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Page 1: multiplicative theory of ideals
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Multiplicative Theory of Ideals

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This is Volume 43 in PURE AND APPLIED MATHEMATICS A Series of Monographs and Textbooks Editors: PAUL A. SMITH AND SAMUEL EILENBERG A complete list of titles in this series appears at the end of this volume

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MULTIPLICATIVE THEORY OF IDEALS

MAX D. LARSEN / PAUL J . McCARTHY University of Nebraska University of Kansas

Lincoln, Nebraska Lawrence, Kansas

@ A C A D E M I C P R E S S New York and London 1971

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COPYRIGHT 0 1971, BY ACADEMIC PRESS, INC. ALL RIGHTS RESERVED NO PART OF THIS BOOK MAY BE REPRODUCED IN ANY FORM, BY PHOTOSTAT, MICROFILM, RETRIEVAL SYSTEM, OR ANY OTHER MEANS, WITHOUT WRITTEN PERMISSION FROM THE PUBLISHERS.

ACADEMIC PRESS, INC. 111 Fifth Avenue, New York, New York 10003

United Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD. Berkeley Square House, London W l X 6BA

LIBRARY OF CONGRESS CATALOG CARD NUMBER: 72-137621

AMS (MOS)1970 Subject Classification 13F05; 13A05,13B20, 13C15,13E05,13F20

PRINTED IN THE UNITED STATES OF AMERICA

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To Lillie and Jean

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Contents

Preface

Prerequisites

Chapter I. Modules

1 Rings and Modules

2 Chain Conditions 3 Direct Sums

4 Tensor Products

5 Flat Modules

Exercises

Chapter II. Primary Decompositions and Noetherian Rings

1 2 Primary Submodules

3 Noetherian Rings

4

Operations on Ideals and Submodules

Uniqueness Results for Primary Decompositions

Exercises

Chapter Ill.

1 Definition

2 3

Rings and Modules of Quotients

Extension and Contraction of Ideals

Properties of Rings of Quotients

Exercises

xi ...

Xl l l

1 8

12 15 21 27

36

39

44 48 52

61

66

71 74

V i i

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Vl l l CONTENTS

Chapter IV. Integral Dependence

1 Definition of Integral Dependence

2 3 4 Almost Integral Dependence

Integral Dependence and Prime Ideals

Integral Dependence and Flat Modules

Exercises

Chapter V. Valuation Rings

1 2 3 Vaiuations

4 Prolongation of Valuations

The Definition of a Valuation Ring

Ideal Theory in Valuation Rings

Exercises

Chapter VI.

1 Fractional Ideals

2 Prufer Domains

3 Overrings of Priifer Domains

4 Dedekind Domains

5 Extension of Dedekind Domains

Priifer and Dedekind Domains

Exercises

Chapter VII.

1 The Krull Dimension

2 3 Valuative Dimension

Dimension of Commutative Rings

The Krull Dimension of a Polynomial Ring

Exercises

Chapter VIII. Krull Domains

1 Krull Domains

2 Essential Valuations

3 The Divisor Class Group

4 Factorial Rings Exercises

82 84 88

92 94

99

105 107 114 118

124 126

132 134 140 144

156 161 164 168

171 179 185 190 194

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CONTENTS ix

Chapter IX. Generalizations of Dedekind Domains

1 Almost Dedekind Domains

2 ZPI-Rings

3 Multiplication Rings

4 Almost Multiplication Rings

Exercises

Chapter X. Prufer Rings

1 Valuation Pairs

2 Counterexamples

3 Large Quotient Rings

4 Prufer Rings

Exercises

Appendix: Decomposition of Ideals in Noncommutative Rings

Exercises

Bibliography

201 205 209 216 220

226 232

234 236

244

252

263

266

Subject Index 294

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Preface

The viability of the theory of commutative rings is evident from the many papers on the subject which are published each month. This is not surprising, considering the many problems in algebra and geometry, and indeed in almost every branch of mathematics, which lead naturally to the study of various aspects of commutative rings. In this book we have tried to provide the reader with an introduction to the basic ideas, results, and techniques of one part of the theory of commutative rings, namely, multiplicative ideal theory.

The text may be divided roughly into three parts. In the first part, the basic notions and technical tools are introduced and developed. In the second part, the two great classes of rings, the Prufer domains and the Krull rings, are studied in some detail. In the final part, a number of generalizations are considered. In the appendix a brief introduction is given to the tertiary decomposition of ideals of noncommutative rings.

The lengthy bibliography begins with a list of books, some on commutative rings and others on related subjects. Then follows a list of papers, all more or less concerned with the subject matter of the text.

This book has been written for those who have completed a course in abstract algebra at the graduate level. Preceding the text there is a discussion of some of the prerequisites which we consider necessary.

At the end of each chapter are a number of exercises. They are of three types. Some require the completion of certain technical details-they might possibly be regarded as busy work. Others

xi

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xii PREFACE

contain examples-some of these are messy, but it will be beneficial for the reader to have some experience with examples. Finally, there are exercises which enlarge upon some topic of the text or which contain generalizations of results in the text-the bulk of the exercises are of this type. A number of exercises are referred to in proofs, and those proofs cannot be considered to be complete until the relevant exercises have been done.

We wish to thank those of our colleagues and students who have commented on our efforts over the years. Special thanks goes to Thomas Shores for his careful reading of the entire manuscript, and to our wives for their patience.

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Prerequisites

A graduate level course in abstract algebra will provide most of the background knowledge necessary to read this book. In several places we have used a little more field theory than might be given in such a course. The necessary field theory may be found in the first two chapters of “Algebraic Extensions of Fields ” by McCarthy, which is listed in the bibliography.

One thing that is certainly required is familiarity with Zorn’s lemma. Let S be a set. A partial ordering on S is a relation < on S such that

(i) s l s f o r a l l s E S ; (ii) if s < t and t I s , then s = t ; and (iii) if s < t and t < u, then s < u.

The set S, together with a partial ordering on S, is called a partially ordered set.

Let S be a partially ordered set. A subset T of S is called totally ordered if for all elements s, t E T either s < t or t < s. Let S’ be a subset of S. An element s E S is called an upper‘bound of S’ if s‘ I s for all s’ E S’. An element s E S is called a maximal element of S if for an element t ES, s 5 t implies that t = s. Note that S may have more than one maximal element.

Zorn’s Lemma. Let S be a nonempty partially ordered set. If evuy totally ordered subset of S has an upper bound in S, then S has a maximal element.

... Xl l l

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xiv PREREQUISITES

If A and B are subsets of some set, then A s B means that A is a subset of B, and A c B means that A E B but A # B. If S is a set of subsets of some set, then S is a partially ordered set with 2 as the partial ordering. Whenever we refer to a set of subsets as a partially ordered set we mean with this partial ordering.

Let S and T be sets and consider a mapping f : S+ T. The mapping can be described explicitly in terms of elements by writing sHf(s). If A is a subset of S, we write f (A) = {f(s) \SEA}, and if B is a subset of T , we write f - l ( B ) = {s ISES and f(s) E B } . Thus, f provides us with two mappings, one from the set of subsets of S into the set of subsets of T and another in the opposite direction. We assume that the reader can manipulate with these mappings.

If S, T, and U are sets and f : S-t T and g : T+ U are map- pings, their composition gf : S+ U is defined by (gf) (s) =g( f ( s ) ) for all s E S.

On several occasions we shall use the Kronecker delta a,,, which is defined by

1 if i= j ,

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Multiplicative Theory of Ideals

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C H A P T E R

Modules

1 RINGS AND MODULES

We begin by recalling the definition of ring. A ring R is a non- empty set, which we aiso denote by R, together with two binary operations (a, b) H U + b and (a, b) H ab (addition and multiplication, respectively), subject to the following conditions :

(i) the set R, together with addition, is an Abelian group; (ii) a(bc) = (ab)c for all a, b, c E R ;

(iii) a(b + c) = ab + ac and (b + c)u = ba + ca for all a, b, c E R.

Let R be a ring, The identity element of the group of (i) will be denoted by 0 ; the inverse of an element a E R considered as an element of this group will be denoted by -a ; a+(-b) will be written a - b. The reader may verify for himself such statements as

Oa = a0 = 0 for all a E R, a( -b) = (-a)b = -(ab) for all a, b E R,

for all a, b, c E R.

A ring R is said to be commutative if ab = ba for all a, b E R. An element of R is called a unity, and is denoted by 1, if la = a1 = a for all a E R. If R has a unity, then it has exactly one unity. We shall assume throughout this book that all rings under consideration have unities. By a subring of a ring R we mean a ring S such that the

a(b -c) = ab -uc

1

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2 1 MODULES

set S is a subset of the set R and such that the binary operations of R yield the binary operations of S when restricted to S x S. By our assumption concerning unities, both R and S have unities. We shall consider only those subrings of a ring R which have the same unity as R.

By a left ideal of a ring R we mean a nonempty subset A of R such that a - b E A and ra E A for all a, b E A and r E R. By a right ideal of R we mean a nonempty subset B of R such that a - b E B and ar E B for all a, b E B and r E R. A left ideal of R which is at the same time a right ideal of R is czlled an ideal of R. If R is a com- mutative ring, then the left ideals, right ideals, and ideals of R coincide. A left ideal, right ideal, or ideal of R is called proper if it is different from R. The ideal consisting of the element 0 alone is called the zero ideal of R and will be denoted simply by 0; it is a proper ideal if and only if R has more than one element.

Let R be a ring. If a E R, then the set

R a = { r a ( r ER}

aR=(av/r E R }

is a left ideal of R, while

is a right ideal of R. The smallest ideal generated by a, is

Since R has a unity, we have a E Ra and a E aR. of R containing a, called the principal ideal the set of elements of R of the form

c Yi as, 7

where ri , si E R and the sum is finite. This ideal will be denoted by (a). If R is commutative, R a = a R = (a).

1.1. Definition. Let R be a ring. A left R-module M is an Abelian group, written additively, together with a mapping (a, x ) t+ ax from R x M into M such that

(2) (3) (ab)x = a(bx),

for all a, b E R and x , y E M . A right R-module N is an Abelian group, written additively, together with a mapping (x , a ) t+xa from N x R into N such that

(1) a(x +y> = a x +by, (a + b)x = ax + bx,

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1 RINGS AND MODULES 3

(1’) (2’) (3’) x(ab) = (xa)b,

(x +y)a = xu +ya, x(a + b) = xu + xb,

for all a, b E R and x, y E N.

We shall adopt the custom of referring to left R-modules simply as R-modules. Many of the results concerning modules, left and right, will be stated for R-modules only. Analogous results hold for right R-modules.

Let R be a commutative ring and let M be an R-module. If a E R and x E M , we define xu to be ax. This makes M into a right R-module. It is immediate that (1’) and (2’) hold. As for (3’), we note that for all a, b E R and x E M we have x(ab) = (ab)x = (ba)x = b(ax) = (xa)b. If R is not commutative, then an R-module cannot necessarily be made into a right R-module in this way.

An R-module M is called unital if l x = x for all x E M. A similar definition applies to right R-modules. We shall assume throughout this book that all modules considered, whether left or right modules, are unital.

An R-module N is called a submodule of an R-module M if N is a subgroup of M and if the module operation R x N + N is the restriction of the module operation R x M - t M to R x N. Suppose that the nonempty set N is a subset of M. Suppose further that x - y E N and ax E N for all x, y E N and a E R. Then N, together with the induced operation of addition, is a subgroup of M . The mapping (a, x) H ax from R x N into N makes this subgroup into a submodule of M . We refer simply to the submodule N .

If L and N are submodules of an R-module M , then their inter- section L n N is also a submodule of M . More generally, if {N, I c( E I } is an arbitrary nonempty family of submodules of M , then

n Na a E I

is a submodule of M . The sum of L and N , denoted by L + N , is defined by

L + N = { x + y l x ~ L and Y E N } .

It is easily seen that L + N is a submodule of M . If N, , . . . , Nk are submodules of M , then

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4 1 MODULES

Nl + a * + Nk = { x l + + xkl xi E N , for each i>

is a submodule of M . We denote this submodule by C:=l N , ; it is called the sum of Nl, . . . , Nk .

The notion of sum of submodules of M can be extended to an arbitrary nonempty family { N , I a E I} of submodules of M . For each finite subset J of I ,

c Na a € J

is a submodule of M . In general, not a submodule of M . However,

U J E I

J finite

the union of submodules of M is

c Na a € J

is a submodule of M . It is this submodule that we call the sum of the family { N , I a E I}, and denote by 2, E, N , .

Let M be an R-module and let x E M . Then

Rx = {ux I u E R)

is a submodule of M. Since M is unital, Rx contains x . If S is a subset of M , we call the smallest submodule of M which contains S the submodule generated by S. This submodule always exists and is unique, for it is precisely the intersection of all of the submodules of M containing S. Note that if S is empty, this submodule consists of the element 0 alone. We call the submodule of M consisting of the element 0 alone the zero submodule of M and denote it simply by 0. If S is not empty, then it is clear that the submodule generated by S is precisely

Rx. X E S

1.2. Definition. Let M be an R-module. If there is a finite subset { x l , . . . , x.} of M such that M = Rx, + - * + Rx, , then we say that M is finitely generated.

Example 1 . We shall denote the ring of integers by 2. Every Abelian group is a 2-module with respect to the mapping (n, x ) H nx given by

if n > O if n = o if n <O.

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1 RINGS AND MODULES 5

Example 2. A ring R is an R-module with respect to its own addition and with its own multiplication as the module operation. Similarly R is a right R-module. The submodules of the R-module R are the left ideals of R, while the submodules of the right R-module R are the right ideals of R.

Example 3. Let K be a field. The K-modules are the vector spaces over K. The submodules of a K-module are its vector sub- spaces.

Example 4. Let M be the set of all ordered n-tuples (al, . . . , a,) of elements of a ring R. If we set

( a l , - * * , a n ) + ( h * . * . v b n ) = ( a l + b l v . - . , a n + b n )

and

a(a,, . . . , a,) = (aa,, . . . , aa,),

then M becomes an R-module. For i= 1, . . ., n, set x2= (0, . . ., 0, 1, 0, . . ., O), where the 1 appears in the ith place. Then

Before proceeding with further definitions, we shall give an important relation between the notions of intersection and sum of submodules.

M = R x l + . * * + R x , .

1.3. Theorem (Modular Law for Submodules). Let K , L, and N be submodules of an R-module M . If K c N , then

K + ( L n N ) = ( K + L ) n N .

Proof. We have K + (L n N ) s K + L. Also, since K c N , we have K + ( L n N ) G N . Hence K + ( L n N ) G ( K + L) n N. Now sup- pose ~ E ( K + L ) n N. Then X E N and x = y + z where y e K and EL. But K s N , so Y E N ; hence z = x - y ~ L n N . Thus x E K + (L n N ) , and we have ( K + L) n N G K + ( L n N ) .

1.4. Definition. A mapping 4 from an R-module M into an R-module N is a homomorphism if +(x + y ) = +(x) + #(y) and +(ax) = a+@) for all x , y E M and a E R. A homomorphism 4 : M - t N is surjective if it maps M onto N , injective if it is one-to-one, and bijective if it is both surjective and injective.

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6 1 MODULES

A bijective homomorphism is also called an isomorphism. In this case, the inverse mapping exists and is also an isomorphism. If there is an isomorphism from M onto N , we say that M and N are isomorphic and write M E N .

It should be clear how to define homomorphism and the related concepts for right R-modules. Note that the relation of being " isomorphic to " between R-modules is reflexive, symmetric, and transitive. If M is an R-module, then the identity mapping of M is an isomorphism of h4 onto itself; we denote it by 1 M .

Let M be an R-module and N a submodule of M . Since M is an Abelian group and N is a subgroup of M , we may consider the factor group M/N. We define a mapping from R x M/N into M / N by (a, x + N ) H ax + N ; this is a well-defined mapping. For, suppose x + N = y + N , then x - y E N and ax - ay = a(x - y ) E N . Hence ax+ N=ay f N . With respect to this mapping, M / N is an R-module, called a factor module of M . The mapping r) from M into M / N defined by +(x) = x + N is a surjective homomorphism. It is called the canonical homomorphism from M onto M / N .

1.5. Theorem. Let M be an R-module and let + be a homomorphism from M into an R-module N . Then

Ker q3 ={XI x E M and r)(x) =0}

is a submodule of M called the kernel of 4. Also,

1n-I r) = M4I x E MI

is a submodule of N called the image of 4. Furthermore,

Im r) M/Ker 4. In fact, i f ?1 is the canonical homomorphism from M onto MjN, then there is a unique isomorphism $ from MlKer + onto Im r) such that r) = $?1*

Proof. We leave it to the reader to show that Ker and Im r) are submodules of M and N , respectively. Define a mapping $ from M/Ker r) into Im r) by $(x + Ker+) = +(x). If x + Ker+ = y + Kerr), then x - y E Ker 4; hence +(x) - +(y) = +(x - y ) = 0. Thus $ is well-defined and is clearly a surjective homomorphism. If

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1 RINGS AND MODULES 7

$(x + Ker $1 = +(x’ + Ker #), then #(x) = #(x’) and #(x - x’) = 0 ; hence x - x‘ E Ker 4 and so x + Ker # = x’ + Ker #. Therefore $ is bijective.

We shall now derive two important corollaries of this theorem. Let L and N be submodules of an R-module M and suppose that L c N . Then L is a submodule of N and the factor module NIL can be imbedded in MIL in the following way. If x E N we identify x + L, as an element of NIL, with x + L , as an element of MIL. Clearly this identification is legitimate; each element of NIL is identified with exactly one element of MIL, and distinct elements of NIL are identified with distinct elements of MIL. Furthermore, after this identification has been made, NIL is a submodule of MIL. The elements of NIL are precisely those cosets of L in M which consist entirely of elements of N.

1.6. Corollary. With M, L, and N as above we have

(M/L) / (N/L) z M / N .

Proof. Define a mapping from MIL into M/N by #(x + L) = x + N. Then # is a well-defined surjective homomorphism. Thus, by Theorem 1.5,

(M/L)/Ker # z M / N .

However #(x+L)=x+N = O if and only if EN. Thus Ker C#J = N/L.

1.7. Corollary. If L and N are submodules of an R-module, then

( L + N ) / N g L/(L n N ) .

Proof. Define a mapping from L into ( L + N ) / N by #(x) = x + N. It is clear that # is a homomorphism. T o show that 4 is surjective we note that if (x +y) + N is a typical element of (L + N ) / N , with X E L and Y E N , then ( x+y)+N=x+N=#(x ) . Thus, by Theorem 1.5,

L/Ker # g (L + N ) / N .

However #(x) = 0 if and only if x E N , that is, if and only if x E L n N . Hence Ker # = L n N .

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8 1 MODULES

If A is a left ideal of a ring R, we can consider the factor module RIA. If A is, in fact, an ideal, we can make RIA into a ring by defining

(a + A)(b +A) = ab +A.

We shall show that this is a well-defined binary operation on RIA and leave it to the reader to show that RIA is a ring with this operation as its multiplication. Suppose that a + A = a’ + A and b + A = b’ +A. Then a - a’ E A and b - b’ E A, and so ab- a’b = a(b - b‘) + (u - a’)b’ E A, since A is both a left ideal and a right ideal of R. Thus ab + A = a’b’ +A. The ring R/A is called a residue class ring of R.

A mapping 4 from a ring R into a ring R‘ is called a homo- morphism if $(a + b) = 4(u) + $(b) and +(ab) = +(a)$(b) for all a, b E R, and +(1) = 1. The modifiers surjective, injective, and bijective retain the same meaning as before when applied to homo- morphisms of modules. A bijective homomorphism is called an isomorphism, and two rings R and R‘ are said to be isomorphic if there is an isomorphism from R onto R’. In this case we write R g R‘.

Let 4 be a homomorphism from a ring R into a ring R . Then Im 4 = {+(a)] a E R} is a ring. The unity of Im 4 is $(1), and Im $ is a subring of R’. Also Ker 4 = {a I a E R and $(a) = 0) is an idealof R, and

Im 4 R/Ker 4. This is proved in the same way as Theorem 1.5. Results like those of the corollaries to Theorem 1.5 also hold for rings and ideals.

2 CHAIN CONDITIONS

Let R be a ring and let M be an R-module. By an ascending chain of submodules of M we mean an increasing sequence

(1)

of submodules of M. By a descending chain of submodules of M we mean a decreasing sequence

Nl c N2 E N3 c * * a

Ni 2 N2 2 N3 2 . * *

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2 CHAIN CONDITIONS 9

of submodules of M. Such a chain (ascending or descending) is called infinite if for each positive integer n the chain has more than n distinct terms; otherwise the chain is called finite. Many of the modules of interest to us satisfy one or both of the following con- ditions :

Ascending chain condition (ACC) : every ascending chain of

Descending chain condition (DCC) : every descending chain submodules of M is finite.

of submodules of M is finite.

1.8. Theorem. (ACC) is equivalent to each of the following conditions:

Finiteness condition (FC): every submodule of M is Jinitely generated.

Maximum condition (MAX) : every nonempty set of submodules of M has a maximal element with respect to set inclusion.

(DCC) is equivalent to the following condition :

of M has a minimal element with respect to set inclusion.

Proof. We shall show that (ACC), (FC), and (MAX) are equivalent and leave it to the reader to show that (DCC) and (MIN) are equiva- lent.

(ACC) +(MAX). Assume that (ACC) holds for M and let Y be a nonempty set of submodules of M . Let Nl E 9’. If Nl is not a maximal element of Y , then there exists N , E Y such that Nl c N , . If N , is not a maximal element of 9, then there exists N , E Y such that Nl c N , c N, . If we continue in this way we will arrive at a maximal element of 9’ in a finite number of steps, for otherwise there exists an infinite ascending chain of submodules of M, contrary to (ACC).

(MAX) i (FC). Assume that (MAX) holds for M and let N be a submodule of M . let Y be the set of all finitely generated sub- modules of M contained in N . Then 9 is not empty since 0 E 9. Therefore Y has a maximal element L. If L # N , choose x E N with x 4 L. Then L c L + Rx G N , which contradicts our choice of L since L + R x is finitely generated. Therefore L = N .

Minimum condition (MIN) : every nonempty set of submodules

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10 1 MODULES

(FC) + (ACC). Assume that (FC) holds for M and consider an ascending chain (1) of submodules of M . Let N = U:=, N , . If x , y E N , then x E N h and y E N k , where we may assume h 5 k . Then x - - y € N k G N . Also,if a E R , then a x E N h s N . Thus N is a submodule of M and consequently has a finite set of generators. Suppose N = Rx, + - * * + Rx, where x i E Nni for i = 1, . . . , m. If no = max{n,, . . ., nm}, then x i E N,, for i = 1, . . ., m and so N G N,, . Therefore N = Nn0 and N , = N,, for all n 2 no . Thus the ascending chain is finite.

1.9. Definition. A sequence of R-modules and homomorphisms

... - M,-, - M , - M,,, - * a *

b n - 1 bn

is an exact sequence i f

Im = Ker 4, for all n.

If, for n > m we have M , = 0, we indicate this by writing

* * + M,-, +M, + 0,

and in a similar way we indicate that all terms from some point on to the left are zero. Thus a sequence

(2) O + L % * M % N + O

is exact if and only if 4 is injective, Im 4 = Ker $, and $ is surjective. An exact sequence of this form is called a short exact sequence.

1.10. Theorem. In the exact sequence (2), M satisJies (ACC) if and only if both L and N satisfy (ACC).

Proof. First, assume that M satisfies (ACC). Let L, E L, G - 0 . be an ascending chain of submodules of L. Then #L,) E 4(L,) s * * is an ascending chain of submodules of M , so there is an integer no such that 4(Ln) = $(L,,) for n 2 n o . Since 4 is injective, we have L, = L,, for n 2 no. Now let N, G N 2 G - - * be an ascending chain of sub- modules of N . Then $-l(N1) E $-'(N2) E * * . is an ascending chain of submodules of M , so there is an integer m, such that $-l(N,)= $-l(N,,) for m 2 m, . Since $ is surjective, we have N , = $($-l(N,))

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2 CHAIN CONDITIONS 11

= +(t,-l(N,,)) = N,,,, for m 2 m, . Therefore, both L and N satisfy (ACC).

On the other hand, suppose that both L and N satisfy (ACC) and consider an ascending chain M1 G M 2 c * * * of submodules of M. Then +-l(+(L) n M,) c +-,(+(I,) n M 2 ) E: * - is an ascending chain of submodules of L and $ ( M I ) G +(MJ G * . is an ascending chain of submodules of N. Hence there is an integer no such that

Since 4 maps L onto +(L), we have +(L) n M,, = +(L) n M,,, for n 2 no . Also M , + Ker + = $J-~(+(M,)) = +-l(+(Mno)) = M,, + Ker i,4 for n 2 no . Therefore, since Ker + = Im + = +(L), we have for n >no,

d-l(+(L) n M,) = +-l(+(L) n Mno) and t,(M,) = +(Mn0) for n L no.

M, = M, n (MI + = M , n (M,o + w>> = Mno + (Mno n +(L)) = Mno.

= Ma0 + W n n +W)) by Theorem 1.3

Thus M satisfies (ACC).

1.11. Definition. A ring R is left Noetherian if it satisjies (ACC) when considered as an R-module.

1.12. Theorem. Let R be a left Noetherian ring. If M is a finitely generated R-module, then M satisfies (ACC).

Proof. The assertion is certainly true when M = 0. Suppose that M # 0 ; we shall prove that M satisfies (ACC) by induction on the number of elements required to generate M . Suppose that M = Rx. Let N be a submodule of M and set A = {a ] a E R and ax E N ) . Then A is a left ideal of R and so A is finitely generated, say, A = Ra, + - * - +Ram. I fy E N , theny = ax for some a E A. Ifa = rlal +.. +rmam, then y = r,alx + * + Y , a, x. Therefore N = Ra,x + - - +Ram x, that is, N is finitely generated. Thus, M satisfies (FC) and so satisfies (ACC).

Now suppose that M requires n > 1 generators and that every R-module which can be generated by fewer than n of its elements satisfies (ACC). Let M = Rx, + * * * + Rx, and set L = Rx, +. . - + Rx, . By the induction assumption, L satisfies (ACC). As we have

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12 1 MODULES

seen, Rx, satisfies (ACC) ; hence every factor module of Rx, satisfies (ACC) by Theorem 1 .lo. Thus, since

MIL = (Rx, + L)/L g Rx,/(Rx, n L),

by Corollary 1.7, MIL satisfies (ACC). Therefore, making use of Theorem 1.10 again, we conclude that M satisfies (ACC).

3 DIRECT SUMS

Let {Ma 1 o! E I> be a family of R-modules indexed by a set I which may be of any nonzero cardinality. Let M be the set of all functions x defined on the set I such that for each a E I we have x, = x(o!) E M a , and subject to the restriction that x, # 0 for only a finite number of a E I. We make M into an R-module by defining

( X + Y ) a = xa + Y a

(ax), = ma, u E R, for all o! E I.

1.13. Definition. The R-module M is the (external) direct sum of the family {Ma I u E I } . We write

If I= (1, . . . , n} we write

M = MI Q @ M,, . Suppose that M = Oao I M a . For each u E I there is a homo-

morphism +a : 1M-t Ma given by

+a(x) = x a

and a homomorphism a+ha : Ma + M given by

if p=o! if p # a .

I t is clear that is surjective and $a is injective. Furthermore,

if a = p if a # / 3

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3 DIRECT SUMS 13

and

(b)

Note that the sum in (b) is finite since +,(x) = 0 for all but a finite number of a E I .

1.14. Proposition. Let M be an R-module and let {M,I a E I } be a family of R-modules. Suppose that for each a E I there exist homo- morphisms +a : M + M , and

(i) for each x E M , +,(x) # 0 for only a jinite number of a E I , (ii) (a) and (b) hold.

: M , + M such that

Clearly @ is a homomorphism. It is surjective since for all x E M , if ya = &(x) for all a E I , then

@(Y) = C G +a(x> = X* U E I

If @(x) = 0, then for each a E I ,

0 = = C #a #AS> = xa t B E I

and so x = 0. Hence @ is an isomorphism.

We now consider an R-module M and a family ( M a ] a E I } of submodules of M . We assume that

M = C Ma a s 1

and

Then each element of M can be written in one and only one way in the form CaE I x, , where x, E M and x, # 0 for only a finite number

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14 1 MODULES

of a E I. For, each element of 11.1 can be written as such a sum in at least one way by (c). Furthermore, if

where xa , y a E M a , then for each a E I,

so that x , =ya . We say that M is the (internal) direct sum of the family of its submodules {Ma 1 a E I ] .

1.15. Proposition. Let {Ma\ a E I > be a family of submodules of an R-module M . Assume that M is the (internal) direct sum of this family of submodules; that is, assume (c) and (d) hold. Then M g 0 a E I M a .

Proof. For every CL E I we define $a: Ma-+ M by $ a ( ~ ) = x for all X E M , . If EM, we write

where x , # 0 for only a finite number of a E I . Then set $,(x) = x , ; 4, is a well-defined homomorphism from M into M a . If X E M ~ , then

if P=a

if /3 # a.

Furthermore, for each x E M , if x = C a E I x , , as above, then

Hence the isomorphism exists by Proposition 1.14.

I n the light of this result, we shall write

M = 0 Ma

whenever {Ma\ a E I ] is a family of submodules of M for which (c) and (d) hold.

Let {R , \ a E I ] be a family of rings. If we consider only the

a f 1

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4 TENSOR PRODUCTS 15

additive structure of each R, we can form the direct sum of the resulting Abelian groups (as 2-modules),

R= @ Ra. a e I

We define a multiplication on R by

for all a E I . (4, = a, b,

Then R becomes a ring, a fact which is easily verified, called the direct sum of the family of rings {Ra I a E I } . The homomorphisms 4, and i ,ha, as defined in the case of modules, are ring homomorphisms. For each a E I, Im J,4, is an ideal of R, and

Ker +a= C Imi,hB. B+a

A result like Proposition 1.14 holds for direct sums of rings. Let R be a ring and let {R , I a E I} be a family of rings, Suppose that for each a E I there are homomorphisms $a : R + R, and t,ba : R, + R such that for each a E R, &(a) # 0 for only a finite number of LY E I ,

and

if a=P

if c t # P

for all U E R.

Then R z B U G , R,.

4 TENSOR PRODUCTS

An R-module M is called free if there is a nonempty subset S of M such that every element of M can be written uniquely in the form

where a, E R and a, # 0 for only a finite number of elements x E S. In this case, we say that the set S freely generates M . Note that

M = @ Rx. 5 E S

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16 1 MODULES

Let S be an arbitrary nonempty set and let M be the set of all formal sums

where a, E R and a, # 0 for only a finite number of elements x E S. Define equality of these sums by

a,x= 1 a,’x Z E S Z E S

if and only if a, = a,’ for all x E S. We make M into an R-module by defining

and a c b ,x= c (ab,)x for all a E R. 2 E S Z E S

If we identify x E S and l x E M, then S is a subset of M. We see immediately that M is a free module freely generated by S. We call M the free module defined on the set S.

Let M be the free module defined on a set S. If 4 is an arbitrary mapping from S into an R-module M’, then there is a unique homomorphism d,’ : M + M’ such that d,‘(x) = d,(x) for all x E S. For, the mapping 4’ from M into M‘ defined by

is a well-defined homomorphism, and is uniquely determined by 4. We say that d,’ is obtained by extending d, by linearity.

Let R be a ring, M a right R-module, and N an R-module. Denote by Z ( M , N ) the free 2-module defined on the Cartesian product set M x N. Thus the elements of Z ( M , N) are formal finite sums

nl(xl, rl) + ’ * ‘ + nk(xk 9 rk>>

where n,, . . ., nk E 2, x l , . . ., xk E M, and yl, . . . , y k E N . Let Y ( M , N ) be the subgroup of Z ( M , N ) generated by the set of all elements of the form

( X I + x2 > Y ) - (x1, Y ) - (x2 ? r), (x, y1 S y 2 ) - (x, n) - (2, r2>,

(xa , y ) - (x, ay),

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4 TENSOR PRODUCTS 17

where x, x,, xz E M , y , y l , y z E N , and a E R. The factor group Z ( M , N ) / Y ( M , N) is called the tensor product of M and N , and is denoted by M O R N . We note that M O R N is not an R-module ; it is simply an Abelian group.

If x E M and y E N , we denote the element (x, y ) + Y ( M , N ) of MORN by x O y ; we call such an element of M o ~ N a simple tensor. We have

(XI + x2) O r = 31 OY + xz O r , 30 (n +YZ) = x OYI t x OYZ,

xu By = x @ uy,

for all x, xl, x2 E M , y , yl , yz E N , and a E R. Thus if we fix y E N , the mapping x w x O y is a (group) homomorphism from M into M 0 N. In particular, if n is an integer, n(x y ) = nx @ y ; similarly, n(x B y ) = x

0 B y = x 0 = 0,

ny. It follows that

(-x) B y = -(x B y ) = x @ (-y)

for a l l x E M a n d y E N . The set of simple tensors {x B y 1 x E M , y E N ) generates the

Abelian group MORN. Indeed, we can make the stronger statement that every element of MORN can be written as a sum of simple tensors. For,

1.16. Definition. Let M be a right R-module and N a n R-module. A mapping 4 from the Cartesian product set M x N into a group G is a bilinear mapping if

+1+ xz 7 Y> = +(% Y ) + 7% Y Y ) ,

+(x, Y1 +YZ) = +& Y1) + +(4 Yz),

+(xu, Y> = d(x, 4, for all x, x,, x2 E M,y ,y l , y2 E N , and aE R.

1.17. Proposition. Let M be a right R-module, N an R-module, and 4 a bilinear mapping from M x N into a group G. Then there is a unique homomorphism II/ : M 0 N --f G such that $(x y ) = $(x, y ) for all x E M andy E N .

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18 1 MODULES

Proof. First, we extend 4 by linearity to a homomorphism 4' from Z(M, N ) into G. Since 4 is bilinear, 4' maps Y ( M , N ) onto the identity element of G. Hence there is a homomorphism $ : M @ R N + G such that $(x@y)=~$(x,y) for all X E M , Y E N . Since every element of M O R N is a sum of simple tensors, $ is uniquely determined by how it acts on the simple tensors.

We shall use this proposition a number of times to obtain homo- morphisms from tensor products into other groups. The procedure is illustrated in the proof of the following result.

1.18. Proposition. If N is an R-module, then

R O R N z N ,

where this is an isomorphism of Abelian groups. Explicitly, there is an isomorphism t,b from R O R N onto N such that $ (a@ y) = ay for all a E R and y E N .

Proof. The mapping 4 from R x N into N given by 4(a, y) = ay is bilinear, so there is a unique homomorphism $ from R O R N into N such that $(a @ y) = ay for all a E R, y E N . It is surjective since #(l 8 y) =y. To show it is injective, let t E Ker $. Since

t = 1 at@yt = 1 la, OYt = 1 1 0 a,yt = 1 0 1 % Y t ,

ify =I ai y i , we have0 = $( t ) = $(lo y) =y.Hencet = 1 @ 0 = 0.

Let M , M ' be right R-modules, let N , N ' be R-modules, and let f : M + M ' and g : N + N ' be homomorphisms. Consider the mapping 4 from M x N into M' BEN' given by

4(% Y ) =f(4 @g(Y).

This mapping is bilinear. We verify one of the requirements; the other two are verified in a similar manner :

+1+ x2 ? Y ) =f(% + x2) 8 g(Y)

= (f(x1) +f(xz))

= f ( 4 8 g ( Y ) + f ( x z ) 8 d Y )

= +(XI, Y ) + 4(% 7 A.

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4 TENSOR PRODUCTS 19

By Proposition 1.17, there is a unique homomorphism from M O R N into M ’ O R N ’ which maps x @ y ontof(x) @ g(y) for all x E M and y E N. We denote this homomorphism b y f o g . The following result is obvious.

1.19. Proposition. If I M : M - t M and l N : N-tN are the identity isomorphisms of M and N , then 1 @ 1, is the identity isomorphism .f M B ~ N . Iff:M+M‘, f ’ : M ’ - t M ” , g : N + N ’ , andg‘:N‘-+N” are homomorphisms, then

(f’ @g’)(f@g) =f’f@g‘g*

1.20. Proposition. Iff: M --f M ‘ and g : N -+ N ‘ are surjective homo- morphisms, then f @ g is surjective.

1.21. Proposition. Iff: M -+ M’ and g : N 3 N’ are surjective homo- morphisms, then Kerf @ g consists of all jinite sums of the form C xi 0 y i , where xi E Kerf or yi E Ker g.

Proof. Let K be the set of all these sums. Then K is a subgroup of M o R N , generated by the set of simple tensors x @ y where x E Kerf ory E Ker g. For sucha simple tensor we have (f@g)(x @ y ) = f ( x ) @ g ( y ) = 0. Hence K G Kerf @ g. Therefore f @ g induces a homomorphism h : M @ NIK -+ M ’ @ N ’ such that

h@@Y +K)=f(x)@gb)

for all simple tensors x @ y in M N . Furthermore,

Ker h = Kerf@ glK.

Define a mapping j from M‘ x N’ into M O R N / K by j (x ‘ , y’) = x@ y + K, wheref(x) = x’ andg(y) =y’. (We use here the fact that

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20 I MODULES

f and g are surjective.) We must show that j is well-defined. Suppose f(xl) = X‘ and g(yl) = y‘. Then x - x1 E Kerf and y --yl E Ker g. Hence

X O Y -x1 oy1= xO ( y -yd + (2 -.1) Or1 E K,

and

X O Y +K=x,Oy1+K.

Now, the mapping j is bilinear ; we shall verify one of the condi- tions, again leaving the verification of the other two to the reader. Let x’, X “ E M ’ , y’ EN’, and let f (xl) = x’, f (x2) = x”, and g(y) =y’. Then f(xl + xa) = x‘ + X ” and

j (x’ + x“, y ’) = (x1+ x z ) 0 Y + K

= (x1O Y + K ) + (xz O r + K )

=&’, Y ’ ) + j ( x ” , y’).

Thus there is a unique homomorphism k : MI @ N ’ +- M @ N / K suchthatk(x’@y’) = x O y +K,wheref(x)= x‘andg(y)=y‘.Then

hk(x’ @ y’) = h(x 0 y + K )

=f(x) Og(y) = 2’ Or’.

Furthermore

0 Y + K ) = k ( f ( x ) 0 d Y ) ) = x @ y + K .

Thus hk and kh are the identity isomorphisms of M’ O R N ‘ and M @ , N / K , respectively. It follows that both h and k are isomor- phisms. Hence Ker h = 0, that is, K = Kerf Og.

1.22. Theorem. Let f 0- N I - ~9- N I I - o

be an exact sequence of R-modules and let M be a right R-module. If 1 is the identity isomorphism of M , then the sequence

1MWf lME3.8 M O R N ’ - M O R N - M @ , N -0

is exact.

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5 FLAT MODULES 21

1.23. Theorem. Let f O-M’-M(I-MW-O

be an exact sequence of right R-modules and let N be an R-module. If 1 , is the identity isomorphism of N , then the sequence

f @ I N %‘@IN M O R N - M O R N - M”O.N-0

is exact.

We shall prove Theorem 1.22, the proof of Theorem 1.23 being similar. By Proposition 1.20, 1 0 g is surjective. By Proposition 1.21, Ker 1 @g consists of all finite sums C xi B y i where either xi = 0 or y t E Kerg = Im f . Hence Ker 1 @ g = I m 1 0 f .

5 FLAT MODULES

1.24. Definition. A right R-module M is flat $ f o r each injective homomorphism f : N ’ +- N from one R-module into another, the homo- morphism

1 M @ f : M@RN’+-M@RN

is injective, where 1 is the identity isomorphism of M.

Thus M is flat if and only if for every exact sequence of R-modules,

0 - N ’ - N L W - 0 , r

the sequence I M @ f l M @ g O - - - - + M O R N ‘ - M @ R N - M O , N “ - O

is exact. Now we need a technical lemma.

1.25. Lemma. Let N be an R-module and F a free right R-module, freely generated by a set S . Then every element of FORN can be written uniquely in the form

where yz # 0 for only a finite number of x E S,

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22 1 MODULES

Proof. Let C x i @ y i E F@RN, where x i E F, yi E N . Let x t = CXE xu:), where a$i) E R. Then

c xt 8Yt = C x 0 (c aPYt). X E S

Hence each element of F 0 N can be written in the required form in at least one way. Note that the sums on i are finite and that for each i, air) # 0 for only a finite number of x E S.

x B y I = 0 implies y z = 0 for all x E S. For each x E S, define 4, : F + xR as follows: if x’ E S, set &(x’) = x when x‘ = x and bZ(x’) = 0 when x‘ # x, and extend 4, to all of F by linearity; that is,

To show the uniqueness, it is sufficient to show that XI

Then, if 1, is the identity isomorphism of N , we have for each x E s,

0 = ( # ’ ~ 8 1 N ) ( ~ ~‘@Yx*) = 1 $ X ( ~ ’ ) ~ Y Z ’ = ~ ~ - Y I * X ’ E S I ’ E S

For each x E S define 7) : XR --f R by q(xa) = a. It is clear that 7 is an isomorphism. Then 7 @I 1,: x R @ , N + R @ RN is an isomorphism. Let #: R 8 EN+ N be the isomorphism of Proposition 1.18. Then, for all x E S,

= #((7) 8 lN)(x 8 Yx)) = #(dx) 8 Yx) = #(l 8 Yz) = Yz *

1.26. Proposition. A free right R-module isflat.

Proof. Let N , N ’ be R-modules and f : N ’ + N an injective homo- morphism. Let 1, be the identity isomorphism of the free right R-module F. Let S be a set of free generators of F . Let t E F O R N ’ and suppose that (1,@ f ) ( t ) = 0. We can write t =IIE x ByX where y X E N ‘ and yX # 0 for only a finite number of x E S. Then

By the statement of uniqueness in the lemma we conclude that f(yX) = 0 for all x E S. Since f is injective, this implies that y X = 0 for all x E S. Thus t = 0.

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5 FLAT MODULES 23

We shall have occasion to consider diagrams of R-modules (left or right) and homomorphisms of the type

MI - M2 MI I

Such a diagram is called a square or triangle, respectively. A square is said to be commutative if g'f= f 'g; a triangle is said to be commutative if hg = f. We shall consider diagrams of R-modules and homomorphisms which involve squares and triangles, and such a diagram is called commutative if every square and triangle in the diagram is commutative.

1.27. Definition. A right R-module P is projective $for every diagram

P

I M - N - 0

of right R-modules, where the row M + N+O is exact, there is a homo- morphism P -+ M such that the diagram

P

M-N-0

is commutative.

1.28. Theorem. Let P be a right R-module. Then the following state- ments are equivalent :

(1) P is projective. (2) P is a direct summand of a free right R-module.

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24 I MODULES

(3) For evevy exact sequence of right R-modules of the form

o + - L ~ M + - P + - o ,

Im f is a direct summand of M . (Such a sequence is called a split exact sequence.)

Proof. (1) + (2) and (3). Let P be a projective right R-module. Let F be the free right R-module defined on the set P. Define (p: F A P by setting d(x)= x for all x E P and extending C$ to all of F by linearity. Then C$ is a surjective homomorphism and we have an exact sequence

0 - L - F - P - 0 ,

where L = Ker 4. We shall show that F r L @P, and thus prove (2).

@

More generally, we shall show that if the sequence f 0 - L - M P P - 0

is exact, then M = (Imf) @ P' where P' r P. This will prove both (2) and (3). Let 1, be the identity isomorphism of P and consider the diagram

P

M - P - 0

Since P is projective there is a homomorphism h : P+ M such that 1, =gh . Let P ' = Im h. Since 1, is injective, so is h, and thus P' 2 P. Let x E (Imf) n P '. Then x E Ker g and x = h(y) , where y E P. Hence x = h(gh(y ) ) = hg(x) =O. Thus (Imf) n P' = 0. Now let z E M ; then

- hg(z)) = g ( 4 - 1 P ( g ( 4 ) = 0,

so z -hg(z) E Imf. Since hg(z) E P', we have

z = ( z - hg(z)) + hg(z) E ( Imf) + P'. Therefore M g (Imf) @ P' .

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5 FLAT MODULES 25

(2) + (1). Assume that P is a direct summand of a free right R-module F. Then there are homomorphisms + : P +F and 4: F + P such that +$ is the identity isomorphism of P, $ being injective and 4 surjective. Consider a diagram

d F-P

If M-N-0

where the row is exact. Let F be freely generated by S. Define j : F+ M by setting j(x) equal to any element of M such that g(j(x)) =f+(x) for each x E S and extending to all of F by linearity. Then the diagram

d F-P

M-N-0

is commutative. Now consider the homomorphism j+: P-t M. We have g(j+) = (gj)$ = (f4)+ = f ( 4 + ) = f. Therefore P is projective.

(3) + (1). Assume (3) holds and let

0-L-F-P-0 * @

be the exact sequence from the first part of the proof that (1) implies (2) and (3). By hypothesis, I m + is a direct summand of F, so that F = ( Im$)@ P' where P' is some submodule of F . Define 7 : P'+P by r)(x)=+(x) for all X E P' . If y E P, then y = 4(x) for some ~ E F . Write x=u+v where u E I m + = K e r + and V E P ' . Then

Y = +@ + v) = 4(4 + $(.> = d(4 = 44. Hence 7 is surjective. Suppose 7(x)= 0 where X E P ' . Then X E Ker 4, so x ~ ( I m $ ) n P'=O, that is, x=O. Thus 7 is an isomorphism and P is isomorphic to a direct summand of a free right

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26 1 MODULES

R-module. Therefore P is projective by the equivalence of (1) and (2).

1.29. Corollary. Every free rkh t R-module is projective.

1.30. Proposition. Let J o - M - M B M” --+ o

be a split exact sequence of right R-modules. Let N be an R-module and 1, the identity isomorphism of N . Then

J @ ~ N g C 3 1 N 0 - M ’ Q R N - M Q R N - M ” Q R N - 0

is exact.

Proof. We must show that f 8 1, is injective. By assumption there is a homomorphism h : M+ M‘ such that hf is the identity isomor- phism of M‘. Then by Proposition 1.19, (h Q I N ) ( f 0 1,) = h f Q 1, is the identity isomorphism of M‘ Q R N . Hence f @ 1, is injective.

1.31. Theorem. A projective right R-module isJEat.

Proof. Let N and N ’ be R-modules and f : N ’ + N an injective homomorphism. Let 1, be the identity isomorphism of the projective right R-module P. By Theorem 1.28 there is a free right R-module F such that F = POP’. Then there is an exact sequence O+P -% F+ P’+O, where g(x ) = x for all x E P. Hence, if 1, and lN, are the identity isomorphisms of N and N ‘ , respectively, both g 9 1, and g 0 l,, are injective by Proposition 1.30. If 1, is the identity isomorphism of F, then the diagram

1 P @ f P Q R N ‘ - P O R N

FQ RN’ F Q RN

is commutative. By Proposition 1.26, F is flat, so 1,Q f is injective. Therefore, lp@ f is injective.

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EXERCISES

EXERCISES

27

1. Isomorphisms. Let M and N be R-modules and suppose there are homo- morphisms +: M - t N and $: N + M such that ++ = 1 and +# = 1,. Show that + and $ are isomorphisms, inverse to one another.

2. Modules and homomorphisms. Let M and N be R-modules and let +: M - t N be a surjective homomorphism. (a) Show that there is a one-to-one correspondence between

the submodules of N and the submodules of M which contain Ker +. In fact, the correspondence is

Ml++ +(Md = {W I x E Ml) ;

the inverse correspondence is

Nl*+-'(N1) = { X I x E M and +(x) E Nl].

Let M I be an arbitrary submodule of M . Show that

+-'(+(MI)) = MI+ Ker 4. (b)

(c) Drop the requirement that + be surjective, and let Ml and N , be submodules of M and N , respectively. Show that

MI(M1 ++-l(N1)) = (+(W + W ( + ( ~ l ) + Nl). 3. Chain conditions.

(a) (b)

(c)

Prove the equivalence of (DCC) and (MIN). Show that the additive group of integers satisfies (ACC) but not (DCC). Let p be a prime. Let G be the additive group of rational numbers and H the subgroup of G consisting of those rational numbers which can be written with denominator prime top. Show that G/H satisfies (DCC) but not (ACC). Let MI, . . . , M , be submodules of an R-module M such that M = Ml + - * * + M,, . Show that M satisfies (ACC) [or (DCC)] if and only if M , does for each i.

(d)

4. Composition series. Let M be an R-module. By a normal series of M we mean a finite chain of submodules of M ,

M = M * 2 M , 2 * . . 2 M r = O .

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28 I MODULES

The integer Y is called the length of the normal series. If Mi-l #Mi for i= 1, . . . , r , the normal series is said to be without repetition. One normal series is called a refinement of a second normal series if every term of the second series is a term of the first series ; it is called a proper refinement if it has some terms not occurring in the second series. A normal series of M is called a composition series of M if it is without repetition and has no proper refinement. The factor modules Mi-,/M,, i = 1, . . . , r , are called the factors of the normal series. Two normal series are called equivalent if they have the same length and if the factors of the two series can be put in one-to-one correspondence in such a way that corresponding factors are isomorphic.

(Zassenhaus' lemma) Let L, L', N , and N ' be sub- modules of M such that L' E L and N ' E N . Show that

(L' + (L n "(L' + ( L n ")) r (N' + ( N n L))/(N ' + ( N n I,')).

Show that any two normal series of M have equivalent refinements. Assume that M has a composition series. Show that any two composition series of M are equivalent, and that every normal series without repetition of M has a refinement which is equivalent to a composition series. Show that M has a composition series if and only if it satisfies both (ACC) and (DCC).

5. The length of a module. If an R-module M has a composition series of length Y, we call r the length of M and write LR(M) = Y. If M has no composi- tion series we write L,(M) = 00.

(a) Show that if 0 4 K-t M-+ N - t 0 is an exact sequence of R-modules then L,(M) = L,(K) + LR(N) (we assume 00

added to 00 or to a nonnegative integer gives 00).

Show that if K and N are submodules of an R-module then (b)

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EXERCISES 29

(c) Let V be a vector space over a field F . Show that LF( V ) = dim, V . Show that V satisfies (ACC) if and only if it satisfies (DCC). Let 0 + Ml +- M , --f * -+ M n -+ 0 be an exact sequence of R-modules with LR(Mi) finite for i = 1, . . . , n. Show that

(d)

n

1 (-l)iLR(Mi) = 0. 1 =1

(e) Let M be an R-module and let 9’ be the family of all finitely generated submodules of M , Show that

L,(M) = sup L,(N). N € Y

(f) Let R be a commutative ring and M an R-module. Let a E R be such that if x E M , then ax = 0 implies x = 0. Show that aM = (ax1 x E M ) is a submodule of M iso- morphic to M . If N is a submodule of M show that M/N r aM/aN. Finally, show that if n is positive integer, then

L,(M/a”M) = nLR(M/aM).

6. Direct sums. (a) Let {Mula E I } be a family of R-modules and let

M = @ u E I M a . Define Ga: M,+M as in Section 3 . Let N be an R-module and for each a t I , let T~ : Mu -+ N be a homomorphism. Show that there is a unique homo- morphism v : M-+ N such that & = vu for all a E I.

(b) Let M‘ be an R-module and, for each c( E I, let y!~: : Ma + M ’ be a homomorphism. Suppose that for each R-module N and for each family of homomorphisms ria: M a + N , one for each M E I , there is a unique homomorphism 7’: M’+N such that q‘#,‘ = qu for all a E I . Show that M‘ 2 M . Let ( M a ] M E 1) and (Nal a E I> be families of R-modules and let M = @ a E I M a and N = O a c 1 N a . For each a E I , let fa: Ma +Nu be a homomorphism. Define f: M+N by f ( ~ ) ~ = fa(xa) for all cc E I; f is called the direct sum of the family of homomorphisms {fal tc E I}. Show that

(c)

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30

Kerf= @ Kerf,, a E I

I MODULES

and

Im f 0 M,/Ker fa. a E I

(d) Continue the notation of (c). Let (L,( 0: E I > be another family of R-modules and for each u E I , let g, : L, --f M , be a homomorphism. Let L = Oa. I L a . Let g be the direct sumof thefamily{g,I ~ E I } . S h o w t h a t L 4 M L + N is exact if and only if La* M a 3 N , is exact for each a c I .

be a one-to-one mapping of the set I onto itself. Show that

(e) Let

0 Ma 0 Mn<a>* a E I a E I

( f ) Let Jand K be nonempty subsets of I such that I = J u K and J n K is empty. Show that

0 Ma z ( 0 Ma)@ ( 0 Ma>- a E I a E J a E K

7. Tensor products. (a) Let M be a right R-module and N an R-module. Let

q(x ,y ) )=x@y for all x e M and T E N . Show that if # is bilinear mapping from M x N into an Abelian group G, then there is a unique homomorphism + : M @I N 3 G such that $7 = 4. For i= 1, 2, let (Ti, .I~) be a pair consisting of an Abelian group Ti and a bilinear mapping qt from M x N into Ti. Assume that for each bilinear mapping 4 from M x N into an Abelian group G there is a unique homomorphism : Ti --f G such that $t 7, = 4. Show that there is a unique isomorphism Q : TI --f T , such that oql = qz . Let R and R‘ be rings and let M be a right R-module and an R’-module such that (a’x)a = a’(xa) for all a’ E R‘, x E M , and a E R. Then M is called an R‘-R-bimodule.

(b)

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EXERCISES 31

Let N be an R-module and define a mapping from R’ x (MORN) into MORNby

Show that this mapping is well-defined and that it makes M O R N into an R‘-module. Let L be a right R’-module, M an R‘-R-bimodule, and N an R-module. Show that there is a unique isomorphism L O R , ( M O R N ) ~ ( L O R , M ) O R N such that x @ ( y @ z ) H ( x @ y ) O Z f o r a l l x E L , y E M , a n d z E N . Let R be a commutative ring and let M and N be R-modules (we may consider them as R-R-bimodules). Show that there is a unique isomorphism MORN+ N O R M such that x B y - y 0 x for all x E M and y E N . Let R be a commutative ring and let M and N be R-modules. Then, according to part (b), M O E N is an R-module. Let Z ’ ( M , N ) be the free R-module defined on the set M x N and let Y’(M, N ) be the submodule of Z’(M, N ) generated by the set of all elements of the

(c)

(d)

(e)

form (x, y1 +YZ) - (x, n) - (x, yz), (x1+ x2 7 Y ) - (Xl, Y ) - ( X Z , y), (xu, y ) - a@, y), and (x, ay) - 4x9 Y ) , where x, xl, x2 E M , y , y l , y z E N , a E R. Show that as R-modules we have

MORN= z ’ ( M , N ) / Y ’ ( M , N ) .

(f) Let { M a [ CL E I } be a family of right R-modules and let ( N , 1 fl E I } be a family of R-modules. Show that

8. Some special tensor products. (a) Let G be an Abelian group and let n be a positive integer.

Let nG = (ax] x E G) and let Z, be the additive group of integers modulo n. Show that 2, OZ G

(b) Let k be the gcd of the positive integers m and n. Show that Z, @,Z, z Z,.

(c) Let U and V be vector spaces over a field K. Show that U OK V is a vector space over K and that

dim, U @ , V = (dim, U)(dim, V ) .

G/nG.

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32 1 MODULES

(d) An Abelian group G is called a torsion group if every element of G has finite order. Show that if either G or H is a torsion group, then G OZ H is a torsion group. An Abelian group G is said to be divisible if nG = G for all positive integers n. Show that if either G or H is divisible, then G QZ H is divisible. Show that if G is a torsion group and His a divisible group, then G &El== 0.

Let M be a right R-module. Show that M is flat if and only if for every injective homomorphism f: N ' L N of a finitely generated R-module N ' into an R-module N, 1 @f is injective.

(b) Let (Ma I CL E 1) be a family of right R-modules. Show that @a I M a is flat if and only if M , is flat for each u E I .

(c) Let M be a right R-module and N and R-R'-bimodule. Show that if M is flat and if N is flat as a right R'-module, then M Q R N is a flat right R'-module. Let R be a subring of a ring R' ; then, in a natural way, R' is an R-R-bimodule. Every R'-module (right or left) is also an R-module. Show that if M is a flat right R'-module, and if R' is a flat right R-module, then M is a flat right R-module.

(e)

(f)

9. Flat modules. (a)

(d)

10. Flat modules and isomorphisms. Let M be a flat right R-module and N an R-module, and let Nl and N 2 be submodules of N . If N' is a submodule of N , and i f f :N'+Nisgivenbyf(y)=yforal lyEN' , then 1,@ f i s injective. We shall identify each element of M Q R N ' with its image under 1 @ f, and thus consider M @ N ' as a subgroup of M @ N . Show that the following isomorphism and equalities hold : (a) MORN/MORN1 2 MOR(N/Nd (b) M O R ( N ~ $ N ~ ) = ( M @ R N ~ ) + ( M O R N ~ ) (c) M @ R ( N l N 2 ) == ( M @ R N I ) ( M @ R N 2 ) *

11. A criterion for flatness. Let M be a right R-module such that if A is a left ideal of R and f : A + R is given by f (u )=u for all a E A , then l , @ f i s injective.

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EXERCISES 33

(a)

(b)

Let

Show that if F is a free R-module, and if g: E+ F is injective, then 1 Show that M is flat.

12. Flatness and exact sequences.

@g is injective.

f 0 - L - M L N - 0

be an exact sequence of right R-modules and assume that N is flat. (a) Prove that

f @1K 9@1K 0- L @ R K - M @ R K - N@RK--O

is exact for every R-module K. Show that M is flat if and only if L is flat. (b)

13. Elementwise criterion for flatness. Let M be a right R-module. (a) Show that M is flat if and only if for every R-module

N and for every pair of finite subsets {xl, . . . , xn} and { y l , . . . ,yn} of M and N , respectively, such that C:=l xi @ y i = 0, there are elements zl, . . . , z k E M and elements bji E R, i = 1, . . . , n, j = 1, . . . , K , such that

n

i = l C biiyi = 0, j = l , ..., k ,

and k

xi = 1 Z j b j i , i = 1, . . . , n, i= 1

(b) Show that M is flat if and only if for every pair of finite subsets (xl, . . . , xn} and {al, . . ., an> of M and R, respec- tively, such that I;= x i a, = 0, there exist elements zl, . . . , zk E M and elements bji E R, i= 1, . . . , n, j = 1, . . . , K, such that

2 !Jjiai=0, j = I, . . . , k , i = l

and

i = l , ..., n.

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34 1 MODULES

(c) Let R be a subring of a ring R'. Show that R is a flat right R-module if and only if for every solution cl, . . . , c, in R' of a system of equations

2 X i f f l h z O , i = 1

where aih E R for each i and h, we have

h = l , ..., r ,

i = 1, . . . , n,

where d,, . . . , d, E R', bji E R for eachj and i, and

j=1 , ..., k, h = l , ..., r.

14. Split exact sequences.

(a) Let O + L L M - % N + O be an exact sequence of R-modules. Show that this is a split exact sequence if and only if there is a homomorphism h : N + M such that

Show that the exact sequence in part (a) is a split exact sequence if and only if there is a homomorphism k : M + L such that kf= 1,. Let L and N be R-modules. Show that there is a split exact sequence 0 -+ L -+ M -+ N -+ 0. If n is a positive integer, let 2, be the additive group of integers modulo n. Construct an exact sequence 0 + 2, --f 2, --f 2, -+ 0 which is not a split exact sequence. Construct other examples of this same type.

g h = 1,. (b)

(c)

(d)

15. Projective modules. (a) Let P be an R-module. Show that P is projective if and

only if there is a subset S of P and for each x E S there exists a homomorphism +z: P - t R such that for all y E P, ~&(y) # 0 for only a finite number of x E S, and

Show that a direct summand of a projective module is projective. Let (P,(a E I> be a family of R-modules. Show that

y = c2. E s ( b z ( Y ) X .

(b)

(c)

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EXERCISES 35

P = O a E I Pa is projective if and only if each Pa is pro- jective. Let P be a projective R-module. Show that if, in the diagram

P

(d)

* A-BLC of R-modules, the row is exact and p+ = 0, then there is a homomorphism 7 : P+ A such that $7 = +. Let P be a projective R-module. Show that if, in the diagram

(e)

6 * P-A-B

CPD-E of R-modules, the square is commutative, the bottom row is exact, and $ + = O , then there is a homomorphism 7 : P+ C such that p7 = a$.

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C H A P T E R

I1 Primary Decompositions and

Noetherian Rings

1 O P E R A T I O N S ON IDEALS AND S U B M O D U L E S

Throughout the remainder of this book, with the exception of the appendix, all rings which are considered will be assumed to be commutative. Thus it will no longer be necessary to distinguish between left and right modules. We shall think of all modules as left modules and use the appropriate notation.

Let R be a ring. If A and B are ideals of R we set

AB=(finite sums C a i b , I a i ~ A , b , ~ B ) .

Then AB is an ideal of R. For, if 1 a, b, and C c j d j are in AB, then

C a, bi - 1 C j d j = C ai bi + C( -Cj)dj E AB,

and if 7 E R, then

r(C a, bl) = C(ra,)b, E AB.

The ideal AB is called the product of A and B. If a E R we write aB or Ba for (a)B; we note that aB = {abl b E B). If A and B are ideals of R, then their sum A + B as submodules of the R-module R is an ideal of R. If A is generated by a,, . . . , a,, we will usually write A = (al, . . . , u,) rather than Ra, + * . * + Ru, .

36

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1 OPERATIONS ON IDEALS AND SUBMODULES 37

2.1. Proposition. Let A , B, and C be ideals of R. Then

(1) A(BC) = (AB)C, (2) A B = BA, (3) (4) (5) (6)

A B s A n B, A(B + C ) = A B + AC, A s B implies that AC s BC, A(B n C ) c A B n AC.

Proof. (l), (2), and (3) are clear. If a E A , b E B, and c E C then a(b + c) = ab + ac E A B + AC, and so sums of elements of this type are in A B + AC. Hence A(B + C ) c AB + AC. It follows from (S), which is clearly true, that A B 5 A(B + C ) and AC G A(B + C ) . Thus A B + AC c A(B + C ) . Finally, A(B n C) G A B and A(B n C) G AC, so that A(B n C) G A B n AC.

Having defined the product of two ideals of R we can define now the nth power of an ideal A of R, where n is a positive integer, in the usual way:

if n > 1.

The rules of exponents hold:

AmAn = Am+" , (A")" = A"", (AB)" = A"B".

It is customary to set Ao = R.

of M , we set Let M be an R-module. If A is an ideal of R and N is a submodule

AN = (finite sums

Then AN is a submodule of M , and the proof of the following proposition is like that of Proposition 2.1.

a, xi I a, E A, xi E N } .

2.2. Proposition. Let A and B be ideals of R and let L and N be sub- modules of M . Then

(1) A(BN) = (AB)N, (2) AN G N, (3) A(L + N ) = AL + A N and ( A + B)N= A N + BN,

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38 11 PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

(4) ( 5 ) (6)

A c B implies that AN c BN, L c N implies that AL G AN, A(L n N ) c A L n A N a n d ( A n B ) N s AN n BN

If a E R we write aN for (a)N, and if x E M we write Ax for

Again, let M be an R-module, A an ideal of R, and N a sub- A(Rx).

module of M . Set

N : A = { x l x E M and Ax c N } .

If x , y € N : A then a(x--)=ax-aycN for all U E A , that is, x - y E N : A. Also, a(rx) = ( m ) x E N for all a E A , where Y is any element of R, that is, YX E N : A, Thus N : A is a submodule of M . If L is any submodule of M , then A L c N if and only if L c N : A.

L : N = ( a l a i z R and U N G L ) .

Then L : N is an ideal of R, and if A is any ideal of R, then A N c L if and only if A E L : N .

The submodule N : A and the ideal L : N are called the residual of N by A and the residual of L by N , respectively.

If L and N are submodules of M , we set

2.3. Proposition. Let A and B be ideals of R and let K , L, and N be submodules of M. Then

A c B implies N : A 2 N : B, ( N : A ) : B = N : AB, (L n N ) : A = (L : A) n ( N : A), N : ( A + B ) =(N: A ) n ( N : B) , L L E N implies that K : L 2 K : N , ( L n N ) : K = ( L : K ) n ( N : K ) , K : ( L + N ) = ( K : L) n (K: N ) .

N implies that L : A G N : A and L : K c N : K,

Proof. We shall prove (l), (Z), (3), and (8) and leave the others for the reader.

(1) We have A(N: B ) E B ( N : B ) E N and so N : B c N : A. (2) We have AB((N: A): B) 2 A(N: A ) c N so that ( N : A ) : B

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2 PRIMARY SUBMODULES 39

c N : AB. Also, AB(N: AB) c N and so B(N: AB) c N : A. Hence N : AB E ( N : A) : B.

(3) We have A((L : A) n ( N : A)) s A(L : A) n A(N: A) c L n N andconsequently(L:A)n ( N : A ) c (LnN) :A .AlsoA( (LnN) :A) E L n N E: L so that ( L n N ) : A E: L : A ; likewise (L n N ) : A G N : A. Thus ( L n N ) : A G L : A n N : A.

(8) We have L G L +N and so by (6), K : (L + N ) E K : L ; likewiseK: ( L + N ) E K : N . Hence K : ( L + N ) G ( K : L) n ( K : N). Let C = ( K : L) n ( K : N ) . Then C s K : L so CL L K. Likewise CN c K, and therefore C(L + N ) = CL + CN E K. Hence C G K : ( L + N ) .

If A and B are ideals of R then the residual A : B is an ideal of R. If C is another ideal of R , then by (2) of the above proposition, ( A : B ) : C = A : B C .

2 PRIMARY SUBMODULES

2.4. Definition. (1) Let M be an R-module. A submodule Q of M is primary if for all a E R and x E M , ax E Q and x 6 Q imply that anM 5 Q for some positive integer n.

(2) A submodule N of an R-module M is irreducible if, for submodules L , and L, of M , N = L, n L, implies that either L , = N or L,= N.

2.5. Proposition. If M satisfies (ACC) then every irreducible sub- module of M is primary.

Proof. Let Q be an irreducible submodule of M . Let a E R and x E M be such that ax E Q but x 6 Q. By (ACC) there is a positive integer n such that Q : ( a n ) = Q : ( a n + l ) = - . * . We have Q c ( Q + ~ ~ M ) ~ (Q+Rx). Let y E ( Q + a n M ) n ( Q + R x ) ; then y=z+anu= z' + bx, where z , z' E Q, u E M , and b E R. Since ay = ax' + abx E Q, it follows that a " + h = ay - az E Q ; hence u E Q : (an+l) = Q : (an). Thus anu E Q and consequently y E Q. Therefore Q = (Q + a"M) n (Q + Rx). Since Q is irreducible and Q # Q + Rx, we must have Q = Q + anM. Thus anM c Q. We conclude that Q is primary.

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40 11 PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

2.6. Proposition. If M satisfies (ACC), then every submodule of M can be written as an intersection of a finite number of irreducible sub- modules of M.

Proof. Let Y be the set of all submodules of M which cannot be written as an intersection of a finite number of irreducible sub- modules of M . If Y is empty, we have finished. Suppose that Y is not empty; then Y has a maximal element N . Since N is not irreducible there are submodules L, and L, of M such that N = L, n L a , N c L,, and N c L,. Then L, $9 and L , $ Y and so there are irreducible submodules K,, . . . , K, and K1‘, . . . , K,’ of M such that L, = K, n * * - n K, and L, = K,‘ n - - n K,’. But then N = K , n * - n K,,, n K,’ n - * n Kn’, which is contrary to the fact that N E 9. Thus, Y must be empty.

Thus, if M satisfies (ACC), it contains many primary submodules. If we combine Propositions 2.5 and 2.6, we obtain the following result.

2.7. Theorem. If the R-module M satisjies (ACC), then every sub- module of M can be written as an intersection of a finite number of primary submodules.

An ideal Q of a ring R is called primary if it is a primary sub- module of R when considered as an R-module. Since R has a unity, Q is a primary ideal if and only if for all a, b E R, ub E Q and b # Q imply that an E Q for some positive integer n.

Since R is commutative we shall call it simply Noetherian if it is left Noetherian. I t follows from Theorem 2.7 that if R is Noetherian, then every ideal of R can be written as an intersection of a finite number of primary ideals.

2.8. Definition. An ideal P of a ring R is prime i f for all elements a, b E R, ab E P implies thut either a E P or b E P.

2.9. Proposition. Let P be an ideal of a ring R. Then the following statements are equivalent

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2 PRIMARY SUBMODULES 41

(1) (2) (3)

P is a prime ideal, I f a , b E R and (a)(b) c P, then either a E P or b E P. If A and B are ideals of R and if AB G P, then either A c P or B s P.

Proof. Clearly (3) implies (2) and (2) implies (1). Now suppose that P is prime and that A B G P , where A and B are ideals of R. If A $ P then there is an element a E A with a $ P. For every b E B, abE AB s P. and so b E P . Thus B E P.

2.10. Definition. ( 1 ) Let A be an ideal of a ring R. Set

Rad(A) = { a I a E R and an E A for some positive integer n},

(2) Let N be a submodule of an R-module M . Set and call Rad(A) the radical of A.

Rad(N) = Rad(N:M),

and call Rad(N) the radical of N .

Let a, b E Rad(A) and r E R ; let am E A and b" E A, where m and n are positive integers. Then (ya)m = rmam E A and

since either m + n - k 2 m or k 2 n. Hence ra, a - b E Rad(A), and we conclude that Rad(A) is an ideal of R. A number of properties of the mapping A i+Rad(A) are given in Exercise 1. We note that a submodule Q of M is primary and if and only if for all a E R and x E M , ax E Q and x $ Q imply that a E Rad(Q).

2.11. Proposition. If Q is a primary submodule of M then Rad(Q) is a prime ideal of R. If A is an ideal of R and if A G P, where P is a prime ideal of R, then Rad(A) E P.

Proof. Let ab E Rad(Q); then for some positive integer il we have anbnM= (ab)"M c Q. If b $ Rad(Q), then bnx 4 Q for some x E M . Since anbnx E Q but bnx $ Q we have ankM G Q for some positive integer k. Thus a E Rad(9)).

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42 11 PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

If a E Rad(A), then an E A ; hence an, and therefore a itself, belongs to every prime ideal of R which contains A.

If Q is a primary submodule of M and Rad(Q) = P, we say that Q is P-primary.

A nonempty subset S of R is said to be multiplicatively closed if ab E S whenever a, b E S. It is clear that a proper ideal P of R is prime if and only if R\P is multiplicatively closed.

2.12. Proposition. Let A be an ideal of R and let S be a multiplica- tively closed set in R such that A n S is empty. Then there is an ideal P of R which is maximal with respect to the properties that A G P and P n S are empty. Furthermore, P is a prime ideal.

Proof. Let 25 be the set of all ideals B of R such that A 5 B and B n S is empty; 25 is not empty since A E %. By Zorn’s lemma, % has a maximal element P. T o show that P is prime, suppose that it is not and let ab E P, a 4 P, b 4 P. Then P c P + (a ) and P c P + (b), and so there are elements s, t E S such that s E P + (a ) and t E P + (b). Hence s = p , + rla, t = p , + rz b, where p , , p , E P and r, , rz E R. Then st = p , p , +p , r , b + rlap, + rlr2 ab E P n S, which contra- dicts the fact that P n S is empty.

2.13. Definition. An ideal P of R is maximal i f P # R and i f there is no ideal A of R such that P c A c R.

If we take S = (1) in Proposition 2.12, we see that each proper ideal of R is contained in at least one maximal ideal of R.

An element a E R is called a zero-divisor if there is an element b E R, b # 0, such that ab = 0. Thus, if R has more than one element, 0 is a zero-divisor. A zero-divisor other than 0 is referred to as proper. An element of R which is not a zero-divisor is called regular. If R has more than one element, then 1 is regular. An ideal of R is called regular if it contains a regular element. An element u E R is called a unit if there is an element v E R such that uv = 1 ; there is only one such element v ; it is denoted by u- l , and it is called the inverse of u. Clearly, 1 is a unit and if R has more than one element then every unit is regular. An element a E R is called nilpotent if

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2 PRIMARY SUBMODULES 43

a E Rad(O), that is, if an = 0 for some positive integer n. If a E R and if a is not nilpotent, then 0 does not belong to the multiplicatively closed set S = {a" In is a positive integer). Hence there is a prime ideal P of R such that a 4 P.

2.14. Proposition. If A is an ideal of R, then

Rad(A) = P,

where the intersection is over all prime ideals of R containing A.

Proof. By Proposition 2.11, Rad(A) is contained in the ideal on the right. Let a $ Rad(A). If S = (a" 1 n is a positive integer) then A n S is empty; hence there is a prime ideal P of R such that A G P and a $ P . Thus a does not belong to the ideal on the right.

Let A be an ideal of R. A prime ideal P of R is called a minimal prime divisor of A if A G P and if there is no prime ideal P' of R such that A c P' c P. If N is a submodule of an R-module M , then the minimal prime divisors of N : M are called the minimal prime divisors of N .

2.15. Lemma. Let A be an ideal of R and suppose that A c P where P is a prime ideal of R. Then P contains a minimal prime divisor of A.

Proof. Let X be the set of all prime ideals P' of R such that A c P' c P. Then X is not empty since P E X. Partially order X by the reverse of inclusion. Then any maximal element of X is a minimal prime divisor of A contained in P . The existence of a maximal element of X will follow from Zorn's lemma if we show that the intersection P' of the ideals in a totally ordered subset {P,} of X is prime. Let ab E P' and suppose that a $ P'. Then a $Pa for some cc and so b e P,. If p # a , then either P , c P, or P, z P , . If P , E P, , then a $ P B and so b E P , . If P, c P B , then b E P B . Hence b E n P , = P ' .

From this lemma and Proposition 2.14 we obtain the following proposition.

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44 11 PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

2.16. Proposition. If N is a submodule of M , then

Rad(N) = (I P, where the intersection is over all minimal prime divisors of N .

3 NOETHERIAN RINGS

I n this section we shall prove several assertions which are of considerable importance in the theory of commutative rings. One consequence of the first of these theorems is that there is an abun- dance of Noetherian rings.

2.17. Theorem (Hilbert Basis Theorem). If R is a Noethmian ring, then the polynomial ring R[X] is a Noetherian ring.

Proof. Let A' be an ideal of R[X] and let A be the set of all a E R such that there is an element of A' of the form axs + c,-lXs-l +. + c, . Then A is an ideal of R and so A = (al, . . . , a k ) . For each i, let a i X S i f - . . €A' . If s =max{s,, . . ., s k ) , then aiXS + - E A'; we denote this polynomial byf, . Let A" = (fl, . . . y f k ) . If we consider R[X] as an R-module, then A' and A" are submodules. Let N be the submodule of R[X] consisting of all polynomials of degree at most s - 1. We shall show that

A'= (A' n N ) +A".

Once this has been done the assertion will follow. For N is finitely generated and so, by Theorem 1.12, A' n N is a finitely generated R-module. Hence there are polynomials f k + 1, . . . , fn E A' n N such that A' = (fl, . . . ,fn).

It remains to verify the equality stated above. Clearly, the right- hand side is contained in the left-hand side. Let

g = b,Xt + - - * + b1 x+ b, E A'.

Then b, E A , so

b, rla1 f * ' ' + r k a k

where rl , . . . . , rk E R. If t < s, then g E A' n N . Suppose that t 2 s.

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3 NOETHERIAN RINGS

Then

g = (rial + - - * + rka,)X + b,-,X t - l + . * - + b,

=rlXt-sdfl -f1') f ' ' ' + rkXt-s( fk - f k ' )

+ b&,X t - l + -. * + b, ,

where f2' = f i - a f X " E N . Thus g =g,' +gl, where g,' E A" and

g, = c,-,Xt-l + * * . +clX+ c, E A'.

45

If we repeat this argument several times, we will obtain finally

g = g l ' f . - . + g ; - s + l +gt-,+,,

whereg,'EAN for i = l , ..., t - s + l a n d g , - , + , E A ' n N . This completes the proof.

2.18. Corollary. If R is a Noetherian ring, then any polynomial ring R[X,, . . . , X,] is a Noetherian ring.

Thus, if K is a field, then any polynomial ring K [ X , , . . . , X,] is a Noetherian ring. The ring 2 of integers is a Noetherian ring (in fact, each of its ideals is principal) so the same is true of any poly- nomial ring 2 [Xl, . . . , X , ] .

2.19. Theorem (Artin-Rees Lemma). Let R be a Noetherian ring and let A, B, and C be ideals of R. Then there is a positive integer r such that

AnB n C = An-'(ArB n C ) for all n > r.

Proof. Let A = (al, . . . , a%). Let X , , . . . , X, be indeterminates and let S, be the set of all homogeneous polynomials f E RIXl , . . . , xk] of degree n such that f (a l , . . . , ak) E A"B n C. (All terms off have degree n.) Let = uFz0 S, and let A' be the ideal of R[X,, . . . , xk] generated by S. By Theorem 2.17, A' is finitely generated, say A' = (fl, . . . , f t ) . I t is clear that we may assume that f i E S for i = 1 , . . . , t . Let di be the degree off, and let r = max{d,, . . . , dt}. Let n > r. If a E AnB n C, then, since a E A", we haveffa,, . . . , ak) = a for some f E s of degree n. Iff = 1 f i g l , where g, E R[X,, . . . , xk],

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46 11 PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

then we may assume that for each i, gt is of degree n - d, and is homogeneous. Then

u=f(' l , *..) uk)=cf t (a l , uk)gt(ul, ak)

E C An-di(AdiB n C ) = C An-'Ar-dt(AdlB n C )

E An-' (A'B n A'-*fC) E AnFr(ArB n C ) ,

and consequently A"B n C E An-'(ArB n C) when n > r . On the other hand, A"-'(A'B n C) E AnB n A"-' C G A"B n C for all n > r , so all is proved.

Now we shall extend the Artin-Rees lemma to modules. Let M be an R-module and let R" = R @ M, the direct sum of R and M as R-modules. Each element of R* can be written uniquely in the form a + x, where a E R and x E M. Define a multiplication in R" by

(u + x)(u' + x') = aa' + u'x + ux'.

Then R" is a commutative ring with unity. If R is Noetherian and M is a finitely generated R-module, then every submodule of the R-module R" is finitely generated; hence R" is a Noetherian ring. The subring {a+OIaE R} of R" is identified with R, and (0 + X I x E M } is an ideal of R" which we identify with M. We have M 2 = 0 and every submodule of M is an ideal of R". Furthermore R*/M E R (as rings).

2.20. Theorem. Let R be a Noetherian ring and M afinitely generated R-module. Let A be an ideal of R and L and N submodules of M. Then there is a positive integer r such that

AnL n N = An-'(A'L n N ) for all n > r.

Proof. Since A + M , L, and N are ideals of R", it follows from Theorem 2.19 that there is a positive integer r such that for all n > r,

A"L n N E ( A +M)"L n N = (A +MY-'(@ +M)'L n N ) c (A"-' + M)((Ar + M ) L n N ) =An-'(A'L n N ) ,

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3 NOETHERIAN RINGS 47

since M 2 = 0. On the other hand,

A"-'(A'L n N ) E A"L n An-" E AnL n N for all n > I,

so the equality is proved.

2.21. Corollary. Let R, A , and M be as in Theorem 2.20. If N = n:=l AnM, then A N = N .

Proof. If r is as in Theorem 2.20, with L = M , then for all n > r,

AnM n N = An-*(ArM n N ) E AN.

Since N E AnM this implies that N G AN. But AN G N , so A = AN.

Now a technical lemma!

2.22. Lemma. Let a , , , b, E R for i, j = 1, . . . , k , and let d = det[a,,]. If C:=l a,, bj = Ofor i = 1, . . . , k, then db, = Ofor j = 1, . . . , k .

Proof. Let di , be the cofactor of a i j in the matrix [a,,]. Then

d f d i j a i h = [o i = l

if j = h if j # h.

Hence

for j = 1, . . . , k.

k k

0 = 1 d,, 1 aihbh = 1=1 h = l

2.23. Theorem (Krull Intersection Theorem). Let R be a Noe- therian ring and M a finitely generated R-module. Let A be an ideal of R. Then

f i A ~ M = O n = l

if and only if l - a g A and ax=0, where a E R and EM, imply that x = 0.

Proof. Suppose that 1 - a E A and ax= 0 imply that x = 0. Let N = (7:=,AnM; then A N = N . Let N = R x , + . . . + R x , . For

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48 11 PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

i= 1 ,..., K , ~ ~ = C F , , a , , x ~ , w i t h a , , ~ A , t h a t i s C ~ = ~ ( 6 , , - a ~ ~ ) x , = O . Let d = det[aij - ui,]. Then 1 -d E A, and by Lemma 2.22, applied to R",wehavedx,=Oforj=l , ..., K.Hencex,=Oforj=l , ..., K , and so N = 0.

Conversely, suppose that for some a E R and nonzero x E M we have 1 - a E A and a x = 0. Then, for each positive integer n we have x = (1 - a).x E AnM. Hence n;= AnM # 0.

2.24. Corollary. Let R, A, and M be as in Theorem 2.23. If A is contained in every maximal ideal of R, then

m

AnM= 0. n = l

Proof. Let a E R and x E M be such that 1 - a E A and a x = 0. If a is not a unit in R then (u) is a proper ideal of R and so is contained in some maximal ideal of R by Proposition 2.12; this implies that 1 is contained in some maximal ideal of R, which is not true. Hence, a is a unit in R, and it follows that x = 0.

4 UNIQUENESS RESULTS FOR

PRIMARY DECO MPOSlTlO N S

Let R be a ring and let M be an R-module. Let N be a submodule of M , and suppose that

N = Q1 r\ ... n Qny

where Ql, . . . , Qn are primary submodules of M ; this is called a primary decomposition of N . Such a decomposition is called reduced if:

(i) No Qi contains the intersection of the remaining Q,; and (ii) Rad(Qi) # Rad(Q,) for i # j .

2.25. Proposition. If a submodule N of M has a primary decomposi- tion, then N has a reduced primary decomposition.

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4 UNIQUENESS RESULTS FOR PRIMARY DECOMPOSITIONS 49

Proof. Let N = Q1 n n Qn where Qi is primary for i = 1, . . . , n. If some Qt contains the intersection of the remaining Qj , simply delete it. Hence we may assume (i) holds. Suppose QLl, . . . , Q,, have the same radical P. By Exercise 6(f), Q = Qll n 1 . . n Qi, is P-primary. In the decomposition for N replace Q,, n . . . n Qlk by Q. Proceeding in this manner we finally obtain a reduced primary decomposition for N .

It can happen that N has no primary decomposition whatsoever [Exercise 13(d)], but if it does have such a decomposition, it may have more than one reduced decomposition [Exercise 11(c)]. In this section we shall determine certain features of primary decompositions which are unique.

Let

be a reduced primary decomposition of N and let P, = Rad(Qi) for i = 1, . . . , k. The prime ideals P,, . . . , Pk are called prime divisors of N. We have

Rad(N) = Rad(Q,) n * * * n Rad(Q,) = PI n - . - n Pk.

Hence, if a E P, . * * P, , then some power of a is in N : M . Conse- quently, if P is any prime ideal of R which contains N: M , then P, * * P, E P and so Pi c P for some i. I t follows that every minimal prime divisor of N is a prime divisor of N and is minimal in the set of prime divisors of N . The next theorem characterizes the prime divisors of N in terms of N itself. We assume throughout this discussion that N # M .

2.26. Theorem. Let N be a submodule of M and assume that N has a primary decomposition. Let N = Q1 Qk be a reduced primary decomposition of N . Let P be a prime ideal of R. Then P = Rad(Q,) for some i if and only if N : Rx is a P-primary ideal of R for some x $ N .

* . *

Proof. Let Pi = Rad(Q,) for i = 1, . . . , k and suppose P= P,. Since the decomposition is reduced there is an element x E Q2 n * * * n Qk with x $ Q1 (simply choose x $ N if k = 1). By Exercise 6(h), Q1 : Rx is a P-primary ideal of R. Furthermore, Qi : Rx = R for i = 2, . . . , k.

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50 11 PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

Hence N: Rx = Q, : Rx and so N : Rx is P-primary ; note that x 4 N . Conversely, suppose that N : Rx is P-primary for some x 4 N . Then

P = Rad(N: Rx) = Rad(Q, : Rx) n * . - n Rad(Qk: Rx).

For each i, Rad(Q,: Rx) = Pi or R, and is equal to Pi for at least one i, since x 4 N . Thus P is the intersection of some of the prime ideals P,, . . . , Pk. By Exercise 5(b), P is equal to Pi for some 2 .

2.27. Corollary. Let N be a submodule of M and assume that N has a primary decomposition. If

N = Q1 n .. . n Qm = Q1' n ... n Qi are two reduced primary decompositions of N , then m = n and the Qt and Qr' can be so numbered that Rad(Q,) = Rad(Q,') for i = 1, . . . , n.

2.28. Corollary. Let R be a Noetherian ring and let P,, . . . , Pk be the prime divisors of the ideal 0. Then the set of zero-divisors of R is precisely PI v * * * v Pk .

Proof. Let a E Pi; then there exists b E R, b # 0, such that an E 0: (b) for some smallest positive integer n. Hence a(a"-'b) = 0 and a is a zero-divisor. Conversely, suppose a is a proper zero-divisor and let ab = 0 where b # 0. Then for somei, b 4 Qi where0 = Q, n n Qk is a reduced primary decomposition of 0. I t follows that a E Rad(Qi) = P i .

2.29. Definition. Let S be a multiplicatively closed set in R. If N is a submodule of M , we set

N , = { X I x E M and sx E N for some s E S } .

Note that N , = M i f ( N : M ) A S is not empty and, in particular, if 0 E S. Let xl, x2 E Ns and a E R. If slxl E N and s2x2 E N, where sl, sz E S, then s,s2 E S and s1s2(x1 - x,) = s2(s1x1) - s,(s, x,) E N ; also s,(ax,) = a(slxl) E N. Hence x, - x2, ax, E N, and we conclude that N , is a submodule of 11.17; certainly N G N , . The submodule N , of M is called the component of N determined by S, or simply the S-component of N .

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4 UNIQUENESS RESULTS FOR PRIMARY DECOMPOSITIONS 51

2.30. Proposition. Let N be a submodule of M which has a primary decomposition, N = Q, n * . n Qk . Let s be a mult+licatively closed set in R. Let Qi be P,-primary and assume that Pi n S is empty for i = 1, . . . , h and that Pi n S is not empty for the remaining i. Then Ns = Q1 n * * * n Qh .

Proof. Let x E N , . Then sx E N = Q1 n * * n Qk for some s E S. For i= 1, . . . , h, sx E Qi but s Pi. Hence x E Qi . Therefore N , s Q1 n n QpL. Now let x E Q1 A - n Qh. If h = k, then x E N. If h # k, choose s, E P, n S f o r j = h + 1, . . . , k. For large enough n we will have

( S ~ + I **.s , )nMc Q h t l n '.. n Q k ,

and so

(Sh+l ' . * Sk)nX E Q1 n * ' 0 Q k = N.

Since (sh+ * . sk)" E S, we have x E N , . Therefore N , = Q1 n - . * n Qh.

Again, let N be a submodule of M which has a primary decom- position

N=Q, n n Q r ,

which we assume to be reduced. Let Pi = Rad(Q,), i= 1, . . . , k. A set of these prime ideals, {Pi , , . . . , PiT}, is called an isolated set of prime divisors of N if each Pi which is contained in at least one of the ideals in this set is itself in this set. If Pi is a minimal prime divisor of N , then {Pi} is an isolated set of prime divisors of N .

2.31. Proposition. r f {Pi , , . . . , Pi,} is an isolated set of prime divisors of N , then Q,, n * - n Qi, depends only on this set and not on the particular reduced primary decomposition of N .

Proof. Let S = R\(Pi, u * * - u Pi,) ; then S is a multiplicatively closed set in R. The proposition will follow if we show that N,=Qi ,n . . * n Qir , and to do this it suffices to show that S n Pi is empty when i = ij for some j, and S n Pi is not empty when i # il, . . . , i, . The former is certainly true. Suppose i # i,, . . . , i, . We have Pi $ P,, f o r j = 1, . . . , r, and SO Pi $ u;=, Pi j by Exercise 5(c). Hence Pi n S is not empty.

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52 11 PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

In the notation of this proposition, Pi, n * * * n Qt, is called an isolated component of N . The proposition asserts that the isolated components of N are uniquely determined by N .

Our principal uniqueness theorem is an immediate corollary to Proposition 2.31 and is stated now.

2.32. Theorem. Let P be a minimal prime divisor of a submodule N of M . Suppose that N has a primary decomposition. If the P-primary submodule Q occurs in a reduced primary decomposition of N , then Q occurs in every reduced primary decomposition of N .

We close this section with a result related to the Krull intersection theorem (Theorem 2.23).

2.33. Proposition. Let R be a Noetherian ring and let M be aJinitely generated R-module. Let A be an ideal of R. Then AnM is a component of the ideal 0.

Proof. Let S = ( l - a ( a E A ) . If a , b E A , then (1 -a ) ( l -b )= 1 - (a + b - ab) E S , so S is a multiplicatively closed set in R. Let N = n:==l A n M ; by Corollary 2.21, we have N = AN. Let x E N and let A x = Q1 n * * n Qk , where Qi is primary and Rad(Q,) = Pi for i= 1, . . . , k. For each i, A x G Qi so that either A c P i or x E Qi [see Exercise 6(c)]. Suppose that A c Pi for somej. Then there is a positive integer m such that A" C_ Pim s Qj: M [see Exercise 6(a)]; that is, A"M s Q j . Hence N c Q,, and consequently x E Q j . Thus x E Qi for i = 1, . . . , k and therefore we have x E A x . Hence x = ax for some a E A. Conversely, if x E M and x = ax for some a E A, then x = a n x € A"M for all n; hence X E N . Thus, X E N if and only if sx = 0 for some s E S ; that is, if and only if x E 0,.

EXERCISES

1. The radical of an ideal. Let R be a ring and let A, B, and C be ideals of R. (a) (b)

Show that Rad(AB) = Rad(A n B) = Rad(A) n Rad(B). Show that Rad(A + B) = Rad(Rad(A) + Rad(B)).

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EXERCISES 53

(c) Show that

Rad(A + BC)= Rad(A +(B n C ) )

= Rad(A + B ) n Rad(A + C).

(d) (e)

Comaximal ideals. Two ideals A, B of a ring R are said to be comaximal if A + B = R. A finite collection A,, . . . , A, of ideals of R is comaximal if each pair is comaximal. Let Al, . . ., A, be comaximal ideals of R and let n,, . . . , n k be positive integers. Prove the following : (a) A;,, . . . , AEk are comaximal. (b) A , n . . . n A , = A , . . . A ,. (c) For a,, . . . , ak E R, there exists x E R such that x - ai E A,

for i = 1, . . . , k. (d) R/(A;’ n . . n A;”) 2 R/A;l @ . . . @ RIA;,.

The ideals B,, . . . , B,, of R are comaximal if and only if Rad(B,), . . . , Rad(B,,) are comaximal.

Show that Rad(Rad(A)) = Rad(A). Show that A + B = Rif andonly if Rad(A) + Rad(B) = R.

(e)

Maximal, prime, and primary ideals. Let R be a ring. (a) Show that an ideal P of R is maximal if and only if RIP is a

field. (b) Show that an ideal P of R is prime if and only if RIP has

no proper zero-divisors. (c) Show that an ideal Q of R is primary if and only if every

zero-divisor of R/Q is nilpotent. (d) Show that if R # 0 and if R and 0 are the only ideals of R,

then R is a field.

2.

3.

4. Ideals in a direct sum of rings. Let R,, . . . , R, be rings and let R = R, @ * @ R, . (a) Show that if A is an ideal of R, then A can be written

uniquely in the form A , @ . @ A,, , where Ai is an ideal of R i for i = 1, . . . , n. Show that A is a maximal ideal of R if and only if for some i, A i is a maximal ideal of Ri and A j = R j for j # i.

(b)

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54 11 PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

(c) Show that the statement of part (b) is true when “maxi- mal ” is replaced by “ prime ” or by “ primary.”

5. Intersections and unions of primary ideals. (a) Let R be a ring, P a prime ideal of R, and Q1, ..., Qk

primary ideals of R. Show that if 8, n - n Qk c P, then Rad(Qi) c P for some i.

(b) Let P,, ..., P k be prime ideals of R. Show that if P = P, n * * n Pk is a prime ideal, then P = Pi for some i and P c Pf for j = 1, . . . , k. Let PI, . . . , P k be prime ideals of R and let A be an ideal of R such that A G P, V * - U pk . Show that A E P, for some i.

(c)

6. Primary submodules. Let R be a ring and let A be an ideal of R such that Rad(A) is finitely generated. Show that there is a positive integer n such that (Rad(A))” c A. Assume that R is Noetherian and let M be an R-module. Let Q be a P-primary submodule of M . Show that for some positive integer n we have P c Q : M . If R is Noetherian, show that a submodule Q of an R-module M is primary if and only if for all ideals A of R and submodules N of M , AN s Q and N $ Q imply that AnM G Q for some positive integer n. Let M be an R-module, P an ideal of R, and Q a sub- module of M . Suppose that: (i) (ii)

Show that P is a prime ideal and Q is P-primary. Show that a submodule Q of M if primary if Rad(Q) is a maximal ideal of R. Let Q1, . . . , Qk be P-primary submodules of M . Show that Ql n Let 4 be a homomorphism from R into R‘. Let Q‘ be a P’-primary ideal of R‘. Show that P = $-‘(PI) is prime and that Q = +-’(Q’) is P-primary. Let M be an R-module, A an ideal of R, P a prime ideal of R, Q a P-primary submodule of M , and N a submodule

Q: M E P G Rad(Q); and If ax E Q and x $ Q, where a E R and x E M , then a E P .

n Qk is P-primary.

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EXERCISES 55

of M . Show that if A $ P then Q: A = Q and that if N $ Q then the ideal Q: N is P-primary. Assume R is Noetherian and M is finitely generated. If N is a submodule of M , show that N : A = N if and only if A is contained in no prime divisor of N . Show that N : A = N if and only if no prime divisor of A is contained in any prime divisor of N .

(i)

( j)

7. The Jacobson radical. Let J(R) denote the intersection of the maximal ideals of the ring R. Then J ( R ) is an ideal called the Jacobson radical of R. (a) Show that J(R) is the set

( a l a ~ R and l f r a isauni t in R forall Y E R } .

Show that J(R/J(R)) = 0. (Nakayama's lemma) Let A be an ideal of R. Show that the following statements are equivalent :

(2)

(3)

(b) (c)

(1) A J(R) . If M is a finitely generated R-module and A M = M , then M = 0. If M is a finitely generated R-module and N is a submodule of M such that M = A M + N , then N = M . If a E A, then 1 f a is a unit in R. (4)

Let R be a Noetherian ring and M a finitely generated R-module. Let N be a submodule of M and A an ideal of R with A s J (R) . Show that n;=l ( N + A n M ) = N .

Let R be a Noetherian ring and M a finitely generated R-module. Let A be an ideal of R and let a E R. Show that for every positive integer n we have a(AnM: (a))= AnM n aM. Show that there is a positive integer Y such that

(d)

8. Corollaries to the Artin-Rees lemma. (a)

(b) AnM: (a) E 0: (a) + An-'IM for all n > r.

Let N be a submodule of M . Show that there is a positive integer Y such that

( N + A"M): (a) E N : (a) + An-'M

(c)

for all n > Y.

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56 11 PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

(d) Let x E M . Show that there is a positive integer r such that

for all TZ > r. (N + A"M): Rx E N : Rx + A"-'

9. A component of 0. Let R be a Noetherian ring and A an ideal of R. (a) Let B = (b I b E R and bAt = 0 for some positive integer t >.

Show that B is an ideal of R. (b) Let 0 = Q1 n n Qk be a reduced primary decompo-

sition of 0 and let Pi =Rad(Qi) for i= 1 , . . . , K. Showthat

B = n Q ~ . A g P i

(c) (d)

(a)

(b)

(c)

Show that there is an integer s such that As n B = 0. Show that B : A = B.

10. The ring of integers. Show that every ideal in the ring of integers is a principal ideal. Determine all of the prime ideals and primary ideals of the ring of integers. Prove the unique factorization theorem for integers as a corollary to results concerning primary decomposition.

11. Ideals in K [ X , Y ] . Let K be a field and consider the ring R = K [ X , Y ] of poly- nomials in two indeterminates X and Y over K. (a) Show that ( X , Y z ) is a primary ideal of R with radical

( X , Y ) . Show that ( X , Y z ) is not a power of ( X , Y ) . This shows that a primary ideal need not be a power of a prime ideal (even in a Noetherian ring). Show that Rad((X2, X U ) ) = ( X ) . Show that ( X 2 , X U ) is not primary. This shows that an ideal with prime radical need not be primary. Show that for each C E K , ( Y - c X , X z ) is a primary ideal of R, and that for c, d E K with c f d we have ( Y - cX, X z ) # ( Y - d X , X z ) . Show that for each c E K, ( X z , X U ) = ( X ) n ( Y - cX , X z ) is a reduced primary decomposition of ( X 2 , X U ) . This shows that an ideal in a Noetherian ring may have infinitely many distinct reducedprimary decompositions. What are the prime divisors

(b)

(c)

of (XZ, XU)?

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EXERCISES 57

12. Ideals in 2 [ X I . Let R = 2 [XI, the ring of polynomials with coefficients in 2. (a) Show that (4, X ) is primary and Rad((4, X)) = (Z,X),but

that (4, X ) is not a power of (2, X ) . (b) Show that (X2, 2 X ) is not primary. Show that (X) is

prime and (X)" G ( X z , ZX) . (c) Show that (4, 2X, X z ) is primary and that (4, ZX, X2) =

(4, X) n (2, X"). This shows that a primary ideal need not be irreducible. Let S be the subring of R consisting of all elements of R having the coefficient of X divisible by 3. Let P be the ideal (3X, X 2 , X 3 ) of S. Show that P is a prime ideal but P2 is not a primary ideal. This shows that apower of aprime ideal need not be primary.

(d)

13. A non-Noetherian ring. Let R be the set of all sequences {a,} ( n 2 1) of elements of the field having two elements, such that for some rn, , depending on the sequence, a, = am,, for all m 2 m , . If we define operations on R by

{a,} + {b,) = {a, + b,,} and {a#),} = {anbn),

then R is a commutative ring with unity. For each i > O , let Pi = {{a,,} 1 a , = 0}, and let Po = {{a,} I for some mo , depending on the sequence, a,,, = 0 for m 2 rn,]. (a) Show that for i 2 0, P , is a prime ideal of R. (b) Show that {Pt I i 2 0) is the set of proper prime ideals of R. (c) Show that an ideal of R is primary if and only if it is prime. (d) Show that nl.= P , = 0, but for every j > 0,

Therefore, 0 is not an intersection of a finite number of primary ideals of R. By part (d) R is not Noetherian. Construct an infinite ascending chain of ideals of R, a nonempty set of ideals of R without a maximal element, and an ideal of R not finitely generated.

(e)

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58 11 PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

14. A theorem of Cohen. (a) Let R be a ring and let A be an ideal of R. Show that if

b E R and if A + (6) and A : (b) are finitely generated, then A is finitely generated. Let Y be the set of all ideals of R which are not finitely generated. Show that if Y is not empty then Y has a maximal element. Assume that Y is not empty and let P be a maximal element of Y . Show that P is a prime ideal. Arguing from this, complete the proof of the following assertion: R is Noetherian i f and only i f every prime ideal of R is jinitely genwated.

(b)

(c)

15. Irreducible ideals. Let R be a ring and A an ideal of R. Let A = B, n n B,, where each Bi is irreducible and no B, contains the intersection of the remaining B, . (a) Show that if A = B1’ n ... n B,’ where B, E Bit for

i = l , ..., K,thenB,=B,’ forZ=l , ..., K. (b) Suppose that we also have A = C, n n C, where

each C, is irreducible and no Ci contains the intersection of the remaining C,. For i = 1, . . . , k, let Di = B, n

n Bi-l n Bii, n - - a n B,. Show that for each i, A = D, n C, for some j . Show that k = n. (c)

16. Primal ideals. Let R be a ring. For each ideal A of R, let

U(A) = { b I b E R and b + A is a regular element of RIA}.

An ideal A of R is said to be primal if R\U(A) is an ideal of R. If A is a primal ideal of R, then R\U(A) is called its adjoint ideal. The adjoint ideal P of a primal ideal A is the unique ideal maximal with respect to the properties A G Pand U ( A ) n P is empty. (a) Show that every irreducible ideal of R is primal. (b) LetKbeafieldandletR= K[X, Y]. Showthat(X2,XY)

is primal with adjoint ideal (X, Y ) . Recall that the radical of (X2, X U ) is (X) [Exercise ll(b)]. By symmetry,

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59

(Y2, Xu) is primal with adjoint ideal ( X , Y ) . Show that ( X U ) = ( X 2 , X U ) n ( Y 2 , X U ) is not primal. For i = 1, . . . , K, let A, be a primal ideal with adjoint ideal Pi. Let B = A, n - * n A, and assume that no A, can be replaced by a strictly larger ideal without changing the intersection. Show that B is primal if and only if some Pj contains every P i , and that in this case Pf is the adjoint ideal of B. Show that every ideal of R is the intersection of all of the primal ideals containing it. A primal decomposition B = A, n - * * n A, is called reduced if no A, contains the intersection of the remaining A,, if no A, can be replaced by a strictly larger ideal without changing the intersection, and if no intersection of two or more of the A, is primal. Show that if R is Noetherian then every ideal of R has a reduced primal decomposition. Let B = A, n * - * n A, = Al' n - n A,' be reduced primal decompositions of B. Show that k = n and that the A, and A,' can be so numbered that A , and A,' have the same adjoint ideal for each i.

17. Prime ideals associated with a module. Let R be a Noetherian ring and M a finitely generated R-module. For x E M set Ann(x) = {a \ a E R and ax = 0}, the annihilator of x. If P is a prime ideal of R such that P = Ann(x) for some x E M , we say that P is associated with M. The set of all prime ideals of R associated with M is denoted by Ass(M).

Show that if P is a prime ideal of R then P E Ass(M) if and only if M has a submodule isomorphic to RIP. Show that if M = 0,. I M a , where each M , is a sub- module of M , then Ass(M) = U d E I Ass(M,). Let P be a prime ideal of R and M a nonzero submodule of RIP. Show that Ass(M) = {P}. Show that every maximal element (with respect to set inclusion) of the set {Ann(x) I x E M and x # 0} belongs to Ass(M). Show that M # 0 if and only if Ass(M) is not empty. Let a E R. Show that ax = 0, for x E M , implies x = 0 if and only if a belongs to no element of Ass(M).

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60 11 PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

18. Properties of Ass(&?). Let R be a Noetherian ring and M a finitely generated R-module. (a) Show that if N is a submodule of M then

Ass(N) E Ass(M) c Ass(N) u Ass(M/N).

(b) Show that if M = I Ma then Ass(M) = Ua. I Ass(M,). (c) Show that if N , n - n Nk = 0, where the N i are sub-

modules of M , then Ass(M) G u:= Ass(M/N,). (d) Let T be a subset of Ass(M). Show that there is a sub-

module N of M such that Ass(M/N) = T and Ass(N) =

Ass(M)\ T .

19. Polynomial rings. Let R be a ring and let X be an indeterminate. (a) Show that R [ X ] is a flat R-module. (b) Show that a, + a , X f . . . + a , X " ~ R[X] is a zero-

divisor in R[X] if and only if there is an element b E R, b # 0, such that bai = 0 for i= 0, . . . , n. Let Q be a P-primary ideal of R. Show that QR[X] is a PR[X]-primary ideal of R[X], and that Q R [ X ] n R = Q. Show that if Al, . . . , A, are ideals of R then

(A, n - * * n Ak)R[X] = A,R[X] n . * . n A,R[X].

(c)

(d)

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C H A P T E R

I11 Rings and Modules of Quotients

1 DEFINITION

Let R be a ring and let M be an R-module. Let S be a multipli- catively closed set in R. Let T be the set of all ordered pairs (2, s) where x E M and s E S. Define a relation on T by

(x, - (x’, s’) if there exists t E S such that t(sx’ - s’x) = 0. This is an equivalence relation on T, and we denote the equivalence class of (x, s) by x/s. Let S - l M denote the set of equivalence classes of T with respect to this relation. We can make S- lM into an R-module by setting

We must show that these are well-defined operations; once this has been done it is easily verified that S - l M is an R-module with respect to these operations.

Suppose XIS = x’/s’ and y / t = y‘/t‘. Then there are elements u, v E S such that u(s’x - sx’) = 0 and v(t’y - yt ’) = 0. It follows that

uv(s’t‘(tx + sy) - st(t’x’ + s)’)) = t’tvu(s‘x - sx’) + s’suv(t’y - ty‘) = 0.

61

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62 111 RINGS AND MODULES OF QUOTIENTS

Hence

( tx + y ) / s t = (t’x’ + s’r’)/s’t’.

Also

u(s‘ax - sax’) = au(s’x - sx’) = 0,

so that

axis = ax’/s’.

The R-module S-’M is called a quotient module, or a module of quotients. Note that if 0 E S, then S- lM = 0. In our discussion of quotient modules we shall always assume that 0 # S. In fact, we will call S a multiplicative system in R if S is a multiplicatively closed subset of R with 0 4 S.

Since R may be considered as an R-module we can form the quotient module S-lR. An element of S- lR has the form als, where a E R and s E S. We can make S-IR into a ring by setting

(a/s)(b/t) = ab/st.

If a/s = a’/s‘ and blt = b’/t’, and if u(s’a - sa’) = 0 and v(t’b - tb’)=O, where u, v E S, then

uv(s‘t’ab - sta’b’) = vt‘bu(s’a - sa’) + usa’v(t‘b - tb’) = 0.

Consequently,

ab/st = a’b’/s’t‘,

and thus we have a well-defined multiplication in S-IR. It is clear that with this operation, S-lR is a ring. The ring S-’R is called a quotient ring of R, or a ring of quotients of R.

3.1. Proposition. Let R and S be as above.

(a)

(b) (c) (d)

If a E R, then as/s is independent of the element s E S, and the mapping 7 : R --f S - lR given by T(a) = asls is a homomorphism. Ker 7 = O,, the S-component of 0. Every element of q(S) i s a unit in S-IR. If 4 : R-tR’ is a homomorphism from R into a ring R such that every element of +(S) is a unit in R’, then there is a unique homomorphism $: S-‘R -+ R’ such that

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1 DEFINITION

R

63

is commutative.

Proof. (a) If a E R and s, t E S, then u(t(as) - s(at)) = 0 for all u E S. Hence asjs = at/t. Thus q(u) = as/s depends only on a. We have

q(a + b) = (a + b)s2/s2 = asis + bs/s

= T ( 4 + V(b)

and

q(ab) = abs2/sz = (as/s)(bs/s) = q(~)q (b ) .

Thus q is a homomorphism. (b) We have a E Ker q if and only if us/s = 0 ; this is equivalent

to ts2a = 0 for some t E S. Since ts2 E S this implies a E 0,. Con- versely, if vu = 0, where z, E s, then av/v = O/v = 0, so that a E Ker q.

(c) Note that s/s = t/t for all s, t E S, and that this element is the unity of S-lR. If s E S then q(s) = s2/s, and (s2/s)(s/s2) = 1, so q(s) is a unit in S-'R.

(d) Define 4: S- lR+ R' by #(.Is) = +(a)+(s)-l. Suppose u/s= b / t and that u(ta - sb) = 0 where u E S. Then 0 = +(u)(+(t)+(u) - +(s)+(b)) and since +(u) is a unit in R', +(t)+(u) - +(s)+(b) = 0. Thus, +(a)+(s)-' = +(h)+( t ) - l . Hence z+h is well-defined. It is a homomorphism since

#(.Is + b / t ) = #((ta + sb)/st) = +(tu + sb)+(st)-l

= ($(t >$(a) + #(s)+(*))+(s) - '+Ct ) = d(.)+(4 + +(b)+(t 1 -l = $(ah) + 4(b/t 1

and

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64 111 RINGS AND MODULES OF QUOTIENTS

Furthermore for all a E R, $y(a) = $(as/s) = $(a)$(s)$(s)-l = $(a). Clearly $ is the only homomorphism from S- lR into R‘ which will make the triangle commutative.

Let R, S, and M be as above. We can make M O R S d 1 R into an R-module by the method described in Exercise 7(b) of Chapter I. Then a(x 0 b/s) = x @ ab/s = ax 0 b/s.

3.2. Proposition. S - l M z M O R S-lR, as R-modules.

Proof. Define 4: S-lM+ M O R S-’R by

$(xis) = x 0 l/s.

x 0 11s = x 0 us’ jsus’ = us’x 0 1 jsus’ = usx’ 0 1 Isus’

If x/s = x’/s’ and if u(s’x - sx’) = 0, where u E S, then

= x’ 0 us/sus’ = x‘ 0 l/s’.

Therefore 4 is well-defined. Since

$(xis + y b ) = $((tx + sy)/st) = ( t x + 9) 0 l/st = tx 0 ljst + sy 0 ljst

= x 0 l/s +y 0 I / t = $(xis) ++bit) and

$(u(xjs)) = $(axis) = ax @ l / s = a(x 0 l / s ) = a+(xjs),

$ is a homomorphism (of R-modules). Now define a mapping $’ from the Cartesian product M x S- lR into S-lM by $’(x, a/s) =ax/s . If a/s=b/t and if v ( ta - sb ) = 0, where v E S, then v(tax - sbx) = 0, and so ax/s = b x / t . Hence $’ is well-defined. After some verification, it follows from Proposition 1.17 that there is a homomorphism (of Abelian groups) $: M ORS-’R+ S- lM such that +(x 0 a/s) = axjs. Furthermore,

$(a(. @I b/t )) = $(x 0 ab/t ) = abx/t

= a[bx/t) = a$(. 0 b / t ) ,

so that Z,4 is, in fact, a homomorphism of R-modules. Since

$#(x 0 a/s) = $(axis) = ax 0 l/s = x 0 a/s

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1 DEFINITION 65

and

$$(XIS) = +(x 0 l/s) = X I S ,

$ and $ are isomorphisms, inverse to one another.

3.3. Theorem. S-IR is a j a t R-module.

Proof. Let N and L be R-modules and let p : L + N be an injective homomorphism. Let $1: S-lL+L @RS-'R and d 2 : S-lN-t NOR S- lR be defined in the same manner as $ in the proof of Proposition 3.2. Define u: S-'L -+ S-lN by u(x/s) = p(x)/s; we verify as usual that a is a well-defined homomorphism. If i= 1,- l R , then the diagram

S-lL & S-IN

L @ , S - ~ R ---+ ~ ~ d s - 1 ~

cp 0 ;)$l(./s) = ( p 0 i)(x 0 1 i s ) = p (4 0 1 i s

o @ i

is commutative, for

= +2(P (x)/s) = $2 4 4 s ) . Now suppose that a E L O R S-lR and that ( p @ i ) (z) = 0. Since +1

is surjective, M = y$(x/s) for some x/s E S-lL, and dZ u(x/s) = 0. Since $ z is injective, p(x)/s = o ( x / s ) = 0. Hence, for some u E S, p(ux) =

up(x) = 0. Since p is injective, ux.= 0. Thus x/s = ux/us = 0. There- fore, M = q$(x/s) = 0, and we conclude that p @ i is injective.

If S is the set of all regular elements of R, then S is a multiplica- tive system in R. The ring S - lR is called the total quotient ring of R. In this case, the homomorphism 7 of Proposition 3.1 is injective, and we shall identify each element of R with its image in S-lR under 7. Thus R is a subring of its total quotient ring. Let T be a multiplicatively closed subset of 5' and let 4 be the homomorphism from R into T-lR given by +(a) = at/t, t E T . Since q ( t ) is a unit in S - l R for all t E T , it follows from Proposition 3.1 that there is a unique homomorphism $: T-'R + S- IR such that the diagram

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66 111 RINGS AND MODULES OF QUOTIENTS

R

T - l R - S-lR * is commutative; in fact t,b(a/t) = alt. It follows that t,b is injective, and we shall identify each element of T-lR with its image in S-lR under $.

If R'has more than one element and no proper zero-divisors, then R is called an integral domain. In this case, the total quotient ring of R is a field, called the quotient field of R.

2 EXTENSION AND CONTRACTION OF IDEALS

I n this section we develop the machinery by which we can compare the ideal structures of a ring R and one of its rings of quotients S I R .

3.4. Definition. Let R be a ring and let S be a multiplicative system in R. Let 77 be the homomorphism 7 : R --f S-IR of Proposition 3.1.

(1) If A is an ideal of R, denote by S-lA the ideal of S-'R generated by y(A). The ideal S-lA is the extension of A to S-lR. If A' is an ideal of S-lR, denote by A' n R the complete inverse image of A' under 7. The ideal A' n R is the contraction of A' to R.

(2)

To reiterate, if A is an ideal of R, S-lA is the set of all finite sums C 7(ai)ri where ai E A, T i E S-lR. In fact, one can easily show that

S-'A = (u /s \ u E A, s E S}.

On the other hand, if A' is an ideal of S-lR, then A' n R= q-l(A' n q(R)). Beware of the notation A' n R for the contraction of A'; for unless 7 is injective, R is not a subset of S-lR, so inter- preting A' n R as the intersection of two sets is nonsense,

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2 EXTENSION AND CONTRACTION OF IDEALS 67

3.5. Proposition. If A is an ideal of R, then S-lA # S-lR if and only if A n S is empty.

Proof. If s E A n S, then ~ ( s ) E S-lA. Hence since ~ ( s ) is a unit in S-lR this means that S-lA= S-lR. Conversely, suppose S- lA = S-lR. Then 1 E S-lA, so there exist a E A and s E S such that 1 = a/s. Hence since 1 = t / t for some t E S, there exist u E S such that u(st - ta) = 0 ; that is, ust = uta which is an element of A n S. Thus A n N is not empty.

3.6. Proposition. If A’ is an ideal of S-lR, then S-’(A’ n R) =A‘.

Proof. The ideal S-l(A’ n R) is the ideal of S-lR generated by .?(A’ n R) = ~[q-l(A’ n q(R))] G A’. Thus S-l(A’ n R) c A‘. On the other hand, if a / s ~ A ‘ for some aER, S E S, then a s / s = (s2/s)(a/s) E A‘ n v(R) and consequently a E v-l(A‘ n q(R)). There- fore a/s= (as/s)(l/s) is in the ideal of S-lR generated by v[~- l (A’ n v(R))] = v(A’ n R ) ; that is, a/s E S-l(A’ n R).

3.7. Proposition. If R is Noethe-rian, then S-lR is Noetherian.

Proof. Let Al‘ G A,’ G As’ c . - . be an ascending chain of ideals of S-lR. Then A,‘ n R 2 A,’ n R c AB’ n R E - * - and so there is an no such that A,‘ n R = AAo n R for all n 2 no. Then, by Proposition 3.6, A,’ = Aho for all n 2 no .

3.8. Proposition. Let {A,‘l a E I } be an arbitrary family of ideals of S-lR. Then

A,’ n R= n (A,’ n R). (G 1 a s 1

(,QI 4 ( G I 1 1 = v-y ,El n (A,’ n m)

Proof. We have A ‘ n R=v-l A,‘ nq(R)

= n T - l ( ~ ’ n ?(R))= n (A‘ n R). , € I , E l

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68 111 RINGS AND MODULES OF QUOTIENTS

I n contrast to the result of Proposition 3.6, if A is an ideal of R, it can happen that A c S-I A n R, even when S-IA # S I R . We shall now examine the process of extending and contracting ideals, in that order. First we consider prime and primary ideals. Recall that an ideal Q of R is primary if and only if ab E Q and b 4 Q, where a, b E R, imply that an E Q for some positive integer n.

3.9. Proposition. Let Q be a primary ideal of R and let P = Rad(Q). Assume Q n S is empty. Then P n S is empty, S-IP is a prime ideal of S-IR, and S-IQ is S-'P-primary. Furthermore S-'P n R = P and S-'Q A R = Q.

Proof. If P n S is not empty, and if s E P n S, then for some positive integer n we have sn E Q n S, contrary to assumption. To show that S-lQ is S-lP-primary, we shall verify that the conditions of Exercise 6(d) of Chapter I1 hold; this will have as a consequence that S-IP is prime. We have S - IQ G S I P . Let a/s E S-IP where a E P and s E S. There is a positive integer n such that an E Q. Then (als)" =

an/sn E S-'Q. Hence S-'P G Rad(S-lQ). Now consider elements als, b/t E S-lR such that (a/s)(b/ t ) = ab/st E S- lQ, and suppose a/s 4 S-IQ. For some c E Q and u E S we have ablst = c/u, so there is an element V E S such that v(uab-sstc)=O. Then avub= vstc E Q, but a 4 Q, and consequently vub E P. Thus b/t =

vub/vut E S-lP. Therefore S-'Q is S-lP-primary. We certainly have Q G S-lQ n R. Let a E S-IQ n R. Then

T(a) = as/s E S-IQ and so as/s = b/t where b E Q and t E S. For some u E S, u(ast - bs) = 0. Thus aust = ubs E Q, but ust $ P since ust ES. Hence a E Q ; thus we have Q = S - l Q n R. Since P is P-primary, we conclude also that S-'P n R = P .

3.10. Corollary. Let P be a prime ideal of R such that P n S is empty. Then there is a one-to-one order preserving correspondence between the P-primary ideals of R and the S-IP-primary ideals of S-'R. This correspondence is QW S- 'Q ; the inverse correspondence is Q'HQ' n R.

Proof. By Proposition 3.9, the mapping Q H S-IQ is one-to-one from the set of P-primary ideals of R into the set of S-lP-primary ideals

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2 EXTENSION AND CONTRACTION OF IDEALS 69

of S-lR. Let Q' be a S-lP-primary ideal of S-IR. By Proposition 3.6, Q' = S-l(Q' n R), and Q' n R is a P-primary ideal of R by Proposition 3.9 and Exercise 6(g) of Chapter 11.

3.11. Corollary. There is a one-to-one correspondence, Pt, S-IP, between the prime ideals of R which do not meet S and the proper prime ideals of S-lR. This correspondence is order preserving.

The proof of this corollary is similar to that of Corollary 3.10 and is left to the reader.

3.12. Proposition. If A is an ideal of R and A, is the S-component of A, then

S-IA n R = A,.

Proof. Suppose a E A,; then as E A for some s E S. Then T(a)r(s) E

r(A) and since ~ ( s ) is a unit in S-*R, r(a) is in the ideal of S - lR generated by q(A); that is, ?(a) E S-lA. Therefore a E S-'A n R. Conversely, suppose a E S-lA n R. Then T(a) E S- IA and so as/s = b/s for some b E A, s E S. Hence there exists u E S such that u(as2 - bs) = 0. From this we see that uas2 = ubs E A. Since us2 E S this proves that a E A,.

If P is a proper prime ideal of R, then S = R \ P is a multiplicative system in R . In this case we denote S- lR by R, and call it the quotient ring of R with respect to P. If A is an ideal of R, we denote the extension of A to R, by AR, rather than S-IA. However for an ideal A' of R, , the contraction of A' to R will still be denoted by A' n R. The ideal PR, of R, is a proper prime ideal of R, . If P' is a proper prime ideal of R,, then by Corollary 3.1 1, P' = P,R, for some prime ideal Pl of P with Pl n S empty. Then PI E P and consequently P' c PR,. It follows that PR, is the unique maximal ideal of R,. Since every nonunit of R, is contained in some maximal ideal, every element of R, not in PR, is a unit in R, . A ring in which there is only one maximal ideal is called a local ring. Hence R, is often called the localization of R at P.

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70 111 RINGS AND MODULES OF QUOTIENTS

3.13. Proposition. Let R be a ring and let A and B be ideals of R. Then A = B if and only if AR, = BR, for every maximal ideal P of R.

Proof. We only need to prove the sufficiency of the condition. Let a E A. Then all E BR, for every maximal ideal P of R. Hence, by the usual argument, for every maximal ideal P of R, there is an element r p E R\P such that T, a E B . Let C be the ideal of R generated by {r , I P a maximal ideal of R}. If C # R, then C is contained in some maximal ideal of R (Proposition 2.12). Since this is not true, we must have C = R. Thus 1 E C and so there are maximal ideals P I , . . . , Pk of R such that 1 = xlrpl + - - * + xk rPk , where xl , . . . , x k E R. Then a = xlrpl a + * . - + x k r p k a E B. Thus A c B, and we show in a similar manner that B c A . Therefore A = B.

Let P be a proper prime ideal of R. If n is a positive integer then P" is not necessarily P-primary [see Exercise 12(d) of Chapter 111. However, every minimal prime divisor of Pn must contain P, so that P is the only minimal prime divisor of P". Denote by P(n) the com- ponent of P" determined by R\P. The ideal P(n) is called the nth symbolic power of P. By Proposition 3.12, Pn) = P"R, n R. If Pn has a primary decomposition, then it follows from Proposition 2.30 and Theorem 2.32 that P(") is the primary ideal with radical P which occurs in every reduced primary decomposition of P".

3.14. Proposition. Let P be a proper prime ideal of a Noetherian ring R and let S = R\P. Then

m n ~ ( n ) = 0,. n = l

Proof. By Exercise 4(a), PnRp = (PR,)" for each positive integer n, and

m

n = l n ( P R , ) ~ = o

by Corollary 2.24. Therefore

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3 PROPERTIES OF RINGS OF QUOTIENTS 71

by Proposition 3.l(b). Hence, by Proposition 3.8, a, m n P)= n (PR, n R )

n = l n = l

- - 0,.

3 PROPERTIES O F RINGS OF QUOTIENTS

In this section we shall obtain several properties of rings of quotients which will prove to be useful. In particular, we shall show that the formation of rings of quotients is transitive and that the formation of rings of quotients commutes with the formation of residue class rings. These facts are consequences of the following universal property of rings of quotients. Let R be a ring and S a multiplicative system in R ; let 7 : R + S- lR be the homomorphism of Proposition 3.1.

3.15. Proposition. Let R' be a ring and let p : R+R' be a homo- morphism such thut ;

(i) (ii)

(iii)

elements of p(S) are units in R ' ; the kernel of p is 0, ; and every element of R' can be expressed in the form p(x)p(s)-l for some x E R, s E S.

Then there is a unique isomorphism t,,!~ : S - lR + R' such that the diagram

R

is commutative.

Proof. By Proposition 3.l(d), there is a unique homomorphism $ : S - l R - t R' such that the diagram is commutative; explicitly, #(a/s) = p(a)p(s)-l. It follows from (ii) and (iii) that $ is injective and surjective.

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72 111 RINGS AND MODULES OF QUOTIENTS

3.16. Theorem. Let T be a multiplicative system in R with S G T. Then

q(T)-lS-'R g T-lR.

Proof. Let 7': R 4 T - l R be the homomorphism of Proposition 3.1. Since every element of q'(S) is a unit in T-'R there is, by Proposition 3.l(d), a homomorphism p : S-lR-t T - lR such that the diagram

R

U

is commutative. We shall show that v( T ) and p satisfy the conditions of Proposition 3.15.

Clearly, elements of p(q( T ) ) = qf( T ) are units in T- lR. Suppose p(a/s) = 0 for some a E R and s E S, that is, q'(u)q'(s)-l = 0. Then there exists an element t E T such that at = 0 ; hence

( a / M t ) = T ( 4 v ( 4 - W ) = q(at)7)W1 = 0.

Thus a/s E On,,, . Conversely, if a/s E O,,,, , then q(at) = 0 for some t E T ; hence there is an element u E S such that aut = 0. This implies that a E 0, , so that p(a/s) = ~ ' ( u ) v ' ( s ) - ~ = 0. Therefore, the kernel of

Finally, if a/t E T-lR, where a E R and t E T, then a/t = q'(a)q'(t)-l = p(q(a))p(q(t))-', which is the form required in (iii) of Proposition 3.1 5 .

p is o,,T) a

3.17. Theorem. Let A be an ideal of R such that A n S is empty. Let (b : R --f RIA be the canonical homomorphism. Then

+(S)-l(R/A) S-lR/S-lA

Proof. Note that +(S) is a multiplicative system in RIA. Let 4': S-lR+ S- lR/S- l A be the canonical homomorphism. Define a mapping p : RIA+ S-lRIS-lA by p(a +A) =q(a) + S - l A ; q is well-defined since A c S-lA n R, it is a homomorphism, and the diagram

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3 PROPERTIES OF RINGS OF QUOTIENTS 73

is commutative. We shall show that +(S) and p satisfy the conditions of Proposition 3.15.

Elements of q(S) are units in S-lR, and since q(S) n S-lA is empty, the elements of +‘q(S) are units in S-lR/S- lA. Hence, the elements of p(+(S)) are units in S I R / S - l A .

Now we determine the kernel of p. The kernel of 4‘ is S-lA and hence the kernel of p+ = +‘TI is q-l(S-lA n q(R)) = S-IA n R = A,, by Proposition 3.12. Hence the kernel of p is ASIA. But ASIA is precisely the +(S)-component of the zero ideal in RIA.

Finally, every element of S-’R/S-lA can be written in the form

4 ’ ( T ( N s ) - l ) = +‘T(414’v(s)-1 = d(+m-l= /-4a + A M + so that (iii) of Proposition 3.15 holds.

3.18. Definition. The set

S = { x l x ~ R and ~ ( x ) isauni t in S-IR}

is the saturation of S.

It is clear that is a multiplicative system in R, and that S G 3.

3.19. Proposition.

S = { x l x ~ R and x y e S forsome Y E R } .

Proof. If xy = s E S then q ( x ) ~ ( y ) = y(s) is a unit in S-IR, and so q(x) is a unit in S-IR. Conversely, suppose ~ ( x ) = xs/s is a unit in S I R ; here s E S . If the inverse of xs/s is zlt, where x E R and t E S, then xsx/st= (xs/s)(x/t)= t / t , so that for some U E S we have x(sxtu) = USt2 E s.

3.20. Theorem. S- lR S- lR; explicitly, there is an isomorphism fyom S-IR onto s- lR which maps a/s E S - ‘R, where s G S, onto a/s E S-lR.

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74 111 RINGS AND MODULES OF QUOTIENTS

Proof. Let p : R + S-IR be the homomorphism defined by p(a) = at/ t , where t E S. Since S c 3, every element of p ( S ) is a unit in S-lR. The kernel of p is 0,- , which contains 0,. If a E 0,- then ax = 0 for some x E 3. Then, if xy E S, we have a(xy) = 0. Hence a E 0,, and we conclude that Ker p = 0,. Consider an element a/ t E S-IR, where a E R and t E 3. If tz E S, then a/t = az/tz = p(az)p(tz)-l. We have verified that S and p satisfy (i)-(iii) of Proposition 3.15. Therefore, there is a unique isomorphism +: S- lR+S- lR such that the diagram

R

S-lR - S-lR @

is commutative. For a E R and s E S we have

+(aN = +(as/s>+(s/s2)

= +(~ /s )+ {sz /s ) -l

= $.l)(a)+.l)(s) -

= p(a)pW - = a/s.

If S consists entirely of regular elements of R, then we have agreed to regard S- lR as a subring of the total quotient ring of R (see the end of Section 1). If, in Theorem 3.16, T also consists entirely of regular elements, it follows that we actually have v(T)-l,S-lR = T-lR. In fact, y ( T ) = T, so that

T-lS- lR = T-IR.

Furthermore, the saturation S of S consists entirely of regular elements of R, and

S-IR = S-1R.

EXERCISES

1. Quotient modules. Let R be a ring, S a multiplicative system in R, and M an R-module.

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EXERCISES 75

(a)

(b)

Show that if we set (a/s)(x/ t ) = ax/st, where a/s E S - lR and x l t E S-lM, then S-lM is an S-lR-module. Show that if M OR S-'R is made into an S-lR-module by the method of Exercise 7(b) of Chapter I, then the iso- morphism of Proposition 3.2, is an isomorphism of S - lR-modules.

2. Isomorphisms of modules of quotients. Let M and N be R-modules and let S be a multiplicative system in R. Verify the following isomorphisms of S-lR-modules. (a) S-l(M @ N ) z S-'M @ S-lN. (b) S-lMBRN 2 S-'(M@RN) z S-'MQS-1,S-lN

E M - OR S-lN.

3. Quotient module mappings. Let M and N be R-modules and let f: M+ N be a homo- morphism. Let S be a multiplicative system in R. Define S-l f : S-IM+ S- 'N by S-'f(x/s) = f ( x ) / s for all x E M , s E S. (a) Prove that S-lf is a S-'R-module homomorphism.

(b) If L is another R-module such that 0 + L 5 M 2 N + 0 is an exact sequence of R-modules, prove that the sequence

0 - S -lL 2 S -lM - S -lN - 0

is exact. Show that if M is a flat R-module, then S- lM is a flat S - lR-module.

s-lf

(c)

Properties of extension of ideals. Let A, B be ideals of R and let S be a multiplicative system in R. Prove the following formulas regarding extension of ideals. (a) S1(AB) = (S-lA)(S-lB). (b) S-l(A n B) = S-lA n S-lB.

(d) S-l(A: B) = S-'A: S-lB if B is finitely generated, or if A and B are contractions of ideals of S - l R [see also Exercise Z(e) of Chapter 1x1.

4.

(c) S-l(A + B) = S-1A + S-1B.

(e) S-l(Rad A) = Rad(S-lA).

(a) 5. Localization properties.

Show that if R is an integral domain and if P runs through the maximal ideals of R, then R = n R, .

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111 RINGS AND MODULES OF QUOTIENTS

Show that if A is an ideal of any ring R and if P runs through the maximal ideals of R, then A = n(ARp n R). If A is an ideal of a ring R such that Rad(A) is a prime ideal Q of R, and if AR, is QRp-primary for each maximal ideal P containing A, prove that A is Q-primary. Show that the assertion of (c) is no longer true if we drop the assumption that Rad(A) is prime. Let x be an element of a ring R such that x is contained in only a finite number of maximal ideals M,, . . . , Mn of R. Prove that if A is an ideal of R containing x and such that ARM, is finitely generated for i = 1, . . . , n, then A is finitely generated. Use (e) to prove that if R is a ring such that R, is Noether- ian for each maximal ideal M of R and each nonzero element of R is contained in only a finite number of maximal ideals of R, then R is Noetherian. If, in a ring R, each ideal with prime radical is a power of its radical, prove that this property holds for R, for each proper prime ideal P of R.

6. Saturation of a multiplicative system. Let R be a ring and let S be a multiplicative system in R. Let s be its saturation. (a) Prove that R \ S is the union of the prime ideals of R which

do not meet S . (b) Let {Mu I a E I} be the set of maximal ideals of S - lR and

let P, = Ma n R. Prove that S-IR = T-lR where T = R \ u P a . If R is an integral domain and P is a prime ideal of R not meeting S, show that R, = S-lRS-,,. If R is an integral domain, show that every ring of quo- tients S - lR of R is an intersection of localizations of R; in particular, S - l R = n RPu where {P,] is the set of ideals of R maximal with respect to not meeting S.

(c)

(d)

7. Transitivity of quotient ring formation. (a) Let R be a ring and let P and Q be proper prime ideals of

R such that P c Q. Prove that

R P E ( R Q ) P R Q *

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EXERCISES 77

(b) Let S be a multiplicative system in R. Let S’ be a multi- plicative system in S - lR and let 5’“ be the multiplicative system in R generated by

5’ u {s” I S“ E R and s“/s E S’ for some s E S}.

Show that S - ’ R 2 S‘-’(S-lR). The formulation of S in (b) is important. Show that the following result is false: Let S be a multiplicative system in R, S’ a multiplicatively closed subset in S I R . Let S” be the multiplicative system in R generated by S u {s” E RI?(s”) E S’}. Then S“ - lR S’-l(S-lR). [Hint. Take R= Z @ 2, S = ((2, 2)”ln 2 01, and S’ =

(c)

8. A tensor product. Let R be a ring, A an ideal of R, and M an R-module. Make RIA O R M into an R-module by setting a((6 +A) @ x) =

(a6 +A) @ x [see Exercise 7(b) of Chapter I]. (a) Show that there is a homomorphism 4: RIA O R M +

MIAM such that $((a + A ) @ x) = ax + A M for all a~ R and X E M .

(b) Show that there is a homomorphism #: M/AM+ RIA O E M such that #(x + A M ) = (1 +A) @ x for all x E M. Show that 4# and $$ are identity isomorphisms, so that 4 and # are group isomorphisms. Verify that they are, in fact, isomorphisms of R-modules . Let S be a multiplicative system in R. Treating R/A as an R-module, show that there is an isomorphism of R-modules S-l(R/A) g S-lR/S-lA. (Compare with Theorem 3.17. Are the isomorphisms the same?)

(c)

9. The prime divisors of an ideal. (a) Let R be a ring and A an ideal of R. Let

U(A) = {b I b E R and b + A is a regular element of RIA}.

Show that U ( A ) is multiplicatively closed and that U(A) n A is empty. An ideal P of R which is maximal

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78 111 RINGS AND MODULES OF QUOTIENTS

with respect to the properties A E P and U(A) n P is empty is called a max ima l p r i m e divisor of A.

(b) A prime ideal P of R is called a prime divisor of A if there is a multiplicatively closed set S in R such that A n S is empty and S-lP is a maximal prime divisor of S-lA. Show that a maximal prime divisor of A is a maximal element in the set of all prime divisors of A. Let P be a minimal prime divisor of A. Show that AR, n R is P-primary ; it is called the isolated P-primary com- ponent of A. Show that if A has a primary decomposition, then the isolated P-primary component of A is the P-primary ideal occurring in every reduced primary decomposition of A. Show that a minimal prime divisor of A is a minimal element in the set of prime divisors of A. Show that if A has a primary decomposition, then the prime divisors of A, as defined in the text after Proposition 2.25, are the same as the prime divisors of A, as defined in this exercise.

(c)

(d)

(e)

10. Families of quotient rings. Let {S, I a E I } be a family of multiplicative systems in a ring R such that for each maximal ideal P of R there is an a E I such that P n S, is empty. (a) Show that if A and B are ideals of R then A = B if and

only if S; lA = S i l B for all a E I . (b) Show that if, for each a E I , S, consists entirely of regular

elements, then R = na I S;lR. (c) Let M and N be R-modules and f: M + N a homo-

morphism. Show that f is surjective (respectively, injective, bijective, the zero homomorphism) if and only if the same is true for S; 'f for all a E I . If P be a proper prime ideal of R, let S ( P ) be the set of regular elements of R\P. Then S ( P ) is a multiplicative system in R, and we denote S(P)- lR by RS(P) . Show that if A and B are ideals of R, then A = B if and only if A&(,, = BRso, for every regular maximal ideal P of R. Show that if P runs over the set of regular maximal ideals of R, then R = 0 Rso,.

(d)

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EXERCISES 79

11. The support of a module. Let R be a Noetherian ring and Ma finitely generated R-module. The support of M is the set Supp(M) = {PI Pis a proper prime ideal of R and S- lM f 0 where S = R\P) .

Show that if M # 0, then Supp(M) is not empty. Show that a proper prime ideal of R is in Supp(M) if and only if it contains an element of Ass(M). Conclude that Ass(M) E Supp(M), and show that these two sets have the same minimal elements. Show that there is a chain of submodules of M , M = Mo x M l =I =I M,,=O, such that for i = O , ..., a - 1 , M,/Mi+lz R/P , for some prime ideal Pi of R. Show that Ass(M) G {Po, . . . , Pn-l} c Supp(M), and that the mini- mal elements of these sets coincide. Show that these minimal elements are precisely the minimal ones among the proper prime ideals of R which contain Ann(M)= n A ~ W .

12. Ass(M) and primary submodules. Let R be a Noetherian ring and M a finitely generated R-module. Let A be an ideal of R. (a) Show that AnM = 0 for some positive integer n if and only

if A c P for every P E Ass(M). (b) Let N be a submodule of M. Show that N is primary

submodule if and only if Ass(M/N) consists of a single prime ideal P, and in this case P = Rad(N). Show that if N is a submodule of M , then Ass(M/N) is the set of prime divisors of N .

(c)

13. Further properties of Ass(M). (a) Let R be a Noetherian ring and M a finitely generated

R-module. Let S be a multiplicatively closed set in R and let F be the prime ideals of R which do not intersect S. Show that the mapping P t--, S-lP is a one-to-one mapping from Ass(M) n F into AssfS-lM) (S- IM being con- sidered as an S-lR-module). Show that if PE F and S - ~ P E Ass(S-lM), then P E

Ass(M). Show that the mapping in part (a) is onto. Show that the mapping x H sx/s from M into S - l M , where

(b)

(c)

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80 111 RINGS AND MODULES OF QUOTIENTS

s E S, is independent of s, and is a homomorphism; let N be its kernel. Let 9' = Ass(M) n Y. Show that N is the unique submodule of M such that Ass(M/N) = Y' and Ass(N) = Ass(M)\Y'.

14. The ring R(X). Let R be a ring and X an indeterminate. Let S be the set of all a, + a,X + - * + a, X" E R[X] such that R = (ao, a,, . . . , a,).

(a) Show that S is a multiplicative system in R[X], consisting entirely of regular elements. Denote the ring S- lRIX]

Show that if A is an ideal of R, then R(X)/AR(X)

Let Q be a P-primary ideal of R. Show that QR(X) is a PR(X)-primary ideal of R(X), and that QR(X) n R = Q. Let A,, . . . , A, be ideals of R. Show that

by R(X).

(R/A)(X). (b)

(c)

(d)

(A, n -. . n A,)R(X) = A,R(X) n * - * n A,R(X).

Show that an ideal M ' of R(X) is maximal if and only if M ' = MR(X) for some maximal ideal M of R. Let A be a proper ideal of R. Show that AR(X) is a proper ideal of R(X). Show that R ( X ) is a flat R-module.

(e)

(f)

(g)

Totally ordered sets of ideals. Let T be an overring of a ring R and let M be a proper prime ideal of T. Let P = M n R and assume that the set of ideals of R, is totally ordered. Prove that the set of ideals of TM is totally ordered. [Hint. T M is an overring of h(Rp), where h : R, -+ T M is given by h(x/y) = x/y.]

15.

16. Symbolic powers of primary ideals. Let P be a proper prime ideal of a ring R and let Q be a P-primary ideal of R . If n is a positive integer then Q(,) = QnRp n R is called the nth symbolic power of Q. (a) Show that Q(,) is P-primary and that if Qn is primary

then Q(n) = 9". (b) Show that P is the unique minimal prime divisor of Qn. (c) Let m and n be positive integers. Show that P is the

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EXERCISES 81

unique minimal prime divisor of Q(m)Q(n), and that the isolated P-primary component of Q(m)Q(n) is Q'm + ").

Rings which are almost Noetherian. A ring R is said to be almost Noetherian if the localization R, is Noetherian for each maximal ideal M of R.

17.

Show that if R is almost Noetherian, then (ACC) holds in the set of prime ideals of R. [H,'nt. Let Pl G P, c be an ascending chain of prime ideals of R and set

Suppose that R is almost Noetherian and that each finitely generated ideal of R has only finitely many minimal prime divisors. Show that if P is a proper prime ideal of R, then P is the unique minimal prime divisor of some finitely generated ideal of R. Suppose that R is almost Noetherian. Show that R is Noetherian if and only if each finitely generated ideal A of R has only a finite number of minimal prime divisors and Rad(A) is finitely generated. Prove that for an almost Noetherian ring R the following statements are equivalent : (1) R is Noetherian ; (2) Each finitely generated ideal of R is an intersection of

a finite number of primary ideals of R ; (3) Each finitely generated ideal of R has only a finite

number of prime divisors. [Hint. To prove that (3) implies (l), use the fact that if P is the unique minimal prime divisor of a finitely generated ideal A and if P R p = ( x l , . . ., xn)RP, then P is the unique minimal prime divisor of B = A + (xl, . . . , x,) and BRp = PR, .] Show that conditions (2) and (3) in (d) and the condition of (c) can hold in a ring R without R being Noetherian. [Hint. Consider the ring of polynomials in a countable number of indeterminates over a field.]

P= UPi . ]

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C H A P T E R

IV Integral Dependence

1 DEFINITION O F INTEGRAL DEPENDENCE

Let R' be a ring and let R be a subring of R'. If a,, . . . , a, E R' we denote by R[a,, . . . , a,] the set of polynomial expressions in a, , . . . , a, with coefficients in R. Thus, if X,, . . . , X, are indeterminates, then

w,, - . - 9 a,] = {f(a,, - * * 9 a,) I f(X1, * - * 9 X,) E R[X,, - * * , X,l>. The mapping f(Xl, . . . , X,)H~( a,, . . . , a,) is a homomorphism from R[X,, . . . , X,] into R' ; consequently its image R[a,, . . . , a,] is a subring of R'. It is clear that R G R[a , , . . . , a,].

4.1. Proposition. Let R be a subring of a ring R' and let a E R'. Then the following statements are equivalent :

(1) There are elements b, , b,, . . . , b, E R (n >_ 1) such that

(2) (3)

b, + b,a+ + bn-,an-l + an = 0. R[a] is afinitely generated R-module. There is a subring R" of R' such that a E R" and R is a finitely generated R-module.

Proof. Suppose (1) holds and let f ( X ) E R [ X ] . Suppose degf(X) = d > n: Iff (X) = c, + c,X + * . + cd Xd, then

f ( a ) = ~ , + c , a + ~ ~ ~ + c , _ , a ~ - ~ +c,ad-n(--bo -b,a--.----b n - 1 an- l )

= c,' + cl'a + . * . + c;-,u"-'.

82

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1 DEFINITION OF INTEGRAL DEPENDENCE 83

If necessary we repeat this argument until we see finally thatf(a) is in the R-module generated by 1, a, . . . , an. Thus, as an R-module, R[a] = R1 + Ra + * + Ran ; hence (2) holds. Clearly, (2) implies (3), taking R" = R[a]. Now suppose (3) holds and let a,, . . . , a, gener- ate R" as an R-module. For i = 1, . . . , n,

n

j = l aat=Cbijaj, bij E R ,

or

C (b , - aij a)a, = 0. , = l

If d = det[bij - aija] then du, = 0 for j = 1, , . . , n by Lemma 2.22. Hence dc = 0 for all c E R . With c = 1 we get d = 0. Since d is a polynomial in a with coefficients in R such that the coefficient of an is

1, we see that (1) holds.

4.2. Definition. Let R be a subring of a ring R' and let a E R'. If the equivalent statements of Proposition 4.1 hold for a, we say that a is integral over R. If every element of R' is integral over R we say that R' is integral over R. If the elements of R are the only elements of R' which are integral over R, we say that R is integrally closed in R'. If R is integrally closed in its total quotient ring, we say simply that R i s integrally closed.

4.3. Proposition. Let R be a subring of a ring R' and let

R, = {a I a E R' and a is integral over R} . Then R, is a subring of R and R c R, .

Proof. Clearly R c R, . Let a, b E R, . Then R[a] is a finitely generated R-module and R[a, b] = R[a][b] is a finitely,generated R[a]-module. Hence R[a, b] is a finitely generated R-modul;; Since a - b, a6 E R[a, b] they are integral over R ; that is, a-b, ab E R,. Therefore, R, is a subring of R'.

In the notation of this proposition, R, is called the integral closure of R in R'. It follows from the next proposition that R, is integrally closed in R'.

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84 Iv INTEGRAL DEPENDENCE

4.4. Proposition. Let R be a subring of a ring R', and R' a subring of a ring R". If R' is integral over R and if a E R" is integral over R', then a is integral over R.

Proof. Suppose b, + b,a + + bn-'an-l + an = 0 where b,, bl, . . . , b,,-' E R'. Then a is integral over R[b,, . . . , b,-,]. Therefore R[b, , . . . , bn-,, a] is a finitely generated R-module. It follows that a is integral over R.

4.5. Proposition. Let R be a subring of R' and let S be a mult+licative system in R. Then S-lR may be considered as a subring of S-IR' and if R' is integral over R, then S - 'R' in integral over S - lR.

Proof. Let 0, and Os' be the S-components of 0 in R and R , respec- tively. We certainly have 0, c 0,' n R. If a E 0,' n R, then sa = 0 for some s E S , and since S G R this implies that a E 0,. Thus 0, =

0 , ' ~ R. Therefore the mapping taking a/s E S - l R onto aIs E S-'R' is an injective homomorphism; if we identify a/s E S-IR with its image als E S-lR', then S-'R may be considered as a subring of S-lR'. Now suppose that R' is integral over R and let aIs E S-IR', a € R', SES. There are elements b o , b,, ..., bn- , E R such that 6 , + b,a + -. + bn-,an-l + an = 0. Then

60/sn + (b,/s"-l)(a/s) +. -. + (bn-l/S>(@/S)n-' + (a/s>n

= (6, + b,a + . * * + b, -,an + a n ) p = 0.

Therefore a/s is integral over S-'R.

2 INTEGRAL DEPENDENCE A N D PRIME IDEALS

Now we are ready to prove some of the principal theorems con- cerning integral dependence. These theorems deal with the relations between the ideals of a ring R and the ideals of a ring R' which has R as a subring and is integral over R. If A is an ideal of R and A' is an ideal of R' such that A =A' n R, then A' is said to lie over A.

4.6. Theorem (The Lying-Over Theorem). Let R be a subring of a ring R' which is integral over R. If P is a prime ideal of R, then there

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2 INTEGRAL DEPENDENCE AND PRIME IDEALS 85

is a prime ideal P' of R' that lies over P. Moreover, if P' and P are prime ideals of R' that lie over P and if P' G P", then P' = P .

Proof. The set of ideals A' of R' such that A' n R s P is not empty, and it follows from Zorn's lemma that this set has a maximal element P'. Then P' n R E P. Suppose P' n R c P and let a E P, a 4 P'. Then P' c P ' f R'a and consequently, by our choice of P', (P' + R'a) n R $ P. Hence there is an element c E P', and an rER', s u c h t h a t c + r a = b $ P b u t b E R . L e t d , ,..., d, , - ,ER be such that do + dlr + - a - + dn-,rn-' + rn = 0. Then

b " + d , - , a b n - l + . . * +d,an-%+d,a"

= (c+ra)"+d,_,a(c+ra)"-' + . . . + d , a n - l ( c + r a ) +doan

= f ( c ) ~ P ' n R c P ;

= j ( c ) + U y P + d,-,rn-l + - * - + d,r +do)

heref(c) is a polynomial in c with coefficients in R'. Hence, since a E P, we have 6" E P, so b E P, a contradiction. Therefore we must have P' n R = P.

Next, we show that P' is a prime ideal. Let S = R\P; then S is a multiplicative system in R'. (If P = R, then P' = R' which is prime, so we may assume P # R.) Let A' be an ideal of R with P' c A'. Then A' n R $ P so A' n R meets S; hence A' n S is not empty. Thus P' is maximal in the set of ideals of R' whose intersection with S is empty. We know that such an ideal is prime (see Proposition

Now suppose that P' and P" are prime ideals of R' that lie over P and that P' c P". Choose a E P" with a 4 P'. Since a is integral over R there is a least positive integer n such that there are elements 6, , . . . , b,,-l E R for which an + bn-,an-l + . - *+ b,a+ b, E P'. Then b, EP" n R = P = P' n R.Hence

2.12).

a(an-l + b, + * . * + b,) E P',

but a $ P', so an-l + 6, choice of n. Thus, if P' c PI', we must have P' = P".

+ 1 * - + b, E P'. This contradicts our

4.7. Corollary (The Going-Up Theorem). Let R and R' be as in Theorem 4.6. Let Po c P, c . . . c P, be a chain of prime ideals of R. If the prime ideal Po' of R' lies over Po, then there is a chain

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86 Iv INTEGRAL DEPENDENCE

Po' c P,' c - - * c P,' of prime ideals of R' such that Pi' lies over P , for i = 0, . . . , r. If, for a given i, there is no prime ideal of R strictly between Pi and Pi +,, then there is no prime ideal of R' strictly between Pi' and Pi+ 1.

Proof. Suppose that 0 2 k < r and that we have shown that there is a chain Po' c c Pk' of prime ideals of R' such that Pi' lies over Pi for i = 0 , . . ., k. Then RIPk may be considered as a subring of RIP,' which, in turn, is integral over RIP, [Exercise 2(a)]. Hence, by Theorem 4.6, there is a prime ideal P ; + , of R' such that Pk' c PLt and Ph + 1/Pk' lies over P, +l /Pk. Then PL + lies over Pk +l. This proves the first assertion of the corollary. Suppose that P' is a prime ideal of R' and that Pi' c P' c Pi+ , . B y Theorem 4.6, P' cannot lie over either Pi or P i t , . Hence the prime ideal P' n R of R is strictly between Pi and Pi +

It is natural to ask whether, in the notation of the corollary, we can start with a prime ideal P,' of R' lying over P, and show the existence of a chain Po' c - * * c P,' of prime ideals of R', such that Pi' lies over Pi for each i. The answer is that we can do this, under somewhat more restrictive hypotheses than those of the going-up theorem.

Let R be an integral domain and let K be the quotient field of R . Let K' be a normal algebraic extension of K and let R' be the integral closure of R in K'. By Proposition 4.3, R' is a subring of K ' and R 5 R'. Let G(K' /K) be the group of K-automorphisms of K'. If a E R' and u E G(K'/K) then u(a) E R'. For, if bo + b,a + * . -

+ b, + an = 0, where 6 0 , . . . , 6 , -1 E R, then

0 = U(b0 + b,a + - * * + bn-lan-l + an) =b,+b,u(a) + - - - + b , - , ~ ( a ) ~ - l +u(a)".

Furthermore a = u ( c l ( a ) ) . Thus u(R') = R' for all u E G(K' /K) . Let P be a prime ideal of R and let P' be a prime ideal of R' lying over P. Then u(P') is a prime ideal of R' and u(P') also lies over P. By Theorem 4.6 either cr(P') = P' or P' $ u(P') and u(P') $ P'.

4.8. Theorem. Let R be an integrally closed integral domain and let K, K', and R' be as above. If the prime ideals P' and P of R' lie over the same prime ideal of R then P = u(P') for some u E G(K'/K).

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Proof. Let P‘ n R = P” n R = P. First, we assume that K’ is a finite extension of K and we let G(K’/K) = {ul, . . . , a,}. Further, we assume that P #u,(P’) for i = 1, . . ., n. Then P $ ui(P’) for i = 1, . . . , n and so, by Exercise 5(c) of Chapter 11, there is an a E P” such that a 4 u,(P’) for i = 1, . . . , n. Then ui(a) 4 P‘ for i = 1, . . . , n. If d is the degree of inseparability of K/K, then

Since Risintegrally closed, R‘ n K = R. Therefore, NKt,,(a) E P” n R = P G P’, a contradiction.

Now let K ’ be of arbitrary degree over K . Let Y be the set of all pairs ( T , 7), where T is a subring of R‘ with R c T , T is integrally closed, the quotient field L of T in K’ is normal over K, and 7 is a K-automorphism of L such that 7(P‘ n T ) =PI‘ n T (note that P’ n T and P n T are prime ideals of T lying over P). The set Y is not empty since (R, i) belongs to Y , where i is the identity isomor- phism of K. Partially order 9 by writing ( T , 7 ) 5 (T’, 7’) if T G T’ and T’(a) =.(a) for all a E T. By Zorn’s lemma Y has a maximal element which we denote henceforth by ( T , 7). If T = R’, we are finished. If T # R’, let a E R’, a 4 T . Let L be the field of quotients of T in K‘. Since K’/K is normal, there is a normal extension L’ of K such that L c L’, a EL‘, and L‘IL is finite. Let T‘ be the integral closure of T in L‘. Then T’ is integrally closed, and T c T’ since a E T’ but a 4 T . The K-automorphism 7 can be extended to a K-automorphism of L’ which we also denote by 7. Then P‘ n T‘ and 7-l(P1’ n T’) are prime ideals of T‘ both lying over P‘ n T, and so by the first part of the proof there is an L-automorphism p of L’ such that p(P’ n T’) = 7-l(P” n T ’ ) ; then 7p (P’ n T’) = P” n T’. Hence (T’, 7p) is in 9, and ( T , 7) < (TI , 7p), which contradicts our choice of ( T , 7). Therefore we must have T = R .

4.9. Theorem (The Going-Down Theorem). Let R be an integrally closed integral domain and let R’ be a ring such that :

(i) R is a subring of R’ ; (ii) R’ is a integral over R ; and (iii) no nonzero element of R is a zero-divisor in R‘.

Let Po c P, c * * c P, be a chain of prime ideals of R and let P,’ be a prime ideal of R’ Iring over P, . Then there is a chain Po’ c PI‘ c - - c

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88 Iv INTEGRAL DEPENDENCE

P,' of prime ideals of R' such that Pir lies over P, for i = 0, . . . , r . If Po = 0 and if P' is a prime ideal of A' with P' n R = 0 and P' G P,', then there is such a chain with P,' = P'.

Proof. We may assume that P, is a proper prime ideal of R. Let S = {abl a E R, a # 0, and b E R'\P,']. Then S is a multiplicative system in R' because of (iii), Let P' be a maximal element in the set of ideals of R' that do not intersect S . Then P' is a prime ideal and P' E F,'. Furthermore, if a E R, a # 0, then a E S and so a $ P'; thus P' n R = 0. Therefore we may replace R' by R'IP' and assume from the start that R' is an integral domain. Then, to prove the theorem, it is sufficient to verify the first assertion under the assump- tion that Po = 0.

Let K and K' be quotient fields of R and R', respectively, with K G K'. We leave it to the reader to show that K' is an algebraic extension of K. Let f(" be a normal algebraic extension of K such that K' 5 K", and let R" be the integral closure of R in K"; then R 5 R' 5 R". By the lying-over theorem, there is a prime ideal P," of R" which lies over P,' and a prime ideal Qo of R" which lies over Po = 0. By the going-up theorem there is a chain Qo c Q1 c * c Qr of prime ideals of R such that Q, lies over P, for i = 0, . . . , r . Since P: and Q,. both lie over P, there is a K-automorphism u of K" such that P:' = ~($3,). Let Pi' = a(Q,) n R' €or i = 0, . . . , r . Then, for each i,

P,' n R = o(Qi) n R = Q, A R = P i .

3 INTEGRAL DEPENDENCE A N D FLAT MODULES

If R is a ring and K is its total quotient ring, then any ring T such that R G T G K is called an overring of R. If R is an integral domain and T is an overring of R, then T is an integral domain and K is the quotient field of T as well as of R.

Let T be an overring of R . If A is an ideal of R then

AT = (finite sums a, t i I uj E A, t , E T }

is an ideal of T . In fact, it is the ideal of T generated by A.

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4.10. Proposition. If T is an overring of an integral domain R, then the following statements are equivalent :

(a) For every prime ideal P of R, either P T = T or T E R,. (b) ( ( y ) : (x))T = T for all x/y E T.

Proof. First, we shall assume that (a) holds. Let x/y E T and suppose that ( ( y ) : (x ) )T # T. Then there is a prime ideal P of R such that ( y ) : ( x ) c P and PT # T . By (a), T G R, and consequently x /yE R,. Thus x /y=a/s where s $ P ; hence xsu=yau for some u E R\P. This implies that su E ( y ) : ( x ) G P, a contradiction. Thus we do have ( ( y ) : (x))T = T .

Conversely, assume that (b) holds. Let P be a prime ideal of R and suppose that PT # T. Let x/y E T. Then ( (y ) : (x))T = T and so ( y ) : (x) $ P. Let s E ( y ) : (x), s $ P. Then sx = ay for some a E R and consequently x/y = a/s E R, . Thus T G R, .

Note that if A,, . . . , A, are ideals of R such that Ai T = T for n A,)T= T . For if 1 < i < R , there are i= 1 , . .., R, then (A, n

elements aij E Ai , t i j E T , f o r j = 1, . . . , a, , such that 1 = ailti, + * + ainitini . Then

and each term on the right-hand side is the product of one element from each of the Ai and R elements of the form t,, . Hence

1 E (A, n - - - n A,)T,

which gives the desired equality.

4.11. Definition. An overring T of a ring R is a flat overring if T is a flat R-module.

4.12. Proposition. A n overring T of an integral domain R is a$at overring i f and only if it satisjies the equivalent conditions of Proposition 4.10.

Proof. Suppose that T satisfies (b) of Proposition 4.10. By Exercise ll(b) of Chapter I and Proposition 1.18 in order to show that T is a

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90 Iv INTEGRAL DEPENDENCE

flat R-module it is sufficient to show that for an arbitrary ideal A of R the homomorphism A 0 T + T given by a 0 b H ab is injective.

Let c E A B R T ; then c = CS=, a, 8 b,, a, E A , 6 , E T . There are elements 6, cl, . . . , c, E R such that 6, = ci /6 for i = 1, . . . , s; hence c = C;=l ai 0 c46. By assumption, ((6): (ci))T = T for i = 1, . . . , s, and so if C = nfZl ( 6 ) : (ci), then CT = T . Now suppose that the image of c is 0, that is, C;=l aici/6 0. If d E C we have dc,/b E R for i = 1, . . . , s; hence

S S

dc = c a, 0 dci/b = c (da, cJb 0 1) t = 1 i = l

= (df$?ci /6) 0 1 = O .

Therefore cC = 0 and consequently 0 = OT = cCT = cT. But c E cT, so c = 0. Thus T is a flat R-module.

Conversely, suppose that T is a flat R-module. We shall show that (a) of Proposition 4.10 holds. I f x/y E T then y(x/y) - xl = 0. Hence by Exercise 13(c) of Chapter I, there are elements zjk E R, j = 1, . . . , r, k = 1, 2, and elements b,, . . . , 6, E T , such that

zj1 Y - Z f Z x = 0, j = 1 , ..., r.

Let P be a prime ideal of R. If zj2 E P fo r j = 1, . . . , r , then PT = T . Suppose zj2 $ P for somej; then (y ) : (x) $ P. Thus, we conclude that either PT = T or (y) : (x) $ P for all x/y E T . Suppose the latter. Let x/y E T ; then there are elements a E R and s E R \ P such that ay = sx, or xly = a/s E R, . Therefore T c R, .

4.13. Proposition. Let T and T' be overrings of an integral domain R with T c T ' .

( 1 ) (2)

If T ' is a flat overring of R, then T ' is a flat overring of T . If T' is a flat overring of T and T is a flat overring of R, then T' is afEat overring of R.

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Proof. Suppose that T' is a flat overring of R. Let a, b E T be such that a/b E T'. Write a = c/s and b = d/s, where c, d , s E R. Then c/d E T' and so ((4 : (c))T' = T' by Proposition 4.12. Hence 1 = tlul + * *

+tkuk, where t , , ..., tkE T', ul,. ,., U k E R, and u , c ~ ( d ) for i= l , ..., K. Then utaE Tb for i = 1, ..., k, so ( T b : Ta)T'= T'. Therefore T' is a flat overring of T, again by Proposition 4.12. The second assertion of the proposition is a consequence of Exercise 9(d) of Chapter I.

4.14. Proposition. If T is an overring of an integral domain R, then the following statements are equivalent:

(1) T is a $at overring of R. (2) T p = R, for all maximal ideals P of T. ( 3 ) T = O R p R , where P runs over all maximal ideals of T .

Proof. (1) + (2). Suppose that T is a flat overring of R and let P be a maximal ideal of T . Certainly we have R, ,, c T p . Let x / y E T p , where x, y E T and y $ P; then there are elements u, v, s E R such that x = U / S and y = v/s. Let C= (s) : (u) n ( s ) : (v). By Proposition 4.12, C T = T ; hence C $ P n R. Let z E C, z $ P n R. Then zx , z y E R and xy f# P, so zy + P n R. Hence x / y = z x / z y E R, R . Thus

(2) + (3). By Exercise 5(a) of Chapter 111, T = 0 T, , where P runs over all maximal ideals of T ; hence, if (2) holds, so does (3).

(3) + (1). Suppose that (3) holds and let Q be a prime ideal of R such that QT # T . Then QT G P for some maximal ideal P of T and Q c P n R . Hence R P n R ~ R P . But T s R p n R SO that T s R , . Thus T is a flat overring of R.

Tp G R P n R .

4.15. Theorem. Let T be an overring of an integral domain R . If T is both integral over R and a $at overring of R, then T = R.

Proof. Let x/y E T ; then ( (y) : (x))T= T by Proposition 4.12. If P is a proper prime ideal of R, then by the lying-over theorem there is a proper prime ideal P' of T lying over P . Since PT 5 P', we must have PT # T . It follows that ( y ) : (x) is not contained in any proper prime ideal of R ; hence (y) : (x) = R. Thus x = ay for some a E R. Then x / y = ay/y = a E R . Thus T = R.

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4 ALMOST INTEGRAL DEPENDENCE

Let R be a subring of a ring R‘ and let a E R’. By the third of the equivalent conditions of Proposition 4.1, a is integral over R if there is a subring R” of R’ such that a E R” and R” is a finitely generated R-module. An example referred to below shows that it is necessary to insist that R be a subring of R’ and not just a submodule of the R-module R‘. However, this weaker condition turns out to be an important one, as we shall see in Chapter VIII, and is worthy of investigation.

4.16. Definition. Let R be a subring of a ring R’. A n element a E R is almost integral over R if there exists a jinitely generated submodule of the R-module R’ which contains all powers of a .

Clearly, an element of R‘ which is integral over R is also almost integral over R. The converse is not true; an example is given in Exercise 11. However, the converse is true if R is Noetherian [see Exercise 1 (c)] .

4.17. Definition. Let R be a subring of a ring R’. The set R, of elements of R‘ which are almost integral over R is the complete integral closure of R in R’. If R, = R, the R is completely integ- rally closed in R. If R is completely integrally closed in its total quotient ring, then we say simply that R is completely integrally closed.

It is immediately clear that the complete integral closure R, of R in R‘ is a subring of R’. However, R, is not necessarily itself com- pletely integrally closed; an example is given in Exercise 14. Further- more, if R, T , and T‘ are rings with R s T c T’, then an element a E T may be almost integral over R as an element of T’ , but not almost integral over R as an element of T ; an example is given in Exercise 13.

Even though the complete integral closure of one ring in another may not be completely integrally closed, we do have the following:

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4 ALMOST INTEGRAL DEPENDENCE 93

4.18. Proposition. Let R be a subring of a ring R' and let R, be the complete integral closure of R in R'. Then R, is integrally closed in R'.

Proof. Let x E R' be integral over R,; say xm + a , , , - , ~ ~ - ~ + * *

+ a, = 0, where a,, . . . , a, E R, . Then x is integral over the ring

R[ao, * - . , a m - , ] *

For i = 0, . . . , m - 1 , a , is contained in some finitely generated sub- module of the R-module R' ; call it Mi = Rxil + - * * + Rxiki , where each xij E R'. Then R[a,, . . . , a,-,] G M,. . * M m - , , which is the submodule of the R-module R' generated by all products

XojoXl j l * ' * Xm-1, im-l

where each j , runs between 1 and k, . Hence x is contained in

R[x] E R[a,, . . . , a,-,, x]

m-1

h = O = C R[a,, . . . , a,-,]xh

which is a finitely generated submodule of the R-module R'. Thus x is almost integral over R, and consequently x E R, . Therefore R, is integrally closed.

4.19. Corollary. Let R, R,, and R, be rings such that R 5 R, c R,. If evevy element of R, is almost integral over R, and if R, is integral over R,, then every element of R, is almost integral over R.

We have the following useful characterization of the complete integral closure of a ring.

4.20. Theorem. The complete integral closure of a ring R with total quotient ring K is

R, = {x I x E K and there exists a regular element r E R such that rxn E R for all positive integers n}.

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94 Iv INTEGRAL DEPENDENCE

Proof. Let x be an element of K which is almost integral over R. Then there exist elements kl, . . . , K, E K such that R[x] G C;= RK, . If r is the product of the denominators of the his, then rxn E R for all positive integers n. Conversely, if x E Ro , say rxn E R for all positive integers n and some regular element r E R, then R[x] 5 Rr-l. There- fore x is almost integral over R.

4.21. Corollary. If R is completely integrally closed, then for all nonunits x E R, n;= (x") consists entirely of zero-divisors.

Proof. Let K be the total quotient ring of R. If x is a regular element of R then x has an inverse in K . If r E nrZl (x") then ~ ( x - l ) " E R for all n. If r is a regular element of R then x-l is almost integral over R and so belongs to R, contrary to assumption. Hence r is a zero- divisor. If x itself is a zero-divisor the assertion is clearly true.

EXERCISES

1. Integral dependence. Let R be a subring of a ring R' and let a E R'. Let A be an ideal of R and suppose that there is a finitely generated sub- module M of the R-module R such that aM G AM, and in addition, if bM= 0 for some b E R[a] then b = 0. Show that there exist a positive integer n and elements c o , c l , . . . , c , - ~ E A such that co + cla + * * + c, + an = 0. Show that a is integral over R if and only if there is a finitely generated submodule M of the R-module R such that aM 5 M and bM= 0 for some b E R[a] implies

Show that if R is Noetherian then a is integral over R if and only if R[a] is contained in some finitely generated sub- module of the R-module R'.

b = 0 .

2. On integral extensions. (a) Let R be a subring of a ring R'. If A' is an ideal of R and

A = A' n R, we may consider RIA as a subring of RIA'. Show that if R' is integral over R, then RIA' is integral over RIA.

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EXERCISES 95

(b) Show that if R and R' are integral domains and if R' is integral over R, then R' is a field if and only if R is a field.

(c) Let R be a subring of R' and assume R' is integral over R. Let M' be a maximal ideal of R' and let M = R n M'. Prove that RL, is not necessarily integral over R , . [Hint. Consider R = K [X - 11 and R' = K [ X ] for a field K. Let M' = (X - l)R'.]

(a) Suppose R' is integral over R and that P' is a prime ideal of R lying over a prime ideal P of R. Without reference to the lying-over theorem, show that P' is maximal if and only if P is maximal. [Hint. Use Exercise 2(b).]

(b) Let R, R', P, and P' be as in (a). Show that there is no ideal A' of R', prime or otherwise, such that A' n R = P and P' c A'.

(c) Show that if R' is integral over R, and if A is a proper ideal of R, then AR' # R'.

(d) Let R' be integral over R and let A be an ideal of R. Show that Rad(AR') = (a I a E R' and there is a positive integer n and elements b, , . . . , b,

(e) Suppose that R is an integral domain and is integral over R. Show that if A' is a nonzero ideal of R' then A' n R # 0.

3. Integral extensions and ideals.

E A such that b, + b,a + * - + bn- lan- l + an = 0).

4. Local integral closure. Let R be an integral domain. Prove the equivalence of the following assertions ; that is, prove that integral closure of an integral domain is a property of localizations.

(1) R is integrally closed. (2) R, is integrally closed for each proper prime ideal P

of R. (3) R, is integrally closed for each maximal ideal M of R.

Let the ring R be a subring of the ring R'. Let P be a prime ideal of R. Show that there is a prime ideal of R' lying over P if and only if P R n R = P .

(b) Let R be an integral domain and F any field having R as a subring. Let P be a prime ideal of R. Show that if a E F,

5. Simple ring extensions. (a)

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96 Iv INTEGRAL DEPENDENCE

a # 0, then either R[a] or R[l/a] contains a prime ideal lying over P.

6. The pseudo-radical of a ring. Let R be an integral domain and let Rad*(R) be the intersection of all of the nonzero prime ideals of R; Rad"(R) is called the pseudo-radical of R. (a) Let R be a subring of an integral domain R'. Show that if

the conclusion of the lying-over theorem holds for R and R', then Rad"(R) # 0 implies that Rad"(R') # 0. Give an example to show that the converse is not true.

(b) Let K be the quotient field of an integral domain R. Show that the following statements are equivalent : (1) Rad*(R) # 0. (2) K = R[a] for some a E K. (3) K = R[a,, . . . , a,] for some a,, . . . , a, E K.

7. Integrally closed domains. Let R be an integral domain. (a) Prove that if R is a unique factorization domain, then R is

integrally closed. (b) Let S be a multiplicative system in R. Show that if R is

integrally closed, then S - lR is integrally closed. (c) Let P be a prime ideal of R. Show by an example that we

may have R integrally closed but RIP not integrally closed.

8. Integral closure and polynomial rings.

(a) Suppose that R is an integrally closed domain. Let K be the quotient field of R and letf(X) be a monic polynomial in R[X]. Show that iff(X) =g(X)h(X), where g(X) and h ( X ) are monic polynomials in K [ X ] , then g(X), h ( X ) E R[X]. [Hint. Take a field L containing K such that in L[X], f(X) and g(X) can be factored into linear factors:

(b) Prove the result of (a) without assuming R is an integral domain.

(c) Let R be a subring of a ring R' and let R, be the integral closure of R in R'. Prove that R,[X] is the integral closure of R[X] in R'[X]. [Hint. Suppose f + r,,-Jn-l + * - .

+ ro = 0 with rl E R[X], and f E R'[X]. Pick an integer

f ( x ) = n ( X - g(X) = JJ(x- bi).]

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EXERCISES 97

m> {n, deg ro , . . . , deg r , Considerg(X) = f ( X ) - X m . Put this in the integral expression forfand simplify. Apply Exercise 8(b).]

9. Integral closure and field extensions. Let R be an integral domain with quotient field K and let L be an algebraic extension of K. Let T be the integral closure of R in L. Let T o = T n K. Prove the following: (a) To is the integral closure of R. (b) L is the field of quotients of T . (c) If x E L is integral over R andf(X) is the minimal polyno-

mial over K which has x as a root, then f ( X ) E T o [ X ] . Furthermore, the ideal of T o [ X ] consisting of those poly- nomials which have x as a root is principal. (Note that this says that if R is integrally closed, x E L is integral over R if and only if f ( X ) E R[X] . )

A proof of the going-down theorem. (a)

10. Let R and R' be as in Theorem 4.9. Let P and Q be proper prime ideals of R with Q c P and let P' be a prime ideal of R' lying over P. Let S = {rsl r E R\Q and s E R'\P'}. Show that S is multiplicatively closed and that S n QR' is empty.

(b) Let %'f = {A'[ A' is an ideal of R', QR' c A', and S n A' is empty}. Let Q' be a maximal element of W (Q' exists by Zorn's lemma). Show that Q' is a prime ideal of R', that Q' c P', and that Q' n R = Q.

11. Almost integral closure. Let R be an integrally closed integral domain with quotient field K. Let T = R + X K [ X ] . (a) Prove that T is integrally closed. ~.

(b) Prove that K [ X ] is the complete integral closure of R in K [ X I *

12. Completely integrally closed rings. (a) Prove that each unique factorization domain is completely

integrally closed. (b) If R is completely integrally closed and S is a multiplicative

system in R, prove that S - lR is completely integrally closed.

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98 Iv INTEGRAL DEPENDENCE

13. On almost integrity. Let R, T,, and T , be rings such that R c Tl E T, . For i = 1,2, let Ri be the complete integral closure of R in Ti. Clearly, Rl G Ra n T,. (a) Show that if T, is a submodule of some Tl-module M such

that Tl is a direct summand of M , then R, = R, n TI, (b) Show that the same conclusion holds if every finitely

generated Tl-module M with TI G M G T2 is a sub- module of a Tl-module of which Tl is a direct summand.

(c) Show that R, = R, n TI if T, is a principal ideal domain. (d) Let K be a field and let X and Y be indeterminates over K.

LetR=K[{XY"In>l}], T,=R[Y],and T 2 = T,[l/X]. With R, and R, as above, show that R, c R, n TI.

The complete integral closure need not be completely integ-' rally closed. Let K be a field and let X and Y be indeterminates. Let

14.

In 2 011. R = K [{XZnfl y n ( Z n + 1)

(a) Show that the quotient field of R is K(X, Y ) . (b) Show that if R = K [{XUn I n 2 O } ] , then R c R' 5 R" G

K[X, Y], where R" is the complete integral closure of R. (c) Show that Y is almost integral over R', and hence is

almost integral over R", but that Y $ R". [Hint. Show that for any element of R, the exponent of Y in any of the monomials of that element is less than or equal to the square of the exponent of X in the same monomial.]

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C H A P T E R

V Valuation Rings

1 THE DEFINITION OF A VALUATION RING

5.1. Definition. A valuation ring is an integral domain V with the property that i f A and B are ideals of V then either A c B or B s A .

5.2. Proposition. For an integral domain V the following statements are equivalent :

(1) (2) (3)

V is a valuation ring. If a, b E V then either (a) c (b) or (b) c (a). If x belongs to the quotient field K of V , then either x E V or x-1 E v.

Proof. That (1) implies (2) is clear. Assume that (2) holds and let x belong to K . Then x = a/b where a, b E V. If (a) G (6) then a = br where r E V ; hence x 7 br/b = r E V. I f (b) c (u) then x f 0 and x - l = b/a E V by the same argument. Finally, suppose (3) holds and let A and B be ideals of V. Suppose A $ B and choose a E A with a $ B. If b E Band b # 0 then a/b $ V , for if a/b E Vthen a E (b) c B. Hence b/a E V and b E (a) E A. Thus B 5 A.

5.3. Corollary. Each overring of a valuation ring is a valuation ring.

99

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100 v VALUATION RINGS

5.4. Proposition. If V is a valuation ring then the nonunits of V form an ideal of V , which is the unique maximal ideal of V.

Proof. Let P be the set of nonunits of V . Let a, b E P and c E V. Then ac is a nonunit of V , that is, ac E P . We may assume that a/b E V . Then a- b= (a/b- l ) b ~ P. Thus P is an ideal of V . If A is a proper ideal of V then every element of A is a nonunit, so A C_ P. Thus P is the unique maximal ideal of V .

5.5. Proposition. Valuation rings are integrally closed.

Proof. Let V be a valuation ring and let K be its quotient field. Let x E K be integral over V , say

X n + an-lXn-l + * * - + a, = 0,

for a,- 1, . . . , a, E V . If x 4 V , then x - l E V . Hence

x = - ( U , , - ~ + ~ , , - ~ X - ~ + * * - + a , ~ - ~ + l ) ,

and thus x E V , a contradiction.

Let K be a field. A partial homomorphism of K is a homo- morphism + from a subring K, of K into an algebraically closed field. Let Y be the set of all such pairs (4, K,). Define a partial ordering - < on 9’ by writing

($5 Kd 2 (h K,)

if K* E K , and t,h is an extension of + to K,, that is, +(a) = +(a) for all a E K,. Zorn’s lemma guarantees that if (I$, Km) E 9, then there is a maximal element of 9’ which is greater than or equal to (9, Km). Maximal elements of Y are called maximal partial homomorphisms. We shall prove below that (+, Km) is a maximal partial homomorphism if and only if K, is a valuation ring with quotient field K .

Let F be the set of all pairs (V, P ) , where V is a subring of the field K and P is a proper prime ideal of V . Define a partial ordering < on F by writing

(VIP PI) G (VZ 9 PZ)

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1 THE DEFINITION OF A VALUATION RING 101

if V , G V 2 and P, n V , = P,. I f this is the case, we say that (V , , P2) dominates (V,, P,). Zorn's lemma guarantees that if (V , P ) E 9, then there is a maximal element of which dominates (V , P ) , Maximal elements of .Y are called maximal pairs. We shall prove below that ( V , P ) is a maximal element of F if and only if V is a valuation ring and P is the maximal ideal of V .

5.6. Lemma. Let x be a nonzero element of a jield K. Let V be a subring of K and let P be the unique maximal ideal of V . Then either P V [ x ] # V [ x ] or P V [ x - l ] # V [ x - ' ] .

Proof. Suppose that P V [ x ] = V [ x ] and P V [ X - ~ ] = V [x-'1. Then there exist a,, . . . , a,,, , b,, . . . , b, E P such that

(1)

and

a, +a,x + * * * + a,xm= 1

(2) b, + b,x-l + . * a + b , x - " = 1.

Choose m and n as small as possible, and suppose m 2 n. Multi- ply (2) by xn. This gives

(1 - b0)x" = blx"-l+ . * . + b, . Now since 6 , E P and P is the unique maximal ideal of V , 1 - 6 , is a unit in V . Thus

~ " = ( 1 -bO)-lb,xn-l + . * * + ( l -bO)-'b,

= clxn-l + * . * + c,,

where c,, . . . , c, E P. Use this in (1) :

1 = a, +a,x + * . . +a, - ,xm-1 + amxm-yclxn-l + ..* + c,) = do + d,x + -. * + dm - 1 x m - 1

and d o , . . . , d,-, E P. This contradicts our choice of m.

5.7. Theorem. Let K be ajield and let V be a subring of K. Then the following statements are equivalent :

( 1 ) ( 2 )

V is a valuation ring with quotient jield K. V has a maximal ideal P such that (V, P ) is a maximal pair.

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102 v VALUATION RINGS

(3) There exists a homomorphism # from V into an algebraically closedFeld such that (4, V ) is a maximal partial homomorphism.

Proof. (1) +(2). Assume that V is a valuation ring with quotient field K . Let P be the unique maximal ideal of V. Suppose that (V, P)< (V', P'). Let x be a nonzero element of V'. If x 4 V, then x - l E P C P', which is absurd since P' is a proper ideal of V'. Thus x E V ; so V' = V and P'= P. Therefore, (V, P ) is a maximal pair.

(2) + ( 3 ) . Assume that P is a maximal ideal of V and that (V, P ) is a maximal pair. Let L be the algebraic closure of the field V / P , and let # be the canonical homomorphism from V into L. If (#I, V ) 5 (#', V'), and if P' is the kernel of 6, then P n V = P, so that (V, P) < (V' , P'). Hence V = V'. Thus, (4, V ) is a maximal partial homomorphism.

(3) + (1). Assume that there exists a homomorphism # from V into an algebraically closed field L such that (4, V ) is a maximal partial homomorphism. Let P be the kernel of #; since #(1) = 1, P is a proper prime ideal of V. If s E V\P then #(s) is a unit in L ; hence # can be extended to a homomorphism #' from V , into L by setting $'(a/s) = #(u)#(s)-' for all a k V, s E V\P. Then (+, V ) 5 (#', Vp) , and so we have V,= V. It follows that P is the unique maximal ideal of V.

Let x be a nonzero element of K . We shall show that either x or x - l is an element of V. By Lemma 5.6, we can assume that PV[x] # V [ x ] . Then there exists a maximal ideal M of V [ x ] such that PV[x] G M ; then M n V = P since P is a maximal ideal of V. Hence the mapping a : V /P+ V [ x ] / M given by a(a + P ) = a + M is an injective homomorphism. Clearly, we have V [ x ] / M = u( V / P ) [ x + MI. Since V [ x ] / M is a field, this implies that x + M is algebraic over a(V/P) . Thus, if s: VIP+ L is given by $(a + P ) = #(a), then 6a- l : a( VIP) + L can be extended to an injective homo- morphism I,!J : V [x]/M -+ L. Let 7~ : V [x] -+ V [x]/M be the canonical homomorphism. Then (#I, V ) 5 (+r, V [ x ] ) , since for all a E V we have

$..(a) = #(a + M ) = $(a + p ) = #(a).

By the maximality of (4, V ) we have V [x] = V. Thus, x E V, which is what we wanted to show. This implies that K is the quotient field of V.

This theorem has an interesting and useful corollary.

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1 THE DEFINITION OF A VALUATION RING 103

5.8. Corollary. Let R be an integral domain with quotient Jield K. The integral closure of R is the intersection of all of the valuation rings of K containing R.

Proof. Since valuation rings are integrally closed by Proposition 5.5, the integral closure of R is certainly contained in the intersection of all valuation rings of K containing R . Conversely, let x be an element of K which is not integral over R . Then x 4 R[x-l] . For, suppose that x E R[x- l ] . Then there is a polynomial f ( X ) E R [ X ] , of degree n, such that x = f ( x - l ) ; then x n + l - x"fx-') = 0, which means that x is integral over R, contrary to assumption. Thus, x-, is not a unit in R[x-'], so there is a maximal ideal P of R[x- l ] such that x - l E P. Let L be the algebraic closure of R[x-']/P. The canoni- cal homomorphism R[x-l] + R[x-l] /P furnishes us with a homo- morphism T: R [ x - l ] + L . Let (+, V ) be a maximal partial homomorphism of K into L such that (T, R[x-l]) I(+, V ) . By Theorem 5.7, V is a valuation ring of K, and R G V . Since + ( x - l ) = 0 we must have x 4 V. Thus x is not in the intersection of all valuation rings of K containing R. This establishes the desired equality.

For Noetherian rings, valuation rings can be characterized by much weaker conditions than for arbitrary rings.

5.9. Theorem. Let V be a Noetherian integral domain which is not a field. Then the following statements are equivalent:

(1) (2) (3)

V is a valuation ring. The nonunits of t ' form a nonzero principal ideal. V is integrally closed and has exactly one nonzero proper prime ideal.

Proof. (1) + (2). Let V be a valuation ring and let P be its ideal of nonunits. Then P # 0 since V is not a field. Since V is Noetherian, P is finitely generated, say .P = ( a l , . . . , u,J. We may assume (a , ) c (a2) c * . * 5 (a,). Then P = (a,).

(2) + ( 3 ) . Assume that (2) holds and let P be the ideal of non- units of V . Then P is the only maximal ideal of V and P = ( a ) where a # 0. By Cdrollary 2.24, n:= Pn = 0. If A is any nonzero proper ideal of V , then A c P, so there is a positive integer n such that

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104 v VALUATION RINGS

A c _ P n a n d A g P " + l . L e t b ~ A , b $ P n + l ; t h e n b=anuwhereu is a unit in V. Now let c E P". Then for some d E V , c = and = h - l d E A. Therefore A = P" = (an). It follows immediately from this that P is the only nonzero proper prime ideal of V.

Now let c be an element of the quotient field of V and write c = r/s, where r , s E V ; assume c # 0. Since every nonzero element of V is a unit times some power of a, we may assume that either r is a unit or s is a unit. Suppose that c is integral over V . Then there are elements b , , . . . , 6,- E V such that 6, + blc + ... + bn-,cn-l + c" = 0. Multiplying by S" we obtain

~ ~ b , + s ~ - ~ b , r + ~ . . + s b , - , r " - ~ + r n = O ,

rn = -s(sn-lbO + sn-%,r + - . . + bn-,rn-l) .

If s is a unit in V, then c E V. If s is not a unit in V , then r" E P, so r E P ; that is, r is not a unit. This is contrary to our assumption. Therefore, c E V and we conclude that V is integrally closed.

(3) + (1). Assume that (3) holds. It is sufficient to show that the unique nonzero proper prime ideal P of V is principal. For once this is done, a repetition of the argument given in the proof that (2) implies (3) will show that every nonzero ideal of V is a power of P. Thus the ideals of V are totally ordered under inclusion, and conse- quently V is a valuation ring.

T o prove that P is principal, we consider P* = (X I x E K and x P E V } where K is the quotient field of V . Then P*P is an ideal of V such that P G P*P E V . Hence P*P = P or P*P = V . We will show below that P*P# P ; hence we have P*P= V. Then there are elements a,, . . . , a, E P and bl, . . . , b, E P* such that a,b, + . * -

+ a, b, = 1. Hence, for some i, ai b, $ P ; that is, there are elements a E P , b E P * such that ab=u , where u is a unit in V. Then a b u - l = 1. Let c E P ; then c = abcu-l. Since bc E V, c E (a). Hence P = (a).

Now we show that P*P# P . Suppose P*P = P . Let P= (a,, . . . , a,). If a E P" then aP E P, so

or

k

j = 1 aai = C r i j aj , i = l , ..., k, T i j E v.

Then If=, (6,,a - ri,)aj = 0, i= 1, . . . , k. Hence, since we are working in the quotient field of V and at least one a, # 0, we have

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2 IDEAL THEORY IN VALUATION RINGS 105

det[&,,a - rt,] = 0.

Thus, a is integral over V and so belongs to V. Therefore P* = V. T o complete the proof we shall show that we cannot have P* = V .

Let a E P, a # 0, and let S = {a"l n is a positive integer}. Then S-IV is the field of quotients of V. For, suppose it is not. Then S-lV has a nonzero maximal ideal P' . Since a is a unit in S-lV, we have a $ P' ; hence P' n V # P and consequently P' n V = 0. However, this is not true, for if clan E P' and clan # 0, then c # 0 and c E P' n V . Thus every element in the field of quotients of V can be written in the form b/an for some b E V.

Now let c E V , c # 0. Then l / c = b/a" and so an = cb E (c). Thus, for each a E P, some power of a is in (c). Since P is finitely generated, it follows that P" G (c) for some smallest positive integer n. Let d E PR-I , d 4 (c). Then dP c (c); that is, ( d / c ) P s V. Thus, d/c E P", but d/c $ V , so that P" # V.

I n the course of this rather long proof we have shown that if V is a Noetherian valuation ring and if P is its ideal of nonunits, then P is principal and every nonzero proper ideal of V is a power of P. Since P is maximal, each such ideal is P-primary [Exercise 6(e) of Chapter 111.

2 IDEAL THEORY IN VALUATION RINGS

In this section we obtain several results concerning the ideals of valuation rings.

5.10. Theorem. Let V be a valuation ring and let A be an ideal of V.

(1) Rad(A) is a prime ideal of V , (2) If B = fl;= A", then B is a prime ideal of V which contains

every prime ideal of V which is properly contained in A.

Proof. (1) Rad(A) is the intersection of the minimal prime divisors of A by Proposition 2.14. But since the set of ideals of V is totally ordered, A has only one minimal prime divisor, which must coincide with Rad(A).

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106 v VALUATION RINGS

(2) Let a and b be elements of V such that a $ B, b $ B . Then a $A", b 6 A"' for some positive integers n and m. Thus A" c (a) and A" c (b), so A"(b) G (a)(b). Actually, we have A"(b) # (a)(b). For, since A" c (a) , there exists x E V such that x a $ A". If A"(b) = (a)(b), then there exists Y E A " such that yb=xab. But this means that xa =y E A", contrary,to our choice of x ; thus A"(b) c (a)(b). Hence A"+"' c A"(b) c (ab), and so ab 4 A"+". Therefore, ab $ B, and we conclude that B is a prime ideal.

If P is a prime ideal of V such that P c A, then P contains no power of A . Hence P c A" for each positive integer n. Thus P E B.

We now turn to the primary ideals of a valuation ring.

5.11. Theorem. Let P be a proper prime ideal of a valuation ring V.

(1) If Q is P-primary and x E V\P, then Q = Q(x). (2) The product of P-primary ideals of V is a P-primary ideal.

i f P # P2, then the only P-primary ideals are powers of P . ( 3 ) The intersection of all P-primary ideals of V is a prime ideal

of V and there are no prime ideals of V properly between it and P.

Proof. (1) Since x $ P, Q c (x). Let K be the quotient field of V and let A = {y 1 y E K and yx E Q). Since Q c (x), A is a subset of V. Furthermore, it is easy to check that A is an ideal of V and Q = A(x). Moreover, since Q is P-primary and (x) $ P , we have A G Q. Thus Q = A and Q = Q(x), as claimed.

(2) Let Ql, Q2 be P-primary ideals of V. Clearly Rad(QlQ2) = P . Letx,ybeelementsof Vwithxy E QlQ2andx $ P . By(l), Q1 = Q1(x). Hence xy E (x)Q1Q2. Since V is an integral domain, this implies that y E QlQ2. Thus Q1Q2 is P-primary.

Now suppose that P # P2 and let Q be a P-primary ideal of V. By Exercise 5(c) , Q contains a power of P2, and so contains a power of P. Thus, there is a positive integer m such that P" E Q but Pm-l $ Q. Let x E Prn-l, x $ Q ; then Q c (x). If we define A as in the proof of (l), then Q = A(x). Since Q is P-primary and x Q, A E P . Therefore, Q = A(x) c P ( x ) E P"', and so we conclude that Q = P".

(3) If P is the only P-primary ideal of V, there is nothing to

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3 VALUATIONS 107

prove. Suppose there is a P-primary ideal Q of V with Q # P, and let {Qa} be the set of all P-primary ideals of V. By (2), 9" is P-primary for each n 2 1 ; hence na Qa c p. However, by Exercise 5(c), each Qa contains a power of Q ; thus na Qa = , Qn. Hence, by (2) of Theorem 5.10, na Qa is a prime ideal of V which contains every prime ideal of V which is properly contained in Q. If P' is a prime ideal of V properly contained in P , then Q $ P ' , so P' c Q; hence P' G na Q a .

3 VALUATIONS

By an ordered Abelian group we mean an Abelian group G on which there is defined a total ordering 5 such that if a, 8, y E G and as B, then a + y 5 B + y . For example, the additive group of real numbers with the natural ordering of the real numbers is an ordered Abelian group. Each subgroup of this group (or, indeed, of any ordered Abelian group), with the induced ordering, is an ordered Abelian group.

Let G1, . . . , G, be ordered Abelian groups and let G = G, @ - *

@ G, . The elements of G may be denoted by n-tuples (al, . . . , a,), where a, E G, for i= 1, . . . , n. If (a1, . . . , a,) and (pl, . . . , B,) are distinct elements of G we write

(ai, * , a,) < (81, . - - , Bn) if ~ , < , 8 ~ or if, for some K > 1 , a i=& for i = l , ..., K - 1 and ak </Ik. We leave it to the reader to show that this is a total ordering on G and that G, together with this ordering, is an ordered Abelian group. We refer to this ordering as the lexicographic ordering of G.

Let G be an ordered Abelian group and let (00) be a set whose single element is not an element of G. Let G* = G w {a} and make G* into a commutative semigroup by defining, for a and in G",

their sum in G if EL, E G, .+a=(, if a = , or /3=co.

We extend the ordering of G to an ordering of G* by defining a 5 CQ

for all a E G". Then G* is an ordered semigroup in the sense that if a, B, y E GX and a I i ? then a+ y iB+ y.

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108 v VALUATION RINGS

5.12. Definition. Let K be a$eld. A valuation on K is a mapping v from K-onto G", where G is an ordered Abelian group, such that

(i) .(a) = co if and onZy if a = 0, (ii) v(ab) = .(a) + v(b) for all a, b E K,

(iii) v(a + b) 2 min{v(a), v(b)) for all a, b E K.

The group G is called the value group of the valuation v. The mapping v from a field K into G" given by .(a) = 0 if a # 0

and v(0) = 00 is clearly a valuation on K, called a trivial valuation. Let v be a valuation on a field K and set

'V={UIUEK and v(u)>O}.

If a, b E V , then v(ab) = .(a) + v(b) 2 0 and

v(a + b) 2 min{v(a), UP)} 2 0,

so that ab, a + b E V. Since v(-1) = v(1) = 0, and hence - 1 E V, we see that V is a subring of K. Let a E K, a # 0 ; if a 4 V then .(a) < 0, so v(l/a) = -v(a) > 0. Thus l/a E V. Therefore V is a valuation ring. Note that K is the quotient field of V. The maximal ideal of V is {a 1 a E K and .(a) > O}. We shall now show that all valuation rings are determined by valuations in this way.

5.13. Proposition. Let V be a valuation ring with quotient $eld K. Then there is a valuation v on K such that

V = { U ~ U E K and v ( u ) > O } .

Proof. Let U be the (multiplicative) group of units of V. Then U is a subgroup of K", the multiplicative group of nonzero elements of K. Let G = K+/U; we write G additively, so that aU + bU = abU. Define a relation on G by bU i aU if a /b E V . This is a well-defined relation. If bU = b'U and aU = a'U, then b'lb E U and a'la E U ; hence a / b E V if and only if a'/b' E V. It is easy to see that we have defined a partial ordering on G; it is a total ordering since if a, b E K", then either alb or bla is in V. Finally, G together with this ordering is an ordered Abelian group. For, if bU 5 aU and if CU is any element of G, then aclbc E V and so bU + cU = bcU 5 acU = aU+ c U .

Now, definev: K+G" by v(0) = co and .(a) = aUif a # 0. Then

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3 VALUATIONS 109

(i) and (ii) from the definition of valuation hold, and it follows that v(1) = 0. Furthermore, v(a + b) 2 min{v(a), v(b)} if either a = 0 or b = 0. Suppose a, b E K* and that b U ( a U . Then alb E V so a/b + 1 E V. Hence v(a/b + 1) 2 v( 1) = 0. It follows that

u(a + b) = v((a/b + 1)b) = v(a/b + 1) + u(b) 2 v(b) = min(u(a), v(b)}.

Therefore, v is a valuation on K , and

V = ( U J U E K and u / ~ E V ) = ( ~ J U E K and v(a)>O).

If V and v are related as in this proposition, we say that v is the valuation determined by V . If v is an arbitrary valuation on K, then {a I a E K and .(a) 2 0} is called the valuation ring of v.

Let v and v’ be valuations on a field K, with value groups G and G’, respectively. We say that v and v’ are equivalent if there is an order-preserving isomorphism + from G onto G’ such that v‘(a) =

+(.(a)) for all a E K*. This relation between valuations is reflexive, symmetric, and transitive. I t is clear that equivalent valuations have the same valuation ring.

Conversely, two valuations on a field K having the same valuation ring are equivalent. To verify this, we shall show that if v is a valuation on K, if Y is the valuation ring of v, and if v’ is the valuation deter- mined by V , then v and v’ are equivalent. Let G be the value group of v and U be the group of units of V. Define 4: G+ K*/U by +(v(u)) = aU. If .(a) = v(b), then v(a/b) = 1, so that a/b E U and aU = bU; thus is well-defined. Since +(.(a)) = v’(a) for all a E K*, it remains to show that + is an order-preserving isomorphism. Now, + is a homomorphism, since

+(.(a) + v(b)) = +(v(ab)) = abU = a u + b u = +(v(.)) + +(v(b)) ;

it is injective since +(v(u)) = 0 implies that a E U and so .(a) = 0, and it is clearly surjective. Finally, if .(a) 5 v(b), then b/a E V and consequently aU I bU, that is, +(.(a)) I +(v(b)).

Let G be an ordered Abelian group. A subgroup H of G is called an isolated subgroup if for each nonnegative element CL of H, 0 I /3 5 GI implies that /3 E H . If H is an isolated subgroup of G and H # G, then H is termed proper.

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5.14. Definition. If an ordered Abelian group G has only a jinite number of isolated subgroups, then the number of proper isolated sub- groups of G is the rank of G.

Thus, G has rank one if and only if G # 0 and G and 0 are the only isolated subgroups of G.

5.15. Proposition. A nonzero ordered Abelian group G has rank one if and only if there is an order-preserving isomorphism from G onto a subgroup of the additive group of real numbers.

Proof. Let G‘ be a nonzero subgroup of the additive group of real numbers; we shall show that G’ has rank one, thus showing the sufficiency of the condition. Let H be a nonzero isolated subgroup of G’. Then H contains a positive element u. If p is a positive element of G’, then by the Archimedian property of the ordering of the real numbers there is an integer n such that 0 < /3 < nu. But na E H , so B E H . Thus, H contains all positive elements of G’ and so must coincide with G’.

Conversely, suppose that G has rank one. We shall show first that if u and j? are positive elements, then there is an integer n such that / 3 s n u . Suppose that this assertion is not true for some pair of positive elements u and ,B of G. Let

S = { y I y E G, y 2 0, and y < na for some integer n}.

Then /3 4 S and if yl, y2 E S, then y1 + y 2 E S. Let H be the subgroup of G generated by S ; then H consists of all elements of G of the form yl- y 2 where yl, y2e S. Consider a positive element of H, say y1 - y2 where yl, y 2 E S , and suppose 0 5 6 I y1 - y z . If y1 5 nu, then since y z 2 0, we have 6 5 6 + y z 5 y1 5 nu; hence 6 E H . Thus H i s isolated, and H # 0, so H = G. But /3 4 H so we have arrived at a contradiction. Therefore, the assertion is true.

Suppose that the set of positive elements of G has a least element GI.

If /3 is any positive element then there is a positive integer n such that ( n - l ) a < , ! ? < n a . T h e n O < / 3 - ( n - l ) a I u , and s o p - ( n - l ) u = t(, or /3 = nu. If y is a negative element of G and if - y = mu, then y = (-m)a. Thus G is an infinite cyclic group, generated by u, and the mapping 4 such that +(nu) = n is the desired order-preserving isomorphism.

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3 VALUATIONS 111

On the other hand, suppose that the set of positive elements of G does not have a least element. Choose, and fix, a positive element a and set +(a) = 1. Let /3 be any other positive element of G and set

I @ ) = {m/n I ma 5 n/3}

.(B> = {.I/. I ma > .PI, and

where m and n represent positivejntegers. There exist positive integers m and n such that a 2 n/3 and /3 <ma. Then l / n E 1(/3) and m / l E u@), so 1(/3) and u(/3) are not empty. Suppose that m/n E 1(/3) and h/k E u@). Then ma 5 n/3 and k/3 < ha; thus mkcl < nha. Hence m% < nh and m/n < h/k. It follows that l (p ) , together with the set of nonpositive rational numbers, and u(/3) are the lower and upper classes, respec- tively, of a Dedekind cut of the rational numbers. Hence, they determine a (positive) real number +(/3). We set + ( O ) = O and if y is a negative element of G, + ( y ) = -$(-y). We leave it to the reader to show that # is the desired order-preserving isomorphism.

5.16. Definition. Let v be a valuation on a field K and let G be its valuegroup. Let V be the valuation ring of v.

(1) (2)

If G has rank n, then v and V have rank n. If G is cyclic, v is a discrete valuation and V is a discrete valuation ring.

Note that if TJ is a discrete valuation, then it has either rank one or rank zero; in the latter case, v is trivial.

It would be surprising if there were not some relationship between the ideal structure of a valuation ring and the group structure of the value group of the valuation determined by the valuation ring. We now explore this relationship.

5.17. Theorem. Let v be a valuation on afield K. Let G be its value group and let V be its valuation ring. Then thereo exists a one-to-one order-reversing correspondence between the isolated subgroups of G and the proper prime ideals of V.

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112 v VALUATION RINGS

Proof. Let 4 denote the set of isolated subgroups of G and let 9 be the set of proper prime ideals of V . For HE 4, set

n ( H ) = {XI x E V and v(x) $ H }

and for P E 9, set

K(P) = {a[ a E G and a $ v(P) and -a $ w(P)}.

Using the fact that a proper ideal of V is a prime ideal if and only if V\P is multiplicatively closed and the fact that

K(P) = {a I a E G and a and -u belong to v(V\P)},

one can easily verify that TI is a mapping of 4 into 9 and K is a mapping of 9’ into 4. Clearly T I and K are both order-reversing. We now show that they are inverse mappings of one another. If x E P E 9, when v(x) E v(P), so v(x) $ K ( P ) . Hence x E TI(K(P)) , which proves that P 2 ~ ( K ( P ) ) . On the other hand, X E V\P implies that v ( x ) $ n(P), and so v(x) E K ( P ) ; that is, x $ TI(K(P)) . Consequently, TI(K(P) ) G P . A similar argument can be used to show that for each H E 4, K(TI(H)) = H .

It is worthwhile to note that this theorem says that if a valuation ring has rank n, then there exists a chain 0 c P, c * * . c P, of proper prime ideals of V, but no longer such chain exists.

We now show that the only Noetherian valuation rings are those which are fields or which have rank one and are discrete.

5.18. Theorem. A valuation ring which is not a field has rank one and is discrete if and only if it is Noetherian.

Proof. Let V be a Noetherian valuation ring which is not a field. If P is the unique maximal ideal of V, then by Theorem 5.9, P is a nonzero principal ideal and P” = 0. Let P = (a). If b E V, b # 0, we can write uniquely b = uan, where u is a unit in V and n is a nonnegative integer. In fact, every element of K , the quotient field of V , can be written uniquely in this way if we allow n to be a negative integer. If U is the group of units of V, then it is easily seen that the mapping b U w n is an order-preserving isomorphism from K*/U onto the additive group of integers. Therefore, V has rank one and is discrete.

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3 VALUATIONS 113

Conversely, suppose that V is a valuation ring that has rank one and is discrete, and let v be a valuation on K having V as its valuation ring and the additive group of integers as its value group. Let A be a nonzero ideal of V. There is an element a E A such that v(a)= min{a(b) 1 b E A}. Let c E A, c # 0 ; then .(a) 5 v(c), so v(c/a) 2 0. Thus c/a E V and c E Va. Therefore, A = Va. It follows that V is Noetherian.

Now we show the role of complete integral closure in the theory of valuation rings. All valuation rings are integrally closed, but those that are completely integrally closed have rank one or are fields.

5.19. Theorem. Let V be a valuation ring which is not equal to its quotient field K. Then V is completely integrally closed if and on& if V has rank one.

Proof. Suppose that V is completely integrally closed. Let P be the maximal ideal of V and let Q be another proper prime ideal of V. If X E P\Q, then Q c (x") for each positive integer n. Hence Q s (x") = 0. Hence Q = 0, and P is the only nonzero proper prime ideal of V. Therefore V has rank one.

Conversely, suppose that V has rank one and let x E K\V. Then x-l E V and by Theorem 5.10, (x-") is a prime ideal of V. It is a proper ideal of V , so either it is the zero ideal or it is the maximal ideal of V. I n the latter case, ( x - ~ ) = (x-l) and it follows easily that x E V , contrary to assumption. Hence n:=l (x-") = 0. Let r be any nonzero element of V; there exists an integer n such that ( r ) $ (x-"), which implies that r $ (x-"), that is, rx" $ V. Therefore x is not in the complete integral closure of V by Theorem 4.20. Consequently V is completely integrally closed.

(x"). By Corollary 4.21,

We close this section with two examples of valuations that will prove useful later.

Let K be a field and let X be an indeterminate. Let p ( X ) be a nonconstant monic irreducible polynomial in K [XI. Iff(X)/g(X) is a nonzero element of the field K ( X ) of rational functions in X over K, we can write

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114 v VALUATION RINGS

f ( X ) / s ( X ) =P(X)"( fdX)/g,(XN,

where f l ( X ) and g l ( X ) are not divisible by p ( X ) and n is an integer which may be positive, negative, or zero, and is uniquely determined by f ( X ) / g ( X ) . Set

ZI(f (X)/g(X)) = n.

If we also set v(0) = co, then v is a valuation on K ( X ) . Clearly v has rank one and is discrete. It is called the p(X)-adic valuation on

Again let K be a field and X an indeterminate. Let w be a valuation K ( X ) .

on K. Iff ( X ) = a, + a,X + * + a, X" E K [XI , we set

w ' ( f ( X ) ) = min{w(a,) I i = 0, . . . , n>.

Then w' is a mapping from K [XI into the value group of w, and w' satisfies (i), (ii), and (iii) in the definition of valuation. Hence w' determines a valuation on K ( X ) [see Exercise 9(d)], which we also denote by w', and call the extension of w to K ( X ) . Note that the value groups of w and w' coincide; hence w and w' have the same rank. In particular, if w has rank one and is discrete, the same is true of w'.

4 PROLONGATION O F VALUATIONS

5.20. Definition. Let v be a valwtion on a jield K and let K' be an extension of K. A valuation v' on K' is a prolongation of v if there is an order-preserving injective homomorphism 4 from the value group of v into the value group of v' such that v '(a) = +(.(a)) for all a E K*, the multiplicative group of nonzero elements of K.

It is clear that if v' is a valuation on K' which is a prolongation of v, and if v'' is a valuation on K' which is equivalent to v', then v" is a prolongation of ZI.

5.21. Proposition. Let v be a valuation on a jield K, let K' be an extension of K , and let v' be a valuation on K' . Let V and V' be the valuation rings of v and a', respectively, and let P be the maximal ideal of V and P' that of V'. Then the following statements are equivalent:

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4 PROLONGATION OF VALUATIONS 115

(1) v' is a prolongation of v . (2) V ' n K = V. (3 ) P'n Y=P.

Proof. (1) 3 (2). Let v' be a prolongation of v ; then there is an order- preserving injective homomorphism t# from the value group of v into that of v' such that v'(a) = +(.(a)) for all a E K". Hence if a E K , then .(a) 2 0 if and only if d(a ) >_ 0. Therefore V' n K = V.

(2) + (3). If V'n K= V , then P'n V = P ' n V ' n K= P ' n K ; we shall show that P' n K = P. If a E P, a # 0, then l / a $ V ; hence l / a $ V' and it follows that a E P' n K. Conversely, if a E P' n K, then l / a $ V ' ; hence l / a $ V and consequently a E P. Therefore P = P' n K.

(1). Assume that P' n V = P . First we shall show that U' n K* = U, where U and U' are the groups of units of V and V', respectively. Let a E U' n K". Then a + P , so .(a) 5 0. Also l / a E U' n K" and it follows that -v(a) = v( l /a) 5 0. Hence .(a) = 0 and so a E U. On the other hand, if a E U, then a + P' so v'(a) 5 0. Likewise -v'(a) = v ' ( l /a ) S O . Consequently v'(a) = 0 and a E U' n KX. Note that we can now conclude that V s V'.

Since U' n K" = U , the mapping 4: K*/U+K'*/U' given by +(aU) = aU' is a well-defined injective homomorphism. If b U 5 aU, then a / b E V and we conclude that a /b E V ; hence bU' 5 aU'. Thus + is order-preserving. There is an order-preserving isomorphism sl, from the value group of v onto K*/U such that t,h(v(a)) = aU for all a E K", and an order-preserving isomorphism 4' from the value group of v' onto K * / U ' such that @(v'(a)) = aU' for all a E K'". Then i,h' -'+,h is an order-preserving injective homomorphism and

(3)

1G'-'t#$(.(u)) = $'-'+(uU) = $'-1(uU') = v'(u)

for all a E K. Thus D' is a prolongation of v.

532 . Theorem. Let v be a vabation on afield K, and Let K be an algebraic extension of K . If v' is a valuation on K which is a prolonga- tion of v, then v' has rank r if and only if v has rank r .

Proof. Let G and G' be the value groups of v and v', respectively. For purposes of this proof we may assume that G is a subgroup of G' and

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116 v VALUATION RINGS

that v'(a) = .(a) for all a E K. Let u E G' and let a E K' be such that d ( a ) = a. There are elements b, , . . . , b, E K, with b, = 1, such that b, + b,a + * - + b, a" = 0. If v'(b, a') # v'(bj ai) whenever i # j, then

CQ = v'(0) = min{o'(b,a')l i= 1, . . . , n> E G .

Hence, for some i and j with i > j we must have v'(b, at) = v'(b,aj), and bi # 0 # b , . Then ( i - j )u = o'(af-j) = o(b,/b,) E G. Thus every element of G'/G is of finite order, and so G' has rank r if and only if G has rank r by Exercise 14(a).

5.23. Corollary. Let V be a valuation ring with quotient field K. Let K be an algebraic extension of K and let R' be the integral closure of V in K . Let V be a valuation ring in K , with maximal ideal P', such that R' s V' and P' n R' is a maximal ideal of R'. Then V' has rank r if and on& if V has rank r .

Proof. If P" = P' n R then P" n V is the maximal ideal of V by Exercise 3(a) of Chapter IV. Hence P' n V is the maximal ideal of V, and the assertion follows from Proposition 5.21 and Theorem 5.22.

If K is an extension of K which is not algebraic over K, then a valuation on K which is a prolongation of a valuation of rank r on K can have rank greater than r . The next theorem considers this situation. It is stated in terms of valuation rings because of later applications.

5.24. Theorem. Let V be a valuation ring of rank r, let K be the quotient field of V , and let K' be the quotient field of the polynomial ring V [X1, . . . , X,]. Then there is a valuation ring V contained in K such that V = V' n K and X1, . . . , X , E V . Every such V' has rank which is 5 r + n, and at least one such V has rank r + n.

Proof. An induction argument shows that it is sufficient to prove the assertions when n = 1. Let v be the valuation determined by V and let v' be a valuation on K ( X ) which is a prolongation. [Note that K ( X ) is contained in the quotient field of V [ X I , and so it must be the quotient field of V [ X ] . ] Let G and G' be the value groups of v and v', respectively; we may assume that G is a subgroup of G'

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4 PROLONGATION OF VALUATIONS 117

and that ~ ’ ( u ) = v(u) for all a E R. If every element of G / G has finite order then v’ has rank 7 by Exercise 14(a). Suppose this is not the case; then there is a positive element a E G‘ such that ka g! G for all positive integers k. Let a E K ( X ) be such that d(u) = a. By the same argument as that used in the proof of Theorem 5.22, it follows that a is transcendental over K . Let 0’’ be the restriction of zl’ to K(u). If we show that w’‘ has rank 5 7 + 1, the same will be true of w’, since K ( X ) is algebraic over R ( u ) and v’ is a prolongation of v”.

Therefore, to prove w’ has rank 5 7 + 1 it is sufficient to do so under the assumption that w’(X) = a. First, we shall show that G’ = G + ( a ) , where ( a ) is the subgroup of G’ generated by a. It is clear that if G‘= G + ( a ) , then the sum is direct. Let f ( X ) = b, + b,X + . * * + b,X“ E K [ X I . Then v‘(biX*) # d(b ,Xy) if i # j . Hence

d ( f ( X ) ) = min((w(bi) + ia) i = 0, . . . , n> E G + (a). Thus, G‘= G + (a) .

Let 0 c H I c - * * c H, be a chain of distinct proper isolated subgroups of G’, and consider the chain 0 c H , n G c * * c H, n G. If H i $ G then H i n G c H , + , n G . For, let y = p + p u ~ H , \ G and y’ = 8’ + qa E H i + l\Hi, where p, 8’ E G and p , q are integers. Since y’ 4 H i we have pf g! H , , and so py‘ ~ q y E Hi+,\Hi . Hence, sincepy’ - q y =pp’ - qj? E G, we have

PY‘ -97 E (Hi+ 1 n G)\(Hi n GI*

If Hi $ G for i= 1, . . . , s, then

O c H , n Gc * * * c H , n G s G .

Otherwise, let j be the largest integer such that H j c G ; then

0 c HI c - * - c H , c H,+, n G c H,,, n G c * . . c H, n G c G.

If H , n G # G, then in either of these two cases we have s 5 7 ; hence s + 1 5 r + 1. Suppose, on the other hand, that Hs n G = G. We claim that H, = G. If H, # G then H,/G is isomorphic to a subgroup (Iza) of ( a ) . Then G‘/H, z (G‘/G)/(H,/G) z (a)/(hcr), which is a finite nontrivial group. Since H, is a proper isolated subgroup of G’ this is impossible. Since H , = G, j = s, and

0 c H , c * .. c H, = G.

Thus s s r , and s f 1 < r + 1 . It follows that G’ has rank < r + 1.

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118 v VALUATION RINGS

To complete the proof of the theorem we shall show that there is a valuation o’ on K ( X ) such that o‘ has rank r + 1, o‘ is a prolonga- tion of o, and o’(X) > 0. Let 2 be the ordered additive group of integers and let G’ = G @ 2. Order G lexicographically; by Exercise 14(b), G‘ has rank r + 1. Let w be the X-adic valuation on the field of quotients of the polynomial ring ( V / P ) [ X ] , where P i s the maximal ideal of V . Letf(X) E V[X],f(X) # 0. Writef(X) = dfl(X), where d E V , fl(X) E V [ X ] , and at least one coefficient of f l ( X ) is a unit in V . Then d is uniquely determined to within a unit factor. Let fl(X) be obtained by replacing each coefficient off,(X) by its canonical image in VIP. Set d ( f ( X ) ) = v ( d ) + w ( f , ( X ) ) . We shall leave i t to the reader to show that v’ is a well-defined valuation on K ( X ) having value group G’. Clearly it is a prolongation of o, and so the theorem is proved.

EXERCISES

1. Overrings of valuation rings. Let V be a valuation ring. (a)

(b)

Show that if V’ is an overring of V , then there is a prime ideal P of V such that V’ = V , . Prove that the set of overrings of V is totally ordered.

2. Valuations and homomorphisms. Let o be a valuation on a field K and let V be its valuation ring. Let M be the maximal ideal of V . (a) Prove that if C# is an isomorphism of a field L onto K, then

vC# is a valuation on L with valuation ring $-I( V ) . (b) Let T be the canonical homomorphism from V onto VIM.

Let V’ be a valuation subring of VIM. Prove that W= r-l( V‘) is a valuation ring and that W , = V. Show that if V has rank Y and V’ has rank r‘, then W has rank r + r‘.

(c)

3. Existence theorem for valuation rings. (a) Let R be a subring of a field K and let P be a nonzero

prime ideal of R. Prove that there exists a valuation ring V of K which contains R and which has a prime ideal M lying over P, that is, with M n R = P . [Hint. Consider the

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EXERCISES 119

set of all subrings T of K which contain R and are such that PT f T , and pick a maximal element of this set.] Extend (a) as follows: let R be a subring of a field K and let 0 c PI c - a - c P,,, be a chain of prime ideals of R. Prove that there is a valuation ring V of K containing R such that V has prime ideals M I , . . . , M,,, which lie over P I , . . . , P,, respectively.

(b)

4. Places and valuation rings. Let K be a field and consider the set K u {a}, where OD

represents an element not in K. Define a + w = + a = OD

for a E K u {m} and am = ma = a for nonzero a E K U {a}. Now let K and I? be fields. A mapping + : K u {a} -+ I? U {a} which preserves addition and multiplication and is such that d( 1) = 1 is called a place of K having values in R. (a) Let 4 be a place of K. If we set 1/0 = OD and 1/.0 = 0, show

that #(O)=O, +(a) = w, and +(l/a) = l/+(a) for all

Let V be a valuation ring with quotient field K. Let P be the maximal ideal of V and set R = VIP. Define +: K u {m} -+K u {a} by setting +(a) = a + P if a E V and +(a) = a if a $ V. Show that + is a place of K having values in R. Let + be a place of K having values in R. Let V = (a I a E K and #(a) E R). Show that V is a valuation ring and that the maximal ideal of V is {a I a E K and #(a) = O}.

Let V be a valuation ring with quotient field K. Let A be an ideal of V . (a) Show that if A is finitely generated, then A is principal. (b) Show that every prime ideal of V which is properly

contained in A is contained in every power of A. (c) Show that if B is an ideal of V with A c Rad(B), then B

contains a power of A.

Primary ideals of valuation rings. Let V be a valuation ring and let P be a nonzero proper prime ideal of V. (a) Show that if there is a finitely generated P-primary ideal

of V , then P is the maximal ideal of V ,

U € K U {a}. (b)

(c)

5. Ideals of valuation rings.

6.

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120 v VALUATION RINGS

(b) Show that the following statements are equivalent : (1) There is an ideal A of V such that A # P and

Rad(A) = P. (2) P is the radical of a principal ideal. (3) P is not the union of the chain of prime ideals

properly contained in P. (4) There is a prime ideal M of V such that M c P and

such that there are no prime ideals M' of V with M c M ' c P.

(5) There exists a P-primary ideal of V distinct from P. A prime ideal which satisfies these conditions is said to be branched. If P is the only P-primary ideal of V, then P is said to be unbranched.

7. Noetherian valuation rings. (a) Let R be a Noetherian integral domain which is not a

field and which has a unique maximal ideal M . Prove that the following statements are equivalent : (1) R is a valuation ring. (2) As a vector space over RIM, M / M 2 has dimension

one. (3) Every nonzero ideal of R is a power of M . Let R be an integral domain and let P= (a) be a nonzero principal prime ideal of R such that n:= P" = 0. Show that R, is a Noetherian valuation ring.

(b)

8. Primary ideals of a rank one valuation ring. Let V be a rank one valuation ring and let P be its maximal ideal. Let Q and QI be P-primary ideals of V . (a) (b)

( c )

(d) (e)

Show that n:==l Q" = 0. If Qn=Qn+l for some positive integer n, show that

If Q G Q1 c P, show that Qln c Q for some positive integer n. Show that if Q c P , then Qz c QP. If Q c Ql, show that Q : Q1 = Q implies that Q1 = P = P2.

Q = Qa = P.

9. Valuations. Let z, be a valuation on a field K .

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EXERCISES 121

(a) (b)

(c)

Show that if .(a) # v(b), then v(a + 6) = min(v(a), v(b)}. Show that if K is a finite field, then .(a) = 0 for all non- zero a E K. Let G be an ordered Abelian group, and let w be a mapping from a field K onto G* such that w(a) = co if and only if a = 0, w(ab) = w(a) + w(b) for all a, b E K , and w(a) >_ 0 implies that w( 1 + a) 2 0. Show that w is a valuation on K. Let R be an integral domain and K its field of quotients. Let G be an ordered Abelian group. Let w’ be a mapping from R into G* which satisfies the conditions in the definition of valuation. Show that there is a unique valuation w on K such that w(a) = w‘(u) for all a E R.

Let p be a prime integer. For each nonzero integer a we can write uniquely a =pna’, where a’ is not divisible by p . Set v,(a) = n, and set v,(O) = co. Show that there is a valuation on the field of rational numbers which coincides with v, when restricted to 2. We denote this valuation by v, ; it is called the p-adic valuation. Show that each nontrivial valuation on the field of rational numbers is equivalent to up for some prime p.

(d)

10. Valuations on the field of rational numbers. (a)

(b)

Valuations and places on K ( X ) . Let K be a field and X an indeterminate.

11.

Let p(X) be a nonconstant monic irreducible polynomial in K [ X ] . Prove that the p(X)-adic valuation on K ( X ) is actually a valuation. (See p. 114 for definition.) Define v on K [XI by v(0) = co and v(f(X)) = -degf(X) if f ( X ) # 0. Show that there is a unique valuation on K ( X ) whose restriction to K [ X ] coincides with v. Show that each nontrivial valuation v on K(X) such that v(a) = 0 for every nonzero a E K is equivalent to some p(X)-adic valuation or to the valuation of part (b). Let w be a valuation on K. Show that the extension of w to K ( X ) is actually a valuation. (See p. 114, for definition.) Let K be a field and let a E K. If X is an indeterminate define 4: K ( X ) u (a> -+ K u (m> as follows: +(a) = co and i f f (X) E K ( X ) then (b ( f (X) ) =f(u) . Show that # is a place on K ( X ) having values in K , and that V , as defined

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122 v VALUATION RINGS

in Exercise 4(c), is given by V = K[X] , , where P = ( X - u).

12. Valuations with prescribed value groups. Let G be an ordered Abelian group. In this exercise a valuation will be constructed having G as its value group. Let K be a field. Let {&lg E G} be a set of indeterminates. Define amappingvfrom K[{X,lg E GI] into G" as follows: v(0) = 00 ; if m = aX,"i * a - Xi: is a nonzero monomial, then v(m) = Cf=, nigr; and if f = m1 + e . 0 + mh , where m,, . . . , m, are distinct nonzero monomials,

then v(f) = min{v(mi)(i = 1, . . . , h). It is now necessary to verify that v satisfies (ii) and (iii) of Definition 5.12; it will follow from Exercise 9(d) that there is a valuation having G as its value group. Verify the following assertions. (a) If m,, . . . , mh are nonzero monomials, then v(m, + * * * + n t h )

2 min{v(mi) I i = 1 , . . . , h}.

v(f2)}. This verifies (iii).

[Hint. Write fi = m, f . . f m, and f2 = m,' + . * + m,' as sums of distinct monomials, where

v(f,)=v(m,)=...=Zl(mh) < v ( m k + l ) s * . * s v ( m , )

(b) If f h f 2 E K[{X,lg E GI19 then .(fi +fi) 2 min{v(f,),

(c) If f i , f 2 E,K[{X,Ig 6 GI19 then 7 4 f J - 2 ) = v(f1) + u(f2).

and

v(fi)=v(m,')=...=v(m,') < v ( m ; + , ) ~ * * - < v(ms'}.

Set f," = m, + -- . + m, and f2* = m,' + consider f1f2 =f1"f2" + ( f1f2 -fl"f2").]

+ m,', and

13. Valuation ideals. An ideal A of an integral domain R with quotient field K is a valuation ideal if there is a valuation ring V of K containing R such that A = B n R for some ideal B of V . If v is the valuation determined by V , then A is a v-ideal. (a) If v is a valuation on K which is nonnegative on R, and if

A is an ideal of R, prove that the following statements are equivalent : (1) A is a v-ideal. (2) (3)

If a, b E R, a E A, and v(b) 2 v(a), then b E A. If V is the valuation ring of D, then AV n R = A.

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EXERCISES 123

Prove that every prime ideal of an integral domain is a valuation ideal. Show that a primary ideal need not be a valuation ideal. {Hint. Consider the ideal ( X z , Yz) of K [ X , Y] . } If A and B are v-ideals of an integral domain R and C is any ideal of R, prove that Rad(A), A n B, and A: C are v-ideals. If A is a v-ideal of an integral domain R, prove that Rad(A) is a prime ideal of R. Show that a valuation ideal need not be primary. {Hint. Consider the polynomial ring K[X, Y] . Let G = 202, ordered lexicographically, and let z, be the valuation on the quotient field of K [ X , Y ] , having G as its value group, con- structed as in Exercise 12. Show that ( X z , X U ) is a v-ideal; it is not primary [see Exercise l l (b) of Chapter 111.)

14. Ordered Abelian groups. (a) Let G be an ordered Abelian group and let H be a sub-

group of G (with the induced ordering). Show that if every element of G/H has finite order, then G has rank r if and only if H has rank Y. [Hint. Show that H' H H n H is a one-to-one mapping from the set of isolated subgroups of G onto the set of isolated subgroups of H.] Let G, and G, be ordered Abelian groups and let G, @ G, be ordered lexicographically. Show that if G, has rank rl and G, has rank r , , then G, 0 G, has rank rl + Y, . Show that the direct sum of Y copies of the additive group of integers, ordered lexicographically, has rank r .

(b)

(c)

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C H A P T E R

VI Priifer and Dedekind Domains

The various aspects of ring theory developed in the first five chapters will now be brought together to study an important class of integral domains, the Prufer domains. The impetus for the study of these domains comes from algebraic number theory. Since the pioneer work of Dedekind and others, including Prufer, an immense literature on Prufer domains and especially on the Noetherian domains in this class, the Dedekind domains, has been developed. In the text of this chapter, we will present a number of fundamental results concerning these domains, with further properties being touched upon in the exercises.

1 FRACTIONAL IDEALS

In this section, we will discuss the notion of fractional ideal relative to an arbitrary ring, although in the remaining sections of the chapter we will be concerned exclusively with integral domains.

6.1. Definition. A fractional ideal of a ring R is a subset A of the total quotient ring K of R such that:

(i) A is an R-module, that is, if a, b E A and Y E R, then a - b, ra E A ; and

(ii) there exists a vegular element d of R such that d A G R .

124

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1 FRACTIONAL IDEALS 125

Note that for (5) to hold it is enough for there to exist a regular element x of K such that xA g R. For, if such an x exists, then x = d/s, where d and s are regular elements of R, and d A = sxA G R.

Each ideal of R is a fractional ideal of R. Some authors call such fractional ideals integral. If x E K , the total quotient ring of R, then Rx is a fractional ideal of R and will be denoted by (x); fractional ideals of this type are called principal.

Let A and B be fractional ideals of R. Their sum

A +B= {a + bl a E A, b E B}

and their product

AB = (finite sums a, bi I a, E A, b, E B>,

as well as their intersection A n B, are fractional ideals of R. Further- more, if B contains a regular element of R, then

[ A : B l = ( x l x ~ K and B x z A }

is a fractional ideal of R. It is clear that [A: B ] is an R-module. Let b be a regular element of R contained in B and let d be a regular element of R such that d A s R. Then bd[A : B ] G d A G R. Note that B contains a regular element of R if and only if it contains a regular elemznt of K. Note also that if A and B are ideals of R, then [A : B ] is not necessarily the same as A : B. In fact, A : B = [A: B ] n R.

Denote by %(R) the set of all nonzero fractional ideals of R.

6.2. Definition. A fractional ideal A of a ring R is invertible $-there exists a fractional ideal B of R such that AB = R.

6.3. Proposition. Let R be a ring with total quoti& ring K.

( 1 )

(2)

(3)

If A E %(R) is invertible, then A is finitely generated (as an R-module) and A contains a regular element of R. If A, B E 9 ( R ) and A c B and B is invertible, then there is an ideal C of R such that A = BC. If A E %(R), then A is invertible if and only if there is a frac- tional ideal B of R such that AB is principal and generated by a regular element of K.

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126 VI PRUFER AND DEDEKIND DOMAINS

Proof. (1) Let B E P ( R ) be such that AB=R. Then there exist a,, . . . , a, E A and b,, . . . , b, E B such that 1 = C?, , a, bi . For each X E A , & , € R,fori= 1, . . ., n ,andx=C?= , a,(xb,).Thus,a,, . . . ,an generate A as an R-module. Let d be a regular element of R such that dB G R. Then d E dR=dAB c AR = A.

(2) If B' E F ( R ) is such that BB' = R, and if C = AB', then C s R and BC= B B A = A.

(3) If x is a regular element of K and B E B ( R ) is such that AB = (x), then A(Bx-l) = R. The necessity of the condition is obvious.

If A is an invertible fractional ideal of R, it follows from (1) of Proposition 6.3 that [R: A] is a fractional ideal of R.

6.4. Proposition. If A E B(R) is invertible and i fB E F(R) is such that AB=R,thrnB=[R:A].

Proof. Since AB = R we have B G [R: A]. Also, A[R: A] c R so that [R: A] = AB[R: A] 5 BR= B.

If A is an invertible fractional ideal of R, we shall denote [R: A] by A-l. It is clear that if B is another invertible fractional ideal of R, then AB is invertible and (AB)-' = A-IB-l. If x is an element of the total quotient ring of R, then (x) is invertible if and only if x is regular, in which case (x)- l= (x-I). Finally observe that a product, A = A, . * - A, of fractional ideals of R is invertible if and only if Ai is invertible for i = 1, , . . , n. If the product A is invertible, then A;l=A-lA, ... An

2 PRUFER DOMAINS

Let R be an integral domain. If A, B E 9 (R) , then each of A + B, AB, A n B, and [ A : B] is in 9(R) . We have seen that invertible fractional ideals of R are finitely generated. I n this section we investi- gate integral domains for which the converse is true.

6.5. Definition. An integral domain R is a Prufer domain ;f each nonzero Jinitely generated ideal of R is invertible.

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2 PRUFER DOMAINS 127

We leave it to the reader to show that if R is a Priifer domain, then each finitely generated fractional ideal of R is invertible. We shall now obtain a number of equivalent conditions for an integral domain to be a Priifer domain.

6.6. Theorem. I f R is an integral domain, then the following state- ments are equivalent :

R is a Priifer domain. Every nonzero ideal of R generated by two elements is invertible. I f AB = AC, where A, B, C are ideals of R and A is finitely generated and nonzero, then B = C. For every proper prime ideal P of R the ring of quotients R, is a valuation ring. A(B r\ C ) = AB r\ A C for all ideals A, B, C of R. (A + B)(A n B) = AB for all ideals A, B of R. I f A and C are ideals of R, with C finitely generated, and if A G C, then there is an ideal B of R such that A = BC. (A+ B): C = A : C + B : C f o r all ideals A, B, C of R with C finitely generated. C : (A n B ) = C : A + C : B for all ideals A, B, C of R with A and B finitely generated. A n (B f C ) = A n B + A n C for all ideals A, B, C of R.

Proof. We begin by showing that (1) and (2) are equivalent. Clearly

(2) + (1). Let C = (c,, . . . , c,) be a nonzero ideal of R ; we shall show that C is invertible by induction on n, assuming that (2) holds. This is true for n = 1, and also for n = 2 by assumption. Suppose n > 2 and that every nonzero ideal generated by n - 1 elements is invertible. We may assume that c,, . . . , c, are all nonzero. Let

(1) =+ (2).

A = ( C ~ , ..., cn-1),

D = f ~ l r cn>,

B = ( c , , ..., c,),

E = c,A-lD-l + ~ ~ 3 - l D - l . Then

CE= ( A + (cn))clA-lD-l + ((cl) + B)c,B-'D-l

= c,D - + c, c, A -1D - 1 + c,c, B - 1D - 1 + c, D - 1 = c , D - ~ ( R + c , B - ~ ) + c,D-'(R+c,A-~).

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128 VI PRUFER AND DEDEKIND DOMAINS

Since c, B-I 5 R and c1A-l E R, this gives

CE= ~ 1 D - l +c ,D- '= ( c ~ , cn)D-'= R.

Thus C is invertible.

nonzero. If A is invertible, then (1) * (3). Suppose AB = AC where A is finitely generated and

B = A-IAB = A-IAC = C.

(3) 4 (4). Suppose that (3) holds for R . If A, B, C are ideals of R with A finitely generated and nonzero, and if AB c AC, then B E C. For we have AC=AB+AC=A(B+C);henceC=B+C and consequently B c C.

Let P be a proper prime ideal of R . We must show that if a/s, b / t E R p , then (a/s) E ( b / t ) or ( b / t ) 2 (a/s). However, since we may assume that s, t if P , and therefore that 11s and l l t are units in R, , it is sufficient to show that either uR, E bR, or bR, E QR,. This is certainly true if either a = 0 or b = 0, so we may assume a # 0 and b # 0. We have (ab)(a, b) c (a2, b2)(a, b), and it follows that (ab) c (a2, b2). Then ab = xu2 + yb2 for some x, y E R. Thus, (yb)(a, 6) E (a)(a, b), and so (yb) _c (a). Let yb = au. Then ab = xu2+ uab or xa2=ab(l -u ) . If u $ P , then a = b ( y / u ) ~ bR,. If U E P , then 1 - u $ P and b = a(x/(l- u)) E aRp .

(4) + (5). Suppose that (4) holds for R and let P be a maximal ideal of R. Then R, is a valuation ring and it follows easily that ( 5 ) holds for ideals of R, . Let A, B, C be ideals of R. Then, using Exercise 4(a) and (b) of Chapter 111,

A(B n C ) R , = (ARp)(BRp n CR,)

= (AB)Rp n (AC)R,

= (AB n AC)R,.

= (AW(BRP) f-l (AW(CR,)

This equality holds for every maximal ideal P of R. Therefore by Proposition 3.13, A(B n C ) = AB n AC.

( 5 ) + (6). If ( 5 ) holds we have for all ideals A and B of R,

(A+B)(A nB)=(A+B)A n ( A + B ) B z A B ,

and the reverse inclusion always holds. (6) + (2). Let C = (cl, c2) be a nonzero ideal of R. If c1 = 0 or

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2 PRUFER DOMAINS 129

c2 = 0, then C is invertible ; hence we shall assume c1 # 0 and c2 # 0. Then A = (c,) and B = (c2) are invertible, and

C(A n B)B-lA-l= ( A + B)(A n B)B-lA-l= ABB-lA-l= R

if (6) holds. Thus, C is invertible. U p to this point in the proof we have shown the equivalence of (1)

through (6). (1) + (7). Assume that R is a Prufer domain. Let A and C be

ideals of R with C finitely generated and with A G C . If C = 0, then A = BC for every ideal B of R. If C # 0, then C is invertible and A = BC for some ideal B of R by Proposition 6.3.

(7) + (4). Let P be a proper prime ideal of R. We must show that, under the assumption of (7), if a, 6 E R, then either aR, 5 bR, or 6R, E aR, . We have (u) c (a, b), so (u) = (a, 6)B for some ideal B of R. Let a= ax+by where x , ~ E B . If X E P, then 1 - x $ P, and so a = b(y/(l - x)) E bR, . Since bB c (a), we have bx E ( U ) so that if x # P, then b E aRp .

(4) + (8). Let A, B, C be ideals of R, with C finitely generated, and let P be a maximal ideal of R. If (4) holds, then the equality in question holds for ideals of R, . Hence, by Exercise 4(c) and (d) of Chapter 111,

((A + B ) : C)Rp = ( A + B)R,: CR,

= (ARp + BR,): CR,

= AR, : CR, -+ BR, : CR,

= (A: C)Rp + (B: C ) R p

=(A: C + B : C)R,.

Therefore, (A + B) : C = A : C + 3 : C by Proposition 3.13.

and for ideals A, B, C of R with A and B finitely generated, (4) + (9). If we assume (4), then for every maximal ideal P of R,

( C : ( A n B))Rp E CR,: ( A n B)R,

= CR,: AR, + CR,: BR,

= ( C : A)R, + ( C : B)R,

= (C: A + C : B)Rp

E ( C : ( A n B))Rp.

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130 VI PRUFER AND DEDEKIND DOMAINS

Thus (C: (A n B))R, = (C: A + C: B)R, for every maximal ideal P of R . Therefore, the desired equality holds.

(4) + (10). This is proved by a similar argument. (8) + (2). Let a, b E R. If (8) holds, then

R = (a, 4: (a, 6) = ((4 + (0 : (Q, b)

= (4 : (a, b) + (4 : (4 b) = (u): (b) + (b ) : (a).

Let 1 = x + y where xb E (a) and ya E (b). Then (xb)b 5 (ab) and (ya)a G (ab). Hence (a, b)(bx, ay) E (ab). But ab = abx + aby, so (ab) = (a, b)(bx, ay). We may assume a # 0 and b # 0. Then (ab) is invertible. Therefore, (a, b) is invertible.

(9) + (2). Let a, b E R. If (9) holds then

R = ((4 n (4): ((4 n (b)) = ((4 n (W: (4 + ((4 (b)): (4 = (b): (a) + (u): (b).

Proceed as above.

and let a, b E R. Since a E (b) + (a - b) we have (10) + (4). Assume (10) holds. Let P be a proper prime ideal of R

(4 = (4 n ((4 + (Q - b)) = ((4 n (4) + ((4 n (a - 0 Let a = t + c(a - b) where t E (a) n (b), c E R, and c(a - 6) E (a). Then c b ~ ( a ) and (l-c)a=t-cbE(b). If c $ P , then b ~ a R , . If c E P , then 1 - c $ P and a E bR, . Thus, R, is a valuation ring.

This completes the proof of Theorem 6.6.

6.7. Corollary. An integral domain R is a Prufu domain if and only if for every maximal ideal P of R the ring of quotients R, is a valuatim ring.

Proof. We need to prove only the sufficiency of the condition. Let P be a proper prime ideal of R and let P' be a maximal ideal of R such that P E P'. Then R\P' c R\P and so R,, G R, . If R,* is a valuation ring, it follows from Corollary 5.3 that R, is also a valuation ring.

To conclude this section we prove some elementary results on the ideal theory in Prufer domains. One result which is immediate, but

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2 PRUFER DOMAINS 131

quite important, is that given two primary ideals A and B of a Prufer domain R, either A and B are comaximal, or they are con- tained in a maximal ideal of R and hence A s B or B G A ; this assertion follows from Corollary 6.7 and Proposition 3.9.

6.8. Theorem. Let R be a Prujer domain and let P be a prime ideal of R.

(1) If Q is P-primary and x E R\P, then Q = Q [Q + (x)]. (2) The product of P-primary ideals of R is P-primary.

Proof. (1) We will show that QRM = [Q2 + Q(x)]RM for each maximal ideal M of R. If Q $ M, then QRM = Q2RM = RM. Let M be a maximal ideal with Q E M. Then QRM is a PRM-primary ideal of the valuation ring R M . Since x $ PRM , QRM = Q(x)RM by Theorem 5.1 1. Since Q2 G Q, Q = Q2 + Q(x).

('2) Let Q,, Q2 be P-primary ideals of R. Then for each maximal ideal M of R such that P c M , QiRM and Qz RM are PR,-primary ideals of the valuation ring RM . By Theorem 5.11, QiRM Q2 RM =

QiQ2 RM is PRM-primary. This proves that QIQz is P-primary by Exercise 5(c) of Chapter 111.

Now let R be a Prufer domain and let P be a prime ideal of R such that P is not the only P-primary ideal of R. We can use the relation between the P-primary ideals of R and the PR,-primary ideals of R, to obtain information about the P-primary ideals of R. For example, the set of P-primary ideals of R is totally ordered, and it follows from Theorem 5.1 1 that if P, is the intersection of the ideals in this set, then P, is prime and there are no prime ideals of R pro- perly between P and P,. Thus, the valuation ring Rp/PIRp has rank one.

If A is an ideal of R let A* be its image under thk composition of the canonical homomorphisms

R --+ R, --f RP)P1 R, . There is a one-to-one order-preserving correspondence between the P-primary ideals Q of R and the P*-primary ideals of R", the cor- respondence being Q t t Q". By Theorem 6.8, this correspondence preserves products. It also preserves residuals ; this follows from

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132 VI PRUFER AND DEDEKIND DOMAINS

Exercise 4(d) of Chapter 111, using the fact the P-primary ideals of R are contractions of ideals of R, (see Proposition 3.9). Hence, if Q1 and Q2 are P-primary ideals of R, then (Q1Q2)* = Ql*Q2* and (Qi : Q2)* = QI* : Q2"-

The correspondence of the preceding paragraph reduces the study of the P-primary ideals of R to the study of the primary ideals of a rank one valuation ring. We can state the following theorem, which follows from Exercise 8 of Chapter V. Note that the assertions hold also when P is the only P-primary ideal of R.

6.9. Proposition. Let Q and Q1 be P-primary ideals of a Prufer domain R. Then:

( 1 ) fi:= Q" is a prime ideal of R ; (2) (3) (4) ( 5 ) ( 6 )

If Q" = Q" + for some positive integer n, then Q = Q2 = P ; If Q c Q1 c P, then Qln c Q for some positive integer n ; If P # P 2, then Q = P" for some positive integer n ; If Q c P, then Q2 c Q P ; If Q c Ql, then Q : Q1 = Q implies that Q1 = P = P2.

3 OVERRINGS OF PRUFER DOMAINS

In this section we give two characterizations of Prufer domains in terms of their overrings.

6.10. Theorem. A n integral domain R is a Prufer domain if and only if mery overring of R is a $at R-module.

Proof. Suppose that every overring of R is a flat overring and let P be a maximal ideal of R. By Proposition 4.13, every overring of R, is a flat overring. Let a and b be elements of R, and suppose that aR, $ bR, . Then bR, : aR, # R, , so bRp: aRp E PR, , the unique maximal ideal of R, . Now consider the ring

R P [ U / b I = { f ( Q / b ) I f ( x ) E RP[XI)*

(We may assume b # 0, for if b = 0, then bR, G aR,, which is what we are trying to show.) This ring is an overring of R, and so is a flat overring of R, . Since a/b E Rp[a/b], we have

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3 OVERRINGS OF PRUFER DOMAINS 133

(bR,: aR,)R,[a/b] = Rp[a/b]

by Proposition 4.12. Thus there are elements x,, . . . , x, E bR, : aR, and b,, . . . , b, E R,[a/b] such that x,b, + - - - + x,b, = 1. There is an integer s and elements a,, E R, , 1 5 i 5 n, 0 5 j 5 s, such that

where d, = xlalj + - . + x, a,, E bR, : a R , . Note that do is not a unit in R, , since bR, : a R , # R, ; hence 1 - do is a unit in R, . If we multiply by (1 - do)s -l(b/a)s, we get

((1 - do)(b/a))s - d,((l - d,)(b/a))s-l- - * - ds(l - do)s-l - 0.

Thus (1 - do)(b/a) is integral over R, . But Rp[(l - do)(b/a)] is a flat overring of R,, so it equals R, by Theorem 4.15. Hence (1 - do)b E

aR, , and since 1 - do is a unit in R, , this implies that b E aR,; that is, bR, c a R p . Therefore R, is a valuation ring. Since P is an arbitrary maximal ideal of R, we conclude that R is a Prufer domain.

Conversely, assume that R is a Prufer domain, Let T be an over- ring of R and let P be a maximal ideal of T. Then T p is an overring of the valuation ring R, ,, and so T , is a valuation ring by Corollary 5.3. Let x E T p . If x 4 R, R , then l/x E R, ; and since it is not a unit in R, ,, R , we have 1 / x E ( P n R)Rp ,, E PT, , which is impos- sible since x E T p . Therefore T p = R, ,, R , and so T is a flat overring of R by Proposition 4.14.

We can immediately draw two corollaries from this theorem, its proof, and Proposition 4.14.

6.11. Corollary. Every overring of a Priifer domain i s a Pri i fu domain.

6.12. Corollary. Let T be an overring of a Priifer domain R. Then there is a set A of prime ideals of R such that

T= R,. P E A

In fact , A is the set of all prime ideals P of R such that PT # T.

We shall now give a second characterization of Prufer domains in terms of their overrings.

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134 VI PRUFER AND DEDEKIND DOMAINS

6.13. Theorem. An integral domain R is a Priifu domain if and only if every overring of R is integrally closed.

Proof. If R is a Priifer domain, then every overring of R is an inter- section of valuation rings (Corollary 6.12). Each of these valuation rings is integrally closed, so the same is true of their intersection.

Conversely, suppose that every overring of R is integrally closed. Let P be a maximal ideal of R; we shall show that R, is a valuation ring. Let a#O belong to the quotient field of R. By hypothesis, Rp[a2] is integrally closed, and since a is integral over Rp[a2] , we conclude that a E R,[u2]. Then there are elements bo , . . . , bn E RP such that a = b, + bla2 + * - * + b,aZn. If we multiply by b2," -l/aZn, we obtain

(bO/u)2n - (bo/U)2?+l + b,bo(bo/a)2n-z + * - . + b,b?-1= 0.

Thus bo/a is integral over R, ; hence bola E R, . If bo/a is a unit in R, , then a E R, . If bo/a is not a unit in R,, then 1 - (bo/a) is a unit in R , . If we multiply the equation expressing a in terms of powers of a2 by l/aZn, we obtain

(1 -(bo/u))(l /u)2~-1 -bb,(l /U)2n-2 -..-- bn= 0.

Since 1 - ( S o / ~ ) is a unit in R,, it follows that l/a is integral over R, . Hence l /u E R, . Therefore R, is a valuation ring. Since P is an arbitrary maximal ideal of R, it follows that R is a Prufer domain.

4 OEDEKIND DOMAINS

This section will be devoted to the study of an important class of integral domains called Dedekind domains. The importance of this class lies in the fact that the ring of integers of a finite algebraic number field is a Dedekind domain. We shall show that the class of Dedekind domains is precisely the class of Noetherian Prufer domains. Then we shall obtain a large number of equivalent conditions for a Noetherian integral domain to be a Dedekind domain; the equiva- lence of some of these conditions will follow from Theorem 6.6. We shall also obtain some results concerning overrings and extensions of Dedekind domains.

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4 DEDEKIND DOMAINS 135

6.14. Definition. An integral domain R is a Dedekind domain if every ideal of R is a product of prime ideals.

6.15. Proposition. For i= 1 , . . ., k, let Pi be an invertible proper prime ideal of an integral domain R . Let A = Pl * Pk . Then this is the only way of writing A as a product of proper prime ideals of R, except for the order of the factors.

Proof. Let A = P1' - * Ph', where Pi' is a proper prime ideal of R for i= 1,. . . , h. Assume Pl is minimal among PI, . . . , P,. Since Pl' Ph' c P,, some Pi' is contained in PI, say Pl' G PI. Since PI -.. Ph G Pl', some Pi is contained in P,'. But then we must have i = 1, and so Pl' = Pl. Since PI is invertible, P2 Ph', and we can repeat the argument. Since the Pi and P,' are proper ideals of R we must have h = l z .

- Pk = P2' -

6.16. Theorem. Let R be a Dedekind domain. Then every nonzero proper ideal of R can be written as a product of proper prime ideals of R in one and only one way, expect f o r the order of the factors.

Proof. Suppose that we know that every invertible proper prime ideal of R is a maximal ideal. We shall show that every nonzero prime ideal of R is invertible. Then the theorem will follow from Proposition 6.15. Let P be a nonzero prime ideal of R. If P = R, then P is inver- tible, so we can assume that P# R. Let a € P, a#., and write (a ) = P, * * * Pk , where each Pi is a proper prime ideal of R. Since (a) is invertible, each Pi is invertible and so maximal. Since P, * - * Pk G P, Pi c P for some i. Hence Pi = P, and P is invertible.

Now let P be an invertible proper prime ideal of R ; we shall show that P is maximal. To do this we shall show that if a E R\P, then P + ( a ) = R. Suppose that for some a E R\P we have P + (a ) # R. Then P + (a ) = Pl * * * P, and P + (a2) = Q1 * - . Qn-, where the Pi and Qi are proper prime ideals of R. Let R'= RIP, Pt'=Pi}P, Q,' = QJP, and a' = a + P. Then

a'R' =PI' * * - Pk'. a'2R' 1 Q,' - - Qn'.

Since a' # 0, the ideals a'R' and aI2R' of R' are invertible, so that the same is true of each Pi' and each Qi'. We have

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136 VI PRUFER AND DEDEKIND DOMAINS

Hence by Proposition 6.15, n = Zk, and we may so number the Qi that for i = 1, . . . , k , Q Z i = Q Z i = Pi . Thus, ( P + (a))z = P + (a2). Then P G (P+ (u ) )~ , and so if b E P, we have b = c + da, where c E P2. Then da E P, but a 4 P, so d E P. Hence P c P2 + Pa. Since P is invertible, there is a fractional ideal A such that PA = R. Then R G P2A + P A a = P + (a). Thus, our assumption that P+ (a ) # R is false.

Since an invertible ideal is finitely generated (Proposition 6.3), and the product of invertible ideals is invertible, we have the following corollary.

6.17. Corollary. If R is a Dedekind domain, then R is Noetherian and every nonzero proper prime ideal of R i s a maximal ideal.

Let R be a Dedekind domain. Recall that F(R) is the set of nonzero fractional ideals of R. As pointed out in Section 6.1, F(R) is closed under multiplication of fractional ideals. I n fact, it is clear that with respect to this multiplication, F(R) is a commutative semigroup with identity element R. By Theorem 6.16 and its proof, every nonzero ideal of R has an inverse in 9 ( R ) . Let A be an arbitrary element of 9 ( R ) ; then d A is an ideal of R for some nonzero d E R. Then, if ( d A ) B = R, we have A ( d B ) = R, so that A has an inverse in F(R). Therefore F(R) is a group. We shall now prove that this is also a sufficient condition for R to be a Dedekind domain.

6.18. Proposition. Let R be an integral domain. Then R is a Dedekind domain if and only if F(R) is a group with respect to multiplication.

Proof. Let Y be the set of all nonzero proper ideals of R which are not products of prime ideals. Under the assumption that F ( R ) is a group, we shall show that Y is empty. Suppose that 9 is not empty. Since every nonzero ideal of R is invertible, R is Noetherian. Hence Y has a maximal element A. Let A c P where P is a maximal ideal of R; then A # P. Let PB = R where B E %(R). Then AB c R and A c AB since R c B. If A c AB, then AB is a product of prime ideals of R

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4 DEDEKIND DOMAINS 137

and the same is true of (AB)P= A. Since this is not true, we have A = A B and so AP= A. However, A is invertible in F(R), and consequently P= R, contrary to our choice of P. Therefore, Y must be empty.

We can now state:

6.19. Theorem. An integral domain R is a Dedekind domain if and only if every nonzero ideal of R is invertible.

I n the next theorem we shall give a number of equivalent condi- tions for a Noetherian integral domain to be a Dedekind domain.

6.20. Theorem. If R is a Noetherian integral domain, then the following statements are equivalent :

R is a Dedekind domain. R is integrally closed and every nonzero proper prime ideal of R is maximal. Every nonzero ideal of R generated by two elements is invertible. If A B = AC, where A , B , C are ideals of R and A # 0, then B = C. For every maximal ideal P of R, the ring of quotients R, is a valuation ring, A ( B n C ) = A B n AC for all ideals A, B, C of R. ( A + B)(A n B) = A B for all ideals A, B of R. If A and C are ideals of R and if A c C, then there is an ideal B of R such that A = BC. ( A + B): C = A : C+ B : C for all ideals A, B, C of R. C : ( A n B)= C : A+ C : B for allideals A, B, C of R. A n ( B + C ) = A n B + A n C for all ideals A, B, C or R. If P is a maximal ideal of R, then there are no ideals of R strictly betwem P and P2. If P is a maximal ideal of R, then every P-primary ideal of R is a power of P. If P i s a maximal ideal of R, then the set of P-primary ideals of R is totally ordered by inclusion. Every overring of R is a j a t overring. Every overring of R is integrally closed.

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138 VI PRUFER AND DEDEKIND DOMAINS

Proof. Let R be a Noetherian integral domain. By Theorem 6.19, R is a Dedekind domain if and only if it is a Prufer domain. Hence, by Theorems 6.6, 6.10, and 6.13, it follows that (l), (3)-(11), (15), and (16) are equivalent.

(2). Assume R is a Dedekind domain. Then R is integrally closed since it is a Prufer domain. By Corollary 6.17, every nonzero proper prime ideal of R is maximal.

(2) 4 (5). Suppose that (2) holds and let P be a nonzero maximal ideal of R. Then R, is Noetherian, and it is integrally closed by Exercise 7(b) of Chapter IV. Furthermore, R, has exactly one nonzero proper prime ideal, namely PR, . Therefore R, is a valuation ring by Theorem 5.9.

(5) + (12). Let P be a maximal ideal of R. If P = 0, then the assertion of (12) holds. Suppose P # 0 and that R, is a valuation ring. Let A be an ideal of R with P2 5 A c P. Then A is P-primary, so AR, n R = A (see Proposition 3.9 and Corollary 3.10). But, either AR, = P2R, or AR, = PR, . Hence either A = P2 or A = P. Note that the same argument shows that for all integers n 2 1, there are no ideals of R strictly between Pn and Pn+l .

(12) + (5). Assume that (12) holds and let P be a nonzero maximal ideal of R. Then PRp is a nonzero ideal of R, and n,"= P"R, = 0 by Corollary 2.24. Hence P2R, # PR, . Also, by (12), there are no ideals of R, strictly between P2R, and PR, . Let P' = PR, and let a E P', a $ PI2. Then P' = aR, + PI2 and by induc- tion we show that P' = aRp + P'" for all positive integers n. Hence

P I = n( aR, +P'").

(1)

m

n = l

But then

PlaR, = ( fi (aR, + P'") n = l

m

= n ( u ~ p + P ' ~ ) / ~ R ~

= n ( P ' / ~ R , ) ~ = 0,

n = 1

m

n = l

by Corollary 2.24, since P'IaR, is the unique maximal ideal of the

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4 DEDEKIND DOMAINS 139

Noetherian ring R,laR,. Therefore P' = aR,, and since P' is pre- cisely the set of nonunits of R, , it follows from Theorem 5.9 that R, is a valuation ring.

(1) + (13). Assume that R is a Dedekind domain and let P be a maximal ideal of R. If P = 0, then P is the only P-primary ideal of R. Suppose P # 0 and let Q be a P-primary ideal of R . Then Q is a product of prime ideals, and since P is maximal, each of these factors must equal P. Hence Q is a power of P.

(13) j (12). This is clear since every ideal between a maximal ideal P and P2 is P-primary.

(5) + (14). This follows from the fact that the set of ideals of a valuation ring is totally ordered by inclusion.

(14) + (12). Assume that (14) holds and let P be a maximal ideal of R. Then P/P2 is a vector space over the field RIP. If P = 0, then the desired conclusion certainly holds. Suppose Pf 0. The set of subspaces of PIP2, each being of the form AIP where A is an ideal of R with P2 s A c P, is totally ordered by inclusion by (14). Hence P/P2 is one-dimensional. Therefore if A is an ideal of R with P2 E A G P, then either A l p 2 = P2/P2 or AIPZ = PIP2; that is, either A= P2 or A = P.

This completes the proof of Theorem 6.20.

It is important to note that there are Prufer domains which are not Dedekind domains. In fact, there are integral domains R which are not Noetherian and which have the property that for each nonzero maximal ideal P of R the ring of quotients R, is a Noetherian valua- tion ring (see Chapter IX).

We now show that overrings of Dedekind domains are Dedekind domains.

6.21. Theorem. Every overring of a Dedekind domain is a Dedekind domain.

Proof. Let R be a Dedekind domain which is not a field and let T be an overring of R other than the quotient field of R. Let M be a maximal ideal of T . Since R is a Prufer domain, TIM = R , by Proposition 4:14. Thus R , A is not a field, and so M n R # 0 . Hence M n R is a maximal ideal of R and therefore R , ,, = TM is a Noetherian valuation ring.

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140 VI PRUFER AMD DEDEKIND DOMAINS

Now let A be a nonzero ideal of T . Then, for each maximal ideal M of T , A T , = &FM)Tnn for some nonnegative integer s (M) . There are only a finite number of maximal ideals of T which contain A ; this follows from Exercise 5(c). Since s (M) > 0 if and only if A c M , we have s ( M ) > 0 for only a finite number of maximal ideals M . By Exercise 5(b) of Chapter 111,

A = AT^ n T ) = fl M ~ ( M ) .

The maximal ideals of Tare pairwise comaximal, so by Exercise 2(b) of Chapter 11,

A = n M S ( M ) ,

This proves that T is a Dedekind domain.

5 EXTENSION O F DEDEKIND DOMAINS

Let R be a Dedekind domain and let K be its quotient field. Let K' be an extension of K and let R be the integral closure of R in K'. Then R' is called an extension of R. We shall now set about to show that if K'IK is finite, then R' is a Dedekind domain. Since K' can be obtained as a purely inseparable extension of a separable extension of K, we can prove this assertion in two steps, first under the assump- tion that K'IK is separable, and then under the assumption that K / K is purely inseparable.

6.22. Proposition. If K'IK is finite and separable, then R' is a Dedekind domain.

Proof. We shall show that R' is Noetherian and integrally closed and that every nonzero proper prime ideal of R' is maximal. By definition, R' is integrally closed, and every nonzero proper prime ideal of R' is maximal by the lying-over theorem. It remains to show that R' is Noetherian.

First, K' is the quotient field of R'. For, let a E K' ; then there are elements b,, , . . . , bk-l E K such that

ak +b,-,a"-l + * . - +b,a + b , = 0.

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5 EXTENSION OF DEDEKIND DOMAINS 141

There are elements co , . . . , ck -1, s E R such that b, = ct/s, i= 0, . . . , K - 1 ; then

(sa)k+c,-,(sa)k-l +.**+(c1s'C-2)(sa) +c0sk--l=0.

Hence sa E R', and a = sals. Now let ul,. . . , u, be a basis of K'/K. Then there are elements

vl, . . . , v,, s E R' such that ut = vi/s, i = 1,. . . , n, and we see from the preceding paragraph that we may choose s E R. Then vl, . . . , v, are linearly independent over K and so form a basis of K'IK. Thus without loss of generality, we may assume that u l , . . . , u, E R'.

Let M = { a , u , + . . . + a , u , J a , , ..., u , E R } . Then M is an R-module and M 5 R'. Let

M* = {bl b E K' and T,.,,(ab) E R for all a E M } ,

where T K , / K is the trace mapping of K'/K. Define R'* in like manner. It follows from properties of the trace

mapping that R'" and M* are R-modules, and we have M c R' G

R'" G M*. If we can show that M" is a finitely generated R-module, then R' and all of its ideals will be finitely generated R-modules by Theorem 1.12. Hence we will be able to conclude that R' is Noethe- rian.

Let wl,. . . , w, E K and consider the following n equations in n unknowns :

n

j = 1 1 TK,/,(uiuj)xj =z wi 9 i = 1, . . . , n.

Since K / K is separable, det[T,.,,(uiuj)] # 0. Hence this system of equations has a unique solution a,, . . . , a, in K. Then a = alul + - *

+ a,u, is the unique common solution of the equations

T K ' / K ( u ~ X) = wi 7 i= 1, . . . , n. Thus, for fixed j , the n equations

T,, /duix) = S i j , i = 1, ..., n, have a unique common solution ui'. Suppose clul' + - * * + c, u,' = 0 where cl,. . . , c, E K. Then, for i = 1, . . ., n,

0 = TK,/K(ut(c1~1' + * * a + cnun')) n

j = l = c cj TK',K(Ut u;) = c* .

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142 VI PRUFER AND DEDEKIND DOMAINS

Therefore ul',. . . , u,I are linearly independent over K, and conse- quently form a basis of K j K .

We shall complete the proof by showing that u,', . . . , u,' E M" and that they generate M" as an R-module. Let a E M and write a = alul + * . * + a,u, where a,, . . . , a, E R. Then for j = 1,. . . , n,

TK,/K(auj') T K , / K ( ( ~ I ~ ~ + * * * + anUn)uj')

n

i = l = c a, T K ' , K ( U , U j ' ) = aj E R ;

thus ui' E M" for j = 1,. . . , n. Finally let b E M" and write b =

blul'+...+b,u,'whereb,, ..., ~ , E K . T h e n f o r i = l , . . . , n,

b, = Txe,K(~l(bl~l' + - + b ,~ , ' ) ) E R.

This completes the proof of Proposition 6.22.

Now suppose that K'IK is finite and purely inseparable. Then K has prime characteristic p , and there is a nonnegative integer e such that ape E K for all a E K'. Iff is a positive integer, set K, = (a 1 a E K' and upf E K}. Then K, is a subfield of K', we have K c Kl G K , G - *

c Ke = K', and up E K,-l for all a E K,. Therefore it is sufficient to prove that R' is a Dedekind domain under the assumption that up E K for all a E K'.

6.23. Proposition. If u p E K for all a E K', then R' is a Dedekind domain.

Proof. First we note that R' = {a I a E K' and up E R}. If a E K' and up E R, then a is a root of x p - up E R [ X ] , so a E R'. Conversely if a E R', then ap E R' n K = R.

Let C be an algebraic closure of K which contains K . Let K" = {cl c E C and cp E K } and let R" be the integral closure of R in K". Then R" = {cl c E K" and cp E R}. The mapping from K" onto K given by c H cp is an isomorphism, and its restriction to R" maps R" isomorphically onto R. Therefore R" is a Dedekind domain. Note that K 5 K" and R' G R .

Let A be a nonzero ideal of R'. Then AR" is invertible ; hence by Proposition 6.4, (AR'')[R": AR"] = R". Let a,, , . . , a, E A and bl, . . , , b, E [R" : AR"] be such that alb, + - * * + a, b, = 1 ; then

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5 EXTENSION OF DEDEKIND DOMAINS 143

alPblP + * - + akPbkP = 1 . For i = 1, . . . , k , biP E K and bipa E R” n K’= R’ for all a E A ; hence biP E [R’: A]. Since aip E A for i= 1 , . . ., k , we conclude that A[R’ : A] = R’ ; that is, A is invertible. Therefore, by Theorem 6.19, R’ is a Dedekind domain.

We summarize all of this as:

6.24. Theorem. Let R be a Dedekind domain, K the quotient field of R, K’ a jn i te extension of K, and R’ the integral closure of R in K’. Then R’ is a Dedekind domain.

6.25. Theorem. Let v be a discrete rank one valuation on afield K and let K be a jn i te extension of K. Then there is a valuation on K’ which is a prolongation of v. Furthermore, up to equivalence of valuations, there is only aJinite number of valuations on K‘ which are prolongations of v . Finally, the valuations on K‘ which are prolongations of v have rank one and are discrete.

Proof. Let V be the valuation ring of v. Then V is a Dedekind domain, so by Theorem 6.24 the integral closure V’ of V in K‘ is a Dedekind domain. If P is the maximal ideal of V , let Pl, . . . , P k be the maximal ideals of V’ containing PV‘. Let Vi’ = V,, and Pi’ = Pi Vi’ for i = 1 ,. . , , k. By condition (5) of Theorem 6.20 each V,’ is a Noetherian valuation ring, and Pi’ is its maximal ideal. Now Pin V = P f o r i = l , ..., k, and Pi’n V ‘ = P , b y Proposition 3.9. Hence for i= 1, ..., k , P i ‘ n V = P i ‘ n V ’ n V = P i n V = P , and consequently, by Proposition 5 2 1 , the valuation vi’ on K‘ determined by Vi‘ is a prolongation of v. Since V,’ is Noetherian, vi’ has rank one and is discrete.

Let v“ be a valuation on K’ which is a prolongation of v and let V” be its valuation ring. Then v”(b) 2 0 for all b E V. Let a E V‘, a f 0; then there are elements b, , bl,. . . , b, -1 E V such that

6, +b,u +... + bn-lan-l +an= 0.

Suppose v’”(a) < 0. Then for i = 0, . . . , n - 1, we have v”(biai) > v”(an). Therefore we have

V ” ( b , + bla + * ‘ + bn-lan-l) > 7J’‘(Un),

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144 VI PRUFER AND DEDEKIND DOMAINS

which is not true. Thus ~ " ( a ) 2 0 ; that is, a E V". Therefore V' E V", and if P" is the maximal ideal of V", we must have P" n I" = Pi for some i, since P = P n V. By condition (15) of Theorem 6.20, V" is a flat overring of V', and therefore V" = V& = Vki = V,' by Proposi- tion 4.14. Thus ZI" and v,' are equivalent for some i, and all is proved.

EXERCISES

1. Criterion for invertibility of ideals. Let R be a ring, let A be an ideal of R, let {Mnl c( E I } be the set of maximal ideals of R, and let S be a multiplicative system in R. (a) If A is invertible, prove that S-'A is invertible. (b) If A is regular and finitely generated and such that ARM,

is invertible for each IX E I , prove that A is invertible. (c) Assume that R is integrally closed. If a, b E R, with a regu-

lar, and if there exists an integer n > 1 such that an- ' b E

(an, b"), prove that (a , b ) is an invertible ideal.

2. Invertible prime ideals. Let R be a ring and let P be an invertible prime ideal of R. (a) Show that i f x E Ps\P+I and y E Pt\Pt+l, then.

xy E Ps + t\Ps + + ' ; therefore, n;= P" is a prime ideal. (b) Show that the powers of P are P-primary and every P-prim-

ary ideal of R is a power of P. (c) If Q is a primary ideal of R with Rad(Q) c P, show that

Q G Pn for each positive integer n. (d) If A is an invertible ideal of R such that P c A , show that

A = R .

3. More on invertible ideals. Let R be a ring with a finite number of maximal ideals M I , . . . , M,, and let A be an invertible ideal of R. (a) Prove that A is principal. [Hint. Find a E A\U:=, A M ,

and show that A = (a).]

(b) If R has a unique maximal ideal and A = (al, . . . , an), show that A = (a,) for some s(1 5 s 2 n).

4. Prufer domains. Let R be a Prufer domain with quotient field K. (a) Let L be an algebraic extension of field of K and let R' be

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EXERCISES 145

the integral closure of R in L. Show that R' is a Prufer domain.

(b) If P is a prime ideal of R, show that RIP is a Prufer domain.

(c) Let {R,} be an ascending chain of Prufer domains all with common quotient field K. Show that u R, is a Prufer domain.

5. Overrings of Prufer domains. Let R be a Prufer domain and let R' be an overring of R. Let A = { P I P is a prime ideal of R and PR' # R'}. (a) Show that if P is a proper prime ideal of R, then P E A if

and only if R' G R, . (b) Let P' be a maximal ideal of R' and P = P' n R. Show that

R, = RL? and P' = PR, n R'. (c) Show that if A is an ideal of R', then A = ( A n R)R'. (d) Show that {PR' I P E A} is the set of proper prime ideals of

R' . 6. Another characterization of Prufer domains.

Let R be an integral domain. Prove that the following are equivalent. [Hint: use Exercise l(c).]

(1) R is a Prufer domain. (2) R is integrally closed and there is a positive integer

n > 1 such that (a , b)" = (an, b") for any a, b E R. (3) R is integrally closed and there exists a positive

integer n > 1 such that an- lb E (a", b") for any a, b E R .

7. Valuation ideals and Prufer domains. Let R be an integral domain. An ideal A of R is a valuation ideal if there is a valuation overring V of R and an ideal B of V such that B n R = A. Refer to Exercise 13 of Chapter V. (a) Prove that if A,, . . ., A, are v-ideals of R and x,, . . ., x,

are elements of R such that xi $ Ai for i = 1, . . . , n, then x1 - * x, $ A, * * . A, . In particular, xs E AIS implies x E A,.

(b) Use (a) to prove that if (a", b") is the intersection of valua- tion ideals, then (a", b") = (a, b)".

(c) Prove that R is a Prufer domain if and only if every ideal of R is the intersection of valuation ideals.

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146 VI PRUFER AND DEDEKIND DOMAINS

8. Completion of ideals and Prufer domains. Let R be an integral domain with quotient field K. Let (V,} be the set of all valuation overrings of R. If A is an ideal of R, A’ = 0 Alr, is called the completion of A and if A = A’, A is called complete. Let A* = 0 (AVa n R); then A* is the intersection of all valuation ideals of R containing A. (a) Prove that the following are equivalent :

(1) R is integrally closed. (2) (3)

(4)

(b) Prove the following conditions are equivalent: (1) (2) (3)

Each principal ideal of R is complete. There exists a nonzero principal ideal of R which is complete. Each principal ideal of R is an intersection of valuation ideals of R.

R is a Prufer domain. Each ideal of R is complete. Each finitely generated ideal of R is an intersection of valuation ideals.

9. Prufer domains, projective modules, and torsion. Let R be an integral domain. (a) Show that a nonzero ideal A of R is invertible if and only

if it is a projective R-module. Thus, R is a Prufer domain if and only if every nonzero finitely generated ideal of R is a projective R-module.

(b) An R-module M is said to be without torsion if ax = 0, where a E R and x E M, implies that either a= 0 or x = 0. Show that a flat R-module is without torsion.

(c) Show that R is a Prufer domain if and only if every finitely generated R-module without torsion is projective.

(d) Let K be the quotient field of R and let M be an R-module. Regarding K as an R-module and R as one of its sub- modules, show that if M is without torsion, then there is an exact sequence

0- t M - t K QR M - t K l R O R M+O.

Let R be a Prufer domain and let M and N be R-modules, each without torsion. Show that M O R N is without torsion,

(e)

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EXERCISES 147

10. Quasi-principal ideals. An ideal M of a ring R is quasi-principal if ( A n B : M ) M = A M n Band ( A + B M ) : M = A : M + B for all idealsA and B of R. (a) (b)

(c)

Show that every principal ideal of R is quasi-principal. Show that the product of quasi-principal ideals of R is quasi-principal. Show that if R is a Noetherian ring with unique maximal ideal, then every quasi-principal ideal of R is principal. [Hint. Use Exercise 7(c) of Chapter 11.1

(d) Show that every quasi-principal ideal of R is finitely generated.

(e) Show that if R is an integral domain, then R is a Prufer domain if and only if every finitely generated ideal of R is quasi-principal.

11. Dedekind domains. Let R be a Dedekind domain. (a) Show that each nonzero element of R is contained in only

finitely many maximal ideals of R. (b) Let A and B be ideals of R with B # 0. Show that there is

an ideal C of R and an element a E R such that B + C = R, AC=(a), and A = AB + (a).

(c) Show that if A is an ideal of R, then there exist elements a, b E A such that A = ( a , 6).

(d) Show that if R has only a finite number of maximal ideals, then R is a principal ideal ring.

(e) Show that if A is a nonzero ideal of R, then RIA is a principal ideal ring.

Integral domains with quotient overrings. Let R be an integral domain. We say R has the QR-property if every overring of R is a ring of quotients of R. (a) Show that if R has the QR-property, then R is a Prufer

domain. (b) Show that if T is an overring of R such that for every

a E T , R[a] is a ring of quotients of R, then T is a ring of quotients of R. Show that if a, b E R, b f 0, and if (a , b)" is a principal ideal for some positive integer n, then R[a/b] is a ring of

12.

(c)

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148 VI PRUFER AND DEDEKIND DOMAINS

quotients of R. Conclude that R has the QR-property if for all a, b E R, some power of (a , b) is a principal ideal. Suppose that R has the QR-property and let A be a finitely generated ideal of R. Show that some power of A is contained in a principal ideal. Suppose that R has the QR-property and let P be a finitely generated prime ideal of R. Show that P = Rad((a)) for some a E R . Let R be a Noetherian integral domain. Show that R has the QR-property if and only if R is a Dedekind domain such that for every ideal A of R, some power of A is a principal ideal.

13. Cancellation of ideals. Let R be a ring, not necessarily an integral domain. We say a fractional ideal A of R can be canceled if AB = AC for frac- tional ideals B, C of R implies that B = C .

Show that if r E R, thin ( r ) can be canceled if and only if r is a regular element of the total quotient ring of R. Show that a fractional ideal A of R can be canceled if and only if for fractional ideals B, C of R, AB c AC implies that B E C. Show that if A,, . . . , A, are fractional ideals of R such that A, - - * Ak can be canceled, then A, can be canceled for each

If A is a fractional ideal of R which can be canceled, if A = A, + - * * + A,, for fractional ideals Al, . . . , A,, of R, and if K is a positive integer, prove that Ak = AIk + * * - Give an example of an ideal which contains a regular element, but which can not be canceled.

i = l , . . . , K .

+ A",.

14. More on cancellation of ideals. Let R be a ring, not necessarily an integral domain. (a) Let A be an ideal of R which can be canceled. Show that if

A can be written as a product of proper prime ideals of R, then it can be so written in only one way, except for the order of the factors.

(b) Suppose that every finitely generated regular ideal of R can be canceled. Show that R is integrally closed.

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EXERCISES 149

(c) Let R be an integral domain and assume that there is a set Y of nonzero proper ideals of R such that every nonzero proper ideal of R can be written as a product of ideals in Y in one and only one way, except for the order of the factors. Show that R is a Dedekind domain and that 9' is the set of nonzero proper prime ideals of R.

15. A construction of Priifer domains. Let K be a field with prime subring k. Let KO be the integral closure of k in K. Let

f ( X ) = X " + an_1xn-1 + - - . + a,X + a,

be a polynomial of positive degree over K which has no roots in K . Let S = { l / f ( a ) I a E K}, let R be the subring of K generated by S , and let R' be the integral closure of R in K. Prove the following statements which constitute a proof that R' is a Priifer domain. (a) If a E K and g ( X ) E K,[X] is such that deg g ( X ) 5 n, then

g(a)lf(u) E R'. Furthermore K is the quotient field of R'. (b) If U E K , then f(a)R'={l, a , ..., an}R'. If r > 0 , then

(c) If a,, a 2 , . . . , a, E K , where s 5 Zt , and if F is a fractional ideal of R' which is generated by a,, a 2 , . . . , a,, then Fnt is principal. Hence R' is a Prufer domain. [Hint. Use Exercise 13(d) and induct on s.]

~ f ( a ) R ' if and only if a E R'.

16. The ideal transform. Let R be a ring with total quotient ring K. If A is an ideal of R, the transform T(A) of A is u,"= (a) Prove that if A is invertible, then AT(A) = T(A). (b) If A, B are ideals such that A" E B for some positive

integer n, prove that T(B) c T(A). (c) If A is invertible and P is a prime ideal of R, show that

PT(A) = T(A) if and only if A c P. (d) Let R' be an overring of R such that R G R' G T(A). Show

that there is a one-to-one correspondence between the prime ideals P' of R' which do not contain AR' and the prime ideals P of R which do not contain A. Show that if P corresponds to P', then P = P'

I x E K and xAn E R}.

R and Rip = R, .

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VI PRUFER AND DEDEKIND DOMAINS

Under the hypothesis of (d), show that if AR' = R', then R' = R,, where {Pa> is the collection of all prime ideals of R which do not contain A. Let A be a finitely generated ideal of R, say

A = (u,, u 2 , . . . , u,,).

Prove that T(A) = n?= T((ui)). Show that if A is a finitely generated ideal of R and {Pa} is the collection of prime ideals of R which do not contain A, then T(A) = n R,, . For u E K, denote by B, the ideal {r I r E R and ru E R}. If R is a Prufer domain, show that T(B,) = R[u] and prove that u is almost integral over R if and only if (I:=, (B,)" # 0. [Hint. Show that if R, and R, are integral domains such that R, c R, and such that { X I x E R, and xR, E R,} # 0, then R, and R, have the same complete integral closure.] Let R be a Prufer domain. Prove that R is completely integrally closed if and only if for each proper finitely generated ideal A of R, , An = 0.

17. The ideal transform and overrings. Let R be an integral domain which does not have a unique maximal ideal. (a) If {x,} is the collection of nonunits of R, prove that

(b) Let x be a nonzero element of R and let R= n T((xa))*

S = { x " l n = l , 2 , ...}.

Show that T((x)) = S-lR. Show that R is integrally closed if and only if the trans- forms of the principal ideals generated by nonzero non- units of R are integrally closed.

(d) Prove that R is a Prufer domain if and only if T((x)) is a Prufer domain for every nonzero nonunit x of R.

(c)

18. Arithmetical rings. A ring R is called an arithmetical ring if for all ideals A, B, C of R we have A n (B+ C ) = ( A n B) + ( A n C ) . (a) Show that R is an arithmetical ring if and only if for all

ideals A, B, C of R we have A + (B n C) = ( A + B) n ( A + C).

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EXERCISES 151

(b) Show that R is an arithmetical ring if and only if for every maximal ideal P of R, the set of ideals of R, is totally ordered by inclusion. Show that the following statements are equivalent : (1) R is an arithmetical ring. (2) (A+ B): C = A : C + B : Cfor allideals A, B, Cof R

with C finitely generated. (3) C: ( A n B) = C : A + C: B for all ideals A, B, C of

R with A and B finitely generated. (d) Show that R is an arithmetical ring if and only if A E C,

where A and C are ideals of R with C finitely generated, implies that there is an ideal B of R such that A = BC.

(c)

19. More on arithmetical rings. - Show that every overring of an arithmetical ring is an arithmetical ring. Let R be an arithmetical ring with only a finite number of maximal ideals. Show that every finitely generated ideal of R is a principal ideal. Show that in an arithmetical ring every primary ideal is irreducible. Let R be an arithmetical ring. Show that every finitely generated regular ideal of R can be canceled. Thus R is integrally closed [see Exercise 14(b)]. Let R be an arithmetical ring and P a proper prime ideal of R. Show that RIP is a Prufer domain.

20. The Grothendieck group. Let R be a ring. Denote by %(R) the set of equivalence classes of finitely generated R-modules under the relation of isomor- phism. Denote the equivalence class containing M by [MI . If [MI, [N] E X ( R ) , set [MI + [ N ] = [ L ] if there is an exact sequence 0 + M + L + N + 0. (a) Show that this is a well-defined binary operation on X ( R ) ,

and that with this operation %(R) is a group: it is called the Grothendieck group of R. Show that if R is Noetherian, then X(R) is generated by the set {[RIP]}, where P runs through the set of proper non- zero prime ideals of R. Show that if R is a Dedekind domain, then X ( R ) z g(R). Show that if R is a field, then X ( R ) 2 2.

(b)

(c) (d)

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152 VI PRUPER AND DEDEKIND DOMAINS

21. Noetherian domains. Let R be a Noetherian integral domain. (a) Assume that R has a unique maximal ideal P. Show that

if P is the only nonzero proper prime ideal of R, then R c [R : PI, that P [R : PI # R implies that P [R: PI" # R for each positive integer n, and that each element of [R: PI is integral over R.

(b) With the same assumption on P, show that if P is not the only nonzero proper prime ideal of R, then P [ R : PI = P and that each element of [R: PI is integral over R. Now assume that not every nonzero proper prime ideal of R is maximal. Let a E R, a # 0, and suppose that (a) has a prime divisor P which is not minimal. Show that for some b E R we have b/a $ R and b/a integral over R. Show also that P is a prime divisor of (c) for every c E P (see Exercise 9 of Chapter 111).

(c)

22. Rings having few zero-divisors. In this exercise, and the four which follow, some of the results which have been obtained for integral domains will be extended to a certain class of rings with zero-divisors. Many of these results will be obtained under more general conditions in Chapter X. A ring R has few zero-divisors if the set of zero- divisors of R is a union of a finite number of prime ideals of R. Thus, Noetherian rings and integral domains have few zero- divisors. (a) Show that a ring R has few zero-divisors if and only if the

ideal 0 has only a finite number of maximal prime divisors (see Exercise 9 of Chapter 111). Show that this is equivalent to the total quotient ring of R having only a finite number of maximal ideals.

(b) Suppose that every ideal of a ring R with prime radical is a power of a prime ideal. Show that R has few zero-divisors if and only if its zero ideal has only a finite number of minimal prime divisors.

For the rest of this exercise, assume the ring R has few zero- divisors. (c) Show that if x, y E R, and if x is regular, then there is an

element u E R such that y + ux is a regular element of R.

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EXERCISES 153

23.

(d) Show that every regular ideal of R is generated by its regular elements. Show that if a regular ideal of R can be generated by n elements, then it can be generated by n + 1 regular elements.

(e) Let T and T' be overrings of R. Show that T G T' if and only if every regular element of T is contained in T'.

A ring having more than a few zero-divisors. In this exercise, an example is given which shows that, in general, the properties enjoyed by rings which have few zero- divisors do not hold for arbitrary rings. The ring of this exercise will be used for counterexamples in Chapter X. Let F be a field and let X and Y be indeterminates. Let G be the set of irreducible polynomials f (X, Y ) in F [X, Y ] such that f ( 0 , O ) = 0 and f ( X , 0) # 0. For each g E G, let 2, be an indeterminate. Let T = F [X, Y , (2, Ig E G}], and let I be the ideal of T generated by

Let R = T/I , and denote the residue classes of X, Y , and 2, by x, y , and z, for all g E G, and let N be the ideal of R generated by y and z, for all g E G. (a) Prove that M is a maximal ideal of R properly containing

the regular prime ideal N, and that every element of M\N is a zero-divisor. Thus, M and N have the same regular elements but are not equal.

(b) L e t f c F [ X , Y ] be such thatf(X, Y ) $ F [ Y ] and let n be the least integer such that f (X, Y ) has a term of the form aY"XP for some nonzero a E F and some positive integer p . Prove thatf(x, y ) is a regular element of R if and only if f (X , Y ) has a term of the form by", where 6 is a nonzero element of F and 0 I m I: n. Let Q be the ideal of R generated by { z g ] g E GI. If f ( X , Y ) E F [ X , Y ] and w E Q, prove that f ( x , y ) + w is regular if and only iff(., y ) is regular.

(c)

24. Quasi-valuation rings. A ring R is a quasi-valuation ring if it has few zero-divisors and if for every pair of regular elements a, b E R, either (a ) s (6 or ( b ) c (a). Let R be a ring which has few zero-divisors.

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154 VI PRUFER AND DEDEKIND DOMAINS

(a) Show that R is a quasi-valuation ring if and only if for every regular element x of the total quotient ring of R, either x E R or x - l E R.

(b) Suppose that R is integrally closed and that R has a unique regular maximal ideal M . Show that if xy E (x2, y2) for every pair of regular elements x, y E M , then R is a quasi- valuation ring.

For the rest of this exercise, assume that R is a quasi-valuation ring. (c) Show that the set of regular ideals of R is totally ordered by

inclusion; conclude that either R has a unique regular maximal ideal or R is its own total quotient ring.

(d) Show that every overring of R is a quasi-valuation ring. (e) Suppose R is not its own total quotient ring K , and let T be

an overring of R with T f K . Let M be the unique regular maximal ideal of T , and let P = M n R. Show that T = Rs(p) . (For the notation used here, see Exercise 10 of Chapter 111.)

(f) Let A be a regular ideal of R. Show that n:==l A" is a prime ideal of R.

(g) Give an example of a quasi-valuation ring which is not a valuation ring.

25. Quasi-valuation overrings of a ring. Let R be a ring which has few zero-divisors. (a) Suppose that is a quasi-valuation ring for every

regular maximal ideal M of R. Show that if T is an overring of R, then T has this same property, and that T = R,(M,, where the intersection is over the set of all regular maximal ideals M of R such that T c RS,,).

(b) Show that a quasi-valuation ring is integrally closed. (c) Show that the integral closure of R is the intersection of all

of the quasi-valuation overrings of R.

26. 2'-rings. A ring R is a P-ring if every overring of R is integrally closed. Throughout this exercise, assume that R has few zero-divisors. (a) Show that R is a P-ring if and only if Rs(M, is a quasi-

valuation ring for every regular maximal ideal M of R.

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EXERCISES 155

(b) Show that the following statements are equivalent : (1) R is a P-ring. (2) Every overring of R is a flat R-module. (3) A(B n C ) = AB n AC for all ideals A , B, C of R

with B and C regular. (4) (A + B)(A n B) = A B for all regular ideals A, B

of R. ( 5 ) Every finitely generated regular ideal of R is invertible. (6) If A B = AC, where A , B, C are ideals of R and A is

finitely generated and regular, then B = C. Show that R is a P-ring if and only if whenever A and B are ideals of R, with B finitely generated and regular, and A c B, then there is an ideal C of R such that A = BC.

(d) Formulate analogs of statements @)-(lo) of Theorem 6.6, which are equivalent to the statement that R is a P-ring, and prove this equivalence.

(c)

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C H A P T E R

VII Dimension of Commutative Rings

1 THE KRULL DIMENSION

Let R be a ring and consider a chain

Po c PI c c p ,

of r + 1 proper prime ideals of R. The length of such a chain is the integer r . Its first term is Po and its last term is P, .

7.1. Definition. The Krull dimension of R is the supremum of the lengths of all chains of distinct proper prime ideals of R. The Krull dimension of R is denoted by dim R.

Either dim R = 00 or dim R = d , where d is a nonnegative integer such that R has a chain of d + 1 distinct proper prime ideals but no chain of d + 2 distinct proper prime ideals. Note that if K is a field, then dim K = 0 ; conversely, if R is an integral domain and dim R = 0, then R is a field.

7.2. Definition. Let P be a proper prime ideal of R. The height of P, denoted by ht P , is the Krull dimension of R,. The depth of P, denoted by dpt P, is the Krull dimension of RIP.

156

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1 THE KRULL DIMENSION 157

Thus the height of P is the supremum of the lengths of all chains of distinct proper prime ideals of R having P as last term. The depth of P is the supremum of the lengths of all chains of distinct proper prime ideals of R having P as first term.

7.3. Definition. Let A be a proper ideal of R. The height of A, denoted by ht A, and the dimension of A, denoted by dim A, are the inJimum and supremum, respectively, of the vatues of ht P as P runs over all of the minimal prime divisors of A. The depth of A, denoted by dpt A, is the Krull dimension of RIA.

Note that if A is a proper ideal of R, then ht A 2 dim A. For example, suppose that R is a Dedekind domain. Since

every nonzero proper prime ideal of R is maximal, dim R = 1. If A is a nonzero prop'er ideal of R, then ht A = dim A = 1 and dpt A = 0. Furthermore, ht 0 = dim 0 = 0 and dpt 0 = 1.

We shall now show that for a ring R, considered as an R-module, an interesting relation exists between (DCC), (ACC), and the property of having zero Krull dimension. I n general, an R-module may satisfy one of (ACC) and (DCC) without satisfying the other (see Exercise 3 of Chapter I). However, if R is considered as an R-module, the two conditions are not independent.

7.4. Theorem. A ring R satisfies (DCC) for ideals if and.only if R is Noetherian and dim R = 0.

Proof. Suppose that the zero ideal of R is the product of maximal ideals of R, say 0 = P, ..- P, . Consider the chain of ideals of R,

0 =PI * - P,, E PI -.. P,-l E * - * c PIP2 E PI c R.

There are no ideals strictly between P, and R. For i = 1, . . . , n - 1, P, Pi+, is a vector space over the field RIP,,,. If R satisfies (ACC), then each of these vector spaces satisfies (ACC) and so is finite dimensional. Thus each of these vector spaces has a composition series; this implies that the chain given above can be refined to a composition series of R, considered as an R-module. This, in turn, implies that R satisfies (DCC). In exactly the same way, we can show that if R satisfies (DCC), then it satisfies (ACC). The facts

PiIP,

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158 VII DIMENSION OF COMMUTATIVE RINGS

which we have used concerning composition series may be found in Exercise 4 of Chapter I.

Now suppose that R is Noetherian and that dim R = 0. Then each proper prime ideal of R is a minimal prime divisor of 0 and it follows that R has only a finite number of proper prime ideals say PI , . . . , Pk, each of which is maximal. For some positive integer r we have

0 = (Rad(0))' = (PI n - - n Pk)' = (P, - - * PkJ,

by Exercise 2 of Chapter 11. Hence R satisfies (DCC). Conversely, suppose that R satisfies (DCC). Let P be a proper

prime ideal of Rand let a E R\P. The set of ideals {P + (an) I n = 1,2, . . .} has a minimal element, so that for some positive integer m we have P+ (a"+') = P f (a'"). Then a" = x + ram+' for some x E P and r E R. Since am(l - ra) = x E P and am 4 P, we have 1 - ra E P, that is, 1 E P + (a). Hence P + (a) = R ; this implies that P is maximal and proves that dim R = 0.

Let P,, P 2 , . . . be distinct maximal ideals of R. Then P, 2 P, n P, 2 P, n P, n P, 2 - . . . Since R satisfies (DCC) this descending chain cannot have infinitely many terms. Hence R has only finitely many maximal ideals, say P,, . . . , Pk . Let A = PI * * * Pk ; if we show that some power of A is the zero ideal then it will follow that R is Noetherian. Since R satisfies (DCC), there is a positive integer r such that A' = A*+,. Assume that AT # 0 and let B be an ideal which is minimal with respect to the properties that BA' # 0 and B 5 A. Note that the set of ideals with these properties is not empty, for A belongs to this set. If P = 0: BA', then P # R since BA' # 0. qe t ab E P and b 4 P. Then abBA' = 0 but bBA' # 0 ; hence bB =!B by our choice of B and aBAr = 0, that is, a E P. Thus P is a proper prime ideal of R. Therefore, A E P and BAT = BA'+I c BA'P= 0, which contradicts our choice of B. Thus we must conclude that A' = 0.

In the remainder of this section we shall obtain results which show that a definite relationship exists between the number of generators of an ideal of a Noetherian ring and the heights of the minimal prime divisors of the ideal. Recall that J(R) denotes the Jacobson radical of R (Exercise 7 of Chapter 11).

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1 THE KRULL DIMENSION 159

7.5. Lemma. Let R be a Noetherian ring and let a E J(R). Let A and B be ideals such that B c A C_ B + (a) and A : (u) = A. Then A = B.

Proof. First assume that B = 0. Since A G (a) we have A =

(A: (.))(a) = Aa. Then A = Aa" for all positive integers n, and consequently

m

A = n ( a ) = ~ . n = 1

Since U E J ( R ) , it follows from Exercise 7(d) of Chapter I1 that A = 0. Now drop the assumption that B = 0. If we set R' = RIB, A' = A/B, and a' = a + B, then a' EJ(R' ) , A' G (a') and A': (a') =A'. Each of these facts is obvious except the last one. If d + B E A': (a') , then da + B E A', so da - c E B for some c E A. But then da - c E A ; hence da E A and d E A : (a ) = A. Thus d + B E A'. From the first part of the proof we conclude that A' = 0. Therefore A = B.

7.6. Theorem (Krull's Principal Ideal Theorem). Let R be a Noetherian integral domain and let a be a nonzero nonunit in R. If P i s a minimal prime divisor of (a), then ht P = 1.

Proof. Since ht P= ht PRp, we may assume that P is the unique maximal ideal of R. Let Q be a prime ideal of R such that Q c P; we must show that Q=0. Since P is a minimal prime divisor of (a), we have dim(R/(a)) = 0. Since R/(a) is Noetherian, it follows from Theorem 7.4 that R/(a) satisfies (DCC). Hence for some positive integer n we have Q ( k ) + (a) = Q(") + ( a ) for all K 2 n. Since Rad(Q("1) = Q and a F$ Q [since P is a minimal prime divisor of (a)], we have Q("): (a) = Q(") by Exercise 6(j) of Chapter 11. I t now follows from Lemma 7.5 that Q ( k ) = Q(") for all K 2 n. Since R is an integral domain, Q(") = 0 by Proposition 3.14. Therefore Q(")= 0 for large n, and since c Q(") we have Qn = 0. Thus Q=O.

7.7. Theorem. Let R be a Noetherian ring. If (al, . . . , a,) is a proper ideal of R, then dim(a,, . . . , a,) 2 r .

Proof. Let P be a minimal prime divisor of (a l , . . . , ar). We shall show that ht P L r , and in doing so we may assume that P is the

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160 VII DIMENSION OF COMMUTATIVE RINGS

unique maximal ideal of R. We must show that if P = Po =I P, 3 ... 3 P, , where each P, is a prime ideal of R, then s 5 r . We may replace R by RIP,, and so assume that R is an integral domain. Furthermore we may assume that there are no prime ideals of R strictly between P and P,. Since P is a minimal prime divisor of (a,, . . . , a,), we have (a l , . . . , a,) $ P1, say a, 4 P,. Then there is no prime ideal P' of R with P1 + (a l ) G P' c P. Hence P is the unique prime divisor of P, + (a,), from which it follows that P= Rad(P, + (a,)). Thus there is an integer t such that ait E P, + (a,) for i= 1, . . . , r. Let att = a,bi + c,, bi E R, ci E P,, and consider the ideal (c,, . . ., c,). Since (c,, . . ., c,) c P,, there is a minimal prime divisor P,' of (c, , . . . , c,) with P,' c P,. Since ai E Rad((a,, c 2 , . . . , c,)) for i= 1, . . . , r, P is the only prime divisor of (a,, c,, . . . , c,). Hence in R/P,', P/Pl' is a minimal prime divisor of the principal ideal generated by a, + P,'. If a, E P,' then (a,, . . . , a,) G Pl', which is not the case since P,' G P,. Hence a, + P,' is a nonzero element of RIP,'; it is also a nonunit in RIP,' since it is contained in P/Pl'. Therefore by Theorem 7.6, ht PIP,' = 1, and we conclude that

We can now prove the theorem by induction on r . If r = 1, the assertion follows from the result of the preceding paragraph. Assume r > 1 and that the assertion is true for any proper ideal generated by fewer than r elements. Then, since PI is a minimal prime divisor of (c2, . . . , c,), we have s - 1 5 r - 1, that is, s 5 r .

P1' = P,.

7.8. Corollary. The dimension, and therefore the height, of a proper ideal of a Noetherian ring is finite.

This does not mean that the dimension of a Noetherian ring is finite. For, the ring may have a sequence P,, P, , . . . of proper prime ideals such that

There is a result which is almost the converse of Theorem 7.7. ht P, = co (see Exercise 3).

7.9. Theorem. Let A be a proper ideal of a Noetherian ring R such that ht A = r 2 1 . Then there are elements a,, . . . , a, E A such that

ht(a,, . . . , a i ) = i for i= 1, . . . , r .

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2 THE KRULL DIMENSION OF A POLYNOMIAL RING 161

Proof. Let P,, . . ., P, be the minimal prime divisors of 0. Since ht A 2 1, A $ Pi for i = 1, . . . , K , and so by Exercise 5(c) of Chapter I1 there is an element a , E A with a, 4 Pi for i = 1, . . . , K. Let P be a minimal prime divisor of (a,). By Theorem 7.7, ht P I 1. But Pi c P for some i, so ht P = 1, and we conclude that ht(a,) = 1. Now suppose that 1 < j 5 r and that we have found elements a,, . . . , u , - ~ E A such that ht(a,, . . . , ai) = i for i = 1, . . . , j - 1. Let Q,, . . ., Qn be those minimal prime divisors of (a l , . . ., u j - l ) such that h t Q i = j - 1 for i = l , ..., n. If A s Q i for some i, then ht A < j - 1 < r , contrary to our assumption. Hence A $ Qi for i = 1, . . . , n, and so there is an element a, E A with aj 4 Qi for i = 1, . . . , n. Let P' be a minimal prime divisor of (a,, . . . , uj) , If no Qi is contained in P', then some minimal prime divisor of (a,, . . . , a,-1) of height greater than j- 1 is contained in P'; hence ht P' 2 j. If Qi c P' for some i, then ht P' 2 ht Qi + 1 = j. By Theorem 7.7, ht P' I j. Hence ht P'= j and therefore ht (a,, . . . , a,) = j .

2 T H E KRULL DIMENSION O F A POLYNOMIAL RING

Let R be a ring. In this section we shall begin an investigation of the relationship between the Krull dimension of R and that of the polynomial ring R [ X ] .

Let A be an ideal of R. Then the ideal A R [ X ] of R [ X ] consists of those polynomials of R [ X ] having coefficients in A ; hence A R [ X ] n R = A. Let 4: R --f RIA be the canonical homomorphism. Define +: R [ X ] 3 ( R / A ) [ X ] by

+(a0 + U l X + * * * + a, X") = +(%) + +(@,)X + - * * + +(an)Xn.

The mapping + is a surjective homomorphism and its kernel is precisely A R [ X ] . Hence,

"IIAWI = (RIA"1

Two special cases are of particular interest. If P is a proper prime ideal of R, then ( R / P ) [ X ] is an integral domain and we conclude that PR[X] is a prime ideal of R [ X ] . If, in addition, P is a maximal ideal of R, then ( R / P ) [ X ] is a Dedekind domain. Thus every nonzero

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162 VII DIMENSION OF COMMUTATIVE RINGS

proper prime ideal of R [ X ] / P R [ X ] is maximal. It follows that dpt P R [ X ] = 1.

7.10. Proposition. Suppose that R is an integral domain. If P is a nonzero prime ideal of R [ X ] such that P n R=0, then for every f ( X ) E R[X]\P we have

( P 3- ( f ( X ) ) ) n R # 0.

Proof. Let K b e the quotient field of R. If S = R\{O}, then S - I R [ X ] =

K [ X ] . Since P n R = 0 , P n S is empty, and consequently S - l P is a nonzero proper prime ideal of K [ X 1. As such, it is a maximal ideal of K [ X ] . Furthermore 5'-'P n R [ X ] = P. Therefore, if f ( X ) E R[X]\P, then f ( X ) $ S-IP; hence f ( X ) and S - l P generate K [ X ] . Now K [ X ] is a principal ideal domain, so S- lP is generated by some g(X) /s , where g ( X ) E P and s E S. Hence there are elements h ( X ) , k ( X ) E R [ X ] and elements u, ZI E S such that 1 = (h(X) /u) (g(X) /s )

uas = wh(X)g(X) + usk(X)f (X) E (P + (f(X))) n R and uws # 0.

+ ( k ( X ) / v ) f ( X ) . Then

7.11. Corollary. Suppose that R is an integral domain. Let P and P' be nonzero ideals of R [ X ] such that P c P'. If P is a prime ideal, then P' n R # 0.

Proof. If P n R f0 , then P' n R # O . If P n R=O and if f ( X ) E

P\P, then

0 # ( P + ( f ( X ) ) ) n R c P' n R.

7.12. Corollary. Let P I , P, , P, be ideals of R [ X ] , with P,prime, such that P , c P, c P,. Then P, n R c P, n R.

Proof. If P, n R c P, n R , we have finished. Suppose that P, n R =

P, n R = P, which is a prime ideal of R. By the remarks at the beginning of the section, R [ X ] / P R [ X ] may be viewed as the poly- nomial ring in one indeterminate over the integral domain RIP. Since P R [ X ] G P, c P, we have 0 # P,/PR[X] c P,/PR[X]. Hence, by Corollary 7.11, P,/PR[X] n R j P f 0. Thus there is an element

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2 THE KRULL DIMENSION OF A POLYNOMIAL RING 163

a E R\P and a polynomial f ( X ) E P, such that a - f ( X ) E PR[X] . If b is the constant term of f ( X ) , then b $ P. However b E P, n R, so P, n R c P3 n R.

7.13. Theorem. If dim R z d, where d is an integer, then

d + 1 <dim R [ X ] 2d+ 1.

Proof. Let Po c P, c c P, be a chain of distinct proper prime ideals of R [ X ] . By Corollary 7.12, Po n R c P, n R c - - * is a chain of distinct proper prime ideals of R. Hence if s is the number of distinct terms in the chain Po n R G P, n R c * * - c P, n R, then we must have s>( r+1) /2 . But s<d+1 , so ( r + 1 ) / 2 < d + l . Therefore r I 2d + 1, and we conclude that dim R [ X ] 5 2d + 1. On the other hand, if Qo c Q1 c . * . c Qd is a chain of d + 1 distinct proper prime ideals of R, then

!& R[X1 C !21R[X1 C ' * * C Qd R[X1 C Qd R[X1 + (x) is a chain of d + 2 distinct proper prime ideals of R [ X ] . Therefore dim R [ X ] > d + 1.

There are large classes of rings for which the lower bound for the Krull dimension of R [ X ] is realized. For example, this is true for Noetherian rings.

7.14. Theorem. If R is a Noetherian ring of finite Krull dimension, then dim R [ X ] = dim R + 1.

Proof. Let dim R = d ; we shall show that dim R [ X ] I d + 1. Let P be a proper prime ideal of R [ X ] . If we show that Pis a minimal prime divisor of an ideal which is generated by d + 1 or fewer elements, then ht P I d + 1 by Theorem 7.7, and the asserted inequality will follow. Let P' = P n R. Then P' is a proper prime ideal of R and so ht P' I d. Hence by Theorem 7.9, there are elements a,, . . . , ad E P' such that if A = (al, . . . , ad). then ht A = ht P'. It follows that P' is a minimal prime divisor of A. Suppose that P, is a prime ideal of R [ X ] such that A R [ X ] G P, G P'R[X] . Then A G P, n R G P' and so P, n R = P' ; hence P , = P'R[X]. Thus P'R[X] is a minimal

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164 VII DIMENSION OF COMMUTATIVE RINGS

prime divisor of A R [ X ] . Since A R [ X ] = a,R[X] + -. + ad R [ X ] , we have ht P’R[X] 5 d. I f P ’R[X] = P, then ht P 5 d. Suppose, on the other hand, that P’R[X] c P. Let f ( X ) E P\P’R[X], and let Q be a minimal prime divisor of A R [ X ] + ( f ( X ) ) such that Q E P. Since A c Q n R s P‘ we must have Q n R = P‘, and thus P’R[X] c Q, equality being ruled out by the fact that f ( X ) E Q. Since P‘R[X] n R = Q n R = P n R it follows from Corollary 7.12 that Q = P. Hence P is a minimal prime divisor of A R [ X ] + ( f ( X ) ) , which is generated by d + 1 elements.

7.15. Corollary. Zf R is a Noetherian ring of Jinite Krull dimension, and ;f X I , . . . , X , are independent indeterminates, then

dim R[X,, * X,,] = dim R + n.

3 VALUATIVE DIMENSION

In this section we shall show that if R is a Prufer domain, then the conclusion of Corollary 7.15 holds for R . Since Prufer domains are characterized in terms of their overrings, it seems natural to study the Krull dimension of Prufer domains by studying the Krull dimension of their overrings. We shall be particularly interested in the Krull dimension of those overrings which are valuation rings, the valuation overrings of R. We note that a valuation ring has finite Krull dimension d if and only if it has rank d (see the remark on p. 112).

Throughout this section, R will be an integral domain.

7.16. Definition. If there is a nonnegative integer k such that dim V 5 k for evety valuation overring V of R, with equality for at least one V , then R has valuative dimension k , and we write dim, R = k . If there is no such k, we write dim, R = 03.

7.17. Proposition. We always have dim R 5 dim, R.

Proof. Let Po c P, c -.. c P, be a chain of distinct proper prime ideals of R. By Exercise 3(b) of Chapter V there is a valuation overring V of R and a chain Po’ c P,’ c * . c P,‘ o f distinct proper

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3 VALUATIVE DIMENSION 165

prime ideals of V such that Pi' lies over Pi for i = 0, 1, . . . , r . The assertion follows immediately from this fact.

7.18. Proposition. If dim, R = k and if X,, . . . , X, are independent indeterminates, then dim, RIXl, . . . , X,] = k + n.

Proof. Let V' be a valuation overring of R[X,, . . . , X,], and let V be the intersection of V' and the quotient field K of R. Then V is a Valuation overring of R and so dim V _< k. Since V n K = V and XI, . , . , X, E V', we have dim V' 5 k + n by Theorem 5.24. Thus, dim, R[X,, . . . , X,] 5 k + n. Now there is a valuation overring W of R with dim W = k , and by Theorem 5.24 there is a valuation overring W' of R[X,, . . . , X,] with dim W' = k + n. Therefore, dim, R[X,, . . . , X,] = k + n.

7.19. Proposition. Suppose that dim R = dim, R. If X,, . . . , X, are independent indeterminates, then dim R[X,, . . . , X,] = dim R + n.

Proof. I t follows from the two preceding propositions that

dim R[X,, . . . , X,] dim, R[X, , . . . , X,] = dim, R + n = dim R + n. The reverse inequality follows from Theorem 7.13.

7.20. Theorem. If R is a Prufer domain and if X1, .. ., X, are independent indeterminates, then dim R[X,, . . . , X,] = dim R + n.

Proof. It is sufficient to show that dim R >dim, R. Let V be a valuation overring of R and let P be the intersection of R and the maximal ideal of V. Then R, E V. Since R is a Priifer domain, R, is a valuation ring. Hence dim V 5 dim R, = ht P 5 dim R. Therefore, dim, R 5 dim R.

The converse of this result is true under the additional hypothesis that dim R = 1. In order to prove this we shall require two prelim- inary results.

7.21. Proposition. Suppose that R is integrally closed and that R has a unique maximal ideal P . Let a belong to the quotient jield of R and

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166 VII DIMENSION OF COMMUTATIVE RINGS

assume that a $ R and l / a $ R. Then PR[a] is a nonmaximal proper prime ideal of R[a].

Proof. Suppose that PR[a] + aR[a] = R[a]. Then there is a positive integer k, elements b,, . . . , bk E R, and an element 6, E P such that 1 = 6, + bla + * * * + bk aka Then 6, - 1 is a unit in R and

( l / u ) k + ( b l / ( b O - l))(l/u)k--' + * * . f bk/(bO - l) = O,

so that l/a is integral over R. Since this is not the case, we conclude that PR[a] + aR[a] # R[a].

Let f ( X ) be a nonconstant monic polynomial in R [ X ] , and let g(X) = f ( X ) - f (u) . If f (u) E R then, since g(X) is monic and g(u) = 0, u is integral over R, which is not true; hence f ( u ) $ R. In particular, f ( u ) #O. Suppose l i f (u) E R. Then l i f (u) E P and so 1 ~ f ( a ) P c PR[a], which is not true; hence lif(a) $ R. Therefore by what was proved above,

Wf(41 +f(a)R[f(a)l f "(41.

Since g(u) = 0, R[a] is integral over R[f(u)]. Hence by Exercise 3(c) of Chapter IV, P andf(a) together generate a proper ideal of R[a], that is,

P R [ 4 +f(a)"l # R[al.

Since 1 + f ( X ) is also nonconstant and monic this implies that

PR[al + (1 +f(.))R[al # R [ 4 .

It follows immediately that f (u) 4 PR[a]. T o summarize, we have shown that PR[a] is a nonmaximal proper

ideal of R[a], and that iff(X) is any nonconstant monic polynomial in R [ X ] , thenf(a) 6 PR[a]. To complete the proof we shall show that under the homomorphism R [ X ] --f R[a] given by h ( X ) H h(a), the inverse image of PR[a] is PR[X] . It will follow that PR[a] is a prime ideal of R[a].

Let h ( X ) E R[X]\PR[X] ; we shall show that h(a) 4 PRta]. Write h(X)=h,(X)+h,(X), where h,(X) has no coefficient in P and h,(X) E PR[X] . Then h , (X) # 0 and its leading coefficient d is a unit in R. Then h,(X)/d is a nonconstant monic polynomial in R [ X ]

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3 VALUATIVE DIMENSION 167

and so h,(a)/d $ PR[a]. Therefore, h,(a) $ PR[a], and consequently $ PR[aI.

7.22. Proposition. If R is integrally closed and if dim,, R = 1, then R i s a Priifer domain.

Proof. If P is a maximal ideal of R, then R, is integrally closed by Exercise 7(b) of Chapter IV, and dim,, R, = 1. Hence, in proving the assertion, we may assume that P is the unique maximal ideal of R and show that R is a valuation ring. Since R is not a field we have dim R = 1 by Proposition 7.17 ; hence dim R [ X ] = 2 by Proposi- tion 7.19.

Assume that R is not a valuation ring. Then there is an element a in the quotient field of R such that a 6 R and l/a 6 R. By Proposition 7.21, dim R[a] 2 2. Consider the homomorphism R [ X ] +- R[a] given byf(X) H f (u ) . If a = b/s, where b, s E R, and if f ( X ) = sX - b, then f ( a ) = 0 ; hence the kernel of this homomorphism is a nonzero prime ideal of R [ X ] . Therefore, dim R [ X ] 2 3, which is a contradiction.

7.23. Theorem. If dim R = 1 and if X is an indeterminate, then dim R [ X ] = 2 if and only if the integral closure of R is a Priifer domain.

Proof. Let R' be the integral closure of R. By Exercise 7, dim R' =

dim R and dim, R'=dim,R. If R' is a Prufer domain, then dim R' = dim, R', and so dim R = dim, R, and dim R [ X ] = 2 by Proposition 7.19.

Conversely, suppose that dim R = 1 and dim R [ X ] = 2. We shall show that this implies that dim, R = 1 ; it follows then that dim,, R' = 1, and that R' is a Prufer domain by Proposition 7.22.

Suppose that dim, R 2 2. Then there is a valuation overring V of R with dim V 2 2. Hence there are proper nonzero prime ideals P, and P, of V such that P, c P, . Choose a E P,\P, and consider the ring R[a]. We have P , n R[u] c P, n R[a] # R[a]. If b/s is a non- zero element of P,, where b, s E R, then b = s(b/s) is a nonzero element of P, n R[a]. Therefore, dim R[u] 2 2. Now, just as in the final paragraph of the proof of Proposition 7.22, this implies that dim R [ X ] 2 3, contrary to assumption.

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168 VII DIMENSION OF COMMUTATIVE RINGS

EXERCISES

1. Heights of prime ideals. Let R be a Noetherian ring and let P be a proper prime ideal of R. (a) Show that ht P i n if and only if P is a minimal prime

divisor of an ideal of R generated by n elements. (b) Suppose that ht P 2 2. Show that there are infinitely many

prime ideals of height one contained in P.

2. Dimension of certain rings. Let R be a Noetherian ring and X an indeterminate. (a) (b)

Show that dim R(X) = dim R. Let a be an element of some ring having R as a subring, and suppose that f(a) = 0 for some nonzero f ( X ) E R[X] . Show that dim R[a] _< dim R.

3. A Noetherian ring of infinite dimension. Let R be a iing such that for every maximal ideal P of R the ring R, is Noetherian and such that each nonzero element of R is contained in only a finite number of maximal ideals. Show that R is a Noetherian ring. Let K be a field and let R = K[X,, X , , . . .] be the ring of polynomials in infinitely many indeterminates XI, X , , . , . over K. Let n,, n,, . . . be positive integers such that 0 < n2 - n, < n3 - n, < - * . Let Pi = (X,, , . . . , X,, + ,). Show that Pi is a prime ideal of R and that ht Pi= %+, - ni *

Let R be the ring of part (b) and let S = R\ Uy= , Pi. Show that S-IR is a Noetherian ring and that dim S-IR = m.

4. Local rings. Recall that R is a local ring if it has a unique maximal ideal. Let R be a Noetherian local ring which is not a field and let P be its maximal ideal. Let d i m R = d , and note that d is finite since d = h t P . (a) Show that there exist d elements a,, . . . , ad E P su,ch that

(u,, ..., ad) is P-primary. Then a,, ..., ad is called a system of parameters of R. Show that no set of fewer than d elements of P generates a P-primary ideal.

(b)

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EXERCISES 169

(c)

(d)

Show that an element a E P belongs to a system of para- meters of R if and only if dim R/(a) = dim R - 1. Show that each of the following is sufficient for an element a E P to belong to a system of parameters of R. (1) dim R/(a) < dim R. (2) a is not a zero-divisor of R. (3) There is no proper prime ideal P' of R with a E P' and

dpt P' = dim R. Let a,, . . . , a, E R be such that ht(a,, . . . , ak) = k. Show that dim R/(a,, . . . , ak) = dim R - k.

(e)

Modules over a local ring. Let R be a Noetherian local ring, let P be its maximal ideal, and let M be a finitely generated R-module. (a) Show that a subset {xal a E I } of M generates M if and only

if {xa + PMI a E I } generates M/PM as a vector space over RIP. Thus {x,, . . . , xk} is a minimal set of generators of M if and only if {xl + PM, . . ., xk + P M } is a basis of MIPM over RIP. Let {xl, . . . , xk} be a minimal set of generators of M . Let A = [aij] be a k x k matrix over R and let yi = C:= aij xj for i= 1, . . . , k. Show that {yl, . . . , yk} is a minimal set of generators of M if and only if det A $ P.

5.

(b)

6. Regular local rings. Let R be a Noetherian local ring which is not a field and let P be its maximal ideal. Then R is called regular if there is a system of parameters of R which generates P.

Show that if k is the"dimension of the vector space PIP" over RIP, then k 2 dim R, with equality if and only if R is regular. Let a,, . . . , a, E P. Show that if R is regular, then these elements belong to a system of parameters which generate P if and only if a, + P2, . . . , a, + P2 are linearly inde- pendent elements of PIP2. Show that this is the case if ht(a1, . . . , an) = n. Let a,, . . . , a, E P be such that ht(a,, . . . , a,) =n. Show that if R/(a,, . . ., a,) is regular, then R is regular. Suppose that R is regular and let A be a proper ideal of R. Show that R/A is regular if and only if A is generated by a subset of a system of parameters which generate P.

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170 VII DIMENSION OF COMMUTATIVE RINGS

7. Dimension of integral extensions. Let R' be integral over a subring R. (a) Show that dim R' = dim R. (b) Assume that R' is an integral domain and show that

dim, R' = dim, R. (c ) Let R be an integral domain and let X,, . . . , X , be inde-

pendent indeterminates. Show that if the integral closure of R is a Priifer domain, then dim R[X,, . . . , X,] = dim R + n.

8. Dimension of other extensions. Let R be an integral domain with quotient field K and let XI, . . . , X,, be independent indeterminates. (a) Let a,, . . . , a, be distinct elements of K and let P be the

kernel of the homomorphism R[X,, . . . , X,] + R[a,, . . . , a,] given byf(X,, . . . , X,) ~ f ( a , , . . . , a,). Show that ht P = n. Show that the following statements are equivalent. (1) If T is an overring of R, then dim T 5 n. (2) If V is a valuation overring of R, then dim V 2 n. (3) If a,, . . . , a, E K, then dim R[a,, . . . , a,] 4 71.

(4) dim R[X,, . . . , X,] 5 2n. Show that the following statements are equivalent. (1) dim, R =n. (2) If T is an overring of R, then dim T s n , and

dim T = n for at least one such T . (3) If a,, .. ., a, E K, then dim R[a,, ..., a,] < n , and

dim R[a,, . . . , a,] = n for some choice of a,, . . . , a , . (4) dim R[X,, . . . , X,] = 2n.

(b)

( c )

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C H A P T E R

VIII Krull Domains

1 KRULL DOMAINS

Let R be a Dedekind domain which is not a field, and let K be the quotient field of R.

(I) There is a farnib {va 1 M E I ] of discrete rank one valuations on K such that ij V , is the valuation ring of v, , then

R = n v,. a e l

In fact, let {Pa 1 a E I> be the family of maximal ideals of R. For each a E I , V , = Rpa is a Noetherian valuation ring. The valuation v, determined by V , has rank one and is discrete by Theorem 5.18, and R is the intersection of the V , by Exercise 5(a) of Chapter 111.

(11) For every a E KX, the set of a in 1 for which v,(a) # 0 is jinite.

For, let a E K", then a = b/c where b, c E R, and both b and c are not zero. For each M E I , v,(a) = v,(b) - v,(c), so it is sufficient to verify the assertion when a E R. However if a E R, then v,(a) # 0 if and only if a E Pa, and we know that a is contained in only a finite number of maximal ideals of R [see Exercise ll(a) of Chapter VI].

171

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172 VIII KRULL DOMAINS

8.1. Definition. An integral domain which is not a field is a Krull domain i f it satisfies ( I ) and ( I I ) .

We have shown that every Dedekind domain which is not a field is a Krull domain. It will be clear from results we shall obtain that there are Krull domains which are not Dedekind domains. Many of the assertions we shall make about Krull domains will also be true of fields. However, we choose to exclude fields from the class of Krull domains.

Let R be an integral domain with quotient field K, and let %(R) be the set of nonzero fractional ideals of R. If A E %(R) we call the fractional ideal

[R: A ] = {at a E K and Aa G R }

the quasi-inverse of A. Two nonzero fractional ideals A and B are said to be quasi-equal if their quasi-inverses are equal; in this case we write A N B. Clearly, quasi-equality is an equivalence relation on the set %(R). We denote the set of equivalence classes of %(R) with respect to this relation by 9(R). The elements of 9(R) are called divisors of R. If A E%(R), the divisor represented by A will be denoted by div A. If A = Ra for some nonzero a E K, we shall write div a instead of div Ra.

8.2. Proposition. If A, €3 E %(R), then A N B i f and only i f A and B are contained in the same principal fractional ideals.

Proof. This follows immediately from the fact that if a E K*, then A G R a if and only if l / a E [R: A].

8.3. Corollary. Let A E %(R) and set

A = n R ~ . A s Ra

Then A E %(R) and A N A.

Proof. Certainly, A is a nonzero R-module. If A d c R where d E R, d # 0, then A c_ R( l /d ) ; hence A s R( l /d ) and so Ad E R. Thus A d E %(R) and it is clear that A N A.

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1 KRULL DOMAINS 173

8.4. Corollary. If A E g(R) , then, in the set of fractional ideals quasi-equal to A, there is one which contains all others, namely, A.

Note that if A, B E F(R), then A N B if and only if A = 8.

8.5. Definition. A fractional ideal A is divisorial if A = A.

Each divisor of R is represented by exactly one divisorial frac-

We can define a partial ordering on 9(R) by

div A < div B i f B s A.

tional ideal.

8.6. Proposition. If A, B, C E 9 ( R ) and if div A 5 div B, then div AC 5 div BC.

- - Proof. We shall show that BC C_ AC. By hypothesis, B _C A; that is, [R: A] c [R: B ] . Then

[ R : AC] = [ [ R : A ] : C] E [ [ R : B ] : C ] = [R: BC],

so that BC G x. If A, B E %(R), we define the sum of the divisors div A and

div B by

div A + div B = div AB.

This is a well-defined operation on 9 ( R ) since

[ R : AB] = [ [ R : A ] : B] = [ [ R : A ] : B] = [ [ R : B ] : A] = [ [ R : B ] : A] = [R: AB]

implies that div A + div B = div A + div 8. With this operation, g(R) is a commutative semigroup with identity element div R. It follows from Proposition 8.6 that if A, B, C E %(R), and if div A 5 div B, then

div A + div C < div B +div C.

Thus, 9(R) is a partially ordered semigroup. We shall show that, in fact, 9(R) is a lattice ordered semigroup; this means that with

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174 VIII KRULL DOMAINS

respect to the given partial ordering, 9(R) is a lattice; that is, every pair of elements in 9 ( R ) has a least upper bound and a greatest lower bound. Lattice ordered semigroups are also called multiplicative lattices.

Let A, B E F(R) . We shall show that div(A n B ) is the least upper bound and div(A + B ) the greatest lower bound of the subset {div A, div B} of B(R). We may assume A and B are divisorial; then A n B is divisorial (for, A n B = A n B = A n B ) and the assertion concerning div(A n B ) follows immediately. Now suppose that C E F(R) and that div C 5 div A and div C 5 div B. Then A L e and B _c c; hence A + B G e and so A + B G e. Therefore div C 5 div(A + B ) . I t follows that div(A + B ) is the greatest lower bound of {div A, div B).

If R is a Dedekind domain, then 9 ( R ) = F(R) , and consequently 9(R) is a group. We shall now determine a necessary and sufficient condition, stated in terms of R itself, for 9(R) to be a group.

8.7. Proposition. 9 ( R ) is a group ;f and only if R is completely integrally closed.

Proof. Let K be the quotient field of R. Suppose that R is completely integrally closed. We shall show that for all divisorial fractional ideals A in F(R) , A[R: A] and R are quasi-equal. Since A[R: A] G R it follows from Proposition 8.2 that it is sufficient to show that if A[R: A] c Ra, then R G Ra; that is, if A[R: A] G Ra, then l /a E R. Now suppose that A E Rb for some b E K ; then l l b E [R: A] and so ( l /b)A c Ra, that is, ( l / a ) A c Rb. Thus ( l /a)A E A = A , and it follows that (l/a)"A s A for all positive integers n. Let d E R, d # 0, be such that d A E R. Then d( l /a ) "A G R, and if c E A, c # 0, we have (dc)(l/a)" E R for all positive integers n. Since R is completely integrally closed, and since dc E R and dc # 0, this implies that l / a E R by Theorem 4.20.

Conversely, suppose that 9 ( R ) is a group. Let a be a nonzero element of K and suppose that there exists d E R, d # 0, such that da" E R for all positive integers n. Then R[a] is a fractional ideal of R and aR[a] c R[a]. But then div R[a] 5 div a + div R[a], and since 9 ( R ) is a group this implies that div a 2 div R ; that is, a E R.

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1 KRULL DOMAINS 175

I n Theorem 5.19, we showed that a valuation ring has rank one if and only if it is completely integrally closed. Clearly the intersection of completely integrally closed integral domains is completely integrally closed. Hence we immediately obtain :

8.8. Corollary. If R is a Krull domain, then 9(R) is a lattice ordered group.

Let R be a Krull domain. Let {v,I a E I} be a family of discrete rank one valuations on the quotient field K of R which satisfy (I) and (11). It is clear that if we replace any of the v, by an equivalent valuation, then (I) and (11) continue to hold. Since each v, is discrete, it is equivalent to a valuation on K having as its value group the additive group of integers. Such a discrete rank one valuation is said to be normed, and we shall assume henceforth that each v, is normed. If A E 9 ( R ) and if a E I, we set

v,(A) = max(v,(u) I A c Ru).

This maximum always exists, for if c E A, c # 0, then v,(c) 2 v,(a) whenever A G Ra.

8.9. Proposition. If R is a Krull domain and if A E F(R), then v,(A) # 0 for only aJinite number of tc in I .

Proof. Let d E R, d # 0, be such that A c Rd. Then v,(A) 2 va(d) . If c E A, c # 0, then v,(c) 2 v,(A). By (11) there is only a finite number of a E I for which either v,(c) # 0 or v,(d) # 0. The assertion follows.

For each CI E I, let 2, = 2, the additive group of integers, and consider the group

Z"'= 0 2,. a E I

For the Krull domain R we can define a mapping C$ from 9(R) into 2") by setting C$(div A), = v,(A) for all a E I . This is a well-defined mapping since it is clear from Proposition 8.2 that if A and B are quasi-equal elements of F(R), then v,(A) = v,(B) for all a E I. It is a consequence of the next proposition that C$ is one-to-one.

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176 VIII KRULL DOMAINS

8.10. Proposition. Let R be a Kru11 domain and let A and B be divisoriat fractional ideals of R. Then A s B if and only ;f v,(A) 2 V, ( B ) for all ci E I .

Proof. We need prove only the sufficiency of the condition. Assume v,(A) >v,(B) for all ci E I. Let a E A ; then for each a E I we have v,(a) 2 va(A). If B G Rb, then v,(B) 2 va(b), so v,(a) 2 v,(b). Hence v,(a/b) 2 0 ; that is, a/b is contained in the valuation ring of v, . Since this is true for each ci E I, it follows from (I) that a/b E R. Hence a E Rb. Therefore a E n,,,, Rb = B. Thus A G B.

If m, n E P), we define m 5 n if m, 5 na for all ci E I. This is a partial ordering on Z(I), and if m,, m 2 , n E 2") and m1 5 m,, then m, + n 5 m2 + n. In fact, with this ordering, 2") is a lattice ordered group; the least upper bound h and the greatest lower bound k of elements m, n E 2'" are given by

ha = max{m, , n,} k, = min{m, , n,} for all c1 E I . An element m E 2'" is called positive if m > 0.

8.11. Proposition. Every nonempty set of positive elements of Z(I) has a minimal element.

Proof. Let S be a nonempty set of positive elements of Z(I) and let n E S. Then n, 2 0 for all t( E I, and there is a finite subset J of I such that n, > O for ci E Jand n, = O for ci E I\J. If m E S a n d m l n , then ma = 0 for c1 E I\ J and 0 5 ma 5 n, for a E I. Hence there is only a finite number of such m and among them there is a minimal element of S.

If R is a Krull domain and if we compose the mapping 4 and the mapping A H div A, where A is divisorial, we obtain a one-to-one mapping + from the set of divisorial fractional ideals of R into 2"). If A G B, then +(A) 2 $ ( B ) ; in particular, if A G R, then $(A) 2 0. Hence if 9 is a nonempty set of divisorial ideals of R, then {$(A) I A E 9) has a minimal element. Therefore Y has a maximal element.

We have verified the necessity of the conditions in the following assertion, which is the principal result of this section.

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1 KRULL DOMAINS 177

8.12. Theorem. An integral domain R which is not afield is a Krull domain if and only if

(a) (b)

R is completely integrally closed, and every nonempty set of divisorial ideals of R has a maximal element.

Let R be an integral domain which is not a field for which (a) and (b) hold. By Proposition 8.7, 9(R) is a lattice ordered group. We shall denote elements of 9(R) by lower case German letters. If a and b are in 9(R) we denote the least upper bound and greatest lower bound of the set {a, b} by a u b and a n 6, respectively.

As a consequence of (b), every nonempty subset of positive elements of B(R) has a minimal element. Let { p a I a E I} be the set of all minimal positive elements of 9(R) ; this set is not empty since R is not a field. For each c1 E I let Pa be the divisorial ideal of R such that pa = div Pa. Then {Pa I a E I } is the set of maximal divisorial proper ideals of R.

Let a E g ( R ) and let b = a u O ; then a=b- (b -a ) , and we have b 2 0 and b - a 2 0. Thus, every element of 9(R) can be writ- ten as the difference of two elements of 9(R) each of which is greater than or equal to zero. We shall show that every element of 9(R) can be written as a linear combination of a finite number of elements of the set ( p a 1 c1 E Z} with integer coefficients. By the remark made above, it is sufficient to show that each positive element of 9(R) can be written as a sum of minimal positive divisors. Suppose this is not true, and that a is minimal among those positive divisors of R which cannot be so written. Then a is not a minimal positive divisor of R and so there is an a E Z such that 0 < p a < a. Then 0 < a - p a < a and so a - p a can be written as a sum of minimal positive divisors. But a = p a + (a - p a ) and so we have contradicted our choice of a.

Thus if a E 9(R), we have

a = c % P a , a E l

where each n, is an integer and na # 0 for only a finite number of a E I. We shall show that this expression is unique.

First of all, suppose that a and b are positive elements of 9 ( R ) and that p a < a + 6. Since p a is a minimal positive element of Q(R), either p a n a = p a or p a n a = 0. I n the former case, p a 5 a.

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178 VIII KRULL DOMAINS

In the latter case, ( p a + b) n (a + b) = b. T o see this, note that b 5 ( p a + 6) n (a + b). If c I p a + b and c I a + 6, then c- b 5 p a and c - b i a. Therefore c - b 5 p a n a = 0 and hence c b. This proves that b 2 ( p a + 6) n (a + 6). Since p a I p a + b and p a I a + b, it follows that p a 5 b. By an induction argument we can show that if p a 5 a1 + * . - + a,, where each ai is positive, then for some i, we have p a 5 a,.

Now suppose that

where each sum has only a finite number of nonzero terms. Assume nu - n,' > 0 for some CI E I and let

J = { a l a E I and n,-n,'>O}.

Then with ma = n, - n,' whenever ci E J and ma = n,' - nu other- wise, we have

Since J is not empty, both sides of this equation are nonzero. By what we have shown in the preceding paragraph, it follows that if a EJ, then p a 5 p, for some f l E I\ J . By the minimality of p , we conclude that p a = p, , which is not true.

Let K be the quotient field of R. If a E K", then

d i v a = 1 napa , a e I

where each n, is an integer uniquely determined by a, and n, # 0 for only a finite number of a E 1. We set va(a) = n, . Then u, is a mapping from K* into the additive group of integers, which we extend to all of K by setting va(0) = a. If b E K" and

div b = n i p a , a e l

then

div ab = div a + div b = C (n, + na')pa.

Hence va(ab) = nu + n,' = u,(a) + v,(b). We know that (div u) n

a s 1

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(div b) = div(Ra + Rb), and R(a + b) G Ra + Rb, so that div(a + b) 2 div(Ra + Rb). Furthermore

Hence either a + b = 0, or

div(a + b) = ma pa 2 C (min{n, , n,")p,. a E I a E I

It follows that ma 2 min{n, , n,'} for each u E I. Therefore v,(a + b) 2 min{v,(u), v,(b)}, even when a+ b = 0 . Thus (ii) and (iii) in the definition of valuation hold for v, when a, b E K*. Since it is clear that they hold when either a=O or b = 0 , we see that for each u E I, v, is a valuation on K.

Consider the family of valuations {v, I cc E I}. For each u E I, elcr

has rank one and is discrete. It is immediate that (11) holds. If a E K* and v,(a) 2 0 for all u E I, then div a 2 0 ; hence Ra c R ; that is, a E R. Thus (I) also holds. Therefore, R is a Krull domain. This concludes the proof of Theorem 8.12.

Since an integrally closed Noetherian integral domain is com- pletely integrally closed [Exercise l(c) of Chapter Iv], and the converse holds even for non-Noetherian domains, we have the

8.13. Corollary. A Noetherian integral domain which is not aJield is a Krull domain if and only if it is integrally closed.

2 ESSENTIAL VALUATIONS

Let R be a Krull domain with quotient field K. We shall continue with the notation of Section 1. In particular, let {pa I 01 E I} be the set of minimal positive elements of 9(R), and for each 01 E I, let Pa be the divisorial ideal of R such that p a = div Pa . Let {v, I u E I} be the family of valuations on K constructed as in Section 1 ; (I) and (11) hold for this family of valuations. The valuations in this family are called the essential valuations of R.

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180 VIII KRULL DOMAINS

We have shown that if a E 9(R), then we can write uniquely

C n u p a , a E I

where n, is an integer for each a E I, and n, # 0 for only a finite number of a E I. If b E Q(R) and

then a i b if and only if n, 2 ma for all a E I. Hence if A is the divisorial fractional ideal of R such that a = div A, then ~ E A if and oniy if va(a) 2 nu for all a E I. Thus every divisorial ideal of R can be described by such a set of inequalities.

Conversely, let {n, I a E I} be a set of integers only a finite number of which are not zero. Then

A = { a l a ~ K and v,(u)>n, for all C L E ~ }

is a divisorial fractional ideal of R; it is the one whose divisor is - L a c rna Pa -

If A is a fractional ideal of R, then by Proposition 8.10, a E A if and only if v,(a) 2 va(k?) = v,(A) for all a E 1.-Therefore

div A = C va(A)pu. a s 1

If for each E J, A, is a divisorial fractional ideal of R, and if A = C, E , A, is a fractional ideal, then div A is the greatest lower bound of the set {div A, I j3 E J}. We leave the proof of this as an easy exercise for the reader.

Ra. Hence div A is the greatest lower bound of the set {div Ra I a E A, a # O}. Therefore

div A = C (min{w,(a) I a E A})p, .

On comparing the two expressions that we have for div A, we con- clude that for all a E I

wa(A) = min{v,(a) I a E A}.

If A is a fractional ideal of R, then A = Ca

a E I

8.14. Proposition. Each of the essential valuations of R is normed.

Proof. Let a E I; then p a < 2p, and so, if A is the divisorial ideal of R such that 2p,=div A, we have A c P a . Let U E P,\A. Then p a 5 div a < Z p , , so 1 5 w,(a) < 2. Therefore v,(a) = 1 .

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2 ESSENTIAL VALUATIONS 181

Let 4: 9 ( R ) -+ 2'') be defined by +(div A), = va(A) for all c i E I . We have seen that r$ is a one-to-one mapping.

8.1 5. Proposition. The mapping 4 is an order-preserving isomorphism from g(R) onto 2")

Proof. First we shall show that 4 is a homomorphism by showing that for divisorial fractional ideals A and B, we have v,(AB) = v,(A) + va(B) for each c1 E I. Let a , b E K be such that

A c Ra, B E Rb, va(A) = va(a), and va(B) = a,@).

Then

AB c_ Rub, and v,(AB) 2 v,(ab) = va(A) + w,(B).

On the other hand, let c E K be such that AB _C Rc and v,(AB) = v,(c). If x E B, x # 0, then Ax G Rc, or A E R(x/c). Hence

un(A) 2 Va(c/x) = va(AB) - va(x);

thus v,(A) + v,(x) 2 v,(AB). By the remarks made above we may choose x so that va(B) = v,(x) ; then v,(A) + va(B) 2 v,(AB). It follows from Proposition 8.10 that 4 is order-preserving. T o show that 4 is surjective we note simply that if n E Z('), then rjh(Cae, n, p a ) = n.

8.16. Proposition. For each a E I , Pa is a prime ideal of R. Further- more if V , is the valuation ring of va , then V , = Rpa .

Proof. If a E R, then v,(u) > 0 if and only if a E P a . Thus Pa is the intersection of R and the ideal of nonunits of V,. Hence Pa is a prime ideal. To prove that V,= RPa, we note that since V , is a Noetherian valuation ring it is sufficient, by Exercise l(a) of Chapter V, to show that Va E Rpa c K. Let a E Va and let ul, . . . , ctk be those elements of I such that v,,(a) < 0 for i = 1, . . . , k. For each i , cli # a, so there is an element si E R\P, such that nai(si) > 0. Then we can choose n so large that vai(asin) 2 0 for i = 1, . . . , k. If s = (sl * * s ~ ) ~ then vB(as) 2 0 for all /3 E I . Hence as = b E R and a = b/s E RPa . Since Pa # 0 we have Rp, # K .

8.17. Proposition. Let P be a nonzero proper prime ideal of R. Then P is divisorial if and only if P = Pa for some c( E I .

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182 VIII KRULL DOMAINS

Proof. Suppose that P is divisorial. Since

div P = C n, pa = C nu div Pa = C div P,"" = div n P,"", ( I E I a E I ( I E I ( I E I

we have nae I P,"" G P. Then Pa G P for some a E I, and since Pa is maximal among the divisorial proper ideals of R, we have P = P a .

Now we are able to describe completely the ideals Pa of R in terms of ideal-theoretic properties of R. An ideal of R is called a minimal prime ideal if it is minimal among the nonzero prime ideals of R.

8.18. Proposition. The family {Pal CL E I } is precisely the family of minimal prime ideals of R.

Proof. We have seen that Pa is prime and that Rpa is a Noetherian valuation ring. Hence by Theorem 5.9, Pa R,, is the only nonzero proper prime ideal of R,, . Then by Corollary 3.1 1, there are no prime ideals of R strictly between 0 and Pa . Then Pa is a minimal prime ideal of R. If P is a minimal prime ideal of R, then by Exercise 5 and Proposition 8.16, there is a subset J of I such that R, = naEJ R,, . Hence if a E J, R, 5 R,, and so Pa G P. Since P is minimal, P = P a .

8.19. Theorem. Let R be a Krull domain with quotient field K . Let R be afinite extension of K and let R' be the integral closure of R in K'. Then R' is a Krull domain. A prime ideal P' of A' is a minimal prime ideal of R if and only if P' n R is a minimal prime ideal of R.

Proof. Let {va[ a E I) be the family of essential valuations of R. Let {v8'(/3 EJ} be the family of valuations on K' determined by the following conditions :

(i) (ii)

Let v,' be the valuation ring of VB' and set R" = n8

Every 08' is a prolongation of some z), . Every prolongation to K' of each v, is equivalent to exactly

one v8'.

v8'. Let a E R', a # 0 ; then there are elements b, , b,, . . . , b,- E R such that

b, + bla + * * + bn-1an-l + a" = 0.

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2 ESSENTIAL VALUATIONS 183

Let /3 E J and suppose that vgr(a) < 0. Then vs'(bi at) > vB'(an) for i = 0, . . . , n - 1, and consequently

vg'(an) < min{v,'(b,), ~g'(b la) , . . . , vg'(bn-lan-l)}

(vs'(b, +b,a+~-*+b,-,a"-l) ,

which is not true. Hence vg'(a) 2 0; that is, a E Vg'. This proves that R' c R".

On the other hand, suppose that a E R". Let f ( X ) be the monic irreducible polynomial in K [ X ] having a as a root. By the lemma proved below, f ( X ) has coefficients in R. Hence a E R', and we have shown that R' = R". In doing so, we have verified that (I) holds for R' and the family of valuations {vgr IS E J } .

Again let a E R , a # 0, and let

bo + bla + - * * + b,-l~"-l + an = 0,

where b, E R for i = 0, . . . , n - 1. For all but a finite number of tl E I , we have v,(b,) = O for i= 0, . . ., n- 1. Choose such an tl E I ; we shall show that if vB' is a prolongation of v, , then v,'(a) = 0. This will show that (11) holds for R' and the family of valuations {vS'1/3 EJ} because, by Theorem 6.25, each v, has only a finite number of prolongations in the family (vgr I /3 €1).

If vgr(a) > 0, then

wg'(bo) < min{v,'(b,a), . . . , vg'(bn- vBr(an)}

<v,'(b,a + - a * +b,-lan-l +a"),

and if v,'(a) < 0, then

vg'(an) < vg'(bO+b,a+...+b,-,a"-'),

both of which are not true. Thus we must have vB'(a) = 0.

over theorem and the going-down theorem. The final assertion of the theorem is a consequence of the lying-

8.20. Lemma. Let V be a valuation ring, K the quotient field of V, and v the valuation on K determined by V . Let K' be a finite extension o f K a n d l e t a E K ' . Let

f ( X ) = co + c,x + - . - + c, - 1 x n - 1 + X" be the monic irreducible polynomial in K [ X ] having a as a root. If v'(a) 2 0 for every prolongation v' of v to K', then c o , cl, . . . , c,-, E V .

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184 VIII KRULL DOMAINS

Proof. Let K be a finite normal extension of K with K' 5 K . Let v" be a prolongation of v to K . Let u E G(K"/K) and define a mapping 8 from K" into the value group of v" by @(b) = v"(u(b)) for all b E K". It is easily seen that 5 is a valuation on K , and that its restriction to K' is a prolongation of v to K . Hence 8(a) 2 0 ; that is, v"(u(a)) 2 0. Now let a,, . . . , a,, be the roots of f ( X ) in K", with repeated roots listed as often as their multiplicity. For each i there is an element u E G(K"/K) such that .(a) = a, . Hence, ~ " ( a , ) >_ 0 for i = 1, . . . , n. Since each coefficient of f ( X ) is an elementary sym- metric function of a,, . . . , a,, we have v"(ci) 2 0; that is, v(ci) 2 0 for i = O , . . . , n - 1.

8.21. Theorem. If R is a Krull domain, then the polynomial ring R [ X ] is a Krull domain. A prime ideal P' of R [ X ] is a minimal prime ideal of R [ X ] if and only if either P' = P R [ X ] for some minimal prime ideal P of R, or P' = Q n R [ X ] where Q is a nonzero proper prime ideal of K [ X ] , where K is the quotient jield of R.

Proof. If p(X) is a nonconstant monic irreducible polynomial in K [ X ] , let up(,) be thep(X)-adic valuation on K ( X ) , and let V,,,, be the valuation ring of vp(x). Let (val c1 E I } be the family of essential valuations of R. For each QEI, let v,' be the extension of v, to K ( X ) , and let V,' be the valuation ring of v,'. Set

It is immediate that R [ X ] 5 R'. On the other hand, let f ( X ) E R', where f ( X ) is a rational function over K . Since f ( X ) E V,,,, for each nonconstant monic irreducible polynomial p ( X ) in K [ X I , it follows that f ( X ) is a polynomial in K [ X I . Then since f ( X ) E V,' for each o! E I, every coefficient of f (X) is in V , for each o! E I . Thus f ( X ) E R [ X ] . We conclude that R' = R [ X ] , and that (I) holds for R [ X ] and the family of valuations consisting of all of the vp(x) and all of the v,'. It is easy to see that (11) holds also, and the details of this argument are left to the reader. Therefore R [ X ] is a Krull domain.

By Exercise 7(b), the various vpCx) and v,' form precisely the family of essential valuations of R [ X ] . Hence a prime ideal of R [ X ] is a minimal prime ideal if and only if it is the intersection with R [ X ] of the ideal of nonunits of some V p c x , or some Va'. The ideal of

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3 THE DIVISOR CLASS GROUP 185

nonunits of V,,,, is QV,,,, , where Q is the principal ideal (p(X)) of K[X] and

Q v p ~ x ) n RXI Q n R[XI*

Furthermore every nonzero proper prime ideal of K[X] is of this form. Let Pa’ be the ideal of nonunits of Va’; we shall show that Pa’ n R[X] = Pa R[X]. We have P,‘ n R = P a , so that Pa R[X] E Pa‘ n R[X]. If f(X) E P,‘ n R[X], then every coefficient of f(X) lies in Pa; hencef(X) E P,R[X]. Therefore, P,‘ n R[X] = P, R[X]. This completes the proof of the theorem.

8.22. Corollary. If R is a Krull domain, then the polynomial ring R[X,, . . . , X,] in k indeterminates is a Krull domain.

3 THE DIVISOR CLASS GROUP

Let R be a Krull domain with quotient field K. We have seen in the preceding section that the family of essential valuations of R is in one-to-one correspondence with the family of minimal prime ideals of R. Thus we may index the essential valuations of R with the minimal prime ideals of R. If P is a minimal prime ideal of R, then z), is the essential valuation of R whose valuation ring is R, (see Proposition 8.1 6).

Let 9(R) be the group of divisors of R, and let 2(R) be the subgroup of 9(R) consisting of all divisors of R of the form diva where a E K, a # 0.

8.23. Definition. The factor group 9(R)/2(R) is the divisor class group of R; we denote it by g(R).

Let R’ be a second Krull domain containing R as a subring. If P is a minimal prime ideal of R and if P’ is a minimal prime ideal of R‘, we shall write P < P‘ if up. is a prolongation of up to the field of quotients of R‘. In this case, there is a positive integer e(P’/P) such that .,,(a) = e(P’/P)vp(a) for all a E K [see Exercise 8(b)]. The integer e(P’/P) is called the ramification index of P’ over P.

We shall assume that the following condition holds, and we shall

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186 VIII KRULL DOMAINS

show that under this assumption there is a homomorphism from W(R) into %(R'):

If P' is a minimal prime ideal of R', then either P' n R = 0

If P is a minimal prime ideal of R and if P' is a minimal prime ideal of R' such that P E P', then P' n R = P. From this it follows that P'R,. n R, = PR, . Hence, by Proposition 5.21, vp . is a prolongation of vp . If a E P, a # 0, then a E P' and so there is only a finite number of minimal prime ideals of R' containing P. Of course, there may be none at all.

(*) or P' n R is a minimal prime ideal of R.

If P is a minimal prime ideal of R, we set

u(div P ) = c e(P'/P) div P' P < P '

if P is contained in at least one minimal prime ideal of R' ; otherwise, we set u(div P ) = 0. It follows from what we have already shown that 9 ( R ) is a free Abelian group, freely generated by {div PIP is a minimal prime ideal of R}. Hence we can extend u by linearity to a homomorphism from 9 ( R ) into 9(R ' ) , which we also denote by u.

8.24. Theorem. The homomorphism 0 maps #(R) into #(R') and so induces a homomorphism u' : W(R) + %(R).

Proof. If a E K, a # 0, then

div u = vp(u) div P P

where the summation is over all of the minimal prime ideals of R. Hence

u(div a) = c vp(u) u(div P ) P

= c zlp(u) c e(P'/P) div P' P P < P '

= 1 ( c e(P'/P)vp(u)) divP', P' P

P < P '

where the first summation is over all P' such that P < P' for some P. Given P', we have P < P' if and only if P' n R = P, and so the inner

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summation is just e(P'/P)vp(a) = vp.(a). If we do not have P < P' for some P, then P' n R = 0 and vp.(c) = 0 for all c E K , c # 0. Therefore

u(div a) = C vp.(a) div P' P'

which is the divisor div a of R'. Thus if (*) holds, we are assured of the existence of the homo-

morphism u'. The two propositions which follow give two important cases when (*) does hold.

8.25. Proposition. If R' is integral over R , then (*) holds.

Proof. Let P' be a minimal prime ideal of R'. By the lying-over theorem (Theorem 4.6), P' n R # 0 . Let P' n R = P I . If P , is not a minimal prime ideal of R, then there is a nonzero prime ideal P, of R such that Pz c P,. Then by the going-down theorem (Theorem 4.9) there is a prime ideal P" of R' such that P" n R = P, and P" c P'. It follows that P" # 0, and thus we have contradicted the fact that P' is minimal among the nonzero prime ideals of R'.

8.26. Proposition. If R' is a j a t R-module, then (*) holds.

Proof. Suppose that there is a minimal prime ideal P' of R' such that P' n R # 0 and P' n R is not a minimal prime ideal of R. Let a E P' n R , a # 0, and let P I , . . . , P , be those minimal prime ideals of R such that vPi(a) > 0 for i = 1, . . . , K . We have P' n R $ Pi for i = 1, . . . , K ; hence, by Exercise 5(c) of Chapter 11, there is an element b E P' n R such that b + Pi for i= 1, . . . , K . Then for each i, vpi(b) = 0, and consequently

vpi(ab) = .,,(a) = max{v,,(a), vpi(b)}.

If Q is a minimal prime ideal of R different from P,, . . ., P , , then .,(a) = 0 v,(ab) = v,(b) = max{v.(a), v,(b)}. Thus div(Ra n Rb) = the least upper bound of {div a, div b} = div ab. Then since both Rub and R a n Rb are divisorial, we have Rub = Ra n Rb. Since Rf is a flat R-module it follows that R'ab =

R'u n R'b (see the remark following the proof). Thus we must have

and

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188 VIII KRULL DOMAINS

u,,(ab) = max{v,,(a),vp.(b)}. However a, b E I", so that .,.(a) > 0 and +(b) > 0. Then vp.(ab) = vp.(u) + vpr(b) > max(u,,(a), z(,.(b)), which is a contradiction.

Remark. I n this proof, we have used the fact that if A and B are ideals of R, then since R' is a flat R-module, we have ( A n B)R' =

AR' n BR'. This is a consequence of Exercise lO(c) of Chapter I, used in conjunction with Proposition 1.18.

Let S be a multiplicative system in R. By Exercise 5, S-'R is a Krull domain, and the minimal prime ideals of S-'R are those ideals of the form S-lP where P belongs to the set F = {PIP is a minimal prime ideal of R and P n S is empty}. Let XI be the sub- group of 9(R) generated by {div P I P E S} and S2 the subgroup generated by {div PI P is a minimal prime ideal of R and P n S is not empty}. Then 9(R) = %, @ X 2 and the mapping T : Q(S-lR) + S, defined by T(div S- lP) = div P is well-defined and is an isomorphism (the reader should verify this). Each element of 9(R) can be written uniquely in the form h, + h, where h, E Sl and h, E X,, and the mapping h, + h, H h, is a surjective homomorphism from 9(R) onto Si . (In the notation of Section 3 of Chapter I , this is the homomorphism $,.) Since for each P E T , e(S-'P/P)= 1, the composition of this homomorphism with 7 - l is precisely the homo- morphism u: 9(R) + 9(S-lR).

0.27. Theorem. Let S be as above. Then ur: GS(R)+%?(S-lR) is defined and is surjective. Suppose there is a subset S' of R such that every element of S is a product of elements of S' and (a) is a prime ideal of R for every a E S' . Then a' is an isomorphism.

Proof. First S - l R is a flat R-module by Theorem 3.3; hence u' is defined by Proposition 8.26. Furthermore, u' is surjective since u is surjective. Now we have

Ker u' = (*(I?) + X2)/2'(l?) S2/(.%', n X(R)).

We will show that if there is a subset S' of R with the stated property, then S2 = 8?, n 2'"(). Consider a divisor div P where P is a minimal prime ideal of R and P n S is not empty. If a E P n S, then a = a, * * ak where ai is a nonzero element of S' for i = 1, . . . , k.

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3 THE DIVISOR CLASS GROUP 189

Then (ai) c P for some i ; since (a,) is a prime ideal this implies that P = (ai). Thus div P E X(R), and we conclude that X 2 c X(R).

8.28. Theorem. If R is a Krull domain and X i s an indeterminate, then u' : V(R) + V ( R [ X ] ) is defined and is an isomorphism.

Proof. First we note that u' is defined, since R [ X ] is a flat R-module [Exercise 19(a) of Chapter 111. If S is the set of nonzero elements of R, then S- l (RIX]) = K [ X ] , where K is the quotient field of R. Let p be the homomorphism from 9 ( R [ X ] ) into 9 ( K [ X ] ) defined as was u above, and let p' : %'(R[X]) -+ %'(K[X]) be the induced homo- morphism. It is well known from elementary abstract algebra that every ideal of K [ X ] is principal. Hence %'(K[X]) = 0 and Ker p' =

%'(R[X]). Let X be the subgroup of 9 ( R [ X ] ) generated by the set of all div P', where P' is a minimal prime ideal of R [ X ] and P' n S is not empty. Then as we have noted already,

W R [ X I ) = Ker p' = (X(R[XI) + X ) / X ( R [ X I )

z %/(A? n X ( R [ X ] ) ) .

By Theorem 8.21, and its proof, it is clear that if P is a minimal prime ideal of R, and P' is a minimal prime ideal of R [ X ] , then P < P' if and only if P' = PR[X] ; in this case, e (P ' /P ) = 1. Thus, for every minimal prime ideal P of R, we have

u(div P ) = div PR[X] .

We conclude that u maps 9 ( R ) onto A?. We shall show that u(H(R)) = X n X ( R [ X ] ) . Then u induces an isomorphism

6: %'(I?) -+ X/ (X n H ( R [ X ] ) ) .

The composite of 6 with the isomorphism from X/ (X n 2 ( R [ X ] ) ) onto ( X ( R [ X ] ) + %)/X is u'.

It is clear that a(%"()) c X n %(R[X]) . Suppose that f ( X ) E

R [ X ] and that R [ X ] f ( X ) E 2. If we show that this ideal of R [ X ] is generated by an element of R, we will be finished. Now f ( X ) is contained in no minimal prime ideal of R [ X ] other than the ones

tained in no proper prime ideal of K [ X ] . Thus f ( X ) is a constant; that is, an element of R.

whose divisors generate 2. Hence by Theorem 8.21, f ( X ) 1s . con-

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190 VIII KRULL DOMAINS

4 FACTORIAL RINGS

8.29. Definition. An integral domain R is a factorial ring if there is a set S of nonzero nonunits of R such that every nonzero element of R can be written uniquely in the form ua, - . * a,, where u is a unit of R and a,, . . . , ak E S, except for the order in which the factors are written.

A factorial ring is also called a unique factorization domain. Familiar examples are any field with S the empty set, the ring 2 of integers with S the set of primes, and the ring of polynomials in one indeterminate over a field with S the set of monic irreducible poly- nomials of positive degree.

Let R be an integral domain. An element p E R is called a prime if ( p ) is a nonzero proper prime ideal of R. An element a E R is called irreducible if it is a nonzero nonunit and if, whenever a = bc where b, c E R, then either b or c is a unit of R. Every prime is irreducible. For, let p be a prime; since ( p ) # 0 we have p # 0 and since ( p ) # R, we conclude that p is not a unit. Suppose p = bc where b, c E R. Then one of b and c is in ( p ) , say b E ( p ) ; write b =pd . Then p = pdc, and since R is an integral domain and p # 0, dc = 1 ; that is, c is a unit of R.

8.30. Proposition. Let R be a factorial ring, and let S be a set whose existence is required in the definition. Then every irreducible element of R is prime, every element of S is prime, and every prime of R is the product of a unit of R and an element of S.

Proof. Let p E S and let ab E ( p ) , where a, b E R. Then ab =pc, where c E R. There are elements p,, . . . , p,, ql , . . . , qm, r l , . . , , r , E S and units u, v , w of R such that

a=up,*. .p, , b = v q , - - * q m , and c = w r l - - - r , .

Then uvp, * -. pkq , qm = wpr, T,.

By the uniqueness of such factorizations, p must be p , for some i or q, for somej. Hence either a E ( p ) or b E ( p ) . Thus ( p ) is a prime ideal and, since p is a nonzero nonunit, p is a prime.

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4 FACTORIAL RINGS 191

Next let p be an arbitrary prime of R. Then p = up, - * p , , where p , , . . . , p k E S and u is a unit of R. One of the factors must be in ( p ) , and since ( p ) f R, some p i must be in ( p ) , say p , E ( p ) . Then p l = p d , where d E R. If d is a unit, then p = d- lp , , which is the assertion. Suppose d is not a unit. Then d = v q , ... qm, where q,, . . . , qm E S and v is a unit of R. Then p , = uvp, . . . p , q1 - - - Qm

which contradicts the uniqueness of such factorizations. Finally, let a be an irreducible element of R and write a = up, * * e p , ,

where p , , . . . , p , E S and u is a unit of R. If K > 1, then a is the product of two nonunits, p , and up, * * p , . Hence we must have k = 1 and a=up , .

8.31. Theorem. An integral domain R which is not aJield is a factorial ring if and only if R is a Krull domain and W(R) = 0.

Proof. Let R be a factorial ring and let S be a set whose existence is required in the definition. If a E R, a # 0, we can write

where u is a unit of R, vp(a) is a nonnegative integer uniquely deter- mined by a for each p E S, and vp(a) = 0 for all but a finite number of p E S. If a and b are nonzero elements of R, then

where w is a unit of R ; that is, vP(a6) = zlp(a) + vp(6) for all p E S. If we set v,(O) = 03, then this equality holds for all a, 6 E R. If either a = 0, b = 0, or a + b = 0, then vp(a + 6) 2 min{vP(a), zlp(b)}.

Suppose a, b and a + 6 are all nonzero. Then

hence vp(a + 6) 2 min{vp(a), vp(b)) for all p E S. Thus for each p E S, there exists a valuation v p on the field of quotients of R (see Exercise l l (d) of Chapter V) such that for all a E R, a # 0,

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192 VIII KRULL DOMAINS

where u is a unit of R . We leave it to the reader to verify that the valuation ring of v p is R,, where P= (p), and that the family of valuations {vplp E S ] satisfies (I) and (11). Therefore R is a Krull domain. Note that for each P E S, v , ( p ) = 1 and v,(q)=O for

T o show %(R) = 0, we shall show that every divisorial fractional ideal A' of R is principal. By Exercise lO(c), A' = Ra' n Rb', where a', b' E K. Let d' E R, d' # 0, be such that a = d'a' and b = d'b' are in R. If A = d'A', then A = (a) n (b) and A' is principal if and only if A is principal. We shall prove that A is principal by induction on the number of elements of S occurring in the factorization of b. If b is a unit of R, then (b) = R and A = (u). Assume that b is a nonunit and that (a) n (f) is principal wheneverfis an element of R requiring fewer elements of S than b in its factorization. Since b is not a unit, we can write b = pc where p E S and c E R . B y our induction assumption, (a) n (c) = (d) for some d E R ; then v,(d) = max(v,(a), v,(c)} for all q E S . Hence v,(d) = v,(pd) = max{v,(a), vq(b)} for all q E S, q # p .

q f s, q f P.

If vp(a) 5 then vp((a) n (pc)) = max{vp(a>, .,(pc)> = .,(pc) =

v,(pd). If %(a) > vp(c), then up(a) 2 .,(pc), and %((a) n (pc)) = vp(a) = v,(d). Thus (a) n (b) = (pd) or (a) n (b) = (d).

Conversely, suppose that R is a Krull domain and that %(R) = 0. Then the minimal prime ideals of R are principal. Choose from each minimal prime ideal a generator and let S be the set of elements of R so chosen. Then the family of essential valuations of R can be indexed by S . If p E S and ( p ) = P, we write v p for v, . Let a be a nonzero element of R. Then vp(u) > 0 for only a finite number of p E S , saypl, * * . , p k * If

then for i = 1, ..

R and q E S , qZP1 , * a * ,

This factorization

. , k, vPi(u) = vp,(a) - v,,(u)v,,(p,) = 0, and for pk, we have also v,(u) = 0. Thus u is a unit of

k

a = u fl p iWa) , i = 1

is unique; for suppose q E S and that q" is the

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4 FACTORIAL RINGS 193

exact power of q occurring in some factorization of a as a product of a unit and elements of S. Then vq(a) = n ; so that if n > 0, we have q = p i for some i and n = vpi(a) . Therefore R is a factorial ring.

Combining the results of Theorems 8.27 and 8.31 we have

8.32. Corollary. Let R be a Krull domain and let S be a muZtiplicative system in R. If R is a factorial ring then S - IR is a factorial ring. On the other hand, if S isgenerated by a set of primes, and if S-IR i s a factorial ring, then R is a factorial ring.

Combining the results of Theorems 8.28 and 8.31, we have:

8.33. Corollary. If R is a Krull domain and X i s an indeterminate, then R [ X ] is a factorial ring if and only if R is a factorial ring.

If R is a Krull domain, then it follows from Theorem 8.12 that R

Maximum condition on principal ideals (MPI) : every non-

satisfies :

empty set of principal ideals of R has a maximal element.

8.34. Theorem. If R i s an integral domain which is not afield, then the following statements are equivalent:

(1) (2) (3)

(4) (5)

R is a factorial ring. R is a Krull domain and every divisorial ideal of R is principal. R is a Krull domain and every prime divisorial ideal of R is principal. R satisfies (MPI) and every irreducible element of R i s a prime. R satisfies (MPI) and the intersection of two principal ideals of R is principal.

Proof. (1) + (2). Assume R is a factorial ring; then R is a Krull domain and 9(R) = &(R). Hence, if A is a divisorial ideal, then div A = div a for some a E R. Then A = (a) .

(2) 3 (3). Clear. (3) + (1). Assume that R is a Krull domain and that every prime

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194 VIII KRULL DOMAINS

divisorial ideal of R is principal. As we have seen, 9(R) is generated by the elements div P, where P runs over the minimal prime ideals of R, and each such P is divisorial. By assumption, div P E #(R). Hence X ( R ) = 9(R), and R is a factorial ring.

(2) j (5). Clear. (5) + (4). Assume that R satisfies (MPI) and that the intersection

of two principal ideals of R is principal. Let p be an irreducible element of R and suppose that ab E (p), where a, b E R. If a = 0, then a E (p) ; assume a # 0. Let (u) n (p) = (c) and set d = ap/c. Since up E (c), d E R. If c = ae, thenp = de and, sincep is irreducible, either d or e is a unit of R. If e is a unit, then a E ( c ) G (p). Suppose d is a unit. Then (p) = (e) and so (c) = (a)(p) = (up), that is, (up) = (a) n (p). Hence ab E (ap) and consequently b E ( p ) . Therefore (p) is a prime ideal.

(4) + (1). Suppose that R satisfies (MPI) and that every irredu- cible element of R is a prime. It follows from (MPI) that every nonzero nonunit of R can be written as a product of a finite number of irreducible elements of R. From each principal ideal which is generated by an irreducible element choose one generator, and let S be the set of elements so chosen. If p E R is irreducible, then there is a unique p' E S such that p = up', where u is a unit of R . Hence every nonzero element of R can be written as a product of a unit and a finite number of elements of S. Suppose that up, . * - p, = nq, * * qn , where u and v are units of R and p,, . . . ,p,, q,, . . ., qn E S. Then UP1 * * - p m E (41) and 4 (ql) , SO some P, is in (qJ, say PI E (qJ. It follows from the definition of S that p1 = ql , and so up2 * - p, = vq, * * - qn . We can then proceed with an argument by induction to show that m = n, u = v, and that the pi and qj can be so numbered that pi = qi for i = 1, . . . , m. Therefore S meets the requirements in the definition of factorial ring.

EXERCISES

1. Residuals of fractional ideals. Let R be an integral domain. (a) Show that if A and B are fractional ideals of R, then

[A : BI = n ~ ( i p ) . b E B b # O

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EXERCISES 195

(b)

(c)

(d)

Show that if A is a fractional ideal of R, then A= [R: [R: A]]. Show that R is a completely integrally closed if and only if [A: A] = R for every divisorial ideal A of R. Show that R is integrally closed if and only if [A: A] = R for every finitely generated nonzero ideal A of R.

2. Primary decomposition in Krull domains. Let R be a Krull domain. (a) Let P be a minimal prime ideal of R, and let n be a positive

integer. Show that

= (a1 u E R and zlp(u) 2 a).

(b) Let a E R, a # 0. Show that there is only a finite number of minimal prime ideals of R which contain a, say PI , . . . , Pk . Show that

(a) = py1) n * - - n ppk),

where ni = vPi(a) for i= 1, . . . , I t , and that this is the unique reduced primary decomposition of the ideal (a).

3. Krull domains. (a) Show that an integral domain R is a Krull domain if and

only if it satisfies the following two conditions: (i) if P is a minimal prime ideal of R, then R, is a

Noetherian valuation ring, and npRp = R ; and (ii) if a E R, a # 0, then a is contained in only a finite

number of minimal prime ideals of R. Now assume that R is a Krull domain. (b) Show that every nonzero prime ideal P of R contains a

minimal prime ideal of R, and if P is not itself minimal, then

Show that if R is Noetherian and if A is a proper ideal of R, then A is divisorial if and only if each prime divisor of A is a minimal prime ideal of R. Suppose that R has a unique maximal prime ideal P. Show that R is a Noetherian valuation ring if and only if P is divisorial.

[R: PI = R. (c)

(d)

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196 VIII KRULL DOMAINS

4. Intersections of Krull domains. (a) Let K be a field. Let I be a set and for each a E I let R, be

a Krull domain which is a subring of K. Let R = naE, R, and assume that for each a E R, a # 0, a is a unit in R, for all but a finite number of a E I . Show that R is a Krull domain.

(b) Let R be a Krull domain with field of quotients K . Let K' be a subfield of K. Show that K' n R is a Krull domain.

5. Krull domains and rings of quotients. Let R be a Krull domain and let (v,l o! E I } be the family of essential valuations of R. Let V, be the valuation ring of v,. Let S be a multiplicative system in R and let J = (a I a E I and v,(s) = 0 for all s E S}. Show that

n va S - 1R = a s f

and conclude that S-lR is a Krull domain.

6. An example. Let R be the ring of holomorphic functions of a complex variable. (a) Show that R is an integral domain (with respect to the

usual addition and multiplication). (b) For the complex number a, let Pa = ( f l f ~ R andf(a) = O}.

Show that Pa is a minimal prime ideal of R. (c) For each complex number a and for each nonzero f~ R,

set v,(f) = n if f ( 2 ) = (Z- u)"g(Z) where g(a) # 0. Define o,(O) = CO. Show that o,, extended to the field of quotients K of R in the usual way [see Exercise 9(d) of Chapter V], is a valuation on K . Show that the valuation ring of v, is R,, . Show that ORpa = R, where the intersection is over all complex numbers. Note that if j is the sine function, then f E P,, for all integers n. Explain how this fact implies that R is not a Krull domain.

(d)

(e)

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EXERCISES 197

7. Essential valuations on polynomial rings. (a) Let R be a Krull domain and let {vaI c1 E I} be a family of

(normed) valuations on the field of quotients of R satis- fying (I) and (11). Assume that for each fs E I we have

where V , is the valuation ring of v=. Show that (vnl CL E I} is the family of essential valuations of R. Show that the family of valuations of R [ X ] consisting of all p(X)-adic valuations and all extensions of essential valuations of R satisfy the assumption of (a).

(b)

8. Prolongations of valuations. Let z, be a valuation on a field K, let K‘ be an extension of K, and let v’ be a valuation on K‘ which is a prolongation of v. Let G,={v‘ (a) lu~K*} . Then Go is a subgroup of the value group G of v’. Denote the index of Go in G’ by e(v’1.u). Let ul’, . . . , at’ E K‘* be such that v’(al’), . . . , ~ ’ ( a , ’ ) represent distinct cosets of Go in G’. (a) Show that q’, . . . , at‘ are linearly independent over K.

Conclude that e(v’/v) 5 [K’: K ] . (b) Assume that z, and u’ are normed rank one and discrete.

Show that for every a E K we have v(u) = e(d/v)z,‘(a). (c) Let P be the maximal ideal of the valuation ring Y of v.

Set R = V / P ; R is called the residue field of K relative to v. Define R‘ in like manner and show that we may regard R as a subfield of I?. Show thatf(v’/v) = [I?’ : R] is finite if [X’ : K ] is finite. Assume that v has rank one and is discrete. Let X’ be a finite extension of K and let vI‘, . . . , vt‘ be the valuations on K’ which are prolongations of zr. Assume that for i= 1, . . ., k, the residue field of I(’ relative to a,’ is a separable extension of R. Show that

(d)

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198 VIII KRULL DOMAINS

9. The approximation theorem. Let al, . . . , a, be normed rank one discrete valuations on a field K such that v, and a, are not equivalent when i # j . (a) Show that there is an element a E K such that vl(a) < 0

and ai(a) > 0 for i= 2, . . . , k. (b) Let n be an integer. Show that there is an element a E K

such that .,(a - 1) > n and .,(a) > n for i = 2, . . . , k. (c) Let a,, . . . , ak E K. Show that if n is an integer then there

is an element aEKsuch that ai(u - ai) > n for i = 1 , . . . , k. 10. The approximation theorem for Krull domains.

Let R be a Krull domain and let {u, I CI E I} be the family of essential valuations of R. Let a l , . . . , ak be (distinct) elements of I and let n,, . . . , nk be integers. (a) Show that there is an element a in the quotient field of R

such that vai(a) = ni for i = 1, . . . , K, and zl,(a) 2 0 for all

(b) Let A, B, and C be divisorial fractional ideals of R such that A _c 3. Show that A = B n Ca for some a in the quotient field of R. Show that every divisorial fractional ideal of R is an inter- section of two principal fractional ideals of R.

a E I\{a1, . e 7 a,>.

(c)

11. Integral domains which are almost Krull. Let R be an integral domain : R is almost Krull if R, is a Krull domain for each proper prime ideal P of R. (a) Let R be almost Krull. Show that R is integrally closed,

and therefore, if R is Noetherian, then it is a Krull domain. (b) Show that if R is almost Krull, then R = n R,, where P

runs over the minimal prime ideals of R. Show that R is completely integrally closed. Show that R is almost Krull if and only if there exists a family {a, I a E I } of discrete rank one valuations on the quotient field of R with the following properties : (1) For each maximal ideal M of R there is a subset I ,

of I such that R , = naEIM V,, where V , is the valuation ring of v, . For each a E I , V , = RPa, where Pa = {xi x E R and

(c)

(2) va(x) > O } .

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EXERCISES 199

12.

(3) For each maximal ideal M of R and each nonzero x in the quotient field of R, v,(x) # 0 for only a finite number of CL E I,,, .

(d) (e)

If R is almost Krull, show that R[X] is almost Krull. If R is almost Krull, show that the integral closure of R in a finite extension of the quotient field of R is almost Krull. Show that if R is almost Krull, and if every nonzero proper ideal of R is contained in only a finite number of maximal ideals, then R is a Krull domain.

(f)

Principal ideal domains. Let R be an integral domain such that every ideal of R is principal; R is called a principal ideal domain. Show that R is a factorial ring. Give an example of a factorial ring which is not a principal ideal domain. An integral domain R is called a Euclidean ring if there is a mapping 4 from R into the set of nonnegative integers such that d(ab) 2 &a) and for nonzero elements a, b E R there are elements q, Y E R such that a = bq + Y and either Y = 0 or 4 ( ~ ) <+(b). Show that a Euclidean ring is a principal ideal domain. Show that the ring of integers and a ring of polynomials over a field (in one indeterminate) are Euclidean rings. Show that the same is true of the ring of all complex numbers a + bi, where a and b are integers.

13. Factorial rings. (a) (b)

Prove directly that (3) implies (4) in Theorem 8.34. Let R be a Krull domain. Show that if every divisorial ideal of R is invertible, then R, is a factorial ring for every maximal ideal P of R. Assume that every divisorial ideal of R is finitely generated and that R, is a factorial ring for every maximal ideal P of R. Show that every divisorial ideal of R is invertible.

(c)

14. Krull domains and the ideal transform. Refer to Exercise 16 of Chapter VI for terminology. Let R be an integral domain with more than one maximal ideal.

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200 VIII KRULL DOMAINS

(a)

(b)

Prove that if R is a Krull domain, then T((x)) is a Krull domain for each nonunit x of R. Let {Pa} be the collection of all minimal prime ideals of R. Suppose that T((x)) is a Krull domain for each nonunit x of R. Prove the following assertions : (1) For each a, RPa is a discrete rank one valuation ring.

(3) If y is a nonzero element of R, then y is contained in only a finite number of the P,’s . [Hint. Assume that y is a nonunit, and show that either y is contained in only a finite number of the P,’s or that y is contained in every maximal ideal of R : do this by considering two cases-whether or not there exists a nonzero nonunit in R\UQB, where {Q8} is the collection of prime ideals of R which contain y.] Now conclude that R is a Krull domain.

(2) R = n R P a .

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C H A P T E R

IX Generalizations of Dedekind Domains

In this chapter we shall study several of the important classes of rings which contain the class of Dedekind domains. The procedure will be to choose a property of Dedekind domains and study the class of rings, or domains, which have the chosen property. The proofs of many of the assertions to be made are similar to those of results in earlier chapters; some will be given in the text but others left as exercises.

1 ALMOST DEDEKIND DOMAINS

9.1. Definition. A n integral domain R is an almost Dedekind domain ;f for each maximal ideal M of R, the ring R, is a Dedekind domain.

Thus, an almost Dedekind domain is a domain which is locally Dedekind. Note that if R, is a Dedekind domain which is not a field, then it follows from the results of Chapter VI that R, is a discrete rank one valuation ring. Hence, if R is an almost Dedekind domain which is not a field, its Krull dimension is one, and R is a Prufer domain. In this section we shall obtain several characteriza- tions of almost Dedekind domains.

201

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202 Ix GENERALIZATIONS OF DEDEKIND DOMAINS

9.2. Proposition. Let R be an integral domain which is not a field. Then the following aye equivalent :

(1) The integral closure of R is an almost Dedekind domain. (2) Each nontrivial valuation ring which is an overring of R hasrank

one and is discrete.

Proof. By Corollary 5.8, every valuation ring which is an overring of R is an overring of the integral closure of R. Hence, there is no loss of generality in assuming R is integrally closed. Thus (1) implies (2) without further ado. Suppose (2) holds. By Exercise 8(c) of Chapter VII and Theorem 7.23, R is a Priifer domain of Krull dimension one. Hence, if M is a maximal ideal of R, then R, is a valuation ring, which has rank one and is discrete. Thus, R is almost Dedekind.

9.3. Corollary. A n overring of an almost Dedekind domain is an almost Dedekind domain.

Proof. Let R be an almost Dedekind domain and let T be an overring of R . Since R is a Priifer domain, T is integrally closed by Theorem 6.13. Furthermore, each nontrivial valuation ring which is an overring of T is also an overring of R, and thus has rank one and is discrete. Therefore, T is almost Dedekind.

9.4. Theorem. If R is an integral domain which is not afield, then the following statements are equivalent :

(1) (2)

(3)

(4)

R is an almost Dedekind domain. R has Krull dimension one and each primary ideal .f R is a power of its radical. If A, B, and C are ideals of R with A # 0, and if AB = AC, then B = C. R is a Priifer domain of Krull dimension one which has no idempotent maximal ideals (an ideal A of R is called idempotent if A2 = A). R is a Priifer domain and for each proper ideal A of R,

Each ideal of R which hasprime radical is a power of its radical.

(5)

(6) n:==l A"=O.

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1 ALMOST DEDEKIND DOMAINS 203

Proof. (1)+(2). Assume that R is an almost Dedekind domain. Since dim RM = 1 for every maximal ideal M of R, dim R = 1 . Let Q be a primary ideal of R, If Q is either 0 or R, then Q is its own radical; hence, we assume Q # 0 and Q # R. Then M = Rad(Q) is a maximal ideal of R and QR, is MR,,.,-primary. Since R, is a Noethe- rian valuation ring, QR, = MnRM for some integer n. Since M is a maximal ideal, Mn is M-primary by Exercise 6(e) of Chapter II. Therefore,

Q=QRM n R = MnRM n R=Mn.

(2) + (1). Assume (2) holds and let M be a maximal ideal of R. Then every proper nonzero ideal Q of R, has MRM as its radical; thus Q is MR,-primary. Therefore, Q n R is an ,'M-primary ideal of R. Hence Q n R = Mn for some integer n, and thus Q = MnRM. Therefore, the only proper nonzero ideals of RM are powers of MR, , and it follows immediately from this that RM is a Noetherian valuation ring.

(1) + (3). Assume that R is an almost Dedekind domain, and let A, B, and C be ideals of R such that A # 0 and AB = AC. If M is a maximal ideal of R then (ARM)(BRM) = (ARM)(CRM) by Exercise 4(a) of Chapter III. Since RM is a Dedekind domain, BRM = CRM . Since this is true for every maximal ideal M of R, it follows by Propo- sition 3.13 that B = C.

(3)+(4). Assume (3) holds. By Theorem 6.6, R is a Prufer domain. If M is a maximal ideal of R, and if M 2 = M = MR, then M = R ; hence M2 # M. It remains to show that dim R = 1. Let P be a proper nonzero prime ideal of R and let x E R\P. Since

(P + (XN3 = ( P + (,))(P2 + (x2)),

we have by (3), (P+(x))~= P 2 + (x2). Thus xPs P2 +(x2). If y E P then xy = a + rx2 for some a E P2 and r E R. Then rx2 E P, and since x 2 $ P, we have r E P. Hence, XP E P2 + (x2)P and it follows by (3) that x E P + (x2). Then x = b + sx2 for some b E P and s E R. Thus x(l - sx) E P, and since x 4 P , we have 1 - sx E P. Therefore, R = P + (x). Since this is true for every x E RjP, P is a maximal ideal of R.

(4) + (2). This implication follows at once from (4) of Proposi- tion 6.9.

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204 Ix GENERALIZATIONS OF DEDEKIND DOMAINS

(4) =+ (5). This implication follows at once from (1) of Proposi-

(5) =+ (4). Assume (5) holds and let M be a maximal ideal of R. tion 6.9.

Then R, is a valuation ring, and

m m

O = M n = n ( M n R M n R)= MnR, n R;

hence n;= MnR, = 0. Thus, by Theorem 5.10, R, has no proper nonzero prime ideal other than MR, . Therefore, dim R, = 1, and since is this true for every maximal ideal of R, dim R = 1. If M is an idempotent maximal ideal of R, then MRM = n;= MnR, , which is not true; hence M2# M.

n = l n = 1 ( n : l )

(2) + (6). Clear. (6) + (2). Assume (6) hoIds and let P be a proper nonzero prime

ideal of R. First of all, we shall show that if P is a minimal prime divisor of a principal ideal (x), then P is a maximal ideal of R. Under this assumption, xR, is PR,-primary. Hence xR, n R is P-primary, and consequently xR, n R = Pn for some integer n. Thus xR, = (PRp)n; this implies that PR, is invertible and so P2R, # PR,. Hence

P 2 c Pc2)= P2R, n R c P.

By (6) , P@) is a power of P, so that we must have P2 = Pc2). Let a E P\Pz and b E R\P. Then Rad(P2 + (ab)) = P, and so P2 + (ab) = P2 or P2+(ab)=P . Since P2 is P-primary, ab$P2. Hence P2 + (ab) = P. Thus a E P2 + (ab), so that a = c +rub for some c E P2 and r E R. Then a( 1 - rb) E P2 and since P2 is P-primary and a $ P2, we have 1 - r b E P. Thus, R = P+ (b), and since this is true for every b E R\P, P is a maximal ideal of R.

If a is an arbitrary nonzero element of P , then P contains a minimal prime divisor P' of (u). By what we have shown, P' is a maxi- mal ideal of R; hence the same is true of P. Therefore, dim R = 1. A primary ideal of R has prime radical, so is a power of its radical. Thus, (2) holds.

We conclude this section by proving that the analog of Theorem 6.24 holds for almost Dedekind domains.

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9.5. Theorem. Let R be an integral domain with quotient field K. Let K' be aJinite extension of K , and let R' be the integral closure of R in K'. If R is almost Dedekind, then R is almost Dedekind.

Proof. Let M be a nonzero proper prime ideal of R'. Let P = M n R and let S= R\P. By Proposition 4.5, S-IR' is integral over R, . Since R' is integrally closed in K', S-lR' is integrally closed in K' by Exercise 7(b) of Chapter IV. Thus, S-lR' is the integral closure in K' of the discrete rank one valuation ring R, . Hence, by Theorem 6.24, S-IR' is a Dedekind domain (since R, is). Therefore, (S-lR')s- lM is a discrete rank one valuation ring. By Exercise 6(c) of Chapter 111, (S-lR')s- lM = RM'. Therefore, R' is an almost Dedekind domain.

2 ZPI-RINGS

If R is a Dedekind domain, then every ideal of R can be written as a product of prime ideals of R. In this section we shall study the class of rings with this property.

9.6. Definition. A ring R is a ZPI-ring i f every ideal of R can be written as a product of prime ideals of R .

The letters ZPI stand for " Zerlegung Primideale." Some authors call a ring a ZPI-ring if every nonzero ideal can be written as a product of prime ideals ; they call the rings of Definition 9.6 general ZPI-rings.

If R is a ZPI-ring, and if P is a proper prime ideal of R, then RIP is a ZPI-ring which is also an integral domain. Hence RIP is a Dede- kind domain and it follows that dim RIP 5 1 . Therefore, dim R 2 1.

A ring is said to be indecomposable if it cannot be written as a direct sum of nontrivial rings. Clearly, every integral domain is inde- composable. We shall show that every ZPI-ring is a direct sum of indecomposable ZPI-rings, and we shall determine all indecompos- able ZPI-rings.

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9.7. Definition. A ring R is a special primary ring if R has a unique maximal ideal M and if each proper ideal of R is a power of M .

T o determine all special primary rings we shall use the following lemma.

9.8. Lemma. Let A be an ideal of a ring R such that there are no ideals of R strictly between A and A2. Then, for every positive integer n, the only ideals between A and A" are A, A2, A3,. . . , A".

Proof. Since the assertion is certainly true when A is idempotent, we shall assume that A 2 c A. Let x E A\A2; then A = A2 + ( x ) and so

( A M 2 = (A2 + (4K4 = 44.

A/(4 = (A/(x))" = (A" + (x))/(4

Therefore, for every positive integer m,

and consequently A = A" + (x). Then A/A" = (A" + (x))/A", and for every positive integer k,

(Ak + A")/A" = (A/A")'" = (A" + (x"))/A".

Therefore Ak = A" + (x") for all positive integers k and m with k s m .

Now let b E An\An+l. Since A" = A"+l+ (%IL), we have b - rxn E A" +l for some rx" $ A" + l. Then rx E A\A2 and therefore A = A2 + (rx). Thus

A" = AA"-'= (A2 + (rx))(A" + (x" - ')) =A"+2+A2(X"-l)+A"(Yx) +(YX") EA"+l+(Yx")

- +(b) . - An+l

Therefore, A"= A"+'+ (b), and since this is true for every b E An\An+l, we conclude that there are no ideals of R strictly between A" and A" + l.

Now consider an ideal 3 of R such that A" E B G A. Choose s to be the largest positive integer such that B G As. If s >_ n then B = A". Suppose that s < n ; then As = A" + (xs). I f c E B\As +l then c - rxs E

A" for some r E R, and rxS $ AS+l since c q? A"'I and A" E AS+'.

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2 ZPI-RINGS 207

Hence rx E A\A2 and so A = A2 + (rx). Therefore, AS = AAS-1- - (A2 +(rx))(A"-l +(."I))

= A n + l + A2(xS-l) + A"-l(rx) + (79) c - A" +A"+' + ( Y X ~ ) = A"' + (YX').

Thus, As = AS+l + (rxs) and it follows that

= (As + (rxs))/(rxs) = (AS+l + (yxs))/(&) = (A/(YX")"+l = (A/(Yx"))"(A/(YX")

( A / ( Y X S ) ) " + 2 = * - - - - ( A / ( Y x s ) ) n .

Consequently,

As E As + ( 7 9 ) =A" + (mS) = An + (c) E B,

and we condude that B = A".

As an immediate consequence of this lemma we have the following result.

9.9. Proposition. If M is a maximal ideal of a ring R such that there are no ideals of R strictly between M and Ma, then for each positive integer n, RIMn is a special primary ring.

9.10. Theorem. The following statements are equivalent:

( 1 ) R is a ZPI-ring. (2)

(3)

R is a Noetherian ring such that for each maximal ideal M of R, there are no ideals of R strictly between M and M a . R is a direct sum of afinite number of Dedekind domains and special primary rings.

Proof. (1) .=$ (2). Let R be a ZPI-ring. We shall show that every prime ideal of R is finitely generated; by Exercise 14(c) of Chapter 11, this implies that R is Noetherian. Let P be a proper prime ideal of R, and let A be an ideal of R such that P c A. If we use the fact that RIP is a Dedekind domain, it follows exactly as in the proof of Theorem 6.16 that for any element a E A\P we have

P= P2 + P(u) G PA E P.

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208 Ix GENERALIZATIONS OF DEDEKIND DOMAINS

Hence P = PA. Therefore, if

o=p;x . . .p;k,

where P,, . . . , P k are distinct proper prime ideals of R and n,, . . . , nk are positive integers, then we may assume that there are no contain- ment relations among P,, . . . , P k . Thus, P,, . . . , Pk are the minimal prime divisors of the ideal 0. We shall show first that these prime ideals are finitely generated ; it is sufficient to show that P, is finitely generated.

If P12 c P, choose x E Pl\(P12 u P, u - * - u PJ ; if P12 = P, choose x E P1\(P2 U - * * U Pk). Let (x) = Q, * * Q,,, , where each Qi is a proper prime ideal of R. Since Q, Qm G P, we have Qi s PI, and so Qi = P,, for some i, say Q, = P,. If m = 1 we are finished; assume m 2 2 , and consider some Q, where ji2. If P12 c P, then Qj#Pi for i= 1, ..., k, so Qi is a maximal ideal of R and not a minimal prime divisor of the ideal 0. If P12 =PI then Qj # P i for i = 2, . . . , k. However, we might have Q j = P,, in which case QIQj = Q1. Hence, we may eliminate every Qj for which this is true, and assume that fo r j = 2, . . . , m, Q j is a maximal ideal of R and not a minimal prime divisor of the ideal 0. If Q1 c Qj , then Q, = QIQ, and we may eliminate Q,. Thus, we may assume P, = Q1 $ Qj for j = 2, . . . , m. Choose y E P1\(Q2 u * . u Qm). We shall show that

Let (x, y ) =N, * * * N , , where each N , is a proper prime ideal of R. Then Ni = P, for some i, say N , = P,. Suppose s >_ 2 and consider N , . Since x E N , we have Qj G N , for some j . If j 2 2 then Q j = N , , which is impossible since y I$ Qj . Hence, P, = Q1 G N , . If P, c N , then P, = PIN, . If P, = N , then x E P12, so P12 = P, and PI = P,N, . Thus, in either case, N , can be eliminated from the product N, * - - N , . Repeating this argument, we come finally to the conclusion that

Now let P be a prime ideal of R which is not a minimal prime divisor of the ideal 0. Then Pi c P for some i, say P, c P . The ideal PIP, of the Dedekind domain RIP, is finitely generated. Since P, is finitely generated it follows that P is finitely generated. This com- pletes the proof that R is Noetherian. Let A be an ideal of R such that P2 G A G P. If we write A as a product of proper prime ideals, then each factor contains P, and so is equal to P since P is maximal. Hence A = Pk for some positive integer K . If k = 1 then A = P ; if k 2 2 then A = P2.

p1= (x, y).

(x, y ) = Pl-

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(2) + (3). Suppose that (2) holds. Let M be a maximal ideal of R. If there exists a prime ideal of R properly contained in M , then by Exercise6(e)and(f),thisprimeidealisP = n;==,Mn. Thus,dimR 5 1. Since M is maximal, all powers of M are M-primary. Since R is Noetherian, each M-primary ideal Q of R contains a power of M . Thus, by Lemma 9.8, Q = M k for some integer K. By Exercise 6(d), P = P M ; thus, by Exercise 6(b), there exists m E M such that p = p m for all p E P . Therefore, if N is a P-primary ideal of R, then p ( 1 - m ) = 0 E N , 1 - m $ P, and so p E N for all p E P. Hence P= N , and we conclude that P is the only P-primary ideal of R.

Now let 0 = Q1 n * * n Qn be a reduced primary decomposition of the ideal 0 ; let QI be P,-primary for i = 1, . . . , n. We claim that Ql, . ~. , Qn are comaximal. For, let i + j and assume Q1 and Qj are contained in the same maximal ideal M . Then P, s M and P, G 111; it follows from the fact that P, # P, that we have, say, P , c P, = M . But then P, G Q, , so that Q, G Q, , and this contradicts the fact that the primary decomposition is reduced. By Exercise 2(d) of Chapter I1 we have

R z R/Ql 0 * * * 0 R/Qn.

For each i , if P, is maximal then QI is a power of P , , while if P, is not maximal then Ql = P, . In the first case, R/QT is a special primary ring. In the second case, R/Q, is a Dedekind domain.

(3) + (1). Each Dedekind domain and each special primary ring is a ZPI-ring. Hence, the assertion follows from Exercise 6(g).

9.11. Corollary. An indecomposable ZPI-ring is either a Dedekind domain or a special primary ring.

3 MULTIPLICATION RINGS

9.12. Definition. A ring R is a multiplication ring if A G B, where A and B are ideals of R, implies that there exists an ideal C of R such that A = BC.

If R is a Dedekind domain, then R is a multiplication ring. The converse is also true when applied to integral domains.

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210 Ix GENERALIZATIONS OF DEDEKIND DOMAINS

9.13. Proposition. If R is an integral domain which is a multiplication ring, then R is a Dedekind domain,

Proof. Let A be a nonzero ideal of R and let a € A, a # 0. Since (a) E A there is an ideal C of R such that (a) = AC. Since (a) is invertible the same is true of A. Therefore, by Theorem 6.19, R is a Dedekind domain.

In this section we shall find some equivalent conditions for a ring to be a multiplication ring, and in the process of doing this, we shall establish some properties of multiplication rings. In particular, we shall show that R is a multiplication ring if and only if it satisfies the apparently weaker condition that A G P, where A is an ideal of R and P is a prime ideal of R, implies that there exists an ideal C of R such that A = PC. Temporarily, we need a name for such rings ; we shall call them weak multiplication rings.

Let R be a weak multiplication ring. If P is a proper prime ideal of R, then RIP is a Dedekind domain (see Exercise 7). Hence d i m R I 1 .

9.14. Proposition. Let R be a weak multiplication ring. If P is a maximal ideal of R, then there are no ideals of R strictly between P and P2.

Proof. Let P be a maximal ideal of R, and assume that there is an ideal A of R such that P2 c A c P. There exists an ideal C of R such that A = PC, and C 9 P. Let c E C\P; since (c) G C we have P(c) G PC = A. Hence P(P + (c)) = P2 + P(c) E A. Since P is maximal, P+ (c) = R. Therefore, P G A, which contradicts the fact that A = P.

9.15. Proposition. Let R be a weak multiplication ring. If M is a maximal ideal of R and if P i s a prime ideal of R such that P c M , then P= Mn and P=MP.

Proof. Since RIP is a Dedekind domain, n:= (M/P)" = 0, and there- fore Mn G P. Since P c M there is an ideal A of R such that

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3 MULTIPLICATION RINGS 21 1

P= MA. Since M $ P, A must be contained in P. Hence, P = M A c MP c P, and consequently

Recall the following definition from Exercise 9 of Chapter III : if P is a minimal prime divisor of an ideal A of a ring R, then AR, n R is called the isolated P-primary component of A. It is the unique minimal P-primary ideal of R which contains A.

9.16. Definition. Let A be an ideal of a ring R. The kernel of A is the intersection of the isolated P-primary components of A , as P runs over the minimal prime divisors of A .

9.17. Proposition. I f R is a weak multiplication ying, then every idealof R i s equal to its kernel.

Proof. Let A be an ideal of R and let AX be the kernel of A. Suppose A # AX and let a E A*\A. Consider the ideal B = A : (a). Let M be a minimal prime divisor of B. Since A E B, M contains a minimal prime divisor P of A, by Lemma 2.15. If P= M then a E ARM n R ; hence, by Proposition 3.12, there exists s E R\M such that as E A. But then s E A : (a) _c M , a contradiction. Thus, we have P c M. Since dim R 5 1, M must be maximal, and it follows from Proposi- tion 9.15 that P= Mn and P=MP. Since B G Mthere is an ideal C of R such that B = MC. If C s B then

W

B = MB= M2B = a . - c n M"=P; n = l

this is not the case since M is a minimal prime divisor of B, and so we conclude that C $ B. Thus, C(a) $ A. Since a E AX s P, we have C(a) c P, and so there is an ideal D of R such that C(a) = PD. Hence,

C(U) = MPD = MC(U) = B(a) G A.

'I'his contradiction implies that the assertion of the proposition is true.

9.18. Theorem. If R is a weak multiplication ring, then every primary ideal of R is a power of its radical.

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212 Ix GENERALIZATIONS OF DEDEKIND DOMAINS

Proof. Let Q be a primary ideal of a weak multiplication ring R, and let P = Rad(Q). If P = R, then Q = R, so we assume that P # R.

We consider two cases. Assume first that P is a maximal ideal of R. If P" # Pn+l for each positive integer n, then Q is not contained in every power of P, since n:= P" is a prime ideal of R [by Exercise 7(d)] which is properly contained in P. Thus, there exists a positive integer k such that Q G P" and Q $ P"+l. By Exercise 7(c), there is an ideal A of R such that Q = PkA and A $ P. If A # R than any maximal ideal of R containing A would contain Q, and hence would equal P. Since this is impossible we have A = R and Q = P".

Continuing to assume that P is maximal, we suppose that P" = Pn+l for some positive integer n. Suppose further that Q 5 P". If a E P" then by Exercise 7(c), there is an ideal A of R such that (a ) = P"A = P2"A = P"(a). Hence, by Exercise 6(b), there exists p E P such that a = p a = p 2 a = * * * , and consequently a E Q since p s E Q for some positive integer s. Thercfore, Q = P". On the other hand, suppose that Q $ P". Then there exists a positive integer k such that Q G P" and Q $ Pk+l and, just as in the preceding paragraph, we have Q = P".

Now assume that P is not a maximal ideal of R. We shall show that Q = P . Suppose that there exists x E P with x $ Q. Let M be a maximal ideal of R containing P , and let B = M(x) + Q. If x E B then x ( 1 - m) E Q for some m E M . Since x q! Q, 1 - m E P c M , so 1 E M , which is not true. Hence x I$ B. By Proposition 9.17, B is equal to its kernel, Let (Pa> be the set of minimal prime divisors of B, and let Qa be the isolated Pa-primary component of B. Since x g B there is an index CI such that x q! Qa . If p E P then for some positive integer n, p" E Q G B c P a , so p E Pa; thus P E P a . However, if y E M\P then yx E B c_ Qa , so y E Pa. Therefore, P c Pa, which contradicts the fact that Pa is a minimal prime divisor of B. Thus, we must have Q = P.

9.19. Proposition. Let R be a weak multiplication ring. If P is a prime ideal of R which is not maximal, then P= P2.

Proof. Suppose that P # P2 and let a E P\P2. Let M be a maximal ideal of R such that P c M . Then P i s the only minimal prime divisor of (Pz + (u))M, and since each ideal of R is equal to its kernel, we have

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(P2 + (a))&'= (P2 + (a))MR, n R

= (P2 + (a))R, n R since MR, = R,

= P2 + (a) .

Hence, a = b + ac for some b E P2 and c E M. Since P2 is equal to its kernel, it is P-primary. Hence, a(1 - c) E P2 and a 6 P2 imply that 1 - c E P c M . Thus 1 E M , which is not true.

9.20. Proposition. Let R be a weak multiplication ring, let A be an ideal of R, and let P be a minimal prime divisor of A. Let P" be the isolated P-primary component of A (such an n exists by Theorem 9.18). If P" # P" i l , then P does not contain the intersection o j the remaining isolated primary components of A.

Proof. Let {P,} be the set of minimal prime divisors of A other than P. For each u, let P,"" be the isolated Pa-primary component of A. Let B = n P,"". By Proposition 9.17, A = P" n B. Since P" # Pn+l, we certainly have P # P 2 ; hence P is a maximal ideal of R. Thus, PnC1 is P-primary. If A 2 Pn+l, then

P n = A R p n R s P n + ' R , n R=Pn+',

contrary to hypothesis ; hence A $ P" +I. By Exercise 7(c) there is an ideal C of R such that A= P"C and C $ P. For each a, P"C c P: and P" $ P,; hence C s P,?. Thus, C c B, and since C $ P, we conclude that B $ P .

Of course, all of the results obtained under the hypothesis that R is a weak multiplication ring hold for multiplication rings. We shall now show that, in fact, every weak multiplication ring is a multiplica- tion ring.

9.21. Theorem. Let R be a ring. The following statements are equiua- lent :

(1) (2) (3)

R is a multiplication ring. R is a weak multiplication ring. R satisjies the following conditions:

(i) (ii)

Each ideal of R is equal to its kernel. Each primary ideal of R is a power of its radical.

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214 Ix GENERALIZATIONS OF DEDEKIND DOMAINS

(iii) If P is a minimal prime divisor of an ideal A of R, ;f n is the least positive integer such that P" is the isolated P-primary component of A, and ;f P" # Pn+I , then P does not contain the intersection of the remaining isolated primary components of A.

Proof. The only part of the theorem that requires proof is the assertion that ( 3 ) implies (1). Assume that ( 3 ) holds. Let A and B be ideals of R such that A G B. Let {Pa} be the set of those prime ideals of R which are minimal prime divisors of both A and B, {P,'} the set of those which are minimal prime divisors of B but not of A, and {Py" } the set of those which are minimal prime divisors of A but not of B. Since A and B are equal to their kernels, we may write

and

where the exponents h a , k, , m a , and ny are chosen to be the least positive integers such that Paha is the isolated Pa-primary component of B, PLkO is isolated P,'-primary component of B, PF is the isolated Pa-primary component of A, and P;"' is the isolated Pipr imary component of A. For each a, we have P,"" G P:a, so ha 5 m a . Let

c= ((7 p y h a ) n ( n py). Clearly, A G C. Let x E BC; then x = C:= bici , where bi E B and ci E C for i = 1 , . . . , n. Therefore, bi E Pta, ci E P,""-ha, and ci E P;"' for each i and each a and y . Consequently, for each i, bici E P,"" and b, ci E P;'"'. Thus x E P,"" and x E P?' for each a and y and, as a result, x E A. Therefore, BC c A.

Let P be a minimal prime divisor of BC. Then P contains either B or C. If B G P, then P is a minimal prime divisor of B. It contains A and so must contain a minimal prime divisor of A ; hence, since BC c A, it must be a minimal prime divisor of A. Thus, P = Pa for some CI. If B $ P then C E P and P is a minimal prime divisor of both A and C. Thus, P= P; for some y. Since BC is equal to its kernel,

BC= ( n ~ 2 ) n ( P';sY),

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where r , and s, are the least positive integers such that P,'" is the isolated Pa-primary component of BC and P;sv is the isolated P;-primary component of BC. Since BC c A we must have m, 5 r a and ny i s , for all CI and y.

For each y , A c_ C G PC""; hence

Pyy= CR,; n R.

Since B $ P;, this implies that

BCR,; n R = P;",;

that is, Pi.'"' is the isolated P;-primary component of BC. Therefore, ny = sy .

If P,"" = P?+l, then P,"" = P z and we conclude that ma = r , . Suppose P,"" #P?+'. Let

and

By (iii), A' $ P, , and by Proposition 9.19, P, is a maximal ideal, so that R = Pza + A'. Thus,

A = P,"" n A' = P,""A'.

Since h, 4 m, , we have Ptu # P,"" + '. Hence, B' $ Pa, R = P,"' + B', and

B=P,h"nB'=P?B'.

Since A' c C' and A' $ P, , we must have C' $ P, . Consequently, if m, - h , > 0, then

R=p;a-hu+C and C=pzu-hu CJ=pma-hnC.

Thus, if m , - h , > 0, we have BC : PFB'C'. If m, - h, = 0, then C= C' and the same equality holds. We note that B'C' $ P,; thus

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216 Ix GENERALIZATIONS OF DEDEKIND DOMAINS

PF is the isolated Pa-primary component of BC. Since ma 5 ra , it follows that ma = r , .

We have shown that BC and A have the same kernels. Therefore, A = BC.

4 ALMOST MULTIPLICATION RINGS

If R is a Dedekind domain, then each ideal of R which has prime radical is a power of its radical. In this section we shall consider those rings which have this property.

9.22. Definition. A ring R is an almost multiplication ring if each ideal of R which has prime radical is a power of its radical.

The terminology stems from the fact that a ring is an almost mul- tiplication ring if and only if it is locally a multiplication ring; we shall prove this later on in this section. Thus, the almost multiplication rings bear the same relation to the multiplication rings as the almost Dedekind domains bear to the Dedekind domains.

If P is a prime ideal of ring R and if S = R\P, we shall denote 0, , the S-component of the zero ideal, by N ( P ) ; that is,

N ( P ) = {X I x E R and xs = 0 for some s E R\P}.

9.23. Theorem. If R is an almost multiplication ring, then for each proper prime ideal P of R, the ring R, is a ZPI-ring.

Proof. Let P be a proper prime ideal of R. If P is a minimal prime divisor of 0, then PR, is the only prime ideal of R,. Hence, by Exercise 5(g) of Chapter 111, every proper ideal of R, is a power of PR, . Thus, R, is a ZPI-ring.

Now assume that P is not a minimal prime divisor of 0. Let Q be a minimal prime divisor of 0 contained in P. Then, R/Q is an almost Dedekind domain by Theorem 9.4. Hence, P / Q is a maximal ideal of R/Q and therefore P is a maximal ideal of R. Thus dim R 5 1. Another consequence of the fact that R/Q is an almost Dedekind domain is that n;==l(P/Q)" = 0 (see Theorem 9.4), that is,

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on"=, P" G Q. Suppose the containment is proper. Then there is a positive integer n such that Q c Pn but Q $ Pn+I. Then, since Rad(Q + P"+I) = P , we have Q + Pn+l = P"; whence (P/Q)"" =

(P/Q)". Since R/Q is an almost Dedekind domain, this is impossible. Thus n:= P" = Q .

I f Q' is any other minimal prime divisor of 0 contained in P, the above argument shows that Q' = ():= P" = Q. Thus, R, has only two proper prime ideals, PR, and QRp . Therefore, every ideal R p has prime radical and consequently is a prime power. This proves that R, is a ZPI-ring.

Our next objective is to prove the converse of Theorem 9.23. In doing so we shall establish some important results concerning the ideal structure of almost multiplication rings.

9.24. Proposition. Suppose that R, is a ZPI-ring for each proper prime ideal P of R . If P is a proper prime ideal of R such that N(P) is not prime, then Rad(N(P)) = P.

Proof. As we have seen, R, has only one minimal prime divisor of 0. Hence, R, is either a Dedekind domain or a special primary ring; since N ( P ) is not prime, R, is not a domain, and so is a special primary ring. Thus, there is a positive integer TZ such that PnR, = 0. Hence, for each p E P there exists s E R \P such that pns = 0, that is, p E Rad(N(P)). Therefore, Rad(N(P)) = P.

9.25. Proposition. Suppose that R, is a ZPI-ring for each proper prime ideal P of R . If P is a proper prime ideal of R such that N ( P ) is not prime, then

(1) (2) (3)

P is maximal and minimal, R, is a special primary ring, and each ideal with radical P is a power of P .

Proof. We have noted already that R, is a special primary ring. Thus, P is a minimal prime ideal of R. Let M be a maximal ideal of R containing P . Then, RM is a Dedekind domain or a special primary ring. If RM is a special primary ring, then M is minimal, so M = P,

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21 8 Ix GENERALIZATIONS OF DEDEKIND DOMAINS

and P is maximal. Suppose R , is a Dedekind domain; then N(M) is prime and M and N ( M ) are the only prime ideals of R contained in M . Hence, either P = M or P = N ( M ) . If P = N ( M ) , then P = N ( P ) which is not true since N ( P ) is not prime. Hence P = M , and con- sequently P is maximal. Finally, if Rad(A) = P then A is P-primary and thus AR, = P"R, for some positive integer. Therefore, A = Pn.

9.26. Proposition. Suppose that R, is a ZPI-ring for each proper prime ideal P of R. If P is a proper prime ideal of R such that N(P) is a prime ideal and P # N(P) , then

(1) (2) (3) P is maximal,

(4)

R, is a discrete rank one valuation ring, N(P) is the only prime ideal of R properly contained in P ,

each ideal with radical P is a power of P, and (5) N(P) = n:=l P".

Proof. Since N(P) is a prime ideal, R, is an integral domain ; hence R, is a discrete rank one valuation ring. Thus, PR, is the only non- zero prime ideal of R, [it is not zero since P # N(P)]. By the one- to-one correspondence between the proper prime ideals of R, and the prime ideals of R contained in P, we see that N(P) is the only prime ideal of R properly contained in P. Let M be a maximal ideal of R containing P. If N ( M ) is not a prime ideal of R, then M is a minimal prime ideal of R by Proposition 9.25; in this case P = M and since P # N ( P ) we have a contradiction. Therefore, N ( M ) is a prime ideal and is the only prime ideal of R properly contained in M . Thus, either N ( M ) =N(P) or N ( M ) = P. Since P # N ( P ) , we must have M = P. Thus, P is maximal.

If Rad(A) = P then AR, = PnR, for some positive integer n ; since P is maximal, this implies that A = P". It follows also from the fact that P is maximal that each power of P is P-primary, and con- sequently N(P) c Pn for each positive integer n. Since R, is a Noetherian valuation ring, ();= PnR, = 0. Therefore, ():= Pn =

"PI.

Now we are ready for the converse of Theorem 9.23.

9.27. Theorem. If R, is a ZPI-ring for each proper prime ideal P of R , then R is an almost multiplication ring.

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4 ALMOST MULTIPLICATION RINGS 219

Proof. Suppose that R, is a ZPI-ring for each proper prime ideal P of R. Let A be a proDer ideal of R such that Rad(A) = P i s a prime ideal of R. By Propositions 9.25 and 9.26, A is a power of P if either N ( P ) is not a prime ideal or if N ( P ) is a prime ideal and P # N(P) . So, we assume that P = N ( P ) . Let Q be a proper prime ideal of R which contains A. Then P c Q, and as in the proof of Proposition

N(Q) z AR, n R L PR, n R=P,

from which it follows that AR, n R = PRQ n R. Thus, by Exercise 5(b) of Chapter 111, A = P. Therefore, inevery case, A isapower of P.

9.26, N(Q) = N ( P ) = P. Thus,

Now we can easily prove our assertion that almost multiplication rings are rings which are locally multiplication rings.

9.28. Theorem. A ring R is an almost multiplication ring i f and only if R, is a multiplication ring fo r each proper prime ideal P of R.

Proof. If R is an almost multiplication ring, then each R, is a multi- plicaion ring by Theorem 0.23. On the other hand, let P be an arbitrary proper prime ideal and assume that R, is a multiplication ring. Then, by Theorem 9.21, primary ideals in R, are prime powers and each ideal in R, is equal to its kernel. Thus, in R, , any ideal with prime radical is primary. Since primary ideals are prime powers, this implies that R, is an almost multiplication ring. Consequently, R, = (Rp)pRp is a ZPI-ring. Therefore, by Theorem 9.27, R is an almost multiplication ring.

9.29. Theorem. A ring R i s an almost mu1tz)lication ring i f and only if whenever an ideal A of R is such that Rad(A) = PI * * * P,, where PI , . . . , P, are distinct prime ideals of R, then A = P:' * - . P,", fo r some positive integers n,, . . . , n, .

Proof. By the definition of almost multiplication ring, any ring which satisfies the condition stated in the theorem is an almost multiplication ring. Conversely, let R be an almost multiplication ring and let Rad(A) = PI . * P, , where PI, . . . , P, are distinct prime ideals of R. First, suppose that P,, . . . , P, are comaximal and that each is a minimal prime divisor of A ; then these prime ideals are all of the

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220 Ix GENERALIZATIONS OF DEDEKIND DOMAINS

minimal prime divisors of A. By Exercise 5(g) of Chapter 111, the isolated Pi-primary component of A is a power, P;l, of Pi for i= 1, . . . , s. Hence, by Exercise 9(b),

A = p;l n n p,"S = p;l . . . Now suppose one of the prime factors of Rad(A), say Pl, is not a

minimal prime divisor of A. Then, P, contains a minimal prime divisor Q of A. Since Pl P, c Q, one of these prime ideals, other than Pl, must equal 9, say P, = Q. Then P, c Pl. Note that this is the only way in which two prime factors of Rad(A) can fail to be comaximal. By Exercise 9(c), PIP, = P, . If A = Pi2 - - - PF, then A = p;ZP;z * * * P,"".

EXERCISES

1. Almost Dedekind domains. Let R be an integral domain. (a) Show that R is an almost Dedekind domain if and only if

R if a Prufer domain with no idempotent nonzero proper prime ideals and for each such prime ideal P, R i scom- pletely integrally closed.

(b) Show that an almost Dedekind domain is integrally closed. (c) Show that if P is a prime ideal of an almost Dedekind

domain, then there are no ideals strictly between P and P2. (d) Show that in an almost Dedekind domain, the set of

primary ideals with the same radical is totally ordered by inclusion.

2. An example. (a) Let (R,l n = 1,2, . . .> be a sequence of almost Dedekind

domains such that R, c R,+l for all n 2 1. Show that R' = u:=l R, is an almost Dedekind domain if and only if it has no idempotent nonzero proper prime ideals.

(b) Let R be an almost Dedekind domain with quotient field K . Let K be an algebraic extension of K and let R' be the integral closure of R in K'. Show that R' is almost Dede- kind if and only if it has no idempotent nonzero proper prime ideals.

( c ) Let K be the field of rational numbers and let K' be

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22 1

obtained by adjoining to K the pth roots of unity for every prime p . Let R' be the integral closure of 2 in K . Show that R' is an almost Dedekind domain. Show that the domain R' of (c) has no finitely generated nonzero proper prime ideals, and hence that R' is not a Dedekind domain. Let R be an almost Dedekind domain which is not a Dedekind domain. Then R has a maximal ideal P which is not finitely generated (by Exercise 14 of Chapter 11). Show that if a is a nonzero element of P then (a) : P = (a), but (a)R, : PRp # (a)R, . With R and P as in (e), show that PR, is invertible for each maximal ideal M of R, but P is not invertible. (Thus, the assertion of Exercise l(b) of Chapter VI is no longer true if the ideal is not required to be finitely generated.)

3. Factorization in almost Dedekind domains. Let R be an almost Dedekind domain and let { M u [ CI E I } be the set of maximal ideals of R. If A is a nonzero ideal of R, for each K E I , let f,(A) be the smallest nonnegative integer k such that ARM, n R = Muk (take Muo = R). For nonzero x E R, write f&) for fa((x>>.

Show that if A is a nonzero ideal of R , then

A = n ~y). Show that if A is a nonzero ideal of R , then for each CI E I there is an integer k 2 0 such that A G Mak and A $ M ; + l ; show that k = f,(A). Show that for nonzero x, y E R and for a E I,

U E I

f a ( 4 = f u W +fdY), and fa(x + Y ) 2 min{f&),fu(r)}.

Hence, for each CI E I , there is a valuation v, on the quoti- ent field of R such that v , ( x ) = f , ( x ) for all nonzero x E R . Show that RMa is the valuation ring of v,. Let x , E Ma\MU2 and let y be a nonzero element of the quotient field of R . Show that there exist a, b E R\M, such that

= aXwJ) a lb.

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222 Ix GENERALIZATIONS OF DEDEKIND DOMAINS

4. Certain types of almost Dedekind domains. Let R be an almost Dedekind domain. (a) Show that R is a Dedekind domain if and only if every non-

zero proper ideal of R is contained in only finitely many maximal ideals of R.

(b) Show that if R has only finitely many maximal ideals then R is a principal ideal domain.

(c) Show that R is a Dedekind domain if and only if R is a Krull domain.

5. Domains with property (#). Let R be an integral domain and let A be the set of maximal ideals of R. The domain R has property (# ) if for every pair of distinct subsets A, and A2 of h we have

0 R P # 0 R P . PEA^ P E A Z

(a) Show that R has property (# ) if and only if for every P E A we have

(b) Show that if for every P E A we have

PSf U (2, Q e A Q # P

then R has property (#). Show that if R has the QR-property (see Exercise 12 of Chapter VI), then R has property ( # ) if and only if each maximal ideal of R is the radical of a principal ideal. Suppose R is a Prufer domain and dim R = 1. Show that if R has property (#), then every overring of R has pro-

Suppose that R is a Prufer domain, that dim R = 1, and that R has property ( #). Let A ={&fa 1 CI E I ). Show that for each aEI, M , is the radical of an ideal with two generators. [Hint. For each /I E 1 let v 4 be the valuation on the quotient field of R determined by RM4 . Show that there exists a, b E R such that

(c)

(d)

perty ( #). (e)

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EXERCISES 223

4 4 < %(@, and %(a) 2 %(b) for all P#a.

Then b/a=s/t, where s E M , and t E R\Ma. Let J = {BIB E I and s E M,}. Let T = R \ u B E , M , and set R' = R, . Show that there is an element t E R such that

t E M , R'\ u M, R'. , € I B f a

Then show that M a =Rad((s, t)).] Show that if an almost Dedekind domain R has property (#), then R is a Dedekind domain.

(f)

6. Some lemmas to Theorem 9.10. Let R be a ring and let A be a finitely generated ideal of R . (a) Let R be a subring of a ring R'. Let N be a finitely generated

submodule of the R-module R' with k generators. Let B and C be ideals of R such that BN 5 CN. Show that for any element b E Bk there exists c E C such that bx = cx for each x EN. [Hint. First, show this for elements of the form b, - * * bk , where b, E B for i = 1, . . . , k, and then for sums of elements of this type.]

(b) Show that if B is an ideal of R and if AB = A, then there exists b E B such that ab = a for all a E A.

(c) Show that if A is idempotent then there is an element a E A such that a2 = a and A = (a).

(d) Show that if B = ():= A", and if AB is an intersection of primary ideals of R, then AB = B.

(e) Assume R is Noetherian and let IM be a maximal ideal of R such that there are no ideals of R strictly between M and M 2 . Show that for each positive integer n we have Mn c

M" if and only if there is a prime ideal P of R such that P C M . Let R and M be as in (e). Assume that for each positive integer n, Mn+I c M". Show that fin"=, M" is the unique prime ideal of R contained in M .

(g) Show that the direct sum of a finite number of ZPI-rings is a ZPI-ring.

(f)

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224 Ix GENERALIZATIONS OF DEDEKIND DOMAINS

7. Weak multiplication rings. Let R be a weak multiplication ring. (a) Show that for every ideal A of R, the ring RIA is a weak

multiplication ring. (b) Show that if R is an integral domain, then R is a Dedekind

domain. (c) Let M be a maximal ideal of R and let A be an ideal of R

with A G Mn. Show that there is an ideal B of R such that A= MnB; if A $ Mn+I then B $ M .

(d) Let M be a maximal ideal of R such that M" # Mn + for

each positive integer n. Show that n:=l Mn is a prime ideal of R.

8. Multiplication rings. (a) Show that every multiplication ring is a subring of a direct

sum of Dedekind domains and special primary rings. (b) Let R be the ring in Exercise 13 of Chapter 11. Show that

R is a multiplication ring which is not a direct sum of Dedekind domains and special primary rings.

9. Some properties of almost multiplication rings. Let R be an almost multiplication ring. (a) Show that for each proper primary ideal Q of R there exists

a maximal ideal M of R such that either Q=N(M) or Q = M" for some positive integer n.

(b) Show that each ideal of R is equal to its kernel. (c) Let P be a prime ideal of R and let A be any ideal of R such

that P c A. Show that PA = P. (d) Show that if A, B, and C are ideals of R, with A regular,

and AB = AC, then B = C. (e) Let S be a multiplicative system in R. Show that S-lR is

an almost multiplication ring.

Show that if R is a multiplication ring, then the following statements are equivalent : (1) The zero ideal of R has finitely many minimal prime

divisors. (2) The zero ideal of R is an intersection of a finite

number of primary ideals. (3) R is Noetherian.

10. Noetherian multiplication rings. (a)

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EXERCISES 225

(b)

(c)

Show that R is a Noetherian multiplication ring if and only if it is a ZPI-ring. Show that a Noetherian almost multiplication ring is a multiplication ring.

11. Almost Dedekind domains and the ideal transform. Refer to Exercise 16 of Chapter VI for terminology. (a) Prove that an integral domain R with more than one

maximal ideal is an almost Dedekind domain if and only if T( (x ) ) is an almost Dedekind domain for each nonunit 31:

of R. Prove that an integral domain R with more than one maximal ideal is a Dedekind domain if and only if T((x)) is a Dedekind domain for each nonunit x of R. [Hint. Use (a) and Exercise 14 of Chapter VIII.] Show that the hypothesis concerning maximal ideals in (a) and (b) is necessary.

(b)

(c)

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C H A P T E R

X Priifer Rings

In this chapter we shall present in detail some extensions of the results concerning Prufer domains to rings which are not integral domains. Under a restrictive assumption, this was done in a series of exercises in Chapter VI. In this chapter we shall show that no such restrictive assumption is necessary.

1 VALUATION PAIRS

In this section a valuation theory for rings is developed which, when restricted to fields, coincides with the theory of Chapter V.

10.1. Definition. A valuation on a ring R is a mapping v from R onto an ordered Abelian group with co adjoined, as in Section 3 of Chapter V , such that for all a , b E R,

(i) v(ab) = .(a) + v(b), and (ii) v (a + b) 2 min{v(a), v(b)} .

It is evident that v( 1) = 0 and v(0) = 00 for every valuation v on R. It is important to note that we have required that v map R onto the ordered Abelian group ; thus, {.(a) I a E R}\{v(O)} is an ordered Abelian group.

226

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1 VALUATION PAIRS 227

10.2. Proposition. Let v be a valuation on R and set

R, = {XI x E R and v(x) 2 0},

P , = { x l x ~ R and v ( x ) > O ) ,

A, = {xi .x E R and v(x) = a}.

Then, R, is a subring of R, P, is a prime ideal of R, , and A, i s a prime ideal of R, . Furthermore, (f R, # R and if A is an ideal of R such that A s R , , t h e n A ~ A , .

Proof. That R, is a subring of R and P, is a prime ideal of R, follow as in the case of a valuation on a field. Clearly, A, is an ideal of R, . If a , 6 E R, and a6 E A,, then v(a) + v(6) = co, so .(a) = co or v(6) = co; hence A, is a prime ideal. Suppose that R, f R, that A is an ideal of R contained in R, , and that A $ A,. Let a E A be such that .(a) # co. Then there exists 6 E R such that v(b) = +a), and since R, # R there exists c E R such that v(c) < 0. Then a6c E A but v(a6c) = n(c) < 0, which contradicts A E R, .

10.3. Definition. A valuation pair of a ring R is a pair (A, P ) , where A is a subring of R and P i s a prope~ prime ideal of A such that for every x E R\A there exists y E P such that x y E A\P.

If (A, P ) is a valuation pair of the total quotient ring of A, then the ring A is called a valuation ring.

10.4. Theorem. If v i s a valuation on a ring R, then (R,, P,) is a valuation pair of R. If (A, P ) is a valuation pair of R, then there exists a valuation v on R such that A = R, and P = P, .

Proof. Let v be a valuation on R. If x E R\R, then U(X) < 0 and there exists y E R such that v(y) = -=a(.) > 0. Then y E P, and v(xy) = 0, so that xy E Rv\P,. Therefore, (R, , P,) is a valuation pair of R.

Conversely, suppose that (A, P ) is a valuation pair of R. If x E R we set

[P : xIR = { X I z E R

and we define a relation - on R by setting x - y if [P :

and xz E P},

= [P : yIR.

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228 X PRUFER RINGS

Clearly, this is an equivalence relation on R; denote by v(x) the equivalence class of an element x of R. On the set of these equivalence classes we define a binary operation by v(x) + v ( y ) = v(xy). This is a well-defined operation, for if x’ E v(x) and y’ E v (y ) , then for any z E R, (xy)z E P if and only if (x ’y’ )~ E P. Thus, v(xy) = ~(x’y’).

Let v(R) = {v(x) 1 x E R). We shall now show that v(R)\{v(O)} is a group with respect to the operation defined above, and that v( 1 ) = A\P. Clearly, v( 1) is an identity element of v(R) with respect to this operation. If x 4 A , then xy E A\P for some y E P. Then y E [P : 1 I R but y I$ [P : xIR, so x rii 1. If x E P, then 1 E [P : xIR; however, 1 4 [P : 1 I R , so x rii 1. This shows that v(1) c A\P. Now suppose that X E A\P and x $ v ( l ) . Since [ P : 1IR= P, we have [ P : 1IR c [P: xIR; hence there exists y $ P such that xy E P. Since P is a prime ideal of A , this implies that y $ A . Hence there exists z E P such that yz E A\P. Thus x(yz) E A\P, while at the same time ( x y ) ~ E P. Since this is impossible we conclude that we do have v(1) = A\P. Now let x 4 v(0). Since [P : 0IR = R we have xy $ P for some y E R. If xy E A , then v(x) + v ( y ) = v(l), so v ( y ) = -v(x). If xy $ A , then (xy))x E A\P for some x E P. Then v(x) + v(yx) = v(l), so v(yx) = -v(x). This proves that v(R)\{v(O)} is a group.

Now define 5 on v(R)\{v(O)} by v(x) 5 v( y ) if [P : xIR 5 [P : y l R . Then v(x) < v(y) is equivalent to the existence of an element z E R such that xz E A\P and y z E P. Clearly, 5 is a well-defined relation. We leave it to the reader to verify that v(R)\(v(O)}, together with the relation 5, is an ordered Abelian group. If we set v(x) 5 v(0) for all x E R, then v(0) plays the role of co. To prove that the mapping x++v(x) from R onto v(R) is a valuation it is sufficient to verify (ii) of Definition 10.1. Let x, y E R and let x E R be such that v (x ) _< v(x) and v(x) 5 v(y). Then [P : zIR G [P : xIR and [P : Z ] R E [P : y I R ; hence [P : zIR c [P : x +yIR and v(x) I v(x + y). It is clear from the definition of 5 that A = R, and P = P, .

10.5. Corollary. Let ( A , P ) be a valuation pair of R. Then the following statements hold:

( 1 ) (2) (3) I f x E R a n d x ” E A , t h e n x E A . (4) (5)

R\A i s closed under multiplication. R\P is closed under multiplication.

A = { x l x ~ R and xy E P for al ly E P}. I f A # R, then P= {xlx E A and xy E A for somey 4 A}.

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1 VALUATION PAIRS 229

For a ring R, let F be the set of all pairs (A, P), where A is a subring of R and P is a proper prime ideal of A. Define a partial ordering < on F by

(4, PI) < (4 9 Pz>

if A , c A, and P, n A, = P, ; if this is the case, we say that (A,, Pz) dominates (A,, P,). Zorn’s lemma guarantees maximal elements of F which dominate any given pair (A, P ) . It follows immediately from the lying-over theorem that, if (A, P ) is a maximal pair in Y, then A is integrally closed in R.

The next theorem shows that the maximal elements of F play the same role here as in the valuation theory of fields ; it is an extension of part of Theorem 5.7 (for the extension of the other part of Theorem 5.7, see Exercise 2).

10.6. Theorem. Let R be a ring and Let F be as above. Then (A, P ) is a valuation pair of R if and only if (A, P ) is a maximal element of F.

Proof. Let (A, P ) be a valuation pair and let (B, Q) E F be such that (A, P)< (B , Q). Assume A c B. If x E B\A, then there exists y E P such that xy E A\P. However, y E Q and so xy E Q n A = P, which is impossible. Thus B = A and Q = P. Therefore, (A, P) is maximal in F.

Conversely, suppose that (A, P ) is a maximal element of F. If A 5 R then (A, P ) is a valuation pair. Suppose that A # R and let x E R\A. Let B = A[x] and Q = PI?. Then Q is an ideal of B and P G Q n A. If P= Q n A then A\P is a multiplicative system in B which does not meet Q. Therefore, there exists a prime ideal M of B containing Q and which does not meet A\P. Then M n A = P and so (A, P ) < (B , M ) . This contradicts the maximality of (A, P ) since x $ A. Thus, we are forced to conclude that P c Q n A.

Let a E Q n A, a 6 P. Then

wherep, , . . . , p , E P, and n is chosen as small as possible. Multiplying by p;-l, we obtain

(xp,), + n&pn)ip:: - - *pi - ap; - 1 = 0. i = O

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230 X PRUFER RINGS

Hence, xp, is integral over A, and therefore xp, E A , If xp, E P, then

which contradicts our choice of n (note that we cannot have n = 0 or n = 1). Hence, xp, E A\P, and so we conclude that (A, P ) is a valuation pair.

For the remainder of this section, let (A , P) be a fixed valuation pair of a ring R and let v be the valuation determined by this valuation pair. Let G, = v(R)\{v(O)}. As in Chapter V, we denote by G," the ordered Abelian group G, with co adjoined.

10.7. Definition. A n ideal I of R,(=A) is v-closed if for all x E I and y E R, v(y) 2 v(x) implies that y E I.

Also recall that a subgroup H of an ordered Abelian group G is isolated if for all nonnegative a E H and /3 E G, 0 5 /3 5 a implies that f i E H . We shall establish a correspondence between the v-closed proper prime ideals of R, and the isolated subgroups of Go.

10.8. Proposition. Let 4 be an order-preserving homomorphism from G, into an ordered Abelian group, and set +( co) = co. Then :

(1) (2) (3)

+v is a valuation on R. Ker + is an isolated subgroup of G, . P,, is a v-closed proper prime ideal of R, .

Proof. (1) I m + is an ordered Abelian group, +v maps R onto (Im +)*, and (i) and (ii) of Definition 10.1 hold for +v.

(2) Let a be a nonnegative element of Ker +and let 0 5/3 5 a. Then 0 5 +(p) 5 +(a) = 0 ; hence +(/3) = 0, that is, /i? E Ker 4.

(3) If x E P,, , then +(v(x)) > 0 , and so v (x) > 0 ; thus P,, c P, . If x E R, , then v(x) 2 0, and so +(v(x)) 2 0 ; thus A, c R,, . There- fore, P,, is a proper ideal of R,, and since it is a prime ideal of R,, , it is also a prime ideal of R,. Now let x E P,, , y E R , and v(y) 2 v(x). Then (~$v)(y) 2 (+v)(x) > 0 and consequently y E P,, . Thus, P,, is v-closed.

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1 VALUATION PAIRS 23 1

10.9. Proposition. Let H be an isolated subgroup of G,. Then there is a surjective order-preserving homomorphism + f rom G, onto an ordered Abelian group such that +v is a valuation on R and H = Ker 9.

Proof. Let + be the canonical homomorphism from G, onto G,/H. Set a + H 5 /3 + H if a 5 8. Since H is isolated, this is a well-defined total ordering of G,/H, and with respect to this ordering, GJH is an ordered Abelian group. The assertion of the proposition follows immediately.

If H and + are as in Proposition 10.9, we set uH = +v. Then we have

P,,=(x]xER and vH(x)>Of

= { X ~ X E R and v(x) + H > H }

= { X ~ X E R and v(x) > c c for all ~ E H } .

On the other hand,

H = {a 1 a E G, and +(a) = +(-a) = 0) =(a] a E G, and max{a, -a} < v(x) for all x E PUH}.

10.10. Theorem. Let v be a valuation on R. Then there is a one-to-one order-reversing correspondence between the isolated subgroups H of C, and the v-closed proper prime ideals Q of R,. The correspondence is given by

H- P U H

and

Qt,{aI CL E G, and max{a, -a>< v(x) for all x E Q)

Proof. I n the light of the remarks made above, all that remains to be shown is that, given Q, the set H = { a ] c1 E G, and max{a, -a} < v (x ) for all x E Q} is an isolated subgroup of G, , and that the correspon- dence is order-reversing.

E H and suppose a = v(x) and /3 = v(y) . By definition of H, -a E H. In showing that a+ /3 E H , it is enough to do so under the assumption that a+ #3 >_ 0. If a + # 3 $ H, then v(xy) = a + #3 2

Let a,

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232 X PRUFER RINGS

v(z) for some z E Q, and this implies that xy E Q since Q is v-closed. Since Q is a prime ideal and x $ Q and y $ Q, we must have x $ R, or y $ R,. Suppose the former. Then there exists u E R, such that v(uxy) = p, contrary to the fact that /3 E H . Thus, a + /3 E H, and we have shown that H is a subgroup of G, . Now suppose that a 2 0 and that y E G, is such that 0 5 y 5 a. If x E Q, then max(y, - y } = y 5 a = max{a, -a } < v (x ) ; hence y E H . Therefore, H i s an isolated subgroup of G, ,

If Q1 and Qz are proper prime ideals of R, such that Q1 c Q z , and if max{a, -a} < v(x) for all x E Q z , then max{a, -u} < v(x) for all x E Q1. Hence, the correspondence is order-reversing.

2 COUNTEREXAMPLES

Although valuation pairs of arbitrary rings possess many of the properties of valuation pairs of fields, they fail to possess others. In this section we shall discuss two examples of this fact.

The first example shows that if (A, P ) is a valuation pair of a ring R, then P may not necessarily be a maximal ideal of A. In this example, the ring A is a valuation ring. A second example of this phenomenon is given in Exercise 3 ; however, in that exercise, A is not a valuation ring.

In this example, R is the ring of Exercise 23 of Chapter VI ; all notation will be as in that exercise.

By Zorn’s lemma, there is a valuation pair (A, P) of the total quotient ring K of R such that R c_ A and P n R = N . Then K is the total quotient ring of A. We shall show that P is not a maximal ideal of A. Let P‘ be the ideal of A generated by P and x. Clearly, P c P‘ E A. If P’ = A then there exist s E A and q E P such that 1 = sx + q. Let q = a/b, where a, b E R and b is regular. Write

b =f’@) + xg’(x, Y ) + w‘, a =f@) + xg(x, Y ) + w, and

wheref(Y),f‘( Y), Xg(X, Y), Xg’(X, Y) E F [ X , Y] and w, w’ E Q. By Exercise 23(c) of Chapter VI, f‘(y) + xg’(x, y) is regular, and consequentlyf’(y) + xg’(x, y) - w’ is regular. If we multiply both a and b by this element, we see that we may assume that w‘ = 0.

If we multiply both sides of 1 = sx + q by bx, we get bx, = az,;

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hence f ' ( y ) z x = f ( y ) z x . This implies that ( f ' ( Y ) -f( Y))Z , E I , which in turn implies thatf'( Y ) =f( Y ) . Thus,f'(y) = f ( y ) , and we have

Since f ( y ) + xg'(x, y ) is regular, and since xg'(x, y)zx = 0, we must have f( Y ) # 0. Write f( Y ) = Y"(Y + h( Y ) ) where Y is a nonzero element of F and h( Y ) E Y F [ Y ] . By Exercise 23(b) of Chapter VI, g'(X, Y ) = Y"g*(X, Y ) , where g"(X, Y ) E F [ X , Y ] . Therefore,

r + h(Y) + (.g(x, Y ) + w)y-" = (. + h(Y) + xg*(x, y))a/b E p.

Next, we show that wy-" E P. Let g,, . . . , g k be those elements of G such that zgl, . . . , xgk occur in the generation of w as an element of Q. Since gi(x, y) E A\P for i = 1, . . . , R , we have nf= g,(x, y ) E A\P. But (nf= gi(x, Y))wY-" = 0 E P, SO WJJ-" E P.

Since h( Y ) E YF [ Y ] , we have h(y) E P ; hence Y + xg(x, y)y-" E

P. Letg(X, Y ) = Y ~ ( Y ' + Yk(X, Y ) ) , where r' is a nonzero element of F [ X ] and k(X, Y ) E F [X, Y ] . Then

r + xyp-"(r' + y k ( x , y)) E P.

If p - rn 2 0, then Y + xyp-"(~' + yk(x, y ) ) E P n R = N , which is not true since Y # 0. If p - m < 0, then X(Y' + yk(x, y ) ) E N , which is not true since r' # 0. Thus, we are forced to conclude that P' # A , and consequently that P is not a maximal ideal of A.

Another way in which valuation pairs of arbitrary rings fail to behave like valuation pairs of fields is the following. There exists a valuation pair (A, P ) and an overring B of A such that there is no ideal Q of B with the property that Q E P and (B, Q) is a valuation pair.

I n fact, let (A, P ) be the valuation pair of the example discussed above. Let p be a regular element of P and let U E A\P. Let B = A [ a / p ] and suppose that there does exist an ideal Q of B such that Q G P and (B, Q) is a valuation pair. Then p $ Q, for if p E Q, then p(a/p) = a E Q G P. Also, 1/p E B. For, suppose l / p $ B ; then there exists q E Q such that qlp E B\Q, and q =p(q/p). Let

1/p = b, + b,(a/p) + * * * + b,(a/p)",

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234 X PRUFER RINGS

where b, , . . . , b, E A and b, # 0. Suppose n 2 2. Multiplying by p n - l givespn-2 = b,a"-l(a/p) + t for some t E A. Sincepn-2 E A, we have b,a"-l(a/p) E A. If z, is the valuation determined by (A, P ) , then z,(un-l) = 0, and since v(b, a"-l(a/p>) 2 0, we must have v(b,a/p) 2 0, that is, 6, alp E A. Then we can write

lip (bn a/P + bn- l)(a/P)n-l + * . * + bo.

Assuming we have chosen n as small as possible, we have arrived at a contradiction. Thus, n 5 1. If n = 0, then 1 / p = b, E A, which is not true. Hence, n = 1 and we have l / p = bla/p + b o . Then, 1 = b,a + bop and so P + (a ) = A. Since P is not a maximal ideal of A, we can choose u E A\P such that P+ (a ) # A. With this choice of a, B has the desired property.

3 LARGE QUOTIENT RINGS

10.11. Definition. Let R be a ring with total quotient ring K, and let S be a multiplicative system in R. The large quotient ring with respect to S, denoted by R,,], is the set of all x E K such that xs E R for some s E S. If S is the complement in R of a proper prime ideal P of R, we write Rip] for Risl .

If P is a proper prime ideal of R, it is not necessarily true that PRipI is a prime ideal of Rip, . In order to study the relationship between the prime ideals of Rip] and those of R, the following opera- tion is introduced.

10.12. Definition. Let S be a mult+licatice system in a ring R. If A is an ideal of R, set

[AIRIsI = {x I x E K and xs E A for some s E S } ;

this is called the extension of A to Ris, .

If P is a prime ideal of a ring R, and if Q is a primary ideal of R contained in P, then [Q]RiPI is a primary ideal of R I P ] and [Q]RLP, n R = Q. The details are left to the reader. However, it is not necessarily true that [PIR,,, is a maximal ideal of Rl,, when P is

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not a maximal ideal of R. Furthermore, if A and B are ideals of R contained in P, it is not the case that [ABIR,,, must equal ([AlR[Pl)([B lR[Pl).

10.13. Definition. Let R be a ring and P a pitoper prime ideal of R. The core C(P) of P in R is the set of all x E R such that f o r each regular element Y E R, there exists an element s E R\P such that xsjr E R.

Some properties of the core of a proper prime ideal are given in Exercise 9. The importance of this concept and the others which have been introduced stems from the following theorem.

10.14. Theorem. Let P be aproper prime ideal of a ring R. The fol- lowing statements are equivalent :

(1) ( R C p 1 , [PIRIP]) is a valuationpair of the totalquotient ring of R. (2) For all x , y E R, either both x and y are in C(P) or there exist

a , b E R, not both in P, such that ax = by. (3) If A and B are ideals of R, not both contained in C(P), then

either AR, G BR, or BR, s AR, .

Proof. Assume (1) holds and let v be the valuation on the total quotient ring of R which is determined by ( R I P ] , [P]R[,]). Let x, y E R and assume v ( x ) 2 v(y) . If v ( y ) = m, then

x, Y E W'IR,,,) n R = [C(p)lR,pl n R = W'), where we have used various parts of Exercise 9. If v(y) # CO, then there exists t E K such that v(ty)=O and v ( t x ) 2 0 ; then ty E Rrp,\[P]Rrp1 . Hence, there exist u, v E R\P such that xtu E R and ytuv E R\P. Now let a = ytuv and b = uxvt. Then a and b are not both in P, and ax = by. Thus, (2) holds.

T o prove the converse, let y be an element of the total quotient ring of R which does not belong to R f p I . There is a regular element r E R such that ry E R. If ry E C(P), then there exists s E R\P such that sy = S Y ~ / Y E R ; this implies that y E Rip , , which is not the case. Hence, ry $ C(P) and so, assuming that (2) holds, there exist a, b E R, not both in P, such that ary = br. Since r is regular, ay = b. Since y $ R,,, , we must have a E P. Thus, b E R\P. Consequently

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a E [PIR,,, , and ay E R,,,\[P]R,,, since [PIR,,, n R = P. There- fore, (RIP, , [P]Rfp l ) is a valuation pair.

Now assume that (2) holds, and let A and B be ideals of R such that A $ C(P). Let a E A , a 4 C(P). If AR, $ BR,, we claim that we may choose a so that as 4 B for all s E R\P. For, assume that this is not the case, that is, assume that there exists q E R\P such that aq E B. Since AR, $ BR, , there exists d E A such that ds 4 B for all s E R\P. Since a 6 C(P), it follows from (2) that there are elements t , z E R, not both in P, such that at = d z . If t E P, then z E R\P and consequently zq E R\P and d(xq) = aqt E B. Since this cannot happen, we must have t E R\P. Since a 4 C(P) , there exists a regular element r E R such that au 4 ( r ) for all u E R\P. Hence a6 = rw for w E R implies that b E P. Therefore if du E (r), then duz = atu E (r) , so tu E P. But t E R\P, so u E P. It follows that d 4 C(P). If we replace a by d we see that we may assume in the first place that as 4 B for all s E R\P. Now apply (2) to a and an arbitrary f E B. There exist x, y E R, not both in P, such that ax = f y . We must have x E P, s o y E R\P. Hence, f y / y = ax/y EAR, . Therefore, BR, E AR, .

Finally, assume that (3) holds, and let x and y be elements of R not both in C(P). Then either xR, 5 yR, or yR, G xR,. If xR, s yR,, then there exists a E R\P such that ax = by for some b E R. Otherwise, there exists b E R\P such that ax = by for some aER.

4 PRUFER RINGS

10.15. Lemma. Let R be a ring in which ( A + B)(A n B) = AB for all ideals A and B of R, at least one of which is regular. Let P be a proper prime ideal of R and let a , b, c E R with a regular. If aR, c bR, , then either bR, G cR, or cR, E bR, .

Proof. Let x E R and y E R\P be such that ay = bx; x and y exist, since aR, c bR, . Now (a , b, c)((a, b) n (c ) ) = (a, b)(c), so bc = xla + x2 b + x3c , where xi E (a , b) n (c ) for i = 1, 2, 3. If xs = au + bv, where u, ZJ E R, then x 3 y = bxu + byv. Therefore,

bc(y - (xu +yv)) = x1ay + x2 by = (x1x + X2Y)b.

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If x = xu + yv $ P, then bR, G cR, because zb = x 3 y E (c). If z E P then y - z $ P since y $ P. Furthermore, x1 and x2 are in (a , b), so

(CRP)(bRP) E ((a, b)RP)(bRP) c (bRP)(bXP).

However, bR, is a regular principal ideal of R,, and so cR, c bR, .

10.16. Lemma. Let R be a ring such that A, B, and C are ideals of R with A Jinitely generated and regular and if AB = A C , then B = C. Let a and b be elements of R and let P be a proper prime ideal of R. If there is a regular element c E R such that cR, G aR,, then either aR, E bR, or bR, c aR, .

Proof. Since (ab)(a, b, c) c (a2, b2, ac)(a, b, c) , we have ab E (a2, b2, ac). Then cR, c aRp implies that abR, c (a2, b2)R,. Therefore, there exists y E R\P such that aby = xu2 + zb2 for some x, z E R. This implies that (zb)(a, b, c) 5 (a , c)(a, b, c), so zb E (a , c). Again using the fact that CR, G aR,, we have (zb)R, c aR,. Hence, there exists v E R\P such that zbv = au for some u E R. Now

abyv = xa2v + zb2v = xa2v + abu,

and therefore (a)(b)( yv - u ) c (a)2. If u $ P, then aR, G bR, .On the other hand, if u E P, thenyv - u $ P since yv $ P. Thus (aRp)(bRp) c (aRp)(aRp). Since aR, is a regular principal ideal of R, , this implies that bR, G aR,.

Now we are ready to define Prufer ring and to prove that Prufer rings are characterized by conditions analogous to those characterizing Priifer domains.

10.17. Definition. A ring Ris aPrufer ring if every Jinitelygenerated regular ideal of R is invertible.

10.18. Theorem. For a ring R, the following statements are equivalent:

( 1 ) R is a Prufer ring. (2) For every maximal ideal P of R, (R IP , , [P]Rcp,) is a valuation

pair of the total quotient ring of R.

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Every overring of R is a j la t R-module. Every overring of R i s integrally closed. A(B n C ) = AB n AC for all ideals A, B, C of R, with B or C regular. ( A f B)(A n B) = AB for all ideals A, B of R, with A or B regular. Every regular ideal of R generated by two elements is invertible. I f AB = A C , where A, B, C are ideals of R and A is finitely generated and regular, then B = C. R is integrally closed, and for any a, b E R, at least one of which is regular, there exists an integer n > 1 such that ( a , b)" = (an, b"). I f A and B are ideals of R, with B jinitely generated and regular, and if A _c B, then there is an ideal C of R such that

(A+ B) : C= A : C + B : C for all ideals A, B, C o f R, with A regdar and C finitely geiaerated. A : ( B n C ) = A : B + A : C for all ideals A, B, Cof R, with C regular and B and CJinitely generated. A n ( B + C ) = A n B + A n C for all ideals A, B, C of R, at least one of which is regular.

A= B C .

Proof. Let K be the total quotient ring of R.

(1) 3 (2). Assume R is a Prufer ring and let P be a maximal ideal of R. Let x E K\R,,, ; then there is a regular element b E R such that xb E R. Then (b, xb) is invertible. Hence there exists a fractional ideal A of R such that (b , xb)A = R. For each a E A, ab E P, for if ab 4 P, then since xba E R we would have x E R F p l . Since (b, xb)A = R, we have xba E R \ P for some a E A. For this element of A, ab E [PIR,,, and xba E R,PI\[P]R~Pl.

(3). Assume (2) holds and let T be an overring of R. Let M be a maximal ideal of T and set P = M n R. Let Q be a maximal ideal of R containing P. Then R,,, c R,,] G T,,,. If x E K\R,,,, then x 6 R,,, , so there exists b E Q such that bx E R\Q. If b 4 P, then bx E R would imply that x E RIP,, which is a contradiction. Now suppose that x E T I M I . Then there exists t E T\M such that xt E T. But b E P c M and bx E R \ P c T\M. Hence bxt E Mn (T\M), which is absurd. Thus x $ TI,, , and we conclude that TIMI = R,,, . Consequently, by Exercise ll(a), T is a flat R-module.

(2)

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( 3 ) + (4). Assume (3) holds, let T be an overring of R, and let T* be its integral closure. By ( 3 ) , T* is a flat R-module, so for any x E T", [R : x]R T* = T" by Exercise Il(a). Therefore,

T" E [ T : xlR T" E IT: XI? TQ G T",

so that [ T : x ] R T " = T *. This means that T * is a flat T-module, and using Exercise ll(b), we conclude that T = T*. (4) +(2). Let P be a maximal ideal of R. Let Z E K \ R [ ~ ] . Since

R L p l [ ~ 2 ] is integrally closed, x E RIp1[z2]. Let

z= a, + alzz + . * * +a,.+, a, E RIP],

where a, # 0 and n is chosen as small as possible. Note that we cannot have n = 0. Suppose n > 1. Multiplying by a?-,, we obtain

(anz)2'L + * * * + up-3al(unz)2 - up-2(Unz) + a, a2-1 = 0,

(a,z2)n + * . * +a:-2a1(a,z2) + (U,u;-l -an-2 n ( n a z > > = O

so a, z E RIP,. Then, multiplying the first equation by a:-', we have

Since R,,, is integrally closed, this implies that a, x 2 E RIP] . But then a,- , + an z2 E RIP,, and

z = a, + a1x2 + * . * +(a, - 1 + unz2)z2'n-1),

which contradicts our choice of n. Thus, we must have n = 1, z = a, + alzz, and a,x E R I P , . Since z $ R I P ] , we also have a, E [P]Rrpl. Since x(1- alz) = a, E R I P , , there exists t E R\P such that tz(1 - q z ) E R. Also, there exists s E R\P such that sa,z E R. Now z $ RcPl, so t (s - sa,x) 4 R\P; but this element is in R, so it must belong to P. Thus 1 - a,x E [P]RLP1, and it follows that

(2) + (5), (1 1)-( 13). Assume that (2) holds and let P be a proper prime ideal of R. If P is regular, then by Exercise 9(b), no regular ideal of R is contained in C(P) . Then by the equivalence of (2) and the third condition of Theorem 10.14, the equalities of (5) and (11)-(13) hold for the extensions of the ideals to R p . The same is true when P is not regular; for, if P is not regular, then in each case at least one of the ideals has its extension to Rp equal to Rp .

a1z E RIPI\[~lRIPl*

(5) j (6) 3 (7). As in the proof of Theorem 6.6. (7) j (2). Same proof as proof that (1) implies (2). (6) + (1). Let A be a finitely generated regular ideal of R, and let

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c be a regular element of A. Let B = [(c) : A]. To prove that A is invertible, it is sufficient to show that AB = (c) ; furthermore, since AB G (c) , all we have to do is to show that cR, E ABR, for every proper prime ideal P of R. Let P be a proper prime ideal of R. By Lemma 10.15, there exist elements a,, . . . , a , E A such that

A = (a1, . . * , a,), a& = c,

and

(U1, . m - 9 a,- 1 ) R p E U k Rp E ' ' * E a, Rp

For i= 1, . . . , n - 1, let x i , y i E R be such that y i 4 P, a,yi = akxi for i = 1, . . . , k - 1, and aiyi = ai+,xi for i= k, . . . , n - 1 . Let b=y1 ...yk-,xk...x,-,.ThenbA c (c),sobEB;hence, u , ~ E A B . But a, b = y1 * * * yk- ,ak and y1 . - * yk- $ P. Therefore, cR, = a, R, G ABR, .

(1) 3 (8). Clear. (8) 3 (5 ) . Assume that (8) holds, and let A, B, C be ideals of R

with B regular. Let P be a proper prime ideal of R. If CR, $ BR, , then there exists c E C such that cR, $ BR, . If b is a regular element of B, then by Lemma 10.16, either bR, E cR, or cR, G bR, . By our choice of c we must have bR, G cR,. Then, if x E B, it follows by Lemma 10.16 that xR, E cR,. Thus, BR, E CR, . We have shown that either CR, c BR, or BR, E CR,. If the former holds, then ACR, G ABR, , and

(AB n AC)R, = ABR, n ACR, = (AR,)(CR,)

= (ARp)(BRp n CR,)= A(B n C)R,.

If the latter holds, we have likewise (AB n AC)R, = A(B n C)R, . Since P is an arbitrary proper prime ideal of R, we have AB n AC =

A(B n C). (4) 3 (9). Assume (4) holds and let a and b be elements of R, at

least one of which is regular. By (4), R is integrally closed. Since (4) implies (8), and since (a, b)(a, b)2 = (a, b)(a2, b2), we have (a, b)2 = (a2, b2).

(9) + (7). Suppose that (a , b)" = (an, b") for a pair of elements a and b of R with a regular, and some n > 1. Then an-lb = xan + yb" for some x,y E R. Let m be the smallest integer greater than 1 for which am-lb = xum + yb" for some x, y E R. Now assume that R is

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integrally closed. Since (yblu)" + xyrn-l - ~ " - ~ ( y b / u ) = 0, we have yb/u = x E R. Then um-lb = xum + zubm-l and since u is regular, the minimality of m implies that m = 2. Hence b = xu + xb. Thus,

(a, b) (y , 1 - z) = (ay, by, a(l - z), b(1 -.)) = (ay, ax, a(1 - z), ax)= (a).

Therefore, (a, b) is invertible. (1) + (10). Take C= B-IA. (10) + (1). Let A be a finitely generated regular ideal of R. If

a is a regular element of A, then (u) G A, so there is an ideal B of R such that (u) = AB. Therefore, A is invertible.

(11)-(13) + (2). In each case, we shall verify the second of the equivalent conditions of Theorem 10.14, and conclude that (R,,, , [P]RLpl) is a valuation pair for each regular maximal ideal P of R . Let P be such an ideal and let x, y E R with x 4 C(P). Then there exists a regular element r E R such that xsjr 6 R for all s E R\P.

Assume that (1 1) holds ; then

= ((4 + (Y)> : ((4 + (4) = (4 : ((4 + (4 + (.> : ((4 + ( r ) ) = (x) : (Y) + (Y) : (x).

Hence, 1 = s + t , where sr E (x) and tx E (r) . Then tx/r E R and consequently t E P. Therefore, s 4 P. Now

R = ((Y) + (x, Y)) : ((Y) + (x, 4 = (x, r> : (Y ) + (Y) : (x, r).

Hence, 1 = u + v, where uy E (x, r ) and vx E ( y ) . If v E R\P then we have what we want. If z, E P then u E R\P and suy E (sx, sr) G (x) and su E R\P. Hence, the desired conclusion holds in this case also. Therefore, (1 1) implies (2).

The proof that (12) implies (2) is the same, for if (12) holds, then

R = ((4 n (9) : ((4 n (9 = (4 : (r) + ( r ) : (4

R = ((x, 4 n (YN : ((x, Y) n (Y ) ) = (x, 4 : (Y) + (Y) : (x, 4.

(4 = (4 n ((Y, r ) + (x -Y)) = (4 n (Y> 4 + (4 n (Y - 4

and

Now assume that (13) holds. Since

= (4 f-3 (Y) + (4 n (I) + (4 n (Y - 4,

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we have x = ux + v r + wx, where ux E ( y ) and wx = z ( y - x ) for some z E R . Since x(l - u - w) = or, our choice of r implies that 1 - u - w E P . Thus, either u E R\P or w E R\P. If u E R\P, the desired result follows from ux E (y). Suppose w E R\P and consider (w + z ) x = zy. Either z E R\P, or z E P in which case w + z E R\P. Thus, once more we draw the desired conclusion. Therefore, (12) implies (2).

This completes the proof of Theorem 10.18.

Because of the equivalence of (1) and (4) in Theorem 10.18, we have the following result.

10.19. Corollary. Every overring of a Prufer ring is a Prufer ring.

The next theorem shows that Prufer rings have an ideal theory similar to that of Prufer domains.

10.20. Theorem. Let R be a Prufer ring and T an overring of R. Let A be the set of regular prime ideals P of R such that P T # T . Then,

(1) for every regular maximal ideal M of T , T[Mj = R,,, , where P = M n R, and M = [PIR,,, n T ,

(2) if P is a regular prime ideal of R, then P E A if and only if

( 3 ) ( A n R)T = A for every regular d e a l A of T , and (4) {PTI P E A} is the set of regular prime ideals of T.

T E , and T = np ,

Proof. B y Corollary 10.19, T is a Prufer ring. Let M be a regular maximal ideal of T. Then (TIMI, [ M ] T , M l ) is a valuation pair. Since R I P ] 5 TI,,, where P = M n R , and R,,] is a Prufer ring, it follows from Exercise 12(b) that

for some regular prime ideal Q 5 P (prove this); in fact, Q is the regular prime ideal such that [Q]R,Pl = [MI T,,] n R I P ] [see Exercise 5(a)]. Then TIMI = R,,, and [ M ] T I M 1 = [QIR,,,. Thus,

Q = [Q]Rrgl n R = [ M ] T I M , n R = [MITIMI n T n R = M n R = P ,

and so TIMI = R I P ] and M = [ M ] T , M 1 n T = [PIR,,, (7 T .

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The first part of (2) follows from Exercise ll(a) since T, being an overring of R, is a flat R-module. By Exercise 6(c), T = n T I M I , where M runs over the set of regular maximal ideals of T. Therefore,

Now let A be a regular ideal of T. By Exercise 6(c),

A = n ( ~ ~ A I T , , ~ n TI,

where M runs over the set of regular maximal ideals of T . For each M , T,,] = RIP] where P = M n R. If x E [A]TrM1, then there exists s E R \ P such that xs E A. But A c RIP, , so there exists t E R\P such that xst E A n R. Consequently,

x E [A n R]RIPI = [(A n R ) T ] T C M I .

Since (A n R)T c A we have, therefore,

[ A I T I M 1 = [(A n R ) T ] T I M I .

The assertion of (3) now follows from another application of Exercise 6(c).

By (3) if Q is a regular prime ideal of T , then Q = (Q n R)T, and Q n R E A. On the other hand, if P E A, then T G RIP,, and hence [PIRIP, n T is a prime ideal of T . Now, by (3) again,

[PIRIP, n T = ([PIRIP, n T n R ) T = P T .

Thus, P T is a regular prime ideal of T.

If we turn now to the primary ideals of a Prufer ring, we can recover most of the results which were obtained in the last part of Section 2 of Chapter VI.

Let R be a Prufer ring and let P be a regular prime ideal of R such that P is not the only P-primary ideal of R. If there exist regular prime ideals of R properly contained in P, let Po be their union. Since (RIP], [PIRIP]) is a valuation pair by Exercise 12(a), the set of ideals of R, containing C(P)R, is totally ordered. Thus, Pi is a prime ideal of R. If there are no regular prime ideals of R properly contained in P, set Po = C(P). In either case, there are no regular prime ideals of R strictly between Po and P . Furthermore, Po is the intersection of the P-primary ideals of R.

Since [Po]R[,I is a prime ideal of RIP], the ring RIP,/[PO]RIPI is a rank one valuation ring. Thus, by Exercise 6(b), R,/Po R, is a rank

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244 X PRUFER RINGS

one valuation ring. There exists a one-to-one correspondence between the P-primary ideals of R and the PR,/P, RR,-primary ideals of R,/P, R, . This correspondence preserves residuals.

In Theorem 6.8 we proved that the product of P-primary ideals of a Prufer domain is a P-primary ideal. Now we shall obtain the same result for regular prime ideals of a Prufer ring.

10.21. Theorem, Let P be a regular prime ideal of a Prufer ring R. Every product of P-primary ideals of R is P-primary.

Proof. We may assume that P is a proper prime ideal of R. Let M be a maximal ideal of R which contains P, and let Q1 and Qz be P-primary ideals of R. It suffices to show that Q,Qz RM is PRM-primary. Since (RcMl, [M]RIMl) is a valuation pair, C ( M ) is contained in every regular ideal of R not meeting R\M. Hence, C ( M ) G Q1Q2. Using various parts of Exercise 9, we can show that C ( M ) is a prime ideal of R. Then, since the set of ideals of R,/C(M)RM is totally ordered, this ring is a valuation ring. Thus, Q1QzRM/C(M)RM is PR,/C(M)R,-primary. Therefore QlQ2 RM is PRM-primary.

We can now prove the following theorem in the same way that we proved a similar theorem in Section 2 of Chapter VI.

10.22. Theorem. Let R be a Prufer ring and let P be a regular prime ideal of R such that P is not the only P-primary ideal of R. If Q and Q1 are P-primary ideals of R, then :

Qn is a prime ideal. Qn = Q"" for some n 2 1 implies that Q = Qz = P. Q L Q1 c P implies that Qln c Q for some n. P2 c P implies that Q = P" for some n. Q c P implies that Qz c QP. Q c Q1 and Q Q1 = Q imply that Q1 = P = P2.

EXERCISES

The valuation determined by a valuation pair. Let (A, P ) be a valuation pair of a ring R, and let v be the valuation on R constructed in the proof of Theorem 10.4.

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EXERCISES 245

2.

(a) Give the details of the proofs of the following facts used in the proof of Theorem 10.4: (1) 5 is well-defined;

(3) If v(x) < ~ ( y ) and v(z) # v(O), then v(xz) < n(yz). Let w be a valuation on R such that A=R,#R and P= P,. Show that there is an order-preserving iso- morphism 4 : w(R)\{w(O)} + v(R)\{v(O)} such that +w = TI.

Maximal partial homomorphisms and valuation pairs. Let R be a ring. A partial homomorphism of R is defined exactly the same way as a partial homomorphism of a field (see Section 1 of Chapter V). Let A be a subring of R and let P be a proper prime ideal of A. Show that (A, P) is a valuation pair of R if and only if there is a homomorphism 9 from A into an algebraically closed field, with kernel P, and such that (4, A) is a maximal partial homomorphism of R. [Hint. When (4, A) is a maximal partial homomorphism and (A, P) is dominated by a maximal pair (B, Q), consider two cases: (i) For each b E B\A, there is a polynomial

(2) If v(.) # +), then .(.I < V(Y) or v(y) < .(.) ;

(b)

f(X)= C a i X ' E A [ X ] i = O

such that f ( b ) = 0 and a, 4 P for some i = 0, 1, . . . , n ; and (ii), the negation of (i).]

3. A counterexample. Let K be the field of rational numbers and let R = K [XI . Let p be a rational prime and let A, be the subring of K consisting of all rational numbers which can be written in the form m/n where p does not divide n. (a) Show that if A = A J X ] and P =PA, then (A, P ) is a

valuation pair. (b) Show that B = P + A X is a proper prime ideal of A, and

P c 3.

4. v-closed prime ideals. Let (R,, PJ be a valuation pair of a ring R, and let ZI be the valuation on R determined by this pair. (a) Show that a prime ideal Q of R, is v-closed if and only if

A, c Q E P, .

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X PRUFER RINGS

Show that the set of v-closed ideals of R, is totally ordered. Show that the set of v-closed ideals of R, is closed under arbitrary unions and intersections. Show that the set of v-closed prime ideals of R, has the same property. Let Q be a v-closed prime ideal of R, and let

H = {a I a E G, and v(x) > max{ct, -a} for all x E Q}.

Show that PvH = Q and R,, = {XI x E R and xQ G Q}.

5. Extensions of ideals to large quotient rings. Let R be a ring and let P be a proper prime ideal of R.

If Q is a primary ideal of R-contained in P , show that [Q]RrPl is a primary ideal of RIP, and that Q = [Q]RLPI n R. Show that every primary ideal of RIP] which is contained in [PIR,,, is of the form [A]RIPI for some ideal A of R. If P is a maximal ideal of R, show that [Pn]RIPI= ([P]RtpJn for each positive integer n. Suppose (R,,, , [P]RIPI) is a valuation pair and that there are no prime ideals of Rrpl strictly between C([P]Rcpl) and [PIRIP]. Show that if A and B are finitely generated regular ideals of R, then

r ~ l ~ ~ P I [ ~ l R ~ P I = [ABIRIPI - Show that it is not true in general that if S is a multiplica- tive system in R, and if A and B are ideals of R, then [A]RISIIB]RrSI = [AB]RlsI. Does it help to require that A and B be regular and/or finitely generated?

6. Large quotient rings. Let R be a ring. (a)

(b)

Let S be a multiplicative system in R. Show that if R is integrally closed, then RtS1 is integrally closed. Let P and Q be prime ideals of R with Q G P. Show that

R~P]/[QIR[PI z RPIQRP * [Hint. Use Proposition 3.15 and Theorem 3.17.1 Let M run over the set of regular maximal ideals of R. Show that if A is a regular ideal of R, then

(c)

A = n AIR[^] R). In particular, R = ORIM1.

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EXERCISES 247

7. Valuation pairs and integral dependence. Let R be a ring with total quotient ring K. Let x be an element of K which is not integral over R.

If z E K is such that zx E R, show that x is not integral over

If P is a proper prime ideal of R such that for all y E K, xy E R implies that y E P, show that x is not integral over

If P is a proper prime ideal of R and if there exists y E R[x] such that, for some w E P we have wy E R\P, show that there exists a valuation pair ( T , Q) of K such that R G T and x $ T. Suppose that for all valuation pairs (T , Q) of K with R 5 T we have x E T. Show that there exists an overring R' of R and a maximal ideal M of R with the following properties : (1) R' is integrally closed. (2) (3) (4) (5) (6)

W l -

RIP, *

Every regular nonunit of R' is in M. If z E K\R', then x is integral over R'[z]. For all y E K , xy E R' if and only if y, xy E M . If y E K\R' and x E R'\M, then xy E K\R. If y E M , then x + y is a zero-divisor in K .

8. Valuation pairs and rings having few zero-divisors. Let R be a ring which has few zero-divisors. (a)

(b)

(c)

Show that if P is a proper regular prime ideal of R, then R,,, = R,(,, (see Exercise 10 of Chapter 111). Show that R is integrally closed if and only if it is an intersection of valuation rings. Show that there exists a prime ideal P of R such that (R, P ) is a valuation pair if and only if R is a quasi- valuation ring.

9. The core of a proper prime ideal. Let P be a proper prime ideal of a ring R. (a) Show that if P is regular, then C(P) G P. (b) Show that if P is regular, then C(P) consists entirely of

zero-divisors. (c) Show that if P is not regular, then C(P) = R and RIP, is

the total quotient ring of R. (d) Show that [C(P)]RIPl = C([P]R,,,).

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248 X PRUFER RINGS

(e) (f)

Show that [C(P)]R[,, n R = C(P). If (R, P) is a valuation pair of the total quotient ring of R, show that C(P) = A,, where z1 is the valuation determined by (R, PI.

10. Large quotient rings and valuation pairs. Let R be a ring and P a maximal ideal of R.

(b) (.) Show that = ( R S I P ) ) [ P R ~ ( p ) l *

Show that (RIP], [P]Rrp,) is a valuation pair if and only if there is a prime ideal Q of R,(,, such that every regular nonunit of R,(,, is contained in Q, and (R,,,,),,, is a valua- tion ring. Show that (RIP], [PIR,,,) is a valuation pair if and only if for every pair of ideals A and B of R, at least one of which is regular, either AR, c BR, or BR, c AR, .

(c)

11. Flat overrings. Let R be a ring and T an overring of R. (a) Show that the following statements are equivalent:

(1) T is a flat R-module. (2) For every x E T, [R: (.)IR T = T , where [R: ( x ) ] R =

{ a l a ~ R and a x ~ R } . (3) For every prime ideal P of R, either PT= T or

T G RIP]. (4) For every maximal ideal M of T, RCM R] = T [ M I . Show that if T is a flat R-module and is integral over R, then T = R.

(b)

12. Prufer rings. Let R be a Prufer ring.

(a)

(b)

Show that for every proper prime ideal P of R, (Rcpl, [P]RLpl) is a valuation pair. Show that if ( T , M ) is a valuation pair of the total quotient ring of R such that R G T , then T = RLM R ] .

13. Generators of an ideal. Let R be a ring which satisfies the hypothesis of Lemma 10.15. Let A be a finitely generated ideal of R, let c be a regular element in A, and let P be a proper prime ideal of R. Show that there are elements a,, . . . , a, E A such that A = (q, . . . , a,), a k = c for some k, and

(a1, . % . , 1)Rp a,Rp E * ' 5 a,R,.

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EXERCISES 249

14. A local characterization of Prufer rings. Let R be a ring and let K be its total quotient ring. Let P be a prime ideal of R and let S = R\P. Then R, can be considered as a subring of S-lK. (a) Prove that ( R I P ] , [P]R[p,) is a valuation pair of K if and

only if (R, , PR,) is a valuation pair of S -IK. [Hint. Define a mapping w on S- lK by w(x/s) = ~(x) for x E K, s E S, where z, is the valuation of (RIP, , [P]R[,,).] Suppose R is a ring with unique maximal ideal P such that (R, P ) is a valuation pair of K with valuation TJ. Show that x E K\R implies that x = l / b where b E R and v(b) < m. Then show that every overring of R is of the form S-'R for some multiplicative system S in R. Let R be a Prufer ring and let T be an overring of R. Let Q be a prime ideal of T and set P = Q n R. Prove that R p g T,.

(b)

(c)

15. A generalization of Theorem 10.20. Let R be a Prufer ring and let T be an overring. Let A be the set of prime ideals of R such that PT # T. Prove the following assertions : (a) If M is a maximal ideal of T , then T , = RN, where

N = M (b) P E A if and only if T c R I P ] .

(d) For any ideal A of T , (A n R)T=A. [Hint. Let Q be a prime ideal of T and let P= Q n R. Show that AT, n R E (A n R)Rp n T (these are contractions and expan- sions) and use the fact that A = n (AT, n T).] The set of prime ideals in T is precisely {PTI P E A}.

R and the contraction of NR, into T is M .

(c) T = ~ , . . R ~ , ~ .

(e)

The ideal transform and large quotient rings. Prove the following results about ideal transforms. (a)

(b)

16.

If x is a regular nonunit of a ring R, then T((x)) = R E N I , where N = {x" I n = 1, 2, . . .}. Let A be an ideal of a ring R and let R' be a ring such that R E R G T (A). Then there is a one-to-one correspondence between the set of prime ideals P of R not containing A and the set of prime ideals P' of R' not containing AR'. Furthermore, if P corresponds to P', then P = P' n R and R[PI= R'[P,l

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17.

X PRUFER RINGS

Let a be a nonnilpotent element of R. Then T((a ) )= nRIPa1 where {Pa} the set of prime ideals of R such that

Let A be a finitely generated nonnilpotent ideal of R. Then T ( A ) = r)RcPal where {P,} is the set of prime ideals of R which do not contain A. [Hint. Use the fact that if

If no maximal ideal of R contains the collection {x,> of regular nonunits of R, then R = If no maximal ideal of R contains all of the regular non- units of R, then R is a Prufer ring if and only if T((x)) is a Prufer ring for each regular nonunit x of R.

(4 $ pa *

A = (al , a 2 , . . -, an), then T(A) = T((a$)).]

T((x,)).

Prufer valuation rings. If (R, P) is a valuation pair and R is a Prufer ring, we call (R, P ) a Prufer valuation pair and R a Prufer valuation ring. (a) Show that if R is a Prufer ring, if A is a finitely generated

regular ideal of R, and if P is a prime ideal of R, then T ( A ) 2 R,,, if and only if A $ P. If M and N are regular prime ideals of a Prufer ring R, prove that N 2 M if and only if R,,, 2 R r N l . Prove that the following are equivalent : (1) (R, P ) is a Prufer valuation pair. (2) R is a Prufer ring with a unique maximal regular

ideal P. (3) (R, P ) is a valuation pair where P is the unique maxi-

mal regular ideal of R. (d) Let ( R , P ) be a Priifer valuation pair, let V be an

overring of R, and let M be a regular prime ideal of 1’. Prove that M G P. Prove that if (R, P ) is a Prufer valuation pair, then every overring of R is a Prufer valuation ring.

(b)

(c)

(e)

18. Generalized transforms. A collection Y of ideals of a ring R is said to be multiplicatively closed if A, B E Y implies that AB E 9. The Y-transform R, of the ring R is the set

R y = {x E K ~ x A G R for some A E Y}.

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EXERCISES 25 1

If B is an ideal of R, denote by B, the set

B , = {x E KI XA c B for some A E Y}.

(a)

(b)

Show that large quotient rings and ideal transforms are generalized transforms. Let Y be a multiplicatively closed collection of ideals of R. Let F be the set of all prime ideals of R such that A $ P for all A E 9’ and let 9 be the set of all prime ideals P‘ of R, such that AR, $ P’ for each A €9’. Prove the following : (1) If P E P and Q is a P-primary ideal of R, then P, is a

prime ideal of R, , Q, n R = Q, and AR, yi Q, for all A E 9.

(2) If P ‘ E 9 and Q’ is a PI-primary ideal of R,, then Q’= (Q’ n R), and A $ Q’ n R for all A E Y .

(3) Pt-, P, is a one-to-one mapping from 9 onto 9.

Let P E F. Show that Q-Q, is a one-to-one corres- pondence between the P-primary ideals of R and the P,-primary ideals of R, provided any of the following conditions hold : (1) Each ideal in Y is finitely-generated. (2) Each P-primary ideal of R contains a power of P. ( 3 ) For each A E Y, AR, = R, . (4) P is a maximal ideal of R. Let T be an overring of R and prove that the following are equivalent : (1) T is a flat R-module. (2) There exists a multiplicatively closed collection Y of

ideals of R such that T = R, and AT= T for all A E Y .

(3) For each proper prime ideal &I of T, T [ M ] =

R I M A R 1 . [Hint. Use Exercise 11.1

(4) For p 9, RfP, = (R,)p,, - (c)

(d)

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Appendix: Decomposition of Ideals

in Noncommutative Rings

It is natural to ask if the primary decomposition theory of com- mutative rings can be generalized to noncommutative rings. As it turns out, the uniqueness results can be generalized, but the existence theorem cannot. This leads to the search for an appropriate decom- position theory. We shall discuss such a theory, the tertiary ideal theory of Lesieur and Croisot.

Let R be a noncommutative ring, that is, a ring which is not necessarily commutative ; we assume, of course, that R has a unity. We shall confine our attention to the ideals of R. Much of what we say can be extended to left ideals or right ideals of R, or to R-modules.

If H and K are subsets of R we set

H K = {finite sums a, bf I ai E H, bi E K}.

If L is another subset of R, then ( H K ) L = H(KL). If H is a left ideal of R, so is H K ; if K is a right ideal of R, so is HK. If H = {a}, we will write a K for H K ; if K = {b), we write Hb for H K . Note that (a) = RaR.

A.1. Definition. An ideal P of R is a prime ideal if AB c P, where A and B are ideals of R, implies that either A G P or B s P.

252

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DECOMPOSITION OF IDEALS IN NONCOMMUTATIVE RINGS 25 3

A.Z. Proposition. Let P be an ideal of R. Then P i s a prime ideal i f and only i f aRb E P, where a, b E R, implies that either a E P or b E P.

This assertion, and certain other assertions made in this appendix, will be left unproved. Their proofs may be considered as exercises.

A nonempty subset M of R is called an m-system if for each pair of elements a, b E M , there is an element x E R such that ax6 E M. A proper ideal P of R is prime if and only if R\P is an m-system.

A.3. Definition. Let A be an ideal of R. The prime radical of A, denoted by Rad(A), i s the set of all a E R such that every m-system of R which contains a meets A.

If A is an ideal of R, then

(i) A c Rad(A), (ii) Rad(Rad(A)) =Rad(A), and

(iii) if P is a prime ideal of R and if A c P, then Rad(A) E P.

A.4. Proposition. Let A be an ideal of R. Then Rad(A) is the inter- section of all of the prime ideals of R that contain A.

Proof. We must show that if a 6 Rad(A), then there is a prime ideal P of R such that A G P and a 6 P. Since a $ Rad(A) there is an m-system M such that a E M and M n A is empty. By Zorn’s lemma, there is an ideal P of R maximal with respect to the properties that A G P and M n P is empty. We shall show that P is prime. Let bRc G P and suppose 6 6 P and c g P. Then A g P C P + ( 6 ) and A E P c P+ (c). Consequently, there are elements x, y E M such that x E P+ (b) and y E P+ (c). For some z E R, we have xzy E M , and

XZY E (P + (b))z(P + (4) = PxP + PZC + 6zP + (b)Z(c)

G P + RbRcR G P ,

which contradicts the fact that M n P is empty. Thus, bRc c P implies b E P or c E P ; hence P is a prime ideal.

A.5. Corollary. If A is an ideal of R, then Rad(A) is an ideal of R.

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254 APPENDIX

A.6. Definition. An ideal Q of R is a primary ideal if AB s Q and B $ Q, where A and B are ideals of R, imply that A E Rad(Q).

A.7. Proposition. Let Q be an ideal of R. Then Q is a primary ideal if and only i f a R b E Q and b $ Q, where a , b E R, imply that a E Rad(Q).

We shall call R a Noetherian ring if the ascending chain condi- tion holds for ideals of R. Note that if R is a left (or right) Noetherian ring, then it is a Noetherian ring.

A.8. Proposition. Let R be a Noetherian ring and let A be an ideal of R. Then there is a positive irzteger n such that (Rad(A))" G A.

This assertion is a consequence of Exercise 2(a) and Proposi- tion A.4.

A.9. Proposition. Let R be a Noetherian ring and let Q be an ideal of R. Then Q i s a primary ideal i f and only i f AB c Q and B $ Q, where A and B are ideals of R, imply that there is a positive integer n such that A" E Q.

This completes our preliminary list of definitions and simple results. If R is a commutative ring, the definitions of prime ideal and primary ideal are equivalent to the ones given in Chapter 11. If R is commutative and Noetherian, then every ideal of R is an intersection of a finite number of primary ideals. The following example shows that this may not be the case when R is not commutative.

Let K be a field and let R be a 3-dimensional vector space over K. Let the elements x, y, z of R form a basis of R over K . We define a multiplication in R in the following manner. First of all, we multiply x, y, and z according to the table

X

Y z

x o z

o z o O Y 0

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DECOMPOSITION OF IDEALS I N NONCOMMUTATIVE RINGS 25 5

We extend this operation to all of R by requiring that the distributive laws hold. Thus,

(ax + by + cz)(dx + ey + f x ) = adx + bey + (af + ce)x.

Then the associative law holds, so that R is a ring. The unity of R is x + y . The mapping a H ax + ay is an injective homomorphism from K into R. We identify each element of K with its image under this homomorphism, and thus consider K as a subring of R.

Each ideal of R is a subspace of the vector space R. Therefore, R is Noetherian. However, the ideal 0 is not an intersection of primary ideals of R. For, let Q be a primary ideal of R. We have

y ( a x + by + cz)x = 0

so that yRz G Q. Hence, either x E Q or y" E Q for some positive integer n. But y" = y for all n, and if y E Q then x = xy E Q. There- fore, in any case, z E Q. Thus, no intersection of primary ideals of R can be the ideal 0.

If A and B are ideals of R, the right residual of A by B is defined to be the set

A : 3 = { x l x ~ R and 3 x s A ) .

The left residual of A by B is defined to be the set

A * . B = ( X I X E R and x 3 s A ) .

Both A . - B and A * . B are ideals of R. If C is an ideal of R, then BC c A if and only if C c A . - B, and CB G A if and only if C c A . 23. The left and right residual operations have properties like those given in Proposition 2.3. Note that ( A . - B) . - C = A . * BC and ( A , . B ) . . C = A * . CB.

A.lO. Theorem. If R is a Noetherian ring, then the following state- ments are equivalent:

(I) Every ideal of R is an intersection of aJinite number of primary ideals.

(2) If A and B are ideals of R, then there is a positive integer n such that A" n B c AB.

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256 APPENDIX

(3) If A and B are ideals of R, then there is a positive integer n such that A = ( A + Bn) n ( A . B").

Proof. (1) + (2). Let A and B be ideals of R and assume that AB = Q1 n - * * n Qk, where Qi is aprimary ideal of Rfor i= 1, . . ., k. For each i, AB G Q + ; hence, either B c Q, or there is a positive integer ni such that A"' c Qi . If B c Qi , set ni = 1, and let n = max(n,, . . . , nk).

Then

A" n B E Q1 n ... n Q k = AB.

(2) + (3). Assume (2) holds and let A and B be ideals of R. We have A . B E A . - B2 5 . . . , and so there is a positive integer t such that A . - Bt= A . * Bt+l= . - * . By (2), there is a positive integer s such that

BSt n ( A . Bt) E Bt(A . Bt) G A.

If n >st , then B" n ( A * B") s BSt n ( A . * Bt) _c A, since A . - B" = A . Bt. Since A G A . - B", we have by the modular law (Theorem 1.3),

( A + B") n ( A . - B") = A + (Bn n ( A . - B")) G A.

The reverse containment always holds, so that we have the asserted equality.

(3) + (1). Assume (3) holds. Since R is Noetherian, every ideal of R is an intersection of a finite number of irreducible ideals ; the proof is exactly the same as in the case of commutative rings (see Proposition 2.6). Thus, to prove (1) it is sufficient to show that an irreducible ideal Q of R is primary. Let AB c Q, where A and B are ideals of R, and assume that An $ Q for every positive integer n. By (3), there is a positive integer n such that Q = (Q + A") n (Q . * A"). Since Q c Q + A", and Q is irreducible, we must have Q . * A" = Q. Since B c_ Q . A" it follows that Q is primary.

Since primary decompositions may fail to exist in a Noetherian ring, it is natural to look for some other class of ideals which have some algebraic significance and such that every ideal of a Noetherian ring is an intersection of a finite number of ideals in this class of ideals. This leads us to the notions of the tertiary radical of an ideal and tertiary ideal.

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DECOMPOSITION OF IDEALS IN NONCOMMUTATIVE RINGS 257

Throughout the rest of the appendix, R is a Noetherian ring. Let A be an ideal of R. The set of all ideals B of R such that for an ideal C of R,

( A . * B) n C E A implies that C 5 A,

is not empty since A belongs to this set. Hence, this set has a maximal element B. Let B' be an arbitrary ideal of this set and suppose that (A . - (B+B')) n C c A. Then (A: B) n ( A : B') n C c A ; hence (A . - B') n C c A and therefore C c A. Thus, 3 + B' belongs to the set of ideals in question; hence B + B' = B, that is, B' E B.

A.ll. Definition. The unique maximal element of the set of ideals described in the preceding paragraph is the tertiary radical of A ; i t is denotedby Ter(A). An ideal A of R is a tertiary ideal if BC _C A and C $ A, where B and C are ideals of R, imply that B G Ter(A).

A.12. Theorem. Every ideal of R is an intersection of ajinite number of tertiary ideals of R.

Proof. We shall show that an irreducible ideal A of R is tertiary. Let B and C be ideals of R such that BC c A and C $ A. Then, since B ( A + C ) = B A + B C c A , we have A c A + C s A : B . Now, suppose that (A . - B) n D c A, where D is an ideal of R. Then, by the modular law.

A = ( (A . * B) n D) + A = ( A . * B ) n (D + A).

Since A is irreducible, this implies that D + A = A, that is, D c A. Therefore, B G Ter(A).

We can now give one more necessary and sufficient for the exist- ence of primary decompositions of ideals of R.

A.13. Theorem. Every ideal of R is an intersection of a$nite number of primary ideals of R i f and only i f every tertiary ideal of R is primary.

Proof. The sufficiency of the condition is obvious in view of Theorem A.12. Conversely, suppose every ideal of R is an intersection of a

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25 8 APPENDIX

finite number of primary ideals of R. Let A be an ideal of R. By The- orem A.lO, there is a positive integer n such that

A = ( A + (Ter(A))") n (A . * (Ter(A))").

Hence

(A . * (Ter(A))) n (A + (Ter(A))n) c A,

and therefore, A + (Ter(A))" G A, that is, (Ter(A))" s A. I t follows, by Proposition A.4, that Ter(A) _C Rad(A). Now suppose that A is tertiary and that BC c A and C $ A, where B and C are ideals of R. Then B z Ter(A), so B c Rad(A). Therefore, A is primary.

It is sometimes convenient to have an elementwise description of the tertiary radical of an ideal ; we now give such a description.

A.14. Proposition. If A is an ideal of R then

Ter(A) = {a 1 a E R and if b 4 A then there is an element c E (b) such that c 4 A and (a)(c) G A}.

Proof. Let a belong to the set on the right and suppose that (A . * (a))n C E A, where C is some ideal of R. If C $ A then there is an element b E C with b A. Hence, there is an element c E (b) such that c 4 A and (a)(c) c A. Then c E (A . * (a)) n C s A, which contradicts the fact that c 4 A.Thus we do have C c A ; we conclude that a E Ter(A).

Conversely, let a E Ter(A) ; then (a) _c Ter(A). If A = R, then a belongs to the set on the right. If A # R, choose b E R so that b $ A. Then (b) $ A and consequently, (A. * (a) ) n (b) $ A. Choose c E (A . * (a)) n (b) so that c $ A. Then c E (b) and (a)(c) c A. There- fore, a belongs to the set on the right.

Let A be an ideal of R. A tertiary decomposition of A is an expression of A as an intersection of a finite number of tertiary ideals of R .

A.15. Proposition. Let A be an ideal of R and let A = TI n - - - n T , be a tertiary decomposition of A such that no Ti contains the intersection of the remaining T i . Then,

Ter(A) = (Ter( TI)) n - - - n (Ter( Tk)).

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DECOMPOSITION OF IDEALS I N NONCOMMUTATIVE RINGS 259

Proof. Let a belong to the set on the right and let b $ A. Then b I$ Ti for some i , say b $ Tl. Since a E Ter( Tl), there is an element b, E (b) such that b, $ T, and (a)(bl) c T,. If b, E T, , then 6 , $ T, n T , and (a)(b,) E T, n T, . If b, I$ T,, then, since a E Ter(T,), there is an element c, E (b , ) such that c1 I$ T, and (a)(c,) G T, . Then c1 $ T, n T , and (a)(c,) c T, n T, . Let b, = b, if b, E T, and b, = c1 if b, I$ T , . Then 6 , E (b) , 6 , $ TI n T,, and (a)(b,) c T, n T , . If we repeat this argument several times we will obtain an element b, E (b) such that b, 4 Tl n *

Conversely, let a E Ter(A). Let b E T , n * * n Tk and b $ TI (if k = 1 the assertion is trivially true). Since b I$ A, there is an element c E (b) such that c $ A and (a)(c) c A. If c E T,, then c E Tl n (b) c T, n n T , = A. Hence, (c) $ TI but (a)(c) E T,. Since Tl is tertiary, this implies that (a) cTer(Tl). If we apply this argument to each of T,, . . . , Tk , we conclude that a E (Ter( T,)) n * - n (Ter( T,)).

n T, = A and (a)(b,) G A. Therefore, a E Ter(A).

A.16. Corollary. Let T, and T, be tertiary ideals of R with Ter( T I ) = Ter( T,). Then T = T, n T , is a tertiary ideal of R with Ter( T) =

Ter( T,) .

Proof. ByPropositionA.15, wehaveTer(T) = (Ter(T,))n(Ter(T,)) =

Ter( T,). Suppose that BC c T and C $ T, where B and C are ideals of R. Then BC G Tl n T,, and either C $ T, or C $ T , . I n either case, B G Ter( T). Therefore, T is a tertiary ideal.

A tertiary decomposition A = T, n . n T, of an ideal A of R is called reduced if:

(i) (ii)

no Ti contains the intersection of the remaining Ti, and Ter( Ti) #Ter( Tj) for i # j .

Because of Corollary A.16, it is evident that every ideal of R has a reduced tertiary decomposition. We shall now obtain a uniqueness result for such decompositions.

A.17. Lemma. Let A be an ideal of R and suppose that A = T n B =

T' n B', where T and T' are tertiary ideals of R, and B and B' are ideals of R. If Ter( T ) # Ter (TI), then A = B n B'.

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260 APPENDIX

Proof. Since Ter(T) # Ter( T’), we may assume that Ter(T) $ Ter(T’). Then, from (Ter(T))(T’ . * (Ter(T))) E T’, it follows that T‘ . * (Ter (7’)) s T’. Thus T‘ . . (Ter( T ) ) = T‘. Then

(T. * (Ter(T))) n (B . * (Ter(T))) = (T n B) . (Ter(T)) = (T’ n B’) . * (Ter(T)) = (T‘ . - (Ter(T))) n (B’ . - (Ter(T))) = T’ n (B’ . * (Ter(T))).

If we intersect each side of this equality with B n B’, and use the fact that B n B’ c B . - (Ter(T)) and B n B’ G B‘ . * (Ter(T)), we obtain

(T. * (Ter(T))) n ( B n B’) = T‘ n (B n B’) = (T’ n B‘) n B = A n B = A E T.

Therefore, B n B’ E T and we have A = T n ( B n B ’ ) = B n B’.

A.18. Theorem. Let A be an ideal of R and let

A= T, n -.- n T,= T,‘ n n T,’

be two reduced tertiary decompositions of A. Then, m = n and the Ti and Ti’ can be so numbered that Ter( Ti) = Ter( Ti‘) for i = 1, . . . , n.

Proof. The assertion is obviously true if A = R ; hence we assume that A # R. Since the decompositions are reduced, this implies that T i # R f o r i = l , ..., mand T , ’ # R f o r i = l , ..., n.Supposethat n = 1. If m = 1, then A = TI = T,‘ and Ter( T,) = Ter( T,‘). Suppose that m > 1. If Ter( TI) # Ter ( T,’), then by Lemma A.17 we have A = T, n * n T, (take B = T, n . * n T, and B‘ = R) which contradicts our assumption that the tertiary decompositions are reduced. Thus Ter( T,) = Ter( TI’). By the same argument, Ter( T,) = Ter(T,’), again a contradiction. Thus, n = 1 if and only if m = 1 .

Now suppose that n > 1. Suppose further that Ter( T,) # Ter( T,’) for i= 1, . . , , n. Then, by Lemma A.17,

A= T, n n T, n T,’ n n T,’ = T, n n T, n T3’ n n T,’

= T, n - - * n T,,, n T,’ = T, n n T , ,

-... -

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DECOMPOSITION OF IDEALS IN NONCOMMUTATIVE RINGS 261

once more a contradiction. Thus Ter( T I ) = Ter( Ti’) for some i. By the same argument, for each j = 2, . . . , m, Ter( Ti) = Ter( Ti’) for some i, and for each j = 1, . . . , n, Ter(T,’) = Ter(Ti) for some i . Hence, m = n, and the assertion is proved.

The remainder of the appendix will consist of a proof that the tertiary radical of a tertiary ideal is a prime ideal. Several exercises expand on the ideas used in this proof.

Let A be a proper ideal of R. If B is an ideal of R such that B $ A, then A * . B is called a proper left residual of A.

A.19. Proposition. A n ideal C of R is a proper left residual of a p r o p e r i d e a l A o f R i f a n d o n l y i f A * . ( A . * C ) = C a n d A c A:C.

Proof. The sufficiency of the condition is clear. Conversely, suppose that C is a proper left residual of A and let C = A . B where B is an ideal of R and B $ A. Then CB 2 A and so B G A . * C. Hence, A * . ( A . * C ) c A * . B = C. On the other hand, since C(A . C) G A, we have C c A - .(A . * C ) . Therefore, A * . ( A . * C) = C. If A =

A . 9 C, then C = R and B G A, contrary to fact. Hence, A c A . C.

A.20. Proposition. If A is a proper ideal of R, then A has a prime proper left residual.

Proof. Since A # R, the set of proper left residuals of A is not empty. Hence, this set has a maximal element P. Let P = A * . B, where B is an ideal of R and B $ A. Suppose that CD G P, where C and D are ideals of R, and suppose that D $ P. We have CDB s PB c A, so C G A * . DB. Since DB E B, we have A * . B E A * . DB, and it follows from the maximality of A * . B that P= A - . DB. Thus, C c P. Therefore, P is a prime proper left residual of A.

A.21. Proposition. Let A be a proper ideal of R. Then A c A . * B if and only if B c P for some prime proper left residual P of A.

Proof. Suppose A c A . B. Since B(A . - B ) c A, we have B c A * . ( A . * B) ; this is a proper left residual of A. It is contained in a maximal proper left residual of A, which is prime by the proof of

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262 APPENDIX

Proposition A.20. Conversely, suppose B c P = A * . C, where C is an ideal of R and C A, and P is a prime ideal of R. From ( A * . C)C c A , it follows that C c A . * ( A . . C ) = A . P c A . - B. If A = A . B, then C c A, which is not true. Hence, A c A . B.

A.22. Definition. Let A be a proper ideal of R. An ideal C of R is an essential left residual of A if there is an ideal B of R with A c B such that C = A . B and such that A c B' G B, where B' is an ideal of R , implies that A * . B = A * . B.

A.23. Proposition. Let A be a proper ideal of R. Every maximal proper left residual of A is an essential left residual of A. Every essential left residual of A is a prime proper left residual of A.

Proof. Let C be a maximal proper left residual of A. By Proposition A.19, C = A * . B, where B is an ideal of R such that A c B. Suppose that A c B' c B, where B' is an ideal of R. Then A - . B G A - . B', so we must have A * . B ' = A . . B.

Now suppose that P is an essential left residual of A ; let P= A * . B where B is an ideal of R satisfying the condition in Definition A.22. Let CD c P, where C and D are ideals of R and D .$ P. Then CDB E PB c A , and so C c A * . DB. Since DB $ A and DB c B, it follows that A c A + DB E B. Consequently,

P = A * . B = A * . ( A + DB) = A * . DB.

Thus, C E P. Therefore, P is a prime ideal of R.

A.24. Theorem. A proper ideal T of R is tertiary i f and only if Ter( T ) is the unique essential left residual of T .

Proof. Let C be an essential left residual of T. Let C = T * . B, where B is an ideal of R satisfying the condition of Definition A.22. We have

Ter( T ) c T * . ( T . * (Ter( T ) ) )

E T - . ( (T . - (Ter(T))) n B)

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EXERCISES 263

and

T E ( T . * (Ter(T))) n B E B.

If T = ( T . * (Ter(T))) n B, then from the definition of Ter(T) it follows that B = T, which is not true. Hence, T c (T. * (Ter( T))) n B, and consequently

Ter(T) E T - . ( ( T . - (Ter(T))) n B ) = T * . B = C.

Note that in this part of the proof we have not used the fact that T is tertiary.

Now, assume that T is tertiary. By Propositions A.23 and A.19, T c T . - C. Since C( T . - C ) 5 T, it follows that C E Ter( T).There- fore, Ter(T) = C, and C is the unique essential left residual of T.

Conversely, suppose that Ter( T) is the unique essential left residual of T. Suppose also that BC G T and B $ Ter( T), where B and C are ideals of R. By Proposition A.23, B is not contained in any prime proper left residual of T. Hence T . * B = T by Proposition A.21. But C 5 T . * B, so C c T . Therefore, T is a tertiary ideal of R.

A.25. Corollary. If T is a tertiavy ideal of R, then Ter( T) is a prime ideal of R.

1.

EXERCISES

Prime ideals and the prime radical. Let R be a ring and let P and A be ideals of R. (a) Show that the following statements are equivalent :

(1) P is a prime ideal of R. (2) (a)(b) E P, where a, b E R, implies that either a E P or

b E P. (3) BC G P, where B and C are left ideals of R, implies

that either B 5 P or C G P. (4) BC s P, where B and C are right ideals of R, implies

that either B E P or C G P. (b) Define minimal prime divisor of A in exactly the same

way as in the commutative case. Show that every prime ideal of R which contains A contains a minimal prime divisor of A ; conclude that Rad(A) is the intersection of all of the minimal prime divisors of A.

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264 APPENDIX

2.

(c)

(d)

Show that if a~ Rad(A), then ah E A for some positive integer n. Show that if a E R, then (a2" 1 n is a positive integer) is an rn-system. Show by example that even when R is commuta- tive, this set need not be multiplicatively closed.

Noetherian rings. Let R be a Noetherian ring. (a) Show that an ideal A of R contains a product of a finite

number of prime ideals of R, each of which contains A. (b) Show that an ideal A of R has only a finite number of

minimal prime divisors. (c ) Suppose that the equivalent conditions of Theorem A.10

hold. Show that if B is an ideal of R and if A = on"=, B", then A = B"A for every positive integer m.

(d) Again, suppose that the equivalent conditions of Theorem A.10 hold. Let P I , . . . , Pk be the minimal prime divisors of an ideal A of R. Show that for every large positive integer, n,

A = (A + P,") n - . n ( A + Pk").

3. Residuals. Let R be a ring and let A, B, and C be ideals of R. (a) Show that (A . B) . . C = (A . * C ) * . B. (b) Show that B = A * . ( A . * B)(or B = A . * (A . B)) if and

only if B = A * . D(or B = A . * 0) for some ideal D of R. (c) Assume A # R and let P be a prime proper left residual of

A. Show that if B $ P, then P is a prime proper left residual of A . * B. (The definition of proper left residual does not really require that R be Noetherian, nor does this result.)

4. Noetherian rings, continued. Let R be a Noetherian ring. (a)

(b)

Show that a proper ideal of R has only a finite number of prime proper left residuals. Show that a proper ideal Q of R is primary if and only if Q has a unique prime proper left residual P ; show that if this is the case, then P is the unique minimal prime divisor of Q, and Rad(Q) = P. Show that every primary ideal of R is a tertiary ideal of R.

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EXERCISES 265

(c) Let Q be a proper ideal of R, and let P be an ideal of R such that P G Rad(Q) and if AB G Q and B $ Q, where A and B are ideals of R, then A G P. Show that P is prime, Q is primary, and Rad(Q) = P. Show that if Q1, . . . , Qk are primary ideals of R with the same prime radical, then Q1 n . * - n Qk is also primary with the same prime radical. Define reduced primary decomposition of an ideal A of R in the same way as in the commutative case. Assume A is an intersection of a finite number of primary ideals of R. Show that A has a reduced primary decomposition, and state and prove a result like Corollary 2.27 for A.

(d)

(e)

5. The tertiary radical and tertiary ideals. Let R be a Noetherian ring. (a) Let A be a proper ideal of R. Show that Ter(A) is the inter-

section of the essential left residuals of A. (b) Let T be a proper ideal of R, and let P be an ideal of R such

that P E Ter(T) and if AB G T and B $ T, where A and B are ideals of R, then A E P. Show that P is prime, T is tertiary, and Ter( T ) = P.

(c) Give an alternate proof of Corollary A.16, without using Proposition A.15, but using part (b) of this exercise.

(d) Show that an ideal 7' of R is tertiary if and only if aRb c T and b $ T, where a, 6 E R, imply that a E Ter(T).

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The list of books which follows contains most of the recent books on the theory of rings and several books on related topics such as the theory of fields and algebraic number theory. Expositions of many aspects of the theory of rings may be found also in the many books on abstract algebra which are available.

Following the list of books is a lengthy list of papers, virtually all of which deal directly with some topic mentioned in the text. The book of Krull contains a list of the older papers on commutative rings.

The book of Lesieur and Croisot contains a more detailed exposi- tion of tertiary decomposition theory than that given in the appendix to this book. The list of papers contains several which are concerned with noncommutative rings.

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Page 311: multiplicative theory of ideals

Subject Index

A

Adjoint ideal, 58 Almost Dedekind domain, 201-205 Almost integral, 92 Almost integral dependence, 92-94 Almost Krull domain, 198 Almost multiplication ring, 216-220 Almost Noetherian ring, 81 Annihilator of element, 59 Approximation theorem, 198 Arithmetical ring, 150 Artin-Rees Lemma, 45 Ascending chain condition, 9

B

Bijective homomorphism, 5 Bilinear mapping, 17 Bimodule, 30 Branched prime ideal, 120

C

Cancellation of ideals, 148 Canonical homomorphism, 6 Chain conditions, 8-12 Cohen’s theorem, 58 Comaximal ideals, 53 Commutative diagram, 23

Complete ideal, 146 Complete integral closure, 92 Completely integrally closed, 92 Completion, of ideal, 146 Component

isolated primary, 78 of submodule, 50

Composition series, 28 Contraction of ideal, 66 Core of proper prime ideal, 235

D

Decomposition primal, 59 primary, 48 tertiary, 258

Dedekind domain, 135-144 Depth of ideal, 157 Descending chain condition, 9 Dimension

of ideal, 157 Krull, 156 valuative, 164

Direct sum, 12-15 external, 12 of homomorphisms, 29 internal, 14

Discrete valuation, 111 Discrete valuation ring, 11 1 Divisible group, 32 Divisor class group, 185-189

294

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SUBJECT INDEX 295

Divisorial fractional ideal, 173 Divisors, 172

Domination, 101, 229 sum of, 173

E

Essential valuations, 179 Euclidean ring, 199 Exact sequence, 10

short, 10 split, 24, 34

of ideal, 66

of valuation, 114

Extension

to large quotient rings, 234

External direct sum, 12

F

Factor module, 6 Factorial rings, 190-194 Few zero-divisors, 152 Finite chain, 9 Finitely generated module, 4 Finiteness condition, 9 Flat modules, 21-26 Flat overring, 89 Fractional ideal(s), 124

divisorial, 173 integral, 125 invertible, 125 principal, 125 product of, 125 sum of, 125

Free module, 15 defined on set S, 16

G

Generalized transforms, 250 Going-down theorem, 87 Going-up theorem, 85 Grothendieck group, 15 1

H

Height of ideal, 157 Hilbert basis theorem, 44 Homomorphism, 5

bijective, 5 canonical, 6 extending by linearity, 16 image of, 6 injective, 5 kernel of, 6 of modules, 5 partial, 100 of rings, 8 surjective, 5

I

Ideal(s), 2 adjoint, 58 comaximal, 53 contraction of, 66 depth of, 157 dimension of, 157 extension of, 66 height of, 157 idempotent, 202 kernel of, 211 powers of, 37 primal, 58 primary, 40 prime, 40 principal, 2 product of, 36 proper, 2 radical of, 41 regular, 42 residual of, 38 sum of, 36 tertiary, 257 transform of, 149, 225, 249 v-closed, 230 zero, 2

Idempotent ideals, 202 Image of homomorphism, 6 Indecomposable ring, 205

Page 313: multiplicative theory of ideals

SUBJECT INDEX

Infinite chain, 9 Injective homomorphism, 5 Integral closure, 83 Integral dependence, 82-91 Integral domain, 66 Integral element, 83 Integral ideal, 125 Integrally closed, 83 Internal direct sum, 14 Inverse, 42 Invertible fractional ideal, 125 Irreducible element, 190 Irreducible submodules, 39 Isolated component, 52 Isolated primary component, 78 Isolated, set of prime divisors, 51 Isolated subgroup, 109 Isomorphic modules, 6 Isomorphism, 6

J

Jacobson radical, 55

K

Kernel of homomorphism, 6 of ideal, 211

Krull dimension,l56 of polynomial ring, 161-164

Krull domain, 172 Krull intersection theorem, 47 Krull’s principal ideal theorem, 159

Large quotient ring, 234-236 Lattice ordered semigroup, 173 Left ideal, 2 Left Noetherian ring, 11 Left R-module, 2 Left residual, 255 Length of module, 28

Lexicographic ordering, 107 Local ring, 69

regular, 169 Localization, 69 Lying-over theorem, 84

M

m-system, 253 Maximal ideal, 42 Maximal pair, 101 Maximal partial homomorphism, 100 Maximal prime divisor, 78 Maximum condition, 9

on principal ideals, 193 Minimal prime divisor

of ideal, 43 of submodule, 43

Minimum condition, 9 Modular law for submodules, 5 Module, 3

finitely generated, 4 free, 15 freely generated by set, 14 length of, 28 projective, 23 without torsion, 146 unital, 3

Module of quotients, 62 Multiplication ring, 209-216 Multiplicative lattices, 174 Multiplicative system, 62 Multiplicatively closed set, 42

N

Nakayama’s lemma, 55 Nilpotent element, 42 Noetherian rings, 44-48, 254 Normal series, 27

0

Ordered Abelian group, 107 Ordered semigroup, 107 Overring, 88

flat, 89

Page 314: multiplicative theory of ideals

SUBJECT INDEX

P

297

R

Radical of ideal, 41 Jacobson, 55 prime, 253 of submodule, 41 tertiary, 257

Ramification index, 185 Rank

of group, 110 of valuation, 11 1

Reduced primary decomposition, 48 Reduced tertiary decomposition, 259 Regular element, 42 Regular ideal, 42 Regular local ring, 169 Residual, 38

left, 255 right, 255

Residue class ring, 8 Residue field relative to valuation, 197 Right ideal, 2 Right R-module, 2 Right residual, 255 Ring, 1

commutative, 1 left Noetherian, 11

Ring of quotients, 62 WX), 80

p-adic valuation, 121 P-primary ideal, 42 P-ring, 154 Partial homomorphism, 100,245 Partially ordered semigroup, 173 Place, 119 Power of ideal, 37 Primal decomposition, 59 Primal ideal, 58 Primary decomposition, 48

reduced, 48 Primary ideal, 40, 254 Primary submodules, 39-44 Prime divisor(s), 49, 78

isolated set of, 51 maximal, 78 minimal, 43

Prime element, 190 Prime ideal(s), 40, 252

associated with module, 59 branched, 120

Prime radical, 253 Principal ideal, 2 Principal ideal domain, 199 Product of ideals, 36 Projective module, 23 Prolongation of valuation, 114 Property (#), 222 Priifer domain, 126-1 34 Priifer rings, 236-244 Priifer valuation pair, 250 Priifer valuation ring, 250 Pseudo-radical of ring, 96 p(X)-adic valuation, 114

Q QR-property, 147 Quasi-equality, 172 Quasi-inverse of fractional ideal, 172 Quasi-principal ideal, 147 Quasi-valuation ring, 153 Quotient field, 66 Quotient module, 62 Quotient ring, 62, 69

S

S-component of submodule, 50 Saturation, 73 Short exact sequence, 10 Simple tensor, 17 Special primary ring, 206 Split exact sequences, 24, 34 Submodule(s), 3

component of, 50 generated by S, 4 irreducible, 39 isolated component of, 52 primary, 39 radical of, 41

Page 315: multiplicative theory of ideals

298 SUBJECT INDEX

residual of, 38 sum of, 3 zero, 4

Subring, 1 SUm

of divisors, 173 of ideals, 36 of submodules, 3

Support of module, 79 Surjective homomorphism, 5 Symbolic power, 70 System of parameters, 168

T

Tensor product, 15-21

Tertiary decomposition, 258-261 Tertiary ideal, 257 Tertiary radical, 257 Torsion group, 32 Total quotient ring, 65 Transform of ideal, 149, 225, 249 Trivial valuation, 108

of homomorphisms, 19

V

v-closed ideal, 230 v-ideal, 122 Valuation@), 106-1 18, 226

determined by valuation ring, 109 discrete, 111 equivalent, 109 essential, 179 extension of, 114 p-adic, 121 prolongation of, 114 p(X)-adic, 11 4 trivial, 108

Valuation ideal, 122, 145 Valuation pair, 227 Valuation ring, 99, 227

discrete, 111 of valuation, 109

Valuative dimension, 164-167 Value group, 108

W

Weak multiplication rings, 210

U

Unbranched prime ideal, 120 Unique factorization domain, 190 Uniqueness of primary decomposition,

Unity, 1 48-52

Zero divisor, 42 proper, 42

ZPI-ring, 205-209

2

Page 316: multiplicative theory of ideals

Pure and Applied Mathematics

A Series of Monographs and Textbooks

Editors

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Metrics. 1953 4 : STEFAN BERGMAN A N D M. SCHIFFER. Kernel Functions and Elliptic Differential

Equations in Mathematical Physics. 1953 5: RALPH PHILIP BOAS, JR. Entire Functions. 1954 6 : HERBERT BUSEMANN. The Geometry of Geodesics. 1955 7 : CLAUDE CHEVALLEY. Fundamental Concepts of Algebra. 1956 8: SZE-TSEN Hu. Homotopy Theory. 1959 9: A. M. OSTROWSK~. Solution of Equations and Systems of Equations. Second

Edition. 1966 10 : J. DIEUDONN~. Treatise on Analysis. Volume I, Foundations of Modern Analy-

sis, enlarged and corrected printing, 1969. Volume 11, 1970. 11 : s. I. GOLDBERG. Curvature and Homology. 1962. 12 : SIGURDUR HELGASON. Differential Geometry and Symmetric Spaces. 1962 13 : T. H. HILDEBRANnT. Introduction to the Theory of Integration. 1963. 14 : SHREERAM ABHYANKAR. Local Analytic Geometry. 1964 15 : RICHARD L. BISHOP A N D RICHARD J. CRITmNnEN. Geometry of Manifolds. 1964 16: STEVEN A. GAAL. Point Set Topology. 1961 17: BARRY MITCHELL. Theory of Categories. 1965 18: ANTHONY P. MORSE. A Theory of Sets. 1965

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19: GUSTAVE CROQUET. Topology. 1966 20: Z. I. BOREVICH A N D 1. R. SRAFAREVICH. Number Theory. 1966 21 : J O S ~ LUIS MASSERA AND JUAN JORGE SCHAFFER. Linear Differential Equations

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1968 30 : L. J. MORDELL. Diophantine Equations. 1969 31 : J. BARKLEY ROSSER. Simplified Independence Proofs : Boolean Valued Models

of Set Theory. 1969 32: WILLIAM F. DONOGHUE, JR. Distributions and Fourier Transforms. 1969 33: MARSTON MORSE AND STEWART S. CAIRNS. Critical Point Theory in Global

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34: EDWIN WEISS. Cohomology of Groups. 1969 35 : HANS FREUDENTHAL AND H. DE VRIES. Linear Lie Groups. 1969 36: LASZLO FUCHS. Infinite Abelian Groups: Volume I. 1970 37: KEIo NACAMI. Dimension Theory. 1970 38: PETER L. DUREN. Theory of Hp Spaces. 1970 39 : BODO PAREICIS. Categories and Functors. 1970 40: PAUL L. BUTZER AND ROLF J. NESSEL. Fourier Analysis and Approximation:

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I n gtepatutiolr WERNER GREUB, STEVE HALPFXIN, A N D JAMES VANSTONE. De Rham Coho- mology of Fibre Bundles : Volume 1 ; Manifolds, Sphere Bundles, Vector Bundles.