Galois Theory and its Algebraic Background [2 ed.] 9781108979184, 9781108838924, 9781108969086

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GALOIS THEORY and Its Algebraic Background

W D.J.H. GARLING

SECOND EDITION

Galois Theory and Its Algebraic Background SECOND EDITION Galois theory, the theory of polynomial equations and their solutions, is one of the most fascinating and beautiful subjects in pure mathematics. Using group theory and field theory, it provides a complete answer to the problem of the solubility of polynomial equations by radicals: that is, determining when and how a polynomial equation can be solved by repeatedly extracting roots using elementary algebraic

operations. This textbook contains a fully detailed account of Galois theory and the algebra that

it needs, and is suitable for both those following a course of lectures and the independent reader (who is assumed to have no previous knowledge of Galois theory). This second edition has been significantly revised and reordered; the first part develops the basic algebra that is needed, and the second part gives a comprehensive account of Galois theory. There are applications to ruler and compass constructions, and to the

solution of classical mathematical problems of ancient times. There are new exercises throughout, and carefully selected examples will help the reader develop a clear understanding of the mathematical theory. D . J . H . GA RLING is Emeritus Reader in Mathematical Analysis at the University of Cambridge and Fellow of St John’s College, Cambridge. He has 50 years’ experience of teaching undergraduate students and has written several books on mathematics, including Inequalities: A Journey into Linear Analysis (Cambridge University Press, 2007) and A Course in Mathematical Analysis (three volumes,

Cambridge University Press, 2013—2014).

Galois Theory and Its Algebraic Background SECOND EDITION

D.J.H. Garling University of Cambridge

“3a CAMBRIDGE 3

UNIVERSITY PRESS

CAMBRIDGE UNIVERSITY PRESS University Printing House. Cambridge CB2 83S, United Kingdom One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamst Road, Port Melbourne, VIC 3207, Australia 314—321, 3rd Floor, Plot 3, Splendor Forum, Jasola District Centre, New Delhi — 110025, India

103 Penang Road. #05-06/07, Visioncrest Commercial. Singapore 238467 Cambridge University Press is part of the University of Cambridge. It furthers the University's mission by disseminating knowledge in the pursuit of education, learning. and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.orgl9781 108838924 D01: 10.1017/9781108979184 First edition © Cambridge University Press 1986 Second edition © DJ.l-l. Gatling 2022 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1986 Reprinted with corrections 1988, 1991. 1993. 1995 Second edition 2022 Printed in the United Kingdom by TJ Books Limited, Padstow, Cornwall. 2022 A catalogue recordfor this publication is availablefrom the British library. Library of Congress Cataloging-in-Publication Data Names: Garling, D. J. H., author. Title: Galois theory and its algebraic background / DJ.H. Garling. Other titles: Course in Galois theory Description: Second edition. | Cambridge ; New York, NY : Cambridge University Press, 2021. | First edition published as Course in Galois theory, Cambridge University Press, 1986. | Includes bibliographical references and index. Identifiers: LCCN 2021002526 (print) | LCCN 2021002527 (ebook) | ISBN 9781108838924 (hardback) | ISBN 9781 108969086 (paperback) | ISBN 9781108979184 (epub) Subjects: LCSH: Galois theory. Classification: LCC QA214 .6367 2021 (print) — LCC QA214 (ebook) | DDC 512/.32—dc23 LC record available at https://1ccn.loc.gov/2021002526 LC ebook record available at https://lccn.loc.gov/2021002527 ISBN 978-1-108-83892-4 Hardback ISBN 978-1-108-96908-6 Paperback Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party intemet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

Contents

Preface

PART I

page ix

THE ALGEBRAIC BACKGROUND

l 1.1 1.2 1.3 1.4 1.5 1.6

Groups

2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11

Integral Domains Commutative Rings with l

Groups Finite Abelian Groups Finite Permutation Groups

Group Series Soluble Groups p-Groups and Sylow Theorems

Polynomials Homomorphisms and Ideals

Integral Domains Fields and Fractions The Ordered Set of Ideals in an Integral Domain

Factorization Unique Factorization Principal Ideal Domains and Euclidean Domains Polynomials Over Unique Factorization Domains More About Fields

2.12 Kronecker’s Algorithm

2.13 Eisenstein’s Criterion 2.14 Localization

ll 17 18 22 25 25 26 27 29 31 35 36 38 41

47 43 50 51

vi

Contents

Vector Spaces and Determinants 3.1 3.2 3.3 3.4

Vector Spaces

The Infinite-Dimensional Case Characters and Automorphisms Determinants PART II

4. 1 4.2 4.3 4.4 4.5 5.1 5.2 6. 1 6.2 6.3 6.4 7.1 7.2 8.1 8.2 8.3 8.4 8.5

THE THEORY OF FIELDS AND GALOIS THEORY

Field Extensions Introduction Field Extensions Algebraic and Transcendental Extensions Algebraic Extensions Monomorphisms of Algebraic Extensions

69 69 70 73 77 80

Ruler and Compass Constructions Some Classical Problems Constructible Points

81 81 81

Splitting Fields Introduction Splitting Fields

85 85 86 89 94

The Extension of Monomorphisms Some Examples

Normal Extensions Basic Properties Monomorphisms and Automorphisms

Separability Basic Ideas Monomorphisms and Automorphisms Galois Extensions

Differentiation Inseparable Polynomials

The Fundamental Theorem of Galois Theory 9.1 9.2 9.3 9.4 9.5

54 54 60 6l 62

Field Automorphisms, Fixed Fields and Galois Groups

Linear Independence The Size of a Galois Group is the Degree of the Extension The Galois Group of a Polynomial The Fundamental Theorem of Galois Theory

98 98 101 103 103 104 106 106 108

112 112 113 115 116 118

Contents

vii

10 The Discriminant 10.1 The Discriminant

122 122

11 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8

126 126 128 129 131 132 134 135 137

Cyclotomic Polynomials and Cyclic Extensions

Cyclotomic Polynomials Irreducibility

The Galois Group of a Cyclotomic Polynomial A Necessary Condition Abel’s Theorem Norms and Traces A Sufficient Condition Kummer Extensions

12 Solution by Radicals 12.1 Polynomials with Soluble Galois Groups 12.2 Polynomials which are Soluble by Radicals

140 140 141

13 13.1 13.2 13.3

Regular Polygons

146 146 147 148

14 14.1 14.2 14.3

Polynomials of Low Degree Quadratic Polynomials Cubic Polynomials

15 15.1 15.2 15.3

Finite Fields Finite Fields Polynomials in Zp [x] Polynomials of Low Degree over a Finite Field

156 156 157 158

16

Quintic Polynomials

161

17 17.1 17.2 17.3

Further Theory Simple Extensions

164 165 166 I68

18 18.1 18.2 18.3 18.4

The Algebraic Closure of a Field Introduction The Existence of an Algebraic Closure The Uniqueness of an Algebraic Closure

Fermat Primes and Fermat Numbers Regular Polygons Constructing a Regular Pentagon

Quartic Polynomials

The Theorem of the Primitive Element The Normal Basis Theorem

Conclusions

149 149 150 153

170 170 171 I75 176

viii

19 19.1 19.2 19.3 19.4 19.5

Contents

Transcendental Elements and Algebraic Independence Transcendental Elements and Algebraic Independence

Transcendence Bases Transcendence Degree The Tower Law for Transcendence Degree

Liiroth’s Theorem Generic and Symmetric Polynomials

20. 1 Generic and Symmetric Polynomials Appendix: Index

The Axiom of Choice

177 177 180 181 182 183 186 186 189 192

Preface

Galois theory is one of the most fascinating and enjoyable branches of algebra. The problems with which it is concerned have a long and distinguished history: the problems of duplicating a cube or trisecting an angle go back to the

Greeks, and the problem of solving a cubic, quartic or quintic equation to the Renaissance. Many of the problems that are raised are of a concrete kind (and

this, surely, is why it is so enjoyable) and yet the needs of the subject have led to substantial development in many branches of abstract algebra: in particular, in the theory of fields, the theory of groups, the theory of vector spaces and the

theory of commutative rings. In this book, Galois theory is treated as it should be, as a subject in its own right. Nevertheless, in the process, I have tried to show its relationship to various topics in abstract algebra: an understanding of the structures of abstract algebra helps give a shape to Galois theory and conversely Galois theory provides plenty of concrete examples which show the point of abstract theory.

The book comprises two unequal parts. In the first part, an account is given of the algebra that is needed for Galois theory. Much of this may well be familiar to the reader, but is included both for completeness and to introduce

the terminology and notation that is used. Much of the algebra (groups, rings, fields and vector spaces) has general interest, and of course the development of Galois theory was responsible for the development of many algebraic ideas. We shall concentrate on presenting those algebraic ideas and results that are

needed for Galois theory. For example, it is important to know that in the right circumstances, the factorization of polynomials with coefficients in a ring is

essentially unique. Group theory plays a large part in Galois theory, but has developed into a huge subject. We shall concentrate on those parts, such as the theory of soluble groups, which are needed in Galois theory.

ix

x

Preface

The second, more substantial, part is concerned with the theory of fields and with Galois theory, and contains the main material of the book; indeed, many readers may wish to start here and refer back to the first part as necessary. Of its nature, the theory develops an inexorable momentum. Nevertheless, there

are many digressions (for example, concerning geometric constructions, finite fields and the solution of cubic and quartic equations): one of the pleasures of Galois theory is that there are many examples which illustrate and depend upon

the general theory, but which also have an interest of their own. The high point of the book is of course the resolution of the problem of when a polynomial is solvable by radicals. I have, however, tried to emphasize (in the final chapter

in particular) that this is not the end of the story: the resolution of the problem raises many new problems, and Galois theory is still a lively subject. The last three chapters have a more abstract nature, and require the use of Zom’s lemma. In full generality, in the uncountable case, this depends upon the Axiom of Choice; this is discussed in the Appendix. Algebra is

principally concerned with finite operations and relations, and is therefore largely concerned with finite or countable sets, and so these chapters have a rather hybrid quality. Two hundred exercises are scattered through the text. It has been suggested

to me that this is rather few: I think that anyone who honestly tries them all will disagree! In my opinion, textbook exercises are often too straightforward, but some of these exercises are quite hard. The successful solution of a challenging problem gives a much better understanding of the powers and limitations of the theory than any number of trivial ones. Remember that mathematics is not a spectator sport!

PART I The Algebraic Background

Groups

It is likely that the reader has already met the concept of a group. It was Galois who first understood the imporance of groups in the study of the roots of a polynomial equation; since then, group theory has blossomed, and developed as a subject in its own right. In this chapter we simply develop those parts of

the theory which we shall need later; one of the main purposes is to explain the notation and terminology that we shall use.

1.1 Groups Suppose that S is a set. A law of composition 0 on S is a mapping from the Cartesian product S x S into S; that is, for each ordered pair (sl, S2) of elements of S there is defined an element s1 0 52 of S.

A group G is a non-empty set with a law of composition 0 : G X G —> G with the following properties:

(i) gl 0 (g2 o g3) = (gl 0 g2) o g3 for all g1, g2,g3 in G — that is, composition is associative; (ii) there is an element e in G (the unit or neutral element) such that e o g =

g oe = g foreachg in G; (iii) to each g in G there corresponds an element g‘1 (the inverse of g) such

thatgog‘l =g—l og=e. Exercise 1.1

Suppose that G is a group. Show that the identity element e is unique, and that for each g e G the inverse element g‘1 is also unique.

4

1 Groups

Two elements g and h of a group commute if g oh = h o g. The commutator

[g,h] ofg and h is the element g‘1 o h‘1 o g o h; thus g and h commute if and only if [g,h] = e. A subset A of a group G is said to be commutative, or

abelian, if and only if any two elements of A commute. The notation that is used for the law of composition varies from situation to situation. Frequently, there is no symbol, and elements are simply placed side by side: goh = gh. When G is abelian, it often happens that the law is denoted by g o h = g + h, the identity element is denoted by 0 and the inverse of an element g is denoted by —g. Let us give some examples of groups. The integers Z (positive, zero and negative) form an abelian group under addition, with identity element 0, but

the non-zero elements do not form a group under multiplication (2 has no multiplicative inverse in Z). The non-zero complex numbers (3* form an abelian group under multiplication, with identity element 1. If S is a non-empty set, a mapping a from S to S is called a permutation of S if it is a bijection: that is, if a(x) = 0(y) then x = y, and if z e S there exists w in S for which a(w) = z. The set 25 of all permutations of S is a group

under the natural composition of mappings. It is not abelian if S has more than two elements. If S = (l, .. . ,n), we write 2,. for 25. We shall consider Sn in more detail in Sections 1.3 and 1.4.

A subset H of a group G is a subgroup of G if it is a group under the law of composition defined on G; that is, if h and h2 are elements of H then so

are h 1 o hz and hf]. If G is a group, {e} and G are subgroups; these are the trivial subgroups of G. If n e Z the set nZ = {nm : m e Z} is a subgroup of Z. The set 11‘ = {z : |z| = l} is a subgroup of the multiplicative group (3*, and ifn > 0 the set Rn = {e2"ik/’I : 0 5 k < n} of nth roots of unity is a subgroup of ']1‘.

A group is cyclic if there is an element g e G such that every element of G is the composition of finitely many copies of g or or finitely many copies of g".

A group is a finite group if it has finitely many elements. The order of a finite group is the number of its elements, and its exponent e(G) is the smallest positive integer n such that g” = e for all g e G. Exercises

1.2

Show that if H is a subgroup of Z, then H is cyclic.

1.3

Show that a subgroup F of a cyclic group is cyclic.

If {Ga}a€A is a family of groups, then the product nae/4 G, is a group, when composition is defined by (g o h)“ = g, 0 ha for a e A.

1.1 Groups

5

The intersection of subgroups of a group G is a subgroup, and so if S is a subset of a group G, there is a smallest group containing S, the subgroup generated by S; this is denoted by (S). It consists of all finite products of elements of S and their inverses, and is called the subgroup generated by S. For example, if G is a group, the derived group 8(G) is the subgroup generated by the set of all commutators [g, h] in G. In the case where S is a singleton {a}, we write (a) for (S); (a) is then an abelian group, the cyclic subgroup generated by a, and consists of {an :n > 0}, e and {a‘" z n > 0}, where a" is a oa o - - - oa (n terms) and a‘” is a“ oa— I o - - - o a"1 (n terms). Suppose that A is a subset of a group G. The centralizer Z (A) is the set {g e G : [g,h] = e, for all h e A} of all elements of G which commute with

every element of A. Exercise

1.4

Suppose that A, A1 and A2 are subsets of a group G, and that A1 9 A2. Show that (i) Z (A) is a subgroup of G

(ii) Z(Az) S Z(Al) (iii) A S Z(Z(A)) (iV) Z(A) = Z(Z(Z(A))) (v) A C Z (A) if and only if A is abelian.

The group Z(G) is called the centre of G; it is an abelian subgroup of G. If G is a finite group (a group with finitely many elements) then the order of G is the number |G| of elements of G. (If G is infinite, its order is 00.) If a e G then the order of a is the order of (a). If a has finite order, the order of

a is the least positive integer n such that a" = e. A mapping ()3 from a group G to a group H is a homomorphism if ¢(gl o g2) = ¢(g1)o¢(g2) for all g1 and g2 in G. A homomorphism which is injective is called a monomorphism, one which is surjective is called an epimorphism

and one which is both is called an isomorphism. If there is an isomorphism of a group G onto a group H, we say that G and H are isomorphic and write G E H. An isomorphism from a group onto itself is called an automorphism.

For example, if k e G, we set g" = k‘1 0 g o k, for each g e G. Then the mapping g —> g" is an automorphism of G (conjugation by k), an inner

automorphism of G. If A is a subset ofa group G, and k e G, we set Ak = {ak : a e A}. Two

subsets A and B of a group G are conjugate if there exists h e G such that B = A”; conjugacy is an equivalence relation on the subsets of G; we denote

6

1 Groups

the equivalence class to which A belongs by conj (A). If {g} is a singleton, we write conj (g); conj (g) is called a conjugacy class. A set A is self-conjugate if conj (A) = {A}. Thus a group is abelian if and only if every singleton is

self-conjugate. A subgroup H of a group G is a normal subgroup if it is self-conjugate; if so, we write H 0) or infinite order (if k = O). 2.12 Suppose that R is an infinite ring such that R/ I is finite for each non-

trivial ideal I. Show that R is an integral domain.

2.5 Fields and Fractions Suppose that R is a ring. An element a of R is invertible, or a unit, if it has a multiplicative inverse; that is, there is an element a“ such that aa‘l = 1. If so,

the inverse is unique, for if a’ is an inverse then a’ = a’ (aa‘l) = (a’a)a‘l = a"1 . If a1 and a; are units, so is am; (with inverse az— 1af'), so that the set of units in R forms a group UR under multiplication.

32

2 Integral Domains

Exercise 2.13 If k e Z", then k e U", the group of units of Zn if and only if the numbers k and n have no common factor.

A ring F is a field if every non-zero element of F is a unit. The study of fields is an essential part of Galois theory. Examples of fields are the sets Q of rational numbers, R of real numbers and C of complex numbers. Proposition 2.9 A ring R is afield if and only if {0} and R are the only ideals in R.

Proof Suppose that R is a field, that J is an ideal in R other than {0} and that

a is a non-zero element of J. Ifb e R then b = a(a‘1b) 6 J, and so J = R. Conversely, suppose that {0} and R are the only ideals in R. If a 75 0 then

= R, so that there exists b such that ab = 1, and a is invertible.

III

Corollary 2.10 If(I) is a ring homomorphism from afield F into a ring R then 45 is a monomorphism. Proof

For the kernel ¢" ({0}) is an ideal in F.

III

Corollary 2.11 If J is an ideal in a ring R, then the quotient R/J is afield if and only if J is a maximal proper ideal in R. Zn is a field if and only ifn is a prime number: Proof For if q : R —> R/J is the quotient map then I is an ideal in R/J if and only if q‘1 (I) is an ideal in R. III Suppose that F is a field. A subfield of F is a subset of F which is a field

under the operations inherited from F. Any subfield must therefore contain 0 and 1. The intersection of subfields is a subfield F0. The intersection of all subfields is therefore a subfield, the prime subfield of F. F is a commutative ring with a 1, and we can therefore consider the ring homomorphism ¢ : Z —> F described earlier in the chapter. Clearly ¢(Z) 9 F0. If ¢ has non-zero characteristic n, then, since F0 is an integral domain, n must be a prime number, and F0 is isomorphic as a field to Zp. On the contrary, if 4) has characteristic 0 then ¢(Z) is isomorphic to Z. If

r/s 6 Q, let ¢(q) = ¢(r)¢(s)‘1. This is well defined, since if q = r’/s’ then

rS’ = r’s so that ¢(r)¢(S’) = ¢(r’)¢(S) and ¢(r)¢(S)" = ¢(r’)¢(S’)". Thus «7) is properly defined, and it is easy to show that ¢(Q) is a subfield of F

and that :15 is a field isomorphism of 0 onto S, (15(0). Clearly every element of ¢(Q) is in F0, so that ¢(Q) = F0. Summing up:

2.5 Fields and Fractions

33

Theorem 2.12 Suppose that F is a field with prime subfield F0. If F has characteristic 0, then F0 is isomorphic as a field to Q. Otherwise, F has characteristic p, where p is a prime number; and F0 is isomorphic to the finite

field Zp. Many integral domains and fields that we shall consider are subsets of C, with the same operations of addition and multiplication. We shall call these numerical integral domains and numberfields.

We now show how to construct a field from an integral domain, in the same way that the field Q of rational fractions is constructed from Z. Suppose that R is an integral domain; let R* = R \ [0}. Intuitively, a fraction is an expression of the form a /b, with a e R and b e R*, but a fraction can be represented by many such expressions. We therefore proceed as follows. We define an equivalence relation ~ on R x R* by setting (a1,b1) ~ (a2,b2) if aibz = a2b1. It is immediate that (a,b) ~ (a,b) and that if (a1,b1) ~ (a2,b2) then (a2,b2) ~ (a1,b1). Finally, if(a1,b1) ~ (a2,b2) and (a2,b2) ~ (a3,b), then a1b2b3 = a2b1b3 = a3b1b2, so that (a1b3 — a3b1)b2 = 0; thus, since R is an integral domain, a1b3 — a3b1 = 0 and (a1,b1) ~ (a3,b3). Consequently ~

is an equivalence relation. Let F be the collection of equivalence classes; we denote the class to which (a,b) belongs by (a /b). We define operations of

addition and multiplication by ai/b1+az/b2 = (a1b2+azb1)/(b1.b2) and(a1/b1)(a2,b2) = (a102)/(b1b2)Exercises

2.14 Check that these do not depend upon the choice of representatives, that under these operations F is a field, with identity 1/1, zero element 0/1 and that the multiplicative inverse of a non-zero a/b equals b/a.

2.15 Show that the mapping a —) a/l is a ring isomorphism of R into F: a/l +b/1 = (a +b)/1 and (a/l).(b/l) = (ab)/l for all a,b e R. F is called the field offractions of R. If R is an integral domain with field of fractions F, the mapping i : R —> F defined by i(r) = r/1 is a monomorphism of R into F. 2.16 Suppose that R is an integral domain with field of fractions F, and that

j is an automorphism of R. Show that there is a unique automorphism j’ ofF such that j’i = ij (where i(r) = r/! ). What is it? 2.17 Identify the field of fractions of the ring Z+ iZ of Gaussian integers with a subfield of (C.

34

2 Integral Domains

To end this section, let us consider some groups that arise in the study of fields. Suppose that F is a field, and that F* is the set of non-zero elements ofF.Ifa e F“ letua(f) = af,forf e F, andifb e F letab(f) = f+b. Then Mp = {flu : a e F*} is an abelian group of permutations of F, with two orbits, the fixed point {0} and F*, and A p = {onJ : b e F} is a transitive abelian

group of permutations of F. The afi‘ine group Wp is then the group generated by Mp and Ap; since abua(f) = of + b = unab/a, every element of Wp can be written as yap), where y(a'b)(f) = af + b. The map w(a'b) —> flu is a monomorphism of Wp onto Mp, with kernel A p, so that A p n. (iii) If S is a non-empty set of proper ideals, there exists Jo in S which is maximal in S; if J e S then J does not strictly contain Jo. Proof Suppose that R is Noetherian and that (In) is an increasing sequence of proper ideals. Then J = U‘JZIJ; is a proper ideal, since 1;; 9! J. It is

therefore generated by a finite subset, and therefore by a subset of some J". Then Jm = J" for m > n, so that (ii) holds. Suppose that S is a non-empty set of proper ideals which does not satisfy

(iii). Let JO 6 S. Then there exists JI e S which strictly contains Jo. Repeating the procedure, we obtain a strictly increasing sequence of proper ideals in S, so that (ii) does not hold. Thus (ii) implies (iii). Suppose that (iii) holds, and that J is a proper ideal in R. Let S be the set of

all finitely generated ideals contained in .1. Then there exists an ideal Jo which is maximal in S. If a 6 J, then C Jo, so that Jo = J. Thus J is finitely generated, and (iii) implies (i). III

Corollary 2.18 (i) A proper ideal in a Noetherian integral domain is contained in a proper maximal ideal.

Exercise 2.21 Show that an integral domain R is Noetherian if and only if every ideal in R is finitely generated: if I is an ideal in R there exists a finite set A such that I = . If R is Noetherian is R[x] Noetherian?

2.7 Factorization In this and the next section, we suppose that R is an integral domain. If a is a non-zero element of R which is not a unit, we say that a factorizes if a = be, where neither b nor c is a unit. If a does not factorize, we say that a

2.7 Factorization

37

is irreducible; if a is irreducible and a = be, then either b or c is a unit. Thus a prime number is an integer greater than 1 which is irreducible in Z. We characterize irreducibility in terms of ideals. A proper principal ideal is an ideal < a >= aR, where a is a non-zero element of R which is not a unit. We say that a and a’ are associates if < a >=< a’ >. Proposition 2.19 Elements a and a’ ofan integral domain R are associates if and only if there exists a unit b such that a’ = ab.

Proof If a and a’ are associates, then a’ = ab for some b and a = a’b’ for some b’, so that a’ = a’b’b, and OR = a’(1R — b’b). Since R is an integral

domain, b’b = IR, and b and b’ are units. Conversely, if a’ = ab, where b is a unit, then a’ e< a >, so that < a’ >g< a >; but a = b‘la’, so that

§< a’ >,too.

El

Let PP denote the collection of proper principal ideals of R. We order PP by inclusion: < a >5< b > if and only if < a >g< b >. It follows from

Proposition 2.19 that < a >5< b > if and only if there is an element c such that a = cb; thus b divides a, which we write as bla.

An element < a > of PP is a maximal element of PP if it is not properly contained in another element of PP. A maximal element does not need to be the largest element of P P. Theorem 2.20 A non-zero element a of R is irreducible if and only if < a > is a maximal element of P P. Proof If a is irreducible, a does not divide 1R, so that < a > 6 PP. If < a > 5 < b >, then bla, so that a = bc; since a is irreducible, c is a unit, and < b >= < a >. Thus < a > is maximal. Conversely, if a is not irreducible and a = be, where b and c are not units, then < a >S< b >,but < b >£< a > ,sothat < a > isnotmaximal. El Corollary 2.21 If every non-zero element of R which is not a unit can be expressed as the product ofa finite number of irreducible elements, then every element of P P is contained in a maximal element of P P. We now introduce a condition which ensures that every non-zero element

of R which is not a unit can be expressed as the product of a finite number of irreducible elements. R is said to satisfy the ascending chain condition for principal ideals (ACCPI) if whenever 11 g 12 C 12 g

is an

increasing sequence of principal ideals then there exists n such that I». = In for all m > n.

38

2 Integral Domains

Exercise 2.22 A Noetherian ring satisfies the ACCPI. Show that Z[X] satisfies the ACCPI, but is not Noetherian. Theorem 2.22 If R satisfies the AC C PI then every non-zero element of R which is not a unit can be expressed as the product of a finite number of irreducible elements. Pmof

Suppose that a = a1 is a non-zero element of R which is not a unit. If

< a] > is maximal, a is irreducible, and there is nothing to prove. Otherwise, there exists < a; > such that < a1 >C< a2 > and < at >7£< a2 >. If < a2 > is maximal, we stop. Otherwise, we repeat the procedure, and con-

tinue. The process must stop after a finite number of steps, by the ACCPI. Thus there exists < an > with < an > maximal, and < an >z< a >. Thus an is irreducible, and a,I la. Hence, writing b0 = an, we can write a = bob]. If be is irreducible, we are done. Otherwise, we repeat, and continue. The process must terminate after a finite number of steps, for otherwise we get a strictly

increasing sequence of principal ideals (170,311 , contradicting the ACCPI. Thus a = bo.b1 . . . bk for some k, where the b,- are irreducible.

El

Exercises

2.23 Show that the polynomial x4 + a2x2 + 6x + 30 is irreducible in zm for all a e Z. 2.24 Find an element of Z[ifi7] which is the product of two irreducible factors and also the product of three irreducible factors.

2.8 Unique Factorization The numerical integral domain Z[i J5] = {a +i «Eb : a, b e Z} is Noetherian, so that any element can be factorized into irreducibles. But 6 = 2.3 = (l +

i~/§)(1 — N5), and 2,3,1 + N5 and 1 — i~/§ are irreducible in mix/5) (why?), so that factorization is not unique. The fact that factorization need not

be unique was not well understood in the beginning of the nineteenth century, leading to fallacious proofs of Fermat’s last theorem.

When is factorization unique? We need a little care. Suppose that r is an element of an integral domain R that factorizes as a product r = r1r2--- rn of irreducible elements of R, that 61,62, ...,6,, are units in R for which 6162 - - - 6,, = IR and that a is a permutation of (1,2, .. . ,n). Let r} = gram

2.8 Unique Factorization

39

for l 5 j S n. Then r = riré - - -r,’, is another factorization of r. We say that this is essentially the same factorization of r, and say that R is a unique factorization domain if every non-zero element which is not a unit can be

factorized as a product of finitely many irreducible elements, and if any two factorizations are essentially the same. If r is a non-zero element which is not a unit, in a unique factorization domain its length l(r) is the number of terms in its factorization into irreducibles. If r is a unit, we set I (r) = 0. Clearly 1 (rs) = l (r) + l(s). In order to characterize unique factorization domains, we need a new

concept. A non-zero element a of an integral domain is a prime if whenever albc then either a|b or alc. In terms of ideals, a is a prime if and only if whenever < bc >§< a > then either < b >§< a > or < c >§< a >.

A simple inductive argument shows that if a is a prime and a|b1.-- - .b,l then there exists j such that albj. Theorem 2.23 A prime element of an integral domain R is irreducible. Proof Suppose that a is a prime, that albc, so that a|b or alc. Suppose that alb, so that b = af, for some f. Then a = afc, so that, since R is an integral

domain, fc = 1 and c is a unit. Similarly, if a|c then b is a unit. Thus a is irreducible. El Corollary 2.24 A prime element of Z is a prime number: Exercise

2.25 Show that a prime number is a prime element of Z (use the fact that (p,a) is cyclic). Deduce the fundamental theorem of arithmetic, that

factorization into prime numbers is unique. Not every irreducible element of an integral domain is prime: in Z(iJ§),

2|(1+i~/§)(1 — N3), but 2+ (1 + N3) and 2 f (1 — N3). Primes are the key to unique factorization. Theorem 2.25 An integral domain R is a unique factorization domain if and only if R satisfies that AC C PI condition and every irreducible element of R is a prime.

Proof Suppose first that R is a unique factorization domain and that (< ai >)§’°l is an increasing sequence of proper principal ideals. Then the sequence ““019; is a decreasing sequence of positive integers, and so there exists n such that l(am) = l(an) for m z n. This means that am and an are associates for m 2 n, so that < am >=< an > for m 2 n. Thus R satisfies

40

2 Integral Domains

the ACCPI condition. Suppose now that r is irreducible and that rlab. We can write ab = rc. If a is a unit, r|b and if b is a unit, rla. Otherwise we factorize a, b and c:

a=S1---Sz. b=ti---tm, c=u1---um so that S] - - ~s1t] - - - t”, = rul - - -u,,. By unique factorization, r is an associate ofan s,- or a tj. Thus r divides a or b. Suppose conversely that r satisfies the ACCPI condition and that every

irreducible element of R is a prime. Then every element can be factorized into a product of irreducible elements, by Theorem 2.22. Suppose that there exist elements which do not have a unique factorization. Then among elements which do not have a unique factorization, there is an element a with a factorization a = b1b2 - - - b,, into irreducible elements, with n as small as

possible; a also has a different factorization a = mg - - ~cm into irreducible elements. Since b] is prime and b1 divides a = clcz - - - cm, there exists c1

such that al Ic]. By rearranging, we can suppose that j = 1, and that c1 = ald. Since c1 and al are irreducible, d is a unit and a1 and c1 are associates. Thus, if

a’ = a2a3 - - -a,., a' = d02C3 - - - cm. But a’ is the product of n — l irreducibles, and so the two products are essentially the same. Multiplying by a], it follows that the two original factorizations are essentially the same, giving the required

contradiction.

El

If R is a unique factorization domain, factorization of elements provides an easy proof of results, as the following theorem shows. Theorem 2.26 Suppose that B is a non-empty subset ofa uniquefactorization domain. Then there exists a e R such that alb for each b in B and such that if

a’ |b for each b in B then a’la. Such an element a is called a highest common factor of B. Any two highest

common factors are associates. Pmof

Pick b0 6 B and factorize it as b1b2---b,,. If J g {1, ...,n} let

p] = “1.6, bj, let .7 = {J : pJIb forallb e B} and let 1’ be a maximal element of .7. Then pylb for all b e B. If clb for all b e B, then clbo, and so c is an associate of bJ for some J e J. Since J’ is maximal, it follows that clpy. Thus p}: is a highest common factor of B. If a and a’ are highest

common factors of B then ala’ and a’ la, so that a and a’ are associates.

El

If the highest common factor of B is 1,2, we say that B is relatively prime. If p is a highest common factor of B then C = {c : Cp 6 B} is relatively prime.

2.9 Principal Ideal Domains and Euclidean Domains

41

Exercises 2.26 Suppose that R is an integral domain. Show that the following are

equivalent: (i) every finite non-empty set of non-zero elements of R has a highest common factor;

(ii) every finite non-empty set of non-zero elements of R has a least common multiple. 2.27 Suppose that R is an integral domain with the property that every nonempty set B of non-zero elements has a highest common factor of the form y1b1+---+ ynbn, with b1, . . . ,bn in B and y1, ...,y,, in R. Show

that R is a principal ideal domain.

2.9 Principal Ideal Domains and Euclidean Domains Recall that a principal ideal < a > in an integral domain is the set aR of all multiples of a by elements of R. An integral domain is a principal ideal domain if every ideal in R is principal. The integral domain Z is a principal ideal domain, since the sets 11% are ideals, and are the only ideals in Z.

Theorem 2.27 If R is a principal ideal domain, it is a unique factorization domain. Proof

We use Theorem 2.25. R is certainly Noetherian, and so it satisfies the

ACCPI condition. Suppose that a is irreducible, that albc and that a ,l’b. Let J =< a,b >. Since < a > is maximal andb ¢< a >, J = R, and there exist d,e such that 1R = da + eb. But then c = dac + ebc. Since albc, it follows

that alc. Thus a is prime.

El

In order to produce examples of principal ideal domains, we introduce

another class of integral domains. A function (13 : R —> Z+ on an integral domain is a Euclidean function if it satisfies the following three conditions: (i) ¢(a) = 0 if and only ifa = 0R; (ii) ifalb then ¢(a) 5 ¢(b); (iii) if a,b e R and b 9E 0 there exist q,r e R such that a = bq + r, where

(#0) < ¢(b). A Euclidean domain is an integral domain on which there exists a Euclidean function.

42

2 Integral Domains

Theorem 2.28 A Euclidean domain R is a principal ideal domain, and is therefore a unique factorization domain.

Proof Let (:3 be a Euclidean function on R. Suppose that J is a non-zero ideal in R, and let J* = J \ {0}. Let a be a non-zero element for which ¢(a) inf{¢(b) : b 6 J*}. Suppose that b 6 1*. Then by (iii) there exist q,r e such thata = qb+r, where ¢(r) < ¢(a). But r = a — qb 6 J, and sor = Thusb=qa,andJ=.

= R 0. El

Let us give some examples. First, Z is a Euclidean domain, with ¢(n) = |n|. Thus, as we have seen, factorization in Z is unique (the fundamental theorem

of arithmetic). Second, let R = Z[w] = z + wZ, where a) = (—1 + i~/§)/2 is a complex cube root of 1. Since 1 + a) + (02 = 0, we can write any element r of Z(w) uniquely as an + dwz. Set ¢(r) = |r|2. Since ¢(r) = c2 — cd + d2,

(1) is a non-negative integer-valued function which certainly satisfies (i) and (ii). Since the points of R form a hexagonal grid in (C with distance 1 between

adjacent points, if a,b e R and b aé 0 there exists a point q in R such that |a/b — q| < 1. Then ifr = a — qb, ¢(r) < ¢(b). Thus Z(w) is a Euclidean

domain. A similar argument shows that the ring Z(i) of Gaussian integers is a Euclidean domain. In contrast Z(2a)) = Z(i J3) is not a Euclidean domain,

since it is not a unique factorization domain (4 = 2.2 = (1 + i~/§)(1 — i #3)). Similarly, Z(2i ) is not a Euclidean domain, since it is not a unique factorization domain (8 = 2.2.2 = (2 + 2i )(2 — 2i )). Note that in these factorizations into irreducibles, the number of factors is different. The next example is important enough to warrant being called a theorem. Theorem 2.29 If F is a field, then the integral domain F [x] ofpolynomials over F is a Euclidean domain. Proof

Let ¢(a) be the degree of a (¢ (Op) = 0). Then (i) and (ii) are satisfied,

and (iii) follows by long division (an easy induction argument). Suppose that a = ao+~e+anx" andb = bo+---+bmx"',wheren andm arethedegrees ofa and b. Ifm < n, take q = 0;: and ifm = n, take q = bm/an. Suppose that the result is true when 0 5 m < n + k, where k z 0. Let q’ = (bm/an)x"

and let b’ = b — aq’: then the degree of b’ is less than m + k, so that there exist q” and r so that b’ = q”a + r, where ¢(r) < ¢(a). Now take q = q +q”. El If a0 and be are two elements of a Euclidean domain R, we can find the highest common factor h of a0 and b0 without having to factorize a0 and b0.

The method is known as Euclid ’s algorithm, and is probably known to the reader when R = Z. We proceed as follows. Suppose that ¢(ao) 5 ¢(bo). We choose qo such that if r0 = b0 — qoao, then ¢(ro) < ¢(ao). If r0 = 0 then

2.9 Principal Ideal Domains and Euclidean Domains

43

aolbo, and a0 is the highest common factor of a0 and b0. Otherwise, we write {a1,b1} = {ro,ao} (so that ¢(a1 S ¢(b1)). Since r0 = b0 — qoao, hlro, so that h is the highest common factor of a1 and b1. But ¢(al) < ¢(ao), and so the

process must terminate after a finite number of steps, with the first possibility. Let us give an example. What is the highest common factor of a0 = 5w + 11w2 and b0 = 13w + 2w2 in Z(w)? Since ¢(a0) = 91 and ¢(bo) = 147, ¢(ao) < ¢(b0). Since the inverse of c = aw + bat2 in the field of fractions

0(a)) is (ba) + aw2)/¢(c), bo/ao = (—770) + 56wz)/91.This suggests that we take qo = —w + (02. Then aoqo = 17w + w2 and r0 = b0 — (1q = —4w + wz. Then ¢(ro) = 21, so that we take 01 = re and bl = a0. Since

b1 /a1 = (—Sw + 2w2)/3, we take q1 = —2w + (02. Then qlal = 7w + l4w2, so that n = b1 —q1a1 = —2w— 3w2. Then (Mn): 7, so that we take a2 = r1

and b2 = a1. But then b2 = (—w + w2)a2, so that r1 = —2w — 3w2 is the required highest common factor. In fact, all these messy calculations are not necessary in this case. For (b is a multiplicative Euclidean function: ¢(ab) = ¢(a)¢ (b), so that h is a highest common factor of a and b if and only if ¢(h) is a highest common factor of ¢(a) and ¢(b). Thus h is a common factor of a0 and be if and only if ¢(h) = 7. We have found one; a simpler associate is w — 2(92.

Besides considering highest common factors, we can also consider lowest common multiples in a principal ideal domain. Proposition 2.30 Suppose that a and b are non-zero elements of a principal ideal domain R. Then there exists an element c, a lowest common multiple of a and b such that ale and blc and such that ifald and bld then cld. Proof The set < a > n < b > is an ideal in R, which is not 0 since ab e< a > n < b >;thusthereexistsc e Rsuchthat< c >=< a > n < b >. Then ale and blc, and ifald and bld then cld. III A lowest common multiple is not unique; if C] and c; are lowest common

multiples of a and b, then they are associates. We can obviously also define a lowest common multiple of a finite subset of R. Exercises 2.28 Using the unique factorization domain Z[i], show that the equations x2 +

2 = y3 and x3 + 1 = y2 each have a unique solution in N, which you should find.

2.29 Using the integral domain Z[i~/2], show that the equation x2 + 2 = y3 has a unique solution in N, which you should find.

44

2 Integral Domains

2.30 Suppose that R is an integral domain. Show that R[x] is a principal ideal

domain if and only if R is a field.

2.10 Polynomials Over Unique Factorization Domains Galois theory is largely concerned with polynomials in one variable, with coefficients in a field K. We shall, however, also need to consider polynomials with integer coefficients, and to consider polynomials in several variables. In

order to deal with both of these, it is convenient to study polynomial rings of the form R[x], where R is a unique factorization domain. In this section, we shall suppose that R is a unique factorization domain, with field of fractions F. If

f=ao+a1x+---+anx'l is a non-zero element of R[x], we define the content of f to be a highest common factor of the non-zero coefficients of f (the fact that this is not

uniquely defined causes no problems). If f has content 1 we say that f is primitive. If y is the content of f then f = yg, where g is primitive. If f is an element of R[x], we can consider f as an element of F[x]. The

next theorem provides a partial converse. Theorem 2.31 Suppose that R is a unique factorization domain. An element of R[x] is a unit if and only if it is a unit in R. If f is a non-zero element of F [x] we can write f = fig, where g is a primitive polynomial in R[x] and

)3 e F. If f = [3’g’ is another such expression then g and g’ are associates in R[x]; there exists a unit a in R such that g = sg’.

Proof The first statement is obvious. Suppose that f is a non-zero element of F [x]. We clear denominators: there exists 8 in R such that 8f 6 R[x]. Let y be the content of 8f. Then 6f = yg,

where g is primitive in R[x], and so f = (8"y)g = fig. Suppose that f = [3'g’ is another such expression. We again clear denominators: there exists a in R such that (#3 and afl’ are in R. Then af = (02,8)g =

(0113’)g’ . As g is primitive in R[x], up? is the content of af : so is afl’ (remember that the content is not uniquely defined!), and so 01/3 and afi’ are associates in R.

This means that there is a unit in R such that afi’ = eafl: aflg = afl’ ’ = saflg’, so that g = eg’ and g and g’ are associates in R[x]. El Theorem 2.32 Suppose that R is a unique factorization domain. If f and g are primitive elements of R[x], so is fg.

2.10 Polynomials Over Unique Factorization Domains

Proof

45

Suppose that

f =ao+a1x+---+anx",

g =bo+b1x+---+bmx"', and

fg = co + c1x + - - - + cm+nxm+". Let d be the content of fg and suppose that d is not a unit. Let r be an

irreducible factor of (1. As R is a unique factorization domain, r is a prime. Since f is primitive, there exists a least i such that r does not divide a.-; similarly there exists a least j such that r does not divide bj. As r is a prime, r does not divide ai b1-. We consider the coefficient

Ci+j = Zakbi+j—k + aibj + Zai+j—lblk ax from K x V into V which satisfies (a) a(x + y) = ax + ay,

(b) (a + fl)x = ax + fix,

(C) (afl)x = 010316), and (d) 1.): = x for all 01, fl in K and x, y in V.

As an example, let S be a non-empty set, and let Kx denote the set of all

mappings from S into K. If f and g are in K‘, define f + g by

(f + g)(s) = f (S) + g(s). fors in S, and, ifa e K, define af by

(aS) = a(f(s)), for sin S. Then it is easy to verify that the axioms are satisfied. If S = {1, . . . ,n}, we write K" for K‘ and, ifx e K", write x = (x1, ...,xn),

where xj is the value of x at j.

54

3.1 Vector Spaces

55

Another example that we shall study in detail is the following. Suppose that L is a field, and that K is a subfield. Then L is a vector space over K, when the algebraic operations are defined in the obvious way. For example, (C is a vector space over R and R is a vector space over Q. A subset W of a vector

space V over K is a linear subspace if, with the same operations of addition and multiplication, it is a vector space over K. If V] and V2 are vector spaces over the same field K, then a {K -linear mapping} (or simply a linear mapping, when the context is clear) from V1 to V2 is a mapping T : V1 —> V2 which satisfies

T(a + b) = T(a) + Nb).

T().a) = AT(a)

for all a,b 6 V1 and all A e K. The set of linear mappings from V] to V2 is denoted by MW, V2): if V1 = V2 it is also denoted by L(V]). When the operations are defined in the obvious way, L(V1, V2) is also a vector space over K: if T 6 MW, V2) then T‘1(0) is the null-space of T, and T(V1) the

image of T. Vector spaces lend themselves to elementary linear geometry. If V is a vector space over K then the line l(a, b) is the set

l(a,b)={(1—A)a+Ab:AeK}=a+{9(b—a):06K}. Exercise

3.1

Suppose that V is a vector space over K, that W is a linear subspace of V and that l(a,b) is a line in V. Then either l(a,b) C W or l(a,b) n W

has at most one point. Theorem 3.1 Suppose that W], . . . , Wn are proper linear subspaces of a vector space V over an infinite field K. Then V ;é U’;=1W,-.

Proof The proof is by induction. The result is certainly true when n = 1. Suppose that it is true for n — 1. Leta be an element of V not in U32: Wj and let b be an element not in W". If a = b there is nothing to prove. Otherwise, l(a,b) meets each W; in at most one point. Since 1 (a, b) has infinitely many points, there is a point e of l(a,b) not in any Wj. El

In fact, we need surprisingly little of the theory of vector spaces. The key is the idea of dimension; as we shall see, this turns out to be remarkably powerful. Suppose that V is a vector space over K. A subset W of V is a linear subspace if it is a vector space under the operations defined on V, for this, it is sufficient that ifx andy are in W anda is in K then x+y e W andax e W. IfA is anon-

empty subset of V, the span of A, denoted by span (A), is the intersection of the

56

3 Vector Spaces and Determinants

linear subspaces containing A; it is a linear subspace of V, and is the smallest linear subspace containing A. If span (A) = V, we say that A spans V. We now turn to linear dependence and linear independence. A subset A of a vector space V over K is linearly dependent over K if there are finitely many distinct elements x1, . . . ,xk of A and elements A1, ... ,Ak of K, not all zero, such that

Alxl+"'+A-kxk=0; if A is not linearly dependent over K, A is linearly independent over K. Note that, even if A is infinite, the sums which we consider are finite. If A is finite and A = {x1, . . . ,xn}, where the x,- are distinct, A is linearly independent over K if it follows from A1x1+---+h,,x,,=0

thatA1=A2=m=An =0. A subset A of a vector space V over K is a basis for V if it is linearly independent and spans V. Exercise

3.2

Suppose that V is a vector space over K. If A is a non-empty subset of V, show that the span of A is the set m

[b E V :b = ZAndn,)\n E K,an E A,m 6N] .

n=1

Our main interest is in finite-dimensional vector spaces; let us consider them

now. A vector space V over K is finite dimensional if there exists a finite subset of V which spans V. First we show that a finite-dimensional space has a finite

basis; this is a consequence of the following theorem. Theorem 3.2 Suppose that A is a finite subset of a vector space V over K which spans V, and that C is a linearly independent subset of A (C may be empty). There exists a basis B of V with C E B E A. Proof

Consider the collection J of all subsets of A which contain C and are

linearly independent. Since |A| < 00, there exists a B in J with a maximum number of elements. B is independent and C Q B Q A; it remains to show that B spans V.

3.1 Vector Spaces

57

Let B = {b}, .. . ,bn}, where the b,- are distinct. Ifa e A\B, B U {a} is linearly dependent (by the maximality of |B|) and so there exist Ao, . . . ,An in K, not all zero, such that hoa+hlb1+-~+hnbn =0. Further, A0 79 0, for otherwise b1, . . . ,bn would be linearly dependent. Thus

a = —).;').1b1— 15112122 — - -- — Ag‘xnbn and a e span (B). Consequently A E span (B), and so span (A) g span (B). As span (A) = V, the theorem is proved. El We would now like to define the dimension of a finite-dimensional vector space as the number of elements in a basis. In order to do this, we need to show that any two bases have the same number of elements. This follows from the next theorem. Theorem 3.3 Suppose that V is a vector space over K. If A spans V and C is a linearly independent subset of V, then |C| S IAI. Proof The result is trivially true if |A| = 00, and so we may suppose that |A| < 00. If |C| = 00, there is a finite subset D of C with |D| > |A|.

As D is again linearly independent, it is sufficient to prove the result when |C| < 00.

El

Theorem 3.3 is therefore a consequence of the following. Theorem 3.4 (The Steinitz exchange theorem) Suppose that C = {ch . . . , or} is a linearly independent subset (with r distinct elements) of a vector space V over K, and that A = {a1, . . . ,as} is a set (with s distinct elements) which spans V. Then there exists a set D, with C g D Q A U C, such that |D| = s and D spans V. Proof We prove this by induction on r. The result is trivially true for r = 0 (take D = A). Suppose that it is true for r — 1. As the set C0 = {c}, . . . ,c,_1} is linearly independent, there exists a set Do with Co Q Do 9 A U Co such that |Do| =s, and D0 spans V. By relabelling A if necessary, we can suppose that D0 = {cls - - ~ ycr—laaraar+lv - ° ‘9aS}'

If s were equal to r — 1, we would have D0 = C0; but c, e span (Do), so that we could write

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3 Vector Spaces and Determinants

r—l

Cr = 2 not. [:1

contradicting the linear independence of C. Thus s 2 r. As c, 6 span (Do), we can write r—l

s

c, = ci + Zajaj. j=r

i=1

Not all aj can be zero, for again this would contradict the linear independence of C. By relabelling if necessary, we can suppose that 0:, 9E 0. Let D = {C}, . . . ,Cr,ar+l, . . . ,as}. Then

r—l

a, =a,—1

s

c, — Zyici — Z ajaj i=1

j=r+l

so that a, e span (D). Thus Span (D) 2 {C}. - ' ' acr—laar,ar+ly - - - vas} = D0

and so span (D) 2 span (D0) = V.

This completes the proof.

El

Corollary (to Theorem 3.3) Any two bases of a finite-dimensional vector space have the some finite number of elements. We now define the dimension of a finite-dimensional vector space V over

K to be the number of elements in a basis. We denote the dimension of V by dim V. If V is not finite dimensional over K we set dim V = 00. Here is one simple but important result: Theorem 3.5 Suppose that U is a linear subspace of a finite-dimensional vector space V over K. Then dimU S dim V, and dimU = dim V if and only if U = V. Proof

Let A be a basis for U, and let C be a finite set which spans V.

Considered as a subset of V, A is linearly independent, and so by Theorem 3.2 there is abasis B of V with A g B Q A U C. Thus

dimU = |A| g |B| = dim V. If dimU = dim V, we must have A = B, so that A spans V and U = V; of course ifU = V,dimU =dim V. [II Corollary 1 Suppose that A is a finite subset of a finite-dimensional vector space V over K. If |A| > dim V, A is linearly dependent.

3.1 Vector Spaces

59

Proof Let U = span (A). If A were linearly independent, A would be a basis for U , so that dim U = |A|. But dim U g dim V, giving a contradiction. El Suppose that V1 and V2 are vector spaces over the same field K. A mapping 4) from V1 into V2 is called a linear mapping if

«#06 +30 = ¢(X) +¢(y). ¢(>~x) = A¢(x) for all x and y in V1 and all A in K. The study of linear mappings is an essential part of the study of vector spaces; for our purposes we shall only need one further corollary to Theorem 3.5. Corollary 2 Suppose that V1 and V2 are vector spaces over K and that d) is a linear mapping of V1 into V2. If dim V] > dim V2, (1) is not one-to-one, and there exists a non-zero x in V] such that 45 (x) = 0. Proof Let n = dim V2. As dim V1 > dim V2, there exist n + 1 linearly independent vectors x], . . . ,xn+1 in V1. Then, by Corollary 1, {¢(x1), ..., ¢(x,,+|)} is linearly dependent in V2, and so there exist in, . . . ,An+l in K, not all zero, such that

11¢(xl) + ---+ An+1¢(xn+1)= 0. But

11¢(x1)+ --- + An+1¢(xn+1)= ¢0~1x1 + - - - + An+1xn+1). since 4) is linear, and

x = 11x1+---+ An+1xn+19'é 0 since {x1, . . . ,xn+1} is linearly independent. As ¢(x) = 0 = 45(0), 4) is not one-to—one. El Suppose that V1 and V2 are finite-dimensional vector spaces over the same field and that T 6 UV], V2). The image T(V1) is a linear subspace of V2; its dimension is the rank r(T) of T. Its null space N (T) = T‘1(0) is a linear subspace of V1; its dimension is the nullity n(T) of T. Theorem 3.6 Suppose that T e L(V1, V2). where V1 is finite dimensional. Then n(T) + r(T) = dim(V1). Proof Let (a1, . . . ,an) be a basis for n(T). Extend it to a basis (a1, . . . ,an, b1, . . . ,bj of VI. Then T(b1), . . . , T(b,-) span T(v|). But they are also linearly independent, for ifAlT(b1)+- - -+AjT(bJ-) = 0, then T(A1b1+- - ~+Ajbj) = O,

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3 Vector Spaces and Determinants

and (Albl + - -- + Ajbj e N(T). Since b1, ...,bj are linearly independent, A; = 0 for l 5 i 5 j; thus b1, ...,bj are linearly independent, and therefore form a basis for T(Vl). Hence r(T) = j. Since dim(V1) = n + j, the result follows. E Corollary 3.7 Suppose that V] and V2 are finite-dimensional vector spaces over K.

(i) There is a linear isomorphism of V1 onto V2 ifand only if V] and V2 have the same dimension. (ii) If T e L(Vl, V2) and dim(V1) > dirn(V2) then there exists x 75 0 such that T(x) = 0. Exercises 3.3

In K", let ej = (0, ...,0,1,0,...,O), where the 1 occurs in the jth position. Let f; = e1+---+e,-. (a) Show that {e1, . . . ,en} is a basis for K". (b) Show that {f}, ...,f,,} isabasis for K". (c) Is {e1,f1,f2, ...,fn} abasis for K"?

3.4

Suppose that S is infinite. For each s in S, let es (t) = 1 if s = t, and let es(t) = 0 otherwise. Is {esz s e S} a basis for K5 ?

3.5

R can be considered as a vector space over Q. Show that R is not finite dimensional over Q. Can you find an infinite subset of R which is linearly independent over Q?

3.6

Suppose that K is an infinite field and that V is a vector space over K. Show that it is not possible to write V = U7=1 Ui, where U1, . . . , Un are proper linear subspaces of V.

3.7

Suppose that V is a finite-dimensional vector space and that T e L(V).

Show thatn(Tj) 5 jn(T).

3.2 The Infinite-Dimensional Case Suppose that V is an infinite-dimensional space. Does it have a basis? We begin with the countable case. Theorem 3.8 Suppose that V is a vector space with countably many points. Then V has a basis.

Proof Suppose that V = (xn):°=l. Let B = {xn : xn ¢ span{x,- : j < n}. Then B is linearly independent, and spans V.

El

3.3 Characters and Automorphisms

61

If V is uncountable, we need Zorn’s lemma (see the Appendix). Theorem 3.9 Suppose that A is a subset of a vector space V over K which spans V and that C is a linearly independent subset of A (C may be empty). There exists a basis B of V with C 9 B Q A.

We have proved this in the case that A is finite in Theorem 3.2 (and made essential use of the finiteness of A). Taking A = V and C the empty set, we see that this theorem implies that every vector space has a basis. Proof

Let S denote the collection of subsets of A which are linearly

independent and contain C. Order S by inclusion. Suppose that T is a chain in S. Let E = UDeT D. E is a subset of A which contains C. Suppose that x1, . . . ,xn are distinct elements of E. From the definition of E, there are sets D1, . . . ,0" in T such that x; e D,~ for 1 g i g n. Since T is a chain, there exists j, with 1 S j g n, such that D; g Dj for 1 S t g n. Consequently

x1, . . . ,xn are all in Dj. As DJ- is linearly independent, {x1, . . . ,x,,} is linearly independent. As this holds for any finite subset of E, E is linearly independent.

Thus E e S. E is clearly an upper bound for T, and so every chain in S has an upper bound. El We can therefore apply Zom’s lemma, and conclude that S has a maximal element B. B is linearly independent and C Q B Q A; it remains to show that span (B) = V. Since span(A) = V, it is enough to show that if a e A \ B then a e span(B). Let Bo = {a} U B. Then B0 is not linearly independent, and so there exist b1, . . . ,bn distinct elements of B and 10,11, . . . ,1", not all 0, in K, such that Aoa + Albl + ,Anbn = 0; since b1, . . . ,b,, are linearly independent, A0 9E O anda = Aaqb] + - - - + hub") e span(B).

3.3 Characters and Automorphisms Let us now establish results about linear independence that we shall need later. Suppose that G is a group and that K is a field. A (K -valued) character on G is a homomorphism of G into the multiplicative group K * of non-zero elements of K. We can think of a character as a K -valued function

on G; recall that the set of all K-valued functions on G is a vector space over K. Theorem 3.10 Suppose that G is a group, that K is afield and that S is a set of K -valued characters on G. Then S is linearly independent over K.

62

3 Vector Spaces and Determinants

Proof If not, there is a minimal non-empty subset {y}, . . . ,yn} of distinct elements of S which is linearly dependent over K. That is, there exist non-zero A1, . . . ,An in K such that

Aiyi(g)+---+Anyn(g)=0

(*)

for all g in G. Each yi is non-zero, since it sends the identity of G to l, and so it 2 2. As yl aé 34,, there exists h in G such that y; (h) qé 31,, (h). Now

A1r1(hg)+---+Anyn(hg) = 0 for all g in G. Using the fact that the y,- are characters, we have that

Aly1(h)y1(g) + - - - + Anyn(h)yn(g) = 0 for all g in G. Now multiply (*) by )4, (h) and subtract:

Mm (h) — yn(h))y1(g) + - - - + An_1(yn—1(h) - yn(h))rn_n(g) = 0 for all g in G. As yl (h) — y,, (h) 5A 0, this means that {y}, . . . , yn_1} is linearly

dependent over K, contradicting the minimality of {1/1, . . . , yn }.

[II

If r is an automorphism of a field K, then the restriction of t to K* is a K-valued character on K*. Spelling the theorem out in detail in this case, we have the following corollary: Corollary Suppose that 1:], . . . , 1'" are distinct automorphisms ofafield K and that k], . . . ,kn are non-zero elements of K. Then there exists k in K such that k|t|(k) + " - +k,.r,,(k) 75 0.

Corollary 3.11 (Dedekind’s lemma) Suppose that S is a set offield homomorphisms from a field F1 into a field F2. Then S is a linearly independent set of Fz-valuedfimctions on F1.

Proof For the restrictions of S to the multiplicative group Ff form a linearly independent set of Fz-valued functions on F1“. CI

3.4 Determinants The theory of determinants is very large, and we shall only consider the basic results that we shall need.

Suppose that K is a field and that n e N. A K-valued n x n matrix A is a function from n x n to K; we write A(ij) as at}. The ith row r,- of A is the element (a,-j) i=1” of K”, and the jth column of is the element (aij)?=l of K”.

3.4 Determinants

63

This defines a linear mapping from K" into itself: if x = (x1, . . . ,x,.) then A(x) = y, where y,- = 233:1 aijxj. For example, let 1,,(ij) = l ifi = j and let 1,, (ij ) = 0 otherwise. 1,, is the unit matrix: it defines the identity mapping on K". We use the determinant of A to determine when A is invertible. Let Mn(K) be the set of n x n matrices; it is a vector space over K of dimension n2. If A e M,l (K), we define the determinant of A to be det(A) = det(cl, . . . ,0") = Z Egalgu) . . . and“), 062,.

where 6,, is the signature of 0.

The determinant has the following properties: Theorem 3.12 Suppose that K is a field and that A = (aij) e K": let A = (c1 . . . c"), where c,- is the jth column of A.

(i) [fl 6 K then det(lc1,c2, . . . ,c,,) = Adet(A). (ii) det(cl,cz, . . . ,0") + det(c’1,cz, . . . ,0") = det(cl + c’l,cz, . . . ,0"). (iii) [ft 6 2,, then det(c,(1), . . . ,Cr(n)) = fir det(c1, . . . ,6"). (iv) det(c1, . . . ,cn) = det(r1, . . . ,rn), where r; is the ith now ofA. (v) det(c1, . . . , en) = 0 if and only if Q, . . . , cn are linearly dependent. Proof

(i), (ii) and (iii) follow easily from the definition.

(iV) det(rl, . . o ,rn) =

Z 6000(1)] n a . aa'(n)n 062,,

= Z 617016—10). . . and—10,) 062,,

= Z €;1010(l)- - - an0(n) 062,,

= Z 60010(1) - - - an0(n) 062,,

= det(c1, . . . ,cn). (v) Suppose first that an, .. . ,an are linearly dependent. By renumbering if necessary, there exist A1 = 1,)»2, . . . ,An such that = 223:] Ma, = 0. But then det(a1, ,an) = det(b1,a2, . . . ,an) = 0.

If a], . . . , an are linearly independent, then, by rearranging if necessary and using (i)—(iii), we can construct a triangular matrix (t1, . . . , t,,) such that ti,- 9e 0

64

3 Vector Spaces and Determinants

for] S i 5 n and ti,- = 0 forj < i, and such that det(tl,...,t,,) = det(al, o o o ,8"). But det(tl, . o . ,tn) = “21:1 tn" # 0-

U

Theorem 3.13 If A, B 6 Mn (K) then det(AB) = det(A). det(B). Proof

Since Abj = Zil=1 aibij,

det(AB) = det(Abl, ...,Abn) n

n

= det (Z aiibill, . . . , Z ainbinn)

i1=1

J'n=1

= Z det(al. ....a,.)e(i1,....i,.>b.-.1. ...,b.-n.. i]...i,,

= det(A). det(B).

[3

Exercise 3.8

Use (iv) and (v) of Theorem 3.4 repeatedly to show that if A 6 Mn (K) there exists T e Mn(K) for which det(T) = det(A) and Ti} = 0 for i > j (T is triangular). What is det(T)? Use this result to give another proof that if A, B e Mn(K) then det(AB) = det(A) det(B).

This means that we can define the determinant of an endomorphism of a finite-dimensional space. Suppose that V is finite dimensional and that S e L(V, V). Let j and j’ be linear isomorphisms of K " onto V.

Theorem 3.14 det(j‘l Sj) = det(j"1Sj’).

Proof LetL = j“j’.Then det(j‘lSj) = det(L‘1(j_'Sj)L) = det(j"'Sj’).

III We can therefore define det(S) = det(j ‘1 Sj ). Theorem 3.15 Suppose that A e L(V), where V is afinite-dimensional vector space over K. Then A has an inverse A‘1 (AA‘1 = A‘IA = I) ifand only if

det(A) 9’: 0. Proof Let j : K" —> V be a linear isomorphism, so that A has an inverse if and only if B = j‘lAj does, and det(A) = det(B). If B = (b1, ,bn), then B ihas an inverse if and only if (b1, .. . ,bn) are linearly independent, which

happens if and only if det(B) 95 0.

El

3.4 Determinants

65

Exercises

3.9

Suppose that d: Mn(K)—> K is linear in each variable aj, that d(e1, ...,e,,) = land that d(al, ,3") = 0 if a,- = aj for some i 95 j. Show that d = det. 3.10 Suppose that A, B, CD 6 Mn(K). Define E e M2,.(K) and show that det(E) = det(A) det(D) — det(B) det(C). 3.11 Suppose that A, B 6 Mn (K) and that A is invertible. Show that A + AB

is invertible for all but finitely many A in K. Suppose that A e Mn(K), thatx e K”, that A e K and that Ax = Ax. Then 1: is an eigenvector of A and A an eigenvalue. The polynomial p(x) = det(A — x1) is the characteristic polynomial of A.

3.12 Suppose that A and B are non-zero elements of M" (K) and that AB = 0. Show that det(A) = det(B) = 0. Suppose that A, B,C e M,.(K) and that BC = In. Show that A and AB have the same characteristic polynomial.

3.13 Suppose that A,P e Mn(K) and that P2 = P. Show that AP and PA have the same characteristic polynomial. 3.14 Suppose that A,B e Mn(K). Show that there is an invertible J 6 Mn(K) such that J A is a projection. Deduce that AB and BA have the

same characteristic polynomial. 3.15 Suppose that A e Mn(K). How would you find the eigenvalues and eigenvectors of A? How many eigenvalues can there be for which the equation A(x) = Ax has a solution in K"? 3.16 Suppose that (bj)’;=l e K”. Define A 6 Mn (A) by setting (1.3.4.1 = 1, for l 5 i < n, a,” = bn_j+1 and aij = 0 for other values ofi and j.

What is the characteristic polynomial of A? What polynomials in K [x] can be characteristic polynomials? 3.17 Let (:3 = (¢1,¢2) be a bijection from (l,...,n]n2) onto (1, ...,n1)x (1, ...,n2). Suppose that A e Mn](K) and B e Mn2(K). Let C e az be defined as Cij = 61¢, (0,,»l (j)b¢2(i).¢2(j). Show that det(C) =

det(A)"l det(B)”.

PART II The Theory of Fields and Galois Theory

4 Field Extensions

4.1 Introduction One of the main topics of Galois theory is the study of polynomial equations. In order to consider how we should proceed, let us first consider some rather trivial and familiar examples.

Polynomials involve addition and multiplication, and so it is natural to consider polynomials with coefficients in a ring R. If we consider the simplest

possible case, when R = Z and p is a polynomial of degree 1, we find there are difficulties: for example, we cannot solve the equation 2x + 3 = 0 in Z. In the case where R is an integral domain, the field of fractions is constructed in order to deal with this problem. Thus, in the example above, if we consider 2 and 3 as elements of Q, the rational field, then the equation has a solution x = —3/2 in Q. Let us now consider a quadratic equation: x2 — 2x — l = 0. We consider this

as an equation with rational coefficients: completing the square, we find that

(x — 1)2 =2. But there is no rational number r for which r2 = 2. For if r = a/b in the lowest terms, then a2 = 2b2, so that a is even, and a = 2c, say. But then

b2 = 202, so that b is also even, giving a contradiction. Instead, the first natural idea is to consider the polynomial as a polynomial with real coefficients: the equation then factorizes as (x — 1 + s/i)(x — 1 — fl) = 0, and we have solutions 1 — fl and 1 + J5. The field R is rather large, however (R is uncountable, while Q is

countable), and it is possible to proceed more economically. The set of all real numbers of the form a + bfi, where a and b are rational, forms a field K: addition and multiplication are obvious, and (a + b~/§)'l = c + d~/§ where

c = a/(az — 21:2) and d = —b/(a2 — 2bA/i). Clearly Q g K g R, and 69

70

4 Field Extensions

K is much smaller than R, since K is countable and IR is not. If we consider x2 — 2x — l as an element of K [x], we can solve the equation in K. Let us express all this in more algebraic language. The polynomial x2 — 2x — l is irreducible in Q[x] (and is therefore irreducible in Z[x]). If, however, it is considered as an element of K [x] or 1R[x], it can be written as a product of linear factors. This suggests the following general programme: given an element f in K [x] (where K is a field), can we find a larger field, L say, such that f considered as an element of L[x] can be written as a product of linear factors? If so, can we do it in an economical way?

4.2 Field Extensions Suppose that we start with a field K. In order to construct a larger field L we frequently have, by some means or another, to construct L, and then find a subfield of L which is isomorphic to K (think of how the complex numbers are constructed from the reals). It is occasionally important to realize that this sort

of procedure is adopted: for this reason we define an extension of a field K to be a triple (i, K, L), where L is another field, and i is a (ring) monomorphism

of K into L. However, much more frequently this is far too cumbersome. If (i, K, L) is an extension of K, the image i(K ) is a subfield of L which is isomorphic to K;

we shall usually identify K with i(K ) and consider it as a subfield of L. In this case we shall write L : K for the extension. Thus C : R is the extension of the real numbers by the complex numbers and IR : Q is the extension of the rational numbers by the real numbers. Very occasionally, when the going gets

rough, we shall need to be rather careful: in these circumstances we shall revert to the notation (i, K, L). Suppose now that L : K is an extension. How do we measure how big the

extension is? It turns out that the appropriate idea is dimension, in the vector space sense. This is an almost embarrassingly simple idea: the remarkable thing is that it is extraordinarily powerful.

To begin with, then, we forget about many of the field properties of L. Theorem 4.1 Suppose that L : K is an extension. Under the operations

(11.12) -> 11 +lzfromL x L toL and (k,l) —> kl from K x L to L, L is a vector space over K.

4.2 Field Extensions

Pmof All the axioms are satisfied.

71

El

Thus C is a real vector space, and R is a vector space over the rationals Q. We now define the degree of an extension L : K to be the dimension of L

as a vector space over K. We write [L : K] for the degree of L : K. We say thatL : K isfiniteif[L : K] < oo,andthatL : K isinfiniteif[L : K] = 00. Thus [C : R] = 2, [1R : Q] = 00, and, ifK is the field ofall r + sfi, with r and s rational, [K : Q] = 2. In this sense, then, K : Q is a more economical extension for solving x2 — 2x — 1 = 0 than R : Q.

The next theorem is very straightforward (there is an obvious argument to try, and it works), but it is the key to much that follows. If M : L and L : K

are extensions, then clearly so is M : K.

Theorem 4.2 Suppose that M : L and L : K are extensions. Then [M : K] = [M : L][L : K]. Proof First suppose that the right-hand side is finite, so that we can write [M : L] =m < oo,and [L : K] =n < oo.Let(x1,...,xm)beabasisforM over L, and let (yl, . . . , y") be a basis for L over K. We can form the products ii (for l S i g m, 1 g j g n) in M. We shall show that the mn elements (ii : l g i g m, 1 < j g n) formabasisforMoverK. First we show that they span M over K. Let 2 e M. As (x1, .. . ,xm) is a basis for M over L, there exist on, . . . ,am in L such that z=a1x1 +---+amxm.

As each at,- is in L, and as (yl, . . . ,yn) is a basis for L over K, for each i there exist fl“, . . . ,flin in K such that

“i = fii1y1+m+ flinyn. Substituting, m

n

z = Z Z flijxi. i=1 j=l which proves our assertion. Second we show that (ii : 1 < i g ml g j S n) is a linearly independent set over K. Suppose that m

n

0 = Z Z Viixi i=1 j=1

72

4 Field Extensions

where the my are elements of K. Let us set It

5i = ZVijG L)

i=1 for1i(K)(B1)

I:

In

K(a1)

'C

K

> M)

We now consider f as an element of K (a1)[x]. We can write f = (x —at1 )h, where h e K(a])[x], and h splits over 2: h=A(x—a2)...(x—a,,). Also 2 = K(a1)(a2, . . . ,an), and so 2 is a splitting field for h over K(ai1).

As degree h = n — l, we can apply the inductive hypothesis: there exists a monomorphism j from E to L such that j|K(a1) = j]. This completes the proof. El Before we establish some corollaries, let us make three remarks. First, like Theorem 6.4, this is an extension theorem: we extend the mapping i. Second, unlike Theorem 6.4, the extension need not be unique: we could map a. to

any of m, .. . , [3,. Third, although the extension need not be unique, there are obviously some limitations on the number of extensions that there can be. This is a topic to which we shall pay much attention later on. Corollary 1 Suppose thati: K —> K’ is an isomorphism and that f e K [x]. Suppose that E : K is a splitting field extension for f, 2’ : K’ a splitting field extension for i (f). Then there exists an isomorphism j: E —> 2" such that

“K = iProof

If we apply the theorem to the mapping i, considered as a monomor-

phism from K to 2’, it follows that there exists a monomorphism j from 2 to 2’ which extends i. We can write f=),(x—a1)...(x—Otn),

94

6 Splitting Fields

witha1,...,at,.inEandAinKJ‘hen

J'(f) = i()»)(x - j(d1))--- (X - j(an)). so that, using Theorem 6.1, it follows that

2' = K'(J'(a1), -.~,J'(an)) S KS), and j is onto.

[II

This leads to the following fundamentally important result: Theorem 6.6 Suppose that f e K [x] is irreducible and that E : K is a splitting field extension for f. If a and 6 are roots of f in 23, there is an automorphism a: )3 —> 2 such that 0(a) = fl and a fixes K. Proof

We may suppose that f is monic: then f is the minimal polynomial

for a and [3 over K. By Corollary 1 of Theorem 6.4, there is an isomorphism r: K(a) —> K03) with 1(a) = ,6 and t(k) = k fork e K. Now 2 : K(a) is a splitting field extension for f over K (a), and 2 : K (,8) is a splitting field

extension for f over K (,3). The result now follows from Corollary 1.

CI

Exercise

6.9

The complex numbers i «5 and 1 + i «5 are roots of the quartic f = x4 — 2x3 + 7x2 - 6x + 12. Does there exist an automorphism o of the splitting field extension for f over Q with o(i J3) = 1 + i J3?

6.4 Some Examples We now consider some examples of splitting fields. First let us consider

polynomials in Q[x]. If f e Q[x] then, as we saw at the beginning of this chapter, f splits over (C[x], and we can, and usually shall, consider the splitting

field of f as a subfield of (C. Such a field is called a number field. Alternatively, we can make the constructions of Theorem 6.2 and 6.3. Corollary 1 to Theorem 6.5 then says that the splitting field that we obtain is essentially the same. Example 6.4.1 f = xl’ — 2 in Q[x] (with p a prime). f is irreducible, by Eisenstein ’s criterion, and there is one real positive root

Zl/Pf is the minimal polynomial of21/p, so that [Q(21/”) : Q] = p. [fa is any root off in C, then (ct/21”)” = cup/2 = 1, so thata = 2l/Pa), where a) is a root ofx” — l. x” — 1 is not irreducible, as

xP—1=(x—1)(xP“+xP‘2+---+1).

6.4 Some Examples

95

Now x”‘1 + x”'2 + - - - + 1 is irreducible over Q (Exercise 2.39), so that ifa) is any root ofx” - 1 other than I then [Q(w) : Q] = p — l. The map n —) a)” is a homomorphism of Z into the multiplicative group C“, with kernel pZ, and so the complex numbers I, w, . . . ,wp’l must be distinct. They are all roots of

x" — 1, so that

xP—l=(x—1)(x—w)...(x—wP-1) and Q(w) : Q is a splittingfield extensionfor x” — 1. Now our original polynomial f splits over Q(w, 21/”) since it has roots

2'/P,w21/P, ...,wP—‘21/P. Further; any splitting field must contain 21/”, and must also contain

or =w21/P/21/p. Thus Q(w,21/p) : Q is the splittingfield extensionfor f. What is [Q(w,21/P) : Q]? In order to answer this, consider the following diagram:

0(0), 2””) Q(21Ip)

QW’)

Q /

Here (and later; when we consider similar diagrams) rising armws represent inclusion mappings.

By the tower law,

[001”) : Q]|[Q(w.2‘/P) : Q] and [Q(w) : Q]|[Q(w.2‘/P) :01. As [Q(21/”) : Q] =p and [Q(a)) : Q] =p — 1, this means that [Q(w,2'/P) : Q] 2 p(p — 1). However if m is the minimal polynomial of 21/P over Q(w), m divides x” — 2 in Q(a)) [x], and so

degree m = [Q(w,21/p) I Q(w)] < P-

96

6 Splitting Fields

Thus, by the tower law,

[000,2‘”) : Q] = [Q(w,2””) : Q(w)][Q(w) : Q] < p(p - 1),

and SO [Q(w,2'/”) 3Q] = P(P - 1)This implies that degree in = p, and so xP — 2 is irreducible over Q(w). This example has many important features. It is perhaps a bit more

complicated than one might imagine. Notice that the pth roots of unity (the roots of x” — 1) played an important role. Notice also that we picked one of them (other than 1): had we picked another, the result would have been the same. (Can you formalize this, using Theorem 6.6?) Notice also that the argument could have been simplified by appealing to Exercise 4.3.

Example 6.4.2 f = x6 — 1 in Qm. f factorizes as

f = (x —1)(x2 +x + l)(x +1)(x2 —x +1). Ifw is a mot 0f + x + 1 then

f = (x —1)(x — w)(x — wzxx +1)(x +w)(x +w2). Thus Q(w) : Q is a splitting field extensionfor f and [Q(w) : Q] = 2.

Example 6.4.3 f = x6 + 1 in Q[x]. The roots off in (C are i, i a), iwz, —i, —iw, —iw2. Thus, arguing as before,

Q(i,w) : Q is the splitting field extension for f, and we have the following diagram:

QU. 60)

\Q(w)

/\

0(1)

Q/ Now we can take a) = (—1 + J3 i ) /2, so that 0) ¢ Q(i) (which consists of all complex numbers oftheform r +is, with r and s in Q). Thus Q(i) 75 (1(0)) and both Q(i) and Q(w) are proper subfields of Q(i , at). It is now easy to conclude

that [Q(i,a)) : Q] = 4.

6.4 Some Examples

97

We now consider examples over more general fields. (To what extent do

we use the fact that we are considering polynomials over the rationals in Examples 6.4.1 to 6.4.3?) Exercises

6.10 Suppose that M : L and L : K are extensions, and that a e M is algebraic over K. Does [L (a) : L] always divide [K (a) : K]?

6.11 Write down all monic cubic polynomials in Zz[x], factorize them completely and construct a splitting field for each of them. Which of these fields are isomorphic?

6.12 Find a splitting field extension K : Q for each of the following polynomials over Q: x4 — 5x2 + 6, x4 + 5x2 + 6, x4 — 5. In each case determine the degree [K : Q] and find a such that K = Q(a).

6.13 Find a splitting field extension K 2 Q for each of the following polynomials over Q: x4 +1, x4 +4, (x4+ 1)(x4 +4), (x4 — l)(x4 +4). In each case determine the degree [K : Q] and find a such that K = Q(a).

6.14 Suppose that L : K is a splitting field extension for a polynomial of degree n. Show that [L 2 K] divides n!

6.15 Find a splitting field extension for x3 - 5 over Z7,l and 213.

7 Normal Extensions

7.1 Basic Properties There are many field extensions. How do we recognize when an extension L : K is a splitting field extension? For this we need the notion of a normal extension. An extension L : K is said to be normal if it is algebraic and whenever f is an irreducible polynomial in K [x] then either f splits over L or f has no

roots in L. Clearly an algebraic extension L : K is normal if and only if the minimal polynomial over K of each element of L splits over L.

The word ‘normal’ is one of the most overworked words in mathematical terminology (normal subgroups, normal topological spaces, . . .). We shall see in Theorem 11.8 that this is a good use of the word. In order to characterize normality, we need to extend the definition of a

splitting field. Suppose that K is a field, and that S is a subset of K [x]. We say that an extension L of K is a splitting field extensionfor S if each f in S splits over L, and if L 2 L’ 2 K and each f in S splits over L’, then L’ = L. If S is a finite set {f1, .. . , fn} then L : K is a splitting field extension for S if and only if it is a splitting field extension for g = fl . . . fn; thus the new definition is only of interest if S is infinite. Theorem 7.1 An extension L : K is normal if and only if it is a splitting field extension for some S 9 K [x].

Proof

Suppose first that L

:

K is normal. L

:

K is algebraic: let

S = {ma: (1 e L} be the set of minimal polynomials over K of elements of L. By hypothesis, each f in S splits over L, and clearly S splits over no proper subfield of L.

98

7.1 Basic Properties

99

Conversely, suppose that L : K is a splitting field extension or S. Let A denote the set of roots in L of polynomials in S. Then clearly L = K (A), and so L : K is algebraic, by Corollary 2 to Theorem 4.7. Suppose that ,6 e L and that m is its minimal polynomial over K. We

must show that m splits over L. First we reduce the problem to one concerning finite extensions. As fl 6 K (A), there exist on, . . . ,an in A such that ,8 e K(al, . . . ,an). There exist f1,...,f,, in S such that 01,-. is a root of f1, for 1 5 i 5 n. Each f; splits over L. Let R be the set of roots ofg = f1...f,,. Then K (R) : K is a splitting field extension for g and fl 6 K (R). We now

consider m as an element of K (R) [x] and construct a splitting field extension H : K (R) for m. Let y be another root of m in H. We must show that in fact

3/ e K (R). We have the following diagram, where upward-pointing arrows denote inclusions:

H

K(R)

TK(fi)

Km. 7)

\ / K(r)

m is the minimal polynomial of both ,3 and 3/ over K, so that [K (,6) : K] = [K (y) : K] = degree m. Also, by the corollary to Theorem 6.4, there is an isomorphism ‘l.’ of K ([3) onto K (y) which sends fl to y and which fixes K. As I fixes K,t(g) = g. Now K (R) 2 K (B) is a splitting field extension for g over K (,6), and K (R,y) : K (y) is a splitting field extension for r(g) = g over K (y), so

that by Corollary 1 to Theorem 6.5 there is an isomorphism a of K (R) onto K(R, y) such that 0|K(fi) = r. This means that [K(R) : K(fl)] = [K(R,y) : K (y)], and so by the tower law

[K(R) : K]=[K(R) I K(fl)][K(I3) I K] =[K(R,y): K(y)][K(}/)1K] =[K(R.V) I K].

100

7 Normal Extensions

But K(R) g K(R,y), and so we must have that K(R): K(R,y). Consequently, y e K (R). CI

The case of finite extensions is particularly important: Corollary 1 A finite extension L : K is normal if and only if L : K is a splitting field extension for some g e K [x]. For if L : K is normal and finite, and a],

,an is a basis for L over K,

then L : K is a splitting field extension for g = mo,l ma,2 . . . mo,” . Let L = K(al, ...,a,,), let m be the minimal polynomial of at,- over K,

and let g = in]... mu. Then L : K is normal if and only if L : K is a splitting field extension for g. Suppose that L 2 K is algebraic. An extension F : L is a normal closure for L : K ifF : K isnormal,andifF : M : L isatowerandM : Kisnormal,

then M = F.

Corollary 2 If L : K isfinite, it has a finite normal closure F : L. With the same notation as in Corollary 1, let F : L be a splitting field extension for 3 over L. Then F : K is a splitting field extension for 3 over K

so that F : K is normal. If F : M : L is a tower and M : K is normal, then each ma splits over M, and so g splits over M; therefore M = F. Corollary 3 If L : K is normal and M is an intermediate field then L : M is normal. For there exists S g K [x] such that L : K is a splitting field extension for S. If we consider S as a subset of M [x], L : M is a splitting field extension

for S. Exercises

7.1

Suppose that L : K and that M and N are normal extensions of K contained in L. Show that M n N and MN (the field generated by M and N) are both normal extensions of K.

7.2

Show that every algebraic extension has a normal closure.

7.3

Suppose that L : K is algebraic. Show that there is a greatest intermediate field M for which M : K is normal.

7.4

Suppose that L : K and that M] and M2 are intermediate fields. Show that if M1 : K and M2 : K are normal then so are K(M1,M2) : K and M1 n M2 : K.

7.2 Monomorphisms and Automorphisms

101

7.2 Monomorphisms and Automorphisms We have just seen that if L : K is normal and M is an intermediate field then L : M is normal. However, there is no reason why M : K should be normal. For example, if to is a complex cube root of 1 then Q(2'/3,w) : Q is normal, since it is the splitting field for f = x3 — 2, while Q(21/3) : Q is not, since f

is irreducible and has one root in Q(2”3) but does not split over (Dal/3). It is important to be able to recognize when M : K is normal. In the next theorem we give necessary and sufficient conditions for this, for finite extensions. Theorem 7.2 Suppose that L : K is a finite normal extension and that M is an intermediate field. The following are equivalent: (i) M : K is normal; (ii) if0' is an automorphism of L which fixes K then 0(M} g M; (iii) ifa is an automorphism of L which fixes K then 0(M) = M. Proof

Suppose first that M : K is normal, and that a is an automorphism of

L which fixes K. Suppose that a e M and let m be the minimal polynomial for a over K. Then m(or ((1)) = a(m(a)) = 0, so that 0(a) is a root ofm. As m splits over M,a(a) e M, and so a(M) g M. Thus (i) implies (ii). El Since [0(M) : K] = [M : K] it is clear that (ii) implies (iii). Suppose now that (iii) holds. As L : K is normal, L : K is the splitting field extension for some g e K [x], by Corollary 1 to Theorem 7.1. Suppose that a e M. Let m be the minimal polynomial for a over K. As L : K is normal, m splits over L. We must show that m splits over M: that is, that all the roots of m

are in M. Let [3 be any root of m in L. By Theorem 6.4, there exists a monomerphism j from K (a) to K, j(g) = g. Now L : j(g) = g. By Corollary which extends j. As a ,3 = 0(a) e M.

K (fl), fixing K, such that j(a) = )3. Since j fixes K (a) and L : K (fl) are splitting field extensions for 1 to Theorem 6.5, there is an isomorphism a: L —> L fixes K ,a(M) = M. In particular, this means that

Exercises

7.5

Suppose that N : L and N’ : L are two normal closures of L : K. Show that there is an isomorphism j of N onto N’ such that j(l) = l for] e L.

7.6

Suppose that L : K is a finite normal extension and that f is an irreducible polynomial in K [x]. Suppose that g and h are irreducible

102

7.7

7 Normal Extensions

monic factors of f in L[x]. Show that there is an automorphism a of L which fixes K such that 0(g) = h. Suppose that L : K is algebraic. Show that the following are equivalent: (i) L : K is normal; (ii) if j is any monomorphism from L to L which fixes K then

1' (L) E L;

(iii) if j is any monomorphism from L to L which fixes K then

1' (L) = L-

8 Separability

8.1 Basic Ideas Normality is a property that an extension may or may not have. Separability is different; most extensions of interest are separable, and we shall have to

work hard to find examples of non-separable extensions. But separability is an

important property; it leads to some very important results in Theorems 8.3 and 8.4. Separability involves several definitions. Suppose first that f is an irreducible polynomial of degree n in K [x] and that L : K is a splitting field extension for f. We say that f is separable (over K) if f has n distinct roots

in L. Suppose next that f is an arbitrary polynomial in K [x]. We say that f is separable (over K) if each of its irreducible factors is separable. Suppose that L : K is an algebraic extension and that a e L. We say that a

is separable (over K) if its minimal polynomial over K is separable, and say that L : K is separable if each a in L is separable over K. Theorem 8.1 Suppose that L : K is separable and that M is an intermediate field. Then L : M and M : K are separable.

Proof It is obvious that M : K is separable.

El

Suppose that a e L. Let m1 be its minimal polynomial over M, m2 its minimal polynomial over K. Let N : M be a splitting field extension for m2, considered as an element of M [x]. Since m2 is separable over K, we can write m2=(x—a1)...(x—a,) where at], . . . ,a, are distinct elements of N. But mllmz in M[x], and so in N[x] m1 = (x —a,-l)... (x —a,-2) for some 1 5 i1 < - - - < is < r. Thus m1 is separable.

103

104

8 Separability

8.2 Monomorphisms and Automorphisms We have already seen that counting dimension leads to some remarkably strong

results. We shall find that counting monomorphisms and automorphisms is equally useful. With this in mind, the results in this section suggest why separability is important. First we consider simple extensions. Theorem 8.2 Suppose that K (at) : K is a simple algebraic extension ofdegree d. Suppose that j : K —> L is a monomorphism. If a is separable over K and if j (ma) splits over L then there are exactly d monomorphisms from K (at) to L extending j; otherwise there are fewer than d such monomorphisms. Proof

By Corollary 2 to Theorem 6.4, there are r such extensions, where

r is the number of distinct roots of j(ma) in L. Now (1 = degree m, = degreej (ma) (Theorem 4.5), so that r 5 d, and r = d if and only if j(ma) splits into d distinct linear factors: that is, if and only if j(ma) is separable

over j (K) and j(ma) splits over L. Clearly a is separable over K if and only if j (mg) is separable over j(K), and so the result is proved. III We now consider the general case. Theorem 8.3 Suppose that K’ : K is a finite extension of degree d, and that j: K —) L is a monomorphism. If K’ : K is separable and j (ma) splits over L for each a in K’, then there are exactly (1 monomorphisms from K’ to L extending j; otherwise, there are fewer than d such monomorphisms. Proof We prove this by induction on d. It is trivially true when d=1. Suppose that it is true for all extensions of degree less than d, and that

[K’ : K] = d.

El

Suppose first that the conditions are satisfied. Leta e K ’\K . By Theorem 8.2 there are exactly [K (a) : K] monomorphisms from K (a) to L extending j. Let k be one of these. We apply the inductive hypothesis to K’ : K (a). First,

[K’ : K(a)] < d. Second, K’ : K(a) is separable, by Theorem 8.1. Iffl e K’, let mp be the minimal polynomial for [3 over K and let up be the minimal polynomial for [3 over K (0:). Then nfi divides m p in K (a)[x] and so k(np) divides k(m,g) in L[x]. But k(mp) splits over L[x], and so k(n,3) splits over L[x], Thus the conditions are satisfied, and so k can be extended in [K ’ : K (01)]

ways. It therefore follows from the tower law that j can be extended in d ways. Suppose next that the conditions are not satisfied. Then there exists a in K’ such that j (ma) has fewer than [K (a) : K] distinct roots in L, and so j can

8.2 Monomorphisms and Automorphisms

105

be extended in fewer than [K (at) : K] ways to a monomorphism from K (a) to

L, by Corollary 2 to Theorem 6.4. Each of these extensions can be extended to a monomorphism from K ' to L in at most [K ’ : K (01)] ways, by the inductive

hypothesis, and so there are fewer than d extensions. Corollary 1 Suppose that L : K is finite and that L = K(a1, . . . ,a,). Ifa,- is separable over K (on, . . . ,ai_1) for 1 5 i 5 r, then L : K is separable. Proof Let F : L beanorrnal closure forL : K. Let K0=K, and let Kj = K(a1, ...,a,-) = Kj_1(aj) for l 5 j 5 r. We assert that there are [Kj : K] monomorphisms from Kj into F which fix K. The result is trivially true for j = 0. Assume that it is true for j — l, and that i is a monomorphism from Kj_1 to F which fixes K. Let n; be the minimal polynomial for 01,- over Kj_1, and let mj be the minimal polynomial for 01,- over K. Then n j |m j in Kj_1[x], and so i(nj)|i(m,-) in i(K,-_1)[x]. But i(mj) = rm, and mi splits in F[x], so that i (nj) splits in F [x]. As aj is separable over Kj_1, i can be extended in

[K,- : Kj_1] ways to a monomorphism from KJ- to F, by Theorem 8.2. The assertion therefore follows inductively, using the tower law. But it now follows from Theorem 8.3 that K,- : K is separable, and so, in particular, L : K is separable. El Corollary 2 Suppose that L : K is finite and that L = K (m, . . . ,a,). If each a,- is separable over K then L : K is separable.

This follows from Corollary 1 and Theorem 8.1. Corollary 3 Suppose that f e K [x] is separable over K and that L : K is a splitting field extension for f. Then L : K is separable.

Apply Corollary 2 to the roots of f in L. Corollary 4 Suppose that L : K is finite, and that L : M : K is a tower. If L : M and M : K are separable, then so is L 2 K. Write M = K(a1, ...,a,), L = M(ar+1, ...,as), and use Corollary 1 and Theorem 8.1.

Exercise 8.1

Suppose that L : K is finite and that L’ : L is a normal closure for

L : K. Show that L : K is separable if and only if there are exactly [L : K] monomorphisms of L into L’ which fix K.

106

8 Separability

8.3 Galois Extensions A separable splitting field extension is called a Galois extension. Theorem 8.4 Suppose that [L : K] is finite. The following are equivalent: (i) L : K is a Galois extension; (ii) I : K is normal and separable; (iii) there are [L : K] automorphisms of L which fix K. Otherwise there are fewer than [L : K] automophisms of L which fix K. Proof This follows from the corollaries to Theorem 7.1 and Theorem 8.3. We shall extend this result in Theorem 9.4. CI

8.4 Differentiation Suppose that f is a non-zero element of K [x] and that L : K is a splitting field extension for f. We say that f has a repeated root in L if there exists a e L and k > 1 such that (x — a)"| f in L[x]. The largest possible value of k is the multiplicity of the root at.

An irreducible polynomial in K [x] is not separable if and only if it has a repeated root in a splitting field. It is therefore important to be able to recognize when a polynomial has a repeated root.

Suppose that f is a non-zero polynomial in C[x], and that at is a root of f. How do we tell if a is a repeated root? We differentiate: a is a repeated root

if and only if f’(a) = 0. Although differentiation has its roots in analysis, the differential operator has strong algebraic properties - in particular, (fg)’ = f’g + fg' — and we can define the derivative of a polynomial in a purely algebraic way.

Suppose that

f=ao+a1x+~~+anxn e K[x]. We define the derivative

Df = a1+ 2a2x + - - - +nanxn'l. Here, asusual, jaj = aj + - ~ . +aj (j times). D is a mapping from K [x] to K [x]. As

D(f + g) = Df + Dg,

D(Otf) = a(Df),

8.4 Differentiation

107

D is a K -linear mapping. Also D(xmxn) = (m +n)m+n—l = mxm—lxn+nxmxn—l = (DxM)xn+xm(Dxn),

and so, by linearity,

D(f8) = (Df)g + f (D8)Notice also that, if K has non-zero characteristic p, then

Dxp = px”_1 = 0. Differentiation provides a test for repeated roots,just as in the case of CD6]. Theorem 8.5 Suppose that is a non-zero element of K [x] and that L : K is a splitting fieldfor f. The following are equivalent: (i) f has a repeated root in L; (ii) there exists a in L for which f (a) = (Df )(a) = 0; (iii) there exists In in K [x], with degree m z 1, such that ml f and mIDf.

Proof Suppose that f has a repeated root at in L. Then f = (x —a)"g, where k >1andg e L[x].Thus Df = k(x — oak—1g + (x — 01)]‘Dg, and so f (a) = Df (a) = 0. Thus (i) implies (ii).

Suppose that (ii) holds. Let m be the minimal polynomial of at over K. Then ml f and mlDf, and so (iii) holds. Suppose that (iii) holds. We can write f = mh, with h in K [x]. As f splits

over L, so does m. Leta be a root ofm in L. We can write f = (x —a)q, with q in L[x]. Then

Df=q+(x—a)Dq. But (x — oz)|Df in L[x], since mlDf, and so (x — a)|q. Thus (x —a)2|f, and

f has a repeated root in L.

El

This theorem enables us to characterize irreducible polynomials which are not separable. Theorem 8.6 Suppose that f e K [x] is irreducible. Then f is not separable ifandonly ifchar K = p > Oand f has theform

f=ao+a1xp+a2x2P +---+anx"”.

108

8 Separability

Proof If f is not separable, there exists m in K [x], with degree m 2 I, such that ml f and mlDf . As f is irreducible, f and m are associates. Thus f IDf; as degree Df < degree f, it follows that Df = 0. This can only happen if char K 7S 0 and f has the form given in the theorem. Conversely, if the conditions are satisfied, Df = O and we can take f = m in Theorem 8.5(iii). CI Corollary 1 Ifchar K = 0, all polynomials in K [x] are separable. Corollary 2 Suppose that charK = p > 0 and that K is perfect. Then K is separable. Proof

Foriff =ao+a1xp +---+anx,’.’ andaj = bf forO E j 5 pthen Cl

f = (120 + blx + - - - + bnx")p, so that f is not irreducible.

Exercises 8.2

Suppose that f is a polynomial in K [x] of degree n and that either char

K = 0 or char K > n. Suppose that a e K. Establish Taylor’sformula:

f = f(a) + Df(a)(x - a) +

8.3

D 2 f(a)(x —ot)2 +-~.+ a( a)(x —“)n° 2! n!

Suppose that f is a polynomial in K [x] of degree n and that either char

K = 0 or char K > n. Show that a is a root of multiplicity r(5 n) if and only if

f(01) = Df(0t) = 8.4 8.5

8.6

= D'_lf(a) = 0 and D’f(ot) 75 0-

Suppose that p is a prime number. By factorizing x”‘l — 1 over Zp, show that (p — 1)! +1 = 0 (mod p) (Wilson’s theorem). Suppose that p is a prime number of the form 4n + 1. Show that there exists k such that k2 + 1 = 0 (mod p). Show that p is not a prime in Z + iZ and show that there exist u and v in Z such that u2 + v2 = p. Suppose that p is a prime number of the form 4n + 3. Show that p is a

prime in Z + iZ.

8.5 Inseparable Polynomials Suppose that

f=ao+a1xp+---+anx"p

8.5 Inseparable Polynomials

109

is in K [x]. We shall write f (x) = g(x"), where

g=ao+a1x+---+anx".

This is a slight abuse of terminology, which does not lead to any difficulties. Theorem 8.7 Suppose that char K = p > 0 and that

f(x) = gov”) =ao+a1x” +---+x"” is manic; then f is irreducible in K [x] if and only if g is irreducible in K [x], and not all of the coefficients a,- are pth powers of elements of K.

Proof If g factorizes as g = g1g2, then f factorizes as f(x) = g1(xp)g2(xp): thus if f is irreducible, so is g.

Suppose next that each a,- is a pth power of an element of K : that is,

a,- = bf, for b.- in K. Then, as before,

f=bg+bfxp+m+b£xw = (b0+b1x+---+b,.x")” and so f factorizes. Thus if f is irreducible, not all the a; can be pth powers of elements of K. Conversely, suppose that f factorizes. We must show that either g factorizes or that all the of are pth powers of elements of K. We can write f as a product of i irreducible factors:

f=fl"'... ,"r where the fi are monic and irreducible in K [x], f,- and f,- are relatively prime, for i 96 j, and n1+-- - + n, > 1. We have to consider two cases. First suppose that r > 1. Then we can write f = h1h2, with h] and h2

relatively prime (take h1 = f1"l ). There exist A1 and Ag in K [x] such that

Mh1+A2h2 =1. Further,

0 = Df = (Dh1)h2 + h1(Dh2). Eliminating h2, we find that

Dh1= A1h1(Dh1) - 12h1(Dh2)

l 10

8 Separability

and so h] tl . As degree Dhl < degree h] , we must have Dh. = 0. Similarly Dh2 = 0. Thus we can write

mm = Co +c:x” + - - - + csxs” = j1(x"). mm = do + dlx" + . . - + dtx‘” = j2(x”) and g factorizes as g = j; jz. Second, suppose that r = 1. Then f = fl", where f] is irreducible, and n > 1. Again there are two cases to consider. If pln, we can write f = k”. If

h =co+c1x+---+csx‘ then

f=hp =cg+cfx”+---+c§’x"’ so that all the ai are pth powers. If p does not divide n,

0 = Df = n(Df1)f1"“ and so Df1 = 0. Thus we can write

f1(x)= do +d1xp + - - - +d1x'p = 810‘”) U

andg = (81)".

Bearing in mind the corollary to Theorem 8.6, this means that if we are to find an inseparable polynomial we must consider fields K of non-zero characteristic which are not algebraic over their prime subfields. With this information, the search is rather short. Let K = Zp (a) be the field of rational expressions in a over Zp. Suppose if possible that —a = fl”, for some )3 in K. Then we can write [3 = f(o:)/g(oz), with f and g in Zp[x]. Thus

-a(g(a))” = (f (00)” and so, since a is transcendental, _xgp = fp.

But p|degree(fP) and p does not divide degree (—xgp). Thus —a is not a pth power in K, and so x” — a is irreducible in K [x], by Theorem 8.7. Let L : K

be a splitting field extension for xP — a, and let y be a root of xl’ — a in L. Then

(x—y)”=x”-y"=x"-a so that xp — (1 fails to be separable in the most spectacular way.

8.5 Inseparable Polynomials

111

Exercises 8.7

Show that a field K is perfect if and only if every finite extension of K

8.8

Suppose that char K = p > 0 and that f is irreducible in K [x]. Show that f can be written in the form f (x) = g(xP"), where n is a non-

is separable.

negative integer and g is irreducible and separable. 8.9

Suppose that char K = p > 0 and that L : K is a totally inseparable

algebraic extension: that is, every element of L\K is inseparable. Show that if 13 e L then its minimal polynomial over K is of the form x”" — a, where a e K. 8.10 Suppose that char K = p 75 0, that f is irreducible in K [x] and that L : K is a splitting field extension for f. Show that there exists a nonnegative integer n such that every root of f in L has multiplicity p". (Hint: Use Exercise 8.5.)

9 The Fundamental Theorem of Galois Theory

9.1 Field Automorphisms, Fixed Fields and Galois Groups Recall that a ring homomorphism from a field L1 to a field L2 is a monomor—

phism, and that if L : K is an algebraic field extension and that t : L —> L is a homomorphism which is fixed on K, then r is an automorphism of L (Theorem 4.9). Galois theory is largely concerned with properties of groups of auto-

morphisms of a field. If L is a field, we denote by Aut L the set of all automorphisms of L. Aut L is a group under the usual law of composition.

Suppose that A is a subset of Aut L. We set ¢(A) = {k e L: o(k) = k for each a in A}.

It is easy to verify that ¢(A) is a subfield of L, which we call the fixed field of A. In this way, starting from A we obtain an extension L : ¢(A). Conversely suppose that L : K is an extension. We denote by F(L : K) the set of those automorphisms of L which fix K: F(L : K) ={a e AutL: a(k) =kforallkinK}. When there is no doubt what the larger field L is, we shall write y(K) for l"(L : K). It is again easy to verify that 1"(L : K) is a subgroup of Aut L; we call l‘(L : K) the Galois group of the extension L: K. In this case, then, starting from an extension we obtain a set of automorphisms. In this chapter we shall study this reciprocal relationship in detail. The operations A —> ¢(A) and L : K —> y(K) establish a polarity between sets of automorphisms of L and extensions L : K. The next theorem is a

standard result for such polarities. Theorem 9.1 Suppose that L : K is an extension, and that A is a subset of Aut L.

112

9.2 Linear Independence

(i) (ii) (iii) (iV)

113

Y¢(A) 2 A; ¢V(K) 2 K; ¢V¢(A) = ¢(A); Mil/(K) = 1/00-

Proof If a e A, a(k) = k for each k in ¢(A), so that a e y¢(A): this establishes (i). If k e K, a(k) = k for each a in y(K), so that k e «by (K): this establishes (ii). If A1 g A2, then clearly ¢(A1) Z_> ¢(A2). Thus it follows from (i) that

¢Y¢(A) S ¢(A); but applying (ii), with ¢(A) in place of K,

¢y¢(A) 2 (MA)This establishes (iii). Similarly if K] g K2, y(K1) Z_> y(K2). Applying this to (ii):

y¢y(K) S y(K); but applying (i), with y(K) in place of A,

Y¢Y(K) 2 YOUThis establishes (iv).

El

Corollary If A is a subset of Aut L, and (A) is the subgroup of Aut L generated by A, then ¢(A) = ¢((A)).

For A S (A) S V¢(A), by (i), and SO ¢(A) 2 ¢((A)) 2 ¢y¢(A) = ¢(A).by (iii)Because of this, we shall usually restrict attention to subgroups of Aut L.

9.2 Linear Independence If r e Aut(L), then the restriction of 1' to the multiplicative group (L"‘, x) is an L-valued character on L*. Thus if n, . . . , tn are distinct automorphisms of

a field L then they are linearly independent L-valued functions on L*. But we need more. Suppose now that G is a subgroup of Aut L. If A e L, we define the trajectory of A, TG (A), to be the element of LG defined by T6 (A) (a) = 0(A).

114

9 The Fundamental Theorem of Galois Theory

LG is a vector space over L; we can also consider it as a vector space over any subfield of L, and in particular as a vector space over ¢(G). The next theorem is particularly important: it takes a rather curious form,

as it is concerned with linear independence over two different fields. Theorem 9.2 Suppose that G is a subgroup of Aut L, that K is the fixedfield of G and that B is a subset of L. Then the following are equivalent: (i) B is linearly independent over K; (ii) {TG(fl): fl 6 B} is linearly independent over K; (iii) {Tg(fi): [3 e B} is linearly independent over L. Proof Clearly (iii) implies (ii). Suppose that B is not linearly independent over K: there exist distinct I31, . . . ,fln in B, and k1, . . . ,kn in K, not all zero, such that

k1fl1+-'-+knfln =0. Thenifor e G, kla(fll) + ' ‘ ' +kna(fin) = “(klfll + ' ' ' +knfin) = 0,

so that lG(B|) + . .. + knTg(,B,,) = 0, and the set {Ta(fl): [3 e B} is not linearly independent over K in LG. Thus (ii) implies (i). Finally, suppose that the set of trajectories {Tc (,8): B e B} is not linearly independent over L in LG. There exist fll, . . . , [3, in B, and non-zero A1, . . . ,A, in L such that

A-1TG(l31)'l""+A-rTG(flr)= 0; further we can find 131, . . . ,fir and Al, .. . ,Ar so that r is as small as possible. In detail, this says that

ll°‘(fl1)+ ---+Ara(fl,) =Oforeacha in G.

(**)

Now ift e G and a e G then r‘la e G, so that Alt—10031) + - - - + Arr—1006,) = 0for each a in G.

Operate on this equation by r: r(A|)a(fi1)+ - - - + 1:(A,)a(fir) = 0 for each a in G.

(**)

Now multiply (*) by 11A,), (**) by 1,, and subtract: (tO‘r)Al _ 7(Avl)A-r)a(fil) + ' ' ' + (IOWM‘r—l _ 7(Ar—1)Ar)a(fir—l) = 0

9.3 The Size of a Galois Group is the Degree of the Extension

115

for each a in G. Thus

(130011 - t(li)lr)TG(fl1)+---+(t(Ar)?~r—1 - t()~r—1)Ar)TG(flr—1) = 0Since there are fewer than r terms in the relationship, it follows from the minimality of r that all the coefficients must be zero: 10.»)..- = 10%))” forl g i < r; in other words,

z().;‘A,-) = 1:1)..- forl g i < r. Now this holds for each 1' in G, and so k,- = Aflli e K, for 1 g i < r. Multiplying (*) by 1:1, we obtain 1610031) + - - - +kr—10(fir—1) +0030 = 0 for each a in G. But as G is a subgroup of Aut L, the identity automorphism is in G. Thus k|fl1+---+kr—lflr—1+ fir = 0 and so B is not linearly independent over K. Thus (i) implies (iii).

II]

9.3 The Size of a Galois Group is the Degree of the Extension When G is finite we can relate the order of G to the degree of L : ¢(G) in a most satisfactory way. Theorem 9.3 Suppose that G is a finite subgroup of Aut L. Then |G| = [L : ¢(G)], G = y¢(G) and L : ¢(G) is a Galois extension. Proof Let K = ¢(G). If B is a subset of L which is linearly independent over K then, by Theorem 9.2, {Tc ([3) : [3 e B} is a subset of LG which is linearly independent over L. But LG has dimension |G|, and so |B| g lGl. Thus L is finite dimensional over K, and [L 2 K] g |G|. On the other hand, by Theorem 8.4, |y¢(G)| g [L : K]. As G g y¢(G), it follows that [L : K] = |G| and that G = y¢(G). Since [L : K] = IGI, it follows from Theorem 8.4 that L : K is a Galois extension. III

What happens if, instead of starting with a group of automorphisms, we start with a finite extension? Here the results are not quite so clear cut. Once again, Theorem 8.4 plays a decisive role.

116

9 The Fundamental Theorem of Galois Theory

Theorem 9.4 Suppose that L : K isfinite. If L : K is a Galois extension, then |y(K)| = [L : K], andK = (In/(K). Otherwise, |y(K)| < [L : K] andK is a proper subfield of (13)! (K ). Proof The relationship between |y(K)| and [L : K] is given by Theorem 8.4. By Theorem 9.3, |y(K)| = [L : ¢y(K)]. Thus, ifL : K is normal and

separable,

[L 1 K] = [L I ¢Y(K)]; as K g ¢y(K), K = ¢y(K). Otherwise

[L : K] > [L I ¢Y(K)] so that K is a proper subfield of ¢y(K).

El

Exercises

9.1

Suppose that L : K is a Galois extension with Galois group G, and that a e L. Show that L = K (a) if and only if the images of a under G are all distinct.

9.2

Suppose that L : K is an extension. If a e l"(L : K), a e EndK(L), the

K-linear space of K -linear mappings of L into itself. Show that l"(L : K) is a linearly independent subset of EndK (L).

9.3

Suppose that L

:

K is a Galois extension with Galois group

G = {01, ...,a,.}. Show that ([31, ...,fl,,) is a basis for L over K if and only if det(oi(flj)) aé 0. 9.4

Suppose that char K =0 and that L : K is a finite extension; let

fil, . . . ,fln be a basis for L over K. Suppose that H is a subgroup of l"(L : K);let yj = 206” aflj,forl g j S n. ShowthatK(y1, . . . ,yn)

is the fixed field for H.

9.4 The Galois Group of a Polynomial The main purpose of the theory of field extensions is to deal with polynomials and their splitting fields. Suppose that f e K [x] and that L : K is a splitting field extension for f over K. Then we call l"(L : K) the Galois group of f; we denote it by PK (f) (or l"(f ), when it is clear what K is). By Corollary 1 to Theorem 6.5, PK (f)

depends on f and K, but not on any particular choice of splitting field. Let us interpret Theorem 9.4 in this setting.

9.4 The Galois Group of a Polynomial

117

Theorem 9.5 Suppose that f e K [x] and that L : K is a splitting field

extensionfor f. Iff is separable then |l"(f)| = [L : K] and K = ¢(l"(f)); otherwise |F(f)| < [L : K] and K is aproper subfield of¢(l‘(f)). An element a of l"( f) is an automorphism of L; it is the action of a on

the roots of f that is all important. The next result shows that we lose no information if we concentrate on this action. Theorem 9.6 Suppose that f e K [x] and that L : K is a splitting field extension for f over K. Let R denote the set of roots of f in L. Each 0 in I‘ (f) defines a permutation of R, so that we have a mapping from l"(f) into the group 22R ofpermutations of R. This mapping is a group homomorphism, and is one-to-one. Proof

Ifa e l"(f), then or( f) = f, since f has its coefficients in K. Thus,

ifa e R,

f(0'(a)) = a(f)(a(a)) = a(f(a)) = 0(0) = 0. Thus 0 maps R into R. Since a is one-to-one and R is finite, aIR is a permutation. By definition, (0102)(a) = 01 (0201)) so that the mapping: a —> a | R is a group homomorphism. Finally, if 0(a) =

r(a) for each a in R, then 0'11: fixes K(R) = L, so that a = r. Notice that Theorem 6.6 states that, if f is irreducible, then 1"(f) acts

transitively on the roots of f: if a and l? are two roots of f in a splitting field, there exists a in l"( f) with 0(a) = 5. Conversely, suppose that f is a monic polynomial of degree n in K [x] which has n distinct roots in a splitting field L, and that l‘(f) acts transitively on the roots of f Let a be a root of f, and let m be the minimal polynomial of (1. Then if )3 is any root of f there exists a in l"(f) such that 0(a) = [3. Thus

m(fi) = m(a(a)) = 0(m)(a(a)) = 0(m(a)) = 0. and so m has at least n roots. Since m divides f,m = f and it follows that f

is irreducible.

El

Exercises 9.5 9.6

Describe the transitive subgroups of 23, E4 and 25. Find the Galois group of x4 — 2 over (a) the rational field Q, (b) the field Z3 and (c) the field Z7.

118

9.7

9 The Fundamental Theorem of Galois Theory

Find the Galois group of x4 + 2 over (a) the rational field Q, (b) the field Z3 and (c) the field 25.

Let us give an example. Theorem 9.7 Suppose that f e Q[x] is irreducible and has prime degree p. If f has exactly p — 2 real roots and 2 complex roots in C then the Galois group

F(f) off overQ is 21,. Proof For l"(f) is transitive and contains the transposition z —) Z, and so the result follows from Theorem 1.20.

CI

As a concrete example, let us consider

f = x5 — 4x + 2. f is irreducible over Q, by Eisenstein’s criterion. The function r —> f (t) on R is continuous and differentiable, and so, by Rolle’s theorem, between any two real zeros of f there is a zero of f’. But

f’ = 5x4 — 4 has only two real zeros, so that f has at most three real zeros. As

f(-2) = -22.

f(0) = 2.

f(1) = -1,

f(2) = 26

f has at least three real roots, by the intermediate value theorem. Thus f has three real roots and two complex roots; by the theorem, F( f) = 25. Notice how useful elementary analysis can be!

Exercise 9.8

Sketch the graph of the polynomial

fr = (x2 + 4)x(x2 — 4)(x2 — 16)... (x2 — 4r2). Show that if k is an odd integer then lfr(k)| 2 5. Show that f, — 2

is irreducible, and determine its Galois group over Q when 2r + 3 is a prime.

9.5 The Fundamental Theorem of Galois Theory The fundamental theorem of Galois theory describes in some detail the polarity that was introduced at the beginning of the chapter. Theorem 9.8 Suppose that L : K is finite. Let G = I‘(L : K), and let Ko = ¢(G). IfL : M : K0, let y(M) = I‘(L : M).

9.5 The Fundamental Theorem of Galois Theory

119

(i) The map ¢ is a one-to-one map from the set of subgroups of G onto the

set offields M intermediate between L and Ko.y is the inverse map. (ii) A subgroup H of G is normal if and only if ¢(H) : K0 is a normal extension. (iii) Suppose that H ~n = 0. and

aolk + aIAk+1 + - - ~ + an—llkw—l +anlk+n = 0 fork =1,2,3,... 10.3 Suppose that f = x" + px + q. Show that

l1=>~2=---=ln—2=0. A01-1: —(n _ UP,

A" = —nq,

An+1=---=)~2n—3=0

10.1 The Discriminant

125

and

lzn—z = (n - 1)P2Show that the discriminant A of f is

A = nn+1n"q"'1 - nn(n - 1)""p" where 11,, = 1 if n (mod 4) = 0 or 1 and 17,, = —1 otherwise.

10.4 Suppose that char K = 2 and that f e K [x] is separable. Show that the discriminant of f always has a square root in K.

10.5 Suppose that f e Q[x] is irreducible, and that its discriminant 8 does not have a square root in Q. If 8 is a square root of A in some splitting field, show that f is irreducible in Q(6).

11 Cyclotomic Polynomials and Cyclic Extensions

We have seen that in order to deal with cubic polynomials it is helpful to have cube roots of unity at our disposal. In this chapter and the next we shall consider splitting fields and Galois groups of polynomials of the form x'" — 1

and x'" — 0 over a field K. Technical problems can arise if char K 9’: 0. Suppose that char K = p > 0 and that m = p’q, where p does not divide q. Then in K [x],

x'" — 1 =(x‘l —1)P’; thus a splitting field extension for x‘l —1 is a splitting field extension for x’" — 1: we need only consider the polynomial x4 — 1. For this reason, in this chapter we shall suppose that char K does not divide m. In this case, D(x"' — 1) =

mx"“1 9e 0, and so x'" — 1 has m distinct roots in a splitting field.

11.1 Cyclotomic Polynomials Suppose that L : K is a splitting field extension for x’" — 1 over K. As x’" — 1 has m distinct roots, L : K is a Galois extension. The set R of roots in L

clearly forms a group under multiplication, and so, by Exercise 10.1, R is a cyclic group of order m. An element a of R is called a primitive mth root of unity if 19 generates R. Thus an element a of L is a primitive mth root of unity ifand ony ife’" = 1 and ej 56 1 for 1 < j < m. For example, in C,i and —i

are the primitive fourth roots of unity: —1 is the only primitive second root of unity and 1 is the only first root of unity. Notice that if e is a primitive mth root of unity then L = K (s). We now define the mth cyclotomic polynomial fiJ-(a) is a homomorphism of FK(,,,) into Um. This is injective, since 0(8) = e if and only if a is the identity in I‘K (,,,). Further, ll} (d>,,,)| = k if and only if there are k images 8““); thus the homomorphism is onto if and only if PK ((a) = a”. Then (I) is an automorphism of F which generates l"(F : Zp).

156

15.2 Polynomials in Zp [x]

Proof

157

Ifa,b e F then p—l

(a +b)p = a” + E (I?)ajb”_j +b” =a” +bp. . J J=| since p divides the intermediate binomial coefficients. Since (ab)? = aP,

(D is a monomorphism, and is therefore an automorphism of F. The set of elements fixed by CD is a subfield G of F. If a e 2,, then (a) = a, so that

Zp g G. But the polynomial x” — x has only p roots in F, so that G = Zp. Thus it follows from the previous theorem that (I) generates I‘(F : Zp). El is called the Frobenius automorphism. Exercises

15.] Suppose that |F| = q, where q = p”. How many elements of F* generate the multiplicative group F*?

15.2 Suppose that IF I — q, where q = p". What are the subfields of F? If G is a subfield of F, what is the Galois group l"(F : G)?

15.2 Polynomials in Z1, [x] We have seen that any finite field is a splitting field for a polynomial x‘l" - 1 over a suitable prime subfield Zp. Let us now go in the opposite direction,

considering polynomials in Zp [x] and their splitting fields. Let us start with the cyclotomic polynomial m over Zp. Let Fm : Zp be a splitting field extension for (1),". Then Fm : Zp is a splitting field extension for xm — 1. Let Em be the set of roots of x’" — 1 in Fm, and let Pm be the set of primitive roots of unity. Recall that there is a monomorphism \II of Pm onto

U,,,, the multiplicative group of units in Zm. Proposition 15.3 Suppose that p does not divide m. Then x’" — 1 is separable, and Em is a cyclic subgroup of (1:3,, x) of order m.

Proof Since D(x"‘ — 1) = mx""1 9e 0, x'" — l is separable. If a,b e Em, then (ab")’" = amb‘m = 1, so that Em is a subgroup of the cyclic group (F,;, x ), and is therefore cyclic

III

What if p divides m?

Proposition 15.4 Suppose that m = lp’, where p does not divide l, l > 1 and r > 0. Then Fm = F[, Em = E] and each element of Em has multiplicity p’.

158

15 Finite Fields

Proof This follows immediately from the fact that x’" — 1 = (x1 — 1)P'. El Suppose that p does not divide m. Let [2 = p (mod m). Then [2 is a unit in Um; let d (11m) be its order. Theorem 15.5 Suppose that p does not divide m. Then the cyclotomic polynomial m is the product of k irreducible polynomials in Zp [x], each of

degree d(p,m), where k = ¢(m)/d(p,m). Fm : Zp is the splitting field for each of these polynomials. Proof

There exists n 6 Pm such that ‘IJ(17) = 13. Then 17 has degree d(p,m).

and its minimal polynomial g,7 has degree d (p,m). Since the Frobenius automorphism generates 1"(F,,,,Zp), it follows that Fm = Z1201), and [Fm : Zp] = d (p,m). Thus Fm : Zp is the splitting field for the polynomial g”. Let Rn be the set of roots of g". If n’ 6 PM \ R07), there exists or e l"(m) such

that 0(1)) = n’, and then Rn’ = o(R,,) and g": = naeRn (x — 0(a)). Thus m the Frobenius automorphism, whose minimal polynomial is mT(x) = x" — 1. HA 6 L let I; = {f e K[x] : f(T)A} = 0. IA is an ideal in K[x], andso I; =< m). >, for some m1 6 K [x], since K [x] is a principal ideal domain. m1 is the minimal polynomial for A, and clearly m Almr. Lemma 17.7 (i) m1 is the lowest common multiple of {m A : A e L}. (ii) Ifu = f(T)A then mulmk.

(iii) Ifm; and mu are coprime, then ml...“ = mim“. Proof (i) Let m be the lowest common multiple of {m1 : A e L}. Then clearly mTIm. But also m(T)A = 0 for all A, so that mlmr. (ii) mA(T)/J. = m).(T)f(T))t = f(T)mA(T))» = 0, so that mulml. (iii) Clearly mHulmlmu. Let v = m1(T)(A + u) = mA(T)/L. Then mvlmHu, by (ii). Then 0 = m,,(T)v = mv(T)mA(T)M, so that mulmvml. Since m; and mmu are coprime,

mulmv and so mulmHu. Similarly mAImHfl. Thus mlmulmlfit, and so mm.” = mAmu. El

We now complete the proof. Suppose that m = p? - - - pz" , where p], . . . , pk are distinct prime numbers. By (ii), for 1 5 i 5 k there exists A,- such that

In)“. = pirifi, so that u,- = fi(T))t; has minimal polynomial pf. Let u = m1+- - -+mk. Then mu(x) = mT(x) = x" — 1. Thus {M.T(/L), . . . , T""(u)} is a normal basis. Exercise

17.5 Show that the primitive nth roots of unity over Q form a normal basis for the splitting field of x" — 1 over Q if and only if n has no repeated prime factors.

18 The Algebraic Closure of a Field

18.1 Introduction If f e Q[x] we can consider f as an element of C[x], and then f splits over C. We therefore have the comforting conclusion that, whenever f e Q[x], we

can find a splitting field extension for f which is a subfield of the fixed field C. In this chapter we shall show that a similar phenomenon occurs for any field K. We must make some definitions. A field L is said to be algebraically closed if every f in L[x] splits over L. Thus the ‘fundamental theorem of

algebra’ states that (C is algebraically closed. An extension L : K is called an algebraic closure of K if L : K is algebraic and L is algebraically closed. Note that C : Q is not an algebraic closure of Q since (C : Q is not algebraic (Exercise 4.17). The next theorem gives two useful characterizations of an algebraic closure: Theorem 18.] Suppose that L : K is an extension. The following are equivalent: (i) L : K is an algebraic closure of K. (ii) L : K is algebraic, and every irreducible f in K [x] splits over L.

(iii) L : K is algebraic, and if L’ : L is algebraic then L = L’. Proof Clearly (i) implies (ii). Suppose that (ii) holds and that L’ : L is algebraic. Then L’ : K is also algebraic (Theorem 4.8). Suppose that a’ e L’. Let m be the minimal polynomial of a’ over K. Then m is irreducible and so, by hypothesis, m splits over L: m=(x—A1)...(x—}.n).

As m(a’) = 0,a’ = A; for some j, and so a’ e L. Thus L = L’ and (iii) holds. Finally suppose that (iii) holds, and that f e L[x]. By Theorem 6.3, there is a

170

18.2 The Existence of an Algebraic Closure

171

splitting field extension L’ for f over L. L’ : L is algebraic, by the corollary to Theorem 6.1 and so, by hypothesis, L’ = L. Thus f splits over L, and so L is algebraically closed. Consequently L : K is an algebraic closure of K.

I]

Corollary Suppose that L : K is an extension and that L is algebraically closed. Let La be the field of elements of L which are algebraic over K. Then La : K is an algebraic closure of K. In particular, if A is the field of complex numbers which are algebraic over Q, then A : Q is an algebraic closure for Q.

18.2 The Existence of an Algebraic Closure We now turn to the fundamental theorem concerning algebraic closures. Theorem 18.2 If K is afield, there exists an algebraic closure L : K. The generality of this statement suggests that we may need to use the axiom

of choice, and the maximal nature of an algebraic closure revealed by Theorem 18.1 reinforces this belief. It is, however, necessary to proceed with some care. Let us begin by giving afallacious argument.

Partially order the algebraic extensions M : K by saying that M1 : K 2 M2 : K if M2 is a subfield of M]. If ‘6 is a chain of extensions M : K, let

N = U{M : M: K e ‘6}. If (1,3 6 N, there exists M: Kin ‘6 such that a and f3 are in M. Define 003,0: + fl and a“ (if 11 9E 0) by the operations in M. This does not depend on M, and so N is a field, and N: K. Ifa e N,

a e M for some M, and so a is algebraic over K. Thus N 2 K is an upper bound for ‘6. By Zorn’s lemma, there is a maximal algebraic extension, and by Theorem 18.1, this is an algebraic closure. What is wrong with this argument? The error comes at the very beginning, when we try to compare extensions. Recall that an extension is really a triple (i ,K ,M), where i is a monomorphism from K into M. Thus in general we

cannot compare extensions in the way that is suggested.

Nevertheless, the fallacious argument has some virtue, and it is possible, by considering fields which, as sets, are subsets of a sufficiently large fixed set,

to produce a correct argument along the lines which the fallacious argument suggests. Exercises 18.1 and 18.2 show one way in which this can be done. We

shall instead give a more ‘ring-theoretic’ argument, which uses the axiom of choice by appealing to Theorem 2.16. We consider a ring of polynomials in very many variables. If f is a nonconstant monic polynomial in K [x] of degree n, then f has at most n roots in

172

18 The Algebraic Closure of a Field

a splitting field extension: we introduce an indeterminate to correspond to each of these possible roots. Let U be the set of all pairs (f, j), where f is a nonconstant monic polynomial in K [x] and 1 S j 5 degree f. For each (f, j) in U, we introduce an indeterminate xj( f ), and consider the polynomial ring K [Xy] of polynomials with coefficients in K and with indeterminates

XU ={Xj(f)1(f,j)€ U}. Now suppose that f is a non-constant monic polynomial in K [x]. We can write

f = x" — a1(f)x"“ + - - ~ + (—1)"an(f) (notice that we have not written monic polynomials in this form before: as we shall see, this can be a very useful form to use). Let g(f) be the element of K[XU][x] that has x1 (f), . . . ,xn(f) as roots:

g(f) = ]'[(x — xiv» j=l

= x" — S1(f)x"" + . . . + (—1)"sn(f), where

SM): 2 xi1(f)---xi,-(f)€K[XU] i| K[XU] 1» M. Then (j,K,M) is an extension of K. Let us set flj(f) = q(xj(f)), for all

u = (f.1) 6 U. Now suppose that

f = x" — a1(f)x"“ + - - - + (—1)"an(f) is a non-constant monic polynomial in K [x]. Then

M) = x" — j(a1(f))x"" + - - - + (—l)"j(an(f)) is the corresponding polynomial in M [x]. But

J'(ak(f)) = q(i(ak(f))) = q(Sk(f)), since sk(f) — i(ak(f)) = tk(f) e I Q J. Thus

M) = x" — q(Sl(f))-’C"_l + - - - + (—1)"q(sn(f» = W” — Si(f)x"" + - - - + (—1)"sn(f» = 4((x - Xi(f))(x -X2(f))- - - ()6 -xn(f))) = ()6 -fl1(f))--- (x - fln(f)).

174

18 The Algebraic Closure of a Field

and j (f) splits over M. Further, each fik (f) is algebraic over j (K) (since it is a root of j ( f)) and the flk( f) generate M over K: thus M : K is algebraic, by Corollary 2 to Theorem 4.7. Consequently (j, K, M) is an algebraic closure of

K. If K is countable, then K (Xy) is countable, so that by using Theorem A.1 we can avoid using the axiom of choice. Exercises 18.1

(i) Suppose that U is a non-empty set, and that P(U) is the set of

subsets of U. Show that if V Q U and f: V —> P(U) is amapping, then f is not onto. (Consider {x: x e V,x ¢ f (x)}.)

(ii) Suppose that U is a non-empty set and that V g W 9 U. Show that if f 2 V —> P(U) is one-to-one then there exists a one-to—one map 3: W —> P(U) such thatglu = f. (Use Zorn’s lemma.)

18.2 Suppose that K is a field. Let U = K [x] x Z+. (i) Show that if (k, K, L) is an algebraic extension, then there exists a

one-to-one mapping of L into U. (Use Zom’s lemma.) (ii) Suppose that (k, K, L) and (l, L, M) are algebraic extensions. Show that if f : L —> P(U) is one-to-one then there exists a one-to-one mapg: M —> P(U) such that f = gl.

18.3 Suppose that K is a field. Let U = K [x] x Z+. (i) Ifa e K, let j(a) = {(x - (1,1)}. Show thatj: K —> P(U) is one-

to-one and that j(K) can be given the structure of a field in such a way that j is a field isomorphism.

(ii) Let 3 be the set of triples (S, + , .) where

(a) 1mg 5 g P(U); (b) (S, + , .) is a field, F(S) say; (c) (i, j (K), F (S)) is an algebraic extension (here i is the inclusion mapping). Define a partial order on 9 by saying that (Sl,+1,.1) 5 (S2, +1.2) if 51 g 52 and (i,F (S1), F (S2)) is an extension (again, i is the

inclusion mapping). Show that under this order, .9? has a maximal element (Zom’s lemma). (iii) Use Theorem 6.2 or 6.3 to show that if (S, +, .) is a maximal element of .9? then (j, K ,F (S)) is an algebraic closure for K. (Here j is

considered as a mapping of K into F (S).)

18.3 The Uniqueness of an Algebraic Closure

175

18.3 The Uniqueness of an Algebraic Closure We now consider problems of uniqueness. First we establish an extension theorem: this uses Zom’s lemma in a very standard way. Theorem 18.3 Suppose that i: K1 —> K2 is a monomorphism, that L : K1 is algebraic and that K2 is algebraically closed. Then there exists a monomorphism j: L —> K2 such that ll = i. Proof Let S denote all pairs (M,0), where M is a subfield of L containing K1, and 0 is a monomorphism from M into K2 such that 0| K1 = i. Partially order S by setting (M1,01) S (M2,02) if M1 9 M2 and 02m, = 01. If‘é is a chain in S, let N = U{M: (M,9) e if}. Ifn e N, then n e M for some (M,0) e ‘6. Set ¢(n) = 9(n). It is now straightforward to verify that (it is

well defined, that «)5: N —> K2 is a monomorphism and that (N,¢) is an upper bound for ‘6'. Thus, by Zorn’s lemma, S has a maximal element (M,0). We

must show that M = L.

If not, there exists a e L\M. a is algebraic over M: let m be its minimal polynomial over M. Then 6 (m) splits over K2, since K2 is algebraically closed. Let

0(M) = (x - fli)...(x - fir)Then 0(m)(/31) = 0, and so by Theorem 6.4 there exists a monomorphism 91: M (a) —> K2 with 01|M = 0. This contradicts the maximality of (M,0). If L is finite or countable, the result follows by a standard induction argument,

without appealing to the axiom of choice. We are now in a position to show that an algebraic closure is essentially

unique.

El

Theorem 18.4 Suppose that (i1,K,L1) and (i2, K, L2) are two algebraic closures for K. Then there exists an isomorphism j : L1 —> L2 such that i2 = ji 1. Proof By Theorem 18.3 there exists a monomorphism j : L1 —> L2 such that i2 = ji 1. L1

1'

K

’ L2

176

18 The Algebraic Closure of a Field

We now use Theorem 18.1. If f is irreducible over K [x], i] (f) splits over LI, and so i2(f) splits over j(L1). As (i2, K,j(L1)) is algebraic, (i2, K,j(L|)) is an algebraic closure for K. Now L2: j (L1) is algebraic, as (i2, K, L2) is, and so L2 = j(L1), by Theorem 18.1(iii). In future, if K is any field, we shall denote by K: K any algebraic closure of K. [II Exercises 18.4 What is the algebraic closure of Q (as a subfield of (C)? 18.5 Show that an algebraically closed field must be infinite.

18.6 Suppose that K (a) : K is a simple extension and that a is transcendental over K. Show that K (a) is not algebraically closed. 18.7 Suppose that K is a countable field. Show how to construct an algebraic closure, by successively constructing splitting fields of the (countably

many) polynomials in K [x]. Is your construction less fallacious than the ‘fallacious proof’ of Theorem 18.2?

18.8 Suppose that L : K is algebraic. In what sense is it true that I = K?

18.4 Conclusions We have now achieved what we set out to do. Some comments are in order. First, the proof of Theorem 18.2 is quite difficult, More to the point, it is quite different from the very special construction of the complex field (C. Here, the hard work is constructing the real number field R from the rational field Q.

C : IR is then a splitting field extension for the polynomial x2 + l, which is irreducible over R. It is then remarkably the case that all polynomials over R split over C. The complex field is a very special one!

Second, the proof uses the axiom of choice in an essential way. This suggests that the theorem should only be used when it is necessary to do so. Third, the existence of an algebraic closure, and the extension theorem

(Theorem 18.3) provide a useful framework in which to work. If one uses this, the theory can be developed more simply in a few places. But the use of the

axiom of choice seems too big a price to pay: for this reason we have not used algebraic closures in the development of the theory.

l9 Transcendental Elements and

Algebraic Independence

19.1 'Ii-anscendental Elements and Algebraic Independence In this chapter we leave the study of algebraic extensions, and consider

problems concerning transcendence. Suppose that L : K is an extension and that a e L. Recall that a is transcendental over K if the evaluation map E, : K [x] —> L is one-to-one; that is, a satisfies no non-zero polynomial relation with coefficients in K. Theorem 19.1 Suppose that L

:

K is an extension and that a e L is

transcendental over K. Then the evaluation map Ea can be extended uniquely to an isomorphism Fa from the field K (x) of rational expressions in x over K onto the field K (a).

Proof The proof should be quite obvious: here are the details. Remember that the field K (x) is obtained by considering an equivalence relation on K [x] x (K [x])* (see Section 2.4). Suppose that (f, g) e K [x] x (K [x])*. As a is transcendental over K,

8(a) 3'5 0, and we can define Ga(f,g) = f (a)(g(0t))"- If (ftg) ~ (n’)

then f8’ = ft: in K [X], so that f (a)g’(a) = f’(a)g(a) and Ga(f.g) = Ga(f’, g’). Thus Go, is constant on equivalence classes: we can therefore

define Fa(f/g) = Ga(f,g). It is straightforward to verify that Fa is a ring homomorphism. Since E, (x) = E, (x) = 0:, Fa (K (x)) 2 K (a). In contrast,

if M? E K(x). Fa(f/g) = f(or)(g(ot))‘l 6 K01), and SO Fa(K(x)) = K (a). Finally if F; is another monomorphism which extends Ed, the set

{V 6 K06): Fa(r) = F&(r)} is a subfield of K (x) which contains K [x]; it must therefore be the whole of K (x), and so E, is unique.

177

178

19 Transcendental Elements and Algebraic Independence

We now generalize the idea of a transcendental element. Suppose that L : K is an extension and that A={a1, ...,a,,} is a finite subset of L (where on, . . . ,an are distinct). Any element f of K [x1, ...,x,.] can be written in the form

f= 2w?” where k1 e K for l

j< m and d, j is a non-negative integer for l

i< n,

1 < j < m. We define the evaluation map EA from K[x1, .. .,x,,] into L by

setting

EA(f)= :kjad'"...and“

It is easy to see that EA is a ring homomorphism. We shall frequently write

EA(f) as f(0!la . . . ,an). We say that A is algebraically independent over K if E,4 is one-to-one: that

is, there is no polynomial relation, with coefficients in K, between the elements on, . . . ,ozn. Thus a one-point set {a} is algebraically independent over K if and

only if a is transcendental over K. We say that an arbitrary subset S of L is algebraically independent over K

if each of its finite subsets is algebraically independent over K. The proof of the next result is exactly similar to the proof of Theorem 19.1: this time we omit the details. El Theorem 19.2 Suppose that L : K is an extension and that A = {(11, .. . ,an} is algebraically independent over K. Then the evaluation map EA can be extended uniquely to an isomorphism FA from the field K (x1, ...,x,.) of rational expressions in x1, . . . ,xn onto the field K (on, . . . ,an).

The next theorem is again very easy: it gives a useful practical criterion for a finite set to be algebraically independent over K. Theorem 19.3 Suppose that L : K is an extension and that at], . . . ,an are distinct elements ofL. Let Ko = K, K,- = K(a1,...,a,-)forl S i g n. Then A = {0:1, . . . ,an} is algebraically independent over K ifand only ifa,- is transcendental over Ki_1, for 1 S i g n.

Proof Suppose that a; is algebraic over Ki_1. Thus f(ai)=ko+k1ai+...

+krair=0

19.1 Transcendental Elements and Algebraic Independence

179

for some non-zero f in K,-_| [x]. We can write each kJ- as

k,- = pj(a1,...,a.~_1)(qj(a1, ...,a.-_1))" where the Pi and q,- are in K[x1, ...,x,-_1] and the qj(a1, . . . ,a;_1) are nonzero. We clear the denominators. Let

lj=Pj

nqk ,f0r0