Group Ring Groups: Volume 1 Orders and Generic Constructions of Units 9783110372946, 9783110372786

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Eric Jespers, Ángel del Río Group Ring Groups De Gruyter Graduate

Also of interest Group Ring Groups. Volume 2: Structure Theorems of Unit Groups Jespers, del Río, 2015 ISBN 978-3-11-041149-2, e-ISBN (PDF) 978-3-11-041150-8, e-ISBN (EPUB) 978-3-11-041275-8

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The Elementary Theory of Groups Fine, Gaglione, Rosenberger, Spellman, 2014 ISBN 978-3-11-034199-7, e-ISBN (PDF) 978-3-11-034203-1, e-ISBN (EPUB) 978-3-11-038257-0

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Journal of Group Theory Christopher W. Parker, John S. Wilson (Editors-in-Chief) ISSN 1433-5883, e-ISSN 1435-4446

Eric Jespers, Ángel del Río

Group Ring Groups

| Volume 1: Orders and Generic Constructions of Units

Mathematics Subject Classification 2010 16S34, 16U60, 20C05, 16H10, 20F05, 20C40, 20E05, 20E34, 16S35, 16G30, 20H10, 20C10, 20H25, 16-02, 11R52, 11R27, 11S45 Authors Eric Jespers Department of Mathematics Vrije Universiteit Brussel Plainlaan 2 1050 Brussel Belgium [email protected] Ángel del Río Departamento de Matemáticas Universidad de Murcia 30100 Murcia Spain [email protected] Acknowledgements The first author has been partially supported by Fonds Wetenschappelijk Onderzoek – Vlaanderen (including project G.015712N) and Onderzoeksraad Vrije Universiteit Brussel. The second author has been partially supported by Ministerio de Economía y Competitividad project MTM2012-35240 and Fondos FEDER. Both authors would also like to thank support by “Instituut ter bevordering van het Wetenschappelijk Onderzoek en de Innovatie van Brussel” (Brusselse hoofdstedelijke regering) for having received a grant “Brains (back) to Brussels” that supported a sabbatical stay of 6 months by the second author at Vrije Universiteit Brussel.

ISBN 978-3-11-037278-6 e-ISBN (PDF) 978-3-11-037294-6 e-ISBN (EPUB) 978-3-11-038617-2 Library of Congress Cataloging-in-Publication Data A CIP catalog record for this book has been applied for at the Library of Congress. Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2016 Walter de Gruyter GmbH, Berlin/Boston Typesetting: PTP-Berlin, Protago-TEX-Production GmbH, Berlin Printing and binding: CPI books GmbH, Leck ♾ Printed on acid-free paper Printed in Germany www.degruyter.com

Preface An explanation of the maybe strange sounding title “Group Ring Groups” with subtitle “Units and Orders” will clearly outline the main topics of this book. A group ring RG of a finite group G over a ring R is the tool that serves as a meeting place for the interplay between group theory and ring theory. In case R is the ring of integers ℤ then, in some sense, the integral group ring is solely based on the defining group G and hence in this case the interplay between G and ℤG should be the strongest. The defining group G is a subgroup of the unit group U(ℤG) of the group ring ℤG and hence, if the interplay between group and ring theory plays its role well, then one should be able to rediscover the group G from the unit group U(ℤG). It is this interaction between groups and group rings and their unit group that is the main theme of this book. This clarifies the title of the book. The integral group ring ℤG is an order in the finite dimensional rational group algebra ℚG. This semisimple algebra contains all the information concerning the rational representations of the group G and one can make use of strong structure theorems to describe this algebra as a direct product of matrix algebras over division algebras. However, since there are many orders in ℚG it is not obvious how to rediscover the integral group ring within this matrix decomposition. Since the unit group of two orders are commensurable, i.e. they have a common subgroup of finite index, it is for the investigation often more convenient to work in the wider context of orders in finite dimensional rational algebras. This explains the subtitle of the book. Instead of starting at once with structural results, we opted to begin with a chapter that contains many examples. It also reflects perfectly that we like to describe the structures as concrete as possible, so that many results can be implemented via the use of software packages, such as GAP or MAGMA. In several sections we will include examples of implementations in GAP. The examples are mainly based on two generic constructions of units: the Bass units, introduced by Bass and based on the idea of cyclotomic units in number theory, and the bicyclic units, introduced by Ritter and Sehgal and, in some sense, these are based on the idea of elementary matrix units. First we compute examples of unit groups in the ring of cyclotomic integers and second we calculate unit groups of integral matrices and orders in quaternion algebras, three important examples of orders. Next we compute the unit group of group rings ℤG of some groups G of small order. It turns out that for most groups G the unit group of ℤG is infinite. Nevertheless, in all these examples, although there are only few Bass and bicyclic units, these units determine a large part of the unit group. We begin the book with introducing the most relevant background. This is done in Chapters 1 to 3. The first main result is that the unit group of an order in a finite dimensional simple algebra is a finitely presented group. This is a classical result and we include a complete proof that also gives a beautiful link with geometric group theory. Although this result can be proved in the even wider context of the arithmetic groups, we opted to stay focused within the context of orders. In the commutative case the

VI | Preface

finite presentation result provides all one may want to know in terms of structure, except that one also would like to determine the rank of the group of units and that one desires to obtain generic constructions of units. Note that if G is abelian then U(ℤG), being finitely generated, has a subgroup of finite index that is a free abelian group of finite rank. An explicit formula for this rank has been obtained by Higman and this goes back to Dirichlet’s Unit Theorem on the unit group of an order in a number field. Moreover, a result of Bass and Milnor, based on properties of cyclotomic units, shows that the Bass units of ℤG generate a subgroup of finite index in U(ℤG). However, for the non-commutative case, the finite presentation property is not as satisfactory and hence we need to introduce some background on K-theory in Chapters 9 and 10. In Chapters 11 to 13 it is shown that for many groups G the Bass and bicyclic units generate a subgroup of finite index in U(ℤG); the exceptions are determined by the existence of exceptional degree two matrix components and non-commutative division algebras in the decomposition of ℚG and also by some special non-commutative epimorphic images of G, namely Frobenius complements. In Chapter 12 we fully describe when such exceptions occur. In Chapter 13 we also give a lot of attention to the group of central units and show that Bass units play a crucial role in this context. The second part of the book is devoted to structure theorems of the unit group and determining units of orders in some of the exceptional simple components of ℚG mentioned above. It was shown by Hartley and Pickel that if G is not abelian then the unit group U(ℤG) contains a non-abelian free subgroup, for most groups G. We will give a survey of the main results and techniques in describing explicit units that generate such a group. In the last chapter of the book we determine when U(ℤG) contains a subgroup of finite index that is the direct product of free-by-free groups. This is the best structural result known. Another large part of the book is devoted to the use of geometric methods to compute the unit group of an order in a quaternion algebra, this via discontinuous actions on hyperbolic spaces. We give a detailed account by first including a new recent proof of the classical result of Poincaré on obtaining presentations from fundamental polyhedra for groups of isometries of Riemann manifolds of constant curvature and, in particular, of hyperbolic spaces. Then we show how to apply this to orders in some quaternion algebras. An outcome of all the above results is that one has now a good, or at least a much better, idea on describing a subgroup of finite index in U(ZG) for all finite groups G, except for those for which ℚG has a non-commutative epimorphic image that is a non-commutative division algebra that is not a totally definite quaternion algebra. Unit groups of orders in such a division algebra are a big unknown and we refer the reader to Kleinert’s book on this topic [137]. In the planning of the book we decided to write a reasonably complete account of the subject of describing large subgroups of the unit group of an integral group ring and an order and the text should be accessible to second-year graduate students. This required the inclusion of lot of background, both old and new, in the theories of groups, rings, representations, as well as in K-theory, number theory and geometric group theory. We have attempted to make the proofs as self-contained as possible

Preface | VII

and we also included many problems, some of which are referred to in order to complete various details in some proofs. The minimum assumed background are basic courses in group theory, ring theory and field theory. A consequence of all this is that the book has become quite extensive and hence it was decided to split it into two volumes. Volume 1 contains all the details on describing generic constructions of units and the subgroup they generate. Volume 2 mainly is on structure theorems and geometric methods. Without being encyclopaedic, we have included what we think are all the main results and techniques used to achieve these results. For other topics we refer the reader to the literature; there are several books devoted to the algebraic structure of group rings: A. Bovdi [31], N. Gupta [78], G. Karpilovsky [126–131], G. Lee [147], I. B. N. Passi [169], W. Plesken [179], C. Polcino Milies and S. K. Sehgal [185], D. S. Passman [171, 172], K. Roggenkamp [197], S. K. Sehgal [201, 202, 204], M. Taylor [217] and A. Zalesskii and A. Mihalev [232]. The present book can be seen as a continuation of [202] and [185] that both deal with units of (integral) group rings. Originally we intended to also include the state of the art on the Isomorphism Problem and the related Normalizer Problem, as well as on the Zassenhaus Conjecture. The latter is one of the fundamental remaining open problems in integral group rings. Since the present book turned out to be more extensive than anticipated, we have decided not to do so. However, there is a need for a separate book on these problems. The impatient reader may consult the following references (among others): [40, 83–86, 149, 151, 183, 184, 202, 203, 223, 224]. The outline of the books is as follows. Volume 1 contains thirteen chapters. Chapter 1 contains some background on group rings and many examples of unit groups of group rings. Chapter 2 contains background on representations of algebras, including semisimple algebras, Brauer groups and crossed products. Chapter 3 is on the Wedderburn decomposition of semisimple algebras including background on representations of groups. Chapter 4 contains background on Dedekind domains, valuations and orders. Chapter 5 contains a proof of the fact that the unit group of an order is finitely presented. Chapter 6 deals with cyclotomic units. Chapter 7 includes a proof of the fact that the group generated by the Bass units generates a subgroup of finite index in the unit group of an integral group ring of a finite abelian group. It also contains an algorithm (yet inefficient) to calculate a set of fundamental central units. Chapter 8 is on generic units. It is determined which polynomials define units in almost all integral group rings of finite cyclic groups. Chapter 9 is on K-theory and it is shown that elementary matrices of degree n over an ideal of an order generate a subgroup of finite index in the special linear group of the order, provided n ≥ 3. Chapter 10 deals with the same issue as Chapter 9 but now for the, somehow surprisingly harder, case n = 2. Chapter 11 contains a proof of the fact that for many groups G the Bass and bicyclic units generate a subgroup of finite index in U(ℤG). Chapter 12 deals with the problem of classifying the cases excluded in Chapter 11. Chapter 13 contains a description of a complete set of primitive idempotents and a description of large subgroups in the

VIII | Preface

central subgroup, including determining a set of fundamental units. The description of a complete set of primitive idempotents is important to describe the Wedderburn components concretely as matrix algebras so that one can apply the K-theory methods. Volume 2 contains six chapters. Chapter 14 is on explicit constructions of units that generate a free subgroup of the unit group. Chapter 15 contains background on hyperbolic geometry and discrete groups. In Chapter 16 we give a proof of the presentation part of Poincaré’s Theorem on discontinuous groups of isometries of a Riemann manifold of constant curvature.. Chapter 17 contains algorithms to determine a fundamental polyhedron, and thus generators (and presentations), of such a group. Chapter 18 contains applications of these algorithms to unit groups of orders in quaternion algebras and integral group rings. Chapter 19 contains a complete classification of when these geometrical methods apply to calculate the unit group of an integral group ring. It turns out that this happens precisely for the finite groups for which the unit group is virtually a direct product of free-by-free groups. We are indebted to O. Broche, M. Caicedo, G. Janssens, A. Kiefer, Á. Pérez Raposo, J. J. Simón and I. Van Gelder for reading parts of the manuscript, to Ann Kiefer for producing some of the 3D pictures of Chapter 18 and to S. Sánchez Pedreño for LATEX assistance. We would like to express our appreciation to our teachers in group rings: D. S. Passman and S. K. Sehgal. They not only introduced us to this wonderful area of mathematics but they constantly have been very inspiring and stimulating. And most of all they are good friends that have proved such wonderful results. Finally we like to mention that we enjoyed the many hours we collaborated on these topics with many people: V. Bovdi, C. Corrales, A. Dooms, A. Giambruno, J. Z. Gonçalves, E. G. Goodaire, R. Guralnick, A. Herman, M. Hertweck, S. O. Juriaans, E. Kleinert, A. Konovalov, G. Leal, L. Margolis, A. Olivieri, G. Olteanu, A. Pita, C. Polcino Milies, M. M. Parmenter, M. Ruiz and P. Zalesskii.

Contents Volume 1 Preface | V 1 1.1 1.2 1.3 1.4 1.5 1.6

Units in group rings: an introduction | 1 Constructions of units: elementary matrices and bicyclic units | 2 Construction of units: cyclotomic units and Bass units | 5 Examples: unit groups of some orders in number fields | 9 Examples: unit groups of some non-commutative orders | 14 Examples: group rings of groups of small order | 20 Finite rings | 28

2 2.1 2.2 2.3 2.4 2.5 2.6

Representations of algebras | 32 Semisimple algebras | 32 Splitting fields | 42 Characteristic polynomial, trace and norm | 46 Brauer group | 55 Cohomology | 57 Crossed products | 64

3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

Wedderburn decomposition of semisimple group algebras | 73 Representations and characters of finite groups | 74 Some operations with characters | 81 Wedderburn components from character tables | 85 Wedderburn components from monomial characters | 92 Strongly monomial characters | 102 Induction theorems | 112 Brauer-Witt Theorem | 114 Examples | 120

4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

Dedekind domains, valuations and orders | 123 Localization and algebraic integers | 123 Dedekind domains | 130 Finitely generated modules over Dedekind domains | 137 Extensions of Dedekind domains | 142 Valuations | 146 Orders | 152 The discriminant | 157 Brauer group of a number field | 160

X | Contents

5 5.1 5.2 5.3 5.4 5.5

The group of units of an order | 170 Lattices in real vector spaces | 170 Hey’s Theorem and Dirichlet’s Unit Theorem | 173 The group of units of an order is finitely generated | 179 The group of units of an order is finitely presented | 194 Subgroups of finite index | 198

6 6.1 6.2

Cyclotomic integers | 205 Cyclotomic fields | 205 Cyclotomic units | 208

7 7.1 7.2 7.3

Central units | 226 The group of central units of an order | 226 Large subgroups of central units: an algorithm | 237 Bass units as generators of large groups of units | 248

8 8.1 8.2 8.3 8.4 8.5

Generic units | 255 Shifted cyclotomic polynomials | 256 The group of generic units | 262 A logarithm function | 265 A basis of generic units for a subgroup of finite index in U(ℤC n ) | 272 Polynomials of small degree defining units | 276

9 9.1 9.2 9.3 9.4 9.5

K-theory | 282 Grothendieck group | 282 The Whitehead group | 284 Stable range condition | 292 Whitehead group and the stable range condition | 295 Applications of K-theory to units | 303

10 10.1 10.2 10.3 10.4

General linear groups of degree 2 | 312 Number theoretical results | 313 Normality of E2 (I1 ≀ I2 ) in SL2 (R, I1 ≀ I2 ) | 314 The factor group SL2 (R, I1 ≀ I2 ) by E2 (I1 ≀ I2 ) | 319 The group E2 (I) is of finite index in SL2 (R) | 326

11 11.1 11.2 11.3 11.4 11.5

Generators of unit groups of group rings | 337 Bass Unit Theorem | 337 Generalized bicyclic units and Bass units I | 341 Bicyclic units and Bass units | 345 Fixed point free groups and Frobenius complements | 348 Group rings of nilpotent groups | 355

Contents | XI

12 12.1 12.2 12.3 12.4 12.5 12.6 12.7

Exceptional simple components | 359 Components of index one | 359 Components of index two | 366 Generalized bicyclic units and Bass units II | 373 Normal closure of the trivial units | 375 Normal complements | 380 Examples: metacyclic groups | 383 Examples with insufficient Bass units and bicyclic units | 390

13 13.1 13.2 13.3 13.4 13.5 13.6 13.7

Idempotents and central units in group rings | 392 Central subgroups and abelian-by-supersolvable groups | 392 Independent units and abelian-by-supersolvable groups | 397 Central subgroups and strongly monomial groups | 400 Independent units and strongly monomial groups | 402 Primitive idempotents and nilpotent groups | 408 Primitive idempotents and strongly monomial groups | 416 Some metacyclic groups | 420

References | 425 Index of Notation | 435 Index | 439

Volume 2 Preface | V 14 14.1 14.2 14.3 14.4 14.5

Free Groups | 1 Ping-Pong Lemma and free groups generated by bicyclic units | 2 Free groups in integral group rings over Hamiltonian groups | 10 Attractors | 14 Free companions | 21 Attractors revisited | 38

15 15.1 15.2 15.3 15.4 15.5

Hyperbolic geometry | 44 Möbius transformations | 44 Riemann varieties of constant curvature | 47 The groups of isometries of ℍ3 and 𝔹3 | 52 Isometric spheres in 𝔹n | 60 Discrete subgroups of PSL2 (ℂ) | 65

XII | Contents

16 16.1 16.2

Poincaré’s Theorem | 69 Polyhedra and tessellations | 69 Group presentations | 80

17 17.1 17.2 17.3 17.4 17.5

Fundamental polyhedra | 95 Dirichlet fundamental polyhedra | 95 Ford fundamental polyhedron | 99 Polyhedra of finite volume | 102 An algorithm to compute a fundamental polyhedron | 106 Symmetries on Dirichlet polyhedra | 113

18 18.1 18.2 18.3

Unit groups of orders in quaternion algebras | 116 Algebras of Kleinian type | 116 Bianchi groups | 119 Calculating fundamental polyhedra for orders in division algebras of Kleinian type | 133 Generators of unit groups of group rings with exceptional components | 142

18.4

19 19.1 19.2 19.3 19.4 19.5 19.6 19.7

Virtually free-by-free groups | 151 Free-by-free groups | 151 Orders and free-by-free unit groups | 158 Virtual cohomological dimension and finite groups of Kleinian type | 162 Examples of groups of Kleinian type | 166 Nilpotent groups of Kleinian type | 172 Groups of Kleinian type that are not nilpotent | 185 Structure theorem | 192

References | 195 Index of Notation | 205 Index | 209

1 Units in group rings: an introduction All rings are associative and have an identity, denoted 1. Occasionally we will also write it as 1R . Moreover, all the modules M are unital, i.e. 1m = m for every m ∈ M. Let G be a group and R a ring. By RG we denote the set of all R-linear combinations of elements of G. The operations in R and G induce a natural sum and product in G which make RG into a ring called the group ring of G with coefficients in R. More concretely, each element a of RG has a unique expression of the form ∑g∈G a g g, with a g ∈ R for every g ∈ G, and a g = 0 for all but finitely many elements g ∈ G. The sum and product in RG are given by the following formulas: ( ∑ a g g) + ( ∑ b g g) = ∑ (a g + b g )g, g∈G

g∈G

g∈G

( ∑ a g g)( ∑ b g g) = ∑ ∑ (a h b h−1 g )g. g∈G

g∈G

g∈G h∈G

For convenience we make the following notation agreement. If a ∈ RG then we denote by a g the coefficient in R of g, that is, a = ∑g∈G a g g. The support of an element a = ∑g∈G a g g ∈ RG is the finite set Supp(a) = {g ∈ G : a g ≠ 0}. By identifying r ∈ R with r1G and g ∈ G with 1R g, we consider R and G as subsets of RG. Under this identification, R is a subring of RG and G a subgroup of the group of units U(RG). By U(T) we denote the group of units, i.e. the invertible elements, of the ring T. Note that RG is a free R-module with basis G. The group ring RG can be characterized by the following property, called the Universal Property of Group Rings: for every ring homomorphism f : R → S and every group homomorphism α : G → U(S) there is a unique ring homomorphism h : RG → S extending both f and α. Clearly h(∑g∈G r g g) = ∑g∈G f(r g )α(g). Applying this property to a ring anti-homomorphism f : R → S and a group anti-homomorphism α : G → U(S) we deduce that there is a unique anti-homomorphism RG → S extending both f and α and this anti-homomorphism maps ∑g∈G r g g to ∑g∈G f(r g )α(g). In particular, if R is commutative then we denote by ∗ the classical involution RG → RG which restricts to the identity on R and inverts the elements of G, i.e. (∑g∈G r g g)∗ = ∑g∈G r g g−1 . If F is a field, then FG is an F-algebra and one usually refers to FG as the group algebra of G over F. The main theme of interest in this book is the unit group U(ℤG) of the integral group ring ℤG of a finite group G. In this introductory chapter, we will use elementary techniques to built up some intuition in the topic. A first relevant step is to give generic constructions of units in group rings. Here the ideas will come from number theory and linear groups. The former allows us to deal with the unit group of some subrings in number fields and the latter allows us to deal with the unit group of some subrings

2 | 1 Units in group rings: an introduction

of matrices over division rings. Recall that a division ring is a ring, not necessarily commutative, in which every non-zero element is invertible. In order to link U(ℤG) with these rings, one will consider ℤG as a subring of the rational group algebra ℚG, and sometimes also as a subring of the real ℝG and complex group algebra ℂG. These algebras are direct sums of matrix rings over division rings and will be discussed in detail in Chapter 2. A second step is to determine whether the generically constructed units describe a large part of the unit group U(ℤG). In this chapter, we will show this for some groups of small order. In Chapters 7 and 11 we will prove some general results in this direction. This chapter is divided into two parts. In the first two sections we will present some generic constructions of units of integral group rings which are inspired by elementary matrices and cyclotomic units. In the subsequent sections we will calculate the group of units of some small rings.

1.1 Constructions of units: elementary matrices and bicyclic units The most obvious units in a group ring RG are those of the form ug, with u ∈ U(R) and g ∈ G. These units are called the trivial units of RG. For example, the trivial units of ℤG are the elements of the form ±g, with g ∈ G. In some cases the trivial units are the only units. If this is the case then we say that the unit group of RG is trivial. The cyclic group of order n is denoted by C n and C∞ denotes the infinite cyclic group. Moreover ⟨g⟩n (respectively, ⟨g⟩∞ ) represents a cyclic group generated by an element g of order n (respectively, of infinite order). Example 1.1.1. Let G be a group. If the unit group of ℤG is trivial then the unit group of ℤ(G × C2 ) is trivial. Hence, for every k ≥ 1, the unit group of ℤ(C2k ) is trivial. Proof. Let x denote the generator of C2 . Clearly, ℤ(G × C2 ) = (ℤG)C2 , the group ring of C2 over the ring ℤG. If u ∈ U(ℤ(G × C2 )) then u = a + bx and u−1 = c + dx for some a, b, c, d ∈ ℤG. From the equality uu−1 = 1 we deduce that ac + bd = 1 and ad+bc = 0. Therefore (a+b)(c+d) = 1 and (a−b)(c−d) = 1. Thus a+b, a−b ∈ U(ℤG). By assumption, a+b = ±g1 and a−b = ±g2 for some g1 , g2 ∈ G. Hence 2a = ±g1 +±g2 . Since every coefficient of a support element of 2a is even, we deduce that g1 = g2 and either a = 0 or a = ±g1 . In the first case u = ±g1 x and in the second case u = ±g1 . In both cases, u is a trivial unit. Recall that an element of a group is said to be torsion (or periodic) if it has finite order. The order of an element g of G will be denoted by |g|. We also use the notation |X| to denote the cardinality of a set X and |z| for the modulus of a complex number z. A group is said to be torsion (or periodic) if every element is torsion and it said to be torsion-free if the only torsion element is 1. Although the focus mainly is on finite groups we mention the open and challenging Unit Conjecture: if G is a torsion-free group and K is a field then every unit of KG is trivial ([172, 202]).

1.1 Constructions of units: elementary matrices and bicyclic units

| 3

For most finite groups G, we will prove that U(ℤG) is not trivial. Hence, for such groups, one would like to construct non-trivial units. Inspiration from linear groups and number theory will provide us with the tools to do so. In this section we will exploit the former. Let R be a ring. The group of units of the matrix ring M n (R) is called the general linear group of degree n over R and it is denoted by GLn (R). In particular, U(R) = GL1 (R). If R is commutative, then GLn (R) = {A ∈ M n (R) : det(A) ∈ U(R)}. In this case, the determinant map det : M n (R) → R induces a group homomorphism GLn (R) → U(R) and the kernel of this map, denoted by SLn (R), is called the special linear group of degree n over R. The diagonal matrices of M n (R) are the matrices of the form x1

0

x2

diag(x1 , . . . , x n ) = (

..

),

.

0

xn

with x1 , . . . , x n ∈ R. Examples of elements of GLn (R) are the diagonal matrices diag(u1 , . . . , u n ) with u1 , . . . , u n ∈ U(R) and the permutation matrices. The permutation matrix, associated to a permutation σ in S n (the symmetric group of degree n), is the n-by-n matrix P σ with 1 in every (σ(j), j) entry, 1 ≤ j ≤ n, and zeros elsewhere. Observe that σ 󳨃→ P σ defines an injective group homomorphism S n → GLn (R) and the determinant of P σ is the sign of σ. For r ∈ R and 1 ≤ i, j ≤ n, we denote by E ij (r) the matrix having r in the (i, j)-entry and zeroed elsewhere. If i ≠ j then E ij (r)2 = 0 and therefore e ij (r) = I n + E ij (r) ∈ GLn (R), with e ij (r)−1 = I n − E ij (r). The matrices of the form e ij (r) are called elementary matrices From elementary linear algebra, one knows that every invertible matrix with entries in a field is a product of elementary matrices, diagonal matrices and permutation matrices. In fact, if F is a field then SLn (F) is generated by the elementary matrices (Problem 1.1.1). This is also true if F is replaced by an Euclidean domain, such as for example the ring of integers ℤ (Problem 1.1.2). It follows that elements of GLn (F) are products of elementary matrices and diagonal matrices. In order to produce some kind of “elementary matrices” in an integral group ring ℤG, we note that if i ≠ j and r ∈ R then E ij (r) = E ii (1)E ij (r)E jj (1)

and

E ii (1)E jj (1) = 0.

More generally, if x and y are elements of a ring R such that yx = 0 then every element of the form 1 + xay is a unit, with inverse 1 − xay. To realize this construction in the integral group ring ℤG, we observe that if H is a finite subgroup of G and h ∈ H then

4 | 1 Units in group rings: an introduction the map x 󳨃→ hx is a bijection H → H. Therefore, if we write ̃ = ∑ h ∈ ℤG, H h∈H

̃ = 0, for every h ∈ H. Hence for every a ∈ ℤG then we have (1 − h)H ̃ ∈ U(ℤG) 1 + (1 − h)a H ̃ Similarly, and its inverse is 1 − (1 − h)a H. ̃ 1 + Ha(1 − h) ∈ U(ℤG) ̃ and its inverse is 1 − Ha(1 − h). If g ∈ G has finite order then we simply write ̃ g̃ = ⟨g⟩. The units of the form ̃ = 1 + (1 − h)g h ̃ b(g, h)

and

̃ g) = 1 + hg(1 ̃ b(h, − h)

(g, h ∈ G),

were introduced by Ritter and Sehgal in [190] and are called the bicyclic units of G. The group generated by the bicyclic units of ℤG is denoted Bic(G). In fact, the group generated by the bicyclic units of G of the form b(x, ̃y) (respec̃ (respectively, b(̃y , x)), with x, y ∈ G, contains all the units of the form 1 + (1 − h)a H ̃ tively, 1 + Ha(1 − h)) with H ≤ G, h ∈ H and a ∈ ℤG (Problem 1.1.3). ̃ (or b(h, ̃ g)) we need that (1 − In order to obtain a non-trivial bicyclic unit b(g, h) ̃ h)g h ≠ 0 (Problem 1.1.4). This happens precisely when g does not belong to N G (⟨h⟩), ̃ (and also b(h, ̃ g)) is a unit of infinite the normalizer of ⟨h⟩ in G. In this case, b(g, h) k ̃ ̃ order because (b(g, h)) = 1 + k(1 − h)g h ≠ 1 for every non-zero integer k.

Problems 1.1.1. Prove that if F is a field then SLn (F) is generated by the elementary matrices. Hint: Calculate (

1 −1

0 1 )( 1 0

1 1 )( 1 a−1

1 0

1 1 )( 1 1

0 1 )( 1 0

a−1

and (

0 1 )( 1 0

−1 1 )( 1 2

−a−1 ) 1

0 ). 1

1.1.2. Prove that if R is a commutative Euclidean domain then SLn (R) is generated by elementary matrices. Extend this to principal ideal domains.

1.2 Construction of units: cyclotomic units and Bass units |

5

1.1.3. Let G be a finite group. Prove that the subgroup of U(ℤG) generated by the bicyclic units of the form b(x, ̃y), with x, y ∈ G, contains all the units of the form ̃ where H is a subgroup of G, h ∈ H and a ∈ ℤG. 1 + (1 − h)a H, ̃ = N G (H), where 1.1.4. Let G be a finite group and H a subgroup of G. Prove CenG (H) ̃ = {g ∈ G : g H ̃ = Hg}, ̃ the centralizer of H ̃ in G, and N G (H) = {g ∈ G : gH = CenG (H) Hg}, the normalizer of H in G. Deduce that the following statements are equivalent for g, h ∈ G: (1) g ∈ ̸ N G (⟨h⟩). ̃ ≠ 1. (2) b(g, h) ̃ (3) b(h, g) ≠ 1. ̃ has infinite order. (4) b(g, h) ̃ (5) b(h, g) has infinite order. 1.1.5. The dihedral group of order 2n is the group given by the following presentation: D 2n = ⟨a, b | a n = b2 = 1, ba = a−1 b⟩. The quaternion group of order 4n is the group given by the following presentation: Q4n = ⟨a, b | a2n = 1, b2 = a n , ba = a−1 b⟩. Calculate all the non-trivial bicyclic units of the dihedral groups D6 , D8 and the quaternion groups Q8 and Q16 . Which of these groups have no non-trivial bicyclic units? Explain why. ̃ is isomorphic to the 1.1.6. Prove that the group generated by the bicyclic units b(g, h) ̃ group generated by the bicyclic units b(h, g).

1.2 Construction of units: cyclotomic units and Bass units If G is a finite abelian group then the bicyclic units introduced in the previous section are all trivial. Hence, for abelian groups, we need to consider other constructions of units. Here, number theory gives us the inspiration. Let ξ be a complex root of unity of order n > 1 and consider the subring R = ℤ[ξ] of ℂ generated by ξ . For every positive integer k let η k (ξ) =

ξk − 1 = 1 + ξ + ξ 2 + ⋅ ⋅ ⋅ + ξ k−1 ∈ R. ξ −1

(1.2.1)

If, additionally, k is coprime with n then k has an inverse l modulo n, that is kl ≡ 1 mod n. In this case η k (ξ)−1 =

ξ −1 ξ kl − 1 = = 1 + ξ k + ξ 2k + ⋅ ⋅ ⋅ + ξ (l−1)k = η l (ξ k ) ∈ R. ξk − 1 ξk − 1

6 | 1 Units in group rings: an introduction

Hence η k (ξ) ∈ U(ℤ[ξ])

and

η k (ξ)−1 = η l (ξ k ),

(if kl ≡ 1

mod |ξ|).

(1.2.2)

The units of this form are called cyclotomic units. Assume now that g is an element of order n > 1 in a group G and let k and l be as above, i.e. positive integers with kl ≡ 1 mod n. We would like to mimic the construction of cyclotomic units inside ℤ ⟨g⟩, where ⟨g⟩ is the subgroup of G generated by g. k −1 Of course, the expression gg−1 does not make sense, but x k (g) = 1 + g + g 2 + ⋅ ⋅ ⋅ + g k−1 is an element of ℤ ⟨g⟩. Unfortunately, x k (g) is not a unit, unless k = 1. To see this we use the augmentation map. The augmentation map of a group ring RG is the R-linear ring homomorphism ω : RG → R that maps g ∈ G to 1. Clearly, a ring homomorphism restricts to a group homomorphism between the respective unit groups. Hence, if u is a unit of RG then ω(u) is a unit of R. In particular, the units of ℤG have augmentation 1 or −1 and thus the units of augmentation 1 of ℤG form a subgroup of index 2 in U(ℤG). The units of augmentation 1 are called normalized units. We now return to our element x k (g). It is not a unit for k > 1 because the augmentation of x k (g) is k. To surpass the augmentation obstacle, we use that k and n are coprime to find a positive integer m such that k m ≡ 1 mod n and put 1 − km g̃ . (1.2.3) n We will see that u k,m (g) is a unit of ℤ ⟨g⟩. These units were discovered by Hyman Bass and are called Bass units or Bass cyclic units[16]. The group generated by the Bass units of ℤG is denoted Bass(G). Observe that u k,m (g) = x k (g)m +r g̃ , with m an integer chosen such that ω(x k (g)m ) ≡ 1 mod ω(g̃ ) and r is the unique integer such that ω(x k (g)m +r g̃ ) = 1. If k ≡ k󸀠 mod |g| then x k󸀠 (g) = x k (g) + s g̃ for some integer s. From the uniqueness of r, we deduce u k,m (g) = (1 + g + g 2 + ⋅ ⋅ ⋅ + g k−1 )m +

u k,m (g) = u k󸀠 ,m (g)

(if k ≡ k󸀠

mod |g|).

(1.2.4)

If, in the expression of u k,m (g) in (1.2.3), the element g is replaced by ζ n , a complex primitive n-th root of unity, then we obtain u k,m (ζ n ) = η k (ζ n )m . (Observe this makes sense because n > 1.) As η k (ζ n )−1 = η l (ζ nk ), the natural candidate for inverse of u k,m (g) is u l,m (g k ). Indeed, u k,m (g)u l,m (g k ) = ((1 + g + g2 + ⋅ ⋅ ⋅ + g k−1 )(1 + g k + g2k + ⋅ ⋅ ⋅ + g(l−1)k ))m + t g̃ = x kl (g)m + t g̃ ,

1.2 Construction of units: cyclotomic units and Bass units |

7

where t is the unique integer so that ω(x kl (g)m + t g̃ ) = 1. From the uniqueness of t and (1.2.4), we deduce u k,m (g)u l,m (g k ) = u kl,m (g) = u1,m (g) = 1. Hence, u k,m (g) is a unit of ℤG and u k,m (g)−1 = u l,m (g k ) (if kl ≡ 1 mod |g|). (1.2.5) A similar argument can be used to prove the following equality u k,m (g)u k,m1 (g) = u k,m+m1 (g).

(1.2.6)

Indeed, u k,m (g) = (1+g+g 2 +⋅ ⋅ ⋅+g k−1 )m +r g̃ and u k,m1 (g) = (1+g+g 2 +⋅ ⋅ ⋅+g k−1 )m1 +r1 g̃ yields that u k,m (g)u k,m1 (g) = (1 + g + g 2 + ⋅ ⋅ ⋅ + g k−1 )m+m1 + t g̃ and the uniqueness of t implies the equality in (1.2.6). Because of (1.2.4), we may assume that for every Bass unit u k,m (g) we have 1 ≤ k < |g|. Moreover, by (1.2.6), all the Bass units of the form u k,m (g) are powers of the Bass unit u k,m k (g), where m k is the order of k in the multiplicative group U(ℤ/nℤ), for n = |g|. So if g ∈ G, then the group generated by the Bass units of the form u k,m (g) is generated by finitely many of them, namely those of the form u k,m k (g) with 1 < k < |g|. We have already observed that the bicyclic units are either trivial or of infinite order (Problem 1.1.4). We now show that almost all non-trivial Bass units also are of infinite order. Lemma 1.2.1. A Bass unit u k,m (g) is torsion if and only if k ≡ ±1 mod |g|. Proof. Let n = |g| and let u = u k,m (g). If k ≡ 1 mod n then u = 1 and the result is clear in this case. So we assume that k ≢ 1 mod n and in particular n > 1. If k = n − 1 2 g̃ = g−2 . Therefore, if k ≡ −1 mod n then and m = 2 then u = (g̃ − g n−1 )2 + 1−(n−1) n m m is a multiple of 2 and u = u n−1,m (g) = u n−1,2 (g) 2 = g−m by (1.2.4) and (1.2.6). This proves that if k ≡ ±1 mod n then u is torsion. Conversely, assume that u is torsion. By the Universal Property of Group Rings, the group isomorphism ⟨g⟩ → ⟨ζ n ⟩, mapping g to ζ n , extends to a ring homomorphism f : ℤ ⟨g⟩ → ℂ. As n > 1, f(g̃ ) = 0 and therefore f(u) = η k (ζ n )m . Since u is torsion, f(u) is a root of unity, hence so is η k (ζ n ). This implies that |ζ nk − 1|= |ζ n − 1|. Thus ζ n and ζ nk are two vertices of a regular polygon with n vertices so that ζ n and ζ nk are at the same distance to 1. This implies that ζ nk is either ζ n or ζ n = ζ n−1 . Then k ≡ ±1 mod n, as desired.

Problems 1.2.1. Let ξ be a complex root of unity of order n > 1. Prove the following assertions for k and k1 integers, coprime with n. ξ k −1 is a cyclotomic unit in ℤ[ξ]. (Observe that we allow ξ −1 −k = −ξ η k (ξ).

(1) η k (ξ) =

(2) η−k (ξ) (3) η k (ξ) = η k1 (ξ) if and only if k ≡ k1 mod n.

k to be negative.)

8 | 1 Units in group rings: an introduction (4) η k (ξ) is a root of unity if and only if k ≡ ±1 mod n. (5) η k (ξ)η k1 (ξ)−1 has finite order if and only if k ≡ ±k1 mod n. (6) U(ℤ[ξ]) is finite if and only if n = 2, 3, 4 or 6. 1.2.2. Let g be a group element of order n > 1. Let k, k1 and m be positive integers such that k m ≡ k1m ≡ 1 mod n. Prove (1) u k,m (g)u k1 ,m (g k ) = u kk1 ,m (g). (2) u k,m (g) = 1 if and only if one of the following conditions hold: (a) k ≡ 1 mod n, (b) n ≤ 2, (c) k ≡ −1 mod n and n | m. 1.2.3. Let N be a normal subgroup of finite group G. Prove that if u is a Bass unit of G/N then some power of u is the image of a Bass unit of G under the natural map ℤG → ℤ(G/N). 1.2.4. Let R be a ring and v ∈ R a unit of finite order n > 1. Assume that n is invertible in R. Let k be a positive integer coprime with n. Prove that the definition of Bass unit u k,m (v) makes sense in R. In case n is odd (and thus −v has order 2n) prove that u k,m (−v) = (1 − v + v2 + ⋅ ⋅ ⋅ + (−1)k−1 v k−1 )m . Hence, also 1 − v + v2 + ⋅ ⋅ ⋅ + (−1)k−1 v k−1 is a unit. Such units are called alternating units of the ring R, in analogy with the definition of alternating units in integral group rings (see [202, Section 17]). 1.2.5. Let g be a group element of order n and let i and j be integers with 1 < i, j and (i, n) = (j, n) = 1. Prove that (1) (1 + g j + ⋅ ⋅ ⋅ + g j(i−1) ) is a unit in ℚ⟨g⟩ (not in ℤ⟨g⟩). (2) If u = h j,i (g) = (1 + g j + ⋅ ⋅ ⋅ + g j(i−1) )(1 + g + ⋅ ⋅ ⋅ + g i−1 )−1 then u = (1 + g j + ⋅ ⋅ ⋅ + ̃ g j(i−1) )(1 + g i + ⋅ ⋅ ⋅ + g(k−1)i ) + 1−ik n g , where k is an integer so that ik ≡ 1 mod n. (3) u ∈ U(ℤ⟨g⟩). (4) u m = u i,m (g j )u i,m (g)−1 , where m is a positive integer so that i m ≡ 1 mod n. Units h j,i (g) as described in part (2) of Problem 1.2.5 are called Hoechsmann units [202, Section 10] (sometimes they are also called constructible units). The above shows that there is a close relationship, in the integral group ring of a finite abelian group, between the group generated by all Bass units and the group generated by all Hoechsmann units. This relationship has been investigated by Hoechsmann and Ritter in [91, 92]. For a survey we refer to [202, Section 10]. In Section 8.2 we give an explicit formula (see 8.2.4) for expressing a Bass unit of the form u k,ϕ(n) (g), with n the order of g, as a product of Hoechsmann. 1.2.6. Let G be an epimorphic image of a finite group G. Let α denote the natural image of α ∈ ℤG in ℤG. Prove the following. (1) If u k,m (g) is a Bass unit in ℤG then there is a Bass unit u k1 ,m1 (g) in ℤG such that u k,m (g)m1 /m = u k1 ,m1 (g) with k1 ≡ k mod |g| and m | m1 . |h| ̃ ̃ |h| . (2) If g, h ∈ G then b(g, h) = b(g, h)

1.3 Examples: unit groups of some orders in number fields |

9

1.3 Examples: unit groups of some orders in number fields In the previous sections we have encountered the following rings: integral group rings ℤG over a finite group G, matrix rings M n (ℤ) and cyclotomic rings of integers ℤ[ξ], with ξ a complex root of unity. These rings R share an important property. Indeed, let ℚG, { { A = {M n (ℚ), { {ℚ(ξ),

if R = ℤG, if R = M n (ℤ), if R = ℤ[ξ]

and G, { { X = {{E ij : 1 ≤ i, j ≤ n}, { 2 k−1 {{1, ξ, ξ , . . . , ξ },

if R = ℤG, if R = M n (ℤ), if R = ℤ[ξ], and k = [ℚ(ξ) : ℚ].

In each case, A is a finite dimensional semisimple rational algebra (that is, A has no non-zero nilpotent ideals) and X is a basis of both the additive group of R and the rational vector space A. In the terminology of Chapter 4 one says that these rings R are orders in A. In Chapter 4, we will show that the groups of units of two orders R and S in a finite dimensional semisimple rational algebra are quite close to each other. More precisely, U(R) and U(S) have a common subgroup of finite index in both. This is one of the main reasons why it is more convenient to consider the more general context of orders in order to study many aspects of the unit group of an integral group ring of a finite group. In this section we calculate the group of units for some orders in number fields of small degree. Recall that a number field K is a finite field extension of the rationals and [K : ℚ] denotes the degree of K over ℚ. The smallest example of an order in a number field is ℤ and clearly U(ℤ) = {1, −1}. Consider now the ring of Gaussian integers ℤ[i] = {a + bi : a, b ∈ ℤ}. The elements of ℤ[i] form a grid in the complex plane as represented in Figure 1.1.

2 1 −2

−1

0

1

2

−1 −2

Fig. 1.1: ℤ[i] as a complex lattice. The units are the points in the unit circle.

10 | 1 Units in group rings: an introduction Observe that {1, i} is a ℤ-basis of ℤ[i] and it also is a ℚ-basis of ℚ(i). Hence ℤ[i] is indeed an order in ℚ(i). The rings ℤ and ℤ[i] have quite a lot in common. For example, the elements of ℤ and those of ℤ[i] form a discrete subset in ℝ and ℂ respectively. Recall that a subset S of a topological space X is said to be discrete if S is discrete with the induced topology. Clearly if S is a discrete subset of X then S ∩ K is finite for every compact subset K of X. The converse holds if S is locally compact and Hausdorff. For example, the closed unit disk only intersects ℤ in 0, 1 and −1 and only intersects ℤ[i] in 0, ±1 and ±i. The discreteness of a group will help us to describe the unit group of some orders, such as ℤ[i]. Indeed, let u, v ∈ ℤ[i] be such that uv = 1. Then |u| |v| = 1 and hence either |u| ≤ 1 or |v| ≤ 1. Hence u or v is a non-zero element in the unit disk. Because of the discreteness there are only finitely many such elements. In the case of ℤ[i] one obtains these easily by looking at Figure 1.1. So we obtain the following example. Example 1.3.1. U(ℤ[i]) = ⟨i⟩ = {±1, ±i} If K is a quadratic field extension of ℚ, i.e. [K : ℚ] = 2, then K = ℚ(√d), for some square-free integer d different from 1. Clearly {1, √d} is an integral basis of ℤ[√d] and a rational basis of ℚ(√d). Hence, ℤ[√d] is an order in ℚ(√d). If d is negative, then it is not difficult to verify that ℤ[√d] is discrete as a subset of ℂ. Hence, as above, one verifies that U(ℤ[√d]) is finite and its elements can be listed (see Problems 1.3.1). However, this method does not apply in the case of real extensions, i.e. when d is positive. The reason being that ℤ[√d] is not a discrete subset of ℝ nor of ℂ. So some other method is needed. The inspiration can already be found in calculating U(ℤ[i]) with the following alternative approach. The Galois group of ℚ(i)/ℚ consists of the identity map and the complex conjugation. Consider the norm map N : ℚ(i) → ℚ defined by N(a + bi) = (a + bi)(a − bi) = a 2 + b2 , with a, b ∈ ℚ. As N(xy) = N(x)N(y), N restricts to a group homomorphism U(ℤ[i]) 󳨀→ U(ℤ)

x = a + bi 󳨃→ N(x) = xx = |x|2 = a2 + b2 Thus, if u = a + bi is a unit of ℤ[i] then necessarily uu = a2 + b2 = 1 and, in this case, the inverse of u is u. Conversely, if a2 + b2 = 1 then u = a + bi ∈ U(ℤ[i]). Solving the Diophantine equation a2 + b2 = 1 we obtain (again) U(ℤ[i]) = {±1, ±i}. We consider this approach for arbitrary real quadratic extensions. So, assume d is a square-free integer with d > 1. Then, Gal(ℚ(√d)/ℚ) = {1, σ} with σ(a + b√d) = a − b√d and the norm map N = Nℚ(√d)/ℚ : ℚ(√d) → ℚ takes the form N(a + b√ d) = (a + b√ d)(a − b√ d) = a2 − db2 , with a, b ∈ ℤ. This map has as good algebraic properties as in the imaginary quadratic case. Namely, u is a unit of ℤ[√d] if and only if N(u) = ±1 and, in this case, u−1 =

1.3 Examples: unit groups of some orders in number fields |

11

±σ(u). Thus, to calculate U(ℤ[√d]) we have to solve two Diophantine equations: x2 − dy2 = 1,

x2 − dy2 = −1.

Unfortunately, as ℤ[√d] is not discrete in ℝ nor in ℂ, this is not that obvious. Somehow, the geometrical intuition of the quadratic imaginary case is not present anymore. One would therefore like to represent ℤ[√d] as a discrete subset in some other Euclidean space. This can be done using both Galois automorphisms of ℚ(√d)/ℚ: 1 and σ. Consider the map Φ : ℤ[√ d] 󳨀→ ℝ2

(1.3.1)

x = a + b√ d 󳨃→ Φ(x) = (x, σ(x)) = (a + b√ d, a − b√ d)

Then, Φ is injective and maps ℤ[√d] onto a discrete subset of ℝ2 (see Problem 1.3.3.) Moreover, every unit u of ℤ[√d] is mapped onto one of the hyperbolas xy = 1 and xy = −1 (because xσ(x) = ±1). In the following example we give the explicit calculations for d = 2.

1 − 2√2 −2√2

2 − √2

1 + √2 2

−1 − 2√2 −√2

−1 − √2

2 + √2

1

xy = 1 xy = −1

0 −2 − √2

1 + √2

−1 √2 1 + 2√2

−2

−1 + √2 2√2 −2 + √2

−1 + 2√2

Fig. 1.2: The plane representation of ℤ[√2] via the map Φ. The units are the points represented in the hyperbolas xy = ±1.

Example 1.3.2. U(ℤ[√2]) = ⟨−1⟩ × ⟨1 + √2⟩. Proof. Set u = 1+ √2. This is a unit of ℤ[√2] because N(u) = (1+ √2)(1− √2) = −1. Let v = a + b√2 ∈ ℤ[√2], with a, b ∈ ℤ, and assume Φ(v) = (a + b√2, a − b√2) belongs 󵄨 󵄨 to the square K = {(x, y) ∈ ℝ2 : |x|, |y| ≤ u}. Then, |a| ≤ |u| < 3 and |b| ≤ 󵄨󵄨󵄨 √u 󵄨󵄨󵄨 < 2. 2

12 | 1 Units in group rings: an introduction

In Figure 1.2 we represent Φ(v) for those points which we simply label by v. Notice that Φ(v) ∈ K for eleven of these points v. If, moreover, v is a unit of ℤ[√2], then a2 − 2b2 = ±1, equivalently Φ(v) is on one of the two hyperbolas xy = 1 and xy = −1. We obtain six units v, namely v = ±1, ±(1 + √2) or ±(1 − √2). Let now v be an arbitrary nontrivial unit in ℤ[√2]. We have to prove that v ∈ ⟨−1, u⟩. Replacing v by −v, if v is negative, one may assume that v > 0. Furthermore, replacing v by v−1 , if v < 1, one also may assume that v ≥ 1. Then u k ≤ v < u k+1 for some non-negative integer k. Therefore w = vu−k is a unit of U(ℤ[√2]) ∩ [1, u). This implies that Φ(w) ∈ K. By the previous paragraph w = 1 and consequently v = u k , as desired. We now consider the cyclotomic rings ℤ[ζ n ]. As any two primitive n-th roots of unity generate the same multiplicative group, the ring ℤ[ζ n ] does not depend on the choice of the particular primitive n-th root of unity. Moreover, if n is odd then −ζ n has order 2n and ℤ[ζ n ] = ℤ[ζ2n ]. We have already calculated U(ℤ[ζ1 ]) = U(ℤ[ζ2 ]) = U(ℤ) = ⟨−1⟩ and U(ℤ[ζ4 ]) = U(ℤ[i]) = ⟨i⟩. We now will calculate U(ℤ[ζ3 ]) and U(ℤ[ζ8 ]). (See Problem 1.3.7 for the case n = 5.) Example 1.3.3. U(ℤ[ζ3 ]) = ⟨ζ6 ⟩. Proof. We may choose ζ3 = e 3 = −1+i2 √3 . As ζ32 = −1 − ζ3 , every element of ℤ[ζ3 ] can be uniquely written as a + bζ3 , with a, b ∈ ℤ. Figure 1.3 represents some elements of ℤ[ζ3 ]. As in Example 1.3.1, ℤ[ζ3 ] is a discrete subset of ℂ and it does not contain non-zero elements of modulus smaller than 1. Therefore all the units have modulus 1. 󵄨 (2a−b)2 +3b2 b√3 󵄨󵄨2 If x = a + bζ 3 , with a, b ∈ ℤ, then |x|2 = 󵄨󵄨󵄨 2a−b = a2 − ab + b2 ∈ ℤ. 2 + 2 i 󵄨󵄨 = 4 2 Thus if x is unit of ℤ[ζ3 ], then |x| = 1 or equivalently (2a − b) + 3b2 = 4. Solving this Diophantine equation (or looking at Figure 1.3) we obtain U(ℤ[ζ3 ]) = {±1, ±ζ3 , ±ζ32 } = ⟨−ζ3 ⟩ = ⟨ζ6 ⟩. 2πi

√3 ζ3 −1

−ζ32 1

ζ32

−ζ3

−√3

Fig. 1.3: ℤ[ζ3 ] as a lattice

1.3 Examples: unit groups of some orders in number fields |

13

Recall that η k (ζ) denotes a cyclotomic unit (see (1.2.1)). Example 1.3.4. U(ℤ[ζ8 ]) = ⟨ζ8 ⟩ × ⟨1 + √2⟩ = ⟨ζ8 ⟩ × ⟨η3 (ζ8 )⟩ ≅ C8 × C∞ , with η3 (ζ8 ) = 1 + ζ8 + ζ82 , a cyclotomic unit. Proof. We may choose ζ8 = e 4 = √22 (1 + i). Then ζ82 = i and (ζ8 + ζ8−1 )2 = ζ4 + 2 + ζ4−1 = i + 2 − i = 2. Thus ℤ[ζ8 ] contains ℤ[√2] and, by Example 1.3.2, u = 1 + √2 ∈ √ U(ℤ[√2]) = ⟨−1⟩ × ⟨u⟩ ⊆ U(ℤ[ζ8 ]). Moreover, v = η3 (ζ8 ) = 1 + ζ 8 + i = (1 + 22 )(1 + i) = √2 √ 2 ( 2 + 1)(1 + i) = ζ 8 u. Therefore ⟨ζ 8 , u⟩ = ⟨ζ 8 , v⟩. πi

Let x ∈ U(ℤ[ζ8 ]). As ζ84 = −1, there are integers a, b, c and d such that x = a + bζ8 + cζ82 + dζ83 . Because ζ8 + ζ8−1 = −ζ83 − ζ8−3 = √2 and ζ82 + ζ8−2 = i − i = 0, we obtain that |x|2 = (a + bζ8 + cζ82 + dζ83 )(a + bζ8−1 + cζ8−2 + dζ8−3 ) = a2 + b2 + c2 + d2 + (ab + bc + cd)(ζ8 + ζ8−1 ) + (ac + bd)(ζ82 + ζ8−2 ) + ad(ζ83 + ζ8−3 ) = a2 + b2 + c2 + d2 + (ab + bc + cd − ad)√2

Thus the map x 󳨃→ |x|2 defines a group homomorphism N : U(ℤ[ζ8 ]) → U(ℤ[√2]) ∩ ℝ+ = ⟨u⟩. Clearly, the kernel of N is ⟨ζ8 ⟩ (see Problem 1.3.8) and N(u) = u2 . An easy argument shows that −1, ±u and ±u−1 are not in the image of N. Hence, the image of N is ⟨u2 ⟩ ≅ C∞ . Since ℤ is a free group, the homomorphism N splits. Thus, U(ℤ[ζ8 ]) = ⟨ζ8 ⟩×⟨w⟩, for every w ∈ N −1 (u2 ). As u and v belong to N −1 (u2 ), the result follows.

Problems 1.3.1. Calculate U(ℤ[√−d]) for d a square-free positive integer. Hint: See Figure 1.4 for d = 2.

√2

−3

−2

−1

0

1

2

3

Fig. 1.4: ℤ[√−2] as a complex lattice. The units are the points in the unit circle.

14 | 1 Units in group rings: an introduction √

1.3.2. Let d ≠ 1 be a square-free integer. Prove that ℤ[ 1+2 d ] is a subring of ℂ if and only if d ≡ 1 mod 4 and that, in this case, ℤ[

a + b√d 1 + √d : a, b ∈ ℤ, a ≡ b ]={ 2 2

mod 2} .

√

Assume that d ≡ 1 mod 4. Prove that ℤ[ 1+2 d ] is a discrete subset of ℂ if and only if d < 0 and calculate the group of units for all the possible values of d < 0 with d ≡ 1 mod 4. 1.3.3. Prove that the map Φ defined in (1.3.1) is injective and its image is discrete in ℝ2 . 1.3.4. Calculate U(ℤ[√3]). 1.3.5. A Diophantine equation of the form x2 − dy2 = m, with d a square-free integer greater than 1 and m an arbitrary integer, is called a Pell equation. This equation can be written in the form N d (z) = m, for z = x + y√d ∈ ℤ[√d] and N d (x + y√d) = x2 − dy2 . Prove that if z ∈ ℤ[√d] is a solution of the Pell equation N d (z) = m and u is a unit of ℤ[√d] then zu is another solution. Prove that there are finitely many solutions of the Pell equation, modulo units. More precisely, prove that there is a finite subset F of ℤ[√d] such that all the solutions of the Pell equation are of the form zu for z ∈ F and u ∈ U(ℤ[√d]). Solve the Pell equation x2 − 2y2 = 7. 1.3.6. Prove that the torsion subgroup of ℤ[ζ n ] is ⟨ζ n ⟩, if n is even, and ⟨ζ2n ⟩ if n is odd. 1.3.7. Prove that ℤ[√5] ⊂ ℤ[ 1+2√5 ] ⊂ ℤ[ζ5 ] and calculate the group of units of these three rings. Prove that the ℤ[√5] has a unit u of infinite order such that ⟨u⟩ has finite index in U(ℤ[ζ5 ]). Prove that η2 (ζ5 ) generates an infinite cyclic group of finite index in U(ℤ[ζ5 ]). 1.3.8. Let n be a positive integer. Prove that the roots of unity of ℚ(ζ n ) form a cyclic group of order n1 = lcm(2, n) generated by ζ n1 . Deduce that if ℚ(ζ n ) = ℚ(ζ m ) with n < m then n is odd and m = 2n. (Hint: Recall that the minimal polynomial of ζ n over ℚ is ∏1≤k≤n,gcd(k,n)=1 (X − ζ k ).)

1.4 Examples: unit groups of some non-commutative orders In this section we determine the structure of the group of units of two non-commutative orders, namely M2 (ℤ) and the Hamilton quaternions ℍ(ℤ) over the integers. Recall that the group of units of M2 (ℤ) is denoted GL2 (ℤ). Because the determi0 ) has order 2 and denant of an invertible integral matrix is either 1 or −1 and ( 10 −1 0 )⟩. Thus, we only need to describe terminant −1, we have GL2 (ℤ) = SL2 (ℤ) ⋊ ⟨( 10 −1 SL2 (ℤ). We will also describe the structure of PSL2 (ℤ) = SL2 (ℤ)/ ⟨−I2 ⟩, the projective

1.4 Examples: unit groups of some non-commutative orders |

15

special linear group of degree 2 over the integers. Here I2 denotes the identity 2-by-2 matrix. As SL2 (ℤ) is generated by the elementary matrices (Problem 1.1.2) and because e ij (n) = e ij (1)n (for i ≠ j), we have SL2 (ℤ) = ⟨(

1 0

1 1 ), ( 1 1

0 )⟩ . 1

(1.4.1)

We first describe the semigroup generated by S = e12 (1) and S󸀠 = e21 (1). Recall that a monoid M is free of rank k and generated by x1 , . . . , x k ∈ M if every e e non-trivial element of M can be expressed uniquely as x i11 ⋅ ⋅ ⋅ x i mm with m, e1 , . . . , e m positive integers and i j ≠ i j+1 for every j = 1, . . . , m − 1. Lemma 1.4.1. The monoid generated by S = e12 (1) and S󸀠 = e21 (1) is free of rank 2. Proof. Let M be the submonoid of SL2 (ℤ) generated by S and S󸀠 . Because S(

a c

b a+c )=( d c

b+d ), d

and

S󸀠 (

a c

b a )=( d a+c

b ), b+d

any A ∈ M has only positive entries and the two diagonal entries of A are non-zero. We now prove the result by contradiction. So, suppose that M is not free. Since M is contained in a group, and thus the cancellation laws hold, there exist ( ac db ) and 󸀠

󸀠

󸀠

󸀠

( ac󸀠 db󸀠 ) in M so that S( ac db ) = S󸀠 ( ac󸀠 db󸀠 ). Hence, a + c = a󸀠 and c = a󸀠 + c󸀠 , so that a + c󸀠 = 0. Since, a, c󸀠 ≥ 0, this implies a = c󸀠 = 0, a contradiction.

Note, however that SL2 (ℤ) is not a free group because it contains the following elements 0 1 0 1 T=( (1.4.2) ) , U = TS = ( ) −1 0 −1 −1 which are of finite order 4 and 3, respectively. The images t of T and u of U in PSL2 (ℤ) have order 2 and 3 respectively. Observe that S󸀠 = TS−1 T −1 and therefore SL2 (ℤ) = ⟨S, T⟩ = ⟨T, U⟩

and PSL2 (ℤ) = ⟨u, t⟩ .

(1.4.3)

Recall that a group G is a free product of subgroups G i , with i ∈ I, if every nontrivial element g of G can be uniquely expressed in the form g1 g2 ⋅ ⋅ ⋅ g r , where r ≥ 1, for every j we have 1 ≠ g j ∈ G i j and if moreover j ≠ r then i j ≠ i j+1 . This expression is called the normal form of g. The empty product is the normal form of 1. Moreover, G is the free product of subgroups G i if and only if g1 g2 ⋅ ⋅ ⋅ g r ≠ 1 for every 1 ≠ g1 ∈ G i1 , . . . , 1 ≠ g r ∈ G i r with i j ≠ i j+1 provided j ≠ r. If G is the free product of G i , with i ∈ I then we write G = ∗i∈I G i . The free product of finitely many groups G1 , . . . , G n is also denoted G1 ∗ ⋅ ⋅ ⋅ ∗ G n . The free group F n of rank n is the free product C∞ ∗ C∞ ∗ ⋅ ⋅ ⋅ ∗ C∞ of n infinite cyclic groups.

16 | 1 Units in group rings: an introduction Theorem 1.4.2. PSL2 (ℤ) is the free product ⟨u⟩3 ∗ ⟨t⟩2 . Proof. We already know that PSL2 (ℤ) = ⟨u, t⟩. Assume that t e1 u f1 t e2 u f2 . . . t e n u f n = 1, for integers e1 , . . . , e n , f1 , . . . , f n . We have to show that some t e i or u f i is 1. Suppose the contrary. Then, t e i u f i is either tu or tu2 , for every i. Hence, I2 or −I2 belongs to the semigroup generated by S = −TU and S󸀠 = TU 2 , yielding a contradiction with Lemma 1.4.1. A famous theorem of Kuroš describes the structure of subgroups of free products (see for example [196, Theorem 6.3.1]). Theorem 1.4.3 (Kuroš’ Theorem). Let H be a subgroup of a free product G = ∗i∈I G i of subgroups G i . Then H is a free product of the form H = H0 ∗ (∗i∈I (∗g∈X i H ∩ (gG i g−1 ))) , where H0 is a free group and X i is a set of (H, G i )-double coset representatives. Corollary 1.4.4. (1) The group SL2 (ℤ) can be given by the following presentation SL2 (ℤ) = ⟨T, V, J | T 2 = V 3 = J, J 2 = 1⟩, where T, V and J correspond to the following matrices: J = −I2 = (

−1 0

0 ), −1

0 T=( −1

1 ) 0

0 1

V=(

−1 ). 1

(2) If G is a torsion-free subgroup of SL2 (ℤ) or PSL2 (ℤ) then G is free. (3) If g is a torsion element of PSL2 (ℤ) then g is conjugate to an element in ⟨t⟩ or ⟨u⟩. In particular, if g is non-trivial then it has order 2 or 3. (4) If g is a torsion element of SL2 (ℤ) then g is conjugate to an element in ⟨T⟩ or ⟨U⟩. In particular, the order of g divides 3 or 4. Proof. (1) As V = −U and T 2 = V 3 = −I2 , the statement is an easy consequence of Theorem 1.4.2 and (1.4.3). (2) Let G be a torsion-free subgroup of SL2 (ℤ). Obviously G ∩ ⟨−I2 ⟩ = {I2 }. Therefore G is isomorphic to its image G in PSL2 (ℤ) and G is torsion-free. Thus it is enough to show that if G is a torsion-free subgroup of PSL2 (ℤ) then G is free. Because of Theorem 1.4.3 and Theorem 1.4.2, G is a free product of a free group and conjugates of ⟨t⟩ and ⟨u⟩. As G is torsion-free, the latter free factors do not occur. So G indeed is a free group. (3) This is an immediate consequence of Theorem 1.4.3. (4) This follows at once from (3).

1.4 Examples: unit groups of some non-commutative orders |

17

If n is a positive integer, then the ring homomorphism ℤ → ℤ/nℤ induces a natural group homomorphism SL2 (ℤ) → SL2 (ℤ/nℤ). Its kernel, denoted SL2 (ℤ, n), is a subgroup of finite index in SL2 (ℤ). It is called the (principal) congruence subgroup of level n (over ℤ). In other words SL2 (ℤ, n) = {(

a c

b ) ∈ SL2 (ℤ) : a − 1 ≡ d − 1 ≡ b ≡ c ≡ 0 mod n} . d

Because SL2 (ℤ, n) is a normal subgroup of SL2 (ℤ), one obtains from Corollary 1.4.4 (4) the following. Corollary 1.4.5. If n > 2 then SL2 (ℤ, n) is a free group and the only torsion elements of SL2 (ℤ, 2) are I2 and −I2 . Corollary 1.4.6. {( ac db ) ∈ SL2 (ℤ, 2) : a ≡ d ≡ 1 mod 4} is a free subgroup of rank 2 generated by e12 (2) and e21 (2) and it has finite index in SL2 (ℤ). Proof. Let H = {( ac db ) ∈ SL2 (ℤ, 2) : a ≡ d ≡ 1 mod 4} and put G = ⟨e12 (2), e21 (2)⟩. Clearly G ⊆ H ⊆ SL2 (ℤ, 2) and −I2 ∈ ̸ H. Thus H and G are free by Corollary 1.4.5 and Corollary 1.4.4 (2). Moreover G has rank 2 because e12 (2) and e21 (2) do not commute. To prove that G has finite index in SL2 (ℤ), we show that its natural image G in PSL2 (ℤ) has finite index. As the images of S = e12 (1) and S󸀠 = e21 (1) in PSL2 (ℤ) are tu and tu2 , G = ⟨(tu)2 , (tu2 )2 ⟩. We claim that every element x of PSL2 (ℤ) is of the form yt i1 u j1 t i2 u j2 , for y ∈ G and integers i1 , i2 = 0, 1 and j1 , j2 = 0, 1, 2. We prove this by induction on the number n(x) of t’s appearing in the normal form of x = t i1 u j1 . . . t i n u j n . Recall that |t| = 2 and |u| = 3. The statement is clear if n(x) ≤ 2. Otherwise, i2 = ⋅ ⋅ ⋅ = i n = 1 and j1 , j2 , . . . , j n−1 = 1 or 2. If i1 = 0 and j1 ≠ j2 then x = (u j1 t)2 y = (tu−j1 )−2 y, with y = tu j2 −j1 tu j3 . . . t i n u j n and the normal forms of x and y have the same number of t’s. Thus, one may assume that either i1 = 1, or i1 = 0 and j1 = j2 . If i1 = 1 then the number of t’s appearing in the normal form of (tu j1 )−2 x is smaller than the number of t’s in the normal form of x, and hence the claim follows. Finally, if i1 = 0 and j1 = j2 then again the number of t’s in the normal form of (tu−j1 )2 x = (u j1 t)−2 x is smaller than the number of t’s in x. Hence, again the claim follows. Finally we prove that G = H. Let n = [H : G] and let k be the rank of the free group H. The Nielsen-Schreier Theorem (see e.g. [196, 6.1.1]) says that a subgroup of index n in a free group of rank k also is free and has rank 1 + n(k − 1). Hence, 2 = 1 + n(k − 1) and thus n = 1 as desired. We now consider the Hamiltonian quaternion algebras over the ring of integers and over the rationals respectively: ℍ(ℤ) = {a + bi + cj + dk : a, b, c, d ∈ ℤ}, ℍ(ℚ) = {a + bi + cj + dk : a, b, c, d ∈ ℚ},

18 | 1 Units in group rings: an introduction where {1, i, j, k} is a ℤ-basis of ℍ(ℤ), respectively a ℚ-basis of ℍ(ℚ), and the multiplication is determined by the rules: i2 = j2 = −1,

ij = −ji = k.

Clearly, ℍ(ℤ) is an order in ℍ(ℚ). Similarly as for the Gaussian integers, we can consider ℍ(ℤ) as a discrete subset of ℝ4 and we have a norm, N : ℍ(ℚ) 󳨀→ ℚ defined by N(a + bi + cj + dk) = ‖(a, b, c, d)‖2 = a2 + b2 + c2 + d2 , where a, b, c, d ∈ ℚ. At first glance there is no apparent reason why N should relate the product in ℍ(ℤ) with that of ℤ. Nevertheless, we shall show that N is product preserving. We do this using two different instructive methods. For the first method, we look again at the norm on ℚ(i) and observe that, for a, b ∈ ℚ, 󵄨󵄨 󵄨 󵄨a −b󵄨󵄨󵄨 󵄨󵄨 |a + bi|2 = a2 + b2 = 󵄨󵄨󵄨󵄨 a󵄨󵄨󵄨 󵄨󵄨b and (

−b ) a

a b

is the matrix associated to the endomorphism R(a + bi) of ℚ(i) given by R(a + bi)(y) = (a + bi)y, with respect to the basis {1, i}. In the terminology of Chapter 2, R : ℚ(i) → M 2 (ℚ) is the matrix representation of the left regular representation of ℚ(i) with respect to the basis {1, i}. In other words R(x) is the matrix associated to the linear map y 󳨃→ xy in the given basis. Moreover |x|2 = det(R(x)). This suggests to consider the matrix representation of the left regular representation R of ℍ(ℚ) with respect to the basis {1, i, j, k}. That is, R(x) is the matrix of the linear map R(x)(y) = xy in the given basis. This gives a homomorphism of ℚ-algebras R : ℍ(ℚ) → M4 (ℚ) and therefore the composition det ∘R restricts to a group homomorphism U(ℍ(ℚ)) → ℚ. If x = a + bi + cj + dk, then a b R(x) = ( c d and

󵄨󵄨 a 󵄨󵄨 󵄨󵄨 󵄨󵄨 b det(R(x)) = 󵄨󵄨󵄨 󵄨󵄨 c 󵄨󵄨 󵄨󵄨d 󵄨

−b a d −c −b a d −c

−c −d a b −c −d a b

It follows that N(xy) = N(x)N(y), as claimed.

−d c ). −b a −d󵄨󵄨󵄨󵄨 󵄨 c󵄨󵄨󵄨󵄨 󵄨 = N(x)2 . −b󵄨󵄨󵄨󵄨 󵄨 a󵄨󵄨󵄨

1.4 Examples: unit groups of some non-commutative orders |

19

For the second method, we represent ℍ(ℚ) as a subring of M2 (ℚ(i)). To do so, we observe that the matrices i1 = (

i 0

0 ), −i

0 −1

1 ), 0

j1 = (

0 k1 = i1 j1 = ( i

i ) 0

satisfy the defining relations of ℍ(ℚ), i.e. i21 = j21 = −I2 ,

i1 j1 = −j1 i1 = k1 .

and

Therefore the ℚ-linear extension of the function that maps 1, i, j and k to I2 , i1 , j1 and k1 respectively is an algebra homomorphism S : ℍ(ℚ) 󳨀→ M2 (ℚ(i)) a + bi −c + di

a + bi + cj + dk 󳨃→ (

(1.4.4) c + di ) a − bi

(1.4.5)

If x = a + bi + cj + dk = (a + bi) + (c + di)j then det(S(x)) = |a + bi|2 + |c + di|2 = N(x). As S is a ring homomorphism, it follows again that N(xy) = N(x)N(y). In the terminology of Section 2.3, N is the reduced norm of ℍ(ℚ) over ℚ. Now, using that N preserves the product we deduce that if x = a + bi + cj + dk ∈ U(ℍ(ℤ)) then 1 = N(xx−1 ) = N(x)N(x−1 ). As N(x) and N(x−1 ) are positive integers, we thus obtain N(x) = a2 + b2 + c2 + d2 = 1. Solving this Diophantine equation for a, b, c and d, we deduce that U(ℍ(ℤ)) = {±1, ±i, ±j, ±k}. (1.4.6)

Problems 1.4.1 (Ping-Pong Lemma). Let H1 and H2 be subgroups of the symmetric group on a set X. Suppose that X contains two subsets X1 and X2 such that h1 (X1 ) ⊆ X2 for every 1 ≠ h1 ∈ H1 and h2 (X2 ) is a proper subset of X1 for every 1 ≠ h2 ∈ H2 . Prove that ⟨H1 , H2 ⟩ is the free product of H1 and H2 . 1.4.2. Let ℂ̂ = ℂ ∪ {∞}, the compactification of ℂ with one point. Consider GL2 (ℂ) acting on ℂ̂ via Möbius transformations, that is az+b

a ( c

, { { cz+d b ) ⋅ z = { ac , { d {∞,

if z ≠ ∞ and cz + d ≠ 0, if c ≠ 0 and z = ∞, otherwise.

If g = ( ac db ) ∈ SL2 (ℂ), with c ≠ 0, then the isometric circle of g is the Euclidean circle in ℂ defined by the equation |cz + d| = 1.

20 | 1 Units in group rings: an introduction (1) Prove that the kernel of the action of GL2 (ℂ) on ℂ̂ by Möbius transformations is formed by the scalar matrices rI2 . (2) Prove that if the isometric circle of g ∈ SL2 (ℂ) is defined, then g maps the isometric circle of g to the isometric circle of g −1 and it interchanges the interior and exterior parts of these circles. (We assume that ∞ is in the exterior part of every circle.) (3) Let z ∈ ℂ, with |z| ≥ 2, and let g = e12 (z) and h = e21 (z). Let B denote the open unit ball of ℂ. Prove that g(B) ⊂ ℂ̂ \ B and h(ℂ̂ \ B) ⊆ B. Conclude that ⟨g, h⟩ is a free group of rank 2. 1.4.3. Let H = ⟨e12 (2), e21 (2)⟩. (1) Prove that H is the only torsion-free subgroup of SL2 (ℤ) containing H. (Hint: use the Nielsen-Schreier Formula; the latter says that if F is a free group of rank k and G is a subgroup of F of index n then G is free of rank 1 + n(k − 1) [196, 6.1.1].) (2) Calculate the index of H in SL2 (ℤ). 1.4.4. Prove that every finite subgroup of GL2 (ℤ) is isomorphic to either the dihedral group D6 of order 6, the dihedral group D8 of order 8 or a cyclic group of order dividing 4 or 6. 1.4.5. Let a and b be non-zero integers and consider the 4-dimensional rational algebra A = ℚ ⊕ ℚi ⊕ ℚj ⊕ ℚk with multiplication given by the following rules: k = ij,

i2 = a,

j2 = b,

ji = −ij.

(1) Prove that R = ℤ ⊕ ℤi ⊕ ℤj ⊕ ℤk is an order in A. (2) Find an injective ring homomorphism f : A → M2 (ℂ) and calculate the composition of f with the determinant map det : M2 (ℂ) → ℂ and the trace map tr : M2 (ℂ) → ℂ. (3) Calculate the group of units of R for the case a = −1 and b = −3.

1.5 Examples: group rings of groups of small order In this section we calculate U(ℤG) for some groups G of small order and prove a theorem of Higman which characterizes the finite groups G for which U(ℤG) is finite. The main tool used here will be the F-representations of G over a field F, namely the group homomorphisms G → GLn (F). Every F-representation of G extends to an F-algebra homomorphism FG → M n (F) and conversely, every such F-algebra homomorphism restricts to an F-representation of G. For example, if n = |G|, then the left regular representation R of ℚG restricts to a rational representation R : G → GLn (ℚ). That is, if G = {g1 = 1, g2 , . . . , g n } and g ∈ G then the (i, j) entry (R(g))i,j of R(g) is 1, 0,

(R(g))i,j = {

if gg j = g i ; otherwise.

1.5 Examples: group rings of groups of small order

|

21

Observe that tr(R(1)) = n and tr(R(g)) = 0 for every non-trivial element g of G, where tr(x) denotes the trace of x. Representation theory of finite groups will be explained in full detail in Chapter 3. However, for groups of small order, the calculations can be done independently. Before calculating some concrete examples, we prove a result of Berman-Higman [24, 89] on torsion units. The result is also true for infinite groups (see e.g. [201, Corollaries 1.3 and 1.7]). Proposition 1.5.1 (Berman-Higman). Let G be a finite group and u a torsion unit of ℤG. If 1 ∈ Supp(u) then u = ±1. If u is central in ℤG then u is a trivial unit. Proof. Let n = |G| and let R : G → GLn (ℚ) be the left regular rational representation of G. Let u be a torsion unit of ℤG. Write u = ∑g∈G u g g, with u g ∈ ℤ. Then R(u) is a torsion n × n matrix and hence it is diagonalizable. Clearly, its eigenvalues ξ1 , . . . , ξ n are roots of unity and the trace of R(u) is ∑ni=1 ξ i = ∑g∈G u g tr(R(g)) = nu1 . As |ξ i | = 1, for every i, one gets that |u1 | ≤ 1 and thus u1 is either 0, 1 or −1. Moreover, if u1 ≠ 0 then either all ξ i ’s must be equal to 1 or all ξ i ’s must be equal to −1. In this case, R(u) = ±I n and hence u = ±1 because R is injective. This proves the first statement. Assume now that u also is central and let g be an element of the support of G. Since, u is a central periodic unit, we clearly get that also ug−1 is a periodic unit, with 1 ∈ Supp(ug−1 ). By the above, ug−1 = ±1 and thus u = ±g, a trivial unit. We now calculate U(ℤC n ) for some cyclic groups C n of small order. Write, C n = ⟨g⟩n . For every divisor d of n, fix a primitive d-th root of unity ζ d in ℂ and define the group homomorphism ρ d : C n → U(ℂ) given by ρ d (g) = ζ d . This is a (linear complex) representation of C n which extends to a surjective homomorphism of rational algebras ρ d : ℚC n → ℚ(ζ d ). n

The kernel K d of ρ d is a maximal ideal of ℚC n . Moreover, (1 + K d ) ∩ C n = ⟨g d ⟩, the only subgroup of order d of C n . If d and d󸀠 are different divisors of n then ⟨g n/d ⟩ ≠ 󸀠 ⟨g n/d ⟩ and so K d ≠ K d󸀠 . By the Chinese Remainder Theorem, the maps ρ d induce an isomorphism Φ : ℚC n / ∩d|n K d → ∏d|n ℚ(ζ d ). By a well-known result, dimℚ (ℚ(ζ d )) = φ(d) (see Proposition 6.1.2) where φ denotes the Euler function. Hence dimℚ ℚC n = n = ∑d|n φ(d) = dimℚ ∏d|n ℚ(ζ d ), and we deduce that ∩d|n K d = 0. Thus we have an isomorphism Φ = ∏ ρ d : ℚC n → ∏ ℚ(ζ d ). (1.5.1) d|n

d|n

In other words ℚC n ≅ ∏ ℚ(ζ d ), d|n

this is the so called Wedderburn decomposition of ℚC n (see Chapter 3). Moreover, Φ maps ℤG into ∏d|n ℤ[ζ d ]. Hence Φ restricts to an injective group homomorphism Φ : U(ℤC n ) 󳨀→ ∏ U(ℤ[ζ d ]). d|n

(1.5.2)

22 | 1 Units in group rings: an introduction Example 1.5.2. If n = 1, 2, 3, 4 or 6 then U(ℤC n ) = ±C n . Proof. Assume that n = 1, 2, 3, 4 or 6. If d | n then ℤ[ζ d ] = ℤ, ℤ[i] or Z[ζ3 ]. From Examples 1.3.1 and 1.3.3 it thus follows that each U(ℤ[ζ d ]) is finite and hence U(ℤC n ) is finite. The result then follows from Proposition 1.5.1. Let R be a ring and G a group. For a finite subset X of G we use the following notation ̃ = ∑ x, X x∈X

an element of RG. If moreover, the cardinality of X is invertible in R then we set ̂ = 1 X. ̃ X |X| ̃ (respectively, ⟨g⟩). ̂ In case X is a If g ∈ G then we simply write g̃ (respectively, ĝ ) for ⟨g⟩ 2 ̃ If moreover |X| is invertible in R then X ̂ is an idempotent ̃ = |X|X. subgroup of G then X ̃ of RG. If X is a normal subgroup in G then X is central in RG. Let N be a normal subgroup of G. Then the augmentation map modulo N (also called a relative augmentation map) is the ring homomorphism ω N,R : RG → R(G/N) ∑ r g g 󳨃→ ∑ r g (gN) g∈G

g∈G

The kernel of ω N,R is called the augmentation ideal of RG modulo N. If the ring R is clear from the context, then we simply write ω N,R as ω N . Observe that ω G = ω, the augmentation map of RG. If an element a = ∑g∈G a g g ∈ RG belongs to ker(ω N ) then, for all g ∈ G, we have ∑n∈N a ng = 0 and thus ∑n∈N a ng ng = ∑n∈N a ng (n−1)g. It follows that ker(ω N ) = ∑ (n − 1)RG = ∑ RG(n − 1). (1.5.3) n∈N

n∈N

̃ = ∑ Assume now that N is finite. Then N n∈N n is a central element of RG and ̃ ̃ = 0 then it is easily N(1 − n) = 0 for all n ∈ N. Conversely, if a ∈ RG is such that a N verified that a ∈ ker(ω N ). Hence ̃ = {a ∈ RG : a N ̃ = 0}, ker(ω N ) = AnnRG (N)

(1.5.4)

̃ in RG. Consequently, we obtain RG-module isomorphisms the annihilator of N ̃ ≅ RG N. ̃ R(G/N) ≅ RG/AnnRG (N) If, furthermore, |N| is invertible in R then ̂= 1 N ̃ N |N|

(1.5.5)

1.5 Examples: group rings of groups of small order

|

23

̃ = AnnRG (N) ̂ = RG(1 − N) ̂ ̃ = RG N ̂ and AnnRG (N) is a central idempotent of RG, RG N We thus obtain that ̂ ⊕ RG(1 − N) ̂ RG = RG N (1.5.6) and the ring isomorphism

̂ R(G/N) ≅ RG N.

(1.5.7)

The map Φ from (1.5.2) relates Bass units with cyclotomic units. Indeed, if d ≠ 1 then ρ d (g̃ ) = 0. Therefore 1, η k (ζ d )m ,

ρ d (u k,m (g)) = {

if d = 1; if d ≠ 1.

(1.5.8)

We now give an example with an infinite unit group. Example 1.5.3. U(ℤC8 ) = ±C8 × ⟨u3,2 (g)⟩∞ . Proof. From (1.5.1) we know that Φ maps ℤC8 injectively into ℤ2 × ℤ[i] × ℤ[ζ8 ] and therefore U(ℤC8 ) is isomorphic to a subgroup of U(ℤ)2 × U(ℤ[i]) × U(ℤ[ζ8 ]). Let b = u3,2 (g) and v = η3 (ζ8 ). By Example 1.3.4, U(ℤ[ζ8 ]) = ⟨ζ8 ⟩ × ⟨v⟩. By (1.5.8) Φ(b) = (1, η3 (−1)2 , η3 (i)2 , η3 (ζ8 )2 ) = (1, 1, −1, v2 ). We claim that ρ8 (U(ℤC8 )) = ± ⟨ζ8 ⟩ × ⟨v2 ⟩. We have ± ⟨ζ8 ⟩ = ρ8 (±C8 ) and ρ8 (b) = v2 . Hence, ± ⟨ζ8 ⟩ × ⟨v2 ⟩ ⊆ ρ8 (U(ℤC8 )) ⊆ U(ℤ[ζ8 ]) = ± ⟨ζ8 ⟩ × ⟨v⟩. Moreover [± ⟨ζ8 ⟩ × ⟨v⟩ : ± ⟨ζ8 ⟩ × ⟨v2 ⟩] = 2. Therefore, to prove the claim it is enough to show that v ∈ ̸ ρ8 (U(ℤC8 )). Let u = ∑7i=0 u i g i ∈ U(ℤC8 ) with ρ8 (u) = v. Therefore, 1 + ζ8 + ζ82 = (u0 − u4 ) + (u1 − u5 )ζ8 + (u2 − u6 )ζ82 + (u3 − u7 )ζ83 . As 1, ζ8 , ζ82 , ζ83 is a basis of ℚ(ζ8 ) over ℚ, we have u0 −u4 = u1 −u5 = u2 −u6 = 1 and u3 = u7 . Consider the augmentation map ω⟨g4 ⟩ : ℤC8 → ℤC4 . Then ω⟨g4 ⟩ (u) = (u0 + u4 )+(u1 + u5 )g +(u2 + u6 )g2 +(u3 + u7 )g3 = (2u0 − 1) + (2u1 − 1)g + (2u2 − 1)g2 + 2u3 g3 is a non-trivial unit of ℤC4 , contradicting Example 1.5.2. Consequently, if u ∈ U(ℤC8 ) then the fourth coordinate of Φ(ub−k ) is a torsion unit, for some k. As the other coordinates are also of finite order, by Example 1.3.1, Φ(ub−k ) has finite order. Therefore, ub−k is a periodic unit in ℤC8 , because Φ is injective. So, by Proposition 1.5.1, ub−k is a trivial unit and the result follows. As examples of non-commutative groups of small order, we will deal with the dihedral group and quaternion group of order 8. As in Problem 1.1.5, we use the following presentation for the dihedral group of order 2n: D2n = ⟨a, b | a n = b2 = 1, bab−1 = a−1 ⟩. There are precisely two group homomorphisms σ : D2n → U(ℂ) with σ(a) = 1. These are σ1 and σ2 defined by σ1 (a) = σ1 (b) = 1,

σ2 (a) = 1, σ2 (b) = −1.

If n is odd these are the unique linear (complex) representations of D2n , i.e. group homomorphisms D2n → U(ℂ). If n is even then D2n has two additional linear complex

24 | 1 Units in group rings: an introduction

representations σ3 and σ4 determined by σ3 (a) = −1, σ3 (b) = 1,

σ4 (a) = σ4 (b) = −1.

The rational linear extension of each σ i gives a surjective map σ i : ℚD2n → ℚ. We can now give a description of the full unit group of ℤD8 . Example 1.5.4. Let B be the subgroup of U(ℤD8 ) generated by the bicyclic units of the ̃ Then U(ℤD8 ) = B ⋊ (±D8 ) and B is free of rank 3. form b(g, h). Proof. Besides the four linear representations σ1 , . . . , σ4 , the dihedral group D8 has a non-linear irreducible (complex) representation given by ρ(a) = (

0 −1

1 ), 0

1 0

ρ(b) = (

0 ). −1

(1.5.9)

Its rational linear extension to ℚD8 we also denote by ρ. As in the cyclic case, the kernel of these homomorphisms are different maximal ideals of ℚD8 and therefore we obtain a ring homomorphism Φ : ℚD8 󳨀→ ℚ4 × M2 (ℚ)

(1.5.10)

defined by Φ(x) = (σ1 (x), σ2 (x), σ3 (x), σ4 (x), ρ(x)). Using the Chinese Remainder Theorem one can prove, as for the map of (1.5.1), that Φ is injective and hence it maps U(ℤD8 ) injectively into U(ℤ)4 × GL2 (ℤ). Consider the following subgroup of U(ℤD8 ) and SL2 (ℤ) respectively: B1 = {1 + (1 − a2 )α ∈ U(ℤD8 ) : α ∈ ℤD8 and ω(α) is even}, 1 + 2a11 Γ = {( 2a21

2a12 ) ∈ SL2 (ℤ) : a11 ≡ a12 ≡ a21 ≡ a22 1 + 2a22

mod 2} .

Observe that B1 is normal in U(ℤD8 ) because a2 is central and the augmentation map ω is a ring homomorphism ℤD8 → ℤ. Let A = e12 (1). It is easy to see that Γ A = {(

a11 a21

a12 ) ∈ SL2 (ℤ, 2) : a11 ≡ a22 ≡ 1 a22

mod 4 and a12 ≡ 0 mod 4}

and this is a proper subgroup of Δ = {(

a11 a21

a12 ) ∈ SL2 (ℤ, 2) : a11 ≡ a22 ≡ 1 mod 4} a22

By Corollary 1.4.6, Δ is a free group of rank 2 that is generated by x = e12 (2) and y = e21 (2). We claim that it is enough to prove the following: (1) U(ℤD8 ) = ±D8 B1 , (2) ρ maps B1 injectively into Γ, (3) B ⊆ B1 and (4) ρ(B)A contains x2 , y, xyx−1 . Then we argue as follows: As Δ is the free group of rank 2 generated by x and y, ⟨x2 , y, xyx−1 ⟩ is a

1.5 Examples: group rings of groups of small order |

25

free subgroup of rank 3 and index 2 in Δ. Hence [Δ : ρ(B)A ] ≤ 2. In fact the index is 2 because Γ A is a proper subgroup of Δ. Thus ⟨x2 , y, xyx−1 ⟩ = ρ(B)A = ρ(B1 )A = Γ A . Then B = B1 and it is free of rank 3, by (2), and U(ℤD8 ) = ±D8 B by (1). So we only have to prove statements (1)-(4). (1) Let N = ⟨a2 ⟩, a central subgroup of order 2. Since D8 /N ≅ C2 × C2 , we obtain from Example 1.1.1 that U(ℤ(D8 /N)) only has trivial units. By (1.5.3) and (1.5.5), ℤ(D8 /N) ≅ ℤD8 / ker(ω N ) and ker(ω N ) = (1 − a2 )ℤD8 . We thus obtain that any unit u of ℤD8 can be written as ±g + (1 − a2 )α, with g ∈ D8 and α ∈ ℤD8 . As ±g + (1 − a2 )α = ±ga2 + (α ± g)(1 − a2 ), we may assume without loss of generality that ω(α) is even. Finally ±g + (1 − a2 )α = ±g(1 + (1 − a2 )(±g−1 α)) ∈ ±D8 B1 . This finishes the proof of (1). (2) Note that B1 is a normal subgroup of U(ℤD8 ) such that σ i (B1 ) = 1 for every i = 1, . . . , 4. Therefore ρ maps B1 injectively onto ρ(B1 ). Assume 1 + (1 − a2 )α ∈ B1 with ω(α) even. Write α = (α0 + α1 a + α2 b + α3 ab) + (α󸀠0 + α󸀠1 a + α󸀠2 b + α󸀠3 ab)a2 , with all α i , α󸀠i ∈ ℤ. B1 ≅ ρ(B1 ). Further, ρ(α) = (

α0 + α2 − (α󸀠0 + α󸀠2 ) −α1 − α3 − (−α󸀠1 − α󸀠3 )

α1 − α3 − (α󸀠1 − α󸀠3 ) ) α0 − α2 − (α󸀠0 − α󸀠2 ).

Because ω(α) is even, it follows that ρ(1 + (1 − a2 )α) belongs to Γ. This finishes the proof of (2). ̃=b ̃ it follows that (3) One of the non-normal cyclic groups of D8 is ⟨b⟩. From b b −1 3 ̃ = 1 + (1 − b)a(1 + b) and u u0 = b(a, b) 0 = 1 + (1 − b)a (1 + b) are the only ̃ bicyclic units of the form b(g, b). Repeating this argument with the remaining nonnormal cyclic subgroups, we see that B = ⟨u0 , u1 , u2 , u3 ⟩, with u k = b(a, ̃ a k b), for 2 k k = 0, 1, 2, 3. Moreover u k = 1 + (1 − a )a(1 + a b) ∈ B1 and we conclude B ⊆ B1 as desired. (4) Because of (1.5.9), we have 1 −4

0 ), 1

3 ρ(u3 ) = ( −2

1 0

4 ), 1

ρ(u1 ) = (

ρ(u0 ) = (

ρ(u2 ) = (

−1 −2

2 ), −1 2 ). 3

Recall that A = e12 (1). Conjugating the above matrices with A we obtain the matrices −3 −4

(

4 ), 5

1 −2

(

0 ), 1

Thus 1 2

y=(

0 ), 1

x2 = (

1 0

4 ), 1

(

1 0

4 ), 1

−3 −2

(

−3 (yx−1 )2 = ( −4

8 ). 5 4 ) ∈ ρ(B)A 5

and so x−1 yx = y−1 (yx−1 )2 x2 ∈ ρ(B)A . This finishes the proof of (4) and of the Example.

26 | 1 Units in group rings: an introduction Recall that if G is a semidirect product N ⋊ H then N is called a normal complement of H in G and H is called a complement of N in G. Dooms and Jespers [49] have shown that ±D 8 has 8 normal complements in U(ℤD8 ) and all turn out to be free. On the other hand, Allen and Hobby discovered in [2] that ±D6 has a normal complement in U(ℤD6 ) that is not torsion-free. There is precisely one such complement and there are three others all of which are free [49]. Both U(ℤD8 ) and U(ℤD6 ) have been calculated in various other ways (see for example [111, 121, 154, 182]). Example 1.5.5 (Higman [88]). U(ℤQ8 ) = ±Q8 . Proof. Recall that Q8 = ⟨a, b | a4 = a2 b2 = 1, bab−1 = a−1 ⟩. Because of (1.4.6), there is an isomorphism ρ : Q8 → U(ℍ(ℤ)) = {±1, ±i, ±j, ±k} which extends to a homomorphism ρ : ℤQ8 → ℍ(ℤ). Let u = ∑g∈Q8 u g g ∈ U(ℤQ8 ), with all u g ∈ ℤ, and assume that u has augmentation 1. Then ρ(u) = (u1 − u a2 ) + (u a − u a3 )i + (u b − u a2 b )j + (u ab − u a3 b )k ∈ {±1, ±i, ±j, ±k}. Multiplying, if necessary, u with a trivial unit of U(ℤQ8 ), one may assume that u1 − u a2 = 1 and u a − u a3 = u b − u a2 b = u ab − u a3 b = 0. Let H = Q8 / ⟨a2 ⟩. The augmentation map ω⟨a2 ⟩ restricts to an augmentation preserving group homomorphism ψ = U(ℤQ8 ) → U(ℤH). As H ≅ C22 , we have U(ℤH) = ±H, by Example 1.1.1. Thus ψ(u) = (u1 + u a2 ) + (u a + u a3 )ψ(a) + (u b + u a2 b )ψ(b) + (u ab + u a3 b )ψ(ab) = −1 + 2(u1 + u a ψ(a) + u b ψ(b) + u ab ψ(ab)) belongs to H. Hence, u1 = 1 and u g = 0, for every 1 ≠ g ∈ Q8 , as desired. An elementary abelian p-group, for p a prime, is an abelian group such that g p = 1, for every g ∈ G. If G is finite then G is elementary abelian p-group if and only if G ≅ C np for some n ≥ 0. The group G is said to be Hamiltonian, if every subgroup of G is normal in G. In Examples 1.1.1, 1.5.2 and 1.5.5 we have encountered some finite groups G for which U(ℤG) is finite. The following result of Higman [88] characterizes the finite groups with this property. The exponent of a group G is Exp(G) = min{n ≥ 1 : g n = 1 for every g ∈ G}, provided such number exists. Otherwise one says that G has infinite exponent. Theorem 1.5.6 (Higman [88]). The following conditions are equivalent for a finite group G. (1) U(ℤG) = ±G. (2) U(ℤG) is finite. (3) G is abelian of exponent dividing 4 or 6 or G ≅ Q8 × A, for A an elementary abelian 2-group. Proof. (1) implies (2) is obvious. (2) implies (3). Assume that U(ℤG) is finite. As non-trivial bicylic units have infinite order, it follows that all the bicyclic units must be trivial. Hence, every subgroup

1.5 Examples: group rings of groups of small order

|

27

of G is normal, i.e. G is a Hamiltonian group. By the Dedekind-Baer classification of Hamiltonian groups ([196, 5.3.7]), G is either abelian or G ≅ Q8 × A × B, where A is an elementary abelian group and B is an abelian group of odd order. Thus it is enough to prove that the order n of each element g of G divides either 4 or 6. We prove this by contradiction. So suppose the contrary, then φ(n) > 2 and therefore there is an integer k coprime with n such that k ≢ ±1 mod n. Choose a positive integer m so that k m ≡ 1 mod n. Because of Lemma 1.2.1, the Bass unit u k,m (g) is a unit of infinite order, yielding a contradiction. (3) implies (1). If G = Q8 × A with A an elementary abelian 2-group then U(ℤG) = ±G by Examples 1.1.1 and 1.5.5. Assume that G is abelian of exponent dividing 4 or 6. Then G ≅ C4k × A or G ≅ C3k × A, with A elementary abelian 2-group. By Proposition 1.5.1 and Example 1.1.1, it is sufficient to prove that U(ℤG) is finite for G = C kd with d = 3 or 4 and k ≥ 1. We argue by induction on k. If k ≤ 1 then U(ℤG) is finite by Example 1.5.2. Assume that k ≥ 2 and U(ℤC k−1 d ) is finite. Write G = H × ⟨g 1 ⟩ × ⟨g 2 ⟩ for some subgroup of G and |g | = |g | H ≅ C k−2 1 2 = d Let g 3 = g 1 g 2 . d Assume first that d = 4. Let K1 = ⟨g12 ⟩, K2 = ⟨g22 ⟩ and K3 = ⟨g32 ⟩ and consider the relative augmentation maps ω i = ω K i : ℤG → ℤ(G/K i ). Then we have a homomorphism 3

3

f = ∏ ω i : ℤG → ∏ ℤ(G/K i ) i=1

i=1

We claim that f is injective. Indeed, let x ∈ ker(f) and write 3

3

j

x = ∑ ∑ x ij g1i g2 i=0 j=0

with x ij ∈ ℤH. For every i = 0, 1, 2, 3 and every j = 0, 1 we have x ij = −x i,j+2 , because x ∈ ker(ω2 ). Similarly x ij = −x i+2,j for every i = 0, 1 and every j = 0, 1, 2, 3, since x ∈ ker(ω1 ). Thus x = (x00 + x10 g1 + x01 g2 + x11 g1 g2 )(1 − g12 )(1 − g22 ). Finally, 0 = ω3 (x) = 2(x00 + x10 g1 + x01 g2 + x11 g1 g2 )(1 − g12 ) and comparing coefficients we obtain x ij = 0 for every i, j. Thus x = 0 as desired. This proves the claim. Therefore f restricts to an injective map f : U(ℤG) → ∏3i=1 U(ℤ(G/K i )). Moreover G/K i ≅ C4k−1 × C2 and hence U(ℤ(G/K i )) is finite, by hypothesis and Example 1.1.1. We conclude that U(ℤG) is finite. The proof for the case d = 3 is similar. We consider four non-trivial subgroups of ⟨g1 , g2 ⟩: K1 = ⟨g1 ⟩, K2 = ⟨g2 ⟩, K3 = ⟨g1 g2 ⟩ and K4 = ⟨g1 g22 ⟩. Repeating the same argument as above, it is enough to show that f = ∏4i=1 ω i : ℤG → ∏4i=1 ℤ(G/K i ) is j injective. Writing an element x of the kernel of f as x = ∑2i=0 ∑2j=0 x ij g1i g2 we deduce from ω1 (x) = ω2 (x) = 0 that x i2 = −x i0 − x i1 and x2i = −x0i − x1i for every i = 0, 1, 2. Thus x = x00 + x10 g1 − (x00 + x10 )g12 + x01 g2 + x11 g1 g2 − (x01 + x11 )g12 g2 − (x00 + x01 )g22 − (x10 + x11 )g1 g22 + (x00 + x10 + x01 + x11 )g12 g22 .

28 | 1 Units in group rings: an introduction From ω3 (x) = 0 one deduces that x10 = x01 = −(x00 + x11 ) and from ω4 (x) = 0 that x00 = x11 = −(x10 + x01 ). The implies x ij = 0 for every i, j. So x = 0 as desired. Recall that a ring (or an algebra) A is said to be simple if it does not have two-sided ideals other than 0 and A. Observe that in Examples 1.5.2 and 1.5.4 we have first calculated an isomorphism Φ : ℚG → ∏ni=1 A i , where each A i is a simple rational algebra. This is called the Wedderburn decomposition of ℚG. Then we observed that Φ maps ℤG injectively into ∏ni=1 Oi , with Oi an order in A i for every i. Using this, we obtain information of U(ℤG) from information on each U(Oi ). In general, ℤG is not a direct product of orders in simple algebras (in fact, it is not a direct product of non-trivial ring, unless G = 1). The main tool to calculate the isomorphism Φ is Representation Theory. In Chapters 2 and 3 we revise the main techniques of Representation Theory and in Chapter 4 we study properties of orders. Other important tools used in some examples of this chapter are the norm and the reduced norm maps. They also will be revised in Chapter 2.

Problems 1.5.1. Let G be a finite group. Prove that if U(ℤG) is torsion then it is finite. 1.5.2. Find a Bass unit of C5 which generates a subgroup of finite index in U(ℤC5 ). 1.5.3. Let G be a finite group and N a normal subgroup of G. Prove that ω N : ℤG → ℤ(G/N) maps Bass units of G to Bass units of G/N and bicyclic units of G to powers of bicyclic units of ℤ(G/N). 1.5.4. Let u0 , u1 , u2 and u3 be the bicyclic units in the proof of Example 1.5.4. Write each u i as a product of the other three. 1.5.5. Show that U(ℤD6 ) = B ⋊ (±D6 ), where B is the group generated by the bicyclic ̃ and show that B is free. units of the form b(g, h), 1.5.6. Find a non-trivial torsion element in Bic(D6 ). 1.5.7. Let g be an element of finite order of a group G and let R be a ring and a ∈ RG. Prove that (g − 1)a = 0 if and only if a ∈ RG g̃

1.6 Finite rings Although we mainly are interested in units of integral group rings and, more generally, in unit groups of orders in finite dimensional rational algebras, for some results we need to know some facts on the unit group of a finite ring. This is the purpose of the present Section. All this is elementary and well known.

1.6 Finite rings |

29

Theorem 1.6.1. If G is a finite subgroup of the multiplicative group of non-zero elements of a field then G is cyclic. In particular, if F is a finite field then U(F) is cyclic. Proof. We prove this by contradiction. So, suppose G is not cyclic. Then it contains a subgroup of the form ⟨a⟩ × ⟨b⟩ with a and b of order p, for some prime p. Then the polynomial X p − 1 has at least p2 roots in the field containing G, a contradiction. Let n be a positive integer. For any integer n we denote by [a]n the congruence class modulo n containing a. Sometimes we will abuse notation and we simply write [a]n as a. As an easy consequence of the Bezout Lemma we have U(ℤn ) = {[a]n : gcd(a, n) = 1}

and therefore U(ℤn ) has φ(n) elements, where φ denotes the Euler function. If gcd(a, n) = 1, then the order of [a]n in U(ℤn ) is called the multiplicative order of a modulo n and it is denoted o n (a). If [a]n generates U(ℤn ) then we say that a is a primitive root modulo n. From Theorem 1.6.1 we know that if p is prime then there exists primitive roots modulo p. We will show that this remains valid for most prime powers. This result will follow at once from the lemma. Lemma 1.6.2. Let p be an odd prime and assume a is a primitive root modulo p e . (1) If e = 1 then either a or a + p is a primitive root modulo p2 . (2) If e ≥ 2 then a is a primitive root modulo p e+1 . Proof. Let a be an integer so that a is a primitive root modulo p e , in particular (a, p) = 1. Clearly o p e (a) divides o p e+1 (a). As also p e−1 (p − 1) = φ(p e ) = o p e (a)

and

o p e+1 (a) | φ(p e+1 ) = p e (p − 1),

we get that o p e+1 (a) is either p e−1 (p − 1) or p e (p − 1). Assume that e = 1 and neither a nor a + p is primitive root modulo p2 . Then p−1 ≡ (a + p)p−1 ≡ 1 mod p2 . Hence a 1 ≡ (a + p)p−1 ≡ a p−1 + (p − 1)a p−2 p ≡ 1 − a p−2 p

mod p2 ,

and so p | a, a contradiction. e−2 Assume now that e ≥ 2. By Euler’s Theorem we know that a p (p−1) = 1 + kp e−1 for some integer k. However p ∤ k, since a is a primitive root module p e . Since e ≥ 2, both 3(e − 1) and 2e − 1 are at least e + 1, and hence ap

e−1

(p−1)

= (1 + kp e−1 )p ≡ 1 + pkp e−1 + (2p)k2 p2(e−1) p − 1 2 2e−1 ≡ 1 + kp e + ≡ 1 + kp e ≢ 1 mod p e+1 . k p 2

Thus o p e+1 (a) ≠ p e−1 (p − 1) and hence, by the first part of the proof, o p e+1 (a) = p e (p − 1) = φ(p e+1 ), i.e. a is a primitive root modulo p e+1 .

30 | 1 Units in group rings: an introduction Theorem 1.6.3. If n is an odd prime power then U(ℤn ) is cyclic. Lemma 1.6.2 also provides an obvious method to calculate primitive roots modulo odd prime powers. Example 1.6.4. Since U(ℤ3 ) = {1, −1}, −1 is a primitive root modulo 3. However −1 is not a primitive root modulo 9. Then 2 = −1 + 3 is a primitive root modulo 9 and hence 2 is a primitive root modulo 3n for every n. For the prime 2 the situation is different. In general ℤ2n is not cyclic. Clearly U(ℤ2 ) = {1} and U(ℤ4 ) = {1, −1} are cyclic. However U(ℤ8 ) = {±1, ±3} and 32 ≡ 1 mod 8. Thus a2 ≡ 1 mod 8 for every odd integer a. This shows that U(ℤ8 ) ≅ C2 × C2 . If e ≥ 3 then 8 divides 2e and hence there is a surjective group homomorphism U(ℤ2e ) → ℤ8 (see Problem 1.6.3). This shows that U(ℤ2e ) is cyclic if and only if e ≤ 2. The same argument shows that if U(ℤn ) is cyclic then n is not multiple of 8. Theorem 1.6.5. Let n be a positive integer. Then U(ℤn ) is cyclic if and only if n is either 1, 2, 4, p e or 2p e with p an odd prime and e ≥ 1. Proof. Assume that p is an odd prime. We already have proved that ℤn is cyclic for n = 1, 2, 4 and p e . On the other hand, by the Chinese Remainder Theorem ℤ2p e ≅ ℤ2 × ℤp e and thus U(ℤ2p e ) ≅ U(ℤ2 ) × U(ℤp e ) = U(ℤp e ). This proves one implication in the statement of the result. Conversely, assume that U(ℤn ) is cyclic. We have to show that n is either 1, 2, 4, p e or 2p e with p an odd prime and e ≥ 1. This is equivalent to showing that n is not divisible by pq with p an odd prime and q either 4 or another odd prime. By Problem 1.6.3, if n is divisible by such product pq, then U(ℤpq ) ≅ U(ℤp ) × U(ℤq ) is cyclic. However, U(Z p ) and U(ℤq ) are cyclic of even order and hence both have cyclic subgroups of order 2. Thus U(ℤpq ) contains at least two different subgroups of order 2 and this implies that it is not cyclic. The proof of Lemma 1.6.2 can be modified to prove the following fact. Lemma 1.6.6. If e ≥ 3 then o2e (5) = 2e−2 . Proof. By Theorem 1.6.5 the group U(ℤ2e ) is non-cyclic. Since φ(2e ) = 2e−1 , we get o2e (5) = 2x e for some x e ≤ e − 2. As o2e−1 (5) divides o2e (5), the sequence (x e )e is nondecreasing. We have to show that x e is exactly e − 2. We prove this by contradiction. So assume that this is not the case and let e be the minimum integer greater than 1 e−4 with x e < e − 2. Clearly e ≥ 4. Then 52 = 1 + k2e−2 with k odd, because x e−1 = e − 3. e−3 e−1 Since e ≥ 4, we have 2(e − 2) ≥ e and so 52 = 1 + 2k2e−2 + k2 22(e−2) ≡ 1 + k22 mod 2e . Therefore e − 3 < x e < e − 2, a contradiction. Theorem 1.6.7.

⟨−1⟩2e−1 , ⟨−1⟩2 × ⟨5⟩2e−2 ,

U(ℤ2e ) = {

if e = 1 or 2; if e ≥ 3.

1.6 Finite rings |

31

Proof. The case e ≤ 2 is obvious. Assume that e ≥ 3. Then, by Lemma 1.6.6, o2e (5) = 2e−2 . Since o2e (−1) = 2, we get that the subgroups of U(ℤ2e ) generated by −1 and 5 have orders 2 an 2e−2 respectively. Since 5i ≡ 1 ≢ −1 mod 4 for every i, we have that −1 ∈ ̸ ⟨5⟩ ⊆ U(ℤ2e ). Hence the result follows. e

e

Let n be a positive integer with prime factorization n = p11 . . . p kk . By the Chinese Remainder theorem we have that U(ℤn ) is isomorphic with the direct product of the groups U(ℤp ni ), for i = 1, . . . , k. As the latter are described in Theorem 1.6.3 and Thei orem 1.6.7, we obtain a description of an arbitrary U(ℤn ). e

e

Theorem 1.6.8. If n = 2e0 p11 . . . p kk is the prime factorization of a positive integer n with e0 ≥ 0 and e i ≥ 1 for every i ≥ 1. Then U(ℤn ) ≅ A × C p e1 −1 (p1 −1) × ⋅ ⋅ ⋅ × C p e1 −1 (p1 −1) ,

with 1, { { A ≅ {C2 , { {C2 × C2e−2 ,

if e0 ≤ 1; if e0 = 2; otherwise.

Problems 1.6.1. If F is a finite field then a primitive element of F is a generator of U(F). Calculate primitive elements for the fields with at most 9 elements. 1.6.2. Let F be an algebraic extension of a finite field. Show that U(F) is periodic. Deduce the converse of the second part of Theorem 1.6.1. 1.6.3. Let n and m be positive integers. Prove that if n divides m then the natural homomorphism U(ℤm ) → U(ℤn ) is surjective. In particular, if U(ℤm ) is cyclic then so is U(ℤn ). 1.6.4. Prove that if p is an odd integer then there is an integer a such that a is a primitive modulo p e for every e. Calculate an integer a such that a is primitive modulo 5e for every e. 1.6.5. Write a computer program that calculates for a given odd prime p an integer a which is a primitive root modulo p e . (Hint: The probability that a random element of U(ℤp ) generates U(ℤp ) is not that small. Estimates this probability.)

2 Representations of algebras This chapter is devoted to the necessary background on representation theory of algebras and the Brauer group of a field. Only results essential for later use are stated, and most will be proved. All material is classical and is based on standard references. Topics on semisimple modules and algebras can be found mainly in [57, 175]. Relevant background on cohomology and the Brauer group can be found in [37, 175, 196, 226] and topics on crossed products in [173, 175].

2.1 Semisimple algebras Throughout this section R is a ring. Recall that all rings are associative and have an identity, denoted 1 (sometimes denoted 1R ). Most results will be written in terms of left modules (which we hence often will simply call modules). If we need to work with right modules then this will be specifically stated. In that case we will sometimes write N R to emphasize that N is a right R-module or R N if N is a left R-module. The endomorphism ring of an R-module M is denoted EndR (M). When we want to emphasize that M is a left (respectively, right) module then we write End(R M), (respectively End(M R )). Similarly, we will use HomR (M, N), Hom(R M, R N) or Hom(M R , N R ) to denote the set of module homomorphism from M to N. Throughout this section M is a left R-module. One says that M is simple if M ≠ 0 and M is the only non-zero submodule of M. The module M is said to be semisimple if it is the sum of simple submodules. The zero module is a semisimple module because the empty sum of modules is interpreted as the zero module. Lemma 2.1.1 (Schur’s Lemma). If M is a simple module then EndR (M) is a division ring. Proof. If 0 ≠ f ∈ EndR (M) then M ≠ ker(f) and 0 ≠ Im(f). Thus, because M is simple, ker(f) = 0 and Im(f) = M. So, f is an isomorphism and hence is invertible in EndR (M). Lemma 2.1.2. If M is a semisimple module then the following properties hold. (1) M is a direct sum of simple modules. (2) Every submodule of M is semisimple and a direct summand of M. (3) Every quotient of M is semisimple. Proof. Let M = ∑i∈I S i , with S i simple for every i and let N be a submodule of M. Let X = {J ⊆ I : N + ∑ S j = N ⊕ (⊕j∈J S j )}. j∈J

Clearly, X is closed under taking unions of totally ordered subsets. Hence, by Zorn’s Lemma, X has a maximal subset, say J. We leave it to the reader to prove that M = N ⊕ (⊕j∈J S j ). The result now follows easily.

2.1 Semisimple algebras | 33

A ring R is said to be semisimple if it is semisimple as left R-module. We denote by R o the opposite ring of R, that is, as an additive abelian group R o = R and the product ⋅ in R o is defined as a ⋅ b = ba, for a, b ∈ R; here the product ba is taken in the ring R. Clearly, right (respectively, left) multiplication yields an isomorphism R o ≅ End(R R) (respectively, R ≅ End(R R )). Theorem 2.1.3 (Wedderburn-Artin Theorem). A ring is semisimple if and only if it is isomorphic to a finite direct product of matrix rings over division rings. Proof. Assume that R = ∏ki=1 M n i (D i ), with D i a division ring. For i = 1, . . . , k and j = 1, . . . , n i let S ij denote the subset of R formed by the k-tuples in R having 0 everywhere except in the j-th column of the i-entry. Then S ij is a simple left R-module and R = ni S ij . Thus R is semisimple as left R-module. ∑ki=1 ∑j=1 Conversely, suppose that R is semisimple. By Lemma 2.1.2, R is a direct sum of simple R-modules and as R = R1, this direct sum is finite (see Problem 2.1.1). Hence, nj n n R ≅ ⊕ki=1 S i i , with each S i a simple R-module and S i ≇ S j if i ≠ j. Then HomR (S i i , S j ) = n

0, if i ≠ j (see Problem 2.1.2). Therefore R o ≅ EndR (R) ≅ ∏ki=1 EndR (S i i ) ≅ ∏ki=1 M n i (EndR (S i )). By the Schur Lemma (Lemma 2.1.1), each EndR (S i ) is a division ring D i . As M n i (D i )o ≅ M n i (D oi ) (Problem 2.1.3), the result follows. It follows from the right-left version of the previous result that a ring R is semisimple if and only if R is semisimple as right R-module. In the following theorem we state some other useful characterizations of semisimple rings. Theorem 2.1.4. The following conditions are equivalent for a ring R. (1) R is semisimple. (2) Every left (respectively, right) R-module is semisimple. (3) Every short exact sequence of left (respectively, right) R-modules splits. (4) Every left (respectively, right) R-module is projective. (5) Every left (respectively, right) R-module is injective. (6) R is Artinian and the Jacobson radical of R is 0. Recall that an R-module P is said to be projective if it is isomorphic to a direct summand of a free R-module. Equivalently, for every homomorphism f : P → N and every surjective homomorphism g : M → N, there is a homomorphism h : P → M such that gh = f . An R-module E is said to be injective if for every homomorphism f : N → E and every injective homomorphism g : N → M there is a homomorphism h : M → E such that hg = f . The Jacobson radical of the ring R is the ideal J(R) of R defined by any of the following equivalent descriptions: J(R) = Intersection of all maximal left ideals of R = Intersection of all maximal right ideals of R

34 | 2 Representations of algebras = {x ∈ R : 1 − xr ∈ U(R), for all r ∈ R} = {x ∈ R : 1 − rx ∈ U(R), for all r ∈ R} = {x ∈ R : 1 − rxs ∈ U(R), for all r, s ∈ R} The following theorem characterizes semisimple group rings. Theorem 2.1.5. Let R be a ring and G a group. The group ring RG is semisimple if and only if R is semisimple, G is finite and the order of G is invertible in R (i.e. |G|x = 1, for some x ∈ R). Proof. Assume first that RG is semisimple and consider R as a left RG-module via the trivial action: g ⋅ r = r, for every g ∈ G and r ∈ R. Then R is semisimple as RG-module by Theorem 2.1.4. Clearly, an additive subgroup of R is an R-submodule if and only if it is an RG-submodule. Hence R is semisimple. As the augmentation map ω : RG → R is a homomorphism of left RG-modules, and because R is projective as left RG-module, there is a homomorphism f : R → RG such that ωf = 1R . Then gf(1) = f(g ⋅ 1) = f(1), for every g ∈ G. This implies that all the coefficients of f(1) are equal and, as f(1) ≠ 0, G is finite and f(1) = ∑g∈G ag for some a ∈ R. Thus 1 = (ωf)(1) = a|G|. Therefore |G| is invertible in R. Conversely, assume that R is semisimple, G is finite and |G| is invertible in R. Because of Theorem 2.1.4, to prove that RG is semisimple it is enough to show that if f : M → N is an injective homomorphism of left RG-modules then f is a split monomorphism. Considering f as a homomorphism of R-modules, the hypothesis that R is semisimple implies that there is a homomorphism of R-modules p : N → M such ̄ that pf = 1M . Let p̄ : N → M be given by p(n) = |G|−1 ∑g∈G g−1 p(gn). Then p̄ is a homomorphism of RG-modules and pf̄ = 1M , as desired. Let R be a semisimple ring. From Theorem 2.1.3 we know that R = B1 × ⋅ ⋅ ⋅ × B k , where each B i is isomorphic to M n i (D i ), for some positive integer n i and some division ring D i . This expression of R is called the Wedderburn decomposition of R and the factors B1 , . . . , B n are called the simple components or the Wedderburn components of R. Each Wedderburn component B i is a simple ring. Therefore the identity e i = 1B i of B i is the unique non-zero central idempotent of B i . Moreover, B i has a unique simple left n B i -module S i , up to isomorphism, B i ≅ S i i both as B i -modules and as R-modules and B i ≅ M n i (EndR (S i )). The idempotents e1 , . . . , e n are called the primitive central idempotents of R. They can be intrinsically described as the non-zero central idempotents which cannot be written as the sum of two non-zero orthogonal central idempotents, or as the central idempotents e such that Re is a simple ring. Every simple R-module of R is isomorphic to the unique simple module of exactly one Wedderburn component of R. Thus there are one-to-one correspondences between isomorphism classes of simple R-modules S i , Wedderburn components B i of R and primitive central idempotents e i of R.

2.1 Semisimple algebras |

35

Recall that if R is a ring or a group then its center is Z(R) = {x ∈ R : xy = yx, for all y ∈ R}. Let F be a field and let A be an F-algebra. Then F1A is a subfield of Z(A) which is isomorphic to F. When there is no danger of confusion, F will be identified canonically with F1A . If F1A = Z(A) then one says that A is a central F-algebra. If, moreover, A is a simple algebra, then one says that A is a central simple F-algebra. A semisimple F-algebra is an F-algebra which is semisimple as a ring. If A is such an algebra then every Wedderburn component B of A is a simple F-algebra and, in particular, the center of B is a field extension of F. Lemma 2.1.6. If A is a finite dimensional algebra over F, then the following conditions are equivalent. (1) A is semisimple. (2) A is a direct product of simple algebras. (3) J(A) = 0. In particular, if A is a simple finite dimensional algebra then it is semisimple. Proof. (1) implies (2) is a consequence of the Wedderburn-Artin Theorem (Theorem 2.1.3); (2) implies (3) is a consequence of the fact that J(A × B) = J(A) × J(B); (3) implies (1) is a consequence of Theorem 2.1.4. Example 2.1.7. Let F be a field of characteristic different from 2. An F-algebra A is said to be a quaternion algebra over F if there exist a, b ∈ U(F) and an F-basis {1, i, j, k} that is subjected to the relations i2 = a,

j2 = b,

k = ij = −ji.

This algebra is denoted ( a,b F ). It is called the quaternion algebra over F defined by = ℍ(ℝ), the classical Hamilton quaternion algebra. a, b ∈ U(F). Clearly, ( −1,−1 ) ℝ The following hold for a quaternion algebra A = ( a,b F ). (1) A is a central simple F-algebra. (2) The map x = x0 + x1 i + x2 j + x3 k 󳨃→ x = x0 − x1 i − x2 j − x3 k (with x0 , x1 , x2 , x3 ∈ F) defines an involution on A, called the quaternion conjugation. Moreover, the map N : A 󳨀→ F defined by x = x0 + x1 i + x2 j + x3 k 󳨃→ xx = x20 − ax21 − bx22 + abx23 is multiplicative, that is, N(xy) = N(x)N(y) for every x, y ∈ A. (3) A is either a division algebra or isomorphic to M 2 (F) and the following conditions are equivalent:

36 | 2 Representations of algebras (a) A ≅ M2 (F), (b) N(x) = 0 for some 0 ≠ x ∈ A, (c) u2 = av2 + bw2 for some (u, v, w) ∈ F 3 \ {(0, 0, 0)}. Proof. (1) and (2) are straightforward. (3) As A is simple and finite dimensional over F, it is semisimple, by Lemma 2.1.6. Therefore A = M n (D) for a division F-algebra D. As 4 = n2 dimF (D), either n = 1 and A is a division algebra or n = 2 and D ≅ F. This proves the first statement. (a) implies (b). If A ≅ M 2 (F) then x2 = 0 for some 0 ≠ x ∈ A. Then N(x)2 = N(x2 ) = 0 and so N(x) = 0. (b) implies (c). Assume that 0 = N(x) with x = x0 + x1 i + x2 j + x3 k ≠ 0. Then x20 − bx22 = a(x21 − bx23 ) and hence a(x21 − bx23 )2 = (x20 − bx22 )(x21 − bx23 ) = (x0 x1 + bx2 x3 )2 − b(x0 x3 + x1 x2 )2 . We have to show that u2 = av2 + bw2 for some (u, v, w) ∈ F 3 \ {(0, 0, 0)}. If x3 = 0 then we take (u, v, w) = (x0 , x1 , x2 ); if x21 − bx23 ≠ 0 then we take (u, v, w) = (x0 x1 + bx2 x3 , x21 −bx23 , x0 x3 +x1 x2 ). Finally, if x3 ≠ 0 and x21 −bx23 = 0 then we take (u, v, w) = (x1 , 0, x3 ). (c) implies (a). If (u, v, w) satisfies the conditions of (c) and x = u + vi + wj then xx = N(x) = u2 − av2 − bw2 = 0 and x and x are non-zero elements of A. Thus A is not a division algebra. Then A ≅ M2 (F). If A and B are two arbitrary F-algebras then the tensor product A ⊗F B is an F-algebra with product given by (a1 ⊗ b1 )(a2 ⊗ b2 ) = (a1 a2 ) ⊗ (b1 b2 ). The maps a 󳨃→ a ⊗ 1 and b 󳨃→ 1 ⊗ b are injective ring homomorphisms f A : A → A ⊗F B and f B : B → A ⊗F B and the elements of f A (A) and f B (B) commute. The algebra A ⊗F B, together with the maps f A and f B , can be characterized, up to isomorphism, by the Universal Property of Tensor Products: for any two algebra homomorphisms g A : A → C and g B : B → C such that the elements of g A (A) and g B (B) commute there is a unique algebra homomorphism h : A ⊗F B → C such that hf A = g A and hf B = g B . Making use of this characterization one easily can prove the following natural isomorphisms: M n (A) ⊗F B ≅ M n (A ⊗F B), A[X] ⊗F B ≅ (A ⊗F B)[X], where A[X] denotes the ring of polynomials in one variable with coefficients in A. The algebras A and B can be considered as subalgebras of A ⊗F B by identifying a ∈ A with a ⊗ 1 and b ∈ B with 1 ⊗ b. Because the tensor product commutes with direct sums, one also has that if a1 , . . . , a n ∈ A are F-linearly independent then, as elements of A ⊗F B, they are linearly independent over B. The following proposition collects properties of the tensor product with one of the factors a simple or central simple algebra.

2.1 Semisimple algebras | 37

Proposition 2.1.8. Let F be a field, A a central F-algebra and B an arbitrary F-algebra. The following properties hold. (1) Z(A ⊗F B) = Z(B). In particular, if B is a central F-algebra then so is A ⊗F B. (2) If A is simple then every ideal of A ⊗F B is of the form A ⊗F I, for some ideal I of B. In particular, if A and B are simple then so is A ⊗F B. Proof. (1) Clearly Z(A ⊗F B) ∩ B = Z(B) ⊆ Z(A ⊗F B). For the reversed inclusion, let 0 ≠ x ∈ Z(A ⊗F B). Write x = ∑ti=1 a i ⊗ b i ∈ Z(A ⊗F B), with t minimal. Then b1 , . . . , b t are F-linearly independent and therefore they are linearly independent over A. For every a ∈ A, we have t

t

∑ a i a ⊗ b i = x(a ⊗ 1) = (a ⊗ 1)x = ∑ aa i ⊗ b i . i=1

i=1

As b1 , . . . , b t are linearly independent over A, we obtain a i a = aa i for every i. Therefore a i ∈ Z(A) = F. Hence, x = 1 ⊗ ∑ti=1 a i b i ∈ B. (2) Let J be a non-zero ideal of A ⊗F B. Put I = J ∩ B. We first show that I ≠ 0. Let 0 ≠ x = ∑ti=1 a i ⊗ b i ∈ J with t minimal. As in the proof of (1), it follows also that b1 , . . . , b t are linearly independent over A. Because, by assumption A is simple, and a1 ≠ 0, we have ∑kj=1 x j a1 y j = 1 for some x j , y j ∈ A. Then k

k

t

t

k

t

∑ (x j ⊗ 1)x(y j ⊗ 1) = ∑ ∑ x j a i x j ⊗ b i = ∑ ( ∑ x j a i y j ) ⊗ b i = ∑ a󸀠i ⊗ b i j=1

j=1 i=1

i=1 j=1

i=1

with a󸀠1 = 1. Then x󸀠 = ∑ti=1 a󸀠i ⊗ b i is a non-zero element of J. Now, for any a ∈ A. t

(a ⊗ 1)x󸀠 − x󸀠 (a ⊗ 1) = ∑ (aa󸀠i − a󸀠i a) ⊗ b i ∈ J. i=2

By the minimality of t, this element is 0. Hence, because of the A-independence of b2 , . . . , b t , we get that aa󸀠i = a󸀠i a for all i. Consequently, all a󸀠i ∈ Z(A). Since, by assumption Z(A) = F, we thus have that all a i ∈ F. Therefore, x󸀠 = 1 ⊗ ∑ti=1 a󸀠i b i ∈ I. So we indeed have shown that I = J ∩ B ≠ 0. Clearly A ⊗ I ⊆ J and if the equality does not hold then the image J of J in (A ⊗F B)/(A ⊗F I) ≅ A ⊗F (B/I) is a non-zero ideal. Applying the previous paragraph in this case, we deduce that 0 ≠ J ∩ (B/I) = (J ∩ B)/I = 0, a contradiction. Theorem 2.1.9 (Noether-Skolem Theorem). Let F be a field. Assume A is a finite dimensional central simple F-algebra and B is a simple subalgebra of A. If f : B → A is an F-algebra homomorphism then there is u ∈ U(A) such that f(b) = ubu−1 for every b ∈ B. Proof. Let f : B → A be an F-algebra homomorphism and set C = A o ⊗F B. We define a left C-module A f as follows. As additive groups A f = A and the action of a ⊗ b ∈ C on a1 ∈ A f is (a ⊗ b)a1 = f(b)a1 a. By Proposition 2.1.8 (2), C is a simple and finite dimensional F-algebra. Therefore, every two left C-modules of the same F-dimension

38 | 2 Representations of algebras are isomorphic. Hence the C-module A f is isomorphic to A i , where i : B → A is the inclusion map. Let ϕ : A i → A f be an isomorphism of C-modules and set u = ϕ(1). Then, for some a ∈ A, 1 = ϕ(a) = ϕ((a ⊗ 1)1) = (a ⊗ 1)u = ua. Thus u ∈ U(A). Moreover, for every b ∈ B, we have ub = (b ⊗ 1)u = ϕ((b ⊗ 1)1) = ϕ(b) = ϕ((1 ⊗ b)1) = (1 ⊗ b)u = f(b)u. Hence the result follows. For a subset X of a ring or a group R we denote by CenR (X), the centralizer of X in R, i.e. CenR (X) = {r ∈ R : rx = xr for all x ∈ X}. Theorem 2.1.10 (Double Centralizer Theorem). Let A be a finite dimensional central simple F-algebra and let B be a simple subalgebra of A. The following properties hold. (1) The centralizer CenA (B) is a simple F-algebra, dimF (B) dimF (CenA (B)) = dimF (A) and CenA (CenA (B)) = B. (2) If B also is central as F-algebra then so is CenA (B) and A ≅ B ⊗F CenA (B). Proof. (1) Because of Proposition 2.1.8 (2), C = B o ⊗F A is simple and, as it also is finite dimensional over F, it is semisimple (Lemma 2.1.6). Let S be the unique, up to isomorphism, simple left C-module and let D = EndC (S). Then C ≅ M n (D), for some positive integer n. Furthermore, dimF (S) = n dimF (D) and dimF (C) = n2 dimF (D). Consider A as a C-module for the product (b ⊗ a) ⋅ a1 = aa1 b, where a, a1 ∈ A and b ∈ B. As left C-modules, A ≅ S k for some positive integer k. Hence, n dimF (S) = dimF (C) = dimF (B) dimF (A) = dimF (B)k dimF (S) and therefore n = k dimF (B). On the other hand, C ≅ M n (D), EndC (A) ≅ M k (D) and the natural isomorphism A ≅ EndA (A) restricts to an isomorphism CenA (B) ≅ EndC (A). This implies that CenA (B) is simple and dimF (B) dimF (CenA (B)) = nk dimF (D) = k dimF (S) = dimF (A), as desired. Applying this to the simple F-subalgebra CenA (B) we obtain dimF (CenA (B)) dimF (CenA (CenA (B))) = dimF (A) = dimF (B) dimF (CenA (B)). Thus dimF (CenA (CenA (B)) = dimF (B). As B ⊆ CenA (CenA (B)), we deduce that B = CenA (CenA (B)). (2) Assume B also is central as F-algebra. Then, by part (1), Z(CenA (B)) = CenA (B) ∩ CenA (CenA (B)) = CenA (B) ∩ B = Z(B) = F. Thus, CenA (B) is central as F-algebra. From the Universal Property of Tensor Products, there is a homomorphism of F-algebras B ⊗F CenA (B) → A. This is an isomorphism because B ⊗F CenA (B) is simple, by Proposition 2.1.8 (2), with the same F-dimension as A. The following proposition describes some behavior of tensor product of semisimple algebras. Recall that if E1 and E2 are subfields of a common field L then the compositum of E1 and E2 in L is the smallest subfield of L containing both E1 and E2 . This field is denoted E1 E2 . A separable F-algebra is a finite dimensional semisimple F-algebra A such that Z(B)/F is a separable field extension for every Wedderburn component B of A.

2.1 Semisimple algebras | 39

Recall that if A is a finite dimensional algebra over a field F then the minimal polynomial of a ∈ A over F is the monic polynomial f of minimal degree for which f(a) = 0, i.e. the monic polynomial of F[X] generating {f ∈ F[X] : f(a) = 0}. Proposition 2.1.11. Let F be a field and A a finite dimensional F-algebra. (1) The following conditions are equivalent. (a) A is separable as F-algebra. (b) A ⊗F E is semisimple for every finite field extension E of F. (c) A⊗F B is semisimple for every semisimple F-algebra B which is finitely generated as module over its center. (2) If A is simple and separable over F and B is a finite dimensional simple F-algebra such that Z(B) is a normal field extension of F then all the Wedderburn components of A ⊗F B are isomorphic as F-algebras. Proof. We first prove (1a) implies (1c) of statement (1) and, at the same time, we prove (2). As the tensor product commutes with direct sums, we may assume without loss of generality that A and B are simple. Let E = Z(A) and L = Z(B). We consider several cases of increasing generality. Assume first that A = E and B = L. By assumption, E/F is a finite separable extension and therefore, by the Primitive Element Theorem, E = F(α) for some α ∈ E. Let f = MinF (α), the minimal polynomial of α over F. Then E ≅ F[X]/(f) and A ⊗F B ≅ F[X]/(f) ⊗F L ≅ (F[X] ⊗F L)/(f ⊗ 1) ≅ L[X]/(f). As E/F is separable, all roots of f in an algebraic closure have multiplicity one and thus we obtain that f = f1 ⋅ ⋅ ⋅ f k , for f1 , . . . , f k coprime irreducible elements of L[X]. By the Chinese Remainder Theorem, A ⊗F B ≅ L[X]/(f) ≅ ∏ki=1 L[X]/(f i ), a product of fields. So (1c) follows in this case. Assume, furthermore that L/F is a normal field extension. Let E and L be algebraic closures of E and L respectively. Let α i ∈ L be a root of f i and set K i = L(α i ). By the previous paragraph, E ⊗F L ≅ ∏ki=1 K i and we need to prove that all the K i ’s are Fisomorphic. For every i = 1, . . . , k, there is an F-isomorphism τ i : E → F(α i ) such that τ i (α) = α i . Since E and L are both algebraic closures of F they are isomorphic. Each τ i extends to an isomorphism E → L, which we keep denoting by τ i . Then σ i = τ i ∘ τ−1 1 ∈ Gal(E/F) and σ i (F(α 1 )) = F(α i ). Because, by assumption L/F is a normal extension, it follows that σ i (K1 ) = σ i (L(α1 )) = L(α i ) = K i . Thus σ i restricts to an Fisomorphism K1 → K i . This proves (2) in this case. Observe that K i is the compositum of τ i (E) and L in F. Using this, we can consider K i as an (E, L)-bimodule with product a ⋅ x ⋅ b = τ i (a)xσ i (b) for a ∈ E, b ∈ L and x ∈ K i . This defines a structure of E-algebra and of L-algebra in K i . Clearly σ i : K1 → K i is an isomorphism of (E, L)-bimodules and hence it is an isomorphism of E-algebras and of L-algebras. Second, assume that B = L. By the previous case E⊗F L ≅ ∏ki=1 K i , with K1 , . . . , K k fields. Then A⊗F B = A⊗E E⊗F L ≅ A⊗E (∏ki=1 K i ) ≅ ∏ki=1 (A⊗E K i ). As A is central simple E-algebra, the algebra A ⊗E K i is a central simple K i -algebra, by Proposition 2.1.8. It also is finite dimensional over K i , because A is finite dimensional over F. Thus each

40 | 2 Representations of algebras A ⊗E K i is semisimple, by Lemma 2.1.6, and we deduce that A ⊗F L is a semisimple L-algebra. If, additionally, L/F is normal then, by the previous case, all the K i are isomorphic as E-algebras and therefore all the C i = A ⊗E K i are isomorphic. Furthermore, the Lalgebra structure of K i induces a structure of L-algebra on C i and the isomorphism C1 → C i is an isomorphism of L-algebras. We now prove the general case. By the previous case A ⊗F L ≅ ∏ki=1 C i , with each C i a finite dimensional simple L-algebra. Clearly, A⊗F B = A⊗F L⊗L B = ∏ki=1 C i ⊗L B. As B is a finite dimensional central simple L-algebra, each C i ⊗L B is a finite dimensional simple L-algebra, by statement (2) of Proposition 2.1.8. So, by Lemma 2.1.6, each C i ⊗L B is semisimple. If, additionally, L/F is normal then all C i ’s are L-isomorphic and therefore all the C i ⊗L B are isomorphic. (1c) implies (1b) is obvious. (1b) implies (1a) We may assume that A is simple. Assume that A is not separable over F. Then the center K of A contains an element α that is not separable over F. Let f be the minimal polynomial of α over F. Then f = (X − α)m g for some g ∈ K[X] with g(α) ≠ 0 and m ≥ 2. Hence, F(α) ⊗F K ≅ (F[X]/(f)) ⊗F K ≅ K[X]/(f) ≅ K[X]/(X − α)m ⊗K K[X]/(g). As X − α + (X − α)m is a non-zero nilpotent element of K[X]/(X − α)m , it follows that α ⊗ 1 − 1 ⊗ α is a non-zero nilpotent element in F(α) ⊗F K. As this element is central in F(α) ⊗ A we obtain that F(α) ⊗ A is not semisimple. Example 2.1.12. The proof of statement 1 of Proposition 2.1.11 indicates how to calculate the Wedderburn decomposition of A ⊗F B, under the assumption that A is a separable F-algebra. We emphasize here the cases where B is a field L and A is a field or, more general, a simple algebra. Assume A is a field E. Then E = F(α) for some α ∈ E. We first calculate f = MinF (α) and then calculate the expression of f = g1 . . . g k as a product of monic irreducible polynomials with coefficients in L. Then E ⊗F L ≅ L[X]/(g1 ) × ⋅ ⋅ ⋅ × L[X]/(g k ). Observe that g1 , . . . , g k are relatively prime in L[X], because f is separable. Moreover, if α i is a root of g i in an algebraic closure L of L then L[X]/(g i ) = L(α i ). On the other hand for every i = 1, . . . , k there is an F-homomorphism σ i : E → L determined by σ i (α) = α i . Therefore, E⊗F L ≅ ⊕ki=1 σ i (E)L, where σ i (E)L denotes the compositum of σ i (E) and L in L. The composition of this isomorphism with the projection onto the i-th component is given by a ⊗ b 󳨃→ σ i (a)b. Assume now that A is simple with center E and B is still the field L. Then we calculate first E ⊗F L = ∏ki=1 σ i (E)L. The i-th component of this decomposition is considered as an E-algebra via the homomorphism σ i : E → σ i (E)L. Then A ⊗E σ i (E)L is a central simple σ i (E)L-algebra in which ax ⊗ b = a ⊗ σ i (x)b, for every a ∈ A, x ∈ E and b ∈ L.

2.1 Semisimple algebras | 41

A complex embedding of a field F is a homomorphism F → ℂ. The complex embeddings of F with image contained in ℝ are called real embeddings of F. If σ is a complex embedding of the field F then σ denotes the composition of σ with complex conjugation. A set of representatives of complex embeddings modulo conjugation of F is a set X of complex embeddings of F such that every complex embedding is either equal or conjugate to exactly one element of X. Example 2.1.13. Let F be a number field. Let σ1 , . . . , σ r be the real embeddings of F and let σ r+1 , . . . , σ r+s be a set of representatives of the complex non-real embeddings of F. Then, by the method explained in Example 2.1.12 these embeddings yield an ℝalgebra isomorphism F ⊗ℚ ℝ ≅ ℝ r × ℂ s and the composition of this isomorphism with the projection onto the i-th component is given by a ⊗ b 󳨃→ σ i (a)b, for a ∈ F and b ∈ ℝ. More generally, if A is a finite dimensional simple rational algebra with center F, then F is a number field. Then A ⊗ℚ ℝ ≅ ∏r+s i=1 A ⊗F L i , where L i = ℝ if i ≤ r and L i = ℂ otherwise. Moreover, L i is considered as an F-algebra via σ i . That is in A ⊗F L i , we have ax ⊗ b = a ⊗ σ i (x)b, for a ∈ A, x ∈ F and b ∈ L. If i > r then A ⊗F L i ≅ M n (ℂ) for some positive integer n. If i ≤ r then A ⊗F L i is isomorphic to either M n (ℝ) or M n (ℍ(ℝ)) (See Problem 2.2.1.) Example 2.1.14. As mentioned earlier, if E1 and E2 are subfields of a common field L then the compositum of E1 and E2 in L is the smallest subfield of L containing both E1 and E2 . Let F be a field and let E1 , E2 and E be field extensions of F. One says that E is a compositum of E1 and E2 over F if there are F-homomorphisms σ1 : E1 → E and σ2 : E2 → E such that E is the compositum of σ1 (E1 ) and σ2 (E2 ) in E. In that case, by the Universal Property of the Tensor Products, there is a surjective homomorphism of F-algebras σ : E1 ⊗F E2 → E. In particular, if E1 ⊗F E2 is semisimple, then E is Fisomorphic to one of the Wedderburn components of E1 ⊗F E2 and every Wedderburn component of E1 ⊗F E2 is a compositum of E1 and E2 over F. 3 2). Observe the relevance of the role of F. For example, take E1 = ℝ and E2 = ℚ(√ Then E1 ⊗ℚ E2 ≅ ℝ×ℂ and E1 ⊗E2 E2 ≅ ℝ. Thus every compositum of E1 and E2 over E2 is E2 -isomorphic to ℝ. However, one compositum of E1 and E2 over ℚ is isomorphic to ℝ and another is isomorphic to ℂ.

Problems 2.1.1. Let M = ⊕i∈I M i , with M a finitely generated module. Prove that M i = 0 for all but finitely many i ∈ I.

42 | 2 Representations of algebras 2.1.2. Let M = ⊕i∈I S i and N = ⊕i∈J T j with S i and T j simple R-modules for every i and j. Prove that if HomR (M, N) ≠ 0 then S i ≅ T j for some i ∈ I and j ∈ J. 2.1.3. Let R be a ring and n ∈ ℕ. Prove M n (R o ) ≅ M n (R)o . 2.1.4. Prove Theorem 2.1.4. 2.1.5. Prove that if A is a finite dimensional simple F-algebra then Z(A) is a field. 2.1.6. Let F be a field of characteristic different from 2 and let a, b, x ∈ U(F). Prove 2 that the quaternion algebras ( axF ,b ) and ( a,b F ) are isomorphic. 2.1.7. Prove that if D and D󸀠 are division F-algebras and n and n󸀠 are positive integers such that M n (D) ≅ M n󸀠 (D󸀠 ) as F-algebras then n = n󸀠 and D and D󸀠 are isomorphic as F-algebras. 2.1.8. Let F be a field containing a primitive n-th root of unity ζ n . Let a, b ∈ U(F). Prove that the F-algebra given by the following presentation (

a, b ) = F[x, y | x n = a, y n = b, yx = ζ n xy] F n

is a central simple F-algebra of dimension n2 . (These algebras are called symbol Falgebras.) a,bc n Prove that if c ∈ U(F) then ( a,b F )n ≅ ( F )n . 2.1.9. Prove that if G and H are groups and R is a commutative ring then R(G × H) ≅ RG ⊗R RH. 2.1.10. Let F be a field of characteristic p and let G be a finite group having a normal Sylow p-subgroup P. Prove J(FG) = ker(ω P ). Hint: argue by induction on |P| to prove that ker(ω P ) is nilpotent.

2.2 Splitting fields Let E/F be a field extension. If A is an F-algebra, then E ⊗F A is an E-algebra and if M is a left A-module then E ⊗F M is a left E-module. One says that a semisimple F-algebra A is split if every simple component of A is isomorphic, as F-algebra, to a matrix algebra over F. Equivalently, A is split if EndA (M) is isomorphic to F as F-algebra, for every simple A-module M. If E/F is a field extension and either E or A is separable over F then A ⊗F E is semisimple. This follows from Proposition 2.1.11 (1); in the first case one uses the equivalent condition (c) and in the second case one uses the equivalent condition (b). If, moreover, A ⊗F E is split as E-algebra then one says that E splits A or that E is a splitting field of A as F-algebra. For example, if E is algebraically closed then every finite dimensional semisimple E-algebra is split and therefore, E splits every finite dimensional separable F-algebra.

2.2 Splitting fields |

43

Lemma 2.2.1. If A is a separable F-algebra then A is split by some finite field extension of F. Proof. Let A be a separable F-algebra. We know that an algebraic closure F of F splits A. Hence there is an isomorphism f : A ⊗F F ≅ ∏ki=1 M n i (F), for some positive integers n1 , . . . , n k . Since A is finite dimensional over F, F contains a finite field extension E of F so that f(A) ⊆ ∏ki=1 M n i (E). Then f restricts to an injective homomorphism A ⊗F E → ∏ki=1 M n i (E). This map is an isomorphism, because dimE (A ⊗F E) = dimF (A) = dimF (A ⊗F F) = ∑ki=1 n2i = dimE (∏ki=1 M n i (E)). So, E is a splitting field of A. In Theorem 2.2.5 we will show that every separable F-algebra is split by a finite Galois extension of F. Observe that if A = A1 × ⋅ ⋅ ⋅ × A k is the Wedderburn decomposition of a separable F-algebra A and E is an extension of F then E splits A if and only if E splits every A i . Thus to study splitting fields one can concentrate on the case of simple algebras. Moreover, if A is simple and A = M n (D), with D a division algebra, then E splits A if and only if E splits D. If A is a finite dimensional central simple F-algebra and E is a splitting field of A then, by Proposition 2.1.8, A ⊗F E ≅ M d (E) for some positive integer d. Therefore dimF (A) = dimE (A ⊗F E) = d2 . This shows that the dimension of a finite dimensional central simple algebra is a square. The positive integer √dimF (A) is called the degree of A, and it is denoted by Deg(A). Moreover, A is semisimple and hence A = M n (D), for some n ≥ 1 and some finite dimensional division F-algebra D. The division algebra D is unique, up to isomorphisms of F-algebras (see Problem 2.1.7). The index of A, denoted Ind(A), is by definition Deg(D). The reduced degree of A is n = Deg(A)/Ind(A). The following proposition characterizes splitting fields of finite dimensional central simple algebras. Proposition 2.2.2. Let A be a finite dimensional central simple F-algebra of degree d and E/F a field extension. Then the following conditions are equivalent. (1) E splits A. (2) There is an F-algebra homomorphism A → M d (E). (3) There is an F-algebra homomorphism f : A → M m (E), for some positive integer m such that f(A)E = M m (E). Proof. (1) implies (2) is obvious. To prove the other implications, suppose f : A → M m (E) is a homomorphism of F-algebras. By the Universal Property of Tensor Products, f extends to a homomorphism of E-algebras f ⊗ 1 : A ⊗F E → M m (E). Because of Proposition 2.1.8, the algebra A ⊗F E is simple. Hence f ⊗ 1 is injective. As dimE (A ⊗F E) = d2 it follows that d ≤ m. If d = m then f is surjective because dimE (A⊗F E) = d2 = dimE (M d (E)). This proves (2) implies (3). Finally, if f(A)E = M m (E) then f ⊗ 1 is surjective and m2 = dimE (M m (E)) ≤ dimE (A ⊗F E) = d2 . Hence m = d and therefore f is an isomorphism. This proves (3) implies (1).

44 | 2 Representations of algebras

Example 2.2.3. Let F be a field of characteristic different from 2. Every quaternion ala,b √ gebra ( a,b F ) has degree 2 and F(√a) and F( b) are splitting fields of ( F ). Conversely, if A is a central simple F-algebra of degree 2 then A is a quaternion algebra over F. Proof. Clearly, every quaternion algebra has degree 2. If E is a field extension of F then a,b 2 2 2 ( a,b F ) ⊗F E = ( E ). Since the equation ax + by = z has a non-trivial solution in both √ F(√a) and F( b), it follows from Example 2.1.7 that both fields are splitting fields of ( a,b F ). Let A be a central simple F-algebra of degree 2. If A is split then A ≅ M2 (F) ≅ ( 1,1 F ). Otherwise A is a division algebra. Let α ∈ A \ F. Then E = F(α) is a subfield of A and 4 = dimF (A) = [E : F] dimE (A). This implies that [E : F] = 2. Hence E = F(i) for some i ∈ A with i2 ∈ F. Clearly E is a maximal subfield of A. Therefore E = CenA (E) = CenA (i). Set H = {x ∈ A : xi = −ix}. If x ∈ A then let x+ =

x + ixi−1 2

and

x− =

x − ixi−1 . 2

Clearly x = x+ + x− , x+ ∈ E and x− ∈ H. This shows that A = E ⊕ H. In particular, H ≠ 0. If 0 ≠ β ∈ H then [F(β) : F] = 2 and F(β)∩E = F. Thus F(β) = F(j) for some j ∈ A\E with j2 ∈ F. Moreover, j = u + vβ for some u, v ∈ F with v ≠ 0. Hence iji−1 = u − vβ ∈ F(β). Then iji−1 ≠ j and (iji−1 )2 = ij2 i−1 = j2 . This implies that iji−1 = −j, i.e. j ∈ H. Thus the map x 󳨃→ xj defines a bijection E → H. We conclude that A = E ⊕ Ej = F ⊕ Fi ⊕ Fj ⊕ Fij 2 2 and A ≅ ( a,b F ), with a = i and b = j , as desired. Two finite dimensional central simple F-algebras A and B are said to be equivalent, or Brauer equivalent, if A ≅ M n (D) and B ≅ M m (D) for some division algebra D and some positive integers n and m, or equivalently, if M m (A) ≅ M n (B) for some n and m (see Problem 2.1.7). Proposition 2.2.4. Let A be a finite dimensional central simple F-algebra and let E be a subfield of A containing F. The following properties hold. (1) CenA (E) is a central simple E-algebra and Deg(A) = [E : F] Deg(CenA (E)). (2) If [E : F] = Deg(A) then E is a maximal subfield of A. (3) CenA (E) is Brauer equivalent to A ⊗F E. (4) If E is a maximal subfield of A then CenA (E) and A ⊗F E are split as E-algebras. (5) If A is a division algebra then E is a maximal subfield of A if and only if [E : F] = Deg(A). Proof. Put C = CenA (E). (1) From the Double Centralizer Theorem (Theorem 2.1.10) we know that C is a simple E-algebra, dimF (A) = [E : F] dimF (C) and E ⊆ Z(C) = C ∩ CenC (C) = C ∩ E = E. Therefore CenA (E) is a central simple E-algebra and dimF (C) = [E : F] dimE (C) = [E : F]Deg(C)2 . Hence, Deg(A) = [E : F]Deg(C). (2) is a direct consequence of (1).

2.2 Splitting fields | 45

(3) Consider A as a left A ⊗F E-module for the product (e ⊗ a)a1 = aa1 e. Then C o ≅ EndA⊗F E (A). Let M be a simple A ⊗F E-module. Because of the Schur Lemma (Lemma 2.1.1), D = EndA⊗F E (M) is a division algebra. Since A⊗F E is a simple E-algebra, we get that, for some positive integers n and m, A ≅ M n and A ⊗F E ≅ M m , as A ⊗F Emodules. Therefore, C o ≅ EndA⊗F E (A) ≅ M n (D) and (A ⊗F E)o ≅ EndA⊗F E (A ⊗F E) ≅ M m (D). Thus A ⊗F E and C are Brauer equivalent. (4) Assume E is a maximal subfield of A. Because C is a central simple E-algebra, we know that C = M n (D), with D a division algebra containing E. Because E(α) is a field containing E for every α ∈ D and since, by assumption, E is a maximal subfield of A, we get that D = E. Therefore, C is split as E-algebra and hence so is A ⊗F E by (3). (5) One implication is an immediate consequence of (2). Conversely, assume A is a division algebra and E is maximal subfield of A. Then, by (4), C ≅ M k (E) for some positive integer k. Because C is contained in the division algebra A, we must have k = 1. Thus C = E and therefore Deg(A) = [E : F], by (1). Theorem 2.2.5. Every separable F-algebra is split by a finite Galois extension of F. Proof. It is enough to prove the theorem for a separable division F-algebra D. We claim that we also may assume that D is central. Indeed, assume that the theorem holds for central division algebras. Let K be the center of D. As, by assumption D is separable as F-algebra, K/F is a separable extension. Let L be the normal closure of K/F. By part 2 of Proposition 2.1.11, D ⊗F L ≅ A n for some central simple L-algebra. By assumption, there is a finite Galois extension E of L which splits A over L. Then E/F is a finite Galois extension and D ⊗F E = D ⊗F L ⊗L E ≅ (A ⊗L E)n , a split E-algebra. This proves the claim. So assume that D is a finite dimensional central division separable F-algebra. Let K be a maximal separable extension of F contained in D and set D󸀠 = CenD (K). Then, by Proposition 2.2.4, D󸀠 is a central division K-algebra. We claim that D󸀠 = K. Suppose the contrary and let L be a maximal subfield of 󸀠 D containing K. Then L/K is a purely inseparable extension and hence [L : K] = p m , with p > 0 the characteristic of F and m ≥ 1. From Proposition 2.2.4 (5) we know that Deg(D󸀠 ) = [L : K] = p m . Let E be a splitting field of D󸀠 over K (Lemma 2.2.1) and fix an isomorphism of E-algebras ϕ : D󸀠 ⊗F E → M p m (E). Let α ∈ L \ K. The minimal k k polynomial of α over K is of the form X p − a p , for some a ∈ K and k > 0, and therefore k (ϕ(α) − aI p m )p = 0. Consequently, the trace of ϕ(α) − aI p m is zero. Hence ϕ(α) has trace zero too. As M p m (E) is generated by ϕ(D󸀠 ) as a vector space over E, it follows that tr(x) = 0 for every x ∈ M p m (E), a contradiction. This finishes the proof of the claim. Therefore K is a maximal subfield of D and hence, by Proposition 2.2.4 (4), K splits D. Let E be a normal closure of K over F. Then E/K is a finite Galois extension of F which splits D. This proves the theorem.

46 | 2 Representations of algebras

Problems 2.2.1. Prove that every finite dimensional division ℝ-algebra is isomorphic to either ℝ, ℂ or ℍ(ℝ) = ( −1,−1 ℝ ). 2.2.2. Let F be a field and a, b ∈ U(F). Prove that if ab is not a square in F then the 2 2 quaternion algebra ( a,b F ) is split if and only if the equation ax +by = 1 has a solution in F. 2.2.3. Let A be a quaternion F-algebra and E a subfield of A containing F-properly. Prove that E splits A. 2.2.4. Let A be a central simple F-algebra of prime degree and E a subfield of A containing F properly. Prove that E splits A. 2.2.5. Let F be a subfield of ℝ and A a central simple F-algebra of odd degree. Prove that ℝ splits F. 2.2.6. Let A be a separable F-algebra and let E and L be splitting fields of A. Let n 1 ≤ ⋅ ⋅ ⋅ ≤ n k and m1 ≤ ⋅ ⋅ ⋅ ≤ m l be the degrees of the different simple components of A ⊗F E and A ⊗F L respectively. Prove that k = l and n i = m i for every i. 2.2.7. Let A be a separable F-algebra and E a field extension of A. Prove that n is the degree of a simple component of A if and only if n is the degree of a simple component of A ⊗F E. n √a, an n-th 2.2.8. Let A = ( a,b F )n be a symbol algebra as in Problem 2.1.8. Let α = root of a in a field extension of F. Prove that there is an F-algebra homomorphism ρ : A → M n (F(α)) given by

0 ρ(a) = diag(α, ζ n α, ζ n2 α, . . . , ζ nn−1 α),

( ρ(b) = ( ( (b

1 0

1 0

..

.

..

.

) ) ) 1 0)

n and F[ √b] are splitting fields of A. Deduce that both F[ √a] n

2.3 Characteristic polynomial, trace and norm In this section F is a field and A is an arbitrary F-algebra. A representation of A is a homomorphism of F-algebras A → M n (F), for n a positive integer, called the degree of the representation. Two F-representations ρ and ρ󸀠 of A are said to be equivalent if they have the same degree, say n, and there is U ∈ GLn (F) such that Uρ(a) = ρ󸀠 (a)U, for every a ∈ A.

2.3 Characteristic polynomial, trace and norm

| 47

There is a one-to-one correspondence between equivalence classes of representations of A and isomorphism classes of left A-modules of finite dimension over F. Indeed, let M be a left A-module that is finite dimensional over F. Then the map ρ : A → EndF (M) a 󳨃→ ρ a given by ρ a (m) = am, for a ∈ A and m ∈ M, is a homomorphism of F-algebras. For every F-basis B of M and for every a ∈ A let ρ B (a) = Matrix associated to ρ a with respect to the basis B. Then ρ B is a representation of A of degree dimF (A), called the representation of A associated to M with respect to the basis B. If B󸀠 is another F-basis of M then ρ B and ρ B󸀠 are equivalent. Conversely, every representation of A of degree n is of the form ρ B for some A-module M of dimension n over F and some basis B of M over F. Moreover, if M and M 󸀠 are left A-modules, of finite dimension over F, and B is an F-basis of M and B󸀠 an F-basis of M 󸀠 then M and M 󸀠 are isomorphic as left A-modules if and only if ρ B and ρ B󸀠 are equivalent. For example, if A is finite dimensional over F then the left regular representation is the representation associated to the regular module A A with respect to some basis. Similarly, there is a one-to-one correspondence between isomorphism classes of right A-modules of finite dimension over F and representations. In this case, the matrices describing the representation, with respect to a basis, are the transposed of the ones used above. The right regular representation is then defined similarly. We will often simply say the regular representation if it is clear from the context whether we are considering left or right modules. Let M be a left A-module that is finite dimensional over F. For every a ∈ A, let ρ a be as above and set MinM/F (a) = Min(ρ a ), minimal polynomial of ρ a , CharM/F (a) = char(ρ a ), characteristic polynomial of ρ a , TrM/F (a) = tr(ρ a ), trace of ρ a , NrM/F (a) = det(ρ a ), determinant of ρ a . MinM/F (a), CharM/F (a), TrM/F (a) and NrM/F (a) are called respectively the minimal polynomial, characteristic polynomial, trace and norm of a with respect to M. Observe that Min M/F (a), CharM/F (a) ∈ F[X] and TrM/F (a), NrM/F (a) ∈ F and they can be calculated using the representation of M in any arbitrary basis over F. Furthermore, TrM/F (a) and NrM/F (a) are coefficients of CharM/F (a), up to a sign. More precisely, if

48 | 2 Representations of algebras n = dimF (M) then CharM/F (a) = X n − TrM/F (a)X n−1 + ⋅ ⋅ ⋅ + (−1)n NrM/F (a).

(2.3.1)

The following properties are easily checked, for a, b ∈ A, α ∈ F and M a left A-module that are finite dimensional over F, with n = dimF (M): TrM/F (a + b) = TrM/F (a) + TrM/F (b), TrM/F (αa) = αTrM/F (a), NrM/F (ab) = NrM/F (a)NrM/F (b), NrM/F (αa) = α n NrM/F (a),

(2.3.2)

TrM/F (1) = n, NrM/F (1) = 1, TrM/F (ab) = TrM/F (ba). If N is another left A-module of finite dimension over F then CharM⊕N/F (a) = CharM/F (a)CharN/F (a), TrM⊕N/F (a) = TrM/F (a) + TrN/F (a),

(2.3.3)

NrM⊕N/F (a) = NrM/F (a)NrN/F (a). Using the canonical form of ρ a , considered as an endomorphism of F-vector spaces, one can easily see that CharM/F (a) is a multiple of the minimal polynomial Min(ρ a ) of ρ a (this is the so called Cayley-Hamilton Theorem) and in fact MinF (ρ a ) and CharM/F (a) have the same irreducible factors in F[X] (see e.g. [145]). In particular, if A is finite dimensional over F then MinA/F (a), CharA/F (a), TrA/F (a) and NrA/F (a) denote, respectively, the minimal polynomial, the characteristic polynomial, the trace and the norm of a with respect to the regular left A-module. Since ρ : A → EndF (A) is injective, MinA/F (a) is the minimal polynomial of a over F. Example 2.3.1 (Separable field extensions). Let E/F be a finite separable field extension of degree n, let F be an algebraic closure of F. Consider E as a finite dimensional semisimple F-algebra. Let σ1 , . . . , σ n be all the distinct F-homomorphisms σ : E → F. Then, for a ∈ E, n

CharE/F (a) = ∏(X − σ i (a)) i=1

and, therefore, n

TrE/F (a) = ∑ σ i (a)

n

and NrE/F (a) = ∏ σ i (a),

i=1

the classical trace and norm for finite field extensions.

i=1

2.3 Characteristic polynomial, trace and norm

| 49

Proof. Let f = MinF (a). Then f is an irreducible polynomial of degree [F(a) : F] in F[X]. By assumption, the roots of f are the different values in {σ1 (a), . . . , σ n (a)}. Therefore CharE/F (a) = f k , for k = [E : F(a)], and the list σ1 (a), . . . , σ n (a) contains each root of f precisely k times. Hence the claim follows. Example 2.3.2 (Simple case). Let F be a field and suppose that A is a simple and finite dimensional F-algebra. Then, up to isomorphism, A has a unique simple left Amodule, say S, and hence the regular left A-module is isomorphic to S n , for some n. Therefore, CharA/F (a) = CharS/F (a)n ,

TrA/F (a) = nTrS/F (a) and

NrA/F (a) = NrS/F (a)n .

Assume, additionally, that A = M n (F) and identify S with the column vectors of dimension n with entries in F. Then the matrix associated to ρ a in the standard basis is precisely a and therefore CharS/F (a) = char(a),

TrS/F (a) = tr(a)

and

NrS/F (a) = det(a).

Hence CharA/F (a) = char(a)n ,

TrA/F (a) = ntr(a)

and

NrA/F (a) = det(a)n .

Let E/F be a field extension. Assume M is an A-module that is finite dimensional over F and let B be a basis of M over F. Then 1 ⊗ B = {1 ⊗ x : x ∈ B} is a basis of E ⊗F M over E. For a ∈ A, we have that (1 ⊗ ρ)1⊗B (1 ⊗ a) = 1 ⊗ ρ B (a). This implies that CharM/F (a) = Char(E⊗F M)/E (1 ⊗ a), TrM/F (a) = Tr(E⊗F M)/E (1 ⊗ a),

(2.3.4)

NrM/F (a) = Nr(E⊗F M)/E (1 ⊗ a). Suppose that [E : F] = n and fix a basis B of E as vector space over F. Then B is also a basis of E[X] as F[X]-module. Let R : E → M n (F) be the representation of E as vector space over F with respect to B. Then the representation of E[X] as F[X]-module with respect to B, which we also denote by R, is given by n

n

f = ∑ f i X i 󳨃→ ∑ R(f i )X i ∈ M n (E[X]). i=0

i=0

Define NrE/F (f) = det R(f)

(f ∈ E[X]).

Lemma 2.3.3 (Change of base field). Let E/F be a finite field extension, let A be a finite dimensional E algebra and let M be a left A-module of finite dimension over E. If a ∈ A then CharM/F (a) = NrE/F (CharM/E (a)) TrM/F (a) = TrE/F (TrM/E (a)) and NrM/F (a) = NrE/F (NrM/E (a)).

50 | 2 Representations of algebras

Proof. By writing M as a direct sum of E[a]-modules we may assume, without loss of generality, that M is indecomposable as an E[a]-module. Then, we may assume that M = E[X]/f(X) for some monic polynomial f = ∑ki=0 f i X i ∈ E[X] (see for example [145, Chapter XIV, Theorem 2.1]) and a acts on M by left multiplication by X. Let B be a basis of E considered as an F-vector space. Then, {bX i : b ∈ B, i = 0, 1, . . . , k − 1} is a basis of the F-vector space M and, with an appropriated ordering of the basis elements, the representation v of a over this basis takes the following form in blocks of n×n matrices (with n = [E : F] and R : E → M n (F) the representation of E as vector space over F with respect to B): 0 In v=( 0 ... 0

0 0 In ... 0

... ... ... ... ...

0 0 0 ... In

−R(f0 ) −R(f1 ) −R(f2 ) ) ... −R(f k−1 )

Then, 󵄨󵄨XI 󵄨󵄨 n 󵄨󵄨 󵄨󵄨 −I n 󵄨󵄨 CharM/F (a) = det(XI nk − v) = 󵄨󵄨󵄨󵄨 . . . 󵄨󵄨 󵄨󵄨󵄨 0 󵄨󵄨 󵄨󵄨 0

0 XI n ... 0 0

... ... ... ... ...

0 0 ... XI n −I n

󵄨󵄨 R(f0 ) 󵄨󵄨 󵄨󵄨 󵄨󵄨 R(f1 ) 󵄨󵄨 󵄨󵄨 ... 󵄨󵄨 󵄨 R(f k−2 ) 󵄨󵄨󵄨󵄨 󵄨 XI n + R(f k−1 )󵄨󵄨󵄨

Adding successively X times each row block to the one just above it, and starting from the last row, we obtain 󵄨󵄨 0 󵄨󵄨 󵄨󵄨 󵄨󵄨−I n 󵄨󵄨 CharM/F (a) = 󵄨󵄨󵄨󵄨 . . . 󵄨󵄨 󵄨󵄨 0 󵄨󵄨 󵄨󵄨 0 󵄨

0 0 ... 0 0

... ... ... ... ...

0 0 ... 0 −I n

R(f)󵄨󵄨󵄨󵄨 󵄨 ∗ 󵄨󵄨󵄨󵄨 󵄨 . . . 󵄨󵄨󵄨󵄨 = det R(f) = NrE/F (f) 󵄨 ∗ 󵄨󵄨󵄨󵄨 󵄨 ∗ 󵄨󵄨󵄨

= NrE/F (CharM/E (a)). A similar argument gives the formulas for the trace and norm. For the rest of the section we assume that A is separable over F. Let E be a splitting field of A (see Lemma 2.2.1 for the existence) and fix an isomorphism of E-algebras h : E ⊗F A → M n1 (E) × ⋅ ⋅ ⋅ × M n k (E). Let h i : E ⊗F A → M n i (E) be the composition of h with the projection onto the i-th component. Then, by definition, the reduced characteristic polynomial the reduced trace

2.3 Characteristic polynomial, trace and norm

| 51

and the reduced norm of a ∈ A are respectively k

RCharA/F (a) = ∏ char(h i (1 ⊗ a)), i=1 k

RTrA/F (a) = ∑ tr(h i (1 ⊗ a)),

(2.3.5)

i=1 k

RNrA/F (a) = ∏ det(h i (1 ⊗ a)). i=1

We now prove that these definitions do not depend on the splitting field E and the chosen isomorphism h. By similarity, we only consider the reduced characteristic polynomial. Let h󸀠 : E ⊗F A → M m1 (E) × ⋅ ⋅ ⋅ × M m l (E) be another isomorphism of E-algebras and let h󸀠i denote the composition of h󸀠 with the projection on the i-th coordinate. By the uniqueness of the Wedderburn decomposition k = l and, permuting the compo󸀠 nents if necessary, we may assume that n i = m i , for each i. Then each h−1 i h i is an −1 󸀠 E-automorphism of M n i (E). By the Noether-Skolem Theorem (Theorem 2.1.9), h i h i is inner and hence h i (1 ⊗ a) and h󸀠i (1 ⊗ a) have the same characteristic polynomial. This shows that RCharA/K (a) does not depend on the choice of h. Next, we prove that it does not depend on the splitting field E. Let E󸀠 be another splitting field of A. Then, one of the simple components K of E󸀠 ⊗F E contains both E and E󸀠 (see Example 2.1.12). As K is a splitting field of A over F, we may assume, without loss of generality, that E󸀠 contains E. Then tensoring by E󸀠 over F we obtain an isomorphism of E󸀠 -algebras k

k

i=1

i=1

h󸀠 : E󸀠 ⊗F A ≅ E󸀠 ⊗E E ⊗F A ≅ E󸀠 ⊗E ∏ M n i (E) ≅ ∏ M n i (E󸀠 ) and h󸀠 (1 ⊗ a) = h(1 ⊗ a). Thus RCharA/F (a) does not depend on the splitting field E. As RChar, RTr and RNr are independent of the splitting field E, one can take E to be a finite Galois extension of F (Theorem 2.2.5). Replacing the isomorphism h by the composition with an element of Gal(E/F), and using the independence of the isomorphism, we deduce that RChar, RTr and RNr are invariant under the action of Gal(E/F). Therefore, RCharA/F (a) ∈ F[X], RTrA/F (a), RNrA/F (a) ∈ F. The reduced trace and norm have similar properties as the standard trace and norm. For example, up to a sign, they are coefficients of the reduced characteristic polynomial: RCharA/F (a) = X m − RTr(a)X m−1 + ⋅ ⋅ ⋅ + (−1)m RNr(a),

(a ∈ A)

(2.3.6)

where m is the sum of the degrees of the simple components of A ⊗F E-modules, for E a splitting field of A over F. Moreover, if a, b ∈ A and α ∈ F then

52 | 2 Representations of algebras RTrA/F (a + b) = RTrA/F (a) + RTrA/F (b), RTrA/F (αa) = α RTrA/F (a), RTrA/F (1) = m, RTrA/F (ab) = RTrA/F (ba),

(2.3.7)

RNrA/F (ab) = RNrA/F (a)RNrA/F (b), RNrA/F (αa) = α m RNrA/F (a), RNrA/F (1) = 1. Lemma 2.3.4. Let A be a separable F-algebra and let L be a subfield of the center of A, containing F. Then RCharA/F = NrL/F ∘ RCharA/L , RTrA/F = TrL/F ∘ RTrA/L ,

(2.3.8)

RNrA/F = NrL/F ∘ RNrA/L . Furthermore, if A is simple of degree n and a ∈ A then CharA/F (a) = RCharA/F (a)n , TrA/F (a) = n RTrA/F (a),

(2.3.9)

NrA/F (a) = RNrA/F (a) . n

Proof. Because of (2.3.1) and (2.3.6), it suffices to prove the statements on characteristic polynomials. Without loss of generality, we also may assume that A is simple, say of degree n. Let K = Z(A) and let E be a splitting field of A over K. Fix an isomorphism h : E ⊗K A ≅ M n (E). If M is the unique simple left E ⊗K A-module then E ⊗K A ≅ M n . Therefore, CharA/K (a) = CharE⊗K A/K (1 ⊗ a) = CharM/A (1 ⊗ a)n = RNrA/K (a)n .

(2.3.10)

The first equality is (2.3.4), the second is a consequence of (2.3.3) and the third one is a consequence of the definition of reduced norm and the fact that h is the representation associated to M with respect to some basis. Let R󸀠A/F = NrK/F ∘ RCharA/K . Let a ∈ A. As NrK/F preserves multiplication, using Lemma 2.3.3 and the equalities in (2.3.10), we have R󸀠A/F (a)n = NrK/F (RCharA/K (a)n ) = NrK/F (CharA/K (a))

(2.3.11)

= CharA/F (a). Let f : K⊗F A ≅ ∏ki=1 A i , with A i simple for every i and let f i denote the composition of f with the projection onto A i . Each A i is a central simple K i algebra of degree n, with K i = σ i (K) for {σ1 , . . . , σ k } the F-embeddings of K in an algebraic closure of K.

2.3 Characteristic polynomial, trace and norm

| 53

Applying (2.3.11) and (2.3.9) for A = A i and F = K i , we have k

R󸀠A/F (a)n = CharA/F (a) = CharK⊗F A/K (1 ⊗ a) = ∏ CharA i /K i (f i (1 ⊗ a)) i=1 k

= ∏ RCharA i /K i (f i (1 ⊗ a)) = RCharK⊗F A/K (1 ⊗ a)n . n

i=1

R󸀠A/F (a)

As both and RCharK⊗F A/K (1 ⊗ a) are monic polynomials, we deduce from uniqueness of factorization that R󸀠A/F (a) = RCharK⊗F A/K (1 ⊗ a).

(2.3.12)

Moreover E ⊗F A ≅ E ⊗K K ⊗F A ≅ ⊕ki=1 E ⊗K A i . If g i denotes the composition of this isomorphism with the projection onto the i-th component then, by (2.3.12), we have k

R󸀠A/F (a) = RCharK⊗F A/K (1 ⊗ a) = ∏ det(g i (1 ⊗ a)) = RCharA/F (a). i=1

Combining this with (2.3.11) we deduce (2.3.9) and we have RCharA/F = NrK/F ∘ RCharA/K . Applying this to F and L we finally conclude RCharA/F = NrK/F ∘ RCharA/K = NrK/L ∘ NrL/F ∘ RCharA/K = NrK/L ∘ RCharA/L , as desired. The equalities (2.3.9) imply the second part of the following lemma. The first part was already observed before. Lemma 2.3.5. Let A be a finite dimensional algebra over a field F and let a ∈ A. Then, MinF (a) divides CharA/F (a) in F[X] and MinF (a) and CharA/F (a) have the same irreducible divisors in F[X]. If, moreover, A is separable then MinF (a) divides both CharA/F (a) in F[X] and RCharA/F (a) has also the same irreducible divisors than MinF (a). If E/F is a finite field extension then (x, y) → TrE/F (xy) is an F-bilinear form of E. This form is non-degenerate if and only if E/F is separable [101, Theorem I.5.2]. This can be generalized as follows. Proposition 2.3.6. The reduced trace of a separable F-algebra A induces a symmetric non-degenerate F-bilinear form τ:A×A→F (a, b) 󳨃→ τ(a, b) = RTrA/F (ab).

54 | 2 Representations of algebras

Proof. Everything but the non-degeneracy is obvious. It is enough to prove that τ is non-degenerate under the assumption that A is simple. Let I = {a ∈ A : τ(a, x) = 0, for every x ∈ A}. We need to show that I = 0. In fact I is a two-sided ideal of A because if a ∈ I then τ(ab, x) = τ(a, bx) = 0 and τ(ba, x) = τ(x, ba) = τ(xb, a) = τ(a, xb) = 0 for every b, x ∈ A. As A is simple, if I ≠ 0 then 1 ∈ I and hence RTrA/F (a) = 0 for every a ∈ A. Let L be the center of A. Then, applying (2.3.8), we have TrL/F (kRTrA/L (a)) = TrL/F RTrA/L (ka) = RTrA/F (ka) = 0 for every k ∈ L and a ∈ A. Since L/F is separable, TrL/F : L → F is surjective (see e.g. [145]) and this implies that RTrA/L (a) = 0 for every a ∈ A. Fix an isomorphism h : E ⊗L A → M n (E) of E-algebras and let E11 (1) = h(∑ki=1 α i ⊗ a i ), with α i ∈ E and a i ∈ A for every i. Then 1 = tr(E11 (1)) = ∑ki=1 α i tr(h(1 ⊗ a i )) = ∑i=1 α i RTrA/F (a i ) = 0, a contradiction. By Proposition 2.3.6, if {x1 , . . . , x n } is a basis of A over F, for A a separable algebra then there are y1 , . . . , y n ∈ A such that RTrA/F (x i y i ) = δ ij . Such y1 , . . . , y n are called a dual basis of x1 , . . . , x n and we have n

a = ∑ RTrA/F (ay i )x i ,

(a ∈ A).

i=1

Problems 2.3.1. Calculate the characteristic polynomial, trace and norm and the reduced versions of an element of a quaternion algebra. 2.3.2. Let F be a field and f : A → B an F-isomorphism of finite dimensional algebras. Prove that NrA/F (a) = NrB/F (f(a)) for every a ∈ A. 2.3.3. Calculate the reduced characteristic polynomial, reduced norm and reduced trace of a symbol algebra of degree 3. Calculate the reduced trace of a symbol algebra of degree n. 2.3.4. Suppose that the field F has a primitive n-th root of unity and let a, b ∈ U(F) be such that X n − a is irreducible in F[X]. (1) Prove that E = F[x] (with x as in Problem 2.1.8) is a maximal subfield of the symbol algebra ( a,b F )n . (Compare with Proposition 2.2.4 and Problem 2.2.8.) a,bN

(α)

E/F (2) Prove that if α ∈ U(E) then ( a,b )n . (Hint: Consider αy.) F )n ≅ ( F (3) Prove that if E contains an element α with N E/F (α) = b then ( a,b F )n is split.

2.3.5. Let A be a simple separable F-algebra and let S be the (unique up to isomorphisms) simple left A-module. Prove that there is a positive rational number q such that RNrA/F (a) = NrS/F (a)q for every a ∈ A.

2.4 Brauer group

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55

2.3.6. Let F be a number field and let A be a separable F-central simple algebra. Let σ be field homomorphism from F to ℂ and let E σ = ℝ, if σ(F) ⊆ ℝ and E σ = ℂ otherwise. Prove the following properties: (1) There is a unique simple component A σ of ℝ ⊗ℚ A with center E σ and having a ring homomorphism f σ : A → A σ extending σ. (2) RNrA σ /E σ (f σ (a)) = σ(RNrA/F (a)). (3) There is a ℝ ⊗ℚ A-module V σ and a positive rational number q σ such that |σ(RNrA/F (a))| = |NrV σ /ℝ (a)|q σ . 2.3.7. Let A be separable F-algebra. Prove that the coefficients of a dual basis of a basis c1 , . . . , c n of A in the given basis are the columns of the inverse of the matrix (TrA/F (c i c j )). 2.3.8. Let E/F be an extension of finite fields. Prove that TrE/F (E) = F and NrE/F (U(E)) = U(F).

2.4 Brauer group Recall that two central simple finite dimensional F-algebras A and B are said to be equivalent if A ≅ M n (D) and B ≅ M m (D) for some division algebra D and some positive integers n and m. The equivalence class containing A is denoted by [A]. The set of equivalence classes is denoted Br(F). It is equipped with the following binary operation: [A][B] = [A ⊗F B] (see Proposition 2.1.11 and Problem 2.4.2). This makes Br(F) into a group, called the Brauer group of F. The identity of Br(F) is the class [F] formed by the split algebras and the inverse of [A] in Br(F) is the class [A o ] containing the opposite algebra A o of A (see Problem 2.4.1). For example, if F is algebraically closed then Br(F) is the trivial group. The only finite dimensional central simple real algebras, up to isomorphisms, are the reals ℝ and the Hamiltonian quaternion algebra ℍ(ℝ) (see Problem 2.2.1). Thus Br(ℝ) is cyclic of order 2. The Brauer group of ℚ is infinite. Let E/F be a field extension. Then we have a group homomorphism ⊗F E : Br(F) → Br(E) [A] 󳨃→ [A ⊗F E] The kernel of ⊗F E is the relative Brauer group Br(E/F) = {[A] ∈ Br(F) : E splits A}. Theorem 2.2.5 can be rephrased as follows Br(F) = ⋃ Br(E/F). E/F, finite Galois

(2.4.1)

56 | 2 Representations of algebras

Proposition 2.4.1. Let A be a finite dimensional central simple F-algebra. If E/F is a finite field extension that splits A then A is Brauer equivalent to an algebra B which contains a subfield isomorphic to E and such that [E : F] = Deg(B). Proof. Suppose E/F is a finite field extension that splits A. Let n = [E : F]. Via the regular representation, we can consider E as an F-subalgebra of M n (F). Hence, E also can be considered as a subfield of A1 = A ⊗F M n (F). Clearly, [A] = [A1 ]. Thus, from the assumption, we obtain that A1 ⊗F E is split. Proposition 2.2.4 yields that C = CenA1 (E) is a central simple algebra over E, which is Brauer equivalent to A1 ⊗F E. Hence, C also is split. Moreover, from Proposition 2.2.4, Deg(A)n = Deg(A1 ) = nDeg(C). Hence, C ≅ M d (E) = E ⊗F C1 ⊆ A1 , with C1 ≅ M d (F), and d = Deg(C) = Deg(A) = Deg(C1 ). Let B = CenA1 (C1 ). Clearly, E ⊆ B. As C1 is a central simple F-subalgebra of A1 , B is also central simple F-algebra and A1 ≅ C1 ⊗F B, by Theorem 2.1.10. Then [A] = [A1 ] = [B] and Deg(B) = Deg(A1 )/Deg(C1 ) = Deg(A1 )/d = Deg(A)n/Deg(A) = [E : F]. Corollary 2.4.2. Let F be a field and E/F is a finite field extension. Assume A is a finite dimensional central simple F-algebra. The following properties hold. (1) If E splits A then Ind(A) divides [E : F]. (2) Ind(A) = min{[K : F] : [A] ∈ Br(K/F), K/F finite field extension}. (3) Ind(A ⊗F E) divides Ind(A) and Ind(A) divides [E : F]Ind(A ⊗F E). Proof. (1) Assume that E splits A. Because of Proposition 2.4.1, there is an algebra B that is Brauer equivalent to A and that contains a field E with [E : F] = Deg(B). It follows that Ind(A) = Ind(B) divides Deg(B) = [E : F]. (2) We know that A = M n (D), with D a division ring. Let E be a maximal subfield of D. Then, by Proposition 2.2.4, [E : F] = Ind(D) = Ind(A) and E splits A. Hence (2) now follows from (1). (3) Again A = M n (D), with D a division algebra. Clearly Ind(A ⊗F E) = Ind(D ⊗F E) and hence Ind(A ⊗F E) divides Deg(D ⊗F E) = Deg(D) = Ind(D) = Ind(A). This proves the first part. Because of (2), there is a finite field extension L/E that splits A ⊗F E and [L : E] = Ind(A ⊗F E). Hence, Ind(A) divides [L : F] = [E : F] [L : E] = [E : F] Ind(A ⊗F E).

Problems 2.4.1. Let F be a field. Prove that if A is a finite dimensional central simple F-algebra of degree d then A ⊗ A o ≅ M d2 (F). (Hint: apply the universal property of the tensor product to the regular left and right representations of A.) 2.4.2. Let F be a field. Prove that if A1 , A2 , B1 and B2 are central simple algebras such that A1 and A2 are Brauer equivalent and B1 and B2 are Brauer equivalent then A1 ⊗F B1 and A2 ⊗F B2 are Brauer equivalent.

2.5 Cohomology

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57

2.4.3. Prove that ℍ(ℝ) ⊗ℝ ℍ(ℝ) ≅ M4 (ℝ).

2.5 Cohomology In this section we give some cohomological background. A positive complex is a sequence of homomorphisms of abelian groups ∂1

∂3

∂2

∂ n−1

∂ n+1

∂n

(M, ∂) : M0 󳨀→ M1 󳨀→ M2 󳨀→ ⋅ ⋅ ⋅ 󳨀→ M n−1 󳨀→ M n 󳨀→ . . . such that ∂ n+1 ∘ ∂ n = 0 for every n ≥ 1. A homomorphism ϕ : (M, ∂) → (M󸀠 , ∂󸀠 ) of positive complexes (M, ∂) and (M󸀠 , ∂󸀠 ) is a commutative diagram of homomorphisms ∂1

∂2

∂ n−1

∂n

∂ n+1

∂󸀠1

∂󸀠2

∂󸀠n−1

∂󸀠n

∂󸀠n+1

M : M0 󳨀→ M1 󳨀→ . . . 󳨀→ M n−1 󳨀→ M n 󳨀→ . . . ↓ ϕ1 ↓ ϕ n−1 ↓ ϕn ↓ ϕ0 M󸀠 : M0󸀠 󳨀→ M1󸀠 󳨀→ . . . 󳨀→ M 󸀠n−1 󳨀→ M 󸀠n 󳨀→ . . . This defines a category of positive complexes. The homology groups of the positive complex (M, ∂) are the abelian groups Hn (M) = ker(∂ n+1 )/Im(∂ n ), with n ≥ 1. If ϕ : (M, ∂) → (M󸀠 , ∂󸀠 ) is a morphism of positive complexes then each ϕ n induces a natural a homomorphism: Hn (ϕ) : Hn (M) → Hn (M󸀠 ). This defines a functor H from the category of positive complexes to the category of abelian groups. Let G be a group, U a left ℤG-module and let −∗ = HomℤG (−, U) denote the contravariant functor that maps a left ℤG-module M to M ∗ = HomℤG (M, U) and maps a ℤG-module homomorphism f : M → N to f ∗ : N ∗ → M ∗ with f ∗ (ϕ) = ϕ ∘ f , for ϕ ∈ HomℤG (N, U). We consider ℤ as a ℤG-module via the trivial action of G, that is gn = n, for g ∈ G and n ∈ ℤ. Associated to a ℤG-projective resolution of ℤ δ n+1

δ3

δ n−1

δn

δ2

δ1

δ0

(P, δ) : ⋅ ⋅ ⋅ 󳨀→ P n 󳨀→ P n−1 󳨀→ ⋅ ⋅ ⋅ 󳨀→ P2 󳨀→ P1 󳨀→ P0 󳨀→ ℤ 󳨀→ 0, that is an exact sequence with all P i projective ℤG-modules, there is a positive complex δ∗1

δ∗2

δ∗3

δ∗n−1

δ∗n

δ∗n+1

P∗ : P∗0 󳨀→ P∗1 󳨀→ P∗2 󳨀→ ⋅ ⋅ ⋅ 󳨀→ P∗n−1 󳨀→ P∗n 󳨀→ . . .

58 | 2 Representations of algebras

Then Hn (G, U) = Hn (P∗ ) = ker(δ∗n+1 )/Im(δ∗n ) is the n-th cohomology group of G with coefficients in U. These groups are, up to isomorphism, independent of the chosen projective resolution. More precisely, if (Q, ϵ) is another projective ℤG-resolution of ℤ then, because of the projectivity of the P i ’s, there are ℤG-morphisms ϕ n : P n → Q n that yield a commutative diagram δ2

δ1

δ0

ϵ2

ϵ1

ϵ0

(P, δ) : ⋅ ⋅ ⋅ 󳨀→ P2 󳨀→ P1 󳨀→ P0 󳨀→ ℤ 󳨀→ 0 ↓ ϕ2 ↓ ϕ1 ↓ ϕ0 ↓1 (Q, ϵ) : ⋅ ⋅ ⋅ 󳨀→ Q2 󳨀→ Q1 󳨀→ Q0 󳨀→ ℤ 󳨀→ 0 Applying −∗ we obtain a homomorphism of positive complexes ϵ∗1

ϵ∗2

δ∗1

δ∗2

Q∗ : Q∗0 󳨀→ Q∗1 󳨀→ Q∗2 󳨀→ . . . ϕ∗ ↓ ↓ ϕ∗0 ↓ ϕ∗1 ↓ ϕ∗2 P∗ : P∗0 󳨀→ P∗1 󳨀→ P∗2 󳨀→ . . . and Hn (ϕ∗ ) : Hn (Q∗ ) → Hn (P∗ ) is an isomorphism. Let H be a subgroup of G. Then, by restriction of scalars, every ℤG-module also is a ℤH-module. Because ℤG is projective as a ℤH-module, restriction of scalars preserves projective modules and projective resolutions. Therefore, if P is a projective resolution of ℤ as ℤG-module then it also is a projective ℤH-resolution. Moreover, for every left ℤG-module U, one has a homomorphism of positive complexes (the vertical maps are inclusions) HomℤG (P, U) : HomℤG (P0 , U) 󳨀→ HomℤG (P1 , U) 󳨀→ HomℤG (P2 , U) 󳨀→ . . . R GH ↓ ↓ ↓ ↓ HomℤH (P, U) : HomℤH (P0 , U) 󳨀→ HomℤH (P1 , U) 󳨀→ HomℤH (P2 , U) 󳨀→ . . . The corresponding cohomology homomorphisms ResGH = Hn (R GH ) : Hn (G, U) → Hn (H, U) are called the restriction maps. For left ℤG-modules M and N, the group G acts on Homℤ (M, N) by f g (m) = g−1 f(gm), with f ∈ Homℤ (M, N), g ∈ G and m ∈ M. Observe that for any subgroup H of G, HomℤH (M, N) consists of the H-invariants elements of Homℤ (M, N). If H also is of finite index in G then the transfer map T HG : HomℤH (M, N) → HomℤG (M, N)

2.5 Cohomology

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59

is defined by T HG (f) = ∑ f s ,

(2.5.1)

s∈S

for f ∈ HomℤH (M, N)) and where S is a right transversal for H in G. This definition is independent of the chosen transversal. We return to the projective ℤG-resolution P of ℤ and we assume that H is of finite index in G. Because of the transfer map, one can define a homomorphism of positive complexes in the opposite direction than that of the restriction R GH : HomℤH (P, U) : HomℤH (P0 , U) 󳨀→ HomℤH (P1 , U) 󳨀→ HomℤH (P2 , U) 󳨀→ . . . ↓ ↓ ↓ T HG ↓ HomℤG (P, U) : HomℤG (P0 , U) 󳨀→ HomℤG (P1 , U) 󳨀→ HomℤG (P2 , U) 󳨀→ . . . The corresponding cohomology homomorphisms CorGH = Hn (T HG ) : Hn (H, U) → Hn (G, U) are called the corestriction maps. As f s = f for f ∈ HomℤG (P i , U), we have (CorGH ∘ ResGH )(f) = [G : H]f.

(2.5.2)

Observe that a ℤG-module is just an abelian group M together with an action of G on M. The ℤG-modules that we will encounter are multiplicative abelian groups U, with an action α : G → Aut(U) of G on U. This action determines the following ℤG-module structure: gu = α g (u), where α g = α(g) for g ∈ G and u ∈ U. In this multiplicative notation (2.5.2) takes the following form. (CorGH ∘ ResGH )(f) = f [G:H] .

(2.5.3)

If G is finite and H = 1 then ResGH (f) = 1 and thus we obtain the following. Proposition 2.5.1. If G is a finite group then f |G| = 1 for every f ∈ Hn (G, U), that is, the exponent of Hn (G, U) is a divisor of |G|. The next goal is to obtain a handy set theoretical description of H2 (G, U) and concrete formulas for ResGH and CorGH within this context. In the remainder of the section U is a multiplicative abelian group and α : G → Aut(U) is an action. So we consider U as a left G-module. The formula for the transfer map (2.5.1) hence takes the following form T HG (f) = ∏ f s . s∈S

The standard resolution of ℤ is the following exact sequence δ n+1

δn

δ n−1

δ3

δ2

δ1

ω

⋅ ⋅ ⋅ 󳨀→ ℤG[G n ] 󳨀→ ℤG[G n−1 ] 󳨀→ ⋅ ⋅ ⋅ 󳨀→ ℤG[G2 ] 󳨀→ ℤG[G] 󳨀→ ℤG 󳨀→ ℤ 󳨀→ 0,

60 | 2 Representations of algebras where ω is the augmentation map, G n is the direct product of n copies of G, ℤG[G n ] is the free left ℤG-module with basis [G]n = {[g1 , . . . , g n ] : g1 , . . . , g n ∈ G} and the maps δ i are given by the following recursive formulas δ1 ([g]) = g − 1, n−1

δ n ([g1 , . . . , g n ]) = g1 [g2 , . . . , g n ] + ∑ (−1)i [g1 , . . . , g i−1 , g i g i+1 , . . . , g n ] i=1

+ (−1)n [g1 , . . . , g n−1 ]. Observe that if G n is identified with [G]n in the obvious way, then the restriction to [G]n gives a group homomorphism Ψ n : HomℤG (ℤG[G n ], U) → M(G n , U), where M(G n , U) is simply the set of maps f : G n → U with the product f1 f2 of f1 , f2 ∈ M(G n , U) defined by (f1 f2 )(x) = f1 (x)f2 (x), for x ∈ G n . The elements of Ψ n (ker(δ∗n+1 )) are the maps f : G n → U satisfying the so called cocycle condition n

α g1 (f(g2 , . . . , g n , g n+1 )) ⋅ (∏ (f(g1 , . . . , g i−1 , g i g i+1 , . . . , g n+1 ))(−1) ) i

i=1

⋅ (f(g1 , . . . , g n ))(−1)

n+1

= 1.

Such maps f are called n-cocycles. The elements of Ψ n (Im(δ∗n )) can be described as follows. For n ≥ 2, these are the maps of the form f t : G n → U,

(2.5.4)

with f t (g1 , . . . , g n ) = α g1 (t(g2 , . . . , g n )) n−1

⋅ ( ∏ (t(g1 , . . . , g i−1 , g i g i−1 , g i+1 , . . . , g n ))(−1) ) ⋅ (t(g1 , . . . , g n−1 ))(−1) , i

n

i=1

where t :

G n−1

→ U. For n = 1, the elements of Ψ1 (Im(δ∗1 )) are the maps f u : G → U,

with u ∈ U, and

f u (g) = α g (u)u−1 ,

for g ∈ G. The maps f t and f u are called n-coboundaries. We denote by Z n (G, U) and B n (G, U) the groups of n-cocycles and n-coboundaries respectively. Considering Ψ n as an identification we have that Hn (G, U) = Z n (G, U)/B n (G, U).

2.5 Cohomology

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61

For example, for n ≤ 2, we have Z 1 (G, U) = { f : G → U : f(gh) = f(g)α g (f(h)), g, h ∈ G}, B1 (G, U) = { f u : G → U : u ∈ U and f u (g) = α g (u)u−1 for all g ∈ G}, Z 2 (G, U) = { f : G × G → U : f(g1 , g2 )f(g1 g2 , g3 ) = f(g1 , g2 g3 )α g1 (f(g2 , g3 )) for all g1 , g2 , g3 ∈ G} B (G, U) = { f t : G × G → U : f t (g, h) = t(g)α g (t(h))t(gh)−1 for all g, h ∈ G 2

and t : G → U}. We now will determine formulas for ResGH and CorGH in terms of cocycles. The difficulty in obtaining these formulas is that the description of Hn (H, U) in terms of cocycles and coboundaries uses the standard resolution of ℤ as ℤH-module and the restriction and corestriction are defined in terms of the standard resolution of ℤ as ℤG-module interpreted as a resolution of ℤ as ℤH-module. Thus to connect the two expressions we need to relate the two resolutions. In one direction this is easy because the inclusions H n → G n induce a commutative diagram for the two standard resolutions: δ2 δ1 ω ⋅ ⋅ ⋅ 󳨀→ ℤH[H 2 ] 󳨀→ ℤH[H] 󳨀→ ℤH 󳨀→ ℤ 󳨀→ 0 ↓ ϕ2 ↓ ϕ1 ↓ ϕ0 ↓1 δ2

δ1

ω

⋅ ⋅ ⋅ 󳨀→ ℤG[G2 ] 󳨀→ ℤG[G] 󳨀→ ℤG 󳨀→ ℤ 󳨀→ 0 Applying HomℤH (−, U) to this diagram we obtain the lower part of the following diagram (the vertical arrows in the upper part are the inclusions): HomℤG (ℤG, U) 󳨀→ HomℤG (ℤG[G], U) 󳨀→ HomℤG (ℤG[G], U) 󳨀→ . . . ↓ ↓ ↓ HomℤH (ℤG, U) 󳨀→ HomℤH (ℤG[G], U) 󳨀→ HomℤH (ℤG[G], U) 󳨀→ . . . ϕ∗ ↓ ↓ ↓ HomℤH (ℤH, U) 󳨀→ HomℤH (ℤG[H], U) 󳨀→ HomℤH (ℤH, U) 󳨀→ . . .

R GH

The upper and lower positive complexes are those obtained from the standard resolutions of ℤ as left ℤG-module and as left ℤH-module respectively; so the elements in the kernels can be interpreted as 2-cocycles. The homology homomorphism Hn (ϕ∗ ) are isomorphisms, because they are obtained from two different projective resolutions of the same module. As we are not distinguishing the cohomology groups obtained by distinct resolutions, we can consider Hn (ϕ∗ ) as an identification. Then ResGH = H(R GH ) = H(ϕ∗ R GH ) (we simply denote this also as R GH ) and therefore we have ResGH (f)(h1 , . . . , h n ) = Ψ n (R GH (Ψ n−1 (f) ∘ ϕ))(h1 , . . . , h n ) = f H (h1 , . . . , h n ), in other words ResGH (f) is simply the restriction of f to H n : ResGH (f)(h1 , . . . , h n ) = f(h1 , . . . , h n ),

62 | 2 Representations of algebras for f ∈ Hn (G, U), h1 , . . . , h n ∈ H. This also explains the use of the terminology “restriction”. We now are going to prove that the corestriction map can be given by the following formula −1 −1 −1 CorGH (f)(g1 , . . . , g n ) = ∏ s−1 c f(s c g 1 s cg1 , s cg1 g 2 s cg1 g2 , . . . , s cg1 ...g n−1 s cg1 ...g n ), (2.5.5) s∈C

where C denotes the set consisting of right H-cosets of H in G and s c is a representative of the right coset c ∈ C. Here we consider G as a group acting on the right on the set C via right multiplication. Observe that s c g belongs to cg and therefore s c gs−1 cg ∈ H. n n We define ℤH-module homomorphisms ψ n : ℤG[G ] → ℤH[H ]. Because {s c : c ∈ C} is a basis of ℤG as ℤH-module and the elements of the form s c [g1 , . . . , g n ], with c ∈ C and g1 , . . . , g n ∈ G, form a basis of ℤG[G n ] as left ℤH-module, it is enough to define ψ n on the elements of this form. We set ψ0 (s c ) = 1 −1 −1 ψ n (s c [g1 , . . . , g n ]) = [s c g1 s−1 cg1 , s cg1 g 2 s cg1 g2 , . . . , s cg1 ...g n−1 g n s cg1 ...g n ].

We claim that the following diagram is commutative δ2

δ1

ω

δ2

δ1

ω

⋅ ⋅ ⋅ 󳨀→ ℤG[G2 ] 󳨀→ ℤG[G] 󳨀→ ℤG 󳨀→ ℤ 󳨀→ 0 ↓ ψ2 ↓ ψ1 ↓ ψ0 ↓1

(2.5.6)

⋅ ⋅ ⋅ 󳨀→ ℤH[H 2 ] 󳨀→ ℤH[H] 󳨀→ ℤH 󳨀→ ℤ 󳨀→ 0 The commutativity of the right square is clear and the commutativity of the second right square can be proved as follows: ψ0 δ1 (s c [g]) = ψ0 (s c (g − 1)) = ψ0 (s c gs−1 cg s cg − s c ) −1 = s c gs−1 cg ψ 0 (s cg ) − ψ 0 (s c ) = s c gs cg − 1

= δ1 ([s c gs−1 cg ]) = δ 1 ψ 1 (s c [g]). The commutativity of the other squares (thus, for n ≥ 2) can be proved as follows: ψ n−1 δ n (s c [g1 , . . . , g n ]) n−1

= ψ n−1 (s c (g1 [g2 , . . . , g n ] + ∑ (−1)i [g1 , . . . , g i−1 , g i g i+1 , . . . , g n ] + (−1)n [g1 , . . . , g n−1 ]))

i=1 n−1

i = ψ n−1 (s c g1 s−1 cg1 s cg1 [g 2 , . . . , g n ] + ∑ (−1) s c [g 1 , . . . , g i−1 , g i g i+1 , . . . , g n ]

+ (−1)n s c [g1 , . . . , g n−1 ])

i=1

= s c g1 s−1 cg1 ψ n−1 (s cg1 [g 2 , . . . , g n ]) n−1

+ ∑ (−1)i ψ n−1 (s c [g1 , . . . , g i−1 , g i g i+1 , . . . , g n ]) + (−1)n ψ n−1 (s c [g1 , . . . , g n−1 ]) i=1

2.5 Cohomology

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63

−1 −1 = s c g1 s−1 cg1 [s cg1 g 2 s cg1 g2 , . . . , s cg1 ...g n−1 g n s cg1 ...g n ] n−1

−1 −1 + ∑ (−1)i [s c g1 s−1 cg1 , . . . , s cg1 ...g i−1 g i g i+1 s cg1 ...g i+1 , . . . , s cg1 ...g n−1 g n s cg1 ...g n ] i=1

−1 + (−1)n [s c g1 s−1 cg1 , . . . , s cg1 ...g n−2 g n−1 s cg1 ...g n−1 ]

−1 −1 = δ n ([s c g1 s−1 cg1 , s cg1 g 2 s cg1 g2 , . . . , s cg1 ...g n−1 g n s cg1 ...g n ])

= δ n ψ n (s c [g1 , . . . , g n ]). Applying HomℤH (−, U) to this diagram, we obtain the upper part of the following commutative diagram HomℤH (ℤH, U) 󳨀→ HomℤH (ℤH[H], U) 󳨀→ HomℤH (ℤH[H 2 ], U) 󳨀→ . . . ↓ ↓ ↓ HomℤH (ℤG, U) 󳨀→ HomℤH (ℤG[G], U) 󳨀→ HomℤH (ℤG[G2 ], U) 󳨀→ . . . T HG ↓ ↓ ↓ HomℤG (ℤG, U) 󳨀→ HomℤG (ℤG[G], U) 󳨀→ HomℤG (ℤG[G2 ], U) 󳨀→ . . .

ψ∗

Using the set-theoretical description of the homological groups, we finally obtain the desired formula CorGH (f)(g1 , . . . , g n ) = (T HG Hn (ψ∗ )) (f)(g1 , . . . , g n ) ∗ = ∏ s−1 c ψ n (f)(s c [g 1 , . . . , g n ]) c∈C −1 −1 −1 = ∏ s−1 c f(s c g 1 s cg1 , s cg1 g 2 s cg1 g2 , . . . , s cg1 ...g n−1 s cg1 ...g n ). c∈C

We finish this section by showing how extensions of groups yield elements of H 2 (G, U). Let U be a group and G an arbitrary group. An extension of G by U is an exact sequence of groups j

p

1→U→H→G→1

(2.5.7) j

󸀠

p

󸀠

This extension is said to be equivalent to the extension 1 → U → H 󸀠 → G → 1 if there is a group homomorphism ϕ : H → H 󸀠 making the following diagram j

p

1 → U 󳨀→ H 󳨀→ G → 1 󸀠 󸀠 ↘j ↓ϕ p ↗ H󸀠

(2.5.8)

commutative. It follows that such a ϕ is necessarily an isomorphism. Assume now that U is an abelian group and consider an extension (2.5.7). For every g ∈ G, select u g ∈ H such that p(u g ) = g. Then we have two maps α : G → Aut(U) g 󳨃→ α g

f:G×G→U (g, h) 󳨃→ f(g, h)

64 | 2 Representations of algebras

defined by the following formulas u g j(a) = j(α g (a))u g ,

u g u h = j(f(g, h))u gh ,

(2.5.9)

g, h ∈ G, a ∈ U. A straightforward calculation shows that the associativity of the product of H implies that α is a group homomorphism and f ∈ Z 2 (G, U), where U is considered as a left ℤG-module via the action of α. Such an action α is independent of the choice of the u g ’s. Even more, two equivalent extensions of G by U induce the same action of G on U. The 2-cocycle f depends on the u g ’s but only up to a 2-coboundary. More precisely, if we have two equivalent extensions as in (2.5.8) and one selects families u g ∈ H and u󸀠g ∈ H 󸀠 , with p(u g ) = p󸀠 (u󸀠g ) = g then, for every g ∈ G, there is t g ∈ U such that u󸀠g = j󸀠 (t g )ϕ(u g ). If f and f 󸀠 are the 2-cocycles defined by {u g : g ∈ G} and {u󸀠g : g ∈ G} respectively, then u g u h = j(f(g, h))u gh and u g u h = j(f 󸀠 (g, h))u gh and a straightforward calculation shows that f 󸀠 = ff t , where f t ∈ B2 (G, U) (see (2.5.4)). This process can be reversed. Indeed, assume we are given a group homomorphism α : G → Aut(U) and a 2-cocycle f ∈ Z 2 (G, U). Let H = U × G. For a ∈ U, g ∈ G, write (a, g) as j(a)u g . Then there is a unique associative product on H such that the map a 󳨃→ j(a) is an injective group homomorphism and such that the equalities f stated in (2.5.9) are satisfied. This defines an extension of G by U, denoted U ⋋α G, and it provides a bijection between equivalence classes of extensions of G by U and 2-cocycles f ∈ H2 (G, U), with respect to some action of G on U (or equivalently, with respect to a ℤG-module structure on U).

2.6 Crossed products In this section we will give a cohomological interpretation of the Brauer group Br(F) of a field F. To do so, we need to recall the notion of a classical crossed product. Because of its later use in the investigation of group rings, we recall a more general notion of crossed product. Let R be a ring and G a group. A crossed product R ∗ G of G over R is an associative ring which has a set of invertible elements {u g : g ∈ G}, a copy of G, such that R ∗ G = ⊕g∈G Ru g , a free left R-module, and Ru g = u g R

Ru g Ru h = Ru gh ,

and

for all g, h ∈ G. A crossed product R ∗ G determines two maps α : G → Aut(R)

and

f : G × G → U(R),

given by the following rules: u g r = α g (r)u g ,

u g u h = f(g, h)u gh ,

(2.6.1)

2.6 Crossed products |

65

for g, h ∈ G, r ∈ R; where α g = α(g). The map α is called the action and the map f is called the twisting of the crossed product. The associativity of R ∗ G is equivalent to the assertions that, for all g, h, k ∈ G, α g α h = ι f(g,h) α gh

and

f(g, h)f(gh, k) = α g (f(h, k))f(g, hk),

(2.6.2)

where ι u denotes the inner automorphism ι u (x) = uxu−1 for u ∈ U(R). So a crossed product is uniquely determined by the given ring R, the group G and maps α and f satisfying (2.6.2). A better notation for R ∗ G hence would be (R, G, α, f). Although being ambiguous, we will stay with the simpler notation R ∗ G. A diagonal change of basis in a crossed product R ∗ G is a replacement of the Rbasis {u g : g ∈ G} by an alternate R-basis {v g = t g u g : g ∈ G}, where t g ∈ U(R) for every g ∈ G. Of course, this results in new maps α and f . It is easy to see that the identity of R ∗ G is of the form 1 = au1 for some a ∈ U(R). Via a diagonal change of basis, we may and will assume that 1 = u1 . The ring R will be considered as a subring of R ∗ G via the embedding given by r 󳨃→ ru1 . Certain special cases of crossed products have their own names. If the action and twisting are trivial, that is if α g = 1 and f(g, h) = 1 for all g, h ∈ G, then R ∗ G = RG, the (ordinary) group ring. If the action is trivial, then R ∗ G = R f G is called a twisted group ring. If the twisting is trivial then R ∗ G is called a skew group ring. In this case the basis elements u g are simply denoted g and clearly they form a subgroup of U(R ∗ G) isomorphic to G. A group ring RG only can be a simple ring provided the ring R is simple and the group G is trivial. Crossed products, however, can be simple. As for group rings, the support of an element a = ∑g∈G a g u g ∈ R ∗ G is denoted and defined as Supp(a) = {g ∈ G : a g ≠ 0}. Lemma 2.6.1. Let R ∗ G be a crossed product with action α and assume that R is simple. If α g is not inner in R for every non-trivial g ∈ G, then R∗G is a simple ring and Z(R∗G) = Z(R)G , the fixed subring of Z(R), that is Z(R)G = {r ∈ Z(R) : α g (r) = r for all g ∈ G} Proof. Let I be a non-zero ideal of R ∗ G and let a = ∑g∈G a g u g be a non-zero element of I with Supp(a) of minimal cardinality. Multiplying, if necessary, with u−1 g for some g ∈ Supp(a), we may assume that a1 ≠ 0. Because R is simple, we have then have that 1 ∈ Ra1 R. Hence, it easily is seen that we also may assume that a1 = 1. Consequently, for any r ∈ R, we get |Supp(ra − ar)| < |Supp(a)|. The minimality condition therefore implies that ra = ar for all r ∈ R. It follows that for any g ∈ Supp(a) and r ∈ R we have ra g = a g α g (r). Hence, Ra g = a g R. As R is simple this yields that a g ∈ U(R) and α g (r) = a−1 g ra g . So, α g is inner. The assumption therefore implies that g = 1 and thus 1 = a ∈ I. The first part of the result thus follows. For the second part, note that obviously Z(R)G ⊆ Z(R ∗ G). To prove the converse inclusion, assume a = ∑g∈G a g u g ∈ Z(R ∗ G) and g ∈ Supp(a). Then, as before, ra g = a g α g (r) for all r ∈ R. So α g is inner. Hence, again by the assumption, g = 1. Thus, a ∈ R. Because a commutes with all u h , with h ∈ G, it follows that a ∈ Z(R)G .

66 | 2 Representations of algebras If R ∗ G is a crossed product and N is a subgroup of G then {a ∈ R ∗ G : Supp(a) ⊆ N} = R ∗ N is a natural sub-crossed product. In the next lemma we show that crossed products show up naturally in the context of group rings, Lemma 2.6.2. Let N be a normal subgroup of a group G. If R ∗ G is a crossed product then R ∗ G = (R ∗ N) ∗ (G/N), where the latter is some crossed product of the group G/N over the ring R ∗ N. In particular, for group rings, RG = (RN) ∗ (G/N). Proof. Let T be a right transversal for N in G. Then R ∗ G = ⊕t∈T (R ∗ N)u t . Because N is normal in G, we have (R ∗ N)u t = u t (R ∗ N) and also (R ∗ N)u t (R ∗ N)u t󸀠 = (R ∗ N)u t󸀠󸀠 , where t󸀠󸀠 is such that Nt󸀠󸀠 = (Nt)(Nt󸀠 ). Hence the result follows. As a second example of crossed products, we now recall the notion of a classical crossed product. Let E/F be a finite Galois extension and let G = Gal(E/F), the Galois group of the extension E/F. Hence one has a natural action α : G → Aut(E). Restriction yields an action of G on U(E). This makes U(E) into a ℤG-module. Let f ∈ Z 2 (G, U(E)), that is, f satisfies the cocycle condition: f(σ, τ)f(στ, ρ) = f(σ, τρ)σ(f(τ, ρ)), where σ, τ, ρ ∈ G. So the maps α and f satisfy (2.6.2). The classical crossed product with cocycle f , denoted (E/F, f), is the crossed product E ∗ G = ∑σ∈G Eu σ . So the product is determined by the following rules u σ a = σ(a)u σ ,

u σ u τ = f(σ, τ)u στ ,

(2.6.3)

where σ, τ ∈ G and a ∈ E. Note that a diagonal change corresponds to replacing f by ff t for some f t ∈ B2 (G, U(E)) (see (2.5.4)). As agreed before, we will assume that 1 = u1 and thus E is a subfield of (E/F, f). Clearly the center of (E/F, f) is F. Hence (E/F, f) is a central F-algebra. Note that Deg(E/F, f) = [E : F] = |G|. So, because of Proposition 2.2.4, E is a maximal subfield of (E/F, f) and it is a splitting field. The following theorem provides the cohomological description of Br(E/F) announced above. Theorem 2.6.3. Let E/F be a finite Galois extension and let G = Gal(E/F). Suppose f, g ∈ Z 2 (G, U(E)). The following properties hold. (1) (E/F, f) is a finite dimensional central simple F-algebra. (2) The following map is a group isomorphism θ E/F : H2 (G, U(E)) → Br(E/F) [f]

󳨃→ [(E/F, f)]

(3) (E/F, f) is split if and only if f ∈ B2 (G, U(E)). (4) (E/F, f) and (E/F, g) are isomorphic as F-algebras if and only if f −1 g ∈ B2 (G, U(E)).

2.6 Crossed products |

67

Proof. (1) This follows from Lemma 2.6.1. (2) Let θ = θ E/F . To prove that θ is injective, suppose θ([f]) = θ([g]). Then A = (E/F, f) = ⊕σ∈G E1A u σ and B = (E/F, g) = ⊕σ∈G E1B v σ are Brauer equivalent and they have the same dimension over F. Hence, they are isomorphic F-algebras. Let ϕ : A → B be an F-algebra isomorphism. Then the map ϕ(e1A ) 󳨃→ e1B is an isomorphism from ϕ(E1A ) to E1B . By the Noether-Skolem Theorem (Theorem 2.1.9), this isomorphism extends to an automorphism of B. Composing ϕ with this automorphism, we may assume that ϕ(e1A ) = e1B for all e ∈ E. Then ϕ(u σ )−1 e1B ϕ(u σ ) = −1 ϕ(u σ )−1 ϕ(e1A )ϕ(u σ ) = ϕ(u−1 σ e1A u σ ) = ϕ(σ(e)1A ) = σ(e)1B = v σ e1B v σ . Therefore, −1 ϕ(u σ )v σ ∈ CenB (E1B ) = E1B . So ϕ(u σ ) ∈ (E1B )v σ . Thus ϕ defines precisely a diagonal change of basis. It follows that f and g are homologically equivalent, that is [f] = [g]. So, indeed, θ is injective. To prove the surjectivity of θ, let [D] ∈ Br(E/F). Then, by Proposition 2.4.1, [D] contains an element A that on its turn contains E as a subfield and [E : F] = Deg(A). Every σ ∈ G yields a natural homomorphism E → A. Hence, by the Noether-Skolem Theorem (Theorem 2.1.9), there is u σ ∈ U(A) such that σ(x) = u−1 σ xu σ for every x ∈ E. We prove that the set {u σ : σ ∈ G} is E-linearly independent. Assume the contrary. Hence ∑σ∈G a σ u σ = 0 for some a σ ∈ E and non-empty X = {σ ∈ G : a σ ≠ 0} of minimal cardinality. Then ∑σ∈G a σ (ρ(b) − σ(b))u σ = ρ(b)(∑σ∈G a σ u σ )b = 0, for every b ∈ E and every ρ ∈ G. By the minimality of X we obtain that ρ(b) = σ(b), for every ρ, σ ∈ X and every b ∈ E. This implies that X has a unique element and this yields a contradiction. So, indeed, {u σ : σ ∈ G} is E-linearly independent. As dimE A = [E : F] = |G| we get that {u σ : σ ∈ G} is an E-basis for A. Consequently, A is a crossed product (E/F, f) for some 2-cocycle f . We now prove that θ is a homomorphism. Let A = (E/F, f) = ⊕σ∈G Eu σ , B = (E/F, g) = ⊕σ∈G Ev σ and C = (E/F, fg) = ⊕σ∈G Ew σ . Let M = A ⊗E B (note that the tensor product is taken over E). So, ae ⊗ b = a ⊗ eb for a ∈ A, b ∈ B and e ∈ E. We endow M with a structure of (A ⊗F B, C)-bimodule as follows: (a ⊗F b)(a󸀠 ⊗E b󸀠 ) = aa󸀠 ⊗E bb󸀠 , (a󸀠 ⊗E b󸀠 )ew σ = a󸀠 u σ ⊗E b󸀠 ev σ , for a, a󸀠 ∈ A, b, b󸀠 ∈ B, e ∈ E and σ ∈ G. This induces an F-algebra homomorphism A ⊗F B → EndC (M C ). This homomorphism is injective because A⊗F B is simple. We now show that it is an isomorphism by proving that the two algebras have the same dimension over F. Indeed, dimE (M) = n2 and dimE (C) = n. As C is simple, the right C-modules are determined, up to isomorphism, by their dimension over F. Therefore M ≅ C n , as right C-module. Hence, EndC (M) ≅ M n (C) and thus dimF (EndC (M)) = n2 dimF (C) = n4 = dimF (A⊗F B). So A ⊗F B ≅ M n (C) and we conclude that A ⊗F B is Brauer equivalent to C. (3) and (4) are re-formulations of the injectivity of θ.

68 | 2 Representations of algebras From Theorem 2.6.3 and Proposition 2.5.1 it follows that [A][E:F] = 1 for every [A] ∈ Br(E/F). Therefore Br(F) is a torsion group. The exponent of a finite dimensional central simple F-algebra A, denoted Exp(A) is the order of [A] in Br(F). In the following theorem we collect some properties on the index and exponent of A. Theorem 2.6.4. Let A be a finite dimensional central simple F-algebra. The following properties hold. (1) Exp(A) divides Ind(A). (2) Exp(A) and Ind(A) have the same prime divisors. Proof. By Theorem 2.6.3 and because A is split by a finite Galois extension E/F (see (2.4.1)), one may assume that A = (E/F, f) = ⊕σ∈G Eu σ with G = Gal(E/F) and some f ∈ Z 2 (G, U(E)). Let A = M n (D), with D a division algebra and m = Ind(A) = Ind(D) (1) Consider V = D n as a left A-module in the obvious way. Then [E : F] = Deg(A) = mn. As E ⊆ A, we also can consider V as a vector space over E. Moreover, nm2 = dimF (V) = dimE (V)[E : F] = dimE (V)mn. Therefore dimE (V) = m. We fix a basis v1 , . . . , v m of V as vector space over E and for every a ∈ A let ϕ(a) be the matrix associated to the map v 󳨃→ av, with respect to this basis. The elements of E give scalar matrices and ϕ(ea) = ϕ(e)ϕ(a) for every e ∈ E and a ∈ A. We let G act on M m (E) in the obvious way. We claim that ϕ(u σ a) = σ(ϕ(a))ϕ(u σ ) for every a ∈ A and σ ∈ G. Indeed, if ϕ(a) = (a ij ), with a ∈ A and a ij ∈ E then u σ av i = u σ ∑j a ji v j = ∑j σ(a ji )u σ v j and the claim follows. Then, ϕ(f(σ, τ))ϕ(u στ ) = ϕ(f(σ, τ)u στ ) = ϕ(u σ u τ ) = σ(ϕ(u τ ))ϕ(u σ ). Taking determinants we obtain f(σ, τ)m t στ = σ(t τ )t σ , for t σ = det(ϕ(u σ )). So, m ∈ B 2 (G, U(E)). Therefore [A]m = 1, by Theof(σ, τ)m = t σ σ(t τ )t−1 στ and thus f rem 2.6.3. Hence, Exp(A) divides m = Ind(A). (2) Let p be a prime divisor of Ind(A). Hence p divides Deg(A) = [E : F] = |G|. Let G p be a Sylow p-subgroup of G and let E p = E G p , the fixed subfield of E under G p . Then p does not divide [E p : F] and hence A E p = A ⊗F E p is not split, by Corollary 2.4.2. On the other hand E splits A, by Theorem 2.2.4 (4), and therefore it splits A E p . Hence, again by Corollary 2.4.2, Ind(A E p ) divides [E : E p ], which is a power of p. Thus 1 ≠ Ind(A E p ) and Ind(A E p ) is a proper power of p that divides Ind(A), by Corollary 2.4.2. By part (1), Exp(A E p ) is a proper power of p. As [A E p ] ∈ Br(E p ) is the image of [A] ∈ Br(F) under the homomorphism ⊗F E, it follows that the Exp(A) is divisible by p. Hence the result follows. We continue with the cohomological interpretation of the Brauer group. Let E/F and L/F be finite Galois extensions with E ⊆ L. Let Φ : Z n (Gal(E/F), U(E)) → Z n (Gal(L/F), U(L)), be given by Φ(f)(σ1 , . . . , σ n ) = f(σ1 |E , . . . , σ2 |E ). This is a group homomorphism such that Φ(B n (Gal(E/F), U(E))) ⊆ B n (Gal(L/F), U(L)).

2.6 Crossed products |

69

Thus it lifts to a group homomorphism Inf : H n (Gal(E/F), U(E)) → H n (Gal(L/F), U(L)), called the inflation map. At the level of the Brauer group, inflation is simply inclusion. More precisely, the following diagram is commutative, Inf

H 2 (Gal(E/F), U(E)) → H 2 (Gal(L/F), U(L)) ↓ θ L/F θ E/F ↓ Br(E/F) 󳨅→ Br(L/F)

(2.6.4)

Combining the equality Br(F) = ⋃E/F, finite Galois Br(E/F) (see (2.4.1)) and the commutativity of diagram (2.6.4) we obtain the homological description of Br(F) that we were seeking: Br(F) = lim H2 (Gal(E/F), U(E)), →

E/F

where the direct limit is taken over the finite Galois extensions E/F and the linking homomorphisms between the cohomology groups are the inflation maps. Proposition 2.6.5. Let E and L be subfields of a given field and let F be a subfield of E ∩ L. Let (E/F, f) be a classical crossed product. Then (E/F, f) ⊗F L = M d ((LE/L, f 󸀠 )), with d = [L ∩ E : F] and f 󸀠 (σ, τ) = f(σ|E , τ|E ) for every σ, τ ∈ Gal(LE/L). Proof. Let G = Gal(E/F), H = Gal(E/L ∩ E), K = Gal(EL/L), A = (E/F, f) = ⊕σ∈G Eu σ and B = (LE/L, f 󸀠 ) = ⊕σ∈K LEv σ . The extensions EL/L and E/L ∩ E are Galois and the map σ 󳨃→ σ|E gives an isomorphism K → H. By Proposition 2.2.4, A ⊗F (L ∩ E) is Brauer equivalent to C = CenA (L ∩ E) = ⊕σ∈H Eu σ = (E/L ∩ E, Res(f)), where Res(f) is the map f restricted to H × H. Therefore, A ⊗F L = (A ⊗F (L ∩ E)) ⊗L∩E L is Brauer equivalent to (E/L ∩ E, Res(f)) ⊗L∩E L and the map eu σ|E ⊗ l 󳨃→ elv σ (with σ ∈ G, e ∈ E and l ∈ L) defines an isomorphism (E/L ∩ E, Res(f)) ⊗L∩E L → B. Hence, A ⊗F L and B are Brauer equivalent. Since Deg(A ⊗F L) = d Deg(B) we conclude that A ≅ M d (B). Corollary 2.6.6. Let E/F be a Galois extension and let G = Gal(E/F). Let F ⊆ K ⊆ E be an intermediate field and let H = Gal(E/K). Then the following diagram is commutative ResGH

H2 (G, U(E)) 󳨀→ H2 (H, U(E)) θ E/F ↓

↓ θ E/K ⊗F K

Br(E/F) 󳨀→ Br(E/K) Corollary 2.6.6 suggests to use the notation ResFK = ⊗F K : Br(E/F) → Br(E/K) and to introduce the corestriction map CorFK : Br(E/K) 󳨀→ Br(E/F)

70 | 2 Representations of algebras

for the unique homomorphism making commutative the diagram CorGH

H2 (H, U(E)) 󳨀→ H2 (G, U(E)) θ E/F ↓

↓ θ E/K CorFK

Br(E/K) 󳨀→ Br(E/F) In this notation (2.5.3) takes the form (CorFK ∘ ResFK )([A]) = [A][K:F] .

(2.6.5)

We next show that if a finite Galois extension E/F is such that G = Gal(E/F) is a cyclic group (we simply say that the extension is cyclic) then the crossed product (E/F, f) has a simpler form. To do so, we perform a diagonal change of basis as follows. Select a generator σ of G and let n = |G|. Put u = u σ and replace the original basis by the new basis consisting of the elements u σ i = u i , with i = 0, 1, . . . , n − 1. Then a = u n ∈ E and the crossed product is completely determined by a and the action of σ on E. To emphasize this dependence this crossed product is usually denoted (E/F, σ, a). In case σ is clear from the context, this is simply denoted (E/F, a). A crossed product of this type is called a cyclic algebra. The arithmetic of cyclic algebras has some nice features: Proposition 2.6.7. Let E/F be a finite cyclic Galois extension and let σ be a generator of Gal(E/F). The following properties hold for a, b ∈ U(E). (1) (E/F, σ, a) ≅ (E/F, σ i , a i ) for every integer i coprime with n. (2) (E/F, σ, a) ≅ M n (F) if and only if a ∈ NrE/F (E). In particular, (E/F, σ, 1) ≅ M n (F). (3) (E/F, σ, a) ≅ (E/F, σ, b) if and only if ab−1 ∈ NrE/F (E). (4) [(E/F, σ, a)][(E/F, σ, b)] = [(E/F, σ, ab)]. (5) If L/F is a field extension then Gal(LE/L) ≅ Gal(E/L ∩ E) = ⟨σ k ⟩ and [(E/F, σ, a) ⊗F L] = [(LE/L, σ k , a)], for k = [L ∩ E : F]. The following example will be used latter. Example 2.6.8. (ℚ(ζ5 )/ℚ, σ, −1) ≅ M2 ( −2,−5 ℚ ). Proof. Let A = (ℚ(ζ5 )/ℚ, σ, −1) = ⊕3i=0 ℚ(ζ5 )u i , with u−1 ζ5 u = σ(ζ5 ) = ζ52 and u4 = −1. Let α = ζ5 + ζ5−1 and β = ζ5 − ζ5−1 . By straightforward calculations we obtain α2 = 1 − α,

β2 = −3 − α

uα = −α−1 u = −(α + 1)u,

(2α + 1)2 = 5, and

uβ = αβu.

1 Let x = 2α+1 (βu + αβu3 ). Then x2 = 1 and hence e = 1+x 2 is a non-central idempotent 2 of A. Moreover x2α+1 = x u = −x and hence conjugation by u2 gives an isomorphism eAe ≅ (1 − e)A(1 − e). Let i = e(u + u 3 )e and j = e(2α + 1)u2 e. Then i2 = −2e, j2 = −5e and ji = −ij. Therefore eAe contains the subalgebra ℚ(i, j) ≅ ( −2,−5 ℚ ). By Example 2.1.7, ℚ(i, j) is

2.6 Crossed products |

71

a rational division algebra of dimension 4. This implies that dimℚ (eAe) = dimℚ ((1 − e)A(1 − e)) ≥ 4. Because the left ℚ(i, j)-module eA(1 − e) is non-zero, its dimension over ℚ is also at least 4 and hence also dimℚ (1 − e)Ae ≥ 4. It follows that eAe = ℚ(i, j) and hence the result follows. The following deep classical theorem states that many central simple algebras are cyclic (for proof we refer to [175, Theorem 18.6]). Theorem 2.6.9. If F is a number field, then every finite dimensional central simple Falgebra is a cyclic algebra (and in particular a crossed product). We finish this section with constructing a crossed product from a group extension 1 → U → H → G → 1, where G is a finite group and U is a finite cyclic group, such that the action of G on U is faithful. Let f ∈ Z 2 (G, U) be a 2-cocycle representing this extension. Without loss of generality, one may assume that U is generated by a complex root of unity ξ and that G is a group of automorphisms of U. The restriction to U = ⟨ξ ⟩ of an automorphism of ℚ(ξ) is an automorphism of U and this provides a group isomorphism Ψ : Gal(ℚ(ξ)/ℚ) → Aut(U). Let G = Ψ −1 (G) and let F be the fixed field of G. Hence G = Gal(ℚ(ξ)/F) and there is a group homomorphism Ψ : Z 2 (G, ⟨ξ ⟩) → Z 2 (G, U(ℚ(ξ))) f

→

f

given by f (σ, τ) = f(Ψ(σ), Ψ(τ)), for σ, τ ∈ G. Furthermore, Ψ(B2 (G, ⟨ξ ⟩)) ⊆ B2 (G, U(ℚ(ξ))) and henceforth Ψ lifts to a group homomorphism H2 (G, ⟨ξ ⟩) → H2 (G, U(ℚ(ξ))). Thus the extension defined by f yields a classical crossed product (ℚ(ξ)/F, f ). Observe that if the extension is split (i.e. f ∈ B2 (G, U)) then the crossed product is split, but the converse is not necessarily true.

Problems 2.6.1. Let U be an abelian group. Prove that a short exact sequence 1 → U → H → G → 1 is split if and only if one (and thus any) 2-cocycle representing the sequence is a 2-coboundary.

72 | 2 Representations of algebras

2.6.2. Prove some of the results given without a proof. 2.6.3. Calculate the equivalence classes of extensions of C2 by C4 . 2.6.4. Calculate the identity of (E/F, f) for an arbitrary cocycle f . 2.6.5. Is a quaternion algebra a crossed product? Is it a cyclic algebra? 2.6.6. Let A = (E/F, f) be a crossed product. Prove that E is a maximal subfield of A. j

p

2.6.7. Let 1 → B → H → G → 1 be an exact sequence of arbitrary groups and for every g ∈ G, let u g ∈ H be such that p(u g ) = g. Construct maps α : G → Aut(B) and f : G × G → B given by the following equalities u g j(b) = j(α g (b))u g ,

u g u h = j(f(g, h))u gh

(g, h ∈ G, b ∈ B),

(2.6.6)

and prove that the associativity in H is equivalent to the conditions in (2.6.2). Investigate how α and f are affected by replacing the list u g , by new elements v g ∈ p−1 (g). Reverse the process, that is, starting from maps α and f as above construct a group H j

p

and an exact sequence 1 → B → H → G → 1. 2.6.8. Prove that the reduced characteristic polynomial, reduced trace and reduced norm do not depend on the splitting field E nor the isomorphisms h : A E → M n (E). (Hint: Noether-Skolem Theorem.) 2.6.9. Let E/F be a Galois extension of dimension n. By Theorem 2.6.3, the skew group ring (E/F, 1) is isomorphic to M n (F) and hence it is isomorphic to EndF (E). In this problem one gives a concrete isomorphism. If x ∈ E then let x󸀠 : E 󳨃→ E denote the left multiplication by x map. Prove that xu σ 󳨃→ x󸀠 σ defines an isomorphism of F-algebras (E/F, 1) → EndF (E).

3 Wedderburn decomposition of semisimple group algebras In this chapter F is a field of characteristic p ≥ 0 and G is a finite group of order coprime with p. From Theorem 2.1.5 we know that FG is a semisimple algebra and hence, by the Wedderburn-Artin Theorem (Theorem 2.1.3), FG is a direct product of matrix rings over division algebras, called the Wedderburn components of FG. In order to study the unit group of integral group rings, it is crucial to develop an algorithmic method that gives a precise description of each of the Wedderburn components of the rational group algebra ℚG and, more generally, of FG. The Wedderburn components of FG are in one-to-one correspondence with the primitive central idempotents of FG. In the first section we revise the basic representation theory of a finite groups G. In particular, we introduce the absolutely irreducible characters of G. It turns out that if FG is split then the Wedderburn components are in one-to-one correspondence with the absolutely irreducible characters of G and it is easy to compute the primitive central idempotents and the Wedderburn components of FG. In general, there is a surjective map (not necessarily injective) associating each absolutely irreducible character χ of G with one of the primitive central idempotents of FG, denoted e F (χ). In this case A F (χ) = FGe F (χ) is the Wedderburn component of FG associated to e F (χ). In Section 3.3 we will show how to compute e F (χ) using Galois descent. Our approach is taken from [231]. Unfortunately, a description of the central idempotents does not provide much information on the Wedderburn components associated to each idempotent. However, in case F has positive characteristic or G is abelian, one can easily describe A F (χ). Most of the present chapter is dedicated to describe A F (χ) in characteristic zero. Because of the applications to integral group rings, we pay special attention to the case F = ℚ. In Section 3.4 we calculate the primitive central idempotents of ℚG that are determined by a monomial character but without actually calculating the monomial irreducible character of G. It turns out that such an idempotent is determined by a Shoda pair (H, K) of subgroups H and K of G. In Section 3.5 we will give a description of simple components determined by such a Shoda pair provided it satisfies some additional conditions; these will be called strong Shoda pairs. Without calculating the character table of G, this will yield a full description of the Wedderburn decomposition of ℚG for a large family of groups, called the strongly monomial groups. This class includes for example all finite abelian-by-supersolvable groups. These results are due to Olivieri, del Río and Simón [165]. In Section 3.6 we prove weak versions of two classical representation theory results: Brauer’s Theorem on Induced Characters and the Berman-Witt Theorem. In Section 3.7 we prove a result of Brauer-Witt giving more specific information on the structure of the Wedderburn components. The proof given follows the approach of Olteanu [167]: it is a computational oriented variation

74 | 3 Wedderburn decomposition

of the classical proof as given in [231]. Even if G is not a strongly monomial group, this method allows the description of many simple components of ℚG. In the final section we outline how this method has been implemented in GAP and we explain its language via some examples.

3.1 Representations and characters of finite groups In this section F is a field of characteristic p ≥ 0 and G is a group. ̃ =∑ Recall that for a finite subset X of G, we write X x∈X x ∈ FG and, if, moreover 1 1 ̂= X ̃= |X| is not divisible by p, then X x ∈ FG. ∑ |X| |X| x∈X Straightforward calculations yield the following result. ̃ where X runs Proposition 3.1.1. If R is a commutative ring then the class sums X, through the finite conjugacy classes of G, form a basis of the center of RG over R. In particular, if F is a field then the dimension of the center of FG is the number of finite conjugacy classes of G. Semisimple group algebras are characterized by Maschke’s Theorem which is a direct consequence of Theorem 2.1.5. Corollary 3.1.2 (Maschke’s Theorem). If F is a field and G is a group then the group algebra FG is semisimple if and only if G is finite and its order is not a multiple of the characteristic of F. The following Theorem is a direct consequence of Maschke’s Theorem and Proposition 2.1.11. Corollary 3.1.3. Every semisimple group algebra is separable. An F-representation of the group G is a group homomorphism ρ : G → GLn (F). The positive integer n is called the degree of the representation. Two F-representations ρ and ρ󸀠 of G are said to be equivalent if they have the same degree n and there is U ∈ GLn (F) such that Uρ(g) = ρ󸀠 (g)U, for every g ∈ G. We now make a link between group algebras and group representations. Clearly, an F-representation ρ of G of degree n extends uniquely to an F-algebra homomorphism ρ : FG → M n (F). Moreover, if ρ󸀠 is another representation then ρ and ρ󸀠 are equivalent if and only if ρ and ρ󸀠 are equivalent. This shows that an F-representation of G and a representation of FG (or equivalently, a left FG-module, see Section 2.3) define the same mathematical notion. In the sequel, we also will denote ρ simply as ρ. We will identify equivalent classes of F-representations of G and isomorphism classes of FG-modules of finite dimension over F and we will use this to transfer module theoretical notions to representation theoretical notions. For example, if ρ : G → GLn (F) and τ : G → GLm (F) are F-representations of G then the direct sum of ρ and τ is the

3.1 Representations and characters of finite groups

|

75

representation ρ ⊕ τ : G → GLn+m (F) given by (ρ ⊕ τ)(g) = (

ρ(g) 0

0 ), τ(g)

with g ∈ G. The reader can easily check that, with the appropriate choice of basis, the module and representation theoretical notions of direct sum agree. The trivial representation 1G : G → U(F) of G is simply the map defined by g 󳨃→ 1. It is the representation of the left module FG F, with multiplication given by gx = x for g ∈ G and x ∈ F. If G is finite, then the left regular representation of G simply means the F-representation of FG considered as left FG-module. The customary terminology for modules and representations differs for some notions. For example, the F-representation associated to a simple FG-module is said to be irreducible, instead of simple. Let ρ be an F-representation of G. The character afforded by ρ is the map χ:G→F g 󳨃→ χ(g) = tr(ρ(g)). An F-character of G, by definition, is the character afforded by an F-representation. In the language of FG-modules, if M is an FG-module then the character afforded by M is the character afforded by its representation with respect to a chosen (and thus any) basis. Hence, the character χ afforded by ρ is the same as the character afforded by the FG-module M associated to ρ and this is precisely the restriction of TrM/F to G (see Section 2.3). The F-character afforded by an irreducible F-representation is called an irreducible F-character. Similar as we agreed for representations, if χ is the character afforded by an Frepresentation of G then we also denote by χ the linear extension of χ to FG. For example, if G is finite and σ is the character of the regular left FG-module then σ( ∑ a g g) = |G|a1 ,

(3.1.1)

g∈G

where a g ∈ F for g ∈ G. An F-class function of G is a map f : G → F which is constant on each conjugacy class of G. The set of F-class functions has an obvious structure of F-algebra and its dimension equals the number of conjugacy classes of G. Lemma 3.1.4. Let ϕ and ψ be the characters afforded by the F-presentations ρ and τ of G respectively. The following properties hold. (1) If n is the degree of ρ then ϕ(1) = n1F . In particular, if F has zero characteristic then ϕ(1) is the degree of ρ. (2) ϕ is an F-class function of G. (3) The character of ρ ⊕ τ is ϕ + ψ. Assume that FG is semisimple, let S1 , . . . , S k be representatives of the isomorphism classes of the simple FG-modules and let χ i denote the character afforded by S i . Then

76 | 3 Wedderburn decomposition m

m

every FG-module that is finite dimensional over F is isomorphic to S1 1 ⊕ ⋅ ⋅ ⋅ ⊕ S k k , for some non-negative integers m1 , . . . , m k , and the character afforded by this module is m1 χ1 + ⋅ ⋅ ⋅ + m k χ k . Thus the F-characters of G are the functions of the form m1 χ1 + ⋅ ⋅ ⋅ + m k χ k , with each m i a non-negative integer. If G is finite and the characteristic of F is not a divisor of |G|, then the following formula defines a symmetric bilinear form on the F-vector space consisting of the Fclass functions of G: 1 [ϕ, ψ] = ∑ ϕ(g)ψ(g −1 ). |G| g∈G From Lemma 2.2.1 and Corollary 3.1.3 we know that every semisimple group algebra FG has a splitting field that is a finite extension of F. If FG is semisimple and split then we simply call F a splitting field of G. We now relate the primitive central idempotents of such an FG with the irreducible F-characters of G. Theorem 3.1.5. Let F be a splitting field of a finite group G. Let e1 , . . . , e n be the primitive central idempotents of FG and let χ i denote the character of the unique (up to isomorphism) simple module of FGe i . The following properties hold. (1) n is the number of conjugacy classes of G. (2) If d i is the degree of FGe i then |G| = ∑ni=1 d2i . In particular, if F has characteristic zero then |G| = ∑ni=1 χ i (1)2 . i (1) (3) e i = χ|G| ∑g∈G χ i (g −1 )g. χ i (1) ≠ 0, (4) χ i (e j ) = { 0,

if i = j; otherwise.

(5) [Generalized Orthogonality Relations] For all h ∈ G, χ j (1) ∑ χ i (gh)χ j (g −1 ) = δ ij |G|χ i (h). g∈G

(6) χ1 , . . . , χ n is an orthonormal basis of the vector space consisting of the F-class functions of G with respect to the bilinear form [ , ]. (7) χ1 , . . . , χ n are linearly independent over any field extension of F. (8) For every g, h ∈ G we have n

|CenG (g)| 1F , 0,

∑ χ i (g)χ i (h−1 ) = { i=1

if g and h are conjugate in G; otherwise.

(9) If E is another splitting field of G such that E and F are subfields of a common field then χ1 , . . . , χ n also are the irreducible E-characters of G. Proof. (1) Since, by assumption, FG is semisimple and split, we get that Z(FG) ≅ F n , where n is the number of primitive central idempotents of FG. Hence, from Proposition 3.1.1, we obtain that n = dimF (Z(FG)) is the number of conjugacy classes of G. (2) Put A i = FGe i . Then FG = A1 ⊕⋅ ⋅ ⋅⊕A n is the Wedderburn decomposition of FG. Because of the split assumption, each A i is isomorphic with M d i (F) for some positive

3.1 Representations and characters of finite groups

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77

integer d i . Then |G| = dimF (FG) = ∑ni=1 dimF (A i ) = ∑ni=1 d2i . The second statement is a consequence of Lemma 3.1.4. (3) If S i is the left ideal of A i mapped via some isomorphism A i ≅ M d i (F) to the first column of M d i (F) then S1 , . . . , S n are representatives of the isomorphism classes d d of simple A-modules and d i = dimF (S i ) and A i ≅ S i i for every i. Thus FG ≅ ⊕ni=1 S i i . Reordering the simple components one may assume that χ i is the character afforded by S i , so that χ i (1) = d i 1F . Hence, if σ represents the character of the regular representation then n

σ = ∑ χ i (1)χ i .

(3.1.2)

i=1

Let ρ i denote the representation associated to S i with respect to some basis. Then ρ i (e j ) = δ ij I d i . Write e i = ∑g∈G a g g. From (3.1.1) and (3.1.2) we thus obtain n

n

a g |G| = σ(e i g−1 ) = ∑ χ j (1) χ j (e i g−1 ) = ∑ χ j (1) tr(ρ j (e i g−1 )) j=1

n

j=1 −1

= ∑ χ j (1) tr(ρ j (e i )ρ j (g )) = χ i (1) tr(ρ i (g −1 )) = χ i (1)χ i (g −1 ). j=1 χ (1)

i Hence e i = |G| ∑g∈G χ i (g −1 )g (note that |G| is invertible in F because FG is semisimple). (4) As e i ≠ 0, the formula in (3) implies that χ i (1) ≠ 0. χ i (1) χ i (h) (5) Let h ∈ G. By (3), the coefficient of h−1 in e i (as an element of FG) is |G|

χ (1)χ (1)

while the coefficient of h−1 in e i e j is i |G|2j ∑g∈G χ i (gh)χ j (g −1 ). Because e i e j = δ ij e i and χ i (1) ≠ 0 (by part (4)), we thus obtain (5). (6) Applying (5) with h = 1 we obtain [χ i , χ j ] = δ ij , i.e. χ1 , . . . , χ n are orthonormal. Hence χ1 , . . . , χ n are F-linearly independent and thus form a basis of the ndimensional space of F-class functions of G. (7) This is an immediate consequence of (6). (8) Let g1 , . . . , g n be representatives of the conjugacy classes of G. Let m i = [G : CenG (g i )], the cardinality of the conjugacy class containing g i . Consider the n × n matrices A = (a ij ) and B = (b ij ) with a ij = m j χ i (g j )

and

b ij =

χ j (g −1 i ) . |G|

Then the (i, j)-entry of AB is 1 n 1 −1 ∑ m k χ i (g k )χ j (g −1 k ) = |G| ∑ χ i (g)χ j (g ) = [χ i , χ j ] = δ ij , |G| k=1 g∈G by (6). Therefore AB = 1 and hence BA = 1. In other words n 1 1 n )χ k (g j ) = ∑ χ k (g −1 ∑ m j χ k (g −1 i i )χ k (g j ) = δ ij . |CenG (g j )| k=1 |G| k=1

78 | 3 Wedderburn decomposition

(9) We have to prove that χ1 , . . . , χ n also are the irreducible E-characters of G. Let K be a field containing both F and E. Since K also is a splitting field of G, we obtain from (1) that n also is the number of primitive central idempotents of KG. Then the primitive central idempotents of FG, EG and KG coincide. Because of (3), the irreducible E-characters of G are determined by the coefficients of these primitive central idempotents of EG. Therefore the irreducible E-characters coincide with the irreducible F-characters of G. Assume that G is finite and p = 0 or p is a prime number that does not divide the order of G. The absolutely irreducible characters of G, in characteristic p, are by definition the irreducible F-characters of G where F is a splitting field of characteristic p of G. A representation is absolutely irreducible if it affords an absolutely irreducible character. Because of Theorem 3.1.5 (9) these definitions are independent of the chosen splitting field within a fixed algebraic closure of the prime field (of characteristic p). Throughout, we will implicitly assume that we fix a field containing the algebraic closure of the prime field in characteristic p and that all absolutely irreducible characters are calculated within this field. For example, in zero characteristic we fix the field of complex numbers and hence the absolutely irreducible characters in zero characteristic are the irreducible complex characters. If χ is an absolutely irreducible character of G then we set e(χ) =

χ(1) ∑ χ(g−1 )g, |G| g∈G

(3.1.3)

which, by Theorem 3.1.5, is the unique primitive central idempotent e of any split semisimple group algebra over G of characteristic p, such that χ(e) ≠ 0. The set of the irreducible complex characters of G usually is denoted Irr(G). We denote by Irrp (G) the set of the absolutely irreducible characters in characteristic p. If Irr(G) = {χ1 , . . . , χ k } then the character table of the group G is a table of the form g1 c1

g2 c2

... ...

gk ck

χ1

χ1 (g1 )

χ1 (g2 )

...

χ1 (g k )

χ2 ... χk

χ2 (g1 ) ... χ k (g1 )

χ2 (g2 ) ... χ k (g2 )

... ... ...

χ2 (g k ) ... χ k (g k )

The first row is formed by representatives g1 = 1, g2 , . . . , g k of the conjugacy classes of G and the second row is formed by the cardinalities of these conjugacy classes, that is c i = [G : C G (g i )]. Let E/F be a field extension. By considering GLn (F) as a natural subgroup of GLn (E), we may consider every F-representation ρ as an E-representation, denoted ρ E . If ρ is the F-representation of G associated to the left FG-module M with respect

3.1 Representations and characters of finite groups |

79

to a basis B, then ρ E is the E-representation of the EG-module E ⊗F M with respect to the basis 1 ⊗ B = {1 ⊗ b : b ∈ B}. Therefore ρ and ρ E afford the same character and the character values of both ρ and ρ E are contained in F. Note that an E-character can have values in F while not being an F-character. For example, the quaternion group Q8 = ⟨a, b | a4 = a2 b2 = 1, ba = a−1 b⟩ of order 8 has the following ℚ(i)representation ρ defined by i 0

ρ(a) = (

0 ), −i

0 −1

ρ(b) = (

1 ). 0

Clearly, the character values of the character χ afforded by ρ are contained in ℚ. However, χ is not a ℚ-character (see Problem 3.1.2). Assume that F is a field of characteristic p, G is a finite group and suppose FG is semisimple. Let Irrp (G) = {χ1 , . . . , χ n } be the set of absolutely irreducible characters of G in the characteristic of F. One may assume that χ1 , . . . , χ n are E-characters, for E a field containing F. Let ρ i be an irreducible E-representation affording the character χ i . If ρ is an F-representation affording the character χ then ρ E is equivalent to m1 ρ1 ⊕ ⋅ ⋅ ⋅ ⊕ m k ρ n for some unique non-negative integers m1 , . . . m n . Then χ = m1 χ1 + ⋅ ⋅ ⋅ + m n χ n . The constituents of ρ are by definition the absolutely irreducible characters χ i with m i ≠ 0. By Theorem 3.1.5 (7), m1 1F , . . . , m n 1F are uniquely determined by the character χ. In particular, if p = 0 then m1 , . . . , m n are determined by χ. This proves the following theorem. Theorem 3.1.6. Let F be a field of characteristic 0. If G is finite group then two Frepresentations of G are equivalent if and only if they afford the same character. Because of Theorem 3.1.6, in zero characteristic, one can transfer notions of representations, which are preserved by equivalence, to notions of characters. For example, the degree, the constituents and the kernel of a character χ in characteristic 0, are respectively the degree, the constituents and the kernel of any representation ρ affording χ. They can be calculated directly from χ. Indeed, obviously the degree of χ is χ(1) and the constituents of χ are the irreducible complex characters ϕ such that [χ, ϕ] ≠ 0. Finally the kernel of χ is ker(χ) = {g ∈ G : χ(g) = χ(1)}. Indeed, let ρ be a complex representation affording the character χ. As ρ(g) has finite order it is diagonalizable and the eigenvalues of ρ(g) are roots of unity. Let d = χ(1), the degree of ρ, and let ξ1 , . . . , ξ d be the eigenvalues of ρ(g). Then χ(g) = ξ1 + ⋅ ⋅ ⋅ + ξ d and hence |χ(g)| ≤ d. Thus χ(g) = χ(1) if and only if ξ1 = ⋅ ⋅ ⋅ = ξ d = 1, or equivalently, if ρ(g) = 1. In positive characteristic, however, this is not possible because non-equivalent representations ρ1 and ρ2 may have the same characters but different constituents (see Problem 3.1.3). However if ρ1 and ρ2 are irreducible E-representations of G for E

80 | 3 Wedderburn decomposition

a splitting field of G, then ρ1 and ρ2 are equivalent (Problem 3.1.5). Thus there is no ambiguity in defining the degree and kernel of an absolutely irreducible character χ of G in characteristic p, as the degree and kernel of an irreducible E-representation affording ρ, for E a splitting field of G of characteristic p. Lemma 3.1.7. Let F be a field of characteristic p ≥ 0. If FG is semisimple, then the constituents of the characters of the irreducible F-representations of G form a partition of the set of the absolutely irreducible characters (in characteristic p) of G. Moreover, if χ is an irreducible F-character with constituents ϕ1 , . . . , ϕ k and the unique Wedderburn component A of FG with χ(A) ≠ 0 is of the form M m (D), where D is a division algebra, then all ϕ i have the same degree, say n, m | n and χ = mn ∑ki=1 ϕ i . Proof. Let E be a splitting field of G containing F. By Theorem 2.2.5, we may assume without loss of generality that E is a Galois extension of F. If FG = A1 ⊕ ⋅ ⋅ ⋅ ⊕ A k is the Wedderburn decomposition of FG then EG = (E ⊗F A1 ) ⊕ ⋅ ⋅ ⋅ ⊕ (E ⊗F A k ) and every Wedderburn component of EG is contained in exactly one of the E ⊗F A i ’s. Therefore, if EG = B1 ⊕ ⋅ ⋅ ⋅ ⊕ B n is the Wedderburn decomposition of EG, then there is a partition {J1 , . . . , J k } of {1, . . . , n} such that E ⊗F A i = ⊕j∈J i B j . Let S i be a simple FG-submodule of A i and T j a simple EG-submodule of B j . By Proposition 2.1.11 (2) all the B j ’s, with j ∈ J i , are isomorphic as F-algebras, and hence, if ϕ j is the character afforded by the EG-modules T j then all ϕ j with j ∈ J i have the same degree, say n i . Then, as FGm modules, A i ≅ S i i , if A i = M m i (D i ) with D i a division algebra, and as EG-modules ni n B j ≅ T j . Therefore ⊕j∈J i T j i ≅ ⊕j∈J i B j = E ⊗F A i ≅ (E ⊗F S i )m i as EG-modules. Then ni m

m i divides n i and ⊕j∈J i T j i ≅ E ⊗F S i . Since the F-character χ i afforded by S i is the E-character afforded by E ⊗F S i , the constituents of χ i are {ϕ j : j ∈ J i }. Since the T j ’s are representatives of the simple EG-modules, the ϕ j ’s are the different absolutely irreducible characters and χ i = mn ii ∑j∈J i ϕ j for every i. Proposition 3.1.8. Let χ be an F-character of a finite group G, where F is a splitting field of G. If χ is absolutely irreducible then [χ, χ] = 1F . The converse holds in zero characteristic. Proof. The first statement is a particular case of Theorem 3.1.5 (5). Furthermore, if χ1 , . . . , χ n are constituents of χ and χ = m1 χ1 + ⋅ ⋅ ⋅ + m n χ n then [χ, χ] = ∑ni=1 m2i 1F . Hence, in characteristic zero we get that if [χ, χ] = 1, then χ = χ i for some i, as desired. Note that the proof of the previous result also implies that [χ, χ] is a positive integer for any character of G in characteristic 0. In characteristic p > 0, [χ, χ] is an element of the field 𝔽p with p elements.

3.2 Some operations with characters

|

81

Problems 3.1.1. Prove Proposition 3.1.1. 3.1.2. Calculate the character tables of D8 and Q8 . Prove that ℚ is a splitting field of D8 . Every irreducible complex character of Q8 takes rational values. Show that ℚ is not a splitting field of Q8 . 3.1.3. Prove the following strong opposite of Theorem 3.1.6. If F is a field of positive characteristic and ρ an F-representation of a finite group G then G has an Frepresentation not equivalent to ρ affording the same character as ρ. 3.1.4. Give examples of non-equivalent F-representations with the same character and different constituents. 3.1.5. Let E be a splitting field of a finite group G and let ρ and σ be irreducible Erepresentations of G. Prove that if ρ and σ afford the same character then they are equivalent. 3.1.6. Let ρ be an F-representation of a finite group G and χ the character afforded by ρ. Prove that χ is absolutely irreducible if and only if ρ E is irreducible for every field extension E of F. 3.1.7. Let G be a finite group and let p be either 0 or a prime integer. Prove that G is abelian if and only if every absolutely irreducible character in characteristic p has degree 1.

3.2 Some operations with characters Assume G and H are groups, M is an FG-module and L is an FH-module. Then M ⊗F L is an F(G × H)-module with multiplication given by (g, h)(m ⊗ n) = gm ⊗ hn, for g ∈ G, h ∈ H, m ∈ M and n ∈ L. If M and L are finite dimensional over F and χ and θ are the characters afforded by M and L respectively, then the character afforded by M ⊗F L as F(G × H)-module is denoted χ ⊗ θ and is given by the following formula (χ ⊗ θ)(g, h) = χ(g)θ(h).

(3.2.1)

The following equality, for characters χ 1 and χ2 of G and characters θ1 and θ2 of H, is easily verified: [χ1 ⊗ θ1 , χ2 ⊗ θ2 ] = [χ1 , χ2 ] [θ1 , θ2 ] In case G = H, then one can consider G as a subgroup of G × G via the diagonal embedding g 󳨃→ (g, g), for g ∈ G. Restricting scalars to FG, one has that M ⊗F L also

82 | 3 Wedderburn decomposition

can be considered as an FG-module. By (3.2.1), the character of G afforded by this FGmodule M ⊗F L is precisely the pointwise multiplication χ ⋅ θ of χ and θ: (χ ⊗ θ)(g) = (χ ⋅ θ)(g) = χ(g)θ(g). This shows that the class of F-characters of G is closed under products. Let f : G → G 1 be a group homomorphism. If ρ is an F-representation of G1 then ρ ∘ f is an F-representation of G. If χ is the character afforded by ρ then the character afforded by ρ ∘ f is χ ∘ f . If χ ∘ f is absolutely irreducible, then χ is absolutely irreducible. There are three relevant particular cases of this construction that we mention now. If H is a subgroup of G and ρ is an F-representation of G then the restriction ρ H of ρ to H is ρ ∘ j, for j : H → G the inclusion map. If χ is the character afforded by ρ then the character afforded by ρ H is χ H , the restriction of χ to H. For the second example, consider a subgroup H of a group G and g an element of G. We use the exponential notation for conjugation, i.e. h g = g −1 hg and H g = {h g : h ∈ H} for h ∈ G and H ≤ G Then h 󳨃→ ghg−1 defines an isomorphism H g → H. If ρ is an F-representation of H affording the character χ then the composition with ρ provides an F-representation of H g which affords the character χ g (x) = χ(gxg−1 ), for g ∈ G. Clearly χ is absolutely irreducible if and only if χ g is absolutely irreducible and in this case e(χ g ) = e(χ)g = g −1 e(χ)g. For the third example we consider a normal subgroup N of G and π : G → G/N the natural group epimorphism. Then every F-representation σ of G/N provides an F-representation ρ = σ ∘ π of G such that N ⊆ ker(ρ). Conversely, if ρ is an F-representation of G such that N ⊆ ker(ρ) then ρ = σ ∘ π for a unique F-representation σ of G/N. Thus there is a one-to-one correspondence between equivalence classes of F-representations of G/N and equivalence classes of F-representations of G with kernel containing N. This one-to-one correspondence preserves irreducible F-representations. Therefore, if χ is a character of G/N then it is absolutely irreducible if and only if so is χ ∘ π and in this case π(e(χ ∘ π)) = e(χ). If furthermore, FG is semisimple ̂ are the Wedderburn components then by (1.5.7) the Wedderburn components of FG N of FG corresponding to equivalence classes of representations ρ with N ⊆ ker(ρ). In ̂󸀠 are the commutative Wedderburn particular, the Wedderburn components of FG G components of FG. An F-representation of a group G is said to be linear if it has degree 1. A linear character of G is the character afforded by a linear F-representation (for some field F). Clearly, every linear F-representation is irreducible and every linear character is absolutely irreducible. Let σ : F → E be a field homomorphism. Then σ induces a natural ring homomorphism σ : M n (F) → M n (E). If ρ is an F-representation of G of degree n, then σ ∘ ρ is an

3.2 Some operations with characters

|

83

E-representation of G of degree n. If χ is the character afforded by ρ then the character afforded by σ ∘ ρ is σ ∘ χ. Moreover, via the action on the coefficients, σ defines a ring homomorphism σ : FG → EG. If χ is an absolutely irreducible character then so is σ ∘ χ, and in this case e(σ ∘ χ) = σ(e(χ)). Assume now that ρ is an F-representation of a subgroup H of finite index in G. Then we define the induced representation ρ G as follows. Take a left FH-module L such that ρ is the representation of L (as FH-module) with respect to some basis over F and let ρ G be the representation of FG ⊗FH L (as FG-module) with respect to some basis over F. Of course the specific form of ρ G , depends on the chosen bases of L and FG ⊗FH L respectively, but, this only up to equivalence. Therefore the character afforded by ρ G is independent of this choice. If ϕ is the character afforded by ρ then the character afforded by ρ G is denoted ϕ G and it is called the character of G induced by ϕ. The following formula provides an expression for ϕ G in terms of ϕ (see Problem 3.2.2): ϕ G (g) = ∑ ϕ(g t ),

(3.2.2)

t∈T,g t ∈H

where T is a left transversal of H in G. If furthermore, H is finite and |H| is not multiple of the characteristic of F then ϕ G (g) =

1 ∑ ϕ(g x ), |H| x∈G,g x ∈H

(3.2.3)

Using (3.2.2) it is easy to prove the following equalities for χ a character of G and ϕ a character of H. [χ, ϕ G ] = [χ H , ϕ] (Frobenius Reciprocity). χϕ = (χ H ϕ) . G

G

(3.2.4) (3.2.5)

To characterize when an induced character is absolutely irreducible we will need Mackey’s Theorem. To state this result, we recall the definition of double coset. Let G be a group and let H and K be subgroups of G. The (K, H)-double coset of G are the sets of the form KgH = {kgh : k ∈ K, h ∈ H}. They form a partition of G. Proposition 3.2.1 (Mackey’s Theorem). Let G be a finite group, H and K subgroups of G, T a set of representatives of double (H, K)-cosets of G and ϕ a character of H. Then (ϕ G )K = ∑ (ϕ tH t ∩K )K . t∈T

Proof. Let L be a left FH-module affording the character ϕ and let B be an F-basis of L. Note that T −1 = {t−1 : t ∈ T} is a set of representatives of the double (K, H)-cosets of G. Every double coset Kt−1 H is a disjoint union of some left H-cosets of G. More precisely, Kt−1 H = ∪k∈T t kt−1 H, where T t is a set of representatives of left cosets of H t ∩ K in K. Then {kt−1 : t ∈ T, k ∈ T t } is a set of representatives all the left H-cosets of G. Thus {kt−1 ⊗ b : t ∈ T, k ∈ T t , b ∈ B} is an F-basis of FG ⊗FH L. For every t ∈ T, x ∈ t−1 Ht

84 | 3 Wedderburn decomposition and n ∈ L, we have x(t−1 ⊗ n) = t−1 ⊗ (txt−1 )n. This shows that t−1 ⊗ L is a left F(H t )module and the character afforded by this module is precisely ϕ t . Let L t be the F-span of {kt−1 ⊗ b : k ∈ T t , b ∈ B}. Then L t is an FK-submodule of FG ⊗FH L and FG ⊗FH L = ⊕t∈T L t . Furthermore L t = ∑k∈T t k(t−1 ⊗ L) = FK ⊗F(H t ∩K) (t−1 ⊗ L) and therefore, the t K character of K afforded by L t is (ϕ tH t ∩K )K . It follows that (ϕ G )K = ∑t∈T (ϕ(H t ∩K) ) . Corollary 3.2.2 (Mackey). Let ϕ be a complex character of a subgroup H of a finite group G. Then ϕ G is absolutely irreducible if and only if ϕ is absolutely irreducible and g [ϕ H g ∩H , ϕ H g ∩H ] = 0, for every g ∈ G \ H. Proof. By Frobenius Reciprocity (3.2.4) and the Mackey Theorem (Proposition 3.2.1), we have [ϕ G , ϕ G ] = [ϕ, (ϕ G )H ] = ∑t∈T [ϕ, (ϕ tH t ∩H )H ] = ∑t∈T [ϕ H t ∩H , ϕ tH t ∩H ], where T is a set of representatives of double (H, H)-cosets. If t ∈ H then ϕ tH t ∩H = ϕ and hence one of the summands of the previous sum is [ϕ, ϕ] > 0. Thus, from Proposition 3.1.8 we know that ϕ G is absolutely irreducible if and only if [ϕ G , ϕ G ] = 1, or equivalently [ϕ, ϕ] = 1 and [ϕ H t ∩H , ϕ tH t ∩H ] = 0 for every t ∈ T \ H. Hence the result follows. The specialization of Corollary 3.2.2 to the case when ϕ is linear is due to Shoda [211] and can be rephrased as follows. Corollary 3.2.3. Let ϕ be a linear complex character of a subgroup H of a finite group G and let K = ker(ϕ). Then ϕ G is absolutely irreducible if and only if for every g ∈ G \ H there is h ∈ H such that h g ∈ H and (h, g) ∈ ̸ K. Proof. This follows from Corollary 3.2.2 by observing that, as ϕ is a linear character, g g [ϕ H g ∩H , ϕ H g ∩H ] = 0 if and only if ϕ H g ∩H ≠ ϕ H g ∩H , or equivalently there is x ∈ H ∩ H g such that ϕ(x) ≠ ϕ(gxg−1 ). Of course, the latter means that h g ∈ H for some h ∈ H such that (h, g) ∈ ̸ K. A character (respectively, representation) of G is said to be monomial if it is the character (respectively representation) afforded by a representation induced from a linear character. One says that G is monomial if every irreducible complex character of G is monomial.

Problems 3.2.1. Let G and H be finite groups. Prove (1) Irr(G × H) = {χ ⊗ θ : χ ∈ Irr(G), θ ∈ Irr(H)}. (2) If χ ∈ Irr(G) and ϕ ∈ Irr(H) then e(χ ⊗ ϕ) = e(χ)e(ϕ) (here we identify FG and FH as subalgebras of F(G × H) in the obvious way). 3.2.2. Prove the formulas (3.2.2), (3.2.4) and (3.2.5). (Hint for (3.2.2): Fix a basis of an FH-module L and use a left transversal of G modulo H to obtain a basis of FG ⊗FH L.)

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85

3.2.3. Let H be a subgroup of finite index in a group G and F a field. Suppose L is a left FH-module that is finite dimensional over F and let B be an F-basis of L. Let ρ be the representation of H associated to L with respect to the basis B. Assume T is a left transversal of H in G and let ρ G be the representation of G associated to the FGmodule FG ⊗FH L with respect to the basis consisting of the elements t ⊗ b with t ∈ T and b ∈ B. Prove that ρ(g) is a |T| × |T|-block matrix with ρ(t−1 1 gt) in the position (t 1 , t) if t1 H = gtH and zeros in all other positions. Also use this to prove (3.2.2).

3.3 Wedderburn components from character tables In this section F is a field of characteristic p ≥ 0 and FG is a semisimple group algebra. If FG is split then the Wedderburn decomposition of FG is determined by the degrees of the absolutely irreducible characters of G in characteristic p. Indeed, from Theorem 3.1.5, we know that there is a one-to-one correspondence between the absolutely irreducible characters of G and the Wedderburn components of FG. This correspondence associates χ ∈ Irrp (G) with the algebra A F (χ) = FGe(χ), where (see (3.1.3)) e(χ) =

χ(1) ∑ χ(g−1 )g. |G| g∈G

Furthermore, if d is the degree of χ then A F (χ) ≅ M d (F), e(χ) is the unique primitive central idempotent e of FG such that χ(e) ≠ 0 and A F (χ) is the unique Wedderburn component A of FG such that χ(A) ≠ 0. If FG is a not necessarily split group algebra then for every χ ∈ Irrp (G) there is a unique primitive central idempotent e of FG such that χ(e) ≠ 0. Indeed, let e1 , . . . , e k be all the distinct primitive central idempotents of FG. If E is a splitting field of FG (its existence follows from Lemma 2.2.1 and Corollary 3.1.3) then every e i is the sum of some primitive central idempotents of EG and, as 1 = e1 + ⋅ ⋅ ⋅ + e k and the e i ’s are orthogonal, every primitive central idempotent of EG appears in the expression of exactly one e i . Therefore, there is exactly one i, with 1 ≤ i ≤ k, such that χ(e i ) = χ(e(χ)) = χ(1) ≠ 0 and χ(e j ) = 0, for every j ≠ i. In other words, χ vanishes on all but one of the primitive central idempotents of FG and hence it also vanishes on all but one of the Wedderburn components of FG. We fix the following notation for each χ ∈ Irrp (G): e F (χ) = Unique primitive central idempotent e of FG with χ(e) ≠ 0, A F (χ) = FGe F (χ) = Unique Wedderburn component A of FG with χ(A) ≠ 0, m F (χ) = Ind(A F (χ)), called the Schur index of χ over F.

86 | 3 Wedderburn decomposition

If χ is a character of G (not necessarily absolutely irreducible) then the field of characters of χ over F (also called the field of character values of χ over F) is F(χ) = F(χ(g) : g ∈ G). The field of character values of χ is an abelian Galois extension of F (Problem 3.3.1). If σ ∈ Gal(F(χ)/F) then σ ∘ χ is a character of G (see Section 3.2). We can also make Gal(F(χ)/F) acting on the group algebra F(χ)G by setting σ( ∑ a g g) = ∑ σ(a g )g, g∈G

g∈G

for σ ∈ Gal(F(χ)/F) and ∑g∈G a g g ∈ F(χ)G. If χ is absolutely irreducible then e(χ) ∈ F(χ)G, σ ∘ χ is absolutely irreducible and we have e(σ ∘ χ) = σ(e(χ)). By (3.1.3), for σ, τ ∈ Gal(F(χ)/F) we have that e(σ ∘ χ) = e(τ ∘ χ) if and only of σ = τ. The following notion of trace map is handy. Let E be a finite Galois extension of F. For a ∈ EG we set σ(a) trE/F (a) = ∑ σ∈Gal(E/F)

and for an E-class function χ of G we put trE/F (χ) =

∑

σ ∘ χ.

σ∈Gal(E/F)

Then trE/F (a) ∈ FG and trE/F (χ) is an F-class function of G. In particular, trF(χ)/F (e F (χ)) ∈ FG. The conjugacy class of an element g in a group G is denoted by g G . Theorem 3.3.1. Let F be a field of characteristic p ≥ 0. Assume FG is a semisimple group algebra and χ ∈ Irrp (G). The following properties hold. (1) e F (χ) = trF(χ)/F (e(χ)) = ∑σ∈Gal(F(χ)/F) e(σ ∘ χ). (2) If ψ ∈ Irrp (G) then A F (ψ) = A F (χ) if and only if e F (ψ) = e F (χ), or equivalently ψ = σ ∘ χ for some σ ∈ Gal(F(χ)/F). (3) If E is a field extension of F and ρ is an irreducible E-representation affording χ then A F (χ) ≅ ρ(FG) = F-span of the image of ρ in M d (E), where d is the degree of χ. (4) The center of A F (χ) is F-isomorphic to F(χ). (5) If L/F is a field extension then A L (χ) and L(χ) ⊗F(χ) A F (χ) are isomorphic F(χ)-alge[F(χ):F] bras, and L ⊗F A F (χ) is isomorphic (as an F-algebra) to a direct product of [L(χ):L] copies of A L (χ). In particular, the degree of A L (χ) as L-algebra is the degree of χ. (6) The character afforded by the unique, up to isomorphisms, simple FG-module contained in A F (χ) is m F (χ)trF(χ)/F (χ) = m F (χ) ∑σ∈Gal(F(χ)/F) σ ∘ χ.

3.3 Wedderburn components from character tables |

87

Proof. Put G F = Gal(F(χ)/F). We fix a splitting field E of FG and a representation ρ : G → GLn (E) affording χ. Recall that n1F = χ(1) ≠ 0 (Theorem 3.1.5 (4)). (1) Set e = trF(χ)/F (e(χ)). Then e ∈ FG and it is a sum of different primitive central idempotents of EG (and thus orthogonal idempotents). Hence e is a central idempotent of FG. If e is not primitive as central idempotent of FG then G F contains a nonempty proper subset I containing 1 such that e1 = ∑σ∈I e(σ ∘ χ) ∈ FG. For any α ∈ G F , we have e1 = α(e1 ) = ∑σ∈I e((ασ) ∘ χ), therefore I = {ασ : σ ∈ I} and hence α ∈ I. This proves that I = G F , contradicting the fact that I is a proper subset of G F . (2) This is a direct consequence of (1). (3) As ρ(e F (χ)) = 1 and ρ vanishes on all the other Wedderburn components of FG, ρ induces a non-zero homomorphism from A = A F (χ) to ρ(A). Because A is simple, this yields the desired isomorphism. (4) Set B = M n (E). Note that B is the E-span of ρ(G). As ρ(A) = F(ρ(G)), we have that the center of ρ(A) is contained in Z(M n (E)) = E1B . We will prove that Z(ρ(A)) = F(χ)1B . Indeed, if x ∈ E and a g ∈ F for g ∈ G are such that x1B = ∑g∈G a g ρ(g) ∈ Z(ρ(A)) then, taking traces, we obtain that nx = ∑g∈G a g χ(g) ∈ F(χ). As n1F = χ(1) ≠ 0, by Theorem 3.1.5 (4), we deduce x ∈ F(χ). Therefore Z(ρ(A)) ⊆ F(χ)1B . G is a central element of FG, there is x ∈ E with Conversely, let g ∈ G. Since g̃ ̃ ρ(g G ) = x1B . Taking traces we get that |g G |χ(g) = χ(1)x and hence G ) = |g G | χ(g)1 = [G : Cen (g)]χ(g)1 . χ(1)ρ(g̃ B G B

(3.3.1)

As n = χ(1) and |G| is coprime with the characteristic of G, we have χ(g)1B = n G [G:CenG (g)] ρ(∑(g )) ∈ Z(ρ(A)). So it follows that also F(χ)1B ⊆ Z(ρ(A)). (5) Enlarging E if necessary, we may assume that the splitting field E contains L. Because of (4), A F (χ) = ρ(FG) is a central simple F(χ)-algebra and ρ(LG) ≅ A L (χ) is a central simple L(χ)-algebra. Hence, from the Universal Property of Tensor Products, we obtain an algebra homomorphism L(χ) ⊗F(χ) ρ(FG) → ρ(LG). Since ρ(FG) is a simple algebra, we know from Proposition 2.1.8 that L(χ)⊗F(χ) ρ(FG) also is a simple algebra. As ρ(G) is a generating set for ρ(LG), considered as L(χ)-vector space, it follows that A L (χ) ≅ L(χ)⊗F(χ) A F (χ), an isomorphism as F(χ)-algebras. Moreover LG ≅ L ⊗F FG and thus we obtain from (1) that L ⊗F A F (χ) is the direct product of the Wedderburn components of LG of the form A L (σ ∘ χ) with σ ∈ G L = Gal(L(χ)/L). Consider the restriction map R : G L = Gal(L(χ)/L) → G F = Gal(F(χ)/F) and let T be a right transversal of R(G L ) in G F . As R is injective, |T| = [G F : R(G L )] =

[F(χ) : F] . [L(χ) : L]

Moreover e F (χ) = ∑ ∑ e((R(σ)τ) ∘ χ) = ∑ e L (τ ∘ χ) τ∈T σ∈G L

τ∈T

and each summand e L (τ ∘ χ) is a primitive central idempotent of LG contained in L ⊗F A F (χ). Hence, L ⊗F A F (χ) ≅ ∏τ∈T A L (τ ∘ χ).

88 | 3 Wedderburn decomposition For every τ ∈ T, let σ τ be an extension of τ to a homomorphism defined on the domain E (and with image some field E τ ). The componentwise action of σ τ gives an automorphism M n (E) → M n (E τ ), which we also denote by σ τ . Hence σ τ ∘ρ is an irreducible E τ -representation of G affording the character τ ∘ χ. Then σ τ : ρ(LG) → σ τ (ρ(LG)) is an isomorphism. From (3) we also know that ρ(LG) ≅ A L (χ) and σ τ ρ(LG) ≅ A L (τ ∘ χ). Hence, we obtain that A L (χ) ≅ A L (τ∘χ) for every τ ∈ T. Hence, L⊗F A F (χ) is isomorphic [F(χ):F] (as an F-algebra) to a direct product of [L(χ):L] copies of A L (χ). Since A L (χ) and L(χ) ⊗F(χ) A F (χ) are isomorphic as central simple algebras, we thus obtain that Deg(A L (χ)) = Deg(A F (χ)). So to prove the final part of statement (5) one may assume that L is a splitting field of FG. In this case A L (χ) ≅ M n (L). Thus n = Deg(A L (χ)). (6) We may assume that E is a splitting field of G containing F. We recycle the notation of the proof of nLemma 3.1.7: FG = A1 ⊕ ⋅ ⋅ ⋅ ⊕ A k , EG = B1 ⊕ ⋅ ⋅ ⋅ ⊕ B n , E ⊗F j A i = ⊕j∈J i B j and ⊕j∈J i T j mi ≅ E ⊗F S i . Assume that ϕ is the character afforded by S i and χ is the character afforded by T j with j ∈ J i . Then A F (χ) ≅ A i = M m i (D i ), for some division algebra D i . By (1), the constituents of ϕ are the characters of the form σ ∘ χ with σ ∈ Gal(F(χ)/F), hence we can parameterize the T j with j ∈ J i as T σ with σ ∈ Gal(F(χ)/F) where σ ∘ χ is the character afforded by T σ . Then n j = n, the degree of m (χ) χ, for every j ∈ J i .By (5), n = Deg(A F (χ)) = m i m F (χ) and E ⊗F S i = ⊕σ∈Gal(F(χ)/F) T σ F and we conclude that ϕ = m F (χ) ∑σ∈Gal(F(χ)/F) σ ∘ χ. Theorem 3.3.1 provides a method to calculate the Wedderburn components of FG from the character table of G. Unfortunately, the description obtained does not provide much information on the Wedderburn components A F (χ) = ρ(FG), except in some special cases, as for example when F has characteristic p or G is abelian. In the rest of the section we will show how to calculate the Wedderburn decomposition in these two cases. Later in the chapter we will develop more sophisticated methods to calculate the Wedderburn decomposition of FG in other more general cases. We start by showing how to calculate the Wedderburn decomposition of a semisimple group algebra FG if the characteristic of F is positive. Recall that Wedderburn’s Theorem states that every finite division algebra is a field. Therefore every finite simple algebra is a matrix algebra over its center. In particular, if F has characteristic p, χ is an absolutely irreducible character of degree n in characteristic p and 𝔽p is the prime subfield of F, then A𝔽p (χ) ≅ M n (𝔽p (χ)), for some positive integer n, (and 𝔽p (χ) is a finite field of order a power of p). From Theorem 3.3.1 (5) we then obtain that A F (χ) ≅ F(χ) ⊗𝔽p (χ) A𝔽p (χ) ≅ M n (F(χ)). Moreover, F ⊗𝔽p A𝔽p (χ) ≅ M n (F ⊗𝔽p 𝔽p (χ)) and F ⊗𝔽p 𝔽p (χ) is a product of [𝔽 (χ):𝔽 ]

[𝔽p (χ):𝔽p ] [F(χ):F] -copies of the field F(χ). In other words, F ⊗𝔽p A 𝔽p (χ)

p p is isomorphic to [F(χ):F] -copies of M n (F(χ)). We now consider the case when G is abelian. In this case the absolutely irreducible characters are the group homomorphisms χ : G → U(F), with F an algebraic closure of F. By Theorem 3.3.1 (3), A F (χ) ≅ χ(FG) = F(χ) in this case. Moreover, if N = ker(χ) then χ(G) is a cyclic group of order d = [G : N] and therefore A F (χ) = χ(FG) = F(ζ d ).

3.3 Wedderburn components from character tables |

89

Conversely every linear character of G is obtained in this fashion. If F = ℚ then one can give a precise description of the primitive central idempotent eℚ (χ) in terms of G and N; so a character free description. For this we need the following notation. For a normal subgroup N of the finite group G and a field F whose characteristic does not divide |G|, we consider the following element in FG ̂ G, ε(G, N) = { ̂−̂ D), ∏D/N∈M(G/N) (N

if G = N; otherwise,

(3.3.2)

where M(G/N) denotes the set consisting of the minimal non-trivial normal subgroups D/N of G/N, with D a subgroup of G containing N. We know that if D is a normal subgroup of G then ̂ D is a central idempotent in FG. If N is a normal subgroup contained ̂ ̂ ̂ ̂−̂ in D then N D = D and hence also N D is a central idempotent of FG. So, ε(G, N) is a central idempotent in FG. Lemma 3.3.2. If χ is a complex linear character of G and N = ker(χ) then G/N is cyclic, e ℚ (χ) = ε(G, N)

and

ℚG ε(G, N) ≅ ℚ(ζ d ),

where d = [G : N]. Proof. As ℚGeℚ (χ) ≅ ℚ(χ) = ℚ(ζ d ), it only remains to prove that eℚ (χ) = ε(G, N). ̂ Furthermore, χ(N) ̂ = 1 and therefore eℚ (χ) is a primitive cenClearly ε(G, N) ∈ ℚG N. ̂ ̂ ≅ ℚ(G/N) of (1.5.7) maps ε(G, N) to tral idempotent of ℚG N. The isomorphism ℚG N ε(G/N, 1) and eℚ (χ) to eℚ (χ), where χ is the natural faithful linear character of G/N induced by χ. Thus, using this isomorphism, one may assume that N = 1. So G is a cyclic group of order d and χ is a faithful linear character of G. Let ε = ε(G, 1). Because ε is a (central) idempotent of ℚG, and thus of ℂG, we have ε = ∑ψ∈Irr(G),ψ(ε)=0̸ e(ψ). On the other hand eℚ (χ) = ∑σ∈Gal(ℚ(χ)/ℚ) e(σ ∘ χ), by Theorem 3.3.1 (1). Furthermore, as G is cyclic of order d and ℚ(χ) = ℚ(ζ d ), the absolutely irreducible characters of G of the form σ ∘ χ, with σ ∈ Gal(ℚ(χ)/ℚ) are precisely the faithful linear characters of G. Thus, it is enough to show that if ψ ∈ Irr(G) then ψ(ε) ≠ 0 if and only if ψ is faithful. This is clear if G = 1. So, we assume that G ≠ 1. Let p1 , . . . , p k be the different prime divisors of d and for each i = 1, . . . , k let g i be a generator of the only subgroup of order p i of G. Then k

ε = ∏(1 − i=1

p −1

j

i gi ∑j=0

pi

).

If ψ ∈ Irr(G) then ψ is linear and p −1

p −1

i i pi , j ψ( ∑ g i ) = ∑ ψ(g i )j = { 0, j=0 j=0

if g i ∈ ker(ψ); otherwise.

Thus ψ(ε) ≠ 0 if and only if g i ∈ ̸ ker(ψ) for every i = 1, . . . , k, or equivalently, ψ is faithful.

90 | 3 Wedderburn decomposition Corollary 3.3.3. Let G be a finite abelian group. The primitive idempotents of ℚG are the elements ε(G, N) with N a subgroup of G so that G/N is cyclic. Proof. Since every complex character of G is linear (Problem 3.1.7), the result follows at once from Lemma 3.3.2 Using Theorem 3.3.1 (1) and Lemma 3.3.2 one can easily prove the following lemma. Lemma 3.3.4. If χ and ϕ are two complex linear characters of a finite group G then the following conditions are equivalent. (1) ker(χ) = ker(ϕ). (2) ϕ = χ i for some i coprime with [G : ker(ϕ)]. (3) ϕ = σ ∘ χ for some σ ∈ Gal(ℚ(χ)/ℚ). (4) eℚ (χ) = eℚ (ϕ). (5) ε(G, ker(χ)) = ε(G, ker(ϕ)). Let G be a finite abelian group and thus every irreducible complex character of G is linear. Furthermore, the pointwise multiplication of functions endows Irr(G) with a group structure: (χ ⋅ ϕ)(g) = χ(g)ϕ(g), (χ, ϕ ∈ Irr(G), g ∈ G). Proposition 3.3.5. If G is a finite abelian group then Irr(G) ≅ G. Proof. Assume first that G = ⟨g⟩n , a cyclic group of order n generated by g and let χ : G → ⟨ζ n ⟩ be an isomorphism. Then Irr(G) = ⟨χ⟩n ≅ G. In the general case G = G1 × ⋅ ⋅ ⋅ × G k , with G i cyclic. Then Irr(G) = Hom(G, U(ℂ)) ≅ ∏ki=1 Hom(G i , U(ℂ)) ≅ ∏ki=1 G i = G. Theorem 3.3.6 (Perlis-Walker [174]). Let G be a finite abelian group and let F be a field of characteristic 0. Let k d denote the number of cyclic subgroups of G of order d. Then FG ≅ ∏ F(ζ d )

kd

[ℚ(ζ d ):ℚ] [F(ζ d ):F]

d, d|n

and, in particular, ℚG ≅ ∏ ℚ(ζ d )k d d, d|n

Proof. As every irreducible complex character of G is linear, Lemma 3.3.2 implies that every Wedderburn component of ℚG is of the form ℚG ε(G, H) ≅ ℚ(ζ d ), for H a subgroup of G with G/H cyclic of order d. Moreover, Corollary 3.3.3 and Lemma 3.3.4 imply that H 󳨃→ ℚG ε(G, H) defines a bijection from the set S of subgroups H of G with G/H cyclic to the Wedderburn components of ℚG. Lemma 3.3.4 also implies that ⟨χ⟩ 󳨃→ ker(χ) defines a one-to-one correspondence between the set of cyclic subgroups of Irr(G) and S. Using this bijection and Proposition 3.3.5 one deduces that the number k d of cyclic subgroups of order d of G coincides with the number of elements H of S with [G : H] = d. Hence it follows that ℚG ≅ ∏d, d||G| ℚ(ζ d )k d . From Theorem 3.3.1 (5)

3.3 Wedderburn components from character tables |

91

we then obtain that 𝔽G ≅ 𝔽 ⊗ℚ ℚG ≅ ∏ (𝔽 ⊗ℚ ℚ(ζ d )) d, d|n

kd

[ℚ(ζ d ):ℚ]

kd

≅ ∏(ℚ(ζ d ) [F(ζd ):F] ) , d, d|n

as desired. Perlis and Walker [174] also obtained a description of the Wedderburn decomposition of semisimple abelian group algebras in positive characteristic. In this case the primitive central idempotents can be described too. See Problem 3.3.9.

Problems 3.3.1. Let G be a finite group of exponent n, F a field of characteristic p ≥ 0 and ζ n a primitive n-root of unity in an extension of F. Let χ be an irreducible F-character of G. Prove that χ(g) ∈ F(ζ n ). Conclude that F(χ)/F is a finite abelian Galois extension. 3.3.2. Let F be a field and let χ be an irreducible F-character of a finite group G. Prove that if E is a splitting field of FG then χ is the direct sum of [F(χ) : F] different irreducible E-representations of G, all of the same degree, say d. Let ψ be one of these irreducible E-representations. Show that [F(ψ) : F]d2 = dimE (E ⊗F A F (ψ)) = dimF (A F (ψ)) = k χ(1), where k is the reduced degree of A F (ψ). ̂ − H). ̂ 3.3.3. Proof that if N is a proper normal subgroup of G then ε(G, N) = ∏N 1 and assume that every finitely generated torsion-free module of rank m − 1 is isomorphic to a direct sum of m − 1 non-zero ideals of R. We consider M as an R-submodule of M (0) . Fix a non-zero element m of M and let N = Fm ∩ M. As M is Noetherian and N is a submodule of M we also get that N is finitely generated. Furthermore, N is a non-zero R-submodule of Fm and hence rkR (N) = 1. If rx ∈ N, with r ∈ R and x ∈ M then rx = am, for some a ∈ F. Thus r = 0 or x = ar m ∈ N. It follows that M/N is torsion-free of rank m − 1. By the induction hypothesis, M/N is isomorphic to a direct sum of m − 1 non-zero ideals of R. Because of Corollary 4.2.9 we know that every ideal of R is projective as an R-module. Hence M/N is a projective R-module. Consequently, M ≅ N ⊕ (M/N) and we conclude that M is isomorphic to a direct sum of m ideals. Secondly, we assume that M is torsion and different from 0. Then AnnR (M) = {r ∈ n n R : rM = 0} is a non-zero ideal of R. Let AnnR (M) = P11 . . . P t t be the factorization of AnnR (M). So each P i is a maximal ideal of R. For every i = 1, . . . , t, let M i = {x ∈ nj n M : P i i x = 0} and let Q i denote the product of the ideals P j with j ≠ i (for example, ni if t = 1, then Q1 = R). Then Q1 + ⋅ ⋅ ⋅ + Q t = P i + Q i = R and Q i M j = 0, if i ≠ j. n Thus Q i M ⊆ M i , because P i i Q i M = AnnR (M)M = 0 and Q i ∑j=i̸ M j = 0. This shows n that M = M1 ⊕ ⋅ ⋅ ⋅ ⊕ M t and AnnR (M i ) = P i i . Using that the class of modules which are of the form given in statement (1) is closed under direct sums, we may assume that AnnR (M) = P n for some maximal ideal P and some positive integer n. In particular, if x ∈ R \ P then Rx + AnnR (M) = R. We claim that the natural map M → M P is bijective. Indeed, to show that it is injective, assume m 1 = 0. Then xm = 0 for some x ∈ R \ P. Hence Rm = (Rx + AnnR (M))m = 0 and thus m = 0. To prove that the map is surjective, assume mx ∈ M P , with m ∈ M and x ∈ R \ P. Because Rx + AnnR (M) = R,

4.3 Finitely generated modules over Dedekind domains

| 139

rm there is an r ∈ R so that rx − 1 ∈ AnnR (M). Hence mx = rxm x = 1 . So the map indeed is surjective and thus bijective. Therefore, M P is a finitely generated torsion R P -module. As R P is a principal ideal domain and every non-zero ideal of R P is a power of P P , we e have that M P ≅ ⊕ui=1 R P /P Pi , an isomorphism of R P -modules, for some positive integers e1 , . . . , e u . Lemma 4.2.3 therefore implies that, as R-modules, M ≅ ⊕ui=1 R/P e i . Finally, let M be an arbitrary finitely generated R-module. Then t(M) is torsion and M/t(M) is torsion-free. By the above, M/t(M) ≅ I1 × ⋅ ⋅ ⋅ × I k , for some k ≥ 0 and e e some I1 , . . . , I k non-zero ideals of R, and t(M) ≅ R/P11 ⊕ ⋅ ⋅ ⋅ ⊕ R/P uu for some u ≥ 0, some positive maximal ideals P1 , . . . , P u of R and positive integers e1 , . . . , e u . Thus M/t(M) is projective by Corollary 4.2.9. Therefore M ≅ (M/t(M)) ⊕ t(M) ≅ I1 ⊕ ⋅ ⋅ ⋅ ⊕ I m ⊕ e e R/P11 ⊕ ⋅ ⋅ ⋅ ⊕ R/P uu as desired. f f e e (2) Let M = I1 ⊕⋅ ⋅ ⋅⊕ I m ⊕ R/P11 ⊕⋅ ⋅ ⋅⊕ R/P uu and N = J1 ⊕⋅ ⋅ ⋅⊕ J n ⊕ R/Q11 ⊕⋅ ⋅ ⋅⊕ R/Q vv . f f e e Then t(M) = R/P11 ⊕ ⋅ ⋅ ⋅ ⊕ R/P uu , t(N) = R/Q11 ⊕ ⋅ ⋅ ⋅ ⊕ R/Q vv , I1 ⊕ ⋅ ⋅ ⋅ ⊕ I m ≅ M/t(M) and J1 ⊕ ⋅ ⋅ ⋅ ⊕ J n ≅ N/t(N). This shows that it is enough to prove (2) assuming that both M and N are either torsion-free (that is u = v = 0) or torsion (m = n = 0). Assume first that M and N are torsion-free. If M ≅ N then m = rkR (M) = rkR (N) = −1 n. Let M i = a−1 i I i and N i = b i J i , with 0 ≠ a i ∈ I i and 0 ≠ b i ∈ I j . Then M i and N i are fractional ideals of R containing R and M i ≅ I i and N i ≅ J i as R-modules. Hence M1 ⊕ ⋅ ⋅ ⋅ ⊕ M m ≅ N1 ⊕ ⋅ ⋅ ⋅ ⊕ N m , M1 ⋅ ⋅ ⋅ M m = (a1 ⋅ ⋅ ⋅ a m )−1 (I1 ⋅ ⋅ ⋅ I m ) ≅ I1 ⋅ ⋅ ⋅ I m and N1 ⋅ ⋅ ⋅ N m = (b1 ⋅ ⋅ ⋅ b m )−1 J1 ⋅ ⋅ ⋅ J m ≅ J1 ⋅ ⋅ ⋅ J m . Thus to prove that I1 ⋅ ⋅ ⋅ I m and J1 ⋅ ⋅ ⋅ J m are isomorphic as R-modules, it is enough to prove that M1 ⋅ ⋅ ⋅ M m and N1 ⋅ ⋅ ⋅ N m are isomorphic as R-modules. Fix an isomorphism f : M1 ⊕ ⋅ ⋅ ⋅ ⊕ M m → N1 ⊕ ⋅ ⋅ ⋅ ⊕ N m . Considering M i as a submodule of (M i )(0) ≅ F, one may consider f as the restriction of an F-automorphism g of the F-vector space F m . Furthermore, as R ⊆ M i and R ⊆ N i for every i, both M1 ⊕ ⋅ ⋅ ⋅ ⊕ M m and N1 ⊕ ⋅ ⋅ ⋅ ⊕ N m contain the standard basis e1 , . . . , e m of F m . Let (a ij ) be the matrix of g with respect to e1 , . . . , e m , that is f(e i ) = ∑m j=1 a ji e j . If x ∈ M i then ∑m j=1 xa ji e j = f(xe i ) ∈ f(M 1 ⊕ ⋅ ⋅ ⋅ ⊕ M m ) = N 1 ⊕ ⋅ ⋅ ⋅ ⊕ N m and thus xa ji ∈ N j for every j. Therefore, for every permutation σ ∈ S m (the symmetric group of degree m), we have

M1 ⋅ ⋅ ⋅ M m a1σ(1) ⋅ ⋅ ⋅ a mσ(m) = (M σ(1) a1σ(1) ) ⋅ ⋅ ⋅ (M σ(m) a1σ(m) ) ⊆ N1 ⋅ ⋅ ⋅ N m This implies that M1 . . . M m det((a ij )) ⊆ N1 ⋅ ⋅ ⋅ N m . Similarly, one also has N1 ⋅ ⋅ ⋅ N m det((a ij ))−1 ⊆ M1 . . . M m . We conclude that multiplication by det(a ij ) is an isomorphism of R-modules M1 . . . M k → N1 . . . N k and therefore I1 ⋅ ⋅ ⋅ I m ≅ J1 ⋅ ⋅ ⋅ J m as Rmodules. To prove the converse (for the torsion-free case) it is enough to show that I1 ⊕ ⋅ ⋅ ⋅ ⊕ I m ≅ R m−1 ⊕ (I1 ⋅ ⋅ ⋅ I m ). As this is clear for m = 1 and the general case follows by induction from the case m = 2, we prove it for m = 2. Let r be a non-zero element of R such that rI1−1 ⊆ R and write rI1−1 = P11 . . . P kk e

e

and

f

f

I2 = P11 . . . P kk ,

with P1 , . . . , P k different maximal ideals of R and e i , f i non-negative integers. By Lemma 4.3.1, for every i = 1, . . . , k, there is a non-zero ideal J i of R and an element

140 | 4 Dedekind domains, valuations and orders e

e

a i ∈ R such that Ra i = P i J i and J i ⊈ P j for every j = 1, . . . , k. Let a = a11 ⋅ ⋅ ⋅ a kk and e e e e J = J11 . . . J k k . Then Ra = P11 . . . P kk J = rI1−1 J and J is not contained in any P i . Thus −1 −1 ar I1 = J and ar I1 + I2 = R. Consequently, there exists an R-module isomorphism f : I1 ⊕ I2 → ar−1 I1 ⊕ I2 and an R-module epimorphism g : ar−1 I1 ⊕ I2 → R. Then g ∘ f : I1 ⊕ I2 → R is an epimorphism and hence I1 ⊕ I2 ≅ R ⊕ I for some R-module I. Then rkR (I) = 1 and hence one may assume that I is a non-zero ideal of R. Using the part of (2) which has been already proved, we deduce that I ≅ I1 I2 . This finishes the proof of the torsion-free case. Assume now that M and N are torsion. Then clearly (b) implies (a). Conversely, assume that f : M → N is an isomorphism of R-modules. Let R1 , . . . , R t be the different elements of {P1 , . . . , P u , Q1 , . . . , Q v }. As in the proof of (1) we can write M = M1 ⊕ ⋅ ⋅ ⋅ ⊕ M k , where M i is formed by the elements of M annihilated by a power of R i and similarly N = N1 ⊕ ⋅ ⋅ ⋅ ⊕ N k . Then f restricts to an isomorphism M i → N i for every i. This shows that one may assume that AnnR (M) = P e and AnnR (N) = P f for some maximal ideal P of R and some positive integers e and f . In fact, as M and N are isomorphic e = f . Hence M ≅ R/P e1 ⊕ ⋅ ⋅ ⋅ ⊕ R/P e u and N ≅ R/P f1 ⊕ ⋅ ⋅ ⋅ ⊕ R/P f v . After reordering the summands, one may assume that e = e1 ≥ ⋅ ⋅ ⋅ ≥ e u and e = f1 ≥ ⋅ ⋅ ⋅ ≥ f v . If e k+1 < e = e k = f l > f l+1 then (P e−1 /P e )k ≅ P e−1 M ≅ P e−1 N ≅ (P e−1 /P e )l and therefore, by Lemma 4.2.3, k = dimR/P (P e−1 M) = dimR/P (P e−1 N) = l. This shows that {i : e i = e} and {i : f i = e} have the same cardinality. Repeating the same argument for decreasing values of e i we deduce that e i = f i for every i and in particular u = v. The following corollary of Theorem 4.3.2 follows easily. Corollary 4.3.3. Let R be a Dedekind domain. Every finitely generated torsion-free Rmodule is projective. Every finitely generated torsion R-module has finite length. Let M be a finitely generated non-zero torsion module over a Dedekind domain R. By e e Theorem 4.3.2, M ≅ R/P11 ⊕ ⋅ ⋅ ⋅ ⊕ R/P kk , with P1 , . . . , P k maximal ideals of R and e1 , . . . , e k positive integers. Then the ideal order of M is e

e

ordR (M) = P11 ⋅ ⋅ ⋅ P kk . Observe that, if P is a maximal ideal of R then e P (ordR (M)) = ∑ e i P i =P

and, by Lemma 4.2.3, this is the number of composition factors isomorphic to R/P in any composition series of M. Using Lemma 4.1.1 one can easily prove that ordR commutes with localization, in other words ordR P (M P ) = ordR (M)P .

(4.3.1) e

Example 4.3.4. If R is a PID and M is a non-zero torsion R-module, then M ≅ R/Rp11 ⊕ e ⋅ ⋅ ⋅ ⊕ R/Rp kk with p1 , . . . , p k irreducible elements of R and thus e

e

ordR (M) = Rp11 ⋅ ⋅ ⋅ p kk .

4.3 Finitely generated modules over Dedekind domains

| 141

More generally, if M ≅ R/Rd1 ⊕ ⋅ ⋅ ⋅ ⊕ R/Rd k with d1 , . . . , d k non-zero elements of R then ordR (M) = Rd1 . . . d k . In particular, if R = ℤ then for d1 , . . . , d k one can take positive integers and |M| = d1 ⋅ ⋅ ⋅ d k . Hence, the ideal order over ℤ of a finite abelian group is the ideal generated by its cardinality. Example 4.3.5. Let R be a PID, M a finitely generated torsion-free R-module and N a submodule of M with rkR (M) = rkR (N). Fix an R-basis m1 , . . . , m k of M and an R-basis n1 , . . . , n k of N. Then n i = ∑ki=1 r ij m j for some r ij ∈ R and ordR (M/N) = R det(r ij ). Proof. The matrix (r ij ) can be converted into a diagonal matrix performing R-elementary operations: reordering the rows or the columns, adding to one row (column) an R-linear combination of the other rows (columns), or multiplying rows or columns by invertible elements of R. These operations do not affect the determinant up to a multiplication by unit of R. Furthermore, the row operations can be considered as a change of basis in N and the column operations as a change of basis in M. So one may assume that (r ij ) is diagonal and suppose that M = R n and N = r11 R × ⋅ ⋅ ⋅×r kk R. Then M/N ≅ (R/r11 R)×⋅ ⋅ ⋅×(R/r kk R) and therefore ord(M/N) ≅ Rr11 ⋅ ⋅ ⋅ r kk = R det(r ij ). Lemma 4.3.6. Let R be the ring of integers of a number field F and I a non-zero ideal of R. Then R/I is finite. Proof. Since R contains a ℚ-basis of F, the rational vector space generated by I is a non-zero ideal in F, and so it is F. Thus I contains a basis x1 , . . . , x n of F over ℚ. If y1 , . . . , y n is an integral basis of R then y i = ∑nj=1 a ij x j for some a ij ∈ ℚ. Then there is a non-zero integer m such that ma ij ∈ ℤ and therefore my i ∈ ∑nj=1 ma ij x i ∈ I. Thus mR ⊆ I and hence |R/I| ≤ |R/mR| ≤ ∞, as desired. If I is a non-zero ideal of the ring of integers R of a number field then the norm of I in R is by definition N(I) = |R/I|. The following lemma collects some properties of this norm. Lemma 4.3.7. Let R be the ring of integers of a number field F and let I and J be non-zero ideals of R. Then (1) N(IJ) = N(I)N(J), for non-zero ideals I and J of R. (2) Let a be a positive integer. The number of non-zero ideals I of R with N(I) = a is finite. (3) ordℤ (R/I) = N(I)ℤ. (4) If 0 ≠ r ∈ R then N(Rr) = |NrF/ℚ (r)|. Proof. (1) If P a maximal ideal of R then dimR/P (R/P n ) = n, by Lemma 4.2.3. Thus, if I = P n and J = P m then N(IJ) = N(P n+m ) = |R/P n+m | = |R/P|n+m = |R/P|n |R/P|m = N(I)N(J). If I and J are relatively prime then N(IJ) = N(I)N(J) by the Chinese Remainder

142 | 4 Dedekind domains, valuations and orders

Theorem. Thus, for arbitrary ideals non-zero I and J we have N(IJ) = N(∏ P e P (IJ) ) = ∏ N(P e P (I)+e P (J) ) P∈Max(R)

P∈Max(R)

= ∏ N(P e P (I) ) ⋅ ∏ N(P e P (J) ) = N(∏ P e P (I) ) ⋅ N(∏ P e P (J) ) = N(I)N(J). P∈Max(R)

P∈Max(R)

P∈Max(R)

P∈Max(R)

(2) If N(I) = a then aR ⊆ I. Since R/aR is finite, aR is contained in only finitely many ideals of R. Thus N(I) = a for only a finite number of ideals I. (3) follows from Example 4.3.4. (4) If x 1 , . . . , x n is a ℤ-basis of R and rx i = ∑nj=1 r ij x j with r ij ∈ ℤ then N(Rr)ℤ = ordℤ (R/Rr) = det(r ij )ℤ = NrF/ℚ (r)ℤ, by Example 4.3.5. Hence, N(Rr) = |NrF/ℚ (r)|.

Problems 4.3.1. Prove that the following conditions are equivalent for a finitely generated Rmodule M over a Dedekind domain R. (1) M is projective. (2) M is torsion-free. (3) M ≅ R k ⊕ I for some k ≥ 0 and some ideal I of R.

4.4 Extensions of Dedekind domains Throughout this section R is a Dedekind domain, F is its field of fractions, E is a finite field extension of F and S is the integral closure of R in E. By Proposition 4.2.6, S also is a Dedekind domain. The goal is to relate the ideal theory of R and S. A relevant example of this situation occurs when F and E are two number fields and R and S are their ring of integers. Let Q be a non-zero prime ideal of S and P = Q ∩ R. By Theorem 4.1.11, P is a non-zero prime ideal of R. The ramification index of Q over R is e(Q/R) = e Q (SP) and the residue degree of Q over R is f(Q/R) = [S/Q : R/P] = dimR/P (S/Q), where we consider R/P as a subfield of S/Q via the natural map R/P → S/Q. In Theorem 4.4.2 we will show that f(Q/R) is finite. Observe that e(Q/R) > 0 because Q is one of the maximal ideals of S containing SP.

4.4 Extensions of Dedekind domains

|

143

Conversely, if P is a non-zero prime ideal of R then the non-zero prime ideals of S (i.e. maximal ideals of S) containing P are called the primes of E above P. If Q1 , . . . , Q k are these primes then R ∩ Q i = P and e(Q1 /R)

SP = Q1

e(Q k /R)

. . . Qk

is the factorization of SP. One says that Q is ramified with respect to R if e(Q/R) > 1 or S/Q is not separable over R/P. Otherwise one says that Q is unramified with respect to R. One says that P is ramified in E, or that P ramifies in E if some prime of E above P is ramified with respect to R. Otherwise one says that P is unramified in E. If F is a number field then so is E. Then both R/P and S/Q are finite fields and therefore S/Q is separable over R/P. Therefore, in this case, Q is ramified with respect to R if and only if e(Q/R) > 1. Similarly, P is ramified in E if P is contained in Q 2 for some maximal ideal Q of S. Lemma 4.4.1. If L is a subfield of E containing F, T is the integral closure of R in L and Q is a maximal ideal of S then (1) e(Q/R) = e(Q/T)e(Q ∩ T/R). (2) f(Q/R) = f(Q/T)f(Q ∩ T/R). e

e

Proof. (1) Let P 1 = Q ∩ T and P = Q ∩ R = P1 ∩ R. Let PT = P11 . . . P kk and P1 S = 󸀠 󸀠 Q1e1 . . . Q te t be the factorizations of PT and P1 S in T and S respectively, with Q = Q1 . Hence e1 = e(P1 /R) and e󸀠1 = e(Q/T). We have to prove that e(Q/R) = e1 e󸀠1 . If Q󸀠 is a maximal ideal of S containing P i then Q󸀠 ∩ T = P i . Hence P j ⊈ Q i for every i = 1, . . . , t and j ≥ 2. This implies that the factorization of PS in S is of the form e1 e󸀠1

PS = (P1 S)e1 . . . (P k S)e k = (Q1

e1 e󸀠t

. . . Qt

e󸀠

e󸀠

t+1 )Q t+1 . . . Q mm ,

for some maximal ideals Q t+1 , . . . , Q m and some non-negative integers e󸀠t+1 , . . . , e󸀠m . Thus indeed e(Q/R) = e1 e󸀠1 . (2) f(Q/R) = [S/Q : R/P] = [S/Q : T/P1 ][T/P1 : R/P] = f(Q/T)f(T ∩ Q/R). Theorem 4.4.2. Let R be a Dedekind domain, F the field of fractions of R and E a finite separable field extension of F. Assume P is a non-zero prime ideal of R. If e1 , . . . , e k and f1 , . . . , f k are the ramification indices and residue degrees over R of the different primes of R above P, then ∑ki=1 e i f i = [E : F]. e

e

Proof. Let S be the integral closure of R in E. Let PS = Q11 . . . Q kk be the factorization e e of PS. Then, by the Chinese Remainder Theorem, S/PS ≅ S/Q11 × ⋅ ⋅ ⋅ × S/Q kk . For every ei i = 1, . . . , k we have dimS/Q i (S/Q i ) = e i , by Lemma 4.2.3. It follows that k

e

dimR/P (S/PS) = ∑ dimR/P (S/Q i i ) i=1 k

e

= ∑ dimR/P (S/Q i ) dimS/Q i (S/Q i i ) i=1 k

= ∑ ei fi . i=1

144 | 4 Dedekind domains, valuations and orders It remains to show that dimR/P (S/PS) = [E : F]. As R P is a local PID, P P = pR P for some p ∈ R. Then PS P = (PS)P = pS P . By Proposition 4.1.6, S contains a basis of E over F. Because, by assumption, E/F is a finite separable field extension, we obtain from Theorem 4.2.6 that S is finitely generated as R-module. Hence, S P is a finitely generated R P -module and it contains a basis of E over F. Let x1 , . . . , x n be a generating system of S P as R P -module of minimal cardinality. If x i ∈ PS P then x i = p v y i for some y i ∈ S P \ pS P . Replacing x i by y i if necessary, we may assume that x i ∈ ̸ (PS)P for every i. We claim that x1 , . . . , x n is a basis of E over F. Since S P contains a basis of E over F, clearly x1 , . . . , x n generate E over F. Assume that x1 , . . . , x n are linearly dependent over F. Then they are linearly dependent over R. Therefore ∑ki=1 a i x i = 0 with all a1 , . . . , a k ∈ R and a i ≠ 0 for some i. Let u = min{e P (a j R) : a j ≠ 0}. Multiplying the equality with p−u and reindexing if necessary we may assume that a i ∈ R p for every i and a1 ∈ U(R p ). Then {x2 , . . . , x n } generates S P over R P , contradicting the minimality. So, indeed x1 , . . . , x n is a basis of E over F. Hence, S P = R P x1 ⊕ ⋅ ⋅ ⋅ ⊕ R P x k . Clearly, as R/P-modules, S P /PS P ≅ (R P /P)x1 ⊕ ⋅ ⋅ ⋅ ⊕ (R P /P)x k , with x i the natural image of x i in S P /PS P . So k = dimR P /P (S P /PS P ). Because S P /PS P and S/SP are isomorphic as R/P-modules (see Lemma 4.1.1), the result follows. Let P be a non-zero prime ideal of R. If σ is an F-automorphism of E then σ(R) = e e R, σ(P) = P and σ(S) = S. Hence, if PS = Q11 ⋅ ⋅ ⋅ Q gg is the factorization of PS in S then each σ(Q i ) is a maximal ideal of S over P and PS = σ(Q1 )e1 ⋅ ⋅ ⋅ σ(Q g )e g . Thus σ permutes the Q i ’s and e(σ(Q i )/R) = e(Q i /R) for every i. If E/F is a Galois extension, then the decomposition subgroup of a non-zero prime ideal Q of R is D Q = {σ ∈ Gal(E/F) : σ(Q) = Q}. Clearly, the cardinality of the orbit of Q under the action of Gal(E/F) is [Gal(E/F) : D Q ]. Theorem 4.4.3. Let R be a Dedekind domain with ring of fractions F, E/F a finite Galois extension. For every non-zero prime ideal P of R the following properties hold. (1) Gal(E/F) permutes transitively the primes of E above P, (2) there are integers e, f and g such that e = e(Q/R) and f = f(Q/R) and g = [Gal(E/F) : D Q ], for every maximal ideal Q of S containing P and (3) [E : F] = efg. Proof. Let G = Gal(E/F) and let S be the integral closure of R in E. (1) By means of contradiction, let Q and Q󸀠 be two non-zero prime ideals of S containing P which are not in the same G-orbit. Let I be the product of the primes in the G-orbit of Q󸀠 and let Q = Q1 , . . . , Q g be all the elements of the orbit of Q. Then I i = I ∏j, j=i̸ Q j ⊈ Q i for every i = 1, . . . , n and hence, for every i, there exists a i ∈ I i \Q i . Then a = ∑ni=1 a i ∈ I \ Q i for every i. Thus σ(a) ∈ I for every σ ∈ G and therefore ∏σ∈G σ(a) ∈ F ∩ I = R ∩ I ⊆ R ∩ Q󸀠 = P ⊆ Q, because the elements of I are integral over R and R is integrally closed in F. As Q is prime in S, σ(a) ∈ Q for some σ ∈ G and therefore a ∈ σ−1 (Q) = Q i , for some i. This contradicts with the choice of a and finishes the proof of (1)

4.4 Extensions of Dedekind domains

|

145

(2) and (3) These are an immediate consequence of (1), because of Theorem 4.4.2. Let R be Dedekind domain with quotient field F and assume that E is a finite Galois extension of F. Let P be a prime ideal of R and Q a prime above P of the integral closure of R in E. Then we use the notation [E : F] e(E/P) = e(Q/R), f(E/P) = f(Q/R) and g(E/P) = . e(F/P)f(F/P) By Theorem 4.4.3, these numbers do not depend on the prime Q. Moreover, Gal(E/F) acts transitively on {Q1 , . . . , Q g }, the set of all primes above P, and therefore the decomposition groups of Q1 , . . . , Q g are conjugate. In particular, if E/F is an abelian extension then all these decomposition groups are equal. We use the bar notation for reduction modulo Q (in S) and modulo P (in R). Every σ ∈ D Q induces an element σ : s 󳨃→ σ(s) of Gal(S/R). This defines a group homomorphism α Q : D Q → Gal(S/R). The kernel T Q of α Q is called the inertia group of Q over R. Observe that g = g(E/P) and |D Q | = e(E/P)f(E/P). Lemma 4.4.4. Let E/F be a Galois extension of number fields and let R and S be the ring of integers of F and E respectively. Then α Q : D Q → Gal(S/R) is surjective. Moreover, if L is the fixed subfield of T Q and S1 is the ring of integers of L then e(Q/S1 ) = e(Q/R) = [E : L] and therefore e(Q ∩ S1 /R) = f(Q/S1 ) = 1. Proof. Write S = R(a) for some a ∈ S. Let G = Gal(E/F) and P = Q ∩ R. By Proposition 4.2.5, we may assume without loss of generality that a belongs to every prime of S above P that is different from Q. In other words, σ(a) ∈ Q for any σ ∈ G \ D Q . Then the characteristic polynomial P = ∏σ∈G (X − σ(a)) ∈ R[X] projects to a polynomial P = ∏σ∈G (X − σ(a)) ∈ R[X] and the non-zero roots of P are the elements σ(a) with σ ∈ D Q . This implies that if τ ∈ Gal(S/R) then τ(a) = σ(a) for some σ ∈ D Q . Hence τ = α(σ). This proves that α is surjective. In particular, [E : L] = |T Q | = |D Q |/[S : R] = e(E/P). Since T Q = Gal(E/L) is contained in D Q , Q is the only maximal ideal of S containing S∩Q. Hence g(Q/S1 ) = 1. The decomposition group of Q with respect to the extension E/L is T Q and the corresponding map T Q → Gal(S/S1 ) ⊆ Gal(S/R) is the trivial map because it is the restriction of α Q to T Q . Applying the surjectivity of α Q to the extension E/L we deduce that f(Q/S1 ) = 1. Therefore e(Q/S1 ) = [E : L]. Problems 4.4.1. Let R = ℤ[i], the ring of Gaussian integers. (1) Calculate the primes of R above 2ℤ and their ramification index and residue degree over ℤ.

146 | 4 Dedekind domains, valuations and orders (2) Prove that pR is a product of two different prime ideals of R if and only if p ≡ 1 mod 4. (Hint: Recall that p ≡ 1 mod 4 if and only if p is the sum of two squares. (see e.g. [101]) 4.4.2. Let p be a prime integer and R the ring of integers of ℚ[√p]. Calculate the primes of P in R and the residue degree and ramification index of these primes over ℤ. Change R by the ring of integers of ℚ[√−p] and repeat the problem.

4.5 Valuations In this section we revise the basic properties of valuations on division rings. For more information we refer the reader to [101], [175], [188] and [208]. Let D be a division ring. A valuation of D is a map v : D → ℝ+ ∪ 0 satisfying the following conditions, for every a, b ∈ D. (1) v(a) = 0 if and only if a = 0. (2) v(ab) = v(a)v(b). (3) v(a + b) ≤ v(a) + v(b). Let v a valuation of a division ring D and let D1 be a division ring that contains D as a subring. A prolongation of v to D1 is a valuation w of D1 such that w(x) = v(x) for every x ∈ D. Let v be a valuation of the division ring D. The v-metric is the metric of D defined by d(x, y) = v(x − y) and the v-topology is the topology of D associated to the v-metric. If w is another valuation of D then v and w are said to be equivalent if the v-topology and wtopology of D coincide. We leave it to the reader to prove the following characterization of equivalent valuations. Lemma 4.5.1. The following properties are equivalent for two valuations v and w of a division ring D: (1) v and w are equivalent. (2) v and w define the same unit ball. (3) w = v a for some positive real number a. We say that v is complete (or that (D, v) is complete) if D is complete with respect to the v-topology, that is, every Cauchy sequence is convergent in D. Let ̂ D v denote the completion of D with respect to the v-topology. Sometimes we will omit the subindex and write simply ̂ D instead of ̂ D v . The elements of ̂ D are equivalence classes of Cauchy sequences with respect to the v-topology. Then, ̂ D is a division algebra for the operations defined in the obvious way: (x n ) + (y n ) = (x n + y n ), (x n )(y n ) = (x n y n ).

4.5 Valuations | 147

Furthermore, v has a prolongation to a valuation ̂v on ̂ D by setting ̂v((x n )) = lim v(x n ). n→∞

Using standard methods, the reader easily may check the details. If a valuation v satisfies the following condition (which implies condition (3) in the definition of a valuation) then one says that v is non-Archimedean. (3󸀠 ) v(a + b) ≤ max{v(a), v(b)}. Otherwise, one says that v is Archimedean. The following lemma gives a criterion to distinguish Archimedean from nonArchimedean valuations. Lemma 4.5.2. The following properties are equivalent for a valuation v of the division ring D. (1) v is non-Archimedean. (2) v(n1D ) ≤ 1 for every n ∈ ℤ. (3) {v(n1D ) : n ∈ ℤ} is bounded. Proof. (1) implies (2) and (2) implies (3) are obvious. Assume that v(n1D ) ≤ a for every n ∈ ℤ. Clearly a ≥ v(1) = 1. Let x ∈ D and set b = max{v(x), 1}. Then, for every positive integer n, we have n

v(x + 1)n = v((x + 1)n ) ≤ ∑ v ((nk)1D ) b k ≤ (n + 1)ab n k=0

and therefore 1

1

v(x + 1) ≤ (n + 1) n a n b. Taking limits, when n tends to infinite, we obtain v(x + 1) ≤ b = max{1, v(x)}. If 0 ≠ y ∈ D then v(x + y) = v(y)v(y−1 x + 1) ≤ v(y) max{v(y−1 x, 1)} = max{v(x), v(y)}, as desired. From Lemma 4.5.1 and Lemma 4.5.2 it follows at once that if v and w are equivalent valuations of D then v is non-Archimedean if and only if so is w. Examples 4.5.3. (1) The trivial valuation of a division ring D is the valuation v defined by v(x) = 1, for every x ∈ D. Clearly D is complete with respect to the trivial valuation. (2) The standard absolute value is a valuation of the complex numbers ℂ. It restricts to an Archimedean valuation on every subfield of ℂ. Clearly, ℝ and ℂ are the only subfields for which this valuation is complete. More generally, let F be a field and σ a complex embedding of F. Clearly v σ (x) = |σ(x)| defines a valuation of F. Moreover F is complete with respect to v σ if and only if σ(F) = ℝ or σ(F) = ℂ. More ̂ → ℂ, also denoted σ, such that generally, σ extends to a field homomorphism F ̂ v(x) = |σ(x)| for every x ∈ F and σ(F) = ℝ, if σ(F) ⊆ ℝ, and otherwise σ(F) = ℂ. Let

148 | 4 Dedekind domains, valuations and orders σ denote the composition of σ with complex conjugation. Clearly v σ = v σ . For example, if F is a number field and σ1 , . . . , σ k are representatives of the complex embeddings of F, up to complex conjugation, then v σ1 , v σ2 , . . . , v σ k are Archimedean valuations of F. In Corollary 5.2.8 we will see that if i ≠ j then there is x ∈ F such that v i (x) > 1 and v j (x) < 1. Hence v σ1 , v σ2 , . . . , v σ k are pairwise non-equivalent, by Lemma 4.5.1. (3) Let R be a Dedekind domain, with quotient field F, and let P be a maximal ideal of R. Recall that e P denotes the exponential P-adic valuation. Fix 0 < ϵ < 1 and set v P (x) = ϵ e P (x) for x ∈ F (including v P (0) = ϵ+∞ = 0). Then v P is a nonArchimedean valuation, called the P-adic valuation valuation on F. If ϵ is replaced by another real number 0 < ρ < 1 then the valuations v󸀠 (x) = ρ e p (x) is equivalent to v. We will abuse terminology by using the expression “the P-adic valuation” independently of the base ϵ used. In case P = Rp, for some prime element p of R, then we simply write e p for e Rp and v p for v Rp and v p also is called the p-adic valuation. If P and Q are different maximal ideals of R then there is x ∈ P \ Q and hence v P (x) < 1 while v Q (x) = 1. Therefore v P and v Q are not equivalent. We quote, without proof, the following well know theorem of Ostrowski which characterizes the complete Archimedean valuations of fields (see e.g. [101, Theorem II.4.1]). Theorem 4.5.4 (Ostrowski’s Theorem). If v is complete Archimedean valuation on a field F then there is a complex embedding σ of F such that σ(F) = ℝ or ℂ and v is equivalent to v σ . The following result shows that Example 4.5.3 collects all the valuations of a number field up to equivalences. Theorem 4.5.5. Let F be a number field and let R be the ring of integers of F. Every Archimedean valuation of F is equivalent to v σ for some complex embedding σ of F. Every non-trivial non-Archimedean valuation is equivalent to v P for some maximal ideal P of R. Proof. If v is an Archimedean valuation of F then, by Ostrowski’s Theorem, there is a ̂ v is equivalent to v σ . Thus v is ̂ v such that the valuation of F complex embedding σ of F equivalent to v σ (restricted to F). See Problem 4.5.5 for the non-Archimedean case. Let F be a number field. A prime or place of F is an equivalence class of non-trivial valuations. An infinite prime (also called infinite place) is an equivalence class of Archimedean valuations and a finite prime (or finite place) is an equivalence class of non-trivial non-Archimedean valuations. By Example 4.5.3 and Theorem 4.5.5, the infinite primes are in one to one correspondence with the embeddings of F in ℂ, up to complex conjugation, and the finite primes are in one-to-one correspondence with the maximal ideals of the ring of integers of F. Let σ be a complex embedding of F. If σ(F) ⊆ ℝ then the infinite prime containing v σ is said to be a real prime (real place) and

4.5 Valuations | 149

otherwise it is said to be a complex prime (complex place). If E is a finite extension of F and P is a prime of F then the primes of E above P are those represented by valuations v of E such that v|F belongs to P. The completion of an Archimedean valuation v σ of F is either ℝ or ℂ (see Example 4.5.3). We now describe the completion of a non-Archimedean valuation v P . We do this in the general setting of discrete valuations on division rings. We need some preparation. Let D be a division ring and let v be a non-Archimedean valuation on D. Then R v = {x ∈ D : v(x) ≤ 1} is a subring of D called the valuation ring of v, and P v = {x ∈ D : v(x) < 1} is a two-sided ideal of R v such that U(R v ) = R v \ P v . This implies that P v is the unique maximal left ideal and the unique maximal right ideal of R v . Hence R v /P v is a division ring called the residue division ring, or residue field if it is commutative. Most of the time we will use the bar notation for the residue division ring. In other words, if R v and P v are as above then we set R v = R v /P v . Moreover, if a ∈ R v and f = a0 + a1 X +⋅ ⋅ ⋅+ a n X n ∈ R v [X] then we denote a = a + P v ∈ R v and f = a0 + a1 X + ⋅ ⋅ ⋅ + a n X n ∈ R[X]. The image of a valuation is a multiplicative subgroup of ℝ+ . A discrete valuation of D is a non-trivial non-Archimedean valuation v on D such that v(U(D)) is cyclic. For example, the P-adic valuations are discrete. Let v be a discrete valuation of a division ring D. Then v(U(D)) is an infinite cyclic subgroup of (ℝ+ , ⋅). Since D is dense in ̂ D, also v(U(D)) is dense in ̂v(U(̂ D)). Further+ more, v(U(D) is a discrete subset of ℝ with respect to the Euclidean topology and this implies that ̂v(U(̂ D)) = v(U(D)). Therefore, ̂v also is discrete. A uniformizer of v is an element p ∈ D such that v(D) = ⟨v(p)⟩ and v(p) < 1. It is clear that p is a uniformizer of v if and only if v(p) = max{v(D) ∩ (0, 1)}. Let R be the valuation ring of v and P the unique maximal ideal of R. If p is a uniformizer of v then P = pR = Rp and every left ideal of R is of the form p n R = Rp n , for some n ≥ 0. Moreover, right multiplication by p n induces an isomorphism R → P n /P n−1 of R-modules. In particular, if D = F is a field then R is a discrete valuation ring, as defined in Section 4.1. Then, every element of F has a unique form up k for some k ∈ ℤ and u ∈ U(R) and if ϵ = v(p) then v(up k ) = ϵ k . We leave it to the reader to complete the details. We now are in a position to describe the completion ̂ D v . Select a set T of representatives of the elements of R/Rp containing 0. Then every non-zero element a of D can i be written uniquely as a truncated convergent series a = ∑∞ i=i0 t i p , for some i 0 ∈ ℤ and i −i each t i ∈ T. Indeed, let v(a) = v(p) 0 . Then, v(ap 0 ) = 1 and hence b0 = ap−i0 ∈ U(R). By the definition of T, there is a unique 0 ≠ t i0 ∈ T with b0 −t i0 ∈ Rp. Thus b0 −t i0 = b1 p with b1 ∈ R. Therefore, there is a unique t i0 +1 ∈ T and b2 ∈ R with b1 = t i1 + b2 p. Hence, b0 = t i0 + t i0 +1 p + b2 p2 . Repeating this process, we obtain two sequences t i0 , t i0 +1 , . . . , and b0 , b1 , . . . with b0 = t i0 + t i0 +1 p + ⋅ ⋅ ⋅ + b i0 +j p j + b j+1 p j+1 . Then, a = t i0 p i0 + t i0 +1 p i0 +1 + ⋅ ⋅ ⋅ + t j p j + b j+1 p i0 +j+1

150 | 4 Dedekind domains, valuations and orders i and hence the series ∑∞ i=i0 t i p converges to a, with respect to the v-topology. Let ∞ i ∑i=i0 t i p be an arbitrary series with t i ∈ T. The partial sums give a Cauchy sequence but the series may not converge in D. The completion ̂ D v of D is formed by all series i ̂ ̂ of this form. The valuation ring R v of D v is formed by the series of the form ∑∞ i=0 t i p ∞ i and its maximal ideal by the series of the form ∑i=1 t i p , with t i ∈ T.

Example 4.5.6. Let R be a Dedekind domain with field of quotient F and let P be a maximal ideal of R. Then R v P = R P , the localization of R at P, P v P = P P and the natural embedding R ⊆ R P induces an isomorphism R/P ≅ R P /P P (Lemma 4.2.3). Therefore, one can take for T the set consisting of representatives of R modulo P. If F is a number field with ring of integers R and P is a maximal ideal of R then we ̂ v ) as F ̂ P (respectively R ̂ P ). (Observe that in some texts ̂ v (respectively R simple write F P P these are denoted F P and R P , but this is in conflict with the notation for localization ̂ (p) is the ring of p-adic ̂ (p) is the field of p-adic numbers and ℤ at P.) For example, ℚ ̂ p respectively. ̂ p and ℤ integers, which we simply denote by ℚ The following theorem is a deep result of Class Field Theory. A proof can be found in [101]. Theorem 4.5.7 (Hasse Norm Theorem). Let E/F be a cyclic extension of number fields and let x ∈ F. Then x is a norm of the extension E/F if and only if for every valuation v ̂ v , where w is some (any) prolongation of v to a ̂ w /F of F, x is a norm of the extension E valuation of E. Observe that the Hasse Norm Theorem states that to check that x ∈ F is a norm of the extension E/F it is enough to select one valuation v P for every prime P of F and one ̂ w /F ̂ v for each P. prolongation w P of v P and check whether x is a norm in E P P We close this section with some applications of the Norm Hasse Theorem. We start ̂ p , for p a prime integer. For with a result which describes the multiplicative group of ℚ every n ≥ 1, let ̂p , Un = 1 + pn ℤ ̂ p ) ≅ U(ℤ/pℤ). Then every element ̂ p ) → U( ℤ ̂ p /p n ℤ the kernel of the natural map U(ℤ n ̂ p ). ̂ of ℚp has a unique expression as p u with n ≥ 0 and u ∈ U(ℤ Proposition 4.5.8 ([207, Chapter II]). Let p be a prime integer. ̂ p ) = ⟨p⟩ × U1 × ⟨ζ p−1 ⟩ and U1 is a subgroup of ̂ p ) = ⟨p⟩ × U(ℤ (1) If p is odd then U(ℚ ̂ p ) isomorphic to the additive group of ℤ ̂p. U( ℤ ̂ 2 = ⟨p⟩ × U2 × ⟨−1⟩ and there is an isomorphism f from U2 to the additive group (2) ℚ ̂ 2 such that f(U3 ) = 2ℤ ̂2. of ℤ Given a field F we let F 2 denote the set of squares of U(F). ̂ p and hence every element of U1 Suppose that p is odd. Then 2 is invertible in ℤ 2 2 2 ̂ is a square, by Proposition 4.5.8 (1). Thus ℚp = ⟨p ⟩ × U1 × ⟨ζ p−1 ⟩. In other words, if n 2 ̂ ̂ x = p u with n ≥ 0 and u ∈ U(ℤp ) then x ∈ ℚp if and only if n even and the image of

4.5 Valuations | 151

u in ℤ/pℤ is a square. Hence ̂ p )/ℚ ̂ 2p ≅ ⟨p⟩/⟨p2 ⟩ × ⟨ζ p−1 ⟩/⟨ζ p−1 ⟩ ≅ C2 , U(ℚ 2 2

and if u is an integer which is not a square modulo p then {1, p, u, pu} is a set of rep̂ p ) module its squares. resentatives of U(ℚ ̂ 2 and the preimage of 2ℤ ̂ 2 under the isomorphism However, 2 is not invertible in ℤ 2 ̂ ̂ U2 → ℤ2 is U3 , a subgroup of index 2 in U2 . Therefore ℚ2 = ⟨4⟩× U3 , that is, if x = 2n u ̂ 2 ) then x ∈ ℚ ̂ 2 if and only if n is even and u ≡ 1 mod 8. Thus with n ≥ 0 and u ∈ U(ℤ 2 ̂ 2 )/ℚ ̂ 2 ≅ ⟨2⟩ / ⟨4⟩ × U2 /U3 × ⟨−1⟩ ≅ C3 , U(ℚ 2 2 and {±1, ±2, ±5, ±10} is a set of representatives of U(ℚ2 ) module squares. The first item of next lemma is a direct consequence of Proposition 4.5.8. The other two can be found in [142, Examples XI.2.14]. ̂p. Lemma 4.5.9. Let p be a prime integer and let F be a finite field extension of ℚ ̂ p if and only if p ≡ 1 mod 4. (1) −1 is a square in ℚ (2) If p is odd then −1 is a sum of two squares of F. ̂ 2 ] is even. (3) If p = 2 then −1 is a sum of two squares of F if and only if [F : ℚ Proposition 4.5.10. Let F be a number field. Then −1 is a sum of two squares in F if and ̂P : ℚ ̂ 2 ] is even for every prime ideal P of the ring of only if F is totally complex and if [F integers of F with 2 ∈ P. ̂ 2] Proof. If i ∈ F then obviously −1 is a squares of F, F is totally complex and [F P : ℚ ̂ 2 , by Proposition 4.5.8. This proves the result in this case. is even because −1 ∈ ̸ ℚ 2 Assume that i ∈ ̸ F. Then NrF(i)/F (x + yi) = x2 + y2 . Therefore −1 is a sum of two squares in F if and only if −1 is a norm in F and this holds if an only if −1 is a sum of two ̂ v for every valuation v of F, by the Hasse Norm Theorem (Theorem 4.5.7).By squares of F ̂ v unless v represents a real prime or v is a Lemma 4.5.9, −1 is a sum of two squares in F ̂v : ℚ ̂ 2 ] is odd. The result then follows prolongation of the 2-adic valuation of ℚ and [F from Theorem 4.5.5. Theorem 4.5.11. The following conditions are equivalent for square-free integer d. (1) ℍ(ℚ(√d)) is not a division ring. (2) −1 is a sum of two squares of ℚ(√d). (3) d < 0 and either d is even or d ≢ 1 mod 8. Proof. The equivalence of (1) and (2) is an immediate consequence of Example 2.1.7. That also (2) and (3) are equivalent can be shown as follows. Write d = 2n u with u an odd integer. By Proposition 4.5.10, −1 is a sum of two squares in ℚ(√d) if and only if ̂ 2 . The latter holds if and only if then n is odd or u ≢ 1 mod 8. d is negative and d ∈ ̸ ℚ 2 As d is square free, this holds if and only if d is even or d ≢ 1 mod 8.

152 | 4 Dedekind domains, valuations and orders

Problems 4.5.1. Prove Lemma 4.5.1. 4.5.2. Prove that the v-metric of a the P-adic valuation, v P (x) = ϵ e P (x) is independent of ϵ. 4.5.3. Let v be a non-Archimedean valuation of a field F and let R be a subring of the valuation ring of v. Let E be a finite field extension of F and let S be the integral closure of R in E. Prove that if w is a prolongation of v to a valuation in E then w is non-Archimedean and S is contained in the valuation ring of w. 4.5.4. Let R be a Dedekind domain with field of fractions F, let E/F be a finite field extension of F, let S be the integral closure of R in L and let Q be a maximal ideal of S and P = R ∩ Q. Let e = e(Q/R) and let v P and v Q respectively be the P-adic valuation in F and the Q-adic valuation in E. (1) Prove that if v P and v Q have been constructed with the same 0 < ϵ < 1 then v Q (a) = v P (a)e , for every a ∈ F. (2) Show how one can define v P and v Q with different ϵ’s so that v Q is a prolongation of v P . (3) Prove that if w is a prolongation of v P in E then w is equivalent to v Q . 4.5.5. Let F be a number field. Prove that every non-trivial non-Archimedean valuation of F is equivalent to a P-adic valuation, for P a maximal ideal of the ring of integers of F. (Hint: Use Problem 4.5.4 to reduce to the case F = ℚ. Let v be a non-trivial nonArchimedean valuation of ℚ and let P = pℤ = ℤ ∩ P v . Prove that v is the P-adic valuation of ℚ, constructed with ε = v(p).)

4.6 Orders In this section R is a commutative Noetherian domain with field of fractions F and A is a finite dimensional F-algebra. If V is a finite dimensional vector space over F then an R-lattice in V is a finitely generated R-submodule of V. A full R-lattice in V is an R-lattice in V that contains a basis of V over F. An R-order in A is a subring of A which also is a full R-lattice in A. A ℤ-order will be simply called an order. Because ℤ is a PID, an order contains a ℤ-basis and this obviously also is a ℚ-basis of A. Examples 4.6.1. Let n be a positive integer and G a finite group. (1) M n (R) is an R-order in M n (F). (2) RG is an R-order in FG

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(3) If R is a Dedekind domain and E is a finite separable extension of F then the integral closure of R in E is an R-order in E (see Proposition 4.1.6 and Theorem 4.2.6). The proof of the following lemma is left to the reader. It provides more examples of R-orders. Lemma 4.6.2. Let A and B be finite dimensional F-algebras, n a positive integer and G a finite group. If O is an R-order in A and O󸀠 is an R-order in B then O × O󸀠 is an R-order in A × B, M n (O) is an R-order in M n (A) and OG is an R-order in AG. From the definitions and Proposition 4.1.3 it is clear that if O is an R-order in A then O is integral over R. Hence the following Lemma is a direct consequence of Proposition 4.1.8. Lemma 4.6.3. Assume that R is integrally closed and O an R-order in A. If a ∈ O then (1) MinF (a), CharA/F (a) ∈ R[X] and TrA/F (a), NrA/F (a) ∈ R. (2) If A is separable then RCharA/F (a) ∈ R[X] and RTrA/F (a), RNrA/F (a) ∈ R. We give one more example of an order. Example 4.6.4. The Hurwitz quaternions Hu(ℤ) (see Example 4.1.10) is a maximal order in ℍ(ℚ), i.e. it is an order in ℍ(ℚ) that is not strictly contained in any order of A. Moreover, it is the unique maximal order containing ℍ(ℤ). Proof. Clearly Hu(ℤ) is an order in ℍ(ℚ) strictly containing ℍ(ℤ). Suppose O is any order in ℍ(ℚ) strictly containing ℍ(ℤ). To prove the result it is sufficient to show that if 2 j+a 3 k α ∈ O \ ℍ(ℤ) then α ∈ Hu(ℤ). Write α = a0 +a1 i+a ∈ ℍ(ℚ), with a1 , a2 , a3 , a4 , q ∈ q ℤ. By adding elements of ℍ(ℤ) we may assume that − 2q < a r ≤ 2q for 0 ≤ r ≤ 3 and a r ≠ 2 2 2 2 2 0 for some r. Thus ( aqr ) ≤ 14 for every r and hence 0 < ( aq0 ) + ( aq1 ) + ( aq2 ) + ( aq3 ) ≤ 1. Furthermore, as α is an integral element, it follows as in the proof of Example 4.1.10) 2 2 2 2 2 2 2 2 that ( aq0 ) + ( aq1 ) + ( aq2 ) + ( aq3 ) is an integer. Hence ( aq0 ) + ( aq1 ) + ( aq2 ) + ( aq3 ) = 1. ai 1 So q = ± 2 for every i. This proves that α ∈ Hu(ℤ), as desired. The following easy observation will be used frequently. Lemma 4.6.5. If L is a full R-lattice in a finitely dimensional F-vector space V and X is a finite subset of V then there is 0 ≠ r ∈ R such that rX ⊆ L. If L1 is another full R-lattice in V then there is 0 ≠ r ∈ R such that rL ⊆ L1 . Proof. Fix a basis v1 , . . . , v n of V contained in L. For every x ∈ X write x = ∑ni=1 a x,i v i , with a x,i ∈ F. Then there is 0 ≠ r ∈ R such that ra x,i ∈ R for every x ∈ X and i = 1, . . . , n. Therefore rx = ∑ni=1 (ra x,i )v i ∈ L. The second statement is a consequence of the first one and the fact that L is finitely generated as R-module. The following lemma can be proved easily using Lemma 4.6.5.

154 | 4 Dedekind domains, valuations and orders

Lemma 4.6.6. (1) If O1 and O2 are R-orders (respectively, full R-lattices) in A then O1 ∩ O2 is an Rorder in A (respectively, full R-lattices). (2) If B is a subalgebra of A and O is an R-order in A then O ∩ B is an R-order in B. In particular, Z(O) is an R-order in Z(A). If L is an R-lattice in A then let Ol (L) = {x ∈ A : xL ⊆ L}

and Or (L) = {x ∈ A : Lx ⊆ L}.

The following proposition proves that R-orders always exists in A. Proposition 4.6.7. If R is a commutative Noetherian domain with field of fractions F, A is a finite dimensional algebra and L is a full R-lattice in A then Ol (L) and Or (L) are R-orders in A. In particular, A has an R-order. Proof. By symmetry, we only prove that O = Ol (L) is an R-order in A. Clearly O is a subring of A. Let v1 , . . . , v n be a generating set of L as R-module. Applying Lemma 4.6.5 to X = {1A } ∪ {v i v j : 1 ≤ i, j ≤ n}, we obtain an element 0 ≠ d ∈ R such that d1A ∈ L and dv i v j ∈ L for every i, j. Clearly, Od = Od1A ⊆ OL ⊆ L and dv i L = ∑nj=1 Rv j ⊆ L. Hence, O ⊆ d−1 L and dv i ∈ O, for every i. The former implies that O is finitely generated over R, because L is Noetherian as an R-module, and the latter implies that O contains a basis of A over F. Thus O is an R-order in A. The following lemma is a partial generalization of the separability case of Theorem 4.2.6. Lemma 4.6.8. Assume that R is integrally closed and A is separable over F. Let O be a subring of A such that O is integral over A and contains a basis of A over F. Then O is an R-order in A. Proof. We only need to prove that O is finitely generated as R-module. Let v1 , . . . , v n be an F-basis of A contained in O. Let w1 , . . . , w n be the dual basis of v1 , . . . , v n in A, that is RTr(v i w j ) = δ ij . If x = ∑ni=1 a i w i ∈ O, with a i ∈ F, then a i = RTrA/F (v i x) ∈ R, by Lemma 4.6.3. This proves that O ⊆ Rw1 +⋅ ⋅ ⋅+ Rw k and therefore O is finitely generated over R, because R is Noetherian. The following lemma collects some elementary properties of the units of an R-order. Lemma 4.6.9. Assume that R is integrally closed. Let O be an R-order in A and a ∈ O. (1) NrA/F (a)1A ∈ aO ∩ Oa. If A is separable over F then RNrA/F (a)1A ∈ aO ∩ Oa. (2) a ∈ U(O) if and only if NrA/F (a) ∈ U(R). If A is separable over F, then the latter is equivalent with RNrA/F (a) ∈ U(R). (3) Let O1 be another order in A and a ∈ O ∩ O1 . Then a ∈ U(O) if and only if a ∈ U(O1 ). (4) If, moreover, R is a Dedekind domain and R/I is finite for every non-zero ideal I of R then U(O ∩ O1 ) has finite index in U(O). More precisely, if 0 ≠ r ∈ R and rO ⊆ O1 then [U(O) : U(O ∩ O1 )] ≤ [O : rO].

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155

Proof. (1) Because of Proposition 4.1.3, CharA/F (a) = X n + r n−1 X + ⋅ ⋅ ⋅ + r1 X + r0 ∈ R[X]. Then (a n−1 + r n−1 a n−2 + ⋅ ⋅ ⋅ + r1 1A )a = a(a n−1 + r n−1 a n−2 + ⋅ ⋅ ⋅ + r1 1A ) = −r0 1O = ±NrA/F (a)1A . This implies that NrA/F (a)1A ∈ aO ∩ Oa. A similar argument shows RNrA/F (a)1A ∈ aO ∩ Oa. (2) If ab = 1 with b ∈ O then NrA/F (a)NrA/F (b) = 1 and NrA/F (a), NrA/F (b) ∈ R, by Lemma 4.6.3. This shows that if a ∈ U(O) then NrA/F (a) ∈ U(R). Assume conversely that NrA/F (a) ∈ U(R). Then 1 ∈ NrA/F (a)R ⊆ aO ∩ Oa and hence a ∈ U(O). If A is separable over F and simple then it follows from (2.3.9) that NrA/F (a) ∈ U(R) if and only if RNrA/F (a) ∈ U(R). Using (2.3.3) and the definition of reduced norm this extends easily to arbitrary separable algebras. (3) is a direct consequence of (2). (4) By Lemma 4.6.6, we may assume that O1 ⊆ O. Because of Lemma 4.6.5, there is an r ∈ R such that rO ⊆ O1 . On the other hand, O/rO is a finitely generated torsion R-module. By Theorem 4.3.2, as an R-module, O/rO is isomorphic to a finite direct sum of R-modules of the form R/I, with I a non-zero ideal of R. By hypothesis, each such R/I is finite and thus O/rO is finite. Let x, y ∈ U(O) with x − y ∈ rO. Then xy−1 − 1 ∈ rO ⊆ O1 and therefore xy−1 ∈ O1 ∩ U(O) = U(O1 ), by (2). This proves [U(O) : U(O1 )] ≤ |O/rO| < ∞. Two subgroups H1 and H2 of a group H are said to be commensurable when their intersection has finite index in both H1 and H2 . Often the group H is clear from the context and hence will not be specifically mentioned. For example, Lemma 4.6.9 (4) can be rephrased saying that the group of units of two orders in A are commensurable. Remark 4.6.10. The above lemma is a practical tool for the investigations of the unit group of an integral group ℤG of a finite group G. It allows to reduce the study of some problems to units of linear groups over orders in division rings. This is done as follows. Let E be the set of primitive central idempotents of ℚG. Then ℤG ⊆ ∏ (ℤG)e ⊆ ∏ ℚGe = ℚG e∈E

e∈E

and, by the Wedderburn-Artin Theorem, ℚGe = M n e (D e ) for some positive integer n e and some division algebra D e . Clearly, ∏e∈E (ℤG)e is an order in ℚG and ℤGe is an order in ℚGe. From the lemma we know that U(ℤG) is of finite index in U(∏e∈E ℤGe) = ∏e∈E U(ℤGe). For each e ∈ E, one can choose an order Oe in D e . Again by the lemma, GLn e (Oe ) and U(ℤGe) have a common subgroup of finite index. It follows, for example, that U(ℤG) is finite if and only if GLn (Oe ) is finite for each e ∈ E. It is this technique that is implicitly behind the proof of Theorem 1.5.6. Later, in Chapter 11, we will construct a finite set of units in U(ℤG) that generates a subgroup of finite index for many finite groups G. In order to prove that a set B has this property, it is sufficient to show that ⟨B⟩ contains a subgroup of finite index in GLn e (Oe ) for each e ∈ E. In Chapter 11 we will discuss virtual structure theorems of U(ℤG), by this we mean descriptions of some structure of a subgroup of finite index in U(ℤG). As a first step we will show that the problems can be reduced to orders in ℚGe.

156 | 4 Dedekind domains, valuations and orders

We close this section with the following lemma which is needed in later chapters. Lemma 4.6.11. Let O be an order in a d-dimensional semisimple rational algebra A and let n be a non-zero positive integer. Then there exists a finite subset X of O with |X| ≤ n d and such that {x ∈ O : |N(x)| = n} = U(O)X. Proof. The cardinality of O/On is precisely n d . Thus, to prove the lemma, it is enough to show that if x, y ∈ O with |NrA/ℚ (x)| = |NrA/ℚ (y)| = n and x − y ∈ nO then x = uy for some u ∈ U(O). Indeed, let x and y be such elements. Because |NrA/ℚ (y)| = n ≠ 0, we have that y is invertible in A. Moreover, by Lemma 4.6.9 (1), ny−1 ∈ Oyy−1 = O. Let u = xy−1 . Then u − 1 = (x − y)y−1 ∈ Ony−1 ⊆ O. Hence u ∈ O and |NrA/ℚ (u)| = 1. By Lemma 4.6.9 (2), u ∈ U(O), i.e. x and y are associates in O. Therefore x = y as desired. Proposition 4.6.12. Let R be a commutative Noetherian domain with field of fractions F and let A be a finite dimensional semisimple algebra A. Let O be an R-order in A and I an ideal of O. Then the following conditions are equivalent. (1) I is a full R-lattice in A. (2) I is essential in R, i.e. I ∩ J ≠ 0 for every non-zero ideal J of O. (3) I intersects non-trivially every Wedderburn component of A. Proof. Clearly I is an R-lattice of A. (1) implies (2) is a consequence of Lemma 4.6.5. (2) implies (3). Suppose that I is essential in R. Let e be a primitive central idempotent of A. As O is a full R-lattice, by Lemma 4.6.5, there is r ∈ R such that 0 ≠ re ∈ O. Then I∩Ore ≠ 0 and hence I∩Ae ≠ 0. Thus I intersects non-trivially every Wedderburn component of A. (3) implies (1). Let e1 , . . . , e n be the primitive central idempotents of A. Clearly Ie i = I∩Ae i and Ae i = F Oe i . Assume that I∩Ae i ≠ 0 for every i. Then FIe i = F(I∩Ae i ) is a non-zero ideal of Ae i = F Oe i and hence FIe i = Ae i . Thus FI = ∑ni=1 FIe i = ∑ni=1 Ae i = A. Therefore, I is a full R-lattice in A.

Problems 4.6.1. Let L be a full ℤ-lattice in a finite dimensional vector space V over ℚ and let L1 be a lattice of V contained in L. Prove that L1 is a full lattice in L if and only if L/L1 is finite. 4.6.2. Let O be an order in a finite dimensional semisimple rational algebra A. Let I be an ideal of O. Prove that R/I is finite if and only if I is essential in O (i.e., I intersects non trivially every Wedderburn component of A) and show that then the centralizer of O in A is the center of A.

4.7 The discriminant

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4.6.3. Let R be a commutative Noetherian domain with field of fractions F and A a finite dimensional semisimple F-algebra. Let O be a subring of A. Prove the following statements. (1) If e1 , . . . , e n are the primitive central idempotents of A then O is a R-order in A if and only if Oe i is an R-order in A for every i. (2) If O is a order in A then M n (O) is a order in M n (A) for every n ∈ ℕ. (3) If P is a prime ideal of R and O is an R-order in A then OP is an R P -order in A. 4.6.4. Consider a quaternion algebra A = ( −a,−b ℚ ) with a and b positive integers. Calculate the norm map NrA/ℚ . Prove that A is a division ring. Show that O = ℤ[i, j] is an order in A and calculate U(O) and U(Z(O)).

4.7 The discriminant In this section F is a field and A is an n-dimensional separable F-algebra. The discriminant over F of an ordered sequence x1 , . . . , x n of elements of A is Δ F (x1 , . . . , x n ) = det((RTrA/F (x i x j ))1≤i,j≤n ). Observe that if x1 , . . . , x n are linearly independent over F then (RTrA/F (x i x j ))1≤i,j≤n is the matrix with respect to this basis of the bilinear form induced by the reduced trace (see Proposition 2.3.6). Then Δ F (x1 , . . . , x n ) ≠ 0. Given a new list y1 , . . . , y n with y i = ∑ni=1 a ij x j and a ij ∈ F, we have Δ F (y1 , . . . , y n ) = det(a ij )2 Δ F (x1 , . . . , x n ).

(4.7.1)

Therefore, in general Δ F (x1 , . . . , x n ) ≠ 0 if and only if x1 , . . . , x n is a basis. Notice that if A is commutative then the reduced trace and the trace coincide and therefore Δ F (x1 , . . . , x n ) = det((TrA/F (x i x j ))1≤i,j≤n ). In the proof of Theorem 4.7.3 we will use the latter as definition of discriminant in an arbitrary commutative ndimensional F-algebra (not necessarily separable). Assume now that F is the field of fractions of an integrally closed Noetherian subring R and A is a separable algebra over F. Let O be an R-order in A. By Lemma 4.6.3, Δ F (x1 , . . . , x n ) ∈ R for every x1 , . . . , x n ∈ O. The discriminant of O over R is the ideal Δ(O/R) of R generated by the elements of the form Δ F (x1 , . . . , x n ), for x1 , . . . , x n with each x i ∈ O (it is sufficient to consider only lists in O that form a basis of A over F). If F is a number field then we call the discriminant of F the positive generator of Δ(R/ℤ), where R is the ring of integers of F. The following lemma can be proved using Proposition 2.3.6, (4.7.1) and the elementary properties of reduced trace (2.3.7). Lemma 4.7.1. Let R be an integrally closed commutative Noetherian domain with field of fractions F, A a separable F-algebra of dimension n and O an R-order in A.

158 | 4 Dedekind domains, valuations and orders (1) Δ(O/R) ≠ 0. (2) If x1 , . . . , x n is an R-basis of O then Δ(O/R) = RΔ F (x1 , . . . , x n ) (3) If P is a prime ideal of R then Δ(OP /R P ) = Δ(O/R)P . Example 4.7.2. Let E/F be a finite field extension. If E is not separable over F then TrE/F = 0 and therefore all the discriminants are 0. Assume otherwise that E is separable over F. Let x1 , . . . , x n be a basis of E over F and let σ1 , . . . , σ n be the F-homomorphisms of E to an algebraic closure of F. Consider the n × n matrix A = (σ i (x j )). Then the (i, j)-entry of A T A is ∑nk=1 σ k (x i x j ) = TrE/F (x i x j ) = RTrE/F (x i x j ). Therefore, Δ F (x1 , . . . , x n ) = det(σ i (x j ))2 . (4.7.2) In particular, if E = F(α) and we set α i = σ i (α) then 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 α1 󵄨󵄨 󵄨󵄨 2 Δ F (1, α, . . . , α n−1 ) = 󵄨󵄨󵄨 α1 󵄨󵄨 . 󵄨󵄨 . 󵄨󵄨 . 󵄨󵄨 n−1 󵄨󵄨α 󵄨 1

1 α2 α22 .. . α2n−1

... ... ... ...

󵄨2 1 󵄨󵄨󵄨 󵄨 α n 󵄨󵄨󵄨󵄨 󵄨 α2n 󵄨󵄨󵄨󵄨 󵄨 .. 󵄨󵄨󵄨 . 󵄨󵄨󵄨 󵄨󵄨 󵄨 α n−1 n 󵄨󵄨

= ∏(α i − α j )2 = ±NrE/F (f 󸀠 (α)), 1≤i 1 or S/Q i is not separable over R/P. By symmetry we only have to prove this for i = 1. Let Q = Q1 , e = e1 and Δ = Δ1 . Set R = R/P and S = S/Q. If e = 1 and S is separable over R, then 0 ≠ Δ(S/R) = RΔ R (u1 , . . . , u k1 ) = R det(Δ), by Lemma 4.7.1. Conversely, if e > 1 we may choose the basis u1 , . . . , u n so that Ψ(u1 ) ∈ Q/Q e . Then (u1 u i )e = 0 for every i. Therefore Tr(S/Q e )/R (u1 u i ) = 0 for every i. Hence

the first column of Δ is zero and hence det(Δ) = 0. If e = 1 but S is not separable over R, then the trace map of the field extension S/R is the zero map. Therefore Δ is the zero matrix and so det(Δ) = 0. We leave it to the reader to prove the following lemma. Lemma 4.7.4. Let F and E be number fields contained in ℂ and let FE be the compositum of F and E in ℂ. Then the following conditions are equivalent: (1) F ∩ E = ℚ. (2) [FE : ℚ] = [F : ℚ][E : ℚ]. (3) For every complex embedding σ of F and every complex embedding τ of E there is a complex embedding of FE extending both σ and τ. When the conditions of Lemma 4.7.4 hold then one says that F and E are algebraically independent. The following result can be found in [162]. Proposition 4.7.5. Let F and E be number fields contained in ℂ. Let R F , R E and R EF be the ring of integers of F, E and EF respectively. Let k

R F R E = { ∑ x i y i : x i ∈ R F , y i ∈ R E }, i=1

160 | 4 Dedekind domains, valuations and orders

the subring of FE generated by R F and R E . Then (1) R F R E ⊆ R FE and (2) if F and E are algebraically independent and d is a greatest common divisor of the discriminants of F and E then dR FE ⊆ R F R E . Proof. (1) is obvious. Assume that the hypothesis of (2) hold and let α1 , . . . , α n and β1 , . . . , β m be basis of R F and R E over ℤ, respectively. Then {α i β j : i = 1, . . . , n, j = 1, . . . , m} is a basis of FE over ℚ. Let x be an arbitrary element of R FE . After some simplification we may write a ij x=∑ αi βj , r i,j with a ij , r ∈ ℤ and r coprime with the greatest common divisors of the a ij ’s. Let x i = a ij ∑m j=1 r β j for every i = 1, . . . , n. Let σ1 , . . . , σ n be the different complex embeddings of F. As F and E are algebraically independent, for every i = 1, . . . , n there is a complex embedding of FE that acts as σ i on F and as the identity on E. This extension we also denote by σ i . Then n

σ k (x) = ∑ σ k (α i )x i . i=1

Let D = det(σ k (α i )). Observe that σ k (x) ∈ R FE and σ k (α i ) ∈ R F for every k and i. Solving for the x i ’s by Cramer’s Rule we obtain Dx i = D i , with D ∈ R F and D i ∈ R FE . Let Δ be the discriminant of F. By (4.7.2), Δ = D2 and hence m Δa ij β j = Δx i = DD i ∈ R FE . ∑ r j=1 In other words, this element of E is integral over ℤ. As β1 , . . . , β m is a basis of R E over Δa ℤ, this implies that r ij ∈ ℤ. Therefore r divides Δa ij for every i and j. As r is coprime with the greatest common divisor of the a ij ’s we deduce that r divides Δ. Similarly r divides the discriminant of E. Hence r divides d and we conclude that dx ∈ R F R E , as desired.

Problems 4.7.1. Prove Lemma 4.7.1 and Lemma 4.7.4.

4.8 Brauer group of a number field In this section we describe the Brauer group of a local field and of a number field. We only include the proofs that help to understand the main concepts. The interested reader can find complete proofs in [175] and [188].

4.8 Brauer group of a number field

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A local field (F, v) is a complete discrete valuation field with finite residue field. As usual we often simply use F to denote (F, v). By an extension of local fields E/F we not only mean that F and E are local fields but also that the valuation of F is the restriction to F of the valuation of E. Examples of local fields of major interest come from non-Archimedean valuations of number fields. More precisely, if F is a number field with ring of integers R and v is a non-Archimedean valuation of F then v is equivalent to v P for a maximal ideal of R (Theorem 4.5.5) and hence the residue field of F with respect to v, as well as the ̂ v , is isomorphic to R/P. As R/P is a finite, F ̂ v is a local residue field of its completion F field. We start with a classical result that relates the roots of polynomials over a valuation ring with those over a residue field with respect to a complete non-Archimedean valuation. Lemma 4.8.1 (Hensel’s Lemma). Let D be division algebra over a field F and let v be a complete non-Archimedean valuation on D. Let R and O be the valuation rings of F and D with respect to v and let Rand D be the residue field of F and D with respect to v. Let 󸀠 f ∈ R[X] with gcd(f , f ) = 1. If f has a root α in O then f has a root x in O such that x = α. (We use the bar notation for the natural maps R 󳨃→ R and O 󳨃→ O and extend this notation to polynomials with coefficients in R.) Proof. The idea of the proof is based on Newton approximation of root of a polynomial. We start selecting x0 ∈ O such that x0 = α. We consider x0 as a first approximation to a root of f . Define recursively z n = −f 󸀠 (x n )−1 f(x n )

and

x n+1 = x n + z n .

We have to make sure that this makes sense, that is f 󸀠 (x n ) ≠ 0. In fact we will show the following stronger facts: v(f 󸀠 (x n )) = 1

and

n

v(f(x n )) ≤ v(f(x0 ))2 ,

(n ≥ 0).

(4.8.1)

󸀠

For this we argue by induction on n. As f (x0 ) = 0 and gcd(f , f ) = 1, one has that 󸀠 f (x0 ) ≠ 0 and this implies that v(f 󸀠 (x0 )) = 1. This gives (4.8.1) for n = 0, since 0 v(f(x0 )) = v(f(x0 ))2 . Assume that n is positive and (4.8.1) has been proven for n. Then n v(z n ) = v(f(x n )) ≤ v(f(x0 ))2 < 1. We use the Taylor expansion f(X + Z) = f(X) + f 󸀠 (X)Z + h(X, Z)Z 2 for some polynomial h ∈ R[X, Z] and taking derivatives with respect to Z we have f 󸀠 (X + Z) = f 󸀠 (X) + h1 (X, Z)Z. for some h1 ∈ R[X, Z]. Replacing X by x n and Z by z n we have f(x n+1 ) = h(x n , z n )z2n

and

f 󸀠 (x n+1 ) = f 󸀠 (x n ) + h1 (x n , z n )z n .

Then v(f(x n+1 )) ≤ v(z n )2 ≤ v(f(x0 ))2 and v(f 󸀠 (x n+1 )) = v(f 󸀠 (x n )) = 1. (Here we have used the dominant principal: v(x) < v(y) implies v(x + y) = v(y).) This finishes the proof of (4.8.1). n+1

162 | 4 Dedekind domains, valuations and orders As f (x0 ) = f (α) = 0, or in other words, v(f(x0 )) < 1, the inequality of (4.8.1) implies that (f(x n )) converges to 0. Using the equality in (4.8.1) we obtain v(x n+1 −x n ) = v(z n ) = v(f(x n )). Furthermore, h−1

v(x n+h − x n ) = v( ∑ x n+i+1 − x n+i ) i=0

≤ max{v(x n+i − x n+i−1 ) : i = 0, . . . , h − 1} = max{v(f(x n+i )) : i = 0, . . . , h − 1} ≤ max{v(f(x0 ))2 = v(f(x0 ))

n+i

: i = 0, . . . , h − 1}

2n

This shows that (v(x n )) is a Cauchy sequence because v(f(x0 )) < 1 as f(x0 ) ∈ O. Since x0 ∈ O and f ∈ R[X], the completeness of D yields that (x n ) converges in O, say to x. Then f(x) = limn→∞ f(x n ) = 0 and x = α, because x n = α for every n. Theorem 4.8.2. Let (F, v) be a complete discrete valuation field with valuation ring R. If D is a finite dimensional division F-algebra of dimension n then (1) v has a unique extension to a discrete valuation w on D given by n w(a) = √ v(NrD/F (a)) (a ∈ D).

(2) [w(D) : v(F)] < ∞. (3) Let O be the valuation ring of D with respect to w. Then O = {a ∈ D : NrD/F (a) ∈ R} = {a ∈ D : a is integral over R}. (4) If Q is the unique maximal ideal of O then P = Q ∩ F = Q ∩ R is the unique maximal ideal of R. Proof. We start proving that if a ∈ D then a is integral over R if and only if NrD/F (a) ∈ R. As v is discrete, R is a PID and hence it is integrally closed, by Example 4.1.5. Therefore, if a ∈ D is integral over D then NrD/F (a) ∈ R by Proposition 4.1.8. Conversely, assume that NrD/F (a) ∈ R and let f = MinF (a) = X n +a n−1 X n−1 +⋅ ⋅ ⋅+a1 X+a0 , the minimal polynomial of a over F. As f and CharD/F (a) have the same irreducible divisors and f is irreducible, NrD/F (a) = ±a0k for some k and therefore v(a0 ) ≤ 1. We have to prove that a i ∈ R for every i. Otherwise, v(a i ) > 1 for some i = 1, 2, . . . , n − 1. Let i be minimal n n−1 + ⋅ ⋅ ⋅ + b X + b . Then with v(a i ) ≥ v(a j ) for every j and set g = a−1 1 0 i f = b n X + b n−1 X g ∈ R[X], 1 ≤ i < n and, using the bar notation for reduction modulo the maximal ideal of R, we have g = X i (1 + b i+1 X i+1 + ⋅ ⋅ ⋅ + b n X n ). By the Hensel Lemma (Lemma 4.8.1), f has a root in R, contradicting the fact that f is irreducible in F[X]. Let Nr = NrD/F . Clearly w satisfies axioms (1) and (2) of the definition of valuation. Let x, y ∈ U(D) with w(x) ≤ w(y). Then v(Nr(xy−1 ))n = w(x)w(y)−1 ≤ 1. Therefore Nr(xy−1 ) ∈ R and hence xy−1 is integral over R, by the previous paragraph. Thus 1 + xy−1 is integral over R, hence Nr(1 + xy−1 ) ≤ 1 and therefore w(x + y)n = Nr(1 + xy−1 )Nr(y) ≤ Nr(y) = max{w(x), w(y)}n . This proves that w is a non-Archimedean valuation on D. That w(U(D)) is discrete and of finite index in v(U(E)) easily follows

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| 163

from the inclusion w(U(D))n ⊆ v(U(F)). This proves (2). If w󸀠 is any valuation of D such that w󸀠 |F = v, then w󸀠 -topology makes D into a topological vector space over F. This implies that the w󸀠 -topology is the product topology with respect to any basis of D [30, Ch. 2 § 2, Theorem 2]. Therefore w󸀠 = w a for some a ∈ ℝ+ . Then v(x) = v(x)a for every x ∈ F, and as v is non-trivial we deduce that a = 1. Hence w󸀠 = w. This finishes the proof of (1). The first equality of (3) is obvious and the second has been proven in the first paragraph of this proof. (4) is obvious. Let F be a local field and let D be a finite dimensional division F-algebra. Let w be the unique valuation of D extending v. The ramification index of D over F is e(D/F) = [w(U(D)) : v(U(F))] and the residue degree is f(D/F) = dimF (D), where F and D are the residue fields of F an D with respect to v and w, respectively. The following example shows that these definitions match with the definitions of ramification index and residue degree over Dedekind domains. Example 4.8.3. Let R be a Dedekind domain with quotient field F, let E/F be a finite separable extension and let S be the integral closure of R in E. Let Q be a maximal ideal of S, P = Q∩R and v the Q-adic valuation on E. Then the valuation rings of v and v|F are S Q and R P respectively and their maximal ideals are Q Q and P P . As R/P ≅ R P /P P and S/Q ≅ S Q /Q Q (Lemma 4.1.2), the residue degree of v over F is dimR/P (S/Q). Moreover, [v(U(E)) : v(U(F))] = [e Q (U(E)) : e Q (U(F))], where e Q is the exponential valuation. If Q Q = aS Q , P P = bR P and e = e Q (PS) then PS Q = bS Q = (Q Q )e , e Q (a) = 1 and e Q (b) = e. Hence, e Q (E) = ℤ and e Q (F) = eℤ and we conclude that e(E/F) = e Q (PS), the exponent of Q in the factorization of PS. The following result is a consequence of Theorem 4.4.2 in case the division algebra D is commutative. A proof in the non-commutative case can be found in [188, Theorem 13.3 and Theorem 14.3]. Corollary 4.8.4. Let D be a finite dimensional division algebra over local field F. Then dimF (D) = e(D/F) f(D/F) (in particular e(D/F) and f(D/F) are finite) and if F = Z(D) then e(D/F) = f(D/F) = Ind(D). Under the assumptions of Corollary 4.8.4, we say that D/F is unramified if e(D/F) = 1 (or equivalently dimF (D) = f(E/F)). Otherwise D is said to be ramified over F. In particular, D is unramified over its center if and only if D is commutative. Let (F, v) be a local field, with valuation ring R, unique maximal ideal P and residue field R = R/P. Let q = |R|. As q is a power of the characteristic of R, if we fix an algebraic closure L of F then L contains a q n − 1 root of unity for every n, which we

164 | 4 Dedekind domains, valuations and orders

denote ζ q n −1 . We let F nu = F(ζ q n −1 ) and R un = integral closure of R in F nu . Corollary 4.8.5. Let (F, v) be a local field, R the valuation ring of v and assume that the residue field R has order q. (1) F nu /F is unramified of degree n. (2) The restriction from F nu to R un induces an isomorphism Gal(F nu /F) → Gal(R un /R). (3) Every unramified extension of F of degree n is F-isomorphic to F nu . Proof. Let P be the unique maximal ideal of R and set m = q n − 1. (1) and (2) Let Q be the unique maximal ideal of R un and w the unique extension of v to a valuation in F nu . If σ ∈ G = Gal(F nu /F) then w ∘ σ is a valuation on F nu , extending v and therefore w(σ(x)) = w(x) for every x ∈ F nu . This implies that σ(R un ) = R un and σ(Q) = Q and shows that the restriction from F nu to R un gives a homomorphism ϕ : Gal(F nu /F) → Gal(R un /R). If g = MinF (ζ m ) then g ∈ R[X], by Lemma 4.6.3. Furthermore, g = ∏σ∈Gal(F nu /F) (X − σ(ζ m )) and g = ∏σ∈Gal(F nu /F) (X − ϕ(σ)(ζ m )). As gcd(g, g 󸀠 ) = 1, g does not have multiple roots in R un and this implies that ϕ is injective. Therefore, f(F nu /F) = |Gal(R un /R)| ≥ |Gal(F nu /F)|= [F nu : F] ≥ f(F nu /F). Hence F nu /F is unramified and ϕ is an isomorphism. (3) Let E/F be finite unramified extension of degree n and let S be the integral closure of R in E. Then [S : R] = f(E/F) = [E : F] = n and therefore S has order q n . This implies that S = R(a) for a a primitive m-th root of unity. By the Hensel Lemma (Lemma 4.8.1), E contains an m-root of unity x such that x = a. As the order of a is m, the order of x is also m. Then n = [S : R] = [R(a) : R] ≤ [F(x) : F] ≤ [E : F] = n and we conclude that E = F(x) which is F-isomorphic to F nu . We use the notation of Corollary 4.8.5. As R un /R is a extension of finite fields, Gal(R un /R) is cyclic generated by the Frobenius automorphism x 󳨃→ x q . The unique element σ n ∈ Gal(F nu /F) that maps to the Frobenius automorphism of Gal(R un /R) is also called the Frobenius automorphism of F nu over F. It is determined by the following property v(σ n (x) − x q ) < 1 for every x ∈ R un . Now we are ready to describe the Brauer group of a local field F. See [175, 17.10] for a proof. Theorem 4.8.6. Let (F, v) be a local field with uniformizer p, valuation ring R and residue field of cardinality q. For every n ∈ ℕ, let F nu = F(ζ q n −1 ), for ζ q n −1 a primitive root of unity in a fixed algebraic closure of F, and let σ n be the Frobenius automorphism of F nu over F. Then a cyclic algebra (F nu , σ n , a) is split if and only if n divides e pR (a) and the map θ : ℚ/ℤ → Br(F) r n

is a group isomorphism.

+ ℤ 󳨃→ [(F nu , σ n , p r )]

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| 165

Let F be a local field and let A be a finite dimensional central simple F-algebra. Then there is a unique rational number which is either 0 or an irreducible fraction nr , with 1 ≤ r < n, representing θ−1 ([A]). This number is called the invariant Inv(A) of A. It is clear that the index of A is the denominator of Inv(A). In particular, Inv(A) = 0 if and only if Ind(A) = 1. This gives a group isomorphism Inv : Br(F) → ℚ/ℤ, which is the inverse of the isomorphism θ of Theorem 4.8.6. Recall that Br(ℝ) = {[ℝ], [ℍ(ℝ)]} and Br(ℂ) = {[ℂ]}. Therefore there is also a group monomorphism Inv : Br(ℝ) → ℚ/ℤ [ℝ] 󳨃→ 0 + ℤ [ℍ(ℝ)] 󳨃→

1 2

+ℤ

as well as the trivial group homomorphism Inv : Br(ℂ) → ℚ/ℤ. As in the case of local fields, the invariant of a central simple algebra A over ℝ or ℂ, denoted Inv(A), is the representative of Inv([A]) in the interval [0, 1). Corollary 4.8.7. Let F be either ℝ, ℂ or a local field. If A is a finite dimensional central simple F-algebra then Exp(A) = Ind(A). Proof. The corollary is obvious if Ind(A) = 1. If A has index n > 1 then Inv(A) = nr for nr an irreducible fraction. Then Exp(A) is the order or nr module ℤ which is precisely n. We now describe the Brauer group of a number field F. Given a prime v of F and a finite ̂ v ⊗F A. Then A v is central simple F ̂ v -algebra. dimensional central F-algebra A let ̂ Av = F This gives a group homomorphism ̂v) Br(F) → Br(F [A] 󳨃→ [̂ Av ] ̂ P , for P a maximal ideal of the ring of ̂ v is either ℂ, ℝ, or a local field F Moreover, F ̂ v ) → ℚ/ℤ. The local integers of F. Therefore we have a group isomorphism Inv : Br(F index Indv (A) and the local invariant Invv (A) of A at v are the index and the invariant of  v , that is Indv (A) = Ind( v ) and Invv (A) = Inv( v ). We say that A ramifies at v, or that A is ramified at v if Indv (A) ≠ 1. In case v is a finite prime represented by the P-adic valuation for a maximal ideal P of the ring of integers of R then Indv and Invv are also denoted IndP and InvP , respectively, and we say that A ramifies at P if it ramifies at v. Similarly, if v is an infinite prime represented by an embedding σ of F in ℂ, then Indv and Invv are denoted Indσ and Invσ , respectively, and we say that A ramifies at σ if it ramifies at v.

166 | 4 Dedekind domains, valuations and orders

Putting together all the homomorphisms given by the local invariants we obtain a group homomorphism INV : Br(F) → ℚ/ℤ(Pl(F)) ̂ v∈Pl(F) [A] 󳨃→ (Invv (A)) where Pl(F) denotes the set of primes of F. The list INV(A) = INV([A]) is called the list of local invariants of A. Observe that ℚ/ℤ(Pl(F)) denotes a direct sum of copies of ℚ/ℤ. This implicitly says that Invv (A) is 0 for all but finitely many places of A. In fact, as in Theorem 4.7.3, the finite primes at which A ramifies are determined by the discriminant. More precisely, all the maximal orders of A have the same discriminant ([188, Theorem 25.3]), say Δ, and A ramifies at a maximal ideal P of the ring of integers of F if and only if P contains Δ ([188, Theorem 25.7]). Moreover there is an exact sequence [175, 18.5] INV

+

1 󳨃→ Br(F) → ℚ/ℤ(Pl(F)) → ℚ/ℤ → 0. In other words, every element [A] of Br(F) is completely determined by its list of invariants and an element of ℚ/ℤPl(F) is the list of invariants of a central simple F-algebra if and only if the sum of its coordinates belongs to ℤ (i.e. is 0 in ℚ/ℤ). This is the so called Albert-Hasse-Brauer-Noether Theorem. The injectivity of INV implies Ind(A) = lcm{Ind(̂ A v ) : v ∈ Pl(F)}.

(4.8.2)

As obviously Exp(A) = lcm(Exp(̂ A v ) : v ∈ Pl(F)}, by Corollary 4.8.7, we conclude wit the following result. Theorem 4.8.8. If A is a central simple algebra over a number field then the index and the exponent of A coincide. For the applications, we are specially interested in the list of invariants of central simple algebras which occur in the Wedderburn decomposition of a finite group algebra over a number field. These are precisely the central simple algebras which are generated over its center, say F, by a finite group and are usually called Schur algebras over F. The following theorem collects some information on Schur algebras. Theorem 4.8.9 (Benard-Schacher Theorem [21, 22]). Let F be a number field and let A be a Schur algebra over F with index m. Then the following properties hold. (1) F has an m-th root of unity. (2) Let v be a prime of F and let σ ∈ Gal(F/ℚ). If σ(ζ m ) = ζ ma then Invv∘σ (A) = a InvP (A). Let F be a finite Galois extension of ℚ and let A be a Schur algebra over F with degree m. Let p be a prime (finite or infinite) of ℚ. By Theorem 4.4.3, if p is a finite prime then Gal(F/ℚ) permutes transitively the primes of F over p. Clearly, this is also true for the infinite primes, where we consider Gal(F/ℚ) acting on the complex embedding of F

4.8 Brauer group of a number field

|

167

by translation. By Theorem 4.8.9 (2), the local invariant at one of the primes of F over p, say P0 , determines the local invariants at all the primes P of F over p. Indeed, there r ar is a σ ∈ Gal(F/ℚ) such that σ(P0 ) = P and thus if InvP0 (A) = mpp then InvP (A) = m pp ar p if σ(ζ m ) = ζ ma . Moreover, a is coprime with m and m P divides m. Therefore, m p is a reduced fraction (if it is not 0). Therefore, the local index is the same at all the primes of F over p. This shows that if A is a Schur algebra over F, then one can describe A up to isomorphisms giving the following numerical data: A positive integer, n and a finite list of pairs (p i , mr ii ), with i = 1, . . . , k, each p i is either ∞ or a prime integer, p i ≠ p j , if i ≠ j. Moreover, if p i = ∞ then mr ii = 12 and otherwise mr ii is an irreducible fraction with 1 ≤ r i < m i . This information describes A as a central simple algebra M n (D), with D a division algebra with center F. Moreover, there are valuations v1 , . . . , v k of F satisfying the following conditions: (1) {v i ∘ σ : i = 1, . . . , k, σ ∈ Gal(F/ℚ)} is a set of representatives of the primes of F for which Ind(A)v ≠ 0; (2) if p i = ∞ then v i (x) = |x| and Invv i (F) = 12 . This implies that F ⊆ ℝ and ℝ ⊗ D is a direct product of copies of M d (ℍ(ℝ)), where d = Deg(D) = Ind(A). 2 (3) if p i is a prime integer, then v i is the P i -adic valuation for some maximal ideal P i of R, containing p i and InvP i = mr ii . If p is a prime of ℚ and P is a prime of F over p then the local index of A at p is, by definition, the local index of A at P. We say that A ramifies at p if its local index at p is greater than 1. The Benard-Schacher Theorem ensures that this does not cause any ambiguity. By the Brauer-Witt Theorem (Theorem 3.7.1), if A is a Schur algebra over F then A is Brauer equivalent to a cyclotomic algebra. Thus, we would like to have information on the local invariants of cyclotomic algebras. As a consequence of Theorem 4.8.6 and Proposition 6.1.2 (1) we have the following property. Corollary 4.8.10. Let F be number field and let A = (F(ζ n )/F), τ) be a cyclotomic algebra and p a prime integer. If A ramifies at p then p divides n. In Section 3.8 we have seen that the output of the Wedderga command WedderburnDecompositionInfo ,

when applied to a group algebra FG, is a list of tuples with numerical data as in (3.8.1). Each of these tuples describes one of the Wedderburn components A of FG. If the length of the tuple is greater than three, then A is an n × n matrix algebra over a cyclotomic algebra (K(ζ m )/K, τ), where n, K and m are the first three entries of the tuple and τ is described by the other entries. In particular, by Corollary 4.8.10, if A ramifies at a prime integer then p divides m. More information is provided by the Wedderga command WedderburnDecompositionWithDivAlgParts .

168 | 4 Dedekind domains, valuations and orders

It gives the reduced degree, center, local indices and index of the Wedderburn components of a group algebra over an abelian number field. We explain the output of this command with the following example. (See the paper [82] of Herman and the manual of Wedderga [36] for details and other related commands.) gap> QG:=GroupRing(Rationals,SL(2,5));; gap> WedderburnDecompositionWithDivAlgParts(QG); [ [ 1, Rationals ], [ 4, Rationals ], [ 1, rec( Center := NF(5,[ 1, 4 ]), DivAlg := true, LocalIndices := [ [ infinity, 2 ] ], SchurIndex := 2 ) ], [ 2, rec( Center := Rationals, DivAlg := true, LocalIndices := [ [ 3, 2 ], [ infinity, 2 ] ], SchurIndex := 2 ) ], [ 5, Rationals ], [ 3, rec( Center := Rationals, DivAlg := true, LocalIndices := [ [ 2, 2 ], [ infinity, 2 ] ], SchurIndex := 2 ) ], [ 3, NF(5,[ 1, 4 ]) ] ]

The output is a list with seven pairs [n, x], representing a central simple algebra M n (D) with D a division algebra. Part of the information of x is the center F of D. If there and is no more information, then D = F. This is the case of four of the entries of the output: [ 1, Rationals ], [ 4, Rationals ], [ 5, Rationals ], [ 3, NF(5,[ 1, 4 ]) ].

They represent ℚ, M4 (ℚ), M5 (ℚ) and M3 (ℚ(ζ5 + ζ54 )) = M3 (ℚ(√5)). In case D ≠ F, then the x contains also the non-trivial local indices of D, and the global index. For example, [ 1, rec( Center := NF(5,[ 1, 4 ]), DivAlg := true, LocalIndices := [ [ infinity, 2 ] ], SchurIndex := 2 ) ],

represents a quaternion division algebra over ℚ(√5) with local index 2 at infinity. Meaning that the local invariants at the two infinite primes of ℚ(√5) are 12 . −2,−5 This algebra is isomorphic to the quaternion algebra ( ℚ( ). Indeed, the entry √5) of the output of WedderburnDecompositionInfo corresponding to this algebra is [ 1, NF(5,[ 1, 4 ]), 10, [ 2, 9, 5 ] ]. This represents the cyclic cyclotomic algebra (ℚ(ζ10 )/ℚ(√5), σ, −1) ≅ ℚ(√5) ⊗ℚ (ℚ(ζ5 )/ℚ(√5), σ, −1) ≅ ℚ(√5) ⊗ℚ ( −2,−5 ℚ )≅ −2,−5 ( ℚ(√5) ) (see Example 2.6.8). The fourth entry of the output represents M2 (D), where D is a quaternion division algebra over ℚ, with local indices 2 at 3 and ∞. Using WedderburnDecompositionInfo

we see that this algebra is also described by the tuple [ 2, Rationals, 6, [ 2, 5, 3 ]] .

4.8 Brauer group of a number field

|

169

Hence D ≅ (ℚ(ζ6 )/ℚ, σ, −1) ≅ ( −1,−3 ℚ ). Finally [ 3, rec( Center := Rationals, DivAlg := true, LocalIndices := [ [ 2, 2 ], [ infinity, 2 ] ], SchurIndex := 2 ) ]

represents M3 (D), where D is a quaternion division algebra over ℚ, with local indices 2 at 2 and ∞. This algebra is isomorphic to M3 (ℍ(ℚ)).

Problems 4.8.1. Prove Corollary 4.8.10. 4.8.2. Let G = ⟨a⟩7 ⋊ ⟨b⟩9 with a b = a2 . Calculate the Wedderburn decomposition of ℚG and the local indices of its simple components show that one of the simple components is a division algebra of degree 3. Prove that this is the smallest group for which the rational group algebra has a simple component of index greater than 2. 4.8.3. Prove that two quaternion Schur algebras are isomorphic if and only if they have the same local index at every rational prime. Give an example of two non-isomorphic Schur algebras of degree 3 with the same local indices at every rational prime.

5 The group of units of an order In this chapter we prove some fundamental properties on the structure of the unit group of a general order. Recall that an order is simply a ℤ-order. In Section 1 we begin with recalling some basic properties of lattices L in a real vector space V. In particular, we prove Minkowski’s Theorem, a crucial result that states that some bounded subsets B of V with large enough volume contain a nonzero element in L. A reference for this classical material is the book of Janusz [101]. In Section 2 we prove Hey’s Theorem and Dirichlet’s Unit Theorem. The latter states that the unit group of the ring of integers in a number field F is finitely generated with rank determined by the number of real and complex embeddings of F. In Section 3 we prove that the unit group of an order in a finite dimensional semisimple rational algebra is finitely generated, a result due to Siegel, and in Section 4 we prove that this group is even finitely presented. These results can be proved in the more general context of arithmetic groups. Since this monograph mainly deals with integral group rings and because of completeness’ sake, we elected to restrict results and proofs within this context. In Section 5 some structural results are proved on the unit group of an order in a finite dimensional semisimple rational algebra. It is characterized when it is finite and when it is central-by-finite. Applications to integral group rings are included.

5.1 Lattices in real vector spaces Throughout this section V is an n-dimensional vector space over ℝ. We consider V as a topological space equipped with the Euclidean topology. More precisely, if we fix a vector space isomorphism f : V → ℝn and consider on ℝn the topology given by any norm then the topology in V is such that f is an homeomorphism. Because of the uniqueness of norm topologies on the n-dimensional Euclidean space ℝn , the topology on V does not depend on the chosen isomorphism f . Observe that the choice of f is equivalent to the choice of the ℝ-basis B = {v1 , . . . , v n } of V given by f(v i ) = e i , where e1 , . . . , e n is the standard basis of ℝn . We refer to B as “reference basis”. We consider ℝn as a metric space with respect to the standard norm in ℝn , i.e. ‖(x1 , . . . , x n )‖ = √ x21 + ⋅ ⋅ ⋅ + x2n . Via the isomorphism f , one can also transfer metric notions from ℝn to V, but these notions depend on the choice of f . In other words, the metric notions depend on the reference basis. For example, if x, y ∈ V, then one defines the distance d(x, y) between x and y and the norm ‖x‖ of x in the given reference basis as the Euclidean distance d(f(x), f(y)) and the Euclidean norm ‖f(x)‖. If X is a subset of V then we say that X is measurable if the Riemann integral ∫f(X) dx exists and in that case we say that the volume of X with respect to f (or with respect to the reference basis) is vol(X) = ∫ dx. f(X)

5.1 Lattices in real vector spaces |

171

The measurability of X is independent on the reference basis but the value of the volume is dependent of the reference basis. If T : V → V is a linear map then, for any subset X of V, vol(T(X)) = |det(T)|vol(X), (X ⊆ V). (5.1.1) (This equality implicitly states that X is measurable if and only if so is T(X).) The validity of (5.1.1) is independent of the choice of the reference basis. Moreover (5.1.1) implies that if B and B󸀠 are two reference bases then volB󸀠 = |det(A)|volB , where A is the change of basis matrix from B to B󸀠 . If A is a real algebra, M is a left A-module, finite dimensional over ℝ and a ∈ A, then (5.1.1) implies vol(aX) = |NrM/ℝ (a)|vol(X),

(X ⊆ M).

(5.1.2)

A lattice L of rank k in V is by definition a finitely generated additive subgroup ℤv1 + ⋅ ⋅ ⋅ + ℤv k of V with v1 , . . . , v k ∈ V linearly independent over ℝ. In that case, we say that v1 , . . . , v k is a basis of L. A full lattice in an n-dimensional real vector space V is a lattice of rank n. (The reader is warned not to confuse the notion of a lattice with that of a ℤ-lattice.) Proposition 5.1.1. Let V be a finite dimensional real vector space. An additive subgroup of V is a lattice in V if and only if it is discrete in V with respect to the Euclidean topology on V. Proof. First assume that L is a lattice. Enlarging L if necessary, one may assume without loss of generality that L is a full lattice. Thus L = ℤv1 ⊕ ⋅ ⋅ ⋅ ⊕ ℤv n , with v1 , . . . , v n an ℝ-basis of V. If K is a compact subset of V, then there is a positive real number ϵ, such that K is contained in {∑ni=1 a i v i : a i ∈ ℝ, |a i | ≤ ϵ}. If x ∈ L ∩ K then x = ∑ki=1 a i v i , with a i ∈ ℤ ∩ [−ϵ, ϵ]. Thus L ∩ K is finite. For the converse, assume that L is a non-zero discrete subgroup of V. We argue by induction on the dimension n of V. Of course the case n = 0 is obvious. So assume that n ≥ 1 and the result holds for all finite dimensional vector spaces of dimension less than n. Hence, we may assume that L is not contained in any proper subspace of V. Consequently, L contains an ℝ-basis v1 , . . . , v n of V. Let W = ℝv1 + ⋅ ⋅ ⋅ + ℝv n−1 and L0 = L ∩ W. By the induction hypothesis, L0 is a lattice in W. Moreover, v1 , . . . , v n−1 ∈ L0 and therefore L0 is a full lattice in W. Thus L0 = ℤw1 ⊕ ⋅ ⋅ ⋅ ⊕ ℤw n−1 for some basis w1 , . . . , w n−1 of W. Therefore, w1 , . . . , w n−1 , v n is a basis of V. Let K = {a1 w1 + ⋅ ⋅ ⋅ + a n−1 w n−1 + a n v n : 0 ≤ a i ≤ 1, for every i = 1, 2, . . . , n}. Clearly K is a compact subset of V and v n ∈ K ∩ L \ W. By hypothesis, K ∩ L \ W is finite and therefore there is a minimal positive real number ϵ such that x0 = a1 w1 + ⋅ ⋅ ⋅ + a n−1 w n−1 + ϵv n ∈ L ∩ K for some 0 ≤ a1 , . . . , a n−1 < 1. Let L1 = ℤv1 +⋅ ⋅ ⋅+ℤv n−1 +ℤx0 . Clearly L1 ⊆ L. Let x = b1 w1 + ⋅ ⋅ ⋅ + b n−1 w n−1 + b n v n be an arbitrary element of L, with each b i ∈ ℝ. Write b n = m n ϵ + c n with m n ∈ ℤ and 0 ≤ c n < ϵ and b i − m n a i = m i + c i

172 | 5 The group of units of an order with m i ∈ ℤ and 0 ≤ c i < 1 for i = 1, . . . , n −1. Let y = m1 w1 +⋅ ⋅ ⋅+ m n−1 w n−1 + m n x0 = (m1 + m n a1 )w1 + ⋅ ⋅ ⋅ + (m n−1 + m n a n−1 )w n−1 + m n ϵv n ∈ L1 . Then x − y = c1 w1 + ⋅ ⋅ ⋅ + c n−1 w n−1 + c n v n . The choice of ϵ implies that c n = 0. Hence x − y ∈ L ∩ W = L0 ⊆ L1 . Thus x ∈ L1 . This proves that L = L1 = ℤw1 + ⋅ ⋅ ⋅ + ℤw n−1 + ℤx0 . So, L is a lattice in V. Let L be a full lattice in V and B = {v1 , . . . , v n } a basis of L. Then B also is an ℝ-basis of V. The fundamental polyhedron of B is by definition n

F B = { ∑ x i v i : 0 ≤ x i < 1}.

(5.1.3)

i=1

Clearly, the sets of the form v + F B = {v + x : x ∈ F B } form a partition of V. Although the fundamental polyhedron F B depends on the chosen basis B its volume does not. Lemma 5.1.2. The volume of a fundamental polyhedron F B of a basis B of a full lattice in a finite dimensional real vector space is independent of the basis B. Proof. Let B = {v1 , . . . , v n } and B1 = {w1 , . . . , w n } be two bases of a full lattice in V. Let f be the isomorphism of V mapping v i to w i . As ℤv1 +⋅ ⋅ ⋅+ℤv n = ℤw1 +⋅ ⋅ ⋅+ℤw n , the matrix A associated to f in the basis B belongs to GLn (ℤ). Thus det(f) = det(A) = ±1. Moreover, f maps the fundamental polyhedron F B of B to the fundamental polyhedron F B1 of B1 . By (5.1.1), vol(F B1 ) = |det(f)|vol(F B ) = vol(F B ), as desired. Let L be a lattice in V. Then we set vol(F B ), ∞,

vol(V/L) = {

if L is a full lattice in V and B is a basis of L; if L is not a full lattice in V.

Lemma 5.1.3. Let L be a full lattice in V with basis B = {v1 , . . . , v n }. Let w1 , . . . , w n ∈ L and write w i = ∑ni=1 a ij v j , with a ij ∈ ℤ for 1 ≤ i, j ≤ n. Set M = ℤw1 + ⋅ ⋅ ⋅ + ℤw n . Then M is a lattice in V. Moreover, M is a full lattice in V if and only if det((a ij )) ≠ 0, or equivalently, M has finite additive index in L and, in this case, [L : M] = |det((a ij ))| =

vol(V/M) . vol(V/L)

Proof. By Proposition 5.1.1, L is discrete in V. Hence so is M. Then, by the same proposition, M is a lattice in V. Clearly, M is a full lattice in V if and only if det((a ij )) ≠ 0, or equivalently, if and only if rkℤ (L) = rkℤ (M). The latter holds precisely when the factor group L/M is torsion as ℤ-module, or equivalently, when L/M is finite. In this case [L : M] = |det((a ij ))|, by Examples 4.3.4 and 4.3.5. Let f be the isomorphism of V mapping v i to w i . Then f maps the fundamental polyhedron F of B to the fundamental polyhedron F 󸀠 of {w1 , . . . , w n } and the matrix associated to this isomorphism in the basis B is (a ij ). Therefore, vol(V/M) = vol(F 󸀠 ) = |det((a ij ))|vol(F) = |det((a ij ))|vol(V/L), by (5.1.1).

5.2 Hey’s Theorem and Dirichlet’s Unit Theorem | 173

Observe that the formula in Lemma 5.1.3 is independent of the reference basis used to calculate the volume. Also condition (1) of the next theorem is independent of the reference basis. Theorem 5.1.4 (Minkowski’s Theorem). Let V be an n-dimensional real vector space and L a full lattice in V. Let X be a bounded subset of V satisfying the following conditions: (1) vol(X) > 2n vol(V/L). (2) if x, y ∈ X then x−y 2 ∈ X. Then, X ∩ L contains a non-zero element. Proof. Fix a basis B = {v1 , . . . , v n } of L. Then 2B = {2v1 , . . . , 2v n } is a basis of 2L. Let F be the fundamental polyhedron of 2B. By the first hypothesis and Lemma 5.1.3 we have vol(F) = vol(V/2L) = 2n vol(V/L) < vol(X). Clearly every element of V has a unique expression as x + y with x ∈ 2L and y ∈ F. This defines a map f : V → F which associates x ∈ V to the unique f(x) ∈ F such that x − f(x) ∈ 2L. Moreover, for every a ∈ 2L, the restriction of f to a + F is the map f(x) = x − a. As translations preserve volume, vol(f(Y)) = vol(Y) for every set Y contained in a + F. Because, by assumption, X is bounded there exist different elements a1 , . . . , a k ∈ 2L so that X ⊆ ∪ki=1 (a i + F). Therefore X is the disjoint union ∪ki=1 (X ∩ (a i + F)) and hence vol(F) < vol(X) = ∑ki=1 vol(X ∩ (a i + F)) = ∑ki=1 vol(f(X ∩ (a i + F))). Because each f(X ∩(a i + F)) ⊆ F, this implies that the sets f(X ∩(a1 + F)), . . . , f(X ∩(a k + F)) cannot be mutually disjoint. Therefore the restriction of f to X is not injective. Hence there exist distinct elements x, y in X with f(x) = f(y). Therefore, x − y = (x − f(x)) + (f(y) − y) ∈ 2L. The second hypothesis therefore yields that 0 ≠ x−y 2 ∈ X ∩ L, as desired. Remark 5.1.5. In many references condition (2) of Minkowski’s Theorem is replaced by the following: X is convex and 0-symmetric. In other words, if x, y ∈ X then the segment joining x and y is contained in X and −x ∈ X. Clearly these conditions imply condition (2) of Minkowski’s Theorem. One can easily find subsets of V satisfying the conditions of Minkowski’s Theorem. For example, take a (closed or open) ball centered at 0 of sufficiently large radius.

5.2 Hey’s Theorem and Dirichlet’s Unit Theorem The results of Hey and Dirichlet presented in this section are classical. Our presentation follows the one given by Kleinert in [135] (or [137]). Dirichlet’s Unit Theorem is one of the most useful tools to study units of an order in a finite dimensional semisimple rational algebra, in particular for the integral group ring of a finite group. Let G be a group acting on the right by homeomorphisms on a topological space X. Let X/G denote the set elements of which are the orbits of this action. Equip X/G with the quotient topology induced by the natural map X → X/G.

174 | 5 The group of units of an order

For example, if V is a finite dimensional real vector space then the underlying additive group of V acts by translations on V. Assume L is a lattice in V. Then the action restricted to L gives an action of L on V. The set consisting of the orbits is the quotient group V/L. Observe that if L has rank k then V/L is homeomorphic to (ℝ/ℤ)k × ℝn−k . In particular V/L is compact if and only if L is a full lattice in V. Let A be a finite dimensional semisimple rational algebra and fix an order O in A. Then A ⊗ℚ ℝ is a finite dimensional ℝ-algebra. Consider the natural injective map A → A ⊗ℚ ℝ associating a ∈ A with a ⊗ 1. Observe that v1 , . . . , v k ∈ A are linearly independent over ℚ in A if and only it v1 ⊗ 1, . . . , v k ⊗ 1 are linearly independent over ℝ in A ⊗ℚ ℝ. This implies that the identification of A as a subset of A ⊗ℚ ℝ makes O into a full lattice in A ⊗ℚ ℝ. Recall from Proposition 2.1.11 that A ⊗ℚ ℝ is a semisimple algebra as well. We consider A ⊗ℚ ℝ as an Euclidean topological space. Let B = {x ∈ A ⊗ℚ ℝ : |NrA⊗ℚ ℝ/ℝ (x)| = 1}, a subgroup of the group of units of A ⊗ℚ ℝ. If a ∈ A then NrA/ℚ (a) = NrA⊗ℚ ℝ/ℝ (a ⊗ 1), by (2.3.4). Combining this with Lemma 4.6.9 (2) we have U(O) = {a ∈ O : NrA/ℚ (a) = ±1} = O ∩ B

(5.2.1)

Thus U(O) is a subgroup of B. If a ∈ U(O) then multiplication by a on the right is a homeomorphism on B. This defines a right action by homeomorphisms of U(O) on B. The set of orbits is B/U(O) = {b U(O) : b ∈ B}, the set of left U(O)-cosets of B. We now prove Hey’s result along the line given by Kleinert in [137, Theorem 1.1]. The proof on its turn is based on that given by Zassenhaus in [233]. Theorem 5.2.1 (Hey’s Theorem [87]). Let D be a finite dimensional rational division algebra. Assume O is an order in D. Let B = {x ∈ D ⊗ℚ ℝ : |NrD⊗ℚ ℝ/ℝ (x)| = 1} and consider the topology of B induced by the Euclidean topology on D ⊗ℚ ℝ. Then B is a subgroup of the group of units of D ⊗ℚ ℝ, U(O) = O ∩ B and B/U(O) is compact. Proof. Because of the comments given before the theorem, it remains to prove that B/U(O) is compact. Let A = D⊗ℚ ℝ, a semisimple ℝ-algebra of dimension n = dimℝ (A). We consider O as a full lattice in A, via the map a 󳨃→ a ⊗ 1. Because of Remark 5.1.5, there exists a bounded subset C of A satisfying the hypotheses of Minkowski’s Theorem (Theorem 5.1.4) with respect to O. In other words, condition (1) takes the form vol(C) > 2n vol(A/O). Note that if a ∈ B then Ca also satisfies the hypothesis of Minkowski’s Theorem. Indeed, clearly Ca satisfies condition (2) and it is bounded because multiplication by a is a linear map. Furthermore, Ca satisfies condition (1) because vol(Ca) = vol(C) > 2n vol(A/O), by (5.1.2) and the fact that a ∈ B. Minkowski’s Theorem therefore guarantees the existence of a non-zero x a ∈ Ca ∩ O for every a ∈ B. Let X = B/U(O) be equipped with the quotient topology. As B is a subset of an Euclidean topological space, B has a countable basis of open subsets. Then, using

5.2 Hey’s Theorem and Dirichlet’s Unit Theorem | 175

that U(O) acts by homeomorphisms on B, it is easy to show that X also has a countable basis of open subsets. Therefore, to prove that X is compact, it is enough to show that every sequence in X has a convergent subsequence (see e.g. [132, Ch. 5, Theorem 5]). So, let (X n ) be a sequence in X. It is enough to show that one can find representatives a n ∈ X n such that the sequence (a n ) has a convergent subsequence. We start selecting a representative a n for every n which will be modified along the proof to obtain the desired conclusion. By the previous paragraph, for every n, there is 0 ≠ c n ∈ C such that x n = c n a n ∈ O. Observe that |NrD/ℚ (x n )| = |NrA/ℝ (c n )|, since a n ∈ B. As C is bounded and the norm is a continuous map, the sequence (NrA/R (c n )) of real numbers is bounded. Moreover, by Proposition 4.1.8, we know that each NrD/ℚ (x n ) is an integer. Therefore, (NrD/ℚ (x n )) is a bounded sequence of integral numbers and hence it only takes finitely many values. Replacing the original sequence by a suitable subsequence, we thus may assume that (NrD/ℚ (x n )) is a constant sequence (and different from 0 because D is a division algebra). Recall that O is a full lattice in A and hence O contains a ℚ-basis of D. Of course one may compute the norm NrD/ℚ using such a basis. Also recall that the norm NrD/ℚ (x n ) is the determinant of the map ρ x n (left multiplication by x n ). From Lemma 5.1.3 it then follows that [O : x n O] = |det(ρ x n )| = |NrD/ℚ (x n )|. Thus [O : x n O] = [O : x1 O] for every n. As O has only finitely many subgroups of a given finite index, we may replace the original sequence by a subsequence in which x n O = x1 O for every n. Thus c n a n = x n = x1 u n for some u n ∈ O with NrD/ℚ (u n ) = 1. In particular, u n ∈ U(O), by Lemma 4.6.9 (2). Replacing a n by a n u−1 n (another representative of X n ), we may assume that c n a n = x 1 for every n. On the other hand, c n ∈ C and C is bounded. Thus (c n ) has a convergent subsequence. Hence one may assume that (c n ) is convergent with limit c, say. Furthermore, |NrA/ℝ (c n )| = |NrA/ℝ (x n )| = |NrA/ℝ (x1 )| > 0 and therefore |NrA/ℝ (c)| = |NrA/ℝ (x1 )| ≠ 0. −1 Hence each c n is invertible and also the sequence (c−1 n ) is convergent with limit c . −1 Consequently, (a n ) is convergent with limit x1 c , as desired. Remark 5.2.2. Hey’s Theorem can be stated in the following equivalent form: B1 /O1 is compact, where B1 = {x ∈ B : NrD⊗ℚ ℝ/ℝ (x) = 1} and O1 = O ∩ B1 . The equivalence follows from the fact that both [B : B1 ] and [O : O1 ] are at most 2. We now prove Dirichlet’s Unit Theorem as an application of Hey’s Theorem. First we need to show a lemma. Lemma 5.2.3. Let F be a number field and let R be the ring of integers of F. Then the roots of unity of F form a finite cyclic subgroup of U(F) formed by the elements u ∈ R such that |σ(u)| = 1 for every complex embedding σ of F. Proof. Let σ 1 , . . . , σ n be the complex embeddings of F and set T = {u ∈ R : |σ i (u)| = 1 for every i}. If u is a root of unity then σ i (u) is a root of unity for every i. Thus u ∈ T. Conversely, assume that u ∈ T. Then the coefficient of X i in the characteristic polynomial χ F/ℚ (u) = ∏nj=1 (X − σ j (u)) is an integer which is, up to a sign, the sum of all the terms that are products of n − i different σ j (u)’s. Since |σ i (u)| = 1 for every i, this coef-

176 | 5 The group of units of an order ficient is bounded by (ni). Therefore, there are finitely many polynomials P1 , . . . , P k ∈ ℤ[X] such that every u ∈ T is a root of some P i . This implies that T is finite. In particular, T is a finite subgroup of F and thus T is cyclic and every element of T is a root of unity. Let F be a number field. Assume that F has precisely r real embeddings σ1 , . . . , σ r and s complex non-real embeddings up to conjugation σ r+1 , . . . , σ r+s . There is an isomorphism of ℝ-algebras f : F⊗ℚ ℝ → ℝr ×ℂs determined by f(x⊗1) = (σ1 (x), . . . , σ r+s (x)), for x ∈ F (see Example 2.1.13). We identify F with its natural image in F ⊗ℚ ℝ. We will use the so called logarithm map: l : U(ℝ)r × U(ℂ)s → ℝr+s (x1 , . . . , x r+s ) 󳨃→ (log|x1 |, . . . , log|x r |, log|x r+1 |2 , . . . , log|x r+s |2 ).

(5.2.2)

This is an epimorphism from a multiplicative group to an additive group. Let B = {x ∈ F ⊗ℚ ℝ : |NrF⊗ℚ ℝ/ℝ (x)| = 1}. By Problem 2.3.2, the composition of l with f|B is an isomorphism from B to the hyperplane V = {(x1 , . . . , x r+s ) : x1 + ⋅ ⋅ ⋅ + x r+s = 0}. The term “pair of complex non-real embeddings of a number field F” refers to a pair {f, f } with f : F → ℂ a field homomorphism with f(F) not contained in ℝ and f is the composite of f with the complex conjugation. Theorem 5.2.4 (Dirichlet’s Unit Theorem). Let F be a number field and assume that F has r real embeddings and s pairs of complex non-real embeddings. If R is the ring of integers of F then U(R) = T × A, where T is a finite group formed by the roots of unity in F and A is a free abelian group of rank r + s − 1. Proof. We use the notation introduced before the theorem. Observe that R is an order in F, by Theorem 4.2.6. As R is a lattice in F ⊗ℚ ℝ, it is a discrete subset of F ⊗ℚ ℝ, by Proposition 5.1.1. Hence, U(R) is a discrete subset of B. By Lemma 5.2.3, ker(l|U(R) ) is a finite cyclic group consisting of all roots of unity of F. It follows that also L = l(U(R)) is a discrete subgroup of V. Thus, again by Proposition 5.1.1, L is a lattice in V. By Hey’s Theorem, V/L is a compact and therefore L is a full lattice in V (see the comments given in the beginning of this section). This implies that L is a free abelian group of rank dimℝ (V) = r + s − 1. Therefore U(R) ≅ f(U(R)) ≅ ker(l ∘ f|U(R) ) ⊕ L, as desired. In the remainder of the section we present some easy consequences of Dirichlet’s Unit Theorem. The first two are direct consequences of Theorem 5.2.4 and statement (4) of Lemma 4.6.9. Corollary 5.2.5. Let F be a number field and assume that F has r real embeddings and s pairs of complex non-real embeddings. If O is an order in F then U(O) = T × A, where T is a finite cyclic group and A is a free abelian group of rank r + s − 1.

5.2 Hey’s Theorem and Dirichlet’s Unit Theorem | 177

Corollary 5.2.6. Let O be an order of a number field F. Then U(O) is finite if and only if F is either ℚ or a quadratic imaginary extension of ℚ. A number field is said to be totally real (respectively, totally complex) if every complex embedding is real (respectively, not real). Corollary 5.2.7. Let F ⊂ E be number fields and let R be the ring of integers of F and S the ring of integers of E. Then, [U(S) : U(R)] < ∞ if and only if F is totally real and E is totally complex and a quadratic extension of F. Proof. Let n = [E : F]; let r and r󸀠 be the number of real embeddings of F and E respectively; and let s and s󸀠 be the number of pairs of complex non-real embeddings of F and E respectively. Then [F : ℚ] = r + 2s, r󸀠 + 2s󸀠 = [E : ℚ] = n[F : ℚ] and, by Dirichlet’s Unit Theorem, the rank of U(R) is r + s − 1 and the rank of U(S) is r󸀠 + s󸀠 − 1. Thus [U(S) : U(R)] < ∞ if and only if r + s = r󸀠 + s󸀠 . If F is totally real and E totally complex and a quadratic extension of F then n = 2, s = 0 and r󸀠 = 0. Then r󸀠 + s󸀠 = s󸀠 = [E:ℚ] = [F : ℚ] = r = r + s. Thus [U(S) : U(R)] < ∞. 2 Conversely, assume that [U(S) : U(R)] < ∞, so that r + s = r󸀠 + s󸀠 . By classical field theory we know that every complex embedding of F has n extensions to an embedding of E. Moreover all the extensions to E of a non-real embedding of F are non-real too. Thus s󸀠 ≥ ns and each of the r homomorphisms F → ℝ extends to one of a list of r󸀠 + s󸀠 − ns homomorphisms E → ℂ. Therefore r ≤ r󸀠 + s󸀠 − ns = r + (1 − n)s and hence (n − 1)s ≤ 0. Then s = 0 because n > 1 and s ≥ 0. Thus F is totally real. Let σ1 , . . . , σ r be the real embeddings of F. Assume that σ i has u i real extensions and v i pairs of complex extensions. Thus r󸀠 = u1 + ⋅ ⋅ ⋅ + u r , s󸀠 = v1 + ⋅ ⋅ ⋅ + v r , and n = u i + 2v i for every i. The latter implies that u i + v i ≥ 1, so that r󸀠 + s󸀠 = r ≤ ∑ri=1 (u i + v i ) = r󸀠 + s󸀠 . Hence u i + v i = 1 for every i. Combining this with n = u i + 2v i > 1, we deduce that n = 2, u i = 0 and v i = 1 for every i. Therefore E is a quadratic extension of F and r󸀠 = 0, i.e. E is totally complex. Corollary 5.2.8. Let F be a number field, R an order in F and let σ1 , . . . , σ k be representatives of the complex embeddings of F up to conjugation. If k > 1, then for every i = 1, . . . , k there is x ∈ U(R) such that |σ i (x)| < 1 and |σ j (x)| > 1 for every j ≠ i. Proof. The notation is as used in the proof of Dirichlet’s Unit Theorem (Theorem 5.2.4). In particular, consider the homomorphism f1 = l ∘ f : B → V = {(x1 , . . . , x r+s ) ∈ ℝr+s : x1 + ⋅ ⋅ ⋅ + x r+s = 0}, where l is the logarithm map and f is the algebra isomorphism F ⊗ℚ ℝ → ℝr × ℂs . Observe that f1 (x) = (log|σ1 (x)|, . . . , log|σ r (x)|, log|σ r+1 (x)|2 , . . . , log|σ r+s (x)|2 ) (see Example 2.1.13). It is shown in the proof of Dirichlet’s Unit Theorem that f1 (U(R)) is a full lattice of V. Let π i : V → ℝr+s−1 be the map that drops the i-coordinate.

178 | 5 The group of units of an order Clearly, π i is a bijective linear map and therefore π i (f1 (U(R)) is a full lattice in ℝr+s−1 . Hence, there is x ∈ U(R) such that all the coordinates of π i (f1 (x)) are positive. In other words log|σ j (x)| > 0 for every j with 1 ≤ j ≤ k and j ≠ i. Then |σ j (x)| > 1 for every j ≠ 1 and therefore |σ i (x)| = ∏j=i,̸ 1≤j≤k |σ j (x)|−1 < 1. Remark 5.2.9. Let F be a number field and let O be an order in F. Let m be the rank of U(O). By Dirichlet’s Unit Theorem (Theorem 5.2.4), F has m+1 complex embeddings of F modulo complex conjugation. Let u1 , . . . , u m ∈ U(O). Then ⟨u1 , . . . , u m ⟩ has finite index in U(O) if and only if u1 , . . . , u m are multiplicatively independent if and only if l(u1 ), . . . , l(u m ) are linearly independent over ℝ. The first equivalence is a direct consequence of Dirichlet’s Unit Theorem and the second is a consequence of its proof. In the next Proposition a criterion is given for this condition to hold. Proposition 5.2.10. Let F be a number field and O an order in F. Let m be the rank of U(O) and let {σ1 , . . . , σ m+1 } be a set of representatives of the complex embeddings of F modulo conjugation. The following conditions are equivalent for u1 , . . . , u m ∈ U(O): (1) ⟨u1 , . . . , u m ⟩ has finite index in U(O). (2) The matrix log|σ1 (u1 )| log|σ2 (u1 )| ( ... log|σ m (u1 )| log|σ m+1 (u1 )|

log|σ1 (u2 )| log|σ2 (u2 )| ... log|σ m (u2 )| log|σ m+1 (u2 )|

... ... ... ... ...

log|σ1 (u m )| log|σ2 (u m )| ) ... log|σ m (u m )| log|σ m+1 (u m )|

(5.2.3)

has rank m. (3) The determinant of any m × m submatrix of the matrix in (5.2.3) is non-zero. Proof. For every i = 1, . . . , m + 1, let δ i = 1, if σ i is real, and δ i = 2, otherwise. Consider the following matrix δ1 log|σ1 (u1 )| δ2 log|σ2 (u1 )| A=( ... δ m+1 log|σ m+1 (u1 )|

δ1 log|σ1 (u2 )| δ2 log|σ2 (u2 )| ... δ m+1 log|σ m+1 (u2 )|

... ... ... ...

δ1 log|σ1 (u m )| δ2 log|σ2 (u m )| ) (5.2.4) ... δ m+1 log|σ m+1 (u m )|

The columns of A are the images of u1 , . . . , u m by the logarithmic map l : U(O) → V, where V is the hyperplane {(x1 , . . . , x m+1 ) ∈ ℝm+1 : x1 + ⋅ ⋅ ⋅ + x m+1 = 0}. By Remark 5.2.9, ⟨u1 , . . . , u m ⟩ has finite index in U(O) if and only if u1 , . . . , u m are multiplicatively independent, if and only if l(u1 ), . . . , l(u m ) generate a full lattice in V if and only if the rank of A is m if and only if the determinant of one of the m × m-submatrices of A is non-zero. The sum of the rows of A is zero. This implies that the determinants of all the m × m submatrices of A can only differ in the sign and thus its absolute value is independent of the row excluded. This proves that either all the m × m-submatrices of A have zero determinant or all have non-zero determinant. The same holds for the

5.3 The group of units of an order is finitely generated |

179

m × m-submatrices of the matrix in (5.2.3), because the determinant of a square submatrix of the matrix in (5.2.3) differs by a non-zero integer with the determinant of the corresponding submatrix of A. The proof of Proposition 5.2.10 shows that the determinants of all the maximal submatrices of the matrix A of (5.2.4) are equal up to a sign (see Problem 5.2.2).

Problems 5.2.1. Let ξ be a complex root of unity. Prove that U(ℤ[ξ + ξ −1 ]) has finite index in U(ℤ[ξ]). 5.2.2. Let F be a number field with ring of integers R. Let m be the rank of U(R). Let u1 , . . . , u m ∈ U(R) and let A be the matrix of (5.2.4). The regulator of u1 , . . . , u m , denoted R(u1 , . . . , u m ), is the absolute value of the determinant of an m × m submatrix of A. One says that u1 , . . . , u m is a list of fundamental units of F if U(R) = ⟨u1 , . . . , u m ⟩ × T, where T is the set of root of units of F. Prove the following statements: (1) ⟨u1 , . . . , u m ⟩ has finite index in U(R) if and only if R(u1 , . . . , u m ) ≠ 0. (2) The regulator of every list of fundamental units is the same. This is called the regulator of F, denoted R(F). ,...,u m ) (3) If ⟨u1 , . . . , u m ⟩ has finite index in U(R) then [U(R) : ⟨u1 , . . . , u m ⟩] = |T| R(u1R(F) . (Hint: The regulator is the volume of the fundamental polyhedron of a lattice!) 5.2.3. Prove that every Bass unit of ℤD2n has finite order modulo the center of ℤD2n .

5.3 The group of units of an order is finitely generated As an application of Dirichlet’s Unit Theorem we know that the group of units of an order in a commutative finite dimensional semisimple algebra is finitely generated. In this section we prove that this remains valid for non-commutative orders. More precisely we will prove the following result. Theorem 5.3.1 (Siegel [212]). The group of units of an order in a finite dimensional semisimple rational algebra is finitely generated. This result was proved for the orders of the form M n (R) with R a commutative order by Hurwitz [96] and the generalization to Hasse domains was given by O’Meara. The result has been generalized to arithmetic groups by Borel and Harish-Chandra [28]. Here we present a proof that follows the approach of Weyl [227, 228] combined with its revision by Kleinert in [135] and [137]. Kleinert calls this the “Reduction Theory Method” as it is based on what Weyl calls “Minkowski’s Theorem of reduction for quadratic forms”.

180 | 5 The group of units of an order We start by reducing the proof to the case where the order is of the form M n (R), for R an order in a division algebra. Let A be a finite dimensional semisimple rational algebra. Write A = M n1 (D1 ) × ⋅ ⋅ ⋅ × M n k (D k ), the Wedderburn decomposition, with each D i a division algebra. Let O be an arbitrary order in A and Oi an order in D i for each i. Then, by Lemma 4.6.2, O󸀠 = ∏ki=1 M n i (Oi ) is another order in A. Moreover, by Lemma 4.6.6, O ∩ O󸀠 is an order in A and U(O ∩ O󸀠 ) has finite index in U(O) and U(O󸀠 ), by Lemma 4.6.9 (4). Recall (see for example [172, Lemma 4.1.7]) that a subgroup N of finite index in a group G is finitely generated if and only if so is G. Hence it follows that if any of the following groups is finitely generated then so are all others: U(O), U(O ∩ O󸀠 ), U(O󸀠 ), U(M n i (Oi )) = GLn i (Oi ) for every i. Thus, to prove Theorem 5.3.1, we only have to prove that GLn (O) is finitely generated for an order O in a finite dimensional rational division algebra D. To do so, we will construct an action of GLn (O) on a connected topological space T that contains a subset F satisfying the conditions of Lemma 5.3.2 below and such that the set X given in the lemma is finite. If a group G defines a right action on a set X, then we denote by x g the action of the element g ∈ G on x. Lemma 5.3.2. Let G be a group acting on the right by homeomorphisms on a connected topological space T, F a subset of T and let X = {g ∈ G : F ∩ F g ≠ 0}. If T = ⋃g∈G F g and F is included in the interior of ⋃x∈X F x then X generates G. Proof. Let H be the subgroup of G generated by X and let S be a right transversal of H in G. For every s ∈ S, let U s = ⋃h∈H F hs . By assumption, T = ⋃s∈S U s . If s, t ∈ S and a ∈ U s ∩ U t then a = x hs = y kt , for some x, y ∈ F and h, k ∈ H. Clearly, y ∈ −1 −1 F ∩ F hst k and thus hst−1 k−1 ∈ X. This implies that st−1 ∈ H and we conclude that s = t. Consequently, the sets U s are mutually disjoint. We claim that U s is open, for every s ∈ S. Indeed, if x ∈ U s , then x = a hs for some h ∈ H and some a ∈ F. Since, by assumption, F is contained in the interior of ⋃x∈X F x , there is an open neighborhood U of a contained in ⋃x∈X F x . Therefore U hs is an open neighborhood of x contained in ⋃x∈X F xhs ⊆ U s . This proves the claim. Summarizing, T = ⋃s∈S U s and the U s ’s are open and mutually disjoint. Since T is connected this implies that |S| = 1. Hence, G = H. Before describing the topological space T and the subset F from Lemma 5.3.2 which will be used to prove Theorem 5.3.1, we recall some basic facts about real quadratic forms. Let V be a vector space over the reals. If f is a symmetric bilinear form on V and q : V → ℝ is the associated quadratic form (that is q(x) = f(x, x) for each x ∈ V) then f(x, y) =

q(x + y) − q(x) − q(y) 2

(x, y ∈ V).

5.3 The group of units of an order is finitely generated |

181

Therefore, f is determined by q and in the remainder we identify symmetric bilinear forms of V and quadratic forms of V. One says that the quadratic form q (or its corresponding bilinear form) is positive definite if q(x) > 0 for every x ≠ 0. If q is a positive definite quadratic form on V then it defines a norm on V and the unit ball for this norm is Bq = {x ∈ V : q(x) < 1}. Assume that k = dimℝ V and fix a basis B = {v1 , . . . , v k } of V. The map associating to a bilinear form f of V the matrix (f(v i , v j )) ∈ M k (ℝ) is an isomorphism of real vector spaces from the set of bilinear maps of V to M k (ℝ). It restricts to a bijection from the subspace of symmetric bilinear maps to the subspace of symmetric matrices. Moreover, if Q is a matrix associated to the quadratic form q then q is positive definite if and only if Q is a positive definite symmetric matrix (i.e. x T Qx > 0 for every non-zero column vector x ∈ ℝk ) if and only if Q = U T U for some U ∈ GLk (ℝ) (see the proof of Lemma 5.3.5 below). For example, if we identify ℝk with the vector space of column vectors with k coordinates, and B is the standard basis of ℝk then the bilinear form in ℝk associated to the matrix Q is given by (x, y) 󳨃→ x T Qy. In particular, the standard Euclidean norm of ℝk is the norm defined by the quadratic form q0 associated to the identity matrix. Then Bq0 = {x ∈ ℝk : x T x < 1} is the standard k-dimensional unit ball. Let Vk = vol(Bq0 ), computed with respect to the standard basis. Writing Q = U T U we have Bq = {x ∈ ℝk : x T Qx < 1} = {x ∈ ℝk : Ux ∈ Bq0 }. Therefore, for an arbitrary vector space V and an arbitrary reference basis B of V, Vk is the volume of the unit ball Bq0 for the quadratic form on V associated with the identity matrix. If q is an arbitrary quadratic form and Q is the matrix associated to q with respect to some basis using (5.1.1) we have vol(Bq ) =

Vk

|det(U)|

=

Vk

√det(Q)

.

(5.3.1)

Using this one easily deduces vol(Bαq ) =

vol(Bq ) α k/2

(α > 0).

(5.3.2)

Moreover, if V = V1 ⊕ V2 , with k i = dim(V i ), q i = q|V i and V1 and V2 orthogonal subspaces with respect to q, then vol(Bq ) = vol(Bq1 ) vol(Bq2 )

Vk . Vk1 Vk2

(5.3.3)

We consider on M k (ℝ) the Euclidean topology and hence the identification of M k (ℝ) with the bilinear forms of V induces a topology on the vector space of bilinear forms. For n ≥ 1 let f n : [0, n] → ℝ be defined by f n (x) = {

x, n−x n−1 x ( n−1 , )

if n = 1; otherwise

182 | 5 The group of units of an order The function f n increases from 0 to 1 in the interval [0, 1] and decreases from 1 to 0 in the interval [1, n]. Therefore, for every 0 < c ≤ 1, there exist real numbers 0 < x n,c ≤ y n,c ≤ n that are uniquely defined by the equality (see Figure 5.1) [x n,c , y n,c ] = f n−1 [c, 1].

1

c x4,c

y4,c

1 3

Fig. 5.1: Plot of f4 (x) = x( 4−x 3 ) .

Lemma 5.3.3. Let q be a positive definite quadratic form in a k-dimensional vector space V and let Q ∈ M k (ℝ) be the matrix associated to q with respect to some reference B and for every x ∈ V let x B denotes the coordinates of x in the basis B. Then (1) If r and r∗ are the minimum and maximum eigenvalues of Q then rx TB x B ≤ q(x) ≤ r∗ x TB x B . k

(2) det(Q) ≤ ( tr(Q) k ) . √k det(Q) then every eigenvalue of Q belongs to the interval (3) Let 0 < c ≤ 1. If tr(Q) c k ≤ t ≤ [x k,c t, y k,c t]. Proof. Let a1 , . . . , a k denote the eigenvalues of Q (some of which may be repeated). Then det(Q) = a1 . . . a k and tr(Q) = a1 + ⋅ ⋅ ⋅ + a k . (1) Since Q is symmetric, there is an orthogonal matrix U such that U T QU = diag(a1 , . . . , a k ). Then x B = Ux B1 , where B1 is the basis of V formed by the elements whose coordinates in the basis B are the columns of U. Then x TB x B = x TB1 x B1 . Henceforth, we may assume without loss of generality that Q = diag(a1 , . . . , a n ). Thus r = min{a1 , . . . , a k } and r∗ = max{a1 , . . . , a k }. Hence, for x B = (x1 , . . . , x k ) ∈ ℝk we have rx TB x B ≤ ∑ni=1 a i x2i = q(x) ≤ r∗ x TB x B . (2) This follows from the well-known inequality between geometric and arithmetic a1 +⋅⋅⋅+a k k means: √a . 1 . . . ak ≤ k (3) Replacing Q by Q/t we may suppose that t = 1. So we are assuming that a=

a1 + ⋅ ⋅ ⋅ + a k a1 . . . a k ≤ 1 ≤ √k k c

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183

and we have to show that each a i ∈ [x k,c , y k,c ], or equivalently that f k (a i ) ≥ c. Then c ≤ ai ∏ aj ≤ ai ( j=i̸

∑j=i̸ a j k−1

k−1

= ai (

)

ka − a i k−1 k − a i k−1 ) ≤ ai ( ) = f k (a i ), k−1 k−1

as desired. Lemma 5.3.4. If A is a finite dimensional semisimple real algebra then A contains a basis B such that if R = R B denotes the regular representation of A with respect to B then R(A)T = R(A), i.e. the image of R is closed under taking transposes. Proof. Without loss of generality, we may assume that A is simple. Indeed, assume that A = A1 × A2 and let B i be a basis of A i such that R B i (A i )T = R B i (A i ). Then B = B1 ⋃ B2 satisfies the required condition. Thus (and because of Problem 2.2.1) we may assume that A = M n (D) with D = ℝ, ℂ or ℍ(ℝ) for some n. Let E = {E ij : 1 ≤ i, j ≤ n}, where E ij denotes the matrix having 1 at the (i, j)-th entry and zeros elsewhere. Put B = E if D = ℝ; B = E ⋃ Ei if D = ℂ and B = E ⋃ Ei ⋃ Ej ⋃ Ek if D = ℍ(ℝ). It is easily verified that B satisfies the required condition. We fix some notation that will be used throughout this section. By D we denote an mdimensional rational division algebra, 𝔻 = ℝ ⊗ℚ D, O is an order in D, n is a positive integer and E = {e1 , . . . , e n } is the standard basis of D n (considered as a right D-vector space). We identify 𝔻n with the column vectors, so that 𝔻n is seen as a (M n (𝔻), 𝔻)bimodule. We fix an injective homomorphism of real algebras R : 𝔻 → M m (ℝ) such that R(𝔻)T = R(𝔻) (see Lemma 5.3.4) and we define an involution x 󳨃→ x on 𝔻 by the following rule R(x) = R(x)T ,

(x ∈ 𝔻).

We consider both the involution ⋅ and the homomorphism R acting componentwise on matrices with entries in 𝔻. In other words, for M = (a ij ) ∈ M k,l (𝔻) we set M = (a ij ) ∈ M k,l (𝔻) and Then

T

T

MN = N M

T

and

R(M) = (R(a ij )) ∈ M mk,ml (ℝ). R(MN) = R(M)R(N)

for compatible matrices M and N. In particular, R : M n (𝔻) → M nm (ℝ) T

is an injective ring homomorphism, M 󳨃→ M defines an involution on M n (𝔻) such that T R(M ) = R(M)T

184 | 5 The group of units of an order and, identifying 𝔻n with M1,n (𝔻) (i.e. column matrices), both the ⋅ and R are defined in 𝔻n . Clearly, it follows that if M ∈ M n (𝔻) then R(M) is symmetric if and only if T

M = M. For every M ∈ M n (𝔻), let tr(M) and nr(M) denote respectively the trace and determinant of R(M). Note that we have such a trace and norm map for every positive integer n. It will be clear from the context which n is used. Observe that if n = 1 then tr and nr are Tr𝔻/ℝ and Nr𝔻/ℝ respectively. However, if n ≥ 2 then tr and nr are different from TrM n (𝔻)/ℝ and NrM n (𝔻)/ℝ respectively. In a similar way as we have identified elements of M k (ℝ) with bilinear forms of a k-dimensional vector space over ℝ, we identify M n (𝔻) with the set consisting of the maps f : 𝔻n × 𝔻n → 𝔻 satisfying the following conditions f(x, ya + zb) = f(x, y)a + f(x, z)b, f(xa + yb, z) = af(x, z) + bf(y, z)

(x, y, z ∈ 𝔻n , a, b ∈ 𝔻).

(5.3.4)

This identification associates M ∈ M n (𝔻) with the map f M : (x, y) 󳨃→ x T My. The topology on M mn (ℝ) induces a topology on M n (𝔻), via the injective homomorphism R, which also can be described as the topology given by the norm ‖M‖ = T

tr(M M). Alternatively, one can describe the topology on M n (𝔻) using the following norm |tr(x T Mx)| : 0 ≠ x ∈ 𝔻n }. (5.3.5) N(M) = max{ tr(x T x) T

Clearly f M (x, y) = f M (y, x) for every x, y ∈ 𝔻n if and only if M = M, or equivalently R(M) is symmetric. The matrices satisfying this condition form a subspace V of M n (𝔻)ℝ . The topological space of interest for the proof of Theorem 5.3.1 (making use of Lemma 5.3.2) is T = {M ∈ M n (𝔻) : R(M) is a positive definite symmetric real matrix}.

(5.3.6)

Observe that T is the intersection of the subsets of V given by the inequalities x T R(M)x > 0, where x runs through all the non-zero elements in ℝnm . Also, for a fixed x, the expression x T Ax with A ∈ M nm (ℝ) is a linear combination of the entries of A and hence the matrices A satisfying x T Ax > 0 form an open half-space of M nm (ℝ). Recall that an open half-space in ℝn is a subset defined by a linear inequality ∑ni=1 a i X i > b. As every open half-space is convex and the intersection of convex subsets is convex, T is convex and hence it is connected. In the sequel we will represent an element of T as either a matrix Q, the associated bilinear form f or the quadratic form q defined by q(x) = f(x, x). Observe that every quadratic form q ∈ T defines a norm x 󳨃→ tr(q(x)) n on 𝔻ℝ . The following lemma provides an alternative description of T .

5.3 The group of units of an order is finitely generated |

185

Lemma 5.3.5. T = {U T U : U ∈ GLn (𝔻)}. Proof. If A = U T U with U ∈ GLn (𝔻) then R(A) = R(U)T R(U) is symmetric and x T R(A)x = (R(U)x)T (R(U)x) > 0 for every x ∈ ℝnm \ {0}. Conversely, assume that R(M) is symmetric and positive definite with M ∈ M n (𝔻). It is well known that then there exists a real and orthogonal matrix W so that R(M) = W T HW, with H a diagonal matrix, say H = diag(r1 , . . . , r nm ) and r1 , . . . , r nm > 0. Put H1 = diag(√r1 , . . . , √r nm ) and let U1 = W T H1 W, an invertible symmetric real matrix with R(M) = U1T U1 . By Lagrange’s interpolation, there is a polynomial f with real coefficients such that f(H) = H1 and hence U1 = W T f(H)W = f(W T HW) = f(R(M)) = R(f(M)). Let U = f(M) ∈ M n (𝔻). Then R(U T U) = R(U)T R(U) = U1T U1 = R(M). Since R is injective, M = U T U as desired. Observe that as M is positive definite, it is invertible and hence from M = U T U one deduces that U ∈ GLn (𝔻). We now define the action of GLn (O) on T which will be used in our application of Lemma 5.3.2. By definition, an element U ∈ GLn (O) acts on Q ∈ T by Q U = U T QU. In terms of the bilinear form f or the quadratic form q the action takes the form f U (x, y) = f(Ux, Uy),

and

q U (x) = q(Ux),

for x, y ∈ 𝔻n . We still need to define the subset F of Lemma 5.3.2. This requires the introduction of new terminology. Let L be a finitely generated right O-submodule of D n containing a ℚ-basis of D n ; i.e. L is a ℤ-lattice in D n that is also a right O-module. Note also that L is a full lattice in 𝔻n . Hence Proposition 5.1.1 is applicable. In particular, for a given q ∈ T , there exists 0 ≠ x ∈ L with tr(q(x)) minimal. A semibasis B in L is an ordered D-basis of the right D-space D n so that B ⊆ L. For a semibasis B = {v1 , . . . , v n } of L, considered as an ordered set, we set Z(L, B) = {q ∈ T : tr(q(v k )) ≤ tr(q(x)) for every x ∈ L \ v1 D ⊕ ⋅ ⋅ ⋅ ⊕ v k−1 D, 1 ≤ k ≤ n}. Note that Z(L, B) is a closed subset of the topological space T , as it is the intersection of closed subsets. For any two real numbers ϵ ≥ 1 and μ ≥ 0, let Z ϵ,μ (L, B) denote the set consisting of the elements q ∈ T satisfying the following two conditions, for every k with 1 ≤ k ≤ n:

186 | 5 The group of units of an order (1) tr(q(v k )) ≤ ϵ tr(q(x)), for every x ∈ L \ (v1 D ⊕ ⋅ ⋅ ⋅ ⊕ v k−1 D). (2) tr(q(v k )) ≤ tr(q(v k − y)) + μ tr(q(v l )), for every y ∈ L ∩ (v1 D ⊕ ⋅ ⋅ ⋅ ⊕ v l D) with l < k. Clearly, for 1 ≤ ϵ ≤ ϵ󸀠 and 0 ≤ μ ≤ μ󸀠 , we have Z(L, B) = Z1,0 (L, B) ⊆ Z ϵ,μ (L, B) ⊆ Z ϵ󸀠 ,μ󸀠 (L, B). Lemma 5.3.6. If ϵ > 1 and μ > 0 then Z(L, B) is contained in the interior of Z ϵ,μ (L, B). Proof. Let B = {v1 , . . . , v n } and q ∈ Z(L, B). Let N be the norm defined in (5.3.5). To prove the result it is sufficient to show that the N-ball centered in q and with radius δ is contained in Z ϵ,μ (L, B) for some δ > 0. Because of Proposition 5.1.1 we have C = min{tr(x T x) : 0 ≠ x ∈ L} > 0. As both x 󳨃→ tr(x T x) and x 󳨃→ tr(q(x)) define (equivalent) norms on the Euclidean n space 𝔻ℝ , there is a positive real number 𝛾 such that

𝛾 tr(x T x) ≤ tr(q(x)). for every x ∈ 𝔻n . Let α = max{1 + C−1 tr(v k T v k ) : 1 ≤ k ≤ n}. So tr(x T x) + tr(v k T v k ) tr(x T x)

≤α

(5.3.7)

for every 0 ≠ x ∈ L. Choose δ > 0 and so that δ≤𝛾

ϵ−1 . ϵ−1+α

(5.3.8)

We are going to check that this choice of δ implies that if q󸀠 ∈ T with N(q󸀠 − q) < δ then q󸀠 satisfies the first condition of the definition of Z ϵ,μ (L, D). Indeed, for such q󸀠 we have for every 0 ≠ x ∈ 𝔻n , |tr(q󸀠 (x)) − tr(q(x))| tr(x T x)

≤δ≤𝛾

tr(q(x)) ϵ−1 0. Let δ < min {𝛾, C1 β−1 } .

(5.3.10)

As tr(q(x)) defines a norm on 𝔻n and L is a discrete subset of 𝔻n , the set Y = {y ∈ L : tr(q(v k − y)) ≤

κ + 𝛾β(1 + μ) } 1 − δ𝛾−1

is finite. Let ν = max{tr((v k − y)T (v k − y)) : y ∈ Y ∩ (v1 D ⊕ ⋅ ⋅ ⋅ ⊕ v l−1 D)} and impose also that δ≤

C1 μ . ν + β(1 + μ)

(5.3.11)

Assume N(q󸀠 − q) < δ. Then we have tr(q󸀠 (v l )) ≥ tr(q(v l )) − |tr(q󸀠 (v l )) − tr(q(v l ))| ≥ tr(q(v l )) − δ tr(v l T v l ) ≥ C1 − βδ > 0. Moreover, if y ∈ L ∩ (v1 D ⊕ ⋅ ⋅ ⋅ ⊕ v l D), then v k − y ∈ L \ v1 D ⊕ . . . v k−1 D (because l < k) and hence tr(q(v k )) ≤ tr(q(v k − y)) (because q ∈ Z(L, B)). Therefore tr(q󸀠 (v k )) − tr(q󸀠 (v k − y)) tr(q󸀠 (v k )) − tr(q(v k )) + tr(q(v k − y)) − tr(q󸀠 (v k − y)) ≤ tr(q󸀠 (v l )) C1 − βδ T

δ(tr(v k T v k ) + tr((v k − y) (v k − y)) ≤ C1 − βδ T

≤

δ(β + tr((v k − y) (v k − y)) C1 − βδ

188 | 5 The group of units of an order If y ∈ Y then

tr(q󸀠 (v k )) − tr(q󸀠 (v k − y)) δ(β + ν) ≤ ≤ μ, tr(q󸀠 (v l )) C1 − βδ

by (5.3.11). Otherwise tr(q󸀠 (v k )) − μ tr(q󸀠 (v l )) ≤ tr(q(v k )) + δ tr(v k T v k ) + δμ tr(v l T v l ) − μ tr(q(v l )) ≤ κ + 𝛾β(1 + μ) < tr(q(v k − y))(1 − δ𝛾−1 ) ≤ tr(q(v k − y)) − δ tr((v k − y)T (v k − y)) ≤ tr(q󸀠 (v k − y)). We conclude that, if δ satisfies conditions (5.3.8), (5.3.10) and (5.3.11) then the N-ball centered in q with radius δ is contained in Z ϵ,μ (L, B). Recall that E = {e1 , ⋅ ⋅ ⋅ , e n } is the standard D-basis of D n . We fix an integral basis {b1 , . . . , b m } of O. Then {b i e j : i = 1, . . . , m; j = 1, . . . , n} is a basis of D n over ℚ and of 𝔻n over ℝ. We consider this basis as a reference basis so that all the volumes in 𝔻n are calculated with respect to this basis. In particular the volume of any fundamental polyhedron of On is 1. An admissible lattice is a finitely-generated free right O-submodule of 𝔻n containing On . Note that the standard basis E of D n is a semibasis for any admissible lattice. Lemma 5.1.3 yields that the volume of the fundamental polyhedron of an admissible lattice L is 1/[L : On ]. Consider the quadratic form q0 : 𝔻ℝ → ℝ given by q0 (x) = tr(xx). Let d0 be the determinant of the matrix (tr(b i b j )). By (5.3.1), the volume of the unit ball of q0 is vol(Bq0 ) = Vm /√ d0 . (5.3.12) Lemma 5.3.7. Let q ∈ T and B = {v1 , . . . , v n } a basis of 𝔻n as right 𝔻-module. Then there is another basis {w1 , . . . , w n } of 𝔻n such that n

n

q( ∑ w i x i ) = ∑ x i a i x i , i=1

((x1 , . . . , x n ) ∈ 𝔻n ),

(5.3.13)

i=1

where a i = q(w i ) and w i − v i ∈ v1 𝔻 ⊕ ⋅ ⋅ ⋅ ⊕ v k−1 𝔻,

(1 ≤ k ≤ n).

(5.3.14)

If moreover, L is a finitely generated right O-submodule of D n , B is a semibasis of L and q ∈ Z(L, B) then for every k ≤ n we have (Vkm )2 d0k 4km

k

k

[L ∩ (v1 D ⊕ ⋅ ⋅ ⋅ ⊕ v k D) : v1 O ⊕ ⋅ ⋅ ⋅ ⊕ v k O]2 ∏ tr(q(v i ))m ≤ ∏ nr(q(w i )). i=1

i=1

5.3 The group of units of an order is finitely generated |

189

Proof. For the first part we argue by induction on n. The case n = 1 is trivial. Let f : 𝔻n × 𝔻n → 𝔻 be the map satisfying (5.3.4) associated to q and let Q = (q ij = f(v i , v j )). As R(Q) = (R(q ij )) is positive definite, R(q11 ) is positive definite and, in particular q11 is invertible. Let w1 = v1 and w i = v i − v1 q−1 11 q 1i , for i > 1. Then f(w 1 , w i ) = 0 for every i > 1 and v k − w k ∈ v1 𝔻 ⊕ ⋅ ⋅ ⋅ ⊕ v k−1 𝔻 for every k. Now we can apply the induction hypothesis to the restriction of f to w2 𝔻 ⊕ ⋅ ⋅ ⋅ ⊕ w n 𝔻. For the second part, assume that q ∈ Z(L, B) and let k ≤ n. Replacing q by its restriction to v1 𝔻⊕⋅ ⋅ ⋅⊕v k 𝔻 and replacing L by L∩(v1 D⊕⋅ ⋅ ⋅⊕v k D), we may assume that k = n. Transferring q via the isomorphism of right 𝔻-modules 𝔻n → 𝔻n that maps e i to v i , we may assume that B is the standard basis of D n . Hence, L is admissible. For every i ≤ n, set α i = tr(q(e i )) and consider the following quadratic form on 𝔻n : n

n

i=1

i=1

q( ∑ w i x i ) = ∑ α−1 i tr(q(w i x i )),

(x1 , . . . , x n ∈ 𝔻).

Recall that Bq = {x ∈ 𝔻n : q(x) < 1}. We claim that L ∩ Bq = {0}. Indeed, assume that 0 ≠ x = ∑ni=1 w i x i ∈ L ∩ Bq (with all x i ∈ 𝔻) and k = max{i : x i ≠ 0}. Because, by assumption, q ∈ Z(L, B), we have that α1 ≤ α2 ≤ ⋅ ⋅ ⋅ ≤ α n . Furthermore, as x ∈ L \ w1 D + ⋅ ⋅ ⋅ + w k−1 D = L \ e1 D + ⋅ ⋅ ⋅ + e k−1 D, we obtain that tr(q(x)) ≥ α k . Hence, k

k

k

i=1

i=1

i=1

−1 −1 1 > q( ∑ w i x i ) = ∑ α−1 i tr(q(w i x i )) ≥ α k ∑ tr(q(w i x i )) = α k tr(q(x)) ≥ 1,

a contradiction. Thus the claim follows. Because of Minkowski’s Theorem (Theorem 5.1.4), the claim implies that the volume of Bq is less than or equal to 2nm /[L : On ]. By (5.3.2) and (5.3.3) we have n

vol(Bq ) = Vnm ∏

vol(Bq i )√α m i vm

i=1

where q i is the quadratic form on 𝔻 given by q i (x) = tr(xa i x). By Lemma 5.3.5, we can write each a i = b i b i with b i ∈ U(𝔻). Hence, vol(Bq i ) =

dx =

∫

1 |det(R(b i ))|

tr(b i xb i x) 0. Suppose s ∈ GLn (O) is such that Z(On , B)s ∩ Z ϵ,μ (On , B󸀠 ) ≠ 0. Let B∗ = {v∗1 , . . . , v∗n }, with v∗i = s−1 (v󸀠i ). Then B∗ is a semibasis of On and 0 ≠ Z(On , B) ∩ −1 Z ϵ,μ (On , B󸀠 )s = Z(On , B) ∩ Z ϵ,μ (On , B∗ ). Let q ∈ Z(On , B) ∩ Z ϵ,μ (On , B∗ ) and set t k = tr(q(v k )). For a given k there exists i ≤ k such that v i ∈ ̸ v∗1 D⊕⋅ ⋅ ⋅+⊕v∗k−1 D. As q ∈ Z ϵ,μ (On , B∗ ), this implies that tr(q(v∗k )) ≤ ϵt i . Because q ∈ Z(On , B), we also have that t n ≥ t n−1 ≥ ⋅ ⋅ ⋅ ≥ t1 and thus tr(q(v∗k )) ≤ ϵt k , (5.3.15) for 1 ≤ k ≤ n. Let l1 < l2 < ⋅ ⋅ ⋅ < l r = n be all the elements l ∈ {1, . . . , n} satisfying v1 D ⊕ ⋅ ⋅ ⋅ ⊕ v l D = v∗1 D ⊕ ⋅ ⋅ ⋅ ⊕ v∗l D. Put also l0 = 0. If k ≠ l i , for every i, then v∗j ∈ ̸ v1 D ⊕ ⋅ ⋅ ⋅ ⊕ v k D for some j ≤ k and hence, by (5.3.15), t k+1 ≤ tr(q(v∗j )) ≤ ϵt j ≤ ϵt k . Using this, one deduces t k2 ≤ ϵ k2 −k1 t k1 ,

(if for some i : l i−1 < k1 < k2 ≤ l i ).

(5.3.16)

5.3 The group of units of an order is finitely generated |

191

By Lemma 5.3.7, there is a 𝔻-basis {w1 , . . . , w n } of 𝔻n satisfying (5.3.13) and (5.3.14) and for such basis we have the following inequality for every k = 1, . . . , n: k

ck ∏ ( i=1

k ti m ) ≤ ∏ nr(a i ), m i=1

(5.3.17)

for a i = q(w i ) and ck =

m mk v2km d0k 4km

[L ∩ (v1 D ⊕ ⋅ ⋅ ⋅ ⊕ v k D) : v1 O ⊕ ⋅ ⋅ ⋅ ⊕ v k O]2 . m

i) Because of Lemma 5.3.3 (2) we know that nr(a i ) ≤ ( tr(a m ) . Moreover, v k − w k ∈ w 1 D ⊕ . . . w k−1 D. Hence t k = tr(a k ) + tr(v k − w k ) ≥ tr(a k ).

Thus k

0 < ck ≤ ∏

nr(a i )

ti i=1 ( m )

m

k

≤∏ i=1

nr(a i ) i) ( tr(a m )

m

≤ 1.

(5.3.18)

Combining this with (5.3.17) and (5.3.18) we have k−1 m k−1 m √nr(a i ) √nr(a i ) tr(R(a k )) tr(a k ) t k m nr(a k ) m nr(a k ) √ ≤ = ≤ ≤ √ ∏ ∏ ti m m m c k i=1 c k i=1 tr(a i ) m

m ≤ √

m

nr(a k ) m nr(R(a k )) = √ . ck ck

k tk Therefore, by Lemma 5.3.3 (3), the eigenvalues of R(a k ) are in the interval [ μm , with μ k = x m,c k and ν k = y m,c k . Using Lemma 5.3.3 (1) we have

νk tk μk tk tr(xx) ≤ tr(xa k x) ≤ tr(xx), m m

(x ∈ 𝔻).

νk tk m ],

(5.3.19)

Write v∗k = ∑ni=1 w i x i , with each x i ∈ 𝔻. Claim. There is a positive real number C, independent of q, such that tr(x i x i ) ≤ C for every i. Combining (5.3.15) and (5.3.19), and using that n

n

i=1

i=1

tr(q(v∗k )) = ∑ tr(q(w i x i )) = ∑ tr(x i a i x) ≥ tr(x i a i x i ), we have tr(x i x i ) ≤

m m m tr(x i a i x i ) ≤ tr(q(v∗k )) ≤ ϵt k . ti μi ti μi ti μi

Thus tr(x i x i ) ≤

mϵ , μi

if i ≥ k,

(5.3.20)

192 | 5 The group of units of an order

and, by (5.3.16),

m k−i+1 ϵ , μi

tr(x i x i ) ≤

if l h−1 < i ≤ k ≤ l h .

(5.3.21)

It remains to consider the case when i ≤ l h−1 < k ≤ l h for some h. So, l = l h󸀠 −1 < i ≤ l h󸀠 < k for some h󸀠 with 0 < h󸀠 ≤ h − 1. Let U be a fundamental domain of O in 𝔻. Then every element of 𝔻 has a unique expression of the form r + u, with r ∈ O and u ∈ U. We now write i

i

n

j=1

j=1

j=i+1

v∗k = ∑ w j u j + ∑ v j y j + ∑ w j x j , with each u j ∈ U and each y j ∈ O. To do so we first write x i = y i + u i , with y i ∈ O and u i ∈ U, and write w i = v i + ∑i−1 j=1 w j α j with each α j ∈ 𝔻. Then i−1

n

j=1

j=i+1

v∗k = ( ∑ w j x j ) + w i x i + ( ∑ w j x j ) i−1

n

= (∑ wj xj ) + wi ui + wi yi + ( ∑ wj xj ) j=1

j=i+1

i−1

i−1

n

= (∑ wj xj ) + wi ui + vi yi + (∑ wj αj yi ) + ( ∑ wj xj ) j=1

j=1

i−1

n

j=1

j=i+1

j=i+1

= ( ∑ w j x󸀠j ) + w i u i + v i y i + ( ∑ w j x j ), for some x󸀠j ∈ 𝔻. Repeating the argument several times we obtain the desired expression. Let y = ∑ij=1 v j y j and z = ∑nj=i+1 w j x j . Clearly, y ∈ L. By (5.3.13), i

n

j=1

j=i+1

i

n

j=1

j=i+1

tr(q(v∗k )) = tr(q( ∑ w j x j )) + tr(q( ∑ w j x j )) = tr(q(v∗k − z)) + tr(q(z)) and tr(q(v∗k − y)) = tr(q( ∑ w j u j )) + tr(q( ∑ w j x j )) = tr(q(v∗k − y − z) + tr(q(z)). Moreover, from the second condition defining the elements of Z ϵ,μ (On , B∗ ) and having in mind that l < k we get tr(q(v∗k )) − tr(q(v∗k − y)) ≤ μ tr(q(v∗l )). Hence, tr(q(v∗k − z)) − tr(q(v∗k − z − y)) = tr(q(v∗k )) − tr(q(v∗k − y)) ≤ μ tr(q(v∗l )) ≤ μϵt l ≤ μϵ l−i+1 t i .

5.3 The group of units of an order is finitely generated

|

193

Combining this with (5.3.19) we have μi ti tr(x i x i ) ≤ tr(x i a i x i ) ≤ tr(q(v∗k − z)) ≤ tr(q(v∗k − y − z)) + μϵ l−i+1 t i m i i νj tj = ∑ tr(u j a j u j ) + μϵ l−i+1 t i ≤ ∑ tr(u j u j ) + μϵ l−i+1 t i m j=1 j=1 ≤ ti

ρ ∑ij=1 ν j + mμϵ l−i+1 m

,

were ρ = max{tr(uu) : u ∈ U}. We conclude that tr(x i x i ) ≤

ρ ∑ij=1 ν j + mμϵ l−i+1 μi

,

if i ≤ l h−1 < k ≤ l h .

(5.3.22)

Now the claim follows from (5.3.20), (5.3.21) and (5.3.22). As v∗k ∈ O, we deduce from Proposition 5.1.1 that there is a finite subset of L which contains B󸀠

s−1

for every s ∈ GLn (O) such that Z(On , B)s ∩ Z ϵ,μ (On , B󸀠 ) ≠ 0.

Since s is completely determined by B󸀠 Z ϵ,μ (On , B󸀠 ) ≠ 0} is finite.

s−1

, we conclude that {s ∈ GLn (O) : Z(On , B)s ∩

We are ready for the Proof of Theorem 5.3.1. Let L1 , . . . , L r be a full list of admissible lattices L which are free as O-modules and such that Z(L, E) ≠ 0. It is finite by Lemma 5.3.8. For each i = 1, . . . , r we choose an isomorphism of O-modules g i : L i → On . Their obvious linear extensions to D, respectively 𝔻, are also denoted by g i . Then B i = {g i (e1 ), . . . , g i (e n )} is a semibasis of On and we set r

F = ⋃ Z(On , B i ).

(5.3.23)

i=1

As each Z(On , B i ) is closed, so is F. Let X = {s ∈ GLn (O) : F ∩ F s ≠ 0}. By Lemma 5.3.9, X is finite. We now prove that F satisfies the first hypothesis of Lemma 5.3.2, i.e. T = ⋃s∈GLn (R) F s . Indeed, if q ∈ T then for every positive real number ϵ, the set B q (x) = {x ∈ 𝔻n : tr(q(x)) ≤ ϵ} is bounded. Hence, by Proposition 5.1.1, it intersects On in a finite set. Thus, one can find consecutively v1 , . . . , v n with v k ∈ On \ v1 D ⊕ ⋅ ⋅ ⋅ ⊕ v k−1 D and tr(q(v k )) = min{tr(q(x)) : x ∈ On \ v1 D ⊕ ⋅ ⋅ ⋅ ⊕ v k−1 D}. Therefore B = {v1 , . . . , v n } is a semibasis of On such that q ∈ Z(On , B). Let g : 𝔻n → 𝔻n be the automorphism of 𝔻-modules given by g(e i ) = v i . Then L = g−1 (On ) is an admissible lattice which is free as O-submodule of 𝔻n and the following rule defines an element q1 ∈ Z(L, E): q1 (x) = q(g(x)), for all x ∈ 𝔻n . Thus L = L i for some i. The following rule defines an element q2 ∈ Z(On , B i ) ⊆ F: q2 (x) = q1 (g −1 i (x)),

194 | 5 The group of units of an order for x ∈ 𝔻n . Hence

g g −1

q = q 2i

,

gg−1 i

∈ GLn (O). This shows that indeed T = ⋃s∈GLn (O) F s . with Fix ϵ > 1 and μ > 0 and let W = ⋃ri=1 Z ϵ,μ (On , B i ). By Lemma 5.3.6, F is contained in the interior of W and, by Lemma 5.3.9, the set Y = {s ∈ GLn (O) : F s ∩ W ≠ 0} is finite. Clearly, X ⊆ Y As the topology on T is normal and F is closed, for every s ∈ Y \ X there is an open subset U s of T with F ⊆ U s and F s ∩ U s = 0. Then the interior of W 󸀠 = W ∩ ⋂s∈Y\X U s also contains an open subset containing F and the set Y 󸀠 = {s ∈ GLn (O) : F s ∩ W 󸀠 ≠ 0} is contained in X. As T = ⋃s∈GLn (O) F s , we have W 󸀠 ⊆ ⋃s∈Y 󸀠 F s ⊆ ⋃s∈X F s . Therefore ∪x∈X F s contains an open subset containing F. By Lemma 5.3.2, we conclude that X is a finite set of generators of GLn (O). Thus GLn (O) is finitely generated, as desired.

Problems 5.3.1. Let A ∈ M k (ℝ). Prove that the bilinear form (x, y) 󳨃→ x T Ay is symmetric and positive definite if and only if A = U T U for some U ∈ GLk (ℝ). (Hint: proof of Lemma 5.3.5.)

5.4 The group of units of an order is finitely presented In this section we show how to modify the proof of Theorem 5.3.1 to prove the following stronger result. Theorem 5.4.1. The group of units of an order in a semisimple finite dimensional rational algebra is finitely presented. This result is classical and originates from the work of Poincaré (see for example [62]). The present formulation is adapted from Swan [216] and the proof presented follows the line of the proof of Theorem 2.7.1 in [54]. The same reduction argument as used in the proof of Theorem 5.3.1 shows that, in order to prove Theorem 5.4.1, it is enough to show that if O is an order in a division algebra D and n is a positive integer then GLn (O) is finitely presented. In the proof of Theorem 5.3.1, we have constructed a connected subset T of an Euclidean space, an action of GLn (O) on T and a subset F satisfying the conditions of Lemma 5.3.2. Moreover, F is a finite union of convex subspaces of T . Indeed, the Euclidean space containing T is the set of symmetric real matrices of a given size (depending on n and D) and T is the intersection of a family of open half-spaces (5.3.6). An open half-space in ℝn is a subset defined by a linear inequality ∑ni=1 a i X i > b, for a i , b ∈ ℝ and a closed subspace is defined similarly using ≥ instead of >. Then F is the union of subsets of the form Z(On , B), for B a semibasis of On (5.3.23). This set Z(On , B) is the intersection of some closed half-spaces with T . Clearly every mentioned open or closed subspace

5.4 The group of units of an order is finitely presented |

195

is convex and therefore T and each Z(On , B) is convex. In particular, T is simply connected, that is, T is path connected and whenever f : [0, 1] → T and g : [0, 1] → T are two paths (i.e. continuous maps) with f(0) = g(0) and f(1) = g(1) then f and g are homotopic relative to {0, 1}. The latter means that there is a continuous map (called a homotopy) h : [0, 1] × [0, 1] → T with h(0, t) = f(t) and h(1, t) = g(t) for all t ∈ [0, 1]. If, furthermore, F is open and connected then we can apply the next proposition to show that GLn (O) is finitely presented. Proposition 5.4.2. Let T be a connected and simply connected topological space and let G be a group acting by homeomorphisms on T . Let U be a path connected open subset of T satisfying the following conditions: (1) T = ⋃g∈G U g . (2) X = {x ∈ G : U ∩ U x ≠ 0} is finite. Then the group given by the following presentation is isomorphic to G: ⟨[x] | x ∈ X, [xy] = [x][y] for x, y ∈ X with U ∩ U y ∩ U xy ≠ 0 ⟩. Before proving Proposition 5.4.2 we show how to use it to prove Theorem 5.4.1. The set F mentioned above satisfies the hypothesis (1) and (2) of Proposition 5.4.2. But, unfortunately, F might be not connected or not open. We will replace F by a subset U of T satisfying the hypothesis of Proposition 5.4.2 . First observe that the set F is the union of finitely many convex subsets of T and satisfies the hypotheses of Lemma 5.3.2. In particular conditions (1) and (2) of Proposition 5.4.2 are satisfied for U = F. If F is replaced by its interior V = F ∘ , then V is the union of finitely many open convex subsets and still satisfies the hypotheses imposed on F in Lemma 5.3.2. Indeed, ⋃x∈X V x is the interior of ⋃x∈X F x and therefore F ⊆ ⋃x∈X V x . Thus T = ⋃g∈G F g ⊆ ⋃x∈X,g∈G (V x )g = ⋃g∈G V g . The remaining hypothesis of Lemma 5.3.2 is obviously inherited from F to V. Unfortunately V is not necessarily connected. (Recall that as F is a subset of an Euclidean space, it is connected if and only if it is path connected.) Now we show how to replace V by a subset U of T satisfying all the conditions of Proposition 5.4.2. Write V = V1 ∪ ⋅ ⋅ ⋅ ∪ V k , with V1 , . . . , V k open convex subsets of T . Then T = ⋃ki=1 ⋃g∈G (V i )g . We select a point x i ∈ V i for every i = 1, . . . , k. Let C = ∪ki=2 C i , where C i is the line segment joining x1 with x i . As C is compact (and contained in T ), C ⊂ U1 ∪ ⋅ ⋅ ⋅ ∪ U l , where each U j is of the form (V i )g , for some g ∈ G and some i = 1, . . . , k, and U j ∩C ≠ 0 for all j. Of course, we may assume that all V1 , . . . , V k are in the list U1 , . . . , U l . Hence, we obtain a connected open subset U = ⋃li=1 U i satisfying conditions (1) and (2) of Proposition 5.4.2. This finishes the construction of U. To finish the proof of Theorem 5.4.1 it thus remains to prove Proposition 5.4.2. Proof of Proposition 5.4.2. Let H = ⟨ [x] | x ∈ X, [xy] = [x][y] for x, y ∈ X with U ∩ U y ∩ U xy ≠ 0 ⟩ . Let x, y, y1 , z ∈ G. We claim that −1 if U x ∩ U y ∩ U y1 ≠ 0 ≠ U z ∩ U y ∩ U y1 then [xy−1 ][yz] = [xy−1 1 ][y 1 z].

(5.4.1)

196 | 5 The group of units of an order −1

−1

−1

−1

Indeed, U (xy1 )(y1 y ) ∩ U ∩ U y1 y = U xy ∩ U ∩ U y1 y −1 −1 U ∩ U yz ∩ U y1 z ≠ 0 and hence

−1

−1

≠ 0 and U ∩ U yz ∩ U (y1 y

−1

)(yz−1 )

=

−1 −1 −1 −1 −1 [xy−1 ][yz−1 ] = [(xy−1 1 )(y 1 y )][yz ] = [xy 1 ][y 1 y ][yz ] −1 −1 −1 = [(xy−1 1 )][(y 1 y )(yz )] = [xy 1 ][y 1 z],

as desired. Let f : {[x] : x ∈ X} → G be the map defined by f([x]) = x. Clearly, f preserves the defining relations of H, that is f([xy]) = f([x])f([y]), if U ∩ U y ∩ U xy ≠ 0. Therefore f extends to a group homomorphism f : H → G. As the set U satisfies the hypotheses of the set F of Lemma 5.3.2, we have that X generates G and hence f is surjective. We need some preparation in order to prove that f also is injective. Let α : [0, 1] → T be a continuous function and fix g, h ∈ G such that α(0) ∈ U g and α(1) ∈ U h . A (U, α)-partition is an increasing sequence A = (a0 = 0, a1 , . . . , a n = 1) of elements of [0, 1] such that, for every i = 1, . . . , n, there is s i ∈ G such that α([a i−1 , a i ]) ⊆ U s i and s1 = g and s n = h. Using that T is the union of the sets of the form U s and the compactness of the image of α, it is easy to see that there is a (U, α)-partition. Given a (U, α)-partition and S = (s1 = g, s2 , . . . , s n−1 , s n = h) as above, we set −1 θ(A, S) = [s1 s−1 2 ] ⋅ ⋅ ⋅ [s n−1 s n ]. −1

Observe that α(a i ) ∈ U s i ∩ U s i+1 = (U ∩ U s i s i+1 )s i+1 for every i = 1, . . . , n − 1 and therefore s i s−1 i+1 ∈ X. We claim that θ(A, S) is independent of S. To prove this, it is enough to show that if T is obtained by replacing s i with t for some i = 2, . . . , n − 1 then θ(A, S) = θ(A, T). To prove this, observe that α(a i−1 ) ∈ U s i−1 ∩ U s i ∩ U t and α(a i ) ∈ U s i+1 ∩ U s i ∩ U t . Then, −1 −1 −1 −1 −1 θ(A, S) = [s1 s−1 2 ] ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ [s i−2 s i−1 ][s i−1 s i ][s i s i+1 ][s i+1 s i+2 ] ⋅ ⋅ ⋅ [s n−1 s n ] −1 −1 −1 −1 −1 = [s1 s−1 2 ] ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ [s i−2 s i−1 ][s i−1 t ][ts i+1 ][s i+1 s i+2 ] ⋅ ⋅ ⋅ [s n−1 s n ]

= θ(A, T), by (5.4.1). If A is a (U, α)-partition and B is obtained from A by inserting one number b then B is a (U, α)-partition and it is easy to see that θ(A, S) = θ(B, T), where T the list obtained from S by inserting an appropriate group element. Thus θ(A, S) also is independent of A because every two (U, α)-partitions have a common refinement and hence we can write θ g,h (α) = θ(A, S). We claim that if α is a closed path and α(0) ∈ U g then θ g,g (α) = 1. To simplify the notation put θ = θ g,g . Since T is simply connected, there is an homotopy h : [0, 1] × [0, 1] → T from α to the constant path β mapping every t ∈ [0, 1] to α(0). Using that the open squares (a1 , a2 ) × (b1 , b2 ), with a i and b i rational, form a basis of open subsets of [0, 1] × [0, 1], it is easy to see that there is a positive integer m such that j−1 j i for every square J of the form [ i−1 m , m ] × [ m , m ] with i, j ∈ {1, . . . , m}, there is s J ∈

5.4 The group of units of an order is finitely presented |

197

G such that h(J) ⊆ U s J . Consider the set P of paths in [0, 1] × [0, 1] which join the points (0, 0) and (0, 1) by concatenating line segments of one of the following forms: j j j j i i i−1 j i j−1 i 󸀠 H i,j = [( i−1 m , m ), ( m , m )]; H i,j = [( m , m ), ( m , m )]; V i,j = [( m , m ), ( m , m )]; or D i,j = j−1 j i [( i−1 m , m ), ( m , m )]. It is easy to see that there is a sequence C 0 , C 1 , . . . , C k of elements in P such that C0 is the line segment [(0, 0), (0, 1)], C k is the concatenation of the segments [(0, 0), (1, 0)], [(1, 0), (1, 1)] and [(1, 1), (0, 1)] and C i is obtained from C i−1 by one of the following operations: – Replacing the concatenation of a vertical segment V i,j , with i < m, followed by a horizontal segment H i+1,j by the diagonal segment D i+1,j (see the blue arrow in Figure 5.2). – Replacing one diagonal segment D i,j by the concatenation of one horizontal H i,j−1 followed by a vertical segment V i,j (see the blue arrows in Figure 5.2). – Replacing one vertical segment V i,m , with i < m, by the concatenation of one 󸀠 diagonal segment D i+1,m followed by a backward horizontal segment H i+1,m (see the red arrows in Figure 5.2).

Let β j be the closed path of T given by restricting h to the path C j . Then α = β0 and β k is the constant path onto α(0). In order to finish the proof of the claim it is enough to show that θ(β i−1 ) = θ(β i ) for every j = 1, . . . , k, because then θ(α) = θ(β k ) = 1. Indeed, there are continuous bijective functions σ : [0, 1] → C i−1 and τ : [0, 1] → C i and 0 < b < c < 1 such that σ([b, c]) and τ([b, c]) is contained in one of the squares J j−1 j i of the form [ i−1 m , m ] × [ m , m ] and σ and τ coincide in [0, 1] \ (b, c). Therefore there is a partition A of the form 0 = a0 < a1 < ⋅ ⋅ ⋅ < a j−1 = b < a j = c < ⋅ ⋅ ⋅ < a n = 1 which is both a (U, β i−1 )-partition and a (U, β i )-partition and such that there is a list S = (s1 , . . . , s n ) of elements of G with both β i−1 ([a j−1 , a j ]) and β i ([a j−1 , a j ]) contained in U s j . Then θ(β i−1 ) = θ(A, S) = θ(β i ), as desired. H󸀠4,6

V3,6

D4,6

H3,4 V2,4 D3,4

D2,2

V2,2

H2,1 Fig. 5.2

198 | 5 The group of units of an order We finally are ready to prove that f is injective. We have to show that if x1 , . . . , x n ∈ X ϵ ϵ and ϵ1 , . . . , ϵ n ∈ {−1, 1} satisfy x11 ⋅ ⋅ ⋅ x nn = 1 in G then [x1 ]ϵ1 ⋅ ⋅ ⋅ [x n ]ϵ n = 1 in H. We claim that one may assume that ϵ i = 1 for each i. Indeed, clearly 1 ∈ X and if −1 x ∈ X then U ∩ U x ∩ U 1x ≠ 0 ≠ U ∩ U x ∩ U x x . Hence [1] = 1 and [x]−1 = [x−1 ]. ϵ ϵ This proves that [x1 ]ϵ1 ⋅ ⋅ ⋅ [x n ]ϵ n = 1 if and only if [x11 ] ⋅ ⋅ ⋅ [x nn ] = 1. Now the claim is clear. So, let x1 , . . . , x n ∈ X be such that x1 ⋅ ⋅ ⋅ x n = 1. Let c0 = 1, c1 = x n , c2 = x n−1 x n , . . . , c n = x1 ⋅ ⋅ ⋅ x n = 1. Then U c i−1 ∩ U c i = (U ∩ U x n−i+1 )c i−1 ≠ 0, for every 1 ≤ i ≤ n. Let C = (c0 , c1 , . . . , c n ). Choose points P0 = P, P1 , . . . , P n , P n+1 = P with P i ∈ U c i−1 ∩ U c i , for 1 ≤ i ≤ n. As U c i is path connected and P i and P i+1 are both contained in U c i for every 0 ≤ i ≤ n, there is a path contained in U c i connecting P i and P i+1 . Joining these paths we obtain a loop α in T . Choosing a (U, α) partition A we have [x1 ] [x2 ] ⋅ ⋅ ⋅ [x n ] = [c n (c n−1 )−1 ][c n−1 (c n−2 )−1 ] ⋅ ⋅ ⋅ [c1 c−1 0 ] = θ(C) = θ 1,1 (α) = 1, as desired.

5.5 Subgroups of finite index In this section we consider subgroups of finite index in U(O), for O an order in a finite dimensional semisimple rational algebra A. We will prove that U(O) has a torsionfree subgroup of finite index. If A is commutative then this can be proved using either Dirichlet’s Unit Theorem (Theorem 5.2.4) or Theorem 5.3.1 and the Structure Theorem of finitely generated groups. Moreover, we will show that if A is simple then U(O) is generated up to finite index by central units and units of reduced norm 1 over the center. By Lemma 4.6.9 (2), we have U(O) = {u ∈ O : RNrA/ℚ (u) = ±1} = {u ∈ O : NrA/ℚ (u) = ±1}.

(5.5.1)

More generally, if A is an F-algebra and R is the ring of integers of F then U(O) = {u ∈ O : RNrA/F (u) ∈ U(R)} = {u ∈ O : NrA/F (u) ∈ U(R)}.

(5.5.2)

To prove this we cannot apply Lemma 4.6.9 (2) because O is not necessarily an R-order. (For example, take A to be a field F and O an order in F different from R.) However the R-submodule O󸀠 of A generated by O is an R-order in A. Then applying Lemma 4.6.9 (2) for O󸀠 and Lemma 4.6.9 (3) for O and O󸀠 considered as ℤ-orders we have U(O) = {u ∈ O : u ∈ U(O󸀠 )} = {u ∈ O : NrA/F (u) ∈ U(R)} = {u ∈ O : RNrA/F (u) ∈ U(R)}. Assume that A is simple and let F = Z(A) and R be the ring of integers of F. As both NrA/F and RNrA/F are multiplicative morphisms, they define group homomorphisms U(O) → U(R). Therefore O1 = {u ∈ O : RNrA/F (u) = 1}

is a normal subgroup of U(O).

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199

Proposition 5.5.1. If O is an order in a finite dimensional semisimple rational algebra A then ⟨O1 , U(Z(O))⟩ has finite index in U(O). If, moreover, A is simple then O1 ∩ U(Z(O)) is cyclic and finite. Proof. Because of Lemma 4.6.9, it is easy to see that, without loss of generality, we may assume that A is simple. Let N = RNrA/F . As Z(O) is an order in F = Z(A) (Lemma 4.6.6) we get that Z(O) ⊆ R and U(Z(O)) has finite index in U(R), by Lemma 4.6.9 (4). If x ∈ U(Z(O)) then N(x) = x d , where d = Deg(A). This implies that the elements of O1 ∩ U(Z(O)) are contained in the torsion subgroup of U(Z(O)). As F is a number field, O1 ∩ U(Z(O)) is cyclic and finite. Moreover, N(U(Z(O))) contains U(Z(O))d , a subgroup of finite index in U(R) and, in particular, [N(U(O)) : N(Z(O))] < ∞. Therefore, there are u1 , . . . , u n ∈ U(O) such that for every u ∈ U(O), there is x ∈ U(Z(O)) such that 1 N(u) = N(u i x). This implies that (u i x)−1 u ∈ O1 and hence u−1 i u ∈ ⟨O , U(Z(O))⟩. So, [U(O) : ⟨O1 , U(Z(O))⟩] ≤ n. Theorem 5.5.2. The group of units of an order O in a finite dimensional semisimple rational algebra A has a torsion-free subgroup of finite index. Proof. First we prove that there is a positive number m such that if n is the order of a torsion element of U(A) then n divides m. Indeed, let ρ denote the left regular representation of A. Let Y be the set formed by the orders of the eigenvalues of ρ(u) where u runs through the torsion units of A. If n ∈ Y and ξ is an eigenvalue of ρ(u) of order n then ξ is a root of CharA/ℚ (u). Thus the minimal polynomial of ξ over ℚ divides CharA/ℚ (u). Therefore φ(n) ≤ dimℚ (A). Using the well-known formula for φ(n), one deduces that Y is finite. Let m be the least common multiple of the elements of Y. Because ρ is injective it is clear that if u is a torsion unit of A of order n then n is the least common multiple of the orders of the eigenvalues of ρ(u). Hence n divides m, as desired. Choose a prime integer p with p > m. The natural ring homomorphism O → O/pO induces a group homomorphism f : U(O) → U(O/pO). As O/pO is finite, N = ker(f) is a subgroup of finite index in U(O). If g is a non-trivial element of N then g = 1 + p k x for some positive integer k and x ∈ O \ pO. If moreover g has order n then 1 = g n = 1 + p k (nx + p k ∑ni=2 (ni)p k(i−2) x i ). As the additive group of O is torsion-free, we have that nx ∈ pO. This implies that p divides n, in contradiction with n ≤ m < p. Therefore N is a torsion-free subgroup of finite index in U(O), as desired. Corollary 5.5.3. Let A be a finite dimensional semisimple rational algebra and let O be an order in A. Let A = A1 × ⋅ ⋅ ⋅ × A n be the Wedderburn decomposition of A. For every i = 1, . . . , n, let Oi be an order in A i . Then U(O) has a subgroup of finite index of the form U(Z(O)) × V1 × ⋅ ⋅ ⋅ × V n , with each V i a subgroup of finite index in O1i .

200 | 5 The group of units of an order Proof. Let O󸀠 = O1 × ⋅ ⋅ ⋅ × On and, for every i = 1, . . . , n, let R i be the ring of integers of Z(A i ). Then Z(O) and R = R1 × ⋅ ⋅ ⋅ × R n are both orders in Z(A) and O and O󸀠 are both orders in A, by Lemma 4.6.2 and Lemma 4.6.6. Clearly Z(O) ⊆ R. It follows that U(Z(O)) has finite index in U(R) = U(R1 ) × ⋅ ⋅ ⋅ × U(R n ) and U(O󸀠 ) has a subgroup of finite index contained in U(O), by Lemma 4.6.9 (4). This implies that each O1i has a subgroup V i of finite index and V1 × ⋅ ⋅ ⋅ × V n is contained in U(O). Moreover, by Theorem 5.5.2, one may assume that each V i is torsion-free. Then, by Proposition 5.5.1, each V i ∩ U(R i ) is a finite torsion-free group. So, V i ∩ U(R i ) = 1 for every i. Let V = V1 × ⋅ ⋅ ⋅ × V n . Then, V ∩ U(Z(O)) = 1. This implies that ⟨V, U(Z(O))⟩ = U(Z(O)) × V. On the other hand, ⟨O1i , U(R i )⟩ has finite index in U(Oi ), by Proposition 5.5.1. Using that V i has finite index in O1i and U(R i ) is contained in the center of U(Oi ), it easily follows that ⟨V i , U(R i )⟩ = V i × U(R i ) has finite index in U(Oi ). Hence, ⟨V, U(R)⟩ has finite index in U(O󸀠 ). As U(Z(O)) has finite index in U(R) we deduce that ⟨V, U(Z(O))⟩ has finite index in U(O󸀠 ) and thus also in U(O). If A is a finite dimensional semisimple rational algebra, then A = A1 × ⋅ ⋅ ⋅ × A n with each A i = M k i (D1 ) and D i a division algebra. For every i, choose an order R i in D i and let Oi = M k i (R i ). Clearly Oi is an order in A i . The group O1i is by definition the special linear group of degree k i over R i . It will be studied in Chapter 9. Corollary 5.5.3 shows that, up to finite index, the study of the unit group of an order in A is reduced to the study of the group of units of the center and that of the special linear group of orders in division algebras. The group of central units is described by Dirichlet’s Unit Theorem. However, the special linear group of an order in a division algebra is much more complicated. In Chapter 9, for most cases, we will give generators for a subgroup of finite index. We finish this section by describing when the unit group of an order in a finite dimensional semisimple rational algebra is finite. To do so we recall the following definition. A quaternion algebra ( a,b F ) is said to be totally definite if F is totally real and a and b are totally negative. In other words, ( a,b F ) is totally definite if every complex embedding σ of F is real and satisfies σ(a), σ(b) < 0. We need the following group theoretical lemma. Lemma 5.5.4. Let U be a group, V a normal subgroup of U and B an abelian subgroup of finite index of V. Then V has a characteristic abelian subgroup C of finite index. If, moreover, B is maximal among the abelian subgroups of V then [U : N U (B)] < ∞ and C can be taken so that C ⊆ B. Proof. Let n = [V : B] and let W be the subgroup of V generated by all the abelian subgroups of V of index n. Since [W : B] < ∞ there are finitely many abelian subgroups A1 = B, A2 , . . . , A k such that W = ⟨A1 , . . . , A k ⟩. Then ∩ki=1 A i is a central subgroup of W of finite index in V and thus C = Z(W) is an abelian subgroup of finite index in V. As W is a characteristic subgroup of V, so is C. If B is maximal among the

5.5 Subgroups of finite index |

201

abelian subgroups of V then C ⊆ B. Moreover the U-conjugates of B are contained in V and contain C. As [V : C] < ∞, the set of U-conjugates of B is finite and hence [U : N U (B)] < ∞. The following proposition is a particular case of a theorem of Malcev (see e.g. [172, Theorem 12.1.3]). Proposition 5.5.5. Let D be a division algebra of finite dimension over its center. If U is a solvable subgroup of U(D) then U contains an abelian subgroup of finite index. Proof. We argue by induction on the derived length of U. This is obvious if U has derived length 1. By the induction hypothesis, U 󸀠 has an abelian subgroup of finite index A. By Zorn’s Lemma U has a maximal abelian subgroup B containing A. Let V = U 󸀠 B. Then V is normal subgroup of U (because U 󸀠 ⊆ U), B is a maximal abelian subgroup of V (because it is maximal abelian in U) and B has finite index in V (because [V : B] = [U 󸀠 : B ∩ U 󸀠 ] ≤ [U 󸀠 : B] < ∞). Let N = N U (B), F = Z(D) and G the group of F-automorphisms of F(B), the subfield of D generated by F and B. As F(B) is a finite field extension of F, the group G is finite. By Lemma 5.5.4, [U : N] < ∞. Moreover, N acts on F(B) by conjugation and the kernel of this action is CenN (B) = B, by the maximality of B. Therefore [N : B] ≤ |G| < ∞ and hence [U : B] < ∞, as desired. Proposition 5.5.6 (Kleinert [136]). Let A be a non-commutative finite dimensional simple rational algebra and O an order in A. Then the following are equivalent. (1) O1 is finite. (2) U(Z(O)) has finite index in U(O). (3) U(O) has an abelian subgroup of finite index. (4) U(O) has a solvable subgroup of finite index. (5) A is a totally definite quaternion algebra. Proof. (1) implies (2) is a consequence of Proposition 5.5.1. (2) implies (3) and (3) implies (4) are obvious. To complete the proof we will prove (5) implies (1), (4) implies (3) and (3) implies (5). (5) implies (1) Assume that A = ( a,b F ) is a totally definite quaternion algebra. Let R be the ring of integers of F. Replacing i and j by an R-multiple, one may assume without loss of generality that a, b ∈ R. This implies that ( a,b R ) = R ⊕ Ri ⊕ Rj ⊕ Rk is an order in A. As [O1 : (O ∩ O1 )1 ] < ∞ for any order O1 in A, it is enough to show that O1 is finite for a chosen order and we choose O = ( a,b R ). The (reduced) norm in A is given by RNrA/F (x0 + x1 i + x2 j + x3 k) = x20 − ax21 − bx22 + abx23 . Thus O1 = {x0 + x1 i + x2 j + x3 k : x20 − ax21 − bx22 + abx23 = 1, x0 , x1 , x2 , x3 ∈ R}. As F is totally real, all the embeddings of F in ℂ are real. Let σ1 , . . . , σ n be these embeddings.

202 | 5 The group of units of an order

Because of Example 2.1.13, there is an isomorphism n

f : A ⊗ℚ ℝ 󳨀→ B = ⨁ ( u=1

σ u (a), σ u (b) ) ℝ

(x0 + x1 i + x2 j + x3 k) ⊗ 1 󳨃→ (σ u (x0 ) + σ u (x1 )i + σ u (x2 )j + σ u (x3 )k)nu=1 . Then f(O) is a lattice in the real algebra B and f(O1 ) is contained in K = {(x0u + x1u i + x2u j + x3u k)nu=1 : x20u − σ u (a)x21u − σ u (b)x22u + σ u (ab)x23u = 1, for every u}. Because σ u (a), σ u (b) < 0 for every u = 1, . . . , n, the set K is compact. Proposition 5.1.1 therefore implies that f(O1 ) is finite. Hence, O1 is finite as desired. Before continuing with the remaining implications we prove that if U(O) has a solvable subgroup U of finite index then A is a division algebra. Indeed, assume that U(O) has a solvable subgroup U of finite index. Set A = M n (D), with D a division ring and let R be an order in D. Then M n (R) is an order in A by Lemma 4.6.2. Hence GLn (R)∩ U(O) has finite index in GLn (R), by Lemma 4.6.9 (4). Therefore GLn (R)∩ U has finite index in GLn (R). Assume that n > 1. Then the elementary matrices e12 (1) and e21 (1) belong to GLn (R) and therefore e12 (m) = e12 (1)m , e21 (m) = e21 (1)m ∈ U, for some m ≥ 3. Because of Corollary 1.4.5, the group ⟨e12 (m), e21 (m)⟩ is free. However, this is impossible as U is solvable. So, n = 1 and thus A = D, a division ring. (4) implies (3). Is a direct consequence of the previous paragraph and Proposition 5.5.5. (3) implies (5). Assume that U(O) has an abelian subgroup U of finite index. Then, by the comments before the proof of (4) implies (3), A = D, a division algebra. Put F = Z(D). Let Σ be the set of subfields L of D containing F such that [U(O) : U(OL ) ∩ U(O)] < ∞, where OL denotes the ring of integers in L. Observe that F(U) ∈ Σ. Claim 1. If L ∈ Σ, E is a subfield of D containing F properly such that E ⊄ L and K = L ∩ E then K is totally real and E is a totally imaginary quadratic extension of K. Indeed, U(OL ) ∩ U(OE ) is a subgroup of finite index of U(OE ). Hence [U(OE ) : U(OK )] < ∞ because U(OL ) ∩ U(OE ) ⊆ U(OK ). It follows from Corollary 5.2.7 that K is totally real and E is a totally complex quadratic extension of K. This proves Claim 1. Claim 2. Let L, E and K = E∩L as in Claim 1 and let C = CenD (K). Assume that E is a maximal subfield of D and either L = F or E and L do not commute. Then C = CenD (K) is a totally definite quaternion algebra over K. By Claim 1, E = K(α) for some α such that α2 ∈ K. Moreover, by the assumptions on E and L, we have that K ⊆ Z(C) ⊆ CenC (L) ∩ CenC (E) ⊆ E = CenC (α). If E and L do not commute then the latter inclusion is strict so that K = Z(C) ≠ E. If L = F then of course K = F = Z(C) = Z(D). Hence we obtain in both cases stated in

5.5 Subgroups of finite index |

203

Claim 2 that K = Z(C) ≠ E. Therefore C is a central simple K-algebra and E is a maximal subfield of C. Then, by Proposition 2.2.4 (5), Deg(C) = [E : K] = 2. Thus C is a quaternion algebra ( a,b K )= K+ Ki + Kj + Kij, with i2 = a, j2 = b and ij = −ji. Without loss of generality we may assume that j ∈ ̸ L. By Claim 1, K(j) = K[√b] is totally complex. Hence b is totally negative. We now show that also a is totally negative. We prove this by contradiction. So, assume that a is not totally negative. Hence K(i) is not totally complex and therefore, by Claim 1, K(i) ⊆ L and thus K(i) = L. Let σ be a (real) embedding of K such that σ(a) > 0. Then, there is x ∈ ℤ such that σ(a)x2 > −σ(b). Consequently, K(xi + j) ≅ K(√ax2 + b) is a subfield of D containing F and that is not contained in L. So, again by Claim 1, K(xi + j) is totally complex, or equivalently ax2 + b is totally negative. However σ(ax2 + b) > 0. This yields the desired contradiction. So both a and b are totally negative and hence C is a totally definite quaternion algebra over K. This finishes the proof of Claim 2. Now we prove by contradiction that if L ∈ Σ and L is not totally complex then L = F. So, suppose L ∈ Σ and L is not totally complex and different from F. Then, D contains a maximal subfield E that is not contained in the centralizer of L in D. It follows that K = L ∩ E is totally real and L is a non-central subfield of CenD (K). By Claim 2, L is totally complex, yielding a contradiction. We are ready to prove that D is a totally definite quaternion algebra over F. We consider two cases. First assume that U ⊆ F. Then F = F(U) ∈ Σ and so D = CenD (F) is a totally definite quaternion algebra over F, by Claim 2. Second, assume that U ⊈ F. Let L = F(U), an element in Σ, and let E be a maximal subfield of D not commuting with L. Then, by Claim 2, K = L ∩ E is a totally real subfield and C = CenD (K) is a totally definite quaternion algebra over K. In particular, L is totally complex. Because of Corollary 5.2.7 we also know that [U(OL ) : U(OK )] < ∞. Hence K ∈ Σ. By the previous paragraph, K = F and hence C = CenD (K) = D is a totally definite quaternion algebra over F. The well-known Tits Alternative [218] states that if F is a field of characteristic 0 then every subgroup of GLn (F) either contains a solvable subgroup of finite index or it contains a free subgroup of rank two. Hence, the unit group U of an order in a finite dimensional simple rational algebra A does not contain a free group of rank two if and only if A is either a field or it is a totally definite quaternion algebra, or equivalently, U is abelian-by-finite (which on its turn is equivalent with U is central-by-finite). Corollary 5.5.7. Let A be a finite dimensional semisimple rational algebra and let O an order in A. Then the following conditions are equivalent. (1) U(Z(O)) has finite index in U(O). (2) U(O) has an abelian subgroup of finite index. (3) U(O) has a solvable subgroup of finite index. (4) Every non-commutative simple component of A is a totally quaternion algebra.

204 | 5 The group of units of an order

The following result is an easy consequence of Dirichlet’s Unit Theorem (Theorem 5.2.4) and Kleinert’s Proposition (Proposition 5.5.6). The proof is left to the reader. Corollary 5.5.8. Let A be a finite dimensional semisimple rational algebra and O an order in A. Then U(O) is finite if and only if every simple component of A is either ℚ, an imaginary quadratic extension of ℚ or a totally definite quaternion algebra over ℚ. As an application of the previous results one can easily give another proof of Theorem 1.5.6 (see Problem 5.5.2).

Problems 5.5.1. Let F be a finite Galois extension of ℚ and R the ring of integers of F. Let a ∈ R \ {0} and for every σ ∈ Gal(F/ℚ) set μ σ = σ(a) a . (1) Prove that if F has a real embedding, then all the complex embeddings of F are real. (2) Prove that NrF/ℚ (μ σ ) = 1 and conclude that μ σ ∈ U(R) (3) Prove that μ−1 τ(μ σ ). Compare with cyclotomic units. σ = σ(μ σ−1 ) = ∏1=τ∈Gal(F/ℚ) ̸ (4) Let X be a set of representatives of non-trivial elements of Gal(F/ℚ) modulo complex conjugation. Prove that ⟨μ σ : σ ∈ Gal(F/ℚ)⟩ has finite index in U(R) if and only if det((log|ρ(μ σ−1 )|)ρ,σ∈X ) ≠ 0. 5.5.2. Prove Higman’s result (Theorem 1.5.6) as an application of Corollary 5.5.8. 5.5.3. Let A be a finite dimensional simple rational algebra. Prove that the following properties are equivalent. (1) A is a totally definite quaternion algebra over its center. (2) Every Wedderburn component of A ⊗ℚ ℝ is isomorphic to ℍ(ℝ). (3) Deg(A) = 2 and A is ramified at every infinite prime of its center.

6 Cyclotomic integers The examples in Section 1.4 suggest that cyclotomic fields ℚ(ζ n ) and their ring of integers ℤ[ζ n ] are crucial to understand the unit group of an integral group ring of a finite cyclic group. Hence, in this chapter we give the necessary background on this topic. Most of the material is taken from [222] and for some classical results we also use [101] and [208]. In [222] the focus is on the real part ℤ[ζ n + ζ n−1 ] of the cyclotomic integers while we are mostly interested in the full ring of integers. Observe that U(ℤ[ζ n + ζ n−1 ]) has finite index in U(ℤ[ζ n ]) (see Problem 5.2.1).

6.1 Cyclotomic fields Recall that for every positive integer n, ζ n denotes a primitive complex n-th root of unity. The field ℚ(ζ n ) is called the n-th cyclotomic field. The n-th cyclotomic polynomial is Φ n = Minℚ (ζ n ). Observe that the field ℚ(ζ n ) does not depend on the particular chosen primitive n-th root of unity ζ n (only on n). Indeed, if ξ1 and ξ2 are two complex primitive n-th roots of unity then ξ2 = ξ1r and ξ1 = ξ2s for some integers r and s such that rs ≡ 1 mod n. This and Proposition 6.1.1 (2) also show that the polynomial Φ n does not depend on the choice of the primitive n-th root of unity. The assignment i 󳨃→ ζ ni defines a bijection from ℤ/nℤ to the complex n-th roots of unity. It restricts to a bijection from U(ℤ/nℤ) to the set of complex primitive n-th root of unity. The following proposition collects the elementary properties of cyclotomic fields. A proof can be found in many references, such as for example [145]. Proposition 6.1.1. Let n and m be positive integers. (1) If n < m then ℚ(ζ n ) = ℚ(ζ m ) if and only if n is odd and m = 2n. (2) Φ n = ∏r∈U(ℤ/nℤ) (X − ζ nr ) ∈ ℤ[X] and X n − 1 = ∏d|n Φ d . (3) ℚ(ζ n )/ℚ is a Galois extension of degree φ(n) (where φ denotes the Euler map). (4) The map U(ℤ/nℤ) → Gal(ℚ(ζ n )/ℚ), associating r with the automorphism σ r of ℚ(ζ n ) defined by σ r (ζ n ) = ζ nr , is a group isomorphism. By Proposition 6.1.1 (1), to study cyclotomic fields ℚ(ζ n ) we may assume without loss of generality that n ≢ 2 mod 4. The following proposition describes the ring of integers of ℚ(ζ n ) and the integral primes which ramify in ℚ(ζ n ). Proposition 6.1.2. Let n be a positive integer with n ≢ 2 mod 4. Then the following properties hold. (1) The primes of ℤ ramified in ℚ(ζ n ) are precisely those dividing n. (2) If p is a prime integer and n = p e n󸀠 with gcd(p, n󸀠 ) = 1 then e(ℚ(ζ n )/pℤ) = φ(p e ). (3) ℤ[ζ n ] is the ring of integers of ℚ(ζ n ).

206 | 6 Cyclotomic integers Proof. We use the following notation throughout the proof: ξ = ζ n , Nr = Nrℚ(ξ)/ℚ and m = φ(n). We also let R denote the ring of integers of ℚ(ξ). As ξ is integral over ℤ, we have ℤ[ξ] ⊆ R. (1) We start by proving that if a prime p ∈ ℤ ramifies in ℚ(ξ) then p divides n. By Proposition 6.1.1.(3, [ℚ(ξ) : ℚ] = n. Thus 1, ξ, ξ 2 , . . . , ξ m−1 is a basis of ℚ(ξ) over ℚ and Δℚ (1, ξ, ξ 2 , . . . , ξ m−1 ) = ±Nr(Φ󸀠n (ξ)), by Example 4.7.2. We have X n − 1 = Φ n (X)P(X), with P(X) = ∏d|n,d=n̸ Φ d ∈ ℤ[X]. Taking derivatives and observing that Φ n (ξ) = 0 we deduce that nξ n−1 = Φ󸀠n (ξ)P(ξ). As ℤ[ξ] is an order in ℚ[ξ] and ξ ∈ U(ℤ[ξ]), we have Nr(ξ) = ±1 (Lemma 4.6.9 (2)) and Nr(P(ξ)) ∈ ℤ (Lemma 4.6.3). This implies that Nr(Φ󸀠n (ξ)) divides Nr(n) = n m . Hence Δℚ (1, ξ, ξ 2 , . . . , ξ m−1 ) divides n m . Since ℤ is a PID, R has a ℤ-basis, say x1 , . . . , x m . Then, ξ i = ∑m j=1 a ij x j , for some a ij ∈ ℤ. By (4.7.1), Δℚ (1, ξ, ξ 2 , . . . , ξ m−1 ) = det(a ij )2 Δℚ (x1 , . . . , x m ) and hence both det(a ij ) and Δℚ (x1 , . . . , x m ) belong to ℤ and divide n m . If p ramifies in ℚ(ξ) then pℤ contains Δ(R/ℤ), by Theorem 4.7.3. By Lemma 4.7.1 (2), Δ(R/ℤ) = ℤΔℚ (x1 , . . . , x m ). Hence p divides Δℚ (x1 , . . . , x m ). Thus p divides n as desired. This proves “half” of (1). For the proofs of both the “other half” and of (2) we fix a prime divisor p of n and set n = p e n󸀠 with p coprime with n󸀠 . Let S denote the ring of integers of ℚ(ζ p e ) and T denote the ring of integers of ℚ(ζ n󸀠 ). We claim that (1 − ζ p e )S is the unique maximal ideal of S containing p and e(ℚ(ζ p e )/pℤ) = φ(p e ). Indeed, by Proposition 6.1.1, e

Φ

pe

e−1 e−1 e−1 Xp − 1 = 1 + X p + X 2p + ⋅ ⋅ ⋅ + X (p−1)p = p e−1 X −1 = ∏ (X − ζ pi e ).

(6.1.1)

i∈U(ℤ/p e ℤ)

Therefore p = Φ p e (1) =

∏

(1 − ζ pi e ).

(6.1.2)

i∈U(ℤ/p e ℤ)

As η i =

1−ζ pi e is a 1−ζ p e i − ζ p e ) for

(cyclotomic) unit in ℤ[ζ p e ], we also know that ℤ[ζ p e ](1 − ζ p e ) =

ℤ[ζ p e ](1 every i ∈ U(ℤ/p e ℤ). Hence, pS = ((1 − ζ p e )S)φ(p ) . Thus, if Q is a prime ideal of S containing p then e(Q/ℤ) ≥ φ(p e ) = [ℚ(ζ p e ) : ℚ] ≥ e(Q/ℤ), by Proposition 6.1.1 (3) and Theorem 4.4.2. This proves the claim. The assumption n ≢ 2 mod 4, implies that p e ≠ 2 and therefore e(ℚ[ξ]/pℤ) ≥ e(ℚ(ζ p e )/pℤ) = φ(p e ) = (p − 1)p e−1 ≠ 1. This proves the “other half” of (1). (2) Let P be a prime ideal of R containing p and set Q = (1 − ζ p e )S. By the claim above, P ∩ S = Q, φ(p e ) = e(Q/ℤ) and e(Q ∩ T/ℤ) = 1. Lemma 4.4.1 and Theorem 4.4.2 therefore yield that φ(p e ) = e(Q/ℤ) ≤ e(P/S)e(Q/ℤ) = e(P/ℤ) = e(P/T)e(P ∩ T/ℤ) = e(P/T) ≤ [ℚ[ξ] : ℚ(ζ n󸀠 )] = φ(p e ). Thus e(ℚ(ξ)/pℤ) = e(P/ℤ) = φ(p e ) as desired. e

6.1 Cyclotomic fields |

207

(3) We argue by induction on the number of primes dividing n. Assume first that n = p e , with p prime and e ≥ 1. Let α = 1 − ξ . By the claim in the proof of part (1), αR is a maximal ideal of R and Rp = Rα m , with m = φ(n) = [ℚ[ξ] : ℚ] = e(αR/ℤ). By Theorem 4.4.2, f(αR/ℤ) = 1. In other words dimℤ/pℤ (R/αR) = 1. This implies that the inclusion ℤ[ξ] ⊆ R induces an isomorphism ℤ[ξ]/(αR ∩ ℤ[ξ]) ≅ R/αR. Hence R = ℤ[ξ]+ αR. If ℤ[ξ]+ α t R = R then αℤ[ξ]+ α t+1 R = αR and therefore R = ℤ[ξ]+ αR = ℤ[ξ] + αℤ[ξ] + α t+1 R = ℤ[ξ] + α t+1 R, because α ∈ ℤ[ξ]. This proves, by induction on t, 2 that R = ℤ[ξ]+ α t R for every positive integer t. In particular, R = ℤ[ξ]+ α em R = ℤ[ξ]+ p em R. We again make use of the integral basis x1 , . . . , x m of R and the expressions ξ i = ∑m j=1 a ij x j , for 0 ≤ i < m mentioned in part (1) of the proof. Also recall that the entries of the matrix (a ij ) are integers and det(a ij ) divides n m = p em . By Cramer’s Rule, p em x i ∈ ℤ[ξ]. Hence p em R ⊆ ℤ[ξ]. Thus R = ℤ[ξ] + p em R ⊆ ℤ[ξ] ⊆ R and the proof is finished. Assume now that n = p e n󸀠 , with p prime, e ≥ 1 and gcd(p, n󸀠 ) = 1. By the previous paragraph the ring of integers of ℚ(ζ p e ) is ℤ[ζ p e ] and by the induction hypothesis the ring of integers of ℚ(ζ n󸀠 ) is ℤ[ζ n󸀠 ]. By Proposition 6.1.1 (3), [ℚ[ξ] : ℚ] = φ(n) = φ(p e )φ(n󸀠 ) = [ℚ(ζ p e ) : ℚ][ℚ(ζ n󸀠 ) : ℚ]. Then ℚ(ζ p e ) and ℚ(ζ n󸀠 ) are algebraically independent because ℚ[ξ] is the compositum of ℚ(ζ p e ) and ℚ(ζ n󸀠 ) in ℂ. Furthermore, the above discriminant calculations show that Δ(ℤ[ζ p e ]/ℤ) and Δ(ℤ[ζ n󸀠 ]/ℤ) are coprime. Hence, by Proposition 4.7.5, the ring of integers of ℚ[ξ] is the subring of ℂ generated by ℤ[ζ p e ] and ℤ[ζ n󸀠 ], and this ring is precisely ℤ[ξ].

Problems 6.1.1. Calculate the exact value of Δ(ℤ[ζ n ]/ℤ). (See the proof of Theorem 6.1.2.) 6.1.2. Let n be a positive integer. Prove that the ring of integers of ℚ(ζ n + ζ n−1 ) is ℤ[ζ n + ζ n−1 ]. 6.1.3. Prove that if n and m are positive integers then ℚ(ζ n ) ∩ ℚ(ζ m ) = ℚ(ζ d ), where d = gcd(n, m). 6.1.4. Let n be a positive integer and p a prime integer not dividing n. Let R = ℤ[ζ n ] and let Q be a maximal ideal of R containing p. Let π : R → R/Q be the natural homomorphism. Prove the following statements. (1) The restriction of π to ⟨ζ n ⟩ is injective. (2) The residual degree f(ℚ(ζ n )/pℤ) is the multiplicative order of p modulo n. (3) Calculate f(ℚ(ζ n )/pℤ) in case n is multiple of p. 6.1.5. Let G be a finite group, let F be a number field and let R denote the ring of integers of F. Let p be a prime integer not dividing the order of G and let P be a maximal ideal of R containing p. Let k = R/Q and identify k with R Q /Q Q under the natural

208 | 6 Cyclotomic integers isomorphism (see Lemma 4.2.3). For every r ∈ R P , let r = r + Q Q ∈ k. Let M be an FG-module which is finitely generated over F. Prove the following statements. (1) M contains a full R-lattice which is an RG-submodule of M. (2) Let N and N 󸀠 be full R-lattices of M which are RG-submodules of M. Then N Q /Q Q N Q ≅ N Q󸀠 /Q Q N Q󸀠 as k-modules. (3) There is an F-basis of M such that the representation of M with respect to this basis has entries in R P . (4) Every F-character of G of degree n is afforded by a representation ρ with image in GLn (R P ). (5) If F splits G then k splits kG and the map χ 󳨃→ χ (with χ(g) = χ(g)) is a one to one correspondence between the absolutely irreducible characters of G in characteristic 0 to the absolutely irreducible characters of G in characteristic p. (6) Let ρ be an absolutely irreducible representation affording the character χ and ρ an absolutely irreducible representation affording the character χ. Then reduction modulo Q Q maps bijectively the set of complex n-th roots of unity to the n-roots of unity in characteristic p and this restricts to a bijection from the eigenvalues of ρ(g) to the eigenvalues of ρ(g), preserving multiplicities.

6.2 Cyclotomic units In this section we fix a positive integer n and we will show that the cyclotomic units of ℤ[ζ n ] generate a subgroup of finite index in U(ℤ[ζ n ]). We assume that n ≢ 2 mod 4 and that n ≠ 1. This does not cause loss of generality by Proposition 6.1.1 (1). Recall that the cyclotomic units based on ζ n are the elements of the form η a (ζ n ) = 1 + ζ n + ζ n2 + ⋅ ⋅ ⋅ + ζ na−1 =

ζ na − 1 ζn − 1

with a ∈ U(ℤ/nℤ). We set G = Gal(ℚ(ζ n )/ℚ). By Proposition 6.1.1 (4), G = {σ a : a ∈ U(ℤ/nℤ)} with σ a (ζ n ) = ζ na . Therefore we can parametrize the cyclotomic units based on ζ n by the elements σ ∈ G and set σ(ζ n − 1) η σ (ζ n ) = . ζn − 1 This suggests an alternative proof of the fact that η σ (ζ n ) is a unit in ℤ[ζ n ]. (See Problem 5.5.1.) The inverse of η σ (ζ n ) is of course η−1 σ (ζ n ) =

ζn − 1 = σ(η σ−1 (ζ n )). σ(ζ n − 1)

Observe that there are φ(n) cyclotomic units based on ζ n . However, η1 (ζ n ) = 1,

and

η σ (ζ n )η σσ−1 (ζ n )−1 =

σ(ζ n − 1) σ(ζ n−1 − 1)

= σ(−ζ n ).

6.2 Cyclotomic units |

209

Therefore ⟨−ζ n , η σ (ζ n ) : σ ∈ G⟩ = ⟨−ζ n , η σ (ζ n ) : σ ∈ X⟩ where X is a set of representatives of the non-trivial elements of G = G/ ⟨σ−1 ⟩. Observe that σ−1 is complex conjugation in ℚ(ζ n ). Hence, by Dirichlet’s Unit Theorem (Theorem 5.2.4), |X| coincides with the rank of U(ℤ[ζ n ]). It is tempting to conjecture that the cyclotomic units based on ζ n generate a subgroup a finite index in ℤ[ζ n ]. This is true in some cases, for example if n is prime power, but not always, (see Problem 6.2.1). In general, to generate a subgroup of finite index of U(ℤ[ζ]), we should use cyclotomic units based on different roots of unity. More precisely we will prove the following theorem. Theorem 6.2.1. Let n be a positive integer with n ≢ 2 mod 4. Let D be the set of divisors m of n with 1 < m and gcd(m, mn ) = 1. For each such m fix a primitive m-th root of unity ζ m and for every σ ∈ Gal(ℚ(ζ n )/ℚ) define ν σ = ∏ η σ (ζ m ). m∈D

If X is a set of representatives of the non-trivial elements of Gal(ℚ(ζ n )/ℚ) modulo complex conjugation then {ν σ : σ ∈ X} is a basis of a free abelian subgroup of finite index in U(ℤ[ζ n ]). Theorem 6.2.1 can be rephrased as follows. Theorem 6.2.2. Let n be a positive integer with n ≢ 2 mod 4. Let D be the set of divisors m of n with 1 < m and gcd(m, mn ) = 1. For each such m fix a primitive m-th root of unity ζ m . For every integer a with gcd(a, n) = 1 let νa = ∏ m∈D

ζ ma − 1 . ζm − 1

Then {ν a : 1 < a < 2n , gcd(a, n) = 1} is a basis of a free abelian subgroup of finite index in U(ℤ[ζ n ]). Observe that each ν a (respectively, ν σ ) is a product of cyclotomic units based on roots of unity of order m ∈ D. Therefore, Theorem 6.2.1 implies the following corollary, which is the main result of this section. Corollary 6.2.3. If ξ is a complex root of unity then the cyclotomic units of ℤ[ξ] generate a subgroup of finite index in U(ℤ[ξ]). If n is a prime power then D = {n} and hence ν σ = η σ (ζ n ) for every σ ∈ G. Thus the next result follows at once. Corollary 6.2.4. If p is prime and k ≥ 1 then the cyclotomic units based on ζ p k generate a subgroup of finite index in U(ℤ[ζ p k ]). As G is an abelian group, Irr(G) is the set of linear characters of G. Recall that Irr(G) has a natural group structure and it is isomorphic to G (Proposition 3.3.5).

210 | 6 Cyclotomic integers

We will use the following notation for a positive integer m: G m = Gal(ℚ(ζ m )/ℚ). In particular G = G n . If m divides n then Resm : G → G m denotes the restriction map. Let ϕ ∈ Irr(G). The conductor of ϕ is the minimum positive divisor c of n such that Gal(ℚ(ζ n )/ℚ(ζ c )) ⊆ ker(ϕ). By Problem 6.1.3, ℚ(ζ c ) is the unique minimal cyclotomic extension containing the field corresponding to ker(ϕ) by the Galois correspondence. If c is the conductor of ϕ and m is a positive integer with c | m and m | n then ker(Resm ) = Gal(ℚ(ζ n )/ℚ(ζ m )) ⊆ Gal(ℚ(ζ n )/ℚ(ζ c )) ⊆ ker(ϕ). As Resm is surjective, we get that ϕ = ϕ m ∘ Resm (6.2.1) for a unique linear character ϕ m of Gal(ℚ(ζ m )/ℚ). The Dirichlet character of ϕ is the map χ = χ ϕ : ℤ → ℂ given by ϕ c (σ a ), χ(a) = { 0,

if gcd(a, c) = 1; otherwise.

If f is a complex valued map we let f denote the composition of f with complex conjugation, that is f (x) = f(x). Clearly the Dirichlet character of ϕ is χ. Outline of the proof of Theorem 6.2.1. We set G = G/ ⟨σ−1 ⟩ and fix a set X of representatives in G of the non-trivial elements of G. Observe that {σ−1 : σ ∈ X} is also a set of representatives of elements of G \ {1}. Because of Proposition 5.2.10, it is clear that Theorem 6.2.1 is equivalent with the non-vanishing of the determinant of the following matrix: A = (log|x(ν y−1 )|)x,y∈X (6.2.2) To show that this determinant is non-zero we will proceed through the following four steps. Step I. For every ϕ ∈ Irr(G) we define α ϕ = ∑ ∑ ϕ(ρ) log|1 − ρ(ζ m )|. m∈D ρ∈G

We will show that if det(A) = 0 then α ϕ = 0 for some ϕ ∈ Irr(G) \ {1}. Step II. We will prove the following equality for every non-trivial linear character ϕ of Irr(G) with conductor c and Dirichlet character χ = χ ϕ : ∞

α ϕ = −τ(ϕ)β ϕ ∑ a=1

χ(a) , a

6.2 Cyclotomic units

|

211

where τ(ϕ) = ∑ ϕ c (ρ)ρ(ζ c ), ρ∈G c

the Gauss sum of ϕ, and β ϕ = ∑ (φ( c|m∈D

n )(∏(1 − χ(p)))). m p|m

Step III. We will prove that if ϕ ∈ Irr(G) \ {1} then τ(ϕ) ≠ 0 and β ϕ ≠ 0. Step IV. Finally we will show that if χ is the Dirichlet character of a non-trivial χ(a) linear character ϕ of G then ∑∞ a=1 a ≠ 0. To complete the proof of Theorem 6.2.1 is now obvious. Indeed, Let ϕ be a nontrivial character of G and χ the Dirichlet character of ϕ. By Step III, τ(ϕ) ≠ 0 and χ(a) β ϕ ≠ 0. As ϕ is a non-trivial linear character of G, ∑∞ a=1 a ≠ 0, by Step IV. Then α ϕ ≠ 0, by Step II. This proves that det(A) ≠ 0, by Step I, and the proof is finished. In the remainder of the section we prove the statements made in Steps I–IV. Step I. We start diagonalizing the matrix A of (6.2.2). Let f : G → ℂ be the map given by f(x) = ∑ log|1 − σ(ζ m )|, m∈D

where σ ∈ G is a representative of x. The value of f(x) does not depend on the representative σ ∈ G because 1 − σ−1 σ(ζ m ) is the complex conjugate of 1 − σ(ζ m ). With a slight abuse of the notation, we can rewrite the matrix A of (6.2.2) as A = (f(xy−1 ) − f(x))x,y∈G\{1} . Consider W = {ϕ : G → ℂ : ∑ ϕ(x) = 0} x∈G

as a complex vector space. Let T be the endomorphism of W given by T(ϕ)(x) = ∑ f(y)ϕ(yx),

(ϕ ∈ W, x ∈ G).

y∈G

We select two bases of W. The first one is formed by the maps ψ x : G → ℂ, with x ∈ G \ {1} given by 1 ψ x (y) = δ y,x − , (y ∈ G). |G| Observe that for y ∈ G we have T(ψ x )(y) = ∑ f(z)ψ x (zy) = f(xy−1 ) − z∈G

1 |G|

∑ f(z) z∈G

212 | 6 Cyclotomic integers

= f(xy−1 ) − f(x) − −1

= f(xy ) − f(x) −

1 |G|

∑ (f(xz−1 ) − f(x)) z∈G

1

∑ (f(xz−1 ) − f(x)) |G| 1=z∈G ̸

= ∑ (f(xz−1 ) − f(x)) ψ z (y). 1=z∈G ̸

So A is the matrix associated to T in the basis formed by the ψ x ’s with x ∈ G \ {1}. The second basis of W is formed by the non-trivial linear characters of G. The fact that the non-trivial linear characters of G belong to W is a consequence of Theorem 3.1.5 (6). If θ is a non-trivial linear character of G and x ∈ G then T(θ)(x) = ∑ f(y)θ(yx) = ∑ f(y)θ(y)θ(x). y∈G

y∈G

This shows that θ is an eigenvector of T with eigenvalue ∑y∈G f(y)θ(y). As the linear characters of G are linearly independent (Theorem 3.1.5 (7)), we conclude that the nontrivial linear characters of G form a basis of eigenvectors of T. Therefore det(A) = ∏ ∑ f(y)θ(y), θ y∈G

where θ runs through the set of non-trivial linear characters of G. Therefore, if det(A) = 0 then ∑y∈G f(y)θ(y) = 0 for some non-trivial linear character θ of G. Let ϕ be the composition of θ with the natural projection map G → G which maps σ to σ. Then ϕ ∈ Irr(G) \ {1} and α ϕ = ∑ f(σ)ϕ(σ) = 2 ∑ f(y)θ(y) = 0. σ∈G

y∈G

This completes Step I. Step II. We need two lemmas. Lemma 6.2.5. Let ϕ be a non-trivial character of G with conductor c and let χ be the Dirichlet character of ϕ. If m ∈ D then ∑ ϕ(ρ) log|1 − ρ(ζ m )| ρ∈G

φ ( n ) (∏p|m (1 − χ(p))) (∑ρ∈G c ϕ c (ρ) log|1 − ρ(ζ c )|) , ={ m 0,

if c | m; if c ∤ m,

where p always represents a positive prime integer. Proof. Note that c > 1 as ϕ is non-trivial by assumption. If Gal(ℚ(ζ n )/ℚ(ζ m )) ⊆ ker(ϕ) then ℚ(ζ c ) ⊆ ℚ(ζ m ) and hence c | m (because c ≢ 2 mod 4). Therefore, if m is not a

213

6.2 Cyclotomic units |

multiple of c then there is ρ0 ∈ Gal(ℚ(ζ n )/ℚ(ζ m )) \ ker(ϕ). Then ∑ ϕ(ρ) log|1 − ρ(ζ m )| = ∑ ϕ(ρ) log|1 − ρρ−1 0 (ζ m )| = ϕ(ρ 0 ) ∑ ϕ(ρ) log|1 − ρ(ζ m )|. ρ∈G

ρ∈G

ρ∈G

and from ϕ(ρ0 ) ≠ 1 we deduce that ∑ρ∈G ϕ(ρ) log|1 − ρ(ζ m )| = 0. We will use the following notation for two positive integers u and v: [u]v = {a ∈ {0, 1, . . . , u − 1} : gcd(a, v) = 1}. In particular, we identify ℤ/uℤ with [u]1 and U(ℤ/uℤ) with [u]u . Let k, m and t be positive integers and assume that m is a multiple of both c and k. We claim that a χ(a) log|1 − ζ mt | = ∑ χ(b) log|1 − ζ mb |.

∑ a∈[mt]k

(6.2.3)

b∈[m]k

Indeed, first observe that [mt]k = {b + dm : b ∈ [m]k , d ∈ [t]1 }. Furthermore, if b+dm b ∈ [m]k then χ(b + dm) = χ(b), for every d ∈ ℤ, and 1 − ζ mb = ∏d∈[t]1 (1 − ζ mt ), since b+dm t b X − ζ m = ∏d∈[t]1 (X − ζ mt ). So, ∑ a∈[mt]k

󵄨󵄨 󵄨 a b+dm 󵄨󵄨 χ(a) log|1 − ζ mt | = ∑ χ(b) log󵄨󵄨󵄨 ∏ 1 − ζ mt 󵄨󵄨 󵄨 󵄨 b∈[m]k

d∈[t]1

= ∑ χ(b) log|1 − ζ mb | b∈[m]k

as claimed. Assume now that c divides m. As ρ(ζ m ) is constant for ρ running through a fixed coset modulo Gal(ℚ(ζ n )/ℚ(ζ m )) and because the restriction map Resm : G → G m is surjective and has kernel Gal(ℚ(ζ n )/ℚ(ζ m )), we have ∑ ϕ(ρ) log|1 − ρ(ζ m )| = |Gal(ℚ(ζ n )/ℚ(ζ m ))| ∑ ϕ m (ρ) log|1 − ρ(ζ m )| ρ∈G

ρ∈G m

n = φ ( ) ∑ χ(a) log|1 − ζ ma | m a∈[m]

(6.2.4)

m

(Note that for the last equality we need m > 1. This is true because c > 1.) Then ∑ χ(a) log|1 − ζ ma | = a∈[m]m

∑ χ(a) log|1 − ζ ma | − ∑ ∑ χ(a) log|1 − ζ ma | a∈[m]1

+

∑

p|m a∈[m]p

χ(a) log|1 − ζ ma | + . . .

∑

p,q|m,p=q ̸ a∈([m]p ∪[m]q )

∑ χ(a) log|1 − ζ ma | − ∑ χ(p)

=

a∈[m]1

+

∑ p,q|m,p=q ̸

p|m

χ(pq)

∑ a∈[m/pq]1

χ(a) log|1 − ζ am |

∑ a∈[m/p]1

χ(a) log|1 − ζ am | + . . . . pq

p

214 | 6 Cyclotomic integers =

∑ χ(a) log|1 − ζ ma | − ∑ χ(p) ∑ χ(a) log|1 − ζ ma | a∈[m]1

+

∑

p|m

a∈[m]1

χ(pq) ∑ χ(a) log|1 − ζ ma | + . . .

p,q|m,p=q ̸

a∈[m]1

= (∏(1 − χ(p))) ∑ χ(a) log|1 − ζ ma | p|m

a∈[m]1

In the second last equality we use (6.2.3) with t = p1 ⋅ ⋅ ⋅ p k , where p1 , . . . , p k are different prime divisors of m not dividing c (note that c divides mt ). This takes care of the summands labeled by lists of different primes not dividing c. The remaining summands are zero because χ(p1 ⋅ ⋅ ⋅ p k ) = 0 if some p i divides c. Combining this with (6.2.4) and applying once more (6.2.3) with k = 1 and t = mc we obtain n ∑ ϕ(ρ) log|1 − ρ(ζ m )| = φ ( ) (∏(1 − χ(p))) ∑ χ(a) log|1 − ζ ma | m ρ∈G p|m a∈[m] 1

n = φ ( ) (∏(1 − χ(p))) ∑ χ(a) log|1 − ζ ca | m p|m a∈[c] 1

n = φ ( ) (∏(1 − χ(p))) ∑ ϕ c (ρ) log|1 − ρ(ζ c )|, m p|m ρ∈G c

as desired. Lemma 6.2.6. If ϕ is a linear character of G c with conductor c then τ(ϕ)τ(ϕ−1 ) = ϕ(σ−1 )c

(6.2.5)

∑ ϕ(ρ)ρ(ζ c )a = χ(a)τ(ϕ).

(6.2.6)

and for every integer a we have ρ∈G c

Proof. If c1 is a proper divisor of c then the restriction of ϕ to the group H c1 = Gal(ℚ(ζ c )/ℚ(ζ c1 )) is a non-trivial linear character of H c1 . Therefore ∑ρ∈H c ϕ(ρ) = 0. 1

If a is coprime with c then ϕ(σ a )−1 = ϕ(σ a ) = χ(a) and ∑ ϕ(ρ)ρ(ζ c )a = ∑ ϕ(ρ)σ a ρ(ζ c ) = ∑ ϕ(σ−1 a ρ)ρ(ζ c ) ρ∈G c

ρ∈G c

ρ∈G c −1

= ϕ(σ a ) τ(ϕ) = χ(a)τ(ϕ). Otherwise c1 = c/ gcd(a, c) is a proper divisor of c, and hence ∑ρ∈H c ϕ(ρ) = 0. On the 1 ac other hand, ζ c 1 = 1 and hence ζ ca ∈ ℚ(ζ c1 ). Thus ρ(ζ c )a = ζ ca , if ρ ∈ H c1 . If T is a set of representatives of G c modulo H c1 then ∑ρ∈G c ϕ(ρ)ρ(ζ c )a = ∑σ∈T,ρ∈H c ϕ(σρ)σρ(ζ c )a = 1

∑σ∈T σ(ζ c )a ϕ(σ) ∑ρ∈H c ϕ(ρ) = 0 = χ(a)τ(ϕ). This proves (6.2.6). 1 In order to prove (6.2.5), we claim that we may assume that c is a prime power. Indeed, assume that c = c1 c2 with gcd(c1 , c2 ) = 1 and (6.2.5) holds for characters of G c i with conductor c i , with i = 1, 2. Then G c = H c1 × H c2 . If {i, j} =

6.2 Cyclotomic units |

215

{1, 2} then let ϕ i ∈ G c be given such that ϕ c i (H i ) = 1 and ϕ i |H j = ϕ|H j . Then ϕ = ϕ1 ϕ2 with ker(ϕ i ) ⊇ H c i . Clearly, ψ i = (ϕ i )c i is a character of G c i such that ϕ i = ψ i ∘ Resc i (see (6.2.1)). We claim that the conductor of ψ i is c i . To show this, assume ker(ψ i ) ⊇ Gal(ℚ(ζ c i )/ℚ(ζ d )) for some positive divisor d of c i . Then Gal(ℚ(ζ c )/ℚ(ζ dc j )) ⊆ ker(ϕ i ). As Gal(ℚ(ζ c )/ℚ(ζ dc j )) ⊆ Gal(ℚ(ζ c )/ℚ(ζ c j )) ⊆ ker(ϕ j ), we deduce that Gal(ℚ(ζ c )/ℚ(ζ c j d )) ⊆ ker(ϕ). By assumption c j d = c and therefore d = c i , as desired. This finishes the proof of the claim. By assumption, τ(ψ i )τ(ψ−1 i ) = ϕ i (σ−1 )c i for i = 1, 2. Then taking ζ c = ζ c1 ζ c2 and having in mind that Resc i restricts to an isomorphism H c i → G c j we have τ(ϕ) =

∑

ϕ1 (ρ2 )ϕ2 (ρ1 )ρ1 (ζ c2 )ρ2 (ζ c1 )

ρ1 ∈H1 ,ρ2 ∈H2

= ( ∑ (ψ1 ∘ Resc1 )(ρ2 )ρ2 (ζ c1 ))( ∑ (ψ2 ∘ Resc2 )(ρ1 )ρ1 (ζ c2 )) ρ2 ∈H2

ρ1 ∈H1

= ( ∑ ψ1 (ρ1 )ρ1 (ζ c1 ))( ∑ ψ2 (ρ2 )ρ2 (ζ c2 )) = τ(ψ1 )τ(ψ2 ) ρ1 ∈G c1

ρ2 ∈G c2

−1 and similarly τ(ϕ−1 ) = τ(ψ−1 1 )τ(ψ 2 ). Then −1 τ(ϕ)τ(ϕ−1 ) = τ(ψ1 )τ(ψ−1 1 )τ(ψ 2 )τ(ψ 2 ) = ψ 1 (σ −1 )c 1 ψ 2 (σ −1 )c 2 = ϕ(σ −1 )c,

as desired. So assume c = p t for some prime p. First we notice that τ(ϕ)τ(ϕ−1 ) = ∑ ϕ(ρ)ϕ(σ)−1 ρ(ζ c )σ(ζ c ) ρ,σ∈G c

= ∑ ϕ(σ−1 σρ)ϕ(σ)−1 σρ(ζ c )−1 σ(ζ c ) ρ,σ∈G c

= ϕ(σ−1 ) ∑ ϕ(ρ) ∑ σ(ρ(ζ c )−1 ζ c ) ρ∈G c

= ϕ(σ−1 )

∑ a∈U(ℤ/cℤ)

σ∈G c

ϕ(σ a ) ∑ σ(ζ c1−a ) σ∈G c

Write 1 − a = bp r with gcd(b, p) = 1. Then ζ c1−a = ζ pbt−r = σ b (ζ p t−r ). Therefore ∑ σ(ζ c1−a ) = Trℚ(ζ c )/ℚ (ζ p t−r ) σ∈G c

p −p { { = [ℚ(ζ c ) : ℚ(ζ p t−r )]Trℚ(ζ pt−r )/ℚ (ζ p t−r ) = {−p t−1 , { {0, t

t−1 ,

if r = t; if r = t − 1; otherwise.

The case r = t only holds for a = 1 and the case r = t − 1 holds for a = 1 + sp t−1 with s = 1, . . . , p − 1. The elements of G c of the form σ1+sp t−1 , with s = 0, 1, . . . , p − 1, form the subgroup H p t−1 = Gal(ℚ(ζ c )/ℚ(ζ p t−1 )) of G c . By the first paragraph of the proof,

216 | 6 Cyclotomic integers ∑σ∈H pt−1 ϕ(ρ) = 0. Consequently, p−1

τ(ϕ)τ(ϕ−1 ) = ϕ(σ−1 )(p t − p t−1 − ∑ ϕ(σ1+sp t−1 )p t−1 ) s=1

= ϕ(σ−1 )(p − p t

∑

t−1

ϕ(ρ)) = ϕ(σ−1 )c.

ρ∈H p t−1

as desired. Let ϕ be a non-trivial linear character of G. Using Lemma 6.2.5 and Lemma 6.2.6 and the power series expansion ∞

log|1 − z| = − ∑ a=1

za , a

(|z| ≤ 1, z ≠ 1)

we have αϕ = ∑ φ ( c|m∈D

n ) ∏(1 − χ(p)) ∑ ϕ c (ρ) log|1 − ρ(ζ c )| m p|m ρ∈G c

= −∑φ( c|m∈D

= −∑φ( c|m∈D

∞

∑ρ∈G c ϕ c (ρ)ρ(ζ c )a n ) ∏(1 − χ(p)) ∑ m p|m a a=1 ∞ n χ(a) ) ∏(1 − χ(p))τ(ϕ) ∑ m p|m a a=1 ∞

= −τ(ϕ)β ϕ ∑ a=1

χ(a) . a

This completes Step II. Step III. This is the easiest step. Let ϕ be a non-trivial character of G and let χ = χ ϕ . Clearly τ(ϕ) ≠ 0 by (6.2.5). Write n = ∏p|n p e p and let P denote the set of primes dividing n but not dividing c. By the definition of the Dirichlet character, if p ∈ P then χ(p) = 0. Moreover, the map I 󳨃→ ∏p∈P p e p is a bijection from the set of subsets of P to {m ∈ D : c ∤ m} = { mn : m ∈ D, c | m}. Then βϕ =

∑

φ(

m∈D, c|m

=

n ) (∏(1 − χ(p))) m p|m p e p )( ∏ (1 − χ(p)))

φ( ∏

∑ m∈D, c|m

p|n, p∤m

p|m, p∤c

= ∑ φ( ∏ p )(∏(1 − χ(p))) ep

I⊆P

=

p∈I

p∈P\I

∏ (φ(p ) + 1 − χ(p)) ep

p|n, p∤c

and each factor of the previous product is non-zero because φ(p e p ) + 1 ≥ 2 and χ(p) is either 0 or a root of unity. Hence β ϕ ≠ 0 and this completes Step III.

6.2 Cyclotomic units

|

217

Step IV. For this step we will make use of the following proposition of which a proof can be found in [101, Proposition IV.2.1]. Proposition 6.2.7. Let f : ℕ → ℂ be a map. Consider the following complex analytic f(a) function F(z) = ∑∞ a=1 a z . (1) Suppose that there are positive real numbers A and B such that, for every real num󵄨 󵄨 ber x ≥ 1, 󵄨󵄨󵄨󵄨∑a∈ℕ, a B, y ∈ ℝ}. ∑

f(a)

(2) If limx→∞ a∈ℕ, xa 1}. (Observe that the limit assumption implies that |∑a 1}.

We also will show that lim (z − 1)ζ K (z) ≠ 0,

z→1

(K, a number field),

(6.2.8)

where the previous limit is calculated in H, and ζℚ(ζ n ) (z) = ∏ L(z, χ),

(z ∈ H)

(6.2.9)

χ

where the latter product runs through the Dirichlet characters χ of the different elements of Irr(G). This will complete Step IV. Indeed, both ζℚ(ζ n ) and ζℚ (z) have a simple pole at z = 1, by (6.2.8). Moreover, ζℚ (z) = L(z, χ0 ), where χ0 is the trivial Dirichlet ζ n ) (z) character, i.e. χ0 (a) = 1 for all a ∈ ℤ. Hence ∏χ=χ̸ 0 L(z, χ) = ℚ(ζ does not vanish ζℚ (z)

at z = 1. Therefore, for every non-trivial Dirichlet character χ, we get ∑∞ a=1 a ≠ 0, as desired. If the function f of Proposition 6.2.7 is a Dirichlet character χ of a linear character ϕ of G then the function F is L(z, χ). In this case, the hypothesis of the proposition is satisfied for A = B = 1. If, moreover, ϕ ≠ 1 then the assumption is satisfied for B = 0 χ(a)

218 | 6 Cyclotomic integers and A = φ(c), where c is the conductor of ϕ. Indeed, ∑ca=1 χ(a) = ∑a∈U(ℤ/cℤ) (ϕ c )(ρ a ), the sum of the elements of the image of ϕ c . As the image of ϕ c is a non-trivial finite subgroup of U(ℂ), i.e. the set of k-th roots of unity for some k, we deduce that ∑ca=1 χ(a) = 0. Let x be a positive real number, let m be the greatest integer smaller than x and set m = cq + r with 0 ≤ r < c and q ∈ ℤ. Then ∑a∈ℕ, a. Because rkR (I) = 1, (equivalently F ≅ I(0) ) one easily verifies that if I and J are non-zero ideals of R then < I >=< J > if and only if I ≅ J as R-modules. It is well-known that Cl(R) is finite (see e.g. [101]) and it usually is denoted Cl(F). Proof of (6.2.8) and convergence of ζ K . For every positive real number ρ ≥ 1 and every C ∈ Cl(K) we set S(ρ, C) = {I ∈ C : I ⊆ R, N(I) ≤ ρ}

and

s(ρ, C) = |S(ρ, C)|.

That is s(ρ, C) is the number of ideals of R in the class C with norm at most ρ. Select an ideal I of R contained in C −1 , the inverse of C in Cl(K). Let v1 , . . . , v d and A be as in Lemma 6.2.8. We are going to relate s(ρ, C) with the volume of A. Set Ω = {a ∈ I \ {0} : |NrK/ℚ (a)| ≤ ρN(I)} , H = {Ra : a ∈ Ω}, 1 ϵρ = d and X ρ = A ∩ ϵ ρ ℤd . √ρN(I)

6.2 Cyclotomic units |

221

By Lemma 4.3.7, Ω = {a ∈ I \ {0} : N(Ra) ≤ ρN(I)}. Then |H| = s(ρ, C) because the map S(ρ, C) → H J 󳨃→ JI is a bijection. Assume that 0 ≠ x = (x1 , . . . , x d ) ∈ ϵ ρ ℤd and let f(x) =

1 ϵρ

∑di=1 v i x i , an element

of I. Then N K/ℚ (f(x)) = μ(f(x)) = ρN(I) ∑di=1 v i x i . Thus x satisfies (6.2.10) if and only if f(x) = ∑di=1 v i x d ∈ Ω. Therefore we have a map g : Xρ → H x 󳨃→ Rf(x) Recall that T is the set of roots of unity of K. We claim that g is surjective and |g−1 (J)| = |T| for each J ∈ H. To prove the claim we only need to show that for every J ∈ H, there are exactly |T| elements in the set Ω J formed by the elements y ∈ Ω such that J = Ry and l(y ⊗ ϵ ρ ) = αw + ∑r+s−1 α i w i with α ∈ ℝ and 0 ≤ α i < 1 for every i = i=1 1, . . . , r + s − 1. Of course there is at least one y0 ∈ Ω with J = Ry0 . The elements y ∈ Ω so that Ry = J are precisely the elements of the form uy0 with u ∈ U(R). Write a r+s−1 a l(y0 ⊗ ϵ ρ ) = αw + ∑r+s−1 α i w i with α, α i ∈ ℝ. If we write u = ξu11 ⋅ ⋅ ⋅ u r+s−1 with ξ ∈ T i=1 r+s−1 and a i ∈ ℤ then l(uy0 ) = αw + ∑i=1 (a i + α i )w i . For each i = 1, . . . , r + s − 1 there is a r+s−1 a exactly one integer a i such that 0 ≤ a i + α i < 1. Then Ω J = {ξu11 . . . u r+s−1 y0 : ξ ∈ T}. This finishes the proof of the claim. s(ρ,C) d Thus |T|s(ρ, C) = |T||H| = |X ρ |. Therefore ρ = N(I) |T| |X ρ |ϵ ρ . To relate s(ρ, C) with the volume of A we consider v1 , . . . , v d as a reference basis, so that we calculate volumes with respect to this basis. For every x ∈ ϵ ρ ℤd consider a cube C x centered in x of side ϵ ρ . More precisely, d

C x = { ∑ v i ⊗ λ i : |λ i − x i | ≤ i=1

ϵρ }. 2

These cubes form a tiling of ℝd . Let X 󸀠ρ be the set formed by the elements x ∈ ϵ ρ ℤd d with C x completely contained in A. Let X 󸀠󸀠 ρ be formed by the x ∈ ϵ ρ ℤ with C x ∩ A ≠ 0 󸀠 󸀠󸀠 d then X ρ ⊆ X ρ ⊆ X ρ . Each cube C x has volume ϵ ρ . Therefore vol(A) = limρ→∞ |X 󸀠ρ |ϵ dρ = d d limρ→∞ |X 󸀠󸀠 ρ |ϵ ρ = limρ→∞ |X ρ |ϵ ρ . Hence lim

ρ→∞

s(ρ, C) N(I) N(I) = lim |X ρ |ϵ dρ = vol(A). ρ |T| ρ→∞ |T|

This is a positive real number by Lemma 6.2.8. Observe that ∑a∈ℕ,a≤ρ f K (a) = ∑C∈Cl(K) s(ρ, C). Hence lim

x→∞

∑ a∈ℕ.a≤ρ

f K (a) s(ρ, C) = ∑ lim = b > 0. ρ→∞ ρ ρ C∈Cl(K)

From Proposition 6.2.7 (2) we deduce that ζ K (a) is convergent in H and limz→1 (z − 1)ζ K (z) = b > 0, as desired.

222 | 6 Cyclotomic integers

Before starting with the proof of (6.2.9) we give some generalities. We begin with some information on the Dirichlet L-function and the Riemann zeta function. We first rewrite L(z, χ) as an infinite product. Let ℙ denote the set of positive prime integers. χ(p a ) χ(p) If p ∈ ℙ and z ∈ ℂ then (1 − p z ) ∑∞ a=0 p az = 1, or equivalently (1 −

χ(p) −1 ∞ χ(p a ) ) = ∑ az . pz p a=0

Fix a positive integer m and let p1 , . . . , p k be the primes ≤ m. Then k

∏(1 − i=1

∞ χ(p11 . . . p k k ) χ(a) χ(p i ) −1 ∑ z ) = ak z = ∑ a1 az pi a∈ℙm a1 ,...,a k =0 (p 1 . . . p k ) a

a

where ℙm is the set of positive integers which are only divisible by p1 , . . . , p k . Therefore k χ(a) χ(p i ) −1 L(z, χ) − ∏(1 − = ∑ . z ) az p i a∈ℙ ̸ m i=1 If z ∈ H then the series defining L(z, χ) is absolutely convergent. Therefore, the right term in the previous equality converges to 0. Thus ∏ (1 −

L(z, χ) = lim

m→∞

p∈ℙ,p≤m

χ(p) −1 χ(p) −1 = − ∏ ) (1 ) . pz pz p∈ℙ

(6.2.12)

Now we use a similar rewriting argument to describe the Riemann zeta function of ℚ(ζ n ) as an infinite product. Letting I run through the non-zero ideals of ℤ[ζ n ] and a a P through the different maximal ideals of ℤ[ζ n ] and setting I = P11 ⋅ ⋅ ⋅ P k k , we have (using Lemma 4.3.7) ζℚ(ζ n ) (z) = ∑ I

1 = N(I)z P ∞

= ∏( ∑ P

a=0

a N(P11 1 ,...,P k ,a 1 ,...,a k

1 a . . . P k k )z

−1 1 1 ) = ∏(1 − ) z N(P a )z N(P) P

= ∏ ∏ (1 − p∈ℙ p∈P

∑

1 −1 1 −g p ) = ∏ (1 − f z ) , f z pp pp p∈ℙ

where f p = f(ℚ(ζ n )/pℤ) (see Section 4.4) and g p denotes the number of maximal ideals a of R containing p. Using the equality 1 − X m = ∏m−1 a=0 (1 − ζ m X) we have f p −1

ζℚ(ζ n ) (z) = ∏ ∏ (1 − p∈ℙ a=0

ζ fap p

−g p

) z

.

(6.2.13)

Recall that G = Gal(ℚ(ζ n )/ℚ). We now review some relations between the subgroups of G and Irr(G) and one application on the calculation of the ramification index of a prime integer over a subfield of a cyclotomic field. For a subgroup H of G we set H ⊥ = {ϕ ∈ Irr(G) : H ⊆ ker(ϕ)};

6.2 Cyclotomic units |

223

and for a subgroup M of Irr(G) we let M ⊥ = ⋂ ker(ϕ). ϕ∈M

If π : G → G/H is the natural homomorphism then the map ψ ∈ Irr(G/H) 󳨃→ ψ ∘ π ∈ Irr(G) induces an isomorphism Irr(G/H) ≅ H ⊥ . It is an easy exercise to check that H = (H ⊥ )⊥ , M = (M ⊥ )⊥ and ⊥ inverts inclusions. Composing this bijection with the Galois correspondence between subgroups of G and subfields of ℚ(ζ n ) we obtain an inclusion preserving bijection, denoted X, between subgroups of Irr(G) and subfields of ℚ(ζ n ). More precisely this bijection associates a subfield F of ℚ(ζ n ) with X(F) = Gal(ℚ(ζ n )/F)⊥ . Then |X(F)| = [G : Gal(ℚ(ζ n )/F)] = [F : ℚ]

(6.2.14)

Let p be a prime divisor of n and set n = p k n󸀠 , with n󸀠 coprime with p. Then we have natural isomorphisms G = Gal(ℚ(ζ n )/ℚ(ζ p k )) × Gal(ℚ(ζ n )/ℚ(ζ n󸀠 )) Irr(G) = X(ℚ(ζ p k )) × X(ℚ(ζ n󸀠 )). Recall that if ϕ ∈ Irr(G) then χ ϕ denotes the Dirichlet character of ϕ. Let ϕ p denote the projection of ϕ into X(ℚ(ζ p k )) along the above direct product decomposition. If M is a subgroup of Irr(G) then let M p = {ϕ p : ϕ ∈ M}. Lemma 6.2.9. Let F be a subfield of ℚ(ζ n ). Then e(P/pℤ) = |X(F)p |. Hence p is unramified in F if and only if χ ϕ (p) ≠ 0 for every ϕ ∈ X(F). Proof. Let L = F(ζ n󸀠 ) = F ⋅ ℚ(ζ n󸀠 ). Therefore X(L) = X(F)X(ℚ(ζ n󸀠 ) = X(F)p × X(ℚ(ζ n󸀠 )). Let F1 be the subfield of ℚ(ζ n ) with X(F1 ) = X(F)p . Then X(ℚ(ζ n󸀠 )) is the group of characters of G with conductor coprime with p and L is the compositum of F1 and ℚ(ζ n󸀠 ). Since p is unramified in ℚ(ζ n󸀠 ) and e(pℤ/ℚ(ζ p k )) = [ℚ(ζ p k ) : ℚ] (Proposition 6.1.2 (2)), we have e(F/pℤ) = e(F1 /pℤ)) = [F1 : ℚ] = |X(F)p |. This proves the first statement. For the second statement observe that pℤ ramifies in F if and only if ϕ p ≠ 1 for some ϕ ∈ X(F). This happens if and only if ϕ ∈ ̸ X(ℚ(ζ n󸀠 )) if and only p divides the conductor of ϕ if and only if χ ϕ (p) = 0. Proof of (6.2.9). Let p be a prime integer. As G is abelian, the decomposition and inertia groups over ℤ of a maximal ideal Q of ℤ[ζ n ] containing p do not depend on the

224 | 6 Cyclotomic integers

maximal ideal Q chosen (see Section 4.4). So denote these decomposition and inertia groups by D p and T p respectively. By Lemma 4.4.4, the fix subfield of T p in ℚ(ζ n ) is the greatest subfield F of ℚ(ζ n ) such that e(F/pℤ) = 1. Therefore F = ℚ(ζ n󸀠 ), by Proposition 6.1.2 (2). By Lemma 6.2.9, T p⊥ = X(ℚ(ζ n󸀠 )) = {ϕ ∈ Irr(G) : χ ϕ (p) ≠ 0}. Set e p = e(ℚ(ζ n )/pℤ) = |T p |,

f p = f(ℚ(ζ n )/pℤ) = [D p : T p ]

and g p = [G : D p ] = |D⊥p |. Observe that this notation is consistent with the one used in (6.2.13). Since T p ⊆ D p , we have D⊥p ⊆ T p⊥ . Furthermore, D p is generated by T p and an element σ ∈ G which projects to a generator of Gal(ℤ[ζ n ]/ℤ)). (Now we are using the bar notation for reduction modulo Q in ℤ[ζ n ] and for reduction modulo pℤ in ℤ.) We can take the Frobenius automorphism x 󳨃→ x p as generator of Gal(ℤ[ζ n ]/ℤ) and hence we can choose σ = σ p . Therefore, D⊥p = {ϕ ∈ T p⊥ : χ ϕ (p) = 1}. In other words, ϕ 󳨃→ χ ϕ (p) defines a group homomorphism T p⊥ → ℂ∗ with kernel D⊥p and image the set of f p -th roots of unity. Then T p⊥ = ⟨D p , ψ⟩ for some ψ with χ ψ (p) = ζ f p . Therefore 0, if ϕ ∈ ̸ T p⊥ ; χ ϕ (p) = { i ζ f , if ϕ ∈ ψ i D⊥p . In other words, when χ runs on the different Dirichlet characters, each f p -th root of unity is obtained as χ(p) for g p = |D⊥p | Dirichlet characters χ. This gives the non-zero values of χ(p) for the Dirichlet characters of the f p g p elements of T p⊥ . The other e p f p g p − f p g p Dirichlet characters vanish on p. By (6.2.12) and (6.2.13) we have f p −1

ζℚ(ζ n ) (z) = ∏ ∏ (1 − p∈ℙ a=0

ζ fap p

−g p

) z

= ∏ ∏(1 − p∈ℙ χ

χ(p) −1 ) = ∏ L(z, χ). pz χ

This proves (6.2.9), completes step IV and finishes the proof of Theorem 6.2.1.

Problems 6.2.1. The aim of this problem is to discover a positive integer n so that V n = ⟨η σ (ζ n ) : σ ∈ Gal(ℚ(ζ n )/ℚ)⟩ does not have finite index in U(ℤ[ζ n ]). The guideline is to go in greater detail through the Steps I–IV in this section. Let X be a set of representatives of the non-trivial elements of Gal(ℚ(ζ n )/ℚ)/ ⟨σ−1 ⟩.

6.2 Cyclotomic units |

225

(1) Calculate the determinant of (log|x(η y (ζ n )|)x,y∈X . (2) Prove that [U(ℤ[ζ n ]) : V n ] < ∞ if and only if χ ϕ (p) = 1 for some p | n and some non-trivial linear character ϕ of G such that ϕ(σ−1 ) = 1. (3) Prove that [U(ℤ[ζ30 ]) : V30 ] < ∞. (4) Prove that V55 does not have finite index in U(ℤ[ζ55 ]). (5) Prove that if n is divisible by four different primes then V n does not have finite index in U(ℤ[ζ n ]). 6.2.2. The aim of this problem is proving Franz Independence Lemma: If {a i : i ∈ U(ℤ/nℤ)} is a list of integers satisfying ∏i∈U(ℤ/nℤ) (ζ di − 1)a i = 1 for every 1 ≠ d | n then a i + a−i = 0 for every i. Here the elements of U(ℤ/nℤ) are identified with integers 1 ≤ i < n with gcd(i, n) = 1. (1) Prove that ∏i∈U(ℤ/nℤ),i< n (ζ di − 1)a i +a n−i is a root of unity, for every d | n. 2 (2) Prove that ∑i∈U(ℤ/nℤ) a i = 0. (Hint: Use (1) with d prime and take norms.) (3) Prove that a i + a−i = 0 for every i. (Hint: Consider first the case n ≢ 2 mod 4 and use Theorem 6.2.2.)

7 Central units The subject of this chapter is the group of central units of U(ℤG) for a finite group G and, more generally, the group of central units of an order O in a finite dimensional semisimple ℚ-algebra A. The main tool is Dirichlet’s Unit Theorem which provides a straightforward method to compute the rank of Z(U(O)) in terms of the center of A. In the particular case where O = ℤG, for G a finite group, one can describe this rank in terms of the arithmetics of G. Having described the rank r of Z(U(ℤG)), each set formed by r multiplicatively independent central units generates a free abelian subgroup of finite index in Z(U(ℤG)). We will prove a result of Bass-Milnor which provides such r multiplicatively independent units; and this is done by making use of Bass units. The original proof of Bass-Milnor uses K-theory. We present a K-theory free approach due to Jespers, del Río and Van Gelder [108].

7.1 The group of central units of an order Recall that if X is either a ring or a group then Z(X) denotes the center of X. Let A be a finitely dimensional semisimple rational algebra and O an order in A. As O contains a basis of A over ℚ, Z(O) = O ∩ Z(A). Moreover, by Lemma 4.6.6, Z(O) is an order in Z(A). Note that U(Z(O)) ⊆ Z(U(O)). In general, the former inclusion is not an equality (see Problem 7.1.1). However, if G is a finite group then it is easily verified that U(Z(ℤG)) = Z(U(ℤG)). The following result is an immediate consequence of Lemma 4.6.9 (3) and Corollary 5.2.6. Corollary 7.1.1. Let A be a finite dimensional semisimple rational algebra and O an order in A. Then U(Z(O)) is finite if and only if the center of every Wedderburn component of A is either ℚ or a quadratic imaginary extension of ℚ. Using Theorem 3.3.1 (4), the specialization of Corollary 7.1.1 to integral group rings takes the following form. Corollary 7.1.2. Let G be a finite group. Then the group of central units of ℤG is finite if and only if for every irreducible complex character χ of G there is an imaginary quadratic extension of ℚ containing χ(G). In general U(Z(O)) is a finitely generated abelian group and hence it is a direct product of a finite abelian group and a free abelian group of finite rank. Dirichlet’s Unit (Theorem 5.2.4) describes the rank for the case when A is a number field. It can also be used to describe the rank of U(Z(O)) for A arbitrary. For this we introduce the following notation. If F is a field of characteristic 0 then r F (A) denotes the number of simple components of F ⊗ℚ A. (Recall that F ⊗ℚ A is semisimple by Proposition 2.1.11.)

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227

In particular, if G is a finite group then r F (ℚG) is the number of simple components of FG, equivalently the number of irreducible F-characters of G. Theorem 7.1.3. Let A be a finite dimensional semisimple rational algebra and O an order in A. Then U(Z(O)) = T × F, where T is a finite group and F is a free abelian group of rank rℝ (A) − rℚ (A). Proof. As Z(O) is an order in Z(A), by Theorem 5.3.1, U(ℤ(O)) is finitely generated and by Lemma 4.6.9 (4) the rank α(A) of U(Z(O)) is independent of O. Moreover, α(A1 × A2 ) = α(A1 ) + α(A2 ) and r F (A1 × A2 ) = r F (A1 ) + r F (A2 ), for every field F. Thus it is enough to prove the theorem under the assumption that A is simple. In that case rℚ (A) = 1, Z(A) is a number field and R = Z(O) is an order in Z(A), by Lemma 4.6.6. By Example 2.1.13, r F (A) = r F (Z(A)) and this number coincides with the number of complex embeddings of F up to complex conjugation. This proves that it is enough to prove the corollary for A a number field and O its ring of integers. In this case, the statement of the corollary is precisely Corollary 5.2.5. Proposition 7.1.4 (Berman-Higman). Let G be a finite group. (1) If u is a torsion unit in ℤG such that 1 is in the support of u then u = ±1. (2) The central torsion units of ℤG are the elements in ±Z(G). Proof. (1) Let u = ∑g∈G u g g be a torsion unit of ℤG with u1 ≠ 0. Assume n = |G|, let ρ : G → GLn (ℂ) be the regular representation of G and let χ be the character afforded by ρ. Then χ(u) = nu1 . Since ρ(u) is a matrix of finite order, it can be diagonalized over ℂ and its eigenvalues ξ1 , . . . , ξ n are roots of unity. This implies that nu1 = χ(u) = ξ1 + ⋅ ⋅ ⋅ + ξ n . As u1 is a non-zero integer and |ξ i | = 1 for every i, we deduce that either all the ξ i ’s are 1 or all are −1. Then ρ(u) = ±I n = ρ(±1). As ρ is injective, we obtain that u = ±1, as desired. (2) Clearly every element of ±Z(G) is a torsion central unit. Conversely, let u be a torsion central unit of ℤG and let g be an element of the support of u. Then v = ug −1 is a torsion unit (not necessarily central) and 1 belongs to the support of v. By (1), v = ±1 and therefore u = ±g. Since u is central in ℤG, necessarily g ∈ Z(G). Hence u ∈ ±Z(G). By Proposition 7.1.4, Theorem 7.1.3 takes the following form for integral group rings. This result has been proven by Ritter and Sehgal in [195] and also by Ferraz in [60]. Corollary 7.1.5. Let G be a finite group. Let rℚ denote the number of irreducible rational characters of G and rℝ the number of irreducible real characters of G. Then the group of central units of ℤG is ±Z(G) × F, for F a free abelian group of rank rℝ − rℚ . Proof. As ℤG is an order in ℚG, by Theorem 7.1.3, U(Z(ℤG)) = T ×F, with T finite and F a free abelian group of rank rℝ (ℚG)−rℚ (ℚG). That this rank is rℝ −rℚ follows from the fact that for a field K of zero characteristic, the number of Wedderburn components of

228 | 7 Central units

KG coincides with the number of irreducible K-characters of G. Clearly, T is the set of torsion central units of ℤG and hence T = ±Z(G), by Proposition 7.1.4. As an application we now easily obtain the description of the unit group of a commutative integral group ring. This is a result due to Higman and has been reproved by Ayoub and Ayoub in [7]. Theorem 7.1.6 (Higman [88, 89]). Let G be a finite abelian group of order n. Then U(ℤG) = ±G × F, where F is a free abelian group of rank φ(d) n + 1 + k2 − 2c = ∑ kd ( − 1) , 2 2 d|n,d>2 where c is the number of cyclic subgroups of G and k d is the number of cyclic subgroups of G of order d. Proof. By Corollary 7.1.5, U(ℤG) = ±G × F were F is a free abelian subgroup of rank rℝ − rℚ . From Theorem 3.3.6 we know that ℚG ≅ ∏d|n ℚ(ζ d )k d and hence rℚ (ℚG) = ∑d|n k d = c. If d ≤ 2 then ℚ(ζ d ) ⊗ℚ ℝ ≅ ℝ. Otherwise, because of Example 2.1.13, φ(d) φ(d) ℚ(ζ d ) ⊗ℚ ℝ ≅ ℂ 2 . Thus, ℝG = ℝ ⊗ℚ ℚG ≅ ℝ × ℝk2 × ∏d|n,d=2̸ ℂk d 2 . Therefore φ(d) rℝ (ℚG) = 1 + k2 + ∑2 a2 > ⋅ ⋅ ⋅ > a k , (3) n ≡ k mod 4, and (4) a1 ⋅ ⋅ ⋅ a k is not a square. Proof. For every g ∈ A n we denote the conjugacy classes of g in S n and A n by g S n and g A n respectively. We also use the notation g󸀠 = g (1,2) . For example, if a = [a1 , . . . , a k ]

234 | 7 Central units is a partition of n with a1 > 1 and b i = a1 + ⋅ ⋅ ⋅ + a i then (1,2)

g󸀠a = g a

= (2, 1, 3, . . . , a1 )(b1 + 1, . . . , b2 ) ⋅ ⋅ ⋅ (b k−1 + 1, . . . , b k ).

As {1, (1, 2)} is a transversal of A n in S n , we have g S n = g A n ∪ g󸀠A n , for every g ∈ A n . If g and g󸀠 are conjugate in A n then g S n = g A n = g 󸀠A n . Otherwise, g S n is the disjoint union of g A n and g󸀠A n ; and in this case we say that g S n splits. Assume that g S n splits. Then either g−1 ∈ g A n or g−1 ∈ g 󸀠A n , since g−1 ∈ g S n . In A 󸀠A the former case g A n = gℝn ≠ gℝ n = g 󸀠A n and we say that g S n splits over ℝ. Similarly, A 󸀠A m A if g ∈ g n for every m coprime with the order of g then g A n = gℚn ≠ gℚ n = g 󸀠A n and we say that g S n splits over ℚ. If we know an h ∈ S n such that g−1 = g h then it is easy to decide whether g S n splits over ℝ or not. Indeed, if h ∈ A n then g S n splits A over ℝ. Otherwise, g−1 ∈ g󸀠(1,2)h ∈ g󸀠 n and hence g does not split over ℝ. Similarly, if {h m : m ∈ U(ℤ/|g|ℤ)} is a family of elements of S n such that g h m = g m for every m ∈ U(ℤ/|g|ℤ) then g S n splits over ℚ if and only if h m ∈ A n for every m. By Corollary 7.1.5 and Theorem 7.1.11, the rank of Z(U(ℤA n )) coincides with the S number of partitions a of n such that g an splits over ℝ but not over ℚ. Observe that condition (1) implies that a represents an element of A n . Thus to prove the result it is enough to prove the following two claims: S Claim 1. g an splits over ℝ if and only if a satisfies conditions (1)–(3). S Claim 2. g an splits over ℚ if and only if a satisfies conditions (1)–(3) but does not satisfy condition (4) (i.e. a1 . . . a k is a square). S In fact we first prove that g an splits if and only if a satisfies conditions (1) and (2). This is clear if a1 = 1, so we assume that a1 ≥ 2. S Assume first that g an splits. Then g󸀠a is not a conjugate of g a in A n . Let i = 1, . . . , k. (b i−1 +1,...,b i )(1,2) 󸀠 Then g a = g a and therefore (b i−1 + 1, . . . , b i ) ∈ A n . Hence a i = b i − b i−1 is odd, i.e. (1) holds. If a i = a i+1 with 1 ≤ i < k then (b i−1 + 1, b i + 1)(b i−1 + (b +1,b i +1)(b i−1 +2,b i +2)⋅⋅⋅(b i ,b i+1 )(1,2) 2, b i + 2) ⋅ ⋅ ⋅ (b i , b i+1 )(1, 2) ∈ A n , by (1). Then g 󸀠a = g a i−1 is conjugate to g a in A n , contradicting the hypothesis. This proves (2). Conversely, assume that a satisfies conditions (1) and (2) and write σ i = (b i−1 + 1, . . . , b i ) and σ󸀠1 = (2, 1, 3, . . . , a1 ). Hence g a = σ1 ⋅ ⋅ ⋅ σ k and g󸀠a = σ󸀠1 σ2 ⋅ ⋅ ⋅ σ k . Recall that the support of a permutation σ ∈ S n is Supp(σ) = {x : σ(x) ≠ x}. Assume S that g an does not split. Then g󸀠a = hg a h−1 for some h ∈ A n . Then h permutes the supports of the σ i ’s. By (2) these supports have pairwise different cardinality and hence the supports of each σ i is invariant under h. Thus h = h1 . . . h k , with Supp(h i ) ⊆ −1 = σ 󸀠 = (2, 1, 3, . . . , a ) Supp(σ i ), (h1 (1), h1 (2), . . . , h1 (a1 )) = h1 σ1 h−1 1 1 1 = hσ 1 h and h i commutes with σ i for every i > 1. The latter implies that h i ∈ ⟨σ i ⟩ ⊆ A n if i > 1. Hence h1 ∈ A n . If h1 (1) = 1 then h1 = (2, 3, 4, . . . , b1 ) ∈ ̸ A n , by (1). If h1 (1) = 2 then h1 = (1, 2) ∈ ̸ A n . Therefore, h1 (1) = i ∈ {3, . . . , a1 } and thus h1 (2) = i + 1, . . . , h1 (a1 − i + 1) = a1 , h1 (a1 − i + 2) = 2, h1 (a1 − i + 3) = 1,

7.1 The group of central units of an order

| 235

h1 (a1 − i + 4) = 3, . . . , h1 (a1 ) = i − 1. The sign of this permutation is (−1)1+(a1 −i+1)(i−1) . Then 0 ≡ 1 + (a1 − i + 1)(i − 1) ≡ 1 + i(i + 1) = 1 + i2 + i ≡ 1 mod 2, because a1 is odd S by (1). This yields a contradiction. Thus g an splits as desired. S Now we prove Claim 1. Assume that g an splits (equivalently (1) and (2) hold). For every i = 1, . . . , k let h i = (b i−1 +1, b i )(b i−1 +2, . . . , b i −1) ⋅ ⋅ ⋅ (b i−1 + a i2−1 , b i−1 + a i2+3 ). Observe that h i ∈ A n if and only if a i ≡ 1 mod 4. Let r be the number of a i ’s with a i ≡ 3 mod 4. Since each a i is congruent to either 1 or 3 modulo 4 we have n = ∑ki=1 a i ≡ Sn 3r + (k − r) ≡ 2r + k mod 4. Moreover (h1 ⋅ ⋅ ⋅ h k )g a (h1 ⋅ ⋅ ⋅ h k )−1 = g−1 a . Therefore g a S splits over ℝ if and only if h1 ⋅ ⋅ ⋅ h k ∈ A n , or equivalently r is even. Consequently g an splits over ℝ if and only if n ≡ k mod 4, as desired. Before proving Claim 2, we associate to each cycle σ = (x1 , . . . , x h ) ∈ S n and each positive odd prime p a permutation τ σ,p ∈ S n . If m ∈ ℤ then m mod h denotes the unique integer i ∈ {1, . . . , h} with m ≡ i mod h. Let r = v p (h) = max{i ≥ 0 : p i | h} and q = phr . By Theorem 1.6.5, U(ℤ/p r ℤ) is cyclic. Fix t ∈ ℤ such that t ≡ 1 mod q and t represents a generator of U(ℤ/p r ℤ). Then τ σ,p ∈ S n is defined as follows: τ σ,p (x) = {

x ti x,

mod h ,

if x = x i for some i ∈ {1, . . . , v}; if x ∈ ̸ {x1 , . . . , x h }.

In other words, τ σ,p permutes the x i ’s according to multiplication by t modulo h on the subindex’s. Let σ and p be as above. Moreover, assume that h is odd, and set τ = τ σ,p . Observe that σ t = τστ−1 because if i ≡ tj mod h then τστ−1 (x i ) = τσ(x j ) = τ(x j+1 ) = x i+t = σ t (x i ), where all induces l have to be interpreted as l mod h. For every j = 1, . . . , h, let O(j) denote the ⟨τ⟩-orbit containing j. By the definition of τ and t we have O(j) = {x i : j ≡ t x i = {x i : j ≡ i

mod h, for some x} mod q and gcd(j, p r ) = gcd(i, p r )}

= {x uj : u ≡ 1 mod q and p ∤ u}. If gcd(j, p r ) = p v then the projection ℤ/hℤ → ℤ/p r−v ℤ restricts to a bijection {u ∈ ℤ/hℤ : u ≡ 1 mod q and p ∤ u} → U(ℤ/p r−v ℤ) and hence 1, p r−v − p r−v−1 ;

|O(j)| = φ(p r−v ) = {

if v = r; if v < r.

Thus the partition associated to τ σ,p is formed by (p − 1)p r−1 , (p − 1)p r−2 , . . . , (p − 1)p, (p − 1), each entry appearing q times, and n − q ones. As p is odd, (p − 1)p v is even for every 0 ≤ v < r. Hence τ is a product of rq cycles of even length and hence τ ∈ A n if and only if either h or r is even.

236 | 7 Central units Let now σ = σ1 . . . σ k with σ1 , . . . , σ k disjoint cycles of different odd lengths a1 , . . . , a k and p an odd prime. For every i = 1, . . . , k, let r i = v p (a i ) and q i = parii . Set r = max{r1 , . . . , r k } and fix t ∈ ℤ such that t ≡ 1 mod q i for every i and t represents a generator of U(ℤ/p r ℤ). Then t also represents a generator of U(ℤ/p r i ℤ) for every i. Let τ = τ σ1 ,p . . . τ σ k ,p . By the previous paragraph τ ∈ A n if and only if ∑ki=1 r i is even. In that case σ t = τστ−1 is a conjugate of σ in A n . Otherwise σ t = τστ−1 = τ(1, 2)σ󸀠 (1, 2)τ−1 is a conjugate of σ󸀠 is A n and hence σ t is not a conjugate of σ in A n . Assume that a = [a1 , . . . , a k ] satisfies (1)-(3). Let p1 , . . . , p m be the prime integers dividing some a i . For every i = 1, . . . , m, let t i be the t of the previous paragraph for t σ = g a and p = p i . Then g ai is conjugate to g a in A n if and only if ∑ki=1 v p j (a i ) is even. tj

If ∏ki=1 a i is a square then ∑ki=1 v p j (a i ) is even for every j = 1, . . . , m and hence g a tj

is conjugate to g a in A n for every j = 1, . . . , k. Note that each g a has the same cycle structure as g a . Hence we may repeat the previous to this element and we obtain that if t is obtained by multiplying the t i ’s then g at is a conjugate of g a for j, l ∈ {1, . . . , m}. Let d = |g a | = lcm(a1 , . . . , a k ). By the Chinese Remainder Theorem t1 , . . . , t m generate S U(ℤ/dℤ). Thus g an splits over ℚ. However, if ∏ki=1 a i is not a square then ∑ki=1 v p j (a i ) is tj

odd for some j = 1, . . . , m. Then g a is not conjugate to g a in A n for some j = 1, . . . , m. S We conclude that g an does not split over ℚ. This finishes the proof of Claim 2.

Problems 7.1.1. Let O be an order in a finite dimensional rational algebra. Prove that U(Z(O)) ⊆ Z(U(O)) and find an example in which the inclusion is strict. 7.1.2. Let g and h be elements of order n of a finite group G and F a field of characteristic not dividing the order of G. Prove that g ∼F h if and only if g is a conjugate of h m for some m ∈ UF (n). 7.1.3. Let G be a finite group and g, h ∈ G. Prove that (1) g and h are ℚ-conjugate in G if and only if the cyclic groups ⟨g⟩ and ⟨h⟩ are conjugate in G. (2) The number of ℚ-conjugacy classes of G coincides with the number of conjugacy classes of cyclic subgroups in G. (3) g and h are ℝ-conjugate in G if and only if g is a conjugate of h or h−1 . 1 (4) The number of ℝ-conjugacy classes of G is c+c 2 , where c is the number of conjugacy classes of G and c1 is the number of conjugacy classes of G which are closed under inverses. 7.1.4. Prove that Z(U(ℤS n )) = {±1}. 7.1.5. Calculate the rank of the group of central units of U(ℤD2n ).

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237

7.1.6. Prove that Z(U(ℤA n )) is finite if and only if n = 1, 2, 3, 4, 7, 8, 9 or 12 (Theorem 4.6 in [60]). 7.1.7. Calculate the positive integers n such that Z(U(ℤA n )) has rank 1. (Aleev, Kargapolov and Sokolov in [1] also investigated the rank of Z(U(ℤA n )).) 7.1.8. Let G be a finite group and let ω denote the augmentation map ℤG → ℤ. Prove the following properties. (1) (gh − 1) ≡ (g − 1) + (h − 1) mod ker(ω)2 . (2) Define a homomorphism G → ker(ω)/ ker(ω)2 . (3) Prove that G/G󸀠 ≅ ker(ω)/ ker(ω)2 . (4) Prove that G ∩ (1 + ker(ω)2 ) = G󸀠 .

7.2 Large subgroups of central units: an algorithm Let O be an order in a finite dimensional semisimple rational algebra A and let T be the group of torsion central units of O. By Theorem 7.1.3, T is finite, it has a complement in U(Z(O)) (i.e. a subgroup H of U(Z(O)) such that U(Z(O)) = T × H) and every complement of T in U(Z(O)) is free abelian. Moreover, the rank of each free abelian subgroup of finite index in U(Z(O)), and in particular of each complement of T in U(ℤ(O)), is r = rℝ (A) − rℚ (A). A system of fundamental central units of O is by definition a basis of a complement of T in U(Z(O)). In case O is commutative we simply call this a system of fundamental units. If K is a number field then a system of fundamental units of the ring of integers of K is usually called a system of fundamental units of K. Calculating the torsion central units and a system of fundamental central units of O provides a complete description of the group of central units. In the particular case of an integral group ring of a finite group G we have T = ±Z(G) by Proposition 7.1.4. In this section we present an algorithm for calculating a system of fundamental central units of an order in a finite number of calculations. This algorithm has been adapted from [29, Section 2.5.3] where the algorithm is presented for the case that A is a number field. The interest of the algorithm is more theoretical than practical because the number of calculations needed to complete the algorithm is finite but usually too large. So, one has an effective algorithm that in general is not efficient. In some cases however it is not very difficult to calculate a basis {v1 , . . . , v k } for a subgroup of finite index of U(ℤO). We will illustrate the method by calculating a system of fundamental central units of ℤG for G = ⟨a⟩5 ⋊ ⟨b⟩4 , with a b = a−1 . Before presenting the algorithm it is convenient to settle the known information about O and A. We assume that we know an integral basis c1 , . . . , c n of O; the Wedderburn decomposition of Z(A), i.e. an isomorphism Z(A) ≅ F1 ×⋅ ⋅ ⋅×F k , where each F i is a number field; and a list of representatives σ i1 , . . . , σ it i of the complex embeddings of F i up to conjugation, for each i = 1, . . . , k. Without loss of generality, we may consider the isomorphism in Z(A) ≅ F1 × ⋅ ⋅ ⋅ × F k as an equality.

238 | 7 Central units

Example 7.2.1. These data are easy to provide for our main example, namely the integral group ring of a finite group G. In this case A = ℚG; the elements of G form an integral basis of ℤG; Z(ℚG) ≅ ∏ki=1 ℚ(χ i ), for some irreducible complex characters χ1 , . . . , χ k of G (see Theorem 3.3.1); and the complex embeddings of ℚ(χ i ) are the elements of Gal(ℚ(χ i )/ℚ). The isomorphism Z(ℚG) ≅ ∏ki=1 ℚ(χ i ) can be precisely described as follows. We say that two irreducible complex characters χ and ψ are equivalent if χ = σ ∘ ψ for some σ ∈ Gal(ℚ(ψ)/ℚ). Then χ1 , . . . , χ k is a set of representatives of the equivalence classes of irreducible complex characters of G. For each i = 1, . . . , k, let n i = χ i (1) and π i the ℚ-linear extension of nχ ii to ℚG → ℚ(χ i ). Then π = ∏ki=1 π i : Z(ℚG) → ∏ki=1 ℚ(χ i ) is an isomorphism of ℚ-algebras (see the proof of Theorem 3.3.1 (4)). In fact, without loss of generality, we may assume that A is commutative. Indeed, the data mentioned above are common for A and Z(A) except for the integral basis of O. To calculate an integral basis of Z(O) from an integral basis c1 , . . . , c n of O we observe that Z(O) = {α = m1 c1 + ⋅ ⋅ ⋅ + m n c n : m1 , . . . , m n ∈ ℤ and αc i = c i α for all i = 1, . . . , n}. Hence, one can effectively calculate an integral basis of Z(O) by solving a system of Diophantine linear equations. For our main example this is not even needed because ̃ =∑ the class sums X x, with X running through the conjugacy classes of G, form x∈X

an integral basis of Z(ℤG) (Proposition 3.1.1). So, in the remainder of the section we assume that we know effectively the following information about an order O in a finite dimensional semisimple commutative rational algebra A: (1) the Wedderburn decomposition A = F1 × ⋅ ⋅ ⋅ × F k , with F i a number field for each i; (2) an integral basis c1 , . . . , c n of O; and (3) a list of representatives σ i1 , . . . , σ it i of the complex embeddings of F i up to conjugation, for each i = 1, . . . , k. For every i = 1, . . . , k, let π i : A → F i denote the projection onto the i-th component. Let {σ1 , . . . , σ t } = {σ ij ∘ π i : i = 1, . . . , k, j = 1, . . . , t i }. The σ i ’s are representatives up to complex conjugation of the homomorphism from A to ℂ. By Example 2.3.1, the trace TrA/ℚ and norm NrA/ℚ , which in this case coincide with the reduced trace and reduced norm respectively, are given by TrA/ℚ (x) = ∑ σ(x) σ∈{σ i ,σ i ,i=1,...,t}

7.2 Large subgroups of central units: an algorithm |

239

and NrA/ℚ (x) = ∏ σ(x), σ∈{σ i ,σ i ,i=1,...,t}

for x ∈ A. This provides an effective method to calculate traces and norms and, in particular, to decide whether a given element x ∈ O is a unit, namely, if NrA/ℚ (x) = ±1 (see Lemma 4.6.9 (2)). It also gives an effective procedure to calculate a dual basis c∗1 , . . . , c∗n of c1 , . . . , c n as explained in Problem 2.3.7. For every i = 1, . . . , k, let r i be the number of real embeddings of F i and s i = t i −r i . Let r = r1 + ⋅ ⋅ ⋅ + r k and s = s1 + ⋅ ⋅ ⋅ + s k . Then t = r + s and the algebra homomorphisms σ1 , . . . , σ t induce an isomorphism (see Example 2.1.13) σ = σ1 × ⋅ ⋅ ⋅ × σ t : ℝ ⊗ℚ A → ℝr × ℂs = (ℝr1 × ℂs1 ) × ⋅ ⋅ ⋅ × (ℝr k × ℂs k ). For every i = 1, . . . , t and a ∈ A such that σ i (a) ≠ 0 we set log|σ i (a)|, l i (a) = { log|σ i (a)|2 = log|σ i (a)| + log|σ i (a)|,

if σ i (A) ⊆ ℝ; otherwise.

As O is an order in A, we know that σ(O) is a full lattice of ℝ ⊗ℚ A and, using Lemma 5.1.3, we can calculate vol(ℝ ⊗ℚ A/O) by calculating the determinant of the matrix formed by the image of σ(c i )’s. We use the standard ℝ-basis of ℝr × ℂs as a reference basis to compute volumes. We also will use the logarithm map (see Section 5.2) l : U(ℝ)r × U(ℂ)s → ℝt (x1 , . . . , x r+s ) 󳨃→ (log|x1 |, . . . , log|x r |, log|x r+1 |2 , . . . , log|x r+s |2 ) We consider σ as an identification. We also consider A as a subset of ℝ ⊗ℚ A, by identifying a ∈ A with 1 ⊗ a (equivalently, we identify a ∈ A with (σ1 (a), . . . , σ t (a))). Via this identification we have l(a) = (l1 (a), . . . , l t (a)) for every a ∈ U(A). We reindex the coordinates of ℝt in block form. For each h = 1, . . . , k one has a block of indices i which are such that σ i = σ hj ∘ π h for some j = 1, . . . , t h . We transfer the block form to the functions l i by denoting l i as l hj with h and j as above. Then l(U(O)) is contained in the real vector space (see Section 5.2) x11 + ⋅ ⋅ ⋅ + x1t1 { { { .. r+s V = {(x11 , . . . , x1t1 , . . . , x k1 , . . . , x kt k ) ∈ ℝ : . { { + ⋅ ⋅ ⋅ + x kt k x k1 {

= =

Clearly V = V1 × ⋅ ⋅ ⋅ × V k , where V i = {(x1 , . . . , x t i ) ∈ ℝt i : x1 + ⋅ ⋅ ⋅ + x t i = 0} = l ({x ∈ U(ℝ ⊗ℚ F i ) = U(R)r i × U(ℂ)s i : Nrℝ⊗ℚ A/ℝ (x) = ±1}) .

0} } } . } } } 0}

240 | 7 Central units V is a subspace of ℝt of codimension k and l(U(O)) is a full lattice of V, because the rank of U(O) is t − k and U(O) ∩ ker(l) is the group of torsion units of O (Lemma 5.2.3). Hence u1 , . . . , u t−k generates a subgroup of finite index in U(O) if and only if l(u1 ), . . . , l(u t−k ) are linearly independent (equivalently, form a basis of V). The strategy to calculate a system of fundamental units of O has three steps. Step 1. Calculate an upper bound for vol(V/l(U(O))). Step 2. Use this bound to calculate a positive real number ρ such that the ball of V with center 0 and radius ρ contains a basis of l(U(O)). Step 3. Use ρ to calculate a system of fundamental units of O. Step 1. Before determining an upper bound for the volume, it is convenient to introduce some notation. For every integer m ≥ 1 and 0 < α ∈ ℝ we consider the simplex S m,α = {(x1 , . . . , x m ) ∈ ℝm : x1 + ⋅ ⋅ ⋅ + x m = 0 and x1 , . . . , x m ≤ α}.

(7.2.1)

As a subset of ℝm , the volume of S m,α is 0. We let vol(S m,α ) denote the volume of S m,α as a subset of the hyperplane with equation x1 + ⋅ ⋅ ⋅ + x m = 0. By convention, vol(S1,α ) = 1. If m ≥ 2 then, by Problem 7.2.1, vol(S m,α ) =

(αm)m−1 √ 2. (m − 1)!

Let C i = vol(S t i ,1 ) = √2

vi

(7.2.2)

t −1

ti i , (t i − 1)!

where v i = 0 if F i = ℚ and v i = 1 otherwise. To calculate an upper bound for the volume vol(V/l(U(O))), we first reduce this problem to the case that k = 1, i.e. we may assume that O an order in a field. To do so, we first calculate for every 1 ≤ i ≤ k an order Oi in F i such that O ⊆ O󸀠 = O1 × ⋅ ⋅ ⋅ × Ok . For example, for a given i, we can effectively calculate a basis of the additive group Oi generated by all π i (c j ). As Oi = π i (O), the group Oi is an order in F i and O ⊆ O1 × ⋅ ⋅ ⋅ × Ok . Then, one can calculate a positive integer m such that mO󸀠 ⊆ O. To do so, one expresses the elements of the calculated basis of Oi as a rational linear combination of the c i ’s. Then, for m one can take the greatest common divisor of the denominators of the coefficients in these linear combinations. By Lemma 4.6.9 (4), [U(O󸀠 ) : U(O)] ≤ [O󸀠 : mO󸀠 ] = mdimℚ (A) . Therefore, by Lemma 5.1.3, we have vol(V/U(O)) ≤ mdimℚ (A) vol(V/U(O󸀠 )) ≤ mdimℚ (A) ∏ki=1 vol(V i /U(Oi )). So, to execute Step 1, it is sufficient to find an upper bound for each vol(V i /U(Oi )). We fix an index i with 1 ≤ i ≤ k. As was mentioned above, we can effectively compute vol(ℝ ⊗ℚ F i /Oi ) and hence we can effectively calculate a real number Q i satisfying π s Q i > 4s vol(ℝ ⊗ℚ F i /Oi ).

(7.2.3)

7.2 Large subgroups of central units: an algorithm |

241

Then, an upper bound for vol(V i /l(U(Oi ))) is given by the following inequality t −1

vol(V i /l(U(Oi ))) ≤ C i log(Q i )t i −1 (∑ i[F i :ℚ]) )t i i

.

(7.2.4)

1≤i≤Q i

Proof of (7.2.4). To simplify notation in the proof, put O = Oi , A = F = F i , t = t i , r = r i , s = s i , Q = Q i , V = V i , C = C i and l j = l ij . Let R be a set of representatives, up to associates in O, of the elements of q ∈ O with 0 < |N F/ℚ (q)| ≤ Q. In other words if a ∈ O and 0 < |NrF/ℚ (a)| ≤ Q then a = qu for unique q ∈ R and u ∈ U(O). Recall from Lemma 4.6.9 that NrF/ℚ (a) ∈ ℤ for a ∈ O and from Lemma 4.6.11 that |R| ≤ ∑Qi=1 i[F:ℚ] . Consider the set 1

X = {x = (x1 , . . . , x t ) ∈ ℝr × ℂs : |x j | < Q t ,

if σ j (F) ⊆ ℝ;

1 t

otherwise.}.

|x j | < Q , 2

Clearly, if x ∈ X then |Nrℝ⊗ℚ A/ℝ (x)| < Q. Then X satisfies the hypothesis of Minkowski’s Theorem (Theorem 5.1.4) for O, considered as a lattice in ℝ ⊗ℚ F = ℝr × ℂs . Indeed, condition (2) obviously holds. Moreover by the choice of Q in (7.2.3) we have vol(X) = 2r π s Q > 2r+2s vol(ℝ ⊗ℚ F/O) and r + 2s = dimℝ (ℝ ⊗ℚ F). For every a ∈ U(F), we set D(a) = {(x1 , . . . , x t ) ∈ V : x j ≤

log(Q) − l j (a)}. t

We now calculate the volume of D(a) as subset of V. For every j = 1, . . . , t, let log|NrF i /ℚ (a i )| . Then y = (y1 , . . . , y t ) ∈ V and y + D(a) = S t,c with y j = l j (a) − ti c=

1 t

Q log ( |NrF/ℚ (a)| ). It then follows from (7.2.2) that t−1 1 Q vol(D(a)) = vol(y + D(a)) = vol(S t,c ) = C ( log ) . t |NrF/ℚ (a)|

Let Y = {x ∈ U(ℝ)r × U(ℂ)s : l(x) ∈ V and σ(q)x ∈ X for some q ∈ R}. − l j (q) for every If y ∈ Y and l(y) = (y1 , . . . , y t ) then there is q ∈ R such that y j < log(Q) t j = 1, . . . , t. Thus l(Y) ⊆ ∪q∈R D(q). Clearly U(O)Y ⊆ l−1 (V). We now prove that the equality holds. Indeed, let y ∈ −1 l (V). Then vol(y−1 X) = vol(X), by (5.1.2). Hence y−1 X also satisfies the conditions of Minkowski’s Theorem for the lattice O. Thus X ∩ yO ≠ {0}, that is, there exists a ∈ O with 0 ≠ x = yσ(a) ∈ X ∩ yO. Then, by (2.3.4), we have Q > |Nrℝ⊗ℚ F/ℝ (x)| = |NrF/ℚ (a)|. Then, y = σ(u)xσ(q−1 ) and xσ(q−1 ) ∈ Y because x ∈ X and l(xσ(q−1 )) = l(yu−1 ) ∈ V. This proves the claim.

242 | 7 Central units Thus V = l(U(O)) + l(Y), in other words l(Y) contains a set of representatives of V/l(U(O)). Therefore, Q 1 vol(V/l(U(O))) ≤ vol(l(Y)) ≤ ∑ vol(D(q)) = C ∑ ( log ) t |Nr (q)| F/ℚ q∈R q∈R ≤ C(log(Q))t−1 |R|

t−1

1 1 ≤ C(log(Q))t−1 (∑ i[F:ℚ] ) t−1 , t t−1 t 1≤i≤Q

where for the last inequality we have used Lemma 4.6.11 and the fact that N F/ℚ (q) ∈ ℤ for each q ∈ R. This finishes the first step for the calculation of a system of fundamental units of O. In the remainder of the section we will frequently use the expression “X is effectively described” for a subset X of a given set Y. By this we mean that there is a finite procedure to decide whether a given element of Y belongs to X. Note that this does not mean that this effective procedure is an efficient one for implementation. If X is finite we also use the expression “the elements of X can be effectively computed” to say that there is a finite procedure to exhibit the list of elements of X. Observe that if X is finite and can be effectively computed and X 1 is a subset of X which can be effectively described then X1 can be effectively computed by simply running on the list of elements of X and selecting those belonging to X1 . Step 2. For the second step we need two lemmas. Lemma 7.2.2. (1) If X is a bounded subset of ℝ ⊗ℚ A which is effectively described and assume one knows ρ > 0 such that |σ i (x)| ≤ ρ for all x ∈ X and for every i = 1, . . . , t then X ∩ O and X ∩ U(O) can be effectively computed. (2) If X is a bounded subset of V which is effectively described and one knows ρ > 0 such that X ⊆ B(0; ρ) then X ∩ l(U(O)) can be effectively computed. Proof. (1) Recall that c1 , . . . , c n is an integral basis of O. Let c∗1 , . . . , c∗n be the dual 󵄨 󵄨 basis of c1 , . . . , c n and let β be an upper bound for {󵄨󵄨󵄨TrA/ℚ (c∗i )󵄨󵄨󵄨 : i = 1, . . . , n}. If x = x1 c1 + ⋅ ⋅ ⋅ + x n c n ∈ X with all x i ∈ ℤ then 󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨 |x i | = 󵄨󵄨󵄨TrA/ℚ (xc∗i )󵄨󵄨󵄨 = 󵄨󵄨󵄨󵄨 𝛾(xc∗i )󵄨󵄨󵄨󵄨 ≤ β ρ. ∑ 󵄨󵄨𝛾∈{σ ,σ ,i=1,...,t} 󵄨󵄨 i i To calculate the elements of X ∩ O we first select the elements x ∈ O of the form x1 c1 + ⋅ ⋅ ⋅ + x n c n with x1 , . . . , x n integers of absolute value at most βρ. This provides a finite list containing all the elements of X ∩ O. As, by assumption X is effectively described, we then simply select those elements of the list which belong to X. This results in the list of elements of X ∩ O. To calculate X ∩ U(O), we select the x ∈ X ∩ O satisfying NrA/ℚ (x) = ±1 (see Lemma 4.6.9).

7.2 Large subgroups of central units: an algorithm

|

243

(2) Using ρ one can calculate a positive integer N such that X is contained in Y = {x ∈ V : |x i | ≤ log(N), for i = 1, . . . , r and |x i | ≤ log(N 2 ) for i = r + 1, . . . , t}. Furthermore Y ∩ l(U(O)) = l(Z) for Z = {u ∈ U(O) : |σ i (u)| ≤ N, for every i = 1, . . . , t}. By (1), Z can be effectively computed. As X∩l(U(O)) ⊆ l(Z), we conclude that X∩l(U(O)) can be effectively computed. Lemma 7.2.3. Let L be a full lattice in a real vector space V. Assume that an upper bound of vol(V/L) is known. Furthermore, suppose that for every effectively described bounded subset X of V with a known ρ1 > 0 such that X ⊆ B(0; ρ1 ), the intersection X ∩ L can be effectively computed. Then, there is an effective way to compute a positive real number ρ such that B(0; ρ) contains an integral basis of L. Proof. Let C be the known upper bound of vol(V/L). We argue by induction on n = dim(V). If n = 1 then ρ = C satisfies the required condition. Assume n > 1 and choose a non-zero u1 ∈ L of minimal norm. Such an element can be effectively calculated by applying the hypothesis of the lemma to a ball centered at 0 and with volume greater than 2n C. By Minkowski’s Theorem (Theorem 5.1.4) such a ball contains a non-zero element in L and u1 is any non-zero element of minimal norm that belong to the intersection of the ball and L. Let V1 be the subspace of V orthogonal to u1 and let π : V → V1 be the orthogonal projection. We claim that L1 satisfies the assumptions of the lemma as a lattice in V1 . First of all L ∩ ker(π) = ℤu1 and hence L1 is full lattice of V1 . Let u󸀠2 , . . . , u󸀠n be a basis of L1 . Then u1 , u󸀠2 , . . . , u󸀠n is a basis of a full lattice L󸀠 of V containing L. 󸀠 ) ≤ vol(V/L) ≤ |uC1 | . Thus Since u1 is orthogonal to L1 , we have vol(V1 /L1 ) = vol(V/L |u1 | |u1 | L1 satisfies the first hypothesis of the Lemma. Let now X1 be a bounded subset of V1 effectively described and X1 ⊆ B(0; ρ1 ) for some known ρ1 > 0. Let X = {x1 + 𝛾u1 : x1 ∈ X1 , |𝛾| ≤ 12 }. Clearly X is effectively described and it is contained in B(0, √ ρ1 + |u41 | ). By assumption, X ∩ L can be effectively computed. For every x1 ∈ X1 ∩ L1 there is 𝛾 ∈ ℝ such that x1 + 𝛾u1 ∈ L. As x1 + (𝛾 + n)u1 ∈ L for every n ∈ ℤ, replacing 𝛾 by some 𝛾 + n, we may assume without loss of generality that |𝛾| ≤ 12 and hence x = x1 + 𝛾u1 ∈ X ∩ L. Thus, x1 = π(x) ∈ π(X ∩ L). This proves that X1 ∩ L1 ⊆ π(X ∩ L). As the reverse inclusion is clear, we have shown that X1 ∩ L1 = π(X ∩ L) and hence X1 ∩ L1 can be effectively computed. This shows that the second hypothesis of the lemma also is satisfied. Thus we have proved the claim. By the induction hypothesis one can effectively compute a positive real number ρ1 such that L1 has a basis u󸀠2 , . . . , u󸀠n formed by elements of norm at most ρ1 . We claim that |u1 |2 ρ = max(|u1 |, √ + ρ21 ) 4 satisfies the desired condition, i.e. B(0; ρ) contains a basis of L. For every i = 2, . . . , n select u i ∈ L such that π(u i ) = u󸀠i . For every i = 2, . . . , n, there is k i ∈ ℤ such that 2

244 | 7 Central units u i = k i u1 + u󸀠i . Replacing k i by k i − n i for a suitable integer n i , we may assume that |k i | ≤ 12 for every i = 2, . . . , n. Then | u1 |≤ ρ and |u i |2 = k2i |u1 |2 + |u󸀠i |2
0 and suppose that U 󸀠 ⊆ ⟨U⟩. Write u󸀠i = ∏j u j with each a ij ∈ ℤ. Let A = (a ij ). Then B U 󸀠 = AB U and hence R U 󸀠 = |det(A)| R U . By Lemma 5.1.3 R [⟨U⟩ : ⟨U 󸀠 ⟩] = |det(A)| = RUU󸀠 . Thus U is a system of fundamental units of O if and only if l(U) = (l(u1 ), . . . , l(u r )) form a basis of l(U(O)), or equivalently the regulator of U is minimal among all the non-zero regulators of subsets of X ∩ U(O) of cardinality t − k. We conclude that we can find one system of fundamental units of O by calculating the

7.2 Large subgroups of central units: an algorithm |

245

regulator of each subset of t − k elements of X ∩ U(O) and selecting one with minimal positive regulator. This finishes the explanation of the program for calculating a system of fundamental units of O. It is important to mention that, although an effective algorithm has been described in order to obtain a set of fundamental units, in general this is not efficient as a large amount of computations are required, even for the case that the algebra A is a number field. We refer the reader to [29, Remark, page 118] and [44, Conclusion 4.9.3] for more detail and comments on this matter. In some cases some of these computations can be avoided and simplified. We give an example. Example 7.2.4. Let G = ⟨a⟩5 ⋊ ⟨b⟩4 with a b = a−1 . Put i = √−1, α = ζ5 + ζ5−1 =

and α󸀠 = ζ52 + ζ5−2 =

−1−√5 2 .

−1+√5 2

The character table of G is

1

b2

b

b3

a

ab2

a2

a2 b2

χ1

1

1

1

1

1

1

1

1

χ2 χ3 χ3 χ4 χ󸀠4 χ5 χ󸀠5

1 1 1 2 2 2 2

−1 −1 −1 2 2 −2 −2

−1 i −i 0 0 0 0

−1 −i i 0 0 0 0

1 1 1 α α󸀠 α α󸀠

1 −1 −1 α α󸀠 −α −α󸀠

1 1 1 α󸀠 α α󸀠 α

1 −1 −1 α󸀠 α −α󸀠 −α

Using this table, we calculate the primitive central idempotents of ℚG, as explained in Theorem 3.3.1. G e1 = eℚ (χ1 ) = ̂ ̂̂ e2 = eℚ (χ2 ) = a b2 − ̂ G ̂−a ̂̂ e3 = eℚ (χ3 ) = e(χ3 ) + e(χ3 ) = a b2 ̂) e4 = eℚ (χ4 ) = e(χ4 ) + e(χ󸀠4 ) = ̂ b2 (1 − a ̂) e5 = eℚ (χ5 ) = e(χ5 ) + e(χ󸀠5 ) = (1 − ̂ b2 )(1 − a We have Z(ℚG)e1 ≅ ℚ(χ1 ) = ℚ = ℚ(χ2 ) ≅ Z(ℚG)e2 , Z(ℚG)e3 ≅ ℚ(χ3 ) = ℚ(i), Z(ℚG)e4 ≅ ℚ(χ4 ) = ℚ(α) = ℚ(√5) = ℚ(χ5 ) ≅ Z(ℚG)e5 . So, the center of ℚG is isomorphic to ℚ × ℚ × ℚ(i) × ℚ(√5) × ℚ(√5) (see Theorem 3.3.1). An isomorphism f : Z(ℚG) → ℚ × ℚ × ℚ(i) × ℚ(√5) × ℚ(√5) is determined by its action on the class sums and, by Example 7.2.1, it can be described by setting G ) = |g G | (χ (g), χ (g), χ (g), f(g̃ 1 2 3

χ4 (g) χ5 (g) , ), 2 2

with g ∈ G. For example, ̃G ) = (5, −5, 5i, 0, 0) f(b

and

̃G ) = (2, 2, 2, α, α) . f(a

246 | 7 Central units The ring of integers of ℚ is ℤ and U(ℤ) = {±1}. The ring of integers of ℚ(i) is ℤ[i] and U(ℤ[i]) = {±1, ±i}, by Example 1.3.1. Because of Example 4.1.9, the ring of integers of ℚ(√5) is a + b √5 ℤ[α] = { : a, b ∈ ℤ, a ≡ b mod 2}. 2 By Dirichlet’s Unit Theorem, U(ℤ[α]) has rank 1 and therefore a system of fundamental units of ℤ[α] is formed by the unique minimal unit of ℤ[α] greater than 1, say β = a+b√5 with a, b ∈ ℤ. As −1 = Nrℚ(α)/ℚ (α) = Nrℚ(α)/ℚ (±β m ) = Nrℚ(α)/ℚ (β)m , for some 2 integer m, we deduce that Nrℚ(α)/ℚ (β) = −1. Moreover, 0 < α < 1 and hence 1 < √5 and α−1 = 1+2√5 . Thus 1 < β ≤ α−1 < 2 and −1 < −β−1 < − 12 . Then β−1 = −a+b 2 −1 −1 0 < a = β − β < 2. Therefore a = 1 and hence b = 1. We conclude β = α , that is α−1 is a fundamental unit of ℤ[α]. Thus α is also a fundamental unit. χ ̃G onto α, it follows that u = a ̃G e is a fundaBecause 24 : ℤG → ℚ(√5) maps a 1 4 ̃ mental units of the maximal order of Z(ℚG)e . Similarly u = a G e is a fundamental 4

2

5

unit of the maximal order of Z(ℚG)e5 . Thus ̂ )̂ b2 v1 = (1−e4 )+u1 = 1+(c5 −1)(1− a

and

̂ )(1−̂ v2 = (1−e5 )+u2 = 1+(c5 −1)(1− a b2 )

form a system of fundamental units of the maximal order of Z(ℚG). In order to calculate the order of v1 and v2 modulo U(ℤG) we perform the following calculations in an interactive GAP session. # G=Semidirect product of C_5 by C_4 F:=FreeGroup("a","b"); a:=F.1; b:=F.2; rel:=[a^5,b^4,b*a*b^-1*a]; G:=F/rel; # QG = group algebra QG := GroupRing(Rationals,G); # f:G-->QG canonical embedding f:=Embedding(G,QG); # cb = canonical basis of QG, i.e. the elements of G as elements of QG cb := CanonicalBasis(QG); # Group algebra elements one := One(QG); a :=G.1; b:=G.2; A:=a^f; B:=b^f; ahat:=(one+A+A^2+A^3+A^4)/5; bhat:=(one+B^2)/2; c5:=A+A^4; v1:=one+(c5-one)*(one-ahat)*bhat; v2:=one+(c5-one)*(one-ahat)*(one-bhat); # Calculating the orders of v1 and v2 modulo ZG: Initiate x=vi and # multiply x by vi until the coefficients of x are integers x:=v1; n:=1; while not ForAll(Coefficients(cb,x), z -> z in Integers) do x:= x*v1; n:=n+1; od;

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247

Print("The order of v1 modulo U(ZG) is ", n,"\n"); x:=v2; n:=1; while not ForAll(Coefficients(cb,x), z -> z in Integers) do x:= x*v2; n:=n+1; od; Print("The order of v2 modulo U(ZG) is ", n,"\n");

The answer of the interactive session is the following: The order of v1 modulo U(ZG) is 12 The order of v2 modulo U(ZG) is 12 12 We conclude that v12 1 and v 2 form a basis for a subgroup of finite index in Z(U(ℤG)). We can now refine this basis and obtain a system of fundamental units of Z(U(ℤG)) as n n follows. Let u = v11 v22 ∈ Z(U(ℤG)) with n1 , n2 ∈ ℤ. If r i is the remainder of n i divided r1 r2 by 12 then v1 v2 ∈ Z(U(ℤG)). The output of the following calculation shows that if r1 and r2 are not multiples of 12 then either r1 = r2 = 4 or r1 = r2 = 8.

Filtered(Cartesian([1..11],[1..11]), x-> ForAll(Coefficients(cb,v1^x[1]*v2^x[2]), z -> z in Integers));

is [ [ 4, 4 ], [ 8, 8 ] ]

Thus w2 = v41 v42 ∈ ℤG and ⟨v1 , v2 ⟩ ∩ ℤG = ⟨v12 1 , w 2 ⟩. Consequently, 12 w1 = v12 1 = (1 − e 4 ) + c 5 e 4

= 65 + 64b2 − 52(a + a−1 ) − 52(a + a−1 )b2 + 20(a2 + a−2 ) + 20(a2 + a−2 )b2 and ̂ + (a + a−1 )4 (1 − a ̂) w2 = v41 v42 = (1 − e4 − e5 ) + c45 (e4 + e5 ) = a = 3 − 2(a + a−1 ) + (a2 + a−2 ) form a fundamental set of central units of ℤG. By Theorem 7.1.5, we conclude Z(U(ℤ(C5 ⋊ C4 ))) = ± ⟨b2 ⟩ × ⟨w1 ⟩ × ⟨w2 ⟩ ≅ C22 × C2∞ .

Problems 7.2.1. Let 2 ≤ m ∈ ℤ. For every positive real number α let S m,α be as in (7.2.1) and T m,α = {(x1 , . . . , x m ) : x1 + ⋅ ⋅ ⋅ + x m = α and x i ≥ 0}. The volume of S m,α and T m,α are calculated in the hyperplanes ∑m i=1 x i = 0 and m ∑i=1 x i = α respectively. Prove (1) vol(T m,α ) = α m−1 vol(T m,1 ) and vol(S m,α ) = α m−1 vol(S m,1 ).

248 | 7 Central units (2) vol(T m,α ) = (3) vol(S m,α ) =

α m−1 √ (m−1)! 2, where the volume is m−1 √2. m m−1 vol(T m,α ) = (mα) (m−1)!

calculated in the hyperplane ∑ x i = 1.

3 7.2.2. Let O be the ring of integers of ℚ(√ 2). Prove that U(O) has rank one and compute a fundamental unit.

7.2.3. Compute a set of fundamental units of ℚ(ζ7 + ζ7−1 ) and for ℚ(ζ7 ). 7.2.4. Let G = D16 , Q16 or A5 . Prove that Z(U(ℤG)) has rank one and compute a set of fundamental units. 7.2.5. Let G = C13 ⋊ C4 = ⟨a⟩13 ⋊ ⟨b⟩4 with the (faithful) action of C4 on C13 defined by a b = a5 . Compute a set of fundamental units of Z(U(ℤG))

7.3 Bass units as generators of large groups of units: abelian group rings Let G be a finite group. The method to calculate a basis for a subgroup of finite index in Z(U(ℤG)) introduced in Section 7.2 is unsatisfactory, not only because in some cases it includes a large amount of calculations but also because even in the cases where it can be completed the units obtained do not provide sensible information on their connection with the group G. Furthermore, and more important, this method does not give a construction of independent units that works for many (preferably all) finite groups G. So, alternatively, we would like to provide a straightforward method to produce a basis for a subgroup of finite index in Z(U(ℤG)) directly from the group elements. In case G is a finite abelian group such a method can be worked out. It is based on the techniques outlined in Remark 4.6.10 and the facts that one knows a concrete description of the Wedderburn decomposition of ℚG and that finitely many concrete generators are described for a subgroup of finite index in the unit group of the ring of integers of these simple components. Indeed, by the Theorem of Perlis and Walker (Theorem 3.3.6), there is an isomorphism f : ℚG → ∏d|n ℚ(ζ d )k d , where k d denotes the number of cyclic subgroups of G of order d. As ℤ[ζ d ] is the ring of integers of ℚ(ζ d ) (Proposition 6.1.2 (3)), this isomorphism maps ℤG into ∏d|n ℤ[ζ d ]k d . Moreover, by Lemma 4.6.9 (4), the image f(U(ℤG)) has finite index in ∏d|n U(ℤ[ζ d ])k d . By Corollary 6.2.3, the cyclotomic units generate a subgroup of finite index in each U(ℤ[ζ d ]). Therefore, there is a positive integer n such that η n ∈ f(ℤG) for every η ∈ ∏d|n U(ℤ[ζ d ])k d which is a cyclotomic unit in one of the factors and 1 in all the other factors. If u η ∈ U(ℤG) is such that f(u η ) = η n then the finitely many units u η of ℤG generate a subgroup of finite index in U(ℤG). In this section we prove a theorem of Bass and Milnor which says that if G is a finite abelian group then Bass(G), the group generated by the Bass units, has finite index in U(ℤG). We present a proof that is different from the original proof as it does not make

7.3 Bass units as generators of large groups of units

249

|

use of K-theoretic methods. Namely, we follow the approach given in [108] where, as well as proving Bass-Milnor Theorem, one provides a basis, formed by Bass units, of a free abelian subgroup of finite index in U(ℤG). The latter generalizes a result of Bass who obtained such basis in case G is a cyclic group [16]. The proof presented only uses elementary methods and a double induction argument to precisely discover the units u η explained above. It turns out that u η belongs to Bass(G), and thus the Bass-Milnor result follows at once. We start recalling from (1.2.1) the notation for cyclotomic units η k (ξ) =

ξk − 1 = 1 + ξ + ξ 2 + ⋅ ⋅ ⋅ + ξ k−1 , ξ −1

where ξ is a root of unity different from 1 and k is a positive integer coprime with the order of ξ . We extend this notation by setting η k (1) = 1. Throughout this section G is a finite abelian group. We also recall the notation for Bass units (1.2.3): u k,m (g) = (1 + g + g 2 + ⋅ ⋅ ⋅ + g k−1 )m +

1 − km g̃ , |g|

where g ∈ G and k and m are positive integers such that k m ≡ 1 mod |g|. We need to fix an isomorphism f : ℚG → ∏d|n ℚ(ζ d )k d . To do so, let H = H(G) denote the set of subgroups C of G such that G/C is cyclic. For every subgroup C ∈ H we fix a linear representation ρ C of G with kernel C. We also denote by ρ C the linear extension of ρ C to ℚG. If d = [G : C] then ρ C (ℚG) = ℚ(ζ d ). Then f = ∏ ρ C : ℚG 󳨀→ ∏ ℚ(ζ[G:C] ) C∈H

C∈H

is an isomorphism (see the proof of Theorem 3.3.6). Recall the following equalities from (1.2.4), (1.2.6), Problem 1.2.2 and (1.5.8): (if k ≡ k󸀠

u k,m (g) = u k󸀠 ,m (g),

mod |g|),

(7.3.1)

u k,m (g) u k,m1 (g) = u k,m+m1 (g),

(7.3.2)

u k,m (g) u k󸀠 ,m (g ) = u kk󸀠 ,m (g), k

(7.3.3)

ρ C (u k,m (g)) = η k (ρ C (g)) , m

(7.3.4)

with C ∈ H, g ∈ G, k, k1 and k󸀠 positive integers so that k m ≡ (k󸀠 )m ≡ k m1 ≡ k1 1 ≡ 1 mod |g|. Furthermore, we have m

n−1

∏ (1 − Xζ ni ) = 1 − X n i=0

(7.3.5)

250 | 7 Central units

Let ξ be a root of unity and assume that k is coprime with n and the order of ξ . If ξ n ≠ 1 then, using (7.3.5) we obtain n−1

n−1

∏ η k (ξζ ni ) = ∏ i=0

1 − ξ k ζ nki 1−

i=0

=

ξζ ni

k i ∏n−1 i=0 (1 − ξ ζ n )

∏n−1 i=0 (1

−

ξζ ni )

=

1 − ξ kn = η k (ξ n ). 1 − ξn

j

Otherwise, i.e. if ξ n = 1, then ξζ n = 1 for some j = 0, 1, . . . , n − 1. Then, using that k is coprime with n, we deduce that {ζ nki : i = 0, 1, . . . , n − 1} = {ζ ni : i = 0, 1, . . . , n − 1} and hence n−1

n−1

∏ η k (ξζ ni ) = ∏ i=0

k(i−j)

1 − ζn

i=0, i=j̸ 1

i−j

− ζn

=

ki ∏n−1 i=1 (1 − ζ n ) i ∏n−1 i=1 (1 − ζ n )

= 1 = η k (ξ n ).

This proves the following equality for every root of unity ξ : n−1

∏ η k (ξζ ni ) = η k (ξ n ) (gcd(k, n|ξ|) = 1).

(7.3.6)

i=0

Lemma 7.3.1. Let C ∈ H and let H be an arbitrary subgroup of G. Set s = |C ∩ H|, t = [H : C ∩ H] and let g ∈ G and k and m be positive integers such that (k, t) = 1 and k m ≡ 1 mod |gh| for every h ∈ H. Then ∏ ρ C (u k,m (gu)) = η k (ρ C (g)t )ms . h∈H

Proof. As C = ker(ρ C ) and CH/C ≅ H/C ∩ H is cyclic of order t, if h runs through the elements of H then ρ C (h) runs through the t-th roots of unity and each t-th root of unity is obtained as ρ C (h) for precisely s elements of H. Therefore, by (7.3.4) and(7.3.6), we have m

t−1

∏ ρ C (u k,m (gh)) = ( ∏ η k (ρ C (g)ρ C (h))) = (∏ η k (ρ C (g)ζ ti )) h∈H

h∈H

ms

= η k (ρ C (g)t )ms ,

i=0

as desired. For an integer k, let Bassk (G) denote the subgroup of U(ℤG) generated by the Bass units of the form u k,m (g) with g ∈ G and k m ≡ 1 mod |g|. The proof of the main result of this section mainly relies on the following lemma. Lemma 7.3.2. Let H ∈ H with d = [G : H] and let k, j ∈ ℕ be such that k is coprime j with d. Set η = η k (ζ d ). Then, there exist a positive integer m and b ∈ Bassk (G) such that ρ H (b) = η m and ρ C (b) = 1 for every C ∈ H \ {H}. Proof. Without loss of generality, we may assume that k is coprime with n = |G|. Indeed, by an easy Chinese Remainder argument there is an integer k󸀠 coprime with n j j such that k ≡ k󸀠 mod d. Then η k (ζ d ) = η k󸀠 (ζ d ), by Problem 1.2.1 (3).

7.3 Bass units as generators of large groups of units |

251

We argue by a double induction, first on n and second on d. The cases n = 1 and d = 1 are trivial. We denote by P(G, H) the statement of the lemma for a finite abelian group G and an H ∈ H. Hence the induction hypothesis includes the following statements: (IH1) P(M, Y) holds for a proper subgroup M of G and Y ∈ H(M). (IH2) P(G, H1 ) holds for H1 ∈ H with [G : H1 ] < [G : H] = d. We consider two cases, depending on whether j is coprime with d or not. Case 1: Suppose j is not coprime with d. Let p be a common prime divisor of d j j/p and j. Then H is contained in a subgroup S of G with [G : S] = p and ζ d = ζ d󸀠 󸀠 with d = [S : H]. For every C ∈ H, let λ C denote the restriction of ρ C to ℚS. Clearly λ C is the ℚ-linear extension of a linear representation of S with kernel S ∩ C. Since S/(S ∩ C) ≅ CS/C and CS/C is a subgroup of G/C we deduce that S/(S ∩ C) is cyclic. Thus C 󳨃→ S ∩ C defines a map H → H(S). This map is surjective, but maybe not injective. Indeed, let C1 ∈ H(S). If C1 ∈ H then clearly the map associates C1 with itself. Otherwise, G/C1 is not cyclic but S/C1 is a cyclic subgroup of index p in G/C1 . Hence p divides [S : C1 ] and G/C1 = S/C1 × L/C1 for some subgroup L of G containing C1 with [L : C1 ] = p. Then G/L ≅ S/C1 , so that L ∈ H and L ∩ S = C1 . Therefore H(S) = {C ∩ S | C ∈ H}. For every Y ∈ H(S) we choose a C Y ∈ H such that C Y ∩ S = Y in such a way that C Y = Y if Y ∈ H. Then ∏ λ CY Y∈H(S)

: ℚS → ∏ ℚ(ζ[S:Y] ) Y∈H(S)

is an algebra isomorphism. By (IH1) there is b ∈ Bassk (S) such that λ H (b) = η m for some positive integer m and ρ C (b) = λ C (b) = 1 if C ∈ H with C ∩ S ≠ H. If C ∈ H satisfies C ∩ S = H then either C = H or p ∤ [S : H] and C = H1 , where H1 /H is the only subgroup of G/H of order p. Hence, if p ∤ [S : H] then b and m satisfy the desired property. Otherwise, ρ H1 (b) is a product of cyclotomic units, by (7.3.4). By (IH2) there is c ∈ Bassk (G) such that ρ C (c) = 1 for every C ∈ H \ {H1 } and ρ H1 (c) = ρ H1 (b)m1 for some positive integer m1 . Therefore ρ H (b m1 c−1 ) = η mm1 and ρ C (b m1 c−1 ) = 1 for every C ∈ H \ {H}. This finishes the proof for this case. j Case 2: j is coprime with d. Then G = ⟨a, H⟩ and ρ H (a) = ζ d for some a ∈ G. As k is coprime with the order of G, there is a positive integer m such that k m ≡ 1 mod |au| for every u ∈ H. Hence η m = ρ H (u k,m (a)), by (7.3.4). Let b = ∏ u k,m (ah). h∈H

For every C ∈ H(G), set d C = [G : C], d󸀠C = [G : ⟨a, C⟩], s C = |H ∩ C| and t C = [H : H ∩ C].

252 | 7 Central units

Then, by Lemma 7.3.1, we have t C d󸀠C u C m s C

ρ C (b) = η k (ρ C (a)t C )m s C = η k (ζ d C

)

,

for some integer u C coprime with d C . Suppose t C d󸀠C is not coprime with d C . Then, by Case 1, there is b C ∈ Bassk (G) such that ρ C (b C ) = ρ C (b)m C for some positive integer m C and ρ C1 (b C ) = 1 for every C1 ∈ H \ {C}. By (IH2), the same holds if d C < d. Let H󸀠 = {C ∈ H : t C d󸀠C is not coprime with d C or d C < d}.

For each C ∈ H󸀠 fix b C ∈ Bassk (G) and m C ∈ ℤ as above. Let m1 = lcm(m C : C ∈ H󸀠 ) and −m 1

b1 = (∏ b C C )b m1 . m

C∈H󸀠

Then b1 ∈ Bass(G), ρ C (b1 ) = 1 if C ∈ H󸀠 and ρ C (b1 ) ∈ ⟨η k (ρ C (a)t C )⟩ if C ∈ H \ H󸀠 . Observe that t H = d󸀠H = 1 and d H = d. Hence H ∈ ̸ H󸀠 . Therefore, ρ H (b1 ) ∈ ⟨η⟩, because η k (ρ H (a)t H ) = η. To finish the proof we prove that H󸀠 = H \ {H}. Suppose the contrary, that is, assume C ∈ H \ {H} with d C ≥ d and gcd(t C d󸀠C , d C ) = 1. The latter implies that d󸀠C = 1, or equivalently G = ⟨a, C⟩, and t C = [CH : C] is coprime with d C = [G : C]. Consequently, t C = 1, or equivalently H ⊆ C. Hence, the assumption d C = [G : C] ≥ [G : H] = d implies that H = C, a contradiction. Theorem 7.3.3 (Bass-Milnor, Jespers-del Río-Van Gelder). Let G be a finite abelian group. For every cyclic subgroup C of G choose a generator a C of C and for every k coprime with the order of C choose m k,C with k m k,C ≡ 1 mod |C|. Then { u k,m k,C (a C ) : 1 < k
|g| 2 can be excluded, that is B1 has finite index in U(ℤG). Finally we have to prove that the elements of B1 are multiplicatively independent. To prove this it is enough to show that the rank of U(ℤG) coincides with the cardinality of the generating set of B1 . For this observe that this cardinality is ∑d k d t d , where d runs through the divisors of |G|, k d is the number of cyclic subgroups of G of order d and t d is the cardinality of {k : 1 < k < 2d , gcd(d, k) = 1}. Obviously t1 = t2 = 0 and t d = φ(d) 2 − 1 for every d > 2. Therefore, the cardinality of the generating set of k φ(d) 2) d = 1+k2 +n−2c , where h d B1 equals ∑d>2 ( d 2 − k d ) = ∑d>2 ( h d −2k ) = n−1−h2 −2(c−1−k 2 2 2 denotes the number of elements of G of order d (so that h1 = 1 and h2 = k2 ) and c is the number of cyclic subgroups of G. By Theorem 7.1.6, this number coincides with the rank of U(ℤG). This finishes the proof. For convenience we will use the function ι : ℂ → ℂ with ι(z) = z for all z ≠ 0 and ι(0) = 1 As an application of the Bass-Milnor result we prove the Bass Independence Theorem (see [202, Theorem 11.8]). Theorem 7.3.4 (Bass Independence Theorem). Let n be a positive integer. Assume i and κ i be integers so that 0 < i < n and ∑i, (i,n)=d κ i = 0 for all divisors d of n. Let i κi C n = ⟨x⟩. If for all complex linear characters χ of C n one has ∏n−1 i=1 ι(χ(x ) − 1) = 1 then κ i + κ n−i = 0 for all i. Proof. Let m = φ(n). For every 0 < i ≤ n let d i = gcd(i, n) and k i =

i di .

Observe that

n n |x d i | = dni , gcd (k i , dni ) = 1 and k m i ≡ 1 mod d i because φ ( d i ) divides φ(n). Thus d i u i = u k i ,m (x ) is a Bass unit of ℤC n . We claim that m ι(χ(x i ) − 1) . χ(u i ) = ( ) ι(χ(x d i ) − 1)

This is clear if χ(x d i ) = 1, because in this case χ(u i ) = 1 and χ(x i ) = 1. Assume otherwise that χ(x d i ) ≠ 1. Then χ(x i ) ≠ 1. Hence χ(x d i ) is a root of unity, say ξ , different n n d −1 ξ d i −1 from 1 and its order divides n . Thus χ(̂ x di ) = ∑ i ξ j = = 0. Therefore ξ −1

j=0

di

χ(u i ) = (1 + χ(x d i ) + χ(x2d i ) + ⋅ ⋅ ⋅ + χ(x(k i −1)d i ))m = ( m

=( as claimed.

m

χ(x i ) − 1 ι(χ(x i ) − 1) = ) ) , ( χ(x d i ) − 1 ι(χ(x d i ) − 1)

χ(x k i d i ) − 1 ) χ(x d i ) − 1

m

254 | 7 Central units κ

i Let u = ∏n−1 i=1 u i . Then, for every linear character χ of C n , the assumption yields

that n−1

χ(u) = ( ∏ i=1

ι(χ(x i ) − 1)

κi

ι(χ(x d i ) − 1)

κi

m

) =(

i ∏n−1 i=1 ι(χ(x ) − 1)

∏d|n ι(χ(x d ) − 1)

κi

∑i, gcd(i,n)=d κ i

m

) = 1.

Thus u = 1 and thus ∏ ∏ u k,m (x d )κ kd = 1. d|n 1 1}.

We have to prove that R = R0 . Let a denote the leading coefficient of f and let d be the degree of f . We claim that if α ∈ R and p is an integral polynomial of degree m then a m p(α) is an algebraic integer. To prove this it is enough to show that aα is an algebraic integer. d−1 d−i−1 i d ̂ Indeed, write f = aX d + ∑d−1 a i X i . Then i=0 a i X with a i ∈ ℤ and let f = X + ∑i=0 a f ̂ is an integral monic polynomial and f ̂(aα) = a d−1 f(α) = 0. This proves the claim. If x is a group element of order n then ℤC n ≅ ℤ[X]/(X n − 1). Therefore, for every positive integer n such that f defines a unit on order n, there exist p n , q n ∈ ℤ[X] such that f p n + (X n − 1) q n = 1. We claim that p n and q n can be chosen such that the degree of p n is smaller than n and the degree of q h is smaller than d. Indeed, as X n −1 is monic, p n = r n +(X n −1)t n with r n and t n integral polynomials such that r n has degree smaller than n. Let s n = q n + ft n . Then, fr n + (X n − 1)s n = 1. If h is the degree of r n and k is the degree of s n then d + h = n + k. As h < n, we have k < d. Replacing p n and q n by r n and s n we obtain the desired conclusion. Let h n denote the degree of q n . By the previous paragraph, we may assume that h n < d. For every α ∈ R we have that a n (α n − 1) and a h n q n (α) are algebraic integers and (α n − 1)q n (α) = 1.

260 | 8 Generic units

Therefore, (∏ a n (α n − 1))(∏ a h n q n (α)) = a d(n+h n ) . α∈R

α∈R

Clearly, the two factors of the left side of the equation are invariant under the action of the Galois group of the splitting field of f over ℚ. Hence these factors are rational numbers. Hence they are non-zero integers as they also are algebraic integers. We conclude that 2 (8.1.4) ∏ |α n − 1| ≤ |a|dh n ≤ |a|d . α∈R

If α ∈

R−

then limn→∞

αn

= 0 and hence there is a positive real number ρ such that ∏ |α n − 1| > ρ.

(8.1.5)

α∈R−

Let r = |R0 | and let δ=

1 min{|α j − 1| : 1 ⩽ j ⩽ rD, α ∈ R0 } 2

if r ≠ 0 and put δ = 1 otherwise. If δ = 0 then R0 contains a root of unity, as desired. So assume δ > 0. Let T = {n : f defines a unit on order n and |α n − 1| ≥ δ for all α ∈ R0 }. We claim that T is infinite. Otherwise there is a positive integer N such that T does not contain any number n greater than N. Condition (3) ensures that there is an increasing sequence of integers N < n0 < ⋅ ⋅ ⋅ < n r such that f defines a unit on order n i for each i = 0, 1, . . . , r, and n i − n i−1 ≤ D for each i = 1, . . . , r. For every i = 0, 1, . . . , r, n we have n i ∈ ̸ T and hence there is α i ∈ R0 such that |α i i − 1| < δ. Since |R0 | = r, necessarily α i = α j for some 0 ≤ i < j ≤ r. Therefore, there exists α ∈ R0 and there are positive integers n and m with n < m ≤ n + Dr and |α n − 1|, |α m − 1| < δ. As |α| = 1, we have 2δ ≤ |α m−n − 1| = |α m − α n | ≤ |α m − 1| + |α n − 1| < 2δ, a contradiction. Thus, indeed T is infinite and hence it contains an infinite increasing sequence (n k ). So, (8.1.6) ∏ |α n k − 1| ≥ δ r α∈R0

Applying (8.1.4), (8.1.5) and (8.1.6) for each n k we obtain 2

∏ |α n k − 1| ≤ α∈R+

|a|d . ρδ r

If R+ ≠ 0 then the left side of the previous inequality diverges, yielding a contradiction. Thus R+ = 0.

8.1 Shifted cyclotomic polynomials

|

261

Fix is a group element x of order n with n > 2d and such that f defines a unit on order n. It exists by the assumption. Because of Lemma 7.1.8, there exists an integer j n such that − 2n < j n ≤ 2n and f(x−1 )f(x)−1 = x j n . Put a d = a. Clearly, a d x−d + a d−1 x−(d−1) + ⋅ ⋅ ⋅ + a1 x−1 + a0 = a d x d+j n + a d−1 x d−1+j n + ⋅ ⋅ ⋅ + a1 x1+j n + a0 x j n . Obviously, x d+j n and x j n belong to the support of the element on the right side and hence they also belong to the support of the left hand side. Consequently, j n ≤ 0 and d + j n = 0. So, j n = −d and a d x−d + a d−1 x−(d−1) + ⋅ ⋅ ⋅ + a1 x−1 + a0 = a d + a d−1 x−1 + ⋅ ⋅ ⋅ + a1 x−d+1 + a0 x−d . In other words, a i = a d−i , for all i. This symmetry yields that if α is a complex root of f(X), then so is α−1 . Thus 0 = |R+ | = |R− |. Consequently, every root of f has modulus 1 and thus each root of f is a root of unity, as desired. Let K be a number field and let S be a finite set of maximal ideals of the ring of integers R of K. Let R S = ∩P∈Max(R)\S R P . The S-Unit Theorem says that the set of elements x ∈ U(R S ) such that 1 − x ∈ U(R S ) is finite [56]. The proof of this theorem ultimately relies on the celebrated Mordell-Weil Theorem and is beyond the scope of this book. Using this theorem, Broche and del Río proved in [34] that Theorem 8.1.6 can be extended as follows. Theorem 8.1.7 (Broche and del Río). If f ∈ ℤ[X] then f defines generic units if and only if f defines units on infinitely many orders. Proof. Assume that f define units on infinitely many orders. Because of Theorem 8.1.6, we have to prove that every root of f is either 0 or an root of unity. Without loss of generality, we also may assume that f is irreducible and different from ±X. Let α be a complex root of f . Let M be the set of positive integers n such that f defines units on order n. As in the proof of Theorem 8.1.6, if n ∈ M then α n − 1 is a unit in ℤ[α]. Consider the number field K = ℚ(α) and let R denote its ring of integers. Let S be the set of maximal ideals P of R such that v P (α) ≠ 0. Clearly, S is finite. Let R S = ∩P∈Max(R), P∈S̸ R P . Then α is a unit of R S and hence ℤ[α, α−1 ] ⊆ R S . If n ∈ M then α n and 1 − α n are invertible inR S . By the S-Unit Theorem, the set {α n : n ∈ M} is finite. As M is infinite, this implies that α is a root of unity.

262 | 8 Generic units

Problems 8.1.1. Let i and j be positive integers. Prove that (1+ X j +⋅ ⋅ ⋅+ X j(i−1) )(1+ X +⋅ ⋅ ⋅+ X i−1 )−1 is a integral (or rational) polynomial if and only if i and j are relatively prime. 8.1.2. Prove that if f ∈ ℤ[X] defines generic units then f(1) is the leading coefficient of f .

8.2 The group of generic units Let G be a finite group. By Gen(G) we denote the group generated by the units of the form Φ m (x), with m a positive integer and x ∈ G. In other words, by Proposition 8.1.3, we have m Gen(G) = ⟨Φ m (x) : x ∈ G, m ∈ ℕ and is not a prime power⟩. gcd(m, |x|) This group is called the group of generic units of ℤG. Because of Proposition 8.1.3 and Theorem 8.1.6, ⟨G, Gen(G)⟩ is formed by the units of the form f(x) with f an integral monic polynomial defining generic units and x ∈ G. By Lemma 8.1.2 (3) we obtain at once the following statement. Proposition 8.2.1. If G is a finite group then all the elements in Gen(G) have augmentation 1. In this section we compare the group Gen(G) with the group generated by the Bass units of ℤG, denoted Bass(G), and with Hoe(G), the group generated by the Hoechsmann units of ℤG. Recall from Problem 1.2.5 that the Hoechsmann units of ℤG are the elements of the form h j,i (g) = (1 + g j + ⋅ ⋅ ⋅ + g j(i−1) )(1 + g + ⋅ ⋅ ⋅ + g i−1 )−1

1 − ik g̃ , d with g ∈ G, say of order d, and i, j, k ∈ ℤ with gcd(ij, d) = 1 and ik ≡ 1 mod d. = (1 + g j + ⋅ ⋅ ⋅ + g j(i−1) )(1 + g i + ⋅ ⋅ ⋅ + g(k−1)i ) +

Proposition 8.2.2. Let n be a positive integer and C n = ⟨x⟩n . Then Gen(C n )∩C n = ⟨x2 ⟩. Proof. By Dirichlet’s Theorem on Primes in an Arithmetic Progression there exist distinct prime numbers p and q with p + 1, q + 1 ∈ nℤ. As gcd(pq, n) = 1, Corollary 8.1.4 yields that Φ pq (x) ∈ Gen(C n ). Because of (8.1.3), Φ pq (X) (X p −1)(X q −1) = (X−1)(X pq − 1). By the choice of p and q and as x n = 1, we obtain that Φ pq (x) (x−1 − 1)2 = (x − 1)2 and thus (x−2 Φ pq (x) − 1)(1 − x)2 = 0. ̂n )2 = 1 − C ̂n = (1 − x)β, for some β ∈ ℚC n , we thus obtain that As (1 − C ̂n ) = 0. (x−2 Φ pq (x) − 1)(1 − C

8.2 The group of generic units

|

263

̂n = 0 and we conSince ω(x−2 Φ pq (x)) = 1, we get from (1.5.4) that (x−2 Φ pq (x) − 1)C 2 2 clude that Φ pq (x) = x . This proves that ⟨x ⟩ ⊆ Gen(C n ). If n is odd then we get that C n = ⟨x2 ⟩ ⊆ Gen(C n ) and thus (2) follows. On the other hand, if n is even then we claim that x ∉ Gen(C n ), and thus also in this case (2) follows. Indeed, assume n is even and x ∈ Gen(C n ). Let ρ denote the character of C n defined by ρ(x) = −1. Then, x = ∏ Φ m i (x d i )k i , for some integers m i , d i and k i with each m i > 0 and gcd(|xmdii |,m ) i

not a prime power. If d i is even then ρ(Φ m i (x d i )) = Φ m i ((−1)d i ) = Φ m i (1) = 1, by mi Lemma 8.1.2 (3). Otherwise 2 divides |x d i | and hence gcd(2,m is not a prime power, i) because n is even. From Lemma 8.1.2 (4) we know that also in this case ρ(Φ m i (x d i )) = k k Φ m i ((−1)d i ) = Φ m i (−1) = 1. We conclude that −1 = ρ(∏ Φ mi i (x)) = ∏ Φ mi i (ρ(x)) = ki ∏ Φ m i (−1) = 1, a contradiction.

Proposition 8.2.3. If G is a finite group then Gen(G) is generated by the units Φ pq (x), where x ∈ G and p and q are distinct primes with gcd(pq, |x|) = 1. Proof. It is sufficient to show that any unit Φ m (x) (with m a positive integer such that m/ gcd(n, m) is not a prime power) is a product of units of the type Φ pq (x d ) and their α α inverses. Write m = p11 ⋅ ⋅ ⋅ p s s , its prime power factorization. Lemma 8.1.2 (1) yields α −1 α −1 that also Φ m (x) = Φ p1 ⋅⋅⋅p s (x e ) is a unit, where e = p11 ⋅ ⋅ ⋅ p s s . Because of Proposition 8.1.3, at least two of the primes, say p1 and p2 , do not divide the order of x e . Hence, applying the first part of Lemma 8.1.2 (2) with e = p1 p2 and n = p3 . . . p s we deduce that Φ m (x e ) is a product of units of the form Φ p1 p2 (x de ) and their inverses. Theorem 8.2.4 (Marciniak-Sehgal). If G is a finite group then Gen(G) = Hoe(G) Proof. First we show that Gen(G) ⊆ Hoe(G). Because of Proposition 8.2.3, it is sufficient to show that each Φ pq (x) ∈ Hoe(G), for x ∈ G, say of order n, and distinct primes p and q with gcd(pq, n) = 1. By Lemma 8.1.2 (2) Φ pq (x) = Φ q (x p )Φ q (x)−1 = (1 + x p + ⋅ ⋅ ⋅ + x p(q−1) ) (1 + x + ⋅ ⋅ ⋅ + x q−1 )−1 = h p,q (x), a Hoechsmann unit in ℤG. Hence, Φ pq (x d ) ∈ Hoe(G). Conversely, let u = (1 + x j + ⋅ ⋅ ⋅ + x j(i−1) )(1 + x + ⋅ ⋅ ⋅ + x i−1 )−1 be a Hoechsmann unit in ℤG, with x and element of G of order n, say, and i and j positive integers with gcd(ij, n) = 1. As 1 + X + ⋅ ⋅ ⋅ + X i−1 = ∏m|i,m=1̸ Φ m , u is a product of units of the form Φ m (x j )Φ m (x)−1 with gcd(n, jm) = 1. From Proposition 8.1.3 we know that these belong to Gen(G) provided m is not a prime power. To deal with the latk−1 ter, suppose m = p k with p a prime and k a positive integer. Replacing x by x p and using Lemma 8.1.2, the problem reduces to the case Φ p (x j )Φ p (x)−1 . If j = j󸀠 j󸀠󸀠 󸀠 󸀠󸀠 󸀠 󸀠 then Φ p (x j )Φ p (x)−1 = Φ p (x j j )Φ p (x j )−1 Φ p (x j )Φ p (x)−1 . Hence, it is enough to deal with the case that j is a prime number. So, suppose q is prime. If q ≠ p then, by Lemma 8.1.2 (2), Φ p (x q )Φ p (x)−1 = Φ pq (y) ∈ Gen(G). This proves that Φ p (y j )Φ p (y)−1 ∈ Gen(G) provided j is coprime with n. In particular, Φ p (x n−p )Φ p (x)−1 ∈ Gen(G), as gcd(p, d) = 1. Also x p(p−1) ∈ Gen(G), by Proposition 8.2.2. Finally, for j = p we have

264 | 8 Generic units Φ p (x p ) = (x p − 1)(x p − 1)−1 = x p(p−1) (1 − x(n−p)p )(1 − x n−p )−1 = x p(p−1) Φ p (x n−p ) and hence Φ p (x p )Φ p (x)−1 = x p(p−1) Φ p (x n−p )Φ p (x)−1 ∈ Gen(G), by Proposition 8.2.2. 2

Recall that by Bass(G) we denote the group generated by all Bass units of ℤG, that is, Bass(G) = ⟨u k,m (g) : g ∈ G, 1 < k < |g|, k m ≡ 1 mod |g|⟩. By Bassm (G) we denote the group generated by all Bass units of the type u k,m (g), with g ∈ G. Proposition 8.2.5 (Hoechsmann [91]). If G is a finite abelian group of exponent e then Bassφ(e) (G) ⊆ Hoe(G). Proof. Let x ∈ G and assume n = |x|. Since, n divides e, the exponent of G, φ(n) divides φ(e)

φ(e). Hence, by (1.2.6), u k,φ(e) (x) = u k,φ(n) (x) φ(n) for every k coprime with n. Therefore, it is enough to show that if m = φ(n) then u k,m (x) is a product of Hoechsmann units. In order to prove this, it is convenient to consider the natural action of A = Aut(C n ) on ℚC n . Each element of A is of the form σ i for a unique i ∈ U(ℤ/nℤ), where the action is defined as follows: x σ i = x i . This action fixes each subgroup of ⟨x⟩ and hence, from the description of the primitive central idempotents of ℚ ⟨x⟩ (Corollary 3.3.3), we deduce that this action respects the simple components of ℚC n = ⊕d|n ℚ(ζ nd ). Therefore this action naturally defines a ℤA-module structure on U(ℚC n ) and on U(ℚ ⟨x⟩ e) for every idempotent e of ℚ ⟨x⟩. We extend the exponential notation of the action of A to the action of ℤA. More precisely, if ∑σ∈A a σ σ with each a σ ∈ ℤ and y ∈ ℚ ⟨x⟩ then y∑σ a σ σ = ∏σ (y a )σ . Note that x − 1 projects to ζ nd − 1 in the component ℚ(ζ nd ). Therefore (1 − x̂)(x − 1) is a unit in ℚ ⟨x⟩ (1 − x̂). For every Bass unit u k,m (x) we have (1 − x̂)u k,m (x) = (1 − x̂)(x − 1)(σ k −1)m

(8.2.1)

and for each Hoechsmann unit h j,i = h j,i (g) we get (1 − x̂)h j,i = ((1 − x̂)(x − 1)) = ((1 − x̂)(x − 1))

σ j (σ i −1)

−1

((1 − ̂x)(x − 1)σ i −1 )

(σ j −1)(σ i −1)

(8.2.2)

A∈ Let ω A : ℤA → ℤ be the augmentation map. As |A| = φ(n), we have φ(n) − ̃ ̃ ̃ ker(ω A ). Moreover, for every a ∈ A, one has A = Aa, and therefore φ(n) (1 − a) = (φ(n) − ̃ A)(1 − a) ∈ ker(ω A )2 . Hence, φ(n) ker(ω A ) ⊆ ker(ω A )2 . Thus

s

φ(n) (σ k − 1) = ∑ z r (σ p r − 1)(σ q r − 1),

(8.2.3)

r=1

for some p r , q r ∈ ℤ with gcd(p r q r , n) = 1 and z r ∈ ℤ. Applying (8.2.1), (8.2.2) and (8.2.3) for m = φ(n) we get

8.3 A logarithm function |

s

(1 − x̂)u k,φ(n) (x) = (1 − x̂)(x − 1)(σ k −1)φ(n) = ∏ ((1 − x̂)(x − 1)(σ pr −1)(σ qr −1) )

265

zr

r=1

s

= ∏(h p r ,q r ) (1 − x̂) zr

r=1

Clearly s

u k,φ(n) (x)x̂ = x̂ = ∏(h p r ,q r )z r x̂ . r=1

Consequently s

u k,φ(n) (x) = u k,φ(n) (x)x̂ + u k,φ(n) (x)(1 − x̂) = ∏(h p r ,q r )z r ,

(8.2.4)

r=1

as desired. Proposition 8.2.5 and Theorem 7.3.3 (with m k,C = φ(Exp(G))) at once yield the following result. Corollary 8.2.6. If A is a finite abelian group then Hoe(A) = Gen(A) is of finite index in U(ℤA). In [91] the index of Hoe(C n ) in U(ℤC n ) is investigated. In general this index is much smaller than that of Bassφ(n) (C n ). For more details we refer the reader to [91] and [202]. Earlier work on this topic was done by Cliff, Hoechsmann, Ritter, Sehgal, Weiss in [90, 92–95].

Problems 8.2.1. Let x be a generator of C n . Prove that both {Φ m (x) :

m is not a prime power} gcd(m, n)

and {Φ pq (x d ), Φ pqr (x d ) : p, q, r distinct primes with gcd(pqr, n) = 1 and d | n} generate Gen(C n ).

8.3 A logarithm function In [155] Marciniak and Sehgal introduced a logarithm function on the group of generic units Gen(C n ). This is a tool to obtain a multiplicative independence criterion of these units. In this section we cover these results.

266 | 8 Generic units Throughout this section C n = ⟨x⟩. Observe that the roots of Φ m (X d ) are roots of unity and thus Φ m (X d ) = ∏i Φ m i (X), for some positive integers m i . It follows at once that n Gen(C n ) = ⟨Φ m (x) : m ∈ ℕ, is not a prime power⟩. gcd(n, m) The following lemma helps to understand the action of linear characters on generic units. Let ι : ℂ → ℂ be the function defined by ι(z) = z for z ≠ 0 and ι(0) = 1. Lemma 8.3.1. Let χ be a complex linear character of C n = ⟨x⟩. If power then m χ(Φ m (x)) = ∏ ι(χ(x d ) − 1)μ( d ) .

m gcd(m,n)

is not a prime

d|m

Proof. By assumption, m is not a prime power and thus, by Lemma 8.1.2, Φ m (1) = 1. m m We argue by induction on n. If n = 1 then ∏d|m ι(χ(1) − 1)μ( d ) = ∏d|m ι(0)μ( d ) = m ∏ 1μ( d ) = 1 = χ(Φ m (1)), as desired. d|m

So assume that n > 1. We first deal with the case that the character χ is not faithful. Then χ = χ ∘ π with π : C n → C n /⟨x k ⟩ ≅ C k the natural epimorphism, for some proper divisor k of n, and χ a linear character of C k . Write π(x) = x. Because k < n, the induction hypothesis yields that ̄ = ∏ ι(χ(x̄ d ) − 1)μ( d ) = ∏ ι(χ(x d ) − 1)μ( d ) , χ(Φ m (x)) = χ(Φ m (x)) m

m

d|m

d|m

again as desired. Next we deal with the case that χ is faithful. Hence χ(x) = ζ n , a primitive n-th root of unity. Assume first that n is not a divisor of m. Then, for all d | m, we have that ζ nd − 1 ≠ 0. Hence, because of (8.1.3), we obtain: χ(Φ m (x)) = Φ m (ζ n ) = ∏(ζ nd − 1)μ( d ) m

d|m

=

∏ ι(ζ nd d|m

− 1)μ( d ) = ∏ ι(χ(x d ) − 1)μ( d ) , m

m

d|m

as desired. Second, assume that n is a divisor of m. Then, again from (8.1.3), we get Φ m (X) = ∏(X d − 1)μ( d ) ∏(X d − 1)μ( d ) m

d|m,n|d

m

d|m,n∤d 󸀠

= ∏(X nd − 1)μ(

m/n ) d󸀠

d󸀠 |m/n

∏(X d − 1)μ( d ) m

d|m,n∤d

= Φ m/n (X n ) ∏(1 − X d )μ( d ) . m

d|m,n∤d

Now m/ gcd(m, n) = m/n is not a prime power. Hence, by Lemma 8.1.2, Φ m/n (1) = 1. Therefore, χ(Φ m (x)) = Φ m (ζ n ) = Φ m/n (1) ∏(ζ nd − 1)μ( d ) = ∏ ι(χ(x d ) − 1)μ( d ) . m

d|m,n∤d

m

d|m

8.3 A logarithm function

|

267

In order to introduce the promised logarithm function, it is convenient to recall the definition of contracted semigroup ring. Let S be a semigroup and R a ring. The semigroup ring of S with coefficients in R is defined in the same way as the group ring. More precisely, each element a of RS has a unique expression of the form ∑s∈S a s s, with a s ∈ R for every s ∈ S, and a s = 0 for all but finitely many elements s ∈ S. The sum and product in RS are defined by (∑ a s s) + (∑ b s s) = ∑ (a s + b s )s, s∈S

s∈S

s∈S

(∑ a s s)(∑ b s s) = ∑ ∑ (a u b v )s. s∈S

s∈S

s∈S u,v∈S,uv=s

An element θ of S is called a zero element if sθ = θs = θ for all s ∈ S. In this case Rθ = {rθ : r ∈ R} is an ideal in RS and the ring RS/Rθ is called the contracted semigroup ring. It is denoted R 0 S. So, in this ring, θ ∈ S is identified with 0 ∈ RS. By Mn we denote the set ℤ/nℤ considered as a multiplicative monoid. Let [a]n denote the element of Mn represented by the integer a. If n is clear from the context then we simply write [a] for [a]n . We define the following equivalence relation in Mn : a ≡ b if and only if a = b or a = −b

(a, b ∈ Mn ).

The set of equivalence classes is denoted by M∗n . The product in Mn induces a product in M∗n which endows M∗n with a structure of semigroup. We will abuse the notation and use [a] both for an element of Mn and for the element in M∗n containing it. Observe that θ = [0] = [n] is the zero of Mn and M∗n . For each positive integer m, consider the following element in ℤ0 Mn : L(m) = ∑ μ ( d|m

m ) [d]. d

Notice that if n | d then [d] = 0 in ℤ0 Mn and ℤ0 M∗n . Hence terms indexed by such a d do not contribute to the sum. The image of L(m) under the natural homomorphism ℤ0 Mn → ℤ0 M∗n will be denoted L∗ (m). Such map induces an isomorphism ℤ0 M∗n ≅ ℤ0 Mn /([−1] − [1])ℤ0 Mn . The following shows that the operator L resembles the Euler φ-function. Lemma 8.3.2. (1) If m and m󸀠 are relative prime positive integers then L(mm󸀠 ) = L(m) ⋅ L(m󸀠 ). α

α

α

(2) If m = p11 p22 ⋅ ⋅ ⋅ p s s is the prime factorization (with α i > 0) then L(m) = ([p1 ] − 1) ⋅ ⋅ ⋅ ([p s ] − 1)[p1 ]α1 −1 ⋅ ⋅ ⋅ [p s ]α s −1 .

(8.3.1)

268 | 8 Generic units Proof. (1) Assume gcd(m, m󸀠 ) = 1. Clearly μ(mm󸀠 ) = μ(m)μ(m󸀠 ). Hence, L(mm󸀠 ) = ∑ μ ( d|mm󸀠

=

mm󸀠 ) [d] d μ(

∑ e|m,f|m󸀠

= ∑μ( e|m,f|m󸀠

mm󸀠 ) [ef] ef

m m󸀠 ) [e] ⋅ μ ( ) [f] e f

= (∑ μ ( e|m

m m󸀠 ) [e]) ⋅ ( ∑ μ ( ) [f]) e f f|m󸀠

= L(m)L(m󸀠 ). (2) By (1) one may assume, without loss of generality, that m = p α with p prime and α a positive integer. Since μ(p r ) = 0 for r ≥ 2, we have L(p α ) = μ(1)[p α ] + μ(p)[p α−1 ] = ([p] − 1)[p]α−1 .

(8.3.2)

Lemma 8.3.3. Let n and m be positive integers and d a positive divisor of n. Then μ(

∑ e|m,gcd(e,n)=d

μ (m) , m )={ d e 0,

if d | m and m | n; otherwise.

Proof. If d ∤ m then the sum ∑e|m,gcd(e,n)=d μ ( me ) is empty and hence it is 0. So, suppose that d | m. Assume that m divides n. Then for e | m one has that gcd(e, n) = d if and only if e = d. Therefore ∑e|m,gcd(e,n)=d μ ( me ) = μ ( md ). Assume that m ∤ n (and d | m). Let e be a divisor of m. As gcd(e, n) = d if and only if gcd(e, gcd(m, n)) = d, replacing n by gcd(n, m), we may assume without loss of generality that n divides m. As we are assuming that n is not divisible by m, we have that n is a proper divisor of m. Write d = d1 d2 , n = d1 n2 n3 and m = m1 m2 m3 m4 such that factors with different indexes are coprime, factors with the same index are divisible by exactly the same primes and dn22 is divisible by all the prime divisors of d2 . Then gcd(e, n) = d if and only if e = d1 e1 d2 e4 with e1 | md11 and e4 | m4 . Hence (recall that, for a non-negative integer x, ϵ(x) = 1 if x = 1 otherwise ϵ(x) = 0), ∑ e|m,gcd(e,n)=d

μ(

m )= e

∑ m e1 | d 1 1

= ϵ(

,e4 |m4

μ(

m1 /d1 m4 m2 )μ( ) μ(m3 )μ ( ) e1 d2 e4

m1 m2 )μ( ) μ(m3 )ϵ(m4 ). d1 d2

The second equality follows from (3.4.9). Assume that this is non-zero. Then m1 = d1 , m4 = 1 and md22 and m3 are square free. If p is a prime divisor of mn22 then p divides d2 and thus dn22 , contradicting the fact that md22 is square-free. Similarly, if p is a prime divisor of mn33 then p is a prime divisor of n3 contradicting the fact that m3 is squarefree. Thus m2 = n2 and m3 = n3 , so that n = m, a final contradiction.

8.3 A logarithm function

|

269

The following lemma is a crucial step in the definition of the logarithm map and in the construction of an independent set of generic units. Lemma 8.3.4. Let m1 , m2 , . . . , m s and n be positive integers such that each is not a prime power For any k1 , . . . , k s in ℤ, the following are equivalent: k k k (1) Φ m11 (x) Φ m22 (x) ⋅ ⋅ ⋅ Φ mss (x) ∈ C n , (2) ∑sj=1 k j L∗ (m j ) = 0 In particular, if Φ m1 (x) = Φ m2 (x) then L∗ (m1 ) = L∗ (m2 )

mi gcd(m i ,n)

Proof. For 1 ≤ i < n, let s

κi = ∑ kj j=1

Let d1 be a divisor of n. As

mj gcd(m j ,n)

μ(

∑ d|m j ,d≡i mod n

mj ). d

is not a prime power, in particular n is not divis-

ible by m j and hence ∑d|m j ,gcd(d,n)=d1 μ ( mod n, then gcd(d, n) = gcd(i, n). Let

mj d )

X j = {(d, i) : d | m j , d ≡ i = {(d, i) : d | m j , d ≡ i

= 0, by Lemma 8.3.3. Moreover, if d ≡ i

mod n, gcd(i, n) = d1 } mod n, gcd(d, n) = d1 }.

Thus s

∑

κi = ∑ kj

i, gcd(i,n)=d1

j=1

∑ μ( (d,i)∈X i

s mj mj μ ( ) = 0. ) = ∑ kj ∑ d d j=1 d|m ,gcd(d,n)=d j

1

We show that each condition (1) and (2) is equivalent to: (3) κ i + κ n−i = 0 for every 1 ≤ i ≤ n. k

k

k

We first show that (1) and (3) are equivalent. Let u = Φ m11 (x) Φ m22 (x) ⋅ ⋅ ⋅ Φ mss (x). Because of Lemma 8.3.1, we get that, for any character χ : C n → ℂ, s

χ(uu∗ ) = ∏ ∏ ι(χ(x d ) − 1) j=1 d|m j

k j μ(

mj d

)

ι(χ(x−d ) − 1)

k j μ(

mj d

)

n−1

= ∏ ι(χ(x i ) − 1)κ i +κ n−i . i=1

So, by the Bass Independence Theorem (Theorem 7.3.4), all κ i + κ n−i = 0 if and only if χ(uu∗ ) = 1 for all linear characters χ of C n , or equivalently uu∗ = 1. By (7.1.2), the latter is equivalent with u ∈ ±C n , and thus with u ∈ C n (because u has augmentation 1, by Proposition 8.1.2 (3). It follows that indeed (1) and (3) are equivalent. Next we show that (2) and (3) are equivalent. To do so, consider the ℤ-submodule V of ℤC n generated by the set {x i + x−i : 0 < i ≤ ⌊n/2⌋}. The semigroup M∗n acts on V as follows: x di + x−di , if n ∤ di; [d](x i + x−i ) = { 0, otherwise.

270 | 8 Generic units The zero element of M∗n , that is [0], acts as the zero operator on V. Hence V is a ℤ0 M∗n module. The annihilator of x + x−1 in ℤ0 M∗n is equal to 0. Hence (2) is equivalent with ∑sj=1 k j L∗ (m j )(x + x−1 ) = 0. Now s

s

j=1

j=1

∑ k j L∗ (m j )(x + x−1 ) = ∑ k j ∑ μ ( d|m j

mj ) (x d + x−d ) d

n−1 s

= ∑ ∑ kj ( i=1 j=1

μ(

∑ d|m j ,d≡i mod n

mj )+ ∑ d d|m ,d≡−i j

μ( mod n

mj )) x i d

n−1

= ∑ (κ i + κ n−i )x i , i=1

It follows that indeed (2) is equivalent to (3). We now prove the existence of a logarithmic function. Theorem 8.3.5 (Marciniak-Sehgal). Let x be a generator of C n . The map Lx : Gen(C n ) 󳨀→ ℤ0 M∗n defined by s

Φ m11 (x) ⋅ ⋅ ⋅ Φ mss (x) 󳨃→ ∑ k j L∗ (m j ) k

k

j=1

is well defined. Furthermore, Lx (u ⋅ v) = Lx (u) + Lx (v),

for all u, v ∈ Gen(C n )

and ker(Lx ) = C n ∩ Gen(C n ) = ⟨x2 ⟩ . In particular, elements u1 , . . . , u s in Gen(C n ) are multiplicatively independent provided that the elements Lx (u i ) are additively independent in ℤ0 M∗n . Proof. Recall that Gen(C n ) is generated by all Φ m (x) ∈ U(ℤC n ). Therefore the result follows at once from Lemma 8.3.4 and Proposition 8.2.2. Observe that the map Lx defined in Theorem 8.3.5 depends on the chosen generator x. Any other generator is of the form x e , with e an integer relatively prime with n. By the definition of Lx and Lx e , we have Lx (f(x)) = Lx e (f(x e )),

for every e coprime with n and every polynomial f ∈ ℤ[X] such that f(x) ∈ Gen(C n ). We end this section describing the image of Lx . For that we first prove the following lemma. Lemma 8.3.6. Let m and e be positive integers. If Φ m (x) ∈ U(ℤC n ) and gcd(e, m) = 1 then Lx (Φ m (x e )) = Lx (Φ m (x)) ⋅ [e].

8.3 A logarithm function

|

271

Proof. First assume e = p α with p a prime. Since gcd(e, m) = 1, Lemma 8.1.2 (2) yields that α

α

Φ m (x p ) = ∏ Φ mp i (x) ∈ Gen(C n ). i=0

Thus, by Theorem 8.3.5 and Lemma 8.3.2 (adapted to L∗ ) we have α

α

α

i=0

i=0 α

Lx (Φ m (x p )) = ∑ Lx (Φ mp i (x)) = ∑ L∗ (mp i ) = L∗ (m) ∑ L∗ (p i ) α

i=0 α

= Lx (Φ m (x)) ∑ L∗ (p i ) = Lx (Φ m (x))(1 + ∑ ([p]i − [p]i−1 )) i=0

i=1

= Lx (Φ m (x)) ⋅ [p] . α

The general case easily follows by induction on the number of prime factors of e. We introduce some notation Un = U (ℤ/nℤ)

U∗n = U (ℤ/nℤ) / ⟨[−1]n ⟩ .

and

For every positive divisor d of n consider the injective map ρ d : Un/d → Mn [k] dn 󳨃→ [dk]n The image of this map is [d]n Un/d = {[k]n : gcd(k, n) = d}. Notice that Un/d is not a subset of Mn , hence [d]Un/d is just a convenient abuse of notation which will be extended to [d]X for subsets X of ℤ Un/d . We have the following decomposition of Mn as a disjoint union Mn = ⋃[d]n Un/d .

(8.3.3)

d|n

Hence ℤ0 Mn = ⨁ [d]n ℤ Un/d d|n

and

ℤ0 M∗n = ⨁ [d]n ℤ U∗n/d , d|n

with the obvious abuse of notation. Let ω d : ℤ Un/d → ℤ and ω∗d : ℤ U∗n/d → ℤ denote the corresponding augmentation maps. Then [d] ker(ω d ) (respectively, [d] ker(ω∗d )) is the additive subgroup of ℤ0 Mn (respectively, ℤ0 M∗n ) generated by the elements of the form [d]([k] dn − 1), with 1 < k < dn and gcd (k, dn ) = 1. The last lemma of this section describes the image of Lx in terms of the “augmentation kernels”.

272 | 8 Generic units Proposition 8.3.7. Lx (Gen(C n )) = ⊕d|n [d]n (ker(ω∗d ))2 . Proof. The additive group ⊕d|n [d](ker(ω∗d ))2 is generated by elements [d]([k] dn − 1)([l] dn − 1), where d | n and gcd(kl, n) = 1. By Dirichlet’s Theorem on Primes in an Arithmetic Progression, there exist distinct primes p and q such that p − k ∈ dn ℤ and q − l ∈ dn ℤ and gcd(pq, n) = 1. So [p] dn = [k] dn and [q] dn = [l] dn . Hence, by Lemma 8.3.2 and Lemma 8.3.6, Lx (Φ pq (x d )) = Lx (Φ pq (x))[d] = [d]([p] − 1)([q] − 1)

= [d]([p] dn − 1)([q] dn − 1) = [d]([k] dn − 1)([l] dn − 1) So, ⊕d|n [d] ker(ω∗d )2 ⊆ Lx (Gen(C n )). From Proposition 8.2.3 we know that Gen(C n ) is generated by elements of the form Φ pq (x e ), with p and q distinct primes, gcd(pq, n) = 1 and e ∈ ℤ. Let d = gcd(e, n) and write e = de1 . Then gcd (e1 , dn ) = 1 and, by the Chinese Remainder Theorem, there is a positive integer r such that e1 ≡ r mod dn and gcd(r, pq) = 1. Hence, by Lemma 8.3.2 and Lemma 8.3.6, Lx (Φ pq (x e )) = Lx (Φ pq (x dr )) = [dr]([p] − 1)([q] − 1) = [d]([pr] − [r])([q] − 1) ∈ [d] ker(ω∗d )2 . Consequently we also obtain that L∗ (Gen(C n )) ⊆ ⊕d|n [d] ker(ω∗d )2 .

8.4 A basis of generic units for a subgroup of finite index in U(ℤC n ) In this section we describe the Marciniak and Sehgal method which provides an explicit basis of a free abelian subgroup of finite index in the unit group of ℤC n . The general idea is to define “stem” units Φ m j (X) for each prime divisor of n and to construct subsets ℑj ⊆ ℤ/nℤ such that the union of partial orbits {Φ m j (x a ) : a ∈ ℑj } forms the desired basis. First we define the stem units. For this we write n = q1 . . . q t with q1 , . . . , q t relatively prime prime powers different of 1. By Theorem 1.6.3 and Theorem 1.6.7, Uq j is cyclic, unless 8 divides q j , in which case, Uq j = ⟨[−1]q j ⟩ × ⟨[5]q j ⟩. We may assume without loss of generality that if n is even then 2 | q1 . Thus we may choose integers a0 , a1 , . . . , a t such that ⟨a0 ⟩ × ⟨a1 ⟩ , ⟨a j ⟩ ,

Uq j = {

if 8 | n and j = 1; otherwise.

Moreover, using the Chinese Remainder Theorem we may assume that a j ≡ 1 mod for every j. Then Un = ⟨a0 ⟩ × ⟨a 1 ⟩ × ⋅ ⋅ ⋅ × ⟨a t ⟩ .

n qj

By Dirichlet’s Theorem on Primes in Arithmetic Progression for every 0 ≤ j ≤ t there is a pair of different primes p j and p󸀠j so that n < p j , p󸀠j and a j ≡ p j ≡ p󸀠j

mod n.

8.4 A basis of generic units for a subgroup of finite index in U(ℤC n ) |

273

Then the stem units are the cyclotomic polynomials Φ p j p󸀠j (X), for j = 0, 1, . . . , n. For every d | n and j = 0, 1, . . . , t let b i = natural projection of a i in U dn , B i = ⟨b i ⟩ t d = max {j = 0, 1, . . . , t : B j ≠ 1} , r d = |⟨B t d ⟩|, T d = {b it d : 0 ≤ i
2 2 2. Prove that Un / ⟨[−1]n ⟩ is cyclic if and only if n is of one of the following forms: (1) a prime power, (2) 2α q with α = 1 or 2 and q an α α odd prime power, (3) p11 p22 with p1 and p2 different odd primes such that gcd(p1 − α1 α2 1, p2 − 1) = 2, (4) 2p1 p2 with p1 and p2 as in (3). Furthermore, prove that [−2] generates Un / ⟨[−1]n ⟩ if and only if either n = p k with p odd or n = 3p k with p a prime and p > 3. 8.4.2. Let n be an odd positive integer and let C n = ⟨x⟩n . Recall from Problem 1.2.4 that i the elements of the form ∑c−1 such that gcd(c, 2n) = 1, i=0 (−x) , with c a positive integer i = Φ2 (x c ) . are called alternating units. Prove that ∑c−1 (−x) i=0 Φ2 (x)

276 | 8 Generic units 8.4.3. Let n be a positive integer and let C n = ⟨x⟩n . For positive integers m and c with (n, cm) = 1 one gets units (m)

uc

(m)

= u c (x) = Φ m (x c )/Φ m (x).

These units were introduced by Marciniak and Sehgal in [155] (see also [202, Section 17]) and they are called alternating-like units. Prove the following results of Marciniak (m) and Sehgal, where A(m) = ⟨u c (x d ) : c, d ∈ ℕ, gcd(c, n) = 1⟩. (1) Each expression Φ m (x c )/Φ m (x) defines a unit in Gen(C n ). (2) The group A(m) is of finite index in U(ℤC n ) if and only if [q]n generates U∗n for every prime q dividing m. (3) For every cyclic p-group C p α there exists a prime q such that the alternating-like units of the form Φ q (x k )/Φ q (x) generate a finite index subgroup in U(ℤC p α ). (4) The group A(m) has finite index in U(ℤC n ) if and only if the following properties are satisfied (a) n is either (1) a prime power, (2) 2α q with q an odd prime power and α = 1 or 2, α α (3) p11 p22 with p1 and p2 different odd primes such that gcd(p1 −1, p2 −1) = 2 α α and α1 , α2 ≥ 1, or (4) 2p11 p22 with p1 and p2 as in (3), (b) for each prime factor q of m the element [q]n generates U∗n . In particular, if m = 2 then A(2) is of finite index in U(ℤC n ) if and only if n = p k with p an odd prime or n = 3p k for some prime p > 3 and, moreover, [2]n generates U∗n .

8.5 Polynomials of small degree defining units Let C n = ⟨x⟩n . We know from (8.0.1) that every unit u in ℤC n can be expressed as f(x) for some f(X) ∈ ℤ[X]. The degree of u is by definition min{deg f(X) : f(X) ∈ ℤ[X], u = f(x)}. Of course trivial units have degree 1 and are defined by the polynomial ±X. Hence the attention is focused on non-trivial units u. Furthermore, without loss of generality, we may assume that u is normalized (i.e. ω(u) = 1) and thus f(1) = 1. In this section we describe the polynomials f ∈ ℤ[X] of degree at most three that define a unit in ℤC n for some cyclic group C n . These descriptions are due to Marciniak and Sehgal [155]. We first show that X is the only integral polynomial f of degree one with f(1) = 1 that defines a unit in ℤC n for some n ≥ 2. Consequently, the only units of degree one are the trivial units. Indeed, let f be a polynomial of degree 1 with integral coefficients such that u = f(x) is a unit in ℤC n with n ≥ 2 and 1 = f(1). By Corollary 7.1.9, u = gv where g is a trivial unit and v is a symmetric unit. If f ≠ X then the support of v has exactly two elements. Hence, v = c(x k + x−k ) for some c ∈ ℤ and thus ±1 = ω(v) = 2c, a contradiction. Next we consider the quadratic polynomials. Because of Proposition 8.1.3, Φ6 (X) = X 2 − X + 1 defines a normalized unit in the group ring ℤC n whenever gcd(n, 6) = 1.

8.5 Polynomials of small degree defining units

|

277

It turns out that this is the only polynomial of degree two defining a non-trivial normalized unit. The following lemma is a relevant step for the proof. Lemma 8.5.1. Let a ∈ ℤ, μ1 and μ2 the roots of the polynomial X 2 + (1 − 2a)X + a2 and for every non-negative integer m let W m (a) = μ1m + μ2m . Then (1) W0 (a) = 2, W1 (a) = 2a − 1 and for m ≥ 1 we have W m+1 (a) = (2a − 1)W m (a) − a2 W m−1 (a).

(8.5.1)

In particular, each W m (a) ∈ ℤ. (2) W m (a) ≡ (−1)m (1 − 2ma + m(2m − 3)a2 ) mod a3 , for every m ≥ 2. (3) If p is a prime integer and W p (a) = 2a p ± 1 then a is either 0 or 1. Proof. Parts (1) and (2) easily follow by induction on m. (3) By (8.5.1) W2 (a) = 2a2 −4a +1, which is clearly different from 2a2 ±1 for a ≠ 0, and W3 (a) = 2a3 − 9a2 + 6a − 1 which is also different from 2a3 ± 1 for a ≠ 0. So in the remainder of the proof we assume that W p (a) = 2a p ± 1 with a ≠ 0 and p a prime with p ≥ 5 and prime. We have to prove that a = 1. First assume a < 0. Then μ1 and μ2 are real and negative because they are the roots of X 2 + (1 − 2a)X + a2 . Hence μ1 /a and μ2 /a are positive. By the well known mean inequality, we get p p μ1 + μ2 1 p (μ 1 /a) + (μ 2 /a) √ ≥ =1− ≥ 1, 2 2a 2a

and thus 2±

W p (a) μ1 p μ2 p 1 p p 1 = = + ≥ 2 − ( ( (1 ) ) ) ≥2− . p p a a a a 2a a

Hence ±

1 p ⩾− . ap a

As a p < 0 this implies ±1 ≥ pa p−1 ≥ p, a contradiction. Hence a > 0. Then the discriminant of X 2 + (1 − 2a)X + a2 is negative and hence μ1 and μ2 are complex conjugate with |μ1 | = |μ2 | = a. Thus p

p

p

p

|μ1 + μ2 | ⩽ |μ1 | + |μ2 | = 2a p . So W p (a) ⩽ 2a p , and thus W p (a) = 2a p − 1. Using part (2) of the Lemma and because p ≥ 5, we get −1 ≡ 2a p − 1 = W p (a) ≡ −1 + 2pa

mod a2

and −1 ≡ 2a p − 1 = W p (a) ≡ −1 + 2pa − p(2p − 3)a2

mod a3 .

278 | 8 Generic units The former implies that a divides 2p. If p | a, then the latter implies that p2 | a, a contradiction. Thus a = 1 or 2. Assume that a = 2 and set W m = W m (2), for every m ≥ 0. Then, W p = 2p+1 − 1

and

W0 = 2, W1 = 3, W m+1 = 3W m − 4W m−1 ,

for m ≥ 1. By induction on m it easily follows that if m ≥ 1 then W2m ≡ 1 mod 16 and W2m+1 ≡ 7 mod 16. Hence, 7 ≡ W p = 2p+1 − 1 ≡ 15 mod 16, a contradiction. We conclude that a = 1, as desired. Proposition 8.5.2 (Marciniak-Sehgal). The polynomial f(X) = X 2 − X + 1 is the only quadratic polynomial in ℤ[X] that defines a non-trivial normalized unit in ℤC n for some n. Proof. Let f(X) = aX 2 + bX + c ∈ ℤ[X] be a quadratic polynomial. Suppose it defines a non-trivial normalized unit f(x) in ℤC n , with C n = ⟨x⟩. Then ac ≠ 0, as otherwise f(x) would be a trivial unit. Further, f(x) being normalized yields that f(1) = 1 and thus a + b + c = 1. Because of Example 1.5.2 we also know that n ∈ ̸ {1, 2, 3, 4, 6} and hence n ≥ 5. Suppose n is the smallest positive integer such that f(x) is a non-trivial normalized unit, with ⟨x⟩ = C n . We claim that n is prime. Indeed, let p be a prime divisor of n. Then f(x p ) = ax2p + bx p + c is a unit in ℤ⟨x p ⟩, since x 󳨃→ x p lifts to a ring homomorphism ℤ⟨x⟩ → ℤ⟨x p ⟩. By the minimality of n we have that f(x p ) is a trivial unit. Consequently, either x2p = x p , x2p = 1 or x p = 1. In the former and latter case n = p as desired. If x2p = 1 then n | 2p and thus p ≥ 5. This proves that every prime divisor of n is at least 5 and from n | 2p we deduce that n = p, as desired. From Lemma 7.1.8 we obtain an integer j with −n < j ≤ 0 such that (ax2 + bx + c)∗ = j x (ax2 + bx + c) and thus ax−2 + bx−1 + c = ax j+2 + bx j+1 + cx j . As 1 belongs to the support of the left hand side, we get that j ∈ {0, −1, −2}. If j = 0 or j = −1 then {1, x−1 , x−2 } contains x2 or x, both in contradiction with n ≥ 5. So j = −2 and thus c = a and b = 1 − 2a. Consequently, f(X) = aX 2 + (1 − 2a)X + a, with a a non-zero integer. As f(x) is a unit in ℤC p , there exist polynomials r(X), q(X) in ℤ[X] such that (aX 2 + (1 − 2a)X + a) ⋅ r(X) = 1 + (X p − 1)q(X). Let α1 and α2 be the complex roots of unity of f(X), and let μ1 = aα1 and μ2 = aα2 . Observe that the notation agrees with Lemma 8.5.1 because μ1 and μ2 are the roots of X 2 + (1 − 2a)X + a2 . So we will use the notation W m (a) = μ1m + μ2m in order to apply Lemma 8.5.1 (2).

8.5 Polynomials of small degree defining units

|

279

We get that p

−1 = (α i − 1)q(α i ) for i = 1, 2. Hence

p

p

1 = (α1 − 1)(α2 − 1)q(α1 )q(α2 ). Let N = deg(q(X)). Then p

p

a2N+2p = (μ1 − a p )(μ2 − a p ) ⋅ a N q(α1 ) ⋅ a N q(α2 ). As a2 = μ1 μ2 = a2 α1 α2 , one gets p

p

(μ1 − a p )(μ2 − a p ) = 2a2p − a p W p (a) = a p (2a p − W p (a)), and therefore a2N+p = (2a p − W p (a)) a N q(α1 ) ⋅ a N q(α2 ). Notice that a N q(α i ) ∈ ℤ[μ1 ] = ℤ[μ2 ] and μ1 and μ2 are algebraic integers. Thus, a N q(α1 ) a N q(α2 ) is an algebraic integer in ℚ, because it is invariant under the action of the Galois group Gal(ℚ(μ1 , μ2 )/ℚ). So, a N q(α1 ) a N q(α2 ) ∈ ℤ. Therefore, 2a p − W p (a) is a divisor of a2N+p in ℤ. Furthermore, Lemma 8.5.1 (2) implies that W p (a) ≡ −1 mod a. Consequently, gcd(a, 2a p − W p (a)) = 1 and thus 2a p − W p (a) = ±1. So W p (a) = 2a p ±1 and therefore a = 0 or 1, by Lemma 8.5.1 (3). Since a ≠ 0, we conclude that a = 1, so that f = X 2 − X + 1, as desired. Consider the Laurent polynomial X + X −1 − 1. Note that X(X + X −1 − 1) = X 2 − X + 1 = Φ6 (X). Hence, by Proposition 8.1.3, X + X −1 − 1 defines a unit in ℤC n if gcd(n, 6) = 1. One can consider X + X −1 − 1 as a symmetric version of Φ6 (X), the unique quadratic polynomial defining non-trivial normalized units. Let C n = ⟨x⟩ and suppose that gcd(n, 6) = 1. Clearly x+x−1 −1 = 1−(x−1)(x−1 −1) ∈ 1+ker(ω)2 . Hence f(x) = x+x−1 −1 has infinite order by Proposition 7.1.7. We also know from Theorem 7.1.6 that U(ℤC n ) = ±C n × V, with V a finitely generated free abelian group. If one would like to construct a set of independent generators of V then, first of all, one needs to find a non-trivial element v ∈ V that is not a power of any other element of V. We will show that x + x−1 − 1 is such an element. To do so we first need the following lemma. Recall that for a group ring element a ∈ RG, we denote by a1 the coefficient of the identity element 1 ∈ G. Lemma 8.5.3. Let G be a finite group. (1) The real group ring ℝG is equipped with the inner product (a, b) = (a∗ b)1 . In particular, ‖a‖ = (a, a)1/2 , defines a norm on ℝG. (2) ‖u m ‖ ≥ ‖u‖m , for any symmetric unit u ∈ ℤG and any positive integer m. Proof. The first part is obvious. To prove the second part, set n = |G| and let u be a symmetric unit in ℤG. Further, let σ : ℝG → ℝG denote left multiplication by u. Then (σ(a), b) = ((ua)∗ b)1 = (a∗ (ub))1 = (a, σ(b)) for all a, b ∈ ℝG. This implies that the matrix associated to σ, in an orthonormal basis of ℝG with respect to the

280 | 8 Generic units

given inner product, is symmetric. Hence σ is diagonalizable with real eigenvalues. Therefore σ2 is diagonalizable with positive real eigenvalues, say θ1 , . . . , θ n and more generally σ2m is diagonalizable with eigenvalues θ1m , . . . , θ m n . From (3.1.1) we obtain m 2 m 2 m m that n((u ) )1 = TrℝG/ℝ ((u ) ) = θ1 + ⋅ ⋅ ⋅ + θ n . Hence, by the Generalized Mean Inequality, m + ⋅ ⋅ ⋅ + θm θ1 + ⋅ ⋅ ⋅ + θ n 2m θ n 2m m √ ‖u m ‖ = √((u2 )m )1 = √ 1 ≥√ = √(u2 )1 = ‖u‖, n n

as desired. Proposition 8.5.4. Let G be a group. If x is a periodic element in G of order relatively prime to 6 then u = x + x−1 − 1 ∈ ℤG is a symmetric unit which is not a proper power of any other symmetric unit. Proof. Because a unit has augmentation ±1, it is clear that |Supp(v)| ≥ 3 for any symmetric non-trivial unit v ∈ ℤG and thus ‖v‖ ≥ √3. Assume x ∈ G is a periodic element in G with order relatively prime to 6. Then we already know that x + x−1 − 1 is a symmetric unit in ℤG. Notice that ‖u‖ = √3. Suppose u = v m for some symmetric unit v and some positive integer m. Then, by Lemma 8.5.3, √3 = ‖u‖ = ‖v m ‖ ≥ ‖v‖m ≥ (√3)m . So m = 1 and the result follows. As an application we now easily give a description of U(ℤC5 ). Example 8.5.5. Let C5 = ⟨x⟩5 Then U(ℤC5 ) = ±C5 × ⟨x + x4 − 1⟩. Proof. We know from Theorem 7.1.6 that the rank of U(ℤC5 ) is one. Hence, by Corollary 7.1.9, U(ℤC5 ) = ±C5 × ⟨u⟩ for some non-trivial symmetric unit u of infinite order. Proposition 8.5.4 yields at once that u = x + x4 −1, satisfies the required condition. Combining Example 8.5.5 and some techniques from the proof of Proposition 8.5.2 we can obtain all the units of degree 3. Proposition 8.5.6 (Marciniak). The only normalized units of degree 3 in ℤC n , with C n = ⟨x⟩ are x3 − x2 + x for gcd(n, 6) = 1 and x3 + x2 − 1 and −x3 + x + 1 for n = 5. Proof. Let u = ax3 + bx2 + cx + d be a unit of degree 3. In particular a ≠ 0 and u is not a trivial unit, so that n ≠ 1, 2, 3, 4 or 6. If d = 0 then ux−1 = ax2 + dx + c is a unit of degree 2. Then aX 2 + bX + c = X 2 − X + 1 and gcd(n, 6) = 1 by Propositions 8.5.2 and 8.1.3. Assume otherwise that d ≠ 0. By Lemma 7.1.8 there is −n ≤ j ≤ 0 such that ax−3 + bx−2 + cx−1 + d = ax j+3 + bx j+2 + cx j+1 + dx j . Since n ≥ 5, x, 1, x−1 , x−2 and x−3 are pairwise different and x3 ≠ x−3 . In particular 1 and x−3 belong to the support of the left side part of the previous equality. We deduce that 1, x−3 ∈ {x j+3 , x j+2 , x j+1 , x j }. Thus j = 0, −1, −2 or −3. If j = −3 then ax−3 + bx−2 + cx−1 +d = a+bx−1 +cx−2 +dx−3 and therefore a = d and b = c. Then 1 = ω(u) = 2(a+b),

8.5 Polynomials of small degree defining units

|

281

a contradiction. If j = −2 then ax−3 + bx−2 + cx−1 + d = ax + b + cx−1 + dx−2 , again a contradiction because x belongs to the support of the right size part and not to the support of the left side part. Assume that j = −1. Then ax−3 + bx−2 + cx−1 + d = ax2 + bx + c + dx−1 . As x−3 belongs to the support of the left side we deduce that x−3 = x2 , i.e. x has order 5, and hence b = 0 and c = d. Then 1 = ω(u) = a + 2c and hence u = (1 − 2c)x3 + c(x + 1). Assume that j = 0 then ax−3 + bx−2 + cx−1 + d = ax3 + bx2 + cx + d. As x−3 belongs to the support of the left side we deduce that x−3 = x2 , i.e. x has order 5. Thus ax2 + bx3 + cx4 = ax3 + bx2 + cx and therefore c = 0 and a = b. Thus 1 = ω(u) = 2a + d, so that u = a(x3 + x2 ) + (1 − 2a). So we are left with n = 5 and two types of possible units of degree 3: u1 (a) = (1 − 2a)x3 + a(x + 1)

and

u2 (a) = a(x3 + x2 ) + (1 − 2a).

Actually u2 (a) = x2 u1 (a) and so it is enough to prove that u2 (a) is a unit of ℤC5 of degree 3 if and only if a = 1. The sufficiency follows from u2 (1)(x + x−1 − 1) = 1. In fact we will prove that it is a unit if and only if a = 0 or 1. This will finish the prove because u2 (0) = 1 is a trivial unit and u2 (1) = x3 + x2 − 1 has degree 3. Let v = x + x4 − 1 and for every integer k write v k = (1 − 2p k − 2q k ) + p k (x + x4 ) + q k (x2 + x3 ) with p k , q k ∈ ℤ. The sequences (p k )k∈ℤ and (q k )k∈ℤ satisfy the following recursive relations p k+1 = 1 − 3p k − q k , q k+1 = p k which implies p k−1 = q k ,

q k−1 = 1 − p k − 3q k .

Then p−1 = p0 = 0,

p k+2 = 1 − 3p k+1 − p k ,

p k−2 = 1 − p k − 3p k−1 .

Using this it is easy to prove that p−k−1 = p k for k ≥ 0. We claim that if k ≥ 1 then |p k+1 | > |p k | > 0 and p k p k+1 < 0. This is clear for k = 1. Assume that the claim is true for k. Then, if p k > 0 then −p k+1 > p k ≥ 1 and hence p k+2 = 1 − 3p k+1 − p k > 1 − 2p k+1 > −p k+1 = |p k+1 |. Thus p k+2 = |p k+2 | > |p k+1 |. Similarly, if p k < 0 then p k+1 > −p k ≥ 1 and hence p k+2 = 1 − 3p k+1 − p k < 1 − 2p k+1 < −p k+1 = −|p k+1 |. Therefore |p k+2 | = −p k+2 > |p k+1 |. This prove the claim. Observe that u2 (a)∗ = u2 (a). Then, by Example 8.5.5, we deduce that u2 (a) is a unit of ℤC5 if and only if it is a power of v. If u2 (a) = v k then p k = 0 and hence k = 0 or k = −1, by the previous paragraph. If k = 0 then a = 0 and if k = −1 then a = 1. This finishes the proof.

9 K-theory In this chapter we introduce basic notions of K-Theory. These provide techniques to study the unit group of a ring. A more extended introduction to K-Theory can be found in [15] and in [46, Chapter 5]. In the first two sections we introduce the Grothendieck group K0 (R) and the Whitehead group K1 (R) of a ring R. In the third section we introduce the stable range of a ring. It turns out that the stable range of an order is two and this has important consequences on its group of units. For an order O, we obtain in the last section a relation between its Whitehead group and the group of central units of O. The contents of this chapter is mainly based on [46, 80, 215].

9.1 Grothendieck group Let R be a ring. For a left R-module M, let (M) denote the isomorphism class of M. We denote by F the free abelian group with basis the set of isomorphism classes of finitely generated projective R-modules and by F0 the subgroup of F generated by the elements (M ⊕ N) − (M) − (N), with M and N running through all finitely generated projective R-modules. The group K 0 (R) = F/F0 is called the Grothendieck group of R. The image of (M) in K0 (R) is denoted [M]. If f : R → S is a ring homomorphism then [P] 󳨃→ [S⊗R P] defines a group homomorphism K0 (f) : K0 (R) → K0 (S). The assignments R 󳨃→ K0 (R) and f 󳨃→ K0 (f) define a functor from the category of rings to the category of abelian groups. Lemma 9.1.1. Let M and N be finitely generated projective left R-modules. Then [M] = [N] in K0 (R) if and only if M ⊕ P ≅ N ⊕ P for some finitely generated projective left R-module P, or equivalently M ⊕ R n ≅ N ⊕ R n for some n ≥ 0. Proof. The only non-trivial statement is that if [M] = [N] then M ⊕ P ≅ N ⊕ P for some finitely generated projective module P. We are going to prove an apparently n stronger result but actually equivalent statement, namely if ∑m i=1 [M i ] = ∑j=1 [N j ] then m n P ⊕ ⊕i=1 M i ≅ P ⊕ ⊕j=1 N j for some finitely generated projective module P. Indeed, the hypothesis means that m

r

s

i=1

k=1

l=1

∑ (M i ) + ∑ (P k ⊕ Q k ) + ∑ ((P󸀠l ) + (Q󸀠l )) n

r

s

j=1

k=1

l=1

= ∑ (N j ) + ∑ ((P k ) + (Q k )) + ∑ (P󸀠l ⊕ Q󸀠l )

(9.1.1)

for some finitely generated projective modules P k , Q k , P󸀠l and Q󸀠l . Since [0] = 0, we may assume without loss of generality that m, n ≥ 1. We argue by induction on r + s.

9.1 Grothendieck group

| 283

If r + s = 0 then the lists (M1 ), . . . , (M m ) and (N1 ) . . . , (N n ) are formed by the same n elements, maybe in different order and hence ⊕m i=1 M i ≅ ⊕j=1 N j . Assume that r + s > 0. Then, after some reordering and renaming, one may assume that one of the following conditions hold. (1) (M1 ) = (P1 ) and (N1 ) = (P󸀠1 ). (2) (M1 ) = (P1 ) and (N1 ) = (P1 ⊕ Q1 ). (3) (M1 ) = (P󸀠1 ⊕ Q󸀠1 ) and (N1 ) = (P1 ⊕ Q1 ). We consider each case separately: (1) Assume (M1 ) = (P1 ) and (N1 ) = (P󸀠1 ). Then, after some cancellation, (9.1.1) becomes m

r

s

i=2

k=2

l=2 n

(P1 ⊕ Q1 ) + (Q󸀠1 ) + ∑ (M i ) + ∑ (P k ⊕ Q k ) + ∑ ((P󸀠l ) + (Q󸀠l )) r

s

k=2

l=2

= (Q1 ) + (P󸀠1 ⊕ Q󸀠1 ) + ∑ (N j ) + ∑ ((P k ) + (Q k )) + ∑ (P󸀠l ⊕ Q󸀠l ). j=2

By induction hypothesis, there is a finitely generated projective module P such that m

m

P ⊕ Q1 ⊕ Q󸀠1 ⊕ ⨁ M i ≅ P ⊕ P1 ⊕ Q1 ⊕ Q󸀠1 ⊕ ⨁ M i i=1 n

n

i=2

j=2

j=1

≅ P ⊕ Q1 ⊕ P󸀠1 ⊕ Q󸀠1 ⊕ ⨁ N j ≅ P ⊕ Q1 ⊕ Q󸀠1 ⊕ ⨁ N j . (2) Suppose (M1 ) = (P1 ) and (N1 ) = (P1 ⊕ Q1 ). Then (9.1.1) reduces to m

r

s

i=2

k=2

l=1

∑ (M i ) + ∑ (P k ⊕ Q k ) + ∑ ((P󸀠l ) + (Q󸀠l )) n

r

s

j=2

k=2

l=1

= (Q1 ) + ∑ (N j ) + ∑ ((P k ) + (Q k )) + ∑ (P󸀠l ⊕ Q󸀠l ). By induction hypothesis there is a finitely generated projective module P with P ⊕ n ⊕m i=2 M i ≅ P ⊕ Q 1 ⊕ ⊕j=2 N j . Then m

m

n

n

P ⊕ ⨁ M i ≅ P ⊕ P1 ⊕ ⨁ M i ≅ P ⊕ P1 ⊕ Q1 ⊕ ⨁ N j ≅ P ⊕ ⨁ N j . i=1

i=2

j=2

j=1

(3) Finally suppose (M1 ) = (P󸀠1 ⊕ Q󸀠1 ) and (N1 ) = (P1 ⊕ Q1 ). Then m

r

s

i=2

k=2

l=2

(P󸀠1 ) + (Q󸀠1 ) + ∑ (M i ) + ∑ (P k ⊕ Q k ) + ∑ ((P󸀠l ) + (Q󸀠l )) n

r

s

j=2

k=2

l=2

= (P1 ) + (Q1 ) + ∑ (N j ) + ∑ ((P k ) + (Q k )) + ∑ (P󸀠l ⊕ Q󸀠l ).

284 | 9 K-theory m n n 󸀠 󸀠 So, P ⊕ ⊕m i=1 M i ≅ P ⊕ P 1 ⊕ Q 1 ⊕ ⊕i=2 M i ≅ P ⊕ P 1 ⊕ Q 1 ⊕ ⊕j=2 N j ≅ P ⊕ ⊕j=1 N j for some finitely generated projective module P.

Using Lemma 9.1.1 and the results in Section 2.1, it is easy to describe the Grothendieck group of a semisimple ring. Example 9.1.2. If R is a semisimple ring and (M1 ), . . . , (M k ) are the isomorphism classes of simple left R-modules then K0 (R) is a free abelian group with basis [M1 ], . . . , [M k ]. Hence, in this case, K0 (R) is free abelian with rank the number of Wedderburn components of R. If R is commutative ring then K0 (R) has a commutative ring structure with multiplication given by [M][N] = [M ⊗R N]. In this case the identity of K0 (R) is [R]. If R is a field then K0 (R) ≅ ℤ as ring.

Problems 9.1.1. Let R1 and R2 rings and R = R1 × R2 . Prove that K0 (R) ≅ K0 (R1 ) × K0 (R2 ). 9.1.2. Let R be a Dedekind domain. Prove that the following map f : ℤ × Cl(R) → K0 (R) (n, < I >) 󳨃→ (n − 1)[R] + [I], is a ring isomorphism where the addition and product in ℤ × Cl(R) are given respectively by (n, a) + (m, b) = (n + m, ab)

and (n, a)(m, b) = (nm, a m b n ).

Describe the zero divisors and the units of K0 (R). Prove that U(K0 (R)) contains a subgroup of index 2 isomorphic to Cl(R).

9.2 The Whitehead group In this section R is an arbitrary ring, I is a two-sided ideal of R and n is a positive integer. Recall that M n (R) denotes the n × n matrix ring over R, I n denotes the identity of M n (R) and GLn (R) = U(M n (R)). Recall also that for 1 ≤ i, j ≤ n and r ∈ R, Eij (r) denotes the n × n matrix having r at the (i, j)-entry and zeros at all the other entries and that the matrices of the form e ij (r) = I n + Eij (r), with i ≠ j, are called elementary matrices. In the following lemma we collect some easy to verify formulas involving elementary matrices.

9.2 The Whitehead group | 285

If g and h are elements of a group then the commutator of g and h is (g, h) = g−1 h−1 gh. If X and Y are subgroups of a group then (X, Y) denotes the subgroup generated by the commutators (x, y) with x ∈ X and y ∈ Y. Lemma 9.2.1. Let r, s ∈ R and n a positive integer. For 1 ≤ i, j, k, l ≤ n with i ≠ j and k ≠ l the following equalities hold. (1) e ij (r)e ij (s) = e ij (r + s). In particular, e ij (r) ∈ GLn (R) and e ij (r)−1 = e ij (−r). e (s), if j ≠ k, and i ≠ l; { { kl (2) e kl (s)e ij (r) = {e il (−rs)e kl (s) = e kl (s)e il (−rs), if j = k and i ≠ l; { if j ≠ k and i = l. {e kj (sr)e kl (s) = e kl (s)e kj (sr), 1, if j ≠ k, and i ≠ l; { { (3) (e kl (s), e ij (r)) = {e il (−rs), if j = k and i ≠ l; { if j ≠ k and i = l. {e kj (sr), We introduce some subgroups of GLn (R) associated to the ideal I of R: GLn (R, I) = {A ∈ GLn (R) : A − I n ∈ M n (I)}, En (I) = ⟨e ij (r) : 1 ≤ i ≠ j ≤ n, r ∈ I⟩ , En (R, I) = normal closure of En (I) in En (R). Observe that GLn (R, I) is the kernel of the natural homomorphism GLn (R) → GLn (R/I) and, in particular, GLn (R, I) is a normal subgroup of GLn (R). Remark 9.2.2. In general, En (I) ≠ En (R, I). Indeed, consider G = {a = (a ij ) ∈ GLn (R, I) : a ii − 1 ∈ I 2 }, a subgroup of GLn (R, I) containing En (I). If r ∈ R and i ≠ j then the (j, j) entry of e ji (r)e ij (1) is 1 + r. Thus, if r ∈ I \ I 2 then e ji (r)e ij (1) ∈ ̸ G and hence e ji (r)e ij (1) ∈ En (R, I) \ En (I). Recall that a unipotent matrix is an upper or lower triangular matrix having 1 at every diagonal entry. Lemma 9.2.3. En (I) contains the unipotent matrices in GLn (R, I) Proof. Let a = (a ij ) ∈ GLn (R, I) and assume that a is unipotent and upper triangular. Let b = ae12 (−a12 ) ⋅ ⋅ ⋅ e1n (−a1n ). Then b is also unipotent and upper triangular and b1i = 0 for every i > 0. By an easy induction argument one shows now that a ∈ En (I). If a ∈ M n (R) and b ∈ M m (R) then a ⊕ b denotes the block (n + m)-matrix ( 0a 0b ). Assume that n ≤ m. Then we consider GLn (R) as a subgroup of GLm (R) by identifying

286 | 9 K-theory a ∈ GLn (R) with the m × m block matrix a ⊕ I m−n . It is clear that GLn (R, I) ⊆ GLm (R, I), En (I) ⊆ Em (I) and En (R, I) ⊆ Em (R, I). We set GL(R) = ⋃ GLn (R),

GL(R, I) = ⋃ GLn (R, I),

n≥1

n≥1

E(I) = ⋃ En (I),

E(R, I) = ⋃ En (R, I).

n≥1

n≥1

Then GL(R) has an obvious group structure and the remaining subsets are subgroups of GL(R). Observe that we can identify GL(R) with the ℕ × ℕ invertible matrices which differ from the identity in only finitely many entries. In this way GLn (R) is identified with the subgroup of GL(R) formed by the matrices (a ij ) satisfying that if a ij ≠ δ ij then 1 ≤ i, j ≤ n. Lemma 9.2.4. Let I and I 󸀠 be ideals of R. (1) If n ≥ 3 then En (II 󸀠 ) ⊆ (En (I), En (I 󸀠 )) and En (R, I 2 ) ⊆ En (I). In particular, En (I 2 ) ⊆ (En (I), En (I)) and En (I) ⊆ (En (R), En (I)). (2) E(II 󸀠 ) ⊆ (E(I), E(I 󸀠 )) and E(R, I 2 ) ⊆ E(I). In particular, E(I 2 ) ⊆ (E(I), E(I)) and E(I) ⊆ (E(R), E(I)). Proof. (2) is an obvious consequence of (1) and the first inclusion of (1) is an obvious consequence of Lemma 9.2.1. It remains to show that En (I) contains every element of the form e kl (xy)e ij (r) with x, y ∈ I and r ∈ R. By Lemma 9.2.1 (2), this is clear unless j = k and i = l. Since n ≥ 3 there is m ≤ n different from k and l. Then, using the formulas in Lemma 9.2.1.((2) and (3)) we have e kl (xy)e lk (r) = (e km (x), e ml (y))e lk (r) = (e km (x)e lk (r) , e ml (y)e lk (r) ) = (e km (x)e lm (−rx), e ml (y)e mk (yr)) ∈ En (I). We now introduce a ring which will be useful to reduce a proof of a result on GLn (R, I) or En (R, I) to the case with I = R. It is the following subring of the direct product R × R: R ∝ I = {(a, b) ∈ R × R : a − b ∈ I}. We use the two natural projections R × R → R to define ring homomorphisms π 1 , π2 : M n (R ∝ I) → M n (R). More precisely, if a = (a ij ) ∈ M n (R ∝ I) and a ij = (b ij , c ij ) then π1 (a) = (b ij ) and π2 (a) = (c ij ). So, we obtain induced homomorphisms π i : GLn (R ∝ I) → GLn (R). Observe that a → (π1 (a), π2 (a)) defines a ring isomorphism ϕ : M n (R ∝ I) → M n (R) ∝ M n (I). Hence, we have the exact sequence u

π2

1 → GLn (R, I) → GLn (R ∝ I) → GLn (R) → 1

9.2 The Whitehead group |

287

where u is the mapping such that (ϕu)(a) = (a, I n ) for a ∈ GLn (R, I). Let d : GLn (R) → GLn (R ∝ I) be the map such that (ϕ ∘ d)(a) = (a, a). Then π2 ∘ d = 1 and therefore the above sequence splits. Thus we have GLn (R ∝ I) = u(GLn (R, I)) ⋊ d(GLn (R)) ≅ GLn (R, I) ⋊ GLn (R), where GLn (R) acts via conjugation on GLn (R, I). We claim that En (R ∝ I) ∩ u(GLn (R, I)) = u(En (R, I)).

(9.2.1)

(9.2.2)

Indeed, if e = e ij (r) with r ∈ I and a = e kl (s) ∈ En (R) then u(e) = e ij ((r, 0)) ∈ En (R ∝ I) and d(a) = e kl ((s, s)) ∈ En (R ∝ I). Then u(e a ) = u(e)d(a) ∈ En (R ∝ I) ∩ u(GLn (R, I)). Hence, u(En (R, I)) ⊆ En (R ∝ I) ∩ u(GLn (R, I)). To prove the converse inclusion, let a ∈ En (R ∝ I). Then, a = e i1 j1 (r1 , s1 ) ⋅ ⋅ ⋅ e i k j k (r k , s k ), with each i t ≠ j t , r t , s t ∈ R and r t − s t ∈ I. Let x t = e i t j t (s t ) ∈ En (R) and y t = e i t j t (r t − s t ) ∈ En (I) for t = 1, . . . , k. Then e i t j t (r t , s t ) = d(x t )u(y t ) and thus a = d(x1 )u(y1 ) ⋅ ⋅ ⋅ d(x k )u(y k ). If moreover a ∈ u(GLn (R, I)) then 1 = π2 (a) = x1 ⋅ ⋅ ⋅ x k . and so a = d(x1 )u(y1 ) ⋅ ⋅ ⋅ d(x k )u(y k ) = [d(x1 )u(y1 )d(x1 )−1 ] ⋅ [d(x1 x2 )u(y2 )d(x1 x2 )−1 ] ⋅ ⋅ ⋅ [d(x1 x2 ⋅ ⋅ ⋅ x k−1 )u(y k−1 )d(x1 x2 ⋅ ⋅ ⋅ x k−1 )−1 ] ⋅ d(x1 ⋅ ⋅ ⋅ x k ) ⋅ u(y k ) −1 = u(x1 y1 x−1 1 ) ⋅ u((x 1 x 2 )y 2 (x 1 x 2 ) )

⋅ ⋅ ⋅ u((x1 x2 ⋅ ⋅ ⋅ x k−1 )y k−1 (x1 x2 ⋅ ⋅ ⋅ x k−1 )−1 ) ⋅ u(y k ) ∈ u(En (R, I)). This finishes the proof of (9.2.2). Observe that (9.2.2) is equivalent to the exactness of the first row of the following commutative diagram, where the vertical maps are inclusions: u

π2

1 → En (R, I) → En (R ∝ I) → En (R) → 1 ↓ ↓ ↓ π2 u 1 → GLn (R, I) → GLn (R ∝ I) → GLn (R) → 1 As d(En (R)) ⊆ En (R ∝ I), (9.2.1) restricts to En (R ∝ I) = u(En (R, I)) ⋊ d(En (R)) ≅ En (R, I) ⋊ En (R).

(9.2.3)

Lemma 9.2.5 (Whitehead’s Lemma). Let I be an ideal of R. The following properties hold. (1) (GLn (R), GLn (R, I)) ⊆ E2n (R, I). (2) E(R, I) = (GL(R), GL(R, I)) = (E(R), E(R, I)). (3) GL(R)󸀠 = E(R).

288 | 9 K-theory Proof. (1) If a ∈ GLn (R) then a ⊕ a−1 ∈ E2n (R) because a ⊕ a−1 = (

In − In

a−1

In 0 )( In 0

In In )( a − In In

In 0 )( In 0

−a−1 ). In

(9.2.4)

Hence, if a, b ∈ GLn (R) then (a, b) = (a, b) ⊕ I n = (a−1 ⊕ a)(b−1 ⊕ b)(ab ⊕ (ab)−1 ) ∈ E2n (R). This shows that (GLn (R), GLn (R)) ⊆ E2n (R)

(9.2.5)

for every ring. If x ∈ GLn (R) and y ∈ GLn (R, I) then u((x, y)) = (d(x), u(y)), where the maps u and d are as defined just before (9.2.1). Therefore u (GLn (R), GLn (R, I)) ⊆ (GLn (R ∝ I), GLn (R ∝ I)). Combining this with (9.2.2) and (9.2.5) applied to R ∝ I we obtain u (GLn (R)), GLn (R, I)) ⊆ (GLn (R ∝ I), GLn (R ∝ I)) ∩ u(GLn (R, I)) ⊆ E2n (R ∝ I) ∩ u(GL2n (R, I)) = u(E2n (R, I)). As u is injective (GLn (R), GLn (R, I)) ⊆ E2n (R, I). (2) Using (1) and Lemma 9.2.4 (2) we have E(I) ⊆ (E(R), E(I)) ⊆ (E(R), E(R, I)) ⊆ (GL(R), GL(R, I)) ⊆ E(R, I). As the last three groups are normal in E(R), we deduce E(R, I) = (E(R), E(R, I)) = (GL(R), GL(R, I)), as desired. (3) is a special case of (2). Corollary 9.2.6. If a ∈ GL(R) and b ∈ GL(R, I) then ab ≡ ba ≡ a ⊕ b ≡ b ⊕ a

mod E(R, I).

Proof. By Lemma 9.2.5, (ba)−1 (ab) = (a, b) ∈ E(R, I). From (9.2.4) we know that b ⊕ b−1 = UAVB, with U, V unipotent lower triangular matrices, A and B unipotent upper triangular matrices such that U, V, AB ∈ GL(R, I) and A, B ∈ E(R). Hence, by Lemma 9.2.3, U, V, AB ∈ E(I) and hence b ⊕ b −1 = U(AVA−1 )AB ∈ E(R, I). Therefore ba(a ⊕ b)−1 = b ⊕ b−1 ∈ E(R, I). Hence the result follows. By Whitehead’s Lemma (Lemma 9.2.5), E(R, I) is a normal subgroup of GL(R). The group K1 (R, I) = GL(R, I)/E(R, I) is called the Whitehead group of the ring R relative to the ideal I. The Whitehead group of R is K 1 (R) = K1 (R, R) = GL(R)/E(R) = GL(R)/GL(R)󸀠 .

9.2 The Whitehead group |

289

The inclusion GL(R, I) ⊆ GL(R) induces a group homomorphism K 1 (R, I) → K1 (R). On the other hand, if f : R → S is a ring homomorphism then the componentwise extension of f induces a group homomorphism GL(R) → GL(S) which in turn induces a group homomorphism K1 (f) : K1 (R) → K1 (S). It is easy to verify that this defines a functor K1 from the category of rings to the category of abelian groups. Our next aim is to present a homological description of K1 (R, I). Let CI denote the category whose objects are pairs (M, f) formed by a finitely generated projective left Rmodule M and an automorphism f : M → M such that the image of f − 1M is included in IM. A morphism ϕ : (M, f) → (M 󸀠 , f 󸀠 ) in CI is a homomorphism ϕ : M → M 󸀠 of R-modules so that the following diagram is commutative f

M 󳨀→ M ϕ↓ ↓ϕ f󸀠

M 󸀠 󳨀→ M 󸀠 . We identify each automorphism f of R n with the matrix associated to f in the standard basis of R n , in other words interpreting an element x ∈ R n as a column matrix, then the automorphism identified with a ∈ GLn (R) is defined by x 󳨃→ ax. Clearly, a ∈ GLn (R, I) if and only if (R n , a) is an object of CI . For each object (M, f) in CI let {M, f} denote the isomorphism class of CI containing (M, f). Let A I denote the free abelian (additive) group with basis the set of isomorphism classes of CI . Definition 9.2.7. Let (M, f) be an object in CI . Fix an isomorphism g : M ⊕ N → R n for some left R-module N and let α be the automorphism of R n defined by α(g(m + n)) = g(f(m) + n), for m ∈ M and n ∈ N. Observe that α ∈ GLn (R, I) because (α − 1R n )(R n ) = g((f − 1M )(M)) ⊆ g(IM) ⊆ I (n) . (Here we use I (n) instead of I n to represent the direct sum of n copies of I to avoid confusion with the ideal I n .) Furthermore α is the only element in GLn (R) for which g is an isomorphism g : (M, f) ⊕ (N, 1N ) ≅ (R n , α). We show that the class of α in K1 (R, I) is independent of the choice of the isomorphism g : M ⊕ N ≅ R n . To do so, assume (R n1 , α1 ) ≅ (M, f) ⊕ (N1 , 1N1 ) and (R n2 , α2 ) ≅ (M, f)⊕(N2 , 1N2 ). Then (R n1 , α1 )⊕(R n2 , 1R n2 ) ≅ (M, f)⊕(N1 , 1N1 )⊕(M, 1M )⊕(N2 , 1N2 ) ≅ (R n1 , 1R n1 ) ⊕ (R n2 , α2 ). That is, there is h ∈ GLn1 +n2 (R) such that h(α1 ⊕ 1) = (1 ⊕ α2 )h.

290 | 9 K-theory Then (1 ⊕ α2 )(α1 ⊕ 1)−1 = h(α1 ⊕ 1)h−1 (α1 ⊕ 1)−1 ∈ (GL(R), GL(R, I)) = E(R, I), by Lemma 9.2.5. Therefore α 1 and α2 represent the same class in K1 (R, I), as desired. Hence, we obtain a group homomorphism Ψ : A I → K1 (R, I) defined by mapping the class {M, f} onto the class of α ∈ GLn (R) in K1 (R, I), with α as in Definition 9.2.7. Let N I be the subgroup of A I generated by the elements of the following two types: – {M, h1 } + {M, h2 } − {M, h1 h2 }, for each (M, h1 ), (M, h2 ) ∈ CI . – {M, f} − {M1 , f1 } − {M2 , f2 } with 0 → (M1 , f1 ) → (M, f) → (M2 , f2 ) → 0 an exact sequence in CI . The class of {M, f} in A I /N I is denoted [M, f]. Theorem 9.2.8. The assignment α 󳨃→ [R n , α], with α ∈ GLn (R, I), defines a group iso−1 morphism Φ : K1 (R, I) → A I /N I . The inverse is given by Φ ([M, f]) = Ψ({M, f}). Proof. Let (M, h1 ) and (M, h2 ) be objects of CI and fix an automorphism g : M ⊕ N → R n . Let α i ∈ GLn (R) be such that g defines an isomorphism (M, h i ) ⊕ (N, 1N ) ≅ (R n , α i ) for each i = 1, 2. Then (α1 α2 g)(m + n) = g((h1 h2 )(m) + n). Therefore Ψ({M, h1 } + {M, h2 }) = α1 α2 = Ψ({M, h1 h2 }) Now let

u1

(9.2.6)

p2

0 → (M1 , f1 ) → (M, f) → (M2 , f2 ) → 0 be an exact sequence in CI and fix isomorphisms g1 : (M1 , f)⊕(N1 , 1) → (R n1 , α1 ) and g2 : (M2 , f) ⊕ (N2 , 1) → (R n2 , α2 ) in CI . As M2 is projective the exact sequence splits in the category of left R-modules and hence there are homomorphisms p1 : M → M1 and u2 : M2 → M such that p i ∘ u i = 1M i and u1 p1 + u2 p2 = 1M . Therefore there is an isomorphism g : M ⊕ N1 ⊕ N2 ≅ R n1 × R n2 = R n1 +n2 with g(m + n1 + n2 ) = (g1 (p1 (m)+n1 ), g2 (p2 (m)+n2 )). Let α be the automorphism of R n1 +n2 in Definition 9.2.7 corresponding to f and g, i.e. such that g is an isomorphism (M, f) ⊕ (N1 ⊕ N2 , 1) ≅ (R n1 +n2 , α). Let x ∈ R n1 and y ∈ R n2 then x = g1 (m1 + n1 ) and y = g2 (m2 + n2 ) for unique m i ∈ M i and n i ∈ N i . Let m = u1 (m1 ) + u2 (m2 ). Then m1 = p1 (m) and m2 = p2 (m). Let β : R n2 → R n1 be defined by β(y) = g1 (p1 (f(u2 (m2 )))). Observe that p1 (f(u2 (m2 ))) = p1 (f(m − u1 (m1 ))) = p1 f(m − u1 p1 (m)) = p1 f(m) − p1 (m) = p1 (f − 1M )(m) ∈ p1 (IM) ⊆ IM1 . Thus β(R n2 ) ⊆ I (n1 ) and hence ( α01 αβ2 ) ∈ GLn1 +n2 (R, I). Moreover, α(x, y) = α(g1 (m1 + n1 ), g2 (m2 + n2 )) = α(g1 (p1 (m) + n1 ), g2 (p2 (m) + n2 )) = α(g(m + n1 + n2 ))

9.2 The Whitehead group

| 291

= g(f(m) + n1 + n2 ) = (g1 (p1 (f(m)) + n1 ), g2 (p2 (f(m)) + n2 )) = (g1 (p1 (f(u1 (p1 (m)) + u2 (p2 (m)))) + n1 ), g2 (f2 (p2 (m)) + n2 )) = (g1 (p1 (u1 (f1 (m1 ))) + p1 (f(u2 (m2 ))) + n1 ), g2 (f2 (m2 ) + n2 ))) = (g1 (f1 (m1 ) + n1 ) + g1 (p1 (f(u2 (m2 )))), α2 (g2 (m2 + n2 ))) = (α1 (g1 (m1 + n1 )) + g1 (p1 (f(u2 (m2 )))), α2 (g2 (m2 + n2 ))) = (α1 (x) + β(y), α2 (y)). Thus α=( I

α1 0

α1 β )=( α2 0

In 0 )( 1 α2 0

α−1 1 β) . I n2

α−1 β

By Lemma 9.2.3, ( 0n1 I1 ) ∈ E(I) and therefore the image of α in K1 (R, I) coincides n2 with the image of α1 ⊕ α2 . In other words Ψ({M, f}) = Ψ({M1 , f1 }) ⊕ Ψ({M2 , f2 }). Using Corollary 9.2.6 we conclude Ψ({M, f}) = Ψ({M1 , f1 })Ψ({M2 , f2 }) = Ψ({M1 , f1 } + {M2 , f2 }).

(9.2.7)

By (9.2.6) and (9.2.7), N I ⊆ ker(Ψ) and hence Ψ induces a group homomorphism Ψ : A I /N I → K1 (R, I). Because

[R n , αβ]

=

[R n , α]

+ [R n , β], we have a group homomorphism Φ : GL(R, I) → A I /N I α

󳨃→ [R n , α].

Every g ∈ GLn (R) defines an isomorphism (R n , f) → (R n , gfg−1 ), where f ∈ GLn (R, I). Therefore E(R, I) = (GL(R), GL(R, I)) ⊆ ker(Φ). Thus, Φ induces a group homomorphism Φ : K1 (R, I) → A I /N I . Clearly, Ψ(Φ(α)) = α for every α ∈ GL(R, I). Thus Φ is injective. Furthermore, if (M, f) is an object in CI then (M, f) ⊕ (N, 1) = (R n , α) for some α ∈ GL(R, I) and some projective R-module N. As [N, 1] = [N, 1] + [N, 1] − [N, 1] = 0 we deduce that [M, f] = [M, f] + [N, 1] = [R n , α] = Φ(α). Therefore, Φ is surjective. Hence, indeed Φ is an isomorphism and the result follows. We freely will use Theorem 9.2.8 and represent elements of K1 (R, I) either as invertible matrices (modulo E(R, I)) or as integral linear combinations of equivalent classes of objects of CI modulo N I . We finish this section with the following lemma that will be used in the proof of Theorem 9.5.1. It can be proven by making use of standard arguments. Lemma 9.2.9. Let R be a commutative ring and S an R-algebra and I an ideal of S. Then the following product endows K1 (S, I) with a structure of K0 (R)-module: [P] ⋅ [M, f] = [P ⊗R M, 1 ⊗ f].

292 | 9 K-theory

Problems 9.2.1. Let S be a subring of a ring R. Let I be an ideal of R contained in S. Prove that GLn (R, I) = GLn (S, I), GL(R, I) = GL(S, I), E(R, I) = E(S, I) and K 1 (R, I) = K1 (S, I). 9.2.2. Prove that if R1 and R2 are rings, I1 is an ideal of R1 and I2 is an ideal of R2 then K1 (R1 × R2 , I1 × I2 ) ≅ K1 (R1 , I1 ) × K1 (R2 , I2 ).

9.3 Stable range condition Let R be a ring and I an ideal of R. The kernel of the natural group homomorphism GLn (R, I) → K1 (R, I) contains En (R, I). We would like to find sufficient conditions for this homomorphism to be surjective and for En (R, I) to coincide with the kernel for some n. The reason being that it would give a finite realization of K1 (R, I) as GLn (R, I)/En (R, I). This is the role of the stable range condition which we define now. Let R be a ring and let M be a right R-module. For m ∈ M let M[m] = {f(m) : f ∈ HomR (M, R)}. Clearly M[m] is a left ideal of R. Furthermore, if r ∈ R then R[r] = Rr. More generally, if (r1 , . . . , r n ) ∈ R n then R n [(r1 , . . . , r n )] = Rr1 +⋅ ⋅ ⋅+Rr n . We say that m is unimodular in M if M[m] = R. In particular, (r1 , . . . , r n ) is unimodular in R nR if and only if ∑ni=1 Rr i = R. Definition 9.3.1. The stable range of R, denoted strR, is the minimum positive integer n (if it exists) satisfying any of the following equivalent conditions: (1) For every k ≥ n and every unimodular element (r, x) of (R × R k )R there is y ∈ R k such that x + yr is unimodular in R kR . (2) For every (r1 , . . . , r n ) ∈ R n and every finitely generated left ideal I of R such that Rr1 + ⋅ ⋅ ⋅ + Rr n + I = R there are x1 , . . . , x n ∈ I such that (r1 + x1 , . . . , r n + x n ) is unimodular in R nR . If such a positive integer n does not exist then we write str(R) = ∞. Clearly fields have stable range one. The following lemma provides a larger class of rings with stable range one. Recall that a ring R is semilocal if R/J(R) is semisimple Artinian, where J(R) denotes the Jacobson radical of R, or equivalently R/J(R) is Artinian. Lemma 9.3.2. If R is semilocal then str(R) = 1. Proof. Let J = J(R). If x ∈ R and (R/J)(x + J) = R/J then yx ∈ 1 + J ⊆ U(R) for some y ∈ R and hence R = Ryx ⊆ Rx ⊆ R. Using this, it follows that it is enough to prove the lemma under the assumption that R is semisimple. Indeed, assume that the lemma holds for semisimple rings and let R be semilocal. Let r ∈ R and I an ideal of R with

9.3 Stable range condition

|

293

Rr + I = R. Then R/J = (R/J)(r + J) + (I + J)/J. By the assumption, there is a ∈ I with R/J(a + x + J) = R/J. Then, by the above, R(a + x) = R, as desired. Thus we may assume that R is semisimple. Then, we can further reduce the proof to the case where R is simple because obviously the stable range of a direct product of two rings is the maximum of the stable ranges of the factors. Thus one may assume, without loss of generality, that R is the endomorphism ring of a finite dimensional right vector space V over a division algebra D. Let a ∈ R and I a left ideal of R so that Ra + I = R. We need to show that R(a + x) = R for some x ∈ I. As R is semisimple, I = Re for some idempotent e of R. Let W = ker(e) and write V = W ⊕ W 󸀠 = a(W) ⊕ U, for some subspaces W 󸀠 and U of V. Then W ∩ ker(a) = 0. Indeed, assume w ∈ W ∩ ker(a). Write 1 = ra + i, with r ∈ R and i ∈ I. Then w = ra(w) + i(w) = 0 + i(w) = 0, as claimed. Hence, a|W is injective. Thus W 󸀠 and U have the same dimension and therefore V D has an automorphism u such that u|W = a|W and u(W 󸀠 ) = U. That is, u is a unit of R such that (u − a)(W) = 0. Then (u − a)(1 − e) = 0 and hence u − a ∈ Re = I. Thus R = R(a + x) with x = u − a ∈ I, as desired. Observe that 3ℤ+5ℤ = ℤ but 3+5ℤ does not contains units of ℤ. In particular str(ℤ) > 1. We will show that if R is an order then str(R) = 2, and in particular str(ℤ) = 2. For the moment we prove. Lemma 9.3.3. If R is an order then str(R) > 1. Proof. Suppose R is an order in the ℚ-algebra A. Let n = dimℚ (A). Fix a basis x1 , . . . , x n of A contained in R and let ρ : A → M n (ℚ) be the regular representation of A with respect to this basis. Fix an integer m such that mR ⊆ ℤx1 +⋅ ⋅ ⋅+ℤx n and choose different prime integers p and q such that p ∤ m and p n ≢ ±1 mod q. As (p, qm) is unimodular in ℤ2 , (p1R , qm1R ) is unimodular in R2R . Assume that str(R) = 1. Then there is x ∈ R such that u = p1R +qmx is invertible in R. Taking norms, Lemma 4.6.9 (2) yields that ±1 = NrA/ℚ (u) = det(pI n + qρ(mx)) ≡ p n mod q, contradicting with the choice of p and q. Let R be a ring. Recall that the Krull dimension of a ring R, denoted Kdim(R), is the supremum of the non-negative integers n such that there is a strongly increasing chain with n + 1 prime ideals of R. The following proposition is crucial for proving that the stable range of an order is at most 2. Proposition 9.3.4 ([15]). Let C be a Noetherian commutative ring of Krull dimension d and let R be a C-algebra which is finitely generated as C-module. Let P and Q be right R-modules such that P is free of rank greater than d and Q is projective. Let p ∈ P, q ∈ Q and let I be a left ideal of R. If P[p] + Q[q] + I = R then P[p + f(q)] + I = R for some homomorphism f : Q → P. Proof. Assume P[p] + Q[q] + I = R. As C is Noetherian, it has finitely many minimal primes ideals, say c1 , . . . , c k . For every i = 1, . . . , k fix a maximal ideal m i of C containing c i and let J = ∩ki=1 m i . Then C/J is semisimple Artinian and R = R/JR is a C/J-

294 | 9 K-theory

algebra which is finitely generated as C/J-module. Thus R/JR is Artinian, and in particular semilocal. Hence str(R) = 1, by Lemma 9.3.2. Consider S = ∩ki=1 (C \ m i ), a multiplicatively closed subset of C. Then the maximal ideals of S−1 C are S−1 m1 , . . . , S−1 m k and hence S−1 J = J(S−1 C). Let n be the rank of P. So, by assumption, n > d. Fix a basis p1 , . . . , p n of P R and write n

p = ∑ pi xi , i=1

∑ni=1

with x i ∈ R. Then R = P[p] + Q[q] + I = Rx i + Q[q] + I and therefore R = ∑ni=1 Rx i + Q[q] + I, where the bar-notation stands for the natural image in R. As str(R) = 1, there are r2 , . . . , r n ∈ R, a homomorphism g : Q → R and c ∈ I such that R = Ru + JR for n

u = x1 + ∑ r i x i + g(q) + c = b + g(q) + c. i=2

S−1 R

S−1 Ru + S−1 JR

= = S−1 Ru + J(S−1 R). So, by Nakayama’s Lemma, S−1 Ru = Hence S−1 R. Therefore there exists t ∈ S ∩ Ru ⊆ S ∩ (R(b + g(q)) + I).

(9.3.1)

For every i = 2, . . . , n let p 󸀠i = p i − p1 r i . Then p1 , p󸀠2 , . . . , p󸀠n is another basis of P R and P1 = p󸀠2 R+⋅ ⋅ ⋅+p󸀠n R is free of rank n−1. So P = p1 R⊕P1 and we define h : P → R by h|P1 = 0 and h(p1 x) = x for every x ∈ R. In other words, x 󳨃→ p1 h(x) is the projection of P onto p1 R along the decomposition P = p1 R ⊕ P1 . Let g1 : Q → p1 R ⊆ P be given by g1 (x) = p1 g(x). Observe that n

n

n

i=1

i=2

i=2

p − p1 b = ∑ p i x i − p1 (x1 + ∑ r i x i ) = ∑ p󸀠i x i ∈ P1 . Then b + g(q) = h(p + g 1 (q))) = h(p1 b + g1 (q)), and thus b + g(q) ∈ P[p + g1 (q)] ∩ (p1 R)[p1 b + g1 (q)].

(9.3.2)

We argue by induction on d. If d = 0 then Spec(C) = {m1 , . . . , m k } and therefore every element of S is invertible in C. In particular Rt = R and hence Ru = R, by (9.3.1). So, by (9.3.2), P[p + g1 (q)] + I = R, as desired. The same argument works for every d provided that t1R ∈ U(R). So, assume that d > 0 and t1R ∈ ̸ U(R). Consider the rings C∗ = C/Ct and R∗ = R/Rt and for every right R-module M let M ∗ = M/Mt, considered as right R∗ -module. Let x∗ denote the natural image of x ∈ M in M ∗ . Then C∗ is a commutative Noetherian with Krull dimension at most d − 1, because t is not contained in any minimal prime ideal of C. Furthermore R∗ is a C∗ -algebra which is finitely generated as module over C∗ , P∗ = P/Pt and P∗1 = P1 /P1 t are free of ranks n and n − 1 respectively, Q∗ = Q/Qt is projective and R = P[p] + Q[q] + I = (p1 R)[p1 b] + P1 [p − p1 b] + Q[q] + I ⊆ P1 [p − p1 b] + Q[q] + Rb + I.

9.4 Whitehead group and the stable range condition

|

295

Hence there exist h1 ∈ HomR (P1 , R), h2 ∈ HomR (Q, R), λ ∈ R and x ∈ I so that 1 = h 1 (p − p1 b) + h2 (q) + λb + x. Therefore, 1 = h1 (p − p1 b) + (h2 − λg)(q) + λ(b + g(q)) + x ∈ P1 [p − p1 b] + Q[q] + R[b + g(q)] + I. Thus R∗ = P∗1 [(p − p1 b)∗ ] + Q∗ [q∗ ] + R∗ [(b + g(q)∗ )] + I ∗ . By the induction hypothesis there is a homomorphism g2󸀠 : Q∗ → P∗1 such that P∗1 [(p − p1 b)∗ + g2󸀠 (q∗ )] + R∗ (b + g(q))∗ + I ∗ = R∗ . As the natural map P1 → P∗1 is surjective and Q is a projective R-module by assumption, g2󸀠 lifts to a homomorphism g2 : Q → P1 . Hence, R = P1 [p − p1 b + g2 (q)] + R(b + g(q)) + I + Rt. Using (9.3.1), (9.3.2) and Problem 9.4.1 we obtain R = P 1 [p − p1 b + g2 (q)] + R(b + g(q)) + I ⊆ P 1 [p − p1 b + g2 (q)] + (p1 R)[p1 b + g1 (q)] + I = P(p + (g1 + g2 )(q)) + I. This finishes the proof. Theorem 9.3.5 (Bass). Let C be a Noetherian commutative ring and R a C-algebra which is finitely generated as C-module. Then str(R) ≤ Kdim(C) + 1. Proof. Let d = Kdim(C), n ≥ d + 1 and (a0 , a1 , . . . , a n ) a unimodular element of R n+1 R . Then R[a0 ] + R n [a1 , . . . , a n ] = R. By Proposition 9.3.4, there is g : R → R n such that R n [(a1 , . . . , a n ) + g(a0 )] = R. If g(1) = (r1 , . . . , r n ) then (a1 + r1 a0 , . . . , a n + r n a0 ) = (a1 , . . . , a n ) + g(a0 ) is unimodular in R nR . If R is an order then R is finitely generated as ℤ-module. From Kdim(ℤ) = 1, Lemma 9.3.3 and Theorem 9.3.5 we obtain the desired result. Corollary 9.3.6. If R is a order then str(R) = 2.

9.4 Whitehead group and the stable range condition In this section we determine some consequences of the stable range of R being finite on the homomorphism GLn (R, I) → K1 (R, I). We begin with two lemmas. Every x ∈ R n is considered as a 1 × n-matrix and hence x T denotes the transposed of x. Lemma 9.4.1. If I is an ideal of R then str(R) = str(R ∝ I). Proof. Let S = R ∝ I. For every x ∈ R n or S n , let x󸀠 denote the n − 1 dimensional vector formed by the first n − 1 coordinates of x and let x󸀠󸀠 denote the last coordinate of x. If x ∈ S n then x1 (respectively, x2 ) denotes the element of R n formed with the first (respectively, second) coordinates of the entries of x.

296 | 9 K-theory The inequality str(R) ≤ str(S) is clear. To prove the converse inequality we assume that n > str(R) and fix a unimodular element x of S n . Then x1 and x2 are unimodular in R n . We have to show that the set x󸀠 +S n−1 x󸀠󸀠 contains a unimodular element of S n−1 . For s ∈ S n , put x̄ = x+sx󸀠󸀠 . Clearly, x̄ 󸀠 +S n−1 x̄ 󸀠󸀠 = (x󸀠 +s󸀠 x󸀠󸀠 )+S n−1 (x󸀠󸀠 +s󸀠󸀠 x󸀠󸀠 ) ⊆ x󸀠 +S n−1 x󸀠󸀠 . Hence, if x̄ 󸀠 + S n−1 x̄ 󸀠󸀠 contains a unimodular element then so does x󸀠 + S n−1 x󸀠󸀠 . Making use of this remark, we will reduce the proof to some special cases by consecutively replacing x by another unimodular element of the form x + sx󸀠󸀠 , with s ∈ S n , until the final element satisfies the required condition. Note that, as x is unimodular in S n , both x1 and x2 are unimodular in R n . First reduction step: We may assume that x󸀠1 is unimodular in R n−1 . Indeed, since n−1 for some r 󸀠 ∈ x1 is unimodular in R n and n > str(R), x󸀠1 + r󸀠 x󸀠󸀠 1 is unimodular in R R n−1 . Let x̄ = x + sx󸀠󸀠 with s󸀠1 = s󸀠2 = r󸀠 and s󸀠󸀠 = 0. Then S n [x]̄ = S n−1 [x̄ 󸀠 ] + S x̄ 󸀠󸀠 = n S n−1 [x󸀠 + s󸀠 x󸀠󸀠 ] + Sx󸀠󸀠 = S n [x] and x̄ 󸀠1 = x󸀠1 + r󸀠 x󸀠󸀠 1 . As x is unimodular in S , it follows n that x̄ is unimodular in S . Hence may replace x by x̄ and thus, indeed, we may assume that x󸀠1 is unimodular in R n . Second reduction step: We may assume that x 󸀠1 is unimodular in R n−1 and x󸀠󸀠 1 = 0. By the first reduction step we may assume that x󸀠1 is unimodular in R n−1 . Thus there T T T T is r ∈ R n−1 such that rx󸀠1 = 1. Then rx󸀠2 ≡ rx󸀠1 = 1 mod I. Let y = rx󸀠2 − 1 and let s ∈ S n−1 with s1 = s2 = r. Then y ∈ I and (1, 1 + y) = sx󸀠T ∈ S n−1 [x󸀠 ]. Furthermore, as x is unimodular in S n , (1, 1) − (u, v)x󸀠󸀠 ∈ S n−1 [x󸀠 ] for some (u, v) ∈ S. Multiplying on the left by (0, y) (an element of S, because y ∈ I) we have (0, y) − (0, yv)x󸀠󸀠 ∈ S n−1 [x󸀠 ], and therefore (1, 1) = (1, 1 + y) − (0, y) ≡ (0, −yv)x󸀠󸀠

mod S n−1 [x󸀠 ].

(9.4.1)

Let z be the element of S n with z󸀠 = 0 and z󸀠󸀠 = (−1, −1 − yv) and set x̄ = x + zx󸀠󸀠 . Then S n [x]̄ = S n−1 [x̄ 󸀠 ] + S x̄ 󸀠󸀠 = S n−1 [x󸀠 ] + S((0, −yv)x󸀠󸀠 ) contains (1, 1), by (9.4.1). Thus x̄ is unimodular in S n . Moreover x̄ 󸀠1 = x󸀠1 is unimodular in R n−1 and x̄ 󸀠󸀠 = x󸀠󸀠 + z󸀠󸀠 x󸀠󸀠 = (0, −yv)x󸀠󸀠 , so that x̄ 󸀠󸀠 1 = 0. By the second reduction step we may assume that x󸀠1 is unimodular in R n−1 and 󸀠󸀠 x1 = 0. Moreover, x2 is unimodular in R n and n > str(R). Thus there is y = x + sx󸀠󸀠 , 󸀠 󸀠 with s ∈ S n such that y󸀠2 is unimodular in R n−1 . As x󸀠󸀠 1 = 0, x 1 = y 1 and hence y 1 = x 1 is unimodular in R n−1 . Hence S n−1 [y󸀠 ] contains an element of the form (1, u). Moreover, as y󸀠2 is unimodular in R n−1 , S n−1 [y󸀠 ] contains an element of the form (v, 1). Then (1, 1) = (1, u) + (0, 1 − u)(v, 1) ∈ S n−1 [y󸀠 ] and hence y󸀠 is unimodular in S n−1 . By the calculations in the first part of the proof y󸀠 ∈ y󸀠 + S n−1 y󸀠󸀠 ⊆ x󸀠 + S n−1 x󸀠󸀠 . This finishes the proof. Lemma 9.4.2. If str(R) < n then every element of En+1 (R) can be written in the following form A 0 In 0 1 0 1 0 In bT (9.4.2) ( )( )( ) )( T )( 0 1 0 B 0 1 c In a 1 for some A, B ∈ En (R) and a, b, c ∈ R n .

9.4 Whitehead group and the stable range condition

|

297

Proof. First observe that if U ∈ GLr (R), V ∈ GLs (R), C ∈ M r,s (R) and D ∈ M s,r (R) then U 0

0 Ir )( V 0

Ir C )=( Is 0

U C󸀠 )( 0 Is

0 ) V

Ir ( D

U 0 )( 0 Is

0 U )=( V 0

0 Ir )( 󸀠 V D

0 ), Is

(

and (9.4.3)

for C󸀠 = UCV −1 and D󸀠 = V −1 DU. We will make use of Lemma 9.2.3 without explicit reference. So we have that E n (R) a t ) and ( I n−1 0 ), for a ∈ R n−1 . contains all the matrices of the forms ( I n−1 a 1 0 1 Let e1 , . . . , e n denote the elements of the standard basis of R n . To prove the lemma, it is enough to show that if x=(

A 0

0 In )( 1 a

0 In )( 0 1

1 bT )( T 1 c

1 0 )( 0 In

0 ), B

as in (9.4.2), then e ij (r)x also is of the form (9.4.2) for every r ∈ R and every i ≠ j, with 1 ≤ i, j ≤ n + 1. This is easy if both i and j are different from n +1 because in that case e ij (r)( A0 01 ) = e ij (r)A 0 ( 0 1 ). If i = n + 1 then, by (9.4.3) we have e(n+1)j (r) (

A 0

0 In )( 1 a

0 In )=( 1 re j

A 0 )( 1 0

0 In )( 1 a

=(

A 0

0 In )( 1 v

=(

A 0

0 In )( 1 v+a

0 In )( 1 a

0 ) 1 0 ) 1

0 ) 1

for some v ∈ R n , and the claim follows. The hard case is for j = n + 1. Let v T = (v1 , . . . , v n )T = A−1 e Ti , the i-th column of −1 A . By (9.4.3), we have (

In 0

e Ti r A )( 1 0

0 A )=( 1 0

0 In )( 1 0

vT r ). 1

(9.4.4)

Moreover v is unimodular in R n , because it is a column of an element of GLn (R). As by assumption str(R) < n, there is a unimodular vector of the form p = (v2 , . . . , v n )−wv1 , for some w ∈ R n−1 . Let X=(

1 wT

0 ) ∈ En (R), I n−1

w1 = (w, 0)

Then (

X 0

0 1 )=( T 1 w1

0 ). In

and

c1 = c − w1 .

298 | 9 K-theory

Hence (

1 cT

0 X )=( 0 In

0 1 )( T 1 c1

0 ). In

(9.4.5)

By (9.4.3) and (9.4.5) we have e i(n+1) (r)x = (

In 0

e Ti r A )( 1 0

(

X 0

0 1 )( T 1 c1

0 1 )( 0 In

A = ( 0

In 0 )( 1 0

vT r X )( 1 0

(

In 0

0 In )( 1 a

0 In )( 1 0

bT ) 1

0 ) B (9.4.6)

b1T 1 )( T 1 c1

0 In )( 1 a1

1 0 )( 0 In

0 ) 1

0 ), B

v

for some a1 , b1 ∈ R n . Observe that X −1 v T = ( p1T ). Since p is unimodular in R n−1 , there is q ∈ R n−1 such that qp T = −v1 . Let A1 = AX and let A2 = A1 ( 10 I−q ). Then n−1 (

A 0

0 In )( 0 1

vT r X )( 1 0

0 A )=( 1 0

0 X )( 1 0

vT r ) 1

=(

A1 0

0 In )( 1 0

X −1 v T r ) 1

=(

A2 0

1 0 ) (0 1 0

q I n−1 0

1 0 0) (0 1 0

=(

A2 0

1 0 ) (0 1 0

q I n−1 0

0 p T r) 1

=(

A2 0

1 0 ) (0 1 0

0 I n−1 0

0 1 p T r) (0 1 0

Observe that A1 and A2 belong to En (R), since A, X ∈ En (R).

0 I n−1 0

v1 r p T r) 1 (9.4.7)

q I n−1 0

0 0) 1

9.4 Whitehead group and the stable range condition

|

299

Applying (9.4.3) we obtain the following rewriting. 1 (0 0

q I n−1 0

0 In 0) ( a1 1

b1T ) 1

In 0 )( 1 0

1 0 ) (0 1 0

In =( a2 1 = (0 v

q

0 1 0) (0 1 0

I n−1 0

1 0 0) (0 1 0

0 I n−1 0

0 I n−1 a󸀠

0

z yT ) 1

I n−1 0

0 1 0) ( 0 1

(9.4.8)

b2 1 )( 0 In

0 ), U

for some z, v ∈ R, a2 ∈ R n , y, a󸀠 ∈ R n−1 and U ∈ En (R). Combining (9.4.6), (9.4.7) and (9.4.8) we have e i(n+1) (r)x = (

(

= (

A2 0

1 0 ) (0 1 0

In 0

b1T 1 )( T 1 c1

A2 0

1 0 ) (0 1 0

1 ( 0

b2 1 )( In 0

0 I n−1 0

0 1 p T r ) (0 1 0

1 0 )( 0 In 0 I n−1 0

0 In 0) ( a1 1

0 ) 1

0 I n−1 0

0 1 0) (0 1 0

0 I n−1 a󸀠

0 ) B

1 0 p T r ) (0 1 v

0 1 )( T U c1

q I n−1 0

0 1 )( In 0

0 0) 1

0 ) B

Rewriting the product of the first two matrices and applying (9.4.3) we obtain e i(n+1) (r)x = (

A2 0

1 (0 0 = (

A3 0

1 0 ) (p T rv 1 0 0 I n−1 a󸀠

0 I n−1 0

0 1 0) ( 0 1

0 In )( 1 a3

with B1 = UB, A3 = A2 ( p T1rv (9.4.3) yields

1 b2 )( T c2 In

1 0 )( 1 0 0 I n−1 ),

1 0 0) (0 1 v

0 1 )( V 0

V = ( I n−1 0

0 I n−1 0

0 1 )( 0 In

0 1 0) (0 1 0

0 I n−1 0

0 p T r) 1

0 ) B1

1 b2 )( T c2 In p T r )( I n−1 0 ). a󸀠 1 1

0 1 )( 0 In

0 ), B1

Some more applications of

300 | 9 K-theory

A3 0

0 In )( 1 a3

1 0 )( 1 0

A3 = ( 0

0 In )( 1 a3

1 0 ) (0 1 0

e i(n+1) (r)x = (

1 b3 )( T c3 In b󸀠

0 1 )( 0 In

t 1 0) ( T c3 1

I n−1 0

0 ) B2

1 0 )( 0 In

0 ) B2

with B2 = VB1 . Factorizing the middle matrix b󸀠

1 (0 0

I n−1 0

1 t 0) = (0 1 0

b󸀠 I n−1 0

1 0 0) (0 1 0

0 I n−1 0

t 0) 1

0 I n−1 0

t 1 0) ( T c3 1

and applying (9.4.3) once more we conclude with 1 A3 0 In 0 e i(n+1) (r)x = ( )( ) (0 0 1 a3 1 0 = (

b󸀠 I n−1 0

1 0 0) (0 1 0

In 0 I n b3T 1 A4 0 )( )( )( T 0 1 a4 1 0 1 c3

1 0 0 )( ) 0 B2 In

1 0 0 )( ) 0 B2 In

󸀠

b ). As B, A , U, V ∈ E (R) we conclude that A , B ∈ E (R). This for A4 = A3 ( 10 I n−1 2 n 4 2 n finishes the proof.

The connection with the finiteness of the stable range and the homomorphism GLn (R, I) → K1 (R, I) relies on the following proposition. Proposition 9.4.3. If I is an ideal of a ring R and n > k = str(R) then (1) GLn (R, I) = En (R, I)GLk (R, I). (2) En (R, I) is normal in GLn (R). (3) En (R, I) = GLn (R, I) ∩ En+1 (R, I). Proof. We first prove the proposition for I = R. (1) By induction, it is enough to prove GLn (R) = En (R)GLn−1 (R). Let a ∈ GLn (R). We have to prove that ua ∈ GLn−1 (R) for some u ∈ En (R). Let b T = (b1 , . . . , b n )T be the last column of a. The proof goes through some reductions. The first reduction consists in showing that we may assume that (b2 , . . . , b n ) is unimodular. Indeed, as (b1 , . . . , b n ) is unimodular and n > str(R), there are r2 , . . . , r n ∈ R such that (c2 , . . . , c n ) = (b2 +r2 b1 , . . . , b n +r n b1 ) is unimodular in R n−1 . Let u = I n +∑ni=2 Ei1 (r i ). By Lemma 9.2.3, u ∈ En (R). Furthermore the last column of ua is (b1 , c2 , . . . , c n ) and this completes the first reduction. In the second reduction we prove that one may assume that b1 = b n − 1. Indeed, by the previous reduction, we may assume that (b2 , . . . , b n ) is unimodular. Hence b n − 1 − b1 = ∑ni=2 r i b i for some r2 , . . . , r n ∈ R. Let now u = I n + ∑ni=2 E1i (r i ), an element of En (R), by Lemma 9.2.3. Then the last column of ua is (b 1 + ∑ni=2 r i b i , b2 , . . . , b n ) = (b n − 1, b2 , . . . , b n ). This completes the second

9.4 Whitehead group and the stable range condition

|

301

reduction. So we assume that b n − b1 = 1. In the third reduction we prove that one may assume that b = (0, 0, . . . , 0, 1). Indeed, let n−1

u = (I n − ∑ Ein (b i ))(I n + En1 (−1)) ∈ En (R). i=1

Then the last column of ua is b1 b2

n−1

b1 b2 b3

n−1

0 0

(I n − ∑ Ein (b i )) (I n + En1 (−1)) ( . ) = (I n − ∑ Ein (b i )) ( ) = ( .. ) .. .. . i=1 i=1 . 1 1+b1 1

Finally, assuming that the last column of a is (0, 0, . . . , 1)T , we have a=(

a1 b

0 ). 1

Taking I n−1 u=( −ba−1 1

0 ) 1

we have u ∈ En (R), by Lemma 9.2.3, and ua ∈ GLn−1 (R), as desired. (2) By (1) for I = R, it is enough to show that if a ∈ GLk (R) and e = e ij (r), with 1 ≤ i ≠ j ≤ n and r ∈ R, then aea−1 ∈ En (R). As k ≤ n − 1, we have a = a1 ⊕ 1, with a1 ∈ GLn−1 (R). If j = n then aea−1 = (

a1 0

0 I n−1 )( 1 0

x a−1 )( 1 1 0

0 I n−1 )=( 1 0

a1 x ) ∈ En (R), 1

as desired. Similarly, if i = n then aea−1 ∈ En (R). Assume that i, j < n. Then e ∈ GLn−1 (R) and aea−1 = ( I n−10+A 01 ) with A = T T a1 Eij (r)a−1 1 . Let u = (u 1 , . . . , u n−1 ) be the i-th column of a 1 and v = (v 1 , . . . , v n−1 ) −1 T the j-th row of a1 . As i ≠ j, vu = 0. Furthermore, the (k, l)-entry of A is u k rv l . Therefore aea−1 = (

I n−1 + A 0

0 I n−1 )=( 1 −rv(I n−1 + A)−1

0 I n−1 + A )( rv 1

I n−1 −rv(I n−1 + A)−1

0 I n−1 )( 1 0

uT I n−1 )( 1 rv

−u T ) 1

I n−1 =( −rv(I n−1 + A)−1

0 I n−1 )( 1 0

uT I n−1 )( rv 1

0 I n−1 )( 1 0

=(

0 ) 1

−u T ). 1

By Lemma 9.2.3, the factors of the last product belong to En (R) and we conclude that aea−1 ∈ En (R), as desired.

302 | 9 K-theory (3) Let X ∈ GLn (R) ∩ En+1 (R). We need to prove that X ∈ En (R). By Lemma 9.4.2, there are matrices A, B ∈ En (R) and a, b, c ∈ R n such that (

X 0

0 A )=( 1 0

0 In )( 1 a

1 bT )( T 1 c

0 In )( 0 1

1 0 )( 0 In

0 ). B

Replacing X by A−1 X, we may assume that A = I n . On the other hand, we use (9.4.3) to rewrite the product of the last two matrices (

1 cT

0 1 )( 0 In

0 1 )=( B 0 1 0

0 1 )( T B c1 1

0 ), In

0

for some c1 ∈ R n . The inverse of ( c T I n ) is ( −c T I n ) = ( cU2 10 ) for some c2 ∈ R n and 1 1 U ∈ En (R). Let X 󸀠 = XU. There exist V ∈ En (R), B󸀠 = BV ∈ En (R), a1 , b1 , b2 , c2 ∈ R n such that (

X󸀠 a1

0 In )=( −a 1 =(

In 0

0 X )( 1 0

0 U )( 1 c2

1 bT )( 1 0

0 1 )=( B 0

0 In )=( −a 1

0 X )( 1 0

0 1 )( T 1 −c1

b1 1 )( V 0

0 1 )=( B 0

b2 ). B󸀠

0 ) In

Hence we get X󸀠 ( a1

1 0 ) = (0 1 0

x M y

0 1 0) = ( 0 1

b2 ), B󸀠

for some x, y ∈ R n−1 and some M ∈ GLn−1 (R). However (

M 0

0 M )=( 1 y

0 I n−1 )( −y 1

0 I n−1 ) = B󸀠 ( 1 −y

0 ) ∈ En (R). 1

0 ) ∈ E (R) because this matrix is a conjugate of By (2) for I = R, we deduce that ( 10 M n M 0 ( 0 1 ) in GLn (R). Then

X󸀠 = (

1 0

0 1 )( M 0

x ) ∈ En (R) I n−1

and therefore X = X 󸀠 U −1 ∈ En (R), as desired. This finishes the proof of the proposition for I = R. To prove the relative versions we apply the non-relative version to R ∝ I. Note that, because of Lemma 9.4.1, str(R) = str(R ∝ I). Recall from (9.2.1) and (9.2.3) that GLn (R ∝ I) = u(GLn (R, I)) ⋊ d(GLn (R)) and En (R ∝ I) = u(En (R, I)) ⋊ d(En (R)). Then u(GLn (R, I)) ⋊ d(GLn (R)) = GLn (R ∝ I) = En (R ∝ I)GLk (R ∝ I) = (u(En (R, I)) ⋊ d(En (R)))(u(GLk (R, I)) ⋊ d(GLk (R))) = u(En (R, I)GLk (R, I)) ⋊ d(En (R)GLk (R)) = u(En (R, I)GLk (R, I)) ⋊ d(GLn (R)).

9.5 Applications of K-theory to units |

303

Hence, GLn (R, I) = En (R, I)GLk (R, I), because u is injective. This proves (1). From (9.2.2) we know that u(En (R, I)) = En (R ∝ I) ∩ u(GLn (R, I)). As En (R ∝ I) is normal in GLn (R ∝ I) we get that En (R, I) is normal in GLn (R, I). This proves (2). Finally, by (9.2.2) we have u(En (R, I)) = En (R ∝ I) ∩ u(GLn (R, I)) = En+1 (R ∝ I) ∩ GLn (R ∝ I) ∩ u(GLn (R, I)) = En+1 (R ∝ I) ∩ u(GLn+1 (R, I)) ∩ u(GLn (R, I)) = u(En+1 (R, I) ∩ GLn (R, I)), which proves (3). For a positive integer n, define K1,n (R, I) = GLn (R, I)/En (R, I), the set of left cosets of GLn (R, I) modulo En (R, I) and consider the natural map ϕ n : K1,n (R, I) → K1 (R, I) xEn (R, I) 󳨃→ xE(R, I) If n ≥ str(R) then ϕ n is surjective by Proposition 9.4.3.(1). In general K1,n (R, I) is not a group, but if n > str(R) then Proposition 9.4.3.(2) implies that K1,n (R, I) is a group and ϕ n is an isomorphism by Proposition 9.4.3.(3). So we have proved the following result. Corollary 9.4.4. Let R be a ring with k = str(R) and let I be an ideal of R. Then, every element of K1 (R, I) is represented by an element of GLk (R, I) and if n > k then En (R, I) is normal in GLn (R, I) and K1 (R, I) ≅ GLn (R, I)/En (R, I). Combining this with Corollary 9.3.6 we have. Corollary 9.4.5. Let O be an order and I is an ideal of O. Then, every element of K1 (O, I) is represented by an element of GL2 (O, I). Furthermore, if n ≥ 3 then En (O, I) is normal in GLn (O, I) and K1 (O, I) ≅ GLn (O, I)/En (O, I) .

Problems 9.4.1. Let M and N be right R-modules and let m ∈ M and n ∈ N. Prove that (M ⊕ N)[(m, n)] = M[m] + N[n].

9.5 Applications of K-theory to units Throughout this section A denotes a finite dimensional semisimple rational algebra and A = A1 × ⋅ ⋅ ⋅ × A k is its Wedderburn decomposition. Put F i = Z(A i ), for each

304 | 9 K-theory i = 1, . . . , k. Let R be the unique maximal order of the center of A and let O be an order in A. Let a ∈ M n (A) and, for each i = 1, . . . , k, let a i ∈ M n (A i ) be the matrix with (m, l)-entry the i-th component of the (m, l)-entry of a. Then a ∈ GLn (A) if and only if a i ∈ GLn (A i ) for every i = 1, . . . , k. We define nr(a) = (RNrM n (A1 )/F1 (a1 ), . . . , RNrM n (A k )/F k (a k )) ∈ Z(A).

(9.5.1)

(Caution: Do not confuse nr with reduced norm.) This defines a map nr : M n (A) → Z(A) which restricts to a group homomorphism nr : GLn (O) → U(R),

(9.5.2)

by (2.3.7) and Lemma 4.6.9 (2). These homomorphisms are compatible with the natural embedding GLn (O) → GLn+1 (O) and hence nr induces a homomorphism nr : K1 (O) → U(R), which we denote with the same symbol. We will use the following notation, where I is an ideal of O: SLn (O) = {x ∈ GLn (O) : nr(x) = 1}, SL(O) = ⋃ SLn (O),

SLn (O, I) = SLn (O) ∩ GLn (O, I) SL(O, I) = SL(O) ∩ GL(O, I).

n≥1

The group SLn (O, I) is called the congruence subgroup of level I. Observe that SL1 (O) has been denoted O1 in Section 5.2, for O an order in a finite dimensional rational simple algebra. By Whitehead’s Lemma (Lemma 9.2.5), E(O, I) ⊆ E(O) = GL(O)󸀠 ⊆ GL(O) ∩ ker(nr) = SL(O), since U(R) is commutative. We define the abelian groups SK1 (O) = SL(O)/E(O),

SK1 (O, I) = SL(O, I)/E(O, I).

Observe that SK1 (O, I) is the kernel of the map K1 (O, I) → U(R), induced by nr. Note that if O is commutative, and thus A is a direct sum of fields, then nr : K1 (O) → U(R) is simply the determinant homomorphism with kernel SK1 (O) and image U(O). Clearly this mapping splits and thus K1 (O) ≅ U(O) × SK1 (O). The following theorem, the main result of this section, shows that SK1 (O) is finite and thus U(O) is a subgroup of finite index in K1 (O). In particular, ℝ ⊗ℤ K1 (O) ≅ ℝ ⊗ℤ U(O). The main theorem proves more and this for arbitrary orders O in finite dimensional semisimple rational algebras A. Recall that an ideal I of such an order is said to be essential if it intersects non-trivially every simple component of A.

9.5 Applications of K-theory to units

| 305

Theorem 9.5.1. Let O be a order in a finite dimensional semisimple rational algebra A, I an essential ideal of O and R the unique maximal order in the center of A. Then both SK1 (O, I) and U(R)/nr(GL(O, I)) are finite. Proof of Theorem 9.5.1 (First part). Proving that U(R)/nr(GL(O, I)) is finite is easy. Inm m deed, if (a1 , . . . , a k ) ∈ Z(O) then, by (2.3.7), nr(a1 , . . . , a k ) = (a1 1 , . . . , a k k ), where m i denotes the degree of A i . Hence, if m = lcm(m1 , . . . , m k ) and x ∈ U(Z(O)) ∩ (1 + I)m m/m m/m then x = (a1 , ..., a k )m = nr(a1 1 , ..., a k k ) ∈ nr(GL(O, I)). Therefore nr(GL(O, I)) m contains U(Z(O)∩(1+ I)) . Because I ∩ Z(O) is an essential ideal of Z(O), we know that Z(O)/(I∩Z(O)) is finite (see Problem 4.6.2) and thus [U(Z(O)) : U(Z(O))∩(1+I)] < ∞, as U(Z(O)) ∩ (1 + I) is the kernel of the natural map U(Z(O)) → U(Z(O))/(I ∩ Z(O)). Hence, because of Lemma 4.6.9 (4) and since U(Z(O)) is finitely generated by Dirichlet’s Unit Theorem (Theorem 5.2.4), we obtain that [U(R) : nr(GL(O, I))] ≤ [U(R) : U(Z(O))][U(Z(O)) : U(Z(O)) ∩ (1 + I)] < ∞. Clearly, a finitely generated abelian group is finite if and only if it is periodic. To prove that SK1 (O, I) is finite, we will show that it is finitely generated and periodic. For the former we use the following proposition which is equivalent to Corollary 5.5.3. Proposition 9.5.2. GLn (O) contains a subgroup of finite index which is isomorphic to a subgroup of finite index in SLn (O) × U(R). Corollary 9.5.3. If O is an order and I is an ideal of O then SLn (O, I) is finitely generated and hence so is SK1 (O, I). Proof. By Theorem 5.3.1, GLn (O) and U(R) are finitely generated. Hence, by Proposition 9.5.2, SLn (O) is finitely generated. To prove that SLn (O, I) is finitely generated we first assume that A is simple. If I = 0 then SLn (O, I) = 1, which is trivially finitely generated. Otherwise O/I is finite, by Problem 4.6.2, and hence so is GLn (O)/GLn (O, I). Thus SLn (O, I) is finitely generated because SLn (O)/SLn (O, I) = SLn (O)/GLn (O, I) ∩ SLn (O) ≅ SLn (O)GLn (O, I)/GLn (O, I) ⊆ GLn (O)/GLn (O, I), a finite group. Assume now the arbitrary case. So A = ∏ki=1 A i . For each i, let Oi be an order in A i and let O󸀠 = ∏ki=1 Oi . If O = O󸀠 then I = ∏ki=1 I i with I i an ideal of Oi for every i and SLn (O, I) = ∏ki=1 SLn (Oi , I i ), which is finitely generated by the previous paragraph. Put O󸀠󸀠 = O ∩ O󸀠 . By Lemma 4.6.6, O󸀠󸀠 also is an order in A. As both O and O󸀠 are full ℤ-lattices in A there is a positive integer m such that mO󸀠 , mO ⊆ O ∩ O󸀠 . Then I 󸀠 = mO(I ∩ O󸀠 )mO is an ideal of both orders O and O󸀠󸀠 , and it is contained in I. Also I 󸀠󸀠 = m2 O󸀠 (I ∩ O󸀠 )m2 O󸀠 is an ideal of both orders O󸀠 and O󸀠󸀠 and I 󸀠󸀠 ⊆ I 󸀠 . By the previous paragraph, SLn (O󸀠 , I 󸀠󸀠 ) is finitely generated. Hence, by Problem 9.5.1, SLn (O󸀠󸀠 , I 󸀠󸀠 )

306 | 9 K-theory is finitely generated. Since I 󸀠 /I 󸀠󸀠 is finite, this implies (with an argument as before) that SLn (O󸀠󸀠 , I 󸀠 ) is finitely generated. So, again by Problem 9.5.1, SLn (O, I 󸀠 ) is finitely generated. As I/I 󸀠 is finite, we obtain that SLn (O, I) is finitely generated. By Corollary 9.4.5, K1 (O, I) ≅ GLn (O, I)/En (O, I), for n ≥ 3 and hence SK1 (O, I) ≅ SLn (O, I)/En (R, I). Thus SK1 (O, I) is finitely generated. So to finish the proof of Theorem 9.5.1 it remains to prove that SK1 (O, I) is periodic. The proof of the periodicity is based on a series of lemmas. Lemma 9.5.4. Consider the following commutative diagram of homomorphisms of abelian groups. f

A → B α↓ ↓β g

C → D Suppose that the kernel and cokernel of both f and g are periodic. Then the kernel (respectively, cokernel) of α is periodic if and only if so is the kernel (respectively, cokernel) of β. The same is true replacing periodic by finite everywhere. Proof. Consider the following commutative diagram with exact rows 1 → A/ ker(f) → B → coker f → 1 α1 ↓ β↓ β1 ↓ 1 → C/ ker(g) → C → coker g → 1 where α1 (a ker(f)) = α(a) ker(g) and β 1 (b Im(f)) = β(b) Im(g). By the Snake Lemma we get an exact sequence 1 → ker(α1 ) → ker(β) → ker(β1 ) → coker α1 → coker β → coker β1 → 1 By assumption, coker f and coker g are periodic and hence so are ker(β1 ) and coker β1 . Therefore ker(α1 ) is periodic if and only if ker(β) is periodic and coker α1 is periodic if and only if coker β is periodic. As ker(f) and ker(g) are periodic, the equality ker(α1 ) = α−1 (ker(g))/ ker(f) implies that ker(α1 ) is periodic if and only if so ker(g) C is ker(α), and coker α1 = Im α C/ ker(g)/ ker(g) ≅ Im α ker(g) , implies that coker α 1 is periodic if and only if coker α is periodic. The same proof works replacing periodic by finite. Lemma 9.5.5. If I is an essential ideal of O then the kernel and cokernel of the natural map K1 (O, I) → K1 (O) are finite. Proof. Let n ≥ 3. By Corollary 9.4.5, K1 (O, I) ≅ GLn (O, I)/En (O, I), and hence the rows of the following commutative diagram are exact 1 → E n (O, I) → GLn (O, I) → K1 (O, I) → 1 ↓ ↓ f ↓ 1 → E n (O) → GLn (O) → K1 (O) → 1

9.5 Applications of K-theory to units | 307

where the left and central vertical homomorphisms are inclusions and the latter is the homomorphism of the statement of the lemma. By the Snake Lemma, there is an exact sequence 1 → ker(f) → En (O)/En (O, I) → GLn (O)/GLn (O, I) → coker f → 1 As I is essential, O/I is finite. Moreover, GLn (O)/GLn (O, I) is isomorphic to a subgroup of GLn (O/I) and therefore it is finite. Then coker f is finite and ker(f) is isomorphic to a subgroup of finite index of E(O)/E(O, I). Clearly, En (O) is generated by the elements of the form e ij (r k ) with 1 ≤ i ≠ k ≤ and O = ℤr1 + ⋅ ⋅ ⋅ + ℤr k . Therefore En (O)/En (O, I) is finitely generated and hence so is ker(f). Let G = GL(O) = ∪n≥1 GLn (O) and N = GL(O, I) = ∪n≥1 GLn (O, I). Then T = G/N is periodic. By the Five-Term Homology Sequence [196, 11.4.17], there is an exact sequence H2 (G) → H2 (T) → N/(G, N) → G/G󸀠 → T/T 󸀠 → 1. Since T is periodic, so is H2 (T) and hence G󸀠 /(G, N), the kernel of the third homomorphism, is periodic. By the Withehead Lemma (Lemma 9.2.5), G󸀠 = E(O) and (G, N) = E(O, I). Therefore E(O)/E(O, I) is periodic, and hence so is ker(f). We conclude that ker(f) = GL(O, I) ∩ E(O)/E(O, I) is a periodic finitely generated abelian group. Hence it is finite, as desired. Lemma 9.5.6. If O ⊆ O󸀠 are orders in A then the kernel and cokernel of the natural homomorphism K1 (O) → K1 (O󸀠 ) are finite. Proof. Let I be a common essential ideal of O and O󸀠 . It exists because mO󸀠 ⊆ O for some positive integer m. By Lemma 9.5.5, the kernel and cokernel of the vertical maps of the following commutative diagram fI

K1 (O, I) → K1 (O󸀠 , I) ↓ ↓ K 1 (O ) → K 1 (O󸀠 ) are finite. So, by Lemma 9.5.4, it is enough to prove that the kernel and cokernel of f I are finite for some essential ideal I of O󸀠 contained in O. Although this is true for every such essential ideal (see Problem 9.5.2) we only need to prove it for I 2 . As in the proof of Lemma 9.5.5, if n ≥ 3 then there is a commutative diagram with exact rows 1 → E n (O, I) → GLn (O, I) → K1 (O, I) → 1 ↓ ↓ fI ↓ 󸀠 󸀠 1 → E n (O , I) → GLn (O , I) → K1 (O󸀠 , I) → 1 Observe that GLn (O, I) = {α ∈ I n + M n (I) : nr(α) ∈ U(R)} = GLn (O󸀠 , I)

308 | 9 K-theory for every n. Thus f I is surjective and the kernel of f I is isomorphic to En (O󸀠 , I)/En (O, I). This is true for every ideal I of O󸀠 contained in O. In particular the kernel of f I 2 is isomorphic to En (O󸀠 , I 2 )/En (O, I 2 ) ⊆ En (I)/En (O, I 2 ) ⊆ En (O)/En (O, I 2 ), by Lemma 9.2.4 (1). By Corollary 9.4.5, En (O)/En (O, I 2 ) is finite and hence En (O󸀠 , I 2 )/En (O, I 2 ) is finite. Therefore ker(f I ) is finite, as desired. Lemma 9.5.7. If O and O󸀠 are orders in A then the kernel (respectively, cokernel) of nr : K1 (O) → U(R) is finite if and only if so is the kernel (respectively, cokernel) of nr : K1 (O󸀠 ) → U(R). Proof. As O ∩ O󸀠 also is an order of A, by Lemma 4.6.6, we may assume without loss of generality that O ⊆ O󸀠 . Consider the following commutative square K 1 (O ) → K 1 (O󸀠 ) nr ↓ ↓ nr U(R) → U(R) where the bottom horizontal map is the identity. By Lemma 9.5.6, the kernel and cokernel of the upper horizontal map is finite. Hence the lemma follows from Lemma 9.5.4. Lemma 9.5.8. Assume that A is simple and let F = Z(A), L a finite field extension of F and S the ring of integers of L. Then the kernel of the natural map α : K1 (O) → K1 (S⊗R O) is periodic. Proof. By Corollary 4.2.7, both R and S are Dedekind domains which are finitely generated as additive groups. The finitely generated R-modules are described in the Theorem of Steinitz (Theorem 4.3.2). We freely use this description. The R-module S R is finitely generated and torsion-free, hence S R ≅ R n ⊕ I for some non-negative integer n and some non-zero ideal I of R. Moreover, S R is projective by Corollary 4.2.9. Consider [S] as an element of K0 (R) and K1 (O) as a K0 (R)-module. Then, the map α : K1 (O) → K1 (S ⊗R O) is multiplication by [S] (see Lemma 9.2.9). The multiplication map I ⊗R I −1 → R is surjective, by Proposition 4.2.8, and it is injective because I −1 is a projective R-module by Corollary 4.3.3. Thus, I ⊗R I −1 ≅ R. We claim that I ⊕ I −1 ≅ R2 . Indeed, let 0 ≠ a ∈ I and J = aI −1 , an ideal of R. Then b 󳨃→ ab defines an isomorphism of R-modules I −1 → J and IJ = a(II −1 ) = aR. Therefore I ⊕ I −1 ≅ I ⊕ J ≅ aR ⊕ R ≅ R ⊕ R, by the Steinitz Theorem. Let x = [R n ⊕ I −1 ] ∈ K0 (R). Then, x[S] = [(R n ⊕ I −1 ) ⊗R (R n ⊕ I)] = [R n ⊕ (R n ⊗R (I ⊕ I −1 )) ⊕ (I −1 ⊗R I)] 2

2

= [R n ⊕ R2n ⊕ R] = [R m ], for m = (n + 1)2 . Therefore, if y ∈ ker(α) then 0 = x[S]y = [R m ]y = my. Hence ker(α) is periodic. Proof of Theorem 9.5.1 (Second part). By Corollary 9.5.3, SK1 (O, I) is finitely generated. As SK1 (O, I) is abelian, to prove that it is finite it is enough to prove that it is periodic.

9.5 Applications of K-theory to units | 309

We first prove that SK1 (O) is periodic. By Lemma 9.5.7 we can change O by another convenient order. For example, we can take O = O1 × ⋅ ⋅ ⋅ × Ok , where each Oi is an order in A i . It is clear that this result holds for O if and only if it holds for each Oi . This shows that, without loss of generality, we may assume that A is simple. Let F be the center of A. We claim that we may replace A by L ⊗F A, where L is a finite extension of F. Hence, in particular, we may assume that A is split. Indeed, if S is the ring of integers of L then we have a commutative diagram f

K1 (O) → K1 (S ⊗R O) nr ↓ ↓ nrS U(R) → U(S) where the right vertical map is induced by the reduced norm of L ⊗F A over L. Assume that ker(nrS ) is periodic. If x ∈ ker(nr) then f(x) ∈ ker(nrS ) and therefore mx ∈ ker(f) for some positive integer m. As ker(f) is periodic, by Lemma 9.5.8, we deduce that x has finite order. So ker(nr) is periodic and the claim is proved. Hence we assume that A = M n (F). Moreover, by the freedom of choosing any order of A, we may assume that O = M n (R). As K1 (O) = K1 (R), we may assume that A = F and O = R. Let x be a representative of an element of ker(nr). By Corollary 9.4.5, we may assume that x ∈ GL2 (R) and, replacing F by a larger field, if needed, one may assume that F contains the eigenvalues of x. So 1 = nr(x) = det(x), with eigenvalues λ, λ−1 ∈ F. If the λ-eigenspace of x has dimension 2, then λ = λ−1 and x = ±1, an element of finite order, as desired. Thus we may assume that both the λ-eigenspace and the λ−1 eigenspace have dimension 1 and we choose eigenvectors u, u󸀠 ∈ R2 with x(u) = λu and x(u󸀠 ) = λ−1 u󸀠 . Let P = R2 ∩ Fu and Q = R2 /P. Clearly, P is a torsion-free R-module and we claim that Q is also torsion-free. Indeed, if 0 ≠ r ∈ R and y ∈ R2 satisfies ry ∈ P then y belongs to the λ-eigenspace of x and hence y ∈ P. This proves the claim. Thus P and Q are finitely generated projective R-modules, by Corollary 4.3.3. Moreover, P is invariant under multiplication by x. Thus x induces an automorphism x of Q. Let y ∈ R2 and write y = au + a󸀠 u󸀠 with a, a󸀠 ∈ F. Let m be a positive integer such that ma, mλ−1 ∈ R. Then maλu = amx(u) ∈ P and mλ−1 u ∈ P, mx(y) = maλu + ma󸀠 λ−1 u󸀠 and mλ−1 y = maλ−1 u + ma󸀠 λ−1 u󸀠 . Therefore mx(y + P) = mλ−1 (y + P). As Q is torsionfree, x(y + P) = λ−1 (y + P). This proves that (Q, x) = (Q, λ−1 ) and hence we have the following short exact sequence in CI : 0 → (P, x|P ) = (P, λ) → (R2 , x) → (Q, x) = (Q, λ−1 ) → 0. Using the homological description of K1 (O), we have x = [R2 , x] = [P, λ] + [Q, λ−1 ]. Moreover P and Q have rank 1 and therefore [P, λ] = [I1 , λ] and [Q, λ] = [I2 , λ−1 ] for some non-zero ideals I1 and I2 of R. Furthermore, [I2 , λ−1 ] = −[I2 , λ]. If n is a multiple of the order of I1 and I2 in the class group of R then the n-th powers of the ideals I1 (n) (n) and I2 are isomorphic to R as R-modules. Hence, the n-th direct sums I1 and I2 are

310 | 9 K-theory isomorphic to R n , as R-modules by Theorem 4.3.2. Thus nx = n([I1 , λ] − [I2 , λ]) = [R n , λ] − [R n , λ] = 0. This proves that x has finite order, as desired. This finishes the proof that SK1 (O) is finite, in other words the kernel of the right vertical map in the following commutative diagram is finite K1 (O, I) → K1 (O) nr ↓ ↓ nr . U(R) = U(R) By Lemma 9.5.5 the kernel and cokernel of the horizontal maps are finite. Thus the kernel of the left vertical map is finite, by Lemma 9.5.4. So SK1 (O, I) is finite. Corollary 9.5.9. If O is a order in A, I is a non-zero two sided ideal of O and n ≥ 3 then En (I) has finite index in SLn (O). Proof. By Theorem 9.5.1, SK1 (O, I) is finite. Moreover, by Corollary 9.4.5, we know that SK1 (O, I) ≅ SLn (O, I)/En (O, I). Thus En (O, I) has finite index in SLn (O, I). Applying this to I 2 and recalling from Lemma 9.2.4 that En (O, I 2 ) ⊆ En (I), we deduce that En (I) has finite index in SLn (O, I). Thus En (I) has finite index in SLn (O), because SLn (O, I) has finite index in SLn (O). Corollary 9.5.10. If O is an order in A then the natural image of U(Z(O)) is of finite index in K1 (O). In particular, rk(K1 (O)) = rk(U(Z(O))) and if T is a torsion-free subgroup of finite index in U(Z(O)) then T is naturally isomorphic with its image in K1 (O). Proof. The first part and the rank statement follow at once from Corollary 9.5.9, Proposition 9.5.2 and Corollary 9.4.5. Since a central unit of norm one is periodic the last statement follows since T is torsion-free. Let O be an order in a finite dimensional semisimple rational algebra A. Let U be a subgroup of U(O). By Lemma 4.6.9 and Proposition 5.5.1, U has finite index in U(OG) if and only if (1) U contains a subgroup of finite index in the center of U(O) and (2) for each simple component S of A the group U contains a subgroup of finite index in the group of reduced norm one units O(S)1 of an order O(S) in S. Note that we implicitly identify O(S)1 with 1−e+O(S)1 , where e is the identity of O(S) and 1 is the identity of A. As a consequence of Theorem 9.5.1 and Corollary 9.5.10, we now show that condition (1) may be replaced by the requirement that the natural projection of U has finite index in K1 (O). Proposition 9.5.11. Let O be an order in a semisimple finite dimensional rational algebra A and write A = A1 ⊕ . . . ⊕ A k , the direct product of its Wedderburn components. For each A i let Oi denote an order in A i and let e i denote the identity of A i . Then, a subgroup U of U(O) is of finite index if and only if both of the following conditions are satisfied: (1) the natural image of U in K1 (O) is of finite index. (2) For each i ∈ {1, . . . , k}, the group U contains a subgroup of finite index in 1−e i + O1i .

9.5 Applications of K-theory to units

| 311

Proof. By Lemma 4.6.9 we may assume without loss of generality that O = ∏ki=1 Oi . Let R = Z(O). If U has finite index in U(O) then it obviously satisfies (2) and contains a subgroup of finite index in U(R). The latter implies condition (1) by Corollary 9.5.10. Conversely, assume that U satisfies conditions (1) and (2). Let f : U(O) → U(R) be the composition of the natural map U(O) → K1 (O) with nr : K1 (O) → U(R). By condition (1), Theorem 9.5.1 and Corollary 9.5.10, f(U) has finite index in U(R). The kernel of f is N = ∏ki=1 (1 − e i + O1i ). Thus (U(O) : UN) = (U(O)/N : UN/N) = (f(U(O) : f(U)) ≤ (U(R) : f(U)) < ∞, and condition (2) implies that (NU : U) = (N : N ∩ U) < ∞. Therefore (U(O) : U) < ∞.

Problems 9.5.1. Let O and O󸀠 be orders in a finite dimensional semisimple rational algebra A with O ⊆ O󸀠 and I and ideal of O󸀠 . Prove that SLn (O, I ∩ O) has finite index in SLn (O󸀠 , I). 9.5.2. Let O ⊆ O󸀠 be orders in a finitely dimensional semisimple rational algebra and let I be an essential ideal of O󸀠 contained in O. Prove that the natural map K1 (O, I) → K1 (O󸀠 , I) has finite kernel and cokernel.

10 General linear groups of degree 2 In this chapter K is a number field such that the unit group of its ring of integers is infinite. By Corollary 5.2.6, this is equivalent with K not being ℚ nor a quadratic imaginary extension of ℚ. The main result of the chapter is a theorem of Vaseršte˘ın [219] and Liehl [148] which states that if R is an order in K and I is a nonzero ideal of R then E2 (I), the group generated by the elementary matrices e ij (r) with r ∈ I, is of finite index in SL2 (R). In [148, 219] one also dealt with the larger class of Dedekind rings of arithmetic type, but this is beyond the scope of this monograph. The proof uses the general theory of Mennicke symbols [17]. We will, however, restrict our focus only on proving precisely what is needed for our purposes. We claim that it is enough to prove the statement for the ring of integers of K. Indeed, if S is the ring of integers of K then nS ⊆ I for some positive integer n. If the statement holds for S then E n (nS) has finite index in SLn (S) and hence E2 (I) has finite index in SLn (R). The condition that U(R) is infinite is essential. For example, if n is a positive integer then E2 (nℤ) has finite index in SL2 (ℤ) if and only if n = 1 or 2. Indeed, we already know that SL2 (ℤ) = E2 (ℤ) (see (1.4.1)) and E2 (2ℤ) has finite index in SL2 (ℤ) (Corollary 1.4.6). On the other hand, if n > 2 then SL2 (ℤ, nℤ) is a free group, by Corollary 1.4.5, and hence E2 (nℤ) = ⟨e12 (n), e21 (n)⟩ is a free group of rank 2. Thus if E2 (nℤ) has finite index in SL2 (ℤ) then necessarily SL2 (ℤ, nℤ) = E2 (nℤ), which is not the case −n because ( 1+n n 1−n ) ∈ SL2 (ℤ, nℤ) \ E2 (nℤ). More generally, the celebrated congruence theorems [18, 148, 206, 219] state that E2 (I) is of infinite index for sufficiently small ideals I of an order in a quadratic imaginary extension of the rationals. The latter, however, is beyond the scope of this monograph. The structure of the proof follows the main lines of [148] and [219], but with the restriction to orders in number fields. So, throughout this chapter R is the ring of integers in a number field K and we assume U(R) is infinite.

For ideals I1 and I2 in an arbitrary commutative ring A we put SL2 (A, I1 ≀ I2 ) = {(

a c

b ) ∈ SL2 (A) : a − 1, d − 1 ∈ I1 I2 , b ∈ I1 , c ∈ I2 } d

and E2 (I1 ≀ I2 ) = ⟨(

1 0

b 1 ),( 1 c

0 ) : b ∈ I1 , c ∈ I2 ⟩ . 1

10.1 Number theoretical results |

313

Clearly E2 (I1 ≀ I2 ) is a subgroup of the group SL2 (A, I1 ≀ I2 ). If I1 = I2 then E2 (I1 ≀ I2 ) = E2 (I1 ). Although it is not essential for the results, it is convenient for the proofs, to work with two ideals I1 and I2 of R. Obviously, E2 (I1 I2 ) ⊆ E2 (I1 ≀ I2 ) ⊆ E2 (I1 + I2 ). Hence, the statement we want to prove is equivalent to the following: if both I1 and I2 are nonzero ideals of R, then E2 (I1 ≀ I2 ) is a subgroup of finite index in SL2 (R).

10.1 Number theoretical results In this section we collect some number theoretical background. Throughout R is the ring of integers of a number field K. We begin by proving four easy lemmas. Lemma 10.1.1. Assume I is a nonzero ideal of R and let R = R/I. If a, b ∈ R are such that aR + bR = R then there is an element z ∈ (a − 1)R such that a + bz ∈ U(R). Proof. Since R is a Dedekind domain, all the prime ideals of the commutative ring R are maximal and there are finitely many of them. Hence, the Jacobson radical J(R) of R is a nil ideal and thus an element z ∈ R is a unit if and only if its natural image in R/J(R) is a unit. Consequently, we may assume that J(R) = {0} and therefore, R is a direct product of finitely many fields. Obviously, we then also may assume that R is one of these fields. The condition aR + bR = R then means that not both a and b are zero. Hence, it is easily verified that there is an element z ∈ R(a − 1) such that a + bz ≠ 0. Lemma 10.1.2. Let F be a finite field such that |F| ≠ 3. If f1 , f2 are nonzero elements in F, then there exist w ∈ F and nonzero u, v ∈ F such that f1 u2 + f2 v2 = w2 . Proof. If −f1 f2 = u2 for some u ∈ F then f1 u2 + f2 f12 = f1 (u2 + f1 f2 ) = 0 and so the property holds. So assume that −f1 f2 is not a square in F. Then U(F) is a cyclic group of even order generated by x, say. Then S = ⟨x2 ⟩ is the set of non-zero squares of F and T = U(F) \ S = xS is the set of non-squares of F. By means of contradiction we assume that f1 X + f2 Y = Z does not have a solution with X, Y, Z ∈ S. Therefore f1 S + f2 S = T. As As −f1 f2 ∈ T we may assume , without loss of generality, that f1 ∈ S and −f2 ∈ T. Then S = f1 S and T = −f2 S. Therefore S − T = T and, in fact S − t = T = tS for every t ∈ T. In particular x(1 + S) = S and so 1 + S = T. Thus −1 ∈ ̸ S, −T = S and S + S = T. Furthermore, S + T ⊆ S ∪ {0}. Indeed, for if S + T ⊈ S ∪ {0} then, as T = −S, there are a, b, c ∈ S with a − b = −c and hence a + c = b ∈ S ∩ (S + S) ⊆ S ∩ T = 0, a contradiction. We now use |F| > 3 to deduce that |S| > 2. Therefore there is s ∈ S \{−x}. Then s + x = s1 ∈ S and hence s = s1 − x ∈ S ∩ (S + S) ⊆ S ∩ T = 0, a contradiction. Lemma 10.1.3. If I is a nonzero ideal of R then R/I is a finite principal ideal ring.

314 | 10 General linear groups of degree 2 e

e

Proof. Let I = P11 . . . P kk be the factorization of I as product of maximal ideals. By the e e Chinese Remainder Theorem R/I ≅ ∏ki=1 R/P i i . Moreover each R/P i i is isomorphic to e i e R P i /(P i P i ) i . As R P i is a PID, R/P i is a PID and hence R/I is a principal ideal ring. Lemma 10.1.4. Let I be a nonzero ideal of R and 0 ≠ r ∈ R. Set X = {(r + i)−1 : i ∈ I with r + i ≠ 0}. Then X contains a ℚ-basis x1 , . . . , x n of K. In particular, ∑ni=1 ℤx i is a full ℤ-lattice of F and thus contains a non-zero ideal of R. Proof. Write K = ℚ(α) for some 0 ≠ α ∈ K. Since I is a full ℤ-lattice of K, mα ∈ I, for some 0 ≠ m ∈ ℤ. Hence, we may assume that α ∈ I \ {−1}. Clearly, K = ℚ((1 + α)−1 ). Hence, for some n, the set {r−1 , r−1 (1 + α)−1 , . . . , r−1 (1 + α)−n } is a ℚ-basis for K, as desired. We also need the following two results, the proofs of which are beyond the scope of this monograph. The first one is a generalization of Dirichlet’s Theorem on Primes in Arithmetic Progression. It is a consequence of the Corollary on page 166 of [144]. Theorem 10.1.5 ([18, A.12, page 84]). If a, b ∈ R are such that aR + bR = R then there are infinitely many maximal ideals of the form zR with z ∈ a + bR. Lemma 10.1.6 ([206, Lemma 3]). Let R be the ring of integers of a number field, I an ideal of R and a0 ∈ R such that a0 R + I = R. Let m be the number of roots of unity in R, p a prime and e = e pℤ (mℤ). Then p e+1 ∤ Exp(U(R/Ra)) for some 0 ≠ a ∈ a0 + I. (Equivalently gcd{Exp(U(R/R(a0 + x))) : x ∈ I, a0 + x ≠ 0} divides m.)

10.2 Normality of E2 (I1 ≀ I2 ) in SL2 (R, I1 ≀ I2 ) Let I1 and I2 be two nonzero ideals of R. In this section we prove that E2 (I1 ≀ I2 ) is a normal subgroup of SL2 (R, I1 ≀ I2 ). We begin with several technical lemmas. Lemma 10.2.1. Let I1 and I2 be nonzero ideals of R. If I is a non-zero ideal of R contained in I1 ∩ I2 , then SL2 (R, I1 ≀ I2 ) = SL2 (R, I ≀ I)E2 (I1 ≀ I2 ). Proof. Let T = R/I 2 . By Lemma 10.1.3, the ring T is a principal ideal ring. Let J1 and J2 be the natural images in T of I1 and I2 respectively. So, J1 = Tt1 and J2 = Tt2 for some t1 , t2 ∈ T. Let A = ( ac db ) ∈ SL2 (T, J1 ≀ J2 ). So, b ∈ J1 , c ∈ J2 , a − 1, d − 1 ∈ J1 J2 and ad − bc = 1. It is sufficient to prove that A ∈ E2 (J1 ≀ J2 ). Write a − 1 = t1 t2 a󸀠 and b = t1 b󸀠 , for some a󸀠 , b󸀠 ∈ T. Clearly Ta + Tb = T and thus also Ta + Tb󸀠 = T. Hence, by Lemma 10.1.1, there exists x, y ∈ T such that y(b󸀠 + xa) = 1. Then, (

a c

b 1 ) ( d 0

xt1 1 ) ( 1 −yt2 a󸀠

0 1 )=( v 1

t1 (b󸀠 + xa) ) w

10.2 Normality of E2 (I1 ≀ I2 ) in SL2 (R, I1 ≀ I2 )

|

315

for some v, w ∈ T. As this product belongs to SL2 (T, J1 ≀ J2 ), we obtain that v ∈ J2 . Consequently, 1 ( −v

0 1 ) ( v 1

1 t1 (b󸀠 + xa) )=( w 0

t1 (b󸀠 + xa) ) ∈ E2 (J1 ≀ J2 ). 1

1 0 ) ∈ E (J ≀ J ), the above yields that indeed A ∈ E (J ≀ J ). As also ( −v 2 1 2 2 1 2 1

Lemma 10.2.2. Let O be an order in a finite dimensional rational division algebra such that U(O) is infinite and let I be a non-zero ideal of O. Then O has a unit u of infinite order such that ( 0u u0−1 ) ∈ E2 (I). Proof. By Problem 4.6.2, O/I is finite and hence I contains a positive integer. Thus, without loss of generality, we may assume that I = nO for some positive integer n. By Theorem 5.5.2, U(O) contains a torsion-free subgroup of finite index. As U(O) is infinite then it has a unit u of infinite order. Then some positive power of u belongs to 1 + n2 O and hence one may assume without loss of generality that u ∈ 1 + n2 O. Thus u = 1 + n2 x for some (non-zero) x ∈ O. Clearly ux = xu. Then (

u 0

0 ) = e12 (n) e21 (nx) e12 (−u−1 n) e21 (−unx) ∈ E2 (I), u−1

as desired. Lemma 10.2.3. If I is a nonzero ideal of R and A ∈ GL2 (K) then there exists a nonzero ideal J of R such that E2 (J) ⊆ A E2 (I) A−1 . Proof. Since GL2 (K) is generated by diagonal matrices ( 0x 0y ) (with x, y nonzero elements in K), ( 10 11 ) and ( 11 01 ), it is sufficient to deal with each of these cases and their inverses separately. Case 1: A = ( 0x 0y ), with nonzero x, y ∈ K. Clearly, AE2 (I)A−1 = ⟨(

1 0

xay−1 1 ), ( 1 ybx−1

0 ) : a, b ∈ I⟩ . 1

Hence, E2 (J) ⊆ A E2 (I) A−1 , with J = x−1 yI ∩ xy−1 I ∩ R. Case 2: A = ( 10 11 ). Then, A E2 (I) A−1 contains all the matrices of the form e12 (a) with a ∈ I. So, it remains to show that there exists an ideal J of R such that A E2 (I) A−1 contains all matrices e21 (a) with a ∈ J. By assumption U(R) is infinite. Thus, by Lemma 10.2.2, U(R) has an element u of infinite order such that u 0 U=( ) ∈ E(I). 0 u−1 Therefore, A E2 (I) A−1 contains 1 0

AU (

1 − u−2 u ) A−1 = A ( 0 1

u(1 − u−2 ) ) A−1 = U. u−1

316 | 10 General linear groups of degree 2 We now select an element y of I with y ≠ 0, −1. Applying Lemma 10.1.4 to the element 1 + y and the ideal Iy, we obtain z1 , . . . , z n ∈ I and an ideal I0 of R such that n

I0 ⊆ ∑ ℤa−1 i ,

for a i = 1 + y + z i y ≠ 0.

(10.2.1)

i=1

Because each Ra i is a non-zero ideal of R, R/Ra i is finite. Thus, there exists a positive integer m such that u2m − 1 ∈ Ra i for each i with 1 ≤ i ≤ n. For each such i, put bi = y + zi y − zi ∈ I

2m c i = a−1 − 1)b i ∈ I. i (u

and

Now, the group A E2 (I) A−1 contains the matrices 1 Bi = ( 0

zi 1 )( 1 0

1 1 )( 1 y

−b i ) 1−y

−1 ai )=( 1 y

0 1 )( 1 0

and the matrices 1 Ci = ( 0

ci ). 1

Hence, AE2 (I)A−1 also contains the matrix U m B i U −m C i B−1 i . Note that (1 − y)a i + yb i = det(B i ) = 1. Hence U m B i U −m C i B−1 i =( =(

ai yu−2m

−u2m b i 1 − y − ci y )( 1−y −y

1 a−1 y(1 − u−2m ) i

bi + ci ai ) ai

0 ) 1

It follows from (10.2.1) that A E2 (I) A−1 contains the matrices of the form e21 (a) with a ∈ J = I0 (u−2m − 1)y. Note that J is nonzero as I0 is nonzero and u is a unit of infinite order. This finishes the proof of Case 2. −1 −1 Case 3: A = ( 10 11 ) , A = ( 11 01 ) or A = ( 11 01 ) . From Cases 1 and 2 we obtain that the result holds for the matrix (

−1 0

0 1 )( 1 0

1 −1 )( 1 0

0 1 )=( 1 0

−1

1 ) 1

.

As E2 (I) is closed under transposition, we therefore easily obtain that the result also −1 holds for A = ( 11 01 ) and A = ( 11 01 ) . Lemma 10.2.4. Let I be a nonzero ideal of R and u ∈ U(R). Put D = ( 0u If A = ( ac db ) ∈ SL2 (R), with b, c ∈ I, then D m/2 AD−m/2 ∈ E2 (I) A E2 (I), where m = |μ(R)|, the number of roots of unity contained in R.

0 ) u−1

∈ SL2 (R).

10.2 Normality of E2 (I1 ≀ I2 ) in SL2 (R, I1 ≀ I2 )

| 317

Proof. Because A ∈ SL2 (R) we know that Ra + Rb = R. Hence also Ra + Rb2 = R. Because D normalizes the group E2 (I), the integers n for which D n AD−n ∈ E2 (I)AE2 (I) form an additive subgroup G of ℤ. We claim that it is enough to show that G contains any n for which u2n − 1 ∈ R(a + xb2 ) for some x ∈ R. Indeed, assume that this holds. Let d be a generator of G. Then for every x ∈ R such that a + xb2 ≠ 0 the assumption implies that Exp(U(R/R(a + xb2 ))) . d divides gcd(2, Exp(U(R/R(a + xb2 ))) Thus d divides gcd {

Exp(U(R/R(a + xb2 )) : x ∈ R, a + xb2 ≠ 0} . gcd(2, Exp(U(R/R(a + xb2 ))))

m Then d divides m 2 , by Lemma 10.1.6. Therefore 2 ∈ G and this is precisely the statement of the lemma. This proves the claim. So, let n ∈ ℤ and let x, r ∈ R be such that u2n − 1 = r(a + xb2 ). Using again that D normalizes E2 (I) we have

D n AD−n ∈ D n A (

1 xb

0 1 ) D−n ( 1 0

−rb ) E2 (I) 1

=(

a + xb2 + dxb)

u−2n (c

b ) E2 (I) −u−2n (c + dxb)rb + d

So, for some r1 ∈ I, r2 ∈ R, D n AD−n ∈ ( Since, ( r1 −ra2 xb

a + xb2 r1

b −1 r2 )A

b 1 )( r2 −xb

a 0 ) E2 (I) = ( r1 − r2 xb 1

b ) E2 (I) r2

= e21 (x) with x ∈ I, we obtain that D n AD−n ∈ E2 (I)AE2 (I),

as desired. Lemma 10.2.5. Let I be a nonzero ideal of R. If A ∈ GL2 (K) then there exists a nonzero ideal J of R such that SL2 (R, J ≀ J) ⊆ A−1 SL2 (R, I ≀ I) A. Proof. Write A = (a ij ) and A−1 = (a󸀠ij ) and let J = {r ∈ R : a ij ra󸀠kl ∈ I 2 for every 1 ≤ i, j, k, l ≤ 2}. Clearly J is a non-zero ideal of R. To prove that J satisfies the required −1 2 11 b 12 condition let B = ( 1+b b21 1+b22 ) ∈ SL2 (R, J ≀ J) and C = (c ij ) = ABA . Then b ii ∈ J and b12 , b21 ∈ J. Thus c11 − 1 = a11 b11 a󸀠11 + a11 b12 a󸀠21 + a12 b21 a󸀠12 + a12 b22 a󸀠21 ∈ I 2 and c12 = a11 b11 a󸀠12 + a11 b12 a󸀠22 + a12 b21 a󸀠22 + a12 b22 a󸀠22 ∈ I 2 . Similarly c21 , c22 − 1 ∈ I 2 .

318 | 10 General linear groups of degree 2 Lemma 10.2.6. If I is a nonzero ideal of R and A ∈ SL2 (K) then there exists a nonzero ideal J of R such that (SL2 (R, J ≀ J), A) ⊆ E2 (I). Proof. Let N be the subset of GL2 (K) that consists of the matrices A such that for every nonzero ideal I of R there exists a nonzero ideal J of R with (SL2 (R, J ≀ J), A) ⊆ E2 (I). We claim that N is a normal subgroup of GL2 (K). We first show that N is closed under conjugation. For this let A ∈ N and X ∈ GL2 (K). Let I be a nonzero ideal of R. Because of Lemma 10.2.3, there exists a nonzero ideal I 󸀠 of R such that E2 (I 󸀠 )X ⊆ E2 (I). Hence, as A ∈ N, there exists a nonzero ideal J 󸀠 of R such that (SL2 (R, J 󸀠 ≀ J 󸀠 ), A) ⊆ E2 (I 󸀠 ). From Lemma 10.2.5 we obtain a nonzero ideal J 󸀠󸀠 of R such that XSL2 (R, J 󸀠󸀠 ≀ J 󸀠󸀠 )X −1 ⊆ SL2 (R, J 󸀠 ≀ J 󸀠 ). Consequently, (SL2 (R, J 󸀠󸀠 ≀ J 󸀠󸀠 ), A X ) = (XSL2 (R, J 󸀠󸀠 , ≀J 󸀠󸀠 )X −1 , A)X ⊆ (SL2 (R, J 󸀠 ≀ J 󸀠 ), A)X ⊆ E2 (I 󸀠 )X ⊆ E2 (I). Hence A X ∈ N as desired. Next we show that N is multiplicatively closed. So, let A1 , A2 ∈ N and let I be a nonzero ideal. Because of Lemma 10.2.3, there exists a nonzero ideal I 󸀠 of R such that E2 (I 󸀠 )A2 ⊆ E2 (I). As A1 , A2 ∈ N, there exist nonzero ideals J1 and J2 of R such that (SL2 (R, J1 ), A1 ) ⊆ E2 (I 󸀠 ) and (SL2 (R, J2 ), A2 ) ⊆ E2 (I). Consequently, for J = J1 ∩ J2 and every C ∈ SL2 (R, J ≀ J) we have (C, A1 A2 ) = (C, A2 )(C, A1 )A2 ∈ E2 (I)E2 (I 󸀠 )A2 ⊆ E2 (I). This proves that A1 A2 ∈ N I . Suppose that A ∈ N and let I be a non-zero ideal of R. Then there is a non-zero ideal J of R such that (SL2 (R, J ≀ J), A) ⊆ E2 (I). By Lemma 10.2.3 there is a non-zero ideal I 󸀠 of R such that AE2 (I 󸀠 )A−1 ⊆ E2 (I). As A ∈ N, there is a non-zero ideal J of A such that (SL2 (R, J ≀ J), A) ⊆ E2 (I 󸀠 ). Then for every B ∈ SL2 (R, J ≀ J), we have (B, A−1 ) = A(A, B)A−1 = A(B, A)−1 A−1 ∈ AE(I 󸀠 )A−1 ⊆ E(I). This proves that N is closed under taking inverses. This finishes the proof of the claim that N is a normal subgroup of GL2 (K). To prove the result we need to show that SL2 (K) ⊆ N. Let u be an element of infinite −m order in U(R) and 2m = |μ(R)|. We claim that A = ( u0 u0m ) ∈ N. Indeed, let I be a nonzero ideal of R and B ∈ SL2 (R, I ≀ I). By Lemma 10.2.3, there exists a nonzero ideal I 󸀠 of R such that E2 (I 󸀠 )B ⊆ E2 (I). Clearly we may assume that I 󸀠 ⊆ I. Because of Lemma 10.2.1, BC ∈ SL2 (R, I 󸀠 ≀ I 󸀠 ) for some C ∈ E2 (I). Because of Lemma 10.2.4, A−1 (BC)A ∈ E2 (I 󸀠 )BCE2 (I 󸀠 ). Hence, as A normalizes E(I) we have (B, A) ∈ B−1 E2 (I 󸀠 )BCE2 (I 󸀠 )A−1 C−1 A ⊆ E2 (I 󸀠 )B E2 (I) ⊆ E2 (I), as desired. So N contains a matrix of the form A = ( 0v v0−1 ) with v a unit of infinite order. As N is normal in GL2 (K), (A, B) ∈ N for every B ∈ GL2 (K). Hence, for every x ∈ K, we have

10.3 The factor group SL2 (R, I1 ≀ I2 ) by E2 (I1 ≀ I2 )

|

319

2

x e12 (x) = (A, e12 ( vxv 2 −1 )) ∈ N and e 21 (x) = (A, e 21 ( 1−v 2 )) ∈ N. As SL2 (K) is generated by elementary matrices (Problem 1.1.1), we have SL2 (K) ⊆ N, as desired.

We are now in a position to prove the normality of E2 (I1 ≀ I2 ) in SL2 (R, I1 ≀ I2 ). Proposition 10.2.7. For nonzero ideals I1 and I2 of R, the group E2 (I1 ≀ I2 ) is normal in SL2 (R, I1 ≀ I2 ). Proof. Let A ∈ SL2 (R, I1 ≀ I2 ). Let r i ∈ I i , for 1 ≤ i ≤ 2. We need to show that A−1 CA ∈ E2 (I1 ≀ I2 ), for C = ( 10 r11 ) and for C = ( r12 10 ). We prove the former (the latter is done similarly). Let I = I1 ∩ I2 and C = ( 10 r11 ). By Lemma 10.2.6, there exists a nonzero ideal J of R such that J ⊆ I and (SL2 (R, J ≀ J), C−1 ) ⊆ E2 (I). Hence, B−1 CB ∈ E2 (I)C ⊆ E2 (I1 ≀ I2 ) for every B ∈ SL2 (R, J ≀ J). Because of Lemma 10.2.1, we can write A = BE with B ∈ SL2 (R, J ≀ J) and E ∈ E2 (I1 ≀ I2 ). Hence, A−1 CA = E−1 B−1 CBE ∈ E−1 E2 (I1 ≀ I2 )E ⊆ E2 (I1 ≀ I2 ) as desired.

10.3 The factor group SL2 (R, I1 ≀ I2 ) by E2 (I1 ≀ I2 ) For nonzero ideals I1 and I2 of R, put W(I1 , I2 ) = {(a, b) ∈ R2 : a − 1 ∈ I1 I2 , b ∈ I1 , aR + bR = R}. Clearly, the first row of any matrix in SL2 (R, I1 ≀ I2 ) is an element of W(I1 , I2 ). Conversely, any element (a, b) ∈ W(I1 , I2 ) is the first row of some matrix in SL2 (R, I1 ≀ I2 ). Indeed, as 1 ∈ I1 I2 + a , we have that Ra + RI2 = R. Hence, from Ra + Rb = R, we obtain that R = Ra + (Ra + I2 )b. So R = Ra + I2 b and there exist c ∈ I2 and d ∈ R with ad − bc = 1. Therefore, d = 1 + bc − (a − 1)d ∈ 1 + I1 I2 and thus ( ac db ) ∈ SL2 (R, I1 ≀ I2 ). This proves the claim. Because of Proposition 10.2.7, E2 (I1 ≀ I2 ) is a normal subgroup of SL2 (R, I1 ≀ I2 ). For (a, b) ∈ W(I1 , I2 ) put ⌈a, b⌉I1 ,I2 = (

a c

b ) E2 (I1 ≀ I2 ) ∈ SL2 (R, I1 ≀ I2 )/E2 (I1 , I2 ), d

where c and d are such that ( ac db ) ∈ SL2 (R, I1 ≀ I2 ). If the ideals I1 and I2 are clear from the context then, for simplicity, we denote ⌈−, −⌉I1 ,I2 as ⌈−, −⌉. Note that the coset is independent of the choice of c and d. Indeed, if also ( ca󸀠 db󸀠 ) ∈ SL2 (R, I1 ≀ I2 ) then (

a c

b a ) ( 󸀠 d c

b ) d󸀠

−1

=(

1 cd󸀠 − dc󸀠

0 ) ∈ E2 (I1 ≀ I2 ). 1

320 | 10 General linear groups of degree 2

We thus obtain a mapping ⌈−, −⌉I1 ,I2 : W(I1 , I2 ) 󳨀→ SL2 (R, I1 ≀ I2 )/E2 (I1 ≀ I2 ) : (a, b) 󳨃→ ⌈a, b⌉. The main property we will prove about this mapping is that it is “multiplicative” with respect to the first coordinate (see Proposition 10.3.8) and, under some restriction, also with respect to the second coordinate (see Proposition 10.3.10). Hence, in the language of [17], ⌈−, −⌉I1 ,R is a Mennicke symbol. In order to prove this property we need a series of lemmas. Lemma 10.3.1. Let I1 and I2 be nonzero ideals of R. If (a, b) ∈ W(I1 , I2 ), x ∈ I1 and y ∈ I2 then ⌈a, b⌉I1 ,I2 = ⌈a, b + ax⌉I1 ,I2 = ⌈a + by, b⌉I1 ,I2 . Proof. The result follows at once from (

a c

b 1 ) ( d 0

x a )=( 1 c

ax + b ) cx + d

and

(

a c

b 1 ) ( d y

0 a + by )=( 1 c + dy

b ). d

Lemma 10.3.2. Let I1 and I2 be nonzero ideals of R. If y ∈ R and (a, b) ∈ W(I1 , I2 y2 ) then ⌈a, b⌉I1 ,I2 = ⌈a, by2 ⌉I1 ,I2 . Proof. If y = 0 then a = 1 and the statement of the Lemma follows from Lemma 10.3.1. y 0 So assume y ≠ 0. Let A = ( 0 y−1 ). By Lemma 10.2.6 there exists a nonzero ideal J such that J ⊆ I1 ∩ I2 y2 and (SL2 (J ≀ J), A) ⊆ E2 (I1 ∩ I2 ) ⊆ E2 (I1 ≀ I2 ). Since, by assumption, (a, b) ∈ W(I1 , I2 y2 ), there exists c, d ∈ R such that C = ( ac db ) ∈ SL2 (R, I1 ≀ I2 y2 ). Because of Lemma 10.2.1, we may write C = BE for some B ∈ SL2 (R, J ≀ J) and E ∈ E2 (I1 ≀ I2 y2 ). As A E2 (I1 ≀ I2 y2 ) A−1 = E2 (I1 y2 ≀ I2 ), we obtain that C−1 ACA−1 = E−1 (B−1 ABA−1 )(AEA−1 ) ∈ E2 (I1 ≀ I2 y2 ) E2 (I1 ≀ I2 ) E2 (I1 y2 ≀ I2 ) ⊆ E2 (I1 ≀ I2 ). Hence C and ACA−1 determine the same element in SL2 (R, I1 ≀ I2 )/E2 (I1 ≀ I2 ). Clearly a by2 ACA−1 = ( y−2 ) and thus ⌈a, b⌉I1 ,I2 = ⌈a, by2 ⌉I1 ,I2 , as desired. c

d

Lemma 10.3.3. Let I be a nonzero ideal of R. If A ∈ SL2 (R, I ≀ I) then 0 A−1 ( −1

1 0 )A( 0 1

−1 ) ∈ E2 (I). 0

Proof. Let C = ( 01 −10 ). Because of Lemma 10.2.6 there exists a nonzero ideal J of R such that (SL2 (J ≀ J), C) ⊆ E2 (I). We may assume that J ⊆ I. Then, by Lemma 10.2.1, A = BE for some B ∈ SL2 (R, J ≀ J) and E ∈ E2 (I). Therefore (A, C) = (B, C)E (E, C) ∈ E2 (I), as desired, because C normalizes E2 (I).

10.3 The factor group SL2 (R, I1 ≀ I2 ) by E2 (I1 ≀ I2 ) |

321

Lemma 10.3.4. Let I1 and I2 be nonzero ideals of R. If x ∈ I1 ∩ I2 , a − 1, d − 1 ∈ x2 R and Ra + Rd = R then ⌈a, dx⌉I1 ,I2 = ⌈d, −ax⌉I1 ,I2 . Proof. If x = 0 then the result is obvious. So assume x ≠ 0. Because of Lemma 10.3.1, for y ∈ R, ⌈a, dx⌉ = ⌈a + ydx2 , dx⌉ = ⌈a + ydx2 , x(d − a) − ydx3 ⌉. By assumption, there exists r ∈ R such that a − 1 = x2 r. Also by assumption, d ∈ 1 + x2 R ⊆ 1 + I1 I2 . So, Rd + I1 I2 = R and therefore r + dy ∈ I1 I2 for some y ∈ R. Consequently, a + ydx2 − 1 ∈ I1 I2 x2 . Moreover x−1 (d − a) − ydx ∈ x−1 (x2 R) + ydx ⊆ xR ⊆ I1 . Hence, (a + ydx2 , x−1 (d − a) − ydx) ∈ W(I1 , I2 x2 ) and thus Lemma 10.3.2 implies that ⌈a, dx⌉ = ⌈a + ydx2 , (x(d − a) − ydx3 ⌉ = ⌈a + ydx2 , x−1 (d − a) − ydx⌉. Then, by Lemma 10.3.1 (with x and yx playing the role of x and y respectively) ⌈a, dx⌉ = ⌈d, x−1 (d − a) − ydx⌉ = ⌈d, x−1 (d − a)⌉. Interchanging the roles of a and d and replacing x by −x, the last equality yields ⌈d, −ax⌉ = ⌈a, (−x)−1 (a − d)⌉ = ⌈a, x−1 (d − a)⌉ Hence, indeed, ⌈a, dx⌉ = ⌈d, −ax⌉. Lemma 10.3.5. Let I1 and I2 be nonzero ideals of R and let b ∈ I1 and a ∈ U(R) ∩ (1 + I1 I2 ). Then (1) ⌈1, b⌉I1 ,I2 = 1 and ⌈a, b⌉I1 ,I2 = ⌈a, 0⌉. (2) If 0 ≠ J = Rx2 ⊆ I1 ∩ I2 and a ∈ U(R) with a − 1 ∈ J 2 then (

a 0

0 ) ∈ E2 (J); a−1

and thus ⌈a, 0⌉I1 ,I2 = 1. Proof. (1) By Lemma 10.3.1, ⌈1, b⌉ = ⌈1, 0⌉ = 1 and ⌈a, b⌉ = ⌈a, b − a(a−1 b)⌉ = ⌈a, 0⌉, as desired. (2) Clearly, a−1 ∈ 1 + J 2 and thus we may write a−1 = 1 + r1 r2 with r1 , r2 ∈ J. Hence, (

a 0

0 1 )=( a−1 0

1 r1 )( 1 −r2 a

0 1 )( 1 0

−a−1 r1 1 )( 2 1 r2 a

0 ) ∈ E2 (J). 1

322 | 10 General linear groups of degree 2 Lemma 10.3.6. Let I1 and I2 be nonzero ideals of R. If 0 ≠ x ∈ I1 ∩ I2 and (a, b1 ), (a, b2 ) ∈ W(x2 R, x2 R) then ⌈a, b1 ⌉I1 ,I2 ⌈a, b2 ⌉I1 ,I2 = ⌈a, b1 b2 ⌉I1 ,I2 . Proof. Put I 󸀠 = Rx2 and let c, d ∈ R such that ( ac yields that d ( −b2

0 −c )=( a −1

1 a )( 0 c

b2 0 )( d 1

b2 ) d

∈ SL2 (R, I 󸀠 ≀ I 󸀠 ). Lemma 10.3.3

−1 a )∈( 0 c

b2 ) E2 (I 󸀠 ). d

So, since E2 (I 󸀠 ) ⊆ E2 (I1 ≀ I2 ), d ⌈d, −c⌉ = ( −b2

−c ) E2 (I1 ≀ I2 ) = ⌈a, b2 ⌉. a

We thus obtain that ⌈a, b1 ⌉⌈a, b2 ⌉ = ⌈a, b1 ⌉⌈d, −c⌉ = ⌈ad − b1 b2 , −ac + b1 a⌉ = ⌈ad − b1 b2 , a(b1 − c)⌉ = ⌈1 − b2 (b1 − c), a(b1 − c)⌉. So it remains to prove that ⌈1 − b2 (b1 − c), a(b1 − c)⌉ = ⌈a, b1 b2 ⌉.

(10.3.1)

In order to do so, we consider three special cases and then prove the statement. Case 1: Suppose b1 − c = z2 for some z ∈ xR. Write z = xr for some r ∈ R. Clearly, 2 b2 z ∈ z2 x2 R ⊆ z2 I1 I2 ⊆ r2 I1 I2 , so that (1 − b2 z2 , ax2 ) ∈ W(I1 , I2 r2 ). Hence, by Lemma 10.3.2, ⌈1 − b2 (b1 − c), a(b1 − c)⌉ = ⌈1 − b2 z2 , az2 ⌉ = ⌈1 − b2 z2 , ax2 r2 ⌉ = ⌈1 − b2 z2 , ax2 ⌉. Because b2 z2 ∈ z2 I1 I2 , Lemma 10.3.4 implies that then ⌈1 − b2 (b1 − c), a(b1 − c)⌉ = ⌈1 − b2 z2 , ax2 ⌉ = ⌈a, −x2 (1 − b2 z2 )⌉. Hence, as b1 = c + z2 , we have ⌈1 − b2 (b1 − c), a(b1 − c)⌉ = ⌈a, −x2 (ad − b2 c − b2 z2 )⌉ = ⌈a, x2 b1 b2 − x2 ad⌉ Lemma 10.3.1 and Lemma 10.3.2 (note that a − 1 ∈ x4 R ⊆ x2 I1 I2 )) therefore yield ⌈1 − b2 (b1 − c), a(b1 − c)⌉ = ⌈a, x2 b1 b2 ⌉ = ⌈a, b1 b2 ⌉. This finishes the proof of Case 1. Case 2: Suppose b2 = z2 for some z ∈ R. Applying Lemma 10.3.2 twice one obtains ⌈1 − b2 (b1 − c), a(b1 − c)⌉ = ⌈1 − b2 (b1 − c), x2 a(b1 − c)⌉ = ⌈1 − b2 (b1 − c), b2 ax2 (b1 − c)⌉

10.3 The factor group SL2 (R, I1 ≀ I2 ) by E2 (I1 ≀ I2 )

|

323

Hence, applying first Lemma 10.3.1 and then Lemma 10.3.4, we obtain that ⌈1 − b2 (b1 − c), a(b1 − c)⌉ = ⌈1 − b2 (b1 − c), ax2 ⌉ = ⌈a, −x2 (1 − b2 (b1 − c))⌉. As 1 = ad − b2 c and applying again Lemma 10.3.4, Lemma 10.3.1 and Lemma 10.3.2 we thus get that ⌈1 − b2 (b1 − c), a(b1 − c)⌉ = ⌈a, −x2 (ad − b1 b2 )⌉ = ⌈a, x2 b1 b2 ⌉ = ⌈a, b1 b2 ⌉, again as desired. Case 3: Suppose b1 u2 + b2 v2 = w2 for some units u, v, w in the ring R = R/Ra. Here we denote by r the element r + Ra ∈ R. Let z, t ∈ R be such that z = (b2 )−1 u (v)−1 and t = w (u)−1 . (Observe that b i ∈ U(R) because aR + b i R = R.) So, 2

b1 b2 z2 − t z2 b2 + 1 = 0.

(10.3.2)

As Ra + Rx = R, we can write z = ar1 + xr2 for some r1 , r2 ∈ R. Let z0 = z − ar1 . So, z0 − z ∈ Ra and z0 ∈ Rx. Similarly, there exists t0 ∈ Rx with t0 − t ∈ Ra. Put c󸀠 = b1 − t20 ∈ x2 R. Then, because of (10.3.2), d󸀠 = (c󸀠 b2 z20 + 1)a−1 ∈ R and ad󸀠 − c󸀠 b2 z20 = 1. Since a − 1 ∈ (I 󸀠 )2 , c󸀠 , b2 z20 ∈ I 󸀠 we also obtain that (

a c󸀠

b2 z20 ) ∈ SL2 (R, I 󸀠 ≀ I 󸀠 ). d󸀠

As b1 − c󸀠 = t20 with t0 ∈ Rx, Case 1 then yields ⌈a, b1 ⌉⌈a, b2 z20 ⌉ = ⌈a, b1 b2 z20 ⌉. Applying Case 2 on both sides of the above equality then implies ⌈a, b1 ⌉⌈a, b2 ⌉⌈a, z20 ⌉ = ⌈a, b1 b2 ⌉⌈a, z20 ⌉. Hence, indeed ⌈a, b1 ⌉⌈a, b2 ⌉ = ⌈a, b1 b2 ⌉. We are now in a position to prove the equality (10.3.1) in the general case. If either b1 or b2 is 0 then a is invertible and a − 1 ∈ Rx4 with x2 ∈ I1 ∩ I2 . Then ⌈a, b1 ⌉ = ⌈a, b2 ⌉ = ⌈a, b1 b2 ⌉ = 1, by Lemma 10.3.5. So, without loss of generality, we may assume that both b1 and b2 are nonzero. Because of Lemma 10.3.5, we also may assume that a ≠ 1. The latter implies that Ra + Rb1 b2 (a − 1) = R. As also 3R is contained in finitely many ideals of R, it follows from Theorem 10.1.5 that there exists y ∈ (a − 1)R such that (a + b 1 b2 y)R is a maximal ideal of R and the field R = R/(a + b1 b2 yR) is different from the field with three elements. Hence, by Lemma 10.1.2, there exists nonzero u, v, w ∈ R such that b1 u2 + b2 v2 = w2 . Consequently, because of Lemma 10.3.1 and Case 3, ⌈a, b1 ⌉ ⌈a, b2 ⌉ = ⌈a + yb1 b2 , b1 ⌉ ⌈a + yb1 b2 , b2 ⌉ = ⌈a + yb1 b2 , b1 b2 ⌉ = ⌈a, b1 b2 ⌉. This finishes the proof.

324 | 10 General linear groups of degree 2 Lemma 10.3.7. Let I1 and I2 be nonzero ideals of R. Let y ∈ R and (a, b) ∈ W(I1 , I2 y) then ⌈a, by⌉I1 ,I2 = ⌈a, b⌉I1 ,I2 . Proof. Because of Lemma 10.3.5, we may assume that y ≠ 0. We first proof the result under the assumption that (a, b) ∈ W(x2 R, x2 yR) for some x ∈ I1 ∩ I2 . Because of Lemma 10.3.2 and Lemma 10.3.6, ⌈a, by⌉ = ⌈a, byx2 ⌉ = ⌈a, b⌉ ⌈a, yx2 ⌉. As a = 1 + yx4 r for some r ∈ R, Lemma 10.3.1 and Lemma 10.3.5 yield ⌈a, yx2 ⌉ = ⌈a − rx2 (yx2 ), yx2 ⌉ = ⌈1, yx2 ⌉ = 1. Therefore we obtain that ⌈a, by⌉ = ⌈a, b⌉, as desired. Now we consider the general case (with y ≠ 0). Let x ∈ (I1 ∩ I2 ) \ {0} and let J = yx2 R. Because, by assumption, (a, b) ∈ W(I1 , I2 yR), there exist c, d ∈ R such that 󸀠 󸀠 A = ( ac db ) ∈ SL2 (R, I1 ≀ I2 y). From Lemma 10.2.1 we obtain matrices B = ( ac󸀠 db󸀠 ) ∈ SL2 (R, J ≀ J) and E ∈ E2 (I1 ≀ I2 y) such that A = BE. Let D = ( 0y 01 ). Then (a󸀠 , b󸀠 ) ∈ W(x2 R, x2 yR) and, by the particular case considered in the previous paragraph, we have DBD−1 E2 (I1 ≀ I2 ) = ⌈a󸀠 , b󸀠 y⌉ = ⌈a󸀠 , b󸀠 ⌉ = B E2 (I1 ≀ I2 ). Hence, as DE2 (I1 ≀ I2 )D−1 ⊆ E2 (I1 ≀ I2 ), we have A−1 DAD−1 = E−1 (B−1 DBD−1 )DED−1 ∈ E2 (I1 ≀ I2 ), and thus ⌈a, b⌉ = A E2 (I1 ≀ I2 ) = DAD−1 E2 (I1 ≀ I2 ) = ⌈a, by⌉.

We are now in a position to prove the multiplicativity of ⌈, ⌉I1 , I2 with respect to the first coordinate. Proposition 10.3.8. Let I1 and I2 be nonzero ideals of R. If (a1 , b), (a2 , b) ∈ W(I1 , I2 ) then ⌈a1 , b⌉I1 ,I2 ⌈a2 , b⌉I1 ,I2 = ⌈a1 a2 , b⌉I1 ,I2 . Proof. Because (a2 , b) ∈ W(I1 , I2 ), there exist c, d ∈ R such that A = ( ac2 db ) ∈ SL2 (R, I1 ≀ I2 ). Because (a1 a2 , b)A−1 = ((a1 − 1)a2 d, (1 − a1 )a2 b), we obtain that ⌈a1 a2 , b⌉⌈a2 , b⌉−1 = ⌈1 + (a1 − 1)a2 d, (1 − a1 )a2 b⌉ = ⌈1 + (a1 − 1)(1 + bc), (1 − a1 )a2 b⌉ = ⌈a1 + (a1 − 1)bc, (1 − a1 )a2 b⌉ Hence, because of Lemma 10.3.7, ⌈a1 a2 , b⌉⌈a2 , b⌉−1 = ⌈a1 + (a1 − 1)bc, (1 − a1 )b⌉. Lemma 10.3.1 thus yields ⌈a1 a2 , b⌉⌈a2 , b⌉−1 = ⌈a1 , b⌉, as desired.

10.3 The factor group SL2 (R, I1 ≀ I2 ) by E2 (I1 ≀ I2 )

|

325

Lemma 10.3.9. Let I1 and I2 be nonzero ideals of R. If (a, b) ∈ W(I1 , I2 ) and b ∈ I1 I2 (I1 + I2 ) then, for all r ∈ R, ⌈a, b⌉I1 ,I2 = ⌈a + br, b⌉I1 ,I2 . Proof. Let (a, b) ∈ W(I1 , I2 ) with b ∈ I1 I2 (I1 + I2 ). Let r ∈ R. Because Ra + I2 = R, there are elements r󸀠 ∈ R and b󸀠 ∈ I2 such that r = ar󸀠 + b󸀠 . Because of Lemma 10.3.1 and Proposition 10.3.8 we obtain that ⌈a + br, b⌉ = ⌈a(1 + br󸀠 ) + bb󸀠 , b⌉ = ⌈a(1 + br󸀠 ), b⌉ = ⌈a, b⌉ ⌈1 + br󸀠 , b⌉. So it remains to prove that ⌈1 + br, b⌉ = 1,

(10.3.3)

for any r ∈ R. Now, note that for b, b󸀠 ∈ I1 I2 (I1 + I2 ), we have 1 −r

⌈1 + (b + b󸀠 )r, b + b󸀠 ⌉ = (

b + b󸀠 1 ) ( 1 r

1 −r

0 1 ) ( 1 0

b 1 ) ( 1 r

1 −r

0 1 ) ( 0 1

1 b󸀠 ) ( 1 r

= (( ((

0 1 ) ( 1 0

0 ) E2 (I1 ≀ I2 ) 1

0 )) E2 (I1 ≀ I2 ) 1 0 )) E2 (I1 ≀ I2 ) 1

= ⌈1 + br, b⌉ ⌈1 + b󸀠 r, b󸀠 ⌉. Hence, to prove (10.3.3), we may assume that b = b1 b2 y with b1 ∈ I1 , b2 ∈ I2 and y ∈ (I1 ∪ I2 ). If y ∈ I1 then (1 + b1 b2 yr, b1 b2 yr) ∈ W(I1 , I2 b1 ) and thus it follows from Lemma 10.3.7 that ⌈1 + b1 b2 yr, b1 b2 y⌉ = ⌈1 + b1 b2 yr, b21 b2 yr⌉. Lemma 10.3.1 and Lemma 10.3.5 therefore yields ⌈1 + b1 b2 yr, b1 b2 y⌉ = ⌈1 + b1 b2 r, −b1 ⌉ = ⌈1, −b1 ⌉ = 1. Similarly, if y ∈ I2 then ⌈1 + b1 b2 yr, b1 b2 y⌉ = ⌈1 + b1 b2 yr, b1 b2 ⌉ = 1. This finishes the proof of (10.3.3) and hence also the proof of the Lemma. Finally, we are now in a position to prove the promised multiplicativity property. Proposition 10.3.10. Let I1 and I2 be nonzero ideals of R. If (a, b1 ), (a, b2 ) ∈ W(I1 , I2 ) and a − 1 ∈ I1 I2 (I1 + I2 ) then ⌈a, b1 ⌉I1 ,I2 ⌈a, b2 ⌉I1 ,I2 = ⌈a, b1 b2 ⌉I1 ,I2 .

326 | 10 General linear groups of degree 2 Proof. Put x = a − 1 ∈ I1 I2 (I1 + I2 ). Because of Lemma 10.3.1, Lemma 10.3.9 and Lemma 10.3.5, we have ⌈1 + rx, rx2 ⌉ = ⌈1 + rx, rx2 − x(1 + rx)⌉ = ⌈1 + rx, −x⌉ = ⌈1, −x⌉ = 1, for any r ∈ R. Therefore, Proposition 10.3.8, Lemma 10.3.9 and Lemma 10.3.1 yield ⌈a, rx2 ⌉ = ⌈a, rx2 ⌉ ⌈1 + rx, rx2 ⌉ = ⌈a + rx + rx(a − 1), rx2 ⌉ = ⌈a + rx + rx2 , rx2 ⌉ = ⌈a + rx, rx2 − x(a + rx)⌉ = ⌈a + rx, −xa⌉.

(10.3.4)

Hence, again by Proposition 10.3.8 and Lemma 10.3.9, ⌈a, b1 x2 ⌉ ⌈a, b2 x2 ⌉ = ⌈a + b1 x, −xa⌉ ⌈a + b2 x, −xa⌉ = ⌈(a + b1 x)(a + b2 x), −xa⌉ = ⌈a2 + ax(b1 + b2 ) + b1 b2 x2 , −xa⌉ = ⌈a2 + b1 b2 x2 , −xa⌉ = ⌈a + ax + b1 b2 x2 , −xa⌉ = ⌈a + b1 b2 x2 , −xa⌉. Because of (10.3.4) we thus obtain ⌈a, b1 x2 ⌉ ⌈a, b2 x2 ⌉ = ⌈a, b1 b2 x3 ⌉.

(10.3.5)

Finally, for any (a, b) ∈ W(I1 , I2 ), Lemma 10.3.1 yields ⌈a, b⌉ = ⌈a, b(1 − a)⌉ = ⌈a, bx⌉ and thus from (10.3.5) we obtain the desired property ⌈a, b1 ⌉⌈a, b2 ⌉ = ⌈a, b1 x2 ⌉ ⌈a, b2 x2 ⌉ = ⌈a, b1 b2 x3 ⌉ = ⌈a, b1 b2 ⌉.

10.4 The group E2 (I) is of finite index in SL2 (R) In this section we prove the main result of the chapter: if J is a non-zero ideal of R then E2 (J) has finite index in SL2 (R). Of course it is sufficient to prove that E2 (J 3 ) has finite index in SL2 (R). In order to prove this, it is for technical reasons convenient to fix throughout the section two non-zero ideals I1 and I2 of R and let I = I1 I2 (I1 + I2 ). Note that if I1 = I2 = J then I = J 3 . Recall from Proposition 10.2.7 that E2 (I1 ≀I2 ) is a normal subgroup of SL2 (R, I1 ≀I2 ). Consider the natural group homomorphism f : SL2 (R, I) 󳨀→ SL2 (R, I1 ≀ I2 )/E2 (I1 ≀ I2 ),

10.4 The group E2 (I) is of finite index in SL2 (R) |

327

that is, (

a b ) 󳨃→ ⌈a, b⌉I1 ,I2 = ( c d

a c

b ) E2 (I1 ≀ I2 ). d

The main effort in this section is dedicated to extend f to a group homomorphism f : SL3 (R, I) → SL2 (R, I1 ≀ I2 )/E2 (I1 ≀ I2 ) whose kernel contains E3 (R, I). We begin by showing that, for every non-zero ideal V of R, each element of SL3 (R, V) can be expressed in a standard form as a product of three matrices. As in Section 9.3, an element x ∈ R n is considered as an 1×n matrix and hence the transpose x T is an n × 1 matrix. We will use the following subgroups of SL3 (R, V): L3 (R, V) = {(a ij ) ∈ SL3 (R, V) : a31 = a32 = 0, a33 = 1}

= {(

bT ) : A ∈ SL2 (R, V), b ∈ V × V} , 1

A 0

R3 (R, V) = {(a ij ) ∈ SL3 (R, V) : a11 = 1, a 21 = a31 = 0}

= {(

1 0

b ) : A ∈ SL2 (R, V), b ∈ V × V} , A

D3 (R, V) = {e31 (t) t ∈ V}.

Observe that D3 (R, V) is isomorphic to the additive subgroup of R. Lemma 10.4.1. Let V be a non-zero ideal of R. Then, SL3 (R, V) = L3 (R, V) D3 (R, V)R3 (R, V). In other words, every element of SL3 (R, V) can be written as a product (

A 0

1 cT ) e31 (t) ( 0 1

b ) D

with A, D ∈ SL2 (R, V), c, d ∈ V × V and t ∈ V. Moreover the element t is unique. Proof. Let X = (x ij ) ∈ SL3 (R, V). Clearly (x11 , x21 , x31 ) is unimodular in R3 , i.e. Rx11 + Rx21 + Rx31 = R. Hence, also Rx11 + Rx21 + Rx231 = R. Because of Corollary 9.3.6, the stable range of R is 2. Hence, we know that there exists c󸀠 ∈ R2 such that a = (x11 , x21 ) − c󸀠 x231 is unimodular in R2 . Put c = c󸀠 x31 . Clearly c ∈ V × V and a ∈ W(V, R). As noticed in the beginning of this section, the latter implies that there exists A ∈ SL2 (R, V) with first column equal to a T . Therefore A1 = (

I2 0

cT A )( 1 0

0 A )=( 1 0

cT ) ∈ L3 (R, V). 1

Moreover, (

I2 0

x11 A −c T ) (x21 ) = ( 0 1 x31

1 0 ) ( 0 ). 1 x31

328 | 10 General linear groups of degree 2

Thus the first column of B1 = e3,1 (−x31 ) (

A−1 0

0 I2 )( 0 1

−c T )X 1

is (1, 0, 0)T . So, B1 ∈ R3 (R, V). Finally, notice that ((

A−1 0

0 I2 )( 0 1

−1

−c T )) 1

= A1 .

So, X = A1 e31 (x31 ) B1 . Hence, X can be expressed in the desired form with t = x31 . The uniqueness of t follows from the fact that t is precisely the entry of X that is in position (3, 1). The lemma allows to extend the homomorphism f to a mapping f : SL3 (R, I) → SL2 (R, I1 ≀ I2 )/E2 (I1 ≀ I2 ) as follows f ((

A 0

cT 1 ) e31 (t) ( 0 1

b )) = f(A) f(D). D

In order to show that f is well defined, the following lemma is useful. Lemma 10.4.2. If ( ac db ) ∈ SL2 (R, I) then −1 ⌈a, b⌉I1 ,I2 = ⌈d, c⌉I1 ,I2 = ⌈d, b⌉−1 I1 ,I2 = ⌈a, c⌉I1 ,I2 .

Hence, if Q = ( 01 10 ) then f(A) = f(A Q ) for any A ∈ SL2 (R, I 3 ). Proof. Because of Proposition 10.3.8 −1 ⌈a, b⌉I1 ,I2 = ⌈ad, b⌉I1 ,I2 ⌈d, b⌉−1 I1 ,I2 = ⌈1 + bc, b⌉I1 ,I2 ⌈d, b⌉I1 ,I2 .

Hence, by Lemma 10.3.1 and Lemma 10.3.5, −1 ⌈a, b⌉I1 ,I2 = ⌈1, b⌉I1 ,I2 ⌈d, b⌉−1 I1 ,I2 = ⌈d, b⌉I1 ,I2

Similarly, by Proposition 10.3.10 and Lemmas 10.3.1 and 10.3.5 we have −1 ⌈d, c⌉I1 ,I2 = ⌈d, b⌉−1 I1 ,I2 ⌈d, bc⌉I1 ,I2 = ⌈d, b⌉I1 ,I2 ⌈d, ad − 1⌉I1 ,I2 −1 = ⌈d, b⌉−1 I1 ,I2 ⌈d, ad − 1 − d(a − 1)⌉I1 ,I2 = ⌈d, b⌉I1 ,I2 ⌈d, d − 1⌉I1 ,I2 −1 = ⌈d, b⌉−1 I1 ,I2 ⌈1, d − 1⌉I1 ,I2 = ⌈d, b⌉I1 ,I2

Finally, applying the above to ( ba dc ) ∈ SL2 (R, I), we also get that ⌈a, c⌉I1 ,I2 = ⌈d, c⌉−1 I1 ,I2 . Lemma 10.4.3. The mapping f is well defined.

10.4 The group E2 (I) is of finite index in SL2 (R) |

329

Proof. Let M ∈ SL3 (R, I). Suppose M=(

1 c1T ) e31 (t) ( 1 0

A1 0

b1 A2 )=( D1 0

1 c2T ) e31 (t) ( 1 0

b2 ) D2

with A1 , A2 , D1 , D2 ∈ SL2 (R, I), b1 , b2 , c1 , c2 ∈ I × I and t ∈ I. We need to prove that f(A1 )f(D1 ) = f(A2 )f(D2 ). Because f is a group homomorphism, this is equivalent with −1 showing that f(A−1 1 A 2 ) = f(D 1 D 2 ). Clearly, (

A1 0 1 ( 0

−1

c1T ) 1

(

c2T A−1 A2 )=( 1 1 0

A2 0

b1 1 )( D1 0

−1

cT ) 1

1 0

b ), D1 D−1 2

1 cT ) e31 (t) = e31 (t) ( 0 1

b ). D1 D−1 2

b2 ) D2

=(

for some b = (b󸀠1 , b󸀠2 ), c = (c󸀠1 , c󸀠2 ) ∈ I × I and ( a11 Write A−1 1 A 2 = ( a21

A−1 1 A2 0

a12 a22 )

d11 and D1 D−1 2 = ( d21

A−1 A2 ( 1 0

Then

a11 + c󸀠1 t cT ) e31 (t) = (a21 + c󸀠2 t 1 t

and 1 e31 (t) ( 0

d12 d22 ).

1 b −1 ) = (0 D1 D2 t

b󸀠1 d11 d21 + tb󸀠1

a12 a22 0

c󸀠1 c󸀠2 ) 1

b󸀠2 d12 ) . d22 + tb󸀠2

Consequently, a11 + c󸀠1 t = d22 + tb󸀠2 = 1 and a21 + c󸀠2 t = d21 + tb󸀠1 = 0, a22 = d11 , c󸀠2 = d12 , b󸀠1 = a12 and b󸀠2 = c󸀠1 . Therefore, A−1 1 A2 = (

1 − c󸀠1 t −c󸀠2 t

and D1 D−1 2 =(

a22 −ta12

a12 ) a22

c󸀠2 ). 1 − tc󸀠1

Thus, because of Lemma 10.4.2 and Proposition 10.3.10, 󸀠 󸀠 󸀠 f(D1 D−1 2 ) = ⌈1 − c 1 t, −ta 12 ⌉I1 ,I2 = ⌈1 − c 1 t, a 12 ⌉I1 ,I2 ⌈1 − c 1 t, −t⌉I1 ,I2

So, by Lemma 10.3.1 and Lemma 10.3.5, −1 −1 f(D1 D−1 2 ) = f(A 1 A 1 )⌈1, −t⌉I1 ,I2 = f(A 1 A 2 ),

as desired.

330 | 10 General linear groups of degree 2

The next step is to show that f is a homomorphism. To do so we again need several lemmas that involve the following subsets of SL3 (R, I): H = {A ∈ SL3 (R, I) : f (AB) = f (A)f (B) for all B ∈ SL3 (R, I)} N = {A ∈ GL3 (R) : f (B A ) = f (B) for all B ∈ SL3 (R, I)} Lemma 10.4.4. The following properties hold. (1) If A ∈ L3 (R, I), B ∈ R3 (R, I) and C ∈ SL3 (R, I) then f (ACB) = f (A)f (C)f (B). In particular L3 (R, I) ⊆ H. (2) H is a subgroup of SL3 (R, I). (3) N is a subgroup of SL3 (R, I) and N normalizes H. (4) If a subgroup K of SL3 (R, I) contains L3 (R, I) and is normalized by E3 (R) then K = SL3 (R, I). Proof. (1) Let C ∈ SL3 (R, I). Because of Lemma 10.4.1, C = A1 D1 B1 for some A1 = T T ( X1 c1 ) ∈ L3 (R, I), D1 ∈ D3 (R, I) and B1 = ( 10 Yb11 ) ∈ R3 (R, I). Hence, if A = ( X0 c1 ) ∈ 0 1

L3 (R, I) and B = ( 10 L3 (R, I) and B 1 B =

T b XX1 c󸀠 Y ) ∈ R3 (R, I), then ACB = (AA 1 )D 1 (B 1 B) with AA 1 = ( 0 1 ) ∈ 󸀠 ( 10 Yb1 Y ), for some c󸀠 , b󸀠 ∈ I × I. Consequently, from the definition

of the map f , we get f (ACB) = f(XX1 ) f(Y1 Y). As, f is a homomorphism this yields f (ACB) = f(X)f(X1 )f(Y1 )f(Y) = f (A)f(X1 )f(Y1 )f (B) = f (A)f (C)f (B). (2) First we show that H is multiplicatively closed. For this let A, A󸀠 ∈ H and let B be an arbitrary element in SL3 (R, I). Then f (AA󸀠 B) = f (A)f (A󸀠 B) = f (A)f (A󸀠 )f (B) = f (AA󸀠 )f (B), by (1). So, AA󸀠 ∈ H. Second we prove that if A ∈ H then also A−1 ∈ H. Indeed, let B ∈ SL3 (R, I). Then, 1 = f (AA−1 ) = f (A)f (A−1 ) and therefore f (A−1 ) = (f (A))−1 . Hence, f (B) = f (AA−1 B) = f (A)f (A−1 B) and thus f (A−1 B) = (f (A))−1 f (B) = f (A−1 )f (B). So, indeed, A−1 ∈ H. (3) That N is a subgroup of SL3 (R, I) is straightforward. To prove that it normalizes −1 −1 H, assume A ∈ N, B ∈ H and C ∈ SL3 (R, I). Then f (B A C) = f ((BC A )A ) = f (BC A ) = −1 f (B)f (C A ) = f (B A )f (C). Hence, B A ∈ H, as desired. (4) Let K be a subgroup of SL3 (R, I) that contains L3 (R, I) and is normalized by E3 (R). Clearly L3 (R, I) contains the set E = {e12 (t) : t ∈ I} and thus E ⊆ K. From Lemma 9.2.1 we obtain that the normal closure of E in E3 (R) contains E3 (I) and hence it is E3 (R, I). So, because of the assumption, E3 (R, I) ⊆ K. Proposition 9.4.3 therefore yields that SL3 (R, I) = E3 (R, I)SL2 (R, I) ⊆ K L3 (R, I). Since, by assumption L3 (R, I) ⊆ K, it thus follows that SL3 (R, I) = K. Lemma 10.4.5. If E3 (R) ⊆ N then f is a homomorphism with E3 (R, I) in its kernel. Proof. Since, by assumption, E3 (R) ⊆ N, Lemma 10.4.4 implies that H = SL3 (R, I), i.e. f is a homomorphism. So, ker(f ) is normalized by E3 (R). Moreover, as E2 (R, I) ⊆ ker(f) and because E3 (R) normalizes ker(f ), we get that ker(f) contains the normal closure of E = {e12 (t) : t ∈ I}. As shown at the end of the proof of Lemma 10.4.4, the latter equals E3 (R, I). So E3 (R, I) ⊆ ker(f ).

10.4 The group E2 (I) is of finite index in SL2 (R) |

Lemma 10.4.6. All matrices of SL3 (R) of the type (

a11 a12 a13 0 1 a23 ) 0 0 a33

331

belong to N.

Proof. Clearly the matrices of the type listed in the statement form a group that is generated by the matrices of the form u D = (0 0

0 1 0

0 0 ), u−1

e12 (t)

and

e23 (t),

with u ∈ U(R) and t ∈ R. We need to show that all such matrices belong to N. So, let M = ABC ∈ SL3 (R, I), with A ∈ L3 (R, I), B = e31 (q) ∈ D3 (R, I) and C ∈ R3 (R, I) (see Lemma 10.4.1). Then MD = AD BD CD and it is readily verified that A D ∈ L3 (R, I), B D ∈ D3 (R, I) and C D ∈ R3 (R, I). Therefore f (M D ) = f (A D )f (C D ). Now, if f (A) = ⌈a11 , a12 ⌉I1 ,I2 then it is easily verified that f (A D ) = ⌈a11 , u−1 a12 ⌉I1 ,I2 . Hence, by Lemma 10.3.7, f (A) = f (A D ). Similarly, we obtain that f (C D ) = f (C). Consequently, f (M D ) = f (A)f (C) = f (M), as desired. Next, let E = e12 (t). Clearly E ∈ L3 (R, I) ∩ R3 (R, I) and so A E ∈ L3 (R, I) and C E ∈ R3 (R, I). Moreover, from Lemma 9.2.1 we have B E = Be32 (qt). Hence, M E = A E B(e32 (qt)C E ) and f (M E ) = f (A E )f (e32 (qt)C E ). Because of Lemma 10.4.4.(1) we thus get f (M E ) = f (A E )f (e32 (qt))f (C E ). a11 a12 c1 a22 c2 ) 0 0 1

Write A = ( a21

1 b1 b2 d12 ). 0 d21 d22

and C = ( 0 d11

f (A E ) = f (

a11 − ta21 a21

Then,

t(a11 − ta21 ) + a12 − ta12 ). a21 t + a22

Hence, by Lemma 10.4.2, f (A E ) = ⌈a21 t + a22 , a21 ⌉I1 ,I2 . Lemma 10.3.9 thus gives f (A E ) = ⌈a22 , a21 ⌉I1 ,I2 = f (A). Similarly, f (C E ) = f (C). Clearly f (e32 (qt)) = 1. So, f (M E ) = f (A)f (C) = f (M), again as desired. Finally we have to deal with conjugations by matrices of the type F = e23 (t). This is done similar as the previous case by noticing that B F = e21 (tq)−1 B and that e21 (tq)−1 ∈ L3 (R, I).

332 | 10 General linear groups of degree 2 001

Lemma 10.4.7. Let Q = ( 0 1 0 ). If A ∈ SL3 (R) and QA T Q ∈ N then A ∈ N. 100

Proof. First we show that f (QM T Q) = (f (M))−1 for M ∈ SL3 (R). To do so, because of Lemma 10.4.1, write a11 M = (a21 0

1 c1 c2 ) e31 (t) (0 1 0

a12 a22 0

b1 d11 d21

b2 d12 ) . d22

Clearly, d22 QM T Q = (d21 0

1 b2 b1 ) e31 (t) (0 1 0

d12 d11 0

c2 a22 a21

c1 a12 ) . a11

Hence, by Lemma 10.4.2, −1 −1 f (QM T Q) = ⌈d22 , d12 ⌉I1 ,I2 ⌈a22 , a12 ⌉I1 ,I2 = ⌈d11 , d12 ⌉−1 I1 ,I2 ⌈a 11 , a 12 ⌉I1 ,I2 = (f (M)) ,

as desired. To prove the result, assume that A ∈ SL3 (R) and QA T Q ∈ N. So, for X ∈ SL3 (R, I), T we have that f (X QA Q ) = f (X). We need to prove that f (X A ) = f (A). Because of the first part of the proof we get that f (X A ) = (f (Q(X A )T Q))

−1

= (f ((QA T Q)(QX T Q)(QA T Q)−1 ))

−1

Since QA T Q ∈ N we thus obtain f (X A ) = (f (QX T Q))

−1

Hence, again by the first part of the proof, f (X A ) = f (X) This finishes the proof. 100

Lemma 10.4.8. Let P = ( 0 0 1 ). If P ∈ N then E3 (R) ⊆ N. 010

Proof. Because of Lemma 10.4.6, e ij (t) ∈ N for t ∈ R and 1 ≤ i < j ≤ 3. Since, P ∈ N by assumption, we thus obtain from Lemma 10.4.7 that e i1 (t) ∈ N for i = 2, 3 and t ∈ R. Because N is a group by Lemma 10.4.4, we obtain from Lemma 9.2.1 that e32 (t) = (e31 (t), e12 (1)) ∈ N. Consequently, E3 (R) ⊆ N. Lemma 10.4.9. Let P be the matrix as in Lemma 10.4.8. Let S = {X ∈ SL3 (R, I) : f (X P ) = f (X)} and let A ∈ SL3 (R, I). (1) If B ∈ R3 (R, I) and C = ( 0c Id2 ) ∈ SL3 (R, I), then A ∈ S if and only if CAB ∈ S. T

(2) There exist C = ( 0c dI2 ) ∈ SL3 (R, I) and B ∈ R3 (R, I) such that CAB = A󸀠 e31 (t), a11 a12 0 a22 c2 ) 0 0 1

with A󸀠 = ( a21

∈ L3 (R, I) and t ∈ I.

10.4 The group E2 (I) is of finite index in SL2 (R) |

333

Proof. To prove the first part, let 1 B = (0 0

b1 d11 d21

b2 d12 ) ∈ SL3 (R, I) d22

and

c C=( 0

d ) ∈ SL3 (R, I). I2

Write d = (d1 , d2 ). Clearly, c C P = P−1 CP = (0 0

d2 1 0

d1 0 ) ∈ L3 (R, I) 1

and 1 B P = (0 0

b2 d22 d12

b1 d21 ) ∈ R3 (R, I). d11

Hence, by Lemma 10.4.4, f ((CAB)P ) = f (C P )f (A P )f (B P ). Because of Lemma 10.3.5 and Lemma 10.4.2, we know that f (C P ) = f(C) and f (B P ) = f(B). Hence, f ((CAB)P ) = f (C)f (A P )f (B). Because of Lemma 10.4.4 we thus obtain that f (A P ) = f (A) if and only if f ((CAB)P ) = f (CAB), as desired. To prove the second part, because of Lemma 10.4.1, write A=(

A1 0

cT ) e31 (t) B1 , 1

with A1 ∈ SL2 (R, I), B1 ∈ R3 (R, I), t ∈ I and c = (c1 , c2 ) ∈ I × I. Let −c1 0 ) 1

and

B = B−1 1 ∈ R3 (R, I).

a11 CAB = (a21 0

a12 a22 0

0 c2 ) e31 (t), 1

1 C = (0 0

0 1 0

Then

as desired. We are now in a position to prove that f is a homomorphism with E3 (R, I) in its kernel. Proposition 10.4.10. The mapping f is homomorphism with E3 (R, I) ⊆ ker(f ). Proof. Because of Lemma 10.4.5 and Lemma 10.4.8, it is sufficient to prove that f (A) = 100 f (A P ), for any A ∈ SL3 (R, I). Recall that P = ( 0 0 1 ). 010

334 | 10 General linear groups of degree 2 So, let A ∈ SL3 (R, I). Because of Lemma 10.4.9, without loss of generality, we may assume that a11 a12 0 A = (a21 a22 c ) e21 (t), 0 0 1 with a11 − 1, a12 , a21 , a22 − 1, c, t ∈ I. Also, because of Lemma 10.4.9, we may replace A by e12 (r)A, for any r ∈ I. Hence, we may assume that a11 ≠ 0. Clearly, a11 A = (a21 + tc t

a12 a22 0

0 c) 1

AP = (

and

a11 t a21 + tc

0 1 c

a12 0 ). a22

As A P has determinant 1, we get that Ra11 +R(tc+a21 )+Rt = R and thus R (tc + a21 )+ R t = R, where R = R/Ra11 . Since a11 ≠ 0, we obtain from Lemma 10.1.1 that there exists r ∈ R such that R (t + r(a21 + tc)) = R. As 1 − a11 ∈ I, we may choose r ∈ I. Put d = rc + 1. Consequently, (a11 , ra21 + dt) ∈ W(I, R). Note that t + r(a21 + tc) = ra21 + td and a11 e23 (r)A P = (t + r(a21 + tc) a21 + tc

a11 a12 ra22 ) = (ra21 + td a22 a21 + tc

0 1 + rc c

0 d c

a12 ra22 ) . a22

Because (a11 , ra21 + dt) ∈ W(I, R), there is a matrix in SL2 (R, I) with first column equal to (a11 , ra21 + dt)T (see the beginning of this section). The inverse of this matrix is of the form (

w11 −(ra21 + dt)

w11 w12 )=( a11 w21

w12 ) ∈ SL2 (R, I). w22

Clearly, (

w11 w21

w12 a11 ) ( ra21 + td w22

0 1 )=( d 0

x a11 d

),

for some x ∈ I. w11 w12 0 w22 0 ) 0 0 1

Put W = ( w21

∈ SL3 (R, I). Then

W e23 (r) A P = (

1 0 a21 + tc

x a11 d c

w11 a12 + w12 ra22 w21 a12 + w22 ra22 ) . a22

Let u = a21 + tc ∈ I and let D = e31 (u)−1 We23 (r)A P . Clearly, 1 D = (0 0

x a11 d c − ux

w11 a12 + w12 ra22 ) ∈ R3 (R, I). w21 a12 + w22 ra22 a22 − u(w11 a12 + w12 ra22 )

Consequently, A P = e23 (r)−1 W −1 e31 (u)D,

(10.4.1)

10.4 The group E2 (I) is of finite index in SL2 (R) |

335

with D ∈ R3 (R, I) and −1

e23 (r) W

−1

1 = (0 0

0 1 0

w22 = (−w21 0

0 w22 −r) (−w21 1 0 −w12 w11 0

−w12 w11 0

0 0) 1

0 −r) ∈ L3 (R, I). 1

The definition of f therefore yields that f (A P ) = f ((

w11 w21

−1

w12 ) ) f(B), w22

where B=(

w21 a12 + w22 ra22 a11 d )=( a22 − u(w11 a12 + w12 ra22 ) b1

a11 d c − ux

r − tda12 ), b2

because w21 a12 + w22 ra22 = −(ra21 + td)a12 + a11 ra22 = r(a11 a22 − a12 a21 ) − tda12 = r − tda12 . Because of the multiplicativity property stated in Proposition 10.3.8 and because of Lemma 10.3.1 and Lemma 10.3.5, we obtain that f(B) = ⌈a11 d, r − tda12 ⌉I1 ,I2 = ⌈a11 , r − tda12 ⌉I1 ,I2 ⌈d, r − tda12 ⌉I1 ,I2 = ⌈a11 , r − tda12 ⌉I1 ,I2 ⌈d, r⌉I1 ,I2 = ⌈a11 , r − tda12 ⌉I1 ,I2 ⌈1 + rc, r⌉I1 ,I2 = ⌈a11 , r − tda12 ⌉I1 ,I2 ⌈1, r⌉I1 ,I2 = ⌈a11 , r − tda12 ⌉I1 ,I2 From Lemma 10.3.1 and Proposition 10.3.10 we also get that (notice that r, 1 − a11 a22 ∈ I) ⌈a11 , r − tda12 ⌉I1 ,I2 = ⌈a11 , r(1 − a11 a22 ) − tda12 ⌉I1 ,I2 = ⌈a11 , −ra21 a12 − tda12 ⌉I1 ,I2 = ⌈a11 , −(ra21 + td)⌉I1 ,I2 ⌈a11 , a12 ⌉I1 ,I2 Consequently, because of Lemma 10.4.2, f (A P ) = f ((

w11 w21

−1

w12 ) ) f(B) w22

= ⌈a11 , −(ra21 + td)⌉−1 I1 ,I2 ⌈a 11 , −(ra 21 + td)⌉I1 ,I2 ⌈a 11 , a 12 ⌉I1 ,I2 = ⌈a11 , a12 ⌉I1 ,I2 = f(A), as desired. We now are in a position to prove the main result of this Chapter.

336 | 10 General linear groups of degree 2 Theorem 10.4.11 ([219] and [148]). Let R be an order in a number field. Assume U(R) is infinite and J is a non-zero ideal of R. The group E2 (J) is of finite index in SL2 (R) Proof. Recall from Corollary 9.3.6 that the stable range of R is 2 and thus, by Proposition 9.4.3, E3 (R, I) is a normal subgroup of GL3 (R). As mentioned early in the chapter, without loss of generality, we may assume that R is integrally closed. Let I1 = J, I2 = J and I = I1 I2 (I1 + I2 ) = J 3 . Let π : SL3 (R, I) → SL3 (R, I)/E3 (R, I) be the natural homomorphism and let h be the restriction of π to SL2 (R, I). Because of Proposition 10.4.10, we know that f : SL3 (R, I) → SL2 (R, J ≀ J) is a group homomorphism with E3 (R, I) ⊆ ker(f ). Hence, f induces a group homomorphism g : SL3 (R, I)/E3 (R, I) → SL2 (R, J ≀ J)/E2 (J). Then g ∘ h = f . By Lemma 10.2.1, f is surjective and therefore so is g. From Theorem 9.5.1 and Corollary 9.4.5 we know that SL3 (R, I)/E3 (R, I) is finite and hence [SL2 (R, J ≀ J) : E2 (J)] < ∞. As [SL2 (R) : SL2 (R, J ≀ J)] < ∞ we deduce that [SL2 (R) : E2 (J)] < ∞, as desired.

11 Generators of the unit group of an integral group ring In this chapter we show that, for many finite groups G, the Bass units together with the bicyclic units generate a subgroup of finite index in the unit group of the integral group ring ℤG. These results are due to Jespers and Leal [112], and Ritter and Sehgal [190–194]. The exceptions are the finite groups G such that their rational group algebra ℚG has exceptional simple components (see Definition 11.2.2) or such that G has nonabelian fixed point free epimorphic images. The basic idea is that the elements of reduced norm 1 together with the central units generate a subgroup of finite index in U(ℤG) (Proposition 5.5.1). So, to obtain a subgroup of finite index one has to “cover” the group of central units and the group generated by the elements of reduced norm 1. Crucial for the proofs are the results proved in Chapter 9 and Chapter 10 on generators for subgroups of finite index in SLn (O) (with n > 1), where O is an order in a simple component of ℚG. The bicyclic units will take care of these groups. To handle the central units of ℤG, we first need to prove the Bass Unit Theorem. As an application, we show a result of Bass and Milnor [16] on the description of finitely many generators of a subgroup of finite index in K1 (ℤG). Several results will be stated and proved in the more general context of orders in finite dimensional semisimple rational algebras. Also, the results on group rings often will be stated in the more general context of group rings over orders R, mainly with R = ℤ[ξ], where ξ is a root of unity. In Section 11.4, we give some classical background on fixed point free finite groups. In particular, we show that these are precisely the Frobenius complements and we state a description of such groups, a result due to Frobenius and some refinements due to Brown. Finally, in Section 11.5, for nilpotent finite groups G such that ℚG does not have exceptional simple components, we show that finitely many generators of the unit group U(ℤG) can be determined, even if G has nonabelian fixed point free epimorphic images. For earlier surveys and related work we refer to [26, 50, 103, 105, 135, 189, 202, 204].

11.1 Bass Unit Theorem Throughout this section O is an order in a finite dimensional semisimple rational algebra A and R is the maximal order (i.e. the ring of integers) in Z(A). For a ring T we know that K0 (T) and K1 (T) are abelian groups, i.e. they are ℤ-modules. Put R0 (T) = ℝ ⊗ℤ K0 (T)

and

R1 (T) = ℝ ⊗ℤ K1 (T),

both are ℝ-vector spaces. As usual, the dual space Homℝ (V, ℝ) of an ℝ-vector space V is denoted by V ∗ .

338 | 11 Generators of unit groups of group rings Let Aℝ = ℝ ⊗ℚ A. In order to define a vector space morphism R0 (Aℝ ) → R1 (O)∗ we need to introduce more notation. For a finitely generated left Aℝ -module V one has a natural ring homomorphism O → Endℝ (V) and hence also a natural group homomorphism natV : K1 (O) → K1 (Endℝ (V)). Note that Endℝ (V) is a central simple ℝ-algebra. Consider the group homomorphism nr : K1 (Endℝ (V)) → U(ℝ), induced by the determinant map, and let f V = nr ∘ natV : K1 (O) → U(ℝ). Observe that if x ∈ K1 (O) is represented by A ∈ GLn (O) then f V (x) = NrV n /ℝ (A), where V n is considered as an M n (Aℝ ) in the standard way. Hence, we obtain the following group homomorphism to the additive group ℝ g V = log|f V | : K1 (O) → ℝ. Its ℝ-linear extension we also give the same name. So, g V : R1 (O) → ℝ. If W also is a finitely generated left Aℝ -module then clearly g V⊕W = g V + g W . We hence obtain a homomorphism K0 (Aℝ ) → R1 (O)∗ and an induced ℝ-linear map G : R0 (Aℝ ) → R1 (O)∗

[V] 󳨃→ g V . Clearly the natural monomorphism K0 (A) → K0 (Aℝ ) yields an injective ℝ-linear map H : R0 (A) → R0 (Aℝ ).

Theorem 11.1.1 (Bass Unit Theorem [16]). If O is an order in a finite dimensional semisimple rational algebra A then the sequence H

G

0 → R0 (A) → R0 (Aℝ ) → R1 (O)∗ → 0 is exact. Proof. Let R be the maximal order of Z(A). First of all observe that dimℝ R1 (O)∗ = dimℝ R1 (O) = rk(K1 (O)) = rk(U(R)) = rℝ (A) − rℚ (A) = dimℝ (R0 (Aℝ )) − dimℝ (R0 (Aℚ )). The third equality is a consequence of Corollary 9.5.10 and the fourth follows from Theorem 7.1.3. As we already know that H is injective, we only have to show that G is surjective and G ∘ H = 0.

11.1 Bass Unit Theorem | 339

We start proving G ∘ H = 0. Let V be a finitely generated left A-module and set Vℝ = ℝ ⊗ℚ V. Then G ∘ H([V]) = g Vℝ = log|f Vℝ | and so, we need to prove that |f Vℝ (x)| = 1 for all x ∈ K1 (O). Let x ∈ K1 (O) be represented by A ∈ GL n (O). Then f Vℝ (x) = NrVℝn /ℝ (A) = NrV n /ℚ (A) ∈ U(ℤ), by (2.3.4) and Lemma 4.6.9 (2). Thus |f Vℝ (x)| = 1, as desired. Next we prove that G is surjective. Because of Lemma 4.6.9 (4), if O1 is another order of A then R1 (O) and R1 (O1 ) are naturally isomorphic to R1 (O ∩ O1 ). So we freely can change O by any other order. If A = A1 × A2 and Oi is an order in A i then G is the direct product of the two maps G : R0 (A iℝ ) → R1 (Oi ), with i = 1, 2. Thus, without loss of generality, we may assume that A is simple. Let F = Z(A), let σ1 , . . . , σ r be the real embeddings of F and let σ r+1 , . . . , σ r+s be representatives up to conjugation of the complex non-real embeddings of F. By the proof of Dirichlet’s Unit Theorem (Theorem 5.2.4), the image of U(R) under the logarithm map x 󳨃→ (log|σ1 (x)|, . . . , log|σ r (x)|, 2 log|σ r+1 (x)|, . . . , 2 log|σ r+s (x)|) is a full lattice in V = {(x1 , . . . , x r+s ) ∈ ℝk : x1 + ⋅ ⋅ ⋅ + x r+s = 0}. Thus, R1 (R)∗ is the ℝ-linear span of the set consisting of the maps x 󳨃→ log|σ i (x)|. By Theorem 9.5.1, the reduced norm over F induces an isomorphism nr : R1 (O) → R1 (R). Therefore R1 (O)∗ is generated by the ℝ-linear span of the maps g i : x 󳨃→ log|σ i (nr(x))|. By Problem 2.3.6 (3) there is a simple Aℝ -module V i and a positive rational number q i such that g V i = q i g i . Thus the image of G contains a generating set of R0 (O)∗ and hence G is surjective. We give an application of the Bass Unit Theorem to the unit group of an integral group ring ℤG of a finite group G. We state the result in the more general context of group rings over an order O in a semisimple finite dimensional ℚ-algebra, say A. (See also the work of Lam [140, 141].) Given a subring R of a ring S we denote by j : K1 (R) → K1 (S) the natural homomorphism induced by the embedding of R in S. Its ℝ-linear extension R1 (R) → R1 (S) we also denote by j. By j∗ we denote the natural induced homomorphism (R1 (S))∗ → (R1 (R))∗ . Clearly (AG)ℝ = Aℝ G. For a finitely generated left Aℝ G-module V and subgroup H of G, we denote by V H the set V regarded as a left Aℝ H-module. Clearly the following diagram is commutative OG

Endℝ (V)

OH

Endℝ (V H )

(the horizontal maps being the natural maps defined by the respective modules). Then, by the definitions of f V and f V H , also the following diagram is commutative

340 | 11 Generators of unit groups of group rings K1 (OG) fV

det

K1 (Endℝ (V))

j

U(ℝ)

f VH K1 (OH) The map V → V H defines a natural restriction homomorphism restr : K0 (Aℝ G) → K0 (Aℝ H). Tensoring with ℝ the previous commutative diagram translates into R0 (Aℝ G)

G

j∗

restr R0 (Aℝ H)

(R1 (OG))∗

G

(11.1.1)

(R1 (OH))∗

Theorem 11.1.2 (Bass-Milnor). Let O be an order in a semisimple finite dimensional rational algebra A and let G be a finite group such that AG is semisimple. Then ⟨j(K1 (Z(O)C)) : C a cyclic subgroup of G⟩ is a subgroup of finite index in the abelian group K1 (OG). Proof. By Corollary 9.5.10, the natural image of U(Z(OG)) in K1 (OG) has finite index in K1 (OG). In particular, j(K1 (Z(OG))) has finite index in K1 (OG) and hence j(K1 (Z(O)G)) has finite index in K1 (OG). Hence, without loss of generality, we may assume that O is commutative. Because K1 ((R1 × R2 )G) is naturally isomorphic to K1 (R1 G) × K1 (R2 G), we also may assume that A is simple. Therefore, let O be an order in a number field, say F. Recall that K1 (OG) is a finitely generated abelian group. So, to prove the result, it suffices to show that the sets j(R1 (OC)), where C runs through the cyclic subgroups of G, generate R1 (OG) as an ℝ-vector space. For this it is enough to prove that if f : R1 (OG) → ℝ is an ℝ-linear map with (f ∘ j)(R1 (OC)) = 0 for all cyclic subgroups C of G, then f = 0. Because of the Bass Unit Theorem and (11.1.1), we first note that the

11.2 Generalized bicyclic units and Bass units I |

341

following diagram (with exact rows) is commutative G

H

0 󳨀→ R0 (FG) 󳨀→ R0 (Fℝ G) 󳨀→ R1 (OG)∗ 󳨀→ 0 restr ↓

H

restr ↓

G

↓ j∗

0 󳨀→ R0 (FC) 󳨀→ R0 (Fℝ C) 󳨀→ R1 (OC)∗ 󳨀→ 0 Let f ∈ R1 (OG)∗ be such that (f ∘ j)(R1 (OC)) = 0, for every cyclic subgroup C of G. By the surjectivity of G there exists x ∈ R0 (Fℝ G) such that G(x) = f . Write x as a real linear combination of Fℝ G-modules M i and let ψ denote the map G → ℝ obtained by replacing in the ℝ-linear combination each M i by TrM i /ℝ . Since j∗ (f) = 0, we have that restr(x) ∈ H(R0 (FC)), for (each) cyclic subgroup C of G. Hence, ψ|C is a real linear combination of F-characters on C. In particular, ψ(g1 ) = ψ(g2 ) for all g1 and g2 that are F-equivalent. The Berman-Witt Theorem (Theorem 7.1.11) implies that also ψ is a real linear combination of F-characters of G. Hence x ∈ H(R0 (G)) and thus f = 0, as desired. For group rings of finite groups G over cyclotomic integers ℤ[ξ] (with ξ a complex root of unity) we obtain the following result. Recall from Problem 7.3.2 that Bass units also have been introduced in group rings ℤ[ξ]G as the units of the form u k,m (ϵg), with ϵ ∈ ⟨ξ ⟩ and g ∈ G. By Bass(ℤ[ξ]G) we denote the group generated by these units. In case ξ = 1 then Bass(ℤG) = Bass(G). Corollary 11.1.3. Let G be a finite group and ξ a root of unity. The natural images of the Bass units in K1 (ℤ[ξ]G) generate a subgroup of finite index. Proof. This follows at once from the Bass-Milnor result (Theorem 11.1.2) and Problem 7.3.2.

11.2 Generalized bicyclic units and Bass units I Let G be a finite group. In this section we describe generators for a subgroup of finite index in U(ℤG). Some of the results are stated in the more general context of group rings over orders in a semisimple group algebra AG, with A a finite dimensional semisimple rational algebra A and with AG semisimple. The following result follows at once from Corollary 11.1.3 and Proposition 9.5.11. Corollary 11.2.1. Let G be a finite group, let ξ be a complex root of unity and let U be a subgroup of U(ℤ[ξ]G) such that for every primitive central idempotent e of ℚ(ξ)G and an order O in ℚ(ξ)Ge, the group U contains a subgroup of finite in 1 − e + O1 . Then ⟨U, Bass(G)⟩ has finite index in U(ℤ[ξ]G). By Corollary 11.1.3, if U contains the Bass units then condition (1) of Proposition 9.5.11 holds. To describe units that will fulfill condition (2), we extend the notion of bicyclic unit.

342 | 11 Generators of unit groups of group rings Recall that a bicyclic unit in an integral group ring ℤG is a unit of the type ̃ = 1 + (1 − h)g h̃ b(g, h)

or

̃ g) = 1 + hg(1 ̃ b(h, − h),

with g and h elements of the finite group G. We now generalize these units as follows. Let A be a semisimple finite dimensional rational algebra such that AG is semisimple. Let O be an order in A. For an idempotent f of AG, let n f be a minimal positive integer such that n f f ∈ OG. It is only for convenience that we take the positive integer minimal; one could replace it by any non-zero integer multiple. Let x1 , . . . , x m be a generating set of O as ℤ-module. Then b(x i g, f) = 1 + n2f (1 − f)x i gf

and

b(f, x i g) = 1 + n2f fx i g(1 − f)

(11.2.1)

are units because ((1 − f)x i gf )2 = 0 and (fx i g(1 − f))2 = 0. These are called generalized bicyclic units. For a given set of idempotents F let GBicF (OG) = ⟨b(x i g, f), b(f, x i g) | f ∈ F , g ∈ G, 1 ≤ i ≤ m⟩ , the group generated by all the generalized bicyclic units defined by the set F. If O = ℤ ̂ then n f = |h| then we simply denote this group GBicF (G). Note that if h ∈ G and f = h and |h|−1

̂ = (1 − h) ∑ (|h| − i)h i . |h|(1 − h) i=0

̂ = 1 + (1 − h)αg h̃ and b(h, ̂ g) = 1 + hgα(1 ̃ Hence, b(g, h) − h) for some α ∈ ℤ ⟨h⟩. So, if ̂ : h ∈ G} then C = {h GBicC (G) ⊆ Bic(G). To prove the relevance of these units, we need to make use of the fundamental results of Bass, Liehl and Vaseršte˘ın, proved in Corollary 9.5.9 and Theorem 10.4.11 respectively, and also of a result of Venkataramana [221] that extends Theorem 10.4.11 to include two-by-two matrices over division algebras that are finite dimensional over the rationals but that are different from a totally definite quaternion algebra with center ℚ (see also the work of Bak and Rehmann [9]). The proof of the latter is beyond the scope of this monograph and hence will be omitted. These fundamental results are collected in Theorem 11.2.3 and are strongly related to the celebrated congruence theorems (for more background we refer to [187]). For convenience we introduce the following terminology. Definition 11.2.2. A simple finite dimensional rational algebra is said to be exceptional if it is one of the following types: (1) a non-commutative division algebra different from a totally definite quaternion algebra, (2) M2 (ℚ), (3) M2 (F) with F a quadratic imaginary extension of ℚ,

11.2 Generalized bicyclic units and Bass units I | 343

a,b (4) M2 ( a,b ℚ ), with a and b negative integers (i.e. ( ℚ ) is a totally definite quaternion algebra with center ℚ). We say that a semisimple finite dimensional algebra A has no exceptional simple components if all its Wedderburn components are not exceptional.

Note that, because of Dirichlet’s Unit Theorem (Theorem 5.2.4) and Proposition 5.5.6, the exceptional simple components that are two-by-two matrices over a division algebra D are precisely those for which an order in D has only finitely many units. Further, all finite dimensional division algebras are exceptional except those for which the unit group of an order has a central subgroup of finite index. Theorem 11.2.3 (Bass, Liehl, Vaseršte˘ın, Venkataramana). Let D be a finite dimensional rational division ring and let O be an order in D. Assume that n is an integer and n ≥ 2. If the simple algebra M n (D) is not exceptional then [SLn (O) : E n (I)] < ∞ for any non-zero ideal I of O. We need one more technical lemma. Recall that a set of matrix units of a matrix algebra M n (D) over a division ring D is a set of elements E i,j ∈ M n (D) with 1 ≤ i, j ≤ n and ∑ni=1 E i,i = 1 and E i,j E k,l = δ j,k E i,l . Lemma 11.2.4. Let D be a finite dimensional rational division algebra and let n be an integer with n > 1. If f is a non-central idempotent in M n (D) then there exist matrix units E i,j , with 1 ≤ i, j ≤ n, such that f = E1,1 + ⋅ ⋅ ⋅ + E l,l , with 0 < l < n. Furthermore, M n (D) = M n (D󸀠 ) with D󸀠 the centralizer of all E i,j . Proof. Put A = M n (D). As fAf is a simple Artinian ring with identity f , there exits orthogonal idempotents E1 , . . . , E l such that f = E1 + ⋅ ⋅ ⋅ + E l and E i AE i is a division ring for each 1 ≤ i ≤ l. Similarly, 1 − f = E l+1 + ⋅ ⋅ ⋅ + E m , a sum of orthogonal idempotents with E j AE j a division ring for l + 1 ≤ j ≤ m. Each AE i is a simple left A-module and all of them are isomorphic, to the unique (up to isomorphisms) simple left A-module, say M. Then, the map ρ : A → EndA (A), associating a ∈ A to right multiplication by a is an isomorphism and EndA (A) ≅ M m (EndA (M)). Hence, m = n and D ≅ EndA (M). Moreover, if u i : AE i → A denotes the inclusion map and p i : A → AE i is the projection along the above decomposition then the elements e i,j = p i u j , with 1 ≤ i, j ≤ n, form a set of matrix units of EndA (A) and ρ(E i ) = e i,i . Therefore, the elements E i,j = ρ−1 (e ij ) form a set of matrix units of A, with E i,i = E i . Moreover, the standard isomorphism EndA (A) ≅ M n (EndA (M)) maps the e ij to the standard matrix units of M m (EndA (M)) and the centralizers of the standard matrix units of M m (EndA (M)) is precisely EndA (M)I n . Thus A ≅ M n (D󸀠 ) with D󸀠 the centralizer of the E ij . Theorem 11.2.5 (Jespers-Leal). Let G be a finite group and R an order in a semisimple finite dimensional rational algebra A. Assume AG is semisimple, e is a primitive central idempotent of AG and O an order in AGe. Assume the simple algebra AGe is not excep-

344 | 11 Generators of unit groups of group rings tional. If f is an idempotent of AG such that ef is non-central (in AGe) then GBic{f} (RG) contains a subgroup of finite index in 1 − e + O1 . Proof. Recall that n f is the minimal positive integer such that n f f ∈ RG. Let x1 , . . . , x m be a generating set of R as a ℤ-module. As AGe = M n (D), for some division algebra D, by Lemma 11.2.4 there is a set of matrix units {E i,j : 1 ≤ i, j ≤ n} of AGe with f = E1,1 + ⋅ ⋅ ⋅ + E l,l for some 0 < l < n. Recall from Lemma 4.6.9 that the unit groups of two orders in AGe are commensurable. Hence, without loss of generality, we may assume that M n (O) is the order chosen in the statement, with O an order in D. Let J = GBic{f} (RG). Note that k

l

[1 + n2f fx i g(1 − f)] [1 + n2f fx j h(1 − f)] = [1 + n2f f(kx i g + lx j h)(1 − f)] , for every k, l ∈ ℤ, g, h ∈ G and 1 ≤ i, j ≤ m. So, the group generated by these units contains all elements of the form 1 + n2f fα(1 − f),

and

1 + n2f (1 − f)αf,

with α ∈ RG. Since {1 + n2f fα(1 − f), 1 + n2f (1 − f)αf : α ∈ RG} ⊆ J, it follows that {1 + n e n2f fα(1 − f)e, 1 + n e n2f (1 − f)αfe : α ∈ RG} ⊆ J. Let i ≤ l and l + 1 ≥ j ≥ n. Then, f OE i,j (1 − f)e = OE i,j . Hence, as O is a finitely generated ℤ-module, there exists a positive integer n i,j such that 1 + n i,j OE i,j ⊆ J ∩ SLn (O). And similarly, 1 + n j,i OE j,i ⊆ J ∩ SLn (O), for some positive integer n j,i . So we have shown the existence of a positive integer x with 1 + xOE i,j ⊆ J ∩ SLn (O)

and

1 + xOE j,i ⊆ J ∩ SLn (O),

for all 1 ≤ i ≤ l and l + 1 ≤ j ≤ n. Now let 1 ≤ i, j ≤ l, i ≠ j and α ∈ O. Then, by Lemma 9.2.1, 1 + x2 αE i,j = (1 + xαE i,l+1 , 1 + xE l+1,j ) ∈ J ∩ SLn (O). Similarly, for l + 1 ≤ i, j ≤ n i , i ≠ j, it follows that 1 + x2 OE i,j ⊆ J ∩ SLn (O). Because of the assumptions, the result now follows from Theorem 11.2.3.

11.3 Bicyclic units and Bass units |

345

Corollary 11.2.6 (Jespers-Leal). Let G be a finite group and ξ a root of unity. Let E be the set consisting of those primitive central idempotents e of ℚ(ξ)G for which ℚ(ξ)Ge is not a division ring. For each e ∈ E let f e be an idempotent in ℚ(ξ)G such that f e e is noncentral and let F = {f e : e ∈ E}. If ℚ(ξ)G does not have exceptional simple components then the group ⟨GBicF (ℤ[ξ]G) ∪ Bass(ℤ[ξ]G)⟩ is of finite index in U(ℤ[ξ]G). Proof. This follows from Corollary 11.2.1 and Theorem 11.2.5. Remark 11.2.7. Note that if ξ is a root of unity of order at least seven then clearly ℚ(ξ)G can not have an exceptional simple component that is a matrix of degree 2. From Theorem 3.6.5 we know that ℚ(ζ|G| ) is a splitting field of ℚG, hence ℚ(ζ|G| )G does not have exceptional simple components if |G| ≠ 6.

11.3 Bicyclic units and Bass units Recall ([170]) that a finite group G is said to be fixed point free if it has an (irreducible) complex representation ρ such that 1 is not an eigenvalue of ρ(g) for all 1 =∈ ̸ g ∈ G. Clearly, subgroups of fixed point free groups are fixed point free. Such groups show up naturally. Indeed, every non-trivial finite subgroup G of a rational division algebra D is fixed point free. To prove this we may assume that G generates D as vector space over ℚ. Fix an isomorphism of ℂ-algebras ρ : ℂ ⊗Z(D) D → M n (ℂ) and let η(x) = ρ(1⊗x) for each x ∈ D. The restriction of η to G is a representation of G. We show that if 1 ≠ g ∈ G then all eigenvalues of η(g) are different from 1. Indeed, suppose v ∈ ℂn is a column vector with η(g)(v) = v. Then, η(g−1)(v) = 0. Because g−1 is invertible in the division ring D, we get that v = 0. So, 1 is indeed not an eigenvalue of η(g) and thus G is fixed point free. In particular, if e is a primitive central idempotent of a group algebra RG of a finite group G such that RGe is a division algebra then Ge is a fixed point free group. This implies the following lemma. Lemma 11.3.1. Let G be a finite group and F a number field. If G does not have a nonabelian epimorphic image that is a fixed point free group then FG does not have a noncommutative division algebra as a simple component. Theorem 11.3.2 (Jespers-Leal). Let G be a finite group and ξ a root of unity. Suppose ℚ(ξ)G does not have exceptional simple components. Let C = {ĝ : g ∈ G}. If G has no nonabelian homomorphic image which is fixed point free, then ⟨GBicC (ℤ[ξ]G) ∪ Bass(ℤ[ξ]G)⟩ is of finite index in U(ℤ[ξ]G). In particular, Bass(G) ∪ Bic(G) generates a subgroup of finite index in U(ℤG).

346 | 11 Generators of unit groups of group rings Proof. Note that because of Lemma 11.3.1, the group algebra ℚ(ξ)G does not have a non-commutative division algebra as a simple component. Because of Corollary 11.2.6, it thus is sufficient to show that if e is a primitive central idempotent of ℚ(ξ)G such that ℚ(ξ)Ge is not commutative then there exists g ∈ G such that ĝ e is a non-central idempotent in ℚGe. Let e be a primitive central idempotent of ℚ(ξ)G with Ge not commutative. By the assumption, Ge is not fixed point free. Thus, there exists a primitive central idempotent e1 of ℂGe such that the non-linear irreducible complex representation ρ : G → (ℂG)e1 mapping x onto xe1 has eigenvalue 1 for some ρ(g), with g ∈ G and ge1 ≠ e1 . Since ρ(g) is diagonalizable one may assume that ρ(g) = (

Ij 0

0 ) with 1 ≤ j < n and D = diag(ξ j+1 , . . . , ξ n ) D

and ξ j+1 , . . . , ξ n are roots of unity different from 1. Consequently ρ(ĝ ) = (

Ij 0

0 ). 0

Hence ĝ e1 is a non-central idempotent of ℂG. It follows that ĝ e is a non-zero idempotent in ℚ(ξ)Ge. Furthermore ĝ e ≠ e, because otherwise ĝ e 1 = ĝ e1 e = e1 e = e1 , a contradiction. The result improves an earlier result of Ritter and Sehgal in [191] where, under the same assumptions, it was proved that the normal closure in U(ℤG) of the group generated by the Bass units and the bicyclic units is of finite index. Suppose G is a finite group such that ℚG does not have exceptional simple components. From Theorem 11.3.2 we obtain at once that if G does not have a nonabelian epimorphic image that is fixed point free then the central units together with the bicyclic units generate a subgroup of finite index in U(ℤG). If one also assumes that ℚG does not have a simple component that is a totally definite quaternion algebra, then the converse of this statement is shown by Ritter and Sehgal (see the proof of Theorem 21.20 in [202]). This can be proved with the method used in the proof of Theorem 11.3.2 (Problem 11.3.1). Recall that ζ n denotes a complex primitive n-th root of unity, for n a positive integer. In case ξ = ζ|G| then we can avoid the assumption that G has no nonabelian fixed point free homomorphic images. The proof is taken from [202, Corollary 28.9]. Proposition 11.3.3. Let G be a finite group and let C = {ĝ : g ∈ G}. If G ≠ D6 (the dihedral group of order 6) then Bass(ℤ[ζ|G| ]G) ∪ BicC (ℤ[ζ|G| ]G) generates a subgroup of finite index in U(ℤ[ζ|G| ]G). Proof. Because of the assumption and Remark 11.2.7, ℚG does not have exceptional simple components. Let ξ = ζ|G| and let e be a primitive central idempotent of ℚ(ξ)G such that ℚ(ξ)Ge is not abelian. We claim that that there exists g ∈ G and ξ1 ∈ ⟨ξ ⟩ such that ̂ ξ1 ge is a non-central idempotent. The result then follows from Corollary 11.2.6. Let

11.3 Bicyclic units and Bass units |

347

g ∈ G be such that ge is not central. Then, ge has at least two different eigenvalues, say ξ1−1 and ξ2 . Clearly these are roots of unity of order a divisor of |G|. Hence they belong to ⟨ξ ⟩. It follows that ξ1 ge has among its eigenvalues 1 and ξ2 ξ1 ≠ 1. As in the proof of Theorem 11.3.2 one gets that ξ̂ 1 ge is a non-central idempotent. If G is group of odd order then, by Problem 4.1.6, we know that ℚ(ξ)G does not have exceptional simple components that are matrix rings of degree 2. Hence the following is an immediate application of Proposition 11.3.3 (for the converse see Problem 11.3.1). Corollary 11.3.4 (Ritter-Sehgal). Let G be a group of odd order and let C = {ĝ : g ∈ G}. If G does not have a nonabelian homomorphic image which is fixed point free and ξ is a root of unity, then Bass(ℤ[ξ]G) ∪ BicC (ℤ[ξ]G) generates a subgroup of finite index in ℤ[ξ]G. The converse holds if ℚ(ξ)G has no simple components that are noncommutative division rings. Often the number of generators constructed can be reduced. For example, if all bicyclic units of ℤG are included in the set of generators, then implicitly one makes use of all idempotents ĝ with g ∈ G. Because of Proposition 9.5.11 and Theorem 11.2.5, one only needs to make use of one non-central idempotent in ℚGe for every primitive central idempotent of ℚG such that ℚGe is a matrix algebra that is not an exceptional component. We consider two examples: the symmetric group and some metacyclic groups. For the first one we give a proof and for the second one we refer to Problem 11.3.2). It is well known that ℚS n is a direct sum of matrices over ℚ (see for example [99, Theorem 5.9]) and there are no matrices of degree 2 if n ≥ 5. Hence Z(U(ℤS n )) is a finite group and thus because of Corollary 11.2.6 one only has to discover a non-central idempotent in orders of simple components of ℚS n , i.e. one does not have to make use of Bass units. Proposition 11.3.5. Let S n be the symmetric group of degree n and let a denote a transposition in S n . If n ≥ 5 then ⟨1 + (1 + a)g(1 − a), 1 + (1 − a)g(1 + a) : g ∈ S n ⟩ is of finite index in U(ℤS n ). Proof. Because of the remarks made before the corollary, it is sufficient to prove that if e is a primitive central idempotent of ℚS n , such that ℚS n e is non-commutative, then ̂ e is central idempotent, i.e. this idempotent is non zero and not equal to e. Choose a ̂ e = 0 then ae = −e and thus ℚS n e = ℚ ⟨b⟩ e is an n-cycle b such that S n = ⟨a, b⟩. If a ̂ e = e then ae = e and again this implies that a commutative ring, a contradiction. If a ̂ e is a non-central idempotent of ℚS n e. ℚS n e is commutative. So, indeed a Proposition 11.3.6. Let G be a split metacyclic group ⟨a⟩m ⋊⟨b⟩n with a b = a r . Suppose that n is odd and r j i ≡ i mod m implies ri ≡ i mod m, (11.3.1)

348 | 11 Generators of unit groups of group rings for each 1 ≤ i ≤ m, 1 ≤ j ≤ s − 1. Then the group generated by the Bass units and the units of the form ̂ and 1 + bg(1 ̂ 1 + (1 − b)g b − b), g ∈ G, is of finite index in U(ℤG).

Problems 11.3.1. Let G be a finite group such that ℚG does not have simple components of exceptional type nor simple components that are totally definite quaternion algebras. Prove that if the central units together with the bicyclic units generate a subgroup of finite index in U(ℤG) then G does not have a nonabelian fixed point free epimorphic image. (Hint: Proposition 5.5.6.) 11.3.2. Let G = ⟨a⟩m ⋊ ⟨b⟩n a split metacyclic group with a b = a r . Prove the following properties: (1) The sum of the dimensions of the commutative simple components of ℚG is n gcd(m, r − 1). (2) If d | m then (⟨a⟩ , ⟨a d ⟩) is a strong Shoda pair if and only if r j ≢ 1 mod md for every j = 1, . . . , s − 1. (3) Suppose that m, n and r satisfy condition (11.3.1). If e is a primitive central idem̂ is a potent of ℚG with ℚGe non-commutative then e = e(G, ⟨a⟩ , ⟨a d ⟩) and be non-central idempotent of ℚGe. (4) Prove Proposition 11.3.6.

11.4 Fixed point free groups and Frobenius complements We note that in the proof of Theorem 11.3.2 the following is shown. Let e be a primitive central idempotent of ℚG and let ρ an irreducible complex representation of G with ρ(e) ≠ 0. If g ∈ G then 1 is an eigenvalue of ρ(g) if and only if ĝ e ≠ 0. Therefore Ge is fixed point free if and only if ĝ e is central in ℚG for every g ∈ G. Let e1 be the sum of the primitive central idempotents e such that ĝ e is not central in ℚG for some g ∈ G. Using the ideas of the previous section one deduces that the group generated by the bicyclic units of G contains a subgroup of finite index in 1 − e + SL1 (ℤGe) for every non-exceptional component of ℚGe1 . However, the bicyclic units do not contribute in the components of ℚG(1 − e1 ) ≅ ℚG/ℚGe1 . Let now e G be the sum of the primitive central idempotents e of ℚG such that ĝ e ≠ 0 for some 1 ≠ g ∈ G. Clearly ℚGe1 ⊆ ℚGe G = ∑1=g∈G ℚG ĝ = ∑1=g∈G ℚG g.̃ ̸ ̸ Brown calls ℚG/ℚGe G the truncated rational group algebra of G and proved that G is fixed point free if and only if 1 − e G is a primitive central idempotent of ℚG (equivalently, the truncated rational group algebra of G is simple) [38, Theorem 1.1]. Thus

11.4 Fixed point free groups and Frobenius complements | 349

the simple components of ℚG(1 − e1 ) are the truncated rational group algebras of the fixed point free epimorphic images of G. Then G is fixed point free if and only if e G ≠ 1. Furthermore, G has a fixed point free epimorphic image if and only if e1 ≠ 1. Hence, the simple components of ℚG(1 − e1 ) are the truncated rational group algebras of the fixed point free epimorphic images of G. This shows that fixed point free finite groups G require special treatment in connection with describing generators for U(ℤG). In this section we recall some background on these groups and we show that they are precisely the Frobenius complements. Recall that a finite group G is said to be a Frobenius group if it contains a proper non-trivial subgroup H such that H ∩ H g = {1} for all g ∈ G \ H. The group H is called a Frobenius complement in G. The following classical result shows that such groups can be decomposed as a semidirect product. For a detailed discussion of Frobenius groups the reader is referred to [98, 170, 196]. Theorem 11.4.1 (Frobenius). Let G be a finite group and H a subgroup of G. Then, G is a Frobenius group with complement H if and only if it contains a normal subgroup N such that G = NH, H ∩ N = {1} and CenG (n) ⊆ N for any non-trivial n ∈ N. Furthermore, in case G is a Frobenius group with complement H, the group N is uniquely determined by H; it is called the Frobenius kernel and it equals (G \⋃g∈G H g )∪ {1}. Also, |N| ≡ 1 mod |H| and therefore |N| and |H| are relatively prime. Proof. We first prove the sufficiency. So, assume G has a normal subgroup N such that G = NH, H ∩ N = {1} and CenG (n) ⊆ N for any non-trivial n ∈ N. Because of the Sylow Theorem and the fact that non-trivial p-groups have non-trivial centers, it follows that |N| and [G : N] are relatively prime. Let g ∈ G \ H and write g = nh with 1 ≠ n ∈ N and h ∈ H. Assume x ∈ H ∩ H g . Then, x = y n for some y ∈ H. Clearly we get that (y, n) = y−1 y n = y−1 x ∈ H. As N is normal in G, this yields (y, n) ∈ H ∩ N = {1} and thus y ∈ CenG (n) ⊆ N. So, y = 1 and thus x = 1. Therefore H ∩ H g = {1}. This shows that G is a Frobenius group with complement H. Conversely, assume G is a Frobenius group with complement H. Let N = (G \ ⋃ H g ) ∪ {1}. g∈G

Because H = N G (H) the subgroup H has [G : H] distinct conjugates. Hence, ⋃g∈G H g \ {1} contains precisely [G : H](|H| − 1) elements. The remaining elements constitute the set N and thus |N| = |G| − [G : H](|H| − 1) = [G : H]. Note that, furthermore, if M is a normal subgroup of G with M ∩ H = {1} then M ∩ H g = {1} for all g ∈ G and thus M ⊆ N. This shows that if a Frobenius kernel exists then it is unique and equal to the set N. If 1 ≠ n ∈ N and x ∈ CenG (n) \ N then x ∈ H g for some g ∈ G and, taking conjugates, we may assume that 1 ≠ x ∈ H. Then x ∈ H ∩ H n = {1}, a

350 | 11 Generators of unit groups of group rings

contradiction. This implies that the action of H on N by conjugation induces an action of H on N \ {1} such that the stabilizer of every point is trivial. In particular, each orbit has the same cardinality than H and hence |N| ≡ 1 mod |H|. The main issue in the proof is to show that N actually is a subgroup of G. To do so we first show that if θ is a ℂ-class function of H with θ(1) = 0 then (θ G )H = θ. To prove this, let 1 ≠ h ∈ H. Then, θ G (h) =

1 |H|

∑ g∈G,

θ(h g ).

h g ∈H

∈ H then ∈ and thus g ∈ H. Therefore θ(h) = θ(h g ). It follows Note that if that θ G (h) = θ(h) if 1 ≠ h ∈ H. Clearly θ G (1) = [G : H]θ(1) = 0. This proves the claim. Now, let 1H ≠ χ ∈ Irr(H) and put θ = χ − χ(1)1H . So, θ(1) = 0. By the above and the Frobenius reciprocity law (3.2.4) we get that hg

hg

H ∩ Hg

[θ G, θ G ] = [θ, (θ G )H ] = [θ, θ]. Since, by assumption, χ is an irreducible non-trivial complex character of H, it follows from the orthogonality relations (Theorem 3.1.5) that [θ G, θ G ] = 1 + χ(1)2 . Again using the Frobenius reciprocity law, we also have that [θ G, 1G ] = [θ, 1H ] = −χ(1). Consequently, χ∗ = θ G + χ(1)1G is a class function of G such that [χ∗, 1G ] = 0. Hence [χ∗, χ∗ ] = 1. Now, θ is a difference of characters and thus so is θ G . Thus χ∗ is an integral linear combination of characters. Since [χ∗ , χ∗ ] = 1, we get from Theorem 3.1.5 (6) that ±χ∗ ∈ Irr(G). Notice that if h ∈ H, then χ∗ (h) = θ G (h) + χ(1) = θ(h) + χ(1) = χ(h).

(11.4.1)

Hence, χ∗ (1) = χ(1) > 0 and thus χ∗ ∈ Irr(G). So, because of (11.4.1), for any non-trivial χ ∈ Irr(H), we have constructed an extension χ∗ ∈ Irr(G). Let M = ∩χ∈Irr(H)\{1H } ker(χ∗ ), a normal subgroup of G. If h ∈ M ∩ H then χ(h) = χ∗ (h) = χ∗ (1) = χ(1) for all χ ∈ Irr(H) \ {1H } and thus h = 1. Thus, by the above, M ⊆ N. On the other hand, if 1 ≠ g ∈ N, i.e. g does not belong to any conjugate of H, then χ∗ (g) − χ(1) = θ G (g) = 0 for any χ ∈ Irr(H). So, g ∈ ker(χ∗ ) and thus g ∈ M. It follows that M = N and hence N is a normal subgroup of G such that |M| = [G : H]. As |MH| = |M| |H| = [G : H] |H| = |G| the result follows. The condition on the centralizers in the Frobenius Theorem (Theorem 11.4.1) can be replaced by several other properties (see Problem 11.4.1). Frobenius groups with complement H also can be described in terms of permutation groups. Indeed, the natural action of G on the right cosets of H yields a faithful transitive action of G in which no non-trivial element of G has more than one fixed point. Indeed, suppose that g ∈ G fixes two distinct right cosets, say Hg1 and Hg2 . Then, Hg1 g = Hg1 and Hg2 g = Hg2 and thus g ∈ H g1 ∩ H g2 . Since g2 g1−1 ∈ ̸ H we obtain that g = 1.

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351

Conversely, let G be a group acting transitively on a set X and assume that no nontrivial element of G has more than one fixed point. Then G is a Frobenius group and its kernel consists of 1 and all elements of G without fixed points. Indeed, choose x ∈ X and let H denote the stabilizer of x. If g ∈ G \ H then H ∩ H g consists of the elements of G that fix the distinct elements x and xg of X. Hence, by the assumption, H ∩ H g = {1} and thus G is a Frobenius group. It easily is verified that the kernel is as described. In this context, we notice that the Frobenius theorem gives at once the following characterization of Frobenius complements. Recall that an action of a group G on a set X is said to be regular if the stabilizer in G of every element of X is trivial. Corollary 11.4.2. A finite group H is a Frobenius complement if and only if there exists a finite group N together with an action (by automorphisms) of H on N such that H acts regularly on the set N \ {1}. In this case, N ⋊ H is a Frobenius group with kernel N and Frobenius complement H. A deep result of Thompson yields the following result (for a complete and detailed proof we refer the reader to [196, Chapter 10]). Theorem 11.4.3 (Thompson). The Frobenius kernel of a Frobenius group is a nilpotent group. This result has a several applications on the structure of Frobenius groups G. For example, one can show that a Frobenius complement in G is uniquely determined up to conjugation (see for example [170, Corollary 17.5]). In order to prove that Frobenius complements are precisely the fixed point free groups we state and prove some basic properties. We start with a group theoretical result. The proof is left to the reader in Problem 11.4.2. Lemma 11.4.4. Let p be a prime integer and G a finite p-group. Then G contains a noncyclic subgroup of order p2 if and only if G is neither cyclic, nor isomorphic to a quatern−1 nion group Q2n , nor a semidihedral group ⟨a⟩2n ⋊ ⟨b⟩2 with a b = a2 −1 . We say that a group G is a disjoint union of subgroups S1 , . . . S n if G = ∪ni=1 S i and S i ∩ S j = {1} for any distinct i and j. Theorem 11.4.5. Let H be a finite group. If H is a Frobenius complement then the following properties hold. (1) For any prime q, the group H does not contain a non-cyclic subgroup of order q2 . (2) Any subgroup whose order is a product of two distinct primes is cyclic. (3) If |H| is even then H contains a unique element of order 2. (4) For any odd prime divisor q of |H|, the Sylow q-subgroups of H are cyclic. (5) The Sylow 2-subgroups of H are either cyclic or quaternion groups. Proof. Let G be a Frobenius group with complement H and kernel N. Because of Thompson’s result we know that N is nilpotent. Furthermore, by Theorem 11.4.1, the

352 | 11 Generators of unit groups of group rings orders of N and H are coprime. Let p be a prime divisor of Z(N) and put V = {x ∈ Z(N) : x p = 1}, an elementary abelian p-group. Let H1 be a subgroup of H. We claim that if H1 is the disjoint union of t + 1 nontrivial subgroups then p | t. Indeed, assume that H1 = S0 ∪ S1 ∪ ⋅ ⋅ ⋅ ∪ S t for some non-trivial subgroups S i with S i ∩ S j = {1} for i ≠ j. Since, V is an abelian group, the point wise product fg of two automorphisms f and g of V defines an endomorphism of V. Denote by c h1 the conjugation of h1 on V and let c S j denote the point wise product (on V) of all the conjugations c s j with s j ∈ S j . Clearly, c S j is an endomorphism of V such that c S j (v) ∈ CenV (S j ) for any v ∈ V. Because H acts regularly on V, it follows that c S j (v) = 1, the constant mapping on the identity, for any 0 ≤ j ≤ t. Similarly, c H1 = 1, i.e. the product of all conjugate maps c h1 , with h1 ∈ H1 , is the constant map onto 1. Because H1 \ {1} = (S0 \ {1}) ∪ ⋅ ⋅ ⋅ ∪ (S t \ {1}), a disjoint union, we obtain that 1 = c S0 (v)c S1 (v) ⋅ ⋅ ⋅ c S t (v) = v t+1 ∏h∈H1 \{1} c h (v) = v t c H1 (v) = v t . Therefore, as the exponent of V is p, one obtains that p | t, as desired. To prove part (1), assume that H1 = C q × C q for some prime q. Then, H1 is the disjoint union of q + 1 subgroups of order q. Hence, by the previous, p | q. But this is impossible as |N| and |H| are relatively prime. So, part (1) indeed follows. Also part (2) easily follows. Indeed, if H 1 is a non-cyclic group of order qq󸀠 , for some distinct primes q and q󸀠 with q > q󸀠 , then H1 is the disjoint union of one subgroup of order q and q subgroups of order q󸀠 . However, this again contradicts with p ∤ q. To prove part (3), let h ∈ H. Because CenH (v) = {1} for any v ∈ V \ {1} it is clear that the mapping V → V defined by v 󳨃→ v−1 v h is injective. Since V is finite, this mapping thus is bijective. Suppose |H| is even and choose h ∈ H of order 2. Then, by the previous, for any x ∈ V there exists v ∈ V such that x = v−1 v h . Because h has order two, we obtain that x h = (v−1 v h )h = (v−1 )h v = (v−1 v h )−1 = x−1 . If also h󸀠 is an 󸀠 󸀠 element of order two in H we thus get that x h = x−1 = x h . So, x h = x h and hence h󸀠 h−1 ∈ CenH (x) = 1. Consequently, h󸀠 = h and (3) follows. To prove the remaining parts, let P be a Sylow p-subgroup of H. By part (1), P does not contain a non-cyclic subgroup of order p2 . By Problem 11.4.2, H is either cyclic or quaternion. So (4) and (5) follow. Proposition 11.4.6. A finite group G is fixed point free if and only if G is a Frobenius complement. Proof. Suppose G is a fixed point free group. Let ρ be an irreducible complex representation such that 1 is not an eigenvalue of ρ(g) for all 1 ≠ g ∈ G. Let n be the exponent of G, and let F = ℚ(ζ n ), R = ℤ[ζ n ], p a prime integer not dividing n, Q a maximal ideal of R containing p and K = R/Q. By Brauer Splitting Theorem (Theorem 3.6.5), we may assume that ρ is an F-representation. By Problem 6.1.5 reduction modulo Q maps bijectively the complex n-th roots of unity to the n-roots of unity in characteristic p and there is an absolutely irreducible representation ρ in characteristic p such that reduction modulo Q maps bijectively the eigenvalues of ρ(g) to the eigenvalues

11.4 Fixed point free groups and Frobenius complements | 353

of ρ(g), preserving multiplicities. Therefore, 1 is not an eigenvalue of ρ(g) for every g ∈ G \{1}. Moreover ρ is a K-representation. and a finite field K with q elements, there is a representation ρ : G → GLn (K) which lifts to ρ and such that the eigenvalues of ρ(g) are the images under θ of the eigenvalues of ρ(g) for any g ∈ G. In particular, the action defined by ρ on K n is such that the stabilizer of any 0 ≠ v ∈ K n is trivial. Therefore, G acts via automorphisms on the abelian group K n and it acts regularly on K n \ {0} and thus G is a Frobenius complement, by Corollary 11.4.2. For the converse, assume H is a Frobenius complement. As in the first part of the proof of Theorem 11.4.5, let G be a Frobenius group with complement H and kernel N. Let p be a prime divisor of the order of the nilpotent group N. Let 𝔽p be the field with p-elements. The group H acts regularly on the elementary abelian p-group V = {x ∈ Z(N) : x p = 1}. Hence, this defines a faithful 𝔽p -representation, say ρ, and ρ(h) does not have eigenvalue 1 for all non-trivial h ∈ H. By ρ we denote the natural extension of ρ to a 𝔽p -representation of H. Clearly, ρ(h) does not have 1 as an eigenvalue for each non-trivial h ∈ H. Since p is not a divisor of |H|, by Maschke’s Theorem, ρ = ρ1 +⋅ ⋅ ⋅+ρ n , for some irreducible 𝔽p -representations of H. Clearly, for each non-trivial h ∈ H, every ρ i (h) does not have 1 as an eigenvalue. It follows (by Problem 6.1.5) that there exists a faithful irreducible ℚ-representation ψ of G such that ψ(h) does not have eigenvalue 1 for any non-trivial h ∈ H. It follows that H is fixed point free. Corollary 11.4.7. A finite nilpotent group is a Frobenius complement if and only if it is either cyclic or isomorphic to Q2n × C q for some positive integers n and q with n ≥ 3 and q odd. Proof. The necessity is a consequence of Theorem 11.4.5. To prove the sufficiency, we will exhibit a representation ρ of G such that 1 is not an eigenvalue of ρ(g) for every g ∈ G \ {1}, both for G = C n and for G = Q2n . Then, because of Proposition 11.4.6 and Problem 11.4.3, it follows that G is a Frobenius complement. For C n n−1 we simply take any faithful linear representation. For Q2n = ⟨a, b | a2 = 1, b2 = n−2 a2 , a b = a−1 ⟩ we observe that Q2n can be realized as a subgroup of the crossed 2 = −1, ζ u product (ℚ(ζ2n−1 )/ℚ(ζ2n−1 + ζ2−1 = ζ2−1 n−1 ), −1) = ℚ(ζ 2n−1 )[u | u n−1 ], by 2n−1 taking a = ζ2n−1 and b = u. Splitting this algebra we obtain a faithful representation ρ : Q2n → M2 (ℂ) with ρ(a) = (

ζ2n−1 0

0 ) ζ2−1 n−1

and

0 1

ρ(b) = (

−1 ). 0

It is now an easy exercise to show that 1 is not an eigenvalue of ρ(g) for any g ∈ Q2n \{1}. We close this section with some results on the classification of subgroups of division rings and Frobenius complements. For proofs the interested reader may consult [3] or [210] for the first part and [170] and [38] for the second part. First of all we recall the definition of Z-groups.

354 | 11 Generators of unit groups of group rings

A Z-group is a finite group with all Sylow subgroups cyclic. Clearly a metacyclic group C m ⋊ C n with gcd(m, n) = 1 is a Z-group and in fact every Z-group is of such a form [170, Proposition 12.11]. Changing the generators if needed, every Z-group has a presentation of the following type: G m,n,r = ⟨a⟩m ⋊ ⟨b⟩n ,

gcd(m, n) = gcd(m, r − 1) = 1, r n ≡ 1 mod m.

(11.4.2)

Recall that if m and r are relatively prime integers then o = o m (r) denotes the multiplicative order of r modulo m. Further, if p is a prime integer then v p (m) denotes the maximum k such that p k divides m. We are ready to state Amitsur’s classification of finite subgroups of division rings and the classification of fixed point free Z-groups. Theorem 11.4.8 ([3]). (Z) A Z-group G m,n,r = ⟨a⟩m ⋊ ⟨b⟩n , with m, n and r as in (11.4.2), is a subgroup of a division ring if and only if one of the following conditions hold: (a) m = 1, (b) n = 4 and r = −1 or (c) if k = o m (r) and for each prime integer p, P p denotes the Sylow subgroup of ⟨a⟩, Q q denotes the Sylow subgroup of ⟨b⟩ and X p = {q | n : (P p , Q q ) ≠ 1} then the following conditions hold for all the primes p and q with q ∈ X p : i. X p1 ∩ X p2 = 0 for every two different primes p1 and p2 ; ii. v q (o |G| (p)) < o q vq (k) (p), with R p = ∏q∈X p Q q ; |P p | |R p |

iii. if q is odd or p ≡ 1 mod 4 then v q (p − 1) ≤ v q (k) and iv. if q = 2 and p ≡ −1 mod 4 then v2 (k) is either 1 or greater than v2 (p + 1). (NZ)The finite subgroups of division rings which are not Z-groups are (a) ⟨s, t | (st)2 = s3 = t4 ⟩ (binary octahedral group), (b) Q m with v2 (m) ≥ 3. (c) Q8 × M with M a Z-group of odd order such that o|M| (2) is odd, (d) SL(2, 3) × M, with o|M| (2) odd, gcd(|M|, 6) = 1 and M a Z-group satisfying the conditions of (Z). (e) SL(2, 5). Theorem 11.4.9. Let G m,n,r be a Z-group with m, n and r as in (11.4.2) and let n0 be the product of the prime divisors of n. Then G m,n,r is a Frobenius complement if and only if n r n0 ≡ 1 mod m. The following result shows that every Frobenius complement has a large normal Zsubgroup and that the class of non-solvable Frobenius complements is very limited. Theorem 11.4.10. Let G be a finite Frobenius complement. Then, (1) G contains a normal subgroup N such that N is a Z-group and G/N is isomorphic to a normal subgroup of S4 or S5 .

11.5 Group rings of nilpotent groups

| 355

(2) If G is not solvable then G = H × N where N is a Z-group of order coprime with 30 and H is isomorphic to either SL(2, 5) or ⟨v, d | d8 = v3 = 1, (d−2 v)5 = d4 , v d = vd2 (v, d−2 )⟩ .

Problems 11.4.1. Let N be a normal subgroup of a finite group G. Assume H is subgroup of G such that G = NH and N ∩ H = {1}. Prove that the following conditions are equivalent. (1) CenG (n) ⊆ N for all non-trivial n ∈ N. (2) CenH (n) = {1} for all non-trivial n ∈ N. (3) CenG (h) ⊆ H for all non-trivial h ∈ H. (4) Every x ∈ G \ N is conjugate to an element of H. (5) If 1 ≠ h ∈ H then h is conjugate to every element of Nh. (6) H is a Frobenius complement in G. 11.4.2. Let p be a prime integer and let G be a non-cyclic finite p-group not having a non-cyclic subgroup of order p2 . Let A = ⟨a⟩ be a maximal normal cyclic subgroup of G. Prove the following statements. (1) A = CenG (A) ≠ G. Hence, G is nonabelian. (2) G/A is abelian. Hence every subgroup of G containing A is normal in G. (3) p = 2 and if b ∈ G \ A then b = a−1 . (4) [G : A] = 2 and G is a quaternion group. (5) Quaternion groups do not have non-cyclic subgroups of order 4. 11.4.3. Prove that the direct product of two fixed point free finite groups of coprime order again is a fixed point free group. 11.4.4. Calculate the truncated rational group algebra of all the nilpotent Frobenius complements.

11.5 Group rings of nilpotent groups In this section we show that Theorem 11.3.2 can be extended for nilpotent finite groups G with no exceptional components of ℚG, by allowing nonabelian fixed point free epimorphic images H that are not embedded in a division algebra. Because of Corollary 11.4.7 such H is of the form Q2n × C q with n ≥ 3 and q odd. From Theorem 11.4.8 it is clear which of these groups can be embedded in division algebras. The crucial step consists in constructing a non-central idempotent in ℚGe for the primitive central idempotents e such that Ge is fixed point free. In a way we give a proof of Theorem 11.4.8 for the fixed point free nilpotent groups.

356 | 11 Generators of unit groups of group rings Let H = Q2n × C q with n ≥ 3 and q odd. If H can be embedded in a division algebra D then we may assume that D is generated by G as ℚ-algebra and hence D is a Wedderburn component of ℚH. Because of Theorem 3.3.6 and Example 3.5.7, one easily deduces that the only Wedderburn component in which H maps injectively is −1 n n−1 n−1 + ζ n−1 }), −1) = ℚH 1−z 2 ≅ A ⊗ℚ ℚ(ζ q ), with Z(Q 2 ) = ⟨z⟩ and A = (ℚ(ζ 2 /ℚ(ζ 2 2 −1 −1 ℍ(ℚ(ζ2n−1 + ζ2n−1 )). Therefore, D ≅ ℍ(ℚ(ζ2n−1 + ζ2n−1 , ζ q )). This shows that Q2n × C q is a subgroup of a division algebra if and only if ℍ(ℚ(ζ2n−1 + ζ2−1 n−1 , ζ q )) is a division algebra. The following theorem characterizes when this happens. Theorem 11.5.1. Let n and q be positive integers with q odd. The following conditions are equivalent. (1) ℍ(ℚ(ζ2n + ζ2−1 n , ζ q )) is not a division ring. (2) The equation X 2 + Y 2 = −1 has a solution in ℚ(ζ2n + ζ2−1 n , ζ q ). (3) q ≠ 1 and also n ≥ 3 or the multiplicative order o q (2) of 2 modulo q is even. Proof. The equivalence between (1) and (2) follows from Example 2.1.7. As ℚ(ζ2n−1 + ζ2−1 n−1 ) ⊆ ℝ, if q = 1 then (2) and (3) do not hold. So assume that q ≠ 1. Then F = ℚ(ζ2n−1 +ζ2−1 n−1 , ζ q ) is totally complex. By Proposition 4.5.10, −1 is a sum of two ̂P : ℚ ̂ 2 ] is even for every prime ideal of the ring of integers squares in F if and only if [F ̂ P /2ℤ ̂ P /2ℤ ̂P : ℚ ̂ 2 )f(F ̂ 2 ) = e(F/2ℤ)f(F/2ℤ) ̂ 2 ] = e(F of F with 2 ∈ P. By Theorem 4.4.2, [F √ and hence −1 is a sum of two squares in ℚ( d) if and only if either e(F/2ℤ) or f(F/2ℤ) is even. If n ≥ 3 then F contains √2, while ℚ2 does not contain √2. Then [F P : ℚ2 ] is even and hence −1 is a sum of two squares in F. So assume that n ≤ 2. Then F = ℚ(ζ q ). As q is odd, e(ℚ(ζ q )/2ℤ)f(ℚ(ζ q )/2ℤ) = f(ℚ(ζ q )/2ℤ) = o q (2), by Proposition 6.1.2 (1) and Problem 6.1.4. Therefore, −1 is a sum of two squares in F if and only if and only if o q (2) is even. This finishes the proof. As a consequence of Theorem 11.5.1 and its preceding comments we obtain the following. Corollary 11.5.2. Let n and q be positive integers with n ≥ 3 and q odd. Then Q2n × C q is a subgroup of a division ring if and only if q = 1 or both n = 3 and o2 (q) is odd. By Theorem 11.5.1, if q is an odd positive integer such that o q (2) is even then ℍ(ℚ(ζ q )) has a non-central idempotent. We now show how to obtain such an idempotent. Note that if x, y ∈ ℚ(ζ q ) with x2 + y2 = −1 then f =

1 (1 + xi + yj) 2

is a non-central idempotent in ℍ(ℚ(ζ q )). Hence it is enough to find explicit solutions in ℚ(ζ q ) of the equation X 2 + Y 2 = −1. Note that o q (2) is even if and only if o p (2) is even for at least one prime divisor p of q and this happens if and only if 2m ≡ −1 mod p for some positive integer m. Consequently, we only have to find solutions for the equation X 2 + Y 2 = −1 in ℚ(ζ p ), for this case. Giambruno and Sehgal have given such a solution in [67].

11.5 Group rings of nilpotent groups | 357

Lemma 11.5.3. Let p and m be positive integers such that p is prime and 2m ≡ −1 mod p. Let m−1

k

a = ∏ (1 + iζ p2 ) k=0

and let ψ denote the unique non-trivial element of Gal(ℚ(i, ζ p )/ℚ(ζ p )). Then the elements a − ψ(a) a + ψ(a) ζ p and y = ζp x= 2 2i satisfy x2 + y2 = −1. Hence

1 (1 + xi + yj) 2 is a non-central idempotent in ℍ(ℚ(ζ p )). f =

Proof. We have m−1

k

k

x2 + y2 = (x + yi)(x − yi) = ζ p2 aψ(a) = ζ p2 ∏ (1 + iζ p2 )(1 − iζ p2 ) k=0

m−1

=

ζ p2

∏ (1 +

k+1 ζ p2 )

=

ζ p2 (1

+

ζ p2 ) (1 + ζ p4 ) ⋅ ⋅ ⋅ (1

m

+ ζ p2 ).

k=0

Then, by an easy induction argument on m one obtains that (1 − ζ p2 )(x2 + y2 ) = ζ p2 (1 − ζ p2

m+1

).

Hence, since p | 2m+1 + 2, (1 − ζ p2 )(x2 + y2 ) = ζ p2 (1 − ζ p−2 ) = ζ p2 − 1. Since ζ p2 ≠ 1, this yields x2 + y2 = −1, as desired. If G is a group then 𝛾 ∈ ℂG can be written uniquely as α + βi with α, β ∈ ℝG. Put Re(𝛾) = α and Im(𝛾) = β. Clearly Re (𝛾) =

𝛾+𝛾 2

and

Im (𝛾) =

𝛾+𝛾 2i

,

where 𝛾 is obtained by taking complex conjugates on the coefficients of 𝛾. Assume now that G is a finite group and N is a normal subgroup such that G/N ≅ ⟨g, h⟩Q8 × ⟨c⟩q with g, h, c ∈ G, q a positive odd integer such that o q (2) is even. Then the only Wedderburn component of ℚG such that G projects with kernel N is ℚGe ≅ ℍ(ℚ(ζ q )), where e = ε(⟨N, g, c⟩ , ⟨N⟩). Fix a prime divisor p of q and m a positive integer such that 2m ≡ −1 mod p. Observe that o p (2) is an even divisor of 2m. Let (o q (2)/2)−1

x = c q/p Re (

∏ k=0

k

(1 + i(c q/p )2 ))

358 | 11 Generators of unit groups of group rings

and (o q (2)/2)−1

y = c q/p Im (

∏

k

(1 + i(c q/p )2 )).

k=0

Both are elements of the group ring ℤ ⟨c⟩. Then the isomorphism ℚGe ≅ ℍ(ℚ(ζ q )) maps xe and ye to the two elements of Lemma 11.5.3. Put 1 f p,N = ε(⟨N, g, c⟩ , ⟨N⟩) (1 + xg + yh). 2

(11.5.1)

Adding the generalized bicyclic units to the Bass units and bicyclic units we obtain the following result proved in [67]. Jespers and Leal in [114] had proved this result partially. Note that the formulation presented differs from the one give in these references. Theorem 11.5.4. Let G be a nilpotent finite group. Assume that o p (2) is even for every odd prime integer p such that G has an epimorphic image isomorphic to Q8 × C p . Let F be the set of idempotents f p,N , where N is a normal subgroup of G such that G/N ≅ Q8 × C q with q odd and p a prime divisor of q. If ℚG does not have exceptional simple components that are two-by-two matrices then ⟨Bass(G), Bic(G), GBicF (G)⟩ is of finite index in U(ℤG). Proof. By Corollary 11.2.1, it is enough to show that for every primitive central idempotent e of ℚG such that ℚGe is non-commutative, the group ⟨Bic(G), GBicF (G)⟩ contains a subgroup of finite index in 1 − e + O1 , for O an order in ℚGe. If Ge is not fixed point free then this follows as in the proof of Theorem 11.3.2. So assume that Ge is fixed point free and let N be the kernel of the map g 󳨃→ ge. Since G is nilpotent, G/N ≅ Ge ≅ Q2n × C q with q odd and n ≥ 3. By assumption M2 (ℚ) is not isomorphic to a simple component of ℚG. This together with (3.5.1) implies that n = 3. If q = 1 then ℚGe ≅ ℍ(ℚ) and the result is obvious. Otherwise, ℚGe ≅ ℍ(ℚ(ζ q )) and, by assumption, o p (2) is even for every prime divisor p of q. As f p,N is a non-central idempotent of ℚGe, this ℚGe is not a division algebra and it is then non-exceptional by assumption. Using Corollary 11.2.6, we obtain the desired conclusion because f p,N ∈ F.

12 Exceptional simple components In Chapter 11 we have shown that if G is a finite group such that G does not have nonabelian fixed point free images and ℚG does not have exceptional simple components then the Bass units together with the bicyclic units generate a subgroup of finite index in U(ℤG). In case G is a nilpotent finite group, it also has been shown that fixed point free images that are not contained in a division algebra can be admitted. So, a remaining problem is to deal with exceptional simple components. In the first two sections of this chapter we show that only restrictive types of exceptional simple matrix components can occur in a rational group algebra ℚG of a finite group G. These results are due to Eisele, Kiefer and Van Gelder [53]. One can also classify the finite groups that are faithfully embedded in such a component (for strongly monomial groups we give complete proofs). In principle all these results could also be deduced from the work of Banieqbal [14], on a classification of the finite subgroups of GL2 (D) with D an arbitrary finite dimensional rational division algebra, or from Nebe’s work [163] on the classification of subgroups G of GLn (D) such that M n (D) is spanned as a ℚ-algebra by G, with D a definite quaternion algebra such that its center has degree d over ℚ. A full classification of the latter is obtained provided nd ≤ 10 and one constructs a representative of the conjugacy classes of the maximal finite such groups. For our purposes, the proof presented is much simpler and more direct. In Section 12.3 we give an algorithm to determine generators for a subgroup of finite index in the unit group of an order of an exceptional simple component that is not a division algebra. These units together with the Bass and bicylic units generate a group that is commensurable with U(ℤG), provided ℚG does not have exceptional simple components that are non-commutative division algebras different from a totally definite quaternion algebra. In Section 12.6 we give several examples of metacyclic groups for which the Bass units together with the bicyclic units generate a subgroup of finite index even though some exceptional components appear in their respective rational group algebras. In Section 12.7 examples of finite groups are given for which the Bass units together with the bicyclic units do not generate a subgroup of finite index in the unit group of their respective integral group ring. Of course, for many finite groups G, the trivial units ±G of ℤG together with the Bass units do not generate a subgroup of finite index in U(ℤG). Nevertheless, in Section 12.4 a result of Kleinert [134] is proven which yields that the normal closure of this group often is of finite index in U(ℤG).

12.1 Components of index one In this section we show that, for a finite group G, the exceptional simple components of ℚG that are two-by-two matrices over a field are very restrictive. These results are due to Eisele, Kiefer and Van Gelder [53].

360 | 12 Exceptional simple components Proposition 12.1.1. Let d be either 0 or a square-free positive integer and let F = ℚ(√−d). The following properties hold for a finite subgroup G of GL2 (F). (1) The exponent of G divides 24 and if d ∈ ̸ {1, 2, 3} then the exponent of G divides 12. (2) G has no elements order 24 and if d ∈ ̸ {1, 3} then G has no elements of order 12. (3) If 8 divides the exponent of G then d = 1 or 2. (4) Every prime divisor of |G| is either 2 or 3. In particular, G is solvable. Proof. Let g ∈ G and let n = |g|. Then, g is conjugate in M2 (ℂ) to a diagonal matrix D such that the least common multiple of the order of the diagonal entries of D is n. This implies that Minℚ (g) = Φ n , the n-th cyclotomic polynomial. Let P = RCharM2 (F)/ℚ (g) and Q = MinM2 (F)/F (g). Then, by Lemma 2.3.5, Φ n divides P in ℚ[X] and Q divides Φ n in F[X]. The first implies that φ(n) = deg(Φ n ) ≤ deg(P) = 2[F : ℚ] and the second implies that F is a subfield of ℚ[ζ n ]. As 2[F : ℚ] ∈ {2, 4}, we have φ(n) ≤ 4 and hence n ∈ {1, 2, 3, 4, 5, 6, 8, 10, 12} and if d = 0 then φ(n) ≤ 2, so that n ∈ {1, 2, 3, 4, 6}. The latter shows that if d = 0 then the exponent of G divides 12 and G has no elements of order 12. We claim that 5 does not divide n. Indeed, if 5 divides n then n = 5 or 10 and F is a quadratic imaginary extension of ℚ contained in ℚ(ζ5 ). However it is well known that the only quadratic subfield of ℚ(ζ5 ) is ℚ(√5). Therefore, n ∈ {1, 2, 3, 4, 6, 8, 12}. If n = 8 then φ(n) = 4 and thus, from the above, F is an imaginary quadratic imaginary extension of ℚ contained in ℚ(ζ8 ). Then d = 1 or 2. If n = 12 then F is a quadratic imaginary extension of ℚ(ζ12 ) and hence d = 1 or 3. This proves (1), (2) and (3). Clearly, (4) is a direct conclusion of (1), Cauchy’s Theorem and then p a q b Burnside Theorem. If n is a positive integer and q is a prime power then 𝔽q denotes the field with q elements and we consider the following groups: GL(n, q) = GLn (𝔽q )

and

SL(n, q) = SLn (𝔽q ).

Corollary 12.1.2 (Eisele, Kiefer and Van Gelder). Let G be a finite group and let e be primitive central idempotent of ℚG such that ℚGe is an exceptional simple component of ℚG of index 1 (i.e. a simple component isomorphic to a 2-by-2 matrix over a field of the form ℚ(√−d) with d a non-negative integer) then Ge is isomorphic to one of the groups in the following table, with ℚGe isomorphic to the algebra indicated in the third column: ID(Ge)

Ge

ℚGe

[6, 1] [8, 3] [12, 4]

S3 D8 D12

M2 (ℚ)

[16, 6] [16, 13]

D+16 = ⟨a⟩8 ⋊ ⟨b⟩2 , a b = a5 D = (⟨a⟩2 × ⟨z⟩4 ) ⋊ ⟨b⟩2 z a = z b = z, a b = z2 a

M2 (ℚ(i))

12.1 Components of index one

ID(Ge)

Ge

ℚGe

[24, 1] [24, 5] [32, 11] [48, 33]

C3 ⋊ C8 S3 × C4 E = (⟨x⟩4 × ⟨y⟩4 ) ⋊ ⟨b⟩2 , x b = y SL(2, 3) ⋊ C2 = (⟨a, b⟩Q8 ⋊ ⟨x⟩3 ) ⋊ ⟨c⟩2 a x = b, b x = ab, c2 = 1, a c = a−1 , b c = b−1 , x c = xa SL(2, 3) : C4 = (⟨a, b⟩Q8 ⋊ ⟨x⟩3 ) : ⟨c⟩4 a x = b, b x = ab, c4 = a2 , a c = b−1 , b c = a−1 , x c = x −1

M2 (ℚ(i))

[16, 8] [48, 29]

D−16 = ⟨a⟩8 ⋊ ⟨b⟩2 , a b = a3 GL(2, 3) = (⟨a, b⟩Q8 ⋊ ⟨x⟩3 ) : ⟨c⟩2 , c2 = a, b c = ab, x c = b−1 x

M2 (ℚ(√−2))

[18, 3] [24, 3]

C3 × S3 SL(2, 3) = (⟨a, b⟩Q8 ⋊ ⟨x⟩3 ), a x = b, b x = ab C3 ⋊ D8 = (⟨x⟩3 ⋊ ⟨a⟩4 ) ⋊ ⟨b⟩2 , x a = x −1 , x b = x, a b = a−1 C3 × D8 C3 × Q8 C3 × Q12 C6 × S3 C3 × SL(2, 3) C3 × (C3 ⋊ D8 )

M2 (ℚ(√−3))

[96, 67]

[24, 8] [24, 10] [24, 11] [36, 6] [36, 12] [72, 25] [72, 30]

|

361

In particular, (1) If ℚGe ≅ M2 (ℚ) then |Ge| ∈ {8, 6, 12}, Exp(Ge) ∈ {4, 6, 12} and Ge has no elements of order 12. (2) If ℚGe ≅ M2 (ℚ(i)) then |Ge| ∈ {16, 32, 24, 96}, Exp(Ge) ∈ {4, 8, 12, 24} and Ge has no elements of order 24. (3) If ℚGe ≅ M2 (ℚ(√−2)) then |Ge| ∈ {16, 48}, Exp(Ge) ∈ {8, 24} and Ge has no elements of order 24. (4) If ℚGe ≅ M2 (ℚ(√−3)) then |Ge| ∈ {18, 24, 36, 72} and Exp(Ge) ∈ {6, 12}. Proof. By assumption, and the definition of exceptional simple algebra we have ℚGe ≅ M2 (F) for F = ℚ(√−d), with d either 0 or a square-free positive integer. Without loss of generality, we may assume that G ≅ Ge, i.e. the map g 󳨃→ ge is injective. This implies that G has a faithful irreducible complex representation of degree 2 with ℚ(χ) = F. We identify G with its image in M2 (F) and thus G is a finite subgroup of GL2 (F). By Proposition 12.1.1, d ∈ {0, 1, 2, 3}, the exponent of G divides 24 but G does not contain elements of order 24, and if d = 0 then the exponent of G divides 12. In particular, 2 and 3 are the only possible prime divisors of |G|. Moreover, as G generates M2 (F), if G has a central element of order n then ℚ(ζ n ) ⊆ F and hence n divides either 4 or 6.

362 | 12 Exceptional simple components Observe that if f is an idempotent of M2 (F) with f ∈ ̸ {0, 1} then there is an automorphism of M2 (F) mapping f to E11 (the matrix having 1 at the (1, 1) entry and 0 elsewhere). Furthermore, f ℚGf ≡ (1 − e)ℚGe(1 − f) and this is algebra is isomorphic to a subfield of F. Hence, the centralizer of f in M 2 (F) is isomorphic to F × F. Let G p be a Sylow subgroup of G, for p = 2 or 3 and let A p = ℚ(G p )e, i.e. the subalgebra of ℚGe generated by G p . We give two observations concerning the algebra A p . Observation 1: Assume A p is not simple. Then, A p has exactly two primitive central idempotents, say e1 and e2 . Clearly, e1 and e2 are primitive idempotents of M2 (F). By the previous, their respective centralizer in M2 (F) is isomorphic to F×F. Consequently, A p e1 and A p e2 are isomorphic to subfields of F. In particular, G p is abelian and hence Ae1 and Ae2 are cyclotomic fields. Furthermore, if A3 is not simple then necessarily F = ℚ(√−3) and G3 ≅ C23 . Also, if A2 is not simple then G2 is isomorphic to a subgroup of C4 × C4 and, if moreover F ⊆ ℚ(√−3) then G2 is isomorphic to a subgroup of C2 × C2 (as F does not contain elements of order 12). Observation 2: Assume A p is a division ring (possibly a field). Then, G p is fixed point free and hence it is either cyclic of order dividing 8 or 3, or isomorphic to Q2n with n ≥ 3 (see Proposition 11.4.6 and Corollary 11.4.7). If G2 = Q2n then it follows from the Wedderburn decomposition of ℚQ2n stated in (3.5.1) that the only division ring in which G2 is embedded is the totally definite quaternion algebra ℍ(ℚ(ζ2n−1 + ζ2−1 n−1 )). A dimension argument easily yields that this algebra can only be contained in M2 (F) if and only if n = 3 and d ≠ 0. Therefore, if A2 is a non-commutative division algebra then G2 = Q8 and d ≠ 0. In order to prove the result it is sufficient to deal with the following cases. (Case 1) Both A2 and A3 are not simple. (Case 2) A2 is a division algebra and A3 is not simple. (Case 3) both A2 and A3 are division algebras. (Case 4) A3 is a division algebra and A2 is not simple. (Case 5) A p is simple but not a division algebra for some p ∈ {2, 3} (in turns out that in this case p = 2). (Case 1): A2 and A3 are not simple. From Observation 1 we get that |G| = 36, G3 ≅ C23 and G2 is isomorphic to either C2 or C22 . Furthermore, ℚGe ≅ M2 (ℚ(√−3)). If G2 ≅ C2 then |G| = 18 and it easily is verified that then G ≅ S3 × C3 . If G2 ≅ C2 × C2 then a computer search using Wedderga [36] shows that then G ≅ C6 × S3 . (Case 2): A2 is a division algebra and A3 is not simple. From Observation 1 we know that then G3 ≅ C23 and d = 3. Hence, by Proposition 12.1.1, G does not have elements of order 8. From Observation 2 we then also get that G2 is isomorphic to either C2 , C4 or Q8 . It is easy to see that if G2 ≅ C2 then G ≅ C3 × S3 , and if G2 ≅ C4 then G ≅ C3 × Q12 . A computer search using GAP and Wedderga shows that there are only three groups G of order 72 with G3 ≅ C23 , G2 ≅ Q8 and having a simple component isomorphic to M2 (ℚ(√−3)). They are the groups C3 ×SL(2, 3), C3 ×(C3 ⋊Q8 ) and C23 ×Q8 . If G is one of the last two then it is strongly monomial and hence, by Corollary 3.5.4,

12.1 Components of index one |

363

e = e(G, H, K) for a strong Shoda pair (H, K) of G. Because G is embedded in ℚGe, the core of K in G is {1} (see Problem 3.5.1). Using Wedderga, one can see that C23 × Q8 does not have such a strong Shoda pair. This excludes this case and hence G is isomorphic to either C3 × SL(2, 3) or C3 × (C3 ⋊ Q8 ). The latter also can be excluded as for every faithful irreducible complex representation χ of G, we have ℚ(χ) = ℚ(ζ12 ). (Case 3): A2 and A3 are division rings. From Observation 2 we know that G3 ≅ C3 and G2 is either C2 , C4 , C8 or Q8 . Moreover, G2 is not isomorphic to C4 for otherwise G ≅ Q12 = ⟨a⟩3 ⋊ ⟨b⟩4 . Then ℚG ≅ ℚ2 × ℚ(i) × M2 (ℚ) × ( −1,−3 ℚ ) and the identity of the component isomorphic to M2 (ℚ) is e = ̂ b2 . Therefore b2 e = e, contradicting the fact that g → ge is injective. If G2 ≅ C2 then G ≅ S3 and d = 0. If G2 ≅ C8 then G ≅ C3 ⋊ C8 and d = 1. Finally, if G2 ≅ Q8 then a computer search shows that d = 3 and G is isomorphic to either C3 × Q8 or Q8 ⋊ C3 ≅ SL(2, 3). (Case 4): A3 is a division ring and A2 is not simple. Then, G3 ≅ C3 and G2 is isomorphic to either C2 , C22 , C2 × C4 or C24 . A computer search shows that there is only one group satisfying this condition with G3 not normal in G, namely (C4 × C4 ) ⋊ C3 . However, this groups does not have exceptional simple components. Thus, G3 is normal in G and hence G is supersolvable and G󸀠 = G3 . Because Deg(ℚGe) = 2 we thus obtain from Corollary 3.5.4, Theorem 3.5.5 and Problem 3.5.1 that e = e(G, H, K) for a strong Shoda pair (H, K) such that [G : H] = 2 and K ∩ K g = {1} if g ∈ G \ H. This implies that K ∩ G3 = {1}. Furthermore, because of Theorem 3.5.12, we may assume that H 󸀠 ⊆ K and H is a maximal subgroup such that H 󸀠 ⊆ K ⊆ H and H/K is cyclic. Hence, H 󸀠 = {1} and thus H is abelian. Because [G : H] = 2 we get that g2 ∈ Z(G) for every g ∈ G. Hence ⟨g2 : g ∈ G⟩ is cyclic and thus G2 can not contain C4 × C4 . Thus, we have shown that G2 is isomorphic to C2 , C2 × C2 or C2 × C4 = ⟨a⟩2 × ⟨b⟩4 . Moreover, in the latter case, a is not central in G. The first and second imply that G ≅ S3 or G ≅ D12 respectively. Suppose, G2 = ⟨a⟩2 × ⟨b⟩4 ≅ C2 × C4 . Write G3 = ⟨x⟩. Because a is not central and G3 is normal in G, we have ⟨a, x⟩ ≅ S3 . Also (b, x) = 1 for otherwise bxb = x2 and thus G ≅ D24 but the only simple component that faithfully contains G −1 is M2 (ℚ(ζ12 + ζ12 )), in contradiction with ℚGe = M2 (F). Hence, G ≅ S3 × C4 . (Case 5): A p is simple but it is not a division ring. Then, A p ≅ M2 (E) with E a subfield of F. Then, as in the proof of (Case 4), e = e(G p , H, K) for a strong Shoda pair (H, K) of G p with [G p : H] = 2. In particular, p = 2. Let n = [H : K] and b ∈ G \ H. We also know that the core of K in G is {1}, i.e. K ∩ K b = {1}. Suppose first that K = {1}. Because of Theorem 3.5.5, we get that H = ⟨a⟩n and M2 (E) is a cyclic algebra (ℚ(ζ n )/E, ζ nr ), with [ℚ(ζ n ) : E] = 2 and a b = a r . If E ≠ F then E = ℚ and hence n = 4. So, G2 ≅ D8 . Otherwise, E = F and n = 8. Moreover, if d = 1 then G2 ≅ D+16 , and if d = 2 then G2 ≅ D−16 . Second, suppose that K ≠ {1}. Then, again by Theorem 3.5.5, H = N G2 (K), F = ℚ(ζ n ) and K b is isomorphic to a subgroup of H/K. This implies that n = 2 or 4. If n = 2 then |K| = 2 and thus H ≅ C22 ; so G2 ≅ D8 . Suppose that n = 4. Then |K| = 4 or |K| = 2. Assume |K| = 4. Then, H = K × K b ≅ C24 = ⟨x⟩ × ⟨y⟩ and x b = y and y b = x. Moreover, b2 ∈ Z(G2 ) ∩ H ⊆ ⟨xy⟩. If b2 = (xy)i , then (bx−i )2 = 1. Hence, replacing b by bx−i , one

364 | 12 Exceptional simple components may assume that b2 = 1 and thus G2 = (⟨x⟩4 × ⟨y⟩4 ) ⋊ ⟨b⟩2 ≅ E, with x b = y. Finally, assume that n = 4 and |K| = 2. Since K is not normal in G2 but K is normal in H, we deduce that H = ⟨x⟩4 × ⟨y⟩2 and y b = x2 y. If ⟨x⟩ is not normal in G then x b = x±1 y 2 and hence x = x b = x−1 , a contradiction. Thus ⟨x⟩ is normal in G2 . Replacing x by xy if necessary, we may assume that x is central in G. Then b2 ∈ Z(G) = ⟨x⟩. If b2 ∈ ̸ ⟨x2 ⟩ then G ≅ D+16 . Otherwise, replacing b by yb if necessary, one may assume that b2 = 1 and hence G ≅ D. So, we have shown that if A p is simple but not a division ring then p = 2 and either d = 1 and G2 is isomorphic to one of the groups D8 , D+16 , D or E or d = 2 and G2 ≅ D−16 . This finishes the proof if G3 = 1. So, assuming that A2 is simple and not a division ring, because of Observation 1 and Observation 2, the remaining cases to deal with are that A3 is simple and not a division ring with G3 ≅ C23 , or A3 ≅ ℚ(√−3) and G3 ≅ C3 . Assume, the former, i.e. G3 ≅ C3 × C3 and A3 is not simple. Then, d = 3 and thus by the above, G2 ≅ D8 . In particular, |G| = 72. A computer search using GAP and Wedderga shows that there are six groups G of order 72 with Sylow subgroups isomorphic to D8 and C23 respectively and having a simple component isomorphic to M2 (ℚ(√−3)). These are the groups G with ID(G) = [72, i], for i ∈ {22, 23, 28, 30, 35, 37, 42}. All of these are strongly monomial. Therefore, e = e(G, H, K) for some strong Shoda pair (H, K) of G and the core of K in G is {1}. Again using Wedderga, one can verify that the last condition is only satisfied if i = 30. This group is isomorphic with C3 × (C3 ⋊ D8 ) = ⟨x⟩3 × (⟨y⟩3 ⋊ (⟨a⟩4 ⋊ ⟨b⟩2 )), y a = y−1 and y b = y. Finally, suppose that G3 ≅ C3 and A3 ≅ ℚ(√−3). Recall that G does not have elements of order 24 and also does not have central elements of order 12. Thus, if g ∈ G has order 8 then (g, G3 ) ≠ 1 and if, moreover, g2 ∈ Z(G) then g does not normalize G3 . In particular, if G is nilpotent then the exponent of G2 is 4 and the exponent of Z(G2 ) is 2, so that G = C3 × D8 . This finishes the proof if G is nilpotent. So, in the remainder of the proof we suppose that G is not nilpotent. Recall from the above that either d = 1 and G2 is isomorphic to one of the groups D8 , D+16 , D or E or d = 2 and G2 ≅ D−16 . Assume that G3 = ⟨g⟩3 is normal in G. Note that h2 acts trivial on G3 , for any h ∈ G. We claim that then G2 can not be isomorphic to D+16 , E, D−16 or D. To show that G2 ≅ E is excluded, assume that G2 = (⟨x⟩4 × ⟨y⟩4 ) ⋊ ⟨b⟩2 with x b = y. Because G3 = C3 , one has that x2 and y2 commute g. As xy−1 ∈ G󸀠2 one also has that xy−1 g = gxy−1 . Thus xy commutes with g and xy is central element of order 4 in G. So F = ℚ(i) and ℚ ⟨x2 , y2 , xy, g⟩ e is a commutative semisimple subalgebra of ℚGe = M2 (F) that properly contains F. As x2 e is a non-central element of Ge of order 2 this algebra contains the idempotent e x̂2 which is non-central in ℚGe. Consequently, ℚ ⟨x2 , y2 , xy, g⟩ e is embedded in F × F = ℚ(i) × ℚ(i). Since, ℚ(i) does not contain a primitive third root of unity, this yields a contradiction. Next we show that G2 ≅ D+16 is excluded as well. Assume G2 ≅ D+16 and thus F = ℚ(i). Write D+16 = ⟨a⟩8 ⋊ ⟨b⟩2 with a b = a5 . Since G does not contain elements of

12.1 Components of index one |

365

order 24 we get that g a = g 2 . Also g b equals g or g2 . If g b = g 2 then ab commutes with g and has order 8. Therefore abg has order 24, a contradiction. Thus g b = g. Then fg(e − f) = (e − f)gf = 0. Moreover, (fgf)3 = f and hence fgf is a third root of unity in F = ℚ(i). Thus fgf = f . Similarly (e − f)g(e − f) = e − f . Thus ge = e, a contradiction. Third we show that G 2 is not isomorphic to D−16 . We prove this by contradiction. So, suppose that G2 is isomorphic to D−16 . Then the element of order 8 in G2 does not centralize G3 , since G does not have elements of order 24. Hence, G ≅ ⟨g⟩3 ⋊ (⟨a⟩8 ⋊ ⟨b⟩2 ) with g a = g−1 and either g b = g or g b = g −1 . So, G is abelian-by-supersolvable and thus, by Theorem 3.5.10, G is strongly monomial. Consequently, by Corollary 3.5.4, e = e(G, H, K) for a strong Shoda pair (H, K) of G. Because Deg(ℚGe) = 2, we obtain from Theorem 3.5.5 that [G : H] = 2. In particular, H ⊃ G󸀠 = ⟨a2 , g⟩ and hence H is either ⟨a, g⟩ ≅ C3 ⋊C8 , ⟨a2 , b, g⟩ or ⟨a2 , ab, g⟩. Since H/K is cyclic, we get that H 󸀠 ⊆ K. As, furthermore, the core of K is trivial and ⟨g⟩ = ⟨(a, g)⟩ and ⟨a4 ⟩ = ⟨(a2 , b)⟩ = ⟨(a2 , ab)⟩ are normal in G, we obtain that g, a4 ∈ ̸ H 󸀠 . This yields a contradiction. Fourth, suppose that G2 ≅ D = ⟨z, a, b⟩, with z central in G2 of order 4 and 2 a = b2 = 1 and (a, b) = a2 . Because G does not contain central elements of order 12, it follows that (z, g) ≠ 1. Hence, G󸀠 = ⟨G3 , z2 ⟩ ≅ C6 . As previously explained in the proof, because G is metabelian, e = e(G, H, K) for a strong Shoda pair (H, K) of G, with [G : H] = 2 and K ∩ K g = {1}, if g ∈ G \ H. In particular, G󸀠 ⊆ H, G3 ⊈ K and z2 ∈ ̸ K. Thus, H 󸀠 ⊆ G󸀠 ∩ K = {1}. So H is abelian and |H| = 24. Then H ∩ G2 is abelian of order 8 and hence z ∈ H. This yields a contradiction, since g ∈ H and (g, z) ≠ 1, Next we show how to deal with G2 ≅ D8 , provided G3 is normal in G. Suppose that G2 ≅ D8 = ⟨x⟩4 ⋊ ⟨y⟩2 . If (G3 , x) = 1 then g y = g−1 because we are assuming that G is not nilpotent. Then G ≅ D24 = ⟨a⟩12 ⋊ ⟨y⟩2 and the only primitive central idempotent e of ℚG with G ≅ Ge is e = e(G, ⟨a⟩ , 1) (see Problem 3.4.5). How−1 ever, ℚGe ≅ M2 (ℚ(ζ12 + ζ12 )) = M2 (ℚ(√3)), contradicting with ℚGe = M2 (√−d). Thus, (g, x) ≠ 1 and replacing y by xy if necessary, G = ⟨g⟩3 ⋊ (⟨x⟩4 ⋊ ⟨y⟩2 ) with g x = g−1 and g y = g. So this finishes the proof in case G3 is normal. Finally, suppose that G3 is not normal in G. If G2 ≅ D8 then a computer search shows that the only group of order 24 with a non-normal Sylow 3-subgroup and a Sylow 2-subgroup isomorphic to D8 and having an exceptional component of index 1 is S4 . This exceptional component is M2 (ℚ) but S4 does not embed faithfully in GL2 (ℚ). Thus G2 is not isomorphic to D8 . Hence, either G has order 48 and G2 is isomorphic to either D+16 , D−16 or D, or G has order 96 and G2 ≅ E. Using GAP one can check that there are 10 groups of order 48 with a nonnormal Sylow 3-subgroup. None of them has a Sylow 2-subgroup isomorphic to D+16 , one has a Sylow 2-subgroup isomorphic to D−16 and another one has a subgroup isomorphic to D. The latter groups are GL(2, 3) and SL(2, 3)⋊C2 = (⟨a, b⟩Q8 ⋊⟨x⟩3 )⋊⟨c⟩2 , with a x = b, b x = ab, c2 = 1, a c = a−1 , b c = b−1 , x c = xa. The first one is faithfully embedded in GL2 (ℚ(√−2)) and the second one is faithfully embedded in GL2 (ℚ(i)). Finally suppose that G2 ≅ E. Then necessarily d = 1 and a computer search shows that there are only two groups G of order 96 with such a Sylow 2-subgroup and a non-

366 | 12 Exceptional simple components normal Sylow 3-subgroup. For these groups ID(G) is either [96, 64] or [96, 67]. The first one has only one exceptional component but the character associated to this component is not faithful. So G should be the second one. This group is SL(2, 3) : C4 = (⟨a, b⟩Q8 ⋊ ⟨x⟩3 ) : ⟨c⟩4 with a x = b, b x = ab, c4 = a2 , a c = b−1 , b c = a−1 and x c = x−1 . It also is shown in [53] that each of the groups showing up in the classification given in Corollary 12.1.2 can be embedded in GL2 (𝔽25 ), were 𝔽25 denotes the field with 25 elements. The proof requires more techniques. We state the result as a curiosity. Proposition 12.1.3. Let K ∈ {ℚ, ℚ(√−1), ℚ(√−2), ℚ(√−3)}. A finite subgroup G of GL2 (K) is embeddable in GL2 (𝔽25 ).

Problems 12.1.1. For every group G listed in Corollary 12.1.2 give a specific isomorphic copy that is a subgroup of GL2 (ℚ(√−d)) and that generates M2 (ℚ(√−d)) as a ℚ-algebra; where d = 0, 1, 2 or 3.

12.2 Components of index two In this section we determine the non-commutative division algebras D that are such that M2 (D) is a simple component of ℚG for some finite group G. As mentioned in the introduction of this chapter, in principle, this can be derived from Banieqbal’s work [14], on the description of finite subgroups of GL2 (D), for D a finite dimensional rational division algebra, and also from Nebe’s work in [163]. These classifications have incredibly tedious and long proofs, and moreover rely partially on the classification of finite simple groups. Again, we present a proof that relies on comparatively elementary techniques and that is self-contained. The classification obtained is due to Eisele, Kiefer and Van Gelder [53]. For nilpotent finite groups such results were obtained earlier by Jespers and Leal in [113]. In this context we mention that Caicedo and del Río, in [39], classified all finite groups G such that ℚG has an exceptional component but ℚ(G/N) does not have exceptional components for any non-trivial normal subgroup N of G. This classification again mainly is based on the work of Amitsur [3] and Banieqbal [14] and it allows to tackle the congruence subgroup problem for integral group rings. Bächle, Caicedo and Van Gelder, in [8], considered this question in the larger context of group algebras 𝔽G over an arbitrary number field. The following abbreviated notation for some quaternion algebras will be used. ℍ1 = (

−1, −1 ), ℚ

ℍ3 = (

−1, −3 ) ℚ

and ℍ5 = (

−2, −5 ). ℚ

(12.2.1)

12.2 Components of index two

|

367

Theorem 12.2.1. Let G be a finite group and let e be a primitive central idempotent of the rational group algebra ℚG. If ℚGe = M2 (D), with D = ( a,b ℚ ) a totally definite quaternion algebra (i.e. a < 0 and b < 0), then D is one of the algebras ℍ1 , ℍ3 or ℍ5 . Proof. Without loss of generality, we may assume that the natural mapping G 󳨀→ Ge ⊆ ℚGe = M2 (D) is injective. We identify G with its image in GL2 (D). Hence, we consider G as subgroup of M2 (D). Moreover, M2 (D) is generated by G. Clearly the center of G has order at most 2. Since the reduced characteristic polynomial over ℚ of g ∈ M2 (D) has degree 4, the minimal polynomial over ℚ of any g ∈ G has degree at most 4. Hence, as in the first part of the proof of Proposition 12.1.1, it follows φ(|g|) ≤ 4 and hence |g| ∈ {1, 2, 3, 4, 5, 6, 8, 10, 12}. Consequently, 2, 3 and 5 are the only possible primes that can divide |G|. We consider two separate cases, depending on whether the primitive central idempotent e is determined by a strong Shoda pair or not. In the first case we are also going to describe the group G and this information will be collected in Corollary 12.2.2. Case 1: Suppose e = e(G, H, K) for a strong Shoda pair (H, K) of G. From Theorem 3.5.12 we know that [G : H] = Deg(M2 (D)) = 4 and the core of K in G is {1}. Let N = N G (K) and put n = |H|. By Theorem 3.5.5 (3) and because the center of D is ℚ, we have H  N and N/H ≅ Gal(ℚ(ζ n )/ℚ). Moreover, N ≠ H, for otherwise ℚGe ≅ M4 (ℚ). Thus, either K = {1} (and thus H is cyclic and H is normal in G) or [G : N] = 2 and K ∩ K g = {1} for every g ∈ G \ N. Case 1.a: Assume K = {1}. Let n = |H| and let h be a generator of H. Then G = N, CenG (H) = H, G/H ≅ Gal(ℚ(ζ n )/ℚ) and hence φ(n) = 4. Because M2 (D) is 16dimensional over ℚ, we also know that |G| ≥ 18. Hence n ∈ ̸ {1, 2, 3, 4, 6}. Therefore, n ∈ {5, 8, 10, 12}. If n = 5 or 10 then G = ⟨a⟩5 ⋊ G2 , where G2 is the Sylow 2-subgroup of G and N/H is cyclic of order 4. Moreover, G2 has a central subgroup M of order 2 with G2 /M ≅ C4 . This implies that G2 is abelian. If n = 5 then H = ⟨a⟩, G2 = ⟨b⟩4 and a b = a2 . In this case ℚGe = ℚGe(G, H, 1) ≅ M4 (ℚ). So, n = 5 is excluded. Suppose that n = 10. If G2 is not cyclic then G ≅ C2 × H with H = ⟨a⟩5 ⋊ ⟨b⟩4 with a b = a2 . From the decomposition of ℚG given in (1.5.6) and (1.5.7) we then obtain that ℚG ≅ ℚH × ℚH. By Theorem 3.5.5 (3), the only non-commutative simple component of ℚH is ℚHe(H, ⟨a⟩ , 1) ≅ (ℚ(ζ5 )/ℚ, −1) ≅ M4 (ℚ). We conclude that if G2 is non-cyclic then ℚG does not have exceptional simple components. Thus G2 is cyclic and G = ⟨a⟩5 ⋊ ⟨b⟩8 with a b = a2 . Then H = ⟨ab4 ⟩. So, again by Theorem 3.5.5 (3), ℚGe ≅ (ℚ(ζ5 )/ℚ, −1) and thus, by Example 2.6.8, ℚGe ≅ M2 (ℍ5 ). Next, assume that n = 8 and thus |G| = 32. A computer search, using GAP and Wedderga, shows that there are only two groups of order 32 with a normal cyclic subgroup of order 8 which is its own centralizer. It turns out that only one of them has

368 | 12 Exceptional simple components an exceptional component isomorphic to M2 (D), with D a quaternion algebra over ℚ. This group is given by the following presentation ⟨a, b, c | a8 = c2 = 1, b2 = a4 , a b = a−1 , a c = a5 , b c = b⟩. Let H1 = ⟨a2 , c⟩ and K1 = ⟨c⟩. Then (H1 , K1 ) is a strong Shoda pair of G with N G (K1 ) = ⟨a2 , b, c⟩. Hence, K1 is a non-normal subgroup of G of order 2 while H is a cyclic normal subgroup of G and K = 1. This implies that 1 = H ∩ K1 = K ∩ H1 and hence e = e(G, H, K) = e(G, H1 , K1 ) by Problem 3.4.3. By Theorem 3.5.5 (3) we have ℚGe = ℚGe(G, H1 , K1 ) ≅ M2 (ℚ(i)/ℚ, −1) ≅ M2 (ℍ1 ). Finally assume that n = 12 and thus |G| = 48. Write H = ⟨a⟩3 ×⟨b⟩4 and let G2 be a Sylow subgroup of G containing b. Then G = ⟨a⟩3 ⋊ G2 and G2 / ⟨b⟩ ≅ G/H. Moreover, H is maximal abelian in G, hence the action of G on H by conjugation is faithful. As [G : H] = Aut(H) = 4 this action induces an isomorphism G/H ≅ Aut(H) ≅ C22 . By Theorem 3.5.5 (3), ℚGe is isomorphic to a crossed product ℚ(ζ12 ) ∗ Gal(ℚ(ζ12 /ℚ)) = ℚ(ζ12 ) ∗ (⟨X⟩2 ⋊ ⟨Y⟩2 ). The structure of this crossed product depends on the structure of G. We have to show that this algebra is isomorphic to M2 (ℍi ) with i ∈ {1, 3, 5}. We do so by considering separately two cases depending on whether G has an element of order 8 or not. Suppose first that G has an element x of order 8. Then, one may assume that b = x2 . Furthermore, because G does not contain elements of order 24, we also get that a x = a−1 . Moreover, since |G2 | = 16, it follows that G2 is isomorphic to one of the following groups: D16 , D+16 , D−16 or Q16 . Also, in the crossed product structure of ℚGe, one may make the following identifications: ζ12 = ax2 e, X = xe and Y = ye for some 5 3 X = ζ12 and X 2 = ζ12 = i. Note that G does not contain y ∈ G 2 \ ⟨x⟩. Therefore ζ12 central elements of order 4 because GL2 (D) does not contain such an element. Since D+16 contains a central element of order 4, say z, that is a square of an element, we obtain that G2 ≇ D+16 . We consider the remaining three cases separately and make use of Problem 12.2.1. Suppose that G2 ≅ D16 . Then G2 = ⟨x⟩8 ⋊ ⟨y⟩2 and we have YX = X −1 Y = −iXY and Y 2 = 1. Moreover, without loss of generality, we also may 7 assume that a y = ζ12 . Then ℚGe ≅ M2 ( −3,2 ℚ ), contradicting the hypothesis. Suppose 7 − Y G2 ≅ D16 = ⟨x⟩8 ⋊ ⟨y⟩2 . Then YX = X 3 Y = iXY. If a y = a then ζ12 = ζ12 and ℚGe ≅ M2 (ℍ1 ). In this case G = ⟨a⟩3 ⋊ (⟨x⟩8 ⋊ ⟨y⟩2 ), −1 Y = = = Otherwise ζ12 = ζ12 and M2 ( 3,−2 with ℚ ), again a contradiction. Finally, suppose that G ≅ Q16 . One may assume without loss of generality that 7 Y a y = a−1 and then we have YX = −iXY, Y 2 = −1 and ζ12 = ζ12 . Hence ℚGe ≅ M2 (ℍ3 ) and G = ⟨a⟩3 ⋊ ⟨x, y⟩Q16 ,

ax

a−1 , a y

a, x y

x3 .

with a x = a−1 and a y = a−1 . Now assume that G2 does not contain elements of order 8. As G2 / ⟨b⟩ ≅ C2 × C2 and ⟨b⟩ is non-central in G2 , G2 is isomorphic to either D8 × C2 = (⟨b⟩ ⋊ ⟨x⟩) × ⟨y⟩,

12.2 Components of index two

|

369

Q8 × C2 = (⟨b⟩ : ⟨x⟩)×⟨y⟩ or D8 ⋊ C2 = (⟨b⟩4 ⋊⟨x⟩2 )⋊⟨y⟩2 with (b, y) = 1 and x y = b2 x. As H = ⟨ab⟩ is maximal abelian in G, a y = a−1 . Moreover, in the first two cases one may assume without loss of generality that (a, x) = 1. So in the first case G ≅ D8 × D6 and in the second case G ≅ Q8 × D6 . In the first case ℚGe ≅ M4 (ℚ), in contradiction with the hypothesis, and in the second case ℚGe ≅ M2 (ℍ1 ). Finally suppose that G2 is as in the third case. Then ℚGe is a crossed product (ℚ(ζ12 )/ℚ, τ) = ℚ(ζ12 )∗(⟨X⟩×⟨Y⟩) 5 Y = ζ12 , X 2 = Y 2 = e and YX = −XY. If a x = a−1 where ζ 12 = abe, X = xe and Y = ye, ζ12 −1,3 −1 X then ζ12 = ζ12 . Hence, ℚGe ≅ M2 ( ℚ ), by Problem 12.2.1, in contradiction with the 7 X = ζ12 and, applying Problem 12.2.1 once more hypothesis. Thus (a, x) = 1. Then ζ12 we have ℚGe ≅ M2 (ℍ3 ). Thus in the last case G is isomorphic to ⟨a⟩3 ⋊ ((⟨b⟩4 ⋊ ⟨x⟩2 ) ⋊ ⟨y⟩2 ) with (a, b) = (a, x) = (b, y) = 1, a y = a−1 and x y = b2 x. Case 1.b: Assume K ≠ {1}. Let n = [H : K]. Then, again by Theorem 3.5.5, ℚGe = M2 (D) with D ≅ ℚ(N/K)e(N/K, H/K, 1)) = (ℚ(ζ n )/ℚ, τ), a totally definite quaternion algebra over ℚ. Then φ(n) = 2 and hence n = 3, 4 or 6. In particular N/K is fixed point free of order 6, 8 or 12, with a cyclic subgroup of index 2. Clearly N/K ≇ S3 nor D12 . Thus n ≠ 3. If n = 4 then N/K ≅ Q8 . If g ∈ G \ N then K and K g are normal subgroups of N with K ∩ K g = {1}. Therefore N contains K × K g and K ≅ K g ≅ (K × K g )/K ≤ N/K ≅ Q8 . Then D is a quaternion algebra in the Wedderburn decomposition of ℚQ8 . Therefore D ≅ ℍ1 . To determine the group G we first suppose that |K| = 8. Then K = ⟨x, y⟩ ≅ Q8 , N = K × K g ≅ Q8 × Q8 . Moreover, as g2 ∈ N and commutes with g, we have g2 = xx g for x ∈ K. Replacing x by gx−1 we may assume that g2 = 1. Thus G is isomorphic to (⟨x1 , y1 ⟩Q8 × ⟨x2 , y2 ⟩Q8 ) ⋊ ⟨g⟩2 , g

g

with x1 = x2 , y1 = y2 . Otherwise K is a cyclic group of order 2 or 4. Using GAP and Wedderga, one discovers the finite groups G with a cyclic subgroup K of cardinality 2 or 4 such that N G (K)/K ≅ Q8 , [G : N G (K)] = 2 and K ∩ K g = 1, for all g ∈ G \ N G (K). Then ID(G) is either [32, 8], [32, 44], [32, 50], [64, 37] or [64, 137]. If n = 6 then N/K = C3 ⋊ C4 and D ≅ ℍ3 . As in the previous case K is isomorphic to a subgroup of N/K and if K is isomorphic to N/K then G is isomorphic to ((⟨x1 ⟩3 ⋊ ⟨y1 ⟩4 ) × (⟨x2 ⟩3 ⋊ ⟨y2 ⟩4 )) ⋊ ⟨g⟩2 , g

g

with x1 = x2 , y1 = y2 . Otherwise K is a cyclic group of order 2, 3 or 6. Again there are several finite groups G with a cyclic subgroup K of order 2, 3 or 6 such that N G (K)/K ≅ C3 ⋊ C4 , [G : N G (K)] = 2 and K ∩ K g = {1}, for all g ∈ G \ N G (K). Thus ID(G) is either [48, 39], [72, 19], [72, 20], [72, 22], [72, 24], [144, 135] or [144, 148]. Case 2: e is not of the form e(G, H, K) for a strong Shoda pair (H, K) of G. Therefore, G is not abelian-by-supersolvable and in particular it is not nilpotent. Without loss of generality, we may assume that G satisfies the following minimal property: for every

370 | 12 Exceptional simple components proper subgroup H of Ge the ℚ-subalgebra of ℚGe generated by H is properly contained in M2 (D). We argue by contradiction. So we assume that D is not isomorphic to neither ℍ1 , ℍ3 nor ℍ5 . We now discuss the structure of the Sylow subgroups of G. Let G p denote a Sylow subgroup of G. We claim that the following properties hold: (R1) either |G2 | ≤ 8 or G2 is isomorphic to one of the groups C24 , D−16 , D+16 , Q16 , D or E; (R2) G3 is either trivial or isomorphic to either C3 or C23 ; (R3) |G5 | ≤ 5. In order to prove the claim, we denote by A p the ℚ-subalgebra of M2 (D) generated by G p and, as in the proof of Corollary 12.1.2, we consider separately the possible structures of A p . Note that, by the minimality assumption, A p is a proper subalgebra of ℚGe. If A p is a division algebra then G p is fixed point free. By Theorem 11.4.7, G p is either cyclic or quaternion. By the constraints on the order of the elements of G, either |G2 | ≤ 8 or G2 is isomorphic to Q16 , |G3 | ≤ 3 and |G5 | ≤ 5. Suppose that A p is not simple. Then A p is the direct product of two division rings isomorphic to subrings of D. If one of them is isomorphic to D then D is a totally definite quaternion algebra over ℚ generated by a p-group Q. Hence, by Theorem 11.4.5, Q is nonabelian and a fixed point free p-group. Therefore, Q ≅ Q8 and D ≅ ℍ1 , a contradiction. Thus A p is a direct product of two fields, say K1 and K2 , and thus G p is abelian and in fact it is the direct product of at most two cyclic groups. Moreover, K1 and K2 are cyclotomic fields of degree at most 2 over ℚ. This implies that p = 2 or 3 and the exponent of G p is either 2, 3 or 4. Thus G p is a subgroup of either C4 × C4 or C3 × C3 . Finally, suppose that A p is simple but not a division ring. Then A p is a two-by-two matrix over a subfield of D. As D is a totally definite quaternion algebra over ℚ, every subfield of D is either ℚ or an imaginary quadratic extension of ℚ. Therefore, G p is isomorphic to one of the nonabelian p-groups listed in Corollary 12.1.2, i.e. p = 2 and G is isomorphic to D8 , D+16 or D−16 , D, of E. This finishes the proof of the claim. Again using the library of small groups of GAP one can calculate the list of all the finite groups G whose Sylow subgroups satisfy the conditions (R1), (R2) and (R3) and that do not have elements of orders that are not allowed. Furthermore, the Wedderburn the decomposition of ℚG should have a simple component A isomorphic to M2 (D), with D a totally definite quaternion algebra over ℚ. This results in the following groups: [48, 28], [120, 5], [144, 124], [144, 125], [144, 127], [240, 89], [240, 90], [720, 409]. However, if ID(G) is either [48, 28], [120, 5] or [720, 409] then every simple component with the given structure has D ≅ ℍ3 ; if ID(G) is either [240, 89] or [240, 90] then the simple component of that form are isomorphic to (ℚ(ζ5 )/ℚ, −1) ≅ M2 (ℍ5 ), by Example 2.6.8. The simple components with the given structure for the remaining groups are either M2 (ℍ3 ) (which is a simple component of ℚG if ID(G) = 5 7 X Y = ζ12 , ζ12 = ζ12 , [144, 124]), or a crossed product ℚ(ζ12 ) ∗ (⟨X⟩2 × ⟨Y⟩2 ) with ζ12 2 Y = −1 such that one of the following conditions hold:

12.2 Components of index two

|

371

3 = i and YX = −iXY, if ID(G) = [144, 124]. (1) X 2 = ζ12 2 (2) X = i and XY = iXY, if ID(G) = [144, 125]. (3) X 2 = −1 and XY = XY, if ID(G) = [144, 127]. By Problem 12.2.1, these crossed products are isomorphic to M2 (ℍ3 ), in the first case, and isomorphic to M2 ( 3,−2 ℚ ) in the second and third case. In all the cases we obtain a contradiction. This finishes the proof.

Corollary 12.2.2. Let G be finite group and D a totally definite quaternion algebra over ℚ. Then G is strongly monomial and isomorphic to a subgroup of GL2 (D) generating M2 (D) over ℚ if and only if G is isomorphic to one of the groups in the Table 12.1 and D is the division algebra appearing in the third column. Proof. All the groups G listed are strongly monomial and have a strong Shoda pair (H, K) such that the core of K in G is 1 and ℚGe(G, H, K) ≅ M2 (D) for D the algebra Tab. 12.1: Strongly monomial groups generating exceptional components of the form M2 (D) with D a totally definite quaternion algebra over ℚ. ID(G)

G

[32, 8]

(⟨a⟩8 ⋊ ⟨b⟩2 )D+ : ⟨c⟩ ,

[32, 44] [32, 50] [48, 16] [48, 40] [64, 37]

(⟨a⟩8 : ⟨b⟩2 )Q16 ⋊ ⟨c⟩2 , a c = a5 , (b, c) = 1 (⟨a⟩2 × ⟨x, y⟩Q8 ) ⋊ ⟨b⟩2 , a b = ax 2 , (b, x) = (b, y) = 1 ⟨a⟩3 ⋊ (⟨x⟩8 ⋊ ⟨y⟩2 ), a x = a−1 , a y = a, x y = x 3 Q8 × D6 (⟨a⟩4 ⋊ ⟨x, y⟩Q8 ) : ⟨b⟩2 , (a, x) = 1, a y = a−1 , b2 = ay −1 , a b = ax −1 , x b = a2 x −1 , y b = xy ((⟨x⟩4 × ⟨y⟩4 ) ⋊ ⟨b⟩2 ) ⋊ ⟨a⟩2 , x b = y, x a = y −1 , b a = x 2 y 2 b g g (⟨x1 , y1 ⟩Q8 × ⟨x2 , y2 ⟩Q8 ) ⋊ ⟨g⟩2 , x1 = x2 , y1 = y2

[64, 137] [128, 937] [48, 18] [48, 39] [72, 19] [72, 20] [72, 22] [72, 24] [144, 135] [144, 148] [288, 389] [40, 3]

D 16

c2

=

a4 ,

ac

= ab,

bc

=b

⟨a⟩3 ⋊ ⟨x, y⟩Q16 , a x = a−1 , a y = a−1 C3 ⋊ (D8 ⋊ C2 ) = ⟨a⟩3 ⋊ ((⟨b⟩4 ⋊ ⟨x⟩2 ) × ⟨y⟩2 ) (a, b) = (a, x) = 1, a y = a−1 , b y = b, x y = b2 x (⟨x⟩3 × ⟨y⟩3 ) ⋊ ⟨a⟩8 , x a = y, y a = x −1 D6 × Q12 ((⟨a⟩3 ⋊ ⟨b⟩2 ) × ⟨c⟩6 ) ⋊ ⟨x⟩2 , a x = a, b x = bc3 , c x = c−1 (⟨x⟩3 × ⟨y⟩3 ) ⋊ ⟨a, b⟩Q8 , (x, a) = (y, b) = 1, x b = x −1 , y a = y −1 (⟨a⟩3 × ⟨b⟩) ⋊ (⟨x⟩8 ⋊ ⟨y⟩)D+ , a x = b, b x = a−1 ,

ℍ1

ℍ3

16

(a, y) = (b, y) = 1 (((⟨a⟩3 × ⟨b⟩3 ) × ⟨x⟩2 ) ⋊ ⟨y⟩4 ) ⋊ ⟨z⟩2 , a y = a−1 , b y = b−1 , (y, z) = 1, x z = xy 2 , (a, z) = 1, b z = b−1 ((⟨x1 ⟩3 ⋊ ⟨y1 ⟩4 ) × (⟨x2 ⟩3 ⋊ ⟨y2 ⟩4 )) ⋊ ⟨g⟩2 g g x1 = x2 , y1 = y2 ⟨a⟩5 ⋊ ⟨b⟩8 , a b = a2

ℍ5

372 | 12 Exceptional simple components

given in the table. (This can be verified using Wedderga [36]). Therefore such G can be embedded in M2 (D) and its image generates M2 (D). Conversely, the assumptions implies that M2 (D) is a simple component of ℚG. If moreover G is strongly monomial then M2 (D) ≅ ℚGe(G, H, K) for (H, K) a strong Shoda pair of G such that the core of K in G is trivial. Then the map g 󳨃→ g e(G, H, K) is injective and the proof in case 1 of Theorem 12.2.1 shows that then Ge and D are as given in the table. As mentioned in the introduction of this section, the classification of the finite subgroups G generating exceptional components of the type M2 (D), with D a division algebra, can be obtained using either [14] or [163]. The table of Corollary 12.1.2 contains those groups for which D is a field and the table in Corollary 12.2.2 contains the strongly monomial groups for which D is not a field. The remaining groups, i.e. the non strongly monomial groups generating an exceptional component of the form M2 (D), for D a totally definite quaternion algebra over ℚ, have been obtained in [53] and it is formed by the groups in Table 12.2. Tab. 12.2: Non-strongly monomial groups generating exceptional components of the form M2 (D) with D a totally definite quaternion algebra over ℚ. ID(G)

D

[96, 190], [96, 191], [96, 202], [144, 128], [160, 199], [192, 989], [320, 1581], [384, 618], [384, 18130], [1152, 155468], [1920, 241003]

ℍ1

[48, 28], [120, 5], [144, 124], [720, 409]

ℍ3

[240, 89], [240, 90]

ℍ5

Problems 12.2.1. Let ξ = ζ12 and consider a crossed product A = (ℚ(ξ)/ℚ, τ) = ℚ(ξ) ∗ (⟨X⟩2 × ⟨Y⟩2 ) with ξ X = ξ 5 . Prove the following statements: (1) If X 2 = i, YX = −iXY and Y 2 = 1 then A ≅ M2 ( −3,2 ℚ ). (Hint: take e = 1+Y and show that eAe = ℚ(ξ ) ⊕ ℚ(ξ3 )(1 − i)X ≅ ( −3,2 3 2 ℚ ).) (2) If X 2 = i, YX = iXY, ξ Y = ξ 7 and Y 2 = 1 then A ≅ M2 (ℍ1 ). (3) If X 2 = i, YX = iXY, ξ Y = ξ −1 and Y 2 = 1 then A ≅ M2 ( 3,−2 ℚ ). (4) If X 2 = i, YX = iXY, ξ Y = ξ 7 and Y 2 = −1 then A ≅ M2 ( 3,−2 ℚ ). (Hint: interchange the roles of Y and XY.) (5) If X 2 = i, YX = −iXY, ξ Y = ξ 7 and Y 2 = −1 then A ≅ M2 (ℍ3 ). (Hint: ((ξ 4 + ξ 3 )x)2 = 1.) (6) If ξ Y = ξ 7 , Y 2 = 1, X 2 = 1 and YX = XY then A ≅ M4 (ℚ). (7) If ξ Y = ξ 7 , Y 2 = 1, X 2 = −1 and YX = XY then A ≅ M2 (ℍ3 ). (8) If ξ Y = ξ −1 , Y 2 = 1, X 2 = −1 and YX = XY then A ≅ M2 ( 3,−2 ℚ ).

12.3 Generalized bicyclic units and Bass units II |

373

(9) If ξ Y = ξ 7 , Y 2 = −1, X 2 = −1 and YX = XY then A ≅ M2 ( 3,−2 ℚ ). (10) If ξ Y = ξ 7 , Y 2 = 1, X 2 = 1 and YX = −XY then A ≅ M2 (ℍ3 ). (11) If ξ Y = ξ −1 , Y 2 = 1, X 2 = 1 and YX = −XY then A ≅ M2 ( −1,3 ℚ ). 12.2.2. Decide which of the groups in Corollary 12.1.2 are strongly monomial. 12.2.3. Calculate the Wedderburn decomposition of ℚG for the finite groups G in Table 12.2 and proof that all of them have a simple component of the form M2 (D) with D the algebra given in the table. Observe that the groups G with ID(G) one of the following [48, 28], [120, 5], [144, 124], [240, 89], [240, 90], [720, 409], appear in the second case of the proof of Theorem 12.2.1. Explain why the other groups in Table 12.2 do not appear in that proof.

12.3 Generalized bicyclic units and Bass units II The results in Section 12.2 show that if M2 (D) is an exceptional simple component of a rational group algebra ℚG of a finite group G, then D is either a field ℚ(√−d), with d ∈ {0, 1, 2, 3}, or a quaternion algebra ( a,b ℚ ), with (a, b) ∈ {(−1, −1), (−1, −3), (−2, −5)}. It is well known that in all four cases with D commutative, the ring of integers is a Euclidean domain (see for example [225] or Problem 12.3.1) and in fact the norm is an Euclidean map, i.e. O is norm Euclidean in the following sense. Let D be a finite dimensional rational division algebra and define N(x) = |RNrA/ℚ (x)| for each x ∈ A. One says that an order O in A is left norm Euclidean order if for every x, y ∈ O, with y ≠ 0, there exist q, r ∈ O such that x = qy + r with N(r) < N(y). Obviously, such an order is a left principal ideal domain and thus it is a maximal order in D (Problem 12.3.3). If the order is left and right norm Euclidean then one simply says it is norm Euclidean. Once can show that each of the three mentioned quaternion algebras contains a norm Euclidean order. More precisely one has the following result. Proposition 12.3.1. Let D be either ℍ1 , ℍ3 or ℍ5 and let ℤ [1, i, j, 1+i+j+k ], { 2 { { 1+j j+k O = {ℤ [1, i, 2 , 2 ] , { { 1+i+j 2+i−k 2+3i+k {ℤ [1, 2 , 4 , 4 ] = ℤ [1, i,

if D = ℍ1 ; if D = ℍ3 ; 1+i+j 2+i−k 2 , 4 ],

if D = ℍ5 .

Then O is norm Euclidean. In [63], Fitzgerald showed that the algebras listed in Proposition 12.3.1 are the only possible totally definite quaternion algebras over ℚ having a left norm Euclidean order O and that O is the unique maximal order. Corollary 12.3.2. Let G be a finite group and let be e a primitive central idempotent of ℚGe. If ℚGe = M2 (D) is an exceptional simple component of ℚGe, with D a division

374 | 12 Exceptional simple components algebra, then D has a unique maximal order, say O. So, O is the ring of integers if D is commutative (see Example 4.1.9) and otherwise (1) O = ℤ[1, i, j, 12 (1 + i + j + ij)] if D = ( −1,−1 ℚ ). −1,−3 1 1 (2) O = ℤ[1, i, 2 (1 + j), 2 (i + ij)] if D = ( ℚ ). (3) O = ℤ[1, 14 (2 + i − ij), 14 (2 + 3i + ij), 12 (1 + i + j)] if D = ( −2,−5 ℚ ). In particular, in each case, the order O is Euclidean. In order to find a finite set of generators for a subgroup of finite index in an order of an exceptional component we will make use of the following lemma. Lemma 12.3.3. Let O be a norm Euclidean order in a quadratic imaginary extension of ℚ or in a totally definite quaternion algebra over ℚ. If B is a ℤ-basis for O then the set 1 X = {( 0

x 1 ),( 1 x

0 ) : x ∈ B} 1

generates a subgroup of finite index in SL2 (O). Proof. Clearly ⟨X⟩ contains E2 (O). Because of Dirichlet’s Unit Theorem (Theorem 5.2.4) and Kleinert’s result (Proposition 5.5.6), we know that U(O) is finite. Hence, the set S formed by the matrices in SL2 (O) of the types ( 0a 0d ) and ( 0c 0b ) is finite. Consequently, it is sufficient to show that if A ∈ SL2 (O) then there exists M ∈ E2 (O) and N ∈ S such that such that A = MN. −1 So, let A = ( ac bd ) ∈ SL2 (O). If c = 0 then d ∈ U(O) and A = ( 10 1bd )( 0a 0d ). If a = 0 then b ∈ U(O) and A = ( 1db−1 01 )( 0c 0b ). Finally, suppose a ≠ 0 and c ≠ 0. Let N denote the norm in O. Then, since O is norm Euclidean, there exists q1 , r1 ∈ O such that a = q1 c + r1 , with N(r1 ) < N(c). Clearly 1 0

(

a −q1 ) ( 1 c

b r1 )=( d c

b − q1 d ). d

Because norms of elements of non-zero elements of O are positive integers, repeating the previous several times, it is clear that we may reduce to the case that a = 0. So the result follows. Because of the previous result one may extend Corollary 11.2.6 on the construction of finitely many generators for a subgroup of finite index in U(ℤG) by allowing also exceptional simple components ℚGe (with e a primitive central idempotent) of the type M2 (D). This provided one can establish concrete isomorphisms M2 (D) → ℚGe. Recall that Bass(G) denotes the group generated by the Bass units of ℤG and Bic(G) the group generated by the bicyclic units of ℤG. Proposition 12.3.4. Let ℚG = ⨁ni=1 ℚGe i = ⨁ni=1 M n i (D i ) be the Wedderburn decomposition of ℚG, with G a finite group. Assume that ℚG does not have exceptional simple components that are division algebras. Also, assume that for each i ∈ {1, . . . , n} such that ℚGe i is not exceptional Ge i is not fixed point free (equivalently, there exists a g i ∈ G such that ĝi e i is not central).

12.4 Normal closure of the trivial units

|

375

For every primitive central idempotent e i with ℚGe i an exceptional simple component, let ψ i : M2 (D i ) → ℚGe i be a ℚ-algebra isomorphism and Oi a norm Euclidean order in D i with ℤ-basis B i . For such i, put 0 b

0 ) : b ∈ B i }. 0

U i = {1 + ψ i ( 00 0b ), 1 + ψ i ( Then, the group

⟨Bass(G) ∪ Bic(G) ∪ ⋃ U i ⟩ i

is commensurable with U(ℤG). Proof. This follows from Corollary 11.2.1, Theorem 11.2.5, Lemma 12.3.3 and Lemma 4.6.9 (4). In Chapter 13 descriptions are given of complete sets of orthogonal idempotents for rational group algebras of finite nilpotent groups G. As a consequence, for such groups, in Chapter 18 a detailed description of the isomorphisms ψ i : M2 (D) → ℚGe i , used in Proposition 12.3.4, will be given.

Problems 12.3.1. Let d be a square-free positive integer, let O be the ring of integers of ℚ(√−d) and let N = Nℚ(√−d)/ℚ . Prove that the following conditions are equivalent. (1) O is norm Euclidean. (2) For every x ∈ ℚ(√−d) there is a ∈ O such that N(x − a) < 1. (3) d = 1, 2, 3, 7 or 11. 12.3.2. Prove Proposition 12.3.1. (For more details see for example [41].) 12.3.3. Let D be a finite dimensional division algebra over G and let O be an order in D. Prove the following statements: (1) If every two sided ideal of O is principal as left ideal then O is a maximal order of D. (Hint: Let O󸀠 be an order of D containing O and let q ∈ O. Then O󸀠 n is a non-zero ideal of O for some positive integer n and qO󸀠 n ⊆ O󸀠 n.) (2) If O left norm Euclidean then every left ideal of O is principal.

12.4 Normal closure of the trivial units In the previous sections it has been shown that, for many finite groups G, the bicyclic units generate a subgroup of finite index in SL1 (ℤG). The Bass units have to be added to get a subgroup of finite index in U(ℤG). Of course, in general, the trivial units ±G are a very small subgroup in U(ℤG). Kleinert in [134] proved that the normal closure

376 | 12 Exceptional simple components in U(ℤG) of the trivial units again is a subgroup of finite index, provided ℚG does not have exceptional simple components (one can even allow many exceptional simple components that are division algebras). We will give a full proof in case there are no exceptional simple components and if M2 (D) is a simple component then D is commutative. The proof is based on results due to Vaseršte˘ın [220] on the normal closure in SLn (D) of a non-central subgroup (with n ≥ 3, or n = 2 and D a field). A note added in proof in [220] states that this result also can be proved for all the non-exceptional algebras. We also state a more general result, due to Margulis [157], which also allows division simple components D with Z(D) a totally imaginary extension of ℚ that is not quadratic. A proof of this result is extremely involved and is beyond the scope of this monograph. We finish with an application on the abelianizer of the unit group U(ℤG). For simplicity reasons we only state applications for integral group rings, there are obvious generalizations to unit group of group rings over larger orders. Recall that if I is an ideal of a ring O and n is a positive integer then En (I) denotes the subgroup of GLn (O) generated by the elementary matrices e ij (x) with x ∈ I. Proposition 12.4.1 (Vaseršte˘ın). Let D be a finite dimensional rational division algebra and O an order in D. Let H be a non-central subgroup of GLn (D) with n ≥ 3. Suppose that H is normalized by En (I) for some non-zero ideal I of O. Then En (J) ⊆ H, for some non-zero ideal J of O. In particular, H contains a subgroup of finite index in SLn (O) (by Corollary 9.5.9). Proof. As mO ⊆ I for some non-zero integer m we may assume without loss of generality that I = mO. Suppose first that H contains an elementary matrix e ij (x) with x ≠ 0 ∈ D. One may assume without loss of generality that x ∈ O. Then using Lemma 9.2.1 and the hypothesis we first obtain that if s ∈ I then e ik (xs) = (e jk (−s), e ij (x)) ∈ H for every k different from i and j. Thus, if r is another element of I then e kl (rxs) = (e ki (r), e il (xs)) ∈ H, provided that i, k and l are different. Interchanging the roles of l and i we also deduce that e ki (r󸀠 rxs) ∈ H for every r󸀠 ∈ I. Hence H contains E n (I 2 xI), as desired. Second, assume that H contains a matrix A = ( zI vn−1 0z ) with z ∈ Z(D) and 0 ≠ v = (v1 , . . . , v n−1 ) ∈ D n−1 . Choose indices i, j such that 1 ≤ i, j ≤ n − 1, v i ≠ 0 and j ≠ i. Then (A−1 , e ij (−m)) = e nj (z−1 v i m) ∈ H. So we are reduced to the first case and thus again the result follows. Third, assume H contains a non-central matrix A of the form ( Bv 0x ) with B ∈ GLn−1 (D). The equality ((

B v

0 ) x

−1

,(

I n−1 w

0 I n−1 )) = ( 1 −xwB−1 + w

0 ) 1

0 shows that either H contains some matrix ( I n−1 ) ≠ I n , or xw = wB for all w ∈ v󸀠 1 n−1 (mO) . The latter implies that B = xI n−1 and x ∈ Z(D). So, either way, we are reduced to the second case and the result follows.

12.4 Normal closure of the trivial units

|

377

Fourth, assume H contains a non-central matrix A = (a ij ) with a1n = 0. Then B = [A, e n1 (−x)] ∈ H for all x ∈ mO and the last column of B coincides with the last column of the identity matrix I n . If B is non-central then we are reduced to the third case with x = 1 and hence we are done. On the other hand, if B is central then we get that B = I n and it follows that a in = 0 for 1 ≤ i ≤ n − 1, so again we are in the third case and the result follows. Finally, we deal with the general case. Because of the third case, we may assume that H contains a matrix A = (a ij ) such that (a1n , . . . , a(n−1)n ) ≠ 0. We claim that there is an elementary matrix e = e ij (d), with d ∈ D \ {0} such that the (1, n)-th entry of e ij (d)Ae ij (−d) is 0. If a1n = 0 one can simply take e = e21 (1). Otherwise a in ≠ 0 for some i ∈ {2, . . . , n − 1} and e = e1i (−a1n a−1 in ) satisfies the desired prop󸀠 erty. Because of Lemma 9.2.1 (2) and because dm ∈ O for some non-zero integer m󸀠 , one gets that e ij (−d)En (m󸀠 mO)e ij (d) ⊆ En (mO). Hence, if B ∈ En (m󸀠 mO) then there B exists C ∈ En (mO) such that (e ij (d)) = e ij (d)C. From the assumption it then follows that e ij (d)He ij (−d) is normalized En (m󸀠 mO). Hence, applying the fourth case to the group e ij (d)He ij (−d), we obtain that e ij (d)He ij (−d) contains En (mJ) for some non-zero ideal J of O. So, e ij (−d)En (mJ)e ij (d) ⊆ H. By assumption and Corollary 9.3.6 we have 4 2 n > ≥2
= str(O). Consequently, by Proposition 9.4.3, we know that En (O, m J ) is a nor4 2 mal subgroup of GLn (O) and, because of Lemma 9.2.4, En (O, m J ) ⊆ En (m2 J). Write e ij (d) = De ij (md)D−1 , with D a diagonal matrix with entry m−1 on the (i, i)-position and 1 elsewhere. We get that e ij (d)En (m5 J 2 )e ij (−d) = De ij (md)D−1 En (m5 J 2 )De ij (−md)D−1 ⊆ De ij (md)En (m4 J 2 )e ij (−md)D−1 ⊆ De ij (md)En (O, m4 J 2 )e ij (−md)D−1 ⊆ DEn (O, m4 J 2 )D−1 ⊆ DEn (m2 J)D−1 ⊆ En (mJ). Hence, En (m5 J 2 ) ⊆ H. We now prove Proposition 12.4.1 for non-exceptional algebras of the form M2 (K) with K a field. Proposition 12.4.2 (Vaseršte˘ın). Let H be a non-central subgroup of GL2 (K) with K a number field that is not ℚ nor a quadratic imaginary extension of ℚ and let O be an order in K. Assume H is normalized by E2 (I), with I a non-zero ideal of O. Then E2 (J) ⊆ H, for some non-zero ideal J of O. In particular, H contains a subgroup of finite index in SLn (O) (by Theorem 10.4.11). Proof. As in the proof of Proposition 12.4.1 we may assume without loss of generality that I = mO for a positive integer m. The assumptions and Corollary 5.2.6 yield that O has infinitely many units. Hence by Lemma 10.2.2, there exists an element u ∈ U(O) of infinite order such that ( 0u u0−1 ) ∈ E2 (mO). Because, by assumption, H contains a noncentral matrix, it easily is verified that H contains a matrix ( ac db ) with a, b, c, d ∈ K

378 | 12 Exceptional simple components and c ≠ 0. Further, let n󸀠 be a positive integer such that n󸀠 ac−1 ∈ O. Because O/n󸀠 mO is finite, replacing if necessary u by a power of u, we also may assume that u − 1 ∈ n󸀠 mO. Hence (u2 − 1)ac−1 ∈ mO. Therefore, there exists t ∈ mO with a + tc = u2 a. Because of the assumption, we obtain that the following matrix belongs to H: 1 (( 0

t a )( 1 c =(

−t )) 1

b 1 )( d 0 u2 a c

−1

x ) y

(

−1

(

a cu−2

u 0

0 a )( u−1 c

x󸀠 u−2 )=( 󸀠 y 0

b u−1 )( d 0

0 ) u

x󸀠󸀠 ) u2

for some x, y, x󸀠 , y󸀠 , y󸀠󸀠 ∈ D. Again by the assumption, for any w ∈ mO, (((

u−2 0

x󸀠󸀠 )) u2

−1

,(

1 0

−w )) = e12 (w(u4 − 1)) ∈ H. 1

Because u has infinite order, we have u4 − 1 ≠ 0. Hence, O(u4 − 1) contains a positive integer, say m󸀠 . It follows that if r ∈ mm󸀠 O then r = (u4 − 1)w for some w ∈ mO and hence e12 (r) = e12 (w(v4 − 1)) ∈ H. Similarly one shows that there is a positive integer m󸀠󸀠 such that e21 (r) ∈ H for every r ∈ mm󸀠󸀠 O. Hence, H contains E2 (mm󸀠 m󸀠󸀠 O) and the results follows. The previous two propositions are special cases of a more general theorem of Margulis. We extract what is relevant for our context. Theorem 12.4.3 (Margulis). Let O be an order in a finite dimensional rational division algebra D. Assume that either n ≥ 2 and M n (D) is not exceptional, or n = 1 and Z(D) is totally complex that is not quadratic over the rationals. If H is a non-central subgroup of SLn (O) which is normalized by a subgroup of finite index in SL n (O) then H has finite index in SL n (O). In particular, SLn (O)/(SLn (O), SLn (O)) is finite. Proposition 12.4.4. Let A be a finite dimensional semisimple rational algebra, let O be an order in A and let U be a normal subgroup of U(O). Let e be a primitive central idempotent of A such that Ae is not exceptional. If Ue contains an element which is not central in A then U contains a subgroup of finite index in 1 − e + SL1 (Oe). Proof. Let Ae = M n (D) with D a division ring and let R be an order in D. By assumption ℚGe ≅ M n (D) is not exceptional. If n = 1 then D is either a field of a totally definite quaternion algebra. Then SLn (R) is finite by Proposition 5.5.6 and hence the results follows. So suppose that n ≥ 2. By Lemma 4.6.9 (4), there is a non-zero ideal I of R such that 1 − e + En (I) ⊆ U(O). By assumption, there is u ∈ Ue such that u is not central in A. We claim that (E n (I), Ue) is not central in Ae = M n (D). Suppose it is, then for every 1 ≤ i, j ≤ n, i ≠ j and y ∈ I, there exists a central element c such that u e ij (y) = c e ij (y) u. As nr(c) = 1 it follows, by Proposition 5.5.1, that c is periodic. Therefore, u e ij (my) = e ij (my) u for some positive integer m. Using that I is finitely generated

12.4 Normal closure of the trivial units

|

379

as ℤ-module, it easily follows that u centralizes E n (kI) for some positive integer k an thus u is central, a contradiction. As E n (I) is of finite index in SLn (R) and E n (I) normalizes (E n (I), Ue), the claim and Theorem 12.4.3 thus yield that (E n (I), Ue) is of finite index in SL1 (Oe). As U is normal in U(O), 1 − e + (E n (I), Ue) = (1 − e + E n (I), U) ⊆ U. Then the result follows. As a consequence of Proposition 12.4.4 we obtain Kleinert’s result. Theorem 12.4.5 (Kleinert). Let G be a finite group and let e be a primitive central idempotent such that ℚGe is not exceptional. Let O be an order in ℚGe. If u ∈ U(ℤG) is such that ue is not central in ℚG then the normal closure of ⟨u⟩ in U(ℤG) contains a subgroup of finite index in 1 − e + SL1 (O). Using a density argument and also the result of Margulis, Kleinert shows in [134] that the previous result remains valid for simple components ℚGe that are division algebras with center a totally complex field that is not quadratic over the rationals. Corollary 12.4.6 (Kleinert). Let G be a finite group such that if an exceptional component of ℚG is a division ring then its center is totally complex but not quadratic over ℚ. Then the normal closure of the group G in U(ℤG) contains a subgroup of finite index in SL1 (ℤG). Proof. Let N denote the normal closure of G in U(ℤG). By the assumption, Theorem 12.4.5 and the comment preceding this corollary, if e is a primitive central idempotent and O is an order in ℚGe then N contains a subgroup of finite index in 1 − e + SL1 (O). Then N contains a subgroup of finite index in SL1 (ℤG). Theorem 12.4.7 (Kleinert). Let G be a finite group such that ℚG does not have simple components that are exceptional two-by-two matrices over a division ring. Assume also that if a non-commutative division algebra is a simple component then it is either a totally definite quaternion algebra or its center is totally complex and not quadratic over ℚ. Then the group generated by the Bass units of ℤG and the normal closure of G in U(ℤG) is of finite index in U(ℤG). Proof. This follows from Corollary 11.2.1 and Corollary 12.4.6. We finish this section with an application to central units, a result due to Jespers, Olteanu, del Río and Van Gelder [120]. Corollary 12.4.8. Let G be a finite group such that ℚG does not have exceptional simple components. Then U(ℤG)/U(ℤG)󸀠 has the same rank as Z(U(ℤG)) and the natural images of the Bass units of ℤG generate a subgroup of finite index in U(ℤG)/U(ℤG)󸀠 . Proof. Let e be a primitive central idempotent of ℚGe such that ℚGe is not a division ring. Let Oe be an order in ℚGe. Since U(ℤG)󸀠 e is not central in SL1 (Oe ) (see Problem 12.4.1), Proposition 12.4.4 and the assumptions easily yield that U(ℤG)󸀠 contains a normal subgroup of finite index in 1− e +SL1 (Oe ). This also holds for the division com-

380 | 12 Exceptional simple components

ponents as they are totally definite quaternion algebras and hence their reduced norm one elements in an order form a finite group by Proposition 5.5.6. Hence, U(ℤG)󸀠 contains a subgroup S = ∏e Se, where e runs through all primitive central idempotents of ℚG, that is of finite index in ∏e SL1 (Oe ) and normal in U(ℤG). Since the elements of S have reduced norm one, S ∩ Z(U(ℤG)) is a finite group. Recall from Proposition 9.5.2 that U(Oe ) has SL1 (Oe )Z(U(Oe )) as subgroup of finite index. Hence the center of the group ∏e U(Oe )/S is of finite index. The first part of the result then follows as Z(U(ℤG)) contains a subgroup of finite index ∏e Z(U(Oe )). For the second part, consider the natural homomorphisms Bass(G) → U(ℤG)/U(ℤG)󸀠 → K1 (ℤG). We know from Corollary 11.1.3 that the natural image of Bass(G) in K1 (ℤG) is of finite index and it has the same rank as Z(U(ℤG)). Hence the natural image of Bass(G) in U(ℤG)/U(ℤG)󸀠 must have the same rank as Z(U(ℤG)) and it thus generates a subgroup of finite index.

Problems 12.4.1. Let A be a finite dimensional semisimple rational algebra, let O be an order in A and let e be a primitive central idempotent of A such that Ae is not a division algebra. Prove that U(O)󸀠 e is not central in SL1 (Oe).

12.5 Normal complements In this section we divert from the main topic of this chapter. We give sufficient conditions on a finite group G in order that the trivial units have a torsion-free normal complement in U(ℤG). The main result will be used in Section 12.6 to describe U(ℤG) for some metacyclic groups G. Throughout this section G is a finite group. Recall that ω N,R denotes the augmentation map of RG modulo N, for R an arbitrary ring R and N a normal subgroup of G. ̂ p the We simply will write ω N,ℤ as ω N . Recall that for a rational prime we denote by ℤ ring of the p-adic integers. 2 Lemma 12.5.1. Let A be a finite abelian p-group and a ∈ A\{1} then a−1 ∈ ̸ ker(ω A,ℤ ̂p ) .

Proof. Since A as a direct product of cyclic groups, it easily is in that, without loss of generality, we may assume that A is cyclic, say of order q. Let g be a generator of A and 2 k 2 let a = g k . Suppose that a − 1 ∈ ker(ω A,ℤ ̂ p ) . Then, by (1.5.3), a − 1 = g − 1 = (g − 1) α ̂ p A. Then (g − 1)(1 + g2 + ⋅ ⋅ ⋅ + g k−1 − (g − 1)α) = 0 and this implies for some α ∈ ℤ 2 ̂ p A (Problem 1.5.7). Applying that 1 + g + ⋅ ⋅ ⋅ + g k−1 − (g − 1)α = b g̃ for some b ∈ Z

12.5 Normal complements |

381

the augmentation map we deduce that k = ω(b)q. Therefore the projection of k in ̂ p /q Z ̂ p ≅ ℤ/qℤ is 0. Thus q | k and hence a = 1. Z Lemma 12.5.2 ([43]). Let A be an abelian normal subgroup of a finite group G. If U(ℤG) ∩ (1 + ker(ω A )) has an element of order p then p divides the order of A. Proof. By means of contradiction, we assume that G has minimal order among the finite groups having an abelian normal subgroup A of order coprime with p for which U(ℤG) ∩ (1 + ker(ω A )) has an element u of order p. If q divides the order of A and B is a Sylow q-subgroup of A then, ω B (u)p = 1. By the minimality assumption, ω B (u) = 1. Hence, replacing A by B, we may assume that A is a q-group. Let 𝔽q = ℤ/qℤ. By Problem 2.1.10, the kernel I of the augmentation ideal of 𝔽q A is nilpotent. As A is normal in G, ker(ω A,𝔽q ) = I𝔽q G = 𝔽q GI and hence ker(ω A,𝔽q ) is a nilpotent ideal of 𝔽q G. n Because α = u − 1 ∈ ker(ω A ), we deduce that u q ≡ 1 mod qℤG for some n ≥ 1. However u p = 1 and hence α ∈ qℤG \ {0}. Thus there is i ≥ 1 with α ∈ q i ℤG \ q2i ℤG. p Then 0 = pα + ∑j=2 (pi)α i ≡ pα mod q2i ℤG, whence α ∈ q2i ℤG, a contradiction. Lemma 12.5.3 ([152]). Let G be a finite group and let A be an abelian normal subgroup of G. Then U(ℤG) ∩ (1 + ker(ω G ) ker(ω A )) is torsion-free. Proof. Observe that it is enough to show that U(ℤG) ∩ (1 + ker(ω G ) ker(ω A )) does not have elements of order p for any prime p. So, we assume that it has an element u of order p. We start showing that one may assume, without loss of generality, that A is a p-group. Otherwise, the p󸀠 -Hall subgroup of A is a non-trivial normal subgroup of G. By induction, ω B (u) = 1, contradicting with Lemma 12.5.2. Therefore, in the remainder of the proof, A is a p-group. Let B = {x1 = 1, x2 , . . . , x k } be a set of representatives of the A-cosets of G. Then B is a basis of ℤG as left and right ℤA-module and the image of α = u − 1 = ∑ki=1 x i α i , by the left regular representation of ℤG expressed in this basis, is a matrix α1 α2 M = (m ij ) = ( . .. αk

x

α12 ..

) ∈ M k (ℤA),

. x

α1k xj

where αx i = ∑kj=1 x j m ji . Then m lj = a ij α i , if x i x j = x l a ij with a ij ∈ A. As α ∈ ker(ω A ), xj

xj

each α i ∈ ker(ω A ). Therefore α i ∈ ker(ω A ) and (a ij − 1)α i ∈ ker(ω A )2 , for every i and j. Let w = (1, . . . , 1) and y = (y1 , . . . , y m ) = wM, the sum of the rows of M. We claim that each y i belongs to ker(ω A )2 . Indeed, α − y1 = α − ∑ki=1 α i = ∑ki=1 (x i − 1)α i ∈ ker(ω G ) ker(ω A ). As, by assumption, also α belongs to ker(ω G ) ker(ω A ) we deduce that xj y1 ∈ ker(ω G ) ker(ω A ) ∩ ℤA = ker(ω A )2 . Thus y1 ∈ ker(ω A )2 . Then y j = ∑kl=1 m lj = xj xj xj ∑ki=1 a ij α i ≡ ∑ki=1 α i = y1 ≡ 0 mod ker(ω A )2 . This finishes the proof of the claim.

382 | 12 Exceptional simple components Moreover, U = I + M has order p and hence, by the Weiss Rigidity Theorem [223], ̂ p A) such that P−1 UP = D = diag(a1 , . . . , a k ) with each a i ∈ A. Let there is P ∈ GLk (ℤ ̂ p A)k . Taking augmentations in the equality w = zP−1 we z = (z1 , . . . , z k ) = wP ∈ (Z k ̂ ̂ p for some i0 = 1, . . . , k. deduce that 1 ∈ ∑i=1 Z p ω(z i ). Thus ω(z i0 ) is invertible in ℤ Moreover, all the coordinates z i (a i − 1) of z(D − I) = wPP−1 MP = yP belong to 2 ker(ω A, Ẑ p ) . Hence, ω(z i )(a i − 1) = (ω(z i ) − z i )(a i − 1) + z i (a i − 1) ∈ ker(ω A, Ẑ p )2 . ̂ p , we deduce that a i − 1 ∈ ker(ω ̂ )2 . From Lemma 12.5.1 As ω(z i0 ) is invertible in Z 0 A, Z p we then obtain that a i0 = 1, Let C1 , . . . , C m be the conjugacy classes of G contained in A and write 1 + α1 = β1 + ⋅ ⋅ ⋅ + β m , where the support of β j is contained in C j for each j. Pick one c j ∈ C j for x x each j and let b j = ∑ki=1 c j i . Then ∑ki=1 β j i = ω(β j )b j . As D and U are conjugate, we get k

k

k

m

x

m

∑ a i = tr(D) = tr(U) = ∑ (1 + α1 )x i = ∑ ∑ β j i = ∑ ω(β j )b j . i=1

i=1

i=1 j=1

j=1

As the supports of the b j ’s are pairwise disjoint, we have ω(β j ) ≥ 0 for every j. Moreover, 1 = ω(1 + α1 ) = ∑m j=1 ω(β j ). Therefore ω(β j ) = 0 for all but one j, say j 0 . Thus ∑ki=1 a i = ω(β j0 )b j0 . This implies that all the a i ’s are conjugate. As a i0 = 1, we conclude that a i = 1 for every i. So U = I and hence u = 1, contradicting the hypothesis. Remark 12.5.4. Zassenhaus Conjecture states that if G is a finite group then every torsion unit of augmentation 1 in ℤG is conjugate in ℚG to an element of G. The Weiss Rigidity Theorem, used in the proof of Lemma 12.5.3, is the main tool of the proof of Zassenhaus Conjecture for finite nilpotent groups [224]. The Zassenhaus Conjecture is one of the most important open problems on units of group rings. It deserves a separate treatment. In [43] one can find a proof of Lemma 12.5.3, not using the Rigidity Theorem, under the additional assumption that G/A is abelian. We are ready to prove the main result of the section. The proof uses the Whitcomb argument which was already used to give a description of U(ℤD8 ) in Example 1.5.4. Proposition 12.5.5 ([43, 152]). Let A be an abelian normal subgroup of a finite group G such that U(ℤ(G/A)) is finite. Then (U(ℤG) ∩ (1 + ker(ω) ker(ω A ))) is a torsion-free normal complement of ±G in U(ℤG). Proof. Let u ∈ U(ℤG) with ω(u) = 1. Because of the assumptions on G/A, Higman’s result (Theorem 1.5.6) yields that u = g + v, for some g ∈ G and v ∈ ker(ω A ). Write v = ∑a∈A v a (a − 1), with each v a ∈ ℤG. Clearly v a − ω(v a ) ∈ ker(ω) and thus v ∈ ∑a∈A ω(v a )(a − 1) + ker(ω) ker(ω A ). Because, xy − 1 = (x − 1) + (y − 1) + (x − 1)(y − 1),

12.6 Examples: metacyclic groups | 383

for every x, y ∈ G, we get that ∑a∈A ω(v a )(a − 1) ∈ (b − 1) + ker(ω A )2 for some b ∈ A. Again applying the previous identity, it follows that g + b − 1 ∈ gb + ker(ω) ker(ω A ). Hence u ∈ gb (1 + ker(ω) ker(ω A )). So we have shown that U(ℤG) = (U(ℤG) ∩ (1 + ker(ω) ker(ω A )))(±G). The result now follows because U(ℤG) ∩ (1 + ker(ω) ker(ω A )) is normal in U(ℤG) and torsion-free, by Lemma 12.5.3. Remark 12.5.6. In [43] it has been proved also that if G has a normal abelian subgroup A with G/A abelian of odd order then ±G has a torsion-free normal complement in U(ℤG).

12.6 Examples: metacyclic groups Throughout this section G is a nonabelian split metacyclic group with a cyclic subgroup of index 2. In other words G = ⟨a⟩n ⋊ ⟨b⟩2 = ⟨a, b | a n = 1, b2 = 1, ba = a i b⟩.

(12.6.1)

Note that i2 ≡ 1 mod n. Set d = gcd(n, i − 1). Clearly Z(G) = ⟨a ⟩ and G󸀠 = ⟨a i−1 ⟩ = ⟨a d ⟩. Also, the non-central conjugacy classes are either of the form a j bG󸀠 , with 0 ≤ j ≤ d − 1, or of the form {a r , a ri } with a r ∈ ̸ Z(G). We claim that every fixed point free epimorphic image H of G is abelian. To prove this we first note that H ≅ C m ⋊ C2 for some m. If m is even then the Sylow subgroup of H is neither cyclic nor quaternion, in contradiction with Theorem 11.4.5 (5). Hence m is odd. Let p be a prime divisor of m. By Theorem 11.4.5 (2) the elements of order 2 and p commute. This implies that the order of the action of C2 on the p-Sylow subgroup of C m divides both 2 and a power of p. Hence this action is trivial. Therefore, H is abelian, as desired. The aim of this section is to prove that in some cases, including all dihedral groups, the Bass and bicyclic units generate a subgroup of finite index in U(ℤG). By the previous paragraph, this is the consequence of Theorem 11.3.2, provided ℚG does not have exceptional simple components. However, it turns out that often ℚG does have exceptional components. In such cases some additional work is needed. Crucial for the proofs is to determine the unit groups of the integral group rings ℤD2n with n ∈ {3, 4, 6, 12}. The results in this section are due to Jespers, Leal, Polcino Milies, Ritter and Sehgal, [104, 117, 121, 191, 202]. We begin with calculating the Wedderburn decomposition of the rational group algebra ℚG and its exceptional simple components. For that, we use the results on strong Shoda pairs in Chapter 3. The Wedderburn decomposition of the rational group algebra of an arbitrary metacyclic group has been obtained in [166]. We have ̂󸀠 ⊕ ℚG(1 − G ̂󸀠 ). ℚG = ℚG G n d

384 | 12 Exceptional simple components ̂󸀠 ≅ ℚ(G/G󸀠 ), a commutative algebra and, by Problem 3.3.5, each simple By (1.5.7), ℚG G ̂󸀠 ) is non-commutative. By Theorem 3.5.12, the primitive central component of ℚG(1− G ̂󸀠 ) are the elements of the form e m = e(G, ⟨a⟩ , ⟨a m ⟩) = idempotents of ℚG(1 − G ε(⟨a⟩ , ⟨a m ⟩) with m | n and m ∤ d. In this case (⟨a⟩ , ⟨a m ⟩) is a strong Shoda pair of G and, because of Theorem 3.5.5, ℚGe m is isomorphic to the cyclic algebra (ℚ(ζ m )/ℚ(λ G ), σ m , 1), where σ m is the automorphism of ℚ(ζ m ) defined by ζ m 󳨃→ ζ mi and λ is a linear character of ⟨a⟩ with kernel ⟨a m ⟩. Thus, the center of ℚGe m is ℚm,i = ∑ ℚ(a j + a ij )e m ≅ ℚ(ζ m + ζ mi , ζ m2 + ζ m2i , . . .). j

Furthermore, from Proposition 2.6.7 one obtains that ℚGe m = M2 (ℚm,i ). Hence the first part of the following result has been proved. Proposition 12.6.1. Let G be the group given in (12.6.1). Let e be a primitive central idempotent of ℚG with ℚGe non-commutative. Then e = e m = ε(⟨a⟩ , ⟨a m ⟩) with m a positive divisor of n such that m is not a divisor of d. Furthermore, for such m, ℚGe m ≅ M2 (ℚm,i ), (a − a i )2 e m is a non-zero central element of ℚGe m and the elements e11 = (

1+b ) em 2

1+b 1−b )a( ) em 2 2 1−b =( ) em 2

e12 = ( −2

e21 = 4 ((a − a i )e m )

(

1+b 1−b )a( ) em 2 2

e22

form a set of matrix units of ℚGe m . Proof. Let R = ℚGe m with m | n and m ∤ d. Then R is a simple algebra of dimension 4 over its center. In particular, rRs ≠ 0 for every r, s ∈ R \ {0}. Clearly, 1+b 1+b 1−b 1+b )R( )⊕( )R( ) 2 2 2 2 1−b 1+b 1−b 1−b ⊕( )R( )⊕( )R( ). 2 2 2 2

R=(

and hence the summands in the previous expression are of dimension 1 over Z(R). Since ℚGe m is not abelian, we get be m ≠ −e m and be m ≠ e m . Thus ( 1+b 2 )e m ≠ 0 and 1−b ( 2 )e m ≠ 0. 1−b Note that ( 1+b 2 )a( 2 )e m ≠ 0. Indeed, for if not, then ae m = [(

1+b 1+b 1−b 1+b 1−b 1−b )a( )+( )a( )+( )a( )] e m . 2 2 2 2 2 2

1+b 1−b r 1−b Hence, for any r ≥ 1, ( 1+b 2 )a ( 2 )e m = 0. Consequently ( 2 )R( 2 ) = {0}, a contra1−b 1+b diction. Similarly, ( 2 )a( 2 )e m ≠ 0. Also,

(a − a i )2 1 − b 1−b 1+b 1+b 1−b ( ) e m = [( )a( )] [( )a( )] e m ≠ 0. 4 2 2 2 2 2

12.6 Examples: metacyclic groups | 385

1−b 1−b 1+b Indeed, for if not, then, because ( 1+b 2 )R( 2 ) and ( 2 )R( 2 ) are one dimensional over Z(R) and because of the previous,

α [((

1+b 1−b 1−b 1−b )R( )) + (( )R( ))] = {0}, 2 2 2 2

1+b 1−b 1−b where α = ( 1−b 2 )a( 2 ). So αR( 2 ) = {0}. Therefore αe m = 0 or 2 e m = 0, a contradiction. Since (a − a i )2 e m is central, the above yields that (a − a i )2 e m has an inverse in Z(R). Hence the defined elements make sense. The result now follows by verifying the identities e11 + e22 = e m and e uv e kl = δ vk e ul .

In the following lemma the exceptional simple components are determined. Of course this also follows from Corollary 12.1.2. However, for the metacyclic groups under consideration, this can be proved very easily. Lemma 12.6.2. Let e m be a primitive central idempotent of ℚG with ℚGe m noncommutative, i.e. m | n but m ∤ d. (1) ℚGe m ≅ M2 (ℚ) if and only if m ∈ {3, 4, 6}. Furthermore, in this case Ge m ≅ D2m . (2) ℚGe m is a simple component which is a two-by-two matrix ring over a quadratic imaginary extension of ℚ if and only if one of the following conditions hold: (a) m = 8 and i ≡ 3 or 5 mod 8 (note that ℚ8,3 = ℚ(√−2) and ℚ8,5 = ℚ(√−1)); (b) m = 12 and i ≡ 5 or 7 mod 12 (note that ℚ12,5 = ℚ(√−1) and ℚ12,7 = ℚ(√−3)) . Proof. By Proposition 12.6.1, ℚGe m = M2 (ℚm,i ) and ℚm,i = ℚ(ζ m + ζ mi , ζ m2 + ζ m2i , . . . ). Since [ℚ(ζ m ) : ℚm,i ] = 2, it follows from Proposition 6.1.1 that dimℚ ℚm,i = φ(m) 2 . Hence, ℚGe m = M2 (ℚ) if and only if m = 3, 4 or 6. Hence, part (1) follows. Note that if m is either 5 or 10 then ℚm,i is a subfield of ℚ(ζ5 ) with [ℚm,i : ℚ] = 2. Then ℚm,i = ℚ(√5). Consequently, if ℚm,i is a quadratic imaginary extension of ℚ (in particular φ(m) = 4) then m = 8 or 12. Part (2) now easily follows. ̃ with g, h ∈ D8 , In Example 1.5.4 it is shown that the bicyclic units of the form b(g, h), generate a free subgroup B (of rank 3) such that U(ℤD8 ) = B ⋊ (±D8 ). Furthermore, if a is an element of order 4 of D8 then B = U(ℤD8 ) ∩ (1 + ker(ω)(1 − a2 )) (see the proof of Example 1.5.4). As usual, ω denotes the augmentation map on ℤD8 . Actually, B is generated by any set formed by three of the four bicyclic units of the form b(a, ̃ a k b) (see also Problem 1.5.4). Problem 1.5.5 says that for D6 there is a similar result. We formulate this in greater detail, without proof, as done by Jespers and Parmenter in [121]. To do so we introduce some notation. For sets A, B, C, D, write (

A C

B a ) = {( D c

b ) : a ∈ A, b ∈ B, c ∈ C, d ∈ D} . d

386 | 12 Exceptional simple components

For example, if n is a positive integer then 1 + nℤ SL2 (ℤ, nℤ) = SL2 (ℤ) ∩ ( nℤ

nℤ ). 1 + nℤ

Proposition 12.6.3. Let D6 = ⟨a, b | a3 = 1, b2 = 1, ba = a−1 b⟩ and let ω denote the augmentation map ℤD6 → ℤ. The following properties hold. (1) The function 3ℤ 2ℤ ℤD6 (1 − a) → ( 3 ) 2 ℤ 3ℤ defined by mapping (α0 + α1 a + β0 b + β1 ba)(1 − a), with α0 , α1 , β0 , β1 ∈ ℤ, to 3(α1 + β0 − β1 ) 2 (α 0 − 2α 1 − β 0 + 2β 1 )

(3

2(−α0 + 2α1 + 2β0 − β1 ) ) 3(α0 − α1 − β0 + β1 );

is a ring isomorphism. 2 0 ) gives a isomor(2) Composing the previous isomorphism with conjugation by ( −3 1 phism of non-unital rings 3ℤ 3ℤ

ker(ω)(1 − a) ≅ (

3ℤ ). 3ℤ

(3) U(ℤD6 ) = W ⋊ (±D6 ), where W = (1 + ker(ω)(1 − a)) ∩ U(ℤD6 ) ≅ SL2 (ℤ, 3ℤ) is a free group of rank 3 and is generated by the 3 distinct (up to inverses) bicyclic units ̃ of the form b(g, h). The next lemma will be needed to deal with metacyclic groups G which are such that ℚG has precisely one exceptional component and, moreover, this a component of the type M2 (ℚ). The proof is elementary (see Problem 12.6.1). Lemma 12.6.4. Let B be a subgroup of a direct product of groups, say G 1 × G2 . Suppose B contains a subgroup S1 × {1} with S1 of finite index in G1 . If the natural projection of B onto G2 is of finite index then B is of finite index in G1 × G2 . Theorem 12.6.5. Let G = ⟨a, b | a n = 1, b2 = 1, ba = a i b⟩. Let S1 be the subgroup of U(ℤG) generated by the Bass units of ℤG and the bicyclic units of the form ̃ a v ) = 1 + (1 + b)a v (1 − b) b(b,

and

̃ = 1 + (1 − b)a v (1 + b), b(a v , b)

and let S2 be the subgroup generated by the Bass units and the bicyclic units of the form ̃ u b, a v ) = 1 + (1 + a u b)a v (1 − a u b) b(a

and

̃ u b) = 1 + (1 − a u b)a v (1 + a u b), b(a v , a

where 0 ≤ u, v < n, and n | (i + 1)u. Suppose the following conditions are satisfied: (i) if 8 | n then i ≡ ±1 mod 8; (ii) if 6 | n then i ≡ 1 mod 6.

12.6 Examples: metacyclic groups |

387

(iii) if 12 | n then i ≡ 1 mod 12. (iv) If n = 2h n󸀠 with n󸀠 odd, h ≥ 2 and i ≡ −1 mod 4 then i ≡ −1 mod 2h . Then (1) S2 has finite index in U(ℤG). (2) Assume additionally that (a) 4 ∤ n or 4 | d; and (b) 3 ∤ n or 3 | d. Then S1 has finite index in U(ℤG). Proof. We use the notation of Proposition 12.6.1. Let e m be a primitive central idempotent of ℚG such that ℚGe m is not abelian. So m is a positive integer that is a divisor of n but it is not a divisor of d. Note that, because of hypothesis (i), (ii) and (iii) and because of Lemma 12.6.2, there is at most one exceptional component and it is of the type M2 (ℚ). Such a component occurs precisely when m = 3 or 4, or equivalently either condition (a) or (b) fails. From Theorem 11.2.5 it follows that S1 contains a subgroup of finite index in SL2 (Om,i ), where Om,i is the ring of integers in ℚm,i whenever m is not 3 or 4. In this case, the group S1 satisfies the assumptions of Corollary 11.2.1, and thus statement (2) follows. On the other hand, assume (a) or (b) fails, i.e. ℚG has a primitive central idempotent e m with m either 3 or 4. Then, by Lemma 12.6.2, ℚGe m ≅ M2 (ℚ) is the only non-commutative exceptional simple component of ℚD2m . From the first part of the proof, we know that S1 (and hence S2 ) contains a subgroup of finite index in SL1 (O), where O is an order in ℚG(1 − e m ). By Lemma 12.6.4 and Corollary 11.2.1, it is enough to show that S2 e m contains a subgroup of finite index in SL2 (Om,i ) = SL2 (ℤ). From Proposition 12.6.3 and Example 1.5.4 we know that in each case the projection of the group generated by the bicyclic units of ℤD2m has finite index in SL2 (ℤ). Note also that the bicyclic units for the groups D6 and D8 are of the type mentioned in the definition of S2 . Now if the bicyclic units of the integral group ring ℤ(Ge m ) are images of such kind of bicyclic units of ℤG, then it follows that S2 e m contains a subgroup of finite index in the reduced norm one elements of an order in ℚGe m , and this for m | n with m ∤ d. So it remains to show that the bicyclic units in ℤD m can be lifted to bicyclic units of ℤG of the type defining S2 . We show the details for non-trivial bicyclic units on the ̃ g) of D2m = ⟨a, b⟩. Every such bicyclic unit is of the form u = b(̃ type b(h, a k b, a j ), with j = ±1 and 0 ≤ k < m. If v is an integer such that v(i +1) ≡ 0 mod n and v ≡ k mod m then b(̃ a v b, a j ) projects to u. Observe that n | i2 − 1 and i ≡ −1 mod m. Therefore, if m = 3 then one can take v = k(i − 1). Assume m = 4 and write n = 2h n󸀠 with n󸀠 odd. By assumption, 2h | i + 1. By the Chinese Remainder Theorem, there exists an integer v such that v ≡ 0 mod n󸀠 and v ≡ k mod m. Then v satisfies the required conditions.

388 | 12 Exceptional simple components In case of general dihedral groups D2n (i.e., i = −1), because of Proposition 12.6.1 and Lemma 12.6.2, there can be up to three exceptional simple components, and they all are of the type M2 (ℚ). Such components occur when n is divisible by 3 or 4. If only one such component occurs (for example if n = 3 or 4), then it follows from Theorem 12.6.5 the Bass units together with the bicyclic units generate a subgroup of finite index in the full unit group. If 6 | n then at least two such components occur and and if n = 12 then there are three such components. In [104] Jespers obtained the following description for U(ℤD12 ). Proposition 12.6.6. In U(ℤD12 ), the trivial units ±D12 have a torsion-free normal complement V which is a semi-direct product of a free group of rank 5 by a free group of rank 3. Furthermore, V is contained in the group generated by bicyclic units of the form ̃ with g, h ∈ D12 . b(g, h), Proof. Write D12 = D6 × C2 = ⟨a, b | a3 = 1, b2 = 1, ba = a−1 b⟩ × ⟨c | c2 = 1⟩. Let W denote the torsion-free normal complement of ±D6 in ℤD6 mentioned in Proposĩi ). tion 12.6.3. Hence W is free of rank 3 generated by x0 , x1 and x2 with x i = b(a, ba Because of Proposition 12.5.5, V = (1+ker(ω)(1−a))∩U(ℤD12 ) is a normal complement for ±D12 in U(ℤD12 ). Let π : ℤ(D12 ) → ℤD6 denote the natural epimorphism (mapping c to 1). As π(V) = W and W is free, V is a semidirect product of K = {u = 1 + α(1 − c)(1 − a) ∈ U(ℤD12 ) | α ∈ ℤD6 } by W. Clearly the natural mapping K → K(1 − ̂c) is an isomorphism. Hence K ≅ K(1 − ̂c) ≅ W2 = (1 + 2ℤD6 (1 − a)) ∩ U(ℤD6 ). Let f be the isomorphism of Proposition 12.6.3 (1) and let g be the composition of f with 2 0 ). Then g(W) = SL (ℤ, 3ℤ), conjugation by ( −3 2 1 1 + 6ℤ 3ℤ

f(W2 ) = (

4ℤ ) ∩ SL2 (ℤ) 1 + 6ℤ

and 2 3

g(W2 ) = ( =(

0 ) 1

−1

1 + 6ℤ 6ℤ

1 + 6ℤ 3ℤ

(

4ℤ 2 )( 1 + 6ℤ −3

0 ) ∩ SL2 (ℤ). 1

2ℤ ) ∩ SL2 (ℤ). 1 + 6ℤ

Thus SL2 (ℤ, 6ℤ) ⊆ g(W2 ) ⊆ SL2 (ℤ, 2ℤ). By Corollary 1.4.4 (4), the only torsion element of SL2 (ℤ, 2ℤ) is −I. As −I ∈ ̸ g(W1 ), W2 is torsion-free and hence free, by Corollary 1.4.4 (2). It is easy to see that SL2 (ℤ, 3ℤ)/SL2 (ℤ, 6ℤ) ≅ D6 and g(W2 )/SL2 (ℤ, 6ℤ) ≅ C3 . As SL2 (ℤ, 3ℤ) is free of rank 3, using the Nielsen-Schreier Theorem (see e.g. [196, 6.1.1]), we deduce that SL2 (ℤ, 6ℤ) is free of rank 13 and W2 is ̃ A straightforward calculation shows that x−1 x0 ∈ K free of rank 5. Let x = b(a, bc). −5 8 and g(x) = ( −12 19 ) ∈ ̸ SL2 (ℤ, 6ℤ). Therefore K = ⟨g −1 (SL2 (ℤ, 6ℤ)), x−1 x0 ⟩. As

12.6 Examples: metacyclic groups |

389

W = ⟨x0 , x1 , x2 ⟩ = g−1 (SL2 (ℤ, 3ℤ)), we deduce that g−1 (SL2 (ℤ, 6ℤ)) is generated ̃ and hence so is K. by bicyclic units of type b(g, h) Arbitrary dihedral groups can now be handled. Proposition 12.6.7 (Ritter-Sehgal [191]). The group generated by the Bass units together with the bicyclic units is of finite index in U(ℤD2n ). Proof. Recall that D2n does not have nonabelian fixed point free epimorphic images. n ̂ m . Then Write D = ⟨a, b | a n = 1, b2 = 1, ba = a−1 b⟩. Let m = gcd(n, 12) and f = a 2n

ℚD2n f ≅ ℚD2m and, by Proposition 12.6.1 and Lemma 12.6.2, for every primitive central idempotent e of ℚD2n (1 − f), the simple algebra ℚGe is not exceptional. Because of Theorem 11.2.5, for each such e with ℚGe not abelian, the group Bic(D2n ) contains a subgroup of finite index in 1 − e + SL1 (Oe ), where Oe is an order in ℚGe. Since every bicyclic unit of ℤD2m is the image of a bicyclic unit of ℤD2n , by the natural projection, by Lemma 12.6.4, we thus only have to prove the result for ℤD2m . If m = 3, 4 or 6 then this follows from Proposition 12.6.3 and Proposition 12.6.6. So, one is left to deal with D24 . Again by Proposition 12.6.1 and Lemma 12.6.2, −1 ℚD24 = ℚD24 ̂ )). a6 × ℚD24 e4 × ℚD24 e12 = ℚD12 × M2 (ℚ) × M2 (ℚ(ζ12 + ζ12

We know that Bic(D24 ) contains a subgroup of finite index in the reduced norm one el−1 ements of an order of the non-exceptional component M2 (ℚ(ζ12 +ζ12 )). Now ⟨a3 , b⟩ = ̂ 3 D8 . Clearly a a6 is a central element. Hence the bicyclic units in ℤD8 project trivially on ℚD24 ̂ a6 . So, by Lemma 12.6.4 and the fact that the bicyclic units of ℤD8 generate a subgroup of finite index in U(ℤD8 ), one obtains that the bicyclic units of ℤD24 contain a subgroup of finite index in the reduced norm one elements of an order in ℚD24 e4 . So the result follows. Note that Ritter and Sehgal prove a bit more. Namely the bicyclic units of the type ̃ together with the Bass units generate a subgroup of finite index in U(ℤD2n ). b(g, h) The proof basically comes down to “cover” elementary matrices in the matrix components M2 (ℚm,i ) (compare with the proof of Theorem 11.2.5). With an appropriate representation one can cover the elementary matrices of the form e12 (x). To cover the other elementary matrices one observes that, in such representation, some element of ̃ | g, h ∈ D 2n ⟩ is closed under D2n takes the form of the matrix A = ( 01 10 ). As ⟨b(g, h) −1 conjugation by elements of D2n and A e12 (x)A = e21 (x), the result follows.

Problems 12.6.1. Prove Lemma 12.6.4.

390 | 12 Exceptional simple components

12.7 Examples with insufficient Bass units and bicyclic units We give some examples of finite 2-groups G such that the Bass units together with the bicyclic units do not generate a subgroup of finite index in U(ℤG). The following result of Jespers and Parmenter [122] is a slight extension of a result of Ritter and Sehgal [193, Proposition 6.3]. Theorem 12.7.1. Let D8 = ⟨a, b | a4 = 1, b2 = 1, ba = a3 b⟩. Let G be a finite 2-group and let f : G → D8 be an epimorphism. If at least two of the elements b, ab, a2 b, a3 b do not have preimages in G of order 2, then the Bass units together with the bicyclic units in ℤG do not generate a subgroup of finite index in U(ℤG). Proof. The ℤ-linear extension of f to a ring epimorphism ℤG → ℤD8 , as well as the induced group homomorphism U(ℤG) → U(ℤD8 ), we also denote by f . Since every Bass unit of ℤD8 belongs to D8 , every Bass unit in ℤG must map to an element of D8 . ̃ in ℤG. Then either f(b(g, h)) ̃ = 1 or Next consider a bicyclic unit b(g, h) ̃ = 1 + c(1 − f(g))f(h)f(g) ̃ = (1 + (1 − f(g))f(h)f(g)) ̃ c, f(b(g, h)) where c =

o(g) o(f(g))

(see Problem 1.5.3).

̃ u2 = b(a, ab), ̃ u3 = b(a, ̃ The bicyclic units of ℤD8 are u1 = b(a, b), a2 b) and −1 −1 −1 ̃ 3 u4 = b(a, a b) (see Problem 1.5.4). Further u4 = u3 u2 u1 . It is easily verified that the given condition on G yields that at least two of these bicyclic units are not images of bicyclic units in ℤG. From Proposition 12.5.5 and Example 1.5.4 we know that V = U(ℤD8 ) ∩ (1 + ker(ω)(1 − a2 )) = U(ℤD8 ) ∩ (1 + ker(ω)(1 − a)) is a normal complement of the trivial units ±D8 and it is a free group of rank three, ̃ Let B be the subgroup of U(ℤG) generated by the bicyclic units of the type b(g, h). ̃ Since G is a 2generated by the Bass units and the bicyclic units of the type b(g, h). group, it follows from the remarks above that f(B) is a proper subgroup of V requiring at most 4 generators. Since V is a free group of rank 3, we conclude that f(B) must be of infinite index in V. Indeed, by the Nielsen-Schreier Theorem (see e.g. [196, 6.1.1]) if f(B) has index n in V then f(B) is free of rank 2n + 1. As f(B) is generated by at most 4 elements, necessarily n = 1 and hence f(B) = V, a contradiction. For a positive integer i, let V i denote the subgroup of V consisting of those units which can be written in the form 1 + 2i β(1 − a2 ) for some β ∈ ℤD8 . Because (1 − a2 )2 = 2(1 − a2 ), it follows that each V i ⊆ V . Also note that for all i, V i is a normal subgroup of V and that the groups V/V1 and V i /V i+1 are of exponent 2 and thus abelian. Since U(ℤD8 ) is finitely generated, so is the group V. Consequently, V/V1 and all V i /V i+1 are finite. So, each V/V i is finite. Let K = ker(f). Obviously, |K| = 2l for some l ≥ 1. We claim that V l ⊆ f(U(ℤG)). Indeed, let 1 + 2l β(1 − a2 ) ∈ V l . Choose a1 , β1 ∈ ℤG such that f(a1 ) = a and f(β1 ) = β.

12.7 Examples with insufficient Bass units and bicyclic units | 391

K =̂ K(1 + 2l β1 (1 − a21 )) is a unit in ℤG ̂ K ≅ ℤD8 . Put u = 1 + ̃ Kβ1 (1 − a21 ). Clearly u ̂ ̂ ̂ Since u(1 − K) = 1 − K is a unit in ℤG(1 − ̂ K), we get that u ∈ ℤG is a unit in the order ℤG ̂ K ⊕ ℤG(1 − ̂ K). Hence, because of Lemma 4.6.9, u ∈ U(ℤG). Obviously, f(u) = l 2 1 + 2 β(1 − a ). So, u ∈ f(U(ℤG)) and the claim has been proved. Suppose that f(B) is of finite index in f(U(ℤG)). Since f(B) ⊆ V, this yields f(B) is of finite index in f(U(ℤG)) ∩ V. Because V l ⊆ f(U(ℤG))V and V l is of finite index in V, it follows that f(B) is of finite index V. However this contradicts the earlier fact that f(B) is of infinite index in V. Therefore, we have shown that f(B) is of infinite index in f(U(ℤG)). ̃ g)) ≠ 1 then it is a power of a bicyclic unit To finish the proof we note that if f(b(h, 2 f(h)), we obtain b(̃ f(h), f(g)) = 1+(1+ f(h))f(g)(1− f(h)). Since b(̃ f(h), f(g)) = b(f(g), ã ̃ that f(Bic(G)) = f (⟨b(g, h) | g, h ∈ G⟩). So, from the previous, f (⟨Bic(G)) ∪ Bass(G)⟩)) is of infinite index in f(U(ℤG)) and thus ⟨Bic(G) ∪ Bass(G)⟩ is of infinite index in U(ℤG). Corollary 12.7.2. If G = Q16 , D−16 , C4 ⋊ C4 or (⟨z⟩2 × ⟨a⟩4 ) ⋊ ⟨b⟩2 , with z central and a b = za, then the Bass units together with bicyclic units generate a subgroup of infinite index in U(ℤG). Actually it turns out that the only indecomposable nonabelian group of order 16 for which the Bass units together with the bicyclic units generate a subgroup of finite index in the unit group of its integral group ring is D16 (see [116, 122, 193]).

13 Idempotents and central units in group rings In Chapter 3 a detailed description is given of the primitive central idempotents of ℚG and in Chapter 7 a description is given of an independent set of generators of the unit group of the integral group ring of a finite abelian group. In this chapter we prove extensions of these results to larger classes of groups. Positive answers are obtained for the class of abelian-by-supersolvable finite groups in which every cyclic subgroup of order not a divisor of 4 or 6 is subnormal. By considering generalized Bass units this can also be done for strongly monomial finite groups which have a complete and non-redundant set of strong Shoda pairs (H, K) with the property that [H : K] is a prime power. In the final section a class of metacyclic groups is given which satisfies all these assumptions. The proofs rely on a detailed description of the Wedderburn components of the rational group algebra ℚG of the groups G considered. Hence the relevance of describing the rational representations of the group G and thus the primitive idempotents of these group algebras. This is done for nilpotent finite groups as well as for some strongly monomial groups. As an application, for some classes of finite groups G, one can describe three nilpotent subgroups of U(ℤG) that generate a subgroup of finite index in U(ℤG).

13.1 Large central subgroups and abelian-bysupersolvable groups Let G be a finite group that is abelian-by-supersolvable group G with the property that every cyclic subgroup of order not a divisor of 4 or 6 is subnormal in G. Recall that in Chapter 3 a detailed description is given of the primitive central idempotents of ℚG and in Theorem 7.3.3 a description is given of an independent set of generators of the unit group of the integral group ring of a finite abelian group. Due to these results, in this chapter it is proven that the group generated by the Bass units of ℤG contains a subgroup of finite index in Z(U(ℤG)). Furthermore, one obtains a description for the generators of this subgroup. This result is due to Jespers, Olteanu, del Río and Van Gelder [120] We introduce a new construction for central units based on Bass units in the integral group ring ℤG. The idea goes back to [124] where finitely many generators are given for a subgroup of finite index in the central unit group of ℤG for G a finite nilpotent group. Throughout this section G is a finite abelian-by-supersolvable group such that every cyclic subgroup of order not a divisor of 4 or 6 is subnormal in G. Clearly, examples are the dihedral and quaternion groups. Also, every finite nilpotent group N is an example of such a group. Indeed, denote by Z i = Z i (N) the i-th center of N, i.e. Z0 = 1 and

13.1 Central subgroups and abelian-by-supersolvable groups |

393

Z i /Z i−1 = Z(G/Z i−1 ), for i ≥ 1. Then, for x ∈ N, the series ⟨x⟩⟨Z1 , x⟩⋅ ⋅ ⋅⟨Z n , x⟩ = N is subnormal in N and thus ⟨x⟩ is subnormal in N. Note that every term in this series is normalized by the elements in the normalizer in G of the subgroup ⟨x⟩. It is well known that the largest nilpotent normal subgroup of G, the Fitting subgroup F(G) of G, is precisely the subgroup generated by all the elements g ∈ G such that ⟨g⟩ is subnormal in G (see for example [196, 12.2.6]). It follows that any g ∈ F(G) has a subnormal series from ⟨g⟩ to G in which each term is normalized by the normalizer of ⟨g⟩ in G Let g ∈ G of order not a divisor of 4 or 6 and let N : N0 = ⟨g⟩
N1
N2
⋅ ⋅ ⋅
N m = G

be a subnormal series in G. For u ∈ U(ℤ ⟨g⟩) define cN 0 (u) = u and N h cN i (u) = ∏ c i−1 (u) , h∈T i

where T i is a transversal for N i−1 in N i , i ≥ 1. In the following lemma it is shown that this construction is well defined, in particular the order of the conjugates defining cN i (u) is not important. Lemma 13.1.1. Let g ∈ G, u ∈ U(ℤ ⟨g⟩), N, N i and T i be as above. The following properties hold for i ≥ 1. x (1) cN i−1 (u) ∈ ℤN i−1 for x ∈ N i . N x (2) cN i−1 (u) = c i−1 (u) for x ∈ N i−1 .

(3) cN i (u) is independent of the chosen transversal T i . In particular, cN m (u) ∈ Z(U(ℤG)). Proof. Clearly (1) and (2) imply (3). We prove part (1) and part (2) by induction on i. The case i = 1 is obvious as u ∈ ℤ ⟨g⟩ = ℤN0 and N0
N1 . Assume the formulas hold for i − 1 and i ≥ 2. In particular, the order of the factors defining cN i−1 (u) is irrelevant. N x = ∏ hx . By the induction hypothesis one has (u) c (u) Let x ∈ N i . Then cN h∈T i−1 i−2 i−1 N h hx ∈ ℤN cN i−1 , which proves (1). i−2 (u) ∈ ℤN i−2 . Since N i−2
N i−1
N i , c i−2 (u) N N x hx = ∏ h󸀠 Next, let x ∈ N i−1 . Then c i−1 (u) = ∏h∈T i−1 cN (u) h󸀠 ∈T i−1 x c i−2 (u) . Note i−2 that T i−1 x also is a transversal for N i−1 in N i−2 . Hence, by the induction hypothesis on (3), the latter equals cN i−1 (u). This proves (2). The lemma allows to construct from a Bass unit based on an element g a product of conjugates that is central. This under the assumption that ⟨g⟩ is subnormal in G. This is essential in the proof of the following result. Further one makes use of the fact that the class of abelian-by-supersolvable groups is closed under taking subgroups. This property does not hold for the larger class consisting of the strongly monomial groups. Recall that an abelian-by-supersolvable finite group G is strongly monomial by Theorem 3.5.10. Because of Corollary 3.5.4 the primitive central idempotents of ℚG are

394 | 13 Idempotents and central units in group rings determined by strong Shoda pairs (H, K) of subgroups of G. We say that two Shoda pairs (H1 , K1 ) and (H2 , K2 ) of G are equivalent if they determine the same simple component of ℚG, i.e. if ℚGe(G, H1 , K1 ) = ℚGe(G, H2 , K2 ) (see Problem 3.4.3). In case (H1 , K1 ) and (H2 , K2 ) are strong Shoda pairs this holds if and only if e(G, H1 , K1 ) = e(G, H2 , K2 ). Clearly, it is enough to consider only one strong Shoda pair in each equivalence class. This simply is expressed as a complete and non-redundant set of strong Shoda pairs. Theorem 13.1.2 (Jespers, Olteanu, Van Gelder, del Río [120]). Let G be a finite abelianby-supersolvable group such that every cyclic subgroup of order not a divisor of 4 or 6 is subnormal in G. Then, the group generated by the Bass units of ℤG contains a subgroup of finite index in Z(U(ℤG)). Proof. We prove this by induction on the order of the group G. If |G| = 1 then the result is clear. So, assume that the result holds for groups of order strictly less than the order of G. Furthermore, because of the Theorem 7.3.3, the result clearly holds if G is an abelian group. Hence, one also may assume that G is not abelian, i.e. the commutator ̂󸀠 ) ⨁ ℚG G ̂󸀠 . From Problem 3.3.5 subgroup G󸀠 is not trivial. Write ℚG = ℚG(1 − G ̂󸀠 ) is a direct sum of non-commutative simple rings and we know that ℚG(1 − G 󸀠 ̂ 󸀠 clearly ℚG G ≅ ℚ(G/G ) is a commutative group algebra. Let z ∈ Z(U(ℤG)) and write ̂󸀠 ))) and z󸀠󸀠 ∈ U(ℤG G ̂󸀠 ). Note that z󸀠 z󸀠󸀠 = 0 = z󸀠󸀠 z󸀠 . z = z󸀠 + z󸀠󸀠 , with z󸀠 ∈ Z(U(ℤG(1 − G We will prove that some positive power of z is a product of Bass units. Since z is an arbitrary element of the finitely generated abelian group Z(U(ℤG)), the result follows. First we focus on the commutative component. Since G/G󸀠 is abelian, it follows from Theorem 7.3.3 that the Bass units of ℤ(G/G󸀠 ) generate a subgroup of finite index in U(ℤ(G/G󸀠 )). A power of each Bass unit of ℤ(G/G󸀠 ) is the natural image of a Bass unit m of ℤG (see Problem 1.2.3). Hence, we get that z󸀠󸀠 = ∏ri=1 b i for some positive integer m and some Bass units b i in ℤG, where we denote the natural image of x ∈ ℤG in ℤ(G/G󸀠 ) by x. From Lemma 1.2.1 we know that u k,m (g) has finite order if and only if k ≡ ±1 mod | g |. In particular, there is a Bass unit based on g ∈ G of infinite order if and only if the order of g is not a divisor of 4 or 6. Moreover, if a Bass unit is of finite order then it belongs to G (see e.g. the proof of Lemma 1.2.1). Hence, without loss of generality, we may assume that each b i is based on an element of order not a divisor of 4 nor of 6. By the assumptions on G, we can construct central units in ℤG which project to some power of a b i in ℤ(G/G󸀠 ). Indeed, each cN i (b i ) is central in ℤG, where Ni is a subnormal series from ⟨g i ⟩ to G when b i is based on g i . Since ℤ(G/G󸀠 ) is commutative, mi the natural image of cNi (b i ) is a power of b i , say b i . Hence z󸀠󸀠

m⋅lcm(m i :1≤i≤r)

r

= ∏ cNi (b i ) i=1

lcm{m i :1≤i≤r} mi .

13.1 Central subgroups and abelian-by-supersolvable groups |

395

Consequently, one may assume that for some positive integer m we have z󸀠󸀠

m󸀠

s

= ∏ cNj (b j ), j=1

where b j runs through a set of Bass units of ℤG with possible repetition. Therefore, 󸀠 ̂󸀠 , with z󸀠󸀠󸀠 ∈ Z(U(ℤG(1 − G ̂󸀠 ))). z m (∏sj=1 cNj (b j ))−1 = z󸀠󸀠󸀠 + G Since G is abelian-by-supersolvable and hence also strongly monomial by Theorem 3.5.10, we know from Corollary 3.5.4 and Problem 3.3.5 that every primitive central ̂󸀠 ) is of the form e(G, H, K) for a strong Shoda pair (H, K) of idempotent of ℚG(1 − G G with ℚGe(G, H, K) not commutative. Let (H, K) be such a strong Shoda pair of G. Then it is also a strong Shoda pair of H and e(H, H, K) = ε(H, K) is a primitive central idempotent of ℚH. Moreover H ≠ G, and hence the induction hypothesis yields that there exists a subgroup, say A1 , of the group generated by the Bass units in ℤH such that A1 is of finite index in Z(U(ℤH)). Clearly, ℤH ⊆ ⨁e ℤHe, where e runs through all primitive central idempotents of ℚH. As both ℤH and ⨁e ℤHe are ℤ-orders in ℚH, Lemma 4.6.6 and Lemma 4.6.9 yield that Z(U(ℤH)) is of finite index in Z(U(⨁e ℤHe)). Hence, A1 is of finite index in Z(U(⨁e ℤHe)). Since ℤ(1 − ε(H, K)) + ℤHε(H, K) ⊆ Z(⨁e ℤH e ), we thus get that A = A(H, K) = A1 ∩ (ℤ(1 − ε(H, K)) + ℤHε(H, K)) is of finite index in U(ℤ(1− ε(H, K)) ⨁ ℤHε(H, K)), and each element of A is a product of Bass units of ℤH. Theorem 3.5.5 gives a description of the simple components associated with a strong shod pair. It follows that ℚGe(G, H, K) ≅ M[G:N G (K)] (ℚHε(H, K) ∗ (N G (K)/H)) and its center consists of the scalar matrices with diagonal entry in (ℚHε(H, K))N G (K)/H, the fixed subfield of ℚHε(H, K) under the action of N G (K)/H. Note that if α = 1 − ε(H, K) + βε(H, K) ∈ A, with β ∈ ℤH, then α n = 1 − ε(H, K) + β n ε(H, K), for n ∈ N G (K). Hence, α and α n commute and thus the product ∏n∈N G (K) α n is independent of the order of its factors. Since U(ℤHε(H, K)) is a finitely generated abelian group, it is readily verified that {∏n∈N G (K) u n | u ∈ U(ℤHε(H, K))} is of finite index in U((ℤHε(H, K))N G (K)/H ). It follows that B = B(H, K) = { ∏ α n : α ∈ A ∩ (1 − ε(H, K) + ℤHε(H, K)) } n∈N G (K)

is a subgroup of finite index of U(ℤ(1−ε(H, K))+(ℤHε(H, K))N G (K)/H ) and the elements of B are products of Bass units in ℤH. Let 𝛾 = 1−ε(H, K)+δ ∈ B, with δ ∈ (ℤHε(H, K))N G (K)/H and T be a right transversal 󸀠 of N G (K) in G. Since ε(H, K)t ε(H, K)t = 0 for different t, t󸀠 ∈ T, it follows that 𝛾t and 󸀠 𝛾t commute and ∏ 𝛾t = 1 − e(G, H, K) + ∑ δ t ∈ 1 − e(G, H, K) + ℤGe(G, H, K). t∈T

t∈T

396 | 13 Idempotents and central units in group rings Clearly, ∏t∈T 𝛾t corresponds to a central matrix in M[G:N G (K)] (ℚHε(H, K)∗(N G (K)/H)), via the isomorphism ℚGe(G, H, K) ≅ M[G:N G (K)] (ℚHε(H, K) ∗ (N G (K)/H)), with diagonal entry in (ℤHε(H, K))N G (K)/H . From the previous it follows that C = C(H, K) = {∏ 𝛾t : 𝛾 ∈ B} t∈T

is a subgroup of finite index in Z (U(ℤ(1 − e(G, H, K)) + ℤGe(G, H, K))). As each 𝛾 ∈ B is a product of Bass units in ℤH, so is ∏t∈T 𝛾t a product of Bass units in ℤG. The proof can now be completed as follows. Write the central unit ̂󸀠 = ∑ z󸀠󸀠󸀠 e(G, H, K) + G ̂󸀠 = ∏ (1 − e(G, H, K) + z󸀠󸀠󸀠 e(G, H, K)), z󸀠󸀠󸀠 + G (H,K)

(H,K)

where (H, K) runs through a complete and non-redundant set of strong Shoda pairs ̂󸀠 . By conof G so that ℚGe(G, H, K) is not commutative and ∑(H,K) e(G, H, K) = 1 − G 󸀠󸀠 struction of C(H, K), there exists a positive integer m such that 󸀠󸀠

(1 − e(G, H, K) + z󸀠󸀠󸀠 e(G, H, K))m ∈ C(H, K) ̂󸀠 )m󸀠󸀠 and thus also for each (H, K). Hence (z󸀠󸀠󸀠 + G 󸀠

s

󸀠󸀠

z m m = (∏ cNj (b j ))

m󸀠󸀠

̂󸀠 )m (z󸀠󸀠󸀠 + G

󸀠󸀠

j=1

is a product of Bass units in ℤG. One can construct a concrete set of generators for the central units. Corollary 13.1.3 ([120]). Let G be a finite abelian-by-supersolvable group such that every cyclic subgroup of order not a divisor of 4 or 6 is subnormal in G. For each such cyclic subgroup ⟨g⟩, fix a subnormal series Ng from ⟨g⟩ to G. Then, ⟨cNg (b g ) | b g a Bass unit based on g, g ∈ G⟩ is of finite index in Z(U(ℤG)). Proof. Because of Theorem 13.1.2 and since Z(U(ℤG)) is a finitely generated group, it is sufficient to show that if u = b1 b2 ⋅ ⋅ ⋅ b m ∈ Z(U(ℤG)), with each b i a Bass unit, say based on g i ∈ G, then there exists a positive integer l so that u l is a product of cNg (b g )’s, with b g a Bass unit based on g ∈ G. In order to prove this, for each primitive central idempotent e of ℚG, write ℚGe = M n e (D e ), with n e a positive integer and D e a division algebra. If Oe is an order in D e then, by Lemma 4.6.9, U(ℤG) ∩ ∏e GLn e (Oe ) is of finite index in U(ℤG). From Proposition 5.5.1, the central matrices in SLn e (Oe ) are periodic. Let u = b1 b2 ⋅ ⋅ ⋅ b m ∈ Z(U(ℤG)), with each b i a Bass unit based on g i ∈ G. Then there exists a positive integer m󸀠 such that 󸀠

m

u m (∏ cNgi (b i )) i=1

m󸀠

∈ ∏ GLn e (Oe ). e

13.2 Independent units and abelian-by-supersolvable groups

|

397

Let k i be a positive integer such that each cNgi (b i ) is a product of k i conjugates of b i . Then cNgi (b i )e and b i k i e have the same reduced norm. Hence, 󸀠

m

󸀠

u km ∏ cNgi (b i )−m k/k i e ∈ SLn e (Oe ) ∩ Z(GLn e (Oe )), i=1

for k = lcm(k i : 1 ≤ i ≤ m) and thus m

󸀠

󸀠

u km ∏ cNgi (b i )−m k/k i e i=1

is a periodic element in Z(GLn e (Oe )). Consequently, 󸀠

m

󸀠

(u km ∏ cNgi (b i )−m k/k i )

m󸀠󸀠

=1

i=1

for some positive integer m󸀠󸀠 , i.e. 󸀠

󸀠󸀠

u km m ∈ ⟨cNg (b g ) | b g a Bass unit based on g, g ∈ G⟩. As mentioned earlier, the group generated by the elements g ∈ G with ⟨g⟩ subnormal in G is contained in the Fitting subgroup F(G) of G, a characteristic subgroup of G that is nilpotent, say of class n. Hence, in Corollary 13.1.3, for g ∈ F(G) one can consider the subnormal series Ng : ⟨g⟩  ⟨Z1 (F(G)), g⟩  ⋅ ⋅ ⋅  ⟨Z n (F(G)), g⟩ = F(G)  G.

(13.1.1)

For a given Bass unit b based on g, the corresponding unit cN n (b) we simply denote c(b). So we have the following consequence. Corollary 13.1.4. Let G be a finite nilpotent group of class n. Then ⟨c(b) | b a Bass unit in ℤG⟩ is of finite index in Z(U(ℤG)). In [124] a similar construction of a central unit, denoted b(n) , is given in case G is nilpotent of class n. It turns out that b(n) is a power of c(b). As an immediate application one obtains that the set formed by the elements of the form b(n) , with b a the Bass units of G, has finite index in the center of U(ℤG). This is the main result of [124].

13.2 Independent sets of central units and abelian-bysupersolvable groups In this section, we obtain a basis formed by products of Bass units of a free abelian subgroup of finite index in Z(U(ℤG)), for G a finite abelian-by-supersolvable group G such that every cyclic subgroup of order not a divisor of 4 or 6, is subnormal in G. This work is due to Jespers, Olteanu, del Río and Van Gelder [120].

398 | 13 Idempotents and central units in group rings

Lemma 13.2.1. Let G be a finite abelian-by-supersolvable group such that every cyclic subgroup of order not a divisor of 4 or 6 is subnormal in G. Let u, v be units in ℤ ⟨g⟩ for g ∈ G and let N be a subnormal series N0 = ⟨g⟩
N1
⋅ ⋅ ⋅
N m = G. Assume h ∈ G and h = G. denote by N h the h-conjugate of the series N, i.e. N h : N0h = ⟨g h ⟩
N1h
⋅ ⋅ ⋅
N m Then (1) cN (uv) = cN (u)cN (v), and h (2) cN (u h ) = cN (u). N N Proof. Let u, v ∈ ℤ ⟨g⟩. Clearly cN 0 (uv) = uv = c 0 (u)c 0 (v). Again let T i denote a transversal for N i−1 in N i , for i ≥ 1. An induction argument on i yields that N x N x N x N N cN i (uv) = ∏ c i−1 (uv) = ∏ c i−1 (u) c i−1 (v) = c i (u)c i (v), x∈T i

x∈T i

N x x for i ≥ 1, since cN i−1 (u) and c i−1 (v) commute by properties (1) and (2) of Lemma 13.1.1. This proves (1). h N h h Let u ∈ ℤ ⟨g⟩ and h ∈ G. By induction on i we prove that cN i (u ) = c i (u) , the Nh h N h h result then follows. For i = 0 we have c0 (u ) = u = c0 (u) . Let i ≥ 1 we obtain from the induction hypothesis that h N h x N hx yh h = ∏ cN = cN cN i (u ) = ∏ c i−1 (u ) = ∏ c i−1 (u) i−1 (u) i (u) . h

h

x∈T ih

x∈T ih

y∈T i

Let G be a group. For g ∈ G define S g = {l ∈ U(ℤ|g| ) : g is conjugate with g l in G}. In other words, S g is the image of the homomorphism N G (⟨g⟩) → U(ℤ|g| ) : h 󳨃→ l h where l h is the unique element of U(ℤ|g| ) such that g h = g l h . The kernel of this homomorphism is CenG (g). Put S g = ⟨S g , −1⟩. Throughout we assume that transversals of S g in U(ℤ|g| ) contain the identity 1. Theorem 13.2.2 ([120]). Let G be a finite abelian-by-supersolvable group such that every cyclic subgroup of order not a divisor of 4 or 6 is subnormal in G. Let R denote a set of representatives of the ℚ-conjugacy classes of G. For g ∈ R choose a transversal T g of S g in U(ℤ|g| ) containing 1 and for every k ∈ T g \{1} choose an integer m k,g with k m k,g ≡ 1 mod |g|. For every g ∈ R of order not a divisor of 4 or 6, choose a subnormal series Ng from ⟨g⟩ to G in which each term of the series is normalized by N G (⟨g⟩). Then {cNg (u k,m k,g (g)) : g ∈ R, k ∈ T g \ {1}} is a basis for a free abelian subgroup of finite index in Z(U(ℤG)). Proof. By assumption, each element g ∈ G that is not of order a divisor of 4 or 6 belongs to the Fitting subgroup of G. Hence, one can indeed choose a subnormal series

13.2 Independent units and abelian-by-supersolvable groups

|

399

Ng in which each term is normalized by the elements of N G (⟨g⟩) (for example the series in (13.1.1)). We also make these choices in such a way that Nh is the x-conjugate of Ng when h = x−1 g i x with g ∈ R and i coprime with |g|, that is when g and h are

ℚ-conjugate. By Corollary 13.1.3, the set B1 = {cNh (u k,m (h)) : h ∈ G, k, m ∈ ℕ, k m ≡ 1

mod |h|}

generates a subgroup of finite index in Z(U(ℤG)). Let t = φ(|G|). We first prove that B2 = {cNg (u k,t (g)) : g ∈ R, k ∈ T g \ {1}} generates a subgroup of finite index in Z(U(ℤG)). To do so we sieve gradually the list of units in B1 , keeping the property that the remaining units still generate a subgroup of finite index in Z(U(ℤG)), until the remaining units are the elements of B2 . By equation (7.3.1), to generate B1 it is enough to use the Bass units of the form u k,m (h) with h ∈ G, 1 ≤ k < |h| and k m ≡ 1 mod |h|. Hence one may assume that k ∈ U(ℤ|h| ). By (7.3.2), for every Bass unit u k,m (h) we have u k,m (h)i = u k,t (h)j for some positive integers i and j. Thus, by Lemma 13.2.1 (1), units of the form cNh (u k,t (h)) with k ∈ U(ℤ|h| ) generate a subgroup of finite index in Z(U(ℤG)). By the definition of a ℚ-conjugacy class, each h ∈ G is conjugate to some g i , for g ∈ R and (i, |g|) = 1. Hence, by (7.3.3), Lemma 13.2.1 (1) and Lemma 13.2.1 (2), the list of generators can be further reduced by taking only Bass units based on elements of R. Obviously one can exclude k = 1 as this gives the identity. Let g ∈ G be of order n. We claim that if l ∈ S g and k ∈ U(ℤn ) then cNg (u l,t (g k )) has finite order. As u n−l,t (g k ) = u l,t (g k )g −lkt , one may assume without loss of generality that l ∈ S g . By (7.3.3), Lemma 13.2.1 (1), Lemma 13.2.1 (2) and because of the assumption that each term in the subnormal series is normalized by N G (⟨g⟩) we get that cNg (u l i+1 ,t (g k )) = cNg (u l,t (g k ))cNg (u l i ,t (g kl )) = cNg (u l,t (g k ))cNg (u l i ,t (g k )). Then, arguing inductively, we deduce that cNg (u l,t (g k ))i = cNg (u l i ,t (g k )), and in particular cNg (u l,t (g k ))t = cNg (u l t ,t (g k )) = cNg (u1,t (g k )) = 1, by (7.3.1). This proves the claim. With g and n as above, every element of U(ℤn ) is of the form kl with k ∈ T g and l ∈ S g . Using (7.3.3) one has u kl,t (g) = u k,t (g)u l,t (g k ). By the previous paragraph, cNg (u l,t (g k )) has finite order. Hence the system can be reduced by taking only k ∈ T g \ {1}. The remaining units are exactly the elements of B2 . Thus ⟨B2 ⟩ has finite index in Z(U(ℤG)), as desired.

400 | 13 Idempotents and central units in group rings Let B = {cNg (u k,m k,C (g)) : g ∈ R, k ∈ T G \{1}}. Using (7.3.2) once more, one deduces that ⟨B⟩ has finite index in Z(U(ℤG)), since so does ⟨B2 ⟩. It remains to prove that the elements of B are multiplicatively independent. To do so, it is enough to show that the rank of Z(U(ℤG)) coincides with the cardinality of B. It is easy to see that |B| = (∑g∈R |T g |) − |R| and |R| equals the number of ℚconjugacy classes. By construction, [U(ℤ|g| ) : S g ] equals the number of conjugacy classes contained in the ℚ-conjugacy class of g. Furthermore, [S g : S g ] = 1 when g is conjugated to g −1 and [S g : S g ] = 2 when g is not conjugated to g−1 . Therefore |T g | = [U(ℤ|g| ) : S g ] is exactly the number of ℝ-classes contained in the ℚ-conjugacy class of g. Hence |B| equals the number of ℝ-classes minus the number of ℚ-conjugacy classes in G. Because of Corollary 7.1.5 this is the rank of Z(U(ℤG)).

13.3 Large central subgroups and strongly monomial groups An interesting question is whether Theorem 13.1.2 remains valid for many other groups. Even for metacyclic groups this is yet unknown. Jespers and Parmenter therefore introduced the so called generalized Bass units in [123] and, in [119], Jespers, Olteanu and del Río proved that they generate a subgroup of finite index in U(ℤ(G)) for G a strongly monomial finite group. In this section we will prove this result. Recall from Problem 1.2.4 that the definition of a Bass unit 1 − km u k,m (x) = (1 + x + ⋅ ⋅ ⋅ + x k−1 )m + (1 + x + ⋅ ⋅ ⋅ + x n−1 ) n makes sense for any torsion unit x of order n in an associative ring R, where k and m are positive integers with (k, n) = 1 and k m ≡ 1 mod n. In particular, if G is a finite group, M a normal subgroup of G, g ∈ G and k and m positive integers such that gcd(k, |g|) = 1 and k m ≡ 1 mod |g|, then M + ĝ M) = 1 − ̂ M + u k,m (g)̂ M. u k,m (1 − ̂ is an invertible element of ℤG(1− ̂ M)+ℤG ̂ M. As this is an order in ℚG, for each element b = u k,m (1 − ̂ M + ĝ M) there is a positive integer n such that b n ∈ U(ℤG). Let n b denote the minimal positive integer satisfying this condition. The element u k,m (1 − ̂ M + ĝ M)n b = u k,mn b (1 − ̂ M + ĝ M) is called a generalized Bass unit based on g and M with parameters k and m. We let n G,M denote the least common multiple of the n b ’s with b running through all the elements of the form u k,m (1 − ̂ M + ĝ M). Note that if M is the trivial group then this yields a Bass unit. Hence, the following result generalizes the Bass-Milnor Theorem (Theorem 7.3.3). Theorem 13.3.1 ([120]). If G is a strongly monomial finite group then the group generH󸀠 + ĥ H 󸀠 ) for a strong Shoda ated by the generalized Bass units b n b , with b = u k,m (1 − ̂ pair (H, K) of G and h ∈ H, contains a subgroup of finite index in Z(U(ℤG)).

13.3 Central subgroups and strongly monomial groups |

401

H 󸀠 ) + ℤH ̂ H 󸀠 , a ℤ-order Proof. Let (H, K) be a strong Shoda pair of G. Let O1 = ℤ(1 − ̂ ̂ ̂ 󸀠 󸀠 in B1 = ℚ(1 − H ) + ℚH H , and let O2 = ℤ(1 − ε(H, K)) + ℤHε(H, K), a ℤ-order in B2 = ℚ(1 − ε(H, K)) + ℚHε(H, K). Observe that B1 ⊆ Z(ℚG) and, because ε(H, K)̂ H󸀠 = 󸀠 ε(H, K), we have that B2 ⊆ B1 . Since H/H is abelian, Theorem 7.3.3 yields that the Bass units of ℤ(H/H 󸀠 ) generate a subgroup of finite index in the group of (central) units of ℤ(H/H 󸀠 ) ≅ ℤH ̂ H 󸀠 . As a power of each Bass unit of ℤ(H/H 󸀠 ) is the natural image of a Bass unit in ℤH (see Problem 1.2.3) we have that the group generated by the units of the form b = u k,m (1 − ̂ H󸀠 + ĥ H 󸀠 ), with h ∈ H is of finite index in U(O1 ). So, the group generated by the generalized Bass units b n b also is of finite index in U(O1 ). Let A1 denote this central subgroup of U(ℤH). By Lemma 4.6.6, B2 ∩ O1 is a ℤ-order in B2 . Hence, by Lemma 4.6.9, U(O1 ) ∩ U(O2 ) is of finite index in U(O2 ). So, we have shown that A1 ∩ O2 is of finite index in the abelian group U(O2 ). The group N G (K)/H acts via conjugation on O2 . As in the proof of Theorem 13.1.2, it follows that B = {∏g∈N G (K) α g | α ∈ A ∩ O2 } generates a subgroup of finite index in U(ℤ(1 − ε(H, K)) + (ℤHε(H, K))N G (K)/H ). Furthermore, if T is a transversal of N G (H) in G then C = {∏t∈T 𝛾t | 𝛾 ∈ B} generates a subgroup of finite index in Z (U(ℤ(1 − e(G, H, K)) + ℤGe(G, H, K))) . Note that because O2 is a commutative ring, the definition of B is independent of the order of the factors α g . Also the definition of C is independent of the order. Indeed, if 𝛾 ∈ B and t1 , t2 ∈ G then 𝛾 = 1 − ε(H, K) + 𝛾1 ε(H, K) for some 𝛾1 ∈ ℤH. If t1 ≠ t2 then ε(H, K)t1 ε(H, K)t2 = 0, because (H, K) is a strong Shoda pair. It follows that 𝛾t1 and 𝛾t2 commute. To finish the proof let u be an arbitrary element in Z(U(ℤG)). Because, by assumption, G is strongly monomial we get from Corollary 3.5.4 that u = ∑ ue(G, H, K) = ∏(1 − e(G, H, K) + ue(G, H, K)), (H,K)

(H,K)

where (H, K) runs through a complete and non-redundant set of strong Shoda pairs H󸀠 + ĥ H 󸀠 ) are again of this form, since the of G. Note that conjugates of b n b = u k,m (1 − ̂ class of strong Shoda pairs is closed under conjugation in G. It follows from the above that some power of u is a product of generalized Bass units. Since, Z(U(ℤG)) is finitely generated the result follows. Remark 13.3.2. The proof of Theorem 13.3.1 shows that the result remains valid if each n b is replaced by some multiple. For example, we can replace each n b , with b of the H󸀠 + ĥ H 󸀠 ), by n H,H 󸀠 . form u k,m (1 − ̂

402 | 13 Idempotents and central units in group rings

13.4 Independent sets of central units and strongly monomial groups In this section we consider strongly monomial finite groups G which have a complete and non-redundant set of strong Shoda pairs (H, K) with the property that [H : K] is a prime power. For this class of strongly monomial groups, Jespers, Olteanu, del Río and Van Gelder in [119] constructed a multiplicatively independent set of elements of Z(U(ℤG)) that generates a subgroup of finite index. In this section this result will be proved. Examples of the groups under consideration are metacyclic groups of the type C q m ⋊ C p n , with p and q different primes and such that C p n acts faithfully on C q m . The class of strongly monomial groups G with a complete and non-redundant set of strong Shoda pairs (H, K) with the property that [H : K] is a prime power is a wider class. For example, the alternating group A4 of degree 4 is such a group that is not metacyclic (and not nilpotent). Note however that not all strongly monomial finite groups have only strong Shoda pairs with prime power index. One can show that all strong Shoda pairs of the dihedral group D2n (respectively, the quaternion group Q4n ) have prime power index if and only if n is a power of a prime (respectively, n is a power of 2) (see Problem 13.4.1). For a strongly monomial finite group we begin with determining the torsion-free rank of Z(U(ℤG)) in terms of strong Shoda pairs. Theorem 13.4.1 ([119]). Let G be a finite strongly monomial group. Then the rank of Z(U(ℤG)) equals φ([H : K]) − 1) , ∑( k [N G (K) : H] (H,K) (H,K) where (H, K) runs through a complete and non-redundant set of strong Shoda pairs of G and if H = ⟨h, K⟩ then 1, 2,

k(H,K) = {

if hh n ∈ K for some n ∈ N G (K); otherwise.

Proof. Corollary 3.5.4 and Theorem 3.5.5 give the following description of the Wedderburn decomposition of ℚG ≅ ⨁ ℚGe(G, H, K) ≅ ⨁ M[G:N K (G)] (ℚ(ζ[H:K] ) ∗ N K (G)/H), (H,K)

(H,K)

with (H, K) running through a complete and non-redundant set of strong Shoda pairs of G. Hence, the torsion-free rank of Z(U(ℤG)) is the sum of the torsion-free ranks of the groups of units of orders in the centers of the simple components. Fix a strong Shoda pair (H, K) of G and let N = N G (K), let F = ℚ(ζ[H:K] )N/H , the center of the simple component M[G:N] (ℚ(ζ[H:K] ) ∗ N/H), and let O be an order in F. Clearly, [ℚ(ζ[H:K] ) : ℚ] φ([H : K]) = . [F : ℚ] = [ℚ(ζ[H:K] ) : F] [N : H]

13.4 Independent units and strongly monomial groups |

403

Since F is a Galois extension of ℚ, the field F is either totally real or totally complex. By Dirichlet’s Unit Theorem (5.2.4), the rank of U(O) is kφ([H:K]) − 1 with k󸀠(H,K) = 1 if 󸀠 (H,K) [N:H] 󸀠 F is totally real, and k(H,K) = 2 if F is totally complex. Therefore, it is enough to prove that k(H,K) = k󸀠(H,K) . Observe that F is totally real, if and only if F is contained in the −1 maximal real subfield ℚ(ζ[H:K] + ζ[H:K] ) of ℚ(ζ[H:K] ). This precisely happens when the Galois group N/H contains complex conjugation, or equivalently if hh n ∈ K for some n ∈ N and h is such that H = ⟨h, K⟩. Therefore k󸀠(H,K) = 1 if and only if k(H,K) = 1. Otherwise k󸀠(H,K) = 2 = k(H,K) . Thus the result follows. In the remainder of the section we assume that G is a strongly monomial finite group G which has a complete and non-redundant set of strong Shoda pairs (H, K) with the property that [H : K] is a prime power. As in the proof of Theorem 13.3.1, to discover a multiplicative independent set of units of Z(U(ℤG)) that generates a subgroup of finite index, it is sufficient to do this for each order ℤGe(G, H, K) + ℤ(1 − e(G, H, K)), with (H, K) a strong Shoda pair such that [H : K] is a prime power. Furthermore, the center of ℤGe(G, H, K) + ℤ(1 − e(G, H, K)) and ℤ[ζ[H:K] ]N G (K)/H + ℤ(1 − e(G, H, K)) are both orders in the center of ℚGe(G, H, K) + ℚ(1 − e(G, H, K)), and hence, because of Lemma 4.6.9, their unit groups are commensurable, i.e. their intersection has finite index in both of them. Hence, one needs to determine units in ℤG projecting to units in ℤ[ζ[H:K] ]N G (K)/H and trivially to the other components. By σ r we denote the automorphism of ℚ(ζ p n ) that maps ζ p n onto ζ pr n . Since, by assumption, [H : K] equals a prime power, say p n , we know from Theorem 1.6.3 and Theorem 1.6.7 that Aut(H/K) is cyclic, unless p = 2 and n ≥ 3 in which case Aut(H/K) = ⟨σ5 ⟩ × ⟨σ−1 ⟩. Often we will abuse notation and identify N G (K)/H with Gal (ℚ(ζ p n )/ℚ(ζ p n )N G (K)/H ) and with a subgroup of U(ℤ/[H : K]ℤ). Since Gal(ℚ(ζ p n )/ ℚ(ζ p n )N G (K)/H ) is a subgroup of Aut(H/K), it follows that N G (K)/H is either ⟨σ r ⟩ for some r or ⟨σ r ⟩ × ⟨σ−1 ⟩ for some r ≡ 1 mod 4. For a subgroup A of Gal((ℚ(ζ p n )/ℚ) and u ∈ ℚ(ζ p n ) put π A (u) = Nℚ(ζ pn )/ℚ(ζ pn )A (u) = ∏ σ(u) = ∏ σ i (u). σ∈A

i∈A

Recall that the cyclotomic units based on a primitive root of unity ξ of order n is an a −1 , with a ∈ U(ℤ/nℤ). We have agreed that η a (1) = 1. element of the form η a (ξ) = ξξ −1 Without specific reference, throughout this section we will make use of the properties of cyclotomic units stated in Problem 1.2.1. Lemma 13.4.2. Let p be a prime number and let A be a subgroup of Aut(⟨ζ p n ⟩). Let T be a transversal of ⟨A, σ−1 ⟩ in U(ℤ/p n ℤ) and assume 1 ∈ T. Then, the set {π A (η k (ζ p n )) | k ∈ T \ {1}} is multiplicative independent and generates a subgroup of finite index in the group U(ℤ[ζ p n ]A ).

404 | 13 Idempotents and central units in group rings Proof. From the comments given before the lemma, we know that A = ⟨σ r ⟩ or A = ⟨σ r ⟩ × ⟨σ−1 ⟩. In both cases, set l = |⟨σ r ⟩|. The arguments given at the end of proof Theorem 13.4.1 show that the unit group n−1 (p−1) U(ℤ[ζ p n ]A ) has rank p ld −1 = |T|−1, where d = 2 if −1 ∈ ̸ ⟨r⟩ and d = 1 otherwise. Moreover ld = |⟨A, σ−1 ⟩|. n Let π = π A . From Corollary 6.2.4, we know that {η k (ζ p n ) | 1 < k < p2 , p ∤ k} generates a free abelian subgroup of finite index in U(ℤ[ζ p n ]). Hence, for every unit u of ℤ[ζ p n ]A we get that u m = ∏hi=1 η k i (ζ p n ) for some integers m, k1 , . . . , k h . Then, u m|A| = π(u m ) = ∏hi=1 π(η k i (ζ p n )). So, {π(η k (ζ p n )) : k ∈ U(ℤ/p n ℤ)} generates a subgroup of finite index in U (ℤ[ζ p n ]A ). First assume d = 1 (i.e. −1 ∈ ⟨r⟩). One of the mentioned properties of cyclotomic units says that η r t i (ζ p n ) = η i (ζ p n )η r t (ζ pi n ), t

for i ∈ T and 0 ≤ t ≤ l − 1. Because π(η r t (ζ pi n )) = π(η r t (ζ pirn )), for 0 ≤ t ≤ l − 1, one deduces that t

(π(η r t (ζ pi n )))2 = π(η r t (ζ pi n ))π(η r t (ζ pirn )) = π(η r2t (ζ pi n )). An induction argument and the properties of cyclotomic units yield that π(η r t (ζ pi n )) is periodic. Consequently, {π(η i (ζ p n )) : i ∈ T} generates a subgroup of finite index in U (ℤ[ζ p n ]A ). Second, assume d = 2 (i.e. −1 ∈ ̸ ⟨r⟩). Let J = T ∪ (−T). Then J is a transversal of ⟨r⟩ in U(ℤ/p n ℤ). The same arguments as above yield that {π(η k (ζ p n )) : k ∈ J} generates a subgroup of finite index in U (ℤ[ζ p n ]A ). If i ∈ T then, by the properties of cyclotomic units, π(η−i (ζ p n )) = π(−ζ p−in )π(η i (ζ p n )) and clearly π(−ζ p−in ) is of finite order. Thus {π(η i (ζ p n ))} : i ∈ T} generates a subgroup of finite index in U(ℤ[ζ p n ]A ). Hence, in both cases, {π(η i (ζ p n ))} : i ∈ T} generates a subgroup of finite index in U(ℤ[ζ p n ]A ). Since η1 (ζ p n ) = 1, one may exclude k = 1. Therefore, the cardinality of the set of these generators is |T| − 1 and thus it is equal to the rank of U(ℤ[ζ p n ]A ). So, this set is independent and the result follows. Because of the natural isomorphism ℚ(H/K) ≅ ℚH ̂ K, the following Lemma is a translation of Lemma 7.3.2. It gives a direct link between generalized Bass units and cyclotomic units.

13.4 Independent units and strongly monomial groups |

405

Lemma 13.4.3. Let H be a finite group, K a subgroup of H and g ∈ H such that H/K = ⟨gK⟩. Put H = {L ≤ H | K ≤ L}. For every L ∈ H, fix a linear representation ρ L of H with kernel L and ρ L (g) = ζ[H:L] (its linear extension to ℚL is also denoted by ρ). Assume L ∈ H and M is a subgroup of H. Let l = |L ∩ M|, t = [M : L ∩ M] and let k and m be positive integers such that (k, t) = 1 and k m ≡ 1 mod |gu| for every u ∈ M. Let n H,K be as defined before Theorem 13.3.1. Then, K+1−̂ K)) = η k (ρ L (g)t )lmn H,K . ∏ ρ L (u k,mn H,K (gu ̂ u∈M

Let H be a finite group and K a subgroup of H such that H/K = ⟨gK⟩ is a cyclic group of order p n . It follows that the subgroups of H/K form a chain, hence H = {L | K ≤ L ≤ n−j H} = {H j = ⟨g p , K⟩ : 0 ≤ j ≤ n}. A crucial property to prove the next result. Let k be a positive integer not divisible by the prime number p and let r be an arbitrary integer. For every 0 ≤ j ≤ s ≤ n construct recursively the following products of generalized Bass units of ℤH: c ss (H, K, k, r) = 1, and, for 0 ≤ j ≤ s − 1, c sj (H, K, k, r) = ( ∏ u k,o pn (k)n H,K (g rp

n−s

ĥ K+1−̂ K))

p s−j−1

h∈H j s−1

j−1

l=j+1

l=0

(∏ c sl (H, K, k, r)−1 )(∏ c l

s+l−j

(H, K, k, r)−1 ).

Recall that for a positive integer l and each integer k coprime to l, o l (k) denotes the multiplicative order of k modulo l. Proposition 13.4.4 ([108, 119]). Let H be a finite group and K a subgroup of H such that H/K = ⟨gK⟩ is cyclic of order p n , with p a prime number. Let H = {L ≤ H | K ≤ L} = n−j {H j = ⟨g p , K⟩ : 0 ≤ j ≤ n}. Let k be a positive integer coprime with p and let r be an arbitrary integer. Then ρ H j1 (c sj (H, K, k, r))

η k (ζ pr s−j )o pn (k)p ={ 1,

s−1

n H,K ,

if j = j1 ; if j ≠ j1 .

(13.4.1)

for every 0 ≤ j, j1 ≤ s ≤ n. Proof. This is proved by a double induction, first on s and, for fixed s, on s − j The minimal cases s = 0 and s = j are obvious. So, assume that s > 1, j < s and the result holds for s1 < s and any 0 ≤ j, j1 ≤ s1 and for s fixed and j1 > j. n−s K+1−̂ K). By Lemma 13.4.3, if 0 ≤ i ≤ j then Let u(k) = ∏h∈H j u k,o pn (k)n H,K (g rp h ̂ ρ H i (u(k)) = η k (ρ H i (g p

n−s

p j−i

)p )o pn (k)p n H,K = η k (ζ p s−i )o pn (k)p n H,K , j−i

i

i

(13.4.2)

406 | 13 Idempotents and central units in group rings and if j ≤ i ≤ n then ρ H i (u(k)) = η k (ρ H i (g p

n−s

))o pn (k)p n H,K = η k (ζ p s−i )o pn (k)p n H,K . j

j

(13.4.3)

By (13.4.2) and the induction hypothesis on s, s+i−j

ρ H j1 (c i

η (ζ s−j )o pn (k)p (H, K, k, r)) = { k p 1,

s+i−j−1

n H,K

= ρ H j1 (u(k))p

s−j−1

,

if j1 = i < j; if j1 ≠ i < j. (13.4.4)

By (13.4.3) and the induction hypothesis on s − j, η (ζ s−i )o pn (k)p ρ H j1 (c si (H, K, k, r)) = { k p 1,

s−1

n H,K

= ρ H j1 (u(k))p

s−j−1

,

if j < i = j1 ; if j < i ≠ j1 (13.4.5)

Now, combining (13.4.2), (13.4.4) and (13.4.5) we have ρ H (u(k))p = η k (ζ p s−j )o pn (k)p n H,K , { { { j s−j−1 p s ρ H j1 (c sj1 (H, K, k, r))−1 = 1, ρ H j1 (c j (H, K, k, r)) = {ρ H j1 (u(k)) { s−j−1 { s+j −j ρ (u(k))p ρ H j1 (c j1 1 (H, K, k, r))−1 = 1, { H j1 s−j−1

s−1

if j = j1 ; if j < j1 ; if j > j1 ;

as desired. We are now in a position to state the main result of this section. As before, for a strong Shoda pair (H, K) of G, we identify N G (K)/H with a subgroup of U(ℤ/[H : K]ℤ) . Theorem 13.4.5. Let G be a strongly monomial finite group that has a complete and non-redundant set S of strong Shoda pairs (H, K) with the property that each [H : K] has m(H,K) . For every (H, K) ∈ S, let T K be a right transversal of prime power, say [H : K] = p(H,K) N G (K) in G, let I(H,K) be a transversal for ⟨N G (K)/H, −1⟩ in U(ℤ/[H : K]ℤ), and assume 1 ∈ I H,K . Then, m

{ ∏ ∏ c0 (H,K) (H, K, k, x)t : (H, K) ∈ S, k ∈ I(H,K) \ {1}} t∈T K x∈N G (K)/H

is a multiplicative independent set that generates a subgroup of finite index in Z(U(ℤG)). Proof. Fix (H, K) ∈ S. Set N = N G (K), T = T K , I = I(H,K) , m = m(H,K) and p = p(H,K) . It is sufficient to prove that {∏ ∏ c0m (H, K, k, x)t : k ∈ I \ {1}} t∈T x∈N/H

is an independent set that generates a subgroup of finite index in 1 − e(G, H, K) + Z(U(ℤGe(G, H, K))). Recall that e(G, H, K) = ∑t∈T ε(H, K). Therefore, it is sufficient to show that {u k = ∏ c0m (H, K, k, x) : k ∈ I \ {1}}, x∈N/H

13.4 Independent units and strongly monomial groups |

407

with k ∈ I \ {1}, is an independent set that generates a subgroup of finite index in the center of 1 − ε(H, K) + U(ℤNε(H, K). Indeed, for then each u k is of the form 1 − 󸀠 ε(H, K) + v k ε(H, K) for some v k ∈ ℤN and u−1 k = 1 − ε(H, K) + v k ε(H, K) for some v󸀠k ∈ ℤN. Then, w k = ∏t∈T u tk = 1 − e(G, H, K) + ∑t∈T v tk ε(H, K)t is a unit in the center of 1 − e(G, H, K) + ℤGe(G, H, K), since the elements of {ε(H, K)t : t ∈ T} are mutually orthogonal idempotents that also are orthogonal to 1−e(G, H, K). So, {w k : k ∈ I\{1}} is an independent set because so is {u k : k ∈ I \ {k}} and it generates a subgroup of finite index in the center of 1 − e(G, H, K) + U(ℤGe(G, H, K)). So, without loss of generality, we may assume that N = G, i.e. K is normal in G and T = {1}. First we deal with the case that H = G. Then ℚGe(G, G, K) ≅ ℚ(ζ p m ). Consider the algebra ℚGe(G, G, K) + ℚ(1 − e(G, G, K)) as a subalgebra of ℚG ̂ K + ℚ(1 − ̂ K). m m (k)p m−1 n G,K o p By Proposition 13.4.4, the elements c0 (G, K, k, 1) project to η k (ζ p m ) in the simple component ℚ(ζ p m ) and trivially in all other components. From Lemma 13.4.2 we know that the set {η k (ζ p m ) | k ∈ I \ {1}} is multiplicative independent and generates a subgroup of finite index in U(ℤ[ζ p m ]). Hence, it follows that {c0m (G, K, k, 1) | k ∈ I \ {1}} is a multiplicative independent set that generates a subgroup of finite index in 1 − e(G, G, K) + U(ℤGe(G, G, K)). So, for the remainder of the proof we may assume that G ≠ H. Consider the non-commutative simple component ℚGe(G, H, K) ≅ ℚHε(H, K) ∗ G/H. Its center is (ℚHε(H, K))G/H ≅ ℚ(ζ p m )G/H . Consider the commutative algebra (ℚHε(H, K))G/H + ℚ(1 − ε(H, K)) as a subalgebra of ℚH ̂ K + ℚ(1 − ̂ K). Since H/K is a cyclic p-group, G/H = ⟨σ r ⟩ or G/H = ⟨σ r ⟩ × ⟨σ−1 ⟩ for some r. Set |⟨σ r ⟩| = l. By Lemma 13.4.2, the set {π (η k (ζ p m )) : k ∈ I \ {1}} is a multiplicative independent set that generates a subgroup of finite index in U (ℤ[ζ p m ]G/H ), where π = π G/H .

If G/H is cyclic then, by Proposition 13.4.4, for each k ∈ I \ {1} the element c0m (H, K, k, 1)c0m (H, K, k, r) ⋅ ⋅ ⋅ c0m (H, K, k, r l−1 ) projects to π(η k (ζ p m ))o pm (k)p n H,K in the component ℚ(ζ p m )G/H and trivially in all other components of ℚH. Hence, the set m−1

{c0m (H, K, k, 1)c0m (H, K, k, r) ⋅ ⋅ ⋅ c0m (H, K, k, r l−1 ) : k ∈ I \ {1}} is multiplicative independent and generates a subgroup of finite index in the group Z(U(ℤGe(G, H, K) + ℤ(1 − e(G, H, K)))), as desired. If G/H is not cyclic, then the elements l−1

1

∏ ∏ c0m (H, K, k, r i (−1)j ) i=0 j=0

408 | 13 Idempotents and central units in group rings project to a power of π(η k (ζ p m )) in the component ℚ(ζ p m )G/H and trivially in all other components of ℚH. Hence, also in this case, we find a multiplicative independent set l−1

1

{∏ ∏ c0m (H, K, k, r i (−1)j ) : k ∈ I \ {1}} i=0 j=0

that generates a subgroup of finite index in Z(U(ℤGe(G, H, K) + ℤ(1 − e(G, H, K)))), again as desired.

Problems 13.4.1. Let n be a positive integer. Prove that the dihedral group D2n has a complete set of strong Shoda pairs (H, K) with [H : K] a prime power if and only if n is a prime power and the quaternion group Q4n has this property if and only if n is a power of 2.

13.5 Primitive idempotents and nilpotent groups Let G be a nilpotent finite group. Recall that such a group is strongly monomial and thus, by Corollary 3.5.4, the primitive central idempotents of ℚG are the elements of the form e(G, H, K) with (H, K) a strong Shoda pair. In this section we give a construction of a set of matrix units {E ij : 1 ≤ i, j ≤ n} of ℚGe ≅ M n (D) (with D a division algebra), that is 1 = ∑1≤i≤n E ii and E ij E kl = δ j,k E il . In particular, a complete set of primitive idempotents is described, in particular the irreducible ℚG-modules are obtained. This is work due to Jespers, Olteanu and del Río in [118]. The following lemma gives a method to produce a complete set of orthogonal primitive idempotents of a classical crossed product (L/F, τ) with trivial twisting τ, i.e. τ(g, h) = 1, for every g, h ∈ G. Here L/F is a Galois extension of a field F of characteristic zero and G = Gal(E/F). Recall that (L/F, 1) = ⊕g∈G Lg and from Theorem 2.6.3 we know that (L/F, 1) ≅ M n (F), with n = [L : F]. Therefore, a complete set of orthogonal primitive idempotents of (L/F, 1) contains n idempotents. Lemma 13.5.1. Let A = (L/F, 1), a classical crossed product with trivial twisting. Let 1 G = Gal(L/F) and let e = |G| ∑g∈G g. If x1 , . . . , x|G| are non-zero elements in L then the x x 1 |G| conjugates e , . . . , e form a complete set of orthogonal primitive idempotents of A if and only if TrL/F (x i x−1 j ) = 0 for all i ≠ j. Proof. Clearly e is an idempotent of A. Moreover, ge = e for every g ∈ G. Therefore, if −1 1 1 1 1 TrL/F (x)e. Thus, x ∈ L then exe = |G| ∑g∈G gxe = |G| ∑g∈G x g ge = |G| ∑g∈G x g e = |G| −1 if x ∈ L then e and xex are orthogonal if and only if TrL/F (x) = 0. Hence, the result follows.

13.5 Primitive idempotents and nilpotent groups |

409

If L/F is a cyclic extension of degree n and F contains a primitive n-th root of unity then L is the splitting field over F of an irreducible polynomial X n − a ∈ F[X] for some a ∈ F (see e.g. [145, Theorem VI.6.2]). Let u ∈ L with u n = a. Then x1 = 1, x2 = u, . . . , x n = u n−1 satisfy the conditions of Lemma 13.5.1. Indeed, the minimal polynomial of u i over F for 1 ≤ i < n is of the form X n/d − a i/d for d = gcd(n, i) and therefore TrL/F (u i ) = [L : F(u i )]TrF(u i )/F (u i ) = 0 and similarly TrL/F (u−i ) = 0. So, this yields natural examples satisfying the conditions of Lemma 13.5.1. On the other hand, it is easy to give an example where there are no elements x1 , . . . , x n satisfying the conditions of Lemma 13.5.1. Let L = ℚ(ζ7 ) and F = ℚ(√−7) and consider the classical crossed product (L/F, 1) of degree 3. If x1 , x2 , x3 satisfy the −1 = x x −1 are non-zero elements of conditions of Lemma 13.5.1 then u = x2 x−1 1 2 1 and u L with zero trace over F. It easily follows that then the minimal polynomial of u over F is of the form X 3 − a for some a ∈ F (see Problem 13.5.1). Hence F contains a primitive third root of unity, a contradiction. The second technical lemma needed is to be able to deal with the building blocks in the proof of Theorem 13.5.3. Note that the different cases are not mutually exclusive. Recall that for a integer prime p we denote by v p (n) the valuation of n at p, that is p v p (n) is the maximum p-th power dividing n. Lemma 13.5.2. Let G be a finite p-group. If G has a maximal abelian subgroup which is cyclic and normal in G then G is isomorphic to one of the following groups: n k P1 = ⟨a, b | a p = b p = 1, b−1 ab = a r ⟩, with either v p (r − 1) = n − k or p = 2 and r ≢ 1 mod 4, n k P2 = ⟨a, b, c | a2 = 1, b2 = 1, c2 = 1, bc = cb, b−1 ab = a r , c−1 ac = a−1 ⟩, with r ≡ 1 mod 4, n k n−1 P3 = ⟨a, b, c | a2 = 1, b2 = 1, c2 = a2 , bc = cb, b−1 ab = a r , c−1 ac = a−1 ⟩, with r ≡ 1 mod 4. Proof. By assumption, the finite p-group G has a maximal abelian subgroup A that is cyclic, say it is generated by a, and it is normal in G. Put |A| = p n . Since A is maximal abelian in G, the kernel of the conjugation action of G on A is isomorphic to a subgroup of Aut(A). Recall again from Theorem 1.6.5 and Theorem 1.6.7 that if either p is odd or p = 2 and n ≤ 2 then Aut(A) is cyclic and otherwise Aut(A) = ⟨σ5 ⟩ × ⟨σ−1 ⟩, where σ r is the automorphism of A given by σ r (a) = a r . First we deal with the case that G/A is cyclic. So, G has a presentation of the form G = ⟨a, b | a p = 1, b p = a s , b−1 ab = a r ⟩, n

k

k

(13.5.1)

k

with p n | r p − 1 and p n | s(r − 1). Hence n ≤ v p (r p − 1) and v p (s) ≥ n − v p (r − 1). If p i ≥ 3 then (1 + xp i )p ≡ 1 + xp i+1 mod p i+2 for every i ≥ 1 and x ∈ ℤ. It follows i i−1 that if either p is odd or p = 2 and r ≡ 1 mod 4 then v p (r p − 1) = v p (r p − 1) + 1, for every i ≥ 1. Furthermore, from the assumption that A is maximal abelian in G, one obtains that a r

p k−1

= ab

p k−1

≠ a and hence r p

k−1

k

≢ 1 mod p n . Thus, n ≤ v p (r p − 1) =

410 | 13 Idempotents and central units in group rings v p (r p

k−1

k

− 1) + 1 ≤ n and hence v p (r p − 1) = n and v p (r − 1) = n − k. Consequently, k

v p (s) ≥ n − v p (r − 1) = k = v p (

k rp − 1 ) = v p (1 + r + r2 + ⋅ ⋅ ⋅ + r p −1 ) . r−1

Since (r, p) = 1, there thus exists an integer x such that xr(1 + r + r2 + ⋅ ⋅ ⋅ + r p p k−1

p k−1

k

−1 )

≡ −s

mod p n . Hence, (a x b)p = b p a rx(r +⋅⋅⋅+r+1) = a s+rx(r +⋅⋅⋅+1) = 1. Replacing b by a x b in (13.5.1), one obtains a presentation of the type P1 . Note that it also follows from the previous that v p (r − 1) = n − k unless p = 2 and r ≢ 1 mod 4. Assume p = 2 and k

k

2k

r ≢ 1 mod 4. Then, v2 (s) ≥ n − v2 (r − 1) = n − 1. Because a = b−2 ab2 = a r and 2k k k thus a r −1 = 1 we also have that v2 (r2 − 1) ≥ n. Hence, v2 (1 + r + r2 + ⋅ ⋅ ⋅ + r2 −1 ) = k

2k

k

−1 v2 ( r r−1 ) ≥ n−1. If v2 (s) ≥ n then G ≅ P1 . If v2 (s) = v2 (1+r+r2 +⋅ ⋅ ⋅+r2 −1 ) = n−1 then k (ab)2 = 1. Replacing b by ab we obtain again that G ≅ P1 . Otherwise, v2 (s) = n − 1 k k k−1 and v2 (1 + r + r2 + ⋅ ⋅ ⋅ + r2 −1 ) ≥ n. Therefore v2 (r2 − 1) > n. Recall that v2 (r2 − 1) < n k k−1 because A is maximal abelian in G. Hence, v2 (r2 − 1) > n > v2 (r2 − 1). The first part of the proof applied to ⟨a, b2 ⟩ yields that k = 1. Note that then v2 (r + 1) = n. So, G is the quaternion group of order 2n+1 and thus is of type P3 (with defining k = 0). Second, assume G/A is not cyclic. So p = 2 and G/⟨a⟩ = ⟨b⟩ × ⟨c⟩, with c acting as inversion on ⟨a⟩ and c2 = 1. This provides a presentation of G of the form k

G = ⟨a, b, c | a2 = 1, b2 = a s , ca = a−1 c, n

k

b−1 cb = a i c, b−1 ab = a r , c2 = 1⟩

(13.5.2)

or G = ⟨a, b, c | a2 = 1, b2 = a s , ca = a−1 c, n

k

b−1 cb = a i bc, b−1 ab = a r , c2 = a2

n−1

⟩

(13.5.3)

Note that in both cases, (bc)−1 a(bc) = c−1 a r c = a r . Since r is odd, we have that either r ≡ 1 mod 4 or −r ≡ 1 mod 4. Replacing b by bc if necessary, the first part of the proof applied to the group ⟨a, b⟩ yields that we may assume r ≡ 1 mod 4, v2 (r − 1) = n − k, v2 (r2 − 1) = n and b2 = 1. Then, c = b−2 cb2 = a i(1+r+⋅⋅⋅+r k

k i(1 + r + ⋅ ⋅ ⋅ + r2 −1 )

k

2k

k

k

2k −1

)c

and

2k

−1 −1 i r r−1 . As v2 ( r r−1 )

| = = k, we get v2 (i) ≥ n − k = v2 (r − 1). Hence, there exists an integer j such that j(r − 1) + i ≡ 0 mod 2n . It then is easy to verify that the commutator of b and a j c is 1. So, replacing c by a j c if needed, we may assume that b and c commute and we obtain the presentation of P2 , if c2 = 1, and the n−1 presentation of P3 , if c2 = a2 . thus 2n

Let p be a prime integer. For shortness, the Sylow p-subgroup of a finite nilpotent group G we call the p-part of G and the product of all other Sylow subgroups is called its p󸀠 -part. Theorem 13.5.3 ([118]). Let (H, K) be a strong Shoda pair of a nilpotent finite group G. Set e = e(G, H, K) and ε = ε(H, K). Let a ∈ H be such that H/K = ⟨a⟩ and set N = N G (K). The 2-parts of N/K and H/K are denoted N2 /K and H2 /K = ⟨a2 ⟩ respectively; and their 2󸀠 -parts are denoted N2󸀠 /K and H2󸀠 /K = ⟨a2󸀠 ⟩ respectively.

13.5 Primitive idempotents and nilpotent groups |

411

Then, ⟨a2󸀠 ⟩ has a cyclic complement in N2󸀠 /K, say ⟨b2󸀠 ⟩. A complete set of orthoĝ onal primitive idempotents of ℚGe consists of the conjugates of b 2󸀠 β 2 ε by the elements [N2󸀠 :H2󸀠 ]−1 2 of T2󸀠 T2 T G/N , where T2󸀠 = {1, a2󸀠 , a2󸀠 , . . . , a2󸀠 }, T G/N denotes a left transversal of N in G and β2 and T2 are given according to the cases below. ̂2 . Moreover, if M2 /K is cyclic (1) If H2 /K has a complement M2 /K in N2 /K then β2 = M then there exists b2 ∈ N2 such that N2 /K is given by the following presentation 2k

n

⟨a2 , b2 | a2 2 = b2 = 1, a2 b2 = a2 r ⟩, and if M2 /K is not cyclic, there exist b2 , c2 ∈ N2 such that N2 /K is given by the following presentation 2k

⟨a2 , b2 , c2 | a2 2 = b2 = 1, c2 2 = 1, a2 b2 = a2 r , a2 c2 = a2 −1 , (b2 , c2 ) = 1⟩, n

n−2

with r ≡ 1 mod 4 (or equivalently, a2 2 is central in N2 /K). Then n−2 k (i) T2 = {1, a2 , a22 , . . . , a22 −1 }, if a2 2 is central in N2 /K and M2 /K is cyclic; and k−1 n−2 n−2 k−1 n−2 (ii) T2 = {1, a2 , a22 , . . . , a22 −1 , a22 , a22 +1 , . . . , a22 +2 −1 }, otherwise. (2) if H2 /K has no complement in N2 /K then there exist b2 , c2 ∈ N2 such that N2 /K is given by the following presentation n

⟨a2 , b2 , c2 | a2 2 = b2

2k

= 1, c2 2 = a2 2

n−1

,

a2 b2 = a2 r , a2 c2 = a2 −1 , (b2 , c2 ) = 1⟩, with r ≡ 1 mod 4. Set m = [H2󸀠 : K]/[N2󸀠 : H2󸀠 ]. Then k (i) β2 = ̂ b2 and T2 = {1, a2 , a22 , . . . , a22 −1 }, if either H2󸀠 = K or the order of 2 modulo m is odd and n − k ≤ 2 and n−2 n−2 1+xa22 +ya22 c2 (ii) β2 = ̂ b2 and 2

T2 = {1, a2 , a22 , . . . , a22

k

−1

, c2 , a2 c2 , a22 c2 , . . . , a22 [N2󸀠 :H2󸀠 ]

x, y ∈ ℚ[a2󸀠

k

−1

c2 } with

, a22 + a2−2 ], k

k

satisfying (1 + x2 + y2 )ε = 0, if H2󸀠 ≠ K and either the order of 2 modulo m is even or n − k > 2. (For concrete values of x and y see Lemma 11.5.3.) Proof. Let T = T G/N a left transversal for N in G. Throughout the proof the properties stated in Theorem 3.5.5 will be used. In particular, ℚGe ≅ M n (ℚ(ξ) ∗ (N/H)) ≅ M n (ℚNε), with ξ a primitive root of unity of order [H : K] and e = ∑t∈T ε t , a sum of orthogonal idempotents. Moreover, the conjugates ε t , with t ∈ T correspond to the “diagonal” elements E tt under the isomorphism ℚGe ≅ M G/N (ℚNε). Therefore, it is sufficient to describe a complete set of orthogonal primitive idempotents for ℚNε = ℚH ε ∗ N/H and then add to these their T-conjugates in order to obtain a complete set of primitive idempotents of ℚGe. So, for the remainder of the proof we may assume that N = G, i.e. K is normal in G and thus e = ε and T = {1}. As the natural isomorphism ℚG ̂ K ≅ ℚ(G/K) maps ε

412 | 13 Idempotents and central units in group rings to ε(H/K, 1), we also may assume that K = 1 and hence H = ⟨a⟩ is a cyclic maximal abelian subgroup of G, which is normal in G and e = ε = ε(H, 1). If G = H then ℚGe is a field, T2 = T2󸀠 = {1} and b2󸀠 = β2 = 1; hence the result follows. So, in the remainder of the proof we assume that G ≠ H. The map aε 󳨃→ ξ induces an isomorphism f : ℚHε → ℚ(ξ), where ξ is a primitive |H|-root of unity. From Theorem 3.5.5, one also obtains a description of ℚGe as a classical crossed product (ℚ(ξ)/F, τ), where F ≅ ℚ(ζ|H| )G is the image under f of the center of ℚGe. Before considering the general case we assume G is a p-group, with p a prime integer. Then G and H = ⟨a⟩ satisfy the conditions of Lemma 13.5.2 and thus G is isomorphic to one of the three groups stated in the lemma. Moreover, H has a complement in G if and only if G ≅ P1 or G ≅ P2 and, in these cases, τ is trivial. We claim that in these cases there exists a list of elements x1 , . . . , x p k in ℚ(ξ) (with p k = [G : H]) satisfying the conditions of Lemma 13.5.1 and the elements f −1 (x1 ), . . . , f −1 (x p k ) correspond to the conjugating elements in G given in the statement of the theorem in the different cases. To prove this, one makes use of the following fact for a positive integer i < p k and a subfield L of ℚ(ξ): if ξ i ∈ ̸ L and ζ q ∈ L, with q = p if p is odd and q = 4 if p = 2, then Trℚ(ξ)/L (ξ i ) = 0. To see this, notice that if d is the minimum positive integer such d d d that ξ ip ∈ L then ℚ(ξ i )/L is cyclic of degree p d and ξ i is a root of X p − ξ ip ∈ L[X]. d d p ip i i Then X − ξ is the minimal polynomial of ξ over L. Hence Trℚ(ξ i )/L (ξ ) = 0 and thus Trℚ(ξ)/L (ξ i ) = 0. n−k First assume that G = P1 and v p (r − 1) = n − k (equivalently a p ∈ Z(G), that is, either p is odd or p = 2 and r ≡ 1 mod 4). Then F is the unique subfield of index k [G : H] = p k in ℚ(ξ) and such that if p = 2 then ζ4 ∈ F. Namely, F = ℚ(ζ p n−k ) = ℚ(ξ p ). If we set x i = ξ i , for i = 0, 1, . . . , p k − 1, then x i x−1 = ξ i−j . If i ≠ j then ξ i−j ∈ ̸ F j −1 i−j and hence Trℚ(ξ)/F (x i x j ) = Trℚ(ξ)/F (ξ ) = 0. Thus, by Lemma 13.5.1, the conjugates ̂ by 1, ξ, ξ 2 , . . . , ξ p k −1 form a complete set of orthogonal primitive idempotents of b of (ℚ(ξ)/F, 1). Then the elements f −1 (x i ) form the elements of T2󸀠 if p is odd or the elements of T2 , in case (1i). Note that ℚGε ≅ M[G:H] (F). n−2 Second assume that G = P1 and p = 2 and r ≢ 1 mod 4 (equivalently, a22 is not central). In this case ζ4 ∈ ̸ F and F(ζ4 ) is the unique subfield of ℚ(ξ) containing ℚ(ζ4 ) k−1 and of index [G : H]/2 = 2k−1 . That is, F(ζ4 ) = ℚ(ζ2n−k+1 ) = ℚ(ξ 2 ). Let x i = ξ i and n−2 −1 x2k−1 +i = ξ 2 +i = ζ4 ξ i , for 0 ≤ i < 2k−1 . Hence, if i ≠ j then x i x j is either ζ4±1 or ξ ±v or ζ4±1 ξ ±v , with v = 1, 2, . . . , 2k−1 − 1. As ξ v ∈ ̸ F(ζ4 ), we have Trℚ(ξ)/F(ζ4 ) (ξ v ) = 0. Since Trℚ(ξ)/F (ζ4 ) = TrF(ζ4 )/F Trℚ(ξ)/F(ζ4 ) (ζ4 ) = [ℚ(ξ) : ℚ(ζ4 )]TrF(ζ4 )/F (ζ4 ) = 0, Trℚ(ξ)/F (ξ v ) = TrF(ζ4 )/F Trℚ(ξ)/F(ζ4 ) (ξ v ) = 0 and Trℚ(ξ)/F (ζ4 ξ v ) = TrF(ζ4 )/F Trℚ(ξ)/F(ζ4 ) (ζ4 ξ v ) = TrF(ζ4 )/F (ζ4 Trℚ(ξ)/F(ζ4 ) (ξ v )) = 0, −1 maps these elements to the it follows that Tr(x i x−1 j ) = 0 for every i ≠ j. Then f elements of T2 for case (1ii). Note again that ℚGε ≅ M[G:H] (F).

13.5 Primitive idempotents and nilpotent groups |

413

Third, assume that G = P2 . Then F = ℚ(ζ2n−k + ζ2−1 n−k ). Since r ≡ 1 mod 4 one also has that n − k ≥ 2. Then, the same argument as in the previous case shows that the n/2 2k+1 elements of the form x i = ξ i and x2k +i = ξ 2 +i = ζ4 ξ i , for 0 ≤ i < 2k satisfy the conditions of Lemma 13.5.1. The elements f −1 (x i ) form now the set T2 of case (1ii). Again ℚGε ≅ M[G:H] (F). Fourth, consider the non-splitting case G = P3 . Then the center of ℚGe is isomor2n−2 + Fc + ̂ ̂ ̂ ̂ phic to F = ℚ(ξ)⟨b,c⟩ = ℚ(ζ2n−k + ζ2−1 n−k ) and b ℚG ε b = b ℚ⟨a, c⟩ ε bF + Fa n−2 ̂ is a primitive F(a2 c) ≅ ℍ(F), which is a division algebra, as F is a real field. Then bε idempotent of ℚGe. Hence ℚGe ≅ M2k (ℍ(F)) and from the first case one can provide the 2k orthogonal primitive idempotents needed in this case by taking the conjugates ̂ by 1, a, a2 , . . . , a2k −1 , and this agrees with case (2i). This finishes the proof in of b case G is a p-group. Finally we deal with the general case, that is, G is not necessarily a p-group. For a prime p let G p denote the p-part of G and G p󸀠 its p󸀠 -part. Then G = G2 × G p1 × ⋅ ⋅ ⋅ × G p r = G2 × G2󸀠 , with p i an odd prime for every i = 1, . . . , r. Moreover, ε(H, 1) = ∏i ε(H p i , 1). As (H, 1) is a strong Shoda pair of G it follows that each (H p i , 1) is a strong Shoda pair of G p i (with p0 = 2). So, each ℚG p i ε(H p i , 1) is a simple algebra. A dimension argument shows that the simple algebra ℚGε(H, 1) is the tensor product over ℚ of the simple algebras ℚG p i ε(H p i , 1). From the above we know that, for n k i ≥ 1, ℚG p i ε(H p i , 1) ≅ M p ki (ℚ(ζ p ni −ki )), for p i i = |G p i | and p i i = [G p i : H p i ]. So, i

i

ℚG2󸀠 ε(H2󸀠 , 1) ≅ M[G2󸀠 :H2󸀠 ] (ℚ(ζ m )), with m = |H2󸀠 |/[G2󸀠 : H2󸀠 ] (= [H2󸀠 : K]/[G2󸀠 : H2󸀠 ]) and then a complete set of orthogonal primitive idempotents of ℚG2󸀠 ε(H2󸀠 , 1) can be obtained by multiplying the different sets of idempotents obtained for each tensor factor. Observe that each G p i , with i ≥ 1, takes the form ⟨a i ⟩ ⋊ ⟨b i ⟩ and so G2󸀠 = ⟨a⟩ ⋊ ⟨b⟩, ki

with a = a1 . . . a r and b = b1 . . . b r . Because each a p i is central, one easily verifies that the product of the different primitive idempotents of the factors from the k p i −1 ̂ odd part (i.e. the conjugates of ̂ b i by 1, a i , a2 , . . . , a i ) are the conjugates of bε i

i

by 1, a, a2 , . . . , a[G2󸀠 :H2󸀠 ]−1 . In the notation of the statement of the theorem, a = a2󸀠 and T2󸀠 = {1, a, a2 , ...., a[G2󸀠 :H2󸀠 ]−1 }, as desired. Note that we also obtain that ℚGε = M2 ([G : H])(ℚ(ζ|H| )G ). So, if |G| is odd then the proof is finished. Otherwise one needs to combine the odd and even parts of G. If H2 has a complement in G2 then ℚG2 ε(H2 , 1) is split over its center and hence we can take T2 as in the 2-group case. If H2 does not have a complement in G2 then ℚG2 ε(H2 , 1) = M[G2 :H2 ]/2 (ℍ(ℚ(ζ2n−k + ζ2−1 n−k ))) and hence −1 ℚGε = M[G:H]/2 (ℍ(F)), with F = ℚ(ζ m , ζ2n−k + ζ2n−k )). In case ℍ(F) is not split (equivalently the conditions of (2i) hold) then one can also take T2 as in the 2-group case. In case ℍ(F) is split then one should duplicate the number of idempotents, or equivalently duplicate the size of T2 . In this case −1 is a sum of squares in F, by Theok [N 󸀠 :H 󸀠 ] rem 11.5.1. Because f(a2󸀠 2 2 ) is a primitive m-th root of unity and f(a22 ) is a primitive [N 󸀠 :H2󸀠 ]

2n−k root of unity, one obtains that there are x, y ∈ F = ℚ(a2󸀠 2

ε, (a22 + a−2 2 )ε) k

k

414 | 13 Idempotents and central units in group rings such that (1 + x2 + y2 )ε = 0. Again by Lemma 11.5.3, f =

1 + xa22

n−2

+ ya22 2

n−2

c2

ε

and

1−f =

1 − xa22

n−2

− ya22 2

n−2

c2

ε

are orthogonal idempotents in ℍ(F). Observe that 1 − f = f c2 . Hence, we obtain that a complete set of primitive idempotents of ℚGε consists of the conjugates of ̂ b2 fε by 2k −1 2k −1 1, a2 , . . . , a2 , c2 , a2 c2 , . . . , a2 c2 , as desired. Also ℚGε = M[G:H] (ℚ(ζ|H| )G ). In the proof of Theorem 13.5.3 we also described the Wedderburn components of ℚG for an arbitrary nilpotent finite group G. We collect this information in the following corollary. Corollary 13.5.4 ([118]). If (H, K) is a strong Shoda pair of a nilpotent finite group G then in cases (1) and (2ii) of Theorem 13.5.3 ℚGe(G, H, K) = M|G/H| ((ℚ(ζ[H:K] ))N G (K)/K ), (where (ℚ(ζ[H:K] ))N G (K)/K is the fixed subfield for the natural action of N G (K)/K on the cyclotomic field ℚ(ζ[H:K] ) = ℚHε(H, K)), and ℚGe = M 1 |G/H| (ℍ (ℚ(ζ[H:K] )N/K )) 2

in case (2i) of Theorem 13.5.3. In particular, if ℚGe is a non-commutative division algebra then [G : H] = 2, N = G n and either ℚGe ≅ ℍ(ℚ(ζ2n + ζ2−1 n )) with [H : K] = 2 , or ℚGe ≅ ℍ (ℚ (ζ m )) with m an odd prime such that the order of 2 modulo m is odd and [H : K] = m or 2m. Proof. As mentioned before the Corollary, the first part collects statements obtained in the proof of Theorem 13.5.3. The second part follows from Theorem 11.5.1. As an immediate consequence one obtains a well-known result on the indices of the simple components of group algebras of finite nilpotent groups over fields of characteristic zero. Corollary 13.5.5 (Roquette [198]). If G is a finite nilpotent group and F is a field of characteristic zero, then FG ≅ ⨁i M n i (D i ), where the D i are either fields or quaternion division algebras. Hence, the index of each simple component of FG is at most 2. Moreover, if the index of a simple component of FG is 2 then the Sylow 2-subgroup of G has subgroups K  H such that H/K is a quaternion group. Notice that Theorem 13.5.3 does not remain valid for metacyclic groups. For example, if G = C7 ⋊ C3 = ⟨a⟩ ⋊ ⟨b⟩, with b−1 ab = a2 and ε = ε(⟨a⟩) then there is not a complete ̂ This set of orthogonal primitive idempotents of ℚGε formed by ℚ(a)-conjugates of bε. is a consequence of the remarks given before Lemma 13.5.2. Let G be a nilpotent finite group and (H, K) a strong Shoda pair in G. In order to describe a complete set of matrix units in a simple component ℚGe(G, H, K) of ℚG

13.5 Primitive idempotents and nilpotent groups |

415

we need to introduce some notation that is based on the different cases stated in Theorem 13.5.3. Set e = e(G, H, K). Let T e = T2󸀠 T2 T G/N

and

̂ βe = b 2󸀠 β 2 ε,

where ε = ε(H, K), T G/N denotes a left transversal of N = N G (K) in G. Set [N 󸀠 :H2󸀠 ]−1

T2󸀠 = {1, a2󸀠 , . . . , a2󸀠 2

},

{1, a2 , . . . , a22 −1 }, { { { k−1 n−2 n−2 k−1 n−2 T2 = {{1, a2 , . . . , a22 −1 , a22 , a22 +1 , . . . , a22 +2 −1 }, { { 2k −1 2k −1 {{1, a2 , . . . , a2 , c2 , a2 c2 , . . . , a2 c2 }, k

in cases (1i) and (2i); in case (1ii); in case (2ii);

and ̂2 , M { { { b2 , β 2 = {̂ { { 2n−2 2n−2 ̂ 1+xa2 +ya2 c2 , {b2 2

in case (1); in case (2i); in case (2ii).

Corollary 13.5.6. Let G be a finite nilpotent group and (H, K) a strong Shoda pair. Let e = e(G, H, K), a primitive central idempotent of ℚG and let T e and β e be defined as above. For every t, t󸀠 ∈ T e , let E tt󸀠 = t−1 β e t󸀠 . Then {E tt󸀠 | t, t󸀠 ∈ T e } is a complete set of matrix units in ℚGe, i.e. e = ∑ E tt

and

E t1 t2 E t3 t4 = δ t2 t3 E t1 t4 ,

t∈T e

for t1 , t2 , t3 , t4 ∈ T e . Moreover, E tt ℚGE tt ≅ F, in cases (1) and (2ii) of Theorem 13.5.3, and E tt ℚGE tt = ℍ(F), in case (2i) of Theorem 13.5.3, where F is the fixed subfield of ℚ(a)ε under the natural action of N G (K)/H. Proof. We know from Theorem 13.5.3 that the set {E tt | t ∈ T e } is a complete set of primitive idempotents of ℚGe. From the definition of the E tt󸀠 it easily follows that E t1 t2 E t3 t4 = δ t2 t3 E t1 t4 , for t i ∈ T e , i = 1, . . . , 4. The second statement is already mentioned in Corollary 13.5.4. A description of a set of matrix units in simple component also allows to describe finitely many generators for a subgroup of finite index in SL1 (ℤG) and thus in U(ℤG). Theorem 13.5.7. Let G be a finite nilpotent group of class n such that ℚG has no exceptional components. For every primitive central idempotent e = e(G, H, K), with (H, K) a strong Shoda pair of G, let N = N G (K) and ε = ε(H, K). Let T e be as in Corollary 13.5.6 and fix an order < in T e . Then the following two groups are nilpotent subgroups of U(ℤG): V e+ = ⟨1 + |G|t−1 β e gt󸀠 : g ∈ G, t, t󸀠 ∈ T e , t󸀠 > t⟩, V e− = ⟨1 + |G|t−1 β e gt󸀠 : g ∈ G, t, t󸀠 ∈ T e , t󸀠 < t⟩.

416 | 13 Idempotents and central units in group rings Hence, if E is the set of primitive central idempotents of ℚG then V + = ∏ V e+

and

V − = ∏ V e−

e∈E

e∈E

are nilpotent subgroups of U(ℤG). Furthermore, if B = ⟨c(b) | b a Bass cyclic unit of G⟩, then the group ⟨B, V + , V − ⟩ is of finite index in U(ℤG). Proof. By Proposition 5.5.1 and Corollary 13.1.4, it is sufficient to show that for each primitive central idempotent e of ℚG if ℚGe ≅ M n (D), with D a division ring, then the group ⟨B, V + , V − ⟩ contains a subgroup of finite index in 1 − e + SLn (O), for some order O in D. The elements of the form 1 + |G|t−1 β e gt󸀠 , with g ∈ ⟨a l ⟩ and t, t󸀠 ∈ T e , project trivially to ℚG(1 − e). By Corollary 13.5.6, they project to an elementary matrix e t,t󸀠 (α g ) of M n (O), for some order O in D. The additive group generated by the α g ’s contains a non-zero ideal of O. Hence it follows from Theorem 11.2.3 that these units generate a subgroup of finite index in SLn (O). Remark 13.5.8. Observe that if G and e are as in Theorem 13.5.7 and ℚGe is a division algebra then V e+ = V e− = 1.

Problems 13.5.1. Let F(α)/F be a separable field extension of degree 3 such that TrF(α)/F (α) = TrF(α)/F (α−1 ) = 0. Prove that the minimal polynomial of α over F is of the form X 3 − a for some a ∈ F. Deduce that if F(α)/F is a Galois extension then F contains a third root of unity.

13.6 Primitive idempotents and strongly monomial groups Let G be an arbitrary finite group. In this section it is shown that the results of the previous section can be extended to simple components of the rational group algebra ℚG that are determined by a strong Shoda pair (H, K) with trivial defining cocycle, that is, f(nH, n󸀠 H) = 1 for all n, n󸀠 ∈ N G (K) (with notation as in Theorem 3.5.5). For such a component, a description is given of a complete set of orthogonal primitive idempotents (and a complete set of matrix units). This work due to Jespers, Olteanu, del Río and Van Gelder [119] and it ultimately relies on the isomorphism given in Problem 2.6.9. Composing such isomorphism with its matrix representation we obtain the following proposition.

13.6 Primitive idempotents and strongly monomial groups

|

417

Proposition 13.6.1. If E/F is a finite Galois extension and n = [E : F] then the classical crossed product (E/F, 1) is isomorphic (as F-algebra) to M n (F). Moreover, if B is an Fbasis of E, then an isomorphism is given by Ψ : (E/F, 1) → EndF (E) → M n (F) , xσ 󳨃→ x󸀠 ∘ σ 󳨃→ [x󸀠 ∘ σ]B , where x ∈ E, σ ∈ Gal(E/F) and x󸀠 denotes left multiplication by x on E. The well-known Normal Basis Theorem [145, Theorem VI.13.1] states that if E/F is a finite Galois extension, then there exists an element w ∈ E such that {σ(w) | σ ∈ Gal(E/F)} is an F-basis of E. This is called a normal basis and hence w is called normal in E/F. In particular, ℚ(ζ[H:K] )/ℚ(ζ[H:K] )N G (K)/H has a normal element, say w. Theorem 13.6.2 ([119]). Let G be a finite group and (H, K) a strong Shoda pair of G. Assume f(aH, bH) = 1 for all a, b ∈ N G (K). Let ε = ε(H, K) and e = e(G, H, K). Let F denote the fixed subfield of ℚHε under the natural action of N G (K)/H and let n = [N G (K) : H]. Let w be a normal element of ℚHε/F and let B be the normal basis determined by w. Let Ψ : ℚN G (K)ε → M n (F) denote the isomorphism given in Proposition 13.6.1 with respect to the basis B. Further, let P, A ∈ M n (F) be the matrices 1 1 (1 ( P=(. ( .. 1 (1 Then

1 −1 0 .. . 0 0

1 0 −1 .. . 0 0

⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ .. . ⋅⋅⋅ ⋅⋅⋅

1 0 0 .. . −1 0

1 0 0) ) ..) .) 0 −1)

and

0 1 (0 ( A = (. ( .. 0 (0

0 0 1 .. . 0 0

⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ .. . ⋅⋅⋅ ⋅⋅⋅

0 0 0 .. . 0 1

1 0 0) ) . .. ) .) 0 0)

{x ̂ T1 εx−1 | x ∈ T2 ⟨x e ⟩}

is a complete set of orthogonal primitive idempotents of ℚGe, where x e = Ψ −1 (PAP−1 ), T1 is a transversal of H in N G (K) and T2 is a right transversal of N G (K) in G. Proof. Set N = N G (K). Because of Theorem 3.5.5 and the assumption that the cocycle f is trivial, we know that ℚGe ≅ M[G:N] (ℚNε) ≅ M[G:N] ((ℚHε/F, 1)) . As in the proof of Theorem 13.5.3, without loss of generality, we will assume that K is normal in G. So, N = G and e = ε. Then B = {w gH : g ∈ T1 }. Since G/H acts on ℚHe via the action induced by conjugation, it easily is seen that the action of G/H on B is regular, that is, the action

418 | 13 Idempotents and central units in group rings

is transitive and the stabilizers are trivial. Hence, Ψ(ge) is a permutation matrix for each g ∈ T1 . Furthermore, 1 1 ( 1 (1 Ψ(̂ T1 e) = ( . n ( .. 1 (1

1 1 1 .. . 1 1

⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ .. . ⋅⋅⋅ ⋅⋅⋅

1 1 1 .. . 1 1

1 1 1) ) . .. ) .) 1 1)

Clearly Ψ(̂ T1 e) has eigenvalues 1 and 0, with respective F-eigenspaces V1 spanned by t (1, 1, . . . , 1) and V0 spanned by the elements (1, −1, 0, . . . , 0), (1, 0, −1, . . . , 0), . . . , (1, 0, 0, . . . , −1). Hence, Ψ(̂ T1 e) = PE11 P−1 . Clearly {E11 , E22 , . . . , E nn } and hence also {Ψ(̂ T1 e) = PE11 P−1 , PE22 P−1 , . . . , PE nn P−1 } is a complete set of orthogonal primitive idempotents of M n (F). Let y = Ψ(x e ) = PAP−1 . As E22 = AE11 A−1 , . . . , E nn = A n−1 E11 A−n+1 it follows that

T1 e)y−1 , . . . , y n−1 Ψ(̂ T1 e)y−n+1 } {Ψ(̂ T1 e), yΨ(̂

is a complete set of orthogonal primitive idempotents of M n (F). Applying Ψ −1 yields the complete set of orthogonal primitive idempotents of ℚGe as described in the statement of the Theorem. Corollary 13.6.3. With notations as in Theorem 13.6.2. If (H, K) is a strong Shoda pair of a finite group G such that f(aH, bH) = 1 for all a, b󸀠 ∈ N = N G (K) then {x ̂ T1 εx󸀠−1 | x, x󸀠 ∈ T2 ⟨x e ⟩} is a complete set of matrix units in ℚGe. Moreover (x ̂ T1 εx−1 ) ℚG (x ̂ T1 εx−1 ) ≅ F, where F is the fixed subfield of ℚHε under the natural action of N/H. Proof. This follows at once from Theorem 13.6.2 and the fact that ℚGe ≅ M[G:H] (F). It remains a problem to give an internal description with the rational group algebra ℚG of the element x e = Ψ −1 (PAP−1 ) introduced in Theorem 13.6.2. In case G is a finite nilpotent group then this problem was avoided in the previous section by using the structure of the group; even in the case when f is not trivial. To obtain a generic description

13.6 Primitive idempotents and strongly monomial groups

|

419

for an arbitrary strongly monomial group might be a hard problem as one first has to overcome the obstacle of obtaining a description of a normal basis for ℚ(ζ[H:K] )/F with F = ℚ(ζ[H:K] )N/H . In the literature one finds some partial results on this. For example, in [79] Hachenberger studied normal bases for cyclotomic fields ℚ(ζ q m ) with q an odd prime. Once such a description is obtained then x e can be determined as follows: Denote by T : ℚG → ℚ the map ∑g∈G a g g 󳨃→ a1 . It is easily verified that T(α) = 1 1 χ |G| reg (α) = |G| ∑χ∈Irr(G) χ(1)χ(α), where χ reg denotes the regular character of G. We can consider Ψ as an isomorphism ℚGe ≅ (ℚ(ζ[H:K] )/F, 1) → M n (F) and extend it to a ring homomorphism ℚG → M n (F), mapping the elements of ℚG(1 − e) to 0. We also denote this extension by Ψ. Then Ψ is the ℚ-linear extension of an irreducible complex representation of G. Let ψ denote the character afforded by this representation. Then χ(Ψ −1 (x)y) = 0 for every χ ∈ Irr(G) \ {ψ}, x ∈ M n (F) and y ∈ ℚG. Therefore, T(x e g−1 ) =

1 |G|

∑

χ(1)χ(x e g−1 ) =

χ∈Irr(G)

1 |G|

∑

χ(1)χ(Ψ −1 (PAP−1 )g −1 )

χ∈Irr(G)

n n ψ(Ψ −1 (PAP−1 )g −1 ) = tr(PAP−1 Ψ(g −1 )). = |G| |G| Hence, x e = ∑ T(x e g−1 )g = g∈G

n ∑ Tr(PAP−1 Ψ(g −1 ))g. |G| g∈G

The following example shows that sometimes these problems can be overcome using elementary linear algebra. Example 13.6.4. Let G = C7 ⋊ C3 = ⟨a, b | a7 = 1 = b3 , a b = a2 ⟩, a metacyclic group and thus a strongly monomial group. The pair (⟨a⟩ , {1}) is strong monomial and the associated cocycle is trivial. So, e = e(G, ⟨a⟩ , 1) = ε(⟨a⟩ , 1) is a primitive central idempotent and B = {ae, a2 e, a4 e} is a normal basis of ℚ(ae) over ℚ(ae + a2 e + a4 e). Consider the isomorphism Ψ : ℚ ⟨a⟩ e ∗⟨b⟩ ≅ M3 (ℚ(ae + a2 e + a4 e)) with respect to B. Clearly A = Ψ(be). Hence, in order to describe x e = Ψ −1 (PAP−1 ) in terms of elements of ℚG, it is sufficient to write Ψ −1 (P) in terms of group ring elements. That is, one needs to find α0 , α1 , α2 ∈ ℚ ⟨a⟩ e such that Ψ −1 (P) = α0 + α1 b + α2 b2 . By definition of P, such elements are solutions of the following system of equations: (α + α1 ∘ b + α2 ∘ b2 )(ae) = (a + a2 + a4 )e { { 0 2 2 2 { (α0 + α1 ∘ b + α2 ∘ b )(a e) = (a − a )e { 2 4 4 { (α0 + α1 ∘ b + α2 ∘ b )(a e) = (a − a )e. Write each α i = (x i,0 + x i,1 a + x i,2 a2 + x i,3 a3 + x i,4 a4 + x i,5 a5 )e with x i,j ∈ ℚ. Note that (1 + a + a2 + a3 + a4 + a5 + a6 )e = 0. The above then leads to a system of 18 linear

420 | 13 Idempotents and central units in group rings

equations in 18 variables. It turns out that 4 5 3 1 4 5 5 1 1 − a − a2 − a − a − a )e 7 14 2 14 7 14 2 11 3 2 1 3 1 4 9 5 +( − a− a − a − a − a ) be 7 14 14 2 7 14 9 11 2 9 3 2 4 1 5 2 2 a− a − a + a − a ) b e. +( − 7 14 14 14 7 2

Ψ −1 (P) = (−

A crucial hypothesis to make use of Proposition 13.6.1 is that the twistings appearing in the simple components are trivial. In Section 13.7 it will be shown that metacyclic groups of the type C q m ⋊ C p n with C p n acting faithfully on C q m do satisfy this condition on the strong Shoda pairs. The following example shows that this is not necessarily true if the action is not faithful. Example 13.6.5. The group G = C19 ⋊ C9 = ⟨a, b | a19 = b9 = 1, a b = a7 ⟩ has the strong Shoda pair (⟨a, b3 ⟩, 1). Let ε = ε(⟨ab3 ⟩, 1). The elements 1, b, b2 are coset representatives for ⟨ab3 ⟩ = ⟨a, b3 ⟩ in G. Clearly the associated cocycle f is not trivial as b2 ⟨a, b3 ⟩b2 ⟨a, b3 ⟩ = b⟨a, b3 ⟩ and b3 = (ab3 )19 and thus f(b2 ⟨a, b3 ⟩, b2 ⟨a, b3 ⟩) = 19 ζ57 ≠ 1. Note, however, that f is cohomologically trivial. Theorem 13.6.2 yields a detailed description of a complete set of orthogonal primitive idempotents of ℚG for any G is a strongly monomial group that has a complete and non-redundant set of strong Shoda pairs (H, K) satisfying f(aH, bH) = 1 for all a, b ∈ N G (K). Apart from the examples given in Section 13.7, one easily verifies that the symmetric group S4 and the alternating group A4 of degree 4 have a trivial twisting in all Wedderburn components of their rational group rings (of course they are not metacyclic nor nilpotent). Trivially all abelian groups are examples as well. So are the dihedral groups D2n = ⟨a, b | a n = b2 = 1, a b = a−1 ⟩. On the other hand, the quaternion groups Q4n = ⟨x, y | x2n = y4 = 1, x n = y2 , x y = x−1 ⟩ has (⟨x⟩ , 1) as a strong Shoda pair with a non-trivial twisting.

13.7 Some metacyclic groups Throughout this section G is a metacyclic group of the type G = C q m ⋊ C p n = ⟨a⟩ ⋊ ⟨b⟩ , with p and q distinct primes, and such that C p n acts faithfully on C q m , i.e. C⟨b⟩ (a) = {1}. Let σ denote the automorphism of ⟨a⟩ given by σ(a) = a b . Write σ(a) = a r , with r ∈ ℤ. We first show that such a group has a complete and non-redundant set of strong Shoda pairs with a trivial associated cocycle. Hence Theorem 13.6.2 is applicable and yields a description of the Wedderburn decomposition of ℚG. Next we give a multiplicative independent set of Z(U(ℤG)) that generates a subgroup of finite index. This is work due to Jespers, Olteanu, del Río and Van Gelder [119]. Ferraz and Simón in [61] dealt

13.7 Some metacyclic groups |

421

with the case m = n = 1. Recall that in Section 12.6 semidirect products of the type C n ⋊ C2 have been investigated. Because of the faithfulness of the action, the automorphism σ has order p n . Hence, p n divides |Aut(⟨a⟩)| = (q −1)q m−1 . Since, by assumption, p and q are different primes it follows that q ≡ 1 mod p n , in particular is q odd. j Therefore, by Theorem 1.6.3, Aut(⟨a q ⟩) = Gal(ℚ(ζ q j )/ℚ) = U(ℤ/q j ℤ) is cyclic for every j = 0, 1, . . . , m and ⟨σ⟩ is the unique subgroup of Aut(⟨a⟩) of order p n . So, for every i = 1, . . . , m, the image of r in ℤ/q i ℤ generates the unique subgroup of n j U(ℤ/q i ℤ) of order p n . In particular r p ≡ 1 mod q m and r p ≢ 1 mod q for every j = 0, ..., n − 1. Therefore, r ≢ 1 mod q and hence G 󸀠 = ⟨a r−1 ⟩ = ⟨a⟩. The description of strong Shoda pairs of metacyclic groups given in Theorem 3.5.12 then yields a complete and non-redundant set of strong Shoda pairs of G. Furthermore, because of Theorem 3.5.5 one obtains a description of the associated simple component of ℚG. Theorem 13.7.1 ([119]). Let p and q be distinct prime numbers. Let G = C q m ⋊ C p n be a finite metacyclic group with C p n = ⟨b⟩ acting faithfully on C q m = ⟨a⟩ and with p and q different primes. Let r be such that a b = a r . A complete and non-redundant list of strong Shoda pairs of G = C q m ⋊ C p n = ⟨a⟩ ⋊ ⟨b⟩ consists of the following pairs of subgroups: i

(G, L i := ⟨a, b p ⟩)

j

and (⟨a⟩ , K j := ⟨a q ⟩),

with 1 ≤ i ≤ n and 1 ≤ j ≤ m. For each pair, there is an trivial associated cocycle (as defined in Theorem 3.5.5). The corresponding simple components are: ℚGε (G, L i ) ≅ ℚ (ζ p i ) and ℚGε (⟨a⟩ , K j ) ≅ ℚ (ζ q j ) ∗ C p n ≅ M p n ((ℚ ⟨a⟩ ε (⟨a⟩ , K j )) C

⟨b⟩

) ≅ M p n (F j ) ,

where F j = ℚ (ζ q j ) p , the fixed field of ℚ (ζ q j ) by the action of C p n . Furthermore F j is the unique subfield of index p n in ℚ (ζ q j ). n

Proof. The only remaining part to be proved is that ℚGε (⟨a⟩ , K j ) ≅ M p n ((ℚ ⟨a⟩ ε (⟨a⟩ , K j )) C

⟨b⟩

) ≅ M p n (F j ) ,

where F j = ℚ (ζ q j ) p , the fixed field of ℚ (ζ q j ) by the action of C p n . Furthermore F j is the unique subfield of index p n in ℚ (ζ q j ). Because the associated cocycles are trivial this follows at once from Proposition 13.6.1. n

For the metacyclic groups under consideration one now can describe an independent large set of central units. Theorem 13.7.2 ([119]). Let p and q be distinct prime numbers. Let G = C q m ⋊ C p n be a finite metacyclic group with C p n = ⟨b⟩ acting faithfully on C q m = ⟨a⟩. Let r be such

422 | 13 Idempotents and central units in group rings that a b = a r . For each j = 1, . . . , m, let I j be a set of coset representatives of U(ℤ/q j ℤ) modulo ⟨r, −1⟩. If p = 2, then the set i

U2 = {c0i (G, ⟨a, b2 ⟩, k, 1) : 1 < k < 2i−1 , 2 ∤ k, i = 2, . . . , n} ⋃ m m 2 {c m m−j (⟨a⟩ , 1, k, 1)c m−j (⟨a⟩ , 1, k, r) ⋅ ⋅ ⋅ c m−j (⟨a⟩ , 1, k, r

n

−1

)

: k ∈ I j \ {1}, j = 1, . . . , m} is multiplicative independent and generates a subgroup of finite index in Z(U(ℤG)). Furm thermore, |U2 | = 2n−1 + q 2−1 − n − m. n If p is odd, then i

U p = {c0i (G, ⟨a, b p ⟩, k, 1) : 1 < k