Resolution of Curve and Surface Singularities in Characteristic Zero [2004 ed.] 1402020287, 9781402020285

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Resolution of Curve and Surface Singularities

Algebras and Applications Volume 4 Editors:

F. Van Oystaeyen University ofAntwerp, u/A, Wilrijk, Belgium A. Verschoren University of Antwerp, RUCA, Antwerp, Belgium

Advisory Board: M.Artin Massachusetts Institute of Technology Cambridge, MA, USA

A. Bondal Moscow State University, Moscow, Russia I. Reiten Norwegian University of Science and Technology Trondheim, Norway

The theory of rings, algebras and their representations has evolved into a well-defined subdiscipline of general algebra, combining its proper methodology with that of other disciplines and thus leading to a wide variety of applications ranging from algebraic geometry and number theory to theoretical physics and robotics. Due to this, many recent results in these domains were dispersed in the literature, making it very hard for researchers to keep track of recent developments. In order to remedy this, Algebras and Applications aims to publish carefully refereed monographs containing up-to-date information about progress in the field of algebras and their representations, their classical impact on geometry and algebraic topology and applications in related domains, such as physics or discrete mathematics. Particular emphasis will thus be put on the state-of-the-art topics including rings of differential operators, Lie algebras and super-algebras, groups rings and algebras, C* algebras, Hopf algebras and quantum groups, as well as their applications.

Resolution of Curve and Surface Singularities in Characteristic Zero

by

K. Kiyek Department of Mathematics, University of Paderborn, Paderborn, Germany

and

J.L. Vicente Departamento de Algebra, Universidad de Sevilla, Sevilla, Spain

.....

"

SPRINGER SCIENCE+BUSINESS MEDIA,LLC

A c.I.P. Catalogue record for this book is available from the Library of Congress.

ISBN 978-90-481-6573-5 ISBN 978-1-4020-2029-2 (eBook) DOI 10.1007/978-1-4020-2029-2

Printed on acid-free paper

All Rights Reserved © 2004 Springer Science+Business Media New York Originally published by Kluwer Academic Publishers in 2004 Softcover reprint of the hardcover 1st edition 2004

No part of this work may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, microfilming, recording or otherwise, without written permission from the Publisher, with the exception of any material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work.

Contents Preface

xi

Note to the Reader

xx

Terminology I

1

2

3

4 5

6

7

xxi

Valuation Theory Marot Rings. . . . . . . . . 1.1 Marot Rings . . . . 1.2 Large Quotient Ring 1.3 Rings With Large Jacobson Radical Manis Valuation Rings . . . . . 2.1 Manis Valuation Rings . . . . . . . . 2.2 Manis Valuations . . . . . . . . . . . 2.3 The Approximation Theorem For Discrete Manis Valuations Valuation Rings and Valuations . . . . . . . . . . . 3.1 Valuation Rings . . . . . . . . . . . . . . . 3.2 Sub rings and Overrings of Valuation Rings 3.3 Valuations....... 3.4 Composite Valuations . . . . . . . . . . . . 3.5 Discrete Valuations. . . . . . . . . . . . . . 3.6 Existence of Valuations of the Second Kind The Approximation Theorem For Independent Valuations Extensions of Valuations . . . . . . . . . . . . . . . . . . 5.1 Existence of Extensions . . . . . . . . . . . . . . 5.2 Reduced Ramification Index and Residue Degree 5.3 Extension of Composite Valuations. . Extending Valuations to Algebraic Overfields 6.1 Some General Results . . 6.2 The Formula ef ~ n . . . 6.3 The Formula L: edi ~ n . 6.4 The Formula L: edi = n . Extensions of Discrete Valuations . v

1 1

2 3 4

5 5 9 14 17 17 18 20 23 24 26 27 29 29 30 30 31 31 33 36 38 40

vi

8

9 10 11 II

1

2

3

4

5 III

7.1 Intersections of Discrete Valuation Rings 7.2 Extensions of Discrete Valuations. 7.3 Some Classes of Extensions 7.4 Quadratic Number Fields Ramification Theory of Valuations 8.1 Generalities . . . . . . . . . 8.2 The Value Groups r, rz and rT 8.3 The Ramification Group . . . . . Extending Valuations to Non-Algebraic Overfields Valuations of Algebraic Function Fields Valuations Dominating a Local Domain . . . . . . One-Dimensional Semilocal Cohen-Macaulay Rings Transversal Elements . . . . . . 1.1 Adic topologies . . . . . 1.2 The Hilbert Polynomial 1.3 Transversal Elements. . Integral Closure of One-Dimensional Semilocal Cohen-Macaulay Rings . . . . . . . . . . . . 2.1 Invertible Modules . . . . . . . . . . . . . . . 2.2 The Integral Closure . . . . . . . . . . . . . . 2.3 Integral Closure and Manis Valuation Rings. One-Dimensional Analytically Unramified and Analytically Irreducible CM-Rings . . . . . . 3.1 Two Length Formulae 3.2 Divisible Modules . . . 3.3 Compatible Extensions. 3.4 Criteria for One-Dimensional Analytically Unramified and Analytically Irreducible CM-Rings Blowing up Ideals. . . . . . . 4.1 The Blow-up Ring R" 4.2 Integral Closure 4.3 Stable Ideals Infinitely Near Rings . .

Differential Modules and Ramification Introduction.......... Norms and Traces . . . . . . . . . . . . . . 2.1 Some Linear Algebra. . . . . . . . . 2.2 Determinant and Characteristic Polynomial 2.3 The Trace Form . . . . . . . . . . . . . . 3 Formally Unramified and Unramified Extensions 3.1 The Branch Locus . . . . . 3.2 Some Ramification Criteria . . . . . . . .

1 2

40 40

41 44 46 46

47 49 53 56 59 61 67 67 69 70

74 74 75 77 79 79 80 81 84 88 88 91 92 96 101 101 104 105 106 109 113 113 115

vii

3.3 Ramification for Local Rings and Applications 3.4 Discrete Valuation Rings and Ramification Unramified Extensions and Discriminants Ramification For Quasilocal Rings Integral Closure and Completion . . . . .

120 124 125 131 134

Formal and Convergent Power Series Rings Formal Power Series Rings. . . . Convergent Power Series Rings . . . WeierstraB Preparation Theorem . . 3.1 WeierstraB Division Theorem 3.2 WeierstraB Preparation Theorem and Applications The Category of Formal and Analytic Algebras . . . 4 4.1 Local k-algebras . . . . . . . . . . . . . . . . 4.2 Morphisms of Formal and Analytic Algebras 4.3 Integral Extensions . . . . . . . . . . . 4.4 Noether Normalization . . . . . . . . . 5 Extensions of Formal and Analytic Algebras .

143 143 146 147 147 152 157 157 157 162 163 166

4

5 6

IV

1 2 3

V

1

2 3 4 5 6

VI

Quasiordinary Singularities 169 Fractionary Power Series. . 169 1.1 Generalities . . . . . . 169 1.2 Intermediate Fields. . 171 1.3 Intermediate Fields Generated by a Fractionary Power Series 175 The Jung-Abhyankar Theorem: Formal Case . 177 The Jung-Abhyankar Theorem: Analytic Case 181 Quasiordinary Power Series . . . . 182 A Generalized Newton Algorithm. 190 5.1 The Algorithm . . . . . 190 5.2 An Example . . . . . . 195 Strictly Generated Semigroups 198 6.1 Generalities....... 198 6.2 Strictly Generated Semigroups 202

zq

The Singularity = XYP Hirzebruch-Jung Singularities Semigroups and Semigroup Rings . 2 2.1 Generalities . . . . . . . . . 2.2 Integral Closure of Semigroup Rings Continued Fractions . . . . . . . . . . . . . 3 3.1 Continued Fractions . . . . . . . . . 3.2 Hirzebruch-Jung Continued Fractions Two-Dimensional Cones . . . . . . . . . . . . 4 4.1 Two-dimensional Cones and Semigroups 4.2 The Boundary Polygon of a and the Ideal of Xu 1

205 205 210 210

213 215 216 218 222 222 225

viii

5

Resolution of Singularities . . 5.1 Some Useful Formulae 5.2 The Case p = 1 . . . . 5.3 The General Case .. 5.4 Counting Singularities of the Blow-up

235 235 237 238 244

VII Two-Dimensional Regular Local Rings 1 Ideal 'Transform . . . . . 1.1 Generalities.............. 1.2 Ideal 'Transforms . . . . . . . . . . . 2 Quadratic 'Transforms and Ideal 'Transforms 2.1 Generalities.............. 2.2 Quadratic 'Transforms and the First Neighborhood 2.3 Ideal'Transforms..... 2.4 Valuations Dominating R 3 Complete Ideals. . . . . . . . . . 3.1 Generalities........ 3.2 Complete Ideals as Intersections 3.3 When Does m Divide a Complete Ideal? 3.4 An Existence Theorem . . 4 Factorization of Complete Ideals 4.1 Preliminary Results . 4.2 Contracted Ideals. . . . . 4.3 Unique Factorization. . . 5 The Predecessors of a Simple Ideal 6 The Quadratic Sequence . . . . . 7 Proximity............. 8 Resolution of Embedded Curves.

247 247 247 249 252 252 255 257

VIII Resolution of Singularities 1 Blowing up Curve Singularities . . . . . . . . . . . . 2 Resolution of Surface Singularities I: Jung's Method 3 Quadratic Dilatations . . . . . . . . . . . . . . . . . 3.1 Quadratic Dilatations . . . . . . . . . . . . . 3.2 Quadratic Dilatations and Algebraic Varieties. 4 Quadratic Dilatations of Two-Dimensional Regular Local Rings. 5 Valuations of Algebraic Function Fields in Two Variables .. 6 Uniformization . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Classification of Valuations and Local Uniformization 6.2 Existence of Subrings Lying Under a Local Ring . . . 6.3 Uniformization...................... 7 Resolution of Surface Singularities II: Blowing up and Normalizing 7.1 Principalization . 7.2 Tangential Ideals . . . . . . . . . . . . . . . . . . . . . . ..

303 303 310 313 313 314 316 320 324 324 328 330 334 334 339

260

261 261 263 268 271 273 273 278 281 282 287 292 296

ix 7.3

The Main Result . . . . . . . . . . . . . . . . . . . . . . .. 343

Appendices

345

A

345 345 345 347 348 349 350 351 353 355 357 361 362 365 367 370 373 378 384 385 387 387 389 392 395 398 403

1

2 3 4

5 6 7 8 9 10 11 12 13 14 B

Results from Classical Algebraic Geometry Generalities . . . . . . . . . . . . . . . 1.1 Ideals and Varieties . . . . . . . . 1.2 Rational Functions and Maps . . . 1.3 Coordinate Ring and Local Rings . 1.4 Dominant Morphisms and Closed Embeddings 1.5 Elementary Open Sets . . . . . 1.6 Varieties as Topological Spaces 1. 7 Local Ring on a Subvariety Affine and Finite Morphisms Products . . . . . . . . . . . . . . . Proper Morphisms . . . . . . . . . 4.1 Space of Irreducible Closed Subsets. 4.2 Varieties and the Functor t . . . . 4.3 Proper Morphisms . . . . . . . . . Algebraic Cones and Projective Varieties . Regular and Singular Points . . Normalization of a Variety . . . Desingularization of a Variety . Dimension of Fibres . . . . . . Quasifinite Morphisms and Ramification . 10.1 Quasifinite Morphisms . 10.2 Ramification..... Divisors............ Some Results on Projections. Blowing up . . . . . . . . . . Blowing up: The Local Rings

Miscellaneous Results Ordered Abelian Groups. 1.1 Isolated Subgroups. 1.2 Initial Index. . . . . 1.3 Archimedean Ordered Groups . 1.4 The Rational Rank of an Abelian Group. 2 Localization . . . . . . . . . . . . . . . . . . . 3 Integral Extensions . . . . . . . . . . . . . . . 4 Some Results on Graded Rings and Modules 4.1 Generalities . . . . . . . . . . . . . . . 4.2 M-Graded Rings and M-Graded Modules 4.3 Homogeneous Localization. . . . . 4.4 Integral Closure of Graded Rings . . . . .

1

409 409 409 411 412 415 419 421 423 423 424 426 430

x

5 6

7 8 9 10

Properties of the Rees Ring Integral Closure of Ideals 6.1 Generalities . . . . . 6.2 Integral Closure of Ideals 6.3 Integral Closure of Ideals and Valuation Theory. Decomposition Group and Inertia Group . Decomposable Rings . . The Dimension Formula . . . . . . . . . . Miscellaneous Results ........... 10.1 The Chinese Remainder Theorem . 10.2 Separable Noether Normalization. 10.3 The Segre Ideal . . . . . . . . . . 10.4 Adjoining an Indeterminate . . . 10.5 Divisor Group and Class Group . 10.6 Calculating a Multiplicity 10.7 A Length Formula ... 10.8 Quasifinite Modules ... 10.9 Maximal Primary Ideals . 10.10 Primary Decomposition in Non-Noetherian Rings. 10.11 Discriminant of a Polynomial . . . . . . . . . . . .

433 437 437 437 440 442 448 449 452 452 453 454 456 457 457 458 459 460 461 462

Bibliography

463

Index of Symbols

475

Index

478

Preface The Curves The Point of View of Max Noether Probably the oldest references to the problem of resolution of singularities are found in Max Noether's works on plane curves [cf. [148], [149]]. And probably the origin of the problem was to have a formula to compute the genus of a plane curve. The genus is the most useful birational invariant of a curve in classical projective geometry. It was long known that, for a plane curve of degree n having m ordinary singular pointsl with respective multiplicities ri, i E {1, ... , m}, the genus p of the curve is given by the formula P

= (n -

l)(n - 2) _ ~ "r.(r. _ 1)

2

2 L..,.

••

.

Of course, the problem now arises: how to compute the genus of a plane curve having some non-ordinary singularities. This leads to the natural question: can we birationally transform any (singular) plane curve into another one having only ordinary singularities? The answer is positive. Let us give a flavor (without proofs) on how Noether did it 2 • To solve the problem, it is enough to consider a special kind of Cremona transformations, namely quadratic transformations of the projective plane. Let ~ be a linear system of conics with three non-collinear base points r = {Ao, AI, A 2}, and take a projective frame of the type {Ao, AI, A 2; U}. Let us consider the three conics of ~ having equations

and the map T:IP'~\r--+lP'~

given by the equations

then T is called a quadratic transformation associated with ties: 1. The conics of

~

r.

It has nice proper-

are mapped onto straight lines.

2. T is self-inverse outside AoAI U AoA2 U AIA2' hence it is birational. 1 An ordinary singular point is a point of multiplicity r with r different tangents. 2For details see [190), Th. 7.4., p. 80 or [69], Ch. 7, Th. 2.

xi

xii

Preface 3. The points A o, AI, A2 have no image: they are the fundamental points of the correspondence. 4. The reference lines (with the reference points removed) contract to points.

The algorithm is to apply a sequence of quadratic transformations to the given curve C of degree n (Le. to C and its successive transforms) in such a way that, if P E C is a singular r-fold point, then we take: 1. Ao

= P.

2. The line AoAi , i = 1,2, cuts C at exactly n - r pairwise different points other than Ao, Ai. 3. The line AlA2 cuts C at exactly n pairwise different points other than A l , A2 [by an appropriate choice of A o, AI, A2 this can always be achieved; on says that then C is in excellent position with respect to the points Ao, A l , A2]' In this situation, P "blows up" to a finite number of points on the opposite side of the reference triangle, which are "less complicated" than P. A finite number of steps in this process allows us to reach the desired result.

The First Resolution of Singularities of Curves The first general proof of resolution of singularities for projective curves over a ground field of arbitrary characteristic was performed by Zariski-Muhly [cf. [143]]. Rather than faithfully comment the arguments of Zariski-Muhly, we are going to think about what a resolution of singularities is, and then we shall deal with the curve case in the spirit of these authors, but borrowing from modern algebraic language. The Noether's theorem of the preceding section leads us to the natural question of whether it would be possible to birationally transform a plane curve into another one which is free from singularities. This cannot be done in general. However, the proper question is not that one; it is rather the following: given an irreducible curve C C 1P'~, does there exist another irreducible curve C' C 1P'~/, birationally equivalent to C and free from singularities? The answer to this question is affirmative, as was shown by Zariski-Muhly. Resolution of singularities of curves a la Zariski-Muhly is achieved in the following way. Let k be an algebraically closed field of arbitrary characteristic (the ground field), let C C 1P'~ be an irreducible curve whose ideal in k[Xo, Xl"'" Xn] = k[X] is p, and let us denote by A(C) = k[X]/p = k[x]

where

Xi

:= Xi

+ p for i

E {O, ... ,n}

the ring of homogeneous coordinates on C. By Noether normalization lemma, we may always assume that A(C) = k[ XO,

XI][X2, •.. , Xn

1,

Preface

xiii

where {Xo, Xl, ••• , Xn} are homogeneous of degree 1, A(O) is integral over k[xo, Xl], the field K := k(xt/xo) is a field of rational functions in one variable over k, and the field F = k(xt/xo, ... , xn/XO) is a finite separable extension of K. Note that F is the subfield consisting of elements of degree 0 of the graded ring of fractions of Q+(A) (which consists of all quotients z/w where z, w E A(C) are homogeneous). Now it is not too difficult to show that the integral closure A(O) of A(O) in Q+(A) is a finitely generated k[ Xo, xI]-module. However, there is something far more important, namely that A(0) has a natural structure of a graded ring and of a graded A(O)-module: it is a homogeneous subring of Q+(A). Then there exists a finite set of elements {zo, Zl, ..• , zn'} in A(O), homogeneous of degree 1, such that A(O) = k[ZO,ZI, ... ,Zn']' Then (zo, Zl, ... , zn') are the homogeneous coordinates of the generic point of an irreducible curve 0' C 1P'~', which is birationally equivalent to 0 and is nonsingular. The smoothness comes from the fact that, A(O') being integrally closed, the singular locus of 0' has codimension 2, that is, it is empty3. This proves resolution of singularities for curves. It turns out that, for many purposes, this resolution process is not satisfactory. It is better to have a finite algorithm, i.e., a finite sequence of birational transformations which, when applied to an irreducible curve 0 C 1P'~, produces a nonsingular curve 0'. This will be the main theme of the next section.

The Controlled Process The above resolution of singularities can be considerably refined. In general, we seek not only 0' but a finite sequence of birational transformations that, when successively applied to 0, produces 0'. These transformations must be as simple as possible. If we take this point of view, we need a control of the process of applying transformations that, on the one side, guarantees us the finiteness of the process and, on the other side, allows us to measure the "improvement" at each step. One must never forget this: if we transform, we need a control measure. Overlooking this obvious fact, has driven several early authors to false proofs of resolution of singularities (cf., for instance, a historic case, [198]). The transformations which work are called blowing ups. We will briefly describe a special case: the blowing up of a linear variety in lP'~. This will be enough for our purposes, since a curve has a finite number of singular points, and we will blow up these ones. Let us write n = T + s + 1, n,T,s E No, n ~ 2, consider the projective space 1P'~ and a linear subvariety L of dimension Tj then the radiation P;: / L is a projective space of dimension s [cf. appendix A, section 12]. We also consider the map f: 1P'~ - t IP'UL

that, to each point P E 1P'~, P ¢ L, associates the point (P + L)/L of IP'klL. Let r c 1P'~ x (1P'~ / L) be the graph of f, and let f be its Zariski closure in the product 3This always happens: the singular locus of a projective variety whose coordinate ring is integrally closed has always codimension 2 at least.

xiv

Preface

r

space. The first projection 7r: ~ Wi: is called the blowing up of L in Wi:, or the blowing up of Wi: with center L. It is a straightforward matter to get a set of bi-homogeneous equations for Let V be a projective variety in Wi: containing Lj then the strict transform of V by 7r is the Zariski closure of f -1 (V \ L). If C is a curve and L = P is one of its points, then it is evident that the strict transform C 1 of C is isomorphic to C outside P and that 7r- 1 (C) is a finite set of points, namely one for every tangent line of C at P. We now apply to C the following algorithm. We take a singular point Po E Co := C and blow up Po. If the strict transform C 1 of C has some singular points, we pick one of them, do the same, and so on. In this way, we get a sequence of blowing ups and strict transforms

r.

7I"n Cm-1 ---+ 71"",-1 71"2 C1 --=t 71"1 C, Cm --"+ ... --=t 0,

... 4

and the task is to prove that this sequence is, at any rate, finite 4 . The proof of this fact can be done in two ways. One is purely algebraic, and the idea is very simple. If we start with Po E Co, then we take on C 1 any point PI such that 7r1(Cd = Po, and so on. We may call this "to follow the track" of Po. Passing to local rings, and given the finiteness of the integral closure, this "following the track" ends in a regular local ring, so at a smooth point. Note that, "to follow the track" is enough to prove what we wanted, because, when dealing with one singular point, the others are unaltered, and the total number of simple points is finite. The second way to deal with the proof is analytic and works only for an algebraically closed field of characteristic zero, essentially C. The method is the same: "to follow the track". However, this time we use complex-analytic parametric equations instead of local rings, but with the same aim. To simplify things, we will assume that C is analytically irreducible at Po, Le., it has only one branch. Since we are dealing with one point at each step, we may go affine. If Po is the origin of en, then we may describe a neighborhood of Po by a set of analytic parametric equations Xl

=

t P , X2

=

L

a2i ti , ... ,Xn

=

i>p

L

ani ti

i>p

where p is the multiplicity of Po in Co. The parametric equations of C 1 in a neighborhood of PI are Xl

=

tP, X2

=

La2iti-P, ... ,X n i>p

=

Laniti-p, i>p

and we see that, after a finite number of steps, the multiplicity must drop. This proves everything. In the case that there are several branches at Po (always a 4Remember that it continues one step further only if the last curve has a singular point.

Preface

xv

finite number!), the same process has a step at which either multiplicity drops or two branches separate (hence multiplicity also drops!), so we continue with two sequences with less branches and smaller multiplicities. This fact is a major issue when we solve singularities of surfaces. The complete neighborhood of a singularity of a complex curve can be analytically parametrized. This is in fact false in the case of surfaces. Hence, we can apply these methods only in some very special cases in which analytic parametric equations exist.

The Surfaces J ung Methods When we dealt before with resolution of singularities of curves, we explained how to take advantage of the two key facts: local parametrization and the finiteness of the singular locus. Now that we are dealing properly with the case of surfaces, we find that none of these facts is true. Let us deal first with the difficulties. It is enough to see them in very easy cases, for instance, surfaces embedded in dimension 3. Let S : z2 - 2 y z + x 2 = be a surface with a singularity in (0,0, 0). By solving the second-degree equation in z we get the two solutions

°

which do not make sense, even as meromorphic functions. However, these "expressions" converge for Ixl < Iyl; this gives us a region around (0,0) E C2, which is "one half" of a neighborhood, something of the kind Walker [cf. [189]] called a wedge. The world would be perfect if we could find, for every (embedded) surface (around a singular point, say (0,0,0)), a Puiseux expansion

z= (i,j)EN~

,i+j>O

but this is not true. Nevertheless, Jung [cf. [102]] did something in 1908. He proved (essentially in the complex-analytic case) that, if a neighborhood of a singular point (say o = (0,0,0)) of a surface can be projected onto a plane in such a way that the discriminant locus is a normal crossing divisor 5 , then the complete neighborhood of 0 can be represented by conjugate Puiseux power series in two variables, of a special type called quasi-ordinary, which we are not to describe here. For instance, around the origin, the surface Z4 - 2 xz2 + x 2 - x 2 y = has a discriminant D = - 256 x 6 (-1 + y) y2, hence, by Jung, the complete neighborhood of (0,0,0) is described by the Puiseux power series z = ,;xVI - .,;y.

°

5i.e., the set of zeroes of xaybU(x, y), with U a unit in the ring of convergent power series

xvi

Preface

There are several proofs of Jung's theorem besides Jung's own one, which is of a topological nature. Abhyankar felt the need of having a purely algebraic proof, which he gave in 1955 [cf. [1]] making use of very sophisticated Galois theory. The authors of this book gave a third one by using modern techniques of ramification theory [cf. [108]]. Finally, there are hints for the existence of a purely combinatorial way of proving it [see, for instance, [175] for the two-dimensional case]. Now that we have introduced local parametrization of a quasi-ordinary (or Jungian) surface singularity, the obvious question is the following: how can we apply this to obtain a full resolution of singularities of a surface? We have two problems: 1. Given a projection IT: X -t S where X and S are surfaces and S is smooth, is there a morphism (1): This is clear.

1 Marot Rings

3

Now we assume that A is a Marot ring, and that M is a finitely generated regular submodule of the A-module Q(A). By adapting the proof of (1) ~ (2), it is easy to see that M has a finite system of generators contained in Reg(Q(A)). (1.4) Corollary: Let A have the following property: For every two elements a, b E A with a E Reg(A) there exists c E A such that b + ca is regular. Then A is a Marot ring. Proof: Note that the ideal Aa + Ab can be generated by the regular elements a and b + ca. (1.5) Corollary: (1) If A is a Marot ring, then every subring ofQ(A) containing A is a Marot ring. (2) If every finitely generated regular ideal of A is principal, then A is a Marot ring.

1.2

Large Quotient Ring

(1.6) REMARK: Let A be a ring, and let p be a prime ideal of A. It is easy to see that the set A[p] := {x E Q(A) Isx E A for some sEA \ p} is a subring of Q(A); this subring is called the large quotient ring of A with respect to p. Note that A[p] = Ap if A is a domain. Properties of the large quotient ring are listed in [118] and [96]. In case A is a Marot ring, the ring A[p] is the localization of A with respect to a multiplicatively closed system, as the next result shows; hence in this case all the usual properties concerning localization can be applied. (1.7) Proposition: Let A be a Marot ring, and let p be a prime ideal of A. We set 8 := Reg(A) n (A \ p). Then 8 is a multiplicatively closed system in A, and we have A[p] = 8- 1 A. The ideal q := pA[p] is a prime ideal of A[p], and we have Ap = (A[p])q. Proof: It is clear that 8 is a multiplicatively closed system in A, and that 8- 1 A C A[p]. Let x E A[p], and choose sEA \ P with sx E A. There exists a regular element a E A with ax E A. We set a := Aa + As; then a is a regular ideal of A, and we have ax C A and a ¢. p. If P contains only zero-divisors, then we set t := a, and if p contains regular elements, then there exists a regular element tEa which does not lie in p [since A is a Marot ring]. In both cases we have tx E A, hence we have shown that x E 8- 1 A. Since 8 does not meet p, the extended ideal q = 8- 1 p is a prime ideal of A[p]. Every element of A \ p is invertible in (A[p])q, hence we have a canonical homomorphism Ap -+ (A[p])q of rings, and it is easy to check that it is bijective.

4

1.3

I Valuation Theory

Rings With Large Jacobson Radical

(1.8) Lemma: Let A be a ring; assume that every prime ideal of A is maximal, and that the Jacobson radical of A is the zero ideal. Then, for every a E A, there exists b E A such that ab = 0 and that a + b is a unit of A. Proof: If a E U(A), then we may choose b = O. If a ~ U(A), then Aa f= A and the set n of those maximal ideals of A which contain a is not empty. The set S := A \ (UmEfl m) is multiplicatively closed. The Jacobson radical t(A) and the nilradical n(A) of A are equal-namely equal to the zero ideal of A-j likewise, the Jacobson radical t(S-l A) and the nilradical n(S-l A) of S-l A are equal. Since a/I E t(S-l A) and n(S-l A) = S-ln(A) = {O}, we have a/I = 0 in S-l A, hence there exists b E S such that ba = O. Let m be any maximal ideal of A. If a ~ m, then bE m and if a E m, then b ~ m, hence a + b ~ m, and therefore a + bE U(A). (1.9) Proposition: Let A be a ring, and let t be its Jacobson radical. The following statements are equivalent: (1) Every prime ideal of A containing t is maximal. (2) For each a E A there exists b E A such that, for all d E A and all units u of A, a + ub and 1 + dab are units of A. (3) For each a E A there exists b E A such that ab E t and that a + b E U(A). Proof (1) ::} (2): Let c.p: A ~ A/t be the canonical homomorphism. Byassumption, every prime ideal of A/t is maximal and the Jacobson radical of A/t is the zero ideal. By (1.8) there exists bE A such that c.p(a)c.p(b) = 0 and that c.p(a) +c.p(b) is a unit of A/t. Thus, we have ab E t and there exist c E A, f E t such that c(a + b) = 1 + f. Now, 1 + f is a unit of A [cf. [63], Ex. 4.7], hence a + bE U(A). Since ab E t, we have 1 + dab E U(A) for every d E A [ef. [63], Ex. 4.7]. Let u E U(A). For every maximal ideal m of A we have a + b ~ m and ab Em, hence either a E m and b ~ m, whence ub ~ m and therefore a + ub ~ m, or a ~ m and bE m, whence ub Em and again a + ub ~ m. Therefore we have a + ub E U(A). (2) ::} (3): By assumption, there exists b E A such that a + b is a unit and such that, for every d E A, the element 1 + dab is a unitj hence we have ab E t. (3) ::} (1): Let p be a prime ideal of A containing t and let m a maximal ideal of A containing p. Suppose that m f= p, and let a E m \ p. By assumption, there exists b E A such that ab E t and that a + b E U(A). Now we have b E P since ab EtC pj thus, we have a + b E m, in contradiction with the fact that a + b is a unit. (1.10) DEFINITION: A ring is said to have large Jacobson radical if it satisfies the conditions of (1.9). (1.11) REMARK: (1) A ring in which every prime ideal is maximal is a ring with large Jacobson radical. (2) A quasisemilocal ring is a ring with large Jacobson radical.

2 Manis Valuation Rings

5

(1.12) Proposition: If the ring of quotients of A is a ring having large Jacobson radical, then A is a Marot ring. Proof: Let a E Reg(A) and bE A. By (1.9) we may choose c E Q(A) such that b + uc E Reg(Q(A)) for every u E Reg(Q(A)). We write c = zr- 1 with z E A, r E Reg(A). Now we have b + (ar)c = b + az E Reg(A), hence A is a Marot ring by (1.4).

(1.13) Corollary: A noetherian ring is a Marot ring. Proof: The ring of quotients of a noetherian ring A is a semilocal ring since the set of regular elements of A is the complement of the union of the finitely many prime ideals in Ass(A) [cf. [63], Th. 3.1], hence it is a ring having large Jacobson radical [cf. (1.11)(2)].

2

Manis Valuation Rings

(2.0) NOTATION: In this section K always denotes a ring having large Jacobson radical and which is its own ring of quotients. Every subring of K having K as ring of quotients is a Marot ring [cf. (1.12)].

2.1

Manis Valuation Rings

(2.1) Lemma: Let V be a subring of K with the following property: The product of elements in K \ V lies in K \ V. (By abuse of language, in this section such a set shall be called multiplicatively closed.) Then V is integrally closed in K, and K is the ring of quotients of V. Proof: There is nothing to show if V = K. Therefore, we treat the case that V "I- K. (1) Let a, bE K. If ab E V, then a E V or bE V since K \ V is a multiplicatively closed set. In particular, for every a E Reg(K) we have a E V or a-I E V. (2) Let t E K \ V, and suppose that t is integral over V. Let F = Tn + alTn-1 + ... + an E V[ T] be a monic polynomial ofleast degree n such that F(t) = O. Note that n ~ 2 since t (j. V. Now t(tn-l + altn-2 + ... + an-d = -an E V implies that t n- 1 + altn-2 + ... + an -l E V, contradicting the choice of n. Hence V is integrally closed in K. (3) Since K is its own ring of quotients, the ring of quotients Q(V) of V can be considered as a subring of K. (a) The Jacobson radical t(K) of K is contained in Q(V). Indeed, let a E t(K); then b := 1 + a is a unit of K, hence b E V or b- 1 E V [cf. (1)]. If bE V, then a EVe Q(V), and if b- 1 E V, then a = (1 - b- 1)(b- 1)-1 E Q(V). (b) Let a E K. By (1.9) there exists c E K such that b:= a + c is a unit of K and that ac E t(K). By (a) there exist elements x E V, r E Reg(V) with ac = x/r. Now (ar)2 - br(ar) + xr = 0, hence (ab- 1r)2 - r(ab- 1r) + xb- 2r = 0 since b is a

6

I Valuation Theory

unit of K. If bE V, then ar is integral over V, hence ar E V by (2), and therefore we have a E Q(V). If b- l E V, then ab-lr E V, and therefore we have a E Q(V).

(2.2) Theorem: Let Vi:- K be a subring of K having K as its ring of quotients. The following statements are equivalent: (1) There exists a prime ideal p of V such that pA = A for every subring A of K properly containing V. (2) K \ V is a multiplicatively closed set. (3) For every regular element x E K we have either x E V or X-l E V. (4) The set of regular cyclic V -submodules of K is totally ordered by inclusion. (5) The set of regular V -submodules of K is totally ordered by inclusion. Let these conditions be satisfied. Then the regular non-units of V are contained in a unique maximal ideal m of V, and for every subring A of K properly containing V we have mA = A. Proof (1) => (2): Let t, t' E K \ V, and suppose that tt' E V. Now V is a proper subring of V[ t] and of V[ t']. Since pV[ t] = V[ t] by assumption, there exist mEN and Zo, . .. ,Zm E P with 1 = Zo + Zl t + ... + zmtm, hence 1 - Zo = Zlt+·· ,+zmtm E (V\p)npV[ t]. We choose mEN minimal with the property that there exist Zl, ... ,Zm E P such that a := Zlt+· . +zmtm E (V\p)npV[ t]. Likewise, we choose n E N minimal with the property that there exist Wl, ... ,W n E P such that a' := Wlt' + ... +Wnttn E (V\p) npV[ t']. We may assume that m ~ n. Since a'zmtm = WlZm{tt')t m- l + W2Zm(tt,)2tm-2 + ... + wnzm(tt,)nt m- n , we see that a'a

= a'zlt + ... + a'Zm_ltm-l + a'zmtm + ... + a' Zm_n_ltm-n-l (a'Zm-n + WnZm{tt,)n)t m- n + ... + (a'Zm-l + WlZm(tt'))t m- l .

= a' Zlt

+

Now a'Zl, ... ,a'Zm-n-1 and a'zm-n + wnzm{tt,)n, ... ,a'Zm-1 + WIZm(tt') are elements of p; on the other hand, we have a' a E V \ p since p is a prime ideal, contradicting the choice of m. (2) => (3): Let x E K be regular. Since 1 = xx- I E V, we see that, if x cf. V, then X-I E V. (3) => (4): A regular cyclic V-submodule of K has the form Vc for some regular c E K. Let a, b be regular elements of K. If Va (5): Let M, N be regular V-submodules of K such that M 0, and therefore me Pv [since Av is a Marot ring]. Conversely, let x E Pv be regular. Then x is a non-unit of A v , hence x E m and therefore Pv em. Now we have shown that m = Pv. (3) We have nv C (Av : K)K. Suppose that there exists x E (Av : K)K with vex) < 00. Since v is surjective, there exists y E K with v(y) = - vex). Since Av '" K, there exists z E K with v(z) < O. Now we have x(yz) E Av but v(xyz) < 0, contrary to the definition of Av. (4) Let x E K and vex) '" 00. Then x fI. nv = (Av : K)K, and there exists a E K with ax fI. Av. We choose a regular b E Av with ba E Av [note that Q(Av) = K]. Now vex) < - v(a) ~ v(b). The regular Av-module Avx + Avb is cyclic [cf. (2.4)(2)], hence we have Avx + Avb = AvY for some regular y E K. This yields y E Avx + Avb, hence v(y) ~ mine {v(x), v(b)}) = vex), and x E AvY, hence v(y) ~ vex). Thus, we have shown that v(y) = vex); since y is regular, it follows that we have v(Reg(K)) = r.

(2.13) Proposition: Every Manis valuation ring of K is the ring of a Manis valuation of K.

2 Manis Valuation Rings

11

Proof: Let V be a Manis valuation ring of K, and let m be the regular maximal ideal of V. Let ~ be the group of regular cyclic V-submodules of K (written additively); its zero element is V. We define II := {Val a E V regular}. Then we have II + II c II, lIn (-II) = {V}, and IIu (-II) =~. We define Va ~ Vb if Va-1b C II; then (~, 0, then let Fi E F be such a polynomialj otherwise we set Fi := 1. We set F := 1 + T2FI ···Fn E Z[T]. It is clear that F E Fj we set k:= deg(F). Let i E {I, ... , n}. First, we consider the case Vi(X) ~ OJ then we have vi(F(x» ~ o. Suppose that Vi (F(x» > OJ then Vi (Fi (x» > 0 [by definition of Fi ], hence Vi (x2FI (x) ···Fn(x» > 0, and therefore vi(F(x» = 0 [ef. (2.10)(2)]; this contradiction shows that vi(F(x» = O. Now, we consider the case Vi(X) < O. Then vi(F(x» = Vi(X k ) [ef. (2.10)(1) ], and Vi(X k ) = kVi(X) = (k-1)vi(x)+vi (x) < Vi(X) [since k ~ 2]. (2.19) Proposition: Let VI: K -+ rIco, ... , Vn: K -+ r nco be Manis valuations of K, and set R := AVI n··· n AVn and lUi := R n Pv, for every i E {I, ... , n}. We assume that Av. rt. Av; for all i, j E {I, ... , n}, i i- j. Then {ml, ... , mn } is the set of regular maximal ideals of R, they are all different, K is the ring of quotients of R, and we have R[m.J = Av. and miAv. = Pv. for every i E {I, ... , n}.

Proof: Let t be the Jacobson radical of K, and set qi := {x E R I Vi(X) = co} for every i E {I, ... , n}. Clearly qi is a prime ideal of R. (1) Let i E {I, ... , n}, and assume that t rt. qi. We show: for every 'Y E ri there exists a regular x E lUi with Vi(X) > 'Y. For the proof of this assertion we may assume that 'Y > o. We choose a E t and a i qi· Since Vi is surjective, there exists b E K such that vi(ab) = -'Y. We choose

2 Manis Valuation Rings

15

with respect to ab a polynomial FE Z[T] as in (2.18); then we have vj{F{ab» $ 0 for every j E {I, ... , n} and vi{F{ab)) < -'Y. Now F{ab) is of the form 1 + da for some d E K, and since a E t, the element 1 + da is a unit of K. The inverse (I + da)-1 of 1 + da lies in R, and we have vi{{1 + da)-I) > 'Y. (2) Let a E K, and assume that vi{a) ¥- 00 for every {I, ... ,n}. For every t E K with ta E t there exists a regular u E R with Vi (ut) > vi{a) for every i E {I, ... , n}. Indeed, set J:= {j E {1, ... ,n} I Vj{t) ¥- oo}. First, we consider the case J ¥- 0. For every j E J we have t f/. qj since ta E t. Let j E J; then there exists a regular element tj E mj with Vj{tj) > vj{a) - Vj{t) by (I), i.e., Vj{ttj) > vj{a). We set u := I1 jEJ tj; u is a regular element of R. Since viet) = 00 for every i E {I, ... ,n} \ J, it follows that Vi{ut) > vi{a) for every i E {I, ... ,n}. Now we consider the case J = 0. Then viet) = 00 for every i E {I, ... , n}, and we choose u = 1. (3) For every a E K there exists a regular element b E R with ab E R and with the following property: for every i E {I, ... , n} with vi{a) ~ 0 we have vi{b) = o. To see this, we choose with respect to a a polynomial F E Z[T] as in (2.18) and set y := F{a). Then we have Vi{Y) $ 0 for every i E {I, ... , n}. Since K has large Jacobson radical, there exists d E K such that Y + ud is a regular element of K for every regular element u of K, and 1 + zyd is a regular element of K for every z E K [cf. (1.9)]. In particular, y+d is a regular element of K and yd E t [cf. [63], Ex. 4.7]. By (2) there exists a regular element r E R with vi(rd) > Vi{y) for every i E {I, ... , n}. Therefore we have Vi(y + rd) = Vi(y) $ 0 for every i E {I, ... , n}. Since y + rd E Reg(K), we may set b:= (y + rd)-I; it is easy to check that b has the desired properties. (4) For every i E {I, ... , n} we have At/; = RIm,], hence we have K = Q(R). Indeed, it is clear that RIm,] c At/,. Let a E At/" and let b be chosen as in (3). Then b E R\ m;, ab E R, and a = ab/b E RIm;]. In particular, if Vi(a) > 0, then we have ab E m;, hence m;At/, = Pt/;. Since Q(At/.) = K, it follows that Q{R) = K. (5) {ml, ... , tlln} is the set of pairwise different regular maximal ideals of R. Indeed, let n be a regular maximal ideal of R and let r E n be a regular element. Suppose that n f/. ml U ... U mn • Then there exists an element a E n such that Vi (a) = 0 for every i E {I, ... , n}. Since K has large Jacobson radical, there exists t E K with a + xt E Reg(K) for every x E Reg(K), and with 1 + yat E Reg(K) for every y E K [cf. (1.9)]; in particular, at E t. By (2) there exists u E Reg(K) such that Vi(ut) > 0 for every i E {1, ... ,n}. Now a+urt lies in Reg(K) since ur E Reg(K), and therefore a + urt is a unit of R since vi(a + urt) = vi(a) = 0 for every i E {l, ... , n}. On the other hand, since a and r lie in n and ut lies in R, it follows that a + urt E n. This contradiction shows that n C ml u··· U tlln, hence that n C m; for some i E {I, ... , n} [cf. [63], Lemma 3.3], and therefore that n = m;. Let i, j E {l, ... , n}, i ¥- j; then we have m; f/. mj since RIm;] = At/; f/. At/, = RIm,] by assumption. Let i E {l, ... , n}; suppose that m; is not a maximal ideal of R. Then there exists a regular maximal ideal n of R containing m;, and, by what we have just shown, we have n C mj for some j E {I, ... , n}, i.e., we have mi C mj.

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I Valuation Theory

Therefore we have shown that {ml' ... , Inn} is the set of pairwise different maximal regular ideals of R. (2.20) Theorem: [Approximation theorem for discrete Manis valuations] Let

VI, ... ,Vn be pairwise different discrete Manis valuation rings of K. For every i E {I, ... , n} let Vi: K -t Zoo be the Manis valuation of K defined by Vi, and let PVi be the regular maximal ideal of Vi. We set R := VI n ... n Vn and ~ := R n PVi for every i E {I, ... , n}. Then:

(1) The prime ideals ml, ... , Inn are regular maximal ideals of R which are pairwise different, and they exhaust the set of regular prime ideals of R. Moreover, for every i E {I, ... ,n} we have RlmiJ = Vi and ~Vi = PVi' (2) For any aI, ... ,an E K and ml, ... ,mn E Z there exists a E K such that

Vi{a - ai)

~

mi

for every i E {I, ... , n}.

(3) For any ml, ... ,mn E Z there exists a E K such that

Vi{a) = mi

for every i E {I, ... , n}.

Proof: The assumption on Vi, ... , Vn implies that Vi ct. Vi for i, j E {I, ... , n}, i :I j [cf. (2.15)]. (I) By (2.19) we know that {ml,"" Inn} is the set of regular maximal ideals of R, that they are pairwise different, and that Vi = RlmiJ and ~ Vi = PVi for every i E {I, ... , n}. Let P be a regular prime ideal of R. There exists i E {I, ... , n} such that P c ~, hence pVi = ~Vi = PVi [since pVi is a regular prime ideal of Vi = RlmiJ and since Vi has only one regular prime ideal by (2.15)]. Let x E ~. Then there exist yEp and a regular element s E R \ mi such that x = y / s, hence that sx = yEp, and that therefore x E p. (2) Let al, ... , an E K. Since Q(R) = K by (2.19), there exist elements TI, ... , Tn E R and a regular element s E R such that al Td s, . .. , an Tn/ S. Since s is regular, we have Vi{S) :I 00 for i E {I, ... , n}. Set m~ := mi + Vi(S) for every i E {I, ... , n}. The existence of a E K satisfying (2) follows from the existence of bE K satisfying vi(b - Ti) ~ m~ for every i E {I, ... , n}. To prove (2) we may assume, therefore, that aI, ... ,an E Rj furthermore, we may assume that

=

=

ml, ... ,mn E N. Let i E {I, ... , n} and set qi := {x E R I Vi{X) ~ mil. It is clear that qi is an ideal of R and that m7'i c qi C ~ [since the ideal m7'i is generated by all products XI'''X mi where XI"",X mi E ~]. Now we apply B{lO.I): the ideals mr 1 , ••• , m~" are pairwise comaximal, hence the ideals ql,"" qn are pairwise comaximal, and therefore there exists a E R such that a == ai (mod qi) for every i E {I, ... , n}, hence we have vi{a - ai) ~ mi for every i E {I, ... , n}. (3) For every i E {I, ... , n} there exists ai E K such that vi{ai) = mi. By (2) there exists a E K such that Vi (a - ai) ~ mi + 1 for every i E {I, ... , ti}. Now we have vi(a) = vi{ai) for every i E {I, ... , n} [cf. (2.10)(2)].

17

3 Valuation Rings and Valuations

(2.21) Corollary: We assume, furthermore, that (VI : K)K, ... , (Vn : K)K is the set of prime ideals of K. Then: (1) Let x E K; then x is regular iffvi(x) < 00 for every i E {1, ... , n}. (2) Let ml, ... ,mn E Z. Then there exists a regular x E K with

Vi(X) = mi

for every i E {1, ... , n}.

(3) Every regular ideal of R is a principal ideal. Proof: (1) Let x E K; if x is regular, then Vi(X) < 00 for every i E {1, ... , n}. Conversely, assume that Vi (X) < 00 for every i E {1, ... ,n}, i.e., that x ~ (Vi : K)K for every i E {1, ... , n} [cf. (2.12)]. Then x is a unit of K [since, by assumption, the ideals (VI : K) K , ... , (Vn : K) K are all the prime ideals of K]. (2) follows from (1) and (2.20) (3). (3) Let i E {1, .. . ,n}. By (2) there exists a regular Xi E K such that Vi(Xi) = 1, Vj{Xi) = 0 for every j E {1, ... , n}, j "I i, and therefore we have RXi C mi. Let y E mi. Then Vi(Y) ~ 1, hence Vj{yjxi) ~ 0 for every j E {1, ... , n}, and we have shown that y E RXi, hence that IDi = RXi. Let a C R be a regular ideal, and, for every i E {1, ... ,n}, set mi := min( {vi(a) I a E a}) [mi E No since a is a regular ideal]. First, we consider the case ml = ... = mn = o. Then a ct mj for every j E {1, ... ,n}, hence a = R [since ml, ... ,mn is the set of maximal ideals of R]. In the general case we apply this argument to the regular ideal ax1m1 ••• x~mn of R.

3

Valuation Rings and Valuations

(3.0) NOTATION: In this section K always denotes a field.

3.1

Valuation Rings

(3.1) DEFINITION: A subring V of K is called a valuation ring of K if it is a Manis valuation ring of K or if V = K. (3.2) NOTATION: Let V be a valuation ring of K. Then V is integrally closed and has K as its field of quotients [cf. (2.1)]. Furthermore, it is quasilocal. Let m be the maximal ideal of V; the field Vjm is called the residue field of the valuation ring V. (3.3) NOTATION: Let V be a valuation ring of K, and let m be the maximal ideal of V. Let A be a subring of K with A c V. The prime ideal m n A is called the center of V in A. In particular, if A is quasilocal and m n A is the maximal ideal of A, then we say that V dominates A. Let V dominate the quasilocal ring A. The residue field Aj{m n A) of A is a subfield of the residue field Vjm of V. The valuation ring V is said to be of the first kind (resp. of the second kind) with respect to A if the residue field Vjm of V is algebraic (resp. not algebraic) over the residue field Aj{m n A) of A.

18

I Valuation Theory

(3.4) Proposition: Let A be a subring of K. The set of valuation rings of K containing A is not empty, and its intersection is the integral closure of A in K. Proof: This follows from (2.4) and (2.5). (3.5) Theorem: [Existence theorem] Let A be a subring of K, and let p be a non-zero prime ideal of A. Then there exists a valuation ring V =F K of K which has center p in A. Proof: By replacing A by Ap we may assume that A is quasilocal with maximal ideal m =F {O}. The set B of subrings B of K with A C Band mB =F B is not empty [ since A E B] and it is ordered by inclusion. Let (Bi)iEI be a chain in B, and set B := UiEI Bii then B is a subring of K. Suppose that mB = B. Then there exist Xl, ... , Xn E m and bl , ... , bn E B with 1 = xlb l + ... + xnbn. Since (Bi)iEI is a chain, there exists j E I with bl , ... , bn E B j , hence we get mBj = Bj, contradicting Bj E B. Therefore B E B is an upper bound of (Bi)iEI. Let V be a maximal element of B [Zorn's lemma]; then we have mV =F V, and therefore there exists a maximal ideal n of V containing mVi for every subring C of K properly containing V we have mC = C by maximality of V, hence nC = C, and therefore V is a valuation ring of K [cf. (2.2)], and V =F K since mV =F V.

(3.6) Corollary: Let A be a subring of K, and let a be a proper non-zero ideal of A. Then there exists a valuation ring V of K with V :::> A and aV =F V. Proof: Since a is a proper non-zero ideal of A, there exists a non-zero maximal ideal n of A containing a. By (3.5) there exists a valuation ring V of K which dominates An.

(3.7) Proposition: Let V be a valuation ring of K, and let M be a finitely generated torsion-free V -module. Then M is a free V -module. Proof: Let {Xl, ... , xn} be an irredundant system of generators of M. Suppose that there exist elements al, ... , an of V which are not all zero, such that al Xl + ... + anxn = O. The ideal of V generated by al, ... , an is a principal non-zero ideal [cf. (2.4)]. By relabelling, we may assume that this ideal is generated by al. Then we have al(xi + (a2/adx2 + ... + (an/al)Xn) = 0, and since M is torsion-free, we have Xl + (a2/al)x2 + ... + (an/adx n = 0, hence Xl = -«a2/al)x2 + ... + (an/adxn), and therefore X2, . .. ,Xn also generate M, contradicting the assumption that {Xl, ... , xn} is an irredundant system of generators of M. Thus, {Xl, ... ,Xn } is a free system of generators of M.

3.2

Subrings and Overrings of Valuation Rings

(3.8) Proposition: Let V be a valuation ring of K. Then: (1) Let W be a subring of K containing V. Then W is a valuation ring of K, and its maximal ideal is a prime ideal of V.

3 Valuation Rings and Valuations

19

(2) The map p I-t Vp is a bijective inclusion-reversing map from the set of prime ideals of V to the set of valuation rings of K containing V. The inverse map maps such a valuation ring of K containing V to its maximal ideal. (3) The set of subrings of K containing V is totally ordered by inclusion. Proof: (1) The first statement follows from (2.4)(1). Let n be the maximal ideal of W. If n = {O}, then W = K and n is a prime ideal of V. In the other case, let x E n be non-zero, and suppose that x ¢ V. Then x- 1 E V, hence 1 = xx- 1 E n, contradicting n"l W. Therefore n is an ideal of V; clearly it is a prime ideal of V. (2) Let p be a prime ideal of V, and set W := Vp. Then W is a subring of K, hence a valuation ring of K, and its maximal ideal pVp is contained in V by (1), hence p = pVp n V = pVp. Conversely, let W be a valuation ring of K containing V, and let n be its maximal ideal. We show that Vn = W. Since every element in V \ n is a unit of W, we have Vn C W. Let z E W. If z E V, then z E Vn. If z ¢ V, then w:= liz E V, and since z ¢ n [as z ¢ V], z is a unit of W, hence w is a unit of Wand therefore w ¢ n, hence z = 11w E Vn . (3) The set of prime ideals of V is totally ordered by inclusion [cf. (2.2)]; the assertion follows from (2).

(3.9) DEFINITION: Let V be a valuation ring of K. If V has only finitely many non-zero prime ideals, then the number of those prime ideals is called the rank of the valuation ring V, and it is denoted by rank(V). In the other case, we say that V has infinite rank, and we set rank(V) = 00. (3.10) VALUATION RINGS OF RANK> 1: Let V be a valuation ring of K with rank(V) > 1. Then V is not noetherian. In fact, suppose that V is noetherian. Let p ~ q "I {O} be prime ideals of V [remember that rank(V) > 1]. Since p and q are finitely generated, these ideals are principal ideals p = V S, q = Vt [cf. (2.4)], hence sand t are prime elements, hence irreducible, and since t = us with u E V, u must be a unit of V, hence p = q, contrary to our assumption.

(3.11) VALUATION RINGS OF RANK 1: Let V be a valuation ring of K ofrank 1. Then the maximal ideal m of V is the only non-zero prime ideal of V; moreover, by (3.8), K is the only subring of V which properly contains V. We have by (2.15): V is noetherian iff m is principal. Moreover, in this case V is a principal ideal domain with only one non-zero prime ideal, and it is a one-dimensional regular local ring. (3.12) Proposition: Let V be a valuation ring of K with maximal ideal m, and let cp : V -+ Vim =: /1, be the canonical homomorphism. Then: (1) The map W I-t cp(W) is a bijective inclusion-preserving map from the set of valuation rings of K contained in V to the set of valuation rings of /1,.

20

I Valuation Theory

(2) Let T C V be a quasilocaI subring of K having K as field of quotients. The map in (1) induces a bijective map from the set of valuation rings of K contained in V and dominating T to the set of valuation rings of K. dominating t.p(T).

Proof: (1) Let W be a valuation ring of K contained in V with maximal ideal n. The maximal ideal m of V is a prime ideal of Wand Wm = V [cf. (3.8)]. Let z E V \ m. If z E W, then t.p(z) E t.p(W). If z f/. W, then liz E W, hence t.p(l/z) = 1/t.p(z) E t.p(W). Therefore t.p(W) is a valuation ring of K.. Conversely, let W be a valuation ring of K., and set W := t.p-l(W). Then W is a subring of V, and m is an ideal of W. Let z E KX. If z f/. V, then liz E meW. If z E m, then z E W. If z E V\ m, then z is a unit of V, hence t.p(z) "10. If t.p(z) E W, then there exists z' E W with z - z' E meW, hence z E W. If t.p(z) f/. W, then 1/t.p(z) E W, hence there exists z' E W with z' - 1I z E m, hence 1I z E W. Therefore W is a valuation ring of K. (2) Let p be the maximal ideal of T. Then m n T c p, hence t.p(T) ~ TI(m n T) is a quasilocal subring of K.. It is easy to check that, if W is a valuation ring of K which is contained in V and which dominates T, then t.p(W) dominates t.p(T). Conversely, ifW is a valuation ring of K. which dominates t.p(T) , then W := t.p-l(W) is a valuation ring of K which dominates T.

3.3

Valuations

(3.13) NOTATION: For the rest of this section let (r, O} is the maximal ideal of the quasilocal ring Av [cf. (2.12)]. The field K. v := AvlPv is called the residue field of v. The rank of Av is said to be the rank of v, and we denote it by rank( v). The rational rank rat. rank(r v) of r v [cf. B(1.29)] is called the rational rank of Vj it is denoted by rat. rank(v). (3.16) REMARK: Let v be a valuation of K with value group r. If r is finitely generated, then the Z-modules rand Zd with d:= rat. rank(v) are isomorphic [cf. [32], Ch. VII, § 4, Cor. 2 to Th. 2, and note that r is torsion-free]. (3.17) NOTATION: Valuations v and w of K are called equivalent if Av = Aw.

3 Valuation Rings and Valuations

21

(3.18) NOTATION: Let v be a valuation of K, and let A be a subring of K contained in Av. The center of Av in A is called the center of v in A, and if A is quasilocal and Av dominates A, then we say that v dominates A. (3.19) REMARK: (1) Let v: K X -+ r be a valuation of K, and let V be the valuation ring of v. Let ~ be defined as in (2.14); the map

in an isomorphism of ordered groups. Therefore we always may identify the value group v(KX) of v with the totally ordered group {Va I a E KX} of non-zero cyclic V-submodules of K. (2) Let V be a valuation ring of K. If V :I K, then we constructed in (2.13) a valuation v of K such that V = Av; this valuation shall be called the canonical valuation of K defined by V. If V = K, then V is the ring of the trivial valuation of K, which also shall be called the canonical valuation of K defined by V. (3.20) REMARK: Let A be a subring of K having K as field of quotients, and let v: A -+ roo be a map with v(l) = 0, vex) = 00 iff x = 0, and with

v(a + b)

~

= v(a) + v(b) for all a, bE A. -+ roo such that w is a valuation of K.

mine {v(a), v(b)}), v(ab)

Then v admits a unique extension w: K In fact, setting w(a/b) := v(a) - v(b) for a, b E A, b :I 0, gives a well-defined map w: K -+ roo since, from alb = a'/b' for a, a' E A, b, b' E A \ {O}, we get ab' = a'b, hence v(a) + v(b' ) = vea') + v(b). It is clear that wlKx: KX -+ r is a homomorphism; since elements x, x, E K can be written in the form x = alb, x' = a'/b with a, a', b E A, b:l 0, we have w(x + x') = w«a + a')/b) ~ mine {w(x), W(X')}), hence w is a valuation of K. Let w' be a valuation of K with w'IA = v. For x = alb with a, bE A and b :I 0 we have w'(x) = w'(a) - w'(b) = v(a) - v(b) = w(a/b) = w(x), hence we have w' = w. (3.21) Proposition: Let v: K -+ roo be a valuation of K, and let V be the ring of v. For every 'Y E r with 'Y ~ 0 the sets a'"( (resp. at) of elements of V with vex) ~ 'Y (resp. vex) > 'Y) are ideals of V, and every non-zero ideal of V contains an ideal a'"( for some 'Y E r with 'Y ~ O. Proof: Clearly ~ (resp. at) is an ideal of V. Let a :I {OJ be an ideal of V, and let x E a be a non-zero element. Set 'Y := vex). Then 'Y ~ 0 and a")' c a [let yEa")', then y = zx with z := y/x E V]. (3.22) V-SUBMODULES OF K AND MAJOR SUBSETS OF r: Let v be a valuation of K, let V be the ring of v, and let r be the value group of v. (1) For any V-submodule N of K we set 0 for all j E {I, ... ,n}, j '" i. Proof: (1) We set R := Al n ... n An, ~ := m(Ai) n R. Then {ml,"" mn} is the set of maximal ideals of R, we have Rm; = Ai and ~Ai = m(Ai) [cf. (2.19)]. Since Ailm(Ai) = Rm; I~Rm; = R/~, we may assume that aI, ... ,an E R. Now the assertion follows from the Chinese remainder theorem B(lO.I). (2) This follows from (1) by choosing aj = 1, aj = 0 for j E {I, ... , n}, j '" i.

28

I Valuation Theory

(4.2) DEFINITION: (1) Valuation rings V, W of K are called independent if K is the smallest subring of K which contains V and W j otherwise they are called dependent. (2) Valuations v and w of K are called independent if the valuation rings Av and Aware independent, otherwise they are called dependent. (4.3) DEPENDENCY AND DECOMPOSITION: (1) Let v and w be non-equivalent non-trivial valuations of K with Av ¢. Aw and Aw ¢. Av. If they are dependent, then there exists a subring A :f. K of K with Av c A and Aw C A, hence there exists a valuation ring V' :f. K of K with A c V' [cf. (3.5)]. Let v' be a valuation of K defined by V'. Then we have non-trivial decompositions v = v' 0 v and w = v' ow. (2) Let v and w be valuations of K which admit non-trivial decompositions v = v' 0 v and w = v' 0 w. Then v and w are dependent. (3) Note that two different valuation rings of K of rank 1 are always independent.

(4.4) Theorem: [Approximation theorem for independent valuations] Let Vl, be pairwise independent valuations of K. Let Xl, •.. , Xn E K and, for i E {I, ... , n}, let 'Yi E r v•• Then there exists X E K with Vi(X - Xi) ~ 'Yi for every i E {I, ... , n}. ••• ,Vn

Proof: If Vi is a trivial valuation for some i E {I, ... , n}, then we have 'Yi = 0, and Vi (x - Xi) ~ 'Yi is true for every X E K. Therefore we may assume that none of the valuations Vl, ... , Vn is trivial. In this case we have Av. ¢. Av; for i, j E {I, ... , n}, i :f. j. Just as in the proof of (2.20) we may assume that 'Yi > 0 and that Xi E R := AVt n .. ·nAvn for i E {I, ... ,n}. Let i E {I, ... ,n}. We set ai:= {z E Av. I Vi(Z) ~ 'Yi} and qi := R n ai. Now rad(ai) is a prime ideal of Av. [let x, Y E Av. with xy E rad(ai), hence with (xy)m E ~ for some mEN, and if we assume, as we may, that Vi(X) ~ Vi(Y), then we have Vi (x2m) ~ Vi(Xmy m ), hence x2m E ~, hence X E rad(~)], and therefore Pi := rad(qi) = rad(ai) n R is a prime ideal of R. It is enough to show that the ideals ql, ... ,qn are pairwise comaximalj then the result follows from the Chinese remainder theorem. Let i E {I, ... , n} j since ml, ... , mn are all the maximal ideals of R [cf. (2.19)], and since qi C mi, it is enough to show that qi rt. mj for every j E {I, ... , n}, j :f. i. Thus, suppose that qi C mj for some j E {I, ... , n} with j :f:. i. Now Pi is a prime ideal of R which is contained in ~ and mj, hence Av. = Rm. and Av; = Rm; are contained in Rp.. Since 'Yi > 0, we have ai :f. {O}, and as qiAv. = ~, we have qi :f:. {O}, hence Pi :f:. {O}, and therefore Rp. :f:. K, contradicting our assumption that Vi and Vj are independent. (4.5) Corollary: Under the assumptions of (4.4) we have: For i E {I, ... , n} let 'Yi E r v•• Then there exists X E K with Vi(X) = 'Yi for every i E {I, ... , n}. Proof: We may assume that Av. :f. K for i E {I, ... , n}. Let i E {I, ... , n}, and choose Xi E K with Vi(Xi) = 'Yi and 8i E rv. with 'Yi < 8i . By (4.4), applied to

5 Extensions of Valuations

29

the elements Xi and the elements c5i , there exists X E K with Vi(X - Xi) > Vi(Xi) for i E {I, ... , n}. From X = Xi + (X - Xi) we now get Vi(X) = 'Yi for every i E {1, ... ,n} [ef. (2.10)].

5 5.1

Extensions of Valuations Existence of Extensions

(5.1) NOTATION: Let LjK be an extension of fields. and let w be a valuation of L. Then the restriction v := wlK of w to K is a valuation of K. Let W be the valuation ring of w, and let r w := w(LX) be the value group of w. Then V := W n K is the valuation ring of v, and the value group r 11 := v(KX) of v is a subgroup of r w. In this situation we often say: The valuation w is an extension to L of the valuation v. By abuse of language, we also say that W is an extension of V to L. (5.2) REMARK: Let Lj K be an extension of fields, let W be a valuation ring of L, and let V be a valuation ring of K. Then W dominates V iff W n K = V. In fact, if W dominates V, then V C W n K, and, for any X E K \ V, X-I lies in the maximal ideal of V, hence in the maximal ideal of W, and therefore X ¢ W. This shows that W n K = V. Conversely, if V = W n K, then we have V C W, and if X E K lies in the maximal ideal of V, then we have X-I ¢ V, hence X-I ¢ W, and therefore X lies in the maximal ideal of W. This shows that W dominates V. In particular, this shows that if V = W n K, then the residue field of v can be considered as a subfield of the residue field of W. (5.3) REMARK: Let Lj K be an extension of fields, let W be a valuation ring of L, and set V:= WnK. We set r:= {Va I a E KX} and ~:= {Wb I bE LX}, considered as totally ordered groups [ef. proof of (2.13)]. Now (5.2) easily yields: the map Va t-+ Wa : r ~ ~ is an injective homomorphism of ordered groups. Therefore the canonical valuation of L defined by W is an extension of the canonical valuation of K defined by V. (5.4) Proposition: Let Lj K be an extension of fields, and let v be a valuation of K. Then v admits at least one extension as a valuation of L. Proof: The valuation ring V of v is a quasilocal subring of L. By (3.5) there exists a valuation ring W of L which dominates V, hence V = W n K by (5.2). Let r ~ {Va I a E KX} be the inverse of the isomorphism in (3.19); then we have an injective homomorphism of ordered groups r ~ {Wa I a E LX}; the canonical valuation w of L defined by W is an extension of v to L.

30

5.2

I Valuation Theory

Reduced Ramification Index and Residue Degree

(5.5) NOTATION: Let L/ K be an extension of fields, let v be a valuation of K, let V be the valuation ring of v, and let w be a valuation of L with valuation ring W which is an extension of v. Let r w (resp. r v) be the value group of w (resp. of v), and let "'w (resp. "'v) be the residue field of w (resp. of v). Then e(W/V) = e(w/v):= (rw: rv),

the index of r v in

r w,

is called the reduced ramification index of w over v, and !(W/V) = !(w/v) := ["'w : "'v]

is called the residue degree of w over v. Note that r w/r v must not be a finite group, and that "'w must not be an algebraic extension of "'v, and if "'w is an algebraic extension of "'v, it must not be a finite extension of "'v. We shall show in (6.10) that e(w/v) and !(w/v) are finite if Lis a finite extension of K.

(5.6) REMARK: Let K C L c L' be a tower of field extensions, let w' be a valuation of L', set w := w'IL and v := wiK. Then we have e(w' Iv)

= e(w' /w)e(w/v),

!(w' Iv)

= !(w' /w)!(w/v),

and, in particular, e(w' Iv) (resp. !(w' Iv)) is finite iff e(w' /w) and e(w/v) (resp. !(w' /w) and !(w/v)) are finite. These assertions follow immediately from the multiplicative behavior of degrees in a tower of field extensions and of indices in a tower of groups.

5.3

Extension of Composite Valuations

(5.7) Proposition: Let L/ K be an extension of fields, let v be a valuation of K, and let w be a valuation of L which is an extension of v. Then: (1) Let v = v' 0 v be a decomposition of v where v'is a valuation of K with Av C AV and v is a valuation of "'Vi with valuation ring Av /Pv l . Then there exist an extension w' of v'to L and an extension W ofv to "'Wi such that w = w' 0 W. (2) Let w = w' 0 'ill be a decomposition of w where w' is a valuation of L with Aw C AWl and 'ill is a valuation of "'Wi. We set v' := w' IK; then we have Av C AV "'Wi is an extension field of "'Vi, and setting v := wl"'v" we have v = v' 0 v. (3) If, in the situation of (2), we set w(A~/) =: fj.1, v(A:,) =: r /, then fj.' (resp. r') is an isolated subgroup of r w (resp. of r v), and fj.' n r v = r'o I ,

I ,

Proof: (1) We may consider v'to be the canonical valuation of K defined by the valuation ring of v' and v to be the canonical valuation of "'Vi defined by the valuation ring Av/Pv" Let r' := v(A:/) be the isolated subgroup of r := r v corresponding to v' [cf. (3.26)], and choose an isolated subgroup fj.' of fj. := r w with fj.' n r = r' [cf. B(l.l1)]. Let w' be the canonical valuation of L defined by

6 Extending Valuations to Algebraic Overfields

31

the valuation ring w- I (~+ u~' u {oo}) corresponding to ~'; note that Aw c AW'. Since AVI = V-I (r + ur'u{ oo}) [cf. (3.23)], we have AWl nK = AVI, hence w' is an extension of v' [cf. (5.3)]. Let rp: AV I --t AV'/PV' = "'V', t/J: AWl --t AWI/PW I = "'w' be the canonical homomorphisms. Since AWl dominates AVI [cf. (5.2)], we have t/JIAv l = rp. Since Aw n Avl = Aw n (AWl n K) = Aw n K = A v , we see that the canonical valuation w of "'w' defined by rp(Aw) is an extension of v. (2) Again let rp: AV I --t AVI/PV I = "'V', t/J: AWl --t AW'/PW' = "'w' be the canonical homomorphisms. We have t/JIAwl = rp, Aw = t/J-I(Aw )' and since Awn "'v' = A v , we have Av = rp-I(Av), hence v = v' 0 v. (3) This follows immediately from the considerations above.

6 6.1

Extending Valuations to Algebraic Overfields Some General Results

(6.1) Proposition: Let K be a field, and let L be an algebraic extension of K. Let v be a valuation of K with valuation ring V, and let B be the integral closure of V in L. Then: (1) The map n f--t Bn is a bijective inclusion-reversing map from the set of maximal ideals of B to the set of valuation rings W of L with W n K = V, the inverse map being W f--t m(W) n B. (2) Every valuation of L extending v is equivalent to a valuation of L defined by a ring Bn where n is a maximal ideal of B. (3) B is the intersection of all the valuation rings of L which are extensions of V. Proof: (1) Let W be a valuation ring of L with W n K = V. Since W is integrally closed [cf. (3.2)], it is clear that B C W. We set n := m(W) n B. Since W dominates V [cf. (5.2)], we have n n V = m(W) n V = m(V), hence n is a maximal ideal of B [cf. [63], Cor. 4.17]. Let x E W be a non-zero element, and let aoTn + alT n- 1 + ... + an with ao = 1 be the minimal polynomial of x over K. We choose j E {O, ... , n} with v(aj) ~ v(ai) for i E {O, ... ,n}, and we set bi := ai/aj, i E {O, ... ,n}. The elements bo, .. . , bn lie in V c B n, and Bn is integrally closed in L by [63], Prop. 4.13. We have bj = 1; by the first part of the proof of (3.34) we get x E Bn or l/x E Bn. Suppose that x ~ Bn. Then l/x E Bn is not a unit of B n, hence l/x E nBn C m(W) which implies that x ~ W, contrary to our choice of x. Therefore we have x E B n, hence we have Bn = W. Conversely, let n be a maximal ideal of B. Then we have n n V = m(V). Let W be a valuation ring of L which dominates Bn [cf. (3.5)]. Then W dominates V, hence W n K = V [cf. (5.2)], and therefore Bn = W by the first part of the proof. (2) follows from (1) and (3) follows from (1) and B(2.5). (6.2) Corollary: IE V = K, then W = L is the only extension of V to L. (6.3) Corollary: (1) Every valuation ring of L which contains B also contains the valuation ring of an extension of v to L.

32

I Valuation Theory

(2) Let w, w' be non-equivalent extensions of v to L. Then we have Aw ct AWl and AWl ct Aw· (3) Let n be a maximal ideal of B, and set W := B n, an extension of V to L. We have Bin = Wlm(W). Proof: (1) Let W be such a valuation ring, and set p := B n m(W). Let n be a maximal ideal of B with pen. Then we have Bn C W. (2) We have Aw = B n, AWl = Bnl where n ¥ n' are maximal ideals of B. (3) This is clear. (6.4) Corollary: Let LI K be a finite extension of fields. Let v be a valuation of K, and let B be the integral closure of Av in L. Then B is quasisemilocal, and every extension W of v to L is equivalent to a valuation defined by a valuation ring of the form Bn where n is a maximal ideal of B. In particular, the non-empty set of pairwise non-equivalent extensions of v to L has at most [L : K ]sep elements. Proof: This follows immediately by applying B(7.3). (6.5) DEFINITION: Let LI K be a finite extension of fields, and let v be a valuation of K. A system of extensions {WI, . .. , w g } of v to L is called a complete set of extensions of v to L if the valuations WI, .•. ,Wg are pairwise non-equivalent, and if every extension of v to L is equivalent to a valuation Wi for some i E {I, ... , g}. (6.6) Corollary: Let L I K be a finite extension of fields, and let L' be a subfield of L containing K. Let v be a valuation of K. If a complete set of extensions of v to L is pairwise independent, then any complete set of extensions of v to L' is also pairwise independent. Proof: (1) Let B be the integral closure of Av in L. We show: A complete set of extensions of v to L is pairwise independent iff every non-zero prime ideal of B is contained in exactly one maximal ideal of B. In fact, assume that a complete set of extensions of v to L is pairwise independent. Let p be a non-zero prime ideal of B which is contained in two different maximal ideals nIl n2 of B. Then Bn! and Bn2 are contained in Bp ¥ L, hence the valuations of L defined by Bn! and Bn2 are dependent, contradicting our assumption. Now assume that every non-zero prime ideal of B is contained in only one maximal ideal of B. Let nl ¥ n2 be maximal ideals of B, and suppose that there exists a subring C ¥ L with Bni C C for i = 1,2. Let W ¥ L be a valuation ring of L with W :J C [cf. (3.5)]. Then we have {O} ¥ (m(W) n B) c nl n n2 [cf. (3.8)], contradicting our assumption. (2) Let B' be the integral closure of V in L'; then B' = BnL'. Let p' be a non-zero prime ideal of B', and suppose that there exist two different maximal ideals n~, n~ of B' with p' C n~ n n2. Using [63], Prop. 4.15, we get the existence of prime ideals p, nl and n2 of B with P C nl n n2, Pn B' = p' and ni n B' = n~ for i = 1,2. Since nl and n2 are maximal ideals of B [ef. [63], Cor. 4.17]' this contradicts the condition shown in (1) to be equivalent to the hypothesis of the corollary.

6 Extending Valuations to Algebraic Overfields

33

(6.7) REMARK: Let K be a field, and let L be a finite normal extension of K; let G := Gal(L/ K) be the group of K-automorphisms of L. Remember that Card (G) = [L: K]sep, the degree of separability of L/K [cf. B(7.3)]. Let v be a valuation of K, and let W be an extension of v to L. For every a E G the map x t-t w(a- I (x)) : LX ~ r w is a valuation of L with value group r wand valuation ring a(Aw); it shall be denoted by wO'. Clearly the residue fields of wand wO' are isomorphic; note that wO' also is an extension of v. We say that the valuations w and wO' are conjugate. Note that, by (6.3)(2), if Aw" =f. A w, then we have Aw" ct. Aw and Aw ct. Aw'" (6.8) Corollary: Let K be a field, and let L be a finite normal extension of K. Let v be a valuation of K, and let WI, W2 be extensions of v to L. Then W2 is equivalent to a valuation w' of L such that w' and WI are conjugate valuations. Proof: Let B be the integral closure of AlI in L, and set qi := B n Pw;> i E {1,2}. The ideals ql, q2 are maximal ideals of B, hence there exists a E Gal(L/K) with a(qt) = q2 [cf. [63], Prop. 13.10], and since Aw; = Bq; for i E {1,2} by (6.1), we have a(AW1) = A W2 ' hence W2 and wf are equivalent valuations.

6.2

The Formula ef ::; n

(6.9) Proposition: Let L/ K be a finite extension of fields, let v be a valuation of K, and let WI, ... ,Wh be extensions of v to L. If the valuations pairwise independent, then we have e(wtfv)f(wtfv)

+ ... + e(wh/v)f(Wh/V)

~

WI, ...

,Wh are

[L: K].

Proof: For i E {1, ... , h} we choose natural integers ei, Ii with ei ~ e(wi/v), fi ~ f(wi/v). (1) Let i E {1, ... , h}. We choose elements 'Yil, ... ,'Yie; E r W; which are pairwise incongruent modulo r 11, and elements Zil, ... ,zil; E Aw; whose images in "'w; are linearly independent over "'11' We set 'Yi := max( {'Yil, ... ,'Yie; }) E r W;' (2) Let i E {1, ... ,h}. By (4.5), for every s E {1, ... ,ei} there exists Xis E L with Wi(Xis) = 'Yis, Wj(Xis) > 'Yj for j E {1, ... , h}, j

=f. i,

and, by (4.4), for every t E {1, ... , Ii} there exists Yit E L with Wi(Yit - Zit)

> 0, Wj(Yit) > 0 for

j E {1, ... , h},j

=f. i.

(3) We show: The elements XisYit, i E {1, ... , h}, s E {1, ... , ed, t E {1, ... , fi}, are linearly independent over K. In fact, suppose that there exists a relation h

e;

I;

i=1

s=l

t=1

L L L aistXisYit = 0

34

I Valuation Theory

with elements aist E K which are not all 0. We may assume that all the elements aist lie in A v , and that aU1 1. (a) Let i E {I, ... ,h}. Since Yit == Zit (mod m(Awi)) for t E {I, ... , fd, the images of the elements Yil, ... ,Yif; in "-Wi are linearly independent over "-v. (b) The w1-value of any element fj:= b1yu + ... + bh Y1h where b1 , ... , bh E Av

=

are not all 0, lies in r v • In fact, we choose t' E {I, ... , fd with v(b t ~ v(b t ) for t E {I, ... , It}; then we can write fj = b1yu + ... + bhY1h = btt{C1Yll + ... + ChY1h) l )

with C1, ... ,ch E Av and Ct ' = 1. From (a) we get W1(C1Yll hence W1 (fj) = v(b t E r v' (c) We consider the elements

+ ... + ChY1/J = 0,

l )

If Xs ::fi 0, then we have W1(X s ) E 1'18 + rv by (b), hence the w1-values of those elements Xl , .•• 'X e1 which are different from are pairwise different, and therefore we have [cf. (2.10)]

°

=

Since aU1 1, we have by (a) w1(au1Yu W1 (xu) = 1'u, and therefore

+ ... + aUhY1h) = 0,

hence W1(Xt} =

(t) (d) We have, by the choice made in (2), h

W1

ei

Ii

(~ ~ ~ aist Xi8Yit

)

> 1'1·

(tt)

From (t) and (tt) we get

contradicting (*) in (3). Therefore we have ed1 + ... + ehfh ~ [L: K]. Since It, .. ·,!h where arbitrary natural integers with ei ~ e(wi/v), Ii < f(wi/v) for i E {l, ... ,h}, the inequality of the proposition follows.

el, ... , eh,

(6.10) Corollary: Let L/ K be a finite extension of fields, let v be a valuation of K, and let w be an extension ofv to L. Then we have e(w/v)f(w/v) ~ [L: K].

6 Extending Valuations to Algebraic Overfields

35

Proof: The proof of (6.9) works also if we consider only one extension w of v to L; in this case the approximation theorem and its corollary are not needed. (6.11) Corollary: Let L / K be an algebraic extension of fields, let v be a valuation of K, and let w be a valuation of L which is an extension ofv. Then: (1) The residue field K. w of w is an algebraic extension of the residue field K. v of v, and the quotient group r w /r v is a torsion group. (2) r w and r v have the same rational rank. (3) r w and r v have the same rank. (4) If, moreover, L is a finite extension of K, then r w is discrete iff r v is discrete.

Proof: (1) We write L = UiEI Li where (Li)iEI is the set of finite extensions of K contained in L. For every i E I we set ri := w(Lf), and we let K.i be the residue field of wlLi. Clearly we have r w = uri, K.w = U K.i. Since r i/r v is a torsion group and K.i is a finite extension of K. for every i E I [cf. (6.10)], the assertions follow immediately. (2) Tensoring the exact sequence of Z-modules [note that Q is a flat Z-module]

with Q and noting that r w/rv ®z Q = {O} by (1), proves the assertion. (3) r w/rv is a torsion group by (1), and the result follows from B(1.11)(2). (4) In this case the factor group r w/rv is finite by (6.10), and the result follows from B(1.27). (6.12) Corollary: Let L/ K be an algebraic extension of fields. In the situation of (5.7)(1), there exists only one decomposition w = w' 0 w. Moreover, we have e(w/v) = e(w' /v')e(w/v) , !(w/v) = !(w/v).

Proof: The first part of the assertion follows from B(l.l1). Clearly we have !(w/v) = !(w I v). We set .6. := r w, r := r v , and define .6.', r' as in (5.7). Note that e(w/v) = (.6. : r), and, by (3.26), e(w/v) = (.6.' : r') and e(w'/v') = (.6./.6.' : r/r'). The factor groups (.6./.6.')/(r/r') and (.6./r)/(.6.'/r') are isomorphic; now the assertion follows since (.6. : r) = (.6./r : .6.' /r') . (.6.' : r'). (6.13) NOTATION: Let L/K be an algebraic extension of fields, let v be a valuation of K, and let w be an extension of v to L. Let V (resp. W) be the valuation ring of v (resp. w). We set r(W/V) = r(w/v) := e(w/v)[ K. w

: K.V]iDS;

r(w/v) is called the ramification index of w over v [cf., e.g., [204], Ch. II, § 5, (3), for the notion of degree of inseparability of algebraic field extensions].

36

I Valuation Theory

(6.14) REMARK: Let K C L C L' be a tower of field extension with L' algebraic over K. Let w' be a valuation of L', set w := w'lL and v := w'IK. Then we have

r(w' jv) = r(w' jw)r(wjv) [by (5.6) and by the multiplicative behavior of the degrees of inseparability in a tower of field extensions].

6.3

The Formula

E edi ::; n

(6.15) A PARTICULAR DECOMPOSITION OF A VALUATION: (1) The results of this paragraph should be stated better in terms of valuation rings instead of in terms of valuations. This would make the statements a little bit clumsy. Therefore, when we speak in this paragraph of a valuation of a field K, we always mean the canonical valuation of K which is associated with the valuation ring of the valuation [cf. (2.13)]. (2) Let Lj K be a finite extension of fields. (a) For every non-trivial valuation v of K we denote by E(v) the set of all extensions of v to L [note that, according to our agreement in (1), two different extensions of v to L are not equivalent]. We have 1 ~ Card( E( v)) ~ [L : K] [cf. (5.4) and (6.4)]. (b) For valuations v, v' of K we set v' < v if Av ~ Av'; clearly this gives a total ordering on the set of valuations of K. For a valuation v of K with rank(v) > 1 we denote by I(v) the set of all non-trivial valuations v' of K with v' < v; note that I(v) is not empty. (c) Let v, v' be non-trivial valuations of K with v' V. We define a map 0, and assume that the theorem holds for all pairs (r', s') with 0:::; r':::; r, 0:::; s':::; sand r' +s' < r+s. We set K':= K(XI, ... ,Xr',YI, ... ,Ys') and v' := wlK'; note that now r v ' = rv EB (ZW(YI) + ... + ZW(Ys'))' and K,v' = K,v (Xl, ... , Xr' ). Replacing K by K' and v by v', it is therefore enough to prove the theorem for the two pairs (r, s) = (1,0) and (r, s) = (0,1).

i j :=Xil [ Herexy I

9 Extending Valuations to Non-Algebraic Overfields

55

(a) Let r = 1, 8 = 0: In this case we have L = K(x), and the image of x in the residue field of w is transcendental over the residue field of v. Therefore, by (6.11), the element x is transcendental over K. (b) Let r = 0 and 8 = 1: In this case we have L = K(y), and the image of w(y) in w(LX)/v(KX) is not a torsion element. Therefore, by (6.11), the element y is transcendental over K. In both cases the result follows immediately from (9.2). The case of a general field extension L / K is covered by the following theorem. (9.4) TheoreIn: Let K be a field, let v be a valuation of K, let r" be the value group and K-" the residue field of v. Let L be an overfield of K, let w be a valuation of L which extends v, let r w be the value group and K-w be the residue field of w. Then: (1) We have

H, moreover, L is a finitely generated extension of K, and if we have equality in

(*), then r w/r" is a finitely generated 'I.-module, and K-w is a finitely generated extension of K-" . (2) We have

If, moreover, L is a finitely generated extension of K, if f" is discrete of finite rank and if we have equality in (**), then f w is discrete of rank rank(f,,) + rat. rank(r w/r,,), and K-w is a finitely generated extension of K-". Proof: (1) Let r, 8 E No with r ~ tr.d",JK- w) and 8 ~ rat.rank(rw/f,,); by (9.3) we get r + 8 ~ tr.dK(L). Therefore (*) has been proved. Now we assume that L is a finitely generated extension of K-hence d:= tr. dK(L) is finite-and that we have equality in (*). Setting r := tr.d",JK- w) and 8 := rat.rank(rw/r,,), we have r + 8 = d. Then there exist elements Xl, ... , Xr in the valuation ring of w such that their images Xl, ... ,xr in K-w are algebraically independent over K-", and elements Yl, ... , Ys in LX such that the images of the elements W(Yl),' .. , w(Ys) in f w/f" are free over Z. We set L' := K(Xl"'" Xr, Ylo.'" Ys) and w' := wIL'. By (9.3) we have K-w ' = K-,,(Xl, ... ,Xr ), f Wi = f" €11 (ZW(Yl) + ... + ZW(Ys») , and the r + 8 = d elements Xl, ... , Xro Yl, ... , Ys are algebraically independent over K. Therefore {Xl, ... ,XroYl, ... ,Ys} is a transcendence basis of Lover K, hence L, being a finitely generated extension of K, is a finite extension of L'. This implies, by (6.10), that K-w is a finite extension of K-w' and that r w/rWi is a finite group. Therefore K-w is a finitely generated extension of K-". The 'I.-module r Wi /r" is finitely generated, and from the exact sequence

we see that f w /f" is a finitely generated 'I.-module.

56

I Valuation Theory

(2) By B(1.30)(2) we have rank(r w)

~

rank(r,,)

+ rat. rank(r wjr,,),

and by (1) above we have

(tt) whence (**) holds [this is clear if r w jr" has finite rational rank; in the other case tr.dK(L) is not finite, and therefore (**) holds trivially]. Now we assume that L is a finitely generated extension of K, hence that tr. dK(L) is finite, and that r" is discrete of rank h-hence that rank(r,,) = rat. rank(r,,) = h is finite [ef. B(1.25)]. By (tt) we get that rat.rank(rwjr,,) is finite. Furthermore, assume that we have equality in (**). Then we must have also equality in (t) and (tt), and this means that, in particular, we have equality in (*), hence, by (1), that K,w is a finitely generated extension of K,,,, and that r wjr" is a finitely generated Z-module. We set r := tr. d"JK,w) and s := rat. rank(r wjr ,,), Then there exists a transcendence basis {Xl,,,,, XT) YI, ... , Ys} of Lover K such that, setting L' := K(XI, ... ,Xr,Yb""Ys) and w' := wiL', we have w'(L'X) = rwI = r" EB (ZW(YI) + ... + ZW(Ys»), and {W(YI), .. "W(Ys)} is a Z-free set [ef. proof of (1)]. From rank(r w) = rank(r ,,) + rat. rank(r wjr,,) [ef. (t)] and rat. rank(r wjr ,,) = rat. rank(r w) - rat. rank(r,,) [cf. B(1.30)] we get rank(r w) = rat. rank(r w), and since L is a finite extension of L', we have rank(r w) = rank(r WI) and rat. rank(r w) = rat. rank(r WI) by (6.11). Moreover, we have rat. rank(r wjr ,,) = rat. rank(r w' jr,,) by B(1.30), since r w' jrw is a finite group by (6.10). Now r w' is a finitely generated free Z-module, and we have rat. rank(r WI) = rat. rank(r ,,) + rat. rank(r w' jr ,,) = rank(r ,,) + rat. rank(r wjr,,) = rank(r w) = rank(r w' ), hence rat. rank(r WI) = rank(r w' ), and therefore r w' is discrete of rank h + rat. rank(r wjr,,) [cf. B(1.39)], and the same is true for r w [cf. (6.11)].

10

Valuations of Algebraic Function Fields

(10.0) In this section k is an arbitrary field. (10.1) DEFINITION: A finitely generated extension field K of k is called an algebraic function field over k. We set r := tr. dk(K); then K is said to be an algebraic function field in r variables over k. For the theory of algebraic function fields of one variable we refer to the books of Chevalley [42] and Eichler [62]. (10.2) REMARK: In the following, when we consider a valuation v of an algebraic function field Kover k, we tacitly assume that k C A", i.e., that vlk is the trivial valuation; then k can be considered as a subfield of the residue field K,,, of v. The transcendence degree tr. dk(K,,,) is called the dimension dim(v) of v.

10 Valuations of Algebraic Function Fields

57

(10.3) Proposition: Let K = k(x) be the field of rational functions over k. Then: (1) Let IP' be a set of representatives for the irreducible polynomials of k[ x]. For every p E IP' the local ring Vp := k[ x ](p) is a discrete valuation ring. Let vp be the valuation of K defined by vp. The residue field ofvp is the field k[ x ]/(p), hence an extension of k of degree deg(p). Let IE KX, and let I = 'Y ITqElPqaq with 'Y E k X and a q E IE for every q E IP' be the factorization of I; then vp(J) = ap. If pi-p' are in IP', then vp and vp' are not equivalent. (2) We set Voo := k[ l/x ](1/x); then Voo is a discrete valuation ring. Let Voo be the valuation of K defined by V00. The residue field of Voo is k. Let I E K x, and write I = g/h where g, h E k[x]. Then voo(J) = deg(h) - deg(g). For any p E IP' the valuations vp and Voo are not equivalent. (3) Every non-trivial valuation v of the function field K is either equivalent to vp for some p E IP' or equivalent to Voo. Proof: k[ x] is a principal ideal domain, hence {Vp I p E IP'} is the set of all valuation rings of K containing k[x] [ef. (3.30)(5)], and Vp/Vpp = k[x]/(p). Since p generates the maximal ideal of Vp, it is clear that vp(J) = ap. We have vp(p) = 1 i- 0 = vp' (p), hence vp and vp' are not equivalent. (2) The ring Voo is a discrete valuation ring by (1), and its residue field is k. Set m := deg(g), n := deg(h), and write 9 = ao + alX + ... + amx m = xm(a m + am_lx- 1 + .. ·+aox-m), h = bo+b 1 + .. ·+bnx n = x-n(bn+bn_lX-l+ .. ·+box-n)j by (1) we have voo(g) = -m, voo(h) = -n, hence voo(J) = n - m. Let p E IP'. We have vp(p) = 1 i- - deg(p) = voo(p), hence vp and Voo are not equivalent. (3) If the ring Av of v contains k[ x], then we have Av = Vp for some p E IP' [ef. (3.30)(5)]. If the ring Av of v does not contain k[ x], then v(x) < 0, Av :) k[ l/x], and the center of v in k[ 1/ x] is the prime ideal generated by 1/ x. Just as before we get that v and Voo are equivalent.

(10.4) Corollary: Let K be a function field in one variable over k. There exist non-trivial valuations of K, every such valuation of K is discrete of rank 1, and its residue field is a finite extension of k. Proof: We chose x E K which is transcendental over k. Then K is a finite extension of k(x), and the result follows from (10.3) and (6.10).

(10.5) REMARK: In (10.4) we have classified the valuations of a field of algebraic functions in one variable. We shall classify valuations of a field of algebraic functions in two variables in section 5 of chapter VIII. Applying the results of (9.4) we get

(10.6) Theorem: Let K be an algebraic function field in r variables over k, and let v be a valuation of K. Then:

58

I Valuation Theory

(1) We have dim( v)

+ rat. rank( v)

~ Tj

if we have equality in (*), then "'v is an algebraic function field in T - rat. rank(r v) variables over k, and r v is a finitely generated free Z-module, minimally generated by T - dim( v) elements. (2) We have dim(v) + rank(r v) ~ Tj if we have equality in (**), then rat. rank( v) = rank( v), v is discrete, and algebraic function field in r - rank(v) variables over k.

"'v is an

(10.7) Proposition: Let K be an algebraic function field in r ~ 1 variables over k, and let v be a valuation of K with dim( v) = r -1. Then v is a discrete valuation of rank 1, and is an algebraic function field in r -1 variables over k. Moreover, the valuation ring Av of v is a localization with respect to a prime ideal of height 1 of an integrally closed k-subalgebra R of K of finite type with Q(R) = K.

"'v

Proof: The first assertions follow from (10.6). We prove the last assertion. We choose elements Xl, ... ,Xr-l E Av whose v-residues in "'v are algebraically independent over k. Then Xl, ... , Xr-l are algebraically independent over k by (9.3). Let Xr E K be such that {Xl, ... , x r } is a transcendence basis of Kover k. This property holds also for {Xl, ... , Xr-l, 1I Xr }, hence we may assume that Xr E Av. We set R' := k[ Xl, ... , Xr ), and we let R be the integral closure of R' in K which is a k-algebra of finite type [d. B(3.6)) contained in Av [cf. (3.4)] and with Q(R) = K. Note that K is a finite extension of Q(R'). Set p' := PvnR', P := PvnR, and v' = vlk(XI, ... , xr). The images in R'/p' of the elements Xl, ... , Xr-l are algebraically independent over k, and since tr. dk("'vl) = tr. dk(",v) [cf. (6.10)), we have tr. dk(Q(R' Ip')) = r - 1, hence p' is a prime ideal of R' of height 1 [d. [63), Cor. 13.4). Therefore p is a prime ideal of R of height 1 [cf. [63), Prop. 9.2). Now Rp is a discrete valuation ring of K [cf. B(10.5)] contained in Av, hence we have Rp = Av [cf. (2.15)).

(10.8) Corollary: Let K be an algebraic function field in r ~ 1 variables over k, let R be subring of K which has K as field of quotients and is a k-algebra of finite type, and let p be a prime ideal of R of height 1. The set of pairwise non-equivalent (r - I)-dimensional valuations of K which have center p in R is finite and not empty. For every such valuation, its residue field is an algebraic function field in r - 1 variables over k, and it is a finite extension of the quotient field of Rip. Proof: Let R = k[tl, ... ,tn ). We have dim(Rlp) = tr.dk(Q(Rlp)) = r -1 by [63], Cor. 13.4j we label the elements tl' ... ' tn in such a way that the images of tl, ... , tr-l in Rip are algebraically independent over k. Then the elements tl, ... , tr-l are algebraically independent over k [cf. (9.3)), and we may assume that {tl, ... , t r } is a transcendence basis of Kover k. Set Ro := k[ tl, ... ,tr ],

11 Valuations Dominating a Local Domain

59

a polynomial ring, and Po := P n Ro. We have r - 1 = tr. dk(Q(Rlp)) ~ tr. dk (Q(Ro/po)) ~ r - 1 [since the images of tl, ... ,tr-l are algebraically independent over k], hence dim(Ro/po) = r - 1 and therefore ht(po) = 1 [cf. [63], Cor. 13.4]. Now Ro is integrally closed in k(tl, ... ,tr); therefore V:= (Ro)Po is a discrete valuation ring of k(t1, ... ,tr ) [ef. B(IO.5)]. Any (r -I)-dimensional valuation of K is discrete of rank 1 [ef. (10.7)], and if it has center p in R, it is an extension of V. Now K is a finite extension of k(tl"'" tr), hence the set of such valuations is finite and not empty [ef. (7.4)]. Furthermore, the residue field of such a valuation of K is an algebraic function field in r - 1 variables over k [cf. (10.6)], and since tr. dk(Q(Rlp)) = r - 1, the last assertion also has been proved. (10.9) Proposition: Let K be an algebraic function field in r ~ 1 variables over k, and let R be a subring of K which has K as field of quotients and is a k-algebra of finite type. Then: (1) lim is a maximal ideal of R, then there exists a zero-dimensional valuation v of K with R c Av and with m = R n Pv. (2) Let v be a zero-dimensional valuation of K with R c Av. Then m := Pv n R is a maximal ideal of R.

Proof: (1) The assertion in case r = 1 follows from (10.8) [since dim(R) = 1 by [63], Th. A on p. 286]. Let r > 1, and assume that the assertion is true for algebraic function fields in r - 1 variables over k. Let p C m be a prime ideal of R of height 1, and let w be an (r - I)-dimensional valuation of K having center p in R [cf. (10.8)]; then is an algebraic function field in r - 1 variables over k, and it is a finite extension of the field of quotients of Rip. There exists a subring S of containing Rip which is a finitely generated Rip-module and has "'w as field of quotients [obtained by adjoining to Rip a basis of "'w over Q(Rlp) consisting of elements which are integral over RIp]. Let n be a maximal ideal of S lying over m/p [cf. [63], Prop. 4.15 and Cor. 4.17]. By induction, there exists a zero-dimensional valuation v of having center n in S. Set v := w 0 v; then v is zero-dimensional and has center min R [cf. (3.12) and (3.26)]. (2) We have dim(Rlm) = tr. dk(Q(Rlm)) and tr. dk(",v) = 0, hence dim (Rim) = 0, and therefore m is a maximal ideal of R [cf. [63], Cor. 9.1].

"'w

"'w

"'w

11

Valuations Dominating a Local Domain

(11.0) NOTATION: In this section R is a local domain with maximal ideal m, field of quotients K and residue field", := Rim. Let v be a valuation of K having center m in R, let V be the valuation ring of v, let r := v(KX) be the value-group of v, and let "'v be the residue field of V. We assume in this section as an overfield of "'; we that V f. K. Since V dominates R, we can consider define tr. d(v) := tr. dK(",v). The set ~ := v(R \ {O}) is a subsemigroup of r, and r is generated by ~ [since K is the field of quotients of R]. The element

"'v

I Valuation Theory

60

a := v(m) := min ( {v (x) I x E m}) is the smallest positive element in E [since m is finitely generated, v(m) is well-defined]. (11.1) Proposition: The semigroup E is a well-ordered set. Proof: We have to show that every non-empty subset S of E has a smallest element. We set S' := v- l (S)nRj let a be the ideal of R generated by S'. Let {Xl, ... ,xn } C S' be a system of generators of a, and set 7:= min({v(xl), ... ,v(xn )}). Then we have 7 ~ vex) for every x E a [since v is non-negative on R], and therefore 7 E S is the smallest element of S [since S C {v(x) I x E a}].

(11.2) REMARK: For every 7 E E we define ideals cry and at of R by setting cry := {x E R I v(x) ~ 7}, a; := {x E R I v(x) > 7}. Clearly cry and at are v-ideals [cf. B(6.11) for the definition of a v-ideal]. (Moreover, for every v-ideal a of R, a'! {O}, we have a = aV(Q)') The set {d EEl d > 7} is not empty, hence has a smallest element 7+ by (11.1), and therefore we have at = a'Y+' Note that at ~ cry since there exist elements x E R with v(x) = 7.

(11.3) THE ASSOCIATED GRADED RING: We consider the graded abelian group grv(R) :=

EB a'Y/a;. 'YEI:

Let 7, d E Ej then we have crya6 C cryH, atat C a~H' Let x E cry, Y E a6j we define

(x + a;)(y + at) := xy + a~Hj

it is easily seen that this gives a well-defined product of two homogeneous elements of grv(R), and we define a product of two arbitrary elements of grv(R) by using the distributive law. Therefore grv(R) with this multiplication is a r-graded ring of type E. (11.4) DEFINITION: The semigroup E is called archimedean if, for any 7 and dEE with, '! 0, there exists an integer n with n, > d.

(11.5) REMARK: Note that E is archimedean if r is archimedean, but, in general, the condition for E to be archimedean is weaker, as the following example shows. (11.6) EXAMPLE: Let R be a two-dimensional regular local ring with maximal ideal m = Rx+Ry, field of quotients K and residue field K.. Thengrm(R) = K.[x,y] [we use the notations introduced in VII(2.0)]. Let p E lPR be the homogeneous prime ideal in grm[x, y] which is generated by y, set r := Z EB Z, ordered lexicographically, and let vp: KX -+ r be the valuation of K as defined in VII(7.5). The value group of vp is r, and we have vp(x) = (1,0), vp(Y) = (1,1), hence E:= vp(R\ {O}) = {(m,n) E Z 2 1 m ~ n ~ O}. Clearly E is archimedean, but r is not archimedean [cf. B(1.19)].

61

11 Valuations Dominating a Local Domain (11.7) Proposition: The following statements are equivalent: (1) ~ is an archimedean semigroup. (2) For every isolated subgroup ~ of r we have ~ n ~ = {OJ. (3) Every non-zero v-ideal of R contains a power ofm.

Proof (1) => (2): Let ~ be an isolated subgroup of r and suppose that ~n~ "I {OJ. Let 'Y be the smallest element of the non-empty set (~n~) \ {OJ [cf. (11.1)]. Let 8 E ~; then there exists n E Z with n'Y > 8, hence 8 E ~ [since ~ is a segment] which implies that ~ C ~, hence that ~ = r [since r is generated by ~], in contradiction with ~ "I r. Therefore we have r n ~ = {OJ. (2) => (1): Let 'Y E ~ be a non-zero element. The subset ~ of r consisting of all elements 8 of r such that there exists an integer n (depending on 8) with n'Y > 8 clearly is a segment; since the sum of two positive elements of ~ lies in ~, ~ is a subgroup of r [cf. B(1.5)(3)]. This subgroup must be r, since otherwise it would be an isolated subgroup of r having non-zero intersection with ~. Therefore, for 8 E ~, there exists n E Z with n'Y > 8, hence ~ is archimedean. (1) ¢:} (3): Assume that ~ is archimedean. Let a be a non-zero v-ideal of R, set (3 := v(a), and choose n E N with na > (3. Then we have mn C a. Conversely, assume that every non-zero v-ideal of R contains a power of m. Let 0 "I 'Y, 8 E ~. We have a:::; 'Y. We choose n E N with mn ~ a.s; then we have n'Y ~ na > 8 [since a.s is a v-ideal by (11.3)]. (11.8) Proposition: We assume that ~ is archimedean. Let pEN. For every n E N with n :::; rat. rank(v) there exists mEN such that setting ~p := {(3 E ~ I (3 < pma}, we have

Proof: Let n E N with n :::; rat. rank( v). Since the image of ~ in r QSlz Q generates this Q-vector space, there exist elements 'Y1, ... ,'Yn E ~ which generate a free Zsubmodule r' of r of rank n. Since ~ is archimedean, there exists mEN with 'Yi < ma for every i E {I, ... , n}. We have mpm C apma . Furthermore, setting r~ := {l1'Y1

+ ... + In'Yn II = (h, ... , In)

we have r~ C ~p and Card(r~)

i(R/mpm)

~ iR(R/apma ) =

E ~,Ill :=

h + ... + In

= (p~n), and therefore we have [cf.

L (3EEp

i R(a{3/at)

:::;

p},

(11.3)]

~ Card(~p) ~ Card(r~) =

(P: n).

(11.9) Theorem: [So S. Abhyankar] Let R be a local domain, and let v be a valuation of the field of quotients of R which dominates R. We set d := rat. rank( v).

Then:

62

I Valuation Theory

(1) We have the inequality rat. rank(v)

+ tr. d(v)

~

dim(R).

(2) If in (*) we have equality, then the group r is isomorphic to Zd, and K. v is a finitely generated extension of K.. (3) If rank(v) + tr. d(v) = dim(R), then we have rank(v) = rat. rank(v), equality holds in (*), and r is discrete of rank d. Before proving Abhyankar's theorem, we need three auxiliary results.

(11.10) Lemma: Let A be a noetherian integral domain, and let x be an element of the field of quotients of A. Then we have dim(A[x])

~

dim(A).

Proof: We write x = alb with a, b E A, b =I- O. The surjective A-algebra homomorphism A[ T] -+ A[ x] mapping T to x has non-zero kernel P since bT - a is mapped to 0, and we have A[x] ~ A[T]/p. Now (*) is trivially true if the Krull dimension of A is not finite. In the other case we have

= dim(A[T]/p) ~ dim(A[T]) -1 = dim (A) [since p =I- (0) and dim(A[T]) = dim (A) + 1, cf. [63], Cor. 10.13]. dim(A[x])

(11.11) Lemma: Let A be a local domain with maximal ideal m, let x be an element of the field of quotients of A, let q be a prime ideal of A[ x] with qnA = m, and set B := A[x]q, n:= qB. The local ring B dominates Ai if the image x E Bin of x is transcendental over Aim, then we have dim (B)

~

dim(A[x])-1.

Proof: It is enough to show that q is not a maximal ideal of A[ x]. If q were a maximal ideal of A[ x], then A[ x ]/q = K.[x] would be a field, hence x would be algebraic over K., in contradiction with our choice of x.

(11.12) Lemma: Let R and V be as in (11.0). Then tr. d(v) =: h is finite, and if h ~ 1, then there exists a local domain Rh which dominates R and is dominated by V such that dim(R) ~ dim(Rh) + h, that K. v is algebraic over the residue field K.h of Rh, and that K.h is a purely transcendental extension of K. with tr. d,,(K.h) = h. Proof: There is nothing to show if tr. d(v) = O. Now let K.v be transcendental over K., and choose an element x E V whose image in K. v is transcendental over K.; we have R[x] C V. Let q be the center of V in R[x], and set Rl := R[x]q, ml := qR1 • Then we have by (11.10) and (11.11) dim(R1 )

~

dim(R[x]) -1

~

dim(R)-1.

11 Valuations Dominating a Local Domain

63

Note that Rl dominates R, that V dominates R 1 , and setting ~l := Rdmb that we have ~l = ~(x), hence that tr.dl«~d = 1. If ~v is transcendental over ~l, we can repeat this construction. Thus, we get a finite sequence R =: Ro C Rl C ... C Rh of local domains with strictly decreasing dimensions which are dominated by V [with h 2:: 1], and denoting by ~i the residue field of Ri, i E {I, ... , h}, we have ~ =: ~o C ~l C ... C ~h C ~v, ~i/ ~i-l is a purely transcendental extension of transcendence degree 1 for i E {I, ... , h}, hence ~h/ ~ is a purely transcendental extension of transcendence degree h, and ~v is algebraic over ~h. Clearly we have dim(Rh) :S dim(R) - h. Proof of Abhyankar's theorem: (1) (a) We show: If 1; is archimedean and tr. d(v) = 0, then rat. rank(v) :S dim(R). It is enough to show the following: For every n E No with n :S rat. rank(v) we have n :S dim(R). Thus, let n E No with n:S rat.rank(v). By (11.8) there exists mEN such that fR(R/m pm )

2::

(p: n)

for every pEN.

Now fR(R/m pm ) is for all large p a polynomial in p with rational coefficients of degree dim(R) [cf. [63], Th. 12.4], and therefore we have dim(R) 2:: n. (b) We show that rat. rank(v) + tr. d(v) :S dim(R). We prove this by induction on dim(R). Let us consider the case dim(R) = 1. Then is archimedean and tr. d(v) = O. In fact, suppose that 1; is not archimedean. Then r has an isolated subgroup ~ with ~ n 1; =1= {O} [cf. (11.7)]. Let q := {x E V I vex) fj. ~} be the prime ideal of V corresponding to ~ [cf. (3.23)(2)]. Then p := q n R is different from {O} [since q =1= {O}]. There exists x E R with vex) E 1; n ~, vex) =1= 0, hence x E m and x fj. p, which means that p =1= m, contradicting dim(R) = 1. Moreover, using (11.12), we have tr. d(v) = O. From (a) we see that (*) holds for R. Now let n > 1, and assume that (*) holds for all pairs (R', v') where R' is a local domain, v' is a non-trivial valuation of the field of quotients of R' dominating R', and dim(R') < n. Let dim(R) = n. If h:= tr.d(v) 2:: 1, then we can apply our induction assumption to (Rh, v) [cf. (11.12)], hence rat. rank(v) :S dim(Rh) :S dim(R) - h, and therefore we get rat. rank(v) + tr. d(v) :S dim(R). If h = 0 and 1; is archimedean, then the result follows from (a). If h = 0 and 1; is not archimedean, then we can choose an isolated subgroup ~ of r with ~ n 1; =1= {O}; we define q and p as above, i.e., q is the prime ideal of V corresponding to ~ and p := q n R. Then we have {O} ~ P ~ m. In particular, the local rings S := Rp and R := R/p have dimension strictly less than n. We set W := Vq , V := V/q. Now W is a valuation ring of K with maximal ideal q [cf. (3.8)] and v(WX) = A [cf. (3.23)], and V is a valuation ring of the residue field K = W/q of W [cf. (3.26)]. The valuation v is composite 1;

64

I Valuation Theory

with valuations wand v where w is a valuation of K defined by the valuation ring W, and v is a valuation of K defined by the valuation ring V [cf. (3.27)]. We may assume that w is a valuation with value-group r /~, and that v is a valuation with value-group ~ [cf. (3.26)]. Note that W dominates S. By induction, we have rat. ranker /~) + tr. dew)

~

(i)

dimeS)

where tr. dew) is the transcendence degree of the residue field K of W over the residue field of S. Let K be the field of quotients of R; note that K is a subfield of K. Let v be the restriction of v to K; v dominates R. Note that K also is the residue field of S, hence that tr. dew) = tr. dj«(K). The value group Li of v is a subgroup of ~, and the residue field of R is induction, we have rat. rank(Li) + tr. d(V) ~ dim(R) where tr. d(V) is the transcendence degree of the residue field of v over more, by (9.4) we have

rat.rank(~/Li) +tr.d",;;(l\;v) ~ tr.dj«(K).

1\;.

1\;;

by (ii)

Further(iii)

Adding inequalities (ii) and (iii) we get, using B(1.30)(1), rat.rank(~)

+tr.d(v)

~

tr.dj«(K) +dim(R),

and adding (i) to this inequality, using (1.30)(1) and that tr. dj«(K) = tr. d(w), we get rat. rank(r) + tr. d(v) ~ dim(S) + dim(R) ~ dim(R). Therefore we have proved the inequality (*) in Abhyankar's theorem. (2) We now assume that rat. rank(v) + tr. d(v)

= dim(R).

(iv)

(a) We show: If ~ is archimedean and rat. rank(v) = dim(R) =: n, then r is isomorphic to zn and I\;v is a finite extension of 1\;. We choose elements 'Y1, ... ,'Yn E ~ which generate a free Z-submodule rl of r of rank n, and we define m and for pEN the sets ~p and r~ as in the proof of (11.8). There exists q E N with

[since fR(R/m 2pm ) is for all large enough p a polynomial in p with rational coefficients of degree n = dim(R), cf. [63], Th. 12.4]. (a) We show that r ~ zn [as a group].

11 Valuations Dominating a Local Domain

65

It is enough to show that r /r' is a finite group, since then r is a finitely generated Z-module, hence we get r ~ by (3.16). Since r /r' is a torsion group, for every (3 E 1: there exists h{3 E N with h{3(3 E r', hence -(3 == (h{3 - 1)(3 (mod r'); since r is generated by 1:, it follows that for every 'Y E r there exists (3 E 1: with 'Y == (3 (mod r'). Therefore there exists a system of representatives for r /r' consisting of elements of 1:, hence we can write

zn

r = U((3i + r')

with (3i E 1: for every i E I

iEI

[disjoint union). Let J c I be a finite subset. Then there exists p* E N with (3j < p*ma for every j E J. Let j E J; for every p ~ p* we have (3j + r~ c 1: 2p , and therefore [cf. (11.8)) Card(J)

(P: n) ~

Card(1: 2p )

~ iR(R/m2pm) ~ q

(P: n)

for all p

~ p*,

hence we have Card(J) ~ q. Therefore I is a finite set. ((3) We show that "-v is a finite extension of "-. First, we show that In fact, suppose that there exists 'Y E 1: with iR(cry/a~) ~ q + 1. We choose elements Yl, ... ,Yn E R with V(Yi) = 'Yi for i E {1, ... ,n}. Let l:= (h, ... ,ln) E N() , and set (3 := 'Y + h V(Yl) + ... + lnv(Yn). Then a{3 = yl a-p at = yla~, and the R-modules a{3/at and cry/a~ are isomorphic. Therefore we have i R(a{3/at) ~ q+l for every (3 E 'Y + r~. We choose p* E N with 'Y < p*ma. For every p ~ p* we have 'Y + r~ c 1: 2p , hence

hence q ~ q + 1 which is absurd. Therefore (*) holds. Second, we show that "-v is a finite extension of "-. Let t ~ ["-v : "-) be a positive integer, and choose elements Xl, ..• , Xt in V such that their images Xl, ... , Xt in "-v are linearly independent over "-. For i E {I, ... , t} let us write Xi = ai/b with ai and b E R, b =I O. We set (3 := v(b). The elements al = xlb, ... , at = Xtb lie in a{3 [since v(xd = ... = v(Xt) = 0). Let al, ... ,cit be the images of these elements in the ,,--vector space a{3/at [note a{3/at is annihilated by m). These elements are linearly independent over "-. In fact, suppose that there exist elements Ul, ... ,Ut E R with v(ulal + .. ·+Utat) > (3; then we have V(UlXl + .. ·+utxd > 0, hence UlXl + ... + UxXt = 0, and therefore Ul = ... = Ut = 0 [where Ui is the

I Valuation Theory

66

image of Ui in f'i" i E {I, ... , t}]. This means that t ~ i R (a[3/at), hence that t ~ q by what we have shown above, and therefore we find that [f'i,v : f'i,] ~ q. (b) We prove the assertions in (2) by induction on dim(R). If dim(R) = 1, then ~ is archimedean and tr. d(v) = 0 [ef. the claim in (b) of part (1)], hence the assertions in (2) follow from (a). Now let n > 1, and assume that the assertions in (2) hold for all pairs (R', v') where R' is a local domain, v' is a valuation of the field of quotients of R' dominating R', 1 ~ dim(R') < n and rat. rank(v') + tr. d(v') = dim(R'). Let dim(R) = n; by assumption, we have rat. rank(v) + tr. d(v) = dim(R), i.e., (iv) holds for (R,v). First, we consider the case h:= tr.d(v) ~ 1. Then we can apply (11.12); for the ring Rh we find, using (*) in (1), that dim(Rh)

rat. rank(v) = rat. rank(v) =dim(R) - h ~ dim(Rh), ~

+ tr. d,,(f'i,v)

- h

hence dim(Rh) = rat. rank(v) , and therefore tr.d"h(f'i,) = 0 by (*) in (1), applied to R h. Therefore we have dim(Rh) = rat.rank(v) + tr.d(v). Now we can apply our induction assumption to (Rh' v) since dim(Rh) < dim(R) = n: we get that f'i,v is a finite extension of f'i,h [since it is a finitely generated extension by induction], hence that f'i,v is a finitely generated extension of f'i" and that r ~ Zd. Second, we consider the case h = o. If ~ is archimedean, then the result follows from (a). If ~ is not archimedean, then we use the notations in part (b) of the proof of (1). If one of the inequalities (i), (ii) and (iii) would be a strict inequality, then, by the proof of (*) in part (b) of (1), we would get strict inequality in (*). Therefore (i), (ii) and (iii) are equalities. By induction, using equality in (i) and (ii), we get that r / ~ and E are finitely generated Z-modules, and that K / K and f'i,ij / f'i, are finitely generated extensions. Furthermore, by equality in (iii), we get from (9.4)(1), using the fact that K / K is a finitely generated extension, that the extension f'i,-v/ f'i,ij is finitely generated and that ~/ E is a finitely generated Z-module. Since ~/ E and E are finitely generated Z-modules, ~ is also a finitely generated Z-module, and since r / ~ and ~ are finitely generated Z-modules, r is a finitely generated Z-module, and therefore also a free Z-module of finite rank d [ef. (3.16)]. This ends the proof of the assertions in (2) of Abhyankar's theorem. (3) Now the assumption is rank(v)

+ tr. d(v)

= dim(R).

From rank(v) ~ rat.rank(v) [ef. B(1.30)(2)] we get rank(v) = rat.rank(v) by (1), and rat.rank(v) + tr.d(v) = dim(R), hence rand Zd are isomorphic by (2) [as groups]. From B(1.39) it follows that r is discrete of rank d.

Chapter II

One-Dimensional Semilocal Cohen-Macaulay Rings INTRODUCTION: The local ring of a point on a curve is a one-dimensional local Cohen-Macaulay ring; in this chapter we study this class of rings. After proving some results on transversal elements in section 1, our main interest in section 2 is the integral closure of a one-dimensional local Cohen-Macaulay ring; we use Manis valuations in describing the integral closure. In section 3 we give necessary and sufficient conditions in order to ensure that the completion of a one-dimensional local Cohen-Macaulay ring which is a domain (resp. has no nilpotent elements) again is a domain (resp. has no nilpotent elements). Here the reader is supposed to be acquainted with the notion of the completion of a local ring and its properties. In section 4 we introduce in the context of this chapter that technique which shall be used over and over again: Blowing up. After defining the notion of the blow-up of a one-dimensional semilocal Cohen-Macaulay ring, we show that by a sequence of blow-ups we can reach the integral closure of such a ring, and prove some interesting results on stable ideals. The notion of infinitely near rings, which shall playa decisive role in section 1 of chapter VIII, is introduced in section 5, and some simple properties are proved.

1 1.1

Transversal Elements Adie topologies

(1.1) In this subsection we collect for easy reference some results on adic topologies. (1.2) ADIC TOPOLOGIES: Let A be a ring, let a be an ideal of A, and let M be an A-module. 67

68

II One-Dimensional Semilocal Cohen-Macaulay Rings

(1) The a-adic topology on A is the topology such that, for every a E A, the family (a + an )nENo is a basis for the neighborhoods of a. With this topology, A becomes a topological ring. Note that A is a Hausdorff space in the a-adic topology iff n~=o an = {O}. (2) The aM -adic topology on M is the topology such that, for every x EM, the family (x + an M)nENo is a basis for the neighborhoods of x. With this topology,

M becomes a topological module over the topological ring A. Note that M is a Hausdorff space in the aM -adic topology iff n~=o an M = {O}. A submodule N of M is open iff there exists n E N with an MeN.

(1.3) Lemma: Let A be a ring, a an ideal of A and M an A-module, endowed with the aM -adic topology. Then: (1) For every subset X eMits closure X is equal to n~=o(X + an M). (2) Every open submodule of M is closed. Proof: (1) Let x E X and n E No; there exists an element Xn E X such that Xn E x + an M, hence we have x E Xn + an Me X + an M, hence x E n~=o(X + an M). Conversely, if x E n~=o (X + an M), then there exists, for every n E No, an element Xn of X such that x E Xn + an M, whence Xn E x + an M, and therefore x E X. (2) If N is an open submodule of M, then we have an MeN for some n E N.

(1.4) Let A be a noetherian ring, a an ideal of A which is contained in the Jacobson radical of A, M a finitely generated A-module and N a submodule of M. Then n~=o(N +anM) = N since we have n~=o an (MIN) = {O} by Krull's intersection theorem [cf. [63], Cor. 5.4], and therefore N is closed in the aM-adic topology of M [cf. (1.3)]. (1.5) NOTATION: In this chapter we are mainly interested in one-dimensional noetherian Cohen-Macaulay rings (CM-rings). The general definition [cf., e.g., [63], definition before Prop. 18.8] says in this case: A one-dimensional noetherian ring R is a CM-ring iff every maximal ideal of R contains a regular element of R. (1.6) OPEN IDEALS: Let R be a semilocal ring of dimension d, and let t be the Jacobson radical of R. R always will be endowed with the t-adic topology which is called the natural topology of R. Thus, an ideal a of R is open iff it contains a power of t. Note that, for every n E No, Rltn is an artinian ring [cf. [63], Cor. 9.1]. (1) An ideal a of R is open iff Ria is an R-module of finite length. In fact, if a is an open ideal of R, then Ria is an artinian ring [because Ria is a homomorphic image of Rltn for some n EN]. Conversely, assume that a is an ideal of R such that Ria is an artinian ring. Then t n + a = t n +1 + a for all sufficiently large integers n; on the other hand, a is a closed set of R since R is semilocal [cf. (1.4)], and is therefore the intersection of the ideals t m + a, mEN [cf. (1.3)]. This implies that a contains a power of t.

1 Transversal Elements

69

(2) Now assume that R is a one-dimensional semilocal CM-ring. An ideal a of R is open iff it contains a regular element. Indeed, if a is open, then a contains a power of t, hence contains a regular element of R [since R is a CM-ring]. Conversely, let rEa be a regular element. It is enough to consider the case where r is a non-unit. Let Rr = ql n· .. n qk be an irredundant primary decomposition of Rr. Then nl := rad(qd, ... , nk := rad(qk) are maximal ideals of R, and Rr contains a power of rad(Rr) = nl n ... n nk = nl ... nk [cf. B(10.1) and [63], Ex. 1.13], hence a fortiori a power of t. Since Rr c a, we have proved the assertion.

1.2

The Hilbert Polynomial

(1. 7) NOTATION: Let S := ffin>o Sn be a graded ring; assume that So is an artinian ring, and that S as So-algebra is generated by finitely many elements in SI. Let M = ffinEZ Mn be a finitely generated graded S-module. Then, for every nEZ, Mn is a finitely generated So-module [cf. B(4.32)], hence has finite length. By [63], Prop. 12.2, there exists a uniquely determined polynomial P(M) E Q[ X] such that fSo(Mn)

= P(M)(n)

for all sufficiently large integers n;

P(M) is called the Hilbert polynomial of M.

(1.8) HILBERT POLYNOMIAL AND MULTIPLICITY: Let R be a semilocal ring of dimension d, let t be the Jacobson radical of R, and let {mb ... , mh} be the set of maximal ideals of R. For every i E {I, ... , h} set Ri := R mi ; then we have dim(Ri) ~ d with equality for at least one i E {I, ... , h}. Let a be an open ideal of R. For every n E No the R-module anjan+1 has finite length [cf. (1.6)(1)]. We set H(a, R)(n) := fR(Rja n ) =

n-l

L fR(a" ja"+!)

,,=0

for every n E No·

The function H(a, R): No -+ No is called the Hilbert function of a. (1) Let a be a proper open ideal of R. Then there exists a non-zero polynomial P(a, R) E Q[ X] of degree max( {ht(mi) liE {I, ... , h}, a C ffii}) such that H(a, R)(n)

= P(a, R)(n)

for all sufficiently large integers n.

Proof: If R is local, this is shown in [63], Th. 12.4. Now assume that R is semilocal. Set ai := aRi for every i E {I, ... , h}. By B(10.9) we have f R(an ja n+!)

h

=L i=1

f Ri (aU af+! )

for every n E No,

70

II One-Dimensional Semilocal Cohen-Macaulay Rings

and, for i E {I, ... , h}, ai is m;Ri-primary if a C mi and ai :::: R; otherwise, hence h

H(a,R) = LH(a;,R;), ;=1

and the assertion follows from the corresponding assertion in the local case. The polynomial pea, R) is called the Hilbert polynomial of a. (2) Let a be an open ideal of R that is contained in a maximal ideal of R of height d. By [63], Ex. 1.21, the polynomial pea, R) E Q( X] has the form

Xd

P(a,R)(X) = e(a,R)T!

+...

[terms of lower degree]

where e(a, R) is a positive integer which is called the multiplicity of a in R; in particular, e(r,R) =: e(R) is called the multiplicity of R. Assume that dim(R) = dim(Rm.} for every i E {I, ... ,h}. Then we have h

e(R) =

L e(R;). i=1

(3) Assume that R is a one-dimensional CM-ring; then Ri is a one-dimensional local CM-ring for every i E {I, ... , h}. Let a be a proper open ideal of R. Then aRi is an m;R;-primary ideal for at least one i E {I, ... , h}, hence the Hilbert polynomial P( a, R) is a polynomial of degree 1, i.e., P(a,R)(X)

= p(a)X -

v(a),

pea) E N, v(a) E Z,

where we have defined p(a) := e( a, R) [to simplify notation and to cover also the case that a = R, cf. below]. It shall be shown later that v( a) E No [cf. (4.11) below]. If a R, then Rjan {O} for every n E No, and we define pea) veal O. For all sufficiently large integers n we have pea) = pea, R)(n + 1) - pea, R)(n) = lR{anjan+l), hence we have lR(anjan+l) lR(amjam+l) for all sufficiently large integers m, n.

=

=

=

=

=

1.3

Transversal Elements

(1.9) DEFINITION: Let R be a one-dimensional semilocal CM-ring, and let a be an open ideal of R. Let t E W; an element a E at such that aan an +t for all sufficiently large integers n is called a transversal element for a of order t. If t = 1, then a is called a-transversal; if R is local and m is its maximal ideal, then an m-transversal element is called a transversal element of R.

=

To show the existence of tranversal elements of some order, we need the following four results. We consider graded rings A = $n?O An, and we set A+ := $n>O An.

1 Transversal Elements

71

(1.10) Lemma: Every homogeneous ideal9J1 =I A of A which is maximal in the set of proper homogeneous ideals of A, has the form 9J1 = 9Jto + A+ where 9Jto is a maximal ideal of Ao; in particular, we have 9J1:J A+.

Proof: Let 9J1 =I A be a homogeneous ideal of A. We have 9J1 n Ao f:. Ao, hence 9J1 C (9J1 n Ao) + A+ f:. A. Now we assume that 9J1 is, in addition, a maximal element in the set of proper homogeneous ideals of A. Then 9J1 = (9J1 n Ao) + A+, and 9Jto := 9J1 n Ao is a maximal ideal of Ao. (1.11) Lemma: The following statements are equivalent: (1) A is an artinian ring. (2) For every n E No the Ao-module An has finite length, and An = {O} for all large enough integers n.

Proof (1) => (2): Since A is an artinian ring, the ring Ao ~ A/A+ is also artinian. Since A is an artinian ring, it is also a noetherian ring [cf. [63], Th. 2.14], hence we have A = Ao [ Xl, ••• , xp] with homogeneous elements Xl,"" xp of positive degree [cf. [63], Ex.1A]; set k i := deg(xi) for i E {l, ... ,p}. Let n E No. Now An is a finitely generated Ao-module, hence satisfies both chain conditions, and therefore has finite length. Since A is artinian, the set of prime ideals Ass(A) = {9J11, ... ,9J18 } of A consists of maximal ideals [ef. [63], Th. 2.14], and each of them is homogeneous [cf. [63], Ex. 3.5], hence 9J1i :J A+ for i E {I, ... , s} [ef. (1.10)]. Since the ideals 9J11, ... ,9J18 are pairwise comaximal, we have 9J1l ... 9J18 = 9J1l n ... n 9J1s =: !)1 :J A+ [cf. B(lO.l)], and !)1 is the nilradical of A, hence it is nilpotent since A is noetherian [cf. [63], Cor. 2.12 and Ex. 1.13]. We choose hEN with !)1h = {O}; then we have Ai = {a}. This implies that X~I ... =0 for all (h, ... , ip) ENg with i 1 + ... + i p ~ h. We set k := max{ {kl"'" kp}). Let n ~ kh. Then the Ao-module An is generated by the power products xiI ... x;r with ilkl + ... + ipkp = n ~ kh; for any such p-tuple (iI, ... , ip) we have i l + ... + ip ~ i 1 (kI/k) + ... + ip(kp/k) = n/k ~ h, hence XiI ... x;r = 0, and therefore we have An = {a}. (2) => (1): The assumptions imply that Ao is an artinian ring and that A is an Aomodule of finite length, hence that A is an artinian Ao-module [cf. [63], Th. 2.13], and therefore A is an artinian ring.

xl

(1.12) Lemma: Let A be noetherian with Ao artinian, and let M be a finitely generated graded A-module. Then we have Mn = {a} for all sufficiently large integers n iff every prime ideal in Ass( M) contains the ideal A+.

Proof: First, we remark that A as Ao-algebra is generated by finitely many homogeneous elements [ef. [63], Ex. 1.4], hence, for every nEZ, Mn is a finitely generated Ao-module [cf. B(4.32)] and therefore of finite length, and Mn = {O} for all sufficiently small integers n.

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II One-Dimensional Semilocal Cohen-Macaulay rungs

Assume that Mn = {O} for all sufficiently large integers n. Then the Ao-module M has finite length, hence satisfies both chain conditions, and therefore the Amodule M satisfies both chain conditions, hence has finite length. This implies that every prime ideal in Ass(M) is a maximal [ef. [63], Th. 2.14] and homogeneous [ef. [63], Ex. 3.5] ideal, hence it contains A+ [ef. (1.10)]. Conversely, assume that every prime ideal in Ass(M) contains A+. Let ':p E Ass(M); since A/A+ e! Ao is an artinian ring by assumption, ':p/A+ is a maximal ideal of A/A+ [cf. [63], Th. 2.14], hence ':p is a maximal ideal of A. This implies that M is an A-module of finite length [ef. [63], Cor. 2.17 and Th. 3.1], and setting 2l := AnnA (M), A/21 is a graded artinian ring [ef. [63], Cor. 2.17 and note that 2l is a homogeneous ideal of A, cf. [63], Ex. 3.5]. Now there exists no E N with (A/21)n = {O} for every n ~ no [cf. (1.11)]. The AI21-module M is finitely generated; let {Xl, ... , xp} be a homogeneous system of generators of M, and set k := max ({deg(xl), ... ,deg(xp)}). Then we have Mn = {O} for every n ~ no + k.

(1.13) Lemma: Let A be noetherian with Ao artinian, and let M be a finitely generated graded A-module. Let ':pI, ... , ':ph be those prime ideals in Ass(M) which do not contain A+. Let tEN, 0: E At, and 1, and assume that the assertion is true for all artinian rings with length smaller than I. Let A be an artinian ring with lA(A) = I, and let t be the Jacobson radical of A. If t = {O}, then A = k1 EB· •• EB kt is a direct sum of infinite fields [cf. B(10.1)]; using the case t = 1 it is easy to prove the assertion. Now we consider the case t:f. {O}j we set A := Alt and Ni := (Ni + tM)/tM for every i E {I, ... , h}. Note that, for every i E {I, ... , h}, we have Ni + tM :f. M by Nakayama's lemma [cf. [63], Cor. 4.8], hence N 1, ..• , N h are proper submodules of M. Since Aim is an infinite field for every maximal ideal m of A, ~d since l-:riA) < lA(A) [as t:f. {O} ], we can apply the induction assumption to A and the A-module M and its proper submodules N 1, ... ,Nh. Thus, there exists an element Y E M which does not lie in any of the A-modules N 1 , ••• , N h. Let y EM be a preimage of Yj then y does not lie in any of the A-modules Nt, ... ,Nh. (1.18) Proposition: Let R be a one-dimensional semilocal OM-ring, and assume that Rim is an infinite field for every maximal ideal m of R. Let a be an open ideal of R. For any tEN there exists an element in at which is transversal for a of order t. Proof: Let t be the Jacobson radical of R, and define S as in (1.15)j the maximal ideals of the artinian ring So = Rlt are the ideals mit where m is a maximal ideal of R. Therefore So satisfies the hypothesis in (1.17). Let '.P1, ... , '.Ph be those prime ideals in Ass(S) which do not contain S+ := EBn>O Sn. Since the ideal S+ is generated by the elements of Sl, the So-modules Sl n '.P1' ... ,Sl n '.Ph are proper submodules of Sl, and therefore [cf. (1.17)] there exists a E Sl which is not contained in any of the ideals '.Pt, ... , '.Ph. We choose a E a with a a + tao Then a is tranversal for a of order 1 by (1.15), hence at is transversal for a of order t by (1.16).

=

2 2.1

Integral Closure of One-Dimensional Semilocal Cohen-Macaulay Rings Invertible Modules

(2.1) REMARK: Let A be a ring and K its ring of quotients. (1) Let M, N be A-submodules of K. Remember that MN is the A-submodule of K which is generated by all products xy with x E M and yEN [cf. B(6.0)]. (2) Let M be an A-submodule of K. Then we define M- 1 := (A: M)K

= {x E K I xM c A}j

M- 1 is an A-submodule of K, and M(A : M)K C A. If M, N are A-submodules

2 Integral Closure

75

of K with N eM, then (A: M)K c (A: N)K. In particular, let a be an ideal of A. Then we have A = (A : A)K C (A : a)K' hence we have a C a(A: a)K cA. (3) An A-submodule M of K is called invertible if M N = A for some A-submodule N of K. An invertible A-submodule of K is a regular submodule.

(2.2) Proposition: Let A be a ring and K its ring of quotients. Then: (1) If M is an invertible A-submodule of K, then (A : M)K is the only Asubmodule N of K satisfying M N = A. (2) Every invertible A-submodule of K is finitely generated. (3) If, in particular, A is quasisemilocal, then every invertible A-submodule of K is cyclic. Proof: (1) Let N be an A-submodule of K satisfying M N = A. Clearly N is contained in (A : M)K, hence we have A = MN c M(A : M)K C A, and therefore A = M(A : M)K. Multiplying this equation by N gives N = (A: M)K. (2) Let M be an invertible A-submodule of K. Since M(A : M)K = A by (1), there exist elements Xl, ... ,Xn E M and Y1, ... ,Yn E (A : M) K such that Xl Y1 + ... + XnYn = 1. Let X E M; then X = (XY1)X1 + ... + (xYn)xn. Thus, we have M = AX1 + ... + AX n since XY1, ... ,XYn E A. (3) Let m1, ... , mh be the maximal ideals of A and let i E {I, ... , h}. Let M be an invertible A-submodule of K. Since A = M(A : M)K ct. mi, there exist Xi E M and Yi E (A: M)K such that XiYi ~ mi, and since mi does not contain the intersection of the other maximal ideals of A, there exists Zi E A \ tlli which lies in all the other maximal ideals of A. We set Y := ZlY1 + ... + ZhYh. Then we have Y E (A: M)K, hence yM is an ideal of A. We show that yM is not contained in a maximal ideal of A. Indeed, suppose that YM c mi for some i E {I, ... , h}; then we have yXi E mi. We consider yXi = (ZlY1 + ... + ZhYh)Xi E mi; by our choice we have Zi(XiYi) ~ mi and Zj(YjXi) E mi for every j E {I, ... ,h}, j "I i, hence we get Zi(XiYi) E tlli. This contradiction shows that yM = A. In particular, Y is a regular element of K and M = y- 1 A.

2.2

The Integral Closure

(2.3) REMARK: Let R be a one-dimensional semilocal CM-ring and let Q be its ring of quotients. In this remark we collect at first some properties of R. Let Ass(R) = {C1, ... ,cd; note that Ass(R) is also the set of minimal prime ideals of R. Then E := R \ (C1 U··· U CI) is the set Reg(R) of regular elements of R [cf. [63], Th. 3.1], Q = E- 1 R, and ill := E-1C1, ... ,ill:= E- 1cI are the prime ideals of Q. Each of them is a maximal ideal, and therefore Q is an artinian ring [cf. [63], Cor. 9.1]; moreover, the canonical homomorphism Q -t Q jill X ..• x Q jill is surjective and has kernel ill n ... n ill [cf. B(1O.1)(2)]. Let i E {I, ... , l}; then we have ili n R = Ci [cf. [63], Prop. 2.2], RjCi is a one-dimensional semilocal domain, and it is easy to check that Qjili is the field of quotients of RjCi. Now we specialize to the case of a one-dimensional local CM-ring R with maximal ideal m. Then we have:

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II One-Dimensional Semilocal Cohen-Macaulay Rings

(1) Let r E Reg(R) n m; then Ass(R/Rr) = {m}, hence rad(Rr) = m. Therefore there exists hEN such that mh C Rr [cf. (63), Ex. 1.13]. In particular, the Rr-adic topology and the m-adic topology on R coincide. (2) For every r E Reg(R) n m we have Q = Rr = R[l/r]. In fact, let x E Q; then we write x = a/s for some a E R, s E Reg(R). If s i m, then we have x E R. If s E m, then we may choose n E N such that rn E mn C Rs, hence rn = ts for some t E Reg(R), and we have x = ta/ts = ta/r n ERr.

(2.4) COMPLETION: Let R be a one-dimensional local CM-ring with maximal ideal m and ring of quotients Q. Let R be the completion of R, and let Q(R) be the ring of quotients of Ii. Then we have dim(R) = 1 [cf. (63), Cor. 10.12), and R is a faithfully flat extension of R; in particular, we have (RM) nQ = M for every Rsubmodule M of Q [cf. [34], Ch. I, § 3, no. 5, Prop. 10] and Reg(R) C Reg(R). This implies that R is a one-dimensional local CM-ring. Moreover, for every regular element rEm we have Q(R) = R[ l/r] [cf. (2.3)]' hence we have RQ = Q(R) [RQ is the smallest subring of Q(R) containing Rand Q]. (2.5) Proposition: Let R be a one-dimensional local integrally closed eM-ring. Then R is a discrete valuation ring.

Proof: Let m be the maximal ideal of R, let Q be the ring of quotients of R, and define m' := (R : m)Q. We have m C mm' C R. We show that m' '" R. Let rEm be regular. By (2.3)(1) there exists hEN with m h eRr. We chose h minimal with this property. Then there exists y E mh - 1 with y i Rr, hence we have y/r i R. On the other hand, we have ym eRr, hence we have y/r Em'. Suppose that mm' = m. Then every element of m' is integral over R [cf. [63), Cor. 4.6], hence m' = R, contrary to what we have shown. Therefore we have mm' = R, hence m is generated by one element [cf. (2.2)], and therefore R is a discrete valuation ring [cf. 1(3.29)]. (2.6) Corollary: Let R be a one-dimensional local eM-ring, and let S be the integral closure of R. Assume that S is a finitely generated R-module. Then S is a one-dimensional semilocal eM-ring, and for every maximal ideal n of S the localization Sn is a discrete valuation ring. Proof: Let Q be the ring of quotients of R. We have dim(S) = 1 by [63], Prop. 9.2, and S is semilocal by B(3.8), hence S is a semilocal CM-ring. Let n be a maximal ideal of S. Since S is a noetherian ring, also V := Sn is noetherian [cf. [63], Cor. 2.3], hence V is a one-dimensional local CM-ring. Since S is integrally closed in Q, V is integrally closed in Qn [cf. [63], Prop. 4.13]. Now Qn is an artinian ring since Q is artinian [cf. (2.3)], hence it is its own ring of quotients, and therefore the ring of quotients Q(V) of V is contained in Qn, hence V is, a priori, integrally closed in Q(V). By (2.5) we see that V is a discrete valuation ring.

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77

(2.7) REMARK: Let A be a noetherian ring, and assume that the Jacobson radical of A contains a regular element r. If A is not reduced, then there exist subrings of the ring of quotients K of A containing A which are integral over A and not noetherian; in particular, the integral closure of A is not a finitely generated Amodule. Proof: Let a E A be a non-zero nilpotent element. We may assume that a2 = O. For any n E N we define Bn := A + A(a/rn} c K. Now Bn is a subring of K containing A, and Bn C B n+ 1 for every n E N. Thus, B := U Bn is a subring of K. Suppose that the increasing chain {B(a/rn)} nEN of ideals of B becomes stationary, hence that there exists mEN such that a/rm+l = b· a/r m for some b E B. We may write b = e + d· a/r k for some e, d E A, kEN. Then we have a/rm+l = ca/rm, hence a(1 - re) = o. Now I - re is a unit of A [cf. [63], Ex. 4.7], hence we have a = 0 which is in contradiction with the assumption a i- O. Therefore B is not a noetherian ring. Clearly B is integral over A. Now B is contained in the integral closure A of A; if A would be a finitely generated A-module, then also B would be a finitely generated A-module [cf. [63], Ex. 1.3].

2.3

Integral Closure and Manis Valuation Rings

(2.8) For the rest of this section let R be a one-dimensional semilocal CM-ring, let Q be the ring of quotients of R, and let S be the integral closure of R. (2.9) DEFINITION: The Manis valuation rings of Q which contain R are called the Manis valuation rings belonging to R. Note that, in case R is a domain, a Manis valuation ring belonging to R is a valuation ring of Q which is different from Q and contains R. (2.10) Proposition: Let R be a one-dimensional semilocal integral domain. Then: {I} The set of Manis valuation rings of Q belonging to R is finite and not empty; each such valuation ring is a discrete valuation ring. (2) Let VI' ... ' Vh be the Manis valuation rings of Q belonging to R; then the integral closure S of R is the intersection VI n ... n Vh, S is a principal ideal domain with h maximal ideals nl, ... , nh, and we have Vi = Sn, for i E {I, ... , h}.

Proof: (a) S is a one-dimensional semilocal domain by the theorem of KrullAkizuki [cf. [63], Th. 11.13]. Let nl, ... , nh be the pairwise different maximal ideals of S. Then Snl' ... , Snh are discrete valuation rings of Q [cf. 1{3.29}] and we have S = n~=1 Sn, [cf. B{2.5}]; S is a principal ideal domain by 1(7.2}. (b) Let V i- Q be a valuation ring of Q containing R; then S is contained in V [cf. 1(3.4}], and therefore we have V = Sn, for some i E {I, ... , h} [cf. 1{7.2}]. (2.11) Theorem: Let R be a one-dimensional semilocal eM-ring. Then:

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II One-Dimensional Semilocal Cohen-Macaulay Rings

(1) The set of Manis valuation rings ofQ belonging to R is finite and not empty; each such Manis valuation ring is a discrete Manis valuation ring. (2) Let {VI,"" Vh } be the set of pairwise different Manis valuation rings of Q belonging to R; then {(VI: Q)Q, . .. , (Vh : Q)Q} is the set of maximal ideals of Q. (3) Let II be a maximal ideal of Q. Then Q/ll is a field, R/(ll n R) is a onedimensional semilocal domain with field of quotients Q Ill, V I-t V /ll is a map from the set of Manis valuation rings of Q belonging to R and having conductor II to the set of Manis valuation rings ofQ/ll belonging to R/(llnR), and it is bijective. (4) The integral closure S of R in Q is the intersection VI n ... n Vh, S has h maximal ideals nl, ... , nh, each of them is regular, and every regular ideal of S is a principal ideal. Moreover, nj Vj is the regular maximal ideal of Vj and Vj = S[ nj 1 for every j E {I, ... , h}. Proof: We use the notations introduced in (2.3). Let V be a Manis valuation ring of Q belonging to R, and set II := (V : Q)Q. Then II is a prime ideal of Q [cf. 1(2.4)(3) (a) ], hence II = lli for some i E {I, ... , l}, Q/ll is a field, and V /ll =I Q/'0 is a valuation ring of Q/ll by 1(2.17) (1), and therefore a discrete valuation ring by (2.10). This implies that V is a discrete Manis valuation ring of Q, again by 1(2.17)(1). Conversely, let i E {I, ... , l}, and let V =I Q/'Oi be a Manis valuation ring of Q/lli which contains R/Ci; then V is a discrete valuation ring of Q/lli by (2.10). The preimage V C Q of V is a discrete Manis valuation ring of Q [cf. 1(2.17)(1)] containing R with conductor 'Oi, and there are only finitely many discrete Manis valuation rings belonging to R and having conductor 'Oi [cf. (2.10)]. Now we have proved the assertions in (1)-(3). The assertion in (4) follows from chapter I, (2.5), (2.19), (2.21), and the fact that every maximal ideal of 8 lies over a maximal ideal of R [cf. [63], Cor. 4.17], hence it is a regular ideal. (2.12) THE REDUCED CASE: We use the notations of (2.3). Let R be a reduced one-dimensional semilocal CM-ring. In this case we can describe the Manis valuation rings of Q which belong to R more precisely. Since R is reduced, we have Cl n ... n Cl = {OJ, '01 n ... n III = {OJ. Let i E {I, ... , l}. Then the onedimensional semilocal CM-domain R/Ci has Ki := Q/'Oi as field of quotients; let Si be the integral closure of R/Ci. Note that Si is a one-dimensional semilocal domain [cf. (2.10)]. Let m~I"'" m~li be the maximal ideals of Si; then {Wi! := (Si)m'. , ... , Wi!' := (Si)m''IIi } is the set of Manis valuation rings of Ki belonging to R/Ci. For j E {I, ... , li} we denote by Wij: Ki -t Zoo the valuation of Ki defined by Wij. Set K := Kl x ... X K/ and S := 8 1 x ... X 8 1• We choose i E {I, ... , l} and j E {I, ... , ld. We define a map Vij: K -t Zoo by Vij((Xl,"" ~)) = Wij(Xi); it is easy to check that Vij is a discrete Manis valuation of K having Vij := Kl x ... X K i - l X Wij X KiH X •.• X Kl as its valuation ring; Kl x ... X Ki-l x m~j x KiH x ... X Kl is the regular maximal ideal of ~j, mij := SI x ... X Si-l x m~j x SiH x ... X SI is 11

1

3 Analytically Unramified and Analytically Irreducible CM-Rings

79

the intersection of this ideal with B, we have Blmi ;] = Vi j , and Kl x ... X K i - 1 X {OJ x Ki+1 x ... X K, is the kernel of Vij. The canonical isomorphism rp: Q ~ K maps 8 onto S [cf. B(3.4)]. We set Vi; := rp-l(Vij ); Vi; is a discrete Manis valuation ring of Q containing R, the intersection \llij of the regular maximal ideal of Vij with 8 is a maximal ideal of 8, and Vij = 8Im i;]. Now {Vij liE {l, ... ,I},j E {l, ... ,li}} is the set of Manis valuation rings of Q belonging to R. In particular, we can state: 8 is a one-dimensional integrally closed reduced semilocal ring, and every ideal of 8 is a principal ideal.

3

One-Dimensional Analytically U nramified and Analytically Irreducible eM-Rings

(3.0) In this section R is a one-dimensional local CM-ring with maximal ideal m and ring of quotients Q.

3.1

Two Length Formulae

(3.1) Proposition: (1) Let M be a finitely generated R-submodule of Q containing a regular element of R. Then we have tR(R/Rr)

= tR(M/rM)

for every r E Reg(R).

(2) Let M be an R-submodule ofQ. Then we have tR(M/rM) $ tR(R/Rr)

for every r E Reg(R).

Proof: (1) For every regular element a E R the map x I-t ax : Q ~ Q is an isomorphism of R-modules; thus, we may assume that M is a regular ideal a of R. Consider the following two exact sequences

o ~ a/ra ~ R/ra ~ R/a ~ 0,

0 ~ Rr/ra ~ R/ra ~ R/Rr ~ O.

All the R-modules in these sequences have finite length by (1.6), hence we obtain

The R-modules R/a and Rr/ra are isomorphic, whence tR(a/ra) = tR(R/Rr). (2) Suppose that we have tR(M/rM) > tR(R/Rr) for some r E Reg(R). Then there exists a finitely generated submodule N of M such that t«N +rM)/rM) > tR(R/Rr). Now we get tR(N/rN) ~ tR(N/(N n rM))

which is in contradiction with (1).

= tR«N + rM)/rM) > tR(R/Rr),

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II One-dimensional Semilocal Cohen-Macaulay Rings

(3.2) Proposition: Let a be a regular ideal of R, and let rEm be regular. Then we have lR(Rlra) = lR(RIRr) + lR(Rla). Proof: The R-modules Ria and Rrlra are isomorphic, hence lR(Rla) = lR(Rrlra). On the other hand, we have lR(Rlra) = lR(RI Rr) + lR(Rrlra). This implies the assertion.

3.2

Divisible Modules

(3.3) DEFINITION: Let A be a ring and M an A-module; for a E A let aM: M -+ M be multiplication by a. (1) M is said to be divisible if, for every a E Reg(A), the map aM: M -+ M is surjective. (2) M is said to be torsion-free if, for every a E Reg(A), the map aM: M -+ M is injective. (3) M is said to be a torsion module if {O} is the only torsion-free submodule ofM. (4) M is said to be simple divisible if M is a non-zero torsion divisible module and admits no proper non-zero divisible submodules. (3.4) DIVISIBLE MODULES: Let A be a ring and M an A-module. (1) Let N be a submodule of M. Then MIN is a divisible module iff M = N +aM for every a E Reg(A). If M is divisible, then MIN is divisible, and if N and MIN are divisible, then M is divisible. (2) A sum of divisible sub modules of M is a divisible submodule of M, hence M contains a unique largest divisible submodule dA(M), namely the sum of all divisible submodules; dA(M) is called the divisible submodule of M. Clearly M is a divisible A-module iff M = dA(M); we have dA(MldA(M» = {O} [cf. (1)]. (3) Let M be an A-module. We set N:=

n

aM.

aEReg(A)

Then we have dA(M) c N, and if M is torsion-free, then we have dA(M) = N. Proof: (a) Let x E dA(M); for every a E Reg(A) there exists Ya E dA(M) such that x = aYa E aM, hence we have dA(M) c N. (b) Now we assume that M is torsion-free. We show that N is a divisible submodule of M, hence that dA(M) = N. Let x E N and b E Reg(A). For every a E Reg(A) there exists Ya E M such that x = baYa [since ba E Reg(A)], and we have aYa = a'Ya' for all a, a' E Reg(A) [since M is torsion-free]. We set z := YIA; then we have z = aYa for every a E Reg(A), hence zEN and x = bz, i.e., N = bN. (4) The divisible submodule dA(A) of A is the conductor fA of A in its ring of quotients Q(A).

3 Analytically Unramified and Analytically Irreducible CM-Rings

81

Indeed, for every a E Reg(A) we have afA = fA since a is a unit of Q(A) and fA is an ideal of Q(A), hence we have fA C dA(A). Let x E dA(A). Take any z E Q(A); then z = air for some a E A and r E Reg(A). There exists y E dA(A) such that x = ry. Now xz = ry(alr) = ya E A, and therefore Q(A)x c A, hence x E fA. (3.5) DIVISIBLE R-MODULES: (1) An R-module M is divisible iff M = mM; in particular, {O} is the only finitely generated divisible R-module. Indeed, if M is a divisible R-module, then M = r M for a regular rEm, hence M = mM. Conversely, assume that M = mM; this implies that M = mh M for every hEN. Let r E Reg(R). Then there exists hEN with mh C Rr [cf. (2.3)(1)], and therefore we have M = rM, i.e., M is a divisible R-module. Now, if M is a finitely generated divisible R-module, then M = mM implies M = {O} by Nakayama's lemma [cf. [63], Cor. 4.8]. (2) Let M be an R-submodule of Q containing R. Then M IRis divisible iff M = R + mM iff M = R + rM for every r E Reg(R) [cf. (1)]. Moreover, in this case M is a subringof Q. In fact, let x E M; then we can write x = ylr with y E R and r E Reg(R). From M = R + rM we get xM = Rx + yM c M + M = M, hence the product of two element of M lies in M. (3) The R-module RI R is divisible. In fact, let x E Rand r E Reg(R) n m. Since R is dense in R [as a topological subspace], we may write x = y + rz for some y E Rand z E R [note that the m-adic topology and the rR-adic topology of R are identical, cf. (2.3)]. Hence RI R is a divisible R-module.

3.3

Compatible Extensions

(3.6) DEFINITION: A proper subring A of Q containing R is called a compatible extension of R if Aj R is a divisible R-module. (3.7) REMARK: Let M be a proper R-submodule of Q containing R. If MIR is a divisible R-module, then M is a compatible extension of R by (3.5)(2). (3.8) REMARK: Let A be a compatible extension of R. (1) The only regular maximal ideal of A is mAo In fact, since A # Q, there exist regular elements in A which are not units, hence there exist regular maximal ideals of A. Let n be such an ideal. Since it contains regular elements of R, we have A = R + n [since A is a compatible extension of R], hence Ajn ~ RI(nnR), hence nnR = m [since Ajn is a field], hence mA en. We have A = R + mA [since mA contains regular elements of R]. Let yEn. Then we can write y = r + z with r E R, z E mA, hence r = y - z ERn n = m, and therefore y E mA, hence n C mA; thus, we have shown that mA = n. (2) Since A is a torsion-free R-module, we have dR(A) = mn A [cf. (3.4)(3)], and since mA contains all regular elements of A which are non-units [cf. (1)], we have

n

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II One-dimensional Semilocal Cohen-Macaulay Rings

dA(A) = nmnA. Therefore we have dR(A) A in Q by (3.4)(4).

= dA(A),

and this is the conductor of

(3.9) Proposition: Assume that R is complete. Let D be the set of proper ideals of Q, and let C be the set of compatible extensions of R. Then the map (J M R+ (J : D -+ C is an inclusion-preserving bijective map, B M dB(B) = mn B is the inverse map, and (J is the conductor of R + (J in Q.

n

Proof: (1) Let B be a compatible extension of R, and set (J := nmnB. By (3.8)(2) (J is an ideal of Q. Let r E Reg(R) n m; then we have B = R + rB [cf. (3.5)(2)]. Let x E B. By induction, there exist elements YO,Yl, ... E Rand Zl, Z2, ... E B such that, for every n E No, x = Yo + Ylr + ... + Ynrn + Zn+lrn+l. Set y := E:'oyvrV E R [note that R is complete]. Now we have x - y E mnB for every n E N, hence we have x - Y E (J, and therefore we have B = R + (J. (2) Let (J E D and set B := R + (J. Then we have B ::f; Q. Indeed, suppose that B = Q. Let rEm n Reg(R). Then l/r = x + d for some x E R, dE (J, hence rd = 1 - rx is a unit of R, which is in contradiction with the fact that d is a zero-divisor in Q. Thus, we have shown that B is a proper submodule of Q. For every r E Reg(R) we have R + rB = R + Rr + r(J = R + (J = B, hence B/R is a divisible R-module, and therefore B is a compatible extension of R [cf. (3.7)]. Set (J' := mn B; then (J' is the conductor of Bin Q, and we have B = R+ ()' [cf. (1»). We have (J C (J', and from B/R = R/(R n (J) = R/(Rn (J') we get R n (J = R n (J', hence we have (J = (J' since Q is a localization of R.

n

(3.10) REMARK: In the sequel, we study the set of maximal compatible extensions of a one-dimensional local CM-ring. As we just saw in (3.9), it is easier to work with a complete ring; therefore, the following result shall playa crucial role. (3.11) Proposition: Let M be the set of all R-submodules of Q containing R, and let M* be the set of all R-submodules of Q(R) containing R. Then: (1) The maps M

M

RM: M -+ M*,

M*

M

M* n Q: M* -+ M

are bijective, inclusion-preserving and mutually inverse. If M* E M*, then M* = R+(QnM*). (2) The map AMRA is a bijective inclusion-preserving map between the set of compatible extensions of R and the set of compatible extensions of R.

Proof: (1) By (2.4) we have RM n Q = M for every R-submodule M of Q. Let M* E M*. It is enough to show that M* = R + (M* n Q). Let x E M*, and choose r E Reg(R) such that rx =: Y E R. The R-module R/ R is divisible [cf. (3.5)(3)], hence there exist u E R, v E R such that Y = u + rv. Now we have x - v = u/r E M* n Q, and therefore we have x = v + u/r E R + (M* n Q).

3 Analytically Unramified and Analytically Irreducible CM-Rings

83

(2) Let A be a compatible extension of R, hence A = R + mA; then we have RA = R + mRA = R + mRA, and RA =I Q(R) by (1). Thus, RA is a compatible extension of R. Conversely, assume that B is a compatible extension of R; then we have B = RA where A = B n Q is a proper R-submodule of Q containing R [cf. (1)]. Choose any r E Reg(R); we show that A = R + rA. Let x E A. Then x = y+rz for some y E R, z E B [because B = R+rB]. Since RjR is a divisible R-module, we may write y = Yl +rz' for some Yl E R, z' E R. Set Zl := z+z' E B. Now x = Yl + rZl, Zl = (x - Yl)jr E Q, hence Zl E A, and therefore x E R + rA. Thus, we have shown that A is a compatible extension of R. (3.12) Corollary: We have the following; (1) The set of compatible extensions of R satisfies the ascending and the descending chain condition. (2) The set of maximal compatible extensions of R is finite and not empty. Proof: This follows from (3.11) and (3.9) since Q(R) is an artinian ring.

(3.13) Proposition: Let A a compatible extension of R. Then: (1) Every R-submodule M ofQ containing A is an A-module. (2) Let M =I Q be an R-submodule of Q containing A. If the only compatible extension of R lying between A and M is A itself, then M is a finitely generated A-module, and MjA is an R-module of finite length. Proof: (1) Let a E A and x E M. We write x = yjr for some Y E R, r E Reg(R). Since A = R + rA [cf. (3.5)(2) j, we may write a = 8 + rb for some 8 E Rand bE A. Now we have ax = (8 + rb)x = 8X + by E M + A eM, and therefore M is an A-module. (2) Let r E Reg(R) n m. Then MjrM is a finitely generated R-module by (3.1), hence M = N + r M where N = RXl + ... + RXn is a finitely generated Rsubmodule of M. There exists hEN such that mhxi C R for every i E {I, ... ,n}, hence mhM C R + mh+1M. Set B:= R+ mhM. Then we have B ~ R+ mB = R+mh+1M ~ R+mhM = B, hence B = R+mB, and the R-module B =I Q is a compatible extension of R by (3.5)(2). Since A = R+mhA C R+mh M = Be M, we have A = B, and therefore we have mhMeA. Now M = N + rM = ... = N + rh M, and since rh MeA, we have M = AXl + ... + AX n + A. Thus, we have shown that M is a finitely generated A-module. Since the R-module M j A is annihilated by m h and since it is a homomorphic image of the finitely generated R-module Mjrh M, it has finite length.

(3.14) Theorem: The R-module QjR is artinian. Proof: By (3.12)(1) the set of compatible extensions of R satisfies both chain conditions. Therefore we can construct a chain R = Ao C Al C ... C An = Q with compatible extensions A l , ... , A n - l of R, and there is no other compatible extension of R lying properly between two consecutive ones. Let i E {O, ... ,n-I},

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and let M be an R-submodule of Q with Ai c M ~ A H1 . Then MIAi is an Rmodule of finite length by (3.13), hence Ai+1IAi is an artinian R-module, and the claim follows easily. (3.15) Proposition: Let M be an artinian R-module. Then M is a noetherian R-module iff M has no divisible submodules except {a}. Proof (1) Assume that M is noetherian; then the divisible submodule dR(M) is noetherian. Now we have dR(M) = {a} by (3.5)(1). (2) Assume that M has no divisible submodules except {a}. Suppose that M is not noetherian. Let M' be a minimal element in the non-empty set of submodules of M which are not noetherian; then M' =I {a}. M' is not divisible by assumption; hence there exists r E Reg(R) such that r M' ~ M'. Now, by the choice of M', rM' is a noetherian module, and N := M'lrM' is not noetherian. We choose hEN such that mh eRr [cf. (2.3)(1)]' and consider the decreasing chain N:::> mN:::>···:::> mhN = {a} of submodules of N. Let i E {O, ... ,h -I}. The Rmodule mi NImH 1 N is artinian, and it is annihilated by m, hence it is an artinian RIm-module; since RIm is a field, mi NImH 1 N is a noetherian RIm-module [cf. [21], Prop. 6.10], and therefore it is a noetherian R-module. This implies that also N is a noetherian R-module, which is in contradiction with the choice of N. This contradiction shows that M is noetherian.

3.4

Criteria for One-Dimensional Analytically Unramified and Analytically Irreducible CM-Rings

(3.16) DEFINITION: A local ring is called analytically irreducible (resp. analytically unramified) if its completion is a domain (resp. is reduced). (3.17) Theorem: The following statements are equivalent: (1) R is analytically irreducible. (2) Q IRis a simple divisible R-module. (3) Q is a field, and the integral closure V of R in Q is a discrete valuation ring and a finitely generated R-module. Proof (1) ¢:} (2): We know that QIR is a simple divisible R-module iff Q(R)IR is a simple divisible R-module [cf. (3.11)], and that there is a bijective map between the set of divisible R-submodules of Q(R)I R and the set of ideals of Q(R) [cf. (3.9)]; in particular, Q(R) is a field iff Q(R)I R is a simple divisible R-module. (2) =} (3): (a) By the equivalence of (1) and (2) we know that Q is a field. (b) A proper R-submodule M of Q which contains R is noetherian. In fact, M IRis a submodule of the artinian module QIR [cf. (3.14)], hence it is artinian, and has no proper divisible submodules except {O} by assumption, hence it is noetherian by (3.15), and therefore M is a noetherian R-module. (c) In particular, the integral closure V of R is a finitely generated R-module, hence V is an integrally closed noetherian domain, and we have dim(V) = 1 [cf.

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85

[63], Prop. 9.2]. We show that V is local, hence that V is a discrete valuation ring [cf. 1(3.28)]. In fact, let n be a maximal ideal of V. Then Vn is a proper subring of Q containing

R, hence it is a finitely generated R-module by (a), and therefore integral over R [cf. [63], Cor. 4.5]. It follows that Vn C V, and therefore we have V = Vn. This implies that every element of V which lies not in n is a unit of V, hence V is local, and n is its maximal ideal. (3) => (2): We show that every proper subring A of Q containing R is contained in V. In fact, assume that A ¢. V, and choose a E A, a ~ V. Then I/a lies in the maximal ideal of V, hence we have Q = V[ a] [cf. (2.3)]. There exists r E Reg(R) with rV c R, and therefore Q = rV[a] C R[a] C A, hence we have A = Q, contradicting the choice of A. This result implies, in particular, the following: Let A/ R be a divisible submodule of Q / Rj then A is a subring of Q containing R [cf. (3.5)], hence we have A C V, and therefore A = R [cf. (3.5){I)]. Thus, we have shown that Q/ R is a simple divisible R-module. This ends the proof. As an easy application of this result we prove

(3.18) Proposition: Let V be a complete discrete valuation ring, let K be its field of quotients, and let L be a finite extension of K. Then there exists a unique extension W of V to L, W is a finitely generated free V -module, and we have e(W/V)!(W/V)

= [L: K].

Let v (resp. w) be the valuation of K (resp. of L) defined by V (resp. W). Then we have v(NL/K(X» = !(W/V)w(x) for every x E L.

Proof: We set n := [L : K] and let T be the integral closure of V in L. Then T contains a basis {Xl, ... , xn} of the K -vector space L. The ring S = V[ Xl, •.. , Xn ] is a finitely generated V -module [cf. [63], Cor. 4.5], hence S is a complete semilocal domain [cf. B(3.8)], and therefore it is local [cf. B(3.5)]. Now T is the integral closure of S, hence it is a discrete valuation ring which is a finitely generated Smodule [cf. (3.17)]. Therefore we have W = T, and W is a finitely generated free V-module [cf. [63], Ex. 4.1 for the first assertion and 1(3.7) for the second one]. The assertion e(W/V)!(W/V) = [L: K] now follows from 1(6.22). Let t be a generator of the maximal ideal of V, and let u be a generator of the maximal ideal ofW. It is enough to show that v(NL/K(U» = !(W/V)j this follows from [note that the norm of a unit of W is a unit of V] 0= v(NL/K(Ue(w/V) It)~

= e(W/V)v(NL/K(U» -

=e(W/V)(V(NL/K(U» -

[L : K]

!(W/V».

(3.19) Theorem: (1) The map A t-+ (integral closure of A in Q) is a bijective inclusion-preserving map from the set of maximal compatible extensions of R to

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II One-dimensional Semilocal Cohen-Macaulay Rings

the set of Manis valuation rings belonging to R; the inverse map is given by V t-+ A where AIR = dR{VIR). (2) The map V t-+ V Ris a bijective inclusion-preserving map from the set of Manis valuation rings belonging to R to the set of Manis valuation rings belonging to R; the inverse map is given by W t-+ W n Q. (3) Let S be the integral closure of R. Then SR is the integral closure of R in Q(R). Proof: (l)(a) Let A be a maximal compatible extension of R, and let V be the integral closure of A in Q. Let x E Q be regular. We show that A[ x] '" Q or A[X-l] '" Q. In fact, suppose that A[x] = Q and A[x- l ] = Q. Since mQ = Q, we have relations

with ao, ... ,am and bo , ... ,bn E mAo We may assume that the relations in (*) are of the smallest possible degrees m and n. Since mA '" A [cf. (3.8)(1)], we have m ~ 1 and n ~ 1. Let, say, n ~ m. Multiplying the first equation in (*) by 1 - bo and the second equation by amx m , we get 1- bo = (1- bo)ao + ... + (1- bo)amxm, (1 - bo)amx m = amblx m- l + ... + ambnx m- n ,

hence we get a relation 1

= Co + ... + Cm_lXm - l

with Co, ..• ,Cm-l E mA

which is in contradiction with the choice of m. Let x E Q be regular; by what we have just shown, we may assume that A[ x] '" Q. Then A[ x] is a finitely generated R-module by (3.13), hence x is integral over A [cf. [63], Cor. 4.6], and therefore x E V. Thus, we have shown that V is a Manis valuation ring of Q, and therefore a Manis valuation ring of Q belonging to R. Let B be defined by B I R = dR (VIR); then B is a compatible extension of R. Since AIR is a divisible R-module, we have B ::> A, hence we have B = A since A is a maximal compatible extension of R. (b) Let V be a Manis valuation ring of Q belonging to R, and let A be defined by AI R = dR (VIR); then A is a compatible extension of R, and there is no further compatible extension of R between A and V. The R-module VIA is artinian by (3.14), and, by (3.13), V is a finitely generated A-module and VIA is a noetherian R-module. Therefore V is integral over A, and we have dR{vIA) = {O} [cf. (3.15)]. (c) Thus, we have shown that the maps mentioned in the statement of (1) are mutually inverse; clearly they are inclusion-preserving. (2) Let V be a Manis valuation ring of Q belonging to R, let A be defined by AIR = dR(VIR), and set B := AR. Then A is a maximal compatible extension of R and V is its integral closure by (1), and B is a maximal compatible extension

3 Analytically Unramified and Analytically Irreducible CM-Rings

87

of R by (3.11). Let W be the integral closure of B in Q{R). Clearly we have RV c W. We set U := Q n Wi then we have V C U by (3.11). Since U is a finitely generated A-module [cf. (3.13)], it is integral over A [cf. [63], Cor. 4.5], hence U C V, and therefore we have W = RV by (3.11). (3) Let Vl , ... , Vh be the Manis valuation rings belonging to R. Then V1R, ... , VhR are the Manis valuation rings belonging to R [cf. (2)], S = Vl n ... n Vh is the integral closure of R in Q by (2.10), V1R n ... n VhR is the integral closure of R in Q{R) by (2.11), and SR = V1R n ... n VhR [cf. [34], Ch. I, § 3, no. 5, Prop. 10, and note that R is a faithfully flat R-module]. (3.20) Proposition: Let All ... , Ah be the maximal compatible extensions of R, and, for every i E {I, ... , h}, let Vi be the integral closure of Ai. Set S := Vl n··· n Vh, T:= Al n··· n Ah. Then: (I) The following sequences of R-modules h

h

0-+ T -+ Q -+ EeQ/Ai -+ 0,

0 -+ S -+ Q -+ EeQ/Vi -+ 0 i=l

i=l

are exact. (2) The R-module SIT has finite length; in particular, we have the estimate h

lR{S/T) ~

L lR{Vi/Ai ). i=l

Proof: (I) Since R is a faithfully flat R-module, a sequence M' -+ M -+ Mil of R-modules is exact iff M' ®R R -+ M ®R R -+ Mil ®R R is exact. We have SR = V1Rn·· ·nVhR and TR = A1Rn .. ·nAhR by [34], Ch. I, § 3, no. 5, Prop. 10, Q{R) = RQ by (2.4){1), V1R, ... , VhR are the Manis valuation rings belonging to R by (3.19){2), and A1R, .. . , AhR are the maximal compatible extensions of R by (3.11). We may assume, therefore, that R is complete. Let Pl, ... , Ph be the maximal ideals of Q labelled in such a way that Ai = R + Pi for every i E {1, ... ,h} [cf. (3.9)]. Then, for any i E {1, ... ,h -I}, we have (Al n··· n Ai) + Ai+l = Q since (Pl n··· n Pi) + Pi+l = Q. The assertions in (I) are now a consequence of the lemma below, whose proof is simple and is left as an exercise. (2) The R-module homomorphism S -+ E9~1 Vi/Ai, defined by S -+ Vi -+ Vi/Ai for every i E {I, ... ,h}, clearly has T as its kernel. The R-modules VdA l , ... , Vh/Ah have finite length [cf. (3.13)], hence we get the estimate in (2). (3.21) LeInIna: Let A be a ring, let M be an A-module, and let N ll ... , Nh be submodules of M with the following property: For every j E {I, ... , h - I} we have (Nl n··· n N j ) + NjH = M. Then the canonical homomorphism M -+ M/Nl EEl ... EEl M/Nh is surjective and has kernel Nl n··· n Nh.

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(3.22) Theorem: The following statements are equivalent: (I) R is analytically unramified. (2) The integral closure 8 of R in Q is a finitely generated R-module. Proof: Let AI, ... ,Ah be the maximal compatible extensions of R, and set T := A 1 n·· ·nAh • Then 8/T is an R-module of finite length by (3.20); in particular, 8/T is a noetherian R-module. Now, 8 is a noetherian R-module iff T is a noetherian R-module by [63], Ex. 1.3, and T is a noetherian R-module iff TR is a noetherian R-module by [34], Ch. I, § 3, no. 5, Prop. 10 and no. 6, Prop. 11. Let {PI, ... ,Ph} be the set of maximal ideals of Q{R); then n := PI n··· n Ph is the nilradical of Q(R). Thus, we have to show that n = {O} iff TR is a noetherian R-module. ~et the prime ideals 1'1,. :'.:' Ph be labelled in such a way that Bl := Al..-R = R + PI, ... ,Bh := AhR = R + Ph are the maximal compatible extensions of R [ cf. (3.9)]. Note that TR = Bl n··· n Bh by [34], Ch. I, § 3, no. 5, Prop. 10. We assume that n = {O}. The canonical R-linear map TR = Bl n ... n Bh ~ BI/Pl ffi ... ffi Bh/Ph has kernel n, hence TR is a submodule of the R-module BI/Pl ffi ... ffi Bh/Ph' Now B/Pi = R/{R n Pi) is a cyclic R-module for every i E {I, ... , h}, hence T R is a noetherian R-module. We assume that T R is a noetherian R-module. Since R + neT R, we see that (R+n)/ R is a divisible [cf. (3.9)] noetherian R-module, hence it is the zero module by (3.5){I). Therefore n is contained in R. Suppose that n is not the zero ideal. Then there exists x E n with x i- 0, x 2 = O. Let rEm be regular. Since n is an ideal in Q, we have in R the ascending chain ({x/rn)R)nEN of ideals. This chain becomes stationary, hence there exist n E Nand s E R with sx = x/r, hence x{I - rs) = O. Now 1 - sr is a unit of R since rEm, hence we have x = 0, which contradicts our choice of x. Therefore we have n = {O}.

4

Blowing up Ideals

(4.0) NOTATION: In this section R always is a one-dimensional semilocal CMring, t is its Jacobson radical, Q is its ring of quotients, and 8 is the integral closure of R in Q.

4.1

The Blow-up Ring RO

(4.1) NOTATION: (1) For R-submodules M, N of 8 we define

M: N:= {M: N)s

= {x E 8 I xN C M}

[the notation M : N instead of the more exact notation {M : N)s will be used only in this section].

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4 Blowing up Ideals

(2) Let a be an ideal of R, and let n E N; then, clearly, an : an is a subring of S containing R, and we have an : an Can+! : an+! , showing that R a :=

U(an: an)

nEN

is a union of an increasing sequence of subrings of S, hence it is a subring of S containing R.

(4.2) DEFINITION: Let a be an open ideal of R. (1) The ring Ra is called the blow-up of R with respect to a; we also say that we get Ra by blowing up the ideal a. (2) Let R be local with maximal ideal m. The ring R m is called the quadratic transform of R. (4.3) Proposition: Let a be an open ideal of R. Then: (1) R a = {xja I sEN, a transversal for a of order s, x E as}. (2) There exist tEN and a E at transversal for a of order t such that Ra = ata- 1 • (3) Ra is a finitely generated R-module, and therefore is a one-dimensional semilocal CM-ring. (4) For all sufficiently large integers n we have R a = an : an and an = an R a. (5) There exists a regular r E R a such that aR a = Rar. (6) If A is any subring of S containing Rand aA is a principal ideal in A, then we have R a cA. Proof: (a) The set on the righthand side in (1) clearly is a subring of Q containing R [cf. (1.16)(2)]; call it B. Let bEat be a transversal element for a of order t [for some tEN]; b is a regular element of R by (1.15). We choose kEN so large that ba kt = akt +t . Then we have bka kt = a2kt . We set s := kt and a := bk ; then we have a E as and aa S = a2s , hence a is transversal for a of order s. We set C := aSa- 1 C Q. Clearly we have C· C = C, hence C is a sub ring of B containing R, and, moreover, it is a finitely generated R-module, hence C is integral over R [d. [63], Cor 4.5]. Since atb- 1 C C, we see that every element of atb- l is integral over R. Now, b was an arbitrary element of R being transversal for a [of some order], hence B is integral over R, and a fortiori over C. (b) Keeping the notation introduced in (a), we shall show that B C C-whence B = C. Let y E B; then y = zjc where c is transversal for a, say of order m, and where z E am. Let us write y = (zcs-ljam)(amjcs); clearly, zcS-lja m and cSjam are elements of C. Moreover, cSjam is a unit of B, and since B is integral over C, cSjam is a unit of C [otherwise, there would exist a maximal ideal n of C containing cSjam, and since there exist maximal ideals of B lying over n by [63], Prop. 4.15 and Cor. 4.17, cSjam would not be a unit of B], hence we have y E C. (c) We shall show that R a = aSa- l [= C = B]. Let y ERa, and choose hEN such that y E ahs : ahs; then y E ahsa- h = C h = C. Let y E aSa-l. Then we have yES [cf. (a)], and ya 2S = yaa S C a2s , hence we have y E a2s : a2s C Ra.

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(d) As Ra is a finitely generated R-module, (3) holds [ef. B(3.8) and [63], Prop. 9.2], the first statement in (4) holds, and from as = aRa we get as = aR a C as R a = aRa, hence as R a = aR a and as R a = as, and therefore we have an Ra = an for every integer n ~ s. (e) The statement in (5) is equivalent to aRa being an invertible Ra-module [cf. (2.2)(3), and note that Ra is semilocal]. By (d) we have aR a . as- 1R a = aR a, hence aR a is invertible. (f) Now we prove (6). Assume that aA = Aw for some wE A. Then w is a regular element of A [because a contains regular elements of R]. Let y ERa, and choose n E N such that yan C an. Then we have yan A C an A, hence yw n E wn A, and therefore we have yEA [since wn is a regular element of A]. (4.4) REMARK: Let a be an open ideal of R. (1) In the course of proving (4.3), we have shown the following: There exist sEN and a E as such that a is transversal for a of order s and that aaB = a2s ; in this case we have R a = aSa- 1 = R[aB/a]. (2) Assume that there exists an a-transversal element a; then we have aRa = aRa. In fact, choose n E N with aan = an + 1 . Then we have aan R a = a n + 1 Ra, and since aRa = Rar for some regular r E Ra [cf. (4.3)(5)], we have aR a = rR a = aRa. Moreover, we have anan = a2n and, by (1), we have R a = R[ an /a n ] = R[ a/a]. In particular, we have R a = R[ ada, ... , ah/a] for any system of generators {al, ... ,ah} ofa. (3) Now we assume that a is contained in the Jacobson radical of R, and that there exists an a-transversal element a; then we have R a = R[ a/a] [cf. (2)]. Let R(a, R) = EBn>O anTn be the Rees ring of R with respect to a [cf. B(5.1)]; there exists a canonical isomorphism 1/J: R(a,R)(aT) -+ Ra [ef. B(5.5)]. Let O an+1Tn be its kernel [cf. B(5.2)]; we set E:= V+(J) C Proj(R(a,R» [cf. B(5.7)]. Every homogeneous ideal ~ of gra(R) which contains a mod a2 contains a power of (gra(R»+ [since ~n = (gra(R»n for every n with aan- 1 = an), hence we have Proj(gra(R» = D+(a mod a2 ). (a) The element a is a regular element of R, and we have aR a = aR a [ef. (2»). Moreover, since Ra is integral over R, every maximal ideal of Ra lies over a maximal ideal of R, and over every maximal ideal of R there lies a maximal ideal of Ra [ef. [63], Cor. 4.17) and therefore the set Max(Ra) of maximal ideals of R a is the set V(aRa) of prime ideals of Ra containing a, and there exists a bijective map En D+(aT) -+ Max(Ra) [ef. B(5.7)(2»). Let p be a maximal ideal of Ra, and let q be the homogeneous prime ideal of R(a, R) corresponding to p. Then there exists a canonical isomorphism R(a,R)(q) -+ (Ra)p [ef. B(5.6)(2»). (b) Now we assume that R is local and that a = m, the maximal ideal of R. Since dim(grm(R» = 1 [cf. [63), Ex. 13.8, 3»), and since the maximal ideal (grm(R»+ of grm(R) is its only irrelevant homogeneous prime ideal, we see the following: Proj(grm(R» consists of all homogeneous prime ideals of grm(R) different from (grm(R»+, we have E C D+ (aT) , and there exists a bijective map

4 Blowing up Ideals

91

Proj(grm(R)) --+ Max(Rm). For another description the reader should read VIII(1.3).

(4.5) Corollary: Let T be a one-dimensional semilocal eM-ring with ring of quotients Q(T), and let cP: R --+ T be a homomorphism such that cp(Reg(R)) C Reg(T). Let a be an open ideal of R. Then: (1) The ideal aT is an open ideal ofT, cp admits a unique extension to a homomorphism of rings cpa: Ra --+ TaT, and the induced homomorphism CPT: T ® R R a --+ TaT, defined by cPT( x ® y) = xcpa (y) for all x E T, y E Ra, is surjective. (2) If cp: R --+ T is a flat homomorphism, then CPT is an isomorphism (in this case, the condition cp(Reg(R)) C Reg(T) holds automatically). Proof: (1) Since a is an open ideal of R, a contains a regular element of R, hence aT contains a regular element ofT, and it is therefore an open ideal ofT [cf. (1.6)(2)]. Clearly cp has a unique extension to a homomorphism 'IjJ: Q --+ Q(T). If cpa exists, it must be the restriction of 'IjJ to Ra. Choose s and a as in (4.4)(1); then we have R a = aSa- 1. Furthermore, since aST = (aT)S, we have aasT = a2s T, hence TaT = (a ST)'IjJ(a)-1, and therefore 'IjJ(Ra) = 'IjJ(a)8'IjJ(a)-1 C aB 'IjJ(a)-1T = TaT; this implies that 'IjJ(Ra)T = TaT. Thus, cpa exists, and CPT is surjective. (2) The homomorphism 'IjJ: Q --+ Q(T) induces a homomorphism T ®R Q --+ Q(T) which is easily seen to be injective; furthermore, T ®R R a --+ T ®R Q is injective [since cp is flat], hence CPT is injective, and therefore an isomorphism [since it is surjective by (1)]. (4.6) REMARK: Let a be an open ideal of R. Corollary (4.5) can be used in the following three cases: (1) T is the t-adic completion R of R; then aR = Ii and iF = Ra R C Q(R). (2) T is the ring of fractions ~-1 R of R with respect to a multiplicatively closed system ~ of R such that ~-1 R is a one-dimensional CM-ring [this is the case if ~ is the complement of a union of maximal ideals of R]. Then the ~ -1 R-algebras (~-1 R)E-1a and ~-1(Ra) are canonically isomorphic. (3) T := Ric, where c is an ideal of R such that Ass(Rlc) contains no maximal ideal. Then Ric is a one-dimensional semilocal CM-ring, and Q(R)/cQ(R) is the ring of quotients of Ric [as one checks easily]. Let cp: R --+ Ric be the canonical homomorphism and 'IjJ: Q(R) --+ Q(Rlc) its extension; since (Rlc)a(R/c) is the smallest subring of Q(Rlc) containing Ric and 'IjJ(Ra) [cf. (4.5)(1)], it is equal to the ring R al(cQ(R) n Ra).

4.2

Integral Closure

(4.7) NOTATION: Set Ro := R, and, for every i E ~, let Ri+1 be the blow-up of Ri with respect to its Jacobson radical. We have an ascending chain

R =

Ro

C R1 C ... C

S

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II One-dimensional Semilocal Cohen-Macaulay Rings

of subrings of the integral closure S of R; the sequence (Ri)i>O is called the blowup sequence of R, or the sequence of blow-ups of R. Note that Ri+j = (Ri)j for all i, j E No [(Ri)j is the j-th blow-up of Ri]. The following result allows us to construct the integral closure S of R "step by step"-at least in the case where S is a finitely generated R-module. (4.8) Proposition: Let (Rik~o be the blow-up sequence of R. Then we have

Proof: Let n E N. We assume that for everyone-dimensional semilocal CM-ring T and every t E Reg(T) such that fR(TITt) < n, the following holds true: For every bET such that bit is integral over T, there exists i E N such that bit E Ti, the i-th blow-up ofT. [If fR(TITt) = 0, then t is a unit ofT, and the assumption is true.] Let r E Reg(R) be such that 1 ~ fR(RIRr) ~ n, and let a E R be such that air is integral over R. Let m be a maximal ideal of R such that rEm, and let Zh + a1Z h- 1 + ... + ah E R[ Z] be an equation of integral dependence of air over R. Multiplying this equation by rh shows that also a E m. Note that t = m q for some ideal q of R. By (4.3)(5) there exists a regular element w E Rl such that Rl w = tR l = (mRt) (qRt}. This implies that mR l is a principal ideal of Rl [cf. (2.2)], say mR l = Rl v where v E Reg(R l ). Since Rl is integral over R, the ideal mR l is a proper ideal of Rl [cf. [63], Prop. 4.15], and therefore v is not a unit of Rl. Now, av- l and rv- l lie in R l , and from (3.1) we get [note that Rl is a finitely generated R-module by (4.3)(3)] fR(RIRr) = fR(Rt/rRt};::: f R1 (Rt/rRt}

> f R1 (Rt/Rl (rv- l )).

By the induction assumption, there exists j E N such that av-l/rv- l E (R1)j-l, and since (Rl)j-l = Rj, it follows that air E Rj .

4.3

Stable Ideals

(4.9) DEFINITION: An open ideal a of R is called stable if aR o = a. (4.10) Proposition: Let a be an open ideal of R. Then: (1) There exists n E N such that an is stable. (2) If an is stable for some n E N, then am is stable for every integer m ;::: n. (3) a is a stable ideal of R iff aRm is a stable ideal of Rm for every maximal ideal mofR. Proof: (1) Since aR o = a iff RO = (a: a)s, and since Ran = R O for every n E N, we get the assertion from (4.3)(4). (2) Let n E N, and assume that an is stable; then an RO = an, hence amRa = am for every integer m ;::: n, and therefore am is stable.

93

4 Blowing up Ideals

(3) Assume that a is a stable ideal of R. Then we have aRa = a. Let m be a maximal ideal of R. We have (Ra)m = R~Rm by (4.6), from this we get (aRm)R~Rm = aRm, and therefore aRm is a stable ideal of Rm. Conversely, assume that aRm is a stable ideal of Rm for every maximal ideal m of R. We have to show that the inclusion a O of E- 1R can be canonically identified with the sequence «E- 1R)i)i>O; in particular, E-l S is the integral closure of E-l R in E-1Q [cf. (4.8)]. (5.2) NOTATION: (1) Let (Ri)i~O be the sequence of blow-ups of R [cf. (4.7)]. A local ring A = (Ri)n, where i E No and n is a maximal ideal of Ri, is said to lie in the i-th neighborhood of R; the set of all local rings in the i-th neighborhood of R shall be denoted by ~i(R). A local ring A is said to be infinitely near to R if A E ~i(R) for some i E No, and we denote by ~(R) the set of all rings which are infinitely near to R. (2) Let A E ~(R); then there exists i E No and a maximal ideal n of Ri such that A = (Ri)n. Now Ri is a finitely generated R-module, hence Rn(A)nR -t A is quasifinite [cf. B(10.1O)] and I(A) := [A/n(A) : R/(n(A) n R)] is finite. (5.3) INFINITELY NEAR RINGS: We collect some simple properties of infinitely near rings. Let (Ri)i>O be the sequence of blow-ups of R. (1) A local ring is infiiiitely near to R iff it is infinitely near to Rm for some maximal ideal m of R. In fact, let A := (Ri)n for some i E No and some maximal ideal n of Ri. Set m:= nnR; the ideal m is a maximal ideal of R [cf. [63], Cor. 4.17]. Then (Ri)n is the localization of (Ri)m with respect to the maximal ideal n· (Ri)m of (Ri)m; by (5.1) we have (Ri)m ~ (Rm)i, hence A is infinitely near to Rm. In the same way we see that if a local ring is infinitely near to Rm for some maximal ideal m of R, then it is infinitely near to R. (2) If A is a local ring infinitely near to R, and if B is a local ring infinitely near to A, then B is infinitely near to R. In fact, choose i E No and a maximal ideal n of Ri such that A = (Ri)n. Now B is infinitely near to Ri by (1), and therefore B is infinitely near to R. (3) Let p be a maximal ideal of R. Every local ring A which is infinitely near to RP is also infinitely near to R. In fact, A is infinitely near to (RP)n for some maximal ideal n of RP by (1). Set m:= Rn n. Note that (Rm)PRm and (RP)m are canonically isomorphic [cf. (5.1)], and that (Rm)pR m is equal to Rm if P -:I m, and is equal to the quadratic transform of Rm if P = m; hence (RP)n, which is the localization of (RP)m with respect to

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II One-Dimensional Semilocal Cohen-Macaulay Rings

the maximal ideal n· (RP)m, is infinitely near to the local ring Rm. Now (RP)n is infinitely near to R by (1), hence A is infinitely near to R by (2). (5.4) REMARK: The integral closure S of R has only finitely many maximal ideals, and each of them is a regular ideal [cf. (2.11)]. Let nl, ... , nh be the maximal ideals of S. Then Vl = S[nd' ... ' Vh = S[n!.] are the Manis valuation rings of Q belonging to R, we have S = Vl n· .. n Vh, and, for every j E {I, ... ,h}, Pj := nj Vj is the regular maximal ideal of Vj [cf. (2.11)]. Let (Ri)i'~O be the sequence of blow-ups of R. We have S = Ui>O Ri [cf. (4.8)]i therefore there exists io E N such that, for every integer i ~ io~ the ideals nl n Ri, ... , nh n Ri are pairwise different. Let j E {I, ... , h}. (1) Define ~j := Rio \ (Rio n nj). Every maximal ideal of S different from nj meets ~j, hence the only prime ideal of ~jlS lying over the maximal ideal of the local ring ~jl Rio is the ideal nj(~jl S). Since every maximal ideal of ~jl S must lie over the maximal ideal of ~jl Rio [cf. [63], Cor. 4.17], the ring ~jl S is a quasilocal ring with maximal ideal nj(~t S), hence we have ~jl S = Snj" The same argument shows also that (RI)n;nRI = ~jl Rl for every l ~ io. (2) The ring ~jl S is the integral closure of ~jl Rio in ~jlQ [cf. [63], Prop. 4.13], and ~jlQ, being its own ring of quotients since it is artinian, is also the ring of quotients of ~jl Rio. Therefore the ring Sn; = (Vj )p; [cf. (1.7)], being quasilocal and the integral closure of a one-dimensional local CM-ring, is a discrete Manis valuation ring [cf. (2.11)(4)]. (3) For every i ~ io we have (Ri+1)n; nRi = (Ri+1)n;nRi+l. By (5.1) we see that (Ri+1)njnRi is the blow-up of (Ri)njnRp and therefore, by (4.8), we have (Vj)Pj = Ui>io (Ri)njnRi· (4) Assume-that S is a finitely generated R-module. Then S is a noetherian ring, and therefore (Vjhj is a local ring whose maximal ideal is generated by one element, hence it is a one-dimensional regular local ring, and therefore it is a discrete valuation ring [cf. 1(3.29)]. (5.5) BRANCH SEQUENCE: Let (Ri)i'~O be the sequence of blow-ups of R, let n be a maximal ideal of S, set V := S[n] , and let P be the regular maximal ideal of the Manis valuation ring V. If, in particular, S is a finitely generated R-module, then Vp is a discrete valuation ring [cf. (5.4)(4)]. (1) For every i E No let Ai = Ai(n) be the local ring (Ri)nnRi. The sequence (Ai)i>O is called the branch sequence of R along n or along V. Note that, for all large enough i, the ring Vp is the only Manis valuation ring belonging to the ring Ai. One often says: Blowing up separates the branches. This will get a geometric interpretation in chapter VIII, section 1. (2) The sequence (e(Ai), f(Ai»iEN o is called the multiplicity sequence of R along n or along V. If, in particular, f(Ai) = 1 for every i E No, then the sequence (e(Ai»i~O is called the multiplicity sequence of R along V. In the rest of this section, we prove for later use some length formulae.

5 Infinitely Near Rings

99

(5.6) Proposition: Assume that S is a finitely generated R-module. Then S/ R is an R-module of finite length, and we have iR(S/R)

=

L

f(A)· iR(An(A) jA).

AE~(R)

Proof: (1) Assume, at first, that R is local, and let (Ri)i>O be the sequence of blow-ups of R. Since S is a finitely generated R-module, it is clear that S j R is an R-module of finite length [there exists a regular r E R which annihilates Sj R], and therefore we have iR(Sj R)

=L

i R(Ri+1j Ri).

i~O

Fix i E No, let ml, ... , mh be the maximal ideals of Ri, and set Aj = (Ri)mj' = mjAj for j E {I, ... , h}. Note that, for j E {I, ... , h}, we have a canonical isomorphism (Ri+1j Ri)mj = A;j /Aj [cf. (5.1)]j A? is the quadratic transform of the local ring A j . Therefore we have [cf. B(1O.10) 1

nj

h

i R(Ri+1j Ri) = L

f(Aj)iAj (A? jA j ).

j=1

(2) Now assume that R is semilo cal , and let ml, ... , mh be its maximal ideals. Then we have [cf. B(10.9)] h

iR(RHI/Ri)

= L iRmj «Ri+1jRi)mj)

for every i E

No.

j=1

For every j E {I, ... , h} we know that «Ri)m·, )i>O - can be identified with the blowup sequence of the local ring Rmj [cf. (5.1)]j now we get the assertion from (1).

(5.7) Proposition: Assume that R is local with maximal ideal m, and let rEm be regular. Then we have iR(RjRr) =

L

f(A)iA(AjrA).

AE~l(R)

Proof: We have iR(RjRr) = iR(RI/rRd by (3.1) and iR(RI/rRd

=

L AE~l(R)

by B(10.10)(2).

f(A)iA(AjrA)

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II One-Dimensional Semilocal Cohen-Macaulay Rings

(5.8) Corollary: Assume that R is local. Then we have e(R) ~

L

e(A);

AE~,(R)

in particular, we have e(R) of R.

~

e(A) for every local ring A in the first neighborhood

Proof: Set m := m(R). We have e(R) = fR(R mjmRm) by (4.12) and mRm = Rmr for some regular r E R m by (4.3), hence fR(RmjRmr) ~ fRm(RmjRmr) = I:AE~l(R) fA (AjAr) by B(10.9), and we have fA (AjAr) ~ e(A) for every A E 11 1 (R) by (4.13). (5.9) Corollary: Let R be local, let n be a maximal ideal of the integral closure S of R, and let (Aik~o be the branch sequence of R along n. Then (e(Ai)k~o is a decreasing sequence of positive integers. If, in particular, S is a finitely generated R-module, then 1 is the smallest element of this sequence.

Proof: Only the last assertion needs a proof. In this case Sn is a one-dimensional local ring, and since Sn is a localization of the discrete Manis valuation ring V = S[n] with respect to its maximal ideal m(V), the maximal ideal of Sn is a principal ideal, generated by a regular element. Therefore Sn is a discrete valuation ring by 1(3.28).

Chapter III

Differential Modules and Ramification INTRODUCTION: Differential modules and ramification is the subject of this chapter. With regard to the principal properties of differential modules, we refer to chapter 16 of Eisenbud's book [63], and add some more results in section 1. After dealing with norms and traces in section 2, we introduce the notion of formally unramified and unramified extensions of rings R -+ S in section 3 and study, in particular, ramification of pairs of local rings in subsection 3.3, ramification of pairs of quasilocal rings in section 5. Under additional hypotheses for a ring extension R -+ S, the set of prime ideals of R which are unramified in S can be described as the closure in Spec{R) of an ideal, the Noether discriminant ideal, which is generated by discriminants; this is done in section 4. In the last section 6 we prove a theorem of Chevalley and a theorem of Zariski: Let k be a perfect field, and let A be a local k-algebra essentially of finite type. Then A is analytically irreducible and analytically normal. These two results shall play an important role when proving the uniformization theorem VIII{6.9): in the proof of it we use namely VIII{6.4) and VIII{6.3) which rely on Chevalley's result.

1

Introd uction

(1.1) As a source for the main properties of differential modules, we use chapter 16 of Eisenbud's book [63]. Let A -+ B be a homomorphism of rings; OB/A denotes the B-module of differentials of B lA, and dB/A: B -+ OB/A is the differential map. The following easy results should be kept in mind. (1.2) REMARK: (I) Let a: A -+ B be a homomorphism of rings, let M be a B-module, and let D: B -+ M be an A-derivation of B with values in M. (a) The map D: B -+ M is, in particular, an A-linear map of A-modules. furthermore, we have D{a(a)IB) = a(a)D{IB) and D(a(a)IB) = a{a)D(IB) +

101

102

III Differential Modules and Ramification

IBD(o:(a)) for every a E A, hence we have D(o:(a))

=0

for every a E A.

(b) For every n E Z we have D(nb) = D(o:(nlA» = o. (2) Let 0:: A -+ B be a homomorphism of rings, and let D: B -+ B be an Aderivation. The following result is easily proved by induction:

Dn(bc) =

~ (:)DV(b)Dn-V(C)

for every n E No and all b,c E B.

(3) Let 0:: A -+ B be a homomorphism of rings, let M be a B-module and let D: B -+ M be an A-derivation. For every b E Band n E N we have

For every unit b of B we have D(bb- 1)

(1.3)

EXAMPLE:

Let B

= 0, hence we have D(b- 1) = -b- 2 D(b).

= A[T1 , ••• ,Tn ] be a polynomial ring over A.

i E {I, ... , n} we have the partial derivative Di: B -+ B with Di(Tj)

For every

= oijlB; if

n = 1, then we often write F' = D 1 (F) for any FEB, and call F' the derivative of F. The differential module {lBIA is a free B-module of rank n, generated by the elements dBIA (Td, .. . ,dBIA (Tn). We need the following results on extending derivations.

(1.4) RING OF FRACTIONS: (1) Let 0:: A -+ B be a homomorphism of rings, let S be a multiplicatively closed subset of A, let T be a multiplicatively closed subset of B, and assume that o:(S) CT. Then there exists a unique algebra-homomorphism o:~'~: S-l A -+ T-1 B such that i~ 00: = o:~'~ 0 i~ [cf. B(2.3)]. (2)'Let N be a B-module, and let d: B -+ N be an A-derivation. Then there exists a unique S-l A-derivation d: T-1 B -+ T- 1N such that the following diagram B-_~d::!....-_...... N

ih]

]'1

T-1B---.....• T-1N d is commutative. Proof [Existence]: Let b, b' E Band t, t' E T satisfy bjt = b'jt' in T- 1B. Then we have td(b) - bd(t) t'd(b') - b'd(t') in T- 1 N. =

1 Introduction

103

= O.

Indeed, there exists til E T satisfying t"(bt' - tb')

d(t")(bt' - tb')

Then we have

+ t"(t'd(b) + bd(t') - td(b') - b'd(t)) = 0,

whence, by multiplying with l/tt', we get

d(t") (~ t

!!.) + til (td(b) -2 bd(t) _ t'd(b') t

~

~2

and therefore the claim has been proved. This shows that there exists a well-defined map

d(b/t)

= td(b) ;

bd(t)

b'd(t'))

=0

in T- 1 N,

d: T-1 B --* T- 1N, defined by

for all b E B, t E T.

It is clear that dis 8- 1 A-linearj moreover, for all b, b' E B, t, t' E T we have

d(~!!.) t t'

= tt'd(bb') -

bb'd(tt') (tt')2

= !!.d(~) t'

t

~d(!!.) t"

+t

hence d is an 8- 1 A-derivation. Clearly, the above diagram is commutative. [Uniqueness] Let d': T- 1B --* T-1 N be any 8- 1 A-derivation such that d'(b/l) d(b)/1 for every b E B. Let b E B, t E T. Then we have by (1.2)(3)

d'(b/t)

= d'«b/l)(I/t)) = d'(b/l)/t -

=

(b/l)d'(t/l)/t 2 = (d(b)t - bd(t))/t2,

i.e., we have d'(b/t) = d(b/t). (3) In particular, let A = Z, 8 = {I}, B a domain and T = B \ {OJ. Every derivation d: B --* B has a unique extension to a derivation d: Q(B) --* Q(B) of the field of quotients Q(B) = T-1 B of B [cf. (2)].

(1.5) SEPARABLE EXTENSION OF FIELDS: Let K be a field, and let d: K --* K be a derivation. Let L be a finite separable extension of K. Then there exists a unique derivation D: L --* L which extends d. Proof: We have L = K(x) by the theorem of the primitive elementj let I be the minimal polynomial of x over K, and let I' be its derivative. For every polynomial F = EaiTi E K[T] we define Fd:= Ed(ai)Tij it is clear that F t-+ Fd: K[T] --* K[T] is a derivation. We define D(x) := Id(x)/I'(x). Let y E L, and choose FE K[T] with y = F(x)j then we define D(y) := Fd(x) +F'(x)D(x). The map D is well-defined: let F, G E K[T] with F(x) = G(x)j then we have F - G = HI for H E K[T], and therefore we find that Fd(x) + F'(x)D(x) = Gd(x) + G'(x)D(x). It is immediate that D is a derivation of L which extends d, and since 0 = I(x) = Id(x) + f'(x)D(x), it is clear that D is the unique extension of d to L. (1.6) Let A --* B be a homomorphism of rings. Assume that B is an A-algebra of finite type. By [63], Prop. 16.1 and Prop. 16.3, the module of differentials OB/A is

III Differential Modules and Ramification

104

a finitely generated B-module. More precisely, if B = A[ Xl, ... , Xn ), then we have n BIA = BdBIA (X1) + ... + BdBIA(X n ) where dBIA : B -+ nBIA is the canonical derivation. (1.7) NOTATION: Let a: A -+ C be a homomorphism of rings; we say that C is essentially of finite type over A or that a is essentially of finite type if C is A-isomorphic to an A-algebra of the form S-l B where B is an A-algebra of finite type and S C B is multiplicatively closed. (1.8) Proposition: Let A be a ring. Then: (1) If B is an A-algebra essentially of finite type and C is a B-algebra essentially of finite type, then C is an A-algebra essentially of finite type. (2) If B is an A-algebra essentially of finite type and A' is an A-algebra, then B' := B ®A A' is an A'-algebra essentially of finite type. Proof: (1) Let B = S-l B1 and C = T- 1C 1 where B1 (resp. C 1) is an A-algebra (resp. a B-algebra) of finite type and S (resp. T) is multiplicatively closed in B1 (resp. in Ct). Let us write C 1 = B[X1"",Xn)/b as a homomorphic image of a polynomial ring; now, since B[X1"",Xn] = S-1(BdX1, ... ,Xn])' the ideal b is of the form S-lb 1 where b1 is an ideal in B1 [Xl, . .. , Xn]. Therefore we have C 1 = S-l B2 where B2 = B1 [Xl, ... , Xn ]/b 1 is a B1-algebra of finite type. Now B2 is an A-algebra of finite type, and C = T-l (S-l B 2) is an A-algebra essentially of finite type [since T- 1(S-l B 2) can be considered as a localization of B2 with respect to a multiplicatively closed system, cf. [21], Ch. 3, Ex. 3]. (2) If B = S-lC where C is an A-algebra of finite type and SeC is multiplicatively closed, then B' = S-l(C ®A A'), and C ®A A' is an A'-algebra of finite type. (1.9) Proposition: Let A be a ring. Let B1"'" Bh be A-algebras which are essentially of finite type. Then Bl ®A'" ®A Bh and B1 x ... X Bh are A-algebras essentially of finite type. The proof is easy and is left to the reader. (1.10) Proposition: Let A -+ C be a homomorphism essentially of finite type. Then n elA is a finitely generated C-module. In fact, we have C = 1,;-1 B where B is an A-algebra B of finite type, and 1,; C B is multiplicatively closed. Now we have nel A = 1,;-10 BIA , and since OBIA is a finitely generated B-module by (1.6), also OelA is a finitely generated C-module.

2

Norms and Traces

(2.0) In this section A always denotes a ring, and A[ X) is the polynomial ring over A in an indeterminate X.

2 Norms and Traces

2.1

105

Some Linear Algebra

(2.1) In the first part of this section we collect some results belonging to linear algebra. (2.2) LINEAR MAPS OF FREE MODULES: The proof of the following two results is left to the reader. Let q.

°

(5.2) Lemma: Let k be an infinite field, and let AI,"" Am be quasilocal kalgebras. We set A := Al X ... x Am. For i E {I, ... , m} let mi be the maximal

132

III Differential Modules and Ramification

ideal of Ai and let ~i := Admi be the residue field of Ai. We assume that the fields ~l, ... ,~m are algebraic extensions of k, and that the ideals ml, ... ,mm are nil ideals. For i E {1, ... , m} let ki be the separable closure of k in ~i, and let qi ~ [k i : k] be a positive integer. Then: (1) For every a E A we have dimk(k[aJ) < 00. (2) There exists a E A with dimk(k[aJ) ~ ql + ... + qm. If there exists i E {1, ... ,m} with mi -:j:. {O} or with ~i -:j:. ki' then there even exists a E A with dimk(k[aJ) > ql + ... + qm·

Proof: (1) Let a = (al, ... , am) E A = Al X .•• X Am. For every i E {I, ... , m} let gi E k[T] be the minimal polynomial of ai over k, and set g:= gcd(gl, ... ,gm). The kernel of the canonical k-algebra homomorphism k[ T] -t k[ all X ••. x k[ am J is the intersection of the ideals (gl), ... , (gm), hence the ideal (g), and therefore we have dimk(k[aJ) ~ deg(g). (2) Let i E {1, ... , m}j for ai E Ai we denote by iii the image of ai in ~i. There exists ai E Ai such that, letting gi be the minimal polynomial of iii over k, the minimal polynomial of ai over k has the form g? with ri EN, deg(gri) ~ qi, and deg(gri) > qi if mi -:j:. {O} or if ~i -:j:. ki [cf. (5.1)]. For every Ci E k we have k[iii + Ci] = k[iii]. Since k is an infinite field, we may assume, replacing iii by iii + Ci with Ci E k, and denoting the minimal polynomial of iii + Ci over k again by gi, that the irreducible polynomials gl, ... , gm are pairwise different, hence, in particular, pairwise coprime. Then the polynomials g?, . .. ,g';,,"' are pairwise coprime. Setting a := (al, ... , am), we have, by the Chinese remainder theorem [cf. B(10.1) J, k[ a] = k[ ad x ... x k[ am]. This proves the assertion. (5.3) REMARK: The following theorem shall be used in the sequel only in the case where S is quasilocalj in this case the proof works even if the residue field of R is finite. Nevertheless, we shall prove the theorem in its full generality, making use of an extension of the ground field in case R has finite residue field. (5.4) Theorem: [Krull] Let R be a quasilocal integrally closed domain, let K be the field of quotients of R, and let L be a finite extension of K. Let S be a subring of L which contains R and is integral over R. Let m be the maximal ideal of R, and let nl, ... ,nm be the finitely many maximal ideals of S. Then: (1) We have [Slnl : RlmJsep + ... + [Slnm: RlmJsep ~ [L: KJ. (2) The equality sign in (*) holds iff there exists a K -basis {Xl, ... , xn} of L contained in S with D L / K (Xl, ... , Xn) E R \ m. If this is the case, then S is the integral closure of R in L, and it is a finitely generated free R-module.

Proof: First, we consider the case that k := Rim is an infinite field. For yES we denote by y the image of y in SimS. (1) Set n:= [L : K]. Let S' be the integral closure of R in L. Now S' has at most n maximal ideals by B(7.3), and for every maximal ideal n' of S' the field S'/n'

5 Ramification For Quasilocal Rings

133

is an algebraic extension of k by B(7A). The number m of maximal ideals of Sis therefore ~ n. By B(10.21) we have a primary decomposition mS = ql n ... n qm where qi is ni-primary for i E {I, ... ,m}. We set A := SimS, Ai := Slqi for i E {I, ... , m}. Note that Ai is quasilocal with maximal ideal ni I qi which is a nil ideal, and has residue field Sini which is algebraic over k. Then A = Al x··· x Am as k-algebras [by the Chinese remainder theorem]. For every a E A we have dimk(k[a]) ~ n [ef. B(7A)], hence we get (*) by using the first part of (5.2)(2). (2) First, we assume that we have equality in (*). By the second part of (5.2)(2) we see that nilqi = {O}, and that Ai is a field which is a finite separable extension of k for every i E {I, ... , m}. Therefore Ai is unramified over k [cf. (4.3)], hence A is unramified over k [ef. (3.8)(1)], hence A = k[ a] [cf. (3.13)], and therefore D A/k (1, a, ... , an-I) =F 0 [cf. (4.2)]. Since {I, a, ... , an-I} is a k-basis of A, there exists a unique monic polynomial f E k[T] of degree n with f(a) = 0; remember that dis(f) = DA/k(I,a, ... ,an- I ). We choose xES with x = a, and we let F E K[T] be the minimal polynomial of x over K; we have deg(F) ~ n. Now F E R[ T] since R is integrally closed [cf. (2.12){2)]' hence dis (F) E R. Let F be the polynomial which we get from F by reducing its coefficients modulo m. We have F(a) = 0, hence F = f and deg(F) = n. Using B(1O.22)(3) we see that the image of dis(F) in k is dis(f), hence that dis (F) = DL/K(I, X, ... , x n- l ) is a unit of R. Second, we assume that there exists a K-Basis {Xl' ... ' Xn} of L which is contained in S, and such that DL/K(XI, ... ,xn) is a unit of R. Then S = RXI + ... + Rx n, and S is the integral closure of R in L. In fact, we have RXI + ... + RX n C S. Let x ELlie not in the R-module RXI + .. ·+Rxn. Then we have x = alXI + .. ·+anxn with aI, ... , an E K, and not all of these elements lie in R. By relabelling, we may assume that al f/. R. We have D L / K (X,X2, ... ,Xn) = a~DL/K(XI,X2, ... ,Xn) [ef. (2.13)(3)], hence DL/K(X, X2, .. . , xn) f/. R [since DL/K(Xb X2, . .. , xn) is a unit of R], and this shows that x is not integral over R. Therefore S = RXI + ... + RX n is the integral closure of R in L. In particular, A = SimS is a quasisemilocal kalgebra with dimk(A) = n, and D A/k(XI, ... , xn) is the image of DS/R(XI, ... , xn) [ef. (2.13)(7)], hence D A/k(Xb ... , xn) =F O. This implies that A is unramified over k [cf. (4.2)], hence that A is a direct product of fields which are finite separable extensions of k [ef. (3.13)], hence that Ai is a field which is a finite separable extension of k for i E {I, ... , m} [ef. B(8.2)], and therefore we have [Ai: k ]sep = [Ai: k] for i E {I, ... , m}, hence we have equality in (*). Now we consider the case that k is a finite field. We use the trick of adjoining an indeterminate [cf. B(lOA)]. We set K' := K(X), L' := L(X); note that [L' : K'] = [L: K]. We set ~ := R[X] \ mR[X], R' := ~-IR[X], m' := mR', S' := ~-l S[ X] and ni := niS'. Then R' is integrally closed, and S' is integral over R'; moreover, S is integrally closed iff S' is integrally closed [ef. [63], Ex. 4.18 and Prop. 4.13]. Furthermore, R' is quasilocal with maximal ideal m', and S' is quasisemilocal with maximal ideals n~, ... , n~: if ')1 is a prime ideal of S[ X] with ')1::) niS[X] for some i E {I, ... ,m} and 'J1n~ = 0, then 'J1nR[X] = mR[X], hence 'J1 = niS[X] by [63], Cor. 4.18. Therefore S' is also the localization of

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III Differential Modules and Ramification

S[X] with respect to the complement of nIS[X] U··· UnmS[X]. Since S -+ S' is faithfully flat [cf. B(lOA)], we have mS' = q~ n· .. n q~ [cf. [34], Ch. I, §3, no. 5, Prop. 10] where, for i E {I, ... , m}, q~ := qiS' is n~-primary. We set k' := R' /m'j for i E {I, ... , m} we set li := S/ni, l~ := S' /n~, and denote by ki the separable closure of k in li. Then k' ~ k(X), and, for i E {I, ... , m}, l~ ~ li(X) and ki(X) is the separable closure of k' in l~. Note that [ki : k] = [k~ : k'] for i E {I, ... , m}. Setting A' := S' /m' S, A~ := S{ / q~, ... , A~ := S' / q~, we have A' = A~ x ... x A~ as k'-algebras, and we have A' = A ®k k(X), A~ = Ai ®k k(X) for i E {I, ... , m}. The assertions of the theorem hold for R' and S' by (1), hence also for Rand S. (5.5) Corollary: Hin (*) we have equality, then we have mSni = niSni and S/ni is a finite separable extension of R/ m for every i E {I, ... , m}. In particular, if R is local , then S is unramified over R. Proof: Using the notation of the proof of the theorem, we have, for i E {I, ... , m}, ni = qi, hence mSn, = niSn, and S/Ui is a finite separable extension of R/m. If R is local, then ni is unramified over A [cf. (3.14)], hence S is unramified over R [cf. (3.2)(5)].

6

Integral Closure and Completion

(6.1) NOTATION: Let R be a semilocal ring with maximal ideals ml, ... , mh, and let t = ml n ... n mh be its Jacobson radical. Set Ri := Rm, for i E {I, ... , h}. Let R be the t-adic completion of R. Generalizing the definition in 11(3.16), we say that R is analytically unramified if R is reduced. (1) We have R = RI X ••• X Rh [cf. B(8A)(I)]; hence R is analytically unramified iff the local rings RI, . .. , Rh are analytically unramified. (2) Assume that R is analytically unramified, and let pi, ... , p; be the minimal prime ideals of R. Then we have Q(R) = Q(R)/PiQ(R) x ... x Q(R)/p;Q(R) [cf. B(3.4)]j let eI, ... , er be the elementary idempotents of Q(R) corresponding to this decomposition. Clearly these idempotents are integral over R. If R is inte~ally closed, then we have el, ... ,er E R, hence R = RIPi X· .. x Rlp;j as the rings Rlpi. are local [being complete semilocal integral domains] and R is decomposable [ d. B(8A)], we have r = h and, after relabelling, pi emiR, R/pi = Ri and ~ is an integrally closed domain for i E {I, ... , h} [d. B(8.2)]. Conversely, if RI , ... , Rh are domains which are integrally closed, then R is integrally closed [d. B(3.3)]. (6.2) DEFINITION: A local ring R is called analytically normal if it is an integrally closed domain and if its completion is an integrally closed domain. (6.3) EXAMPLE: Let R be a regular local ring. Then R is analytically normal. In fact, the completion R is regular [since dim(R) = dim(R) by [63], Cor. 10.12, and

135

6 Integral Closure and Completion

emdim(R) = emdim(R) by [63], Th. 7.1], and a regular ring is a factorial domain [cf. [63], Th. 19.19], hence, in particular, is integrally closed [cf. [63], Prop. 4.10]. (6.4) Proposition: Let R be a semiloca.l ring, R its completion, and a an ideal of R. Let aR = qi n ... n qh be an irredundant primary decomposition of aR, and set pi := rad(qi) for every i E {I, ... , h}. Then a = (qi n R) n··· n (qh n R) is a primary decomposition of a. Let i E {I, ... , h}; then we havepinR = rad(qinR), and there exists a prime ideal Pi E Ass(Rla) with pi nRc Pi. Proof: Let i E {I, ... , h}. Obviously qi n R is a primary ideal of R and pi n R is its prime ideal. Since R is a faithfully flat R-module [cf. [34], Ch. III, § 3, no. 5, Prop. 9], we have aR n R = a [cf. [34], Ch. I, § 3, no. 5, Prop. 10], and therefore a = (qi n R) n ... n (qh n R) is a primary decomposition of a. Since RI aR is the completion of Ria [cf. [63], Lemma 7.15], every regular element of Ria remains a regular element in RI aR [note that an element a of a ring A is regular if the multiplication map aA: A -+ A is injective]. Let i E {I, ... , h} and x E pi n R. The image of x in RlaR is a zero-divisor [cf. [63], Th. 3.1], hence the image of x in RI a is a zero-divisor, and therefore x is contained in the union of the prime ideals in Ass(Rla). This implies that pi n R is contained in a prime ideal p E Ass(Rla) [cf. [63], Lemma 3.3]. (6.5) Corollary: If, in (6.4), a is a prime ideal, then a everyi E {I, ... ,h}. Proof: We have a C

= qi n R = pi n R

for

qi nRc pi nRc a [since {a} = Ass{Rla)].

(6.6) Proposition: Let R be a semilocal domain, R its completion, and let p be a prime ideal of R such that Rp is a discrete valuation ring and that Rip is analytica.11j' unramined. Then an irredundant primary decomposition of pR has the form pR = pi n ... n Ph where pi, ... 'Ph are prime ideals of R. Moreover, for every i E {I, ... , h} we have: (a) the ring Rpi is a discrete valuation ring, (b) for every x E R with xRp = pRp we have xRpi = pi Rpi . Proof: Set E := R \ Pi then we have E- 1R = Rp. Since RlpR is reduced, the ideal pR admits an irredundant intersection as stated, and pi n R = P for every i E {I, ... ,h} [cf. (6.5)]. This implies, in particular, that Enpi = 0 for every i E {I, ... , h}. Since Rp is a discrete valuation ring, there exists x E R with xRp = pRp. Using the commutative diagram

R

I I

R

•

E-1R--E-1R

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III Differential Modules and Ramification

yields X(E-l R) = p(E-l R). On the other hand, p(E-l R) = E-lpin·· ·nE-lph is an irredundant primary decomposition of p(E-l R) [cf. [63], Th. 3.10]. Therefore, by localizing, we get xRPi = PiRpi for every i E {I, ... ,h}. Now x is a regular element of Rpi since x is a regular element of R, hence R pi is a local ring whose maximal ideal is generated by a regular element, and therefore R pi is a discrete valuation ring [cf. 1(3.29)].

(6.7) Let R be a ring, let Q(R) be its ring of quotients, and let s, t E R be such that t is not a zero-divisor in R and that (Rt : RS)R = Rt. Let z E Q(R)j if sz E R and tz E R, then we have z E R. Proof: Since szt E Rt, we have zt E (Rt : RS)R = Rt, hence zt = zIt for some Zl E R. Since t is not a zero-divisor of R, we have z = z' E R. (6.8) Lemma: Let R be an integrally closed semilocal domain, R its completion, and Q(R) the ring of quotients of R. Assume that there exists a non-unit t E R\ {OJ such that Rip is analytically unramified for every prime ideal p E Ass(RI Rt). Let z E Q(R) be an element with tz E R. If z is integral over R, then we have z E R. Proof: (1) Let Ass(RI Rt) = {PlI .. " Ph} and set E := R\ (PI U··· UPh). We have ht(Pi) = 1 for every i E {I, ... , h} [cf. [63], Th. 11.2]. Now E- l R is a noetherian integrally closed domain [cf. [63], Prop. 4.13], and E-lpl, ... ,E-lph is its set of prime ideals of height 1 which is also the set of maximal ideals of E-l R. Therefore we have E-l R = RPl n ... n RPh [note that E-l RE-l Pi = RPi for i E {I, ... , h} ] and the rings R p1 , ... ,RPh are discrete valuation rings [cf. B(1O.5)], hence E- l R is a semilocal principal ideal domain having E-lPl,"" E-lPh as set of non-zero prime ideals [cf. 1(7.2)], and therefore there exist elements xL ... , xh E E-l R with (E- l R)x~ = Pi(E- l R) for i E {I, ... , h}. For i E {I, ... , h} we can write x~ = xi/s with Xl, ... ,Xh E Rand sEE. Let, for i E {I, ... ,h}, Vi be the valuation of Q(R) defined by R pi . Since sEE, we have Vl(S) = ... = Vh(S) = 0, hence we have VI(Xi) = du for alII, i E {I, ... , h}. For i E {I, ... ,h} we setei:= viet). We show that X~l · .. x~hlt E R. By [63], Cor. 11.4, it is enough to show that X~l ... X~h It E Rp for every prime ideal P of R of height 1. Thus, let P be a such a prime ideal. If P E Ass(RIRt), then clearly X~l ... X~h It E Rp, and if P ~ Ass(RI Rt), then t ~ P, hence again xp ... X~h It E Rp. Now we can write X~l ... X~h = ts with S E Rand VieS) = 0 for i E {I, ... , h}, which implies that sEE. (2) Let i E {I, ... , h}. From (6.6) we get an irredundant primary decomposition RPi = pil n ... n Pik( i) where pil, ... , Pik( i) are prime ideals of R and pij n R = Pi for every j E {I, ... , k(i)} [note that RlpiR is reduced by assumption]. Moreover, R pij is a discrete valuation ring for every j E {I, ... , k(i)}j let Wij be the discrete valuation of Q(R p!) defined by Rp! .• Let j E {I, ... , k(i)}. The image in RPi; of a regular element of R is a regular element, hence the canonical homomorphism R -t Rp!.., extends to a homomorphism

.

~

6 Integral Closure and Completion

137

'Pij: Q(R) -+ Q(Rpi) of rings of quotients. The map Wij := Wij 0'Pij: Q(R) -+ Zoo clearly is a Manis valuation of Q(R) [cf. 1(2.8), and note that Q(R) is a ring having large Jacobson radical, cf. 1(1.11)(2)]. (3) Let i E {I, ... , h}, j E {I, ... , k(i)}. Since xiRpi = PiRpn we have Wij(Xi) = 1 for every j E {I, ... , k(i)} by (6.6), and for I E {I, ... , h}, If; i, we have Xi ~ PI, hence Xi ~ pij , and therefore we have Wlj(Xi) = 0 for every j E {I, ... , k(l)}. From this we get, by (1), that Wij(t) = ei for all i E {I, ... , h} and j E {I, ... ,k(i)}. (4) Let z E Q(R) be integral over R with tz E R. We have to show that z E R. Now t is a regular element of R. For every s E ~ we have (Rt : RS)R = Rt [cf. [63], Th. 3.1], hence (Rt: Rs)ii. = Rt [note that (Rt: RS)R ~ RsI(RsnRt), and use [34], Ch. I, § 3, no. 5, Prop. 10]. We know that X~l ... x~" = ts with s E ~ [cf. (1) ]; by (6.7) it is enough to show that (s' z)t E RX~l ... x~" for some s' E ~. (a) Let i E {I, ... , h}, j E {I, ... , k(i)}. Now z E Q(R) is integral over R by assumption, hence 'Pij(Z) E Q(Rpi) is integral over 'Pij(R) C R pij , and therefore we have 'Pij(Z) E R pij [note that a discrete valuation ring is integrally closed]. This implies that Wij(Z) ~ o. (b) Since tz E R, we can write tz = X{l ..• x,"y with y E Rand (h, ... , Ih) E ~. We consider the case that there exists i E {I, ... , h} with Ii < ei. For every j E {I, ... , k(i)} we have

Ii + Wij(Y) = Wij(X{l ... x,"y) = Wij(tZ) = ei + Wij(Z),

(*)

hence Wij(Y) ~ 1 [cf. (a)], i.e., we have 'Pij(Y) E pijRpij' and therefore we have

Yi E prj [Since 'Pi;t(pijRpi) = Pi'j]. We have, on the one hand, Xi(~-l R) pi(~-l R), hence Xi(~-l R) = pi(~-l R), and, on the other hand, pi(~-l R) ~-lpil n··· n ~-lp:,k(i) [cf. [63], Th. 3.10]. Therefore there exist s' E ~, Y' E with s'Y = XiY', hence we have S'zt -- xlt I

...

= = R

Xli-lXIi+1xli+l i-I i i+1· .. xl"y' h •

Note that Wij(S') = 0 for all j E {I, ... , k(i)}. If Ii + 1 < ei, we may continue the procedure just described until we arrive at a representation

S"zt -- xlt I

...

XIi-lXlxIHl i-I i i+1

••

·xl"y" h

with I > ei and with elements s" E ~ and y" E R. Therefore we may assume that the;e exist E ~, fj E R and natural integers h, ... ,Ih with Ii ~ ei for every - = XlIt ... x 1,,l. E {I , ... , n } sueh t h at tzs h y.

s

(6.9) Theorem: Let R be an integrally closed local domain, K its field of quotients, L a finite separable extension of K, and S the integral closure of R in L. The ring S is semilocal. Assume that Sip is analytically unramified for every prime ideal p of S of height 1. If R is analytically normal, then S is analytically

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III Differential Modules and Ramification

unramified, the completion S of S is a finite torsion-free R-algebra which is integrally closed in its ring of quotients, and it is the integral closure of R in Q(S). Furthermore, the local rings Sn, n maximal ideal of S, are analytically normal. Moreover, we have dimQ(il) (Q(S)) = [L: K].

Proof: (1) First, note that S is a finitely generated R-module by (4.5)(2), hence it is semilocal, and its natural topology and the mS-adic topology [m being the maximal ideal of R] are identical [cf. B(3.8)]. We choose xES such that L = K(x)j the minimal polynomial f of x over K has coefficients in R [ef. (2.12)(2)]. With d := D L / K (I,x, ... ,xn- 1 ) [n := [L : K)) we have dS c R + Rx + ... + Rxn- 1 = R[x] C S [ef. (4.5)(3)]. From dS C R[x] c S we see that Sand R[x] have the same ring of quotients, hence that Sand R[ x] have the same integral closure in Q(S), say T [note that d is a regular element in S since S ~ S is fiat, cf. [34], Ch. III, § 3, no. 5, Prop. 9, and that S is a finitely generated R-module, hence is integral over R by [63], Cor. 4.5]. Clearly f E Q(R)[ X] is a separable polynomial, hence Q(R)[x] = Q(R)[X]/(f) is an unramified Q(R)-algebra [ef. (3.13)]. Now we have dT C R[x] by (4.5)(3) [because R is an integrally closed domain by hypothesis], hence we have dT C S. Since S is an integrally closed noetherian domain, every prime ideal in Ass(S/Sd) has height 1 [ef. [63], Th. 11.2]; the assumption on S shows that we can apply (6.8), and we get T = S. (2) Let yES be a nilpotent element, and let rEm be a regular element. Then y/r n E Q(S) is integral over S for every n E N, hence we have y/r n E Sand y E mnS. Since S is a finitely generated R-module, we get y = 0 by Krull's intersection theorem [ef. [63], Cor. 5.4]. Thus, S is a reduced ring, and Sn is analytically normal for every maximal ideal n of S [cf. (6.1)(2)]. (3) Since R[ x ] is a free R-module of rank n, we see that R[ x] = R[ x ] ®RR is a free R-module ofrank n. Since Q(R[x)) = R[x] ®il Q(R), we get dimQ(il) (Q(S)) = n since Q(S) = Q(R[x)). (6.10) Corollary: We have nS/RR = ns/il and branch(S/R)

= V(ns/il).

Proof: We set L' := Q(S), K' := Q(R). Clearly we have S ®il K' = L'. From (2.10) we get nS/R C ns/il, hence nS/RR C nS/R- On the other hand, we have R ®R

I't S = 1\n(R ®R S) = 1\n S

[cf. [63], Prop. A2.2], showing that ns/il C

nS/RR. The last assertion follows from (4.12).

(6.11) REMARK: The following two results in (6.12) and (6.13) are needed for the proof of theorem (6.14). In (6.12) we mention only those results on homogenization which are needed in the sequel. For the notion of separable generated fields of algebraic functions and p-bases the reader should consult appendix Al of Eisenbud's book [63].

6 Integral Closure and Completion

139

(6.12) HOMOGENIZATION: Let k be a field, and let R = k[Xl, ... ,Xn ] be an integral k-algebra of finite type. Let K be the field of quotients of R. Set r := tr. dk(K) = dim(R); note that ht(m) = r for every maximal ideal m of R [cf. [63], Th. A on p. 286]. (1) Let Yo be transcendental over K, set Yi := XiYO for i E {I, ... , n} and S := hR:= k[yo, ... ,Yn]; note that dimeS) = tr.dk(Q(S)) = r+ 1 [cf. [63], Th. A on p. 286]. Let p E No, and let Sp be the k-vector space generated by all power products y~Oy~l ... y~" with io + ... + in = p. Then the sum S = L Sp is direct:

In fact, if Lp Lio+",+;,,=p c~:~...• i"y~O ... y~" = 0 with c~:~...•i" E k, then we get . c~p). Xii ... xi" = 0 hence". . c~p). Xii ... xi" = "L..Jp y p ". L..JIO+···+I,,=p '0 ..... 1" 1 n , L..J10+·"+I,,=p 10 ..... 1" 1 n

°

o [since Yo is transcendental over K],

hence Lio+"+i"=pC~:~.... i,,y~Oy~l .. ,y~" = 0 for every p E No. We set S+ := Lp>o Sp; S+ is a homogeneous prime ideal of S. (2) For any 9 E S we set ag := g(I,Xl, ... ,xr ) E R; note that 9 1-+ ag : S -t R is a surjective homomorphism of rings. For every homogeneous ideal b of S we denote by a b the image of b in R under this homomorphism, and for every ideal a of R we denote by ha the homogeneous ideal of S which is generated by all homogeneous elements 9 E S with agE a. Then: (a) For every ideal a of R we have a(ha) = a. (b) Let a, b be ideals of R. If a ~ b, then we have ha ~ hb. (c) If 13 is a prime ideal of R, then hp is a prime ideal of S. (d) If m is a maximal ideal of R, then we have ht(hm) = r. Proof: (a) Clearly we have a(ha) C a. Let I E a; then there exists a homogeneous 9 E S with ag = I, hence we have 9 E ha and therefore we get I E a(ha). (b) This follows immediately from the definition. (c) Let gl, g2 E S be homogeneous with g1g2 E hI'. Then a(glg2) = ag1 ag2 E 13, and therefore gl or g2 lies in hp, hence hI' is a prime ideal [cf. B(4.1)]. Let {O} =: Po C PI C ... CPr := m be a strictly increasing chain of prime ideals of R. Then {O} C hpl C ... C hPr = hm C S+ is a strictly increasing chain of prime ideals of S, and since dimeS) = r + 1, we see that ht(hm) = r.

(6.13) Lemma: Let k be a field, let R be an integral k-algebra DEfinite type, and let m be a maximal ideal DE R. IE K := Q(R) is separably generated over k, then there exists a transcendence basis {ZI, ... zr} DE K / k contained in m such that (i) K is a finite separable extension DE k(ZI,'" ,zr), (ii) R is a finitely generated k[ ZI, ... , Zr ]-module, (iii) mnk[ZI"",Zr] = (ZI,,,.,Zr). Proof: If k has characteristic 0, this is just Noether normalization [cf. [63], Th. 13.3]. We consider the case that k has positive characteristic p. We write R = k[ Xl, ... , Xn]. We adjoin a transcendental element Yo to K, and set L := K(yo), Yi := XiYO for i E {I, ... , n}, S := k[yo, ... , Yn]; note that S is a graded ring by (6.12), and that L is separably generated over k. It is enough to show: There exist r homogeneous elements (1, ... , (r of S such that, setting (0 := Yo, we have the following: {(o, (1, ... ,(r} is a separating transcendence

140

III Differential Modules and Ramification

basis of Ljk, (l" .. ,(r E h m, and S is integral over k[(o, ... ,(r]. In fact, if this is proved, then setting Zi := a(i for i E {I, ... r}, we have: Zl, ... , Zr lie in m, hence m n k[ Zl, ... , zr] = (Zl,"" Zr), R is integral over k[zl, ... , zr] [since 9 1-+ a 9 : S --t R is a surjective homomorphism] and K is a finite separable extension of k(Zl"",Zr) [note that Zi = (;/y~iwith deg((i) =: mi for i E {I, ... ,r}, k((O,(l, ... ,(r) = k(YO,Zl, ... ,zr), and since L = K(yo) is a finite separable extension of k(Zl, . .. , zr)(YO), also K is a finite separable extension of k(Zl, ... , zr)]. Now S(o is a prime ideal of S, ~O,l := S(o is not contained in h m , we have ht(S(o) = 1 [by Krull's principal ideal theorem, cf. [63], Th. 10.2] and {(o} is a p-independent set over k [note that kLP = kKP((C)], We construct, for j E {2, ... ,r + I}, homogeneous elements (1," .,(j-l of hm such that, setting mj-l := S(o + ... + S(j-l, we have: ht(mj-d = j, h m does not contain any of the isolated prime ideals ~j-l,i' i E {I, ... , hj-d, of mj-l, and {(o, ... , (j-d c L is a p-independent set over k. Assume that j E {I, ... , r}, and that (0, (1, ... , (j-l have already been constructed. We choose homogeneous elements u, v E S with u ¢ Ui \Pj-l,i, U E h m [by prime avoidance, cf. [63], Lemma 3.3], v E h m, v E ni ~j-l,i' and v ¢ kLP( (0, (1, ... , (j-d [the existence of v follows from the fact that the elements of every non-zero ideal of S generate Lover k, ]. We write deg(u) = pas, deg(v) = pbt with a, bE No, pf sand pf t. We set (j := {

u tPb _ a u tp

+ VS

+ (C

if a

a+l-b

VS

< b,

if a ~ b.

Then (0,"" (j satisfy also the above conditions. In fact, (j is homogeneous. With L j- l := kLP((o, ... ,(j-d we have [Lj-l[v] : Lj-d = p [cf. [63], p. 559, first paragraph after the definition], hence [L j - l [ V S ] : L j - l ] = p since p f s, and therefore {(o, ... , (j} C L is a p-independent set over k. We have ht(mj) = j + 1 [by Krull's principal ideal theorem, d. [63], Th. 10.2, since (j ¢ Ui ~j-l,i]. We have rad(mj_l) = n ~j-l,i' rad(mj) = n ~j,i' Since mj-l C mj, every ideal ~j,i' i E {I, ... , h j }, contains an ideal ~j-l,il for some i' E {I, ... , hj-d. Therefore none of the ideals ~j,i is contained in h m. We can continue the construction, and find elements (0, (1, ... , (r as desired. Since ht(mr ) = r + 1 = dim(S), mr C S+ and mr is homogeneous, we have rad(mr) = S+, and therefore S is integral over k[(o'''',(r] [ef. B(4.34)], hence {(0, ... , (r} is a transcendence basis of L j k. Since {(o, ... , (r} is a p-independent set of Ljk, it must be a p-basis of Ljk [otherwise we could enlarge this set, in contradiction with the fact that in our case a p-basis is a transcendence basis, ef. [63], Cor A.1.5], and therefore {(o, ... , (r} is a separating transcendence basis of Ljk, again by [63], Cor. A.1.5.

(6.14) Theorem: Let k be a perfect field, and let A be an integral local k-algebra essentially of finite type. Then: (1) [Chevalley] The k-algebra A is analytically unramified. (2) [Zariski] The k-algebra A is analytically normal.

6 Integral Closure and Completion

141

°

Proof: (a) We show (1) and (2) by induction on r := dim(A). The case r = is trivial. Let r > 0, and assume that (1) and (2) hold for every integral local k-algebra essentially of finite type having dimension < r. By (6.1) this means, in particular: If S is a semilocal domain of dimension < r, and if, for every maximal ideal n of S, Sn is an integral local k-algebra essentially of finite type, then S is analytically unramified. (b) Let A be an r-dimensional integral local k-algebra essentially of finite type with maximal ideal m and field of quotients L. We show that there exists a subfield k' of L such that L is separably generated over k' and that A is the localization of a k' -algebra of finite type with respect to a maximal ideal. There exist elements Xl, ... , Xn in A such that setting To := k[ Xl, ... , xn], A is a localization of To with respect to the prime ideal m n To of To. Set s := tr. ddL). Since ht(m n To) = r, we have dim(To/(mn To)) = s - r [cf. [63], Th. A on p. 286 and Cor. 13.2]. Since L is separably generated over k [theorem of F. K. Schmidt, ef. [63], Cor. A1.7], we can apply (6.13) to To (and an arbitrary maximal ideal of To) and get: There exist Ul, ... , Us E To such that, setting Ro := k[ Ul, ... , Us ], To is a finitely generated Ro-module, and L is a finite separable extension of Ko := Q(Ro). Set Ii := Ro/(m n Ro) = k[Ul,"" us] where Ui is the image of Ui in Ro for i E {I, ... , s}. We have dim(Ro) = dim(To/(m n To)) since To/(m n To) is integral over Ro [ef. [63], Prop. 9.2], hence tr. dk(Q(Ro)) = s - r [ef. [63], Th. A on p. 286]. By relabelling, we may assume that {Ul,' .. ,us - r } is a transcendence basis of Q(Ro) over k. We set ~ := k[ Ul,.'" u s- r ]\ {O}. Note that ~n(mnRo) = 0. We set k' := ~-lk[Ul,,,,,Us_r] = k(Ul,""U s- r ). Then R':= ~-lRo = k'[u s- r+l,""us ], and since R' / (mnR') = k' [u s - r +1, ... , us], the ideal mnR' is a maximal ideal of R' [the integral affine k'-algebra R" := R' /(mnR') is a field since tr. dk' (Q(R")) = 0, hence dim(R") = 0, hence R" is artinian]. We set T := ~-lToj then T is a k'algebra of finite type, it is integral over R' [cf. [63], Prop. 4.13], m nTis a maximal ideal of T [ef. [63], Cor. 4.17], and A = TmnT. Note that Ko is a purely transcendental extension of k', hence L is separably generated over k', and that tr. d k , (L) = r. (c) We apply (6.13) to the k'-algebra T and the maximal ideal m n T of T. This yields the existence of elements Zl, ... , Zr E T such that, setting R := k'[Zl, ... , Zr], we have m n R = RZl + ... + Rzr, a maximal ideal of R, L is a finite separable extension of K = Q(R) and T is a finitely generated R-module. Considering a primary decomposition of (mnR)T, we easily see that q := (mnR)A is an m-primary ideal of A, hence that there exists lEN with ml C q. (d) Let S be the integral closure of R in Lj note that T C S. Then S is a finitely generated R-module by (4.5){2), and dim(8) = r [ef. [63], Prop. 9.2], hence dim(Sn) = r for every maximal ideal of S [ef. [63], Th. A on p. 286]. Moreover, S is a k'algebra of finite type, and k' is a k-algebra essentially of finite type, hence S is a k-algebra essentially of finite type [ef. (1.8)]. Let n be a maximal ideal of S. Then n n R is a maximal ideal of R [cf. [63], Cor. 4.17], and since R' := RnnR is a regular local ring [cf. [63], Cor. 19.14], it is analytically normal by (6.3). Set 8' := SnnRj S' is the integral closure of R' in L [ef. [63], Prop. 4.13],

142

III Differential Modules and Ramification

and since it is a finitely generated R'-module, it is semilocal [ef. B(3.8)]. Let n' be any maximal ideal of S'. Then n' n S is a maximal ideal of S by [63], Cor. 4.14 [since (n'nS)nR = (n'nR')nR = nnR], and we have Sn'ns = S~" hence S~, is an r-dimensional integral integrally closed local k-algebra which is essentially of finite type. Let p' be a prime ideal of S' with ht(p') = 1, and let n' be a maximal ideal of S' containing p'. Now dim(S~,/p'S~,) < r, and since S~,/p'S~, is an integral local k-algebra essentially of finite type, it is analytically unramified by our induction hypothesis. Since S~,/p'S~, ~ (S'/p')n'/p" we get from (6.1) and (6.9): S'is analytically unramified, and S~, is analytically normal for every maximal ideal n' of S', hence, in particular, we have dim(Sn) = rand Sn is analytically normal. (e) We set B := A[S], the smallest subring of L containing A and S. Since S = R[ t 1, ... , th] with elements h, ... , th which are integral over R, we see that B = A[ h, ... , th], hence that B is a finitely generated A-module, and therefore we have dim(B) = r [ef. [63], Prop. 9.2]. Moreover, B is a semilocal domain and a power ~ of the Jacobson radical t of B is contained in mB [cf. B(3.8)]. Let n be a maximal ideal of B. Then neBn C mBn C nBn, hence nle Bn C ml Bn C qBn C nBn, and therefore qBn is nBn-primary. The elements tl, ... , th are, a priori, integral over T; setting T := T[ h, ... , th], we see that T is integral over T, dim(T) = r and that B = (T)mnT' We have n n A = m, hence n n T lies over the maximal ideal m n T of T and it is therefore maximal, hence ht( n n T) = r, and since Bn = T nnT , we get dim(Bn) = r. Now (nnS)nR = (nnA)nR = mnR, hence n n S is a maximal ideal of S, and (m n R)S C n n S, hence (m n R)Bn = qBn C (n n S)Bn C nBn, and therefore (n n S)Bn is an nBn-primary ideal of Bn. Note that Snns is analytically normal by (c). We consider the local homomorphism Snns --t Bn; note that (n n S)Bn, the ideal of Bn generated by the maximal ideal (n n S)Snns of Snns, is an nBn-primary ideal of Bn. From B(9.6) we therefore get Snns = Bn. This being true for every maximal ideal n of B, we see that B is analytically unramified [ef. (c) and (6.1)(1)]. Since the completion of A can be considered as a subring of the completion of B with respect to its Jacobson radical, we see that A is analytically unramified. Thus, we have shown (1). From Snns = Bn we see that Bn is integrally closed; this being true for every maximal ideal of B, we get that B is integrally closed [cf. B(2.5)]. Therefore B is the integral closure of A. Hence, if A is integrally closed, we have B = A, hence B is local with maximal ideal n, and since B = Snns, B is analytically normal by (c). This proves (2), and the theorem. (6.15) Corollary: Let k be a perfect field, let A be an integral local k-algebra essentially of finite type, and let A be its completion. If A is integrally closed, then A is integrally closed. Proof: A is a domain and An Q(A) = A [by [34], Ch. I, § 3, no. 5, Prop. 10 since A is a faithfully flat A-module]; therefore A is integrally closed. (6.16) REMARK: For more general results concerning the completion of a local domain the reader should consult chapters 14 and 15 of [116].

Chapter IV

Formal and Convergent Power Series Rings INTRODUCTION: In this chapter we prepare the ground for the proof of the JungAbhyankar theorem in section 2 and the study of quasiordinary power series in section 4 of chapter V. We assume that the reader is acquainted with the notion of power series over a field; in section 1, for the convenience of the reader, we give some background, introduce convergent power series over the real and complex numbers in section 2, and prove WeierstraB division and preparation theorem in section 3. The category of formal (resp. analytic) algebras-these are homomorphic images of formal (resp. convergent) rings of power series-is treated in section 4. The last section considers, in particular, the completion of an analytic algebra; these results are needed to prove the convergent case of the Jung-Abhyankar theorem.

1

Formal Power Series Rings

(1.0) In this section R is a ring. (1.1) THE RING OF FORMAL POWER SERIES: (1) The ring offormal power series in indeterminates T 1 , ••• ,Tn with coefficients in R shall be denoted by R[ T 1 , .•• , Tn ] [for a definition of this ring and some simple properties which shall be used in the following without proof the reader should refer to, e.g., [32], Ch. V, §4, no. 1, or [204], Vol. II, Ch. VII, § 1, and also [63], Ch. 7]. Usually a power series will be written in the form

F= and we call the support of F the set

143

144

IV Formal and Convergent Power Series Rings

Of course, polynomials are exactly power series with finite support. (2) Sometimes we will use the representation of a power series as a sum of forms:

(il, ... ,i,,)ENO il+··+i,,=h

h=O

and we will call Fh the form of degree h of F. If F 1: 0, we will call the initial form of F the form of F of smallest degree; this degree is the order of F and will be denoted o(F). If F = 0, then we define o(F) = 00 [with the usual rules for addition and order in Noo := No U {oo}, cf. 1(2.7)]. The order function has the following properties: o(F+G) ~min({o(F),o(G)}), o(FG) ~o(F)+o(G).

A power series F is a unit of R[ T1 , ••• , Tn] iff Fo is a unit of R. This Fo, whether it is zero or not, will be called the constant term of F. Let m (T1 , ••• , Tn) be the ideal of R[T1 , • •• , Tn) consisting of the power series with zero constant term; the intersection of the powers of m is the zero ideal. Note that for F E R[TI, ... , Tn], F 1: 0, we have o(F) = n iff F E mn but F f/. mn+1. In particular, if R k is a field, then k[ T1 , ••• , Tn] is quasilocal with maximal ideal m and residue field k. If R is a domain, then the initial form of the product of two power series is the product of the corresponding initial forms, so R[ T 1 , ••• , Tn] is also a domain, and the order of the product is the sum of the orders of the factors. (3) The ring R[ TI, ... , Tn] is complete for the m-adic topology. Moreover, it is the completion of R[ T1 , ••• , Tn] for the n := m n R[ T1 , ••• , Tn ]-adic topology-note that n = R[ TI, ... , Tn ]Tl + ... + R[ Tb ... , Tn ]Tn-and it is also the completion of R[ T1 , ••• , Tn]n for the nR[ T 1 , ••• , Tn ]n-adic topology.

=

=

(1.2) SUBSTITUTION: Consider the ring of formal power series R[ U1 , ... , Um ], and let m be the ideal of this ring generated by Ut, ... , Um. Let A be an R-algebra, (l an ideal of A, and assume that A is Hausdorff and complete in its a-adic topology. We choose elements XI, ... , Xm E a. Let G E R[ UI, ... , Um ], and for every hE No we consider the form Gh of degree h of G and the element Gh(XI, . .. , xm) E A. It is clear that Gh(Xl, ... , xrn) E ah, hence that the sum 00

LGh(Xl, ... ,Xrn ) =: G(XI, ... ,Xrn ) h=O

converges for the a-adic topology of A. The map G H G(xI, ... ,xrn }: R[UI, ... ,Urn ] ~ A is an R-algebra homomorphism which is obviously continuous, for mh is mapped into (lh for every integer h ~ O. Since R[ U1 , • •• , Urn] is dense in R[ U1 ,· •• , Urn D,

145

1 Formal Power Series Rings

there exists only one continuous R-algebra homomorphism cp: R[ UI , ... , Um] -t A with cp(Ui) = Xi for every i E {I, ... , m}. This homomorphism is called a substitution. In particular, let A = R[ T I , ... , Tn] and the Xi be power series of positive orderj then the homomorphism will be called a power series substitution. We get the constant term of a power series F by choosing Xl = ... = Xn = 0: F(O) := F(O, ... ,0) = Fo. (1.3) PARTIAL DERIVATIVES: We set Rn:= R[TI , ... ,Tn]; let m be the ideal of Rn generated by TI"'" Tn. (1) Let j E {I, ... , n}. For every F=

set

i.'

.

/'('l, ... ,'n

Til ... I

Tij-lTij-ITij+l ... Tin j-l

j

j+l

n .

(il, ... ,in)EJ'il O

It is easy to check that Dj : Rn -t Rn is an R-derivation of Rn, and that DiDj = DjDi for all i, j E {I, ... , n}. Let j E {I, ... , n}j since Dj(m h ) C mh - l for every hEN, we see that Dj is a continuous map. To emphasize that we are dealing with the indeterminates TI"'" Tn, we often write 8/8Tj instead of Dj; for FERn the power series 8/8Tj (F) is called the j-th partial derivative of F. (2) For every FERn we have [we use the notation of (1) 1 il!···in!'/'il, ... ,i n

= D~l ... D~n(F)(O)

for all (il, ... ,i n ) E N{j.

Assume, in particular, that R is a Q-algebraj then we have the Taylor expansion F=

L

(il, ... ,in)EJ'il O

(1.4) JACOBIAN: Let F(l), . .. , F(m) E R[ T I , . .. , Tn] be power series with constant term equal to zero. Then the (m, n)-matrix with elements in R

) ( 8F(i) 8Tj (0) l~i~m,l~j~n E M(m, n; R) is called the Jacobian matrix of F(1), ... , F(m). We consider the substitution homomorphism cp: R[ UI , . .. , Um ] -t R[ T I , .. . , Tn] defined by cp(Ui) := F(i) for i E {I, ... , m}; this substitution homomorphism is called the substitution homomorphism defined by F(l), ... , F(m). The Jacobian matrix of F(l), ... , F(m) is called also the Jacobian J

O. Then, for

3 Weierstraf3 Preparation Theorem

151

every 9 E k[(X1 , ... , Xn+l}], there exist a polynomial ho E k[(X1 , ... , Xn}][Xn+l] and a power series hI E k[(X b ··., Xn+l}] such that

9

and that ho = 0 or that deg(ho) uniquely by f and g.

~

= ho+hd, m - 1. Moreover, ho and hI are determined

Proof: We shall use the notation of the proof of (3.3) [where Z must be replaced by Xn+l]. First, we consider the case f = X::'+l - p where P = E~o PiX~+I' and Pi E k[ (Xl, ... ,Xn ) 1has constant term zero for every i E No. (1) For any FE k[(Xb ... ,Xn}] and (s, ... ,s) E IR+ write 11F1I(s, ... ,s) =: F(s), and for any G E k[(XI , ..• , Xn+l}] and (s, ... , s, t) E 1R~+l write IIGII(s, ... ,8,t) := G(s, t). In particular, writing G = Ei>O GiX~+l with Gi E k[(X I , ... , Xn)] for i ~ 0, we have G(s, t) = Ei~O Gi(s)ti. (2) Let F = E(il, ... ,in)ENo'il> ... ,inxil ... X~n E k[(XI, ... ,Xn)] have constant term ,0, ... ,0 = O. Let s be a positive real number such that F(s) < 00. For every positive real number t such that t < s we have

(3) Since p and 9 are convergent, there exists s E ~ such that Ns, s) < 00 and g(s,s) < 00. We choose, E k such that ITI > l/s. Consider the linear change of variables defined ~(Xi) = XiI, for every i E {I, ... , n + I}. Then we have ... , X n +1}] be Xn+1-regular of order m > O. Then F can be written, in a unique way, as F = U·P where U E k[{XI"'" X n+1}] is a unit and P E k[{X I , ... , X n }][Xn+1] is a WeierstraB polynomial of degree m. Proof [Existence]: Set Rn := k[{Xb ... ,Xn}], and let m be the maximal ideal of Rn. By WeierstraB division theorem [cf. (3.8)] we find a unique expression

X:+1 where Ho

= Ho + HIF,

=0

Ho E Rn[Xn+1], HI E k[{XI> ... ,Xn+1}]

or deg(Ho) ::; m - 1. Taking residues modulo m we see that

X:+1 = Ho + HIF and F = Fi . X:+1 where Fi is a unit of k[Xn+l ]. This shows that Ho = 0 and that HIFl = 1, hence that HI is a unit of k[Xn+1].

3 WeierstraB Preparation Theorem

153

Therefore HI is a unit of k[{XI"'" Xn+l}] [cf. (1.1)(2) resp. (2.4)(4)], and our result follows by taking U := HII, P := X:+l - Ho. [Uniqueness] Let F = U· P be a representation of F as stated. We set HI := U- I , Ho := X:+l - P. Then we have F = Ho + HIF, which shows [cf. (3.8) and note that Ho = 0 or deg(Ho) ::; m - 1] that U and P are determined uniquely. (3.10) DEFINITION: In the situation of (3.9), the WeierstraB polynomial P is called the WeierstraB polynomial associated with F. We add some consequences of WeierstraB division theorem and WeierstraB preparation theorem. We write Rn = k[{X I , ... ,Xn}]; note that k[(XI , ... ,Xn+l)] is a subring of k[(XI"'" Xn)][Xn+l~' the power series ring over k[(XI , ... , Xn)]. (3.11) REMARK: (1) Let F E Rn[Xn+ l ] be a WeierstraB polynomial of degree m, and let G E Rn[Xn+ l ] be a polynomial. The equation

G = HF+AIX:+jl + ... +Am

[cf. (3.8)] is the division identity in Rn[ Xn+d, as follows from the uniqueness part of (3.8), hence, in particular, HE Rn[ Xn+l] is a polynomial. (2) Let F E Rn[ Xn+l] be Xn+l-regular of order m > O. Then in F = UP [cf. (3.9)] the unit U E k[{XI , ... , Xn+l}] is a polynomial in Rn[Xn+l] of degree deg(F) - deg(P). (3.12) REMARK: Let P E Rn[ X n+ l ] be a WeierstraB polynomial, and assume that P = FI···Fh where FI, ... ,Fh E k[{XI"",Xn+l}] are non-units. Clearly, the power series H, ... , Fh are Xn+l-regular of positive orders, hence there exist units UI , ... , Uh in k[{XI"'" Xn+d] and WeierstraB polynomials PI,"" Ph in Rn[ Xn+l] such that Fi = UiPi for every i E {I, ... , h} [cf. (3.9)]. Since PI ... Ph is a WeierstraB polynomial in Rn[ Xn+l], by (3.9) we get P = PI ... Ph and UI ·· ,Uh = 1. (3.13) Corollary: Let F E Rn[Xn+l] be a WeierstraB polynomial which is irreducible. Then F is also irreducible in k[{XI, ... , Xn+l}]'

Proof: Indeed, suppose that F = FIF2 where F I , F2 E k[{XI"'" Xn+d] are nonunits. Since F is Xn+l-regular of positive order, also FI and F2 are Xn+l-regular of positive orders. By (3.9) we have Fi = UiPi, i = 1,2, where UI , U2 are units of k[{XI"'" Xn+l}] and PI, P2 E Rn[Xn+l] are WeierstraB polynomials. By (3.12) we see that F = PI P2 , which contradicts the irreducibility of F. (3.14) REMARK: Let

be a WeierstraB polynomial, Le., AI"", Am E Rn have positive orders.

154

IV Formal and Convergent Power Series Rings

(1) We show that

k[{XI, ... ,Xn+11] pnRn [Xn+1]

= R n[Xn+1] P.

Indeed, let F E k[{Xb ... , Xn+d] P n Rn[Xn+d, i.e., there exists a formal resp. convergent power series H E k[{XI, ... ,Xn+11] such that F = HP. We choose q E N such that the degree of F as a polynomial in X n +1 is at most m + q. Let us write 00

H = LHiX~+1

with Hi ERn

for every i E

No.

i=O

Then, for every i E N such that i > q, we have

Hi

+ Hi+1AI + ... + Hi+mAm

= 0.

Since AI, ... , Am lie in the maximal ideal m of Rn, we see that Hi Em for every i E N such that i > q. Applying (*) again we conclude that Hi E m2 for every i > q. By induction, we see that Hi E jEN mj for every i > q, hence that Hi = for every i > q, and that therefore H E Rn[ X n+1]. (2) Now R n [Xn+1]/(P) can be considered as a subring of k[{XI , ... , Xn+d]/(P) by (1). Take any G E k[{XI' ... ' Xn+11]. By WeierstraB division theorem there exists H E k[{XI' ... ,Xn+d] such that G - HP E Rn[Xn+I ], which implies that

°

n

k[{XI, ... ,Xn+d]/(P)

= Rn[Xn+1]/(P).

(3) Let x := X n+1 + (P) be the image of X n+1. Then Rn[Xn+d/(P) is a free Rn-module having {1, x, ... , x m - l 1 as a basis. (3.15) REMARK: (1) Let k be an infinite field. Let F E k[{Xb ... ' Xnl] be a non-unit power series different from zero, of order m, and let Fm be its initial form. Since k is infinite, we may find (-YI, ... , 'Yn-l) E k n - l such that F m(-YI, ... ,'Yn-l,1) "# 0. Consider the linear change of variables cp: Rn -+ Rn defined by cp(Xi) = Xi + 'YiXn for every i E {1, ... , n - 11 and cp(Xn) = Xn [cf. (1.4) and (2.4)(8)]. Now cp(Fm)(O, ... ,O,Xn ) = Fm('YI, ... ,'Yn-I,1)X~\ hence cp(F) is Xn-regular of order m. This shows that every non-zero power series can be made regular in Xn of order equal to its own order, by applying a linear change of variables. Since the affine space k n is not a union of finitely many hypersurfaces [since k is infinite], the same is true simultaneously for a finite number of power series. (2) It can be shown that, even in the case of a finite field, every non-zero formal power series in k[XI, ... ,Xn] can be made regular in Xn of order equal to its own order, by applying a suitable k-algebra automorphism of k[ Xl, ... , Xn] [cf. [204], vol. II, Ch. VII, Lemma 3]. We tacitly assume this result. (3.16) Proposition: The power series ring k[{Xl] is a discrete valuation ring, and its maximal ideal is generated by X. In particular, k[{Xl] is noetherian and factorial.

3 WeierstraB Preparation Theorem

155

Proof: The orderfunction 0: k[{X}]-t No U{oo} can be extended in a unique way as a discrete valuation v: k( {X}) -t Zoo and k[ {X}] is the ring of v [cf. chapter I, (3.20) and (3.28)]. (3.17) Proposition: The ring Rn is noetherian.

We prove it by induction on n. The ring RI is noetherian [cf. (3.16)]. Let n > 1, and assume that k[{X I , ... ,Xn-d] is noetherian. Let a =I- Rn be a non-zero ideal of Rn and choose F E a, F =I- O. By applying a k-automorphism [cf. (3.15)] and the WeierstraB preparation theorem [cf. (3.9)], we may assume that F is a WeierstraB polynomial with respect to Xn of degree m ~ 1. By (3.14)(3) we see that Rn/(F) is a finitely generated k[{XI'"'' Xn_d]-module, hence it is a noetherian ring [cf. [63], Cor. 1.3], and therefore the ideal a/ (F) is a finitely generated ideal of Rn/(F). It follows that a is a finitely generated ideal of Rn, hence Rn itself is a noetherian ring. (3.18) Proposition: The ring Rn is factorial.

We prove it by induction on n. The ring RI is factorial [cf. (3.16)]. Let n > 1, and assume that k[{Xl, ... ,Xn-d] is factorial; then k[{Xl,,,,,Xn-dHXn] is also factorial by Gauf3's lemma. Let FERn be a non-unit different from zero. By (3.15) we may assume that F is Xn-regular of positive order, and, by (3.9), that F = UP where U E Rn is a unit and P E k[{Xl,,,,,Xn-d][Xn] is a WeierstraB polynomial. Since k[ {Xl, ... , Xn-d][ Xn] is factorial, we may write, uniquely up to units, P = PI",Ph where PI,,,,,Ph E k[{Xl, ... ,Xn-d][Xn] are irreducible. By (3.13) this is a factorization of P in Rn into a product of irreducible elements. Let P = Fl'" Fm be another factorization of P into irreducible elements in Rn. Then the elements Fl, ... ,Fm are Xn-regular. For every i E {1, ... ,m} we can write Fi = ViPI where Vi is a unit of Rn and PI E k[{XI, ... ,Xn-d][Xn ] is an irreducible WeierstraB polynomial [cf. (3.9)]. By (3.12) we have P = P{··· P:n, hence a factorization of P into a product of irreducible elements of k[ {Xl, ... , X n- 1 }] [ Xn ]. Since the latter ring is factorial, we have m = h and, after relabelling, UiPi = PI for every i E {I, ... , h} where UI , ... , Uh are units of k[{Xl, ... , Xn-d]. Therefore we have Fi = ViUiPi for every i E {I, ... , h} which means that P = PI ... Ph is a factorization of P in Rn into irreducible factors which is unique up to relabelling and units. (3.19) Corollary: Let FE k[{Xl, ... ,Xn+1}] be a power series which is X n +l regular of positive order m. If F is an irreducible element of k[{Xl, ... , Xn+d], then the quotient field of the domain k[{Xl"" ,Xn+1}]/(F) is a simple algebraic extension of degree m of the quotient field of k[{Xl, ... , Xn}].

Proof: Since k[{XI'"'' Xn+d] is factorial by (3.18), the ideal generated by F is a prime ideal. By WeierstraB preparation theorem we have F = UP where

156

IV Formal and Convergent Power Series Rings

U E k[{X1' ... ' X n+1}] is a unit and P E R n [Xn+1 ] is the WeierstraB polynomial associated with F. Now the result follows from (3.14). (3.20) Theorem: The ring Rn is an n-dimensional regular local ring. Proof: Rn is a noetherian ring by (3.17). The ideals generated by Xl' ... ' Xj, j E {O, ... , n}, form a strictly ascending chain of prime ideals, and the maximal ideal of Rn is minimally generated by n elements, hence n ::; dim(Rn) ::; emdim(Rn) = n, hence we have emdim(Rn ) = dim(Rn), and Rn is regular [ef. 11(4.19)]. (3.21) REMARK: From the result in (3.20) we now can deduce the factoriality of Rn [ef. (3.18)] also by applying Th. 19.19 in Eisenbud's book [63]. (3.22) Theorem: Assume that k is algebraically closed. Let F E Rn[Z] be a monic polynomial of positive degree, and assume that F(Z) E k[Z] has at least two different zeroes; then we have F(Z) = I1~1 (Z _"(1)8 1 , where m 2:: 2, "(1,· .. , "(m are pairwise different elements in k and 81, ... , 8 m E N. Then there exist monic polynomials F1, ... ,Fm E Rn[Z] such that F = Fl·· ·Fm and Fl(Z) = (Z _"(1)8 1 for every l E {I, ... , m}. Proof [by recursion]: Set G(Z) := FbI degree as F and

G(Z) = FbI

+ Z)

+ Z)

E Rn[Z]; G is monic of the same

m

= Z8 l

II (Z +

"(1 - "(I)SI.

1=2

Therefore G is Z -regular of order 81. Let H E Rn [ Z] be the WeierstraB polynomial associated with G; then we have G = UH, where U E Rn[Z] is monic [ef. (3.11)] and degz(U) = degz(G) - degz(H). Set F1(Z) := H(Z - "(d and F2 (Z) U (Z - "(1). Then F1 and F2 are monic polynomials in Rn [ Z], and we have

F

= F1F2,

F1(Z)

= (Z - "(d

m

8l ,

F2(Z)

= II(X -

"(j)S;.

1=2

Repeating the above argument with F2 yields the result. (3.23) DEFINITION: Let A be a quasilocal ring with residue field k. Then A is called henselian if for every monic polynomial F E A[T] such that w(F) = G H, where G, H E k[T] are monic and relatively prime, there exist monic polynomials G, HE A[T] with F = GH and with w(F) = F, w(G) = G. (3.24) REMARK: Let A be a henselian ring, and let a be an ideal of A contained in the maximal ideal of A. Then A/a is a henselian ring. (3.25) Corollary: Let k be algebraically closed. Then Rn is a Henselian ring.

4 Formal and Analytic Algebras

4

157

The Category of Formal and Analytic Algebras

(4.0) For the sake of simplicity, we assume in this section that k is algebraically closed.

4.1

Local k-algebras

(4.1) REMARK: Let B be a local k-algebra with maximal ideal n and residue field k. Then we have B = k IB E9 n as a direct sum of k-vector spaces. Let {YI, ... , Yh} be a system of generators of the ideal n, and let T E No. Any y E B admits a representation y=

"1.···.'". yi1I

'Y.

(i1 •...• i,,)EN~

.• • yi" h

+ Zr

i1+··+i ,, el > ... > eg

= 1.

Define ei-l ni := ei

Then we have

f3i

130

for every i E {O, ... , g},

mi

£or every i E { 1, ...} ,g.

for every i E {I, ... , g} ,

190

V Quasiordinary Singularities

and we have fio

= eo = n1 e1 = n1 n2e2 = ... = n1 ... n g. Furthermore, we have mini+1 < mi+1 gcd(mi' ni) = 1,

for every i E {I, ... ,g -I}, ni > 1 for every i E {I, ... , g}.

The 9 pairs (m1' n1), ... , (mg, ng) of coprime integers are called the characteristic pairs of (td,y(t». Define ho :=

l:11J,

l

mi+1 hi:= - J - mi ni+1

for every Z. E {I, ... ,g -I}.

We can write g-l hi

ho

y(t)

=L

Jl.=o

'YJl.tnJl.

+L

L 'Yi,Jl.t(mi+Jl.)ni+l···n9 + L 'Yg,Jl. tmg +

i=l Jl.=o

00

1J

(*)

Jl.=o

where the coefficients 'YJl.' 'Yi,Jl. lie in k and 'Yi,O :I 0 for every i E {I, ... , g} [note that fio = d, that, for every i E {I, ... ,g-I}, we have fii+J.Lei = (mi+J.L)ni+1 ... ng for every J.L E {O, ... , hi}, and that mg = fig]. (3) The integral closure of R[y] is the discrete valuation ring Rd [cf. (4.20)]. Let v be the valuation of Qd defined by Rd. It can be shown that r(y) = v(R[y] \ {O}) [cf., e.g., [40], Ch. IV], hence that r(y) is an invariant of the ring R[ y].

5

A Generalized Newton Algorithm

(5.0) We keep the notations introduced in the earlier sections of this chapter.

5.1

The Algorithm

(5.1) REMARK: Let F be as in (4.2), and assume that there exists y E Qn,d with F(X1 , ••• , X n , y) = O. (This is the case if F is a quasiordinary WeierstraB polynomial, cf. (4.3) and (4.4).) Let y = EiES(y) (XiMi where Mi = X ri / d for every i E S(y), be the normal representation of y. In the following, we will describe an algorithm to find Mi and (Xi for every i E S(y) from the coefficients At, ... , Ad of F. More precisely: we will see that the monomials M 1 , M 2 , ••• are determined uniquely while the coefficients (Xl, (X2, ... are determined only up to factors in k which are roots of unity. (5.2) NOTATION: (1) With respect to d and the sequence (ri)iES(y) we may define the sequence (Vi)iES(y) as in (6.4) [cf. the remark in (1.14)]. (2) We set Qn(O) := Qn and 81 := [Qn(Y) : Qn]j furthermore, we set Qn(i) := Qn(M1, ... , Mi) and 8i+1 := [Qn(Y) : Qn(i)] for i E S(y). (3) By (1.10) we have [Qn(i) : Qn] = V1'" Vi for every i E S(y)j note that 8i = Vi8i+1 for every i E S(y), and, in particular, 81 = d = V182. There exists hE S(y) with Qn(Y) = Qn(h) [cf. (1.20)]. Then we have Vi = 1 for every i E S(y) with i > h, and we have 8i = Vi'" Vh for every i E {I, ... , h}.

5 A Generalized Newton Algorithm (5.3)

DETERMINATION OF

M1

191

AND

a1: Note that

(_l)d A d{X1, ... ,Xn ) = y(1) ... y(d). (I) Clearly, by (*) and (1.I8)(2)(a), the leading monomial of Ad{Xt, ... , Xn) is Mtj thus, T1 can be found by inspecting Supp{Ad), namely, T1 is the smallest element in SUpp{Ad). Furthermore, we can calculate V1 [cf. (1.I3) j and 62 [since d= V162j. (2) Now Qn{M1) is a cyclic extension of Qn of degree V1 and [Qn{Y) : Qn{Mt} j = 62 • Let r 1 c Gal{Qn,d/Qn) be a system of representatives for the cosets of the subgroup Gal(Qn,d/Qn{Mt}) [cf. (1.I3»); then we have V1 = Card{r1). Let e1 E k be a primitive v1-th root of unity. There are V1 pairwise different conjugates of M1 over Qn, namely e{ M1 for I E {I, ... , vd. (3) From (2) we see that the conjugates of y can be ordered in the following way: y

(j) _

-

{a 1 M1 (j) a 1 M1

+ yP) + Y1(j)

for j E {I, ... , 62 }, . for:J E {62 + 1, ... , d}j

here we have

Thus, there are V1 different choices for the leading coefficient of y, and any two such choices differ by a factor which is a v1-th root of unity. Clearly, if yP) ::f. 0, then M2 is the leading monomial of y~j) for every j E {I, ... ,d}. (4) Let 8 1, ... ,8d E Z [Tl, ... , Td j be the elementary symmetric polynomials in d indeterminates T 1, ... ,Td. For I E {O, ... , d - I} we are interested in the leading term of Ad-l{X1, ... ,Xn ) = (_I)d-l 8d_l{y(1), . .. ,y(d». The minimal polynomial of a1M1 over Qn is Y"'l - ar l Mfl [cf. (1.I3) j, hence

{ylll _ arl Mfl )62 =

t (~2)

yllli (-arl Mfl )62 -1

1=0

is the field polynomial of a1M1 E Qn{Y) over Qn. Therefore Mt- 1 is the leading monomial of A d - 1 iff A d - 1 ::f. 0 and V1 I (d - I). Let A E {I, ... , 62 }, and let f3>. be the leading coefficient of A>'1I1 if A>'1I1 ::f. 0 and define f3>. := 0 if A>'1I1 = O. Comparing the coefficients of the leading monomial Mt in 0 = F{y) shows that a1 is a zero of the polynomial

therefore the polynomial n may be used to calculate the leading coefficient of y. (5) Now we make the substitution

Yi

:= y - a1 M1

192

V Quasiordinary Singularities

to get F 1 (X 11/d , ... , X1/d n , y,) 1 : = F( Xl,"" X n , Yi 1 d .1.1 vd + A 1,1 (X1/d F1 --.1.1 1 , ... , X n/ )vd-1

+ a 1M 1) where

+ ... + A I,d (X11/ d, ... , Xl/d) n

E Rn,d [Y1] .

The d zeroes of F1 are [cf. (3)] yP) for j E {I, ... ,82 },

,B~j~, M1 + y~j) for j E {82 + 1, ... ,d} [note that ,B~j~ ,

:=

Q~j)

- Q1

°

=I

for every j E {82

+ 1, ... ,d} ].

(5.4) THE INDUCTION ASSUMPTION: Let i E N, and assume that the terms M 1, ... , Mi, Q1,"" Qi have been calculated. Let VI, ... , Vi, 81 , ... , 8i+ 1 be the natural integers which are defined by [Qn(j) : Qn] = VI .. 'Vj and 8j := 8j+1Vj for j E {I, ... ,i}. Assume, furthermore, that in the course of calculating M 1 ,.·. ,Qi we have got the following data. (1) A system of representatives fi C Gal(Qn,d/Qn) for the cosets of the Galois group Gal(Qn,d/Qn(i»). The set fi has V1"'Vi elements. (2) Order the conjugates of y in such a way that y(l), . .. ,y(Oi+t} are those conjugates of y the first i terms of which are left invariant by the elements of fi' i.e.,

QW, l E {I, ... ,i} and j E {8

+ 1, ... , d}, are non-zero elements of k. y~l) =I 0, 'then the leading monomial of y~j) is MiH for every j = {I, ... , d}. where

iH

If

(3) A polynomial

D._yd r, - i

+ A.',1 (X11/ d, ... 'Xn1/ d)yd-1 i + ... + A·',d (X 11/ d, ... , Xl/d) n

E R n,d [,,...] L, •

(4) For every l E {I, ... ,i} and every j E {8'H + 1, ... ,8,} a sum of monomials

a~j)

L ,B}!2 M i

:=

A,

,B}!2 E k for A E {l, ... , i}, ,B}!t) =I 0,

A=' such that Yi(j)

} for j E { 1, s ... ,: UiH,

a(j) i

+ Yi(j)

£or J. E { UiH s:

+ 1, ... , d}

are the d zeroes of Fi . Note that, for l E {I, ... , i} and j E {8'H,"" 8t}, M, is the leading monomial of ai j ) + y~j) .

5 A Generalized Newton Algorithm

193

(5.5) THE INDUCTION STEP: Starting with the data in (5.4), we will describe the next step. If Ai,d = 0, then y~l) = 0, and we are done: y = 2::=1 QIM, is a zero of F. Now we assume that Ai,d =I 0. (1) It is easy to calculate M iH . Namely, by (1.18)(2), and since 0, - 0'+1 = OIH(VI -1) for every I E {I, ... ,i}, the monomial M 6i+1M6.-6i+1M6.-1-6i ... M61-62 _

i+l

i-I

i

1

-

i 6i+1 ITM 6/+ 1(v -l) Mi+l 1 l

1=1

is the leading monomial of Ai,d(X:/ d, ... , X~/d). Thus, we know Mi+l, and therefore we know ViH and Oi+2 = OiH/ViH' (2) We know that Gal(Qn(i + 1)/Qn(i)) is a cyclic group of order ViH, and using the construction described in (1.14), we may find a representative 'I/J E Gal(Qn,d/Qn(i)) for a generator of the Galois group Gal (Qn(i + 1)/Qn(i)). Then the set ri+l := {ri'I/J' II E {I, ... , Vi+d} is a system of representatives for the cosets of the subgroup Gal (Qn,d/Qn(i + 1)) of Gal(Qn,d/Qn)' (3) Let ci+l E k be a primitive Vi+l-th root of unity. The conjugates y(1), ... ,y(6i +d can be ordered in the following way:

here we have (16i+2+'\) Qi+l

1 Qi+l = ci+l

£,or I E { 1, ... , Vi+l

r}. - 1} and 1\\ E { 1, ... , Ui+2

Thus, there are ViH different choices for the leading coefficient Qi+l of yF), and any two such choices differ by a factor which is a viH-th root of unity. Clearly, if Yi~1 =I 0, then Mi+2 is the leading monomial of Y!~1 for every j E {I, ... , OiH }. The zeroes of Fi can now be written in the following form:

",. .... '+1 M·,+1

Y(j) i

-

+ y(j) i+l

{ ",(j) M· + y(j) .... i+l 1+1 i+l (j)

Gi

(j)

for j E {I, ... , Oi+2}, for j E {0i+2 + 1, ... ,oi+d, (j)

+ Qi+l MiH + YiH

for j E {OiH

+ 1, ... ,d}.

Let 8 1 , ... , 86i+1 E Z[Tb ... , T6i+1] be the elementary symmetric polynomials in OiH indeterminates T 1, ... , ni+1' The coefficient of Z 6i+1 -Vi+1 in the polynomial

(ZVi+1 _ 1)6i+2 =

IT (Z - c~+D6i+2

Vi+1 1=1

V Quasiordinary Singularities

194

is, by binomial expansion, ' , ... ,ci+1 Vi+l- 1, ... ,ci+1 Vi+l- 1) Ut+2 --(_I)Vi+1SVi+l (1 , ... , l,Ct+1,··.,Ct+1,

_~,

[where, on the right hand side, there are di+2 elements 1, di+2 elements ci+1, ... , r Vi+l- 1 ]. Ui+2 eIements ciH · A i,d-oi+l +Vi+l (X11/ d, ... , Xl/d) f yOi+l -Vi+l·In F· . The coe ffi C1ent n 0 i i IS, up to a sIgn, the (d - diH + lIiH)-th symmetric function of the zeroes of Fi . The leading monomial of Ai,d-Oi+l +Vi+l is, therefore, the product of the leading monomial of SVi+l (yP), ... , YlOi+tl) and the leading monomial of a~otl ... a~Oi+1H), hence is equal to M':'i+l M?i-Oi+l M?i-l-Oi ... M 01- 02 t+1 t t-1 1

1 II MO/+ 1(v/-1) = MVi+ t+1 I , i

1=1

and the leading coefficient (:3 of Ai,d-Oi+l +Vi+l is equal to

hence we find that

As ai+1 is determined up to multiplication by a 1Ii+1 -th root of unity, we may calculate QiH by using the last displayed equation; let QiH be an element of k satisfying (*). (4) Using the system of representatives r iH, we may calculate the coefficients aW1,1 for every 1 E {I, ... , i + I}, j E {diH + 1, ... , d}, appearing in the following set of displayed equations

Y (j)

=

I

I:

alMI + 1=1 iH (') L.,;ai~l,IMI 1=1

"'"

Y~~2

for j E {I, ... , 6i+2},

t

+

( ') Yit2

for j E {6i+2

+ I, ... ,d}.

(5) Now we make the substitution

and we get a monic polynomial FiH (Xi/d, ... , X;/d, ViH) E R n ,d[Vi+1] of degree d, having as zeroes the elements (j) + y(j) ai+1 i+1

£0r J. E {~Ui+2

+ 1, ... , d}

5 A Generalized Newton Algorithm where for every j E {8i+2

+ 1, ... , 8i+1}

(j) .- (3(j) G i+1·i+1,i+1 M·1+1,

and

G~~l

195

:= GP) - Qi+1Mi+1

'" ...J. 0 i+1,i+1 ..- ",(j) o, a + b ~ I} be the triangle in LR with vertices 0, Lt, h. Then we have TnL =-{O,Lt,h} iff {Lt,h} is a Z-basis of L. Proof: Assume that {Lt, l2} is a Z-basis of L; then it is clear that Tn L consists only of 0, Lt and l2. Conversely, assume that {it, h} is not a Z-basis of L. Then there exist rational numbers aI, a2 such that l := alit +a2l2 E L, but (aI, a2) i Z2. Set a~ := al - Lad, a2 := a2 - La2J, and If := l- Ladll - La2Jl2. If a~ + a2 ~ 1, then If = a~ II + a2l2 lies in Tn L, and if a~ + a2 > 1, then l" := Lt + l2 - If = (1 - aULt + (1 - a2)l2 lies in Tn L, and these points If and l" are different from the vertices of T.

(4.12) Proposition: Let L be a free Z-module of rank 2, and let {el' e2} be a Z-basis of L. Let r E N, and let al, . .. , ar E N be such that ai ~ 2 for every iE{I, ... ,r}. Define subsets {/o, ... ,/r+d, {e3, ... ,e r +2} ofL by

10 := el, Ii := li-l + ei+1, ei+2 := (ai - 2)/i-l + (ai - I)ei+l

Ir+l

:=

for i E {I, ... , r},

Ir + e r +2·

Then, for everyi E {I, ... ,r + I}, the sets {Ii-l,ei+d and {/i-l,/i} are Z-bases of L. Furthermore, we have li-l+li+l =adi

foreveryiE {I, ... ,r},

(*)

and, for everyi E {O, ... ,r -I} and j E {O, ... ,r+ I} satisfyingi +2 < j, we have j-2 Ii + Ii = (ai+1 - I)/i+l + (a" - 2)1" + (aj-l - I)/j-l. (**) ,,=i+2

L

VI The Singularity zq = Xyp

226

Proof: Since 10 = el, we see that {fo, e2} is a Z-basis of L. Let i E {I, ... , r}, and assume that {fi-l, ei+d is a Z-basis of L. Since h = li-l + ei+1, ei+2 = (ai - 2)/i-l + (ai -l)ei+1 and (ai -1) - (ai - 2) = 1, it follows that also {Ii, ei+2} is a Z-basis of L. For every i E {I, ... , r} we have li-l

+ Ii+l = h-l + Ii + ei+2 = h-l + Ii + (ai - 2)h-l + (ai - l)ei+1 = (ai - l)(h-l + ei+d + Ii = (ai - l)h + h = ai/i.

It is easy to see that {fi-b Ii} is a Z-basis of L for every i E {l, ... , r + I}. The formula (**) can easily be shown by induction on j.

(4.13) We keep the notation introduced in (4.12). (1) We define uniquely integers p, q as in (1.0) by -q- = al

q-p

-fa; - ... -~;

note that p ~ 1. Let AO := q, Al := q- p. By the definition of the Hirzebruch-Jung continued fraction expansion for q/(q - p), there exist integers A2, . .. , Ar+1 such that we have

and that we have

(2) We show that Ir+1 = pel + qe2; in particular, 10 and Ir+1 are linearly independent over JR. Proof: Set e* := pel +qe2. Then we have e* = (Ao-Ad/o+Aoe2. Let i E {I, ... , r}, and assume that we have already shown that e* = (Ai-l - Ai)h-l + Ai-lei+l. We see that

(Ai - Ai+l)/i+Aiei+2 = (Ai - Ai+1)(h-l + ei+d + Ai«ai - 2)h-l = (Ai-l - Ai)/i-l + Ai-lei+l.

+ (ai - l)ei+1)

Therefore we have e* = (Ar - Ar+dlr + Arer+2 = Ir + er+2 = Ir+l. (3) Set I.lQ := 0, JLl := 1, and define JL2,· .. , JLr+1 by JLi+1 = aiJLi - JLi-l for every i E {I, ... , r}. Note that JLo < JLl < ... < JLr+1 [we have JLl > J.Lo, and if we have already shown that J.Li > J.Li-l for some i E {I, ... , r}, then it follows that JLi+l - JLi = (ai - l)JLi - J.Li-l ~ JLi - JLi-l > 0]. (4) We show that we have

h

Ai

= - 10 q

+ -JLiq Ir+1

for every i E {O, ... ,r + I}.

4 Two-Dimensional Cones

227

Since fo = e1 and h = fo + e2 = (l/q)«q - p)fo + fr+1) = (l/q)(Ado + {Ldr+t) , the assertion holds for i = and i = 1. Let j E {I, ... , r}, and assume that the assertion is true for every i E {O, ... , j}. Then, by (4.12)(*), we have

°

f j+1

= aj f j

-

f j-1

= aj).j -q Aj-1 JOI + aj{Lj -q {Lj-t f r+1 --

Aj+1 I {Lj+1 f --JO + - - r+1· q q

(5) From (4) we get {Lr+1 = q. (6) Let i E {O, ... , r + I}. A linear combination O!ofo

+ ... + O!r+dr+1 where are non-negative integers, can represent fi only in the trivial way: we have O!i = 1 and O!j = for all j E {O, ... ,r + I}, j i i. In fact, since the sequence (Ai)Oofr+1j (f is a strongly convex rational polyhedral cone (with respect to L) in iR of dimension 2. Let r := (f n L be the semigroup associated with (f, let () be the convex hull in LR of the set «(f n L) \ {OJ, and let 8() be the boundary polygon of ()j by abuse of language, we call 8() the boundary polygon of (f. (4.15) Proposition: With notations as in (4.14) we have: (1) The elements fo, h, ... , !r+1 in this order are the points of L lying on the compact edges of the boundary polygon 8(). (2) The set {fo, ... , fr+d is a minimal set of generators ofr. (3) Let mEr and m i OJ if m does not lie on one of the halflines JR.>0 Ii, i E {O, ... , r + I}, then there exists exactly one index j E {I, ... , r + I} such-that mE NoIi-1 + No!;. (4) We have !r+1 = e*. For every i E {1, ... ,r} the point fi lies on the line through fi-1 and fH1 iff ai = 2, and it is a vertex of () iff ai > 2. (5) For i E {O, ... ,r} we set (fi := JR.>oli + JR.>oli+1' Then (fi is a nonsingular strongly convex rational polyhedral co~e of dimension 2. The intersection of any two such different cones is either {O} or a halfline JR.~0 Ii for some j E {I, ... , r} . Furthermore, we have (f = (fo U··· U (fr' Proof: (a) By (4.13)(4) the elements fo, ... ,fr+1 lie in (f. Let i E {O, ... ,r}j since {Ii, fHd is a Z-basis of L by (4.12), (fi is nonsingular. Since the sequence (AI)O 0, with

a

a

a

cp(TdD )

•••

cp(TZ~!l) = cp(Tt)cp(Tf!-l)

[if i = t + 1, then we have f3 = 0], and with a + f3 ~ 'Yo + ... + 'YtH. This implies that a k-basis of nV /nv+l is given by all elements cp(TrTf!-l)' i E {O, ... , t + Il, a > 0, a + f3 = 11 [f3 = if i = t + 1]. Counting these elements shows that

°

dimk(nV /nv+l)

= (t + 2) + (t + 1)(11 -

1)

= (t + 1)11 + 1;

235

5 Resolution of Singularities

this implies the remaining assertion in (2). We show: the images in a/rna of the t(t + 1)/2 elements Qij, i, j E {O, ... , t + I}, i < j - 1, are linearly independent over k. Then the assertion in (3) follows from Nakayama's lemma [ef. [63], Cor. 4.8]. Indeed, let 'Yij, i, j E {O, ... , t + I}, i < j - 1, be elements in k such that t+1

L

'YijQij E mo..

i,j=O i0 10 + 1R>0 1s+1 and, for every i E {O, ... , s}, we set ai := 1R>0 1i + 1R>0 1i+1; let () be the convex hull of (a n N) \ {O} and let B() be its boundary [cf. (4.18)].

236

VI The Singularity zq = Xyp

(5.2) For every i E {O, ... , s + I} we define non-negative integers

= lIinl + /-Lin2· = 1, 110 = 1, III = 1, and we have [ef.

/-Li, IIi

by

Ii

Then we have /-Li+1

/-Lo

= 0, /-Ll

= bi/-Li -

/-Li-l,

= billi -

lIi+1

lIi-l

(4.12)(*)]

for every i E {I, ... , s}.

By (4.13)(3) and (4.13)(5) we get

o= /-LO < 1 = /-Ll < /-L2 < ... < /-Ls+1 = q; it is easy to see that [for the last equality sign cf. (4.13) (2) ] 1 = 110

= III

(1) We have

~ 112 ~ ... ~ IIs+1

= p.

=1

for every i E {o, ... , s}. Proof: We have /-LIllO - /-LollI = 1. Let i E {I, ... , s}, and assume that we have shown already that /-LiIIi-l -/-Li-illi = 1; then we have /-Li+1l1i - /-Lilli+1

/-Li+1l1i -/-Li lli+1

= (bi/-Li -/-Li-dlli - (billi = /-Lilli-l - /-Li-l IIi = 1.

lIi-l)/-Li

(2) We have /-Lillj -/-Ljlli

> 0 for all i,j

E {O, ... , s

+ I}

such that i

> j.

Proof: We have, by (1), /-Li

/-Li-l

----=

1

for every i E {I, ... , s + I},

hence we get h

= '" L..J

/-Li+h _ /-Li-l lIi+h lIi-l

and the assertion is proved. (3) In particular, since /-Ll = III qllj ~ P/-Lj

£ or .z E {1, ... , s + I} an d h E {O , ... , s + I-' } z ,

1

j=O lIi+jlli+j-l

= 1 and /-Ls+l = q,

for every j E {O, ... , s + I},

IIs+1

/-Lj ~ IIj

= p, we have

for every j E {I, ... , s

+ I}.

(5.3) Note that ko = m2, kl = ml, hence that {kl,ko } is the basis of M dual to {nl,n2}' Let i E {O, ... ,s}. (1) Let {m~i),m~i)} be the Z-basis of M dual to the Z-basis {li,li+d of N. By (5.2) we get (i) mo

= - lIi+1 ko

+ /-Li+1 kl ,

m 1(i) --

lI·k I/.·k 1, , 0 -,..,

hence, by (5.2)(1), we find that ko

(2) Note that

r

lTi

(i) = /-Limo(i) + /-Li+lml'

=

No mo(i) + No m (i) l •

kl

l . = lIi m o(i) + lIi+1 m (i)

5 Resolution of Singularities

5.2

237

The Case p = 1

(5.4) In this subsection we assume that p = 1. (5.5)

REMARK:

-q-

q-1

(1) Then we have u = lR~o nl

= bl - fb;" - ... - [ba

[cf. (3.9)(2)]. The semigroup generators where

ko

= ml,

kl

ru

where

8

+ lR~o (nl + qn2)

=q -

1 and bl

and

= ... = bs = 2

has the set {ko,kl,k2} as a minimal system of

= ml + m2,

k2

= (q -1)ml + qm2

[cf. (4.18) for p = 1], and the affine variety Xu has one singular point Xu which is an As-singularity [cf. (4.22)]. Note that ko, k2 are linearly independent over Q, and that qkl ko+k2' which shows again that k[ru] k[U, V][W]/(W q -UV). (2) The projective variety BIz .. (Xu) is covered by three open affine subsets Uo, UI , U2 having, respectively, the coordinate rings Ro, RI , R2 where

=

=

R j := k[r u][ e(ko)/e(kj ), e(kd/e(kj), e(k2)/e(kj)]

for j = 0,1,2

[cf. A(14.1)(3)]. We show that Uo and U2 are nonsingular, and that U1 is nonsingular if q $ 3, and that it is (isomorphic to) a toric variety which has a singularity of type A q - 3 if q ~ 4. Indeed, the subsemigroup of M generated by ko, kl' k2' kl - ko, k2 - ko is also generated by ko and kl - ko since k2 - ko = q(kl - ko) + (q - 2)ko, and the subsemigroup of M generated by ko, kl' k2' ko - k2' kl - k2 is also generated by k2 and kl - k2 since ko - k2 = q(kl - k2) + (q - 2)k2; note that {ko, kl - ko} and {k2' kl - k2} are Z-bases of M. The subsemigroup r I of M generated by ko, kl' k2' ko - klo k2 - kl is also generated by ko - kl' k2 - klo kl' and we have (q - 2)kl = (ko - kd + (k2 - kl). If q = 2, then we have r l = Z (ko - kl ), hence UI is an open affine subset of Al [cf. (2.2)]. If q 3, then we have r l No (ko - kl ) + No (k2 - kd; since ko - kl' k2 - kl are linearly independent over Q, the affine variety UI is nonsingular [cf. (4.10)]. If q ~ 4, then UI is (isomorphic to) a toric variety and has one singular point of type A q - 3 •

=

=

Thus, we have shown:

(5.6) Proposition: Assume that p = 1. Then Xu has one singular point, namely a singularity of type A q - l . By blowing up the singular point, one gets a nonsingular variety if q $ 3, and a variety with exactly one singular point if q ~ 4; this singular point is a singularity of type A q - 3 , and it has an open afIine neighborhood which is (isomorphic to) a toric variety. (5.7) Corollary: Assume that p = 1. Then Xu can be desingularized by repeatedly blowing up points. Proof: This follows from (5.6) and A(14.4).

238

5.3

VI The Singularity zq

= Xyp

The General Case

(5.8) NOTATION: If 8 = 1, then we set v := 1 and i(O) := 0, i(l) := 1, i(2) := 2. If 8 > 1, then let v' be the number of those vertices of () which are different from 10 , 11, 18 , Is+1-note that 0 ~ v' ~ s - 2-, set v := v' + 2, and label these vertices as li(2), ... ,li(v-l) where i(2) < ... < i(v - 1). We set i(O) := 0, i(l) := 1, i(v) := 8, i(v + 1) = s + 1. Note that h, Is are, in general, not vertices of (). By (4.18)(1) we have b i (a)+1 = ... = bi (a+1)-1 = 2 for every a E {I, ... , v - I}.

From the recursion formula in (5.2) we find for every a E {I, ... , v - I} Iti(a+l) - lti(a+1)-1 Vi(a+1) - Vi(a+1)-1

If v > 2, then we have of ().

= =

bi (2)

lti(a+1)-1 - lti(a+l)-2 Vi(a+1)-1 - Vi(a+1)-2

lti(a)+1 - Iti(a), Vi(a)+1 - Vi(a)·

> 2, ... , bi (v-l) > 2 since li(2), .•. ,li(v-l) are vertices

(5.9) NOTATION: We define elements

(Ta,li(a»)

= ... = = ... =

TO, ... , Tv E

= (Ta ,li(a)+1) = 1

M by requiring that

for every a E {1, ... ,v -I}

[note that {li,li+d is a Z-basis of N for every i E {O, ... , 8}, cf. (4.18)(1)]. Let a E {I, ... , v-I}; the points li(a), li(a)+1, ... ,li(a+1) lie on a line, hence we have (Ta,

II')

=1

for every

f3

E {i(a), i(a)

+ 1, ... , i(a + I)}.

(1) Using (5.2)(1) and (5.3)(1), we see that

= ko, = -(Vi(a)+1 - vi(a»)ko + (lti(a)+1 -lti(a»)k1 Tv = -Vi(v)+1ko + lti(V)+lkl. TO

Ta

for every a E {1, ... ,v -I},

(2) Let a E {I, ... , v-I}. Using (5.3)(1) and the definition of TO, . .. , Tv, we get T a -- m(i(a» 0

+ m(i(a» l'

T

v -

k

t+ 1 -- m(s) O·

By (5.2) and (5.8) we get Ta -

(5.10) Lemma: The 3 + (b 1 1

+ (bi(a+l)

-

1) m (i(a+1» .

2) + ... + (b s

-

2)

mo(i(a+l» -

b 1 «b1 - (3)TO

+ f3Td,

f3

1

= t + 2 elements

E {O, ... , bd,

239

5 Resolution of Singularities

in case s

= 1, and

1 bl _ 1 (bl - 1- ,B)TO + ,BTl), ,B E {O, ... , bl - I}, 1 b 2 (bi(o) - 2 - ,B)To-l + ,BTo),a E {2, ... ,v -1},,B E {I, ... ,bi(o) - 2}, i(o) 1 bs _ 1 «bs - 1- ,B)Tv-l + ,BTv), ,B E {I, ... , bs - I},

in case s ~ 2, in this order, are the elements ko, kl' ... ,kt+l on the compact edges of the boundary polygon 8(}. Proof: Note that 3 + (b l - 2) + ... + (b s - 2) = t + 2 by (3.10)(*). First, we consider the case s = 1. Here we have bl = q, J.L2 = q, V2 = q - 1, i(O) = 0, i(l) = 1, i(2) = 2, t + 1 = q, TO = ko, Tl = ktH' Cl = ... = Ct = 2 [cf. (4. 18){2) ], hence we have

k J. _- (q - j)ko + jktH q

for every j E {O, ... ,q}

[cf. (4.13)(4)]. This is the formula given above. Now assume that s ~ 2. We use induction with respect to sand bl . We define q', p' as in (3.9)(2)(b); let s' be the length of the Hirzebruch-Jung continued fraction expansion of q'/ (q' - p'), and let 10, ... , l~, H be the data corresponding to q', q' - p'. We define v' and TO, ... , T~' with respect to 10,... , l~, H in the same way as we defined v and TO, ... ,Tv with respect to 10, ... , IsH. Furthermore, let t' be the length of the Hirzebruch-Jung continued fraction expansion of q' / p', and let ko, ... , k~, H be the data corresponding to q', II . (1) bl = 2: then we have Cl ~ 3, s' = s - 1, t' = t, q' = q - p, p' = p, b~ = b2, b~ = b3, ... , b~, = bs , ~ = Cl - 1, ~ = C2' ... '~' = Ct. We set n~ := It = nl + n2, n~ := n2. Then {ni, n~} is a IE-basis of N. We find [cf. (4.13) and (4.18) for notations which are not explained here] that 10 = n~ = It, Ii = ni + n~ = 12 [as bl = 2, we see that n2 = n3], and that Ii = liH for i E {2, ... , s}. We set m~ := ml, m~ := - ml + m2. The IE-basis {mi,m~} of M is dual to {ni.n~}. We set mi := m~ = - m2, m~ := mi - m~ = 2m2 + ml; then we have ko = mi = - m2, ki = ko + m~ = kl' k~ = ki + m~ = kl + m3 = k2 [because m~ = (Cl - 3)ko + (Cl - 2)m~ = m3] and kj = kj for every j E {3, ... , t + I}. (2) bl ~ 3: then we have Cl = 2, s' = s, t' = t - 1, q' = p, p' = 2p - q, bi = bl - 1, b~ = b2, ... ,b~, = bs , C2 =~, ... ,Ct = ~,. We set ni := -n2, n~ := nl +2n2. Then we have 10 = n! = - nl, Ii = n~ + n~ = It, l~ = Ii + n~ = 12, and Ii = Ii for every i E {3, ... ,s+ I}. We set m!:= -m2 +2mt. m~:= mI. The IE-basis {m!,m~} · d U al to {' -I -I o f M IS n l ,n2'} . lIT vve set m + m2, m l := m I2 = ml 2 := m Il - m2I = m2. Then we have ko = m! = ml + m2 = kl' kl = m! + m~ = k2, and kj = kjH for j E {2, ... ,t}. (3) s = 2, bl = 2: then we have v = 2. Now we have ko = TO, kl = Tl by (5.9){1),

VI The Singularity zq = Xyp

240 hence we get

Since r~ = r2 and II case s = 1, we get

(4) s

{1/{b 2

+ l3 = b2l2, we get r~ = (b2rl

- r2)/(b2 - 1), hence, by the

= 2, -

bl > 2: from l~ + l~ = {bl - 1)1~ and lo 1)) ({bl - 2)ro + rd. By (5.9){1) we find that

+ l2 =

bIll we get r~

=

= - {bl - 3)k~ + {bl - 2)k~ = - {bl - 2)ko + {bl -l)kl = rl, r~ = - p'k~ + q'k~ = - (2p - q)kl + p{2kl - ko) = - pko + qkl = r2. We assume that the formulae are true for b~ = bl - 1. Then we have r~

k/3+l

= k~ = bl ~ 2 ({bl 1

-

= bl -1 ({bl -

2-

,B)r~ + ,BrU

2 - ,B)TO

+ {,B + l)Td,

,B E {I, ... ,bl - 2},

and by (5.9)(1) the displayed equation holds for ,B = 0 and ,B we have

= -1; furthermore,

(5) We assume that s ~ 3, and that the assertion is true for s - 1. (a) bl = 2: then we have ko = TO, kl = Tl by (5.9){1). If b2 = 2, then we have v' = v and T~ = Ta for every a E {I, ... , v}, and this case is finished. If b2 > 2, then we have v' = v-I, and T~ = T2, . .. , T~_l = Tv. Since we have (T~,l~) = 0, (T~,lD = 1 by definition, we get T~ = (1/(b2 - 2))((b2 - l)rl - T2), hence, again by induction, we get k/3+l

= k~+l = b2 ~ 1 ({b2 = b2 1_ 2 ({b2 -

2-

,B)T~ + ,BTU

2 - ,B)Tl

+ ,BT2),

,B E {I, ... ,b2

-

2}.

Now it is clear by induction that the assertion is also true in this case. (b) bl ~ 3: then we have v' = v, and T~ = Ta for every a E {I, ... , v}. We assume that the formulae are true for bl - 1. We can use the same argument as in the case s = 2, bl > 2.

5 Resolution of Singularities

(5.11)

NOTATION:

We set

To: := IR~o li(o:)

note that in NR..

TO, .•. ,Tv

241

+ IR~o li(o:+1)

for every a E {O, ... ,v};

are strongly convex rational polyhedral cones of dimension 2

(5.12) Let a E {O, ... , v}. We set Ao: := i(a + 1) - i(a);

then we have Ao = Av = 1. (1) Let Zo, Zl E Z, and define recursively Zj for every j E N by Zj+1 then we have Zj = jZ1 + (1 - j)zo for every j E No. (2) By (1) we have

= 2zj -

Zj-1;

lj = (j -i(a) )li(O:)+1 + (1- (j -i(a)))li(o:) for every j E {i(a), i(a) + 1, ... ,i(a+ I)}.

(3) By (2) we have

Since {li(o:' li(o:)+1 -li(o:)} is a Z-basis of N, the only point of the toric variety Xra which can be singular is a singularity of type A.x a -1 [cf. (5.5)(1)]. (5.13) We set

bi := b1 -

1, b~ := bi(o:) - 2 for every a E {2, ... , v-I}, b~ := bs

-

1.

(5.14) Let a E {I, .. . ,v -I}. We set 1

So: := fj(rO:-1 - ro:). 0:

(1) We have by (5.12)(2), and since li(O:+l)-l

+ li(o:+1)+1

= bi(o:+1) li(o:+l) ,

+ l)li(o:+1) - Ao: li(o:+1) +1 , 1)(bi(O:+l) - 1) + l)li(O:+l) + (1- Ao:)li(o:+1)+1'

li(o:) = (Ao:(bi(O:+1) - 1) li(o:)+1 = ((Ao: -

Now we have (ro, h)

= 1,

(ro, l2)

= b1 -

1, and for a E {2, ... , v - 2} we get

(2) By (1) we find that {To:, So:} is dual to the Z-basis {li(o:),li(o:)+1 -li(o:)} of N. (3) We set m1,0: := So:, m2,0::= To: - m1,0:'

242

VI The Singularity zq

= Xyp

Then we have [cf. (4.18)]

a minimal system of generators for the subsemigroup r T" of Mis {mI,a, m2,a} if Aa = 1, and is {mI,a, mI,a + m2,a, (Aa - l)mI,a + Aam2,a} if Aa > 1. (4) Note that r TO = No m~O) + No m~O), and that r Tv = No m~8) + No m~8).

(5.15) Let a E {I, ... ,v -I}. (1) We have 1

1

Aara = ~(ra+l - ra) + ;;;(ra-I - ra) a+I

a

[evaluate both sides at li(a) and li(a+l)]. (2) From (1) we get mI,I

1

= b* (ro -

rI), (Aa - l)m1,a + Aa m 2,a

1

1

= ~(ra+1 -

ra).

0+1

(5.16) For every j E {O, ... , t + I} let rj be the subsemigroup of M generated by the elements k o , ... , kt+l' ko - k j , ... , k j - 1 - kj, kj+I - k j , ... , kt+l - k j . For every a E {O, ... , v} let j(a) E {O, ... , t + I} be defined by kj(a) = ra [cf. (5.10)]; we have j(O) = 0, j(l) = b1 if s = 1, j(l) = b1 -1 if s ~ 2, and j(v) = t + 1. (1) s = 1: then we have b1 = q = t + 1. By (5.3)(1) we get m~O) = ki - ko, m~O) = ko, and by (5.10) we get k1 - ko = (l/q)(r1 - ro), hence k{3 - ko = /3m~O) • {OJ (0) for every /3 E { 1, ... ,b1 } • Smce r TO = No mo + No m 1 ,we have shown that ro = r TO • Let j E {I, ... , t+ I}. By (5.9)(2) we have m~I) = kt+l = r1 and ro = ko = m~1) + (1) _ ( . ) (1) {} _ (1) . (1) b1m 1 ,hence k{3-kj - J -/3 m 1 for every /3 E 0, ... , b1 and k j - mo - Jm 1 . _ (1) (1) . {} _ Thus,wehaverj-ZmO +Nom1 foreverYJE 1, ... ,t ,andrt+l-r T1 • (2) s ~ 2: then we have v ~ 2. Just as above we see that ro = r TO" Since rv = kt+l = m~8) and m~8) = (1/(b 1 - 1)(rv_1 - rv) = k t - kt+l' we see that rTv C rt+l. Just as above we can show that rt+l C rTv ' and therefore we have rt+1 = rTv • Let a E {I, ... ,v -I}; we show that rj(a) = r T " . (a) First, we show that ro, ... , rv, ro - ra, ... , rv - ro lie in r T". Since

we see that ro-I - ro, ro, and ra+l - ro E r T " , and therefore ro-1 and ra+l lie in r T " . Assume that a < v - 1. Then we have [cf. (5.15)(1)] 1 Ao+lro+l + ~(rO+l - ro) 0+1

1

= ~(ro+2 0+2

ro+l),

5 Resolution of Singularities

243

and therefore Ta:+2 -To:+1 lies in r T", hence To:+2 -Ta = (Ta+2 -Ta+t) + (Ta+1 -Ta) lies in r T,,' Continuing in this way, we find that the elements T a+2, ... , Tv and T a+2 - T a, ... , Tv - To: lie in r T" • Likewise, if 0: ~ 2, then we have AO:-1 Ta-1

1

+ I1(T a -1 a

Ta)

-

1

= ~(Ta-2 a-1

Ta -1),

and therefore T a-2 - T a-1 lies in r T", hence T a-2 -Ta = (Ta -2 - T a-1) + (T a-1 - T a) lies in r T,,' Continuing in this way, we find that the elements To, ... ,Ta-1 and TO - Ta , ... , Ta-1 - Ta lie in r T,,' (b) We show that k{3 - Ta lies in r T" for every [3 E {O, ... , t + I}. Since ko = TO, we know already that ko - Ta E r T,,' Let '1 E {I, ... ,v} and [3 E {I, ... ,b~}; then we have by (5.10) k{3 - To:

= b~ ((b~ -

(3)(T-y-1 - Ta)

-y

+ [3(T-y

- Ta

»,

hence we have k{3 - Ta E r T " since T-y-1 - Ta , T-y - Ta E r T" by (a). Thus, we have shown that rj(a) CrT" for every 0: E {I, ... ,v -I}. (c) By (5.10) and (5.15)(2) we have kj(a)+1 - kj(a)

1

= ~(Ta+1 a+1

m1,a k j (a)-l - kj(o:)

0:

T" C rj(a),

a+1

hence we have

k{3 - kj

for

0:

E {I, ... ,v -I} and j E {j(o:)

In fact, we set rj := Z ma + No [3-j

kj;

+ Aa m 2,a,

+ Aa (m1,a + m2,a)

rj(a)

E {I, ... ,v -

= r T". I}.

+ 1, ... ,j(o: + 1) -I}

we have

we have .

= ~(Ta+1 - Ta) = (J - (3)ma a+1

Therefore we have rj C rj, and the elements

.m r'j '

I)m1,a

I = ~(Ta - Ta+d

ma := ~(Ta - Ta +1) a+1

We show: For

= (Aa -

= Ta = kj(a) ,

+ m2,a

+ Aa (m1,a + m2,a)

Thus, we have shown that r (3) We set 1

Ta)

for [3 E {j(o:), ... ,j(o: + I)}. Ta

= kj (a), •.. ,kj (a+1) = To:+1

lie

VI The Singularity

244

zq = XYP

We assume that a: < v - 2, and let f3 E {j(a: + 1), ... ,j(a: + 2)}. Then we have

hence k{3-kj = (k{3-To 1 or a2r-l = 1 in which case we have q2r-l = q2r-2· (3) Assume that b1 = 2 and that s ~ 2. (a) First, we consider the case where b1 = ... = bs = 2. Then we have s = q - 1 and p = 1 [cf. (3.8)], and we have q ~ 3 and

~1 =1+lq-2+fI, qhence we have r = 2, al = 1, a2 = q - 2, a3 = 1. (b) Now we consider the case where at least one of b2 , ••• , bs is greater than 2, and we choose u E {2, ••• , s} such that b1 = ... = b1£ - l = 2, b1£ > 2. Then we have 71"j-l

hence

71"j

=q -

= 271"j -

jp for every j

71"jH E

{O, ... , u}, and

71"1£-1

where 0

~ 71"1£H

= b1£ 7l"1£ -

71"1£-1 71"1£-1 71"1£

for every j E {I, ... , u - I},

= b1£ 7l"1£ -

71"1£+1

< 71"1£. It is clear that

= b1£ -lb1£+ 1 -

••• -

fb;.

Furthermore, since p < q - p and 0 < q - up < p, we have

q=(q-p)+p,

q-p=(u-l)p+(q-up).

If u = s, then we have 71".-1 = b.7I"., hence we have p = (b. - l)(q - sp), and therefore we obtain r = 2, al = 1, a2 = s - 1 and a3 = b. - 1. If u < s, then we

246

VI The Singularity

zq = XYP

have 1ru -1 = bu1ru -1ru+!, hence we have p = (b u - 2)(q - up) + (1ru - 1ru+!), and therefore we obtain r > 2, al = 1, a2 = u - 1, a3 = bu - 2, and 1ru-I Ca4 + ... +la2r-I' - = a3 + 1 +l~ 1ru

(4) Now we prove the assertion by induction with respect to sand bl . If s = 1, then we have r = v = 1. Let s ;:::: 2, and assume that the assertion is true for all q, p such that q/(q - p) has a Hirzebruch-Jung continued fraction expansion of length smaller than s. Let q/(q - p) have a Hirzebruch-Jung continued fraction expansion of length s. If bi = ... = bs = 2, then we have v = 2, i(l) = 1, i(2) = s = q - 1, hence we get Al = q - 2, and by (3)(b) we find that r = 2, a2 = q - 2. Now we assume that bi = 2, and that at least one of b2 , • .• , bs is larger than 2. We use the notation of (3)(b). Set q' := 1ru-I and p' := 1ru , and let v' (resp. r') be defined with respect to q', p' in the same way as v (resp. r) was defined with respect to q, p. If u = s, then we have v = 2, Al = S - 1, and we have r = 2, a2 = s - 1 by (3)(b). If u < s, then we have v' > 2, v' = v-I, and [cf. (**) 1r' = r - 1. In this case Io, ... ,Iu lie on a line and Iu+! does not lie on this line. Let Ih, . .. ,1~-u+2 (resp. AL ... , A~') be defined as in (4.18) (resp. (5.12)) with respect to q', p'. Then we have Ih = Iu, I~ = lu+1,"" 1~+u-2 = Is+I' and therefore we get A~ = A2,"" A~' = Av. By our induction assumption, applied to q' and p', we get from (*) and (**) that A2 = a4, . .. ,Av-I = a2r. Furthermore, we have Al = i(2) - i(l) = u - 1. If bi > 2, then we have q < 2p, hence we get al > 1, and we have -q- - 1 = bi q-p

-

1-

-q- - 1 = al -1 q-p

[b; - ... - [bs,

+ja;" + ... +la2r-l.

Using the result in (5.10)(2), and by induction, we see that the assertion is true in this case also. (5.20) REMARK: The result in (5.19) can be interpreted in the following way: the continued fraction expansion in (5.19) allows us to determine the number v-I of singular points of BI", .. (Xu) and, moreover, to say that every singular point Ya of BI", .. (Xu ) is a singularity of type A a2Q - I , a E {I, ... ,v - I}. We know that Ao = Av = 1.

Chapter VII

Two-Dimensional Regular Local Rings INTRODUCTION: The theory of complete (= integrally closed) ideals of a twodimensional regular local ring R was introduced by Zariski [cf. [191] and the appendix of [204]]. He used it in his paper [192] to resolve surface singularities. Similarly we shall use facts from this theory in section 7 of chapter VIII. We usually do not assume that the residue field of R is infinite (under this additional assumption we could prove the fact that the product of complete ideals is complete in a much simpler way, cf., e.g., [98], Th. 3.7). We introduce the notion of ideal transform in section 1. In section 2 we define quadratic transforms of a two-dimensional regular local ring and study their properties. Complete ideals are introduced in section 3; one of the main results of the theory, the unique factorization of complete ideals, is proved in section 4. The predecessors of a simple ideal shall play a decisive role in the resolution process in section 7 of chapter VIII; they are studied in section 5. The connection between simple complete ideals of R and valuations of the second kind which dominate R is described in section 6; here we also prove the length formula of Hoskin-Deligne. Some results concerning proximity are proved in section 7. The result of (8.13) in section 8 is a local version of the theorem of resolution of singularities of curves embedded in a regular surface-and shall play a decisive role when proving this theorem in section 1 of chapter VIII.

1 1.1

Ideal Transform Generalities

(1.0) Let ReS be two factorial domains having the same field of quotients. We associate with an ideal a of R an ideal as of S, the transform of a in S, and we study properties of this transform operation. 247

248

VII Two-Dimensional Regular Local Rings

(1.1) Lemma: Let R be an integral domain with field of quotients K, and let p

be a prime ideal of R such that the localization Rp is a discrete valuation ring. Let S be a subring of K containing R, and set pS := pSp n S. The following statements are equivalent: (1) pS f:. S. (2) q := pS is the unique prime ideal of S whose intersection with R is p, and we have Sq = Rp. (3) There exists an ideal b of S such that b n R = p. (3') pS n R = p. (4) Sp f:.K. (5) Sp = Rp. (5') S C Rp. Proof: Since Rp is a discrete valuation ring, we have p f:. {O}; moreover, any subring of K properly containing Rp is equal to K [cf. 1(3.11)]. Set M := R \ p. Then we have Sp = M-1S. From ReS we get Rp eSp, and this implies that either Sp = K or Sp = Rp; for any ideal b of S with b n R = P we have M n b = 0. The implications (2) :::} (1), (2) :::} (3), (2) :::} (5') and (4) :::} (5) are clear. We have (1) :::} (4) since (1) implies that pSp f:. Sp, and we have (4) :::} (3) since from Sp f:. K we get Sp = R p , hence p = pSp n R = (pSp n S) n R = b n R with b := pSp n S. Assume that (3) holds. Then p is contracted from S, hence it is the contraction of its extension pS, and therefore we get (3) :::} (3'). Since (3') :::} (3), we have (3) ¢} (3'). Since Rp eSp, we have (5) ¢} (5'). It remains to show that (5) :::} (2). Assume that (5) holds; then we have S C Sp = Rp. Now q := pSp n S = pRp n S is a prime ideal of S, and q n R = pRp n R = p. Furthermore, since M C S \ q, we have Sp C Sq, and we have Sq f:. K since q is a prime ideal of S different from {O}, hence Rp = Sq. Let q' be any prime ideal of S with q' n R = p; by the same argument we get Rp = Sq., hence Sq = Sq •. This equality implies, as is easily checked, that q = q'. The following result shall be needed later. (1.2) Proposition: Let R be a semilocal domain, and let ffil, ... , ffih be the maximal ideals of R. If the rings Rmi are factorial for i E {I, ... , h}, then R is also factorial.

Proof: It is enough to show that every prime ideal of R of height one is principal [cf. [63], Prop. 3.11 and Th. 10.1]. Thus, let p be a prime ideal of R of height 1. For i E {I, ... , h} we set Ri := R mi . Since Ri is factorial and pRi is either the unit ideal or a prime ideal of Ri of height 1, there exists an element Pi E P with PiRi = pRi. For i E {I, ... , h} there exists an element ai ~ ffii with ai E ffij for j E {I, ... , h} with j f:. i [cf. [63], Lemma 3.3]; note that Piai also is a generator of pR i . We set p:= Plal + ... + Phah; then we have pEp. Let i E {I, ... , h}. Then we have Pjaj E PffiiRi for j E {I, ... ,h}, j f:. i, hence pRi = PiaiRi C pRi + PffiiRi C pRi' and therefore pRi = pRi + PffiiRi; by Nakayama's lemma [cf. [63], Cor. 4.8] we get pRi = pRi, and this implies that p = Rp by B(2.5).

1 Ideal Transform

1.2

249

Ideal Transforms

(1.3) GREATEST COMMON DIVISOR: Let R be a factorial domain with field of quotients K. (1) Let a be a non-zero ideal of R. An element x E R is a greatest common divisor for the elements of a iff the following two conditions are satisfied: (i) We have a C Rx; (ii) For every y E R with aeRy we have Rx CRy. Clearly every non-zero ideal of R admits a greatest common divisor of its elements. Let x be a greatest common divisor for the elements of a, and let Z E R be a non-zero element. Then zx is a greatest common divisor for the elements of za. (2) Let MeR be multiplicatively closed with 0 f/. M, and set S := M- 1 R; the ring S is factorial by 1(3.30)(3). Let a be a non-zero ideal of R, and let x be a greatest common divisor for the elements of a. Then x is also a greatest common divisor for the elements of the ideal as of S. Proof: We have as C Sx. Let yES be an element with as C Sy. We may assume that y = Pl ... Ph with irreducible elements Pl, ... ,Ph E R which satisfy PiS =I S for every i E {I, ... ,h}. Then we have [cf. I(3.30)(3)(b) 1 a C (as) nRc (Sy)

n R = Ry,

hence we have Rx cRy, and therefore we have Sx C Sy. (3) Let a be a non-zero ideal of R, and let x E R be a non-zero element. Then x is a greatest common divisor for the elements of a iff

In particular, 1 is a greatest divisor for the elements of a iff (R : a)K = R. Proof: Assume that x is a greatest common divisor for the elements of a. From a C Rx we get Rx- 1 = (R : RX)K C (R: a)K. Let z E (R : a)K be a non-zero element. Then we can write z = zd Z2 where Zl, Z2 E R are non-zero elements, and we have aZl C Rz2. Now ZlX is a greatest common divisor for the elements of aZl [cf. (1) l, hence RZl x C Rz2, and therefore Z = zd Z2 E Rx- 1 , hence (R: a)K C Rx- 1 • Thus, we have shown that Rx- 1 = (R: a)K. Conversely, assume that (R : a)K = Rx- 1 • Then we have x- 1 E (R : a)K, hence ax- 1 C R and therefore we have a C Rx. Let y E R with aeRy. Then we have Ry-l C (R : a)K = Rx- 1 , hence Rx CRy, and therefore x is a greatest common divisor for the elements of a. (4) Let a be a non-zero ideal of R, and set b:= a(R: a)K.

Then b is an ideal of R, and it is the unique ideal of R with the following properties: (i) (R: b)K = R, Le., 1 is a greatest common divisor for the elements of b; (ii) a = bz for some Z E R; any element Z as in (ii) is a greatest common divisor for the elements of a.

250

VII Two-Dimensional Regular Local Rings

Proof: Let x be a greatest common divisor for the elements of a. Then we have a(R : a)K = ax- l [cf. (3)], hence we have

(R: b)K

= (R: ax-l)K = (R: a)Kx = Rxx- l = R,

and we have a = bx, hence b = ax- l C R [since a C Rx]. Let C be a non-zero ideal of R with (R : C)K = R and a = cz for some z E R. Then we have (R : a)K = (R : C)K z-l = Rz- l , hence we have a(R : a)K = az- l = c. Now we have proved that b satisfies the assertions in (i) and (ii). Let z E R be an element with a = bz. Then we have (R : a)K = (R : b)Z-l = Rz- 1 , and therefore z is a greatest common divisor for the elements of a [cf. (3)]. (5)(a) Let a, b be non-zero ideals of R such that 1 is a greatest common divisor for the elements of a and of b. Then 1 is a greatest common divisor for the elements of abo In fact, we have (R : ab)K = ((R : a)K : b)K = (R : b)K = R; hence the assertion follows from (3). (b) Let aI, a2 be non-zero ideals of R, and let, for i = 1,2, Xi be a greatest common divisor for the elements of ai. Then XlX2 is a greatest common divisor for the elements of ala2. In fact, for i = 1,2 we have ai = biXi where bi is an ideal having 1 as a greatest common divisor for its elements. Then 1 is a greatest common divisor for the elements of blb 2 by (a), and from ala2 = bl b2(XlX2) we get the result by (4). (6) Let a be a non-zero ideal of R. By (4) we have a unique representation where n E No, PI, ... , Pn are pairwise different principal prime ideals of R, el,." ,en are natural integers and b is a non-zero ideal of R with (R: b)K = R.

(1.4) DEFINITION: Let R be a factorial domain with field of quotients K, and let S be a factorial subring of K containing R. Let a be a non-zero ideal of R, and write a = p~l ... p;'nb as in (1.3)(6)(*). We define the transform of a in S to be that ideal as of S which is defined by as := q? ... q~n (bS)(S : bS)K where qi := pr for every i E {I, ... , n} [cf. (1.1), and note that RPi is a discrete valuation ring by 1(3.30)(4)]. In particular, if a is a non-zero principal prime ideal, then the ideal as as defined here is the same as the ideal as defined in (1.1). (Note that aR = a.) Some basic properties of the "transform" operation are given in the next proposition [for the notion of integral closure of an ideal cf. section 6 of appendix B].

(1.5) Proposition: Let R be a factorial domain with field of quotients K, and let SeT be factorial subrings of K containing R. Let a, aI, a2 be non-zero ideals of R. Then:

1 Ideal Transform

251

(1) If a is a principal prime ideal, then either as n R = a, in which case as is a principal prime ideal of S with as n R = a, or as n R =/:. a, in which case we have as = S. (2) If (R : a)K = R, then as = (as)(S : as)K, and therefore (S : as)K = Sand as = za s where z is a greatest common divisor for the elements of as. (3) [Compatibility with products] (ala2)s = arar (4) [Transitivity] (aSf = aT. (5) If, in particular, S is a ring offractions of R, then as = as. (6) If a2 C al C a2 (the integral closure of a2 in R), then a~ C ar C (a~) (the integral closure of a~ in S). (7) We have as = as [integral closure of as in S ].

Proof: (1) Let a =: p be a non-zero principal prime ideal of R with pS n R = p. Then q := pS is a prime ideal of S with q n R = p, and we have Sq = Rp [cf. (1.1)], hence q is a prime ideal of height 1 of S, and therefore is a principal prime ideal [since S is factorial]. If pS n R =/:. p, then we have pS = S by (1.1). (2) If (R : a)K = R, then we have as = (as)(S : as)K by definition; the last assertions follow from (1.3)(2), applied to as. (3) Let i E {I, 2}, and let ai = bixi where Xi is a greatest common divisor for the elements of ai and bi := ai(R : ai)K; let Yi E S be a greatest common divisor for the elements of biS. Then we have ala2 = bX1X2 with b := b1b2. Since 1 is a greatest common divisor for the elements of b1 and b2 , 1 is also a greatest common divisor for the elements of b [cf. (1.3)(3)], hence (R : b)K = 1. Furthermore, we have [cf. (1.3)(1)] (S : b1S)K = SY1 1 and (S : b2 S)K = SY2 1, and since Y1Y2 is a greatest common divisor for the elements of (b 1 S)(b 2 S) = bS [cf. (1.3)(2)], we have (S : bS)K = SY1 1Y2 1, and this yields (S : bS)K = (S : b1S)K(S : b2S)K. Now it is clear that ara~ = (ala2)s. (4) Let p be a non-zero principal prime ideal of R with pS n R = p. Then q := pS is the unique prime ideal of S with q n R = P and with Sq = Rp. Assume that qT n S = q. Then qT is the unique prime ideal of T with qT n S = q and with TqT = Sq. Since qT n R = q n R = p, it follows that pT = (psf. It is easy to check that pT = (pS)T in the other two cases. Let a be a non-zero ideal of R with (R : a)K = R. We have as = as(S : as)K [cf. (2)], hence we have (as)T = (aST)(T : aST)K, again by (2). We write as = as z with some z E S [cf. (2)]. Then we have (as)T = (asT)(T: aST)K = (aT)(T: z-laT)Kz- 1 = (aT)(T: aT) = aT.

Let a be a non-zero ideal of R. We write a = p~l ···p;"b as in (1.3)(6)(*). Then we have as = bSy where Sy = q~l ... q;" with qi := pr. By what we have just shown, and by (2), we get (aSf = aT. (5) Let p be a non-zero principal prime ideal of R. Then pS is either a prime ideal of S, in which case we have pS n R = p, or pS = S; in both cases we get pS = pS. Let a be an ideal of R with (R : a)K = R, i.e., 1 is a greatest common divisor for the elements of a. Then 1 is a greatest common divisor for the elements of as [cf.

252

VII Two-Dimensional Regular Local Rings

1(3.30)(6)]. Therefore we have (8 : a8)K = 8, hence as = (a8)(8 : a8)K = a8. Now the assertion follows from (3). (6) Let i E {1,2}; we use the notation introduced at the beginning of the proof of (3). Let p be an irreducible element in R; then R(p) is a discrete valuation ring [ef. 1(3.30)(2)]. We write ai = Xibi where Xi is a greatest common divisor for the elements of ai and bi := ai(R : ai)K. Then xiR(p) is a greatest common divisor for the elements of aiR(p) [ef. 1(3.30)(6)], hence xiR(p) = aiR(p). Since R(p) is a principal ideal ring, every ideal of R(p) is integrally closed. Every element of Cl2 is integral over a2, hence Cl2R(p) is integral over a2R(p), hence a2R(p) = Cl2R(p), and therefore we have a1R(p) = a2R(p). This implies that x1R(p) = x2R(p). Since this equality holds for every pEP, we get RX1 = RX2 =: Rx for some X E R [cf. 1(3.30) (4)(b) ]. In particular, we have (R : a1)K = &-1 = (R : a2)K, and from a2 C a1 C Cl2 we get b2 C b1 C b2 with b2 = X- 1Cl2 [note that b2 is the integral closure of the ideal X-1a2]. The inclusion b2 C b1 C b2 implies that

[since every element of bl is integral over b2 and b28 is an ideal of 8]. From (*) we get, applying the argument above to the ring 8 and the ideals b18, b2 8 of S, that (8: b18)K = (8 : b28)K. Then we have, by (2), bi

= (b28)(8 : b28)K C (b 1 8)(8 : b18)K = br C b~

[with regard to the last inclusion, note that from b18 C b2 8 we see that every element of (b 18)(8 : b18)K is integral over (b 28)(8 : b28)K = bn. By (3) we now have ~ = (Rx)Sb~ C (Rx)Sbr = and (Rx)Sbr C (Rx)Sb~ = (Rx)Sb~ = (b2)SX

= ~.

ar

(7) This follows immediately from (6).

2 2.1

Quadratic Transforms and Ideal Transforms Generalities

(2.0) In this subsection R is an n-dimensional regular local ring with n ~ 2, = m is its maximal ideal, and KR = K = Rim is its residue field. Note that R is a factorial domain [cf. [63], Th. 19.19]; let K be the field of quotients of R. Let {Xl, ... , xn} be a system of generators of m; then (Xl' ... " Xn) is a regular sequence on R, and if we denote, as usual, by In(z) E grm(R) the leading form of z E R, then we have grm(R) = K[ln(xd, ... , In(xn)], and In(x1), ... , In(xn) are algebraically independent over K [ef. [21], Th. 11.22 or [63], Ex. 17.16]. Note that the ring RIRxl is a regular local ring of dimension n -1 [ef. [63], Cor. 10.15]. For a polynomial f E R[ Z] (resp. a homogeneous polynomial h E R[ Xl, ... , Xn]) we denote by 7 E K[Z] (resp. h E K[Xl, ... ,Xn ]) the polynomial (resp. the homogeneous polynomial) which we obtain by reducing the coefficients of f modulo m (resp. the coefficients of h modulo m). mR

2 Quadratic Transforms and Ideal Transforms

253

We denote by M(R) = M the multiplicative monoid of non-zero ideals of R. For a, b E M we say that a divides b if there exists e E M with b = ae; this implies that b C a. Clearly a divides b iff b = a(b : a) [if we have b = ae for some ideal e of R, then e C b : a, hence b = ae C a(b : a) C b, and therefore b = a(b : a), and if b = a(b : a), then a divides b).

(2.1) REMARK: In particular, if n = 2, then we have the following: Let a be a non-zero ideal of R; we have a = zb where z E R is a greatest common divisor for the elements of a and b is an ideal of R having 1 as a greatest common divisor for its elements [cf. (1.3)]. Since every prime ideal of R of height 1 is a principal ideal [since R is factorial], it follows that a non-zero proper ideal of R is an m-primary ideal iff 1 is a greatest common divisor for its elements. (2.2) THE ORDER FUNCTION: For every nonzero fER we define ordR(f) = ord(f) = s if f E m S , f ~ m S +! [note that the intersection of the powers of the maximal ideal m is the zero ideal by Krull's intersection theorem, cf. [63], Cor. 5.4]. Since grm(R) is a domain, we have ord(fg) = ord(f) + ord(g) for all non-zero f, 9 E R, and if f + 9 '" 0, then we have ord(f + g) ~ mine {ord(f), ord(g)} ). The canonical extension of ord to KX [cf. 1(3.20)] shall be denoted by VR = v; v is a discrete valuation of rank 1 of K, v is non-negative on R and has center min R, hence the residue field", of R can be considered as a subfield of the residue field of v. The valuation ring of VR will be denoted by VR' Let a be a non-zero ideal of R; then ordR(a) = ord(a) := min{ord(a) I a E a} is called the order of a. Set r := ord(a); note that a C mT, but a ct. mT+!. For non-zero ideals a, b of R we have ord(ab) = ord(a) + ord(b). Clearly the powers mT, r E N, are v-ideals, and are therefore integrally closed in R [cf. B(6.16)(2)]. (2.3) Proposition: We set A := R[X2/Xl,'" ,xn/xI} = R[ m/xI}. Let, for i E {2, ... ,n}, Ti be the v-residue of xii Xl. Then: (1) T2, ... ,Tn are algebraically independent over "', and the rational function field "'(T2,"" Tn) is the residue field of v. In particular, v is of the second kind with respect to R, and we have rank(v) + tr. d",(v) = dim(R). (2) v is non-negative on A, has center mA = AXl in A, AXl is a prime ideal of height 1 of A, and AAxl is the ring of the valuation v. (3) We have A/mA = "'[ T2, ... , Tn], hence the prime ideals of A containing AXl correspond uniquely to the prime ideals of the polynomial ring "'[ T2, ... , Tn]. For every maximal ideal q of A with Xl E q the local ring Aq is regular and we have dim(Aq) = n. (4) The ring A is an n-dimensional integrally closed noetherian domain, and we have mffi An R = mffi for every mEN. Proof: (a) Suppose that AXl = A; then there exists yEA with X1Y = 1. We choose a homogeneous polynomial f E R[ X 2, ... , Xn] with Y = f(X2/Xl,"" xn/xd; let d be the degree of f. Then we have xt = xt+!y = Xd(X2,"" xn). This implies

254

VII Two-Dimensional Regular Local Rings

that xt E md +1, hence that d = II(Xt} ~ II (md +1 } = d + 1 which is absurd. Therefore we have rnA AXI, AXI n R m, and II is non-negative on A since II(X2/XI} = ... = lI(x n/xd = o. We show that T2,.·. , Tn are algebraically independent over It. Let 7 E It[ Z2, ... , Zn] be a non-zero polynomial of degree d, and choose a ~olyno­ mial ! E R[ Z2, ... , Zn] of degree d whose image in It[ Z2, ... , Zn] is f. We define the homogeneous polynomial 9 E R[XI, ... ,Xn] by g(XI, ... ,Xn} = XtJ(X2/ X b . .. , Xn/ Xl}; then we have lI(g(Xl, ... , Xn)) = d since not all coefficients of 9 lie in m [cf. [21], Prop. 11.20 or [63], Ex. 17.16], hence the II-residue of g(XI, ... , xn}/xt is not zero, which means that !(T2, ... , Tn) =I- o. We show that It(T2' ... ' Tn} is the residue field ltv of II, hence that tr. d/C(II} = n - 1. In fact, every non-zero element in ltv is the image of a quotient u/v where u, v E R have the same II-value r; there exist homogeneous polynomials g, h E R[ Xl, ... , Xn] of degree r having not all their coefficients in m such that u = g(XI, ... ,Xn}, V h(XI, ... ,Xn}. We have u/v g(xI, ... ,xn}/h(xl, ... ,xn} = g(l, X2/XI, ... ,xn/xd/h(l, X2/XI, ... , xn/xd, and therefore the II-image of u/v is g(l, T2, .. . , Tn}/h(l, T2, .. . , Tn) E It(T2, . .. , Tn). (b) By (a) we have mA n R = m. Let 0:: A ~ A := A/mA be the canonical homomorphism; then we have Ti O:(Xi/XI) for i E {2, ... , n}, o:(R) It and A = 11:[ T2, ... , Tn]. In particular, A is a polynomial ring over a field in n - 1 indeterminates, hence we have dim(A} = n -1 [cf. [63], Cor. 10.13]. (c) Let! E R[Xb ... ,Xn] be homogeneous of degree r; if f(xI, ... ,Xn)/x'i E A lies in the center of II in A, then all the coefficients of! lie in m since T2, . .. , Tn are algebraically independent over It, and therefore we have !(XI, ... ,xn}/x'i E AXI. Since AXI = mA, it follows that AXI is the center of II in A, hence AXI is a prime ideal of height 1 by Krull's principal ideal theorem [cf. [63], Th. 10.1], and AAz 1 is contained in the ring of II. Since AAz 1 is a one-dimensional local domain whose maximal ideal is generated by one element, it is a discrete valuation ring [ cf. 1(3.29)], hence it is the ring of the discrete valuation II. Moreover, since A is an R-algebra of finite type, and since regular local rings are universally catenary [cf. [63], Cor. 18.10], we have dim(A} = ht(Axd + dim(A/Axd = 1 + (n -1) = n. (d) Let q be a maximal ideal of A with Xl E q. Then q := q/AXI is a maximal ideal in A, and q can be generated by n - 1 elements [cf. [63], Ex. 4.27 or [204], vol. II, Ch. VII, § 7, Th. 24], hence q can be generated by n elements. Since A is an R-algebra of finite type, and since regular local rings are universally catenary, we have n = ht(q) + dim(A/q}, hence dim(Aq) = n, and therefore Aq is an ndimensional regular local ring. (e) Since R is integrally closed and all the powers mm are integrally closed in R, the ring A is integrally closed [cf. B(6.9)]. (f) We have mm A = Axf for every mEN. Let z E Axf n R. Then we have II(Z) ~ m, hence we get z E mm [Since mm is a II-ideal].

=

=

=

=

=

(2.4)

REMARK:

cf. 1(11.9).

=

Note that now we have equality in (3) of Abhyankar's theorem,

2 Quadratic Transforms and Ideal Transforms

2.2

255

Quadratic Transforms and the First Neighborhood

(2.5) For the rest of this chapter R is a two-dimensional regular local ring.

(2.6) QUADRATIC TRANSFORMS: (1) We consider the canonical homomorphism of graded rings r.p: R,(m,R)

= EBmnT n -+ n:2:0

EBmnlmnH

= grm(R)

n:2:0

from the Rees ring R,( m, R) c R[ T] to the associated graded ring gr m(R). Note that grm(R) = ~[x,y] [here m = Rx + Ry, x:= In(x) = x mod m2 , y:= In(y) = y mod m2 ]. The kernel of r.p is the ideal J :=

EB mnHTn; n:2:0

it is a homogeneous prime ideal of R,( m, R). Let IP'R = IP' be the set of homogeneous prime ideals of grm(R) of height 1, i.e., lP'is the set of closed points ofProj(grm(R)). Every p E IP' is a principal ideal, generated by an irreducible homogeneous element 7 E ~[x,y]. We set deg(p) := deg(7). Let p = (7) E IP' where 7 E grm(R) is homogeneous of degree m, and choose f E mm with In(l) = 7; then we have ordR(I) = m and n~ := r.p-l(p) = J + fTmR,(m,R). Now n~ is a closed point of Proj(R,(m, R)), and the local ring of this point is Sp := R,(m, R)(n~). Every such ring is called a quadratic transform of R. (2) Let p = (7) and f be as in (1). Either x or y do not lie in p. We consider the case that x does not lie in p. Then we have xT ~ n~, and in R,(m, R)(xT) = R[ mix] =: A we have the following [cf. B(5.7)]: The prime ideal of A determined by J is the ideal Ax, the maximal ideal np of A determined by n~ is the ideal generated by x and f I xm [since, for n E N, the homogeneous elements of n~ of degree n are linear combinations of terms xiyiTn, i + j = n + 1, and xiyi flTn, i + j + lm = n, with coefficients in RJ, and we have R,(m, R)(n~) = An p [cf. B(5.6)], and Sp := An p is a two-dimensional regular local ring [cf. (2.3)(3)] with field of quotients K, and its maximal ideal is generated by x and f I xm. Moreover, mSp = Spx is a prime ideal of Sp, and Sp dominate R. The residue field of Sp is ~[T] I (7(1,17)) [note that np n R = m], hence the residue field of Sp is a finite extension of ~ of degree deg(p). We set [Sp : R] := deg(p). Furthermore, the valuation ring V R of VR contains Sp and has center Spx in Sp [since VR has center Ax in A by (2.3)(2)]. (3) Let, as in (2), p = (7) E IP', let 9 E grm (R) be homogeneous of degree d > 0 and assume that 9 ~ p, or, equivalently, that 7 does not divide g. We choose g E md with In(g) = g. Then n~ determines a prime ideal n~ in R,(m,R)(gTd) = R[mdlg] [cf. B(5.5) and B(5.6) J, and we have Sp = R[ md I g ]nilp . We show: Sp consists of all elements zlw with z, w E R, ord(z) ~ ord(w) and In(w) ~ p. Proof: In fact, let z, w be as stated. Set s := ord(w); the condition 7 f In(w) is equivalent to wTs ~ n~. Upon replacing, if necessary, w by w n and z by

256

VII Two-Dimensional Regular Local Rings

wn-Iz for some n E N, we may assume that ord(w) is a multiple of d, say td. By this replacement the condition wTord(w) rf. n~ is not affected. Now we have zjw = (zjyf)j(wjyf), we have zjgt, wjgt E R[mdjg], and, moreover, wjgt rf. n~. This means that zjw ESp. On the other hand, every element of Sp can be written as (zjgt)j(wjgt) where z, w E R, ord(z) ~ td, ord(w) = td for some tEN and wTtd rf. n~, hence does not divide the element In(w).

7

{2.7} DEFINITION: The set {Sp I P E lP'} of quadratic transforms of R is called the first neighborhood of R, and shall be denoted by NI (R). {2.8} REMARK: Let P E lP' and n E N. Then we have mnsp n R = mn. In fact, let us choose x E m \ m2 with (In (x)) ::/: p. Then Sp is a localization of A := R[ mjx] with respect to a maximal ideal of A containing Ax, and therefore we have Axn = Spxn nA [ef. [63], Th. 3.10, and note that Ass(AjAxn ) = {Ax}]. Now the result follows from (2.3}(4). {2.9} INTERSECTION OF QUADRATIC TRANSFORMS: wise different, and set S := SPl n ... n Sp,.

Let PI, ... ,Pt E lP' be pair-

By prime avoidance [ef. [63], Lemma 3.3] we can choose a homogeneous element 9 E grm(R) of positive degree with 9 rf. PI U··· U Ptj we set d := deg{g), and we choose 9 E md with In(g) = g. We set A:= R[mdjg] = 'R.{m,R)(gTd)j

A is integrally closed by B(6.9). In A we have the maximal ideals n~l' ... ' n~, of height two determined by the prime ideals n~l' ... ,n~, of 'R.( m, R) [note that, for i E {I, ... , t}, we have gTd rf. n~i and SPi = An~i ] and the prime ideal p determined by Jj clearly we have p = mAo Now we have S = An"Pi

n··· nAn"P, .

We set E := A \ (n~l U··· un~,)j then E- I A is a two-dimensional integrally closed . al·d ~-1 np1 " , ••• ,nt := ~ ~-1" ~-IA senn·1ocal· nng WI·th maxlm 1 eals nl := ~ np , • S·!nce ~ is the intersection ofits localizations {E- I A)n pi = An~i' i E {I, ... , t} [ef. B(2.5)]' we have S = E-I A and Sni

= SPi

for i E {I, ... , t}.

(There should be no confusion with regard to the different meanings of the subscripts ni resp. Pi of S.) By (2.6){3) it is easy to check that S is the set of all elements ujv where u, v E R, ord(u) ~ ord{v), and In(v) ¢ PI U·· ·UPt. The localizations Snl' . .. , Sn, are factorial, hence S is also factorial [ef. (1.2)]. Moreover, since mS = pS, mS is a prime ideal of S, and since (mS)Sni = mSni is a principal prime ideal of Sni = SPi for i E {I, ... , t}, the ideal mS is a prime ideal of S of height 1, hence it is a principal ideal mS = Sz for some z E S.

2 Quadratic Transforms and Ideal Transforms

257

Thus, we have proved: (2.10) Proposition: A finite intersection S of quadratic transforms of R is a two-dimensional factorial semilocal domain; mS is a principal prime ideal of S.

2.3

Ideal Transforms

(2.11) IDEAL TRANSFORM: (1) Let p E IP' and set S := Sp; let n be the maximal ideal of S. Since Rand S are factorial rings, for every ideal a of R the ideal transform as [cf. (1.4)] of a in S is defined. In the following, we give an explicit description of as. (2) Let {x,y} be a regular system of parameters of R with In(x) f/:. p. Set A := R[ m/x]; then we have S = An p and mS = Sx [cf. (2.6)(2)]. Let q be a prime ideal of height 1 of A different from Ax. Then x f/:. q, hence we have Aq = (A x )qA Furthermore, we have Ax = Rx [since A C Rx], and (Rx)qR2 = RqnR, hence we have Aq = RqnR, and therefore q n R is a prime ideal of height 1 of R. Let a be an m-primary ideal of R of order n; note that 1 is a greatest common divisor for the elements of a [cf. (2.1)]. The only prime ideal of A of height 1 which contains aA is Ax since mh C a for some hEN, hence xh E aA, and we have aA C Ax n , aA rt Axn+1 [since Axm n R = mm for every mEN by (2.3)(4)]. In particular, no height one prime ideal of A contains the ideal x-naA of A. Furthermore, this implies that Sx is the only prime ideal of S of height 1 which contains as, as C Sx n , as rt Sx n+1 and xn is a greatest common divisor for the elements of as. Therefore we have (S : as)K = Sx- n , hence as = S· a/xn. Moreover, as is either the unit ideal or it is an n-primary ideal of S [cf. (2.1)]. Now let hER be irreducible; note that RRh is a discrete valuation ring. Then we have either ShnR = Rh in which case (Rh)S is a principal prime ideal of S, or we have RhnS :I Rh in which case (Rh)S = S [cf. (1.5)]. Set n := ord(h). If h = ux with a unit u of R, then we have n = 1 and Sh n R = Sx n R = m:l Rh, hence S = (Rh) S = S . h / x. Now we consider the case that x and h are not associated in R. If In(h) f/:. p, then xn /h E S, hence h/xn is a unit of S [cf. (2.6)(3)]. Therefore we have Sh = Sx n = mnS, hence Sh n R = Axn n R = mn:l Rh, and therefore we have S = (Rh)S = S . h/xn. If In(h) E p, then we have (Sh) n R = Rh. In fact, let u, v E R with ord(u) ~ ord(v) and In(v) f/:. p, and (u/v)h =: z E R. From hu = vz and h f v [since In(v) f/:. p and In(h) E p] we get that h I z and therefore that u/v E R. In this case we have SRh = RRh [cf. 1(3.30)]. Furthermore, we have RRhh = RRh . h/xn = SRh . h/xn. Now h/xn is irreducible in SRh and h/xn E S. We show that h/xn is also irreducible in S. Suppose, on the contrary, that h/xn is not irreducible in S. Then we have h/xn = (u/v)(u'/v') where u, v, u' and v' E R, deg(u) = deg(v), deg(u') = deg(v'), In(v) and In(v') lie not in p, and u/v, u'/v' are non-units of S. Since h E In(p), we see that h divides neither v nor v'. From hvv' = xnuu' it follows therefore that h divides exactly one of the elements u, u', say h divides u, hence u = hu" with u" E R. Then we have vv' = xnu'u", and therefore In( u') f/:. p, hence u' /v' is a unit of S, in contradiction with our 2

•

258

VII Two-Dimensional Regular Local Rings

assumption. Hence h/ xn is irreducible in S, and therefore S . h/ xn is a prime ideal of S. Thus, we have S·h/x n = SRh ·h/xnnS, i.e., we have (Rh)S = S ·h/xn , and we have shown: H hER is irreducible and ord(h) = n, then (Rh)s = S· h/xn , and h/xn is either irreducible in S-and this is the case if In(h) E p-or it is a unit of S-and this is the case ifIn(h) ¢ p. (3) Let I, 9 E R be irreducible and not associated, and assume that (Rf)s and (Rg)S are prime ideals of S. Then we have (Rf)s =I (Rg)S and Sx =I (Rf)s. Proof: Suppose that (Rf)s = (Rg)s. Set m := ord(f), n := ord(g), and assume, as we may, that m 2: n. There exist u, v E R with In(u) ¢ p, In(v) ¢ p, deg(u) = deg(v), and I/xm = (g/xn)(u/v). Then we have Iv = guxm- n. Since I and 9 are irreducible and not associated, and x divides neither I nor g, we see that I and u, as well as 9 and v are associated, contradicting In(f) E p and In(g) E p [cf. (2)]. Similarly, one can prove the second assertion in (3). (2.12) RESULTS ON IDEAL TRANSFORMS: We apply the results of (2.11) and get: (1) Since ideal transforms are compatible with products [cf. (1.5)], we have the following: (a) Let I E m, I =I 0, and set n := ord(f). Let p E lP and choose x E m \ m2 with In(x) ¢ pj then we have (Rf)sp = Sp· I/xn. Moreover, we have shown: Let grm(R) In(f)

= II qn U ) q

qED>

be the factorization of the homogeneous principal ideal grm(R) In(f) of grm(R) into a product of homogeneous principal prime ideals of grm(R). Then we have (Rf)sp =I Sp iff np(f) > O. (b) Let a be an ideal of R of order n. Let p E lP' and choose x E m \ m2 with In( x) ¢ p. Then we have aSp = Sp . a/ xn [since every non-principal proper ideal of R has the form gb with 9 E R and an m-primary ideal b of R, cf. (1.3) and (2.1)]. (2) Let PI, ... ,Pt E lP, and let S be the intersection of the rings SPi' ... , SPt. Then S is factorial, and there exists z E S with mS = Sz [cf. (2.10)]. Let a be an ideal of R of order n. Then we have

In fact, let {x, y} be a system of generators of m and let i E {I, ... , t}. Then we have either In(x) ¢ pi-in which case we have (mS)Sp; = Sp;x, hence Sp;z = SiXor In(y) ¢ pi-in which case we have (mS)Sp; = Sp;y, hence Sp;z = Sp;y. By (1) we have aSp; = Sp; . a/z n for every i E {I, ... , t}, and since aSp; = aSSp; for every i E {I, ... , t} [cf. (1.5)] and a/z n c as, we get (*) [cf. B(2.5)]. (2.13) CHARACTERISTIC IDEAL OF AN IDEAL: Let a be an non-zero ideal of R, and set r := ord(a). The homogeneous ideal c(a) of grm(R) which is generated by all the elements In(f) = I mod mr +1 where I E a is of order r, is a principal idealj it is called the characteristic ideal of a. We have c(a) = (9) for some homogeneous

2 Quadratic Transforms and Ideal Transforms

259

9 E grm(R), and we set deg(c(a))

=: deg(g). We have 0 ::::; deg(c(a)) ::::; T. Every generator of c(a) is called a characteristic form of a. Note that c(a) = (1) if a is a power ofm. If b is another non-zero ideal of R, then we have c(ab) = c(a)c(b), and if a C band ord(a) = ord(b), then c(b) divides c(a), and therefore deg(c(b)) ::::; deg(c(a)). In particular, if a = RJ is a non-zero principal ideal of R, then we have c(a) = (In(f)).

(2.14) Proposition: Let pEP, and let S := Sp be the quadratic transform of R defined by p. Let a be a non-zero ideal of R, and let rI' be the highest power of p which divides c(a). Then we have s := ords(aS ) ::::; h, and if h > 0, then we have s > O. In particular, we have as = S iff p f c(a). Proof: We set b := as and T := ord(a). Let x E m \ m2 be such that In(x) £t p. Then we have b = S· ax- r [cf. (2.12)]. Let p = (g), set n := deg(g), and choose 9 E mn with In(g) = g. There exists an element Z E a of order T such that In(z) E c(a) C ph but that In(z) £t ph+1. Then we have In(z) = ghw where w E grm(R) is homogeneous of degree T - hn and 9 f w. We choose w E mr - hn with In(w) = w; then we have z_ghw E m r+1, hence in S we have z/xr _(g/xn)h . (w/x r - hn ) E Sx. From 9 f W we get that w/x r - hn is a unit of S [cf. (2.6)(3)]. Since the maximal ideal n of S is generated by x and 9 / x n , we see that z / xr £t nh+1, and since z/x r E b, we have b ct. n h +1, hence s ::::; h, and therefore b = S if h = O. Now we assume that h > O. If we apply the argument just given to any element z E a of order ~ T, then we get z/x r E Sx, and this means that s > O. Therefore we have b '" S iff p divides c(a).

(2.15) Corollary: Let a be a non-zero ideal of R, and let c(a) =

II qn

q (l1)

qEP

be the factorization of the ideal c(a). For pEP we have aSp", Sp iff np(a)

> O.

(2.16) Lemma: Let x E m \ m2, and let nl, ... , nh be pairwise different maximal ideals of A := R[ m/x] containing x. For i E {I, ... , h} let qi be an ~-primary ideal of A. For every sEN we have Ass(A/XBql ... qh) = {Ax, nl, ... , nh}' Proof: We set q := ql ... qh = ql n ... n qh [note that the ideals ql," . ,qh are pairwise comaximal by B(lO.I)]; then we have {nl,"" nh} = Ass(A/q). Now we have Ax E Ass(A/xBq) since Ax is a minimal prime overideal of xBq [cf. [63], Th. 3.1]. Let i E {I, ... , h}; since ~ E Ass(A/q), there exists Zi E A with q : AZi = ni, hence with x 8q : AXBZi = ~, and therefore we have ni E Ass(A/xBq). No height one prime ideal of A different from Ax contains X8 q. Since A is twodimensional [cf. (2.3)(4)]' it is enough to show that no maximal ideal of A different from the ideals nl, ... , nh lies in Ass(A/xBq). Let n £t {nl. ... , nh} be a maximal ideal of A which contains xBq. Since q ct. n, we have x E n, and therefore we have

260 n

VII Two-Dimensional Regular Local Rings

= Ax + A(g/xn ) where 9 E mn for some n EN and g/x n ¢ lli for i E {I, ... , h} = xBq which implies that n ¢ Ass(A/xBq).

[ef. (2.6)(2)]. We show that xBq : n

Let Z E xBq : n. Then we have z· (g/xn) E xBq C AxB. Since g/x n ¢ Ax, we have z E Axs , hence z = xBw with w E A, and w· (g/xn) E q. Since g/x n ¢ ni for i E {1, ... ,h}, we have wE qi for i E {1, ... ,h}, hence wE q, and therefore we have z E xBq. (2.17) TRANSFORM OF IDEALS: Let x E m \ m2 , and set A:= R[m/x]. Let a be a non-zero ideal of R, and set r := ord(a). We have aA c mr A = Axrj we set a' := x- r aA, and we call a' the transform of a in A. In the following, we study properties of a'. (1) Let P E 1P, and assume that In{x) ¢ pj then we have Sp = Anp and aSp = a'Sp [ef. (2.12)(I)(b)]. (2) In particular, assume that a is an m-primary ideal. We have aA = xra'. There exists hEN with h > r and with mh C a, hence every prime ideal of A which contains a', also contains the principal prime ideal Ax. No height 1 prime ideal of A contains a' [ef. (2.11)(2)]. Let us consider the case that a' is a proper ideal of Aj then Ass(A/a') consists of those maximal ideals of A which contain a', each of them contains x and therefore Ass{A/ a') = {npl' ... , np,,} where P1, ... , Ph E IP are the different prime ideal factors of c{a) different from (In{x» [cf. (2.15) and (2.6)(2)]. Let a' = q1 n ... n qh = q1 ... qh where qi is an np;-primary ideal for i E {I, ... , h}, be the primary decomposition of a'. Then [cf. B(10.1) and [63], Cor. 9.1] A/a' is artinian, and A/a' ~ A/q1 x ...

X

A/qh.

We set E:= A \ (Upl U·· ·Unp,,), S:= E- 1A and Si:= Sp; for i E {I, ... ,h}j then we have S = S1 n···nSh [ef. (2.9)], a'SnA = a' and aSnA = aA [cf. (2.16) and [63], Th. 3.10]. Since mS = Sx, we find that a'S = as and that a'Si = as; for i E {I, ... , h} [ef. (1.5)]. Since Sis semilocal, we have a'Sl n ... n a'Sh = a'S [ef. B(2.5)]. Therefore we have (a'Sl n··· n a'Sh) n A = a', and a similar result holds with a' replaced by xr a' = aA. The image of E in A/a' is the set of units of A/a'j therefore we have A/a' ~ S/as. Let i E {1, ... ,h}. Since xra'Si = aSi = xra s ;, we have as; = a'Si = qiSi, and this implies that A/qi ~ Silas;. In particular, we have S/as ~ St/a S1 x ... x Sh/as".

(3) Now let a = RJ be a non-zero ideal in R, set r := ord(f), and let P1,··· ,Ph be the pairwise different factors of c(a) = (In(f» which are different from (In{x». The only maximal ideals of A of the form np for some P E 1P, P '" (In(x», with J/ xr E np are the ideals np1 , ••• , np" .

2.4

Valuations Dominating R

(2.18) DIRECTIONAL IDEAL OF A VALUATION: Let v be a valuation of K dominating R with valuation ring Vj note that V '" K. Then K, can be considered as a

261

3 Complete Ideals subfield of the residue field K. v of V, and m is a v-ideal with v(m) by Abhyankar's theorem [cf. 1(11.9)], we have

> O. Note that,

rat.rank(v) +tr.dn(K. v ):::; dim(R) = 2, hence we have either tr. dn(K. v ) = I-in this case V is of the second kind with respect to R, K. v is a finitely generated extension of K., rank (v) = rat. rank( v) = 1, v is discrete ofrank 1 [cf. 1(11.9)(3)], or we have tr. dn(K. v ) = O-in this case V is of the first kind with respect to R. If, in particular, the value group of v is archimedean, then every non-zero proper v-ideal of R is an m-primary ideal [cf. 1(11.7) ]. For the rest of this subsection we assume that V i- VR. We choose x, y E m with m = Rx + Ry, and we set x := In(x), y := In(y). We may assume that vex) :::; v(y), hence that v(m) = vex). Clearly we have v(z) ~ VR(z)v(m) for every z E R, z i- O. Since V i- VR , there exists z E R \ {OJ with v(z) > VR(z)v(m). We choose f E R[X,Y] with z = f(x,y). Then we have v(f(l,y/x» > o. This means that the v-residue ( of y/x is algebraic over K.. Let 9 E K.[Z] be the minimal polynomial of ( over K.. Set d := deg(g), and define the homogeneous polynomial 9 E K.[x, y)] = grm(R) by g(x, y) := xdg(ylx)j then 9 is irreducible and g(l, () = O. Every irreducible homogeneous polynomial Ii E grm(R) with h(l,() = 0 has the form h = 'Yg with "I E K. x , and it is called a directional form of v or of V. We set p := (g). We have Pi- (x), and we set p(V) = p(v) := p E IP' and call it the directional ideal of v or of V. Set A := R[ mlx]j we have V :J A and v(np ) > O. Therefore Up is the center of v in A, hence V dominates Sp. If v(y) = vex), then we have B:= R[mly] C V, Pi- (y), and the center of V in B is the maximal ideal of B corresponding to p, hence we have Sp = Bm(V)nB, and therefore Sp is the only quadratic transform of R which is contained in V. Thus, we have shown: (2.19) Proposition: Let v be a valuation of K dominating R, let V be the valuation ring ofv and assume that V i- VR . Let p = p(v) be the directional ideal of V. Then V dominates Sp and Sp is the only quadratic transform of R which is contained in V. Moreover, if rat. rank( v) = 2, then V is of the first kind with respect to R.

3 3.1

Complete Ideals Generalities

(3.0) In this section we keep the notations and hypotheses introduced in section 2. For any finite non-empty subset P C IP' we set

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VII Two-Dimensional Regular Local Rings

Remember that SP is a semilocal factorial domain, that {m(Sp) n Sp I PEP} is the set of maximal ideals of SP, and that we have e = npEP eSp for every ideal e of Sp [cf. (2.10) and B(2.5)]. (3.1) COMPLETE IDEALS: Let S be a two-dimensional regular local ring with maximal ideal n. Let a be an ideal of S. Since S is integrally closed, the integral closure of a in S is also the integral closure of a in the field of quotients of S [cf. B(6.9)]. An ideal a of S which is an integrally closed ideal of S will be called complete following the time-honored tradition set up by Zariski. We use the notion of complete ideals also for the rings Sp. Every principal ideal of S is complete since S is integrally closed. Let a be a non-zero non-principal proper ideal of S. Then we have a = ab where a is a greatest common divisor for the elements of a, and b is an n-primary ideal [cf. (2.1)]. Since S is integrally closed, we know [cf. B(6.16)(6)] that a is complete iff b is complete. Thus, in studying complete ideals of S, it is often enough to consider only complete n-primary ideals of S. The following general result shall be very useful.

(3.2) Proposition: Let a, b be integrally closed ideals of an integrally closed noetherian domain A, and let e be a non-zero ideal of A. If we have ae = be, then we have a= b. Proof: Let a E aj then we have ae C ae = be, hence a is integral over b by B(6.2)(3), hence a lies in b and therefore we have a C b. By symmetry, we also obtain b C a, hence we have a = b. (3.3) REMARK: (1) Let T ::) R be a subring of K, and let b be an ideal of T which is integrally closed in T. Then a := R n b is complete. In fact, if z E R is integral over a, then, considering an equation of integral dependence for z over a, one sees that z is integral over b [since ai C bi for i E No ], hence that z E b n R = a. (Note that this argument works nor only for R but also for any integrally closed subring of T having field of quotients K.) (2) Let a be a non-zero ideal of R, and set r := ord(a). Then we have a C mr , a¢. ar+1 , and since mr is integrally closed [cf. (2.2)], the integral closure of a has the same order as a. (3.4) Proposition: Let a be a complete ideal of R, and let P C non-empty subset. Then aSp and aSp are complete ideals of Sp.

JP>

be a finite

Proof: We set T := Sp. We have aT = zraT where r := ord(a) and z E T with mT = Tz. Since T is integrally closed [cf. (2.10)], we know that aT is complete iff aT is complete [cf. B(6.16)(6)]. We show that aT is complete. Let u E T be integral over aT. Then we have an equation un + alUn - 1 + ... + an = 0 where ai E (aT)i = aiT for i E {1, ... ,n}. This means that there exist bi E ai and w E R with In(w) ¢ p for every pEP and with ord(bi ) ~ ord(w) and

3 Complete Ideals

263

ai = bijw for i E {l, ... ,n} [ef. (2.9)]. Set u' := uw. Then we get an equation u ln + bluln - l + b2 wu ln - 2 + ... + bnwn- l = 0, hence u' is integral over a, and therefore we have uw = u' E a, hence we have u = u' jw EaT. (3.5) Corollary: Let x E m \ m2 , and set A := R[mjx]. Let a be a complete ideal of R. Then aA and the transform a' of a in A are integrally closed ideals ofA. Proof: Since A is integrally closed [ef. (2.3)(4)], the proof given above works in this case also [one uses a power of x instead of w].

3.2

Complete Ideals as Intersections

(3.6) Lemma: Let B be a noetherian domain with field of quotients L, and let P i- {O} be a prime ideal of B. Let {Xl, ... , xn} be a system of non-zero generators ofp, and set B i := B[pjxd fori E {l, ... ,n}. Then there exist i E {I, .. . ,n} and a prime ideal q of height 1 of Bi with q n B = p. Proof: The rings B l , ... , Bn are B-algebras of finite type, and therefore noetherian rings. Let V be a valuation ring of L having center p in B [ef. 1(3.5)], and choose i E {l, ... ,n} with VXi = pV C m(V) [ef. 1(2.4)]. We have Bi C Vj we set Pi := m(V) n B i . Now we have pBi = BiXi C Pi, and clearly Pi n B = p. Therefore we have BiXi n B = p. Now let q C Pi be a minimal prime ideal of BiXi. By Krull's principal ideal theorem [cf. [63], Th. 10.1] we have ht(q) = 1, and clearly we have q n B = p. (3.7) Proposition: Let a be an ideal of R, and let a be the integral closure of a in R. Then: (1) For every valuation ring V of K with V ~ R we have aV = aVo (2) For every z E K \ a there exists a valuation ring V of K being of the second kind with respect to R and such that z i aV. Proof: (1) Since R is integrally closed, the integral closure of a in K is also the integral closure of a in Rj therefore (1) is just a restatement of part of B(6.14). (2) We consider the ring

B:=R[z-la]= U(a+Rz)njznj n~O

B is an R-subalgebra of K of finite type, hence it is noetherian. We set p :=

m + z-laB. We have

therefore p is an ideal of B. We have 1 i p, since otherwise there would exist wE mand n E N with l+w E z-la(a+Rz)njzn, hence (l+w) z n+l would

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VII Two-Dimensional Regular Local Rings

lie in an+1 + zan + ... + zn a, and z would be integral over a [note that 1 + w is a unit of R]. Therefore we have p n R = m. We have B = R + p since (a + Rz)njzn = R + z-la(a + Rz)n-l jzn-l for n EN. Therefore we have Bjp = (R + p)jp ~ Rjm, hence p is a maximal ideal of B. There exists an R-algebra C of finite type with B c C c K and a prime ideal q of C of height 1 with q n B = P [cf. (3.6)]. Since regular local rings are universally catenary [cf. [63], Cor. 18.10], we have ht(q) + dim(Cjq) = dim(C). Now Cjq is an integral Rjm-algebra of finite type, and therefore we have dim(Cjq) = tr. dR / m Q(C jq) [cf. [63], Th. A on p. 286]. Let n be a maximal ideal of C with q C n. Then we have n n B = P and dim(Cn ) = dim(R) [cf. [63], Th. 13.8, and note that Rand C have K as field of quotients, hence K ®R C = K and dim(K) = 0]. We have dim(Cn ) = dim(C) [since C is catenary], and therefore we get at last

tr.dR/m(CqjqCq) = dim(R) -1

=1

[note that Q(Cjq) = CqjqCq]. Let V be a valuation ring of K having center qCq in C q [cf. 1(3.5)]; then V is of the second kind with respect to R. Furthermore, we have z-la C m(V), hence z does not lie in aVo (3.8) Proposition: Let a be an ideal of R. If a is complete, then we have

a=

naSp-

pEP

Proof: The left hand side of (*) is contained in the right hand side. Let z E K \ a; it is enough to show that there exists p E IP' with z ¢ aSp. By (3.7) there exists a valuation ring V of K dominating R with z ¢ aVo If V = VR, then every quadratic transform of R is contained in VR, and if V i- VR, then there exists a quadratic transform Sp of R which is dominated by V by (2.19), and we have z ¢ aSp. (3.9) REMARK: Let {x, y} be a system of generators of m. For every complete ideal a of R we have

a = aR[mjx] n aR[mjy]

=

naSp.

pEP

This follows immediately from (3.8). (3.10) Corollary: Let a be a complete non-zero ideal of R, and let P C IP' be a finite non-empty subset. Set r := ord(a). If P contains all p E IP' which divide c( a), then we have Proof: We have mr S p

nR = (

n

mr Sp)

pEP

n R = mr

3 Complete Ideals

265

[for the first equality sign cf. (3.0), and for the second equality sign cf. (2.8)]. On the other hand, we have aSp C mrSp, hence aSp = aSp n mrSp, and therefore we have aSp n R = aSp n mr. For every P ft P we have aSp = mr Sp [cf. (2.15)], hence we obtain from (3.8) [note that mr is complete] aSp

n mr

=

n nn

mr Sp =

aSp

pEP

pEP

n

aSp = a.

pEl?

(3.11) Corollary: Let a be a complete non-zero ideal of R. We have c(a) iff a is a power of m.

= (1)

Proof: We set r := ord(a). If c(a) = (1), then we have aSp = mr Sp for every p E IP' [cf. (2.15)], hence a = mr by (3.8). Conversely, if a = mr , then we have c(a) = (1). (3.12) Proposition: Let a be a complete m-primary ideal of R which is not a power of m. Set r := ord(a) and P := {p E IP'I p I c(a)}. Then P :I 0, and the canonical map mr /a -t mrSp/aSp is an isomorphism of R-modules.

Proof: Since we have aSp n mr = a [cf. (3.10)], this map is injective. (1) Let {x, y} be a system of generators of m, set A := R[m/x], B := R[ m/y] and G := R[ m 2 /xy] = R[y/x, x/y]. Note that G is a localization of A with respect to the multiplicatively closed system {(y / x)n I n E No}j likewise, G is a localization of B with respect to the multiplicatively closed system generated by x/yo (a) Let a' be the transform of a in A. First, we assume that a' is not the unit ideal. Then the set {PI, ... ,Ph} of prime ideals of IP' which divide c(a) and are different from (In(x)) is not empty and Ass(A/a' ) = {nI, ... , nh} where ni = np • for i E {I, ... , h} [cf. (2.15) and (2.17)(2)]. We consider the commutative diagram

mr A/aA -..!.. 0 for i E {I, ... , l}, and ei 0 for i E {I + 1, ... ,h} [where 0 ~ 1 ~ h j, we have

=

r-sI b · .. bI, C() =m a = PIel ... PIe, , E {I, ... ,l} we have C(bi) = P~' .

a

and for i

Proof: We set Si := Sp, for i E {I, ... , h} and S := Sl n··· n Sh. (1) and (2): We have as = aS1n· . ·naSh [cf. (3.0)], and aSnR = a by assumption, hence we have

a = a1 n··· n ah.

Let i E {I, ... , h}. Since aSi is a complete ideal of Si [cf. (3.4)], ai is also a complete ideal of R [cf. (3.3)]. From a C ai we see that ai contains a power of m, we have ord(ai) ~ r, and since ai C aSi, we have ord(ai) = vR(ai) ~ vR(a) = r, and therefore we obtain ord( ai) r, hence ai is m-primary. We have ai aiSi n R [since ai is contracted from Si, it is the contraction of its extension], and therefore c(ai) =: P~' is a power of Pi, bi is complete and m-primary or bi = R, and we have bimr-s, = ai [cf. (3.26)], hence ord(bi) = 8i and C(bi) = c(ai). (3) We set t := 81 + ... + 8h and b := b1 ... bh. Since c(ai) = P~' divides c(a), also c(a1 ... ah) divides c(a), hence t = 81 + ... + Sh ~ deg(c(a» ~ r. Since mr-tb C bimr - s , = ai for i E {I, ... , h}, we have mr-tb C a1 n··· n ah, hence

=

=

For i E {I, ... , h} the ideal bi is m-primary or bi = R; therefore mr - t b C a is m-primary, hence there exists j E N with m r+i C mr-tb. We have c(b) = p~l ... p~h. Clearly we have ord(mr-tb) = ord(a), hence c(a) divides c(b) [cf. (2.13)]; on the other hand, c(b) = c(a1 ... ah) divides c(b). Therefore we have

c(a1 ... ah)

= c(a) = p? ... p~h,

= =

and, in particular, s t 81 + ... + 8h. Our aim is to show equality in (*). We relabel the ideals Pt. ... ,Ph as in (5). If 1 = 0, then we have s = 0 and a = mr by (3.11), hence a1 = ... = ah = mr by (2.8), and therefore b1 = ... = bh = R. In particular, we have equality in (*). Now we consider the case 1 > o. Then c(a) = p~l ... p~'. Let I E a be of order r. e'

e'

The ideal (In(f» therefore has the form P1 1 ••• PI' q where e~ ~ e1, ... ,e; ~ e/, q is generated by a homogeneous element of grm (R) [q = (1) is allowed] and none of the ideals PI, ... ,PI divides q; moreover, there exists an element I E a of order r with (In(f» = p~l ... p~' . (a) Let I E a be of order r and with (In(f» = p~l ... p~'q. By the lemma above, there exist elements ft, ... , II and 9 in R such that 1- ft··· Ilg E mr +i C a and with (In(h» P~' for i E {I, ... , l} and (In(g» q. Now we have ft ... ihg E a. Let i E {I, ... , l}. We choose x E m \ m2 such that (In(x» '" Pi. Then the element ft·· . h-tfi+1 .. ·llg/xr - s , is a unit of Si [cf. (2.6)(3)], hence h/xs , E as, =

=

=

4 Factorization of Complete Ideals

277

x-raSi, and therefore we have limr-si E aSi n R = ai, hence Ii E ai : m r - Si = bi. Therefore we have h ... II E b. Since ord(g) = deg(In(g)) = r - 8, we have 9 E m r - s , hence h ... Ilg lies in mr-sb. (b) Let l' be an element of a of order r with (In(f')) = P~~ ... p~~q where e~ > ei for at least one i E {I, ... , l}. Then I + l' satisfies the assumptions of (a) for I, hence I + l' E mr-sb, and therefore l' lies in mr-sb. (c) Lastly, let I' E a and ord(f') > r. Then I + l' satisfies the assumptions of (a) for I, and again we find that l' lies in mr-sb. Thus, we have shown that a is contained in mr-sb, hence we have a = mr-sb. (4) and (5) follow immediately.

(4.6) Corollary: If a is a simple complete m-primary ideal different from m, then c( a) is a positive power of P E lJD, and aSp is a simple complete m(Sp)-primary ideal ofSp • Proof: The first assertion follows from (4.5)(5). (Another proof follows from (3.27).) We set S := Sp, and we have to show that as is a simple ideal of S. Suppose that as is not simple, and write as = bl ··· bh with h > 1 where bl , ... , bh are complete m(S)-primary ideals of S [cf. (4.1)(2)). By (3.27)(i) there exist complete m-primary ideals aI, ... ,ah of R, not divisible by m, with af = bi for i E {I, ... , h}. By (3.27)(ii) we find that al ... ah = mta for some t E No which is absurd since a is simple. We want to show that a product of complete m-primary ideals of R is complete, again. We can do this now in a special case.

(4.7) Lemma: Let al, ... ,an be complete m-primary ideals of R. If the ideals c(al), ... ,c(an ) are pairwise coprime, then al ... an is a complete m-primary ideal. Proof: Since the product of a complete ideal with a power of the maximal ideal is complete, again [cf. (3.20)), and since, if ai is not a power of m, we can write ai = mti a~ with ti E No and a complete m-primary ideal a~ which satisfies ord( aD = deg(c(aD) for i E {l, ... ,n} [cf. (3.22)), we may assume, to begin with, that ord(ai) = deg(c(ai)) for i E {I, ... , n}. By (4.5)(5) and the assumption that, for i, j E {I, . .. ,n}, i ;j; j, the ideals c(ai) and c(aj) have no non-trivial common divisor, it is easily seen that it is enough to show the following: Let bl , ... , bh be complete m-primary ideals where, for i E {I, ... , h}, we have C(bi) = p~i with Pi E lJD and ei > 0, with ord(bi) = deg(c(bi)) =: 8i, and with pairwise different PI,.·· ,Ph· Then a' := bl ... bh is complete. We set 8 := 81 + ... + 8h; then we have ord(a') = 8. For i E {I, ... , h} we set Si := SPi> and we set S:= Sl n··· nSh and a:= a'SnR. Then we have as = a'S [since an extended ideal is the extension of its contraction) and a' c a, hence r := ord(a) ~ 8, and a is m-primary or a = R. Let i E {I, ... , h}; we have bjSi = m S; Si for every j E {I, ... ,h} with j ;j; i [cf. (2.14))' hence a'Si = ms-sibiSi, and therefore a'Si is a complete ideal of Si [ef.

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VII Two-Dimensional Regular Local Rings

(3.4) and (3.20)], hence also a'S = a'S1n·· ·na'Sh [cf. (3.0)] is a complete ideal of S [cf. proof of (3.3)(1)], hence a = a'S n R is a complete ideal of R [cf. (3.3)(1)]. We have mr aSi = (mr as) = (as)Si = (a' S)Si = a'Si = mS - Si biSi = m S bfi, hence aSi = ms-rbfi [since mSi is a principal ideal of Sd. Now bfi is an m(Si)-primary ideal of Si [cf. (2.11) and (2.14)], the ideal aSi is either equal to Si or it is also an m(Si)-primary ideal of Si and mr-sSi is a principal ideal of Si; therefore we have r = 8, and a is m-primary. We set ai := aSi n R; then ai is a complete ideal of R, we have aiSi = aSi = a'Si = m S - Si biSi, and since mS - Si bi is contracted from Si by (3.10), and ai is contracted from Si by definition, we have ai = mS - Si bi . This implies that ai = m S - Si (a : mS - Si ) [cf. (2.0)], hence that ai : mS - Si = bi [cf. (3.2)], hence we get a = mr - t b1 ... bh where t := deg(c(a)) by (4.5), hence t = 81 + ... + 8h = 8 = r, and therefore we get a = a'; this means that a' is complete.

4.2

Contracted Ideals

Now we are going to show that a product of complete ideals of R is complete; we need the following four lemmas (with the exception of lemma (4.11) which fits into this context but is needed only later).

(4.8) CONTRACTED IDEALS: Let x Em \ m2 , and set A:= R[m/x]. Let a be an ideal of R which is contracted from A; then a is the contraction of its extension, hence we have a = aA n R. We have mn = Axn n R for every n E ]I\! [d. (2.3)(4)]. A proper ideal of R which is contracted from A is not a power of Rx. (4.9) Lemma: Let x E m \ m2 , and set A := R[m/x]. contracted from A iff a : Rx = a : m.

An ideal a of R is

Proof: (1) We assume that aA n R = a. Let a E a: Rx. For every z E m we have az = (ax)· (z/x) E aA n R = a, hence am C a, showing that a: Rx C a: m. Since x E m, we have a : mea: Rx, hence we have shown that a : Rx = a : m. (2) We assume that a: Rx = a : m. There exists y E R such that m = Rx + Ry. Let a E aA n R; then we can write a = r/xn with n E No and r E amn. If n > 0, then we can write r = byn + cx with b E a and c E amn - 1 ; since r = ax n , we see that b = xd for some d E R [since (x, y) is a regular sequence on R]. This implies that r = x«yd)yn-1 + c). We have b E a, hence d = b/x E a: Rx = a: m, and therefore we have yd E a, hence r' := (yd)yn-1 + c E amn - 1, and we have a = r' /xn-1. By recursion we find that a E a. (4.10) Lemma: Let x E m \ m2 , and set A := R[ m/x]. The product of a finite number of ideals of R contracted from A is also contracted from A.

Proof: By induction, it is enough to prove the lemma for two ideals a, b. Let a and b be proper ideals of R which are contracted from A. Then a and b are not equal to powers of Rx [cf. (4.8)]. By (4.9) it is enough to show that ab : Rx = ab : m.

4 Factorization of Complete Ideals

279

Let z E R with zx E abo Now R/ Rx is a discrete valuation ring, hence a(R/ Rx) is a principal non-zero ideal of R/ Rx. We choose a E a with a(R/ Rx) = a(R/ Rx). Let a' E a; then there exist u, v E R with a' = ua + vx, hence v E a : Rx, and since x(a : Rx) c a, it follows that a = Ra+x(a : Rx). Likewise, there exists b E b with b = Rb + x(b : Rx). Therefore we have ab = Rab + xa(b : Rx) + xb(a : Rx) [note that xa(b : Rx) C ab and xb(a : Rx) c ab]. We have ab ¢ Rx; for every c E R with cx E ab we therefore have c E Rab + a(b : Rx) + b(a : Rx), hence cm Cab [since b : Rx = b : m and a: Rx = a: m by (4.9), we have m(b : Rx) C b and m( a : Rx) c a]. This shows that ab : Rx C ab : m, and therefore we have ab : Rx = ab : m. The following result shall play an important role in section 7 of chapter VIII.

(4.11) Lemma: Assume that R has infinite residue field K,. Then: (1) Let a be an m-primary ideal of R. Then there exists x E m \ m2 such that a is contracted from A = R[ m/x) iff J.t(a) = ord(a) + l. (2) Let aI, ... ,an be complete m-primary ideals of R. Then there exists x E m \ m2 such that aI, ... ,an are contracted from A = R[ m/ x). Proof: (1) We set T := ord(a), and let {Xl, yd be a system of generators of m. Then we have grm(R) = K,[In(xl), In(Yd]. Let I E a be of order Tj since K, is an infinite field, there exists a linear combination ofIn(xd and In(Yl) with coefficients in K, which does not divide In(f) in grm(R), hence there exists x E m \ m2 such that In(x) does not divide In(f) in grm(R). We choose y E R with m = Rx + Ry. This means that I = Ei+i=r aiixiyi with aii E R for all i, j E No with i + j = T, and aOr is a unit of R. Then we have a + Rx = RI + Rx. The ring R = R/ Rx is a discrete valuation ring [cf. (2.0) j, and the image] of I in R has order T in R. We have R/(a + Rx) = R/ R], and therefore we have lR(R/(a + Rx) = T. Since (a + Rx)/a ~ R/(a: Rx), and since R/a is an R-module of finite length, we have lR((a: Rx)/a) = T. Since a: mea: Rx, we have lR((a : m)/a) :::; T with equality iff a is contracted from A = R[ m/x) [cf. (4.9»). Since lR((a : m)/a) = J.t(a) -1 by (3.17), we get the assertion. (2) We have J.t(ai) = ord(ai) + 1 for i E {1, ... ,n} by (3.19). For i E {1, ... ,n} we choose Ii E ai of order ord(ai). Since K, is an infinite field, there exists x E m \ m2 such that In(x) does not divide In(M in grm(R) for every i E {I, ... , n}. By the proof of (1) this implies that the ideals al, ... , an are contracted from R[ m/x]. (4.12) Lemma: Let p E IP', and let al, ... , an be complete m-primary ideals of R. If c(ai) is a power of p for every i E {1, ... , n}, then (al ... an)Sp n R = al ... an. Proof: We set a := al··· an. We choose x E m \ m2 with (In(x)) :j:. p; then S := Sp is a localization of A := R[ m/ x) with respect to a maximal ideal n of A which contains x. Let i E {1, ... , n}; we have aiS n R = ~ [cf. (3.10)), hence aiA n R = ai, hence ai : Rx = ai : m [cf. (4.9)], and therefore we have aAn R = a [cf. (4.10)). Since a is an m-primary ideal, and since c(a) is a power of p, we have as n A = aA [cf. (2.17)(2)), and therefore we obtain as n R = a.

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VII Two-Dimensional Regular Local Rings

Now we can put our results together and prove:

(4.13) Theorem: The product of complete ideals of R is complete. Proof: Clearly it is enough to prove the theorem for a product a = al··· an of complete m-primary ideals al, ... an. We know that every complete m-primary ideal of R is a product of simple complete m-primary ideals of R [ cf. (4.1) (2) ], even if we do not know yet that the factors in such a product are determined uniquely. Nevertheless, it makes sense to prove by induction on i-and this will prove our assertion-: If a is a product of complete m-primary ideals al, ... , an and each of the factors aI, ... ,an is a product of simple complete m-primary ideals having colength ~ i, then a is complete. If the ideals al, ... , an are products of simple complete m-primary ideals of colength 1, then the ideals al, ... ,an are powers of m, hence also a is a power of m, and therefore a is complete [cf. (2.2)]. Let i > 1, and assume that a = al ... an is complete if the ideals aI, ... ,an are products of simple complete m-primary ideals having colength < i. Now we consider a product a = al ... an of complete m-primary ideals where all the ideals al, ... , an can be written as a product of simple complete m-primary ideals having colength ~ i. If all the factors aI, ... , an of a are powers of m, then a is complete. In the other case, for i E {I, ... , n} we write every ideal ai which is not a power of m as a product ai = mt• bil ... bin. where ti E No, where the ideals bil, ... ,bin. are complete mprimary ideals which are not divisible by m, and where C(bij) is a positive power of a prime ideal Pij E 1P', j E {I, ... , ni} [cf. (3.10) and (4.5)]; note that we can write any of the ideals bij as a product of simple complete m-primary ideals having colength ~ 1. In al ... an we group together all those factors bij for which the characteristic ideal is a power of the same prime ideal of 1P', hence we choose pairwise different ql, . •• ,qk E IP' and write a = mtcl ... Ck where, for i E {I, ... , k}, Ci = Cil ... Cik. is a product of complete m-primary ideals none of which is divisible by m, and for which the characteristic ideal is a positive power of qi, and where we can write the ideals Cij as a product of simple complete m-primary ideals having colength ~ i. It is enough to show that the ideals Cl, ... ,Ck are complete [cf. (4.7) and (3.20)]. Let i E {I, ... , k} and set Si := Sq.; let ni be the maximal ideal of Si. Since Ci = CiSi n R [cf. (4.12)], it is enough to show that CiSi is a complete ideal of Si [cf. (3.3)(1)]. We consider the transform CT' of Ci which is the product of the complete ni-primary ideals c~. We write Ci as a product of simple complete mprimary ideals of R having colength ~ l; then the transform in Si of each of these simple m-primary ideals is simple [cf. (4.6)] and has colength < i [cf. (3.14)], hence, by our induction assumption, applied to Si and cf', the ideal cf' of Si is complete, and therefore CiSi is also a complete ideal of Si [cf. B(6.16)(6)].

(4.14) Corollary: The set of non-zero complete ideals of R (resp. the set of complete m-primary ideals of R) is under multiplication a monoid with cancellation law. Every non-zero complete ideal (resp. every complete m-primary ideal) of R

281

4 Factorization of Complete Ideals

has a factorization into a product of simple complete (resp. simple complete mprimary) ideals of R. Proof: The first assertions follow from (4.13) and (3.2), while the last assertion was proved in (4.1){2). (4.15) NOTATION: Let T be a two-dimensional regular local subring of K with field of quotients K; let n be its maximal ideal, and let IP'T be the set of nonzero principal homogeneous prime ideals of grn(T). In the following, we use these notations: MC(T) is the monoid of complete non-zero ideals of T and MC'P(T) is the submonoid of complete ideals of T of finite colength. SC'P(T) is the set of simple complete n-primary ideals of T, and, for p E IP'T, SC'P(T,p) is the subset of SC'P{T) of those ideals whose characteristic ideal is a positive power of p, and MC'P(T,p) is the submonoid of MC'P(T) which is generated by the elements of the set SC'P(T,p). (4.16) Corollary: Let p E IP' and set S := Sp. The map a I-? as : SC'P(R,p) -+ MC'P(S) is injective and SC'P{S) is its image. Proof: We only have to show [cf. (4.6)]: If cp is a simple complete m(S)-primary ideal of S, then the unique complete m-primary ideal a of R with as = cp and aSq = Sq for every q E IP' different from p and which is not divisible by m [cf. (3.27)] is simple. Suppose that a is not simple. Then we have a = al ... ah with h > 1 where al,'." ah are simple complete and m-primary ideals [cf. (4.1){2)]. We label these ideals in such a way that = cp and ~ = ... = a~ = S. Then we have al = mta for some t E No by (3.27)(ii), hence mt a2 ... ah = R which is absurd.

ar

4.3

Unique Factorization

We quote Zariski [cf. [204], vol. 2, p. 385]: "The culminating point of our theory of complete ideals is a theorem of unique factorization of complete ideals into simple complete ideals." These results do not hold in regular local rings of higher dimension. For a certain generalization one may consult Lipman's paper [131]. (4.17) Theorem: Every complete ideal of R has a unique factorization into a product of simple complete ideals of R, i.e., MC(R) is a free monoid which is generated by the simple complete m-primary ideals of R and the prime ideals of height 1 of R. Proof: It is enough to prove the assertion for complete m-primary ideals of R [cf. (3.1)]. Since MC'P(R) has cancellation law [cf. (4.14)], it is enough to show the following: If a simple complete m-primary ideal a of R divides a product bl ... bh of simple complete m-primary ideals bl , ... , bh of R, i.e., if ac = bl ··· bh for some ideal c of R, then we have a = bi for some i E {I, ... , h}. We prove this

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VII Two-Dimensional Regular Local Rings

by induction on lR(Rja). If lR(Rja) = 1, then a = m, and the result follows from (3.25). Assume that l > 1, and that the result is true for simple complete m-primary ideals a of R with lR(Rja) < l. Now let a be a simple complete mprimary ideal of R with lR(Rja) = l. Then c(a) is a positive power of p for some p E IP [cf. (4.16)], and, for i E {I, ... , h}, C(bi) is either a positive power of p or is not divisible by p [cf. (4.16)]. We set S := Sp. We have as,s = br .. , b~ [cf. (1.5)], and, for i E {I, ... , h}, br is a simple complete m(S)-primary ideal of S or br = S [cf. (2.14) and (4.16)]. Now as is a simple complete m(S)-primary ideal of S [cf. (4.16)] with ls(Sja S ) < lR(Rja) [cf. (3.15)]. By our induction assumption we have as = br for some i E {I, ... ,h}j in this case we have a = bi by (3.27).

5

The Predecessors of a Simple Ideal

(5.0) In this section we keep the notations and hypotheses introduced in section 2. In particular, let VR be the valuation of K defined by the order function ordR of R, and let VR be the valuation ring of VR. (5.1) VALUATION ASSOCIATED WITH A SIMPLE COMPLETE IDEAL: (1) Let p f; m be a simple complete m-primary ideal of R. We have c(p) = rf where p E IP and lEN, the ring S := Sp is the only quadratic transform of R with pS f; S, pS is a simple complete m(S)-primary ideal of S, and pSn R = p [cf. (3.10) and (4.16)]. Furthermore, we have lR(Rjp) > ls(SjpS) [cf. (3.15)]. We set Ro := R, p(O) := p, and Rl := S, p(l) := pS. If pS =I- m(S)-and this is the case iff ls(SjpS) > I-then we can repeat the procedure described above, and we get a sequence R = Ro C Rl C ... C Rh of two-dimensional regular local subrings of K where RiH is a quadratic transform of Ri for i E {O, ... , h - I}, and a sequence (p(i»O IIR(z)v(m) iff In(z) E p. Proof: We set r := IIR(Z). We write Z = f(x,y) where f E R[X, Y]. Then we have v(z) = rv(m) + v(f(I, y/x)). Therefore we have v(z) > rv(m) iff ](1, () = 0, hence iffg(In(x),In(y)) divides ](In(x) , In(y)) in K[In(x),In(y)]. (2) Let a be an m-primary v-ideal of R. Then c(a) is a power ofp. Proof: Since a is a v-ideal, it is a complete ideal of R [cf. B(6.16)(2) 1and we have a = aV n R, hence a is contracted from S and we have a = as n R, and therefore c(a) is a power of p [cf. (3.26)]. (3) Let a be an m-primary v-ideal of R and set r := ord(a). Let hEN, and set q := mhaV n R. Then we have mha = mr +h n q. Proof: We set il := mr +h n qj il is an m-primary ideal of R which is contracted from S, and we have mha C il. Now mha C il C mr +h implies that ord(il) = r + h, hence that deg(c(il)) ~ deg(c(a)) ~ ord(a) = r, and therefore we have ord(il) - deg(c(il)) ~ h. This implies that [cf. (3.24)] il = mhe with e := il : mh. We have mha C il C q and v(mha) = v(q), and therefore we have v(il) = v(q). Now we get v(mh) + v(e) = v(il) = v(q) = v(mha) = v(mh) + v(a), hence v(e) = v(a), and therefore e C a [since a is a v-ideal]. Therefore we have il = mhe C mha, and therefore il = mha. (4) Let a be an m-primary v-ideal of R. Then as is a v-ideal of S. Proof: Set r := ord(a). Let z E S, z ¥- 0, with v(z) ~ v(a S). We have to show that z E as. We write xrz = Ut/U2 with Ui, U2 E R, In(u2) rt p, and ord(ui) ~

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VII Two-Dimensional Regular Local Rings

ord(u2) =: h [cf. (2.6)(3)]. Now we have to show that Ut/U2 E as = xra s . We have V(UI/U2) = v(xrz) 2 v(xra S ) = v(aS) = v(a). Since U2 E mh, we have v(ut} 2 v(mha). Now we have VR(Z) 2 0, hence VR(Ut/U2) 2 r, and therefore VR(UI) 2 r+vR(u2) = r+h, hence UI E mr+h. From this and from v(ut} 2 v(mha) we get by (3) that UI E mha, hence that Ut/U2 = (x h jU2)a' with a' E as, hence that Ut/U2 lies in as. (5) Let b be an m(S)-primary v-ideal of S, and let a E MCP(R,p) be the ideal with as = b [cf. (4.16)(2)]. Then a is a v-ideal of R. Proof: We set r := ord(a); we have as = xrb and aSnR = a. We set q := aVnR. Just as in the first part of the proof of(5.1)(2) we get a = mr nq. Since m does not divide a, we have deg(c(a)) = ord(a) [cf. (3.22)]. We use this to show that q C mr. Suppose not; then we have s := ord(q) < rand mr-sq C mr n q = a, and from ord(mr-Sq) = ord(a) we get deg(c(a)) ~ deg(c(q)) ~ ord(q) = s < r = ord(a), contradicting the equality above. (6) Let bl , b2 be m(S)-primary v-ideals of S, and let aI, a2 E MCP(R,p) be the m-primary v-ideals with af = bi for i = 1,2 [ef. (5)]. If al ~ a2, then we have bl ~ b2 . Proof: Let i E {1,2} and set ai := ord(ai); then we have aiS n R = ai and aiS = x ai af = x ai bi. Note that al 2 a2. Suppose that the assertion is not true. Then we have bl :) b2 [ef. B(6.13)(I) and note that bl , b2 are v-ideals]. Since mal-a2a2S = x a1 b2 C x a1 bl = alS, and since m a1 - a2 a2 and al are contracted from S [cf. (3.10)], we get m a1 - a2 a2 Cal. This implies that al = deg(c(at}) ~ deg(c{a2)) = a2 [ef. (3.22)], hence that al = a2, and therefore we have a2 Cal, contradicting al ~ a2. (7) Let a be an m-primary v-ideal of R, and set r := ord{a). If as = S, then we have a = mr , and if as =I- Sand al is the unique complete m-primary ideal with ar = as (cf. (3.27)}, then we have a = mtal for some t E No. Proof: If as = S, then we have as = xr S, hence mr C a, and therefore a = mr, and if as =I- S, then we have mtal = a for some t E No by (3.27).

(5.3) THE PREDECESSORS OF A SIMPLE IDEAL: Let P be a simple complete mprimary ideal of R of rank h. If h > 0, then we have c(p) = pn with p E lP' and n E N. Let (Ri)o h«()S). (2) We prove the assertion by induction on h(a). The assertion is true if h(a) = 0 [cf. (5.6)]. Now let h(a) > 0, i.e., a is not a power of m. Let c(a) = p~l ... pr' be the factorization of c(a)j note that 1 > O. We set Si := SPi for i E {I, ... ,l} and S := Sl n··· n S" and we choose z E S with mS = Sz. Furthermore, we set 1li := m(Si) n Sj note that n1,"" nl are the maximal ideals of S [cf. (2.9)]. We set, := a : b. Since m is either a factor of a or equal to a predecessor of a simple factor of a, it is enough to consider only the case that , is not a power of m [this includes the case that, = R]. We write

here we have ao, Co E ~, and, for i E {I, ... , I}, I1i is a product of simple complete m-primary ideals of SCP(R,pi) and'i is a product of simple complete m-primary ideals of SCP(R,Pi) or 'i = R [cf. (5.5)]. With T := ord(a), 8 := ord(b) and t := ord(,) we have for the ideals as, bS and ,s of S [cf. (5.5)]

287

6 The Quadratic Sequence

For i E {I, ... , I} the ideal af is an lli-primary ideal of 8, and therefore we have [cf. B{lO.l)] ar ... af = ar n ... n af. This implies that cS

= cf ... cf = as : z8H- r bS =

=

n I

I

1=1

(ar : Z8H- r bS)

i=1

(n ar) :z8+t- r bS

I

= II (ar : z8+t- r bS ) i=1

[since, for i E {I, ... , I}, ar : zs+t-rb S is either equal to 8 or it is an ni-primary ideal of 8]. For i, j E {I, ... , I} we have (ar : z'H- r bS )8j = z·+t-rb S; since z·+t-rb S; = 8j if i ¥- j. Therefore we 8j is a localization of 8, and we have Si Si have c = Cfi = afi : z·+t-rb , and by induction, applied to the m{8i )-primary ideal afi of 8i ofrank < h{a) and the ideal z·H-rb Si of 8i, we obtain that every simple complete m{8i )-primary factor of CSi is a predecessor of, or coincides with a simple complete m{8i)-primary factor of afi. This proves the lemma.

a7; :

a7; :

(5.8) Proposition: We assume that ti. is an infinite field. For any two complete m-primary ideals a, b of R with be a, iR{a/b)

= 1 and ord{b) = ord{a) + 1

we have: Every simple complete m-primary factor of b is a predecessor of, or coincides with a simple complete m-primary factor of a. Proof: If a would be a power of m, say a = mr , then b C mrH C mr , and 1 = iR{mr /b) ~ iR{mr /mrH) = r+ 1 which is absurd. Therefore a is not a power ofm. We have b : m = a: m by (3.23){3). It is enough to show the following: Every simple complete m-primary factor of a : m is a predecessor of, or coincides with a simple complete m-primary factor of a. But this is an immediate consequence of (5.7).

6

The Quadratic Sequence

(6.1) In this section we change our point of view. Until now we studied a fixed two-dimensional regular local ring with field of quotients K. Now we start with a field K, and we consider all two-dimensional regular local subrings of K having K as field of quotients. (6.2) NOTATION: Let K be a fixed field. We denote by n the set of all twodimensional regular local subrings of K having K as field of quotients. For 8 E n let ms be its maximal ideal, ti.s = 8/ms its residue field, ords its order function which gives rise to the discrete valuation Vs of K with valuation ring Vs, and let IP's be the set of homogeneous prime ideals of height 1 of grms (8).

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VII Two-Dimensional Regular Local

rungs

(6.3) Lemma: Let 8, TEn. Then: (1) If 8 c T, then T dominates 8. In particular, K.s is a subfield of K.T. (2) If 8 c T and msT mT, then we have 8 T. (3) If 8 ~ T, then we have Vs i: VT.

=

=

Proof: (1) Suppose that T does not dominate 8. Then p := mT n 8 is a prime ideal of 8 of height 1 [we have mT n 8 i: {O} since 8 and T have the same field of quotients], and therefore p = 8x for some x E 8. VT is a discrete valuation ring of K which dominates T, hence m(VT) n 8 = p. Now 8 p is a discrete valuation ring of K, and we have 8 p C VT, hence 8 p = VT. In particular, we have VT(X) = 1, hence x is a regular parameter of T. We choose yET such that mT = Tx + Ty. Since JIT(Y) = 1, we have y = x(z/w) with z, w E 8, and we may assume that neither w nor z is divisible by x in 8. Therefore we have lIT(W) = lIT(Z) = O. The relation wy = xz implies that x divides win T, hence lIT(W) > 0, contradicting JIT(w) = O. Therefore we have proved that T dominates 8. (2) This follows from B(9.6) [note that a regular local ring is analytically normal, hence, in particular, analytically irreducible, cf. 111(6.3)]. (3) Suppose that Vs = VT =: V, hence that liS = liT. Let {x,y} be a regular system of parameters of 8, and let' be the image of y/x in K.v := V/m(V). Then ( is transcendental over K.s and we have K.v = K.s«() [cf. (2.3){1)]. By the same argument we get that K.v is purely transcendental over K.T of transcendence degree 1. Therefore ( is transcendental over K.T. We have JIT(x) = lIT(y) = 1. Suppose that y mod rn} is contained in the subspace of mT/rn} generated by x mod rn}, hence that y = ax+z with a unit aofTand z E rn}. Since lIT(Z/X) ~ 1, this yields that ( E K.T, in contradiction with the fact that ( is transcendental over K.T. Therefore y mod rn} does not lie in the subspace of mT/rn} generated by x mod rn}. Considering x/y instead of y/x, we see that x mod rn} does not lie in the subspace of mT/rn} generated by y mod rn}. Therefore {x,y} is a regular system of parameters of T, and therefore we have msT = mT. From (2) we now get 8 = T, contrary to our assumption. (6.4) Proposition: Let 8, TEn with 8 determined sequence

c

T. Then there exists a uniquely

8 =: 8 0 C 8 1 C ... C 8 n := T

{ with n

~

0] where, for i E {I, ... , n}, 8i is a quadratic transform of 8i-l.

Proof: There is nothing to show if 8 = T; hence we assume that 8 i: T. Then T dominates 8 and Vs i: VT by (6.3); note that VT dominates T. (1) We show that there exists a quadratic transform 8 1 of 8 with 8 1 CT. Let {x, y} be a regular system of parameters of 8, and set u := y / x. (a) If u E T, then we have A := 8[u] C T, and since T dominates 8, the ideal mTnA = m(VT )nA is a prime ideal of A which contains msA = Ax. Suppose that m(VT) n A = Ax. Then we have AAz C VT, and since Vs = AAz [cf. (2.3)(2)], we get Vs = VT, contradicting our assumption. Therefore we have mT n A = np where pEPs, and the quadratic transform 8 1 of 8 defined by p is contained in T.

6 The Quadratic Sequence

289

(b) Likewise, if l/u E T, then there exists a quadratic transform S1 of S with S1 CT. (c) We show that either u or l/u lie in T. Suppose that this is not the case. We set B := T[ u] and q := mTB. The ideal q is a prime ideal of B, the q-residue u of u is transcendental over K.T and B/q = KT[U] [cf. 1(3.34)]. Let V be a valuation ring of K having center q in B [cf. 1(3.5)]. Then KT [u] is a subring of V/m(V). Moreover, V dominates S, and since U is also transcendental over KS, we have V = Vs by (2.19). We have vs(x) = 1, and since vs(mT) > 0, we have vs(mT) = 1, hence x ¢ n?, and since vs(y) = 1, we also see that y ¢ n? Just as in (3) of the proof of (6.3) we obtain that y mod n? does not lie in the subspace of mT In? generated by x mod n? Applying this reasoning to B' := T[ l/u], we see that also x mod n? does not lie in the subspace of mT In? generated by y mod n? Now we have shown that {x, y} is a regular system of parameters of T, i.e., that msT = mT. This implies that S = T by (6.3)(2), contradicting our assumption. Therefore we have u E T or l/u E T (2) If S1 =j:. T, we can continue this process. We have to show that after a finite number of steps we obtain a ring Sn such that Sn = T. Suppose that there is an infinite sequence S =: So C S1 C S2 C ... C T C VT such that, for i ~ 1, Si is a quadratic transform of Si-1, and that T contains all the rings Si. Let u E VT be such that the vT-residue of u is transcendental over K.T. We write u = xo/Yo with Xo, Yo E So; it is clear that Xo, Yo E mso. Now mSoS1 is a proper principal ideal S1Z1 of S1; hence we have Xo = Z1XI, Yo = ZlY1 with Xl> Y1 E Sl, hence u = Xl/Y1, and again we have Xl, YI E ms1 • Note that VT(XO) > VT(Xt}. Continuing, we get a strictly decreasing sequence (VT(Xi»iEN of positive integers; this is absurd. (3) With regard to uniqueness, let S' be any quadratic transform of S which is contained in T. Then VT dominates S' [cf. (6.3)], and therefore S' = S1 by (2.19). By repeating this argument, we see that, for i E {I, ... , n -I}, Si is that quadratic transform of Si-1 which is contained in T. (6.5) DEFINITION: Let S, TEn with S c T. The sequence S =: So C S1 C ... C Sn := T

of (6.4) is called the quadratic sequence between S and T; n is called the length of the sequence. (6.6) NOTATION: Let R E nand n E No. We denote by Nn(R) the set of SEn with S :J R and such that the quadratic sequence between R and S has length n, and set N(R) := UnENo Nn(R). Note that No(R) = {R}, that NI (R) is the first neighborhood of R, and that N(R) is the set of all two-dimensional regular local subrings of K which contain R.

(6.7) Proposition: Let S, TEn with S cT. Then:

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VII Two-Dimensional Regular Local Rings

(1) K.T is a finite extension of K.s. (2) Let a be an ms-primary ideal of 8. Then aT is an mT-primary ideal ofT or

aT=T.

Proof: Both assertions are true if T is a quadratic transform of S [cf. (2.6)(2) and (2.11)(2)]. In the general case, we consider the quadratic sequence between 8 and T and use induction [we use the multiplicative behavior of field degrees and also (1.5)(4)].

(6.8) NOTATION: Let 8, TEn with 8 c T. The degree [K.T : K.s] shall be denoted by [T: 8]. (This agrees with the notation introduced in (2.6)(3).)

(6.9) Corollary: [Hoskin-Deligne] Let a be a complete m-primary ideal of R. Then we have lR(R/a)

=

L

[8:

R]ords(aS)(O~ds(aS) + 1),

SEN(R)

and there exists n E N such that as = S for every 8 E Ui~n Ni(R).

Proof: The formula in (*) follows by repeated application of (3.14); since lR(R/a) is finite, the second assertion follows immediately.

(6.10) SIMPLE COMPLETE IDEALS, QUADRATIC SEQUENCES, VALUATIONS OF THE SECOND KIND: Let R En; we associate with R the following three sets: (1) The set N(R) of rings 8 E n with R c 8; (2) The set SC'P(R) of simple complete mR-primary ideals of R; (3) The set V(R) of valuation rings V of K which dominate R and are of the second kind with respect to R [such a V is discrete of rank 1 by (2.18)]. We show that there are bijective maps between these three sets. THE SETS SC'P(R) AND N(R): To each simple complete mR-primary ideal P of R there corresponds a quadratic sequence R = Ro c ... C Rh = 8 p such that pSp = msp [cf. (5.1)]; we associate with P the ring Sp. Conversely, let 8 E N(R), and let R =: Ro C Ri C ... C Rh := 8 be the quadratic sequence between R and 8. Set p(h) := ms and, for i E {l, ... ,h}, let p(i-i) C Ri-i be that simple complete mRi_l-primary ideal with (p(i-i»Ri = p(i) [cf. (3.27)]; we associate with 8 the simple complete m-primary ideal p := p(O) of R. This sets up a bijective map between the set SC'P(R) of simple complete mR-primary ideals of R and the set N(R) of rings 8 E n with 8 ::::> R; in particular, for p E SC'P(R) the length of the quadratic sequence between R and 8 p is equal to the rank of p. Let p E SC'P(R) be of rank h, let Po ::::> Pi ::::> ••• ::::> Ph := P be the set of simple vp-ideals defined by p [cf. (5.3)], and let R =: Ro C Ri C ... C Rh := 8 p be the quadratic sequence between Rand 8 p; then we have SPi = Ri for i E {O, ... , h}. THE SETS N(R) AND V(R): Let 8 E N(R); we associate with S the discrete valuation ring Vs. Then Vs dominates R and it is of the second kind with respect to

6 The Quadratic Sequence

291

R [since Vs is of the second kind with respect to S, cf. (2.3)(1)], hence Vs E V(R). Conversely, let V E V(R). If V = VR, then we associate R with V. If V f; VR, then V has a directional ideal p(V) E IP' [cf. (2.18)], and setting Rl := Sp(V), then V is still of the second kind with respect to R 1 • If V = VR1' then we associate Rl with V. Otherwise we repeat this process. Suppose that there exists an infinite sequence R =: Ro C Rl C R2 C ... in n such that for i ~ 1 the ring Ri is a quadratic transform of Ri-l, and V dominates all the rings Ri. Just as in (2) of the proof of (6.4) this leads to an absurd result. Therefore there exists h E No such that V = VRh' Now we associate with V the ring Rh. Thus, we have a surjective map N(R) --+ V(R). To show that this map is injective, we have to assert: Let S, T E N(R) and S f; T; then we have Vs f; VT. We consider the quadratic sequences R =: ~ C R~

c ... c

R'm := S, R =:

R:;

C R~

c ... c

R~ :=

T.

We prove the assertion by induction on min(m, n). We may assume that m ~ n. If n = 0, then we have Vs f; VR by (6.3). Now we assume that n > 0, and we choose i E {O, ... ,n} minimal with R~ = R~'. Ifi = 0, then we have R~ f; R~, hence Vs f; VT since Vs and VT contain only one quadratic transform of R by (2.19). If i ~ 1, then we replace R by R~ = R~', and, by induction, we again get that Vs f; VT. (6.11) Corollary: (1) Let a be an ideal of R and let a. Then we have a= aVs·

n

a be the integral closure of

SEN(R)

(2) Let a, b be complete ideals of R. Then we have a C b iff aVs C bVs for every S E N(R). Proof: (1) follows from B(6.15), (3.7) and (6.10), and (2) is a consequence of (1). (6.12) Corollary: Let Ro C Rl C ... be a sequence in n such that Ri+l is a quadratic transform of Ri for every i E No. Then V := URi is a valuation ring of K which is of the first kind with respect to Ro. Moreover, V is the only valuation ring of K which dominates all the rings R i , i 2:: o.

Proof: The ring V is integrally closed and quasilocal with maximal ideal n := UmR,· Clearly V dominates all the rings Ri; in particular, Vln is algebraic over RolmRo [since the residue fields of all the rings Ri, i ~ 0, are finite extensions of the residue field of R]. Suppose that V is not a valuation ring of K. By 1(3.34)(2) there exists a valuation ring W of K which dominates V and which is of the second kind with respect to V. Then W dominates Ro and it is also of the second kind with respect to Ro, and therefore we have W = VR, for some i E No [cf. (6.10)], hence W does not dominate Ri+1 [cf. (2.6)(2)]. This contradiction shows that V is a valuation ring of K, and it is of the first kind with respect to Ro.

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VII Two-Dimensional Regular Local Rings

If WI is any valuation ring of K which dominates all the rings Ri, i ~ 0, then W' dominates V, hence we have W' = V by 1(3.8).

7

Proximity

(7.0) We keep the notations of the last section.

(7.1) DEFINITION: Let R, 8 E 0 with R ~ 8; we say that 8 is proximate to R and we write 8 ~ R or R -< 8 if the valuation ring VR contains 8. (7.2) REMARK: (1) Let REO; then we have R -< 8 for every 8 E Nl(R). (2) Let REO and 8 1 E N l (R). There exists a unique quadratic transform 82 of 8 1 which is contained in VR. In fact, we have mR81 = 8 1 x with x E Rand m(VR) n 8 1 = 8 1 x, and x is part of a regular system of parameters of 8 1 [cf. (2.6)(2)]. We choose u E 81 with (x,u) = ms1 • Then we have VR(U) = 0, and the only quadratic transforms of 8 1 which are contained in VR are localizations of B := 8tl x/u] with respect to maximal ideals of B containing u. It is easy to check that m(VR) n B = B . x/u, hence B . x/u is a prime ideal of B, and therefore BB.x/u C VR is a discrete valuation ring [cf. B(1O.5) and note that B is integrally closed by (2.3)(4)], hence we have BB.x/u = VR. The only maximal ideal of B containing u and x/u is the ideal n generated by u and x/u. Therefore 82 := Bn is the only quadratic transform of 8 1 contained in VR. Note that mS2 = (u, x/u), [82 : 8tl = 1 and m(VR) n 8 2 = 8 2 . x/u. (3) The argument in (2) can be applied to 8 2 [ x/u 2 ] etc., and it leads to the following result. Let REO and p E lPR, and set Rl := 8 p ; then there exists a unique infinite sequence

where RiH is a quadratic transform of Ri for i ~ 0. For i ~ 1 we have mRi = (u, X/U i - l ), m(VR) n Ri = Ri . X/U i - l and [RiH : Rd = 1 for i ~ 1. The union URi is a valuation ring of K which dominates all the rings Ri, i ~ 0, it is contained in VR, and it is of the first kind with respect to R [cf. (6.12)]. In (7.6) below we shall describe the valuation defined by this valuation ring. Note that R -< Ri for every i E N. (4) Let 8 ~ T be rings in 0, and let 8 =: 8 0 C 8 1 C ... C 8 n := T be the quadratic sequence between 8 and T. Then T is proximate to 8 n -1, and if n ~ 2, then T is proximate to at most one other ring 8i, i E {O, ... , n - 2}. Proof: Assume that n ~ 2 and that T is proximate to one of the rings 80, ... , 8 n - 2 • We choose i E {O, ... , n - 2} minimal with 8i -< T. There is nothing to show if i = n - 2; hence we assume that i ~ n - 3. We apply to 8i, 8iH and VSi the construction in (3). Then we obtain x E 8i, U E 8H1 with vsi(u) = and msi +; = (u,x/u j - l ) for j E {1, ... ,n-i}; in particular, we get mT =

°

293

7 Proximity

(u, x/U n - i - 1). For j E {I, ... , n - i - 2} we have VSi+i (u) = VS i+i (x/ui - 1) = 1, hence vSi+i(x/un-i-1) = 1- (n - j - i) ~ -1, and therefore T is not proximate to any of the rings SH1, ... , Sn-2.

(7.3) Proposition: Let S ~ Tin 0, and let S =: So C Sl C ... C Sn := T be the quadratic sequence between Sand T. We have mT = (x, y) with mSn_1T = Tx. For every non-zero ideal a of S we have aT = xCydaT where c:= ordsn _1 (aSn-d, and if there exists i E {O, ... , n - 2} with Si ~ T, then we have d := ordsi (aSi), and d:= otherwise.

°

Proof: The assertion is clear if n = 1. Now let n = 2; we have mSoSl = SlX1, and we choose Y1 E Sl with mS1 = (X1,Y1). If So ~ S2, then ms 1 S2 = S2Y1, and {x := Y1, Y := xl/yt} is a system of generators of mS2 [cf. (7.2)(2)]. With c := ords1 (aSt} and d := ordso(a) we get aS2 = xfaS1 S2 = xCydaS2 • If S2 is not proximate to So, then S2 is not the ring Sp with p = In(xt} E grml (St}, hence S2 is a localization of Sdyl/X1]' Therefore we have mS1S2 = S2X1, and with x := Xl we obtain aS2 = x CaS2 with c := ords 1 (aSt}. Now let n ~ 3, and assume that the assertion holds for all S ~ T such that the quadratic sequence between Sand T has length n - 1. Let S := So C Sl C ... C Sn := T be a quadratic sequence of length n. First, we consider the case that there exists i E {O, ... , n - 2} with Si ~ T. If i = n - 2, then we can argue as above. If i < n - 2, then we choose t, u E Si+1 with mSiSi+1 = SHIt and mSH1 = (t, u). Then we have mSn_1 = (u, t/U n - 2- i ), msn_1Sn = Snu and mSn = (u, t/u n - 1- i ) by (7.2)(3). Set d:= ordsi(aSi), c' := ordsn_2 (aSn-2). Then we have, by induction, aSn-1 = u C' (t/un-2-i)daSn-l, and therefore aSn = uc(t/un-1-i)daSn with c := c' + d + ord sn _1 (a Sn - 1 ) = ord(aSn_t}. Now we consider the case that T is proximate only to Sn-1. If Sn-1 is proximate only to Sn-2, then we choose Xn -1 E Sn-1 with msn_2 S n -l = Sn-1Xn-1; we have, by induction, aSn-l = X~_l aSn - 1 with c' := ordsn _ 2 (aSn -2). Since Sn is not proximate to Sn-2, we have msn_1Sn = SnXn-l, and with x := Xn -1 we get aSn = xCaSn with c:= c' +ordsn_1 (a Sn - 1) = ordsn _ 1 (aSn-1)' Now we assume that there exists i E {O, ... , n-3} with Si ~ Sn-l. In Si+1 we choose t and u as above, and we have msn_2 S n -l = Sn-lU, mSn_1 = (u, t/U n - 2- i ). By induction, we have aSn-l = u C' (t/un-2-i)daSn-l with c' := ordsn_2 (aSn -2) and d := ordsi (aSi). Since Sn is not proximate to Sn-2, we have msn_1Sn = Sn . u/t n - 2- i , and u/(t/u n - 2- i ) is a unit of Sn. Therefore we get with x := t/U n - 2- i that aSn = xCa Sn with c := c' + d + ordsn_1 (a Sn - 1) = ordsn_1 (aSn -1).

(7.4) Corollary: Let S ~ T in 0, let P be a simple complete ms-primary ideal of S, and let ms =: Po :J PI :J ... :J Ph := P be the sequence of simple vp-ideals defined by p. IE pT is a principal ideal, then PiT is a principal ideal for every i E {O, ... ,h-l}.

°

Proof: Let i E {I, ... ,h -I}. If pT is a principal ideal of T, then pT = T by (7.3), hence T ct Sp, the ring in associated with p, and therefore T ct SPi [cf. (6.10)],

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VII Two-Dimensional Regular Local Rings

hence we have

pi = T and therefore PiT is a principal ideal, again by (7.3).

Now we describe the valuation mentioned in (7.2).

(7.5) A

PARTICULAR CLASS OF VALUATIONS:

Let p E lP'R. We define a valuation

with value group Z x Z where Z x Z is ordered lexicographically, in the following way. Let fER be a non-zero element, and let (In(J» = I1 qE iP Rqnq(f) be the factorization of the homogeneous principal ideal (In(J» in grmR(R). We define

Let f, 9 E R be non-zero elements. Since (In(Jg» = (In(J»(In(g», we have

IIp(J g) = IIp(J)

+ IIp(g).

Assume, furthermore, that f + 9 ¥ 0. If ordR(J + g) > min( {ordR(J),ordR(g)}), then we have IIp(J + g) > min( {lIp(J) , IIp(g)}). Now we consider the case that ordR(J + g) = min({ordR(J),ordR(g)}). If ordR(J) ¥ ordR(g), say ordR(J) > ordR(g), then we have In(J + g) = In(g), hence np(J + g) = np(g), which implies that np(J + g) ~ min({np(J),np(g)}). If, on the other hand, ordR(J) = ordR(g), then we have In(J + g) = In(J) + In(g), and again we find that np(J + g) > min( { np (J), np (g)} ). Therefore we have in both cases

The canonical extension of lip to K X [cf. 1(3.20)] shall also be denoted by lip, and it is a valuation of K. Let Vp be the valuation ring of lip. We choose fER with (In(J» = p, and set h := ordR(J) = deg(p). Then we have IIp(J) = (h,I). Let mR = (x,y); we may assume that (In(x» ¥ p. Then (1,0) = IIp(x) ::=:; IIp(Y), hence IIp(mR) = (1,0). Let (m, n) E Z x Z; then we have IIp(Jnxm-hn) = (m, n). Therefore lip is surjective, hence lip is a discrete valuation ofrank 2, and Vp ~ YR. Since (0,1) is the smallest positive element in Z x Z, we have IIp(m(Vp)) = (0,1) < (1,0) = IIp(m) , hence Vp dominates R. Then [lI:vp : II:R] is finite by Abhyankar's theorem [cf. 1(11.9)], hence Vp is of the first kind with respect to R. It is easy to see that p is the directional ideal of Vp, hence that the quadratic transform Sp = R[yjx b:,f/xh) of R is the first quadratic dilatation of R with respect to lip [cf. VIII(3.1) for the notion of quadratic dilatation]. (7.6) Proposition: With notations as in (7.5) we have: (1) Let T E S2 with R ~ T C Vp. Then we have Sp C T. (2) We set f := In(J) , ( := In(y)jln(x) E Q+(grmR(R»o. Then we have II:vR = II:R«(), ](1, () E II:R[ (] is irreducible and denoting by vp the valuation

7 Proximity

295

of the rational function field II:R«() in one variable over II:R defined by the discrete valuation ring II:R[1n(x), 1n(y) hpj = II:R[ (](f(l,()), we have vp = VR 0 v p, and II:R[(]/(f(l,()) is the residue field ofvp; in particular, vp and Sp have the same

residue field. (3) Let TEn with Sp eTc VR. Then we have T C Vp. Moreover, Vp is the only valuation ring of K dominating T and contained in VR, the inclusion map T 2, and that the assertion holds for n - 1. Let S E N n - 1 (R) and let T be a quadratic transform of S. By induction, there exists a regular system of parameters {u, v} of S such that the total transform aof a in S has the form uveas where e E {O, I}. Let n be the maximal ideal of S. If nT = Tu, then we set x := Uj if v/u is not a unit of T, then {x,y := v/u} is a regular system of parameters ofT, and we have n· (uveaS)T = xyeaT, and if v/u is a unit of T, then we choose yET such that {x, y} is a regular system of parameters of T, and we get n· (uveaS)T = xaT . If nT = Tv, then we set x := v. If u/v is not a unit of T, then {x, y := u/v} is a regular system of parameters of T, and we

8 Resolution of Embedded Curves

301

have n· (uveaS)T = xyaT , and if u/v is a unit of T, then we choose yET such that {x, y} is a regular system of parameters of T, and we get n· (uveaS)T = xaT . (2) Now we assume that I E R is reduced (resp. analytically reduced). We show: Any generator of the total transform of RI in T is reduced (resp. analytically reduced). By induction, it is enough to show: Let S E N 1 (R); then any generator of the total transform of RI is S is reduced (resp. analytically reduced). If I is reduced, then this follows from the multiplicative behavior of forming the ideal transform, and from (2.11)(3). Now let I be analytically reduced, and choose x E R with mRS = Sx. Let Is be a generator of (Rf)s. Then xis is a generator of the total transform of RI in S; if Is is not a unit of S, then x does not divide Is in S [cf. (2.11)(3) j, hence Sx + Sis is an ms-primary ideal of S. Then Bx + Bls is an m(B)-primary ideal of B, and therefore xis is a reduced element of B, hence xis is analytically reduced. If Is is a unit of S, then xis is analytically reduced.

(8.12) NOTATION: Let R En. A reduced non-unit I of R is said to define a curve C in R; C is called regular if ord(f) = 1. The irreducible factors of I are called the irreducible components of C. C is called a normal crossing curve if either C is regular or C has two irreducible components C 1 , C 2 , C1 and C2 are regular, and if C1 is defined by It and C2 is defined by 12, then we have £R(RIt,Rh) = 1. Note that in the latter case the condition to be a normal crossing curve is equivalent to {It, h} being a regular system of parameters of R. (8.13) Proposition: Let (Ri)i~O be an increasing sequence in n such that Ri is a quadratic transform of Ri-1 for every i E N. Let IE R be analytically reduced. Then there exists mEN such that the total transform of RI in Rm is a normal crossing curve. Proof: (1) Let SEn and let 9 E S be reduced; we set ords(g) =: n, and we assume that 9 is a product of irreducible factors g1, ... ,gn of order 1. We define the integer ()s(Sg) -_ (ord(g) - 1)2(ord(g) - 2)

+

'" .L.J

(£s(Sgj, Sgt) - 1).

l:Sj2j this is a birational map. Then one can show: Applying a finite number of quadratic

310

VIII Resolution of Singularities

transformations to an irreducible projective curve in ]p2 leads to an irreducible projective curve in ]p2 which as singularities has at most ordinary multiple points [cf. [69], Ch. 7, Th. 2]. Thus, if we want to work only in the projective plane, we can, in general, not resolve singularities of plane projective curves.

2

Resolution of Surface Singularities I: J ung's Method

(2.0) In this section we assume that the ground field k is algebraically closed Let X be an irreducible surface. We shall resolve the singularities of X by using the fact that we can resolve the singularity of a twodimensional toric variety by repeatedly blowing up points [cf. chapter VI, (5.7) and (5.18)]. This method of resolving singularities should bear the name of H. E. W. Jung, as we pointed out in the preface. of characteristic zero.

(2.1) Theorem: Let X be an irreducible surface. Then there exists a desingularization 1T: X' --+ X of X.

Proof: We can consider X as an open subset of an irreducible projective surface. Therefore, by appendix A, (8.2) and (8.3), we may assume, to begin with, that X is an irreducible normal projective surface. There exists a finite surjective morphism cp: X --+ Z:= ]p2 [cf. A(12.4) j. The branch locus D := branch(X/Z) is a pure one-dimensional closed subset of Z [cf. A(1O.1O) and note that X is normal by assumption, and that Z is regular]' i.e., it is a curve. We apply embedded resolution to D c Z, as follows. By (1.9) there exists a proper morphism 1T: Z' --+ Z, obtained as a finite sequence of blowing ups of points, and a finite set FeD such that 1T induces an isomorphism Z' \ 1T- 1 (F) --+ Z \ F, and that 1T- 1 (D), the total transform of Din Z', is a divisor with normal crossings. We set S := cp-l (F); note that S is a finite sets of points [by A(2.7) ]. We consider the projective variety X x z Z'. Let p: X x z Z' --+ X, q: X x z Z' --+ Z' be the projections; p is proper by A(4.23), q is finite by A(3.14)(8), and both maps are surjective and closed. We set W := Z \ F, W' := 1T- 1 (W) and U := cp-l(W); note that U = X \ S is irreducible, and that the restriction of 1T induces an isomorphism W' --+ W. We consider U Xw W' as an open subset of X Xz Z' [cf. A(3.14)(4) j; we have p-l(U) = q-l(W') = U Xw W', p(U Xw W') = U and q(U x w W') = W'. Let X' be the closure of U x w W'. Since p and q are closed maps, we have p(U Xw W') c p(X'), q(U Xw W') c q(X'), hence the restrictions p' := piX': X' --+ X, q' := qIX': X' --+ Z' are surjective. Now p induces an isomorphism U Xw W' --+ U [cf. A(3.14)(7) j, hence X' is an irreducible projective surface; p': X' --+ X is proper by A(4.19), and X' \ p,-l(S) --+ X \ S is an isomorphism. The morphism q': X' --+ Z' is finite by A(2.6). Clearly we have deg(q') = deg(cp) =: m [since p' and 1T are birationalj. Let w' E W' and set w := 1T(W') E W; there lie exactly m points of X over W [cf. A(1O.8)], and these points lie in U; therefore there lie exactly m points of X, over w', hence q' is

2 Resolution of Surface Singularities I: Jung's Method

311

unramified over Wi [cf. A(IO.8)]. Thus, we have branch(q/) C 1I"-1(D); moreover, branch(q/) is a proper closed subset of Zl by A(IO.9). Thus, replacing X' by X and ZI by Z, it is enough to show the following: Let X be an irreducible projective surface, Z a regular projective surface, O R; = A;r, the valuation ring of v. Let e E A;r, and write e = alb with a, b E R~ b '" O. If b rI. m, then we have e E R. Now we consider the case that b E m. Since 0 ~ vee) = v(a/b), we have veal ~ v(b), hence a E m. We can write a = alzl, b = blzl with aI, bl E R I . We have c = at/bl . If bl f/. ml,

320

VIII Resolution of Singularities

then we have c E RI. If bl E ml, then we have v(ad ~ v(b l ), hence al E m, and we can write al = a2z2, bl = b2z 2 with a2, b2 E R 2, and we have a = a2zlz2, b = b2z l Z2 and c = a2/b2. Continuing, we see: If i E Nand c ~ Ri-l, then we have a = aizl ... Zi, b = bizl ... Zi with ai, bi E Ri. For n E {I, ... , i} the center ofv in Rn is mn , hence we have v(zn) > 0, and therefore we obtain v(b) ~ i. Thus, this construction cannot have infinitely many steps. Therefore, there exists i E N with c E Ri, and we have proved the assertion. (d) Let i E No. Now Ri is a one-dimensional local domain. Let Vi be the set of valuations of "'w belonging to Ri ; Vi is a finite non-empty set of discrete valuations of rank 1, and Vi :J Vi+l. Note that v E Vi for every i E No, and that Si := EV; All is the integral closure of Ri. We set {Vl, •.• , Vh} = 170 \ {v}. We consider the case that h ~ 1; let j E {I, ... , h}. There exists aj E Au with aj ~ Alii [since Au ct. Alii]' and by (c) there exists i j E N with aj E Rii . We choose io E N with io ~ i j for j E {1, ... ,h}. Then we have aj E Rio for j E {1, ... ,h}; this implies that Via = {v}. In particular, Au is the integral closure of Rio. Since Rio is essentially of finite type over the field k, Au is a finitely generated Rio-module by B(3.6), hence the increasing chain Ro C RI C ... c Au becomes stationary, and there exists i l E N with Ril = Ril +1 = ... = Au [cf. (c)]. (e) Let i ~ iI, and choose fh E Ri with vCih) = 1. Let Yi E Ri be an element with image fh Then we have V(Yi) = (0,1) [cf. 1(3.33»). Since Pi is a prime ideal of height 1 of the factorial domain Ri [note that Ri is a regular local ring], Pi = RiXi is a principal ideal. Since (Ri)p; = Aw [since (Ri)p; is a discrete valuation ring which is contained in AwJ, we have V(Xi) = (1, ai) with ai E Z. We show that mi = RiXi + RiYi. In fact, let z E ~, and let Z E mi be the image of Zi in Ri; we can write Z = tfh with t E Ri. We choose t E Ri having image t. Then we get Z - tYi E Pi, hence Z E RiXi + RiYi.

nIl

5

Valuations of Algebraic Function Fields in Two Variables

(5.0) In this section k is a field of arbitrary characteristic, and K is an algebraic function field over k. Every valuation v of K is tacitly assumed to be trivial on k; remember that, by definition, dim (v) = tr. dk(",v). (5.1) REMARK: Let K be an algebraic function field in one variable over k. There exist non-trivial valuations of K, and every non-trivial valuation of K is discrete of rank 1 [cf. 1(10.4)].

(5.2)

CLASSIFICATION OF VALUATIONS OF FUNCTION FIELDS IN TWO VARIABLES:

Let K be an algebraic function field in two variables over k, and let v be a valuation of K with value group r. We have dim (v) = 1 or dim(v) = 0 [cf. 1(10.6)]. (1) If dim(v) = 1, then v is a discrete valuation ofrank 1 [cf. 1(10.7)], hence r is isomorphic as an ordered group to the ordered group Z of integers.

5 Valuations of Algebraic Function Fields in Two Variables

321

(2) If dim( v) = 0, then we have the two possibilities rank( v) = 2 or rank( v) = 1 [cf. 1(10.6)]. (a) If rank(v) = 2, then v is discrete ofrank 2 [cf. 1(10.6)], i.e., r is isomorphic to Z x Z, ordered lexicographically. (b) Now we consider the case rank(v) = 1. In this case r is a subgroup of the reals [cf. B(1.21)]. The rational rank of r is either 2 or 1 [cf. 1(10.6)]. (bl) If rat. rank(r) = 2, then there exist rationally independent real numbers Pi, P2 E r such that r = ZPl EB ZP2 [cf. 1(10.6)]. (b2) It remains the case that rat. rank(r) = 1. If r is discrete, then r is isomorphic to Z. If r is not discrete, then we may assume that r is a subgroup of the rationals [ if r contains an irrational number p, then we have Z P =I r, but r ®z Q = QP; for every (J E r\Zpthere exist integers m, n with m =I 0 and (J = {!'jp, and r·l1 pC Q is order-isomorphic to r]. In the rest of this section we show that all the possibilities mentioned above do really occur.

(5.3) REMARK: Let K be an algebraic function field in r 2: 2 variables over k. Then K admits valuations of dimension r - 1 [cf. 1(10.8)]. Such valuations are discrete of rank 1 [cf. 1(10.7)], hence this result takes care of case 1 in (5.2). (5.4) Let r E N, and let K be the field ofrational functions in r variables over k. Let r be a totally ordered abelian group. For any r elements 11, ... "r E r there exists a valuation v: K -t roo with v(KX) = Z,l + ... + Z,r [cf. 1(9.1)]. (1) Let r := zr, endowed with lexicographic ordering; note that rank(r) = rat. rank(r) = r. Let {el' ... , e r } be the canonical Z-basis of r, and let v: K -t roo be a valuation with v(KX) = Zel + ... + Zero Then we have v(KX) = r, hence rank(v) = rat. rank(v) = rand dim(v) = O. This takes care of case (2)(a) in (5.2). (2) Let Pi, ... ,Pr E lR be rationally independent, and let v: K -t lRoo be a valuation with v(KX) = ZPl + ... + Zpri then we have rank(v) = 1, rat.rank(v) = r and dim(v) = o. This takes care of case (2)(bl) in (5.2).

(5.5) NON-DISCRETE VALUATIONS OF RANK 1 AND RATIONAL RANK 1: We have to consider the case mentioned in (5.2)(2)(b2). (1) Let r c Q be a non-discrete subgroup with 1 E r. We write the elements of r always in reduced form min with m E Z, n E N, gcd(m, n) = 1. We divide the prime numbers which occur in the denominators n of elements of r into two classes P and Q: The first class P contains those prime numbers P which occur in the denominators n of elements of r only to a bounded power: For pEP there exists J.tP E N such that PP'P occurs in the denominator n of some element of r, but PP'p+1 does not occur in the denominator of any element of r. In the second class Q we put the remaining primes. If Q is empty, then P is infinite [since r is non-discrete]. We show: r consists of all rational numbers whose denominators are of the form Pil P;2 ... q~l q~2 . .. where Pi , P2, ... E P, ql, Q2, ... E Q, J.tl, J.t2, ...

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VIII Resolution of Singularities

are non-negative integers with JLI ~ JLPl' JL2 ~ JLP2' ..• and VI, V2, ••• are arbitrary non-negative integers. Proof: (a) If alb E r where a E Z, bEN and gcd(a, b) = 1, then there exist integers a', b' with aa' + bb' = 1, hence l/b = b' + a'(a/b) E r [since 1 E r by assumption], and therefore all integral multiples of l/b belong to r. (b) If r contains aI/bl , a2/b2 with ab a2 E Z, bb b2 E Nand gcd(al,bd = gcd(a2, b2) = 1, and if gcd(b l , b2) = 1, then (a l b2 + a2bl)/blb2 E r, and since a l b2 + a2bl and bl b2 are relatively prime, we have 1/(b l b2) E r by (a). Now the assertion follows from (a) and (b). (2) Given a non-discrete subgroup r of Q as in (1), we construct a valuation v with value group r of K := k(x, y), the field of rational functions in two variables over k, as follows: we set

The set P' U Q' is countable; we write P' U Q' assertion in (1), it is clear that

r - {

= {ml, m2, ... }.

Then, by the

ns Ins E Z, SEN} ml···ms

where now the elements ns/ml ... ms in general are not in reduced form. (a) Let (Ci)iEl'I be a sequence in kX. We define a sequence (Xi)iEl'I in K by recursion, as follows. We set Xl := X, X2 := y, and Xi+2 := (Xi -CixH\)/xH\ for every i E N. We set Ri := k[ Xi, XiH], qi := RiXi + RiXiH C Ri for every i E N. Let i E N; since Xi = xH.\ (Xi+2 + Ci) E qi+l C RiH, we have RI C R2 C ... , and we see that {Xi, Xi+d is a transcendence basis of Kover k, hence qi is a maximal ideal of Ri, ht(qi) = 2, qi+l n Ri = qi, Si = (Ri)q; is a two-dimensional regular local ring and {Xi, XiH} is a regular system of parameters of Si, and if ni denotes the maximal ideal of Si, then we have Si C SiH, niH n Si = ni and Sdni = k. We set S := Ui>l Si, n := Ui>l ni; then S is a proper quasilocal subring of K, n is its maximal ideal, and we have Sin = k. There exists a valuation v of K which has center n in S [cf. 1(3.5)]; set Ll := v(KX). Then we have V(Xi - CixH\) > v(xH\) for every i E N [since V(Xi+2) > 0], whence V(Xi) = miv(xi+d and therefore

We have rank(Ll) = 2 or rank(Ll) = 1 [cf. 1(10.6)]. If rank(Ll) = 2, then Ll is isomorphic to Z x Z, ordered lexicographically, and if rank(Ll) = 1 and rat. rank(Ll) = 2, then Ll is isomorphic as a Z-module to Z2 [cf. 1(10.6)]. Since v(xi+d = v(xI)/mlm2··· mi for every i E Nand v(xd > 0, and since the sequence (mI· .. mi)i>l is not bounded above, the two possibilities for Ll just mentioned are excluded. Therefore we have rat. rank(Ll) = 1. Thus, replacing Ll by (l/v(xd)Ll, we can consider Ll as a non-discrete subgroup of Q with V(XI) = 1 E Ll.

323

5 Valuations of Algebraic Function Fields in Two Variables

(b) We show that ~ = r and that Av = S. (i) Let i E N; we show that Si+1 is a quadratic dilatation of Si with respect to v.

In fact, set ~(l) := Xi+l, 1J(1) := Xi, and define recursively ~(j+l) := ~(j), ~(m;)

.,p+1) := 1J(j) /~(j)

for j E {I, ... , mi - 2},

:= ~(m.-l), 1J(m.) := (1J(m.-l) _ Ci~(m.-l)/~(m.).

Then we have ~(j) = Xi+l for j E {I, ... , mil and 1J(j) = Xi/X{+l for j E {I, ... ,mi -I}, whereas 1J(m.) = Xi+2' Note that, for j E {l, ... ,mi}, {~(j},1J(j)} is a transcendence basis of Kover k. We have

v(1J(j) = (mi - j)v(~(j) > v(~(j)

for j E {I, ... ,mi -I}.

Set A(j} := k[~(j),1J(j}), p(j} := ~(j}A(j) + 1J(j)A(j}, T(j} := (A(j)p(i) for j E {I, ... , mil. Then we have A(j+1) = A(j} [1J(j} /~(j)] c T(j)[ 1J(j} /~(j)] C T(j+1) , hence T(j+l) is the first quadratic dilatation of T(j) with respect to v for j E {l, ... , mi - I}, and therefore Si+1 = T(m.) is a quadratic dilatation of Si = T(1) with respect to V. Now S is the valuation ring Av by (4.4). (ii) Let I(xl, X2) E k[ Xl, X2] be a non-zero polynomial with 1(0,0) = 0. We show that there exist hEN and n E N with

l(xl,x2)

= xhlh(xh,xh+d

with h(Xh,Xh+1) E Rh, h(O,O)

::f. 0.

(**)

We choose /-L E N with v(xi) ~ V(f(Xl' X2», and we set z := xi /I(xl, X2)' Then we have z E A v , hence by (i) there exists hEN with z E Sh, i.e., there exist polynomials p(Xh,Xh+1), q(Xh,Xh+1) E Rh with q(O,O) ::f. and with z = p(Xh,Xh+d/q(Xh,Xh+l). Since we have Xl = x hg(Xh,Xh+1) with 1/ E Nand g(Xh' Xh+l) E Rh, g(O, 0) ::f. [as is easily seen by induction], we have

°

°

I(xl, X2)p(Xh' xh+d

= X~" g(Xh' Xh+1)l'q(Xh' xh+d =x~"r(xh,xh+d where r(xh,xh+d E Rh, r(O,O)::f. 0.

Let us consider the factorization of I(xl, X2) in Rh into a product of irreducible factors. The last displayed equation shows that the only irreducible factor g(Xh' xh+d of I(xl, X2) in Rh with g(O, 0) = is a power of Xh; this implies the assertion. (iii) Let l(xl,x2) E k[Xl,X2] with 1(0,0)::f. 0. It is easily seen by induction: For every hEN we can write

°

I(xl, X2) = xh!h(Xh, xh+d

::f. a, n EN. r. By (*) in (a) we have ~ => r. A non-

with h(Xh, Xh+1) E Rh, !hea, a)

(iv) We have to show that ~ = zero element z E K can be written as a quotient z = I(Xl,X2)/g(Xl,X2) with polynomials I(xl, X2), g(Xl' X2) E k[ Xl, X2]' By what we have just shown in (ii) and (iii), there exist hEN, polynomials Ih(xh, Xh+1), gh(Xh, Xh+1) E Rh with h(O, a) ::f. 0, gh(a, 0) ::f. and n E Z with

° z

l(xl,x2)

n

= g(Xl,X2) = xh

h(Xh,Xh+1) gh(Xh,Xh+1)'

VIII Resolution of Singularities

324

We have V(Xh) = 1j(ml" ·mh-I) for hEN [ef. (*) in (a) above], hence v(z) is an integral multiple of 1j(ml ... mh-I), i.e., v(z) E r, hence r => ~, and therefore we have ~ = r.

6

U niformization

(6.0) In this section K is a field; sometimes we specify a subfield k of K. We keep the notation Ineftk(K) introduced in (4.0).

6.1

Classification of Valuations and Local Uniformization

(6.1) CLASSIFICATION OF VALUATIONS: Let R be a two-dimensional regular local ring with maximal ideal m, field of quotients K and residue field K which we assume to be algebraically closed. We classify the valuation of K which dominate R. Let v be a valuation of K dominating R; let V be the valuation ring of v. We have tr.d .. (K v ) = 1 or tr.d .. (K v ) = 0 [ef. 1(11.9)]. (1) If tr. d.. (K v ) = 1, then v is a discrete valuation of rank 1 [cf. 1(11.9)], hence r v is isomorphic as an ordered group to the ordered group Z of integers. (2) If tr. d.. (K v ) = 0, then we have the two possibilities rank(v) = 2 or rank(v) = 1 [ef. 1(11.9)]. (a) Ifrank(v) = 2, then v is discrete ofrank 2 [cf. 1(11.9)], i.e., rv is isomorphic to Z x Z, ordered lexicographically. The isolated subgroup of r v of rank 1 is {O} x Z [ef. B(1.25)]; then we can write v = w 0 v where w is a discrete valuation of K of rank 1 and v is a discrete valuation of Kw [cf. chapter I, (3.23) and (3.26)]. Let W be the valuation ring of w; then V is contained in W, hence W contains R. If W does not dominate R, then p := m(W) n R is a prime ideal of R of height 1, hence Rp is a discrete valuation ring, we have Rp = W, and v is a valuation of the type described in (4.7). Now we consider the case that W dominates R. Let {x,y} be a system of generators of m, and assume that v(x) ~ v(y). Let 9 be a directional form and let p = (g) be the directional ideal of v; note that 9 is linear since K is algebraically closed. If W = VR where VR is the valuation ring of K defined by the order function of R, then we have Kw = K(() where ( is the v-image of yjx [cf. VII(2.3)], and v(g(l, = 0, and therefore we have v = vp where vp is the valuation defined in VII(7.5). In this case we can choose x and y in such a way that v(x) = (1,0) and In(y) = p, hence that v(y) = (1,1). Now let RI be the quadratic dilatation of R with respect to v; note that RI = Sp is a quadratic transform of R [cf. (4.1)], and that {y, y j x} is a regular system of parameters of R I ; in particular, we have v(yjx) = (0,1) and RI C W. Again we have the two possibilities: Either W does not dominate RI in which case W is the localization of RI with respect to a prime ideal of height 1, hence v is a valuation of the type discussed in (4.7), or W dominates R I . Then we have either W = VRt, hence v is of the type discussed in VII(7.5) with respect to R I , or we have W i- VRt' In the latter case we repeat the construction. Taking into account

(»

325

6 Uniformization

VII(6.IO) and (4.8), we can state: There exists a quadratic dilatation S of R with respect to v which is a two-dimensional regular local ring having a regular system of parameters {x,y} with vex) = (I,a) with a E Z and v(y) = (0,1). (b) Now we consider the case rank(v) = 1. In this case rv is a subgroup of the additive group IR [cf. B(1.2I)]. The rational rankofr v is either 2 or 1 [cf. 1(11.9)]. (bI) Ifrat.rank(r v) = 2, then there exist rationally independent real numbers PI, P2 E r v such that r v = ZPI EB ZP2 [cf. 1(11.9)]. (b2) It remains the case that rat. ranker v) = 1. If r v is discrete, then r v is isomorphic to Z. If r v is not discrete, then we may assume that r v is a subgroup of the rationals [cf. (5.2) (2)(b2) ]. (6.2) Theorem: [Local uniformization] Let R E Ineftk(K) be two-dimensional and regular with maximal ideal m and algebraically closed residue field 1\.. Let v be a valuation of K dominating R and with tr. d,,(l\.v) = O. Let fER be nonzero. Then there exist a quadratic dilatation S of R with respect to v which is a two-dimensional regular local ring and a regular system of parameters {x, y} of S such that f = xGyb u where a, bE No, u is a unit of S, and where the following conditions are satisfied: (a) If rank(v) = 2, then rv = Z x Z, ordered lexicographically, and we have vex) = (1, c) with c E Z, v(y) = (0,1); in particular {v(x), v(y)} is a Z-basis of rv. (b) Hrank(v) = 1, then (bI) if rat. rank(v) = 2, then either vex) and v(y) are rationally independentand in this case {v(x),v(y)} is a Z-basis ofrv- or vex) and v(y) are rationally dependent-and in this case we have b = 0, (b2) if rat. rank(v) = 1, then we have b = O. Proof: (a) We assume that rank(v) = 2. Then v is discrete of rank 2 by 1(11.9); we may assume that rv = Z x Z, ordered lexicographically. By (6.I)(2)(a) there exists a quadratic dilatation So of R with respect to v which is a two-dimensional regular local ring and a system of generators {xo, Yo} of the maximal ideal no of So such that v(xo) = (1, c) with c E Z and v(Yo) = (0,1). For every i E N let Si be the i-th quadratic dilatation of So with respect to v. We have v(xo) > iv{yo) for every i E N, hence, for i E N, {Xi := xo/y~, Yi := Yo} is a system of generators for the maximal ideal tli of Si. For /1 f. /1' E No and v, v' E No the v-values of x~Yo and x~' Yo' are not equal, and for /1 E No and v f.. v' E No the v-values of x~Yo and x~Yo' are not equal. Since every non-zero element h E So is a linear combination of monomials x~Yo with coefficients which are units of So, we see that v{h) = n) with m > 0 iff Xo divides h [remember that So is factorial since it is a regular local ring]. In So we write f = x~o 90 with eo E No, 90 E So and Xo does not divide 90. Then we have V(90) = CO, n) with n E Z. By induction, we define elements 9i E Si, i E N, by writing 9i-1 = 9i with ei E No, 9i E Si and Yi does not divide 9i. We have tli-ISi = SiYi for i E N. Let i E N; if 9i-1 E ni-l, then we have ei > O. In particular, if lEN and 9i-1 E ni-l for i E {I, ... , I}, then

em,

y:i

326

VIII Resolution of Singularities

we have v(go) ~ (0, l). Therefore there exists lo E N such that glo is a unit of Sio' Since Xo = XloY:~ and Yi = Yl o for i E {O, ... , lo}, we see that we have I = x't yf U with a, b E No and a unit U E Sio' 0 0 (b) Now we have rank(v) = 1. We set Ro := R, tno := m. Let (Rdi>O be the sequence of quadratic dilatations of R with respect to v. By (4.4) we-have U>o Ri = Av. Let i E No; RiH is a quadratic transform of Ri [cf. (4.1) 1, and we -can choose a system of generators {Xi, yd of the maximal ideal mi of Ri such that miRi+1 = RiHXiH and YiH = (Yi - UiXi)jxi, i.e., we have Xi = Xi+1 and Yi = Xi(Yi+1 + Ui), or that XiH = Yi and Yi+1 = (Xi - UiYi)jYi, i.e., we have Yi = XiH, Xi = Yi(Xi+1 + Ui); here Ui is a unit of Ri-l [cf. proof of VII(2.19) and note that", is algebraically closed, hence that the irreducible polynomials in grmi (Ri) are linear]. We set 10 := I· We define Ii E Ri inductively in the following way: If i E Nand li-l E Ri-l is defined, then we define Ii E Ri by li-l = X~i Ii where Xi does not divide Ii [remember that Ri is factorial since it is a regular local domain]. We consider the case that Ii is not a unit of Ri; let Ii = Ii! ... lihi be a factorization of Ii in Ri into a product of irreducible elements. For j E {I, ... , hd the ideal Rdij is a prime ideal of height 1 of R i ; let Wij be the valuation of K defined by the discrete valuation ring (Ri)(fii)' Let Vi be the set of valuations of K which have center mi in Ri and are composite with Wij for some j E {I, ... , hd. Now Vi is a finite non-empty set by (4.7). Let v E Vi; then we have v = Wij 011 for some Wij with j E {I, ... , hi}, and v is a valuation of rank 2. We have Wij(Xi) = 0. Let Vij be the center of Wij in R i - l ; note that Vij is not the zero ideal [since Wij is not the trivial valuation of K], and that Xi ~ Vij. Therefore we see that Vij "I mi-I, hence we get ht(Pij) = 1. We have li-l E Pij, hence Ii-I is not a unit of Ri-l. Let Ii-I = Ii-l,l .. ·Ii-l,hi_l be the factorization of li-l in Ri-l into a product of irreducible elements. Since Vij is a principal prime ideal [note that Ri-I is a factorial domain 1, there exists j' E {I, ... , hi - d such that Pij = Ri - I!i-l ,j' , hence that the valuation Wi-l,j' of K defined by the discrete valuation ring (Ri-I)(fi_l.i') is the valuation Wij, i.e., we have v E Vi-I. Thus, we have shown that Vi C Vi-I' Suppose that ni>O Vi "I 0, and let w be a valuation in this intersection. Now w dominates Ri for every i E N, hence we get Av = Aw by (4.4). We have rank(v) = 1 and rank(w) = 2. This contradiction shows that the intersection is empty. For every i E N the set Vi is finite; therefore there exists io E N with Vi = 0 for every i ~ i o. This means that lio is a unit of Rio' Therefore we have I = xr; y~; u' where aI, bl E No and u' is a unit of Rio' We consider separately the following two cases. Case (i): rat.rank(v) = 1, or rat.rank(v) = 2 and V(Xio), V(Yio) E rv C rv ®zQ are linearly dependent over Q [we identify r v with its image in r v ®z Q]. If rat. rank(v) = 1, then we may assume, in addition, that rv C Q [cf. (6.1) (2) (b2) ]. Therefore, in both cases we can write V(Xio)jV(Yio) = nojnl with coprime positive integers no, ni. If no = nl = 1, then there exists a system of parameters in RioH of the form {XioH = Xio,YioH = (Yio - u*Xio)jxio} where u* is a unit of Rio, hence we get I = xioH U where a = al + bi and U E RioH is a unit. In the general

327

6 Uniformization

with nl > n2 > ... > nk = 1 be the Euclidean algorithm for the integers no, nl. We set Zi-l £or i E {1, ... , k Zo := Xio, Zl := Yio, Zi+l := -8-' - } 1 j Zi' then we have

V(Zi) ni ( ) =v Zi+l niH

. forzE{O, ... ,k-I},

hence we have V(Zk-l) = V(ZZk). Furthermore, we set ~o:= Xio, 1]0:= Yio, So:= and

°

~81 +'+8i-1 +j := Zi, 1]81 +-+8i-1 +j := Zi-/ for i E {I, ... , k}, j E {I, ... , sd. zi

We set t := Sl + ... + Sk. The sequence of quadratic dilatations of So := Rio with respect to v takes the form

where, for i E {I, ... ,k -I}, j E {I, ... ,si}, and for i = k, j E {I, ... ,sk -I}, {~i,rli} is a regular system of parameters in Si. We have ~t-1 = Zk, 1]t-1 = zk-d ZZk- 1. We choose a unit Ut-1 E St-1 with v(zk-d ZZk - Ut-1) > 0, and we set ~t := ~t-1' 1]t := (1]t-1 - Ut-1~t)/~tj then {~t,1]tl is a regular system of parameters for the quadratic dilatation St of St-1 with respect to v. Note that 1]t-1 =~t(1]t+Ut-1). Now it is easy to check that we have ~o = ~f~l1]LI' 1]0 = ca" b" WIt . h a, ' a, "b' , b" E 1"1). ~L There £ore we h ave Xio = Xio+tU1, a2 b2+ U2 "t-11]t-1 Yio = Xio t where a2, b2 E Nand UI, U2 are units of R io +t , hence we have f = xfo+tu with a := a1a2 + b1 b2 ~ 0, u:= U'U1U2, and U is a unit of Rio+t. Case (ii): rat. rank(v) = 2 and the elements v(Xio)' v(Yio) are linearly independent over Q, i.e., we have r v i8)z Q = QV(Xio) + QV(Yio). It is enough to show that {p:= v(Xio),q := v(Yio)} is a Z-basis of rv. Since p :I q, we may assume that p < q. Let ~ be a system of representatives of". = Rio /mio in Rio. Let Z E Rio and set r := v(z). Since p > 0, there exist lEN with r < lp. There exists a polynomial F = L:i+j9 (J"ijXiyj E Rio[X, Y] of degree ~ 1 with coefficients (J"ij E ~ such that z* := Z - F(Xi o, Yio) E mio. We have v(m~o) = lp > r, hence v(z*) > r, and therefore we have v(z) = v(F(Xio, Yio)) [cf. 1(2.10)]. Since V((J"ij) = if (J"ij :I 0, and since p and q are linearly independent over Q, the elements in the set {(J"ijX!oY!o I i,j E {O, ... ,l},(J"ij :I O} have pairwise different v-values, and therefore there exist s, t E {O, ... ,l} with v((J"stxfoY~o) < v((J"ijx~oYfo) for all i, j E {O, ... , l} with (J"ij :I 0 and (i, j) =I (s, t). This means that r = v(z) = v(F(Xio, Yio)) = v((J"8txfoyfJ = ps + qt [cf. 1(2.10)]. Therefore we have shown that v(Rio \ {O}) C Nop + Noq, hence that v(KX) = Zp+ Zq.

°

328

6.2

VIII Resolution of Singularities

Existence of Subrings Lying Under a Local Ring

(6.3) Proposition: Let K be an algebraic function field in r variables over a perfect field k. Let L be a finite extension of K, let S E Ineftk(L) with dim(S) = r, and let n be the maximal ideal of S. The following statements are equivalent: (1) There exists R E Ineftk(K) such that S lies over R. (2) (K n n)S is an n-primary ideal of S. (3) There exist elements Xl, ... , Xn E K n n such that SXl + ... + SXn is an n-primary ideal of S. If these conditions are satisfied, then R is determined uniquely, namely R = S n K, and we have dim(R) = r.

Proof: Clearly (2) and (3) are equivalent since S is noetherian. (1) => (3): Let m be the maximal ideal of R. Let B be the integral closure of R in L which is a k-algebra essentially of finite type [cf. B(3.6)]. Let nl, ... , nl be the maximal ideals of B labelled in such a way that S = B n1 • Then we have mB = ql n·· ·nql where qi is ni-primary for i E {I, ... , l}. Now we have mS = qlS, hence mS is n-primary; since nn R = m [cf. B(7.3)(2)], we can take for Xl," . ,X n a system of generators of m. (3) => (1): We choose elements Xl, ... , Xn E K n n such that q := SXl + ... + SXn is an n-primary ideal of S. We set B':= k[Xl,""X n]; then K':= k(Xl, ... ,Xn ) is the field of quotients of B', and q' := q n B' = B' Xl + ... + B' Xn is a maximal ideal of B' [since B' / q' = k]. Let A' be the integral closure of B' in K'; A' is a k-algebra of finite type by B(3.6). Note that A' c S. We set p' := A' n n; since p' n B' = n n B' = q, we see that p' is a maximal ideal of A' [cf. [63], Cor. 4.17]. We set R' := A~I; then m' := p'R' is the maximal ideal of R'. Moreover, we have dim(R') = tr. dk(K') =: s [cf. [63], Th. A, p. 286]. Note that R' E Ineftk(K'), and that R' c S. We have q'S c m'S C n, and q'S = q is n-primary; hence m'S is n-primary also. Since dim(R') = s, there exists an m'-primary ideal a in R' which can be generated by s elements [cf. [63], Cor. 10.7], and there exists mEN with m'm cae m'. Also, there exists n E N with nn em'S, hence nnm C m,ms C as C n, and therefore as is n-primary. This implies, by [63], Cor. 10.7, that r = dim(S) ~ s. On the other hand, we have s = tr. dk(K') ~ tr. dk(K) = r, hence r = s, and therefore L is a finite extension of K', and also K is a finite extension of K'. Let B be the integral closure of A' in L; B C S is a k-algebra of finite type by B(3.6), and since (nnB)nA' = p', the ideal nnB is a maximal ideal of B, hence S := BnnB E Ineftk(L) lies over R', ti:= (nnB)S is the maximal ideal of S, and dim(S) = dim(R') = r [cf. B(7.5)]. Now A := B n K is the integral closure of A' in K, it is a k-algebra of finite type, and An(nnB) is a maximal ideal of A, hence R := AAn(nnB) E Ineftk(K). Note that S lies over R, and that R lies over R', hence that S lies over R' [cf. B(7.3)]. Furthermore, S dominates S, both rings have the same field of quotients, and we have dim(S) = dim(S) = r. Note that tinR' = m', hence that m'S C tiS, and that therefore tiS is an n-primary ideal of S. Since k is perfect, S is analytically irreducible by III(6.14). From B(9.6) we

6 Uniformization

329

now get S = S. Since S lies over R, we have R = S n K by B(7.3), hence R is determined uniquely. The last assertion follows from B(7.5). (6.4) Theorem: Let K be an algebraic function field in two variables over an algebraically closed field k. Let L be a finite extension of K and let w be a zerodimensional valuation of the algebraic function field Lover k. Let S E Ineftk(L) be two-dimensional and regular. If w dominates S, then there exist a quadratic dilatation S* of S with respect to wand R* E Ineftk(K) such that S* lies over R*. Proof: Let {~, 1]} be a transcendence basis of Kover k. By replacing ~ by 1/ ~ or 1] by 1/1] if necessary, we may assume that w(~) ~ 0, w(1]) ~ O. Replacing S by a quadratic dilatation with respect to w, we may assume that ~ and 1] lie in S [cf. (4.4)]. Set R:= SnKj clearly R is integrally closed in K [since S is integrally closed in L]. Since ~ and 1] lie in K and S, these elements lie in R, and therefore K is algebraic over the field of quotients of R. Let z E Kj then there exists c E R \ {O} such that cz E K is integral over R, hence cz E R, and therefore K is the field of quotients of R. Set v := wiK. Note that dim(v) = 0 [since L is a finite extension of K, hence K,w is a finite extension of K,v by 1(6.10)]; moreover, we have rank(v) = rank(w), rat. rank( v) = rat. rank( w) and r w /r v is a torsion group [cf. 1(6.11)]. We choose Zl E K with V(Zl) > O. If rat.rank(v) = 1, then we set Z2 := 1, and if rat. rank(v) = 2, then we choose Z2 E K such that V(Z2) > 0 and that V(Zl), V(Z2) are linearly independent over Q in r v ®z Q, then v(zd, V(Z2) are linearly independent in r w ®z Q. We set z := ZlZ2 E K; note that v(z) > O. By (4.4) we may replace S by a quadratic dilatation of S with respect to w which contains Zl and Z2, hence we may assume that Zl and Z2 lie in S; note that z is not a unit of S. By (6.2) there exists a quadratic dilatation S* of S with respect to w such that Zl, Z2 E S* and that z = xaybu where {x,y} is a regular system of parameters of S*, a, b E No, u is a unit of S*, and with the following additional property: either we have a > 0 and b = 0, or the elements w(x) and w(y) are rationally independent. We set R* := S* n K and p* := n* n R* where n* is the maximal ideal of S*; note that R* is integrally closed, that K is the field of quotients of R* [proof as above], and that z E p* since z E n*. We set q* := p* S*. We show that in both cases q* is an n* -primary ideal of S*; then we have R* E lneftk (K) and S* lies over R* by (6.3). Case 1: a > 0 and b = O. Since K is the field of quotients of R* and since y is algebraic over K, there exists a non-zero c E R* such that cy is integral over R*. Let xn + iIX n - 1 + ... + in E K[X] be the minimal polynomial of cy over K; then the coefficients iI, . .. in lie in R* [cf. 111(2.12)(2)], and we have in =i' o. We set io := 1 and gi := in_iCi for i E {O, ... , n}. The elements go, .. ·, gn lie in R*, and from go = _y( gnyn-1 + ... + gl) we see that the non-zero element go E S* is divisible by y. We write go = xlh* with 1 E No, h* E S* and x does not divide h*j note that y divides h*. We set h:= g3/z l . Then we have h E K [since z E K] and h = h*a/ul E S*, hence h E S* n K = R*, and h is not divisible by

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=

x in S·. From h E yS* C n* we get hE n* n R* p*. Suppose that q* is not an n* -primary ideal of S*. Then there exists a non-unit t E S* with q* c S*t [since S* is a two-dimensional regular local ring and q* is not n* -primary, every prime

ideal in Ass(S* jq*) has height 1, and every such prime ideal is a principal ideal]. We have Z = xau E q* C S*t, and therefore we have S*t = S*x a' with 1 ~ a' ~ a. We have h E p* C p* S* = q* C S*t = S*x a' , hence h is divisible by x in S*. This contradiction shows that q* is n* -primary. Case 2: w(x) and w(y) are rationally independent. From Zl, Z2 ESC S* and ZlZ2 = Z = xaybu in S* we see that Zl = xalyblul, Z2 = xa2yb2u2 with al,a2,bl ,b2 E No, a = al + a2, b = bl + b2, and units Ul, U2 E S*. We set bl ,Z3:= Zla2 Z2-al and Z4:= Zlb2 z2-bl ; the e1ement s Z3, Z4 l'le m . K , an d we d := al a2 b2 a have Z3 = y-du3 and Z4 = xdU4 with units U3 := Uf2 u2 t, U4 := U~2U2bl of S*. Suppose that d = O. Then Z3 and Z4 are units of S*, hence a2w(zd -alw(z2) = 0, b2w(zd - bl w(Z2) = O. Since w(zd = V(Zl), W(Z2) = V(Z2) are rationally independent by the choice made above, we obtain al = a2 = bl = b2 = 0, hence a = al + a2 = 0, b = bl + b2 = 0, and therefore Z = ZlZ2 is a unit of S*. Since S* dominates Sand Z E S, we see that Z is a unit of S, and this contradicts the choice of z. Therefore we have d :f: O. From Z3 = y-du3 and Z4 = xdU4, and the fact that w(x) and w(y) are rationally independent, we now see that the elements W(Z3) = V(Z3) and W(Z4) = V(Z4) are rationally independent. We set

I

I

* {Z3 Zl :=

z3

1

ifd 0,

* {Z4 Z2 :=

zi 1

ifd>O, if d < O.

We have zi = yldlu~, zi = xldlu~ with units u~, u~ E S*, hence zi and zi lie in S*. Since Z3, Z4 E K, we have zi, zi E K, hence zi E n* n K = p* for i E {1,2}. Therefore we have n*2l d l C q*, hence q* is n* -primary.

6.3

Uniformization

(6.5) NOTATION: Let K be an algebraic function field in two variables over an algebraically closed field k. Let v be a valuation of the algebraic function field K. We say that v can be uniformized if there exists a regular R E Ineftk(K) which is dominated by v and with tr. dk(l\:) = tr. dk(l\:v) (I\: is the residue field of R). (6.6) REMARK: Let K be an algebraic function field in two variables over an algebraically closed field k. Let A be a subalgebra of K of finite type which has K as field of quotients, let p be a non-zero prime ideal of A, set R := Ap and let I\: = Q(Ajp) be the residue field of R. We have dim( R) = I-and in this case we have tr. dk (I\:) = l-or dim(R) = 2-and in this case we have I\: = k since p is a maximal ideal of A, hence Ajp is a finitely generated field extension of k with tr. ddAjp) = 0, hence an algebraic extension of k [cf. [63], Th. A on p. 286 and Cor. 13.4], hence I\: = k since k is algebraically closed.

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Let v be a valuation of K. We have dim (v) = 1 or dim(v) = O. If dim (v) = 1, then v can be uniformized iff there exists a one-dimensional regular R E lneftk (K) which is dominated by v. The existence of such an R follows from I(10.7). If dime v) = 0, then v can be uniformized iff there exists a two-dimensional regular R E Ineftk(K) which is dominated by v. If k has characteristic 0, we shall prove the existence of such an R in theorem (6.9) below. (6.7) Lemma: Let K be an algebraic function field in two variables over an algebraically closed field k of characteristic O. Let L be a cyclic Galois extension of K of prime degree p. Let v be a valuation of K with tr. d k (v) = 0 and rat. rank( v) = 1 which has only one extension w to L. If v can be uniformized, then w can be uniformized. Proof: By hypothesis, there exists a two-dimensional regular R E Ineftk(K) with residue field k such v dominates R. Let m be the maximal ideal of R. The local rings in qln(L) lying over R are localizations of the integral closure S of R in L with respect to maximal ideals of S. Since v admits only one extension w to L, the valuation ring Aw is the integral closure of Av in L [cf. I(6.1)], Aw dominates Av [cf. I(5.2)], we have S C Aw and mS C n := Snm(Aw); hence n is a maximal ideal of S [cf. [63], Prop. 9.2]. We have a(m(Aw)) = m(Aw) for every a E Gal(LI K), hence a(n) = n for every a E Gal(LI K). Since the maximal ideals of S are conjugate under Gal(LI K) [cf. [63], Prop. 13.10], S has only one maximal ideal n, hence S is the only ring in qln(L) lying over R, and it is therefore the integral closure of R in L. Since S E Ineftk(L) and dimeS) = 2 [ef. B(7.5)], the local ring S has residue field k by (6.6). Since k is algebraically closed of characteristic 0, k contains a primitive p-th root of unity; then there exists a primitive element z for L/ K such that the minimal polynomial for z over K has the form F = ZP - r for some r E KX. Clearly we may assume that r E R. By (6.2), replacing R by a quadratic dilatation of R with respect to v, we may assume that r = xau where {x,y} is a regular system of parameters of R, a E ~ and u is a unit of R; by the argument above, the integral closure S of R in L is local. Suppose that a == 0 (mod p). Then z' := zx- a/ p is a primitive element for LI K, and ZP - u is the minimal polynomial for z' over K. This implies that D L / K (I,z', ... ,z'P-l) = ±ppuP- 1 [cf. I(7.6)] is a unit of R, hence [Sin: Rlm]sep = [L: K] = p [ef. III(5.4)]. On the other hand, we have Rim = Sin = k. This contradiction shows that p does not divide a. We choose a positive integer b with ab == 1 (mod p). Since b is not divisible by p, also z' := zbx(l-ab)/p is a primitive element for LIK, and ZP - xu b is the minimal polynomial of z' over K. Since u b is a unit of R, also {xu b, y} is a regular system of parameters of R. Therefore we may assume: {x, y} is a regular system of parameters for R, and ZP - x is the minimal polynomial of a primitive element z for LIK. We set B' := R[ z]; note that B' C S since z is integral over R, and that Q(B') = K(z) = L. Set q' := B' n n, B := B~I and q := q'Bq/. Then Q(B) = L, and since ReB c S, we see that B is integral over R, hence dim(B) = 2 [ef. [63],

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VIII Resolution of Singularities

Prop. 9.2]. We have x = zP E q', hence z lies in q'. Let b' be the ideal of B' generated by the elements y, z; then we have me b' C q'. Now we have B'/b' = k, and therefore we have q' = b', hence q is generated by y and z. This implies that B is a regular local ring, hence B is integrally closed, and therefore we have B = S, hence W can be uniformized.

(6.8) Lemma: Let K be an algebraic function field in r variables over a perfect field k. Let L be a finite extension of K, let R E Ineftk(K), and let S E Ineftk(L) lie over R. Let m (resp. n) be the maximal ideal of R (resp. of S). We assume that dim(R) = r, and that Rim = Sin = k. IfmS = n and if S is regular, then also R is regular. Proof: The local ring R is analytically irreducible [cf. III(6.9)]. Since dim(R) = dimeS) [cf. B(7.5)], and since S is a quasifinite R-module by assumption, one can show, just as in the proof of B(9.6)(I), that the canonical homomorphism cp: Ii -t S, induced by the local homomorphism R X2} is a transcendence basis of Kover k. (b) If rat. rank{ v) = 1, let {Xl, X2} be an arbitrary transcendence basis of Kover k. By replacing, if necessary, Xl resp. X2 by its inverse, we may assume that V{Xl) ~ 0 and V(X2) ~ O. Since k is algebraically closed and dim(v) = 0, we have K,v = k. Therefore, replacing Xi by Xi - 'Yi for some 'Yi E k, we may even assume that V{Xi) > 0 for i E {1,2}. In both cases we set Kl := k(Xl,X2), Rl := k[Xl,X2kl:l, Z 2) and denote by ml the maximal ideal of Rl; then v has center ml in R l . Note that Rl is a two-dimensional regular local ring with residue field k. (2) Let L be the smallest Galois extension of Kl containing K, and let W be an extension of v to L. Set Vl := vlKl • If rat.rank(v) = 2, then we have rv = r V1 by (1)(a). Let Wl = W, W2, ••• , Wn be a complete set of extensions of Vl to L.

6 Uniformization

333

Then D := AWl n ... n Aw.. is the integral closure of AVl in L, and n := nl := D n m{ AWl)' ... ,On := D n m{ AWn) are the pairwise different maximal ideals of D [cf. 1{6.1)). By B{lO.l) there exists zED such that z E nl, z (j. nj for j E {2, ... , n}. Let xm +alXm- l + .. ·+am E Kl[X] be the minimal polynomial of z over K l ; note that aI, ... ,am lie in AVl [cf. III(2.12) (2)]. Replacing Rl by a quadratic dilatation of Rl with respect to VI, we may assume that the elements aI, ... ,am lie in Rl [cf. (4.4)). Let B be the integral closure of Rl in L; note that z E BcD. We set q := B n m(Aw) = B n n, S := B q • Then S is the local subring of L lying over Rl such that w dominates S; note that dimeS) = 2 by B(7.5) and that S has residue field k by (6.6). From z E B n n = q, z (j. B n nj for j E {2, ... , n}, we get q =I B n nj for j E {2, ... , n}, hence n is the only maximal ideal of D lying over q. Note that Rl c Avl' S CAw, and that AVl dominates R l . (i) Let G l := Gal{L/Kl ) be the Galois group of Lover K l , and let a E G l . If a{q) = q, then we have a{n) = n [since n is the only maximal ideal of D lying over q). Therefore we have Gz{w/vt} ::) Gz{S/R l ). On the other hand, we have Gz{w/vt} C Gz{S/Rt} by B{7.lO){2). Therefore we have Gz{w/vt} = Gz{S/R l ). (ii) We set R := S n K, m := n n R. Then S lies over Rand R lies over Rl [cf. B{7.3))' hence we have R E Ineftk(K) by B{3.6); note that dim(R) = 2 and that R has residue field k. We set G := Gal(L/ K). (iii) Let (Kl)z be the fixed field of Gz(S/ R l ), and Kz be the fixed field of Gz(S/ R). By B(7.8) Kz is the compositum of (Kdz and K and Gz(S/ R) = Gz(S/ R 1 ) n G, Gz(w/v) = Gz(w/vt} n G. Therefore we have Gz(w/v) = Gz(S/ R) by (i). (iv) The residue field "'Vi = k of VI is algebraically closed; by 1(8.3)(4) we have GT(W/Vt} = GZ(W/Vl), and GT(W/Vt} is abelian by 1(8.16). Therefore GZ(W/Vl) = Gz(S/R 1 ) = Gal(L/(Kt}z) is abelian, hence Kz is an abelian Galois extension of (K1 )z. By Galois theory there exists a sequence of fields (K1 )z =: Lo C Ll C ... C Lh := Kz such that Li/ Li-l is a cyclic extension of prime degree for i E {I, ... h}. Let i E {O, ... , h}. We set Si := S n Li and Wi := wlLi. Note that Si lies over R 1 , hence Si E lneftk (Li) [cf. B(3.6)), dim(Si) = 2 [ef. B(7.5)) and Si has residue field k. Now m 1 So is the maximal ideal of So by B(7.7); since Rl is a twodimensional regular local ring, we see that So is also a two-dimensional regular local ring. Therefore Wo can be uniformized. Let i E {I, ... , h}; since w is the only extension of Wo to L [ef. 1(8.3)(1)), Wi is the only extension of Wi-l to Li. If rat. rank(v) = 1, then we can apply (6.7) successively to the extensions L1/ Lo, L2/ L 1 , .•. ,L h/ Lh-l, and get: Wh can be uniformized. If rat. rank( v) = 2, then we have rv = r Vl [ef. above], rv = r Wh and r Vl = rwo by 1(8.7), hence r Wh = r wo ' and since Wh is the only extension of Wo to Lh, we see that Lo is the fixed field of GZ(Wh/WO) by B(7.7)(2). Since dim(v) = 0, we have dim(w) = 0, and since k is algebraically closed, the residue field of w is k. Therefore the residue fields of Wo and Wh are equal to k, hence we have GZ(Wh/WO) = GT(Wh/WO) by B(7.11)(I}. We have r whir Wo = GT(Wh/WO) [ef. 1(8.16)), hence GT(Wh/WO) is trivial, and this yields Lo = L h. Therefore Wh = Wo can be uniformized. (v) Hence in both cases Wh can be uniformized, i.e., there exists a two-dimensional

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VIII Resolution of Singularities

regular local ring S E Ineftk(Kz) with residue field k which is dominated by Who Since Sh C Awh , and Sh is essentially of finite type over k, we can, replacing S by a quadratic dilatation of S with respect to Wh, assume that Sh c S [cf. (4.4)]j note that a first quadratic dilatation of S is two-dimensional and regular by (4.1) and has residue field k by VII(6.7). Let SL E Ineftk(L) be the ring lying over S which is dominated by Wj SL is two-dimensional and has residue field k. Clearly we have S C SL CAw, and SL dominates S. From B(7.10) we get GZ(W/Wh) C GZ(SL/S) c GZ(S/Sh). From B(7.9) we get GZ(W/Wh) = Gz(w/v) n G and GZ(S/Sh) = Gz(S/R) n G. From (iii) we therefore get GZ(W/Wh) = GZ(SL/S) = GZ(S/Sh). Since W is the only extension of Wh to L, the field Lh is the fixed field of GZ(W/Wh) by B(7.7)(2), hence Lh is also the fixed field of GZ(SL/S)j since m(S)SL = m(SL) by B(7.7)(3), we see that SL is a two-dimensional regular local ring with residue field k. Applying (4.1) yields: Every quadratic dilatation of SL with respect to W is two-dimensional and regular and has residue field k. Replacing SL by a quadratic dilatation with respect to W we may assume by (6.4): There exists RK E Ineftk(K) such that SL lies over RKj note that RK = SL n K [ef. B(7.3)]. We have R C RK C A v , dim(RK) = 2, and RK has residue field k. Since RK dominates R and Aw dominates RK, from B(7.1O) we get Gz(w/v) C Gz(SL/RK) c Gz(S/R), hence Gz(w/v) = GZ(SL/RK) = Gz(S/R) [ef. (iii)]. Since Kz is the fixed field of Gz(S/R), we see that K is also the fixed field of Gz(SL/RK), and therefore we have m(RK)SL = m(SL) by B(7.11)(2). Since SL is regular, also RK is regular by (6.8).

7

Resolution of Surface Singularities II: Blowing up and Normalizing

(7.0) In this section k is an algebraically closed field of characteristic zero, and K is a field of algebraic functions in two variables over k. All varieties are varieties over k. All valuations of K will tacitly be assumed to be valuations over k.

7.1

Principalization

(7.1) REES RING AND INTEGRAL CLOSURE: Let R be an integrally closed domain, and let a be an ideal of R. Let b be the integral closure of a in R. We assume that bn is integrally closed in R for every n E Nj then bn is the integral closure of an for every n E N [cf. B(6.16)(4)], hence 'R.(b, R) is the integral closure of'R.(a,R) [ef. B(6.8)]. Under this assumption, we have the following: Let q be a homogeneous prime ideal of 'R.( a, R) which is not irrelevant such that 'R.(a, R)(q) is integrally closed. We choose x E a with xT ¢ q. Then 'R.(a, R)(q) is a localization of 'R.(a, R)(xT) = R[ a/x] with respect to a prime ideal p [cf. B(5.6)(2)]. Now 'R.(b, R)xT is integral over 'R.(a, R)xT [cf. [63], Prop. 4.13], hence also ('R.(b,R)xT)O = 'R.(b,R)(xT) = R[b/x] is integral over ('R.(a,R)xT)O =

7 Resolution of Surface Singularities II: Blowing up and Normalizing

335

R(a,R)(xT) = R[alx] [ef. B(4.24)(1)]. Furthermore, R(b,R)(xT) = R[b/x] is integrally closed [cf. B(4.30)(2)], hence R[ b/x]p = R[ alx]p [since every localization of R[ bI x] with respect to a multiplicatively closed system is integrally closed by [63], Prop. 4.13]. Now R[ bI x]p is the localization of R[ bI x] with respect to a prime ideal p' [ef. B(2.3)(3)], hence R[b/x]p = R[b/x]p' = R(b,R)(q') with a homogeneous prime ideal q' of R(b, R) [ef. B(5.6)(2)], hence we have R(a,R)(q) = R(b,R)(q').

(7.2) Lemma: Let S E Ineftk(K) with dim(S) = 2, and let n be the maximal ideal of S. Let 1:: be the set of zero-dimensional valuations v of K with Av :::> S. Then 1:: is not empty, and we have: (1) Every v E ~ dominates S, and we have S = nVEE A v , n = nVEE Pv· (2) Let w be a one-dimensional valuation of K with Aw :::> S; then there exists v E 1:: with Av cAw. Proof: We can write S = Ap where A is a k-subalgebra of K of finite type and p is a prime ideal of A [cf. B(2.3)(3)]; since dim(S) = 2, we see that p is a maximal ideal of A [ef. B(7.5)], hence 1:: :f. 0 [ef. 1(10.9)]. Since k C S, we know that S is the intersection of all valuation rings of K which contain the ring S [cf. 1(3.4)]. Let v E 1::; then PvnS is the maximal ideal of S since "'v = k, hence v dominates S. Let w be a one-dimensional valuation of K with S CAw; "'w is a field of algebraic functions in one variable over k [cf. 1(10.7)]. Let O qn, with qo := m and qn := an - 1bTn for every n E N, is a homogeneous non-irrelevant prime ideal ofR( a, R). Moreover, writing a = b + k ao with ao E R, the ideal bI ao . S is the maximal ideal of S. Proof: (1) Let n be the maximal ideal of S; note that Sin = k by B(7.5). There exist Y1, ... ,Yh E K such that, setting B:= R[Y1, ... ,Yh], we have S = BnnB and p := n n B is a maximal ideal of B [cf. B(7.5)]; therefore the elements /1 := Y1 mod p, ... ,'h := Yh mod p lie in k. Replacing Yj by Yj - Ij for j E {I, ... , h}, we may assume that Yj E P for j E {I, ... , h}. Since Q(R) = K, there exist elements aO,a1, ... ,ah E R, ao:f. 0, with Yj = ajlao for j E {l, ... ,h}. Clearly we may assume that ao, a1, ... ,ah have greatest common divisor 1 in R. We set

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VIII Resolution of Singularities

a := Rao+·· ·+Rah; then 1 is a greatest common divisor for the elements of a, and we have B = R[a/ao]. Since B t- R [as S t- Rl, we have ao E m. Suppose that a = R; then we have B = R[ a/ao] = R[ l/ao], hence we get mS = S, contradicting the assumption that S dominates R. Therefore a is an m-primary ideal of R [cf. VII(2.1)]. Since S = R[ a/aojp, we have S = n(a, R)(q) where q is a homogeneous prime ideal of n(a, R) with aoT ~ q [cf. B(5.6) j; since powers of complete ideals of R are complete [cf. VII(4.13)]' we may replace a by its integral closure [cf. (7.1) j; hence we may assume, to begin with, that a is a complete m-primary ideal of R, generated by ao, ... , ah, and that B = R[ a/ao j. (2) Let E be the set of zero-dimensional valuations v of K with Av :::> S. By (7.2) we know that v dominates S for every vEE, and that S = nvEE Av. Since S dominates R, every vEE dominates R also. For every vEE we have a/ao c A v , hence a is contained in a~ := {z E R I v(z) ~ v(ao)}. We set a' := nvEE a~. Now a~ is a v'-ideal, hence it is integrally closed, and therefore a' is a complete ideal of R [cf. B(6.16) j which contains a. For every a' E a' we have v(a' lao) ~ 0 for every VEE, hence a'/ao lies in S, and therefore we have B' := R[a'/aol C S, hence S = B~nBI. Replacing a by a' and B by B ' , we see that we may assume that B = R[ a/ ao j where a is complete and that BnnB = S. Likewise, for vEE we define bv := {z E R I v(z) > v(ao)}; then b := nVEE bv is a complete ideal of R. Note that be a, and that v(a) = v(ao) and that v(b) > v(a) for every vEE. We show that alb is a simple R-module which implies that b is an m-primary ideal. In fact, let a E a. We have a/ao E S, hence there exists 'Y E k with a/ao - 'Y E n, whence v(a-'Yao) > v(ao) for every vEE [since the valuations in E dominate S], and therefore we have a E b + k ao, hence a = b + k ao; since ao ~ b, the assertion has been proved. (3) We have

[note that v(an-1b) = (n -l)v(ao) +v(b) > v(al)) for every v EEl; in particular, we have an t- an- 1b for every n E N. We set qo := m, qn:= an-1bT n C n(a,R)n for every n E N. Then q := EDn>o qn is a homogeneous prime ideal in n(a, R) which is not irrelevant, and we ootain S = n( a, R) (q). In fact, it is easy to check that q is a prime ideal of n( a, R) [cf. (*) above and use B(4.1) j. Since an t- an-1b for every n E N, the ideal q is not irrelevant. We have aoT ~ q; then p:= (b/ao)· R[a/aoj is the prime ideal in R[a/aoj corresponding to q. We have b/ao C n since n = {z E S I v(z) > 0 for every VEE} [cf. (7.2) j. Conversely, let z E R[ a/ao j n n, hence v(z) > 0 for every vEE. We can write z = a/a8 for some n E N and a E an; then we have v(a) > v(a8) = v(an ), hence a E an-1b by (*), hence z E (b/ao) . R[a/aoj. Thus, we have shown that p = R[ a/ao j n n, hence that S = R[ a/ao lp and n = (b/ao) . S, and, in particular, that S = n(a,R)(q).

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(7.4) Corollary: Let VR be the valuation of K determined by the order function of R and let v be a zero-dimensional valuation of K which dominates S. If the first quadratic dilatation Rl of R with respect to v is not contained in S, then we have vR(b) = vR(a) + 1. Proof: Let n be the maximal ideal of S. For typographical reasons, if x E m \ m2 and Y E grm(R) is a form of degree 1 not associated with In(x) , then the maximal ideal of A := R[ mix] determined by Y [cf. VII(2.6)] shall be denoted by n(y). Since ma c b c a [cf. (7.3)], we have 1 + vR(a) ~ vR(b) ~ vR(a), hence we have vR(b) = vR(a) + 1 or IIR(b) = vR(a). We show that IIR(b) = vR(a) implies that Rl is contained in S, contradicting our assumption. Thus, suppose that vR(b) = vR(a) =: r. Let Xl, Yl be a regular system of parameters of R with v(xd < V(Yl), Le., we have v(m) = V(Xl). The construction in the proof of VII( 4.11) shows that there exists a linear combination x:= alXl + blYl of Xl, Yl with units aI, bl of R such that every simple complete m-primary factor of a and of b different from m is contracted from A = R[ mix]. Since v dominates R, we still have v(m) = v(x)j hence with Y := Yl we have m = Rx + Ry and v(y) > v(x). Let oj In(x) + f3j In(y), j E {I, ... , l}, be the pairwise different non-associated characteristic forms associated with the various simple complete m-primary factors of a and of b different from m. Then we have f3j :I 0 for j E {I, ... , l}. We include among these forms the homogeneous polynomial y := In(y), also there may be no simple complete m-primary factor of a or of b having characteristic form In(y). We label these forms as Yl := In(y)'Y2 := 02 In(x) + /32In(Y)'···'Yh := ohIn(x) + /3hIn(y)j note that 02, ... ,Oh are different from o. We now write

where ro, So E No, and where, for i E {I, ... , h}, ai is the product of those simple complete m-primary factors of a which have characteristic form Yi [ai = R if a does not have a simple complete m-primary factor with characteristic form Yi]j similarly for band bl , ... , bh. Note that the ideals al, ... , ah and bl , ... , bh are contracted from A by VII(4.IO). We set ri := vR(ai), Si := vR(bi) for i E {I, ... , h}. Then we have vR(a) = ro + rl + ... + rh, vR(b) = So + Sl + ... + Sh· (1) We show that ri ~ Si and bi C m 8 . - r'ai for every i E {I, ... ,h}, and that So ~ roo In fact, let i E {I, ... , h}. There is nothing to show if ai = R. Thus, assume that ai :I R, and let a~ be the transform of ai in A. Since ai is contracted from A, we have ai = aiA n R = x r • a~ n R, a~ is n(Yi)-primary and we have a~An(g.) n A = a~ [cf. VII(2.17)]. Let a' be the transform of a in Aj then we have a' An(g,) = a~An(g.) [cf. VII(2.15)]. Let b' be the transform of b in A. Since vR(a) = vR(b) and b C a, we have b'An(g.) C a'A n (9.). Let b~ be the transform of bi in A. We have b' An(g,) = b~An(g.), hence b~ :I A, and therefore b~ is n(Yi)primary, b~An(9.) n A = b~, X S , b~ n R = bi and b~ C a~. Suppose that ri > Si. Then we have x r • -so X" b~, c x r • a'·t' and since mr .- s• b·t is contracted from A [cf.

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VIII Resolution of Singularities

VII(4.1O)], we have mri - Si bi C Cli, hence m divides ai [cf. VII(3.22)(2)], contrary to the definition of Cli. Therefore we have Si ~ Ti, hence xSib~ C x8i-rixria~, and, as above, bi C mSi-riai. Since vR(b) = vR(a), hence TO + ... + Th = So + ... + Sh, we get So :::; TO' (2) We have a = b+kao, hence dimk«a+mr+l)/mr+l) and dimk«b+mr+l)/mr+l) differ at most by 1. On the other hand, for i E {I, ... , h} the k-vector space (ai +mri+l )/mri+l is generated by uri [since deg(c(ai)) = vR(ai) by VII(3.22)(I)]; similarly for bi. Therefore we have dimk«a+ mr+1)/mr+l) = dimk(mro /mro+l) = TO, dimk«b + mr+l)/mr+l) = So, and since be a, we get TO = So or To = So + 1. (a) If TO = So, then we have Tj = Sj for every j E {I, ... , h}, and bi "I Cli for exactly one i E {I, ... , h} [note that, if e ~ () and e' ~ ()' are complete m-primary ideals of R, then we have ee' ~ ()e' ~ ()()' by unique factorization of complete ideals, cf. VII(4.17)]. (b) If TO = So + 1, then we have Si = Ti + I for one i E {I, ... , h}, while Tj = Sj for every j E {I, ... ,h}, j " l i, hence bi C mai and, as above, bj = aj for j E {I, ... , h}, j "I i. Therefore, in both cases, there exists exactly one i E {I, ... , h} with bi "I ai. (c) We show that b1 "I a1. Suppose that bl = aI, hence that Sl = TI. Let j E {2, ... ,h}. Since v(y) > v(x) and aj"l 0, we have v(ajx+ (3jy)r; = v(xr;) = v(mr;). Since aj + mr;+l = R(ajx + (3jyt; + mr;+l, we get v(aj) = v(mr;), and similarly v(bj) = v(m8;). Since Sl = T1, we have SO+S2+" '+Sh = TO+T2+" +Th. We have v(a) - v(ad = (TO + T2 + ... + Th)v(m) = (So + S2 + ... + sh)v(m) = v(b) - v(bd, hence v(a) = v(b), contradicting the definition of b [cf. (2) in the proof of (7.2) ]. (3) Let v' be a zero-dimensional valuation of K having center n in S. Since S dominates R, we have v'(m) > O. Furthermore, we have v'(a) < v'(b) by definition of b. Suppose that v'(y) < v'(x); then we have v'(m) = v'(y). We have a1 + mr1+l = yrlR + mr1 +1, and therefore v'(ad = T1v'(m), and, similarly, v'(b 1) = slv'(m). If TO = So, then we have Ti = Si for i E {I, ... , h} and ai = bi for i E {2, ... ,h} [cf. (2)(a) and (2)(c)]. From v'(a) < v'(b) we get v'(ad < v'(bd. From T1 = Sl we get v'(ad = v'(b 1). This contradiction shows that in this case we have v'(y) ~ v'(x). If TO = So + 1, then we have Sl = T1 + 1 and a = mSo (madb 2 ••• bh [cf. (2)(b) and (2)(c)]. From v'(a) < v'(b) we get v'(ma1) < v'(bd. This inequality yields (TI + I)v'(m) < slv'(m) = (T1 + I)v'(m), hence T1 + 1 < T1 + 1 since v'(m) > 0 which is absurd. Therefore also in this case we have v'(y) ~ v'(x). Since S is the intersection of all valuation rings of zero-dimensional valuations of K having center n in S [cf. (7.2)], we have y/x E S, hence R[ m/x] C S; since S dominates R, also v dominates R, and the first quadratic dilatation of R with respect to v is RI = R[m/x]R[m/z)nn C S, as we wanted to show. This ends the proof of (7.4). (7.5) Corollary: Let n be the maximal ideal of S. Let v be a zero-dimensional valuation of K which dominates S. If the first quadratic dilatation R1 of R with

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respect to v is not contained in S, and if there exists an l-th quadratic dilatation R, of R with respect to v with S c R" then nR, is a principal ideal of R,.

Proof: By (7.4) we have vR(b) = vR(a) + 1. Using the notation in the proof of (7.3), we have aR[a/aol = aoR[a/aol, and since R[a/aol C S, the ideal as is a principal ideal of S, hence aR, is a principal ideal of R,. From (4.3) and VII(5.8) we get that the ideal bR, is a principal ideal of R,; since n = b/ao . S, it follows that nR, is a principal ideal of R"

7.2

Tangential Ideals

(7.6) TANGENT LINES AND MAXIMAL PRIMARY IDEALS: Let X C An be an affine algebraic variety passing through the origin, let k[ T 1, ... , Tn] be the affine coordinate ring of An, set a := I(X) and let A = k[Tb ... , Tn]/a = k[ h, ... , t n ] be the affine coordinate ring of X. Let R be the local ring of the origin on X, and, for i E {I, ... , n}, let Xi be the image of ti in R. The maximal ideal n of R is generated by the n elements Xl, .. . ,Xn , and we have R = Am where m is the maximal ideal of A generated by t1, ... , tn. Since R/n = k, we know that every element X E n admits a representation X = f31x1 + ... + f3nxn + x, with f3b' .. , f3n E k and x' E n2 • (1) Let LeAn be a line through the origin, and let (-Y1,"" 'Yn) E k n be a nonzero vector determining L, i.e., the line L is the intersection of all hyperplanes Z((a1T1 + ... + anTn» in An where (a1,"" an) E k n is a non-zero vector with 'Y1a1 + .. '+'Ynan = O. The elements a1X1 + .. ·+anxn +X' in n [where a1, ... , an E k and x' E n2 ] with 'Y1 a1 + ... + 'Ynan = 0 form a maximal n-primary ideal of R [cf. B(1O.I6) for the definition of a maximal n-primary ideal]. In fact, it is easy to check that the set q of these elements is an ideal containing n2 and properly contained in n, hence it is an n-primary ideal [since n is a maximal ideal]. We have 'Yi "10 for some i E {I, ... , n}, hence Xi Ft q. For every j E {I, ... , n}, j "I i, we have xj := Xj - (-yj/'Yi)Xi E q, hence q is a maximal n-primary ideal, generated I I I 2 bY Xl" .. , Xi-1, Xi+1" .. ,XIn an d n. (2) Let q be a maximal n-primary ideal of R; then n/q is a one-dimensional kvector space, and we have n2 C q [cf. B(IO.I7)(2)]. Let zEn \ q; then every element X E n has a unique representation X = 'YZ + Y with 'Y E k and y E q. In particular, we have Xi = 'YiZ + Yi with 'Yi E k and Yi E q for i E {I, ... , n}. Since q "I n, we have (-Y1,' .. ,'Yn) "I O. Let X E n; we write x = a1X1 + ... + anXn + x' with a1, ... ,an E k and x' E n2 • Then we have

= (a1 'Y1 + ... + an'Yn)z + a1Y1 + ... + anYn + x'; we see that x E q iff a1 'Y1 + ... + an'Yn = O. The one-dimensional subspace of k n x

with basis {('Y1,'" ,'Yn)} is uniquely determined by q. (3) From (1) and (2) it follows that there exists a bijective map between the set of lines in An passing through the origin and the set of maximal n-primary ideals of R.

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(4) Let q be a maximal n-primary ideal of R, and let LeAn be the line through the origin corresponding to q. We show: L is a tangent line to X at the origin iff

nl

f:. nl - 1 q

for every lEN.

In fact, let q correspond to bl,"" 'Yn) E k n \ {O}, and assume, as we may, that 'Yl f:. O. The elements x~ := Xl'X~:= X2 - b2hl)Xl,""X~:= Xn - bnhl)Xl generate n, and q is generated by x~, ... ,x~ and n2 • After having made a linear change of coordinates, we may assume, therefore, that 'Yl = 1, 'Y2 = ... = 'Yn = 0, and that q is generated by X2, ... , Xn and n2 • Then the line corresponding to q is the line L = Z{{T2, ... , Tn)). We know that L is a tangent line to X at the origin iff In{a) C (T2, ... , Tn) [cf. A{6.13){4)]. We assume that L is not a tangent line to X at the origin. Then there exists F E a with In{F) ¢ (T2, ... , Tn). We set r := deg{F), and we write F = Fr + ... + Fr+s where Fi E k[ T1 , ••• , Tn] is homogeneous of degree i for i E {r, .. . ,r + s}, and In{F) = Fr. We have Fr = O:oT[ + G2T2 + ... + GnTn with 0:0 E k X and G 2, ... ,Gn E k[Tb ... , Tn] homogeneous of degree r - 1. Now we have 0 = Fr{Xl,"" xn)+·· +Fr+s{Xl,'" ,xn) and Fr+i{Xl, ... , xn) = O:iXr+i+Yi with O:i E k and Yi E nr - 1q for i E {I, ... ,s}, we have Xr(O:O+O:IXl + .. '+O:sxn E E nr - 1 q, and therefore we have nr = nr - 1 q. Conversely, assume nr - 1 q, hence that nr = nr - 1 q for some r E N. Then there exist Y2, .. . ,Yn E nr - 1 such that = Y2X2 + ... + YnXn' This implies that there exist homogeneous polynomials G2, ... ,Gn E k[Tb ... ,Tn] of degree r - 1 and a polynomial H E k[T1 , ••• , Tn] with H{O) = 0 such that F := (T[ + G2T2 + ... + GnTn){1 + H) E a, and we have In{F) = T[ + G2T2 + ... + GnTn ¢ (T2,' .. ,Tn), hence L is not a tangent line to X at the origin. (5) A maximal n-primary ideal q of R is called a tangential ideal if the line in An through the origin corresponding to it is a tangent line to X at the origin; note that q is a tangential ideal iff (*) in (4) holds.

xr

xr

(7.7) ANOTHER CHARACTERIZATION OF TANGENTIAL IDEALS: We keep the notations of (7.6) and we assume, in addition, that X is irreducible. Let F be the field of rational functions on X. (I) Let v be a zero-dimensional valuation of F which dominates Rj then we have v{n) = min(v(xl), ... , v(xn)). The v-ideal q := {x E R I v(x) > v{n)} of R is an n-primary ideal [since n2 C q]; it is properly contained in n. Let zEn \ q; for x E n we have v{xJz) ~ 0, hence there exists 'Y E k with v{xJz - 'Y) > 0 [since v is zero-dimensional], i.e., we have x - 'Yz E q. Therefore q is a maximal n-primary ideal [cf. B{1O.16)]. (2) Let q be a maximal n-primary ideal of R; we assume that there exists a zerodimensional valuation v of F dominating R such that q is a v-ideal [i.e., we have qAv n R = q]. Then we have q = {x E R I v{x) > v{n)}. In fact, since q is a v-ideal different from n, we have v(q) > v{n) [since n is a v-ideal]. We choose zEn with v{n) = v{z). Let x E nand v{x) > v(n). We have x = 'Yz + Y with

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341

'Y E k and y E q, and since vex) > v(z), we have 'Y = 0, hence x E q. This proves our assertion. (3) Let q be a maximal n-primary ideal of R. Then q is a tangential ideal iff there exists a zero-dimensional valuation v of F dominating R such that q is a v-ideal. In fact, to prove this assertion, we may assume, as above, that we have Xl ¢ q, and that q is generated by X2, ... , Xn and n2. (a) Assume that q is a v-ideal where v is a zero-dimensional valuation v of F which dominates R. We have v(n) > v(q), and since v(n) v(xd, we have v(Xj) > v(xd for j E {2, ... ,n}, and therefore v(n'-lq) ~ (1-1)v(n) +v(q) > lv(xd = v(xD, hence x~ ¢ n'-lq for every IE Nj this means that (*) in (7.6)(4) holds. (b) Assume that (*) in (7.6)(4) holds. We have Xl =I- 0 since Xl ¢ q. We set A' := A[m/xt] = k[Xl,X2/Xl, ... ,Xn/xt]j let m' be the ideal in A' which is generated by Xl, X2/Xl," ., Xn/Xl' Suppose that m' = A'. Then there exist mEN and elements Wl,' .. , Wn E mm with 1 = (Wt/Xr)Xl + (W2/Xr)· (X2/Xl) + ... + (wn/xr)· (xn/xd, hence x~+l = W1X~ + W2X2 + ... + WnXn E nmq, hence we have nm = nm- 1 q, contradicting (*). Now m' is a maximal ideal of A'j let v be a zero-dimensional valuation of F which dominates A:U, [cf. 1(10.9)]. Since mA' = xlA' C m', we have m' n A = m, hence we see that A:U, dominates Am = R. We have V(Xl) > 0, v(xj/xd > 0 for j E {2, ... ,n}, and therefore v(q) > v(xd = v(n). Let zEn and v(z) > v(n)j then we have z = 0!2X2 + ... + O!nXn + z, with 0!2, ... , O!n E k and z' E n2, hence z lies in q, and therefore q is a v-ideal.

=

(7.8) TANGENTIAL IDEALS AND BLOWING UP: Let X c IP'n be an irreducible projective variety with field of rational functions F, and assume that X passes through p = (1: 0: ... : 0). Then 0 X,p is also the ring of p, considered as a point of the affine variety X n {(~o : ~1 : •.. : ~n) I ~o =I- O} contained in An. Let n be the maximal ideal of OX,p, and let E be the non-empty set of zero-dimensional valuations of F which dominate OX,p [cf. 1(10.9)]. Let 71": X' -+ X be the blowing up of X with center p, and let E' = 71"-1 ({p}) be the exceptional fibre. Now X' C IP'n x IP'n-l is an irreducible projective variety with field of rational functions F [cf. A(13.8)]. In the following, we use the notations introduced in A(13.11). We set fh := Yi + b for j E {1, ... ,n}, Zj:= Zj + b1 for j E {2, ... ,n}. We assume that 'tit =I- OJ then we have Zj =Yj/fh for j E {2, ... ,n}. (1) Let q = (p, (131: ... : f3n» be a point in E'. Let E~ be the non-empty set of zero-dimensional valuations of F which dominate OX',qj note that E~ C E. Let q be the maximal n-primary ideal of OX,p corresponding to (131," . , f3n) [cf. (7.6)(3)]. We show: q is a v-ideal for every v E E~, and therefore q is, in particular, a tangential ideal ofOx,p [cf. (7.7)(3)]. For the proof we assume, as we may, that 131 =I- O. The ideal q is generated by Y2 - (f32/f3dY1"",Yn - (f3n/f31)"Y1 and n 2. The ring OX',q is the localization of k[jit, Z2 - 132/131," ., Zn - f3n/f31] with respect to the maximal ideal generated by Yl' z2-f32/ 131. ... , Zn -f3n/ 131. Let v E E~. Then we have V(Y1) > 0, v(Zj-f3j /f3d =

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v(Yj!fh -f3j/ f3d > 0 for j E {2, ... , n} which means that v(Yj - (f3j/ f3dYl) > V(Yl) for j E {2, ... , n}, hence v(q) > V(Yl) = v(n). Let yEn and v(Y) > v(n). We have Y = 02Y2 + ... + onYn + Y' with 02, ... ,On E k and Y' E n2 , hence Y E q. Therefore q is a v-ideal. (2) Let (f31,' .. , f3n) E k n \ {O}, and let q be the maximal n-primary ideal of OX,p corresponding to (f31,"" f3n). We show: If q is a tangential ideal, then the point q := (p, (f31 : ... : f3n)) E IP'n x IP'n-l lies in E', and q is a v-ideal for every v E :E~. For the proof we assume, as we may, that f31 :f. O. Then q is generated by the elements Y2 - (f32/ f31 WI' ... , Yn - (f3n/ f31Wl and n 2 • There exists v E :E such that q is a v-ideal [cf. (7.7)(3)]; in particular, we have v(Yj - (f3j/f3dYl) > V(Yl) for j E {2, ... , n}. Now p E X is the point of X which is dominated by v, v dominates exactly one point q E X' [cf. (3.4)], and clearly we have q E E'. We have V(Yl) > 0, v(Zj - (f3j/ f31)) > 0 for j E {2, ... , n}, hence (Yl' Z2 - f32/ f31, ... , zn - f3n/ f3d is the center of v in k[Yl' Z2, ... , zn], hence q = (p, (f31: ... : f3n)). (3) We consider the bijective map between the set of lines in An passing through the origin and the set of maximal n-primary ideals of OX,p [cf. (7.6)(3)]. From (1) and (2) we get: If (f31," . , f3n) E k n \ {O} corresponds to q, then q := (p, (f31: ... : f3n)) E IP'n x IP'n-l lies in E' iff q is a tangential ideal of OX,p; in this case q is a v-ideal for every zero-dimensional valuation v of F which dominates OX',q. (4) Let q E E' correspond to the tangential ideal q of OX,p' We consider the Rees ring R(n,Ox,p) = ffim>o nmTm c OX,p[T], and we show: The homogeneous ideal P = ffim>oPm ofn(n,Ox,p) with Po := n and Pm := nm-1qTm for every mEN is a prime ideal which is not irrelevant, and we have OX',q

= R(n,Ox,p)(p)'

Proof: Note that q and n are v-ideals for every zero-dimensional valuation of F which dominates Ox' ,q' There exists zEn with n = k z + q. Just as in (3) of the proof of (7.3) it follows that P is a homogeneous prime ideal in R(n,Ox,p), and from (7.7)(3) it follows that P is not irrelevant. We may assume that q = (p, (1: 0: ... : 0)); then q is the ideal generated by Y2' ... , Yn, n 2 , and Ox' ,q is the localization of k[Yl' Z2, . .. , zn] with respect to the prime ideal generated by Yl,Z2, ... ,zn. An element z' in OX',q has the form , F(Yl,Z2,,,,,Zn). ( rr» G( ) k[T rr> ] z = G(- _ _ ) wIth F T1, ... ,.Ln, Tl'"'' Tn E 1, .. ·,.Ln Yl,Z2,···,Zn and G(O, . .. ,0) :f. O. Making the substitution Zj that we can write

= Yj/Yl

for j E {2, ... , n} shows

z' - !(Yl"" ,Yn) _ !r(Yl"" ,Yn) + !r+l(Yl,'" ,Yn) +... - g(Yl,···,Yn) - gr(Yl,···,Yn)+gr+1(Yll···,Yn)+···

(*)

where !i(TI, ... , Tn), gi(T1, ... , Tn) E k[T1, ... , Tn] are homogeneous of degree i for i = r, r + 1, ... , and where

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343

Conversely, every quotient (*) for which (**) holds is an element of 0 x' ,q. It is clear that the elements IcYl' ... ,fin)' g(fil, ... ,fin) in (*) lie in nr. Let r E N, 9 = gr + gr+l + ... E k[ T1 , ••• ,Tn) where gi is homogeneous of degree i for i = r, r + 1, ... and gr i O. We write gr = aT[ + . .. with a E k. Set w:= g(fil' ... ,fin); then W E nr. We show that a i 0 iffw f/. nr- 1 q. In fact, note that nrH c nr- 1 q [since n2 C q). If a i 0, then w E nr- 1 q would imply that Yl E nr- 1 q, hence nr = nr- 1 q, contradicting (7.7)(3). Conversely, if w f/. nr- 1 q, then a i O. Thus, we have proved the assertion.

7.3

The Main Result

(1.9) Theorem: Let X be an irreducible normal projective surface and let

be a sequence such that, for i ~ 1, Xi is the normalization of a surface obtained by blowing up a singular point of Xi-I. Then (*) is a finite sequence. Proof: Let K be the field of rational functions on X. We set So = {Xo}, and let, for i E No, SiH be the set of all normalizations of all those surfaces which can be obtained by blowing up a singular point of a surface in Si. Since the set of singular points of an irreducible normal surface is finite [cf. A(7.17)(2)), it is easily seen by induction that Sn is a finite set for every n E No. Let n E No, and let Tn be the set of all local subrings Q of K such that there exists a sequence where, for every i E {O, ... ,n}, 8 i = OXi,Xi with Xi E Si and Xi E Sing(Xi ), and, for i E {I, ... , n}, Xi is the normalization of Xi := Bl xi _ 1 (Xi-d where Xi-l E Sing(Xi-d, and Xi lies over a point xi of the exceptional fibre of Xi. Every element of Tn is the local ring of a singular point on one of the finitely many members of Sn; therefore Tn is a finite set. Note that To = {Oxo,x I X E Sing(Xo)}. Suppose that there were infinitely many sequences of the form (**). Since To is finite, there would be infinitely many such sequences beginning with a specific 8 0 E To; then 8 0 = OXo,xo with Xo E Sing(Xo). Since 7i is finite, there would be some 8 1 E 7i such that among those sequences beginning with 8 0, there would be infinitely many which begin with 8 0 ~ 8 1 ~ ...• Since 72 is finite, there would be some 8 2 E 72 such that among the sequences which begin with 8 0 ~ 8 1 ~ ... there would be infinitely many beginning with 8 0 ~ 8 1 ~ 8 2 ~ ...• Continuing in this manner, we define 8 3 ,84 , ..• , and so obtain a strictly increasing infinite sequence (8i)i~0 where, for i ~ 1, 8 i = OXi,Xi with Xi E Sing(Xi ), Xi is the normalization of Xi := Bl xi _ 1 (Xi-d where Xi-l E Sing(Xi- 1 ), and Xi lies over a point xi of the exceptional fibre of XI. In particular, 8i is a normal quadratic dilatation of 8 i- 1 for every zero-dimensional valuation of K which dominates 8i-l [cf. (3.5)]. For every i E No let ni be the maximal ideal of 8i.

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Now V := Ui>O Si is the ring of a zero-dimensional valuation v of K which dominates Si for every i ~ 0 [cf. (4.6»). By the theorem of uniformization (6.9) there exists a two-dimensional regular local k-subalgebra R of K having K as field of quotients and being essentially of finite type over k which is dominated by v. Let (Rik~o be the sequence of quadratic dilatations of Ro := R with respect to v [cf. (3.2»). Since v is zero-dimensional, for every i E N the ring Ri is a two-dimensional regular local ring and (Ri)i?O is a strictly increasing sequence [cf. (4.2)]. Therefore the ring Ui>O Ri is the ring of a zero-dimensional valuation of K which dominates Ri for every i ~ 0 [cf. (4.4»), hence it must be equal to V by (4.4). Since, for every i E No, the k-algebras Ri are essentially of finite type [cf. (3.1))' and also the k-algebras Si are essentially of finite type, for every i E No there exist a(i), b(i) EN with R; c Sa(i), Si C Rb(i)' We choose i E N with Ro c Si; since V f/. Si, not every quadratic dilatation of Ro with respect to v is contained in Si. By replacing Ro by an appropriate quadratic dilatation R j of Ro with respect to v and relabelling, we may assume that Ro ~ Si, but that the first quadratic dilatation Rl of Ro with respect to v is not contained in Si. We choose I ~ 1 with Si c R I . Then the extended ideal tliRI is a principal ideal [cf. (7.5)], i.e., niRI = R/w with w E RI' Let Xi+l be the point of X:+1 = BIz. (Xi) lying under Xi+1 and set S:+1 := OXi+l,Z:+l; then we have Si+l = Sm(v)ns where S is the integral closure of S:+1 [cf. (3.5)(2)]. We show that S:+1 c RI. Let q C Si be the maximal ni-primary ideal of Si corresponding to the point xi+1 E BIz. (Xi); it is a tangential ideal and a v-ideal [cf. (7.8)(3)], and therefore we have v(q) > v(tli). The non-zero elements of S:+1 have the form ZdZ2 with Zl E nrt, Z2 E ni \ ni-1q for some mEN [cf. (7.8)(4»); we have Zl = wmz~, Z2 = wmz~ with zL z~ E RI. Since Z2 ~ ni-1q, we have V(Z2) = mv(ni) = mv(w) [cf. (7.7)(2»)' hence v(z~) = 0, and therefore z~ is a unit of RI. This means that Z = zU z~ E RI, hence that S:+1 c R/, hence that S C RI since Hz is integrally closed. Now, since v dominates R/, we have m(V) n S = m(RI) n S; therefore we see that Si+1 = Sm(R')ns is contained in RI. Repeating this argument we get Sj C RI for every j ~ i, hence V C R/, contrary to the fact that (R;)i?O is strictly increasing. Therefore there exist only finitely many sequences of the form (**), hence (*) is a finite sequence.

(7.10) Corollary: [Resolution of singularities] Let X be an irreducible projective surface. There exists a desingularization 7r: Y -+ X where Y is an irreducible regular projective surface and 7r is the composition of a finite number of morphisms which are either of the form Z -+ Z where Z is an irreducible projective surface and Z is its normalization, or of the form BIz(Z) -+ Z where Z is an irreducible normal projective surface and z E Sing(Z).

Appendix A

Results from Classical Algebraic Geometry In this first appendix we treat only those aspects of classical algebraic geometry which are needed in this book. After introducing the notions of (affine and projective) varieties and stating, mostly without proof, some results on varieties and morphisms of varieties in section 1, we study affine and finite morphisms in section 2. Products and fibre products in the category of varieties is the contents of section 3. When defining the notion of resolution of singularities in section 8, we need the concept of proper morphisms; these are introduced in section 4. Regular and singular points of a variety as well as the classical Jacobian criterion for regular points are treated in section 6. One process of resolution of singularities uses repeatedly blowing up and normalization. Normalizing is the contents of section 7, while blowing up shall be dealt with in sections 13 and 14; section 5 describes the connection between affine algebraic cones and projective varieties which is used when constructing the normalization of a projective variety. Dimension of fibres and ramification for finite maps is treated in sections 9 and 10. In section 11 we introduce the group of Cartier divisors on an irreducible variety X and the group of Weil divisors on an irreducible normal variety X, and we show that these groups coincide if X is an irreducible nonsingular variety.

1 1.1

Generalities Ideals and Varieties

(1.0) NOTATION: In this chapter k is an algebraically closed field of arbitrary characteristic, and m, n E No. We denote by A1 = An (resp. 1P';: = IP'n) the ndimensional affine (resp. projective) space over k. An and IP'n are endowed with the Zariski topology. The coordinates of a point in An are denoted by (6,6, .. ·, ~n), and the homogeneous coordinates of a point in IP'n are denoted by (~o: 6: ... : ~n). 345

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A Algebraic Geometry

The elements F of the polynomial ring k [ Tl , ... , Tn] will be considered as functions F: An ---+ k; for any subset M of k[ T 1, ... , Tn] we denote by Z (M) C An the zero set of M, and for any X C An we denote by I(X) C k[T1, ... , Tn] the ideal of polynomials vanishing on X. For any subset M of the graded polynomial ring k[ To, ... , Tn] we denote by Z+ (M) the zero set in ]p'n of the homogeneous elements of M, and for any X C ]p'n we denote by I+(X) the ideal in k[ To, .. . ,Tn] generated by all homogeneous polynomials vanishing on X; I+(X) is a homogeneous ideal. A good source for the following results from algebraic geometry which are given without proof are the first three sections of the first chapter of Hartshorne's "Algebraic Geometry" [82] or the first two chapters of the first volume of Shafarevich's "Basic Algebraic Geometry" [174].

(1.1) NOTATION: (1) A closed subset X of An is called an affine variety. An open subset of an affine variety is called a quasi-affine variety; equivalently, a quasi-affine variety is a locally closed subset of some An. (2) A closed subset X of ]p'n is called a projective variety. An open subset of a projective variety is called a quasi-projective variety. (3) A variety is an affine or quasi-affine or projective or quasi-projective variety. (4) Note that every locally closed subset of a variety is a variety. (1.2) REMARK: These definitions differ from those used in [82] where a variety is always assumed to be irreducible. (1.3) IDEALS AND VARIETIES: (1) Let X C An be closed. Then we have X Z(I(X)), and the ideal I(X) is a radical ideal. The reduced k-algebra

=

A(X) := k[ T 1, ... ,Tn ]/I(X) is called the affine coordinate ring of X. The variety X is irreducible iff I(X) is a prime ideal. Note that an element f E A(X) can be considered as a function f: X ---+ k; moreover, it is a regular function on X [cf. below], and if F E k[ T 1, ... , Tn] has f as its image, then we have FIX = f. (2) Let X C ]p'n be closed. Then we have X = Z+(I+(X)), and the ideal I+(X) is a homogeneous non-irrelevant radical ideal. The reduced graded k-algebra

S(X) := k[ To, T 1, ... ,Tn ]/I+ (X) is called the homogeneous coordinate ring of X. The variety X is irreducible iff I+(X) is a prime ideal. (3) Let a be an ideal in R := k[T1, ... ,Tn ]. Then we have I(Z(a)) = rad(a) [Hilbert's Nullstellensatz]. Moreover, the map X I--t I(X) is an inclusion-reversing bijective map from the family of closed subsets of An to the family of radical ideals of R; the irreducible closed subsets of An correspond to prime ideals of R.

1 Generalities

347

(4) Let a be a homogeneous ideal in S := k[To, ... , Tn] and assume that 2+ (a) '10. Then we have I+(2+(a)) = rad(a) [homogeneous version of Hilbert's Nullstellensatz]. Moreover, the map X I-t 4(X) is an inclusion-reversing bijective map from the family of non-empty closed subsets of rn to the family of non-irrelevant homogeneous radical ideals of Sj the irreducible closed subsets of rn correspond to non-irrelevant homogeneous prime ideals of S. The empty set of rn is the zero set of every irrelevant homogeneous ideal of S. (1.4) REGULAR FUNCTIONS AND MORPHISMS: (1) Let X be a variety. The definition of a regular function on X is the same as in [82], p. 15 [where it is defined only for irreducible varieties]. The ring of global regular functions on X is denoted by O(X)j it is a reduced k-algebra of finite type. In particular, if X is affine, then we have a canonical injective k-algebra homomorphism A(X) -t O(X). (2) Let X and Y be varieties. A morphism cp: X -t Y is defined as in [82], p. 15 [where it is defined only for irreducible varieties]; cp induces a k-algebra homomorphism cp*: O(Y) -t O(X). (1.5) LOCAL RINGS: (1) Let X be a variety, and let x E X. The local ring of x in X shall be denoted by OX,z, and ez(X) is the multiplicity of the local ring OX,z. (2) Let cp: X -t Y be a morphism of varieties. For each x E X the morphism cp induces a local homomorphism of k-algebras cp;: OY,