Multiplicative Theory of Ideals (Pure & Applied Mathematics)

Multiplicative Theory of Ideals

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

MULTIPLICATIVE THEORY OF IDEALS M A X D. LARSEN / PAUL J . McCARTHY University of Nebraska Lincoln, Nebraska

University of Kansas Lawrence, Kansas

@

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

1971

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

To Lillie and Jean

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Contents xi

Preface

...

Prerequisites

Xlll

Chapter I. Modules 1 2 3 4 5

Chapter II. 1 2 3 4

1 8 12 15 21 27

Rings and Modules Chain Conditions Direct Sums Tensor Products Flat Modules Exercises

Primary Decompositions and Noetherian Rings

Operations on Ideals and Submodules Primary Submodules Noetherian Rings Uniqueness Results for Primary Decompositions Exercises

36 39

44 48 52

Chapter Ill. Rings and Modules of Quotients 1 Definition 2 Extension and Contraction of Ideals 3 Properties of Rings of Quotients Exercises Vii

61 66 71 74

CONTENTS

Vlll

Chapter IV. 1 2 3 4

Integral Dependence

Definition of Integral Dependence Integral Dependence and Prime Ideals Integral Dependence and Flat Modules Almost Integral Dependence Exercises

Chapter V.

Valuation Rings

1 T h e Definition of a Valuation Ring 2 Ideal Theory in Valuation Rings 3 Vaiuations 4 Prolongation of Valuations Exercises

Chapter VI.

Dedekind Domains

5 Extension of Dedekind Domains Exercises

Chapter VII.

156 161 164 168

Krull Domains

1 Krull Domains 2 Essential Valuations 3 The Divisor Class Group

4 Factorial Rings Exercises

124 126 132 134 140 144

Dimension of Commutative Rings

1 T h e Krull Dimension 2 T h e Krull Dimension of a Polynomial Ring 3 Valuative Dimension Exercises

Chapter VIII.

99 105 107 114 118

Priifer and Dedekind Domains

1 Fractional Ideals 2 Prufer Domains 3 Overrings of Priifer Domains 4

82 84 88 92 94

171 179 185 190 194

ix

CONTENTS

Chapter IX. Generalizations of Dedekind Domains 1 2 3 4

Almost Dedekind Domains ZPI-Rings Multiplication Rings Almost Multiplication Rings Exercises

Chapter X.

201 205 209 216 220

Prufer Rings

2 Counterexamples 3 Large Quotient Rings 4 Prufer Rings Exercises

226 232 234 236 244

Appendix: Decomposition of Ideals in Noncommutative Rings

252

1 Valuation Pairs

Exercises

263

Bibliography

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. I n 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

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.

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 e v u y totally ordered subset of S has an upper bound in S, then S has a maximal element.

...

Xlll

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 I S E Sand 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 mappings, 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,

Multiplicative Theory of Ideals

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CHAPTER

Modules

1 RINGS A N D MODULES

We begin by recalling the definition of ring. A ring R is a nonempty 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, T h e 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. T h e reader may verify for himself such statements as Oa = a0 = 0 a( -b) = (-a)b = -(ab)

a(b -c)

= ab -uc

for all a E R, 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 1

1

2

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 commutative 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}

is a left ideal of R, while

aR=(av/r E R } is a right ideal of R. Since R has a unity, we have a E Ra and a E aR. The smallest ideal of R containing a, called the principal ideal generated by a, is 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

(1) (2) (3)

a(x +y>= a x +by, (a b)x = ax bx, (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 (1’) (2’)

(3’)

3

RINGS AND MODULES

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

+

+

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 . T h e mapping (a, x) Hax 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 intersection 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

a E I

Na

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

+ N , is

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 , , submodules of M , then

. . . , N kare

1

4 Nl

+ a

*

+Nk= { x l + +xkl xi E N ,

MODULES

for each i>

is a submodule of M . We denote this submodule by C:=l N , ; it is called the sum of Nl, . . . ,N k. T h e 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 . I n general, the union of submodules of M is not a submodule of M . However,

U

c Na

J E I a€J J finite

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) Hnx given by if n > O if n = o if n m we have M , * *

= Ker

= 0, we

4,

for all n.

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) only if both L and N satisfy (ACC).

if

and

Proof. First, assume that M satisfies (ACC). Let L, E L , G 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 n o . Now let N , G N 2 G * be an ascending chain of submodules 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,)) -

--

0

.

2

11

CHAIN CONDITIONS

= +(t,-l(N,,)) = N,,,, for m 2 m, . Therefore, both L and N satisfy (ACC). O n 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 d-l(+(L) n M,) = +-l(+(L)n Mno)and t,(M,) = +(Mn0) for n L no. 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 = I m = +(L),we have for n >no,

+

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

+

+

=

Ma0

W

n

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 f y 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

-

+ +

+

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 S U M S

Let {Ma1 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

Ya

(ax), = m a ,

u E R,

for all

o! E

I.

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

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

= MI Q

@ M,,

.

Suppose that M = O a o I M a . For each u E I there is a homomorphism +a :1M-t Ma given by +a(x)

=xa

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

It is clear that

is surjective and

$a

p=o!

p#a.

is injective. Furthermore, if a = p if a # / 3

3

13

DIRECT SUMS

and (b)

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

=0

for all but a finite

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 homomorphisms +a :M +M , and :M , +M such that

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

Clearly @ is a homomorphism. It is surjective since for all x E M , if y a= &(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

BEI

#AS> = xa

t

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

as1

and

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

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 {Ma1 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 0a E I M a . Proof. For every CL E I we define X E M , .If EM, we write

$a:

Ma-+ M by

$ a ( ~ )= x

for all

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

4,

if P = a if /3 # a.

Furthermore, for each x

E

M , if x = CaEI 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 af1

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

4

15

TENSOR PRODUCTS

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

R = @ Ra. aeI

We define a multiplication on R by

(4, = a, b,

for all a E I .

Then R becomes a ring, a fact which is easily verified, called the direct sum of the family of rings {RaI a E I } . T h e 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, if if

a=P ct#P

and for all Then R z

UE

R.

B U GR,., 4 T E N S O R 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. I n this case, we say that the set S freely generates M . Note that M = @ Rx. 5ES

1

16

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 Define equality of these sums by a,x= ZES

E

S.

1 a,’x ZES

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

2 ES

for all a E R.

ZES

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 We say that d,’ is obtained by extending d, by linearity.

4.

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

+

r), r2>,

x2 > Y ) - (x1, Y ) - (x2 ? (x, y1 S y 2 ) - (x, - (2, (XI

n)

( x a , y ) - (x,

ay),

4

17

TENSOR PRODUCTS

where x,x,, xz E M , y , y l , y z E N , and a E R. T h e 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 M O R N by x O y ; we call such an element of M o ~ Na simple tensor. We have

+

+

+

(XI x2) O r = 3 1 OY xz O r , 30 (n+YZ) = x OYI t x OYZ, xu B y = x @ uy,

for all x,xl, x2 E M , y , y l , 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 . I n particular, if n is an integer, n(x y ) = nx @ y ; similarly, n(x B y ) = x ny. It follows that 0 B y =x

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

0 = 0,

for a l l x E M a n d y E N . T h e 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 )

+(x, Y1 +YZ)

= +&Y1)

+ 7% Y ) , + +(4Yz), Y

+(xu, Y >= d(x, 4,

for all x,x,, x2 E M , y , y l , y 2 E N , and aE R. 1.17. Proposition. Let M be a right R-module, N an R-module, and

4a

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 a n d y E N .

1

18

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 homomorphisms 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

RORNzN,

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 RN into N such that $(a @ y) = ay for all a E R, y E N . It is surjective since #(l 8 y) =y. T o show it is injective, let t E Ker $. Since

t=

1at@yt = 1 la, OYt = 1

1 0 a,yt = 10 1% Y t ,

ify = ai y i I , we have0 = $ ( t ) = $(loy) =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 ( 48 g ( Y ) + f ( x z ) 8 d Y ) = +(XI,

Y)

+ 4(% A. 7

4

19

TENSOR PRODUCTS

By Proposition 1.17, there is a unique homomorphism from M O R N into M ’ O RN ’ 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 - t N 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 3N’ are surjective homomorphisms, then Kerf @ g consists of all jinite sums of the form C xi 0y 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 RN , generated by the set of simple tensors x @ y where x E Kerf o r y 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 ) @ g b )

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 RN / K by j ( x ‘ ,y’) = K , wheref(x) = x’andg(y) =y’. (We use here the fact that

x@ y

+

I

20

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 o y 1 = x O ( 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 conditions, again leaving the verification of the other two to the reader. Let x’, X “ E M ’ , y’ E N ’ , 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 =&’, Y ’ ) + j ( x ” , y’).

+K )

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) O g ( y )= 2’ Or’.

Furthermore

0Y

+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 isomorphisms. Hence Ker h = 0, that is, K = Kerf Og. 1.22. Theorem. Let

0-

NI-

f

~ 9N II-

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

MORN’ is exact.

- 1MWf

MORN

lME3.8

M @ , N -0

5

21

FLAT MODULES

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 MORN-

f@IN

MORN-

%‘@IN

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 K e r g = Im f . Hence Ker 1 @ g = I m 1 0f . 5 F L A T 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 homomorphism 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 r- 0 , the sequence IM@f

lM@g

O----+MORN‘-M@RN-MO,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 y z # 0 for only a finite number of x E S ,

22

1

Proof. Let

C x i @ y iE F@RN,

CXExu:), where a$i)E R. Then

c

xt 8Yt =

MODULES

where x i E F, y i E N . Let x t =

C

x0

XES

(caPYt).

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. To show the uniqueness, it is sufficient to show that x ByI = 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,

XI

Then, if 1, is the identity isomorphism of N , we have for each xE

s,

0 = ( # ’ ~ 8 1 N ) ( ~ X’ES

~ ‘ @ Y x * )=

1$ X ( ~ ’ ) ~ Y Z ’ = ~ ~ - Y I *

I’ES

For each x E S define 7) : X R--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 8Yx))= #(dx) 8 Yx)= #(l 8 Yz)=Yz *

1.26. Proposition.

A free right R-module isflat.

N , N ’ be R-modules and f : N ’ + N an injective homomorphism. 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 RN ’ and suppose that (1,@ f ) ( t )= 0. We can write t x ByX where y XE N ‘ and y X # 0 for only a finite number of x E S. Then

Proof. Let

=IIE

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.

5

23

FLAT MODULES

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

I

MI

M2

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

M-N-0

I

of right R-modules, where the row M + N+O is exact, there is a homomorphism 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 statements are equivalent :

(1) (2)

P is projective. P is a direct summand of a free right R-module.

24 (3)

I

MODULES

For evevy exact sequence of right R-modules of the form o+-L~M+-P+-o,

I m 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 exact sequence

C$

is a surjective homomorphism and we have an @

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, = g h . Let P ' = I m 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 ( g h ( y ) )= hg(x) =O. Thus (Imf) n P' = 0. Now let z E M ; then - hg(z))=g ( 4 - 1P ( g ( 4 ) = 0,

so z -hg(z)

E

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

Therefore M g (Imf) @ P ' .

+ hg(z) E (Imf) +P'.

5

25

FLAT MODULES

(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

+

F-P

d

M-N-0

If

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 F-P

d

M-N-0

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

(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 FWrite . 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 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

XE

1

26

MODULES

R-module. Therefore P is projective by the equivalence of (1) and (2). 1.29. Corollary. Every free r k h t R-module is projective.

--

1.30. Proposition. Let

o

M

J

M BM” --+

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

gC31N

0-M’QRN-MQRN-M”QRN-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 isomorphism 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.

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 l N ,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 Proof. Let N and

1P@f

PQRN‘-

PORN

FQ RN’

F Q RN

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

27

EXERCISES EXERCISES

1. Isomorphisms. Let M and N be R-modules and suppose there are homo= 1 and morphisms +: M - t N and $: N + M such that +# = 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 ={WI x E Ml) ; the inverse correspondence is Nl*+-'(N1)

={XI

xEM

and +(x)

E Nl].

(b) Let M I be an arbitrary submodule of M . Show that +-'(+(MI))= M I + Ker 4. (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+)

N l ) .

3. Chain conditions. (a) Prove the equivalence of (DCC) and (MIN). (b) Show that the additive group of integers satisfies (ACC) but not (DCC). (c) 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). (d) Let MI, . . .,M , be submodules of an R-module M such * * M,, . Show that M satisfies (ACC) that M = Ml [or (DCC)] if and only if M , does for each i.

+- +

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*2M,2*..2Mr=O.

I

28

MODULES

is called the length of the normal series. If . . ., 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 submodules of M such that L' E L and N ' E N . Show that T h e integer

Y

Mi-l # M i for i= 1,

(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. T h e 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 composition 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). (b) Show that if K and N are submodules of an R-module then

+

29

EXERCISES

(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). (d) 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 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 = (ax1x E M ) is a submodule of M isomorphic 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 aDefine . 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 homomorphism 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 :M a+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 . (c) Let ( M a ]M E 1) and (Nala 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: M a +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 tc E I}. Show that

{fal

30

I

MODULES

Kerf= @ Kerf,, a E I

and Im f

0M,/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 = O a .I L a . Let g be the direct sumof thefamily{g,I ~ ESIh 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 . (e) Let be a one-to-one mapping of the set I onto itself. Show that

0Ma a0 Mn* EI

a E I

(f)

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

K

0Ma z ( a0 Ma)@ ( 0Ma>E J aEK

aEI

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 T i 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 : Ti--f G such that $t7,= 4. a unique homomorphism Show that there is a unique isomorphism Q : T I --f T , such that oql= q z . (b) 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.

31

EXERCISES

Let N be an R-module and define a mapping from R’ x ( M O R N )into M O R N b y Show that this mapping is well-defined and that it makes

M O RN into an R‘-module. (c) 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 Rsuch N that x @ ( y @ z ) H(x@y)OZforallxEL,yEM,andzEN. (d) 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 . (e) Let R be a commutative ring and let M and N be R-modules. Then, according to part (b), M O EN 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 form (x, y1 +YZ) - (x, n)- (x, yz), (x1+ x2 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 7

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, OZG G/nG. (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 O KV is a vector space over K and that dim, U @ , V = (dim, U)(dim, V ) .

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 O ZH is a torsion group. An Abelian group G is said to be divisible if nG = G for (e) all positive integers n. Show that if either G or H is divisible, then G Q ZH is divisible. (f) Show that if G is a torsion group and His a divisible group, then G &El== 0. 9. Flat modules. (a) 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 (MaI 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. (d) 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.

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 iff:N'+Nisgivenbyf(y)=yforallyEN', then 1@ , fis injective. We shall identify each element of M Q RN ' 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/MORN12 MOR(N/Nd (b) MOR(N~$N~)=(M@RN~)+(MORN~) (c) M@R(Nl N 2 ) == ( M @ R N I ) (M@RN2)* 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 b y f ( u ) = u for all a E A , then l , @ f i s injective.

33

EXERCISES

Show that if F is a free R-module, and if g: E+ F is injective, then 1 @g is injective. (b) Show that M is flat. 12. Flatness and exact sequences. Let (a)

f

0-L-MLN-0

be an exact sequence of right R-modules and assume that N is flat. (a) Prove that 0-

L@RK-

f @1K

M@RK-

9@1K

N@RK--O

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

C:=l

n

C biiyi = 0,

j = l , ..., k ,

i = l

and k

xi =

1

i =1

Zjbji,

i = 1, . . . , n,

(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 , respectively, such that x i a, = 0, there exist elements zl, . . . , zk E M and elements bji E R, i= 1, . . . , n, j = 1, . . ., K, such that

I;=

2!Jjiai=0,

i=l

j

= I,

..., k ,

and

i = l , ..., n.

34

1 (c)

MODULES

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

h=l,

XifflhzO,

...,r,

i=1

where aihE R for each i and h, we have

i = 1, . . . , n, where d,,

.. .,d , E R',bji E R for e a c h j and i, and j=1,

..., k, h = l , ..., r.

14. Split exact sequences. be an exact sequence of (a) Let O + L L M - % N + O R-modules. Show that this is a split exact sequence if and only if there is a homomorphism h : N + M such that g h = 1,. (b) 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,. (c) Let L and N be R-modules. Show that there is a split exact sequence 0 -+ L -+ M -+ N -+ 0. (d) 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. 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

y = c2.E s ( b z ( Y ) X . (b) Show that a direct summand of a projective module is projective. (c) Let (P,(aE I > be a family of R-modules. Show that

35

EXERCISES

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

* A-BLC (e)

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 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$.

CHAPTER

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 ) .

1a, b, and C c j d j are in A B , then C a,bi - 1 = C ai +C( E AB,

Then AB is an ideal of R. For, if

bi

Cjdj

and if

7 E R,

-Cj)dj

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 a B 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

1

37

OPERATIONS ON IDEALS AND SUBMODULES

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

A(BC)= ( A B ) C , A B = BA, 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.

(1)

(2) (3) (4) (5) (6)

+

+

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 A C , 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 A C 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. Let M be an R-module. If A is an ideal of R and N is a submodule of M , we set AN =(finite sums

a, xiI a, E A , xiE N } .

Then A N is a submodule of M , and the proof of the following proposition is like that of Proposition 2.1. 2.2. Proposition. Let A and B be ideals of R and let L and N be sub-

modules of M . Then (1) (2)

(3)

A ( B N )= ( A B ) N , AN G N, A ( L + N ) = A L + A N and ( A

+ B ) N = A N + BN,

11

38

(4) (5) (6)

PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

A c B implies that A N c BN, L c N implies that A L G A N , 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 A x for A(Rx). Again, let M be an R-module, A an ideal of R, and N a submodule 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. If L and N are submodules of M , we set L:N=(alaizR

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 . T h e submodule N : A and the ideal L : N are called the residual of N by A and the residual of L by N , respectively. 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 N implies that L : A G N : A and L : K 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 ) .

+

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 A B ( ( N :A ) :B ) 2 A ( N :A ) c N so that ( N :A ) :B

2

39

PRIMARY SUBMODULES

c N : AB. Also, A B ( N : A B ) 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 ( L n N ) : A . A l s o A ( ( L n N ) : 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:BC. 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 + a n u = 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 a n M .Thus anM c Q. We conclude that Q is primary.

+

+

+

+

+

+

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 submodules of M .

Y be the set of all submodules of M which cannot be written as an intersection of a finite number of irreducible submodules 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, $9and 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. Proof. Let

-

-

-

--

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 submodule 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.

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.

2.8. Definition.

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

2

(1) (2) (3)

41

PRIMARY SUBMODULES

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 A B 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}, and call Rad(A) the radical of A. (2) Let N be a submodule of an R-module M . Set Rad(N) = Rad(N:M), and call Rad(N) the radical of N . Let a, b E Rad(A) and r E R ; let amE A and b"

E A,

where m and

n are positive integers. Then (ya)m= rmamE 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 ankMG Q for some positive integer k. Thus a E Rad(9)).

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 multiplicatively 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.

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, rlr2ab E P n S, which contradicts the fact that P n S is empty. Proof. Let 25 be the set of all ideals

+

+

+

+

+

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

2

43

PRIMARY SUBMODULES

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 $ P afor 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.

44

11

PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

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

Rad(N) = (IP,

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 abundance 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 if - . . € A ' . If s =max{s,, . . ., s k ) , then a i X S 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, . . . ,f n E A' n N such that A' = (fl, . . .,fn). It remains to verify the equality stated above. Clearly, the righthand side is contained in the left-hand side. Let g = b,Xt

+ - - + b1 x+ b, E A'. *

Then b, E A , so b,

where rl, .. . . , rk E R . If t

rla1 f* '

'

+

rkak

< s, then g E A' n N . Suppose that t 2 s.

3

45

NOETHERIAN RINGS

Then

+ +

+ +

g = (rial - - * rka,)X b,-,X t - l =rlXt-sdfl -f1') f ' ' ' rkXt-s(fk b&,X t - l -.* b, ,

+ +

+

where f2'

=f i - a f X " E N .

+ . - + b, *

-fk')

Thus g =g,' +gl, where g,'

g, = c,-,Xt-l

+

* * .

+clX+ c,

E A"

and

E A'.

If we repeat this argument several times, we will obtain finally g = g l ' f . - . + g ; - s + l +gt-,+,,

whereg,'EAN for i = l , completes the proof. 2.18. Corollary.

..., t - s + l

a n d g , - , + , E A ' n N . This

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 polynomial 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 R IX l , . . . , xk] of degree n such that f ( a l , . . . , ak) E A " B n C. (All terms off have degree n.) Let = uFz0S, 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 AnBn C, then, since a E A", we haveffa,, ..., ak)= a for some f E of degree n. Iff = f i g l , where g, E R[X,, ..., xk],

s

1

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 k ) = c f t ( a l ,

u=f('l,

uk)gt(ul,

ak)

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

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

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 XI 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 ) ,

+

+

3

47

NOETHERIAN RINGS

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:=lAnM, then A N = N .

N= Proof.

If r is as in Theorem 2.20, with L = M , then for all n > r, AnM n N = An-*(ArMn N ) E AN.

Since N E AnM this implies that N

G

AN. But A N

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 =

Proof. Let

1, . .., k, then db, = Ofor j = 1, .. ., k .

d i , be the cofactor of a i j in the matrix [a,,].Then

f

dijaih

i = l

=

d

[o

if j = h if j # h.

Hence 0=

k

k

1=1

h=l

1d,, 1 aihbh = for j = 1,

...,k.

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

fi A ~ M = O

n=l

if and

only that x = 0.

if l - a g A

and a x = 0 , where a E R and EM, imply

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

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 AnM # 0.

n;=

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

U N I Q U E N E S S RESULTS FOR PRIMARY DECO M P O S l T l O 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=

Q1r\

... 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 .

If a submodule N of M has a primary decomposition, then N has a reduced primary decomposition.

2.25. Proposition.

4

49

UNIQUENESS RESULTS FOR PRIMARY DECOMPOSITIONS

Proof. Let N = Q1 n n Qn where Qi is primary for i = 1, . . . , n. If some Q t contains the intersection of the remaining Q j , simply delete it. Hence we may assume (i) holds. Suppose QLl,. . . , Q,, have the same radical P. By Exercise 6(f), Q = Qll n n Qi, is P-primary. I n the decomposition for N replace Q,, n . . . n Qlkby Q. 1 . .

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. T h e prime ideals P,, . . . , Pk are called prime divisors of N . We have Rad(N) = Rad(Q,) n * * * n Rad(Q,)

-

= PI n - . n P k .

Hence, if a E P, . * * P, , then some power of a is in N : M . Consequently, 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 f o r 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, Q i : Rx = R for i = 2, .. ., k.

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) = P i 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 = Q1n .. . 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 Qiwhere0 = 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 , = {XI 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 N s and a E R. If slxlE N and s2x2E 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, - x 2 , 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 .

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 P i n S is not empty for the remaining i. Then N s = 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 Q i . Therefore N , s Q1 n n Q p LNow . 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,)nMcQ h t l n '.. n Q k ,

(S~+I

and so (Sh+l ' . * Since (sh+

*

Sk)nX E

Q1 n

*'

0

Q k = N.

. sk)" E S , we have x E N , .Therefore N , = Q1n - . * n Qh.

Again, let N be a submodule of M which has a primary decomposition N=Q, n

n Qr,

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 { P i , , . . ., 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. T h e proposition will follow if we show that N , = Q i , 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 P i j by Exercise 5(c). have Pi $ P,, f o r j = 1, . . . , r, and SO Pi $ Hence Pi n S is not empty.

u;=,

11

52

PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

In the notation of this proposition, Pi, n * * * n Qt, is called an isolated component of N . T h e 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 AnM is a generated R-module. Let A be an ideal of R. Then 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:==lA n M ; by Corollary 2.21, we have N = A N . Let x

E

N

and let A x = Q1 n * * n Qk, where Q iis primary and Rad(Q,) = Pi for i= 1, .. .,k. For each i, A x G Q i so that either A c P i or x E Qi [see Exercise 6(c)]. Suppose that A c Pifor 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. T h e radical of an ideal.

Let R be a ring and let A , B, and C be ideals of R. (a) Show that Rad(AB) = Rad(A n B) = Rad(A) n Rad(B). (b) Show that Rad(A B) = Rad(Rad(A) Rad(B)).

+

+

53

EXERCISES

(c)

Show that Rad(A

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

(d) Show that Rad(Rad(A)) = Rad(A). (e) Show that A B = Rif andonly if Rad(A)

+

2.

+ Rad(B) = R.

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,, .. . , a k E R, there exists x E R such that x - aiE A , for i = 1, . . . , k. (d) R/(A;’ n . . n A;”)2 R/A;l @ . . . @ RIA;,. (e) T h e ideals B,, . . . , B,, of R are comaximal if and only if Rad(B,), . . ., Rad(B,,) are comaximal.

+

3. 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.

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 A i is an ideal of R i for i = 1, . . . , n. (b) Show that A is a maximal ideal of R if and only if for some i, A i is a maximal ideal of Riand A j = R j for j # i.

54

11 (c)

PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

Show that the statement of part (b) is true when “maximal ” 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 Pkis a prime ideal, then P = Pi for some i and P c Pf for j = 1, . . . , k. (c) 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.

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 submodule of M . Suppose that: (i) Q: M E P G Rad(Q); and (ii) If ax E Q and x $ Q, where a E R and x E M , then aEP. 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 n Qk is P-primary. 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

55

EXERCISES

of M . Show that if A $ P then Q: A = Q and that if N $ Q then the ideal Q: N is P-primary. (i) 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 . ( j) Show that N : A = N if and only if no prime divisor of A is contained in any prime divisor of N .

7. T h e 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 ~ Rand l f r a isaunitin 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 : (1) A J ( R ) . (2) If M is a finitely generated R-module and A M = M , then M = 0. (3) If M is a finitely generated R-module and N is a submodule of M such that M = A M N , then N=M. (4) If a E A, then 1 f a is a unit in R. (d) 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 A n M )= N .

(b) (c)

+

n;=l

+

8. Corollaries to the Artin-Rees lemma. (a) 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. (b) Show that there is a positive integer Y such that A n M : (a) E 0: (a) + An-'IM for all n > r. (c)

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 for all n > Y.

56

11

PRIMARY DECOMPOSITIONS AND NOETHERIAN RINGS

(d) Let x E M . Show that there is a positive integer r such that ( N A " M ) : Rx E N : Rx A"-' for all TZ > r.

+

+

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

nQ~.

AgPi

(c) Show that there is an integer s such that As n B = 0. (d) Show that B : A = B. 10. The ring of integers. (a) Show that every ideal in the ring of integers is a principal ideal. (b) Determine all of the prime ideals and primary ideals of the ring of integers. Prove the unique factorization theorem for integers as a (c) 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 polynomials 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). (b) 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. (c) 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 - c X , X z )# ( Y - d X , X z ) . Show that for each c E K , ( X z ,X U ) = ( X ) n ( Y - c X , 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 of (XZ, XU)?

57

EXERCISES

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 ,Z X ) . (c) Show that (4,2 X , X z ) is primary and that (4,ZX,X2) = (4,X)n (2,X"). This shows that a primary ideal need not be irreducible. (d) 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 ( 3 X , X 2 , X 3 ) of S. Show that P is a prime ideal but P2is not a primary ideal. This shows that apower of aprime ideal need not be primary.

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 P o = {{a,} I for some m o , 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 {PtI 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. P , = 0, but for every j > 0, (d) Show that

nl.=

(e)

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.

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. (b) 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. (c) 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.

+

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 , t h e n B , = B , ’ f o r Z = l , ..., K. n C, where (b) Suppose that we also have A = C, n 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 n B,. Show that for each i, A = D,n C, for some j . (c) Show that k = n. - - a

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. T h e 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,

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 a x = 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 = I M a , where each M , is a submodule of M , then Ass(M)= U d EAss(M,). I 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).

0,.

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). I M a then Ass(M)= Ua.I Ass(M,). (b) Show that if M = n N k = 0, where the N i are sub(c) Show that if N , n modules of M , then Ass(M)G Ass(M/N,). (d) Let T be a subset of Ass(M). Show that there is a submodule N of M such that Ass(M/N)= T and Ass(N)= Ass(M)\ T .

-

u:=

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 zerodivisor in R[X] if and only if there is an element b E R, b # 0, such that bai = 0 for i= 0, . . . , n. (c) 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. (d) Show that if Al, .. . , A , are ideals of R then (A, n - * * n A k ) R [ X ]= A,R[X] n .* . n A,R[X].

CHAPTER

I11 Rings and Modules of Quotients

1 DEFINITION

Let R be a ring and let M be an R-module. Let S be a multiplicatively 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 - l M 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)

+)’)s sx’) + s’suv(t’y - ty‘) = 0.

- st(t’x’

= t’tvu(s‘x -

61

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 - l M = 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 - l R 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 - t R ’ 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

1

63

DEFINITION

R

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 + = asis + bs/s = T ( 4 + V(b) b)s2/s2

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,. Conversely, 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 - l R + 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

= $(ah)

and

+ +(b)+(t 1

+ 4(b/t1

-l

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 - l R 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 Rinto an R-module by the method described in Exercise 7(b) of Chapter I. Then a(x 0b/s) = x @ ab/s = ax 0b/s.

3.2. Proposition. S - l M Proof. Define

z M O RS-lR, as R-modules.

4:S-lM+ M O RS-’R by $(xis) = x 0l/s.

If x/s = x’/s’ and if u(s’x - sx’) = 0, where u E S , then x 0 11s = x 0us’ jsus’ = us’x 01jsus’ = usx’ = x’0us/sus’ = x‘

Therefore

0l/s’.

4 is well-defined. Since $(xis

+yb)

= $((tx

+ sy)/st)

= tx 0 ljst

01Isus’

=(tx

+ sy 0ljst

= x 0l/s + y

+9)0l/st

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 - l R into S - l M by $’(x, a/s) =ax/s. If a/s=b/t and if v ( t a - s b ) = 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 - l M such that +(x 0a/s) = axjs. Furthermore, $(a(. @I

b/t )) = $(x 0ab/t ) = abx/t = a[bx/t)= a$(.

0b / t ) ,

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

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

1

65

DEFINITION

and $$(XIS)

= +(x

0l/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 N O RS - l R 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

&

L @ , S - ~ R---+

o@i

S-IN

~ ~ d s - 1 ~

is commutative, for

cp 0;)$l(./s)

(401i s 44s).

= ( p 0i)(x 01is) = p

= + 2 ( P (x)/s) = $2

Now suppose that a E L O RS - l R and that ( p @ i ) ( z )= 0. Since +1 is surjective, M = y$(x/s)for some x/s E S-lL, and dZu(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. Therefore, 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 multiplicative system in R. The ring S - l R is called the total quotient ring of R. I n this case, the homomorphism 7 of Proposition 3.1 is injective, and we shall identify each element of R with its image in S - l R 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 - I R such that the diagram

66

111

RINGS AND MODULES OF QUOTIENTS

R

T-lR

*

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. I n this case, the total quotient ring of R is a field, called the quotient field of R.

2 EXTENSION AND CONTRACTION O F 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) (2)

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.

T o reiterate, if A is an ideal of R, S - l A is the set of all finite sums C 7(ai)riwhere 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 interpreting A' n R as the intersection of two sets is nonsense,

2

67

EXTENSION AND CONTRACTION OF IDEALS

3.5. Proposition. If A is an ideal of R, then S - l A # 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 - l A = 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 ( An’ 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)).Therefore a/s= (as/s)(l/s) is in the ideal of S-lR generated by v [ ~ - l ( An’ 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.

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 n o . Then, by Proposition 3.6, A,’ = Aho for all n 2 no . Proof.

3.8. Proposition. Let {A,‘l a

EI

} be an arbitrary family of ideals of

S-lR. Then

(G 1 A,’

Proof.

n R=

We have A ‘ n R=v-l (,QI

4

A,‘ ( G I

= =

n (A,’ n R).

as1

v-y n (A,’ ,El

1

1

nq(R)

n

m)

n T - l ( ~ ’n ?(R))= n (A‘ n R).

,€I

, E l

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. T o 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 - I Q 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/snE 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 - l Q , 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

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. of A, then

If A is an ideal of R and A,

is the S-component

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 - l R 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 - I A 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 - l R 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.

111

70

RINGS AND MODULES OF QUOTIENTS

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

A

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, 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 r P k ,where xl, .. , x k E R . Then a = xlrpla * . x k r p k aE B. Thus A c B, and we show in a similar manner that B c A . Therefore A = B.

I

+- - +

+ -+

.

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 component 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, nR. 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

n m

~ ( n= ) 0,.

n=l

Proof. By Exercise 4(a),

PnRp= (PR,)"for each positive integer n,

and

n(PR,)~ o m

=

n=l

by Corollary 2.24. Therefore

3

71

PROPERTIES OF RINGS OF QUOTIENTS

by Proposition 3.l(b). Hence, by Proposition 3.8,

n

n (PR, m

a,

P)=

n=l

nR)

n=l

- 0,.

3 PROPERTIES O F R I N G S OF Q U O T I E N T S

I n this section we shall obtain several properties of rings of quotients which will prove to be useful. I n 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 - l R be the homomorphism of Proposition 3.1. 3.15. Proposition. Let R' be a ring and let p : R+R' morphism such thut ;

be a homo-

(i) elements of p(S) are units in R ' ; (ii) the kernel of p is 0, ; and (iii) 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.

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 - l R - t T - l R 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 - l R . 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 p is o,,T) Finally, if a/t E T - l R , 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 . 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

+ ( S ) is a multiplicative system in RIA. Let S-lR+ S - l R / 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 - l A n R , it is a homomorphism, and the diagram Proof. Note that

4':

+

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 - l A is empty, the elements of +‘q(S)are units in S - l R / S - l A . 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 - l A 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(Ns)-l)

d(+m-l=/-4a + A M +

= +‘T(414’v(s)-1 =

so that (iii) of Proposition 3.15 holds. 3.18. Definition. The set

S = { x l x ~ R and ~ ( x ) isaunitin S-IR}

is the saturation of S.

is a multiplicative system in R, and that S G

It is clear that

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 Swe have x(sxtu)= USt2 E

s.

3.20. Theorem. S - l R

fyom S-IR onto

a/s E S-lR.

s-lR

S - l R ; explicitly, there is an isomorphism which maps a/s E S - ‘R, where s G S, onto

74

111

RINGS AND MODULES OF QUOTIENTS

Proof. Let p : R + S - I R 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 - l R + S - l R 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 ) + { s z / 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 - l R 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. I n 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.

75

EXERCISES

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

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 homomorphism. 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

-

(c)

4.

2

- s-lf

0 S -lL S -lM S -lN 0 is exact. Show that if M is a flat R-module, then S - l M is a flat S- lR-module.

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) S 1 ( A B ) = (S-lA)(S-lB). (b) S-l(A n B ) = S-lA n S-lB. (c) S-l(A B) = S-1A S-1B. (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. (e) S-l(Rad A)= Rad(S-lA).

+

+

5. Localization properties. (a) Show that if R is an integral domain and if P runs through the maximal ideals of R, then R = R, .

n

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 ( A R p 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 Noetherian 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 be its saturation. (a) Prove that R \ S is the union of the prime ideals of R which do not meet S . (b) Let {MuI a E I}be the set of maximal ideals of S - l R and let P, = Ma n R. Prove that S-IR = T-lR where T=R\u Pa. If R is an integral domain and P is a prime ideal of R not (c) meeting S, show that R, = S-lRS-,,. (d) If R is an integral domain, show that every ring of quotients S - l R of R is an intersection of localizations of R ; in particular, S - l R = RPu where {P,] is the set of ideals of R maximal with respect to not meeting S.

s

n

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 (RQ)PRQ *

77

EXERCISES

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

5’u {s” (c)

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 “ - l R S’-l(S-lR). [Hint. Take R= Z @ 2, S = ((2, 2)”ln 2 01, and S’ =

8. A tensor product. Let R be a ring, A an ideal of R, and M an R-module. Make RIA O RM 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 . (c) 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?)

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

78

111

(b)

(c)

(d) (e)

RINGS AND MODULES OF QUOTIENTS

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

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 I S;lR. elements, then R = (c) Let M and N be R-modules and f:M + N a homomorphism. 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 . (d) 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 ) - l R by R S ( 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 = Rso,.

na

0

79

EXERCISES

11. T h e 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 - l M 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 = M o x M l =I =I M,,=O, such that for i = O , ..., a - 1, M , / M i + l zR / P , for some prime ideal Pi of R. Show that Ass(M) G {Po,. . . , Pn-l}c Supp(M), and that the minimal 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)= A~W.

n

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). (c) Show that if N is a submodule of M , then Ass(M/N) is the set of prime divisors of N . 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 - I M being considered as an S-lR-module). E then P E (b) Show that if P EF and S - ~ PAss(S-lM), Ass(M). Show that the mapping in part (a) is onto. (c) Show that the mapping x H sx/s from M into S - l M , where

80

111

RINGS AND MODULES OF QUOTIENTS

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 A s s ( M / N )= Y' and A s s ( N )= Ass(M)\Y'.

s E S, is

14. T h e 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 = ( a o ,a,, . . . , a,). (a) Show that S is a multiplicative system in R[X], consisting entirely of regular elements. Denote the ring S - l R I X ] by R(X). (b) Show that if A is an ideal of R, then R(X)/AR(X) (R/A)(X). (c) 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. (d) Let A,, . . ., A, be ideals of R. Show that

+

+ +

(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. (f) Let A be a proper ideal of R. Show that AR(X) is a proper ideal of R(X). (g) Show that R ( X ) is a flat R-module. (e)

15. 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 T M 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.]

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(,) = QnRpn 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

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+ ").

17. 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 . 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 P= UPi.] 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. T o 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.]

CHAPTER

IV Integral Dependence

1 D E F I N I T I O N 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,, - .-

I f(X1, - 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,]. 9

a,] = {f(a,,

-

* * 9

a,)

*

* 9

* *

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 b, b,a+ bn-,an-l an = 0. R [ a ] is afinitely generated R-module. (2) (3) There is a subring R" of R' such that a E R" and R is a finitely generated R-module.

+

Proof. =d

+

+

Suppose (1) holds and let f ( X ) E R [ X ] . Suppose degf(X) c,X * . cd Xd, then

> n: Iff (X) = c,

+

+ +

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

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

82

n-1

an-l)

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, generate R" as an R-module. For i = 1, . . . , n,

+ + +

n

aat=Cbijaj,

bij E R ,

j = l

or

C (b, - aija)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 anis 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

Ia

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 finitelygenerated 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'.

Iv

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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 ,respectively. 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 - I R 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', S E S .There are elements b o , b,, ..., b n - , E R such that 6 , b,a - . bn-,an-l an = 0. Then

+ + + 60/sn

+

+ (b,/s"-l)(a/s) +.-.+ = (6,

(bn-l/S>(@/S)n-'

+ (a/s>n

+ 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 concerning 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

2

85

INTEGRAL DEPENDENCE AND PRIME IDEALS

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 dn-,rn-' rn = 0. Then such that do dlr b " + d , - , a b n - l + . . * +d,an-%+d,a" = (c+ra)"+d,_,a(c+ra)"-' + . . . + d , a n - l ( c + r a ) +doan =j ( c ) U y P d,-,rn-l * d,r +do) =f(c)~P'nRcP; 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 2.12). 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

+

+ + + -

+

a

+

-

+

+- - +

+

a(an-l

+

+

+ . + b,)

+ b,

*

*

E

P',

+ -+

* b, E P'. This contradicts our but a $ P', so an-l 6, choice of n. Thus, if P' c PI', we must have P' = P". 1

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 P o , then there is a chain

Iv

86

INTEGRAL DEPENDENCE

--

Po' c P,' c * c P,' of prime ideals of R' such that Pi' lies over P , for i = 0, . . . , r. If,f o r 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 P i + , . 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. T h e 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 * . an = 0, where 6 0 , . . . , 6 , -1 E R, then b,

-

+

+ +

+

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).

2

87

INTEGRAL DEPENDENCE AND PRIME IDEALS

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”nR = 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 isomorphism 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 7 p (P’n T’) = P”n T’. Hence (T’,7p) is in 9, and ( T ,7 ) < ( T I ,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 Poc 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

- -

Iv

88

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 assumption 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 Qoc Q1c * 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 . I n fact, it is the ideal of T generated by A.

3

89

INTEGRAL DEPENDENCE AND FLAT MODULES

4.10. Proposition. If T is an overring of an integral domain R, then

the following statements are equivalent : (a) (b)

For every prime ideal P of R, either P T ( ( y ): (x))T= T for all x/y E T .

=T

or T E R,.

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 / y E 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 i= 1 , . .., R, then (A, n n A,)T= T . For if 1 < i < R , there are elements aijE 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. 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

Proof.

Iv

90

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 0b Hab is injective. Let c E A B RT ; 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= ai 0 c46. By assumption, ((6): (ci))T= T for i = 1, .. ., s, and so if C = ( 6 ) : (ci), then CT = T . Now suppose that the aici/6 0. If d E C we have dc,/b E R image of c is 0, that is, for i = 1, .. ., s; hence

C;=l

nfZl C;=l

c a, 0dci/b c (da,cJb 0 1) S

dc =

S

=

i=l

t=1

= (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 zjkE R,j = 1, . . . , r, k = 1, 2, and elements b,, . . ., 6, E T , such that

zj1Y - Z f Z

x = 0,

j=1,

..., r.

Let P be a prime ideal of R. If zj2E P f o 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.

3

91

INTEGRAL DEPENDENCE AND FLAT MODULES

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 * * + t k u k , where t,, ..., t k E T', ul,.,., U k E R, and u , c ~ ( d )for i = l , ..., K. Then u t a E 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) (2) (3)

T is a $at overring of R. T p= R , for all maximal ideals P of T. 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 z x , z y E R and xy P, so zy P n R. Hence x / y = z x / z y E R, R . Thus f#

Tp G R P n R .

+

0

(2) + (3). By Exercise 5(a) of Chapter 111, T = 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 But P . T s R p n RSO that T s R , . Thus T is a flat overring of 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 = a y / y = a E R . Thus T = R.

92

Iv

INTEGRAL DEPENDENCE

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. T h e 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 integrally 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 completely integrally closed; an example is given in Exercise 14. Furthermore, 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:

4

93

ALMOST INTEGRAL DEPENDENCE

R, be the complete integral closure of R in R'. Then R, is integrally closed in R'.

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

+

Let x E R' be integral over R,; say xm a , , , - , ~ ~ +-* ~* E R, . Then x is integral over the ring a, = 0, where a,, .. ., a,

Proof.

+

R[ao, * - . , a m - , ] *

For i = 0, . . . , m - 1 , a , is contained in some finitely generated submodule of the R-module R' ; call it M i= 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

+- +

XojoXljl

* ' *

Xm-1, i m - l

where each j , runs between 1 and k, . Hence x is contained in R[x] E R[a,,

... , a,-,,

x]

m-1

=

C R[a,, . . ., a,-,]xh

h=O

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, = {xI x E K and there exists a regular element r E R such that rxn E R for all positive integers n}.

Iv

94

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. Therefore x is almost integral over R.

4.21. Corollary. If R is completely integrally closed, then for all nonunits x E R, (x") consists entirely of zero-divisors.

n;=

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 zerodivisor. 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 submodule 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 , cl, an= 0. . . ., c , - ~ E A such that co cla * * c, 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 b M = 0 for some b E R[a] implies b=0. Show that if R is Noetherian then a is integral over R if and only if R[a] is contained in some finitely generated submodule of the R-module R'.

+ + +

+

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.

95

EXERCISES

(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'.]

3. Integral extensions and ideals. (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 a E R' and there is a positive integer n and elements b, , . . . , b, E A such that b, b,a * bn-lan-l an = 0). (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.

I

+

+

+ +-

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. 5.

Simple ring extensions. (a) 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,

Iv

96

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 R I P 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: bi).] g(X) = J(xf(x) =n ( X (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

+

+

+

97

EXERCISES

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) Tois 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 polynomial 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 polynomials 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 ] . )

10. A proof of the going-down theorem. (a) 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 ] . closed. (a) Prove that T is integrally ~. (b) Prove that K [ X ]is the complete integral closure of R in K[XI*

+

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 - l R is completely integrally closed.

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 Ribe the complete integral closure of R in Ti. Clearly, RlG 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 submodule of a Tl-module of which Tlis 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.

14. The complete integral closure need not be completely integ-' rally closed. Let K be a field and let X and Y be indeterminates. Let R = K [{XZnfly n ( Z n + 1) In 2 011. (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.]

CHAPTER

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 BsA. 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

Proof. That x belong to

E

v.

(1) implies (2) is clear. Assume that (2) holds and let

K . Then x = a/b where a, b E V . If (a) G (6) then a = br where r E V ; hence x 7br/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 B a n d 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

v

100

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 Thus P is the unique maximal ideal of V .

C_

P.

5.5. Proposition. Valuation rings are integrally closed. Proof. Let

x

E

V be a valuation ring and let K be its quotient field. Let

K be integral over V , say Xn

for a,-

1,

+ an-lXn-l+ - +a, = 0, * *

. . ., a, E V . If x 4 V , then x - l x=

and thus x

E

- ( U , , - ~ + ~ , , - ~ X - ~

E

V . Hence

+**-+a,~-~+l),

V , a contradiction.

Let K be a field. A partial homomorphism of K is a homomorphism 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 (hK,)

+

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 (+, K m )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 PZ) 9

1

101

THE DEFINITION OF A VALUATION RING

if V , G V 2and 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 o f 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

+ + a,xm=

a, +a,x

(1)

* * *

1

and b,

(2)

+ b,x-l + .

*

a

+b,x-"=

1.

Choose m and n as small as possible, and suppose m 2 n. Multiply (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 . T h u s ~"=(1 -bO)-lb,xn-l + . * * + ( -bO)-'b, l = clxn-l * .* c,,

+ +

where c,, . . . , c,

E P.

Use this in (1) :

+ = do + d,x + -. + dm

1 = a, +a,x

+a,-,xm-1

*..

*

and d o , . . ., d,-,

E P.

+ amxm-yclxn-l + ..*+ c,)

-1 x m - 1

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.

v

102

(3)

VALUATION RINGS

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 (#', V p ) , 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 P V [ 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 ] / Mgiven by a(a P ) = a M is an injective homomorphism. Clearly, we have V [ x ] / M = u( V / P ) [ x M I . Since V [ x ] / Mis a field, this implies that x M is algebraic over a ( V / P ) .Thus, if VIP+ L is given by $(a P ) = #(a), then 6 a - l : a( VIP)+L can be extended to an injective homomorphism I,!J : V [x]/M-+ L. Let 7~ : V [x] -+ V [x]/Mbe 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.

+

+

+ +

s:

+

+

+

This theorem has an interesting and useful corollary.

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.T h e canonical homomorphism R[x-l]+R[x-l]/P furnishes us with a homomorphism 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 nonunits of V . Then P is the only maximal ideal of V and P = ( a ) where a # 0. By Cdrollary 2.24, Pn = 0. If A is any nonzero proper ideal of V , then A c P, so there is a positive integer n such that

n:=

v

104

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 = a n u w h e r e u 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 ,

or 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 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 consequently V is a valuation ring. T o prove that P is principal, we consider P* = (XI 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, aib, $ P ; that is, there are elements a E P , b E P * such that a b = 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

rEP;

+

k

aai =

C r i jaj ,

i=l,

..., k,

TijE

v.

j=1

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

2

IDEAL THEORY I N 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) (2)

Rad(A) is a prime ideal of V ,

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).

106

v

VALUATION RINGS

(2) Let a and b be elements of V such that a $ B , b $ B . Then a $A", b 6A"' 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 y b = x a b . 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) (2) (3)

If Q is P-primary and x E V\P, then Q = Q(x). 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 . 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 1y 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, Q2be 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 # P2and 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

3

107

VALUATIONS

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 Qa c p.However, by Exercise 5(c), each Qa contains a power of Q ; thus Qa = Qn. Hence, by (2) of Theorem 5.10, 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 Qa.

na na

na

,

na

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 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

as

+

+

-

, a,) < (81,. - - , Bn) if ~ , < , 8 ~or if, for some K > 1 , a i = & for i = l , ..., K - 1 and ak .

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. I n 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:

4 (1) (2) (3)

115

PROLONGATION OF VALUATIONS

v' is a prolongation of v . V ' n K = V.

P'n Y=P.

Proof. (1) 3 (2). Let v' be a prolongation of v ; then there is an orderpreserving 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. (3) (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

+

+

+

1G'-'t#$(.(u))

= $'-'+(uU)= $'-1(uU') = v'(u)

for all a E K. Thus D' is a prolongation of v . 5 3 2 . 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 prolongation 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

v

116

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'(bjai) 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 . T h e 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'

4

117

PROLONGATION OF VALUATIONS

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 , X y ) 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 G c * * *Hc, n G s G . Otherwise, let j be the largest integer such that H j c G ; then 0 c HI c - * -Hc, 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 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

s

+

+

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.

v

118

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 prolongation 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 it 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) Show that if V’ is an overring of V , then there is a prime ideal P of V such that V’ = V , . (b) 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 . (c) Show that if V has rank Y and V’ has rank r‘, then W has rank r r‘.

+

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

119

EXERCISES

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.] (b) Extend (a) as follows: let R be a subring of a field K and let 0 c PI c 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. 4. Places and valuation rings. where OD Let K be a field and consider the set K u {a}, 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 +(a) = w, and +(l/a)= l/+(a) for all that #(O)=O, - a -

+

+

+

U€KU

{a}.

(b) 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}-+Ku {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. be a place of K having values in R. Let V = (c) Let (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}. 5. Ideals of valuation rings. 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. 6. 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 ,

+

+

+

v

120

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 . (b) Let R be an integral domain and let P = (a) be a nonzero P" = 0. Show principal prime ideal of R such that that R, is a Noetherian valuation ring.

n:=

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 . Q" = 0. (a) Show that If Q n = Q n + l for some positive integer n, show that (b) Q = Qa = P. ( c ) If Q G Q1 c P, show that Qln c Q for some positive

n:==l

integer n. (d) Show that if Q c P , then Qz c QP. (e) If Q c Ql, show that Q : Q1= Q implies that Q1 = P = P2.

9.

Valuations. Let z, be a valuation on a field K .

121

EXERCISES

+

(a) Show that if .(a) # v(b), then v(a 6) = min(v(a), v(b)}. (b) Show that if K is a finite field, then .(a) = 0 for all nonzero a E K. (c) 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. (d) 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.

+

10. Valuations on the field of rational numbers. (a) 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. (b) Show that each nontrivial valuation on the field of rational numbers is equivalent to up for some prime p.

11. Valuations and places on K ( X ) . Let K be a field and X an indeterminate. 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 iff(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

v

122

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 Xi: is a nonzero monomial, then v(m)= Cf=,nigr; and if f = m 1 mh, where m,, . . . , m, are distinct nonzero monomials, = 1, . . . ,h). It is now necessary to verify then v(f) = min{v(mi)(i 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,, . . . ,mhare nonzero monomials, then v(m, * * * nth) 2 min{v(mi) I i = 1, . . ., h}. (b) If f h f 2 E K[{X,lg E GI19 then .(fi +fi) 2 min{v(f,), v(f2)}.This verifies (iii). (c) If f i , f 2 E,K[{X,Ig6 GI19 then 7 4 f J - 2 ) = v(f1) u ( f 2 ) . [Hint. Write fi= m, f . . f m, and f2 = m,' .* m,' as sums of distinct monomials, where * a -

+ + e . 0

+ +

+ + +

v(f,)=v(m,)=...=Zl(mh)

< v ( m k + l ) s* . * s v ( m , )

and

< v ( m ; + , ) *~* - < v(ms'}. Set f," = m, + --.+ m, and f2* = m,' m,', and consider f1f2 =f1"f2" ( f 1 f 2 -fl"f2").] v(fi)=v(m,')=...=v(m,')

+

+ +

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) If a, b E R, a E A , and v(b) 2 v(a),then b E A. (3) If V is the valuation ring of D, then AV n R = A.

123

EXERCISES

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 = 2 0 2 , ordered lexicographically, and let z, be the valuation on the quotient field of K [ X , Y ] ,having G as its value group, constructed as in Exercise 12. Show that ( X z ,X U ) is a v-ideal; it is not primary [see Exercise ll(b) of Chapter 111.)

14. Ordered Abelian groups. (a) Let G be an ordered Abelian group and let H be a subgroup 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.] (b) 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, . (c) Show that the direct sum of Y copies of the additive group of integers, ordered lexicographically, has rank r .

+

CHAPTER

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

I n 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

1

125

FRACTIONAL IDEALS

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 = s x A 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. Furthermore, if B contains a regular element of R, then [ A : B l = ( x l x ~ Kand 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 b d [ 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. I n 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 A B = 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 a n 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 a n ideal C of R such that A = BC. If A E %(R),then A is invertible if and only if there is a fractional ideal B of R such that A B is principal and generated by a regular element of K.

VI

126

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 = a, bi For each

..

..

C?,,

.

A , & , € R,fori= 1, . . ., n , a n d x = 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. XE

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 f B 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 A i is invertible for i = 1, , . .,n. If the product A is invertible, then A;l=A-lA, ... An

-

2 PRUFER D O M A I N S

+

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 investigate integral domains for which the converse is true. 6.5. Definition. An integral domain R is a Prufer domain ;f each

nonzero Jinitelygenerated ideal of R is invertible.

2

127

PRUFER DOMAINS

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 statements are equivalent :

R is a Priifer domain. Every nonzero ideal of R generated by two elements is invertible. I f A B = 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 ) = A B r\ A C for all ideals A, B, C of R. ( A+ B)(An 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.

(1)

We begin by showing that (1) and (2) are equivalent. Clearly

=+ (2).

(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 A=(C~ ..., , cn-1), D = f ~ l cn>, r

Then

+

B = ( c , , ..., c,), E = c,A-lD-l ~~3-lD-l.

+

+ + +

C E = ( 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-~).

+

+

VI PRUFER AND DEDEKIND DOMAINS

128

Since c, B-I

5

R and c1A-l E R, this gives CE=

+c,D-'=

~1D-l

( c ~ cn)D-'= ,

R.

Thus C is invertible. (1) (3). Suppose AB = AC where A is finitely generated and nonzero. If A is invertible, then

*

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 A C = A B + A C = A ( B + C ) ; h e n c e C = 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 fi 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)(BRpn CR,) = (AW(BRP)f-l(A W (C R , ) = (AB)Rpn (AC)R,

= (AB n AC)R,.

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)(An B ) = ( 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

2

129

PRUFER DOMAINS

c2 = 0, then C is invertible ; hence we shall Then A = (c,) and B = (c2) are invertible,

C ( A n B)B-lA-l= ( A

assume c1 # 0 and c2 # 0. and

+ B)(An 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 BIf .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. (4) + (9). If we assume (4), then for every maximal ideal P of R, and for ideals A, B, C of R with A and B finitely generated, ( C : ( A n B))RpE 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.

+

+

VI

130

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: (4b) = (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 = ((4n (4):((4 n (b))

((4

(4+ ((4 (b)):(4

= n (W: = (b): (a) (u): (b).

+

Proceed as above. (10) + (4). Assume (10) holds. Let P be a proper prime ideal of R and let a, b E R. Since a E (b) (a - b) we have

+

(4= (4 n ((4+ (Q - b)) = ((4n (4)+ ((4n (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

2

131

PRUFER DOMAINS

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 contained 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) (2)

If Q is P-primary and x E R\P, then Q = Q [Q + (x)]. The product of P-primary ideals of R is P-primary.

+

Proof. (1) We will show that QRM = [Q2 Q(x)]RMfor 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-primaryideal 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 QzRM are PR,-primary ideals of the valuation ring RM . By Theorem 5.11, QiRM Q2RM = QiQ2RM 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 properly 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)P1R, . 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 correspondence being Q t t Q". By Theorem 6.8, this correspondence preserves products. It also preserves residuals ; this follows from

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"T h e 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) (2) (3) (4) (5) (6)

fi:= Q" is a prime ideal of R ; 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 O F PRUFER DOMAINS

I n 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:a R p E PR, , the unique maximal ideal of R, . Now consider the ring RP[U/bI

I f ( x E) RP[XI)*

={ f ( Q / b )

(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

3

133

OVERRINGS OF PRUFER DOMAINS

(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

+- - - +

+- +

. x, a,, E bR, : a R , . Note that do is not a where d, = xlalj unit in R, , since b R , : 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 overring of R and let P be a maximal ideal of T . Then T pis 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 impossible 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 is a Priifu

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,. PEA

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.

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 intersection 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

+

+- +

( b O / u ) 2 n- (bo/U)2?+l

+ b,bo(bo/a)2n-z + -.+ b,b?-1= *

0.

Thus bo/ais integral over R, ;hence bola E R, . If bo/ais 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 equivalence of some of these conditions will follow from Theorem 6.6.We shall also obtain some results concerning overrings and extensions of Dedekind domains.

4

135

DEDEKIND DOMAINS

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 - Pk = P2' - Ph', and we can repeat the argument. Since the Pi and P,' are proper ideals of R we must have h = lz.

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 invertible, 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 Q i 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'

Q,'

1

--

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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PRUFER AND DEDEKIND DOMAINS

Hence by Proposition 6.15, n = Zk, and we may so number the Q i that for i = 1, .. ., k , Q Z i = Q Z i = P i . 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 is 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 A B c R and A c A B since R c B. If A c AB, then AB is a product of prime ideals of R

4

137

DEDEKIND DOMAINS

and the same is true of (AB)P= A. Since this is not true, we have A = A B and so A P = 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 conditions for a Noetherian integral domain to be a Dedekind domain. 6.20. Theorem. If R is a Noetherian integral domain, then thefollowing statements are equivalent :

R is a Dedekind domain. R is integrally closed and every nonzeroproper 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 Cfor 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 is 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.

+

+

+

+

VI

138

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.

(1) (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 P n + 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 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 induction we show that P' = aRp P'" for all positive integers n. Hence

n,"=

+

+

n(aR, +P'"). m

PI=

n=l

But then PlaR,

=

( fi (aR, +P'") n=l

n +P ' ~ ) / ~ R ~ n ( P ' / ~ R , ) ~0,= m

=

( u ~ p

n=1 m

=

n=l

by Corollary 2.24, since P'IaR, is the unique maximal ideal of the

4

DEDEKIND DOMAINS

139

Noetherian ring R,laR,. Therefore P' = aR,, and since P' is precisely 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 P2is 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/P2is a vector space over the field RIP. If P = 0, then the desired conclusion certainly holds. Suppose P f 0. The set of subspaces of PIP2,each being of the form A I P where A is an ideal of R with P2s A c P, is totally ordered by inclusion by (14). Hence P/P2is one-dimensional. Therefore if A is an ideal of R with P 2 E A G P, then either A l p 2 = P2/P2or 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. I n 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 valuation 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.

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 =R , by maximal ideal of T . Since R is a Prufer domain, TIM 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 , ,, = T Mis a Noetherian valuation ring.

Proof. Let

VI

140

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

MS(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 assumption 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.

5

141

EXTENSION OF DEDEKIND DOMAINS

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

+c0sk--l=0.

+.**+(c1s'C-2)(sa)

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 Noetherian. Let wl,. .., w,E K and consider the following n equations in n unknowns : n

1 TK,/,(uiuj)xj

=z

wi

i = 1, .. ., n.

9

j=1

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

i= 1, .. ., n.

7

Thus, for fixed j , the n equations T,,/duix)

i = 1, ...,n,

=Sij,

+ - + c, u,'

have a unique common solution ui'. Suppose clul' where cl,. . . , c, E K . Then, for i= 1, . . ., n, 0 = TK,/K(ut(c1~1'

c

+ + *

* a

n

=

j=l

cj TK',K(Ut u;) = c*

cnun'))

.

* *

=0

VI

142

PRUFER AND DEDEKIND DOMAINS

Therefore ul',.. ., u,I are linearly independent over K , and consequently 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 ~ ~*

c

* *

anUn)uj')

n

=

a, T K ' , K ( U ,

Uj')

= aj E R ;

i=l

thus ui'E M" for j = 1,. . . , n. Finally let b E M" and write b = blul'+...+b,u,'whereb,, ..., ~ , E KT. h e n f o r i = l , ..., n,

+ - +b , ~ , ' ) )

E R.

b, = Txe,K(~l(bl~l'

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 apeE 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 upE 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. 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 Proof. First we note that

+- +

5

143

EXTENSION OF DEDEKIND DOMAINS

alPblP + * - + akPbkP= 1 . For i = 1, . . ., k , biP E K and bipaE R” n K’= R’ for all a E A ; hence biP E [R’:A ] .Since aipE 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 quotientfield of R,

K’ a j n i t e 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 j n i t e 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. 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 P i n 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 Proof.

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),

144

VI

PRUFER AND DEDEKIND DOMAINS

which is not true. Thus ~ " ( a2) 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 Proposition 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 {Mnlc( 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 IXE I , prove that A is invertible. (c) Assume that R is integrally closed. If a, b E R, with a regular, 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, P" is a prime ideal. (b) Show that the powers of P are P-primary and every P-primary 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. +

+

+

n;=

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

145

EXERCISES

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 R, is a Prufer common quotient field K. Show that domain.

u

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, bER. 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 $ A i for i = 1, . . . , n, then x1 - * x, $ A, * * . A , . In particular, xsE AISimplies x E A,. (b) Use (a) to prove that if (a",b") is the intersection of valuation 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.

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’ = Alr, is called the completion of A and if A = A’, A is called complete. Let A* = (AVan 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) Each principal ideal of R is complete. (3) There exists a nonzero principal ideal of R which is complete. (4) Each principal ideal of R is an intersection of valuation ideals of R. (b) Prove the following conditions are equivalent: (1) R is a Prufer domain. (2) Each ideal of R is complete. (3) Each finitely generated ideal of R is an intersection of valuation ideals.

0

0

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 submodules, show that if M is without torsion, then there is an exact sequence 0 - t M - t K Q RM - t K l R O RM+O.

(e) 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,

147

EXERCISES

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) Show that every principal ideal of R is quasi-principal. (b) Show that the product of quasi-principal ideals of R is quasi-principal. (c) 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.

+

+

12. 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. (c) 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

148

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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 fractional 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 i=l, ...,K. If A is a fractional ideal of R which can be canceled, if A = A, + - * * + A,, for fractional ideals A l , . . . , A,, of R, and if K is a positive integer, prove that Ak = AIk * * A",. Give an example of an ideal which contains a regular element, but which can not be canceled.

+

+

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.

149

EXERCISES

(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 K O 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 ~ f ( a ) Rif' and only if a E R'. (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.]

16. T h e 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 x E K and xAn E R}. (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' R and Rip= R, .

u,"= I

VI

PRUFER AND DEDEKIND DOMAINS

Under the hypothesis of (d), show that if AR' = R', then = 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

R'

A = (u,, u 2 , . . ., u,,).

n?=

Prove that T ( A )= 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)= 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 {XI 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.

n

,

17. T h e 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 R= T((xa))* (b) Let x be a nonzero element of R and let S = { x " l n = l , 2 , ...}.

n

Show that T((x))= S-lR. (c) Show that R is integrally closed if and only if the transforms of the principal ideals generated by nonzero nonunits 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. 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 (Bn C )= ( A B) n ( A C).

+

+

+

+

151

EXERCISES

(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. (c) 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.

+

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 isomorphism. Denote the equivalence class containing M by [ M I . 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. (b) Show that if R is Noetherian, then X ( R ) is generated by where P runs through the set of proper nonthe set {[RIP]}, zero prime ideals of R. (c) Show that if R is a Dedekind domain, then X ( R ) z g ( R ) . (d) Show that if R is a field, then X ( R ) 2 2.

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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. (c) 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). 22. Rings having few zero-divisors. I n 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 zerodivisors of R is a union of a finite number of prime ideals of R. Thus, Noetherian rings and integral domains have few zerodivisors. (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 zerodivisors. (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.

+

153

EXERCISES

(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'.

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23. A ring having more than a few zero-divisors. I n this exercise, an example is given which shows that, in general, the properties enjoyed by rings which have few zerodivisors 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 mI :n. Let Q be the ideal of R generated by { z g ] g EGI. If (c) 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.

+

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.

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,y 2 ) for every pair of regular elements x, y E M , then R is a quasivaluation 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 = R s ( p )(For . the notation used here, see Exercise 10 of Chapter 111.) (f) Let A be a regular ideal of R. Show that A" is a prime ideal of R. (g) Give an example of a quasi-valuation ring which is not a valuation ring.

n:==l

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 quasivaluation ring for every regular maximal ideal M of R.

155

EXERCISES

(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 ) = A B 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 (c) 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.

+

CHAPTER

VII Dimension of Commutative Rings

1 THE KRULL DIMENSION

Let R be a ring and consider a chain

Poc 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 Poand 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

1

157

THE KRULL DIMENSION

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 is Noetherian and dim R = 0.

if

and.only if R

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 PIP2E PI c R.

There are no ideals strictly between P, and R. For i = 1, . .., n - 1, P, PiIP, 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

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 P I , . . . , 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 A T# 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. q e 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).

1

159

THE KRULL DIMENSION

7.5. Lemma. Let R be a Noetherian ring and let a EJ(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

n m

A=

(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' E J ( 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 is a minimal prime divisor of (a), then ht P = 1. Proof. Since ht P = ht P R p , 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. It 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

VII

160

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

+

+

+

+

+

+

+

P1' = P,. 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 . 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 ht P, = co (see Exercise 3). ideals such that There is a result which is almost the converse of Theorem 7.7. 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.

2

161

THE KRULL DIMENSION OF A POLYNOMIAL RING

P,,. . ., P, be the minimal prime divisors of 0. Since ht A 2 1, A $ Pifor 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 a j 4 Qi for i = 1, . . . , n. Let P' be a minimal prime divisor of (a,, . . . , u j ) , 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' 2j. 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 . Proof. Let


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 Rb = B. Thus A G B.

n,

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

+

+

k, = min{m, , n,} for all c1 E I . An element m E 2'" is called positive if m > 0. ha = max{m, , n,}

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

177

KRULL DOMAINS

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 Pabe 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 written 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

%Pa,

a El

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.

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 Ip 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={alaEI

and n,-n,'>O}.

Then with ma= n, - n,' whenever wise, we have

ci E J

and ma= n,' - nu other-

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 E J , then p a 5 p, for some fl 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 diva=

1 napa, aeI

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 =

nipa, ael

then div ab = div a + div b =

Hence va(ab) = nu

C

as1

(n,

+ na')pa.

+ n,' = u,(a) + v,(b). We know

that (div u ) n

2

179

ESSENTIAL VALUATIONS

+

(div b) = div(Ra Rb), and R(a 2 div(Ra Rb). Furthermore

+

Hence either a

+ b) G Ra + Rb, so that div(a + b)

+ b = 0, or

div(a

+b)

mapa 2

= a E I

C (min{n, ,n,")p,.

aEI

+

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 completely 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.

VIII

180

KRULL DOMAINS

We have shown that if a E 9(R),then we can write uniquely

C

nupa,

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 if A 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, a E I}be a set of integers only a finite number of which are not zero. Then

I

A = { a l a ~ K and v,(u)>n,

for all

CLE~}

divisorial fractional ideal of R ; it is the one whose divisor is -is a rna Pa -

Lac

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 E 1.-Therefore

a

div A =

C

va(A)pu.

as1

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. If A is a fractional ideal of R, then A = Ca Ra. Hence div A is the greatest lower bound of the set {div Ra a E A, a # O}. Therefore div A = C (min{w,(a) I a E A})p, .

I

a E I

On comparing the two expressions that we have for div A, we conclude that for all a E I wa(A)= min{v,(a) I a E A}. 8.14. Proposition. Each of the essential valuations of R is normed.

< 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 . Proof. Let a E I;then p a

2

181

ESSENTIAL VALUATIONS

Let 4: 9 ( R )-+ 2'')be defined by +(div A), = va(A) for all ci E I . We have seen that r$ is a one-to-one mapping. 8.1 5. Proposition. The mapping from g(R) onto 2")

4 is an order-preserving isomorphism

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) v a ( 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 V a ( 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 v a ( 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. Furthermore 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,= R P a ,we note that since V , is a Noetherian valuation ring it is sufficient, by Exercise l(a) of Chapter V, to show that VaE Rpac K. Let a E Vaand 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 = Pafor some c( E I .

182

VIII

Proof. Suppose that

P is divisorial. Since

div P = C n, pa = (IEI

nae

KRULL DOMAINS

C nu div Pa = C div P,""= div n P,"",

aEI

(IEI

(IEI

we have I P,""G P. Then PaG 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, PaR,, is the only nonzero proper prime ideal of R,, .Then by Corollary 3.11, 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, = R,, . Hence if a E J, R, 5 R,, and so Pa G P. Since P is minimal, P = P a .

naEJ

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) Every 08' is a prolongation of some z), . (ii) Every prolongation to K' of each v, is equivalent to exactly one v8'.

n8

v8'. Let Let v,' be the valuation ring of VB' and set R" = a E R', a # 0 ; then there are elements b, , b,, ..., b,- E R such that

b, + bla

+ + bn-1an-l + a" = 0. * *

2

183

ESSENTIAL VALUATIONS

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 l a )., ..,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". O n 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".I n doing so, we have verified that (I) holds for R' and the family of valuations {vgrIS 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'1/3E J } because, by Theorem 6.25, each v, has only a finite number of prolongations in the family (vgr/3 €1). If vgr(a)> 0, then

I

wg'(bo) < min{v,'(b,a), ... , vg'(bnvBr(an)} 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 - I R is a factorial ring, then R is a factorial ring.

Combining the results of Theorems 8.28 and 8.31, we have:

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.

8.33. Corollary. If

If R is a Krull domain, then it follows from Theorem 8.12 that R satisfies : Maximum condition on principal ideals (MPI) : every nonempty set of principal ideals of R has a maximal element.

R is an integral domain which is not afield, then the following statements are equivalent:

8.34. Theorem. If

(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 is 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

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 irreducible 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 U P 1 * * - 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= q l , and so up2 * p, = vq, * * q n . 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 = ~(ip).

n

b E B

b#O

195

EXERCISES

(b) 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. (d) Show that R is integrally closed if and only if [A:A] = R for every finitely generated nonzero ideal A of R.

(c)

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 P I , ...,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. 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 n p R p= 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

(a)

[R:PI = R. 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. (d) 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.

(c)

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 R, a Krull domain which is a subring of K . Let 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

naE,

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 S - 1R =

n va

asf

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 Pais 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,, . (d) Show that ORpa= R, where the intersection is over all complex numbers. (e) 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.

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 satisfying (I) and (11). Assume that for each fs E I we have

where V , is the valuation ring of v=. Show that (vnlCL E I} is the family of essential valuations of R. (b) 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).

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 ) l u ~ 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. (d) 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

VIII

198

KRULL DOMAINS

9. T h e 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 a E K s u c h 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 a E I\{a1, . 7 a,>. (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. (c) Show that every divisorial fractional ideal of R is an intersection of two principal fractional ideals of R. e

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. (c) 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 , = V,, where V , is the valuation ring of v, . (2) For each a E I , V , = R P a ,where Pa = {xi x E R and va(x) > O}.

naEIM

199

EXERCISES

(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) If R is almost Krull, show that R[X]is almost Krull. (e) If R is almost Krull, show that the integral closure of R in a finite extension of the quotient field of R is almost (f)

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.

12. 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 ( ~>. Show that if A is a nonzero ideal of R , then A=

n~ y ) .

U E I

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 M a k and A $ M ; + l ;show that k =f,(A). Show that for nonzero x,y E R and for a E I, fa(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 quotient field of R such that v , ( x ) = f , ( x ) for all nonzero x E R . Show that RMais 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.

222 4.

Ix

GENERALIZATIONS OF DEDEKIND DOMAINS

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 nonzero 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 #

PEA^

0

R P .

PEAZ

(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 (#). (c) 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. (d) Suppose R is a Prufer domain and dim R = 1. Show that if R has property ( # ) , then every overring of R has property ( #). (e) 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 a E I , M , is the radical of an ideal with two generators. [Hint. For each /IE 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

223

EXERCISES

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 a n d s E M,}. Let T = R \ u BE , M , and set R' = R, . Show that there is an element t E R such that t E M , R'\

u

M, R'.

,€I

Bfa

(f)

Then show that M a =Rad((s, t)).] Show that if an almost Dedekind domain R has property (#), then R is a Dedekind domain.

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). A", and if AB is an intersection of (d) Show that if B ==:)( 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 PCM. (f) Let R and M be as in (e). Assume that for each positive integer n, Mn+Ic 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.

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+Ithen B $ M . (d) Let M be a maximal ideal of R such that M" # Mn for each positive integer n. Show that 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. 10. Noetherian multiplication rings. (a) 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.

n:=l

+

EXERCISES

225

(b) Show that R is a Noetherian multiplication ring if and only if it is a ZPI-ring. (c) 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. (b) 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.] (c) Show that the hypothesis concerning maximal ideals in (a) and (b) is necessary.

CHAPTER

X Priifer Rings

I n 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

I n 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

1

227

VALUATION PAIRS

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, is a prime ideal of R, . Furthermore, (f R,# R and if A is an ideal of R such that A s R,,thenA~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 is a p r o p e ~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 is 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=

and we define a relation

-

{XI

zER

and xz E P } ,

on R by setting x

-y

if [P :

= [P :yIR.

X

228

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 : 1IR 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 :1IRc [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 ( y z ) 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))xE 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. T o 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 :zIRG [P : xIR and [P : Z ] R E [P : y I R ; hence [P : zIRc [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)

(4) (5)

R\A is closed under multiplication. R\P is closed under multiplication. IfxERandx”EA,thenxEA. A = { x l x ~ and R xy E P for a l l y E P } . I f A # R, then P= {xlx E A and xy E A for somey 4 A}.

1

229

VALUATION PAIRS

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)< (4Pz> 9

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. T h e 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 5R 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 6P. Then

wherep, , . . . ,p , E P, and n is chosen as small as possible. Multiplying by p;-l, we obtain (xp,),

+n&pn)ip:: i=O

-

-

*pi- ap; - 1 = 0.

X

230 Hence, xp, is integral over A , and therefore xp,

EA,

PRUFER RINGS

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 fi E H . We shall establish a correspondence between the v-closed proper prime ideals of R, and the isolated subgroups of Go .

4 be an order-preserving homomorphism from G, into an ordered Abelian group, and set +( co) = co. Then :

10.8. Proposition. Let

(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, .

+

(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 5/35 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,, . Therefore, 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. Proof.

1

231

VALUATION PAIRS

10.9. Proposition. Let H be an isolated subgroup of G,. Then there is a surjective order-preserving homomorphism f r o m G, onto an ordered Abelian group such that +v is a valuation on R and H = Ker 9.

+

Proof. Let Set a H

+

+ be the canonical homomorphism from G, onto G,/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. T h e assertion of the proposition follows immediately.

If H and have

+ are as in Proposition 10.9, we set uH= +v. Then we

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-

PUH

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 correspondence is order-reversing. Let a, E H and suppose a = v(x) and /3 = v ( y ) . By definition of H , -a E H. I n 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

+

+

X

232

PRUFER RINGS

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. v(z) for some

+

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. I n 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

+

a =f@)

+ xg(x, Y ) + w ,

and

+

b =f’@) xg’(x, Y )

+ w‘,

wheref(Y),f‘( Y), X g ( 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,;

+

+

+

2

233

COUNTEREXAMPLES

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 ) z x = 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/bE p.

Next, we show that wy-" E P. Let g,, . . . ,g k be those elements of G such that zgl, . . . , xgkoccur in the generation of w as an element of Q. Since gi(x, y) E A\P for i = 1, . . . , R, we have g,(x, y ) E A\P. But 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 +' Y k ( X , Y ) ) ,where r' is a nonzero element of F [ X ]and k(X, Y ) E F [X, Y ] .Then

nf=

(nf=

+

+ xyp-"(r' + y k ( x , y ) ) P. If p - rn 2 0, then Y + xyp-"(~'+ yk(x, y ) ) P n R = N , which is not true since Y # 0. If p m < 0, then +yk(x,y ) ) E N , which is r

E

E

-

X(Y'

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/pE 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)", * * *

X

234

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-2E 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

(bna/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 R i s l .

If P is a proper prime ideal of R, it is not necessarily true that PRipIis a prime ideal of Ri p,.In order to study the relationship between the prime ideals of Rip]and those of R, the following operation 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. T h e details are left to the reader. However, it is not necessarily true that [PIR,,, is a maximal ideal of Rl,, when P is

3

235

LARGE QUOTIENT RINGS

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. T h e 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 following statements are equivalent :

(1) (2)

(3)

( R C p[1P, I R I Pis] )a valuationpair of the totalquotient ring of R. 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. 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 , Let x,y E R and ring of R which is determined by ( R I P ][P]R[,]). assume v ( x ) 2 v ( y ) . If v ( y )= m, then

x,Y E W'IR,,,)

n R = [C(p)lR,pln 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 . is a regular element ring of R which does not belong to R f p IThere r E R such that ry E R. If ry E C(P),then there exists s E R\P such Y R ; this implies that y E R i p , ,which is not the case. that sy = S Y ~ / E 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

X

236

PRUFER RINGS

a E [PIR,,, , and ay E R,,,\[P]R,,, since [PIR,,, n R = P. Therefore, ( R I P ,[, P ] R f pis l ) 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 E A R , . 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 y R , or y R , 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) = A B 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 + x 3 c , where xiE (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.

+

4

237

PRUFER RINGS

+

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, .

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 Proof. Since (ab)(a,b, c)

+

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) (2)

R is a Prufer ring. For every maximal ideal P of R, ( R I P ,[P]Rcp,) , is a valuation pair of the total quotient ring of R.

X

PRUFER RINGS

Every overring of R is a j l a t R-module. Every overring of R is 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) = A B for all ideals A, B of R, with A or B regular. Every regular ideal of R generated by two elements is invertible. If 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= BC. ( 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.

+

+

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 lSince . (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. (2) (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 6R,,, , 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 . there exists t E T\M such that xt E T. suppose that x E T I M IThen But b E P c M and bx E R \ P c T\M. Hence bxt E M n (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.

4

239

PRUFER RINGS

( 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? TQG 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 [ is ~ 2integrally ] closed, x E RIp1[z2].Let z= a,

+ alzz +

.**

a, E RIP],

+a,.+,

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

+ + up-3al(unz)2 - up-2(Unz) + a, a2-1 = 0,

(anz)2'L

* * *

so a, z E RIP,. Then, multiplying the first equation by a:-', we have (a,z2)n

+

*.*

+a:-2a1(a,z2)

+

(U,u;-l

- a nn- 2 (a n z> > = O

Since R,,, is integrally closed, this implies that a, x 2 E RIP]. But then an z2 E RIP,, and

a,-

,+

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

+

a1z E RIPI\[~lRIPl*

(2) + (5), (11)-(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 . (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

X

240

PRUFER RINGS

c be a regular element of A . Let B = [(c): A ] .T o prove that A is invertible, it is sufficient to show that A B = (c) ; furthermore, since

A B 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)Rp

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 y 1 . * 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

-

( A B n AC)R, = ABR, n ACR, = (AR,)(CR,) = (ARp)(BRp n CR,)= A(B n C)R,.

If the latter holds, we have likewise ( A B n AC)R, = A ( B n C)R, . Since P is an arbitrary proper prime ideal of R, we have A B 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

+

+

4

241

PRUFER RINGS

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). I n each case, we shall verify the second of the equivalent conditions of Theorem 10.14, and conclude that (R,,, , [ P ] R L pisl )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 (11) 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, (11) implies (2). The proof that (12) implies (2) is the same, for if (12) holds, then R = ((4n (9) : ((4 n (9 = (4 : (r)

+ ( r ) : (4

and R = ((x, 4 n (YN : ((x,

Y)

n ( Y ) )= (x,4 : (Y)

+ (Y): (x,4.

Now assume that (13) holds. Since

(4n (Y>4 + (4n (Y - 4 = (4 f-3 (Y) + (4 n (I +)(4n (Y -4,

(4= (4n ((Y, r ) + (x -Y))

=

X

242

PRUFER RINGS

+ +

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)

(2) (3) (4)

for every regular maximal ideal M of T , T[Mj= R,,, , where

P = M n R , and M = [PIR,,, n T , if P is a regular prime ideal of R, then P E A if and only if TE , and T = , ( A n R)T = A for every regular d e a l A of T , and {PTI P E A} is the set of regular prime ideals of T.

np

Proof. B y Corollary 10.19, T is a Prufer ring. Let M be a regular [ M ] T , M lis) a valuation pair. Since maximal ideal of T. Then (TIMI, R I P5 ] 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 , P= l [MI T,,] n R I P [see ] Exercise 5(a)]. Then TIMI = R,,, and [ M ] T I M =1[QIR,,,. Thus,

Q = [Q]Rrgln R = [ M ] T I Mn, R = [ M I T I MnI T n R = M n R = P , and so TIMI = R I P and ] M = [ M ] T , Mn 1 T = [PIR,,, (7 T .

4

243

PRUFER RINGS

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 = T I M where I, M runs over the set of regular maximal ideals of T. Therefore,

n

Now let A be a regular ideal of T . By Exercise 6(c), A=

n

( ~ ~ A In T ,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 ] T r Mthen 1 , there exists s E R \ P such that xs E A. But A c R I P ,so , there exists t E R\P such that xst E A n R. Consequently, x E [An R]RIPI = [(An R ) T ] T C M I .

Since (A n R)T c A we have, therefore, [(An 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, [AITIM1=

[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 Pobe 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 Poand P . Furthermore, Pois 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

X

244

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 Q1and 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 :

Q n is a prime ideal. Qn = Q"" for some n 2 1 implies that Q = Qz = P. Q L Q1c P implies that Qln c Q for some n. P2c 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.

245

EXERCISES

Give the details of the proofs of the following facts used in the proof of Theorem 10.4: (1) 5 is well-defined; (2) If).(v # +), then (I. < V(Y) or v ( y ) of regular nonunits of R, then R = T((x,)). If no maximal ideal of R contains all of the regular nonunits 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.

17. 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. (b) 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 . (c) 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 maximal 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. (e) Prove that if (R, P ) is a Prufer valuation pair, then every overring of R is a Prufer valuation ring. 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 A B E 9. T h e Y-transform R, of the ring R is the set

R y = {x E K ~ xGAR for some A E Y}.

25 1

EXERCISES

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) Show that large quotient rings and ideal transforms are generalized transforms. (b) 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. (4) For p 9, RfP,= (R,)p,, (c) Let P E F.Show that Q-Q, is a one-to-one correspondence 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. (d) 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 A T = T for all A E Y. (3) For each proper prime ideal & of I T, T [ M ] = RIM R 1 . [Hint. Use Exercise 11.1 A

Appendix: Decomposition of Ideals in Noncommutative Rings

It is natural to ask if the primary decomposition theory of commutative 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 decomposition 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,bfI 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 H K . 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 A B c P, where A and B are ideals of R, implies that either A G P or B s P. 252

DECOMPOSITION OF IDEALS IN NONCOMMUTATIVE RINGS

25 3

A.Z. Proposition. Let P be an ideal of R. Then P is 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), is 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 intersection of all of the prime ideals of R that contain A.

We must show that if a 6Rad(A), then there is a prime ideal P of R such that A G P and a 6P. 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 Proof.

XZY E

(P

+ (b))z(P+ (4)

= PxP G

P

+PZC+ 6zP + (b)Z(c)

+ RbRcR G P ,

which contradicts the fact that M n P is empty. Thus, bRc implies b E P or c E P ; hence P is a prime ideal.

cP

A.5. Corollary. If A is an ideal of R, then Rad(A) is an ideal of R.

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 condition 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 Proposition A.4. A.9. Proposition. Let R be a Noetherian ring and let Q be an ideal of

R. Then Q is 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

x

o

z

Y

O Y

0

z

o z o

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 Hax 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(ax

+ 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 . Therefore, 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 ) .

T h e left residual of A by B is defined to be the set A * . B = ( X I X E Rand 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.

.

-

. - .-

If R is a Noetherian ring, then the following statements are equivalent:

A.lO. Theorem.

(I) (2)

Every ideal of R is an intersection of aJinite number of primary ideals. If A and B are ideals of R, then there is a positive integer n such that A" n B c AB.

256

APPENDIX

If A and B are ideals of R, then there is a positive integer n such that A = ( A Bn)n ( A . B").

(3)

+

(1) + (2). Let A and B be ideals of R and assume that AB = n Qk, where Q i is aprimary ideal of R f o r 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 Q i, set ni = 1, and let n = max(n,, . . . ,nk). Then

Proof.

Q1n

- * *

A" n B

E Q1n

... n Q k =

AB.

(2) + (3). Assume (2) holds and let A and B be ideals of R. We have A B E A . B25 . . . , 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

.

-

-

.

BStn ( A . B t )E Bt(A Bt) G A.

.

If n > s t , then B" n ( A * B") s BStn ( 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 + (Bnn ( 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 A n$ 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.

257

DECOMPOSITION OF IDEALS IN NONCOMMUTATIVE RINGS

Throughout the rest of the appendix, R is a Noetherian ring. Let A be an ideal of R. T h e 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 B') n C c A ; ( A . (B+B')) n C c A. Then ( A : B) n ( 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 ; it is denotedby Ter(A). A n ideal A of R is a tertiary ideal if B C _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

.

*

Since A is irreducible, this implies that D Therefore, B G Ter(A).

+

B ) n (D A).

+ A = A, that is, D c A.

We can now give one more necessary and sufficient for the existence 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

25 8

APPENDIX

finite number of primary ideals of R. Let A be an ideal of R. By Theorem 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 zTer(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. Therefore, 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 Ticontains the intersection of the remaining T i . Then,

Ter(A) = (Ter( TI)) n

- - - n (Ter(Tk)).

259

DECOMPOSITION OF IDEALS I N NONCOMMUTATIVE RINGS

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 Tln * n T, = A and (a)(b,) G A. Therefore, a E Ter(A). Conversely, let a E Ter(A). Let b E T , n * * n Tkand 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,)).

-

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,).

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.

Proof.

A tertiary decomposition A = T, n . n T, of an ideal A of R is called reduced if: (i) no T i contains the intersection of the remaining Ti, and (ii) 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'.

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’

T and we have A = T n ( B n B ’ ) = B n B’.

E

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 T i 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 , ..., m a n d 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) # T e r ( 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,,

261

DECOMPOSITION OF IDEALS I N NONCOMMUTATIVE RINGS

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. T h e 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.

T h e 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 C, then C = R and B G A, contrary to fact. Hence, A c A C. Proof.

. .

.

.

-

.

.

.

.

9

A.20. Proposition.

If A is a proper ideal of R, then A has a prime

proper left residual.

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. Proof.

-

.

A.21. Proposition. Let A be a proper ideal of R. Then A

c

A

if and only if B c P for some prime proper left residual P of A.

.

.-

.B *

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 Proof.

.

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 Ter( T ) is the unique essential left residual of T .

if

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 )

-

-

263

EXERCISES

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).Therefore, 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. ideal of R.

If T is a tertiavy ideal of R, then Ter( T ) is a prime EXERCISES

1. 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.

264

APPENDIX

(c) Show that if a~ Rad(A), then ah E A for some positive integer n. (d) 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 commutative, this set need not be multiplicatively closed. 2. 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 = 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,

on"=,

+

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) Show that a proper ideal of R has only a finite number of prime proper left residuals. (b) 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.

265

EXERCISES

(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. (d) 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. (e) 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.

-

5. T h e 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 intersection 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 A B 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).

Bibliography

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. T h e book of Krull contains a list of the older papers on commutative rings. T h e book of Lesieur and Croisot contains a more detailed exposition 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. Boom Atiyah, M. F., and MacDonald, I. G., “ Introduction to Commutative Algebra.” Addison-Wesley, Reading, Massachusetts, 1969. Bourbaki, N., “ &ments de Mathematique, Algebre,” Chaps. 4 and 5 (Polynomes et fractions rationnelles; Corps commutatifs). Hermann, Paris, 1959. Bourbaki, N., ‘‘ Elements of Mathematique, Algebre Commutative,” Chaps. 1 and 2 (Modules Plats ; Localisation). Hermann, Paris, 1961. Bourbaki, N., “ Elements de Mathematique, Algebre Commutative,” Chaps. 3 and 4 (Graduations, filtrations, et topologies; IdCaux premier associCs et decomposition primaire). Hermann, Paris, 1961. 266

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1964a Classes de diviseurs et dkrivtes logarithmiques. Topology (Suppl. 1 ) 3, 81-96. 1964b Anneaux graduts factoriels et modules rtflexifs. Bull. SOC.Math. France 92, 237-249. 1968 On a construction due to P. M. Cohn. Proc. Cambridge Philos. SOC.64, 249-250. Sato, H. 1953 Zum Teilerkettensatz in kommutativen Ringen. Proc. Japan Acad. 29, 10-12. 1956157 Some remarks on Zariski rings. J. Sci. Hiroshima Univ. Ser. A 20, 93-99. 1957/58a On splitting of valuations in extensions of local domains. J . Sci. Hiroshima Univ. Ser. A 21, 69-75. 1957/58b A note on principal ideals. J . Sci. Hiroshima Univ. Ser. A 21, 77-78. Satyanarayana,M. 1969 Generalized primary rings. Math. Ann. 179, 109-114. Scheja, G. 1965 ifber Primfactorzerlegung in zweidimensionalen lokalen Ringen. Math. Ann. 159, 252-258. Schilling, 0. F. G. 1944 Automorphisms of fields of formal power series. Bull. Amer. Math. SOC. 50, 892-901. Schmidt, F. K. 1936 u b e r die Erhaltung der Kettensatze der Idealtheorie bei beliebigen endlichen Korperenveiterungen. Math. Z . 41, 443-450. Schneider, J. E. 1969 A note on the theory of primes. Pacific J . Math. 30, 805-810. Scholz, A. 1943 Zu Idealtheorie in unendlichen algebraischen Zahlkorpern. J. Reine Angew. Math. 185, 113-126. Scott, W. R. 1954 Divisors of zero in polynomial rings. Amer. Math. Monthly 61, 336. Seidenberg, A. 1945 Valuation ideals in polynomial rings. Trans. Amer. Math. SOL.57, 387-425. 1953 A note on the dimension theory of rings. Pacific J. Math. 3, 505-512. 1954 On the dimension theory of rings 11. Pacific J. Math. 4, 603-614. 1966 Derivations and integral closure. Pucijic J . Math. 16, 167-173. 1967 Differential ideals in rings of finitely generated type. Amer. J. Math. 89, 22-42. Shannon, R. T. 1970 The rank of a flat module. Proc. Amer. Math. SOC.24, 452456. Skolem, T. 1939 Eine Bemerkung uber gewisse Ringe mit Anwendung auf die Produlitzerlegung von Polynom. Nordisk Mat. Tidskr. 21, 99-107. Smith, W. W. 1969 Projective ideals of finite type. Canad. J . Math. 21, 1057-1061. Snapper, E. 1950 Completely primary rings I. Ann. of Math. (2 ) 52, 666-693.

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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

SUBJECT INDEX

295

Divisorial fractional ideal, 173 Divisors, 172 sum of, 173 Domination, 101, 229

E Essential valuations, 179 Euclidean ring, 199 Exact sequence, 10 short, 10 split, 24, 34 Extension of ideal, 66 to large quotient rings, 234 of valuation, 114 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, 151

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

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

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

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

0

Ordered Abelian group, 107 Ordered semigroup, 107 Overring, 88 flat, 89

297

SUBJECT INDEX

P 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

R Radical of ideal, 41 Jacobson, 55 prime, 253 of submodule, 41 tertiary, 257 Ramification index, 185 Rank of group, 110 of valuation, 111 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

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

298

SUBJECT INDEX

residual of, 38 sum of, 3 zero, 4 Subring, 1

V

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 of homomorphisms, 19 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

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, 114 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, 48-52 Unity, 1

2 Zero divisor, 42 proper, 42 ZPI-ring, 205-209

Pure and Applied Mathematics A Series of Monographs and Textbooks Editors

Paul A. Smith and Samuel Eilenberg Columbia University, New 'fork

1: ARNOLD SOMMERFELD. Partial Differential Equations in Physics. 1949 (Lectures on Theoretical Physics, Volume V I ) 2 : REINHOLD BAER.Linear Algebra and Projective Geometry. 1952 3 : HERBERT BUSEMANN A N D PAUL KELLY.Projective Geometry and Projective Metrics. 1953 4 : S T E F A N BERGMAN A N D M. SCHIFFER. Kernel Functions and Elliptic Differential Equations in Mathematical Physics. 1953 5: RALPHPHILIP BOAS,JR. Entire Functions. 1954 6 : HERBERT BUSEMANN. The Geometry of Geodesics. 1955 7 : CLAUDE CHEVALLEY. Fundamental Concepts of Algebra. 1956 8: SZE-TSENHu. Homotopy Theory. 1959 9: A. M. OSTROWSK~. Solution of Equations and Systems of Equations. Second Edition. 1966 Treatise on Analysis. Volume I, Foundations of Modern Analy10 : J. DIEUDONN~. 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. BISHOPA N D RICHARD J. C R I T m N n E N . Geometry of Manifolds. 1964 16: STEVEN A. GAAL.Point Set Topology. 1961 17: BARRYMITCHELL. Theory of Categories. 1965 18: ANTHONYP. MORSE.A Theory of Sets. 1965

Pure and Applied Mathematics A Series of Monographs and Textbooks

19: GUSTAVECROQUET.Topology. 1966 20: Z. I. BOREVICH A N D 1. R. SRAFAREVICH. Number Theory. 1966 21 : J O S ~LUISMASSERA AND JUAN JORGE SCHAFFER. Linear Differential Equations and Function Spaces. 1966 22 : RICHARD D. SCHAFER. An Introduction to Nonassociative Algebras. 1966 23: MARTXNEICELER.Introduction to the Theory of Algebraic Numbers and Functions. 1966 24 : SHREERAM ABHYANKAR. Resolution of Singularities of Embedded Algebraic Surfaces. 1966 TREVES. Topological Vector Spaces, Distributions, and Kernels. 1967 25 : FRANCOIS D.h x AND RALPHS. PHILLIPS. Scattering Theory. 1967. 26 : PETER 27: OYSTEINORE.The Four Color Problem. 1967 28 : MAURICEHEINS.Complex Function Theory. 1968 29: R. M. BLUMENTHAL A N D R. K. GETOOR. Markov Processes and Potential Theory. 1968 30 : L. J. MORDELL. Diophantine Equations. 1969 31 : J. BARKLEY ROSSER.Simplified Independence Proofs : Boolean Valued Models of Set Theory. 1969 32: WILLIAMF. DONOGHUE, JR. Distributions and Fourier Transforms. 1969 33: MARSTONMORSEA N D STEWART S. CAIRNS.Critical Point Theory in Global Analysis and Differential Topology. 1969 34: EDWIN WEISS.Cohomology of Groups. 1969 35 : HANSFREUDENTHAL 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 Categories and Functors. 1970 39 : BODOPAREICIS. 40: PAUL L. BUTZER A N D ROLFJ. NESSEL.Fourier Analysis and Approximation: Volume 1, One-Dimensional Theory. 1971 41 : EDUARD PRUCOVEEKI. Quantum Mechanics in Hilbert Space. 1971 : An Introduction to Transform Theory. 1971 42 : D.V.WIDDEE 43 : MAXD. LARSEN AND PAUL J. MCCARTHY. Multiplicative Theory of Ideals. 1971 B E H ~ N SRing . Theory. 1971 44 : ERNST-AUGUST I n gtepatutiolr

WERNER GREUB,STEVEHALPFXIN, A N D JAMES VANSTONE.De Rham Cohomology of Fibre Bundles : Volume 1 ; Manifolds, Sphere Bundles, Vector Bundles.