One-Dimensional Cohen-Macaulay Rings

Lecture Notes in Mathematics A collection of informal reports and seminars Edited by A. Dold, Heidelberg and B. Eckmann, Z0rich

327 Eben Matlis Northwestern University, Evanston, IL/USA

1-Dimensional Cohen-Macaulay Rings

Springer-Verlag Berlin. Heidelberg • New York 1973

A M S Subject Classifications (1970): 13-02, 13Cxx, 13E05, 1 3 E l 0 , 13F05, 1 3 H x x

I S B N 3-540-06327-7 Springer-Verlag B e r l i n . H e i d e l b e r g . N e w Y o r k I S B N 0-387-06327-7 Springer-Verlag N e w Y o r k " H e i d e l b e r g " Berlin This work is subject to copyright. AII rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the publisher, the amount of the fee to be determined by agreement with the publisher, © by Springer-Verlag Berlin - Heidelberg 1973. Library of Congress Catalog Card Number 73-80869. Printed in Germany. Offsetdruck: Julius Bettz, Hemsbach/Bergstr.

TABLE

Chapter

I

h-Divisible Chapter

and C o t o r s i o n

Chapter

Extensions

11

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

28

IV

Localizations Chapter

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

34

V

Artinian Chapter

Divisible

Modules . . . . . . . . . . . . . . . . . . . .

42

Vi

Strongly Chapter

Unramified

Ring Extensions

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

56

VII

Closed

Chapter

Components

of R . . . . . . . . . . . . . . . . . . . .

68

Modules . . . . . . . . . . . . . . . . . . . . .

79

VIII

Simple

Divisible

Chapter

IX

Semi-Simple Chapter

and U n i s e r i a l

Divisible

Modules

. . . . . . . . . . .

85

X

Integral

Chapter

Closure . . . . . . . . . . . . . . . . . . . . . . .

90

XI

Primary

Chapter

Decomposition

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

96

Ring . . . . . . . . . . . . . . . . . . .

103

XII

The F i r s t Chapter

Neighbourhood

XIII

Gorenstein Chapter

Rings . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter

116

XIV

Multiplicities

The

I

III

Chapter

The

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

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

Compatible

The

Modules

II

Completions

The

OF C O N T E N T S

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

126

Ideal

141

XV

Canonical

of R . . . . . . . . . . . . . . . . . . . . .

References . . . . . . . . . . . . . . . . . . . . . . . . . . . .

153

Index

155

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

To

Margarita Hill Matlis

INTRODUCTION The purpose Artinian ring,

of these notes

modules

is to present

over a 1-dimensional,

a structure

Noetherian,

and to show that m a n y of the properties

derived

from a knowledge

to recapture Macaulay

of this theory.

m a n y of the known

results

rings by these techniques,

Thus we have a new and unifying &mining

a category

of modules

theory for

Cohen-Macaulay

of such a ring may be

In fact, we shall be able

about

1-dimensional

and find

Cohen-

some new ones as well.

way of looking

that previously

at these

rings

has received

by ex-

scant

attention. My original modules

aim was to study the category

over a Noetherian

there was a structure theory of divisible was possible developed

apparent

modules

to apply the general "Cotorsion

the problems

all redefined

in terms

arbitrary

zero divisors). discovery modules

Macaulay

the most

ring;

multiplicities, Gorenstein

important namely, latent

rings,

about

valid.

modules ring

residue

the existence

ring,

became

most

true for an

thing of all was the divisible

and relate to each

a 1-dimenslonal, reduction

ideals,

local,

number,

and so forth.

of canonical

were

then

Thereupon

theory of Artinian

degrees,

re-

(even though it had

surprising

the multiplicity,

striking

(that is, the elements

to really understand

things

domains

and cotorsion

commutative

remained

Cohen-Macaulay

it became possible

torsion

elements

about divisible

integral

that it

After a time it became

of an arbitrary

But perhaps

I found

[12] to yield

that by using the structure

other the most

ring.

Modules"

all of the tool theorems

1-dimenslonal,

the elementary

about

torsion-free,

divisible

l, and to see if

theorems

of the regular

results

generalize

I was considering.

that are not zero divisors)

of the striking

dimension

over a Dedekind

that if divisible,

practically

of Krull

theory that would

in m y paper,

sults about

domain

of Artinian

Cohen-

latent

Theorems analytic

about

ir-

VIII reducibility,

analytically unramified

ponents are all seen to be theorems d i v i s i b l e modules,

rings,

and the analytic

about the structure

and these theorems

are all u n i f i e d

com-

of A r t i n i a n

by this struc-

ture theory. The structure quite

elementary.

direct

of an A r t i n i a n module In the Dedekind

sum of a m o d u l e

divisible m o d u l e s

of finite

submoduies.

ule c h a r a c t e r i z e s

essentially

length and of a finite number of

Since this d e c o m p o s i t i o n

Dedekind

1-dimensional

a local problem,

R-modules

of an Artinian m o d u l e

a satisfactory

W h i l e we shall not be able for a Dedekind

structure

ring, we

theory for Artinian

that is far more complex than for Dedekind

haps m o r e i n t e r e s t i n g

is

that R is a Noetherian,

ring.

that is p o s s i b l e

shall be able to present

ring.

and for the sake of simplicity we shall

Cohen-Macaulay

to attain the p r e c i s i o n

nonzero

of an A r t i n i a n mod-

Cohen-Macaulay

of finding the structure

1-dimensional

gp~ is a

rings, we can not hope to find anything

assume for the rest of this i n t r o d u c t i o n local,

is a

in the sense that it has no proper,

as good for an a r b i t r a r y The problem

ring is

case an Artinian module

of the form Zp~ for various primes p .

simple d i v i s i b l e m o d u l e divisible

over a Dedekind

rings,

but per-

for that reason.

We shall show that an A r t i n i a n R - m o d u l e is the sum of an R-module of finite length and of a d i v i s i b l e A r t i n i a n R-module. A r t i n i a n divisible submodules

R-module

has a c o m p o s i t i o n

such that the factor modules

ules in the sense described previously. exists a J o r d a n - H S l d e r

series of d i v i s i b l e

are simple d i v i s i b l e

R-mod-

We shall prove that there

theorem for these c o m p o s i t i o n

shall have to replace the i s o m o r p h i s m

Every

of the factors

series,

but we

of two compo-

sition series w i t h the notion of equivalence. Two m o d u l e s morphic

are defined

image of the other.

B are equivalent

to be equivalent, Two Artinian,

if each is a homo-

divisible

R-modules

A and

if and only if B ~ A/C, where C is a f i n i t e l y gen-

erated

submodule of A.

Thus, contrary to the usual practice of con-

sidering only finitely generated

R-modules

over a Noetherian

ring,

we actually go to the extreme of "throwing them away". The Jordan-H~lder

theorem for Artinian divisible R-modules

gives

us the concept of the divisible length of an Artinian R-module, function that proves to be most useful.

a

Since modulo its divisible

submodule an Artinian R-module has finite length in the classical sense, we see that we have two numerical

invariants

to describe an

Artinian R-module. We prove that there are only a finite number of equivalence classes of simple divisible correspondence

R-modules;

and these are in one to one

with the prime ideals of rank 0 of the completion of

R, and hence with the analytic components ponent of R corresponds

in the full ring of quotients Q of

(A ring is called a pseudo valuation

submodule,

ring if modulo its divisible

it becomes an ordinary valuation

these pseudo valuation of representatives

Each analytic com-

uniquely with one of the pseudo valuation

rings containing R and contained R.

of R.

rings,

ring.)

If V 1 ..... V n are

then Q/VI,...,Q/V n are a complete

of the equivalence

set

classes of simple divisible

R-modules. The set of isomorphism given simple,

divisible

classes

R-module is in one-to-one

with the ideal class semi-group ponent of R. endomorphism

equivalent

to a

correspondence

of the corresponding

analytic com-

And by analogy with Schur's Lemma we find that the rings of these simple,

of the integral extensions tient field.

of R-modules

divisible R-modules give us all

of this analytic component in its quo-

Hence these endomorphism

rings are complete,

Noetherian,

local domains of Krull dimension i. We define a semi-simple equivalent

to a finite direct

divisible

R-module to be one that is

sum of simple,

We find that every Artinian divisible

divisible R-modules.

R-module is semi-s~iple

if and

only if R is analytically

unramified.

proof of the known theorem only if the integral While there however,

closure

any restrictions

module

to be a P-primary correspond

the completion module

of R.

R-modules

decomposition

We define

divisible

the

"genetic

representative ules appears

to a direct

primes

responding

divisible

of each equivalence

series

maximal

ideal

unramified A/(AM)

divisible

class of simple, series

to it appears

in fact K conR-modules. divisible

ring extension

ring extensions

components

of R.

of the cor-

R-module

ring A o R is called

A/R is the divisible

component

A is a closed component

submodule

valuation

rings containing

divisible

submodules

of V/R,

strongly

ideal of A and

of maximal

We show that a subring of Q is a closed

Equivalently,

if and only

(If M is the

of R in Q and these are called

of R if and only if it has an analytic

R-mod-

of R.

of R.

exist only a finite number

unramified

A

for K, and the num-

over R if AM is the only regular maximal There

P-primary

as a factor in the

A/R of K is a divisible

of R, then a commutative

% R/M.)

pletion.

and,

is equal to the latent m u l t i p l i c i t y

if A is a strongly unramified

R-

P.

R-module;

in a composition

equivalent

submodule

R-

divisible

determined

prime ideal of rank 0 of the completion

A proper

divisible

of R and K is the R-module

code" for all Artinian,

as a factor

R-modules

if all of its composi-

sum of uniquely

for the various

ber of times a m o d u l e composition

divisible

there is,

to the same fixed prime ideal P of rank 0 of

then K is an Artinian

tains

if and

in general,

an Artinian

R-module

If Q is the full ring of quotients Q/R,

unramifled

generated.

for Artinian,

on R.

leads to a n e w

We then see that every Artinian,

is equivalent

divisible

of R is finitely

decomposition

without

tion factors

that R is analytically

is no semi-slmple

a primary

This result

strongly

the closed component

of R as its comof R if and only if

where V is one of the pseudo

R and contained

the closed components

in Q.

Modulo

their

of R are analytically

Jr-

XI reducible

Noetherian

local domains

we have a way of passing using induction The point

of view in w r i t i n g

these notes

as possible,

an elementary

logical

algebra.

and proofs

tried

knowledge

have been given

ature for the bits of machinery

ground

information

modules

that

for this particular of searching

to arbitrary

portance

is the duality

Complete

modules

[12],

in Corollary

rings.

is based

ology.

In the applications

we make,the

ditional

modules

between

of the M-adic

topology

basis

in [16].

for the treatment

in the notes.

Chapter

divisible modules

While

elementary

of strongly

gives

ring may be reduced

We also

R-modules

with

of R.

us the powerful

in nature,

IV shows how the problem

to the local case.

on

ad-

and the

cf R found in [97.

unramified

over a 1-dimensional,

and

coincides

completion

Chapter III deals with the theory of compatible as enunciated

im-

modules

result.

the Artinian

completion

so that

of R in the R-top-

M-adic

of the two topologies

tool of the duality

Noetherian

M-adic

and thus H is the ordinary

reworked

h-divisible

of H, the completion

is based

torsion-free

on this

back-

and complete

Of particular

2.4 between

on the one hand and torsion,

together

We have

the liter-

This material

here the properties

This coming

study.

the n e c e s s a r y

cotorsion,

develop

the R-topology

and homo-

theorems

through

but has been

commutative

Much of the analysis

to

that have to be used.

h-divisible,

Modules"

them

have been pro-

of these notes present

concerning

"Cotorsion

it is applicable

the other.

definitions

we need in this investigation.

on the paper

ring theory

of most of the general

to save the reader the trouble

by

has been to make

of commutative

Most of the relevant

two chapters

domain

and the reader is only assumed

as tools and preliminaries

The first

I, and hence

irreducible

length of K.

possess

needed

to an analytically

dimension

on the divisible

as self contained

vided

of Krull

ring extensions this forms the

ring extensions

later

of considering

Noetherian

Cohen-Macaulay

It also shows that for these

XII rings a divisible module no nonzero

nilpotent

ible module in [I0].

information

Chapters

theory

summand.

injective

This is a reworking

found

of a divis-

of material

all of the necessary in [9] and are pre-

local C o h e n - M a c a u l a y

time.

Chapters

and

reduction

ring,

and carried

concerned

with the study of Artinian

further.

sense.

Chapter XIV combines

the results

Artinian

two different

of development.

Chapter XV considers shows that the existence

here

the

of a 1-dimensional

to the study of Gorenof the

pioneering

work is

these two chapters

modules

The material

develop

here by the concept

Basically

with those concerning streams

number

and some of N o r t h c o t t ~ s

explained

in the classical

XII and XIII

role is played

divisible

and most of it appears

ring and this is applied

A special

first neighborhood

the theory of Artinian,

in this introduction,

of the m u l t i p l i c i t y

structure

together

modules

V through XI present

for the first

stein rings.

submodule

proofs.

as outlined

in print

and that if the ring has

then the torsion

4.5 and 4.6 collect

concerning

sented without

modules

elements

is a direct

Theorems

is h-divisible,

are

that have finite

length

here m a y be found in [15]. of the preceding

divisible

modules,

two chapters

and unites

the question

of canonical

of such ideals

is decided

ideals,

these

and

by the divisible

of K.

An effort about providing the author.

has been made throughout attribution

for those

these notes

to be scrupulous

thsorems not due originally

to

CHAPTER I h - D I V I S I B L E AND C O T O R S I O N MODULES

Throughout

these notes R will be a c o m m u t a t i v e

ring.

An ele-

ment of R that is not a zero divisor in R will be called a regular element

of R.

The set z~

tively closed. R.

of regular elements

The ring Rj is called

of R is m u l t l p l i c a -

the full ring of quotients

We shall c o n s i s t e n t l y use the n o t a t i o n Q = R / and K = Q/R.

always assume that Q J R. module ~

by the symbol AQ.

some basic

come later. concepts

theorems

for integral

tiplications

divisible.

domains

is a g e n e r a l i z a t i o n

is said to be divisible

by the regular elements

C l e a r l y a homomorphic

and if both A and B/A are divisible, submodules

has no divisible

submodule

submodules;

divisible

of B.

if the mul-

of R are all R-module

is

of an R-module B,

then B is also divisible.

of a module

an R-module B has a unique largest tains every divisible

of similar

image of a divisible

On the other hand if A is a submodule

sum of two divisible

and

that may be found in [12].

An R-module

on the m o d u l e

epimorphisms.

definitions

that we shall need as tools for what is to

Most of this m a t e r i a l

Definitions:

B/d(B)

We

If A is an R-module we shall denote the

In this chapter we shall review some elementary prove

of

is again divisible. submodule

d(B)

The Thus

that con-

B is said to be reduced if it

that is, if d(B) = O.

Clearly,

is reduced.

The concept dual to divisible m o d u l e is said to be t o r s i o n - f r e e ule by the regular elements of a t o r s i o n - f r e e R - m o d u l e

is that of torsion-free. if the m u l t i p l i c a t i o n s

of R are all monomorphisms. is torsion-free.

no nonzero

on the modA submodule

If A is a submodule

an R-module B, and if both A and B / A are torsion-free, torsion-free.

submodules.

of

then B is also

An R-module is said to be a torsion R-module

torsion-free

An R-

if it has

An R-module B has a unique

largest

torsion

submodule

of B, and B/t(B)

t(B)

that contains

is torsion-free.

order if there is a regular

submodule

B is said to be torsion of bounded

element

easy to see that an R-module

every torsion

r of R such that rB = 0.

is both torsion-free

It is

and divisible

if

and only if it is a Q-module. A more useful

concept

than divisible

plying homological

methods

is that of h-divisible.

said to be h-divisible free and divisible

if it is a homomorphic

R-module;

image of a Q-module. R is an integral

if it is a homomorphic

A direct Consequently module

h(B)

sum of h-divisible

said to be h-reduced, that is, if h(B) B/h(B)

mental

exact

I.i.

R-modules

B contains

a unique

It follows

is divisible.

If B is an R-module,

largest

h-divisible

submodule

of B.

h-divisible

sub-

B is

submodules;

some necessary

situation

that and

to occur.

then we have three funda-

sequences:

0 -~ B/t(B) ~ Q ®R B -~ K @R B -* O.

(2)

0-*HOmR(K,B)

(3)

0 -- B/h(B) -- Ext~(K,B)

- * H O m R ( Q , B ) -*h(h)

~ 0.

-~ Ext~(Q,B)

we have t(B) ~ Tor~(K,B);

and only if Q D R B = 0.

Furthermore,

-* 0.

and B is a torsion B i__ssh-reduced

R-module

if

if and only if

HOmR(Q,h ) = 0. Proof.

We

later.

We shall examine p r e s e n t l y

(i)

Therefore,

if and only

it is not true in general

for this desirable

if

is again h-divisible.

every h-divisible

Unfortunately,

conditions

Theorem

R-module

if it has no nonzero

= 0.

is h-reduced.

sufficient

R-module.

to say about the converse

that contains

is

(in particular,

is h-divisible

image of an injective

an R-module

An R-module

image of a torsion-

ring

then an R-module

that an h-divisible

shall have something

of ap-

that is, if it is an R-homomorphic

If Q is a semi-simple

domain),

from the definition

for the purposes

As our starting point we take the exact sequence:

(*)

0-*R-*Q-*K-~

0.

If we tensor this sequence with B and identify B with R ®R B, we have an exact sequence:

O-*Tor~(K,B)

- * B a-*Q ~R B - * K

where a(x) = i ~ x for x c B. ker a = t(B).

To establish

Because Q ~R B is a Q-module,

thus t(B) c ker a.

Conversely,

image of Q @R Rx in Q ~R B is O. hence Q ®R Rx = O. that ax = 0.

®R B - * O (i) we have to show that it is torsion-free,

suppose that x ¢ ker a.

and

Then the

But Q @R " is an exact functor and

Thus, there exists a regular element a of R such

Hence x ~ t(B) proving that ker a = t(B), and that

Tor~(K,B) ~ t(B).

We also see that B = t(B) if and only if

Q ®R B = 0. If we apply the functor HomR(.,B ) to exact sequence identify B with HomR(R,B),

(*) and

we obtain an exact sequence:

0 -~ HomR(K,B ) -~ HomR(Q,B ) ~-~ B ~ Ext~(K,B) -* Ext~(Q,B) -~ 0, where B(f) = f(1) for f ~ HomR(Q,B ).

Thus to establish (2) and (3)

it is sufficient to prove that Im B = h(B). Since HomR(Q,B ) is a Q-module, and thus Im B c h(B). mapping onto h(B). maps onto x.

it is torsion-free

On the other hand,

and divisible

there is a Q-module V

Let x E h(B) and choose an element y ¢ V that

Since Qy c V, there is an R-homomorphism f : Q - ~ h ( B )

such that f(1) = x.

Therefore,

x ¢Im

B and hence Im B - h(B).

is now clear that B is h-reduced if and only if HomR(Q,B) Corollary 1.2.

Let B be a torsion R-module.

It

= 0.

Then the natural

map:

: K ~R H°mR(K'B) -+h(B) defined by ~(x ~ f) = f(x) for x ( K and f ¢ HomR(K,B)

is an iso-

morphism. Proof. image

By T h e o r e m

of the Q - m o d u l e

I.I,

Q @R H°mR(K'B)

divisible.

Thus

other

let y ¢ h(B).

hand,

morphism

Define

R-module,

= O.

morphism

f : K-~h(B)

Im ~

such that ~(i)

by v(z)

and v(1)

and hence

Then by T h e o r e m

and hence

v : Q-~h(B)

Therefore,

there

= ~(az)

R c kerv

Im ~ c h(B).

i.i there

= y.

is a regular

then Q/S is

element

for all z ~ Q. and we have

On the

is an R-homo-

If S = ker ~,

such that y ~ f(K).

is h-

a in S N R.

Then v ( a -I) = y

an induced

Hence we

R-homo-

see that

: h(B).

Every

element

x @ f, w h e r e a E R.

of K D R HomR(K,B)

can be w r i t t e n

f E H o m R ( K , B ) and x = a -I + R for some

Suppose

that x ~ f ¢ ker ~;

that

g (HOmR(K,B)

as follows:

is an element

z ¢ K such that y = az,

is any element This

is an R - h o m o m o r p h i c

and hence K @R H°mR(K'B)

Im ~ is h-divisible,

~ : Q -~ h(B)

a torsion

K ~R H°mR(K'B)

shows

and hence

that

if y ( K, then

of K such that

is f(x) since

in the form regular

= O.

Define

K is d i v i s i b l e

and we let g(y)

= f(z).

aw = O, then w ~ Rx and hence

g is a w e l l - d e f i n e d

element

R-homomorphism.

f ~ x = ag @ x = g ® ax = g ® 0 = O.

there If w

f(w)

Clearly

= O.

ag = f,

Thus ~ is an isomor-

phism. Corollary

1.3.

Let B be a t o r s i o n

R-module.

Then we have the

following: (1)

I f x E B, then x ~ h(B)

i_~f and o n l y if there

exists

f ¢ H O m R ( K , B ) such that x E f(K). (2)

B i__ssh-reduced i f an__dd only i f H o m R ( K , B ) = O.

(3)

I__ffA and C are

h-divisible

submodules

of B, then A = C

if and only if H o m R ( K , A ) = H o m R ( K , C ) . Proof.

This c o r o l l a r y

Theorem

1.4.

Assume

h-dlvi______sibl____eeR-module,

then

follows

immediately

from C o r o l l a r y

that Q l_ss ~ s e m i - s l m p l e t(B)

i__{sa direct

rin~.

su___~an_ddofB.

1.2.

If B i_ss an

Proof.

By Zorn's Lemma there

and divisible

submodule

and divisible

submodules,

C)/C.

(t(B) +

Since

C of B.

exists

a maximal

B/C

Then

has no nonzero

and the torsion

submodule

torslon-free

and prove that B is a torsion

R-module.

Let x be a nonzero

then Q/S is isomorphic

simple

ring,

summand

and divisible

B/C

= x.

of B.

Its complementary

I.I,

there is

If S = Ker f, submod-

Since Q is a semi-

of Q, and thus SQ/S is a direct summand

in Q/S is a torsion-free

of B, and hence is equal to O.

R-module,

submodules,

Now the torsion

and SQ is an ideal of Q. summand

is

loss of gen-

and divisible

Then by Theorem

to a submodule

submodule

Q/S is a torsion

of B.

such that f(1)

SQ is a direct

of Q/S.

We thus have

element

f : Q-+B

uie of Q/S is SQ/S,

of

torslon-free

t(B) O C = O, we may assume without

erality that B has no nonzero

an R-homomorphism

torsion-free

and hence x is a torsion

shown that B is a torsion

R-module,

Therefore

element

of B.

and this completes

the proof of the theorem. Definition.

An R-module

if HomR(Q,C ) = 0 and Ext~(Q,C) sion module order,

is h-reduced.

then Ext~(Q,B)

EXt~(Q,B)

establishing Theorem

(1)

R-module

R-module

a cotor-

of bounded

and torsion-free,

Thus a torsion

R-module,

and hence of bounded

and A is any Q-module,

of F.

then

For we have HomR(A,C ) = 0 because

Now we can write A = F/P,

P is a Q-submodule

in particular,

R-module

R-module.

HomR(A,C ) = 0 = Ext~(A,C).

modules.

Thus,

is both torsion

If C is a eotorsion

h-reduced.

= 0.

If B is a torsion

= 0 for all n > 0.

order is a cotorsion

C is said to be a cotorsion

C is

where F is a free Q-module

We then have Ext~(A,C)

~ HomR(P,C )

and

0,

our assertion. 1.5.

Let 0 ~ A - ~ B - ~ C

~0

be an exact

sequence

The____nnw_eehav___eeth__eef__ollowing: If A and C are cotorsion,

then B is also cotorsion.

of R-

(2)

I f B is cotorsion,

then A is cotorsion if and only if C

is h-reduced. Proof.

This theorem is an immediate consequence of the long

exact sequence obtained by applying the functor ExtR(Q,. ) to the given exact sequence. Theorem 1.6. C is cotorslon, Proof.

Let B and C be R-modules.

If B is torsion,

or if

then HomR(B,C ) is cotorsion.

We have the canonical duality isomorphism

HomR(Q,HOmR(B,C)) if B is torsion,

~ HomR(Q D R B,C) by [3, Ch. II, Prop. 5.2]. or C is cotorsion,

then HomR(B,C ) is h-reduced.

Suppose that B is a torsion R-module, module containing C.

Thus

and that E is an injective R-

Then we have an exact sequence of h-reduced

R-modules: O-*HOmR(B,C) We have Ext~(Q,HOmR(B,E))

[3, Ch. Vl, Prop. 5.1]. torsion R-module.

-*HOmR(B,E) -*HOmR(B,E/C).

~ HomR(TOr~(Q,B),E)

= 0 for all n > 0 by

Hence by Theorem 1.5, HomR(B,C)

is a co-

Suppose that C is a cotorsion R-module,

be a free R-module mapping onto B with kernel A.

and let F

Then we have an

exact sequence of h-reduced R-modules. O-~HOmR(B,C)

--~HOmR(F,C) -*HOmR(A,C ).

HomR(F,C ) is isomorphic to a direct product of copies of C and thus is a cotorslon R-module.

Hence HomR(B,C ) is cotorsion by Theorem

1.5. Definition.

The homological dimension of an R-module A (abbrev-

iated hdRA ) is defined to be the smallest integer n such that Ext~(A,B) = 0 for all m ~ n + 1 and all R-modules B. integer n does not exist, then hdRA = ~.) hdRA ~ 1 if and only if Ex~(A,B)

(If such an

It is easy to see that

= 0 for every R-module B that is a

homomorphic image of an inJective R-module. Theorem 1.7.

The following statements are equivalent:

(1)

hdRQ : 1

(2)

If 0 - ~ A - ~

B-~ C-~ 0 is an exact sequence of R-modules and

A and B are cotorsion, then C is cotorsion. (3)

Ext~(K,B) is a cotorsion R-module for all R-modules B.

(4)

Ext~(K,B) = 0 for all h-divisible R-modules B.

(5)

Ext~(K,B) = 0 for all torsion, h-dlvisible R-modules B

(6)

hdRK = I.

Proof. (1) ~ > reduced R-module.

(2).

We see from Theorem 1.5 that C is an h-

Now we have an exact sequence: Ext~(Q,B) -~Ext~(Q,C) -~Ext~(Q,A).

Since B is a cotorslon R-module and hdRQ = I, the end terms of this sequence are zero.

Thus Ext~(Q,C) = O, and hence C is a cotorsion

R-module. (2) ~-> (3). containing B.

Let B be an R-module and E an injective R-module

Then we have an exact sequence:

0 -~ HomR(K,B ) -~ HOmR(K,E ) a_.HOmR(K,E/B ) -~ Ext~(K,B) -- 0 By Theorem 1.6 every term in this sequence, except possibly Ext~(K,B) is a cotorslon R-module. sion R-module.

Hence by (2), Im a is a cotor-

But then by (2) again, Ext~(K,B) is also a cotorsion

R-module. (3) = >

(4).

If B is an h-divislble R-module, then by Theorem

I.I (3) we have Ext~(K,B) ~ Ext~(Q,B). free and divisible.

Thus Ext~(K,B) is torsion-

Since it is cotorsion by (3), we see that

Ext~(K,B) = 0. (4) ~---> (5)-

This is a trivial assertion.

(5) ~ >

Let B be a homomorphic image of an inJective R-

(6).

module.

It is sufficient

to see that an inJective homomorphic

to prove that Ext~(K,B) R-module is h-divisible,

= 0.

It is easy

and hence every

image of an inJective R-module is h-divislble.

We have

an exact sequence: HomR(K,B/t(B))

-~Ext~(K,t(B))

Clearly HomR(K,B/t(B)) Q-module,

= 0.

-~Ext~(K,B)

-~Ext~(K,B/t(B)).

On the other hand,

we have by [3, Ch. VI, Prop. 4.1.3]

Ext~(K,B/t(B))

~ Ext~(Q D R K,B/t(B)).

Ext~(K,B/t(B))

= O.

Therefore,

see that Ext~(K,t(B))

since B/t(B)

that

But Q @ R K = 0, and thus

from the previous

~ Ext~(K,B).

is a

exact sequence we

Thus by (5) it is sufficient

to

prove that t(B) is h-divisible. Let x ¢ t(B); f : Q-~B

then by Theorem I.I there is an R-homomorphism

such that f(1) = x.

If S = Ker f, then S contains

ular element of R and hence Q/S is a torsion R-module. Im f c t(B), and from this it follows (6) ==> (i).

This assertion

is obvious.

The gollowing

(I)

B/h(B)

is h-reduced for every R-module B.

(2)

If 0-*A

two statements

(1) ~=> (2).

then B is also h-divisible.

Let A be a submodule of an R-module B,

and suppose that A and B/A are h-dlvlsible. is a homomorphic

Since B/h(B)

are equivalent:

~ B - ~ C ~ 0 is an exact sequence of R-modules

such that A and C are h-divisible,

B/h(B)

Therefore

that t(B) is h-dlvisible.

Lemma 1.8.

Proof.

a reg-

image of B/A.

is h-reduced

Then A c h(B),

Thus B/h(B)

and hence

is h-divisible.

by (1), we see that B/h(B) = 0.

Therefore

B is h-divlsible. (2) -~-> (1). taining h(B). (2).

Let B be an R-module and A a submodule of B con-

If A/h(B)

But then A = h(B), Theorem 1.9.

is h-dlvisible,

then A is h-divislble

showing that B/h(B)

If hdRQ = I, then B/h(B)

by

is h-reduced. is h-reduced

for every

R-module B.

On the other hand, if Q is a semi-simple

ring, then the

two statements are equivalent. Proof.

If hdRQ = l, then by Theorem 1.7, Ext~(K,B)

torsion R-module. Ext~(K,B)

Since B/h(B)

is isomorphic

is a co-

to a submodule of

by Theorem i.i (3), we see that B/h(B)

is h-reduced.

On the other hand let us assume that Q is a semi-simple ring and that B/h(B)

is h-reduced for every R-module B.

sion h-dlvisible R-module.

To prove that hdRQ = 1 it is sufficient

by Theorem 1.7 to prove that Ext~(K,A) have Ext~(K,A) ~ Ext~(Q,A). Ext~(Q,A)

Let A be a tor-

= 0.

By Theorem 1.1 (3) we

Hence it is sufficient

to prove that

= 0.

Let us consider an exact sequence of the form: O-~A-~C-~Q-~ Then A is the torsion submodule of C. divisible R-modules

and,

O. By Lemma 1.8, C is an h-

since Q is a seml-simple

Theorem 1.4 that A is a direct summand of C. Ext~(Q,A)

ring, we see from

This proves that

= O, and hence hdRQ = I.

Remarks.

Hamsher [4] has proved for an integral domain R that

hdRQ = 1 if and only if every divisible R-module is h-dlvislble. Therefore,

because of Theorem 1.4, both of these conditions

are

equivalent to the condition that the torsion submodule of a divisible module is a direct summand.

I suspect that these results hold true

even if we only assume that Q is a semi-simple ring instead of a field.

However, we shall not need these facts, and thus we shall

not reproduce here the proof of Hamsher's Theorem I.i0.

result.

Let A be an R-module and C a cotorsion R-module

containing A such that C/A is torsion-free

and divisible.

Then the

following statements are true: (I)

If D is a cotorsion R-module,

then every R-homomorphism

from A into D can be lifted uniquely to an R-homomorphism of C into D.

10

(2) sion-free

I f D is a cotorsion and divisible,

Proof.

R-module containing A, and D/A is tor-

then D is isomorphic

to C.

We have an exact sequence:

HomR(C/A,D ) -~ HOmR(C,h ) ~ HomR(A,D ) -~ Ext~(C/A,D). The end terms of this sequence are O, since D is a cotorsion R-module.

Thus HomR(C,D ) ~ HomR(A,D ) .

The first statement of the theorem

follows easily from this, and the second statement follows easily from the first.

CHAPTER

II

COMPLETIONS Definitions:

We shall

say that an ideal of R is a regular

ideal if it contains

a regular

we define a topology

on A, called

submodules

r is a regular

a uniform denoted

of R.

If A is an R-module,

the R-topology

of the form IA, where I is a regular

system of neighborhoods where

element

topology,

by ~.

the inverse

of 0 in A.

element

of A, by taking the ideal of R, as a

The submodules

of the form rA,

of R, give the same topology.

and hence

It is standard

every R-module point

limit of the R-modules

This is

has a unique completion

set topology

that A is equal to

A/rA:

= lira A/rA. Let C be the direct product ranges

over the regular

If we represent

elements

of the R-modules of R.

A/rA,

where

Then A is a submodule

r of C.

an element x of C in the form (x r + rA), where

x r ~ A, then ~ ~ ~ if and only if x r - Xrs c r A

for every regular

r and s in R.

~ : A--A

~(x) where

We have a natural

= (x + rA) for every x ~ A. r ranges

over the regular

Theorem 2.1. bounded

order.

R-homomorphism

given by

The kernel elements

Let A be an R-module

of ~ is equal to N rA, r of R.

whose tcrsion

submodule

has

Then

(1)

Ker N is the divisible

(2)

A is a cotorsion

(3)

t(A)

~ t(A),

submoduie

of A and is torslon-free.

R-module.

and thus if A is torslon-free,

X is also tor-

sion-free.

(4) A/~(A) is Proof. all regular divisible

(i)

torsion-free

and divisible

and

A/~(A) ~ Ext~(Q,A).

It is clear that ker ~ = N rA, where

elements

submodule

of R.

Since t(A)

has bounded

of A and is torsion-free.

r ranges

over

order, A rA is the

i2

(2)

F o r each r e g u l a r

~ s ( ( Z r + rA>) where

=

s ranges

morphism

of C / A into a d i r e c t

Theorem

1.5,

(3)

of c o p i e s

z r c A).

elements

product

of R.

of copies

of C is a c o t o r s i o n

A is a c o t o r s i o n

regular

element

r ¢ R.

can a s s u m e

s ¢ R such that

H e n c e Xrs ¢ r A

without

Xrv - x v ( v A N

element

t(A)

+ t(A);

= O.

(Xrv - x v) = x r - X r v ¢ r A .

= Z(~(A)). (4)

Because

of C.

of t(A).

= O.

Therefore,

~ = ~ ( x v)

( A and s u p p o s e

H e n c e x = sy for some y ~ A, and ~

that A/~(A)

element

of R such that vt(A)

element

r ( R there

Yr - Yrq

¢rA

= <x r + rA>

element

+ t(A),

Therefore

A/D(A)

t(A)

t(~(A))~t(A). for

some regular

¢ ~(A)c~(A).

is t o r s i o n - f r e e . Le~ s be a r e g u l a r

Let v be a r e g u l a r F o r each

regular

Yr ¢ A s u c h that Xru - x u = uy rE urA.

and thus vy r - V Y r q ~ r A .

Hence Let ~ = (vy r + rA);

- uy r + rA> = (x u + rA> = ~(Xu)

c D(A).

is divisible.

W e h a v e an e x a c t

The end terms of this HomR(Q,A/~(A))

of A.

= O, and let u = vs.

q ¢ R, uy r - UYrq

then ~ ¢ A and ~ - s ~ = (Xru

- ~(y)

that A/~(A)

is d i v i s i b l e .

an e l e m e n t

is an e l e m e n t

and h e n c e

sx r - x ( rA for every

and we have p r o v e d

of R a n d ~

r.

Then

sx = ~(x)

r c R.

F o r any r e g u l a r

Then

that

element

element

~ rA, we

for every

and divisible,

s ( R and x c A.

We shall next p r o v e

regular

Thus x r - x v = (x r - X r v )

element

~ ~ ~(A)~

Thus by

T h e n t h e r e is a

that x r ¢ t(A)

regular

Therefore,

C l e a r l y any

and since x r - Xrs

K e r ~ is t o r s l o n - f r e e

Let ~ = (x r + rA)

~ k e r ~s" s Thus we have a m o n o -

R-module.

of R s u c h that vt(A)

= vt(A)

Then ~ =

sx r ~ rA for e v e r y

loss of g e n e r a l i t y

Let v be a r e g u l a r

: C -+ C by

R-module.

Let x = <x r + rA) be an e l e m e n t

element

+

s c R define ~s

(where

over all r e g u l a r

direct product

fixed

element

sequence:

sequence

~ Ext~(Q,D(A)).

are zero b e c a u s e But

since A/D(A)

A is c o t o r s i o n . is t o r s l o n - f r e e

Hence and

13 divisible,

we have H o m R ( Q , A / D ( A ) )

Ext~(Q,~(A)). Ext~(Q,Ker

~ A/~(A).

Thus A/~(A)

Since K e r ~ is a Q - m o d u l e ,

~) = 0 f o r all n > 0 by

we have

[7, Ch. VI,

Ext~(Q,Ker

Prop.

O)

4.1.3].

It

follows from this that

Definition. a*

: K-~K

Let A be an R-module,

@R A b y a*(x)

an R - h o m o m o r p h i s m

Proof. x @ a, w h e r e

~R A) is the d i v i s i b l e

submodule

k o f A onto H o m R ( K , K

Every

element

the r e g u l a r

of A;

and t h e r e

some r e g u l a r

in the f o r m element

r ~ R.

t h e n x @ a = 0 if and o n l y if a ~ rA.

1 ® (a-rb)

It is n o w c l e a r elements

T h e n the k e r n e l

such that k = k~.

of K @R A can be w r i t t e n

= O, t h e n in Q ~R A we have

I.I.

~R A)

a ~ A and x = r -I + R for

some b ~ A, and hence Theorem

= a*.

R-module.

If x ® a is of this form, if x ® a

®R A) by k(a)

We can t h e n d e f i n e

Let A be a t o r s i o n - f r e e

of k : A-*HOmR(K,K is an i s o m o r p h i s m

We d e f i n e

= x ® a for e v e r y x ¢ K.

k : A ~ HomR(K,K

T h e o r e m 2.2.

and a ( A.

= O.

For

r -I @ a = 1 @ b = r -I ® rb for Therefore,

that K e r k = A

a - rb = 0 by

rA, w h e r e

r ranges

of R, and h e n c e K e r X is the d i v i s i b l e

over

submod-

ule of A. Let ~ = be an e l e m e n t p h i s m ~*

: K-~K

@ R A by ~*(x)

regular

element

defined

R-homomorphism.

: A-+H°mR(K'K Suppose

It is not hard to v e r i f y We can then d e f i n e

®R A) by [(~)

the proof,

an R - h o m o m o r -

= ~*.

that ~* is a w e l l -

an R - h o m o m o r p h l s m

It is c l e a r that k = [~.

is an e l e m e n t

a r ~ rA for e v e r y

of K e r ~. regular

T h e n b y the

r c R, and hence

Thus ~ is a m o n o m o r p h i s m .

Let f ~ H o m R ( K , K t h e r e is an element f(r-l+R)

We d e f i n e

= x @ a r if x ~ K and rx = 0 for some

that ~ =

first p a r a g r a p h = 0.

r of R.

of A.

= (c'l+R)

T h e n 0 = (c-l+R)

@R A);

t h e n if r is a r e g u l a r

a r ~ A such that

f(r-l+R)

element

= (r-l+R)

@ b w h e r e b ~ A and c is a r e g u l a r

@ a r.

element

® rb, and h e n c e by the first p a r a g r a p h

of R, For of R.

of the p r o o f

14 there is an element a r ~ A such that rb = ca r . = (c-lr-l+R)

® ca r = (c-lr-l+R)

for any regular element =

f(r-l+R)

~ rb = (c'l÷R) ® b = f(r-l+R).

s ~ R we have:

- sf(s-lr-l+R)

= O.

Thus (r-l+R) ® a r

(r'l+R)

® (ar-ars)

Hence a r - ars ¢ r A ,

= (a r + rA> is an element of A.

Now

and thus

Clearly ~(~) = f, and we have

shown that ~ is an isomorphism.

Corollary 2.3. lowing statements

A is a cotorsion R-module.

(2)

A is complete in the R-topology.

(3)

A ~ HomR(K,K @R A).

(4)

A ~ HomR(K,B)

torsion-free

R-module.

Then the fol-

are equivalent:

(1)

Proof.

1.5.

Let A be a torsion-free

for some R-module B.

(i) -~-> (2).

By Theorem 2.1, A is cotorsion,

and divisible.

But A/A is a reduced R-module by Theorem

Hence A = A, and thus A is complete

in the R-topology.

(2) -~-> (3).

This is an immediate

(3) ~ >

(4).

This is a trivial assertion.

(4) = >

(1).

This follows directly from Theorem 1.6.

Corollary 2.4. equivalence

and A/A is

(Duality).

consequence

of Theorem 2.2.

The functor HomR(K,. ) i__ssa natural

from the category~o_~f

torsion-free,

cotorsion R-modules

onto the category ~ of torsion h-divlsible modules;

and its inverse

and adjoint is given by K ®R "" If A is in ~

and B is in S ,

then

(I)

H°mR(K ~R A,B) ~ HomR(A, HOmR(K,B))

(2)

H°mR(h'K ®R A) ~ HOmR(HOmR(K,B),A).

Proof.

The first statement

Corollary 1.2 and Corollary 2.3. canonical

duality isomorphism

second isomorphism,

of the theorem is a consequence The first isomorphism

[3, Ch. Ii, Prop. 5.2].

let C = HomR(K,B).

of

is the To prove the

Then B ~ K @R C and

15

A ~ HomR(K,K

@R A).

Thus the second isomorphism

reduces

to the

first.

Definitions. R-topology. naturally

We shall let H denote

If we identify

isomorphic

the completion

of R in the

K ~R R with K, then by Theorem 2.2,

to HomR(K,K),

Theorem 2.2 sends an element

H is

and the m a p p i n g ~ of R into H of

of R into m u l t i p l i c a t i o n

by that ele-

ment on K.

Corollary

2.5.

(i)

H is a commutative

(2)

We have an exact sequence 0 -~ d(R)

where d{R)

--

is the divisible

ring,

and is a torsion-free

R i H -~ Ext~(Q,R) submodule

R-module•

0

--

of R and k is a ring homomor-

phism. (3) Ext~(H,C)

Tor~(H,B)

= 0 for every torsion

= 0 for every cotorsion

Proof.

Let f be an element

R-module

R-module

B and

C•

of H, let r be a regular

element

of

R and let y = r -I + R in K. Since

rf(y)

= O, it follows

for some s c R (depending

on y).

that f(y)

~ Ry, and hence f(y) = sy

If g is another

g(y) = vy for some v ~ R.

Thus g(f(y))

= f(g(y)).

showing that H is a commutative

Hence gf = fg,

Since H = HomR(K,K), (2)

The exact

= g{sy)

element

H is a torsion-free

sequence

is a direct

of H, then

= svy = f(vy) ring.

R-module.

consequence

of Theorems

2.1

and 2•2. (3)

This is a fairly

easy consequence

• ®R B and HomR(-,C ) to the exact Remarks.

sequence

of applying of (2).

As we shall see in Theorem 2.7,

ideal of R, then H/HI % H ®R R / 1 %

R/I.

the functors

if I is a regular

It follows

that the comple-

16 tion of H in the R-topology is equal to the completion of R in the R-topology;

that is, H is complete in the R-topology.

Corollary 2.6.

(l)

Assume that R is a reduced R-module.

I f Q is a semi-simple ring,

Then:

then H is a faithfully flat

R-module. (2)

If H is a faithfully flat R-module and A is any R-module,

then H @R A is a finitely generated ~roJectiv ~ H-module if and only if A is a finitely generated projective R-module. (3)

I f R is a Noetherian local ring and hdRQ = l, then H is a

faithfully flat R-module and H @R A is thee completion o f A for every finitely generated R-module A. Proof. Ext~(Q,R)

(1)

Since Q is a seml-simple

is a projective Q-module.

follows that Ext~(Q,R)

ring, the Q-module

Because Q is flat over R, it

is a flat R-module.

(2)

Hence exact sequence

of Corollary 2.5 shows that H is a faithfully flat R-module

(see

[2]). (2)

This is a property of faithfully flat ring extensions of

R (see [2]). (3) sequence:

If A is a finitely generated R-module, Fl-~F0

free R-modules.

~A~O,

we have an exact

where F 1 and F 0 are finitely generated

Hence we have an exact sequence:

H ®R FI a_. H D R F 0 ~ H D R A-~ O, and H @R FI and H D R F 0 are finitely generated free H-modules. fore Im a is an h-reduced R-module,

There-

and hence Ker a is a cotorsion

R-module by Theorem 1.5.

But since hdRQ = l, Im a is a cotorsion

R-module by Theorem 1.7.

Because Im a = Ker S, it follows from

another application of Theorem 1.7 that H @R A is a cotorsion R-module. We have an exact sequence:

17

0 -~ Tor~(H,A) -~ T o r ~ ( H / R , A ) 8-~ A -- H ®R A "-~ (H/R) D R A -- O. Because A is finitely generated, Lemma that A is a reduced R-module. because H/R is a Q-module. phism Tor~(H,A)

it follows from the Nakayama However Tor~(H/R,A)

is divisible

Therefore 8 = 0 and we have an isomor-

~ Tor~(H/R,A),

and an imbedding A c H ~R A such that

(H D R A)/A is torsion-free and divisible.

If A is the completion of

A, then ~ is a cotorsion R-module and A / A is torsion-free and divisible by Theorem 2.1.

Hence b y Theorem I.I0, A is isomorphic to

H D R A.

If I is an ideal of R, then we have an exact sequence: O-*Tor~(H,R/I)

- * H @R I - ~ H - ~ H

But we have shown that Tor~(H,R/I)

~ Tor~(H/R,R/I)

and divisible and H ~R I is cotorsion. = Tor~(H/R,R/I).

®R R/I -~0.

Therefore,

is torsion-free Tor~(H,R/I)

= 0

It follows that H and H/R are flat R-modules.

Thus,

as is well known (see [2]), H is a faithfully flat R-module. Theorem 2.7.

(I)

i_~f T is a torsion R-module,

Thus T has the s~ructure of an H-module, unique.

and this structure is

Every R-submodule of T is also an H-submodule of T.

is any H-module, (2)

then T ~ H D R T.

then HomR(T,A ) = HomH(T,A ) .

I f C is a eotorsion R-module,

has the structure of an H-module, A is any H-module, Proof.

(I)

If A

then C ~ HomR(H,C).

Thus C

and this structure is unique.

If

then HomR(A,C ) = HomH(A,C ) . Because d(R) @R T = 0 = Ext~(Q,R)

@R T, we see that

if we tensor the exact sequence of Corollary 2.5 with T, we obtain an isomorphism: for all x ¢ T.

8 : T-~H

D R T, where % is defined by 8(x) = 1 @ x

Thus T has an H-module structure extending that of R

defined by fx = 8-1(f ® x) for all f c H and x ~ T. Suppose that there is another H-module structure on T extending that of R, and let us denote it by f o x.

There is a regular element

18

r ( R (depending

on x)

there are e l e m e n t s fore,

such that

and hence

fx = sx also

an H - s u b m o d u l e

shows

= ~(sx)

= s~(x)

is an H - h o m o m o r p h i s m , (2)

HA(R)

is divisible,

the two H - s t r u c t u r e

There-

are i d e n t i c a l .

that e v e r y R - s u b m o d u l e

of T is a l s o

of T.

If ~ is an R - h o m o m o r p h i s m ~(fx)

Because

g ~ H and s c R such that f = rg + k(s).

fx = sx = f . x;

The fact that

rx = 0.

= f~(x)

of T into an H - m o d u l e - g~(rx)

= f~(x).

A, then

Thus we see that

and hence H o m R ( T , A ) = H o m H ( T , A ) .

We have an exact

sequence:

HomR(H/~ (R) ,C) -~ HomR(H,C ) ~-~ C -~ Ext~(H/~ (R) ,C) w h e r e ~ is d e f i n e d a cotorsion Corollary

R-module,

2.5,

of this

sequence

C has the

and d i v i s i b l e are zero,

structure

we have D = ~.

C is by

and thus v

of an H - m o d u l e of

= x. H-module

structure

and let us d e n o t e

F o r a fixed x E C, d e f i n e D : H ~ C by ~(f)

= ~(1)

Since

f ( H, x E C and ~ is an e l e m e n t

that C has a n o t h e r

it by f - x. Then ~(~)

where

such that ~(1)

Suppose

for all e ~ H o m R ( H , C ) .

and H A l R ) is t o r s i o n - f r e e

Therefore,

b y fx = ~(f),

HomR(H,C)

= ~(i)

the end m o d u l e s

is an i s o m o r p h i s m . defined

by ~(~)

= x = ~(1) Therefore

= ~(~),

= f • x.

and since ~ is an i s o m o r p h i s m ,

f • x = fx,

and we

see that the H - s t r u c t u r e

on C is unique. Let A be any H - m o d u l e ; the c a n o n i c a l

duality

then

isomorphism

since C ~ H o m R ( H , C ) of

[3, Ch. II,

Prop.

we can u t i l i z e 5.2]

and ob-

tain:

H°mH(A'C)

~ H°mH(A'H°mR(H'C))

Definitions:

A commutative

if it has o n l y one m a x i m a l called

a local

of m a x i m a l

ideal.

~ H°mR(H

ring is c a l l e d A Noetherian,

ring , and a N o e t h e r i a n

ideals

is c a l l e d

~H A,C)

~ HomR(A,C).

a quasi-local quasi-local

ring w i t h o n l y a f i n i t e

a semi-local

ring.

ring ring is number

i9 Theorem 2.8.

There is a one-to-one correspondence between the

set of regular ideals I of R and the set of regular ideals J of H such that k-l(j) (i)

is a regular ideal of R that satisfies the following

I f I is a regular ideal of R, then H ~R I is the completion

off I, and H ~R I ~ HI. H/HI ~ R/I;

and (HI)/k(I)

more, Tor~(H,I) (2)

k-!(HI) = I:

is torsion-free and divisible.

Further-

= O.

If J is an ideal of H such that k-l(J) is a regular ideal of

R then J = H(k-l(J)) (3)

HI is a regular ideal of H;

and H/J ~ R/(X-I(J)).

I f I 1 and 12 are regular ideals of R, then H(I 1 n I2)

= HI 1 A HI 2 and H(I 1 + 12) = HI 1 + HI 2. (4)

I f R is a quasi-local

ring with maximal ideal M, then H i__ss

quasi-local ring w i t h maximal ideal HM. Proof.

(i)

We have a map ~ i : H 8R I ~ HI defined by

~ l ( f ® x) = fx for all f c H and x c I; ~2 : H ~R R - ~ H

and we have a map

defined in a similar fashion.

We also have the

canonical map ~3 : H ~R R/I -~H/HI defined by ~3(f ~ (r+l)) = fr + HI. Thus we have a commutative diagram:

0-~H

@R I - ~ H

0 -* HI

@R R - ~ H

' H

@R R/! - * 0

' H/HI

The top row is exact by Corollary 2.5,

~0

since TorR(H,R/I)

= O.

It is

clear that the vertical maps are epimorphisms and that the middle vertical map is an isomorphism.

It follows from diagram chasing

that the end vertical maps are isomorphisms

H ~R I "-- HI;

also.

Therefore,

and by Theorem 2.7, H/HI is isomorphic

to R/I.

Now k(1) c HI, and since H is a torsion-free R-module, k(1) contains a regular element of H. Because H/HI is isomorphic

Therefore,

HI is a regular ideal of H.

to R/I, we see that H/HI is a torsion R-

module of bounded order, and hence is a cotorsion R-module.

H is

20 also a cotorsion Theorem

R-module,

R-module

by

1.5.

Let i' = k-l(HI); of R.

and thus HI is a cotorsion

then I c I', and hence

I' is a regular

ideal

Thus by what has already been proved we have H/HI' ~ R/I'

H/HI ~ R/I.

But clearly

HI = HI',

and thus R/I' ~ R/I.

and

It follows

from this that I = I' = X-I(HI). Because k-l(HI)

= I, we have HI n X(R)

H/k(R)

is a divisible

H/X(R)

=

R-module,

(Hi+k(R))/k(R)

is a torsion-free

and divisible

If ~ is the divisible a regular

ideal of R.

2.1,

then K e r

~

=

Therefore

~ / ~ is torsion-free have proved

that

By Theorem have H ®R K ~ K.

k(I) % I / ~

Therefore,

and hence Hl/k(1)

c I because

If ~ is the com-

= ~.

By Theorem 2.1,

and I is a cotorsion

R-module.

and divisible

Hence it follows

from Theorem

I.I we have Tor~(H,K)

= 0;

Thus we have an exact

sequence an isomorphism:

and that I.I0 t h a t

~H

all

®R Q ~ K

and by Theorem 2.7 we

sequence: ~0"

n > 0 by Theorem 1.1,

we o b t a i n

from this

Tor~(H,I) ~ Tor~(~ % Q,I). By ~3, Ch. VI,

Prop. ~ . 1 . 2 ] we have Tor~(H % Q,I) ~ Tor~(H % Q,Q % I ) . I is a regular ideal,

It follows

Let J be a regular ideal of R.

H = HI + X(R).

But sinoe

we have Q ®R ! ~ QI = Q, and thus

Tor~(H ®R Q'Q ®R I) = O.

regular

We

to HI.

R Torn(K,! ) = 0 for

(2)

I is

and ~ : I ~ I is the map of Theorem

and divisible

R-module.

~

= ~.

hence ~(I) ~ I / ~

0 ~H Since

of R, then

(HI)/~ is also torsion-free

HI is a cotorsion is isomorphic

, and

~

= HI/X(1),

2.5,

R-module.

submodule

pletion of I in the R-topology

By Corollary

and thus H = HI + k(R).

HI/(HI n k(R))

~

= k(I).

= O.

ideal of H such that I = k-l(J)

Since H/X(R)

Consequently,

that Tor~(H,I)

is a divisible

J = HI + k(I)

we see that H/J = H/HI ~ R/k-l(J).

= HI.

R-module,

is a

we have

Therefore,

by (1)

21

(3)

Let I I and 12 be regular ideals

I 1 + 1 2 are also regular ideals

of R.

Then I I D 12 and

of R, and we have an exact sequence:

0 - ~ I 1 N I 2 - ~ I 1 ¢ 1 2 -~I 1 + 12 -~0. By

(1) we have Tor[(H,I 1 + I2) = 0.

diagram with exact

0-~H

rows:

~R (If N 12) ~

(H ~R If) @

0----~HI 1 N Hi 2 The middle

Hence we obtain a commutative

(H ~R I2) - ~ H ~R

~ HI I ® Hi 2

(If + I2) ~ 0

~ HI I + HI 2

vertical map is an isomorphism

by (I).

'~ 0

Since it is ob-

vious that HI ! + HI 2 = H(I I + 12) , the right hand vertical map is also an isomorphism isomorphism.

by (I).

Because

Thus the left hand vertical map is an

the image of this map is H(I I ~ 12), we have

H(I 1 n 12) = HI 1 ~ HI 2. (%)

Suppose

that R is a quasi-local

By (i) HM is a m a x i m a l of HM, where x r ( R.

ideal of H.

x r is in M.

ideal M.

Let ~ : (x r + rR~ be an element

It is easy to see that if some x s is a unit in

R, then every x r is a unit in R. contradicting

ring with maximal

But then ~ would be a unit in H,

the fact that H}4 is a proper But then every element

1 - ~ is a unit in H.

ideal of H.

Thus every

1 - x r is a unit in R, and hence

This shows that HM is the only maximal

ideal

of H.

Theorem 2.9. let H(S)

Let S be a ring extension

denote the completion

a torsion-free

R-module

of R in Q,

o f S in the S-topology.

and we have an exact

sequence:

O-*HOmR(K,S/R ) -~H ~-~H(S) 8-~Ext~(K,S/R) where ~ i__ssa rin___gghomomorphism. (i)

If S/R has bounded

order the sequence

becomes:

S~Q,

and

Then H(S)

is

22

0 -* H ~--* H ( S ) ~ S / R -~ 0 . (2)

I f hdRQ = I, then 6 is onto.

(3)

I f hdRQ = 1 and S/R i_~s h-divisiblue,

0-~HOmR(K,S/R) Proof.

By T h e o r e m 2 . 2 ,

first that HOms(Q/S,Q/S) s ~ S.

an R-homomorphism x ~ Q/S. R-module. H(S)

H(S) ~ HOms(Q/S,Q/S).

We s h a l l

by ~(x)

Then b~ = 0, and hence

show

Let g ~ HomR(Q/S,Q/S)

a, b ( R and b is regular in R.

~ : Q/S-*Q/S

= sg(x)

- g(sx)

and

Define

for all

~ = 0 since Q/S is a divisible

Thus g ~ HOms(Q/S,Q/S),

is thus a torsion-free

becomes:

--H ~ H(S) -*0.

= HomR(Q/S,Q/S).

Then s = a/b where

the sequence

proving

R-module,

the desired

equality.

since Q/S is a divisible

R-mod-

ule. By Theorem 2.7, = HomH(Q/S,Q/S).

Q/S is an H-module

and HomR(Q/S,Q/S )

Thus we have a ring homomorphism

defined by ~(f)(x)

~ : H-~H(S)

= fx for all f E H and x ~ Q/S.

An argument

sim-

ilar to that given in Theorem 2.7 shows easily that ker

= HomR(K,S/R). Let ker

~ be the

~ = S/R.

canonical

Then from the

R-homomorphism of K onto exact

sequence

Q/S with

0-~S/R-*K--*w

Q/S-~0

we obtain a diagram:

HomR(K,K) HomR(K,Q/S)

°\

xt (K,S/R)

xt (K,K)

HornR (Q/S, Q/S) where ~ and k are induced by ~, and the top row is exact. easily verified

that the diagram

because k is obviously Im ~ = ker 8. theorem.

commutes.

a monomorphism,

This establishes

It is

We let 8 = ~k;

it is easily checked

the first

exact sequence

then, that

of the

2] In fact, k is an isomorphism. is easily verified

For let f ¢ HomR(K,Q/S).

that if q ~ Q, then there exists an element

Sq ~ S such that f(q + R) = Sqq + S. induces an R-homomorphism hence k is onto.

Thus f(S/R)

g : Q/S-~Q/S.

= 0, and hence f

But then k(g) = f, and

Since we have already observed

phism, we see that k is an isomorphism. If hdRQ = l, then Ext~(K,K) shows that ~ is onto, h-divisible,

Then it

This proves that Im ~ = Im 6.

= 0 by Theorem 1.7.

and hence 8 is onto.

then Ext~(K,S/R)

that k is a monomor-

The diagram then

If in addition S/R is

= 0 by Theorem 1.7, and hence

(3) is

established. Suppose now that S/R has bounded order. HomR(K,S/R ) = 0.

Since Ext~(Q,S/R)

that S/R % Ext~(K,S/R). Ext~(K,K)

% Ext~(Q,K)

Then,

of course,

= 0, we see by Theorem 1.1 (3)

We also see by Theorem 1.1 (3) that

and hence Ext~(K,K)

is torsion-free.

The

diagram then shows that v is onto, and hence 8 is onto in this case also.

Thus we have proved that the sequence

O - - H Z H(S) --S/R-- 0 is exact. Theorem 2.10. visible

H is a subrin~ of H ®R Q;

submodule of R, then Q / $

correspondence

and if Z> is the di-

c II ®R Q and there is a one-to-one

between the se ~ of rings B between R and Q and the set

of rings 0 between H and H ®R Q ~-yen b-lf 0 = H ®R B and B/~9 = 0 n Q/~.

Furthermore:

(1)

B/R and (H D R B)/H are isomorphic

(2)

Q/B a_nd (H ®R Q)/(H D E B) are isomorphic

(3)

If B = B / ~ ,

(4)

(H D R B)/B is a torsion-free

Proof. Tor~(H,Q/B)

torsion modules. torsion modules.

then B c H @R B, and H ®R B = HB = H + B. and divisible R-module.

Since Q/B and B/R are torsion R-modules, = 0 and TorR(H,B/R)

we have

= 0 by Corollary 2.5 (3);

and we

24 have H ~R Q/B ~ Q/B and H ~R B/R ~ B/R by Theorem 2.7

(I).

Thus we

have exact sequences: 0 ~ H ~ H @R B ~ B/R ~ 0 and 0 - - H @R B ~ H

®R Q ~ Q / B

~0.

If we let B = Q, then we see that H c H @R Q" H @R B is a ring between

In general we see that

H and H @R Q' and we have proved

(1) and

(2) of the theorem. Now if V is any Q-module, Prop. 4.1.2]

we have Tor~(B,V)

Q ~R B ~ Q, it follows Tor~(B,Ext~(Q,R)) exact

sequence

then Tor~(B,V)

~ Tor~(Q @R B,V);

that Tor~(B,V)

= 0.

Therefore

of Corollary

0 - - ~ ®R B ~

that

2.5(2)

exact

sequence

and since

let R : R / ~

we have

and tensor

with B we obtain an exact

~B

the

sequence

~R B ~ 0.

=~.

Therefore,

Thus we have proved

(H 8R B ) / B is a torsion-free

For by [3, Ch. VI,

In particular,

H ~R B ~ Ext~(Q,R)

~ ~ ~R B.

The previous

= 0.

if we

Now g9 is an ideal of Q, and hence then B = B/X~

= 0.

if B = B / ~ ,

that B c H ®R B and

and divisible

R-module.

shows that if x c B and ~ = x + /9~,

then we may identify ~ w i t h 1 ® x in H ~R Q"

Furthermore,

since

H c H @R Q' we may identify

an element h ~ H with its image h ® 1 in n H @R Q" If k ~ H @RnB, then k = Z h i ® xi, where h i ~ H and i=l x i ~ B. Hence k = Z h i ~ i ~ HB, showing that H ®R B = HB. Suppose i=! that h ¢ H and x ~ B. Then x = a/b, where a ~ R and b is a regular element there

of R.

exist

5 = r + ~ =

Because

elements ¢ R).

(g ® a) +

H / ~ is a divisible

R-module

by Corollary

g ¢ H and r ~ R such that h = bg + 5

2.5,

(where

Thus h @ x = (bg + 5) ® x = bg ® x + (5 ® x)

(I @ rx)

=

(ag ® i) +

(i ® rx) ¢ H + B.

Thus we have

shown that H ®R B = H + B. We identify Q = Q / ~ with B. suppose

w i t h a subring of H @R Q as we have done

Then of course we have B c that ~ ~ (H ®R B) A Q.

(H ®R B) N Q.

Then there

exists

On the other hand, a regular

element

25 b in R such that bk ~ B. torsion-free,

However,

and thus k ~ B.

Let 0 be a ring between

we have proved

This proves

then because regular

it is sufficient

to show that 0 = H + B.

(H ®R Q)/H ~ K is a torsion R-module,

and hence there

h = bg + ~. Therefore,

Thus b(k-g) k = g + ~

exist

elements

Theorem 2.11.

~ H ÷ B, proving

(2)

Let ~

in Q.

Since

Assume

submodule

that

of R and let

every regular

element

Then

The categories eotorsion

exists a

is complete.

be the divisible

of R, is isomorphic

sion-free,

Let k ~ O;

But H/R is a divisible

that 0 = H + B.

Q is the full ring of quotients

completion

the proof

= ~ ¢ R, and hence k - g = r/b ~ D A Q = B.

= R / ~ , Q = Q / ~ , and K = Q/~. of Q is invertible

there

Then

g ~ H and r ~ R such that

H + B = H @R B, the proof of the theorem

(1)

(H ®R B) A Q = B.

R and Q, and to complete

element b ~ R such that bk = h ~ H.

R-module,

(H ®R B ) / B is

H and H ®R Q' and let B = 0 A Q.

= B / ~ , where B is a ring between of the theorem

that

that

K % K;

to H, the completion

of torsion

R-modules

of R;

h-divisible

and H, the

of R.

R-modules

and tor-

are equal to the same categories

of

-modules. {3)

I is a regular

ideal

of R, if and only if ~ : I / ~

regular ideal of R and then the completions topologies, Proof.

respectively, Now ~

an ideal of Q.

of I and ~ in the R and

are isomorphic.

is torsion-free

Therefore

is a

and divisible,

~ is a ring;

and thus ~

and by assumption,

is

Q is equal

to its own full ring of quotients. Let b be a regular in R. since ~

If a ¢ R and ba ~ ~ is a divisible

a = d because element

element

of E.

R-module.

If we let ~ element

the image of b

, then ba = bd for some element

R is torsion-free.

that b is a regular

of R, and b = b + ~ ,

But then b(a-d)

d ~ ~,

= 0, and hence

This shows that b is a regular

be the set of all elements b in R such of R, then we have Q % Rj .

Since

26 is a subset of the regular

elements

of R, it follows

tained in the full ring of quotients its own full ring of quotients ring of quotients Since Q ~

of R.

However,

by assumption,

= ~9~ c R, we have

~K

= Hom~(K,K),

and thus R and R have isomorphic

torsion

h-divisible 1.2,

h-divisible

R-module

R, and hence ~ is a regular ideal of R.

cotorsion

R-module

if and only if

in I.

ideal

of R.

Then there is a

In the first para-

is a regular

Conversely,

element

of

let [ be a

Then there is an ideal I of R such that element

of ~;

Since Q is the full ring of quotients

ment q of Q such that ~

= i.

Now q = q + ~ ,

then ~ = c + ~ ,

where q E Q and hence

element of R.

ca/b - 1 E ~ ,

Therefore

and hence ca - b ~ ~ .

that I is a regular

If I is a regular @R I) ~ H o m ~

of I in the R-topology

where

of R, there is an ele-

q = a/b where a E R and b is a regular

H°mR(K'K

h-divisible

R-module.

that b = b + ~

Let c be a regular

I, showing

Every

2.3, we can prove

ideal of R and let ~ = I / ~ .

graph of the proof we showed

c E I.

using Corollary

is a torsion-free ootorsion

A is a torsion

element b of R that is contained

= I//~.

completions.

On the other hand if A is an R-module,

fashion,

Let I be a regular

regular

HomR(K,K )

is of the form K @R A for some R-module

~-module.

it is a torsicn-free

regular

Hence K is

and clearly K ~R A ~ K D R A % K @ ~ (R ~R A) is a

In similar

that an R-module

+ R)/R = O.

Furthermore,

then R D R A % A, and hence K D R A ~ K ~ R-module.

and thus Q is the full

: (Q~

and of course K ~ K.

A by Corollary

Q is equal to

of R.

an R-module,

torsion

that Q is con-

Thus

b is an element

of

ideal of R.

ideal of R we have K ®R I ~ ~ @ ~ ~, and hence (K,K ~

~).

Thus by Theorem 2.2 the completion

is isomorphic

to the completion

of ~ in the

R-topology. T h e o r e m 2.12.

H is complete

in the H-topology.

Furthermore,

if

27

an R-module cotorsion

is a cotorsion

(or torsion)

Proof. H-module.

(or torsion)

Let k : R - ~ H be t h e ring homomorphism

also a torsion

element

of R, then k(b)

H-module.

H-module,

H-topology.

R-module.

By Corollary

H-module.

Let V be the full

ring of quotients

R-module,

HomH(V,C ) c HomR(V,C ) = O.

element

= 0.

Therefore

Let us consider

where A is an H-module,

R-module,

of H;

is

an exact

on C.

in the

it is suf-

then C is a co-

then V is a torsion-

it is sufficient sequence

Thus to prove

of the form:

and f and g are H-homomorphisms.

Since

s : A - ~ C such that

s

But HomR(A,C ) = HomH(A,C ) by Theorem 2.7,

s is an H-homomorphism.

that

f-~A g-~v--~O

= 0, there is an R - h o m o m o r p h i s m

is the identity

of H

R-module

and hence V is a Q-module.

0--C

splits

is a regular

2.5.

then we will have shown that H is complete

torsion

hence

of Corollary

Hence to finish the proof of the theorem,

free and divisible

is an

2.3 if we prove that H is a

to prove that if C is a cotorsion

Ext~(V,C)

R-module

Thus a torsion

ficient

ExtH(V,C)

then it is also a

H-module.

H is a torsion-free

cotorsion

R-module

By Theorem 2.7 a cotorsion

If b is a regular because

(or torsion)

f

o

and

Therefore, the given exact sequence

over H, showing that Ext~(V,C)

= 0.

CHAPTER COMPATIBLE

Theorem

3.1.

torsion-free

III

EXTENSIONS

Let A be a c o m m u t a t i v e

as an R-module.

ring e x t e n s i o n

Then the f o l l o w i n g

of R that is

statements

are

equivalent: (1)

If J is a r e g u l a r

ideal

of A, then J O R is a r e g u l a r

(2)

If x is a r e g u l a r

element

ideal

of R.

y ( A such that xy = b is a r e g u l a r (3)

A ®R Q is the full

Proof.

(1) = >

is a r e g u l a r (1).

If x is a r e g u l a r

of A.

element

of A, then Ax ideal

of R by

y ~ A such that yx = b is a r e g u l a r

of R. (1).

of J.

By

Let

J be a r e g u l a r

(2) there

J O R is a r e g u l a r (2) = >

(3).

of A, then X = z/x, (2) t h e r e

ideal

is an e l e m e n t

H e n c e k = z y / x y = zy/b.

t h e r e is an e l e m e n t 1/x = y/b.

torsion-free of A, then

3.2.

ring of q u o t i e n t s element

of A.

b o t h b and y are r e g u l a r

expression

ring of q u o t i e n t s

Let x be a r e g u l a r

element

y ¢ A and a r e g u l a r

J A R is a r e g u l a r

in the

for k shows that of A. of A.

element

Then by

(3),

b ~ R such that

(2).

Let A be a c o m m u t a t i v e

as an R-module,

By

element

Since A ~R Q = AQ is c o n t a i n e d

H e n c e xy = b, p r o v i n g

Theorem

of the full

R-module,

of A, this

to the full

(3) ~---> (2).

of R in Ax O R, and

y ~ A such that xy = b is a r e g u l a r

in A.

A @R Q is equal

element

z ¢ A and x is a r e g u l a r

Since A is a t o r s i o n - f r e e

ring of q u o t i e n t s

of A, and x a r e g u l a r

of R.

If X is an e l e m e n t where

ideal

is a r e g u l a r

of R.

full

is an element

of R.

ideal of A and hence Ax O R is a r e g u l a r

(2) = > element

element

ring of q u o t i e n t s

Thus there is an e l e m e n t

element

hence

(2).

of A, then there

and a s s u m e ideal

ring e x t e n s i o n

of R that is

that if J is a r e g u l a r

of R.

Then the R - t o p o l o g y

ideal and

29

the A-topology on any A-module coincide. has one of the following properties

Furthermore,

an A-module

as an A-module if and only if it

has the same property as an R-module: (I) (5)

torsion,

divisible, Proof.

(6)

(2)

h-reduced,

(3)

cotorsion,

(4)

complete,

h-divisible.

It is obvious from Theorem 3.1 that the R-topology and

the A-topology on an A-module coincide

and that (1), (4), ~und (5) are

t rue. Let B be an A-module. Ext~(Q,B)

By [5, Ch. VI, Prop. 4.1.3], we have

~ ExtA(A ~R Q,B) for all n _> 0.

quotients

of A by Theorem 3.1,

have a commutative

Since A ~R Q is the ring of

(2) and (3) are seen to be true.

We

diagram:

H°mA(A ~R Q,B) ~ HomR(Q,B)

B

where ~ is the canonical duality isomorphism, kR(f) = f(1) for f ¢ HOmR(Q,B), Theorem 1.1, B is h-divisible (resp. hA) is an epimorphism. Theorem 3.3.

X R is defined by

and h A is defined similarly.

By

over R (or over A) if and only if

R

Thus we see that (6) is true.

Suppose that A is an integral domain containing R

and tha___~tthe quotient field of A is algebraic over Q. tient field of A is isomorphic

Then the quo-

to A ~R Q' and thus all of the state-

ments of Theorem 3.2 are true for R and A. Proof.

If x is a nonzero element of A, then the m i n i m u m poly-

nomial for x over Q has a nonzero constant term.

From this it fol-

lows readily that if J is a nonzero ideal of A, then J D R ~ 0. Theorem 3.3 is thus a consequence of Theorems 3.1 and 3.2. Theorem 3.4.

Let A be a commutative

is torslon-free as an R-module and a s s ~ e

ring extension of R, that that if J is any regular

3O

ideal

of A, then J O R is a r e g u l a r

statements

ideal

of R.

are equivalent:

(I)

A = J + R for e v e r y

(2)

A / R is a d i v i s i b l e

(3)

A / J "-- R / ( J A R) for e v e r y r e g u l a r

I f any of these regular

Then the f o l l o w i n g

regular ideal

J o_ff A.

R-module.

statements

are true,

ideal

J of A.

then J = A ( J A R) for e v e r y

ideal J of A.

Proof. a regular showing

(i) ~---> (2).

ideal

If b is a r e g u l a r

of A, and hence

A = Ab + R.

that A / R is a d i v i s i b l e

(2) -~-> (I). element

of R, then Ab is

Thus b(A/R)

= A/R,

R-module.

If J is a r e g u l a r

b ~ J A R.

element

Then b(A/R)

ideal

= A/R,

of A, c h o o s e

a regular

and hence A = Ab + R

cJ+RcA. (I) -----> (3).

Let J be a r e g u l a r

ideal

of A;

then

ideal

of A.

Since A / J is a

A/J = ( J + R ) / J ~ R / ( J N R). (3) ==> cyclic

(i).

R-module

A = Rx + J.

Let

J be a r e g u l a r

by assumption,

there

Thus 1 = rx + J, w h e r e

have A : Ar + J.

We also

is an e l e m e n t x ~ A such that r ~ R and j E J.

have Ar = Rxr + Jr,

Therefore,

we

and hence A = Rxr + J.

But Rxr + J = R + J, and thus A = R + J. Assume

that any one of t h e s e

let J be a r e g u l a r

i d e a l of A.

equivalent

If we let I = A(J A R),

regular

ideal of A and I c J.

J = I +

(R A J) c I, and we see that

Definitions: shall

of

R,

Thus by

is true,

and

t h e n I is a

(1) we have A -- I + R.

Hence

J = I.

Let A be a c o m m u t a t i v e

say that A is a c o m p a t i b l e

following

statements

extension

ring e x t e n s i o n

of R.

of R if it s a t i s f i e s

We the

conditions:

(1)

A is t o r s i o n - f r e e

(2)

A is not

(3)

if J is a r e g u l a r

as an R - m o d u l e .

equal to its full

ring of q u o t i e n t s .

i d e a l of A, t h e n J A R is a r e g u l a r

ideal

31

(4)

A = J + R for every

We shall

regular

say that R is a c l o s e d

ideal

J of A.

domain

if R is an i n t e g r a l

domain,

and if J O R ~ 0 for e v e r y n o n z e r o

course,

H is also an i n t e g r a l

module,

and H/R is a d i v i s i b l e

R-module,

we

in this case H is a c o m p a t i b l e

extension

of R.

the s t a t e m e n t s Theorem

3.1,

J of H.

the q u o t i e n t

field

domain,

F o r if D is a proper,

In p a r t i c u l a r ,

J O R = O,

morphism

on K.

since

This

nonzero,

all of

By

of K is

R-submodule

of H by C o r o l l a r y

element

proves

R-

to H @R Q"

h-divisible

ideal

every n o n z e r o

contradiction

R and H.

every p r o p e r R - s u b m o d u l e

of K, then J = H o m R ( K , D ) is a n o n z e r o However,

of

see b y T h e o r e m 3.4 that

of H is e q u a l

then

Then,

Since H is a t o r s i o n - f r e e

of T h e o r e m 3.2 are then true about

If R is a c l o s e d h-reduced.

domain.

ideal

that

1.3.

of R acts as an epievery p r o p e r

submod-

ule of K is h - r e d u c e d . We have p r o v e d domain,

[9, T h e o r e m 4.5]

then the c o n v e r s e

ule of K is h - r e d u c e d ,

Corollary

3.5.

R (A not a s s u m e d

is true;

that if R is a q u a s i - l o c a l namely,

then R is a c l o s e d

Let A be a p r o p e r

to be a ring).

if e v e r y p r o p e r R - s u b m o d -

domain.

R-submodule

Then A is a c o m p a t i b l e

sion of R if and o n l y if A/R is a d i v i s i b l e

Proof.

Suppose

nonzero

element

element

of R.

every r e g u l a r

in A, we ideal

and A is a t o r s i o n - f r e e extension

where

R-module.

see that A x c

of R by T h e o r e m

A.

of A has a r e g u l a r

R-module. 3.4.

Let x be a

a ~ R and b is a r e g u l a r

Then A = Ab + R, and h e n c e Ax = Aa + Rx.

and Rx are c o n t a i n e d Clearly

then x = a/b,

ring exten-

R-module.

that A i R is a d i v i s i b l e

of A;

of Q c o n t a i n i n g

Therefore,

The c o n v e r s e

Since Aa

Thus A is a ring. intersection

with

A is a c o m p a t i b l e statement

R, ring

also f o l l o w s

from T h e o r e m 3.4.

T h e o r e m 3.6.

Let A be a c o m p a t i b l e

ring e x t e n s i o n

of R.

Then:

32

(i)

A @R Q i__ssthe full ring of quotients

statements (2)

of Theorem

3.2 are true for R and A.

If J is a regular

A/J % R/(J D R).

of A, and all of the

ideal of A, then A = J + R and

Furthermore

J = A(J O R) and J D R is a regular

ideal of R. (3)

A finitely

generated (4)

torsion

generated

torsion

(5)

R-module.

ideal of A is finitely

N is a regular maximal

if N O R is a regular maximal (6)

ular ideal of A.

ideal

prime)

If A is a reduced

Let S be an R-submodule

is a proper

over A.

ideal of R.

% R/M.

ideal M, then AM i s

AM contains

R-module,

every reg-

then AM is the only

ring.

of Q containing

R.

and AS/S is a divisible

subring of Q containing

over R,

ideal of A if and only

ring w i t h maximal

of A and A / ~

A ~R Q we have AS = A + S;

generated

generated

prime)

ideal of A and A is a quasi-local

(7)

S.

(resp.

(resp.

If R is a quasi-local

r_egular maximal

ule.

is also a finitely

If every regular ideal of R is finitely

then every regular

maximal

A-module

Then in R-module.

If S

R, then AS/S is a divisible

Thus if AS ~ A ~R Q' then AS is a compatible

S-mod-

ring extension

of

If A c S, then S is also an A-module. (8)

If R is an integral

tion subring Proof. Theorems cyclic follows

of Q containing (i),

(2),

3.1 and 3.4.

torsion

(4) and

Assume

ideal of R.

(3), a cyclic

torsion

A-module

every

regular

of generators.

ideal of A.

of R and hence N O R = M.

ideal of A.

is a

statement

By (5), N A R By (2) we have

AM = A(N O R) = N, and thus AM is the only regular maximal and contains

of

and let M be the maximal

Let N be any regular maximal ideal

consequences

and the general

on the number

that R is quasi-local

is a regular maximal

A + V = Q or A c V.

(5) are immediate

by Theorem 3.4,

easily by induction

(6).

if A c Q, and if V is a valua-

R, then either

As for

R-module

domain,

ideal of A

By (2) we have A/AM ~ R/M.

33

If A is not a q u a s i - l o c a l of A such that I ~ AM.

ring,

then there is a proper ideal I

Choose an element y E I such that y ~ AM.

Since I is not a regular ideal of A by the p r e c e d i n g paragraph, a zero divisor in A, and hence there exists a nonzero such that yz = 0.

Let b be any regular element of R.

is a regular ideal of A and is not contained Ab + Ay = A.

Thus Az = b A z .

If we assume A to be reduced,

this c o n t r a d i c t i o n

quasi-local

ideal AM.

(7).

Let S be an R - s u b m o d u l e

R-module,

AS c A + S.

we have Aq = Aa + Rq.

Since Ab + Ay

submodule

of A.

shows that A is a

of Q c o n t a i n i n g

then q = a/b w h e r e a ¢ R and b is regular in R. divisible

element z ¢ A

in AM, we have

Hence Az is a divisible

ring w i t h m a x i m a l

y is

R.

Let q ~ S;

Because A/E is a

Thus Aq c A + S~ and hence

Since 1 ¢ A N S, we have the reverse inequality.

We

then have AS/S = (A ÷ S)/S ~ A/(S n A), and hence A S / S is a homomorphic image of A/R. Suppose

Therefore,

that S is a proper

AS/S is a divisible R-module. subring of Q c o n t a i n i n g R.

lows from Theorem 3.2 that since AS/S is a divisible also a divisible

S-module.

It fol-

R-module it is

It is obvious that the i n t e r s e c t i o n with

S of a regular ideal of AS is a regular ideal of S. AS ~ A @R Q' then AS is a c o m p a t i b l e

ring extension

Therefore,

if

of S b y Theorem

3.4. If A c S, then AS = A ÷ S = S, and hence S is an A-module. (8).

Assume that R is an integral

V is a v a l u a t i o n

subring of Q c o n t a i n i n g R.

A V = A + V is a p r o p e r V - s u b m o d u l e V-module

of bounded

thus AV = V.

domain,

order.

By

of Q.

that A c Q, and that

If A + V ~ Q, then

Hence

(AV)/V is a torsion

(6) A V / V is a divisible V-module,

Hence we have A c V.

and

CHAPTER

IV

LOCALIZATIONS Definitions:

A commutative

n < ~ if e v e r y chain of p r i m e and there is at l e a s t

ring is said to have K r u l l

ideals

Noetherian

ring of K r u l l

R contains

a regular

mary component

ring of K r u l l

I x = O, or AnnR(x)

AnnR(x ) = [r¢

let T be a t o r s i o n

R

I rx = 0].

w h e r e M S ranges

R-module.

Mi-primary.

decomposition

can w r i t e

1 = rI +

x = rlx +

... + rnX.

hence

ideal

from

closed.

I, and let T be a

of R, d e f i n e

the M a - p r i -

is an M a - p r i m a r y

Clearly

Since

Therefore,

sum of the Ta's.

T a is a s u b m o d u l e

Cohen-Macaulay

of T.

ring and

Ta

element

of I~

(k ~ i),

... + rn, w h e r e

is an M i - p r i m a r y

ideal]

ideals

of R.

Furthermore,

is an RMjmOdUle

If L i = A Jk'

i = 1,...,n.

differs

Then

Let x be a n o n z e r o

Take a p r i m a r y

ideal of

of K r u l l d i m e n s i o n

Such a d o m a i n

dimension

over all of the m a x i m a l

and hence

Proof.

domain

Let R be a 1 - d i m e n s l o n a l

T=Z@

TMj

ring if it is a

that it m a y not be i n t e g r a l l y

If M a is a m a x i m a l

T h e o r e m 4.1.

%

ring.

say

of T by:

T a = [x C T where

We will

i, and if e v e r y m a x i m a l

A Noetherian

ring o n l y in the fact

R-module.

n + I terms.

Cohen-Macaulay

Cohen-Macaulay

Let R be a N o e t h e r i a n torsion

exactly

dimensional

element.

1 is a 1 - d i m e n s i o n a l a Dedekind

in R has at m o s t n + i terms,

one chain w i t h

that a ring R is a 1 - d i m e n s i o n a l

dimension

of T, and let I = AnnR(x).

I = Jl ~

then R = L 1 + r i ~ L i.

ideal,

H e n c e we

Jiri x = 0 for

AnnR(rix ) contains

or rix = O.

... + L n.

Ji is

Thus we h a v e

Jiri c I, we have

either

"'' ~ Jn w h e r e

a power

This p r o v e s

of M i and

that T is the

35

We prove next that the sum is direct. maximal

ideals

of R, and suppose

that x ( T 1 N

integers

M ~k2

kI k2 Since R = M 1 + (M 1

M knx n

= 0.

(j E> 1 Tj).

Then

k1 kl,...,k n such that M 1 x = 0 and

there are positive

...

Let MI,...,M n be distinct

shown that T = Z @ T~. a It is obvious that T a is uniquely

k .. Mnn)

-

we see that x = 0.

,

Hence we have

R - Ma, and hence

(T~)M~ = o.

(Ta)Ma Z T a.

Thus T M ~

Z~

by the elements

On the other hand,

of

if 8 ~ a, we have

(T~)M : ( T ~ ) M ~ T~.

8

Theorem 4.2.

divisible

a

Let R be a 1-dimensional

Cohen-Macaulay

ring.

Then hdRQ = I. Proof.

Case I.

R is a local

Let b be a regular J

= [bn].

Noetherian

rank 0 and is a m a x i m a l

ideals

ideal of R j .

Since R c R j

Let F be a Tree R-module define an R-eplmorphism

Case II.

f : F-*Q

has

every regular

General

that R j

free basis

= i/b n.

= Q. [x n] and

It is then easy

case. h-divisible

R-module.

by Theorem

Ext~(K,B)

an exact

Let us consider

ele-

of f, then P is a free R-module w i t h

that hdRQ = l, it is sufficient

(i)

Hence R~

is equal to its own full

c Q, it follows

by f(Xn)

the

Hence we have hdRQ = I.

Let B be a torsion

= 0.

of R j .

Therefore,

w i t h a countable

to see that if P is the kernel {Xn+l-bXn].

between

ring and every prime ideal of Rj

is a unit in Rj , and hence R /

ring of quotients.

ideal of R, and let

correspondence

of rank 0 in R and the prime

is a semi-local

basis

of the m a x i m a l

Then there is a one-to-one

prime ideals

ment of Rj

element

ring.

In order to prove

1.7 to prove

sequence

that

of the form:

0 -~ B -* A - * K -+ O.

By Theorem 4.1 we have B Z aE ~ BMa,

A Z aZ @ A Ma and K ~ aZ ® KMa ,

36

where M a ranges quence

over all of the maximal

(1) is the direct

sum of the exact

0 If every one of these because

ideals

se-

0.

split,

(2) is an ~ - s e q u e n c e ,

Thus exact

sequences:

- %

sequences

of R.

then

(1) is also split.

it is sufficient

Hence

to prove that

C~

Ext'_ (K M , B M ) Ha G tients of RM, ~ module,

= 0.

If we prove that Q M

is the full ring of quo-

then we see that ~BM~ is a torsion,

h-divisible

and the theorem will follow from Case I and Theorem

Let M be a maximal

ideal of R and let C = Ix ¢ Q

RMa-

1.7.

I sx = 0 for

some s ~ R - M).

Then C is an ideal of Q and we let Q = Q/C.

Q is a semi-local

ring and every prime ideal of Q has rank 0 and is

maximal

in Q, the same is true of Qo

ring of quotients. and M = M/I. quotients

then I c M, and we let ~ = R/I,

that R M = %

- M is a zero divisor ring of quotients Theorem 4.~. Then an R-module Proof.

in the full

ring of

But since Q is equal to its full ring of quotients,

that Q is the full ring of quotients

We observe

full

Thus Q is equal to its full

We have ~ c Q, and Q is contained

of R.

it follows

Let I = R 0 C;

Since

and QM = QM"

Because

no element

of

in R, we have Q ~ = Q, and hence Q = Q M

is the

of R M-

Let R be a 1-dimensional is divisible

Of course

of R.

Cohen-Macaulay

ring.

if and only if it is h-divisible.

an h-divisible

R-module

is divisible,

and

hence we must prove the converse. Case I:

R is a local

Let b be a regular = {bn}.

element

of the maximal

ideal

of R and let

As in the proof of Theorem 4.2 we then see that Q = R ~ .

Let D be a divisible found Xn,

ring.

R-module,

and let x ~ D.

let Xn+ 1 be an element

an R-homomorphism

Let x = xi, and having

of D such that bXn+ 1 = x n.

f : Q - ~ D by f(a/b n) = aXn+l,

Define

where a ¢ R and

37 n > O.

It is easily verified

and that f(1) copies

= x.

Thus D is a homomorphic

of Q and thus D is an h-divisible

Case II:

General

we see that D/h(D)

the torsion M a ranges

I.

R-module.

is h-reduced

that D is h-reduced,

sum of

R-module.

by Theorem 1.9.

of D;

Thus we can assume

that D = O.

then by Theorem 4.1,

over all of the m a x i m a l

divisible

image of a direct

Since hdRQ = 1 by Theorem 4.2,

and we shall prove

submodule

R-homomorphism

Case.

Let D be a divisible

sion,

that f is a well-defined

ideals

of R.

Let T = t(D),

T ~ Z @ T

, where

Since TMa is a tor-

RM -module, TMa is an h - d i v i s i b l e I~a-module by Case

But in the proof of Theorem 4.2 we saw that the full ring of quo-

tients

of RM~ is a homomorphic

divisible free.

R-module.

image

Thus TMa = 0 for all M a, and hence D is torsion-

But then D is h-divisible, Corollary

4.4.

w i t h n o nilpotent module

Q ~ RP1 @

elements

... @ Rp,.~

other than zero. R-module

Let P1,...,Pt

have 0 = P1 0

and therefore

Let R be a 1-dimensional

of every divisible

Proof.

of Q, and thus TMa is an h-

and thus Q is a semi-simple

an immediate

consequence

Remarks. study divisible no nonzero

Corollary modules

nilpotent

when R is a local D = t(D) @ D/t(D). direct

over the fields pletely

elements,

we

that RPi is a field for

ring.

over a 1-dimensional

The theorem

is now

1.4.

Cohen-Macaulay

it is sufficient

For if D is a divisible is a Q-module,

Hence D/t(D)

that are direct

described

Then

4.4 and Theorem 4.1 show that in order to

Now D/t(D)

sum of fields.

of rank 0 in R.

of Theorem 4.3 and Theorem

elements,

ring.

summand.

nilpotent

From this it follows

i = 1,...,t,

ring

Then the torsion-sub-

ideals

Since R has no nonzero

... 0 Pt"

Cohen-Macaulay

is a direct

be the prime

D = O.

by the dimensions

R-module,

the case then

and Q is a finite

is a direct

summands

to consider

ring w i t h

sum of vector

spaces

of Q, and hence can be com-

of these vector

spaces

over these

38

fields.

For the torsion part,

divisible, maximal

torsion

RMa-modules

ideals M S of R.

of the vector

Hence

space components

If R is a 1-dimensional potent

elements,

= Z @ T a is a direct

Ta, where M a ranges

to the local case.

modules

and observe

studied

locally.

of D/t(D),

over all of the

we shall know D completely.

Cohen-Macaulay

ring with nonzero

define Artinian

that they are torsion R-modules,

in the rest of these notes

reasons we shall to local,

nil-

the study of its torsion

We shall presently

For these

sum of

if we k n o w the Ta's and the dimensions

then we can still reduce

modules

rings.

t(D)

But before we do this we still need

and hence can be

restrict

1-dimensional

R-

our attention

Cohen-Macaulay

some results

of a general

nature. Definitions. reducible

An ideal of a commutative

if it is not the intersection

An R-module

of two p r o p e r l y

is said to be indecomposable

of two properly The proof

smaller

ring is said to be irlarger ideals.

if it is not the direct

sum

submodules.

of the next theorem may be found in [9] and we shall

not prove it here. Theorem 4.5. (1)

Let R be a commutative

An injective

C = E(R/P),

where

R-module

E(R/P)

Noetherian

C is indecomposable

is the inJective

envelope

ring.

Then:

if and only if of R/P,

P a prime

ideal of R. (2)

An ideal I o f R i_ss irreducible

where x is a nonzero (3) injective (4) Rp.

element

Every injective

if and only if I = AnnR(x),

of an indecomposable

R-module

is a direct

injective

R-module.

sum of indecomposable

R-modules. Let P be a prime ideal of R and Rp the P-adlc

Then E(R/P)

is also the injective as an ~p-module

envelope

Rp/PRp.

Viewing

E(R/P)

gives

morphism

between

Rp and HomR(E(R/P), E(R/P) ) •

completion

of

over Rp and Rp o f

rise to a natural

iso-

39 (5)

If P is a prime ideal of R, then E = E(R/P) = U knnE(Pn), n where AnnE(Pn ) = {x ( E I P nx = 0]. I_~f P is a maximal ideal of R, then AnnE(Pn ) i_~sa finitely generated R-module. (6)

I_~f I is an ideal of R, and I = I 1 A ... N I n is an irredun-

dant intersection o_~f irreducible E(R/II) ~ ... ~ E(R/In).

ideals Ij, then E(R/I)

Furthermore,

Ij i_~sPj-primary for some

prime ideal Pj and E(R/Ij) ~ E(R/Pj). (7)

Let R be a complete,

ideal M, and E = E(R/M). = It

Noetherian local rin~ with maximal

If I is an ideal of R, then AnnR(AnnE(I))

and if D is a submodule of E, then AnnE(AnnR(D)) Definitions:

If R is a commutative

a set of R-submodules

A1 c ~

ring, A an R-module,

and

of A, then ~- is said to satisfy the

Ascending Chain Condition ules in ~ :

= D.

(ACC) if for every ascending chain of mod-

c ... c A n c ... there is an index n o such that

An = AnO for all n _> n O .

This is equivalent

to the assertion that

every subset of ~- has a maximal element with respect to inclusion. If A has ACC on the set of all of its submodules, be a Noetherian

R-module.

then A is said to

A Noetherian R-module is finitely gen-

erated. In similar fashion we define the Descending Chain Condition (DCC) by reversing the inclusion DCC is equivalent

relations.

to asserting that every subset of ~

element with respect to inclusion. its submodules,

To say that ~

satisfies

has a minimal

If A has DCC on the set of all of

then A is said to be an Artinian R-module.

An

Artinian module over an integral domain or a 1-dimensional Noetherian Cohen-Macaulay

ring is e torsion R-module.

If an R-module is both Artinian and Noetherian, that it has finite length (in the classical

we shall say

sense).

The proof o f the following theorem is obtained by a trivial modification of the proofs of [9, Theorem 4.2] and [9, Corollary

40

4.3]

and shall not be repeated

here.

Theorem 4.6.

Let R be a commutative,

(Duality).

local ring w i t h maximal M-adic

topology,

ideal M, let R be the completion

and let E = E(R/M)

Let A be an R-module.

Noetherian

b_~e the injective

Then A is an Artinian

H o m R ( A , E ) is a Noetherian

R-module.

of R in the

envelope

R-module

of R/M.

if and only if

l__Snthi____sscas____eewehav____~e:

A ~ Hom~(HOmR(A,E),E). I n Particular

A is an Artinian

ule o~f E n, where

R-module

E n is a direct

Conversely,

sum of n > 0 copies

let B be an R-module.

ule i_~f an__~donly if Hom~(B,E)

if and only if A is a submodof E.

Then B is a Noetherian

is an Artinian

R-module.

~-mod-

In this case,

we have:

B % H o m R ( H O m ~ ( B , E ) ,E). The next theorem

is a g e n e r a l i z a t i o n

proof given here is due to Hamsher Theorem 4.7.

Suppose

ules of Q containing ment and non-unit y ~ B/R.

quasi-local

c

R and A ~ Q and B ~ Q. Then

But then regular

ring,

loss of generality

(c -1 + R) = r-lx ~ A/R. element

ele-

(c -1 + R) = x + y, where x E A/R and

cx = 0 = cy, and hence there exist

unit and R is a quasi-local Without

Then K

Let c be a regular

- (r + t)c -1 E R, and thus 1 - (r + t) E R c .

unit.

ring.

that K = A/R @ B/R, where A and B are R-submod-

and t in R such that x = rc -1 + R and y = tc -1 + R. -1

and the

R-module.

of R.

Therefore,

Cor. 4.2],

[4].

Let R be a commutative,

is an indecomposable Proof.

of [13,

it follows

r

But then

Since c is not a

that either

we can assume

elements

r or t is a

that r is a unit.

Thus we have shown that if c is a

of R, then either c -1 ~ A or c -1 ~ B.

Because A ~ Q and B ~ Q, there

exist

elements

e and d in R that

41

are regular elements and not units of R such that c -I ( A and d -1 c B.

Now c-ld -1 ¢ A U B, and hence we can assume that

c-ld -1 ~ A.

Thus d -1 = c(c-ld -1) ~ A A B = R, contradicting

that d is not a unit of R.

Therefore,

the fact

K is an indecomposable

R-mod-

ule. Theorem 4.8.

Let R be a Noetherian domain of Krull dimension I,

M a maximal ideal of R, a n d different

from M.

Proof.

Then~

[M a] a collection of maximal ideals of R

RMa D R h

"-- Q and hence

(Oe RM) + h

= Q"

If x ¢ Q, then x = a/b, where a, b ~ R and b ~ O.

b is contained

Since

in only a finite number of maximal ideals of R, we see

that x ( RMa for all but a finite number of Ma's. an R-homomorphism

~ : Q -~ Z @ K e

by ~(x)

Thus we can define

= E (x + h e )

M~

for all x ~ Q.

e

It is obvious that Ker ~ = n h ' and thus we have a monomorphism e 0 -* Q/(Ne h a ) -~ a7 ~ K Me. By Theorem 4.1, KMe is isomorphic to the Me-primary

component

of K and hence KMe D R h

(E ~ KMa ) D R R M = 0, from which it follows Therefore,

(he h a )

DR RM --- Q"

Thus we have

that Q/(N RMa) D R R M = 0.

It follows from this that

[(Oa RMe) + RM]N = Q for every maximal

(Oe h ~ ) + h : Q"

= 0.

ideal N of R, and thus

CHAPTER V

ARTINIAN DIVISIBLE MODULES Throughout the remainder of these notes R will be a Noetherian, local, 1-dimensional, Theorem 5.0.

(1)

Cohen-Macaulay ring with maximal ideal M.

The following statements are true.

The R-tp~01ogy and the M-adic topology o_nnannR-module are

the same.

(2)

H is complete and is the completion of R in the M-adic

topology.

(3)

H is a complete, Noetherian, local, 1-dimenslonal Cohen-

Macaulay ring with maximal ideal HM. (4)

H is a faithfully flat R-module.

(5)

I_~f A is a finitely generated R-module, then H D R A is the

completion of A in the R (o__~rM-adic) topology. Proof.

(1)

A regular ideal of R is an M-primary ideal of R,

and hence contains a power of M.

It follows easily from this that

the R-topology and the M-adic topology on an R-module are the same. (2)

If I is a regular ideal of R, then by Theorem 2.8, we have

H/HI ~ R/I.

Thus the completion of H in the R-topology is equal to

the completion of R in the R-topology; R-topology.

that is, H is complete in the

Therefore H is complete in the M-adic topology, and is

the M-adic completion of R. (3)

Statements (3), (4), and (5) could of course be established

by a reference to general theorems in any standard textbook on Noetherian rings.

However, in the 1-dimensional case we are con-

sidering, they follow readily from theorems we have already proved. Hence for the sake of completeness, we shall sketch the proofs here. By Corollary 2.5 and Theorem 2.8, H is a commutative, quasi-local ring with maximal ideal HM.

Clearly the (HM)-adic topology on H is

equal to the M-adlc topology on H, and thus H is complete in the

43 (HM)-adic topology.

If b is a regular element of M, then Rb contains

a power of M and hence n (HM) n = n Hb n. But O Hb n is a divisible Rn n n submodule of H, and thus O (HM) n = O. HM is a finitely generated n ideal of H, and is generated by elements bl,...,b kLet G(H) be the graded ring of H w i t h respect to HM;

that is,

G(H) =

~ ~ (HM)n/(HM) n+l w i t h m u l t i p l i c a t i o n defined in the obn=O vious way. If we let Si = bi + (HM)2' then G(H) = (H/HM)~b I .... ,bn ~, and hence G(H) is a commutative,

Noetherian ring.

If h is a nonzero

element of H, then there exists an integer n ~ 0 such that h ~ (HM) n but h # (HM) n+l.

If we let h = h + (HM) n+l in G(H), then h is called

the leading form of h. If J is a nonzero ideal of H, let ~

be the ideal of G(H) gen-

erated by the leading forms of the nonzero elements of J.

Since

G(H) is a Noetherian ring, there is a finite set of elements hl,--°,h t in J whose leading forms generate

~

over G(H).

Since H

is complete in the (HM)-adic topology and n (HM) n = O~ it is not n hard to show that h 1,...,h t generate J over H. Thus H is a NoetherJan ring. Let b be a regular element of M. HM, Hb is an HM-primary ideal. ideal theorem Therefore,

Since Hb contains a power of

It follows from the Krull principal

~18, Ch. III, Theorem 61 that H has Krull dimension 1.

H is a complete,

local,

1-dimensional,

Cohen-Macaulay

ring. (4) and (5).

Since hdRQ = 1 by Theorem 4.2,

these statements

are true by Corollary 2.6 (3). Remarks.

We shall now recapitulate

some of the results of the

earlier chapters as they apply to a local, Macaulay ring R. and K = Q/R.

1-dlmensional~

Cohen-

As before, Q is the full ring of quotients of R

We have hdRQ = 1 and hdRK ~ 1.

R is the R-topology,

and H -= HomR(K,K ) .

divisible R-module isomorphic

H/R is a torsion-free and

to ExtlCQ,R) i ~

H is the completion of

~

6

44

Every divisible nilpotent

elements,

ule is a direct divisible ule,

R-module

is h-divisible.

then the torsion

summand.

submodule,

and B/h(B)

then its completion

to HomR(K,K

is reduced.

then h(B)

is a cotorsion

R-mod-

is its

If B is a torsion

If A is a torsion-free

the maximal

R-module

ideal of R by M.

if and only if it is an M-primary

the M-adic

topology

Noetherian,

local,

on an R-module 1-dimensional,

ideal HM and H/HM ~ R/M. a finitely

of a divisible

R-mod-

R-

and is i s o m o ~ h i e

D R A).

We shall denote regular

submodule

If B is an R-module,

then h(B) ~ K ~R H°mR(K'B)"

module,

If R has no nonzero

generated

ideal.

The R-topology

and

are the same.

H is a complete,

Cohen-Macaulay

ring w i t h maximal

H is a faithfully

R-module,

An ideal of R is

flat R-module.

then its completion

If A is

is isomorphic

to

H ~R A. If E = E ( R ~ )

is the injective

also the inJective HomR(E,E)

~ H.

converse

H-modules An Artinian

is alse true.

Definitions. define

Theorem

and Artinian R-module

Artinian

reduced.

By the Nakayama

Conversely,

generated.

the duality of Theorem 4.6 beR-modules

R-modules

= [x ~ A

Let A be an Artinian

i_ff and only if A is finitely Proof.

By Theorem 4.5 we hate

given by the funcH-module,

suppose

generated

and I is an ideal of R we shall

R-module. (i.e.,

it contains

with respect

to the property

Because

B is reduced,

there exists

= 0] and the

Then A is reduced

has finite generated

that A is reduced

minimal

IrA

I Ix = 0}.

Lemma a finitely

Since A is Artinian,

and the

are torsion R-modules.

of A in R by AnnRA = (r c R

of I in A by AnnAI 5.1.

of R/~ over R, then E is

is an A r t i n i a n

If A is an R-module

the annihilator

annihilator

of H/HM over H.

We also have available

tween Noetherian tor H O m R ( - , E ) .

envelope

envelope

length). R-module

but not finitely

a submodule

B that is

of not being finitely

a regular

is

element

generated.

r ¢ R such that

45

B ~ rB.

By the m i n i m a l i t y

It follows

of B we see that rB is finitely

that B/rB is not finitely

Since B/rB is Artinian, Theorem 4.6. integer

Because

B/rB is finitely finitely

generated

generated,

ideal of R, there exists

an

Thus we have B/rB c Ann nRr of A n nE ~ k , and

sum of n copies

R-module

by Theorem 4.5.

and this contradiction

Therefore,

proves

that A is

generated.

Corollary finitely sense.

Rr is an M-primary

But AnnEn M k is a direct

A n n ~ k is a finitely

generated.

we have B/rB c E n for some n > 0 by

k > 0 such that M k c Rr.

c AnnEnMk .

generated.

5.2.

generated

If A is an Artinian module, and hence has finite

then A/h(A)

is

length in the classical

Thus A = h(A) + B, where B is a finitely

generated

submodule

of A. Proof. generated

A/h(A)

by Theorem

Theorem 5.3. and C a finitely D

=

and hence is finitely

5.1.

Let D be a divisible generated

We have D/B ~

ible and finitely =

and reduced,

R-module,

submod__ul__~e o_~f D.

B a submodule

of D,

I_ff D = B + C, then

B.

Proof.

D

is Artinian

C/(B

generated.

N C), and hence D/B is b o t h divis-

By the Nakayama

Lemma,

we see that

B.

Theorem

5.4.

Given an exact

sequence

O-~A-+B-~

of R-modules:

C-~O

then

(i)

If A and C are reduced,

(2)

If B is reduced and A is Artinian,

Proof. Hence

suppose

B is also reduced.

If A and C are reduced, that B is reduced

then C is reduced.

then clearly B is reduced.

and that A is Artinian.

Then

46 H o m R ( Q , B ) = 0 and hence we have an exact

sequence:

0-*HOmR(Q,C ) ~Ext~(Q,A). By Theorem 5.1, A has finite module.

length,

On the other hand Ext~(Q,A)

Thus Ext~(Q,A)

and thus Ext~(Q,A) is torsion-free

= O, and by the preceding

that HomR(Q,C ) = 0.

exact

Hence C has no nonzero

is a torsion

and divisible.

sequence

this implies

h-divisible

submodules,

and thus C is reduced. Definitions. module

We will say that an R-module

if it is a nonzero

proper nonzero We will tion series

divisible

torsion,

module

that has no

submodules.

say that a torsion of divisible

divisible

is a simple divisible

divisible

modules

R-module

D has a composi-

if it has a chain of divisible

sub-

modules:

0 = DO c D 1 c D2 c such that Di/Di_ 1 is a simple The next theorem series

shows

of divisible modules

divisible

modules.

More

stronger

Definition. K.

A similar Theorem

lowing statements

R-module

that the property is confined

holds

a composition

to the class of Artinlan

it also shows that either the

submodules

condition

for i = l,...,n.

of having

is separately

equivalent

to

of being Artinlan.

We shall let K n denote a direct

statement 5.5.

divisible

surprisingly

ACC or the DCC on divisible the apparently

... c D n = D,

sum of n copies

of

for E n.

If D is a torsion

divisible

module,

then the fol-

are equivalent:

(1)

D has b o t h DCC and ACC on divisible

(2)

D has a composition

(3)

D is a homomorphic

series image

submodules.

of divisible modules.

of K n for some n > 0.

47

(4)

D is a submodule

(5)

D is an Artinian module.

(6)

D has ACC on divisible

submodules.

(7)

D has DCC on divisible

submodules.

Proof.

(4) < = = >

of E n for some n > O.

(5).

This equivalence

has already been stated

in Theorem 4.6. (1) ~-> simple

(2).

divisible

submodules

Since D has DCC on divisible submodule

D 1.

submodule

has ACC on divisible

of steps with a composition (2) ~ >

~ O;

element

series

submodules

of HomR(K,DI)

epimorphism

an epimorphlsm

of D.

1.3, module,

Suppose

gi:K i -~ D i.

.

every

that for a

Choose

1.3 an R - h o m o m o r p h i s m

f : K - * D i + ! such

and using gi and f we can obHence by induction

If we define M -I = [q ¢ Q

a regular

element

generated

submodule

extension

of M-I/R and thus K c E(M-I/R),

direct

submodules

divisible

is an epimorphism.

gi+l:Ki+l-~Di+l

Since D

there is an

gn on K n onto D n = D.

(3) ==> (5).

M-I/R.

with D/D2.

By Corollary

Then Di+ 1 = D i + f(K),

tain an epimorphism

Now D 2

... c D n = D be a composition

of D.

x ( Di+ 1 - Di, and by Corollary

D2/D 1.

stops in a finite number

of divisible

and since D 1 is a simple

given i > 0 we have found

that x ~ f(K).

the process

Let 0 = D O c D I c

of divisible

HomR(K,DI) nonzero

(3).

submodule

of D and we may proceed

submodules,

D has a

D/D 1 has DCC on divisible

and thus has a simple divisible

is a divisible

series

Clearly

submodules,

b ~ R such that bM -I c R. of K.

also K n) is contained K n is Artinian

Hence M-I/R is a finitely

It is easy to see that K is an essential

Since M-I/R is finitely sum of copies

I qM c R], then there is

generated,

of E by Theorem 4.5. in a finite

by Theorem 4.6,

direct

and hence

the injective E(M-I/R)

envelope

of

is a finite

Therefore, sum of copies

K (and hence of E.

every homomorphic

Thus

image of

K n is also Artinian. (5) ~ >

(6).

By Theorem 4.6 we have D c E n for some n > O.

Let

48

D I c D2 c

... c D m c

Then HOmR(K,DI) H-submodules R-module therian

... be a chain of divisible

c HomR(K,D2)

c

... c HomR(K,Dm)

of HomR(K,En ) by Theorem 2.7.

by the proof of (3) ~ > H-module

HomR(K,Dm)

c

of D.

... is a chain of

But K is an Artinian

(5), and hence HomR(K,E n) is a Noe-

by Theorem 4.6.

Thus there is an index m 0 such that

= HomR(K,Dmo ) for all m _> m O.

m ~ m 0 by Corollary

submodules

1.3, p r o v i n g

Therefore

D m = Dm0 for all

that D has ACC on divisible

submod-

ules. (6) -~-> (I).

Since D has ACC on divisible

submodules,

divisible

If x ¢ D, then by

a maximal

Artinian

Corollary

1.3 there is an R - h o m o m o r p h i s m

Since K is Artinian

submodule

by (3) = >

and hence f(K) c A. (7).

This is a trivial

(7) = >

(5).

It is a consequence

R-modules.

submodules,

Suppose

divisible

We assert

divisible

submodule

implication. of (3) = >

R-module

divisible

of D satisfying

that this assertion B 1 of D.

submodule

B 2 of A I.

divisible

submodule

of D.

construct

a divisible

submodule

B of D such that

Then there is a submodule Choose

A 1 of D such

a nonzero Artinian

Then B 1 @ B 2 is a nonzero

Continuing

submodule

Artinian

in this way we see that we can

of D that is an infinite

the f a c t that D has DCC on divisible

hence we have established

the existence

of a submodule

direct

sum.

submodules,

and

B of D with

properties.

We denote the injective respectively; E(D).

is a sum of Artinian

is false and choose a nonzero Artinian

divisible

the desired

(5) and Corollary

A O B = O, then A is reduced.

that A 1 n B 1 = 0 and A 1 is not reduced.

This contradicts

is Artinian,

that since D has DCC on divisible

there is an Artinian

if A is any submodule

such that x ~ f(K).

A = D.

(5) = >

divisible

f:K-~D

(5), we see that A + f(K)

Therefore,

1.3 that every torsion,

A of D.

there is

envelopes

of B and D by E(B)

and E(D),

and we can assume

that E(B)

is a direct

summand

of

Thus there is a submodule

C of E(D)

such that E(D) = E(B) @ C.

49

If we let A = D A C, then A O B = 0, and hence A is reduced. D/A is isomorphic

to a submodule

of E(B),

and E(B)

Theorem 4.6, we see that D / A is an Artinian Since D is a sum of Artinian marked,

it is sufficient

submodules D I c D2 c submodules

of D.

that [(D i + A)/A) modules

of D/A.

However,

ible submodules

(6).

divisible

Hence

and reduced,

divisible

sub-

there is an integer n O such From this it follows

by Theorem

immedi-

Therefore,

Since D n O A is Artinian

(D n N A)/(Dn0 A A) is also reduced

we see

and hence has ACC on divis-

(D n N A) for all n ~ n o .

(D n A A)/(Dno A A).

both divisible

divisible

Let

chain of Artinian

that D n + A = Dno + A for all n ~ n O .

Dn/Dno ~

as we have re-

chain of Artinian

D / A is Artinian

by (5) = >

ately that D n = Dno +

R-modules

(D i + A)/A ~ Di/(D i A A) is divisible

is an ascending

by

divisible module.

that D is Artinian.

... be an ascending

Since

is Artinian

to prove that D has ACC on Artlnian

in order to prove ... c D n c

divisible

Since

5.4.

and reduced,

Hence Dn/Dno

is

and we have D n = Dno for all n > n O •

Thus D is an Artinian module. Corollary submodule

If a torsion

divisible

A such that D/A is Artinian,

Proof. (7) ~ >

5.6.

This assertion

was proved

(5) in the p r e c e d i n g

Corollary

5.7.

ule of D is finitely

D has a reduced

then D is Artinian. as part of proving

that

theorem.

Let D be a torsion

is a simple divisible

R-module

R-module

divisible

R-module.

if and only if every proper

generated.

Then D R-submod-

In this case D is an Artinian

R-

module. Proof.

If D is a simple divisible

DCC on divisible

submodules

by Theorem

Since a proper

finitely

5.5.

generated

R-module,

in a trivial way,

by Theorem

submodule

5.1.

then D satisfies

and hence D is Artinian

of D is reduced,

it is

50 Definition.

We shall say that two R-modules are equivalent if

each is a homomorphic

image of the other.

This is clearly an equiv-

alence relation on the class of R-modules. R-modules,

we shall denote this by A - ~ B ,

If A and B are equivalent and we shall denote the

class of R-modules equivalent to A by [A]. Lemma 5.8.

If A is an Artinian divisible R-module and B is a

reduced submodule of A, then A/B is equivalent simple divisible R-module,

to A.

Thus if A is

every nonzero homomorphic

image of A is

a simple divisible module equivalent to A. Proof. (*)

Let D = A/B;

then we have an exact sequence:

0 -~ HomR(D,B ) -* HomR(D,A ) ~ HomR(D,D ) -* Ext~(D,B)

By Theorem 5.1, B has finite length, annihilator. Ext~(D,B),

and thus Ext~(D,B)

Let r ~ R be a regular element of the annihilator of

let I be the identity mapping on D, and let w : A ~ D

the canonical homomorphism.

Then by exact sequence

there exists g ~ HomR(D,A ) such that ~g = rl. morphism.

has a regular

It follows that A = B + g(D).

be

(.) we see that

Thus ~g is an epi-

Since B is finitely gen-

erated we have A = g(D) by Theorem 5.3, and hence g is an epimorphlsm.

Therefore,

we have proved that A is equivalent to D.

If A is a simple divisible R-module,

then A is Artinian by Cor-

ollary 5.7, and the second statement of the Lemma follows from the first and Corollary 5.7. Lemma 5.9.

Let A be an R-module and B c C submodules of A such

that C/B is a simple divisible R-module.

If S is a submodule of A

such that S O C i__ssArtinian an__ddreduced,

then (S + C)/(S + B) is a

si_~_~le divisible R-module equivalent to C/B. Proof.

Since B c (S + B) N C and (S + C)/(S + B)

C/((S + B) N C), we see that (S + C)/(S + B) is a homomorphic

image

of C/B and is ~Ither 0 or a simple divisible module equivalent to C/B

51

a c c o r d i n g to L e m m a 5.8.

T h e r e f o r e it is sufficient

S + C = S + B and arrive at a contradiction. have C = B +

(S A C), and hence C/B ~

to assume that

But in this case we

(S A C)/(S A B).

Since S A C

is f i n i t e l y generated by T h e o r e m 5.1, and C/B is a n o n z e r o module,

divisible

we have our desired contradiction.

Definition.

If A is an A r t i n l a n d i v i s i b l e R-module,

say that two c o m p o s i t i o n equivalent

series of d i v i s i b l e

if there is a one-to-one

factor m o d u l e s

submodules

correspondence

of A are

between the sets of

of the two series such that c o r r e s p o n d i n g

ules are equivalent.

Of course equivalent

we shall

composition

factor mod-

series then

have the same n u m b e r of terms in their series. Given two d e s c e n d i n g

chains of submodules

that one of the chains is a refinement every term of the other chain somewhere

of A, we shall say

of the other if it contains in its sequence.

Theorem 5.10 (A J o r d a n - H S l d e r type of Theorem). Artinian

divisible

Then

(1) (2)

A has a c o m p o s i t i o n

series of divisible R-modules.

Any two c o m p o s i t i o n

series for A are equivalent.

(3)

Any chain of divisible

composition Proof. (2) divisible of A.

R-module.

Let A be an

submodules

of A can be refined to a

series. (1)

This is part of Theorem 5.5.

Let 0 = A O c A 1 c submodules

... c A n = A be a c o m p o s i t i o n

of A, and let B be a simple divisible

series of submodule

There is an integer i such that B c Ai, but B ~ Ai_ 1.

of an obvious induction

0 c B c B + A1 c is a c o m p o s i t i o n

argument it is sufficient

... c B + Ai_ 1 c Ai+ 1 c

series of d i v i s i b l e

submodules

Because

to prove that

... c An = A of A equivalent

the given series in order to complete the p r o o f of (2).

to

52

Now Ai_ 1 ~ B + ~ - I of A i.

c ~,

and B + Ai_ I is a divisible submodule

Since Ai/Ai_ 1 is a simple divisible R-module,

divisible modules properly between ~ - I

and A i.

Therefore,

B + Ai_ 1 = A i and hence Ai/Ai_ 1 = (B + A i _ l ) / ~ _ 1 % equivalent to B by Lemma 5.8. that (B + Aj)/(B + Aj_I)

there are no

B/(B O q _ l )

is

On the other hand for J < i - 1 we see

is equivalent to Aj/Aj_ 1 by Lemma 5.9 and we

have proved our assertion. (3)

This is an immediate consequence of Theorem 5.5.

Definition.

If A is an Artinian R-module,

then according to

Theorem 5.10 every composition

series of divisible submodules of h(A)

has the same number of terms.

Thus we can unambiguously define

L(A) to be the number of terms in a composition submodules of h(A).

It can be immediately

series of divisible

seen that L(A) = 0 if and

only if A is reduced. Theorem 5.11.

Let O - ~ A - ~ B - ~ C

Artinian R-modules.

- ~ 0 be an exact sequence of

Then: L(B) = L(A) + L(C).

Proof.

We shall assume that C = B/A and that v : B - ~ C is the

canonical map.

By Corollary 5.2, B = h(B) + BI, where B I is a

finitely generated R-module.

We then have C = v(h(B)) + v(B1), and

it follows easily from Theorem 5.3 that vh(B) = h(C).

Thus we have

an exact sequence O-~h(B) Now L(B) = L(h(B))

A A-~h(B)

and L(C) = L(h(C));

have L(h(B) N A) = L(A).

-~h(C) -~0. and since h(A) c h(B) N A we

Thus without loss of generality we can as-

sume that B is a divisible R-module. Now we have an exact sequence:

53

O-+A/h(A) Since A/h(A) consequence L(B/h(A)) erality,

is reduced,

-~B/h(A)

- ~ C -*O.

we have L(A/h(A))

of T h e o r e m 5.10 that because

= L(B)

- L(h(A))

we can assume

= L(B)

= 0.

h(A)

- L(A).

that A is reduced.

of B is a composition

We then have L(B)

is divisible

Thus without

series

series of divisible

= L(C) = L(C) + L(A),

from Lemma

of divisible submodules

sub-

of C.

since L(A) = 0.

i f A and B are equivalent

Co r011ary 5.12.

we have

loss of gen-

It then follows

5.9 that the image under v of a composition modules

It is an immediate

Artinian

R-modules,

then L(A) = L(B). Proof.

L(B)

Since B is a homomorphic

L(A);

image of A, we have

and since A is a homomorphic

image of B we have

~(A) ~ L(B). Corollary Athen

5.13.

If A and B are Artinian

and B are equivalent

divisible

R-modules,

if and only if there exists

a homomor-

phism f of A onto B such that Ker f is reduced. Proof.

If such a homomorphism

alent by Lemma 5.8. we have an exact

On the other hand,

5.11 we have L(A)

we have L(A)

= L(B).

Corollary R-modules,

5.14.

f-~A

Corollary

Remarks. We shall prove

if A and B are equivalent,

f-+B-~0.

= L(B) + L(Ker f), and b y Corollary

5.12

Thus L(Ker f) = 0, and hence K e r f is reduced. If A and B are equivalent

then A and B have equivalent

Proof.

then A and B are equiv-

sequence: 0-~Ker

By Theorem

exists,

Artinian

composition

divisible

series.

5.13 and Lemma 5-9.

The converse

of Corollary

in a later theorem

5.14 is not true in general.

that the converse

of Corollary

5.14

54 is true if and only if the integral generated

following

5.15.

Let f ~ HomR(K,K)

statements

f is an eplmorphism

(2)

C is a finitely

(c)

c ~ R.

generated

Thus if B is an R-submodule Proof.

(I)

(3).

5.5 and hence by Theorem

~ HomR(K,K

Therefore

to H by Theorem 2.2.

Since

C is isomorphic

2.8.

Thus H ~ H D R C, and by Corollary

ule.

However,

ring,

Since C is an R-submodule

module.

we have

2.6,

of C is

to a regular

to H D R C by Theorem C is a projective

and thus projective

R-mod-

R-modules

of Q, it is an indecomposable

are R-

Thus we have C ~ R.

(3) ---~> (2). Suppose

This implication

epimorphism

is trivial.

that B is an R-submodule

out loss of generality

we can assume

of Q such that Q/B ~ K. that R c B.

of K onto Q/B with kernel B/R,

of K onto K with kernel B/R. versely,

R-module

the completion

of C is isomorphic

R is a local

But

Thus K ~ K D R C, and hence

D R C).

ideal of R, the completion

But K is

generated.

and since Q/C is a torsion I.I.

we see

5.1, Ker f is reduced

Since Ker f = C/R, we have f(K) ~ Q/C.

Q/C ~ K @R C by Theorem

isomorphic

- L(f(K)),

if and only if f is an epimorphism.

f(K) = K by (2) --~-> (I);

H ~ HOmR(K,K)

R-module.

Since L(Ker f) = L(K)

if and only if Ker f is finitely (2) = >

Then the

of Q, then Q/B ~ K if and only if B ~ R.

(2).

that Ker f is reduced by Theorem

and let Ker f = C/R.

are equivalent:

(I)

free.

of R in Q is a finitely

R-module.

Theorem

Artinian

closure

not contain

5.16. a proper

By (I) ~---> (3) we see that B ~ R.

If A is a nonzero submodule

Thus we have an

and hence an epimorphism

if B ~ R, then it can be immediately

Theorem

With-

Artinian

isomorphic

Con-

seen that Q/B ~ K. R-module,

to itself.

then A does

Thus every

55 monomorphism Proof.

of A into itself is an isomorphism. Suppose that B is a proper

an i s o m o r p h i s m of A onto B. = L(A)

= L(h(A)).

= h(A). module

of itself,

assume without classical

= h(A),

an i s o m o r p h i s m

n a m e l y B/h(A).

loss of g e n e r a l i t y

and L(h(B)) and f(h(A))

of A/h(A)

Therefore,

of A and that f is = L(B) : h(B)

onto a p r o p e r

sub-

by C o r o l l a r y 5.2 we can

that A has finite

length in the

sense.

We shall let

S(A)

denote the length of a c o m p o s i t i o n

A whose factors

are ordinary

then have

=

~(A)

are isomorphic.

~(B) Hence

tradiction proves Remarks. That is,

Now h(B) c h(A),

Thus we have h(B)

Hence f induces

submodule

+

simple R-modules

•(A/B).

~(A/B)

But

that A cannot be isomorphic

c be regular element B = Rc-1/R.

to R/M.

We

since A and B This con-

to B.

to Theorem 5.16 is false in general.

there do exist i s o m o r p h i s m s of themselves.

= ~(B),

= 0, showing that B = A.

The dual assertion

factor modules

~(A)

isomorphic

series of

of Artinian modules

As an example,

onto proper

we may let A = K.

Let

of R that is not a unit in R and define

Then A/B is a proper factor m o d u l e of A that is isomor-

phic to A.

Theorem 5.17.

Let A be a torsion d i v i s i b l e R-module.

a sum of Artinian divisible Proof. Corollary

R-modules.

This is an immediate

1.3.

Then A is

consequence

of Theorem 5.5 and

CHAPTER VI STRONGLY UNRAMIFIED

Throughout sional,

this chapter R will be a Noetherian,

Cohen-Macaulay

Definition.

ring w i t h m a x i m a l

If A is a c o m m u t a t i v e

say that A is strongly unramified A, AM contains

RING EXTENSIONS

ring.

over R.

tween the set of regular of regular J A R = i.

Proof.

ring extension

There is a one-to-one

(i.e. M-prlmary)

(i.e. HM-primary)

ideals

ideals

We

however,

is an unramified

2.5 that H is a t o r s i o n - f r e e

R-module

Cohen-Macaulay

that J is an H M - p r l m a r y

of R.

strongly u n r a m i f i e d

moreover,

we see from Corollary

By Theorem 5.0,

Hence M n c

H is a

Therefore,

The o n e - t o - o n e statements

H is a compatible

and so H is

correspondence

of the

now follow directly from what

has already been proved and Theorem 2.8. The next theorem

It

(HM) n A R c J 0 R;

By T h e o r e m 2.8, H / H M ~ R/M,

and the remaining

H/HI ~ R/I

ideal of H and hence there exists an

(HM) n c J.

over R.

I of R and the set

ring w i t h maximal ideal HM.

and thus J N R is a regular ideal of R. ring extension

be-

and that H/R is a divisible

Let J be a regular ideal of H.

integer n > 0 such that

correspondence

and divisible.

Since R is a reduced R-module,

1-dimensional

of R and is

J of H given by HI = J and

Thus HI D R = I and J = H(J O R);

(HI)/I is t o r s i o n - f r e e

theorem,

It is clear,

over R, then A(AM)

H is a compatible

strongly unramified

follows

Of course,

of R in the usual sense.

Theorem 6.0.

local,

of R, we shall

ideal of A, AM D R = M, and AM + R = A.

that if A is strongly unramified

R-module.

ring extension

over R if A M is a regular ideal of

note that A need not be a q u a s i - l o c a l

and

1-dimen-

ideal M.

every regular ideal of A, and A/AM ~ R/M.

AM is then a m a x i m a l

extension

local,

shows that there are m a n y other compatible,

57 strongly

unramified

extensions

the theory of Artinian Theorem 6.1. not assume following

divisible

and these play a big role in

R-modules.

Let A be an R-module

in advance statements

c such that R c A ~ Q.

that A is a ring extension

of R).

(We do

Then the

are equivalent:

(1)

A/R is a divisible

(2)

A is a compatible

(3)

A is a strongly

R-module. ring extension

unramified

If A is a strongly unramlfied statements

of R;

of R.

ring extension

ring extension

of Theorem 3.6 are true for R.

of R.

of R, then all of the

O f particular

importance

are the f o l l o w i n g : (a)

If S is an R-module

such that A c S c Q, then S is an A-

module. (b)

A finitely

as an R-module, (c)

generated

torsion

and the same finite

If J is a regular

A-module

has finite

length

length over A.

ideal of A, then J c AM, A = J + R,

A/J ~ R/(J n R), and J = A(J n R). (d)

If R is an integral

domain of Krull dimension Proof. (2) ~ >

domain,

of (I) and

(3) as well as the statements

(3) = >

(1).

If b is a regular

at the end of the theorem are

element

of M, then there is an

Since A = A M +

= A/R from w h i c h it follows

a divisible

(2) and the implication

3.5 and Theorem 3.6.

integer n > 0 such that M n c Rb. M(A/R)

that b(A/R)

R, we have = A/R.

Hence A/R is

R-module.

Remarks.

We can not overemphasize

the importance

ment in Theorem 6.1 that if R is a Noetherian dimension

local

1.

The equivalence

given by Corollary

then A is a Noetherian

l, and A is a strongly

Q, then A is also a Noetherian

unramified

local domain

of the state-

local domain of Krull

ring extension of Krull

of R in

dimension 1.

58

This fact greatly 1-dlmensional

simplifies

domains

the proofs

and explains

of theorems

the difference

about

local

between C o r o l l a r y

6.2 and Theorem 6.9. W i t h the aid of Theorem 6.1 we shall give a short direct proof of the Theorem of Krull-Aklzuki.

(For a different

proof,

see [17,

Theorem 33.2].

C o r o l l a r y 6.2. Noetherian

(Theorem of Krull-Akizuki).

domain of Krull dimension

be a finite algebraic a field)

Let R be a

1 w i t h quotient

field Q, let L

field e x t e n s i o n of Q, and let T be a ring

such that R c T c L.

dimension

I;

of finite

length over R.

Then T is a N o e t h e r i a n

and if J is a nonzero

(not

domain of Krull

ideal of T, then T/J is a m o d u l e

If R is a local domain then T is a semi-

local domain. Proof.

By taking a finite integral

that L = Q.

To prove the theorem it is sufficient

J is a nonzero R.

ideal of T, then T/J is a m o d u l e

Let I = J D R;

contained

extension

then I is a nonzero

of R.

As we have just seen,

and thus it is sufficient R M.

of finite

ideals of R.

JM = TM for all but a finite number of m a x i m a l T/J ~ Z @ TM/JM,

to prove that if length over

ideal of R, and hence I is

in only a finite n u m b e r of maximal

Theorem 4.1,

of R we m a y assume

Thus

ideals M of R.

where M ranges over all m a x i m a l

By ideals

this sum has only finitely m a n y terms

to prove that TM/J M has finite length over

Thus we m a y assume that R is a local domain. Let A/R = h(T/R);

then by Theorem 6.1, A is a strongly unram-

ified extension of R and T/J has finite length over R if and only if it has finite length over A. T/R is a reduced R - s u b m o d u l e

Hence we may assume that R = A. of K.

Since K is an Artinian R-module

by T h e o r e m

5.5, we see that T/R has finite length by T h e o r e m

It follows

that T/(J A R) has finite length.

homomorphic

Thus

Because

5.1.

T/J is an R-

image of T/(J D R), we see that T/J is a m o d u l e of finite

59 length over R. Theorem 6.3.

(Independence of strongly unramified

extensions).

Assume that R is a Noetherian local domain of Krull dimension 1. A I ..... A n b_ees_trongly unramified extensions

Let

of R i__nnQ, N i the maxi-

mal ideal of A i, S = A 1 A ... A A n and M i = N i D S.

I_~f A i + Aj = Q

for i ~ j, then: (1)

M1,...,M n are the only maximal ideals of S;

they are all

distinct and SMi = A i(2)

Q/S ~ Q/A 1 @ ... ~ Q/A N and A i +

Proof.

n

j¢i

Aj = Q.

Let T = A2 O ... O An, and Pj = Nj A T for J ~ 2.

Then

by induction on n, the Pj's are the only maximal ideals of T, and we have Tpj = Aj and Q/T ~ Q/A 2 ¢ ... @ Q/A n . A 1 + T = AIT, and thus A 1 + T is a ring. suppose that this is not the case.

In fact A 1 + T = Q.

For

Then by localizing A 1 + T at a

maximal ideal we obtain a 1-dimenslonal both A 1 and T.

By Theorem 3.6 we have

local ring U that contains

Let N be the maximal ideal of U and P = N A T.

Then

P is a maximal ideal of T by Corollary 6.2, and hence P = Pj for some J ~ 2.

But then Aj = Tpj c U, and thus Q = A 1 + Aj c U.

tradiction Q/S

:

shows that A 1 + T = Q.

This con-

Now S = A 1 O T, and thus

(T + AI)/S = T/S @ AI/S Z (T + AI)/A 1 @ (T + AI)/T = Q/A 1

@ Q/T Z Q/A 1 @ Q / ~

@ ... @ Q/A n .

By Corollary 6.2,

Thus we have proved statement

S is a Noetherian

(2).

semi-local domain of Krull

dimension i, and thus each M i is a maximal ideal of S. Q/SMI Z (Q/S)MI Z (Q/AI)MI @ ... • (Q/An)M1.

We have

However, Q/SMI is an

indecomposable SMl-mOdule by Theorem 4.7 and (Q/AI)MI = (Q/A1). Therefore, we have (Aj)MI = Q for all J ~ 2.

This shows that

M 1 ~ Nj for J ~ 2, and hence M 1 ~ Mj for j ~ 2.

that the Mi's are all distinct. Since SMi c ~

for all i = l,...,n, we have

Thus we have proved

60

S c SM1 A ... A %

c A 1 A ... A An = S.

S = SM1 N ... A SMn.

Therefore

If P is a maximal ideal of S different from

the Mi's then (SM1 A ... O % )

= Q by Theorem 4.8. But then S p = Q, P and this contradiction shows that the Mi's are the only maximal ideals of S. We have SMI + ( S %

O ... O SMn) = Q by Theorem 4.8.

x ¢ A1, we have x = y + z, where y E SM1 and z E ( S ~

Thus if

N ... D SMn )"

Thus z = x - y ~ (A 1 + SM1) = A l, and hence z ~ A1 N S~

0 ... 0 SMn c A 1 D ... A A n = S.

Therefore,

x ~ SM1 + S = SM1, and we have shown that A 1 = SM1.

By similar ar-

guments we have A i = SMi for all i = 1,...,n. Remarks.

The necessity of the condition Q = ~

in Theorem 6.3 is demonstrated

+ Aj for i ~ j

by the fact that if A 1 + ... + A n ~ Q,

then S = A 1 O ... A A n is a local ring.

The similarity of Theorem

6.3 with an analogous theorem concerning independent valuation

rings

[17, Theorem ll.18] will of course not go unnoticed by the careful reader. Theorem 6.4. B/R = h(S/R),

Let S be a ring extension of R in Q (S ~ Q), let

and let H(S) be the completion of S in the S-topology.

Then we have an exact sequence: 0 ~ HomR(K,B/R ) ~ H ~ H(S)

where ~ is a ring homomorphism. as an R-module.

Proof.

Furthermore,

semi-local,

0

S/B has finite length

Noetherian,

1-dimensional

ring.

It is clear that HomR(K,B/R)

exact sequence:

S/B ~

Therefore H(S) is a finitely generated H-module,

and hence is a complete, Cohen-Macaulay

~

~ HomR(K,S/R).

We have an

61

Because

hdRK = l ,

Ext~(K,S/R)

the

end t e r m s

~ Ext~(K,S/B).

of

this

sequence

Now S/B is a reduced,

and hence has finite length by Theorem 5.1. Ext~(K,S/B)

are

~ S/B by Theorem i.i.

O, a n d

Artinian

thus R-module

Therefore,

The theorem now follows

immediate-

ly from Theorem 2.9.

L e m m a 6.5.

If I is an ideal of H, then H/I is a torsion-free

R-module if and only if I is an unmixed Proof.

Suppose that I = Jl O

of I in H, where Ji is P i - p r i m a r y in H.

ideal of rank 0 in H.

... O Jt is a normal d e c o m p o s i t i o n and Pi is a prime ideal of rank 0

If H/I is not a t o r s i o n - f r e e

R-module,

and r regular in R such that rh ~ I.

But there exists an index i

such that h ~ Ji' and hence rh ~ Ji implies Pi D R has no regular

elements,

H/Z is a t o r s i o n - f r e e

R-module.

Conversely, is not unmixed

there exists h ¢ H - I

that

r ~ Pi"

and this c o n t r a d i c t i o n

suppose that H/I is a t o r s i o n - f r e e

of rank 0 in H, then HM belongs

there exists h ~ H - I such that HMh c I. is not t o r s i o n - f r e e

over R.

However,

proves that

R-module.

to I;

If I

and hence

But then Mh c I, and H/I

This c o n t r a d i c t i o n

proves

that I is un-

mixed of rank 0 in H.

Definition. is a t o r s i o n - f r e e

If I is an unmixed

ideal of rank 0 in H, then H/I

R-module by Lemma 6.5,

have a canonical monomorphism:

and hence by Theorem I.! we

K @R I -~ K @R H.

By Theorem 2.7

we can i d e n t i f y K @R H with K, and from now on we shall c o n s i s t e n t l y identify K ~R i w i t h its image in K.

Thus if x c K and f ~ !, we

shall identify x @ f w i t h f(x).

Theorem 6.6. pondence between

There is a one-to-one, the set of unmixed

the set of proper divisible

ideals

submodules

B/R = K @R I and I = HomR(K,B/R).

order preserving,

corres-

{I} of rank 0 in H and

{B/R1

of K given by

62 If I and B/R correspond,

then:

(1)

B is a strongly unramified

(2)

H/I is isomorphic

(3)

AnnHQ/B = I

(4)

K D R H/I ~ Q/B and HomR(K,Q/B ) ~ H/I.

(5)

Let X~ be the divisible

ring o_~f quotients o~f H/I. of h/~,

Q/~

ring extension of R.

as a ring t_~oH(B), the completion of B.

submodule of B and Q(1) the full

Then Q / ~

is the full ring o_~f quotients

c Q(1), and (H/l) N Q/x9 = B/x9 .

Furthermore,

H/I

is the completion of B / ~ . Proof.

Let B/R be a proper divisible

I = HomR(K,B/R). ideal of H.

Since H = HomR(K,K)

submoduie of K, and let

it is clear that I is a proper

By Corollary 1.2, we have K D R I = B/R.

from Theorem 6.1 that B is a strongly unramified R.

It follows

ring extension of

By Theorem 6.4, we have an exact sequence: O-~I

~ H ~-~H(B) ~ 0

and thus H/! is isomorphic as a ring to H(B).

Since H(B) is a tor-

sion-free R-module, we see that I is an unmixed ideal of rank 0 in H by Lemma 6.5. By Corollary 2.5, we have an exact sequence:

0 -~ ~

-~ B -~ H(B) -~ ExtBI(Q,B) -~ 0

where M~ is the divisible submodule of B.

It follows from this that

K D R B -= K D R H(B).

Now K D R B --- Q/B by Theorem 1.1, since Q/B is

a torsion R-module.

Hence we have Q/B -= K D R H(B) --- K D R H/I.

H/I is complete in the B-topology, by Theorem 3.2.

it is complete in the R-topology

Thus by Theorem 2.2 we have

H / I ~ HomR(K,K ~R H / I )

~ HomR(K,Q/B ) .

Because H/I "= HomR(K,Q/B),

we have AnnH(Q/B ) c I.

hand we have Q / B "-- K D R H/I, and thus I c AnnH(Q/B ) . AnnH(Q/B)

= I.

Since

On the other Therefore,

63

Let B = B / ~

and Q = Q / ~ .

a semi-local Noetherian

Now

is the completion

is an ideal of Q, and Q is

ring of Krull dimension 0.

lar element of Q is invertible full ring of quotients

~

of B.

in Q.

Hence by Theorem 2.11, Q is the

We also have by Theorem 2.11 that H(B)

of B in the B-topology.

an imbedding B c H/I c Q(I).

Since H(B) ~ H/I, we have

This gives rise to an imbedding

c Q(I) under which we have B c (H/l) O Q = B I. sion-free B-module

because

Thus every regu-

it is contained

Now BI/B is a

in H(B)/~ = (H/I)/B.

torOn

the other hand BI/B is a torsion B-module because it is contained Q/B.

in

Thus BI/B = 0, and hence B = (H/I) A Q. The only thing remaining to be proved is that if we start with

an unmixed ideal I of rank 0 in H, then I = HomR(K,K @R I). J = H°mR(K'K ~R I);

Let

then by Theorem 2.2, I ~ J and J/l is a torsion-

free and divisible R-module.

Hence

(HM)(J/I)

= J/i.

But J is an

ideal of H, and thus J/I is a finitely generated H-module. is a local ring with maximal

Since H

ideal HM, we see by the Nakayama Lemma

that J/I = 0, and hence J = !. Corollary 6.7. of rank 0 i_~nH. Proof.

Let II, .... I k be a finite set of unmixed ideals

Then K @R ( ~ i j) = h( Q~ (K @R lj)).

We have HOmR(K,h ( O (K @R Ij)) = HomR(K, O

J

(K @R Ij))

J

= 0 H°mR(K'K ~R lj). By Theorem 6.6 we see that J A HomR(K,K @R Ij) = O !j; and hence by the same theorem,

J

J

K ~R ( nJ z j) : h ( NJ (K ~R z j) ) . Remarks. an H-module

Since H is a compatible

is torsion,

or divisible,

if the same is true over R. of quotients

ring extension of R, or complete

over H if and only

By Theorem 3.1, H @R Q is the full ring

of H, and since H @R K ~ K by Theorem 2.7, we see that

(H @R Q)/H ~ K.

By Theorem 2.10 there is a one-to-one

correspondence

between the set of rings B between R and Q and the set of rings ~ between H and H @R Q given by 0 = H ®R B and B = O O Q.

We have

64 B/R ~ (H O R B)/H and Q/B ~ (H ~R Q)/(H ®R B). B c H ®R B and that H ~R B = HB = H + B.

It is also true that

Furthermore,

(H ~R B)/B is

torsion-free and divisible. Theorem 6.8.

(I)

There is a one-to-one correspondence between

the set of ideals ~o_~f H ~R Q and set of the unmixed ideals I o~f rank 0 in H given by (2)

There is a one-to-one correspondence

strongly unramified ideals

~

X9 O H = I and QI = ~0. between the set of

ring extensions 0 of H i~n H ~R Q and the set of

of H @R Q given by 0 = H + x9 and

~

= d(0),

the divisible

submodule of 0. (3)

There is a one-to-one correspondence between the set of

ring extensions B of R i~n Q and the set of ring extensions 0 of H i~n H ~R Q given by 0 = H @R B and B = 0 O Q.

B is strongly unramified

over R if and only if H ~R B is strongly unramified over H. (4)

If B and 0 = H ~R B are corresponding

strongly unramifie_~d

ring extensions of R i_~n Q and H in H ~R Q' respectively,

then

d(0) A H = I is the unmixed ideal of rank 0 in H corresponding to B by Theorem 6.6.

Furthermore ~/d(O)

= H/I is the completion of both

B and ~ in their respective topologies.

Proof.

(1)

H is a 1-dimensional,

local, Cohen-Macaulay ring. n Let P1,...,Pn be the rank 0 prime ideals of H and ~ = H - U Pi" i=l Then J is the set of regular elements of H, and hence H j = H ®R Q" There is clearly a one-to-one correspondence of HE

and the set of unmixed ideals I of rank 0 in H given by

n H = I and Ij (2)

H @R Q"

= ~.

But Ij

= H~I

Let 0 be a strongly unramified

H ®R Q' and let ~ H-module,

between the set of ideals

= d(0).

Then ~

ring extension of H in

is a torsion-free

and hence is an H ®R Q-module; Suppose that 0 ~ H + ~ .

= QHI = QI.

that is ~

and divisible is an ideal of

Let H I = (H + ~)//~9

and

65 Q1 ~ (H @R Q ) / ~ "

Then H I ~ H / ( ~

local, Cohen-Macaulay Let 01 ~ q / ~ ;

N H) is a complete,

1-dlmenslonal

ring and Q1 is its full ring of quotients.

then H 1 c 01 c Q1 and D 1 is a reduced Hi-module.

see that O1/H 1 ~ O/(H +

dQ ) is a homomorphic

is a nonzero divisible Hi-module.

We

image of O/H, and thus

Hence HOmHl(Ql,ql/H l) ~ O.

How-

ever, we have an exact sequence:

H°mHl( QI'QI ) -~ H°mHi( QI'~!/HI ) -* EXtHl( QI' HI ) The end terms of this sequence are 0, since Q1 is reduced and H 1 is a complete ring.

Therefore HomHI(QI,OI/H I) = 0.

This contradiction

shows that Q = H + ~Q. On the other hand if ~

is an ideal of H ®R Q' then ~

divisible H-module and (H + /~)/i~O

~ H/(H n Mg) is reduced.

is the divisible submodule of H + /_0. is a divisible H-module,

H +~

is a

Since (H + X g ) / H ~ / ( H

Thus N ~)

is a strongly unramified extension

of H in H ~R Q" (3)

This statement has been proved in Theorem 2.10.

(4)

Let B be a strongly unramlfled ring extension of R in Q and

let Q = H 8 R B, and I ~ d(D) n H. hence Q/H ~ d(Q)/I. module,

we

By (2) we have n = H + d(0), and

Since d(q) is a torsion-free and divisible R-

see that K ®R I & Tor~(K,d(O)/I) & t(d(~)/I) = d(n)II.

By (3) we have B/R ~ Q/H & d(D)/I, and therefore K ®R I a B/R.

It

follows from Theorem 6.6 that I is the unmixed ideal of rank 0 in H corresponding

to B.

By Theorem 6.6, H/I is the completion of B.

Now

Q/d(Q) = (H + d(O))/d(Q) & H/I, and hence H/I is the completion of O. Remarks.

Theorem 6.8 shows that every proper divisible submod-

ule of K is of the form QI/I, where I is an unmixed ideal of rank 0 in H. The next theorem is a generalization of the Theorem of KrullAklyuki.

It shows that some of the strongly unramified extensions of

R in Q may not he Noetherian.

A different proof of this theorem may

66

be found in [4, Prop.

6.4.13].

! am grateful

to Mrs.

Judy Sally for

drawing it to my attention. T h e o r e m 6.9.

Let R be a 1-dimensional,

local,

Cohen-Macaulay

ring.

Then every ring between R and Q is N o e t h e r i a n

has no

nonzero nilpqtent

Proof.

Suppose

element

elements.

that R has a nonzero nilpotent

out loss of g e n e r a l i t y

we can assume

of M, the maximal

all n > O.

that c 2 = 0.

and S n c Sn+l.

of the S n and I the ideal of S g e n e r a t e d

and an element

Let b be a regular

by the elements

c/b n.

Now there

If I

is an in-

r,t E R such that s = r + tc/b k.

lows that c/b n+l = rc/b n.

Therefore,

in R, we see that c = 0.

Conversely,

With-

We let S be the u n i o n

s E S such that c/b n+l = sc/b n.

S is not a N o e t h e r l a n

c.

ideal of S, then there exists an integer n

tegral k > 0 and elements

is a unit

element

ideal of R, and let S n = R + R c/b n for

Then S n is a ring,

is a finitely g e n e r a t e d

if and only if R

It fol-

c(l-rb) = 0, and since 1 - rb

This

contradiction

proves

that

ring.

suppose that R has no nilpotent

elements

~ 0.

Then

0 = Pl N ... n Pt' where p!,...,p t are the rank 0 prime ideals of R. It follows

that Rpi is a field for all i = l,...,n.

Q ~ Rpl @ ... @ Rpt,

Q is a direct

sum of fields.

Since Thus Q is a semi-

simple ring. Let B be a ring such that R c B ~ Q. is a strongly u n r a m l f l e d is a finitely g e n e r a t e d

ring extension R-module.

If B'/R = h(B/R),

of R.

Let ~

be the divisible

and divisible,

and hence

~

submodule

Since Q is a seml-slmple

e E ~

such that Qe = ~

is a p r i n c i p a l

=

of B.

is a Q-module;

of Q.

~e.

By T h e o r e m 5.1,

Thus without

we can assume that B is a strongly u n r a m i f i e d

ring,

extension Then ~

that

B/B'

loss of generality,

is,

of R.

is t o r s l o n - f r e e x9 is an ideal

there is an idempotent

It follows

then B'

element

that /9 = Be, and hence

ideal of B.

If we let ~ = R/(R N ~ ), B = B / ~ ,

and Q = Q / ~ ,

then B

is a

67

ring e x t e n s i o n of R in Q.

F u r t h e r m o r e R has the same properties

and Q is the full ring of q u o t i e n t s erated ideal of B, it follows is Noetherian. is a reduced

Thus without

submodule of B/R,

tension of R.

~ O.

have proved ring.

that B is N o e t h e r i a n

Because

loss of g e n e r a l i t y we m a y assume that B

on L(K).

If S/R is a proper,

ring.

p r o p e r submodule

L(Q/S)

< L(K),

Thus by T h e o r e m ele-

then B is also a N o e t h e r i a n

we can assume that B / R has no

submodules.

Hence by T h e o r e m 5.1,

of B/R is a finitely g e n e r a t e d

Let J be an ideal of B and I ~ J 0 R. ideal of ~.

ring ex-

it follows by i n d u c t i o n that if we

loss of g e n e r a l i t y

divisible

generated

nonzero,

Of course S has no nilpotent

that S is a N o e t h e r i a n ring,

nonzero,

finitely

if and only if

Since S c B, S is a reduced R-module.

Thus without

proper,

is a finitely gen-

then S is a strongly u n r a m i f l e d

3.6 (6), S is a quasi-local ments

Since ~

R-module.

We p r o c e e d by induction divisible

of R.

as R

every

R-module.

Suppose

If J is a regular

that J is not a

ideal of B,

J = BI by T h e o r e m 3.6 (2), and hence J is finitely g e n e r a t e d

then over B.

Thus J is not a regular ideal of B, and ~ance I is contained in the intersection

of some of the prime

there exists an element

c E R that

prime ideal of R that contains not contained regular

b E I + Rc.

is not contained

ideal of R;

because

every p r o p e r submodule

follows

that J = Jb + I.

erated ring.

ideal of B.

Therefore

in any rank 0 I + Rc is

and hence there is a

Thus b = a + rc where a E I and r E R.

Now Jb is not a finitely generated

hence Jb I Ja c JI.

of rank 0 of R.

I and such that cI = 0.

in any rank 0 prime

element

ideals

R-module.

Hence B = Jb + R

of B/R is finitely generated.

It

Since Ic ~ 0, we see that Jc = 0, and

Therefore

J = BI, and h e n c e J is a finitely gen-

This c o n t r a d i c t i o n proves

that B is a N o e t h e r i a n

CHAPTER VII THE CLOSED C O M P O N E N T S

Throughout Macaulay

this chapter R will be a 1 - d i m e n s i o n a l

local Cohen-

ring.

Definitions.

R is said to be a n a l y t i c a l l y i r r e d u c i b l e

c o m p l e t i o n H is an integral domain. integral

OF R

domain.

if its

Of course R is then also an

We recall that we have defined R to be a closed

domain if R is an integral

domain and J O R J 0 for every nonzero

ideal J of H.

T h e o r e m 7.I.

The f o l l o w i n g

statements

(l) (2)

R is a n a l y t i c a l l y

(3)

Every nonzero R - e n d o m o r p h i s m

(4)

K is a simple divisible module.

R is a closed

are equivalent:

domain. irreducible. of K is an epimorphism.

E is a simple divisible module.

(6)

Q is a field and the integral

disc rete v a l u a t i o n Proof.

closure V of R in Q is a

ring that is a f i n i t e l y generated

(I) ~ >

(2).

Let P be a prime ideal of rank 0 in H.

Then P O R is a p r i m e ideal of R and P n R ~ M. hence P = 0.

Therefore,

(2) ~---> (3)-

pose that f is not an e p i m o r p h i s m

an integral (3) = > B is not

element of H = HomR(K,K).

on K.

hence b y T h e o r e m 5.11, L ( K e r f) / O;

g(K) c K e r f.

Then L(f(K))

element

We then have fg = O, c o n t r a d i c t i n g

(4).

reduced,

< L(K),

and

g ¢ H such that the fact that H is

Thus every element of H is an e p i m o r p h i s m Let B be a proper,

nonzero R - s u b m o d u l e

~hen by C o r o l l a r y 1.3,

of K that is not an epimorphism.

Sup-

that is K e r f is not reduced.

1.3 there is a nonzero

domain.

Thus P N R = 0, and

H is an integral domain.

Let f be a nonzero

Hence b y C o r o l l a r y

R-module.

of K.

on K. If

there is an R - e n d o m o r p h i s m

Hence B is reduced,

and thus K is

69 a simple

divisible

(4) ~=> (6). per nonzero finitely

ideal

generated

~9 + R = Q.

Suppose %~.

Since

R-module

has Krull

Therefore,

is a p r o p e r

submodule

ring.

local domain

~ R/(~

Lemma.

~

is not a

It follows

that

n R) has Krull dimension

is an integral

of R in Q;

then V/R

of K, and hence V is a finitely

5.1.

of Krull

(2).

(2) ~

R-module,

A R), and this is a contradiction

0 and R / ( ~

closure

Since

Therefore,

valuation

(6) ~ >

every localization generated

of V with

R-module,

V is a Noetherian,

dimension

1 by Corollary

generated

R-

respect

to

we see that V

integrally

closed,

6.2 and hence V is a

ring. By Theorem 6.4,

H is a subring

of H(V),

and H(V)

domain.

(i).

Let J be an ideal of H that is maximal

spect to the property ideal

by the Nakayama

ideal of V is a finitely

is a local

discrete

is a divisible

Then Q has a pro-

Q is a field.

by Theorem

a maximal

~

dimension

Let V be the integral

module

that Q is not a field.

But then Q / ~

since Q / ~ 1.

module.

J O R = O.

It is easily

with

re-

seen that J is a prime

of H, and since HM O R = M, we have J = O. (I) ~

(5).

let J = AnnHD. J = 0.

Let D be a nonzero

a simple divisible (2).

R-endomorphism divisors.

R-submodule

Then J is an ideal of H and J A R = O.

By Theorem 4.5,

(5) ~ >

divisible

of E, and

Hence we have

we have D : AnnEJ = AnnEO = E, and thus E is

R-module. Since E is a simple

of E is an epimorphism.

But H Z HomR(E,E)

divisible

R-module,

Thus HomR(E,E)

b y Theorem 4.5,

every

has no zero

and thus H is an inte-

gral domain. Remarks.

We note that it follows

7.1 that R is analytically submodule

irreducible

of K (or of E) is a finitely

from Corollary

5.7 and Theorem

if and only if every proper generated

R-module--in

fact,

70 if and only if it has finite length. Definition.

We shall call a maximal strongly unramified

sion of R in Q a closed component of R.

exten-

By Theorem 6.1, A is a

closed component of R if and only if A/R is a maximal proper divisible submodule of K.

By Theorem 5.5, K has ACC on divisible

submod-

ules, and thus closed components of R exist. Theorem 7 . 2 . RinQ.

Let A be a strongly unramified

Then the following statements

ring extension of

are equivalent.

(1) (2)

Q/A is a simple divisible A-module.

(3)

H(A), the completion of A in the A-topology,

A is a closed component of R.

is an inte-

gral domain. (4)

If ~

is the divisible submodule of A, then A / ~

analytically irreducible,

Noetherlan,

i_~sa__n.n

local domain of Krull dimen-

sion 1. Proof.

(1) (2).

By Theorem 3.2, an A-module is a torsion,

divisible A-module if and only if it is a torsion, ule.

Furthermore,

by Theorem 3.6

Q is also an A-module.

divisible R-mod-

(7), every R-module between A and

Thus A is a closed component of R if and

only if Q / A is a simple divisible A-module. (2) ~ >

(3)-

Since Q/A is a simple divisible A-module,

element of H(A) = HomA(Q/A,Q/A ) is an epimorphism. has no zero divisors.

Since H(A) is a commutative

Therefo~

every H(A)

ring by Corollary

2.5, H(A) is an integral domain. (3) ~ >

(4).

If ~

is the divisible

submodule of A, then X9~

is the kernel of the ring homomorphism A -~ H(A), and thus ~ prime ideal of A. Let R = R / ( ~

~Q

is a

is a Q-module and hence ~Q is an ideal of Q.

D R), A = A / ~ ,

and Q = Q / ~ .

Then ~ is a Noetherian,

local domain of Krull dimension 1 and Q is the quotient field of R.

71 Since A/R "-- A/(R + ~ ) is a homomorphism image of A/R, it follows that A / R is a divisible R--module. ramified

ring extension of R.

Therefore,

A is a strongly un-

By Theorem 6.1, A is a Noetherian

local domain of Krull dimension I.

Since H(A) = H(A) is an integral

domain, A is analytically irreducible. (4) ----->(2). graph.

We shall use the notation of the preceding para-

By Theorem 7.1, Q / A is a simple divisible A-module.

Since

Q / A ~- Q/A, we see that Q/A is a simple divisible A-module. Remarks. divisible = A/~

Let A be a closed component of R, and let @

submodule of A.

Since A = A/~Q

is the full ring of quotients

and thus ~

is a maximal ideal of Q.

be the

is an integral domain and

of A, we see that Q is a field Since the sum of two maximal

ideals of Q is equal to Q, we observe that ~9 is the unique maximal ideal of Q contained Let R = R / ( ~

in A.

n R).

Then ~ is a Noetherlan local domain of

Krull dimension l, and Q is the quotient field of R. Q/A~

Q/A, we see that Q / A is a simple A-module.

Since

Thus A is a closed

component of R. Because Q/A % (H ®R Q)/(H D R A) by Theorem 2.10, we see that H D R A is a closed component of H. ponent of H arises in this way.

Furthermore,

every closed com-

By Theorem 6.8, H D R A = H + L,

where L is an ideal of H D R Q and L is the divisible submodule of H @ R A.

But then by the previous paragraph L is a maximal ideal of

H D R Q.

Because ~Q c L, and g} is a maximal ideal of Q, it follows

that ~

= Q n L.

By Theorem 6.8 the completions

H D R A are all equal to H/(H 0 L).

of A, A / ~ }, and

We shall enlarge upon these re-

marks in the next theorem. Definition.

If P is a prime ideal of rank 0 in H, then H/P is

called an analytic component of R.

H/P is a complete,

Noetherian,

72 local

domain of Krull dimension If B is a ring,

B-topology

we recall

There is a one-to-one

set of closed components

its completion

correspondence

in the

and hence there is a one-to-one

The correspondence

between

A of R and the set of prime ideals

set o fclose_____ddcomponents, o f R R.

that we denote

by H(B).

Theorem 7.3.

0 i__snH;

i.

correspondence

and the set of analytic

the

P of rank

between

the

components" o f

is given by:

K @R P = A/R and HomR(K,A/R)

= P.

We then have: (i)

P = AnnHQ/A

(2)

H(A)

(3)

If

submodule (H/P)

/~

n Q/~

is

an

of

rank

O.

by

Theorem

of K,

preceding

By

of the

and the

corresponding

the

A is

theorem

6.6,

of

A/R

a closed follow

of

by

of

nature

Theorem

6.6

by

necessarily

0 in

the

proper of

7.2,

H,

then

unramified

of

a maximal

H,

rank

A/~.

Theorem then

of

a strongly

component

from

Then

ideal

divisible

and

completion

R.

ideal

A is

is

L the

P = HomR(K,A/R),

a prime

where

the

a prime

order-preserving

Theorem thus

P is

P is

P = A/R,

let

and

A = H + L,

is

component

Hence if

H ®R H/P

If we

of A,

R. and

exten-

one-to-one divisible

The the

sub-

remaining Remarks

theorem.

Definition.

writing

domain.

K ®R

of

L N H = P,

a closed

Conversely, 6.6,

submodule

Furthermore,

~ H/P.

R in Q.

statements

then

A be

H(A)

correspondence module

divisible

= A/~.

integral

6.6,

of

the A,

Let

Theorem

sion

is

of H ®R

Proof. H(A)

= H/P

If P is a prime

closed component

P = PA and A = Ap.

ideal of rank 0 in H and A is the

of R in Q, we shall denote

this by

73 Theorem 7.4.

If P is a prime

ideal of rank 0 in H, then

P = AnnH(AnnKP ) .

Proof.

It is clear that P c AnnH(An~KP ) = I.

to 0 in H, there

is a nonzero

element

f E H such that Pf = 0.

f(K) c AnnKP , and hence AnnKP is not reduced. tains no regular

elements

ideal P' of H of rank O.

Since P belongs

It follows

of H, and hence I is contained

Thus

that I conin a prime

Since P c p', we see that P = P', and hence

P = AnnH(AnnKP ) . Definition. integral

domain

A ring W is called

a valuation

such that if x is a nonzero

field of W, then either x or I/x is in W. local.

A valuation

a Noetherian

ring.

domain of Krull dimension

valuation

divisible

its divisible

submodule

pseudo

valuation

valuation

rings belonging

set of closed

submodu!e

local,

rings between

Theorem 7.5.

ring is quasi-

valuation

ring if it is

ring is a Noetherian

it is a valuation

local

ideal.

ring.

The

ring is thus a prime ideal.

Cohen-Macaulay

ring,

R and Q will be called

then the the pseudo

to R.

There is a one-to-one

components

rings V belonging

A valuation

ring V is a pseudo valuation

of a pseudo valuation

If R is a l-dimensional,

of the quotient

I, and every ideal is a principal

We shall say that a commutative ring if module

element

ring is called a discrete A discrete

ring if it is an

correspondence

between

the

A of R and the set of pseudo valuation

to R given by h(V/R)

= A/R and

(Integral

Closure

of A in Q) = V. Proof. divisible

Let A be a closed component

submodule

is the quotient

Then ~

field of A / ~ .

ically irreducible, Let V / ~

of A.

Noetherian,

be the integral

of R, and let ~

is a maximal

By Theorem 7.2, local

closure

ideal

of Q and Q/z@

A/~

is an analyt-

domain of Krull

of A / ~

be the

in Q / ~ .

ideal of Q, it is easy to see that V is the integral

dimension

!.

Since ~

is an

closure

of A in

74 Q.

By Theorem 7.1, V / ~

is a discrete valuation ring.

Therefore V

is a pseudo valuation ring belonging to R. Conversely, let V be a pseudo valuation ring belonging to R, and let A/R = h(V/R).

Then A is a strongly unramified ring extension of

R in Q by Theorem 6.1.

Since ViA is a reduced R-module, it is a

finitely generated torsion R-module by Theorem 5.1. bounded order.

Thus by Theorem 2.9 we have H(A) c H(V).

the divisible submodule of V. Since V / ~

Hence V/A has

Then H{V) = H ( V / ~ )

Because H(A) c H ( V / ~ ) ,

an integral domain.

be

by Theorem 2.11.

is a valuation ring by definition, H { V / ~ )

valuation ring.

Let ~

is also a

we conclude that H(A) is

Therefore, A is a closed component of R by

Theorem 7.2. Corollary 7.6.

Let V be a pseudo valuation ring belonging to

R, and A the closed component of R corresponding to V. (I)

V/A is a torsion R-module of finite length.

(2)

If~

is the divisible submodule of V, then ~

Then:

is the

divisible submodule of A. (3) ~

is a maximal ideal of Q.

(4)

V/~

is a discrete valuation ring belonging to R/(~

(5)

V/~

i__ssthe integral closure of A / ~ ,

closed component of R / ( ~ Proof. and both ~ '

If

~

and ~

Q, we see that

~'

and A/~4~

O R).

is a

A R).

' is the divisible submodule of A, then ~' c ~ ; are ideals of Q. = ~.

Since ~9' is a maximal ideal of

Now R/(LG O R) is a Noetherian local domain

of Krull dimension 1 and quotient field Q/L~. Noetherian ring by Corollary 6.2, and hence V / ~

Thus V/%~

is a

is a discrete val-

uation ring. The other statements of the corollary were proved in the course of proving Theorem 7.5. Theorem 7.7.

The integral closure of R i__nnQ i_ss equal to the

75 intersection o f all o f the pseudo valuation rings belonging t__ooR. Proof°

If an element of Q is integral over R, then it is inte-

gral over every closed component

of R.

Hence by Theorem 7.5 the in-

tegral closure of R is contained

in every pseudo valuation ring be-

longing to R. Let

~l''''' ~t

be the set of maximal

ideals of Q, and for each

i I l,...,t let W i be the intersection of those pseudo valuation rings of R that contain

~i"

Then W i / ~

is the intersection of all of

the valuation rings belonging to R/(R N ~9,i), and hence W i / ~ i is the integral closure of R/(R N /%i ) in Q / ~ i " Suppose that x is an element of the intersection of all of the pseudo valuation rings belonging to R.

Then x 6 W1, and hence there

is an integer n I > 0 and a polynomial fl(X) of degree < n I with conl efficients in R such that x - fl(x) £ Xg~I. Let Yl = xnl fl (x); then Yl E W2, and hence there is an integer n 9 > 0 and a polynomial n2 f2(X) of degree < n 2 such that Yl - f2(Yl ) E ~ i n ~ 2 " Continuing in this way we see that there is a polynomial < n R nln 2 ... n t such that x n - g(x) 6 element of

Z~ 1 N ... N

integer k > 0 such that

~t

~I

is nilpotent,

[x n - g(x)] k ~ 0.

g(X) of degree

N ... n

~9~t.

But every

and hence there exists an This equation shows that

x is integral over R. Theorem 7.8.

Let AI, .... A n b__~ethe closed components o f R and

V i,...,V n the pseudo valuation rings belonging to R. and S = f] V i. i

Let T = n A i i Then S is the integral closure of R i__n_nQ, and:

(!)

n Q/T /- Z @ Q/A k and A k + n A w Q. kml m~k m

(2)

OJS -" Z @ Q/V t and V k + ~ V m - Q. km I m~k

(3)

S/T is an R-module of finite length,

n

and hence S is a

finitely generated T-module. Proof.

(1)

Let

~l'''''

~t

be the maximal

ideals of Q and

76 for each i = l,...,t ponents

let B i be the intersection

of R that contain

~i"

Since

of those closed com-

~I

+ (X~2 N ... N ~ t ) ~ Q, t we see that B I + (B 2 N ... n Bt) = Q. Since T ~ ~ Bi, we see that i~l Q/T ~ (B1/T) ¢ (B 2 n ... n Bt)/T ~ Q/(B 2 n ... n B t) ~ Q/B I. By induction on t we have Q/(B 2 N ... n Bt) ~ Q/B 2 ¢ ... ~ Q/B t.

Thus

Q/T ~ Q / B 1 ® ... ¢ Q/B tLet Ri = R/(R N Ali,...,Api

~ i ) , Qi ~ Q / ~ i

be the closed components

~Ji = Aji/X~i"

and Bi ~ B i / ~ i " of R that contain

Then Ri is a Noetherlan

sion I, and ~li,...,~pi

Let ~i

and let

local domain of Krull dimen-

are the closed components

of Ri"

Since

~Ji + A--mi ~ Qi for J ~ m, we can apply Theorem 6.3 and obtain Qi/~i ~ Qi/~li @ ... @ ~i/'Api. Therefore

Q/T ~

Let C k ~

But Qi/'Bi ~ Q/B I and Qi/~ji & Q/Aji.

t n Z @ Q/B i ~ Z • Q/Aji ~ Z @ Q/A k. i=l i,J k=l

n Am; m~k

then we have an exact sequence o -. C k / T

-~ Q / T -~ Q / C k ~

0. n

Then by what we have already proved,

Q/T ~

Z ~ Q/A k and k=l

Q/C k ~, Z ¢ Q/Am, and hence L(Q/T) ~ n and L(Q/C k) ~ n - i. m~k L(Ck/T ) m I and hence Ck/T is not h-reduced. But Ck/T = (C k + Ak)/Ak,

and since Q/A k is a simple divisible

Thus

R-module,

we see that C k + A k = Q. (2)

Since V k ~ A k for k ~ i ..... n, we see by (i) that

vk + m kn

Q. Hence

By induction Q/s =

Vk/S •

Vml/S

Q/(m k vml ® Q/Vk"

on n we have Q/( n v m) ~ z ~ Q/Vm, and hence m~k m~k

n z ® Q/v t. k~l (3)

As we observed

in Theorem 7.5,

tion ring of R corresponding Vk/A k is an R-module

if V k is the pseudo valua-

to the closed component

of finite length.

A k of R, then

Hence L(Q/V k) z L ( Q / A k ) =

1.

77

Thus by (1) and (2) we have L(Q/T) ~ n = L(Q/S), Therefore,

and hence L(S/T) - O.

by Theorem 5.1, S/T is an R-module of finite length.

Remarks.

Suppose that R is a Noetherian domain of Krull dimen-

sion 1 so that the pseudo valuation rings of R are in fact bonafide valuation rings.

In this case if S is the integral closure of R in Q,

then S is a semi-local principal

ideal domain by Corollary 6.2.

If

NI,...,N n are the maximal ideals of S, then SNI,...,SNn are exactly the valuation rings belonging to R.

We then have

Q/S & Q/SNi @ ... @ Q/SNn by Theorem 4.1, and this fact could have been used to give a slightly different

proof of Theorem 7.8.

The next theorem was proved in the case where R is an integral domain by Northcott Theorem 7-9.

[19, Th. 5]. There is a one-to-one correspondence between the

set of pseudo valuation rings belonging t__q R and the set of prime ideals of rank 0 in H. Proof.

This is an immediate consequence of putting together

Theorems 7.3 and 7.5. Remarks.

Let V be a pseudo valuation ring of R and let A be

the corresponding

closed component

component of H in H ®R Q"

of R.

Then H ®R A is a closed

To show that H @R V is the corresponding

pseudo valuation ring of H it is sufficient

by Theorem 7.5 to show

that H @R V is the integral closure of H ®R A in H ®R Q" Now on the one hand H ®R V is an integral extension of H @R A in H @R Q because module.

(H ®R V)/(H ®R A) ~ V/A is a finitely generated H-

On the other hand if ~ is the integral closure of H ®R A in

H ®R Q' and B = D N Q, then by Theorem 2.10, Q = H ®R B and B/A ~ (H ®R B)/(H ®R A).

Hence by Corollary 7.6 applied to H, we see

that B/A is an H-module of finite length. over A and hence B c V.

Therefore,

B is integral

Thus Q = H ®R B c H ®R V, proving that

78 H @R V is the integral Let ~

closure of H ®R A in H @R Q"

be the divisible

module of H @R V.

Then ~

and H ®R A, respectively.

submodule

of V, and L the divisible

and L are the divisible

submodules

of A

Hence H ®R A z H + L and ~9- - L N Q.

we let P - L O H, then H/P & (H ®R A)/L is the completion

sub-

If

of both A

and H @R A, and hence P is the prime ideal of rank 0 in H correspondlng to A by Theorem 7.3. By Corollary 7.6, it is the integral Because complete

(H ®R V)/L is a discrete valuation ring, and

closure of (H @R A)/L = H/P in its quotient

(H ®R V)/L is finitely generated discrete

valuation ring.

as an H/P-module,

Furthermore,

V / ~ , and thus (H ®R V)/L is the completion

field.

it is a

(H @R V)/L O Q/Le of V / ~ .

All of these remarks go to show the nature of the one-to-one correspondence

between the set of pseudo valuation rings belonging

R, and the set of complete discrete integral

closures

of the analytic

valuation

components

rings which are the of R.

to

CHAPTER VIII SIMPLE DIVISIBLE MODULES Throughout Macaulay

this chapter R will be a 1-dimensional,

local, Cohen-

ring.

Lemma 8.1.

l_~f T I and T 2 are equivalent

torsion R-modules,

then

AnnHT 1 - AnnHT 2. Proof.

By definition,

g : T 2 - ~ T I.

f : T 1 ~ T 2 and

By Theorem 2.7, T 1 and T 2 are H-modules

H-homomorphisms. -

we have R-epimorphisms

It follows

immediately

and f and g are

from this that AnnHT 1

AnnHT 2. Definition.

class.

Let T be a torsion R-module,

and [T] its equivalence

Because of Lemma 8.1, we can unambiguously

define AnnH([T])

to be AnnHT. Theorem 8.2.

Let D be a simple divisible

R-modul~,

morDhism of K o mto D, C/D = Ker f, and A/R = h(Ker f). (i)

A is a closed component

the equivalence (2)

o_~f R and is uniquely

f a_nm R-hom_~oThen:

determined

by

class of D.

D is Isomorphic t__~oQ/C, and C is isomorphic

to a regular

ideal of A. (3) R-module

If I is a regula r ideal of A, then Q/I is a simple divisible equivalent

R-module equiva!ent

to D.

C qnverselF,

i_~fD' is any simple divisible

to D, then D' ~ Q/I for some regular ideal i o_~f

A.

(4)

D is equivalent

Proof. = L(K) - I.

too Q/A and AnnH D = PA"

By Theorem 5.11 we have L(A/R) = L(C/R) = L(K) - L(D) Thus A is a closed component

is an A-module; sion R-module.

of R.

By Theorem 3.6, C

and by Theorem 5.1, C/A is a finitely generated Therefore,

hence C is isomorphic

C is a finitely generated A-module,

to a regular ideal of A.

tor-

and

Because C/R = Ker f,

80 it follows that D g K/(Ker f) ~ Q/C. Now Q/C ~ (Q/A)/(C/A), 5.8.

and hence D is equivalent

to Q/A by Lemma

If I is a regular ideal of A, then A/I is a cyclic torsion R-

module by Theorem 3.6.

Therefore,

Q/I is also equivalent

to Q/A by

Lemma 5.8. By Lemma 8.1 we have AnnHD ~ AnnH(Q/A), have AnnH(Q/A) uniquely

= PA"

correspond,

Therefore,

and by Theorem 7.3 we

AnnHD = PA"

Because A and PA

and because AnnHD is an invariant

ence class of D, we see that A is independent and of the representative

of the equivalence

of the equival-

of the homomorphlsm class of D.

f

By what we

have already proved this shows that if D' is a simple divisible Rmodule equivalent regular

to D, then D' is isomorphic

to Q/I, where I is a

ideal of A.

Corolla£y 8. 3 .

(i)

the set o_~f equivalence

There is a one-to-one

classes

the set o__ffclosed components

correspondence

[D] of simple divisible

between

R-modules,

A o_~f R i_.nnQ given by Q / A ' v D

and

and

A/R = h(Ker f), where f i_~s any R-homomorphism o f K onto D. (2)

There is also a one rto-one correspondence

o_~f classes

[D] and the set o__ffprime ideals P of rank 0 i__nnH glven by

P = AnnH([D]) (3)

between the set

and D'-~K ®R H/P.

P an d [D] correspond,

i__ffP = PA and Proof.

and A and

[D] correspond

Q/AA~D. If A is a closed component

- 1 and L(Q/A) = L(K) - L(A/R). Q/A is a simple divisible

of R in Q, then L(A/R) ~ L(K)

Therefore,

R-module.

L(Q/A) = l;

mainder of the theorem follows Corollary

8.4.

The re-

from Theorem 8.2.

Let V I ..... V n b_~e the pseudo valuation rin~s o f

Then Q/VI,...,Q/V n are a full set of representatives

equivalence

and hence

If £ is a rank 0 prime ideal o f

H, then by Theorem 5.G and 7.3 we have K ®R (H/P) ~ Q/A£.

R.

i_~fand only

classes of simple divisible

R-modules

of the

and Q/V i is not

81

equivalent to Q/Vj if i ~ J. Proof.

Let A i be the closed component of R corresponding

by Theorem 7.5.

to V i

By Corollary 7.6, V i is a finitely generated A i-

module, and hence V i is isomorphic to a regular ideal of A i.

The

corollary now follows from Theorems 8.2 and Corollary 8.3. Definitions.

If D is a simple divisible R-module we shall let

G([D]) denote the set of isomorphism classes of R-modules D' equivalent to D. If A is any ring we shall let G(A) denote the semi-group

of

isomorphism classes of regular ideals of A. Theorem 8,~. P ~ AnnH(D), o_~f R.

Let D be a simple divisible R-module,

let

and let H = H/P, the corresponding analytic cpmponent

Then there is a one-to-one correspondence

between the sets

G([D]) and G(H) given by Y ~ HomR(K, DI) an d D' ~ K ®R Y' where Y is a nonzero ideal of H and D' is an R-module equivalent t_2oD. Proof.

Let D' be an R-module equivalent to D.

If A is the

closed component of R corresponding to the equivalence class [D], then by Theorem 8.2 there is a regular ideal I of A such that D' Z K @R I.

Let ~ = HomR(K,D' ) ~ HomR(K,K ®R I);

then by Theorem

2.2, I is isomorphic to the completion of I in the R-topology,

and

hence by Theorem 3.2 to the completion of I in the A-topology.

Now

is the completion of A by Theorem 7-3, and hence by Theorem 2.8, T i8 isomorphic to a nonzero ideal of H.

By Corollary 1.2 we have

K ®R I ~ K ®R I ~ D'. On the other hand suppose that T is a nonzero ideal of H. is the divisible submodule of A, then A/~

is an analytically irre-

ducible Noetherian local domain by Theorem 7.2; tion by Theorem 7.3. and thus ~ O A / ~

Hence A/~

If

and H is its comple-

is a closed domain by Theorem 7.1,

is a nonzero ideal of A / ~ •

is a regular ideal I of A such that I / ~

By Theorem 2.11 there

= ~ A A/~.

By Theorems

82 2.8 and 2.11 the completion of I is isomorphic to T, and thus by Theorem 2.2 we have Y ~ HomR(K,K ®R I).

By Corollary 1.2,

K @R ~ ~ K ®R I, and thus by Theorem 8.2, K @R ~ is equivalent to D.

Corollary. 8.6.

Le__.~tD be a simple divisible R-module,

b__eethe close______~dcomponent of R corresponding equivalent

to D.

t__qoD i__ssisomorphic t__ooD if and o n ~

and let A

Then every R-module

i__ffA i_~s~ pseudo val-

uation ring. Proof.

A is

a

pseudo valuation ring if and only if its comple-

tion H is a complete discrete valuation ring.

It is also clear that

is a complete discrete valuation ring if and only G(~) consists of one element.

Thus the corollary

is an immediate consequence of

Theorem 8.5. Remarks.

(I)

Corollary 8.6 shows that modules can be equival-

ent without being isomorphic,

and thus settles any lingering doubts

that the whole concept of equivalence isomorphism.

(2)

This, of course,

is Just a vacuous extension of

could have been seen much earlier.

Schur's Lemma states that the endomorphism ring of

module is a division ring.

a

simple

By analogy we should expect that the

endomorphism ring of a simple divisible R-module should be something special.

That this is indeed the case is shown by the following

theorem. Theorem 8,~.

Let D be a simple divisible R-module and H the

corresponding a nal~tic component o_~fR. plete, Noetherlan,

Then F = HomR(D,D ) is a com-

local domain of Krull dimension i.

F is an exten-

sion ring o_~fH i_.nnits quotient field and i__ss~ finitel~ ~enerated ~module.

Conversely, i_~f F is a rin~ with these properties

is a simple divisible R-module D' equlvalent F ~ HomR(D',D' ). Proof:

then there

t._~oD such that

83 By Theorem 8.5, there is an ideal T of S such that D $ K ®R Y" Since ~ is a complete R-module, we have HomR(D,D) ~ H o m ~ [ , ~ ) Corollary 2.4.

by

By Theorem 2.7, we have HomR(T,T) & HomH(T,T).

cause T is an H-module, we have

HOmH(Y,T) = Hom~(Y,Y).

Be-

Hence the

statements about F follow from Corollary 6.2 and Theorem 7.1. Conversely, theorem.

let F be a domain with the properties defined in the

Let D' = K ®R F;

since F is isomorphic to an ideal of H,

we see by Theorem 8.5 that D' is a simple divisible R-module equivalent to D.

By the same arguments as in the preceding paragraph we

have HomR(D',D') ~ Hom~(F,r) ~ F. Theorem 8.8.

Let D 1 and D 2 b_~esimpl e divisible R-modules.

Then

the following statements are equivalent: (i)

D 1 is equivalent t__ooD 9.

(2)

HomR(DI,D 2) ~ O.

(3)

AnnHD 1 = AnnHD 2.

Proof.

The equivalence of (I) and (2) is due to Lemma 5.8;

and

the equivalence of (i) and (3) is due to Corollary 8.3. Corollary 8. 9 .

l_f_fD is a simple divisible R-modul_.~e, then there

is a submodule of K equivalent t_%oD. Proof.

If P = AnnH(D),

then we have seen in the course of the

proof of Theorem 7.4 that AnnK(P) is not reduced. simple divisible R-module D' contained in AnnK(P). p c AnnHD' = P' we see that P = P'.

Hence there is a Since

But then D and D' are equivalent

by Corollary 8.3. Corollary 8.10. HomR(D,K ) ~ O.

l__ffD is a divisible R-module, then

Thus if B is a ring extension of R in Q, B ~ Q, then

HomR(Q/B,K) 6 0 and HOmR(B,H) ~ 0. Proof.

Since E is a universal inJective R-module, we have

HomR(D,E) ~ O.

Let f be a nonzero element of HomR(D,E).

Then Im f

84 iS an Artinian divisible R-module and HOmR(Im f, K) c HomR(D,K ). Thus we can assume that D is an Artinian divisible R-module.

There

is a submodule C of D such that D/C is a simple divisible R-module. Since HomR(D/C,K) c HomR(D,K ) we can assume that D is a simple divisible R-module.

But then by Corollary 8.9 we have HomR(D,K) ~ 0.

Let B be a ring extension of R in Q and B ~ Q.

Then

H°mR(B'H) ~ H°mR(B'H°mR(K'K)) ~ H°mR(K ®R B,K) ~ 0. Theorem 8.11.

Let AI,...,A n be the closed component s o~f R, and

let D b__ee~ simple divisible submodule of K correspondin 5 to A I. (I)

D c

N

Then

(Aj/R)

j~l (2)

A1/R contains every simple divisible submodule of K that is

not equal t_~oD. Proof.

If J > i, then by Corollary 8.9 there is a simple divis-

ible submodule Dj of K corresponding to Aj.

By Corollary 1.7 there

is an element f £ HomR(K,K) such that f(K) = Dj. f(D) c D, and thus f(D) c D N Dj. reduced.

By Theorem 2.7,

But D ~ Dj, and hence D n Dj is

Thus f(D) = 0 and D c Ker f.

Therefore, D c h(Ker f);

since h(ker f) = Aj/R by Theorem 8.2, we have D c Aj/R.

Dc

~

and

Thus

(A j/R).

j~l Now if D' is any divisible submoduie of K that is not equal to D, we apply the argument of the preceding paragraph with D in the role of Dj and D' in the role of D, and we obtain D' c AI/R.

CHAPTER SEMI-SIMPLE

Throughout Macaulay

this

semi-simple

modules.

ment

divisible

expect

R will

is not

DIVISIBLE MODULES

be a 1 - d i m e n s i o n a l ,

local C o h e n -

Theorem following

with

the theory

but if we r e p l a c e

is

divisible

R-

isomorphism

modules

R-module

R-modules.

see in the next

R-module

seml-slmple

divisible

divisible

Let B be an A r t i n i a n

statements

divisible

of o r d i n a r y

that a s e m i - s i m p l e

as we shall

9.1.

an A r t i n i a n

if it is a sum of simple

sum of s i m p l e

correct,

correct,

say that

module

to find

to a direct

becomes

chapter

We will

By a n a l o g y

we w o u l d morphic

AND UNISERIAL

ring.

Definition. a

IX

This

is isostate-

by e q u i v a l e n c e

it

theorem.

divisible

Then

R-module.

the

are e q u i v a l e n t :

(i

B is a s e m i - s i m p l e

divisible

(2

B i~s e q u i v a l e n t

(3

l_~f C i__ssan F d i v i s i b l e

to a direct

R-module. sum of s i m p l e

divisible

R-mod-

ules. submodule

of B,

then B i_~s e q u i v a l e n t

to c ~ B/C. Proof. B.

(i) ~ >

If B I @ B,

rained uing of B:

in B I.

(2).

there Let

is a s i m p l e

divisible

in this way we o b t a i n an a s c e n d i n g D I ~ D2 ~

every Bj

is a simple

divisible

submodules

that B = D n.

divisible

that

5.5,

divisible

= n.

module

series

submodule

of

B 2 of B not conm then D I @ D 2. Contin-

chain

of d i v i s i b l e

submodules

D i = B I + B 2 + ... + B i, and

submodule

by T h e o r e m

a composition

see that L(B)

D i such

of B.

there

Now D i / D i _ I is a n o n z e r o

is a simple

structed

...

divisible

submodule

D 2 I B I + B 2 and D I = BI;

0

hence

Let B I be a simple

is an

Since B has ACC on i n t e g e r n > 0 such

homomorphic

by Lemma

of d i v i s i b l e

5.8.

image

of Bi,

Thus we have

submodules

and con-

of B, and we

86

We have an exact

sequence:

0 - ~ S - ~ B I @ ... • B n -~ B - ~ 0. Using Theorem

5.11,

thus S is a r e d u c e d B1 @

(I).

This

(i) ---> (3). B has ACC

divisible ty that this

R-module.

... @ B n. (2) m >

Since

n F~ L ( B i) - L(B) ,, n - n = 0, and i-l By C o r o l l a r y 5.13, B is e q u i v a l e n t to

we have L(S) =

is a t r i v i a l

Let C be a d i v i s i b l e

on d i v i s i b l e

submodule

the case.

of B s u c h that

submodule

A of B that

C N A is reduced.

is not

We assert

D ~ A + C.

by T h e o r e m

of B. 5.5,

that A + C = B.

is a s i m p l e

n c)) + L ( D / D

n D)) + L ( C / ( A 1 n c)) < L(C)

to the p r o p e r -

For s u p p o s e

divisible

submodule

that D

of A,

= L((A + C)/A) n (A + C)) = L(C) + 1.

.. L(AI/A)

+ I.

is a

by the m a x i m a l l t y

We have L((A 1 + C)/A)

hand we have L((A 1 + C)/A)

there

with respect

Let A 1 = A + D;

+ L((A 1 + C)/(A + C)) = L ( C / ( A

- L(D/(A

submodule

is m a x i m a l

T h e n there

A 1 O C is not reduced.

On the o t h e r

statement.

This

+ L((A 1 + C)/AI) contradiction

shows

that A + C = B. We have an exact

sequence:

O-~A

Since A N C is reduced, Lemma

5.8.

n c -~A

we see that B is e q u i v a l e n t

Now B/C = (A + C)/C ~ A/(A

ent to B/C

by Lemma

5.8.

Combining

ence we see that B is e q u i v a l e n t (3) m > submodules Hence

R-modules, Kerf

(i). of B.

by C o r o l l a r y

such that

@ C - ~ B - ~ O.

is reduced.

N C), and h e n c e A is e q u i v a l -

this w i t h

the p r e c e d i n g

Then 5.13

by a s s u m p t i o n there

f(A) c A.

B is e q u i v a l e n t

is a h o m o m o r p h l s m Since

f(A)

divisible

to A @ B/A.

f of A @ B/A onto B

is a sum of simple

On the o t h e r hand L(f(A))

Thus we have

equival-

to B/C @ C.

Let A be the sum of all of the s i m p l e

Ker f is reduced. we have

to A @ C by

f(A) ~ A.

divisible

~ L(A)

since

87 Suppose module

that A ~ B;

then there

D of B/A and f(D) c A and

exists

f(D) ~ O.

Let C = f-l(f(D))

then f(C) = f(D) and hence C is not reduced an exact

a simple divisible

by Theorem

5.4.

subD A;

We have

sequence: 0 - ~ S - ~ C ¢ D - ~ f(D) -~ 0

where

S c Ker f, and hence

S is reduced.

+ L(D) ~ L(C) + I, and hence tradiction

shows

L(C) = O.

But then 1 = L(f(D)) Thus C is reduced.

= L(C)

This con-

that B I A, and hence B is a sum of simple

divisible

R-modules. Remarks. divisible

R-module

the integral remarks

We shall prove

the converse

Corollary

9.2.

5.14).

simple divisible are nonzero

R-module.

of divisible Lemma

A is a submodule Artinian

divisible

if and only

R-module.

if

(See the

is true for

images

Artinian

submodule

R-

of B.

R-modules.

homomorphic

By Theorem

divisible

submodules

9.~.

generated

divisible

image of B is a semi-

9.1 we see that both A and B/A

of B.

At the other extreme

say that an A r t i n i a n

R-module

The same a s s e r t i o n

r o ~

every nonzero

homomorphic

Definition.

every Artinian

Let B b__ee~ semi-slmple , divisible,

Then A and B/A are semi Tsimple Clearly

that

corollary.

and let A be ~ nonzero,

Proof°

divisible

of R is a finitely

of the following

Corollary module,

is a semi-simple

closure

following

in a later theorem

from seml-simplicity

R-module

is linearly

is uniserial

we shall

if its lattice

ordered.

I f D i__~s~ u niserlal,

divisible , Artinian

of D, then h(A) and D/A are uniserial,

module and

divisible,

modules.

Proof. ible Artinian

It is of course module.

trivial

that h(A)

Let C/A be a divisible

is a unlserlal, submodule

of D/A.

divisThen

88

by Corollary

5.2,

C = h(C) + B, where B is a finitely

module.

Thus C/A = [(h(C)+ A)/A] + [(B + A)/A],

finitely

generated

From this Artinian

R-module.

it follows

immediately

and

5.3,

?.4.

(B + A)/A

is a

C/A = (h(C) + A)/A.

that D/A is a unlserlal,

Let D be a unlserlal,

Then D i_~s equivalent

to a submodule

L(D) ~ min(L(K),L(E)). o f the uniserial, Proof.

Corollary

Thus

divisible,

A contains

divlsibl____~e, Artlnian

there

is a finite

homomorphic

every homomorphic

of K.

bound on the

modules.

image of K in D. image

of K in D.

Since D Thus by

1.3, A = D.

Let B be the unique have an exact

simple divisible

Hence we have a derived

exact

O-*HOmR(D/B,E)~*~ Since HomR(B,E) f E HomR(D,E) sufficient

submodule

of D.

Then we

sequence: 0 --~ B i

D X D/B--

O.

sequence:

HomR(D,E) i~ HomR(B,E) --0.

~ 0, ~* is not an epimorphism.

such that f ~ Im ~*.

by Lemma

5.8 to prove

Thus we can choose

To prove the theorem

it will be

that Ker f is reduced.

Then B c K e r f

morphism

But then ~*(g) = f, and we have f E I m

g : D / B - * E.

This contradiction Remark.

shows

that Ker f is reduced,

We shall prove

in Theorem

Theorem 9.5.

and hence

Suppose

this is not the case.

statement

module.

of E and to a factor module

divisible , A r t i n i a n

Let A be a maximal

is uniserial,

tains

R-

module.

Theorem

Hence

Thus by Theorem

generated

f induces

proving

later on that L(K) = L(E).

9.4 that L(D) ! min(L(K),L(E))

Let B be an R-module

one and only one uniserial,

equivalent

divisible,

that

an R-homo~*.

the theorem. Thus the

is redundant. to E.

Artinian

Then B con-

R-module

from

$9

each equivalence Proof.

class

Case I:

By Theorem 9.4, Artlnlan

module

of such modules. Assume B = E.

E contains

from each equivalence

that D 1 and D 2 are equivalent u!es of E.

be extended

and by Theorem

D 2 = g(Dl) c D I. Case II:

However,

In similar

General

By Corollary

Case:

5.13 there

that Ker f is reduced. talns at least equivalence

class

2.7,

of E.

Thus

g(D 1 + D 2) = g(Dl). = L(g(DI))

By Theorem

is an R - e p l m o r p h i s m

f can

4.5, Thus

Thus we have

5.8 and Lemma

Artlnian

Suppose

Artinian

9.3,

f of E onto B such

9.3 and Theorem

divisible,

divisible,

that

Therefore,

9.4,

R-module

B confrom each

D 1 and D 2 are

submodules

of B.

By Coro-

g of B onto E such that Ker g is g(D I) and g ( D 2) are equivalent

uniserial,

divisible

by Case I we have g(D I) = g(D2),

D 1 = D 2.

R-module,

D2, and E are H-modules.

Artlnian

sub-

and hence

L(D 1 + D2) ~ n(g(D 1 + D2))

= L(DI) , and thus D 1 = D 1 + D 2.

and therefore,

is an R-

B'-~E.

is an R-eplmorphlsm

By Lemma

there

submod-

fashion we have D 1 c D2, and thus D 1 = D ~

to D 1 and D 2 and are equivalent modules

DI,

Suppose

Artinain

By definition

g(D I) = f(D I) = D 9.

of such modules.

unlserial,

divisible,

g of E.

divisible,

of such modules.

E is an inJectlve

Thus by Lemma

one unlserial,

llary 5.13 there reduced.

Since

to an R - e n d o m o r p h l s m

we have g(D I) c D I.

equivalent

unlserlal,

f of D 1 onto D 2.

= H;

one unlserial,

class

We will show that D 1 = D 2.

homomorphism

HomR(E,E)

at least

Similarly,

D 2 = D 1 + D2,

153

REFERh~CES (I)

H. Bass,

"On the ubiquity of Gorenstein rings," Math. Zeitschr.,

82 (1963), 8-28. (2)

N. Bourbaki,

"Elements de Math@matique, Alg~bre Commutative",

Fascicule XXVII, No. 1290, Hermann, Paris (1961). (3)

H. Cartan and S. Eilenberg,

"Homological Algebra", Princeton

University Press, Princeton, N. J. (1956). (4)

J. Dieudonn@ and A. Grothendieck, Alg@brique I", Springer-Verlag,

"Elements de G6om@trie

Berlin,Heidelberg,

New York

(1971) (5)

D. Ferrand and M. Raynaud, Noeth@rien",

"Fibres Formelles d'un Local

Ann. Scient. Ec. Norm. Sup. 4 e Serie, t.3,

(1970), 295-311. (6)

R. Hamsher,

"On the structure of a one dimensional quotient

field", J. of Algebra, 19 (1971), 416-425. (7)

J. Herzog and E. Kunz, Macaulay-Rings",

"Der Kanonische Modul eines Cohen-

Lecture Notes in Mathematics No. 238, Springer-

Verlag, Berlin, Heidelberg, New York (1971). (8)

J. Lipman,

"Stable ideals and Arf rings", Am. J. of Math.,

Vol. XCIil, No. 3, (1971), 649-685. (9)

E. Matlis,

"Injective modules over Noetherian rings", Pacific

J. Math., 8 (1958), 511-528. (I0) E. Matlis, ~8~J - - J ~al ~ J

"Divisible modules",

Proc. Amer. Math. Soc., II (1960)

*

(ii) E. Matlis,

"Some properties of a Noetherian domain of dimension

i", Canadian J. Math., 13 (1961), 569-586. (12) E. Matlis,

"Cotorsion modules", Memoirs Amer. Math. Soc., No. 49,

(1964). (13) E. Matlis,

"Decomposable modules",

(1966), 147-179.

Trans. Amer. Math. Soc., 125

154

(14) E. Matlis,

"Reflexive Domains", J. Algebra, 8 (1968), 1-33.

(15) E. Matlis,

"The multiplicity and reduction number of a one-

dimensional local ring", Proc. London Math. Soc., (to appear). (16) E. Matlis,

"The theory of Q-rings",

(to appear).

(17) M. Nagata,

"Local rings", Interscience Publishers, New York,

N. Y., (1962). (18) D. G. Northcott,

"Ideal Theory", Cambridge Tracts, No. 42,

Cambridge University Press, London (1953). (19) D. G. Northcott,

"General theory of one-dimensional local rings"

Proc. Glasgow Math. Assoc., 2 (1956), 159-169. (20) D. G. Northcott,

"On the notion of a first neighborhood ring",

Proc. Camb. Phil. Soc., 53 (1957), 43-56. (21) D. G. Northcott,

"The theory of one-dimensional rings", Proc.

London Math. Soc., 8 (1958), 388-415. (22) D. G. Northcott,

"The reduction number of a one-dimensional

local ring", Mathematic&, (23) D. G. Northcott,

6 (1959), 87-90.

"An algebraic relation connected with the

theory of curves'[, J. London Math. Soc., 34 (1959), 195-204. (24) 0. Zariski and P. Samuel,

"Commutative Algebra", Vol. II, Van

Nostrand, Inc., Princeton, N. J. (1960).

155

I NDEX page

(i)

AI

(2)

analytic component

77

(35 (4)

analytically irreducible

74

(55 (6)

analytically unramlfied

96

AnnH([T])

85

(v5 (8)

Artinian module

45

ascending chain condition (ACC)

45

(9) (io)

canonical ideal

150

closed component

76

ill) (12)

closed domain

74

compatible extension

36

(13)

completion

17

(145

composition series of divisible modules

52

(155 (16)

cotorsion module

II

descending chain condition (DCC)

45

(lr)

discrete valuation ring

79

(18)

divisible module

(19) (2o) (21)

E = E(R/M) equivalent

composition

equivalent

modules

(225

first neighborhood ring

(23) (24)

Gorenstein ring

99

H, the completion of R

21

(25) (26)

h-divisible

8

h-reduced

8

(27) (28)

hdRA, the homological dimension of A Hilbert polynomial

(29)

indecomposable module

150

144

analytically ramified

7 5o series

57 56 109

12 I09

44

156

Page 44

(30)

irreducible ideal

(31)

K = Q/R

(32)

Krull dimension

(33)

latent multiplicity

133

(34)

latent residue degree

141

(35)

local ring

24

(36)

L(A), the divisible length of A

58

(37)

2~(A),

(38)

minimal ideal for a prime P

138

(39)

multiplicity

109

(40)

Noetherian module

45

(41)

1-dimensional

4O

(42)

P-primary component

104

(43)

P-primary divisible module

102

(44)

pseudo valuation ring

(45)

Q, the full ring of quotients

(46)

quasi-local

(47)

R-topology

(48)

reduced module

(49)

reduction number

(50)

regular element

(51)

regular ideal

17

(52)

semi-local ring

24

(53)

semi-simple divisible module

91

(54)

simple divisible module

52

(55)

strongly unramified extension

62

(56)

superficial

(57)

torsion module

7

(58)

torsion module of bounded order

8

(59)

torsion-free module

7

7

the classical length of A

Cohen-Macaulay. ring

ring

4O

109

79 7 24 17

element

7 109 7

109

157

Page

(60)

uniserial divisible module

93

(61)

valuation ring

79

CHAPTER X THE INTEGRAL Throughout Macaulay

this

chapter

CLOSURE

R will be a 1-dimensional,

Cohen-

ring.

Theorem

i0.i.

Let S be the intesral

n be the number of distinc~

Proof,

By Theorem

closure

pseudo valuation

L(K) = L(S/R)

fore,

local,

of R in Q and let

rings of R.

+ n.

5.11 we have L(K) = L(S/R)

it is only necessary

be the full set of pseudo

to prove valuation

Then

that L(Q/S) rings

+ L(Q/S).

= n.

of R.

There-

Let V 1 ..... V n

By Theorem 7.7,

n

S =

~ Vi, and by Theorem 7.8, Q/S --- Q/V 1 @ ... @ Q/V n. By Coroli--I lary 8.4 Q / V i is a simple divisible R-module. Thus L(Q/S) = n. Definition.

R is said to be analytically

pletion H has no nonzero course,

to asserting

nilpotent

elements.

that the intersection

unramified This

if its com-

is equivalent,

of the rank 0 prime

of

ideals

of H is O. Remarks.

The equivalent

theorem have been proved ferent way

of statements

for integral

(i) and

domains

(2) of the next

by Northcott

in a dif-

[19, Th. 8].

Theorem

10.2.

The following

(i)

R i__~sanalytically

(2>

The integral

statements

are e~uivalent:

unramified.

closure

S of R in Q is a finitely

generated

R-module.

(3)

L(K) = n, where n i__ssthe number

of distinct

t.ion rin~s o~f R. (4)

K is a semi-simple

divisible

R-module.

(5)

E i_~s ! semi-simple

divisible

R-module.

pseudo

valua-

91

(6)

Every nonzero Artinian divisible

(7)

I_~f C and D are Artinlan divisible

are equivalent Proof.

R-_~modul__~ei_~ssemi-slmpl e. R-modules,

if and only i__ffthey have equivalent

(I) <m,,> (2).

composition

Let V 1 .... ,V n be the distinct

uation rings of R and h(Vi/R) = Ai/R. the closed components

then C and D

of R.

series.

pseudo val-

By Theorem 7.5, A I, .... A n are

Let Pi = H°mR(K'Ai/R);

then P1,...,Pn

are the prime ideals of rank 0 in H by Theorem 7.3.

Since

S = n v i, we have i O Pi = N HomR(K, Ai/R) = N HOmR(K, Vi/R) = HomR(K,N (Vi/R)) i i i i = HomR(K,S/R). Thus by Corollary

1.3, Q Pi = 0 if and only if S/R is reduced. And i by Theorem 5.1, S/R is reduced if and only if S is a finitely generated R-module. (2) (3).

This is an immediate

consequence

of Theorem i0.I

and Theorem 5.1. (2) ,.N> (4).

If V 1 ..... V n are the distinct pseudo valuation

rings of R, then by Theorem 7.8 we have Q/S & Q/V 1 • ... @ Q/V nSince Q/V i is a simple divisible lows that Q/S is a semi-simple finitely generated 5.8;

R-module,

divisible

by Corollary 8.3, R-module.

then K is equivalent

and hence K is semi-simple (4) =.=> (5).

R-module

If S/R is a to Q/S by Lemma

by Theorem 9.1.

By Theorem 5.5, E is a homomorphic

some m > 0, and thus E is a semi-simple (5) ="> (6).

divisible

A nonzero Artinian divisible

image of K m for

R-module.

R-module

is isomor-

phic to a submodule of E m for some m > O, by Theorem 5.5; it is a semi-simple (6) =t> (7).

divisible

5.14.

R-module

by Corollary

and hence

9.2.

Let C and D be Artinian divisible R-modules.

they are equivalent, Corollary

it fol-

then they have equivalent

Conversely,

composition

If

series by

suppose that they have equivalent

compo-

92

sition series.

Then by Theorem 9.1,

they will be equivalent

same finite direct sum of simple divisible modules, will be equivalent (7) m.=> (4).

they

be the simple divisible factors

of a composition series for K.

then K and D have equivalent are equivalent

and hence,

to each other. Let D I , . . . , D

(with repetitions)

to the

composition series,

by assumption.

If D = DI@...~Dm, and hence K and D

Therefore K is a semi-simple divisible

module by Theorem 9.1. (4) ==> (2). ule. D.

Suppose that S is not a finitely generated R-mod-

Then by Theorem 5.1, S/R contains a simple divisible R-module By Corollary

A/R = h(Ker f);

1.3, there is an epimorphism

of K onto D.

Let

then by Theorem 8.2, A is a closed component

By Theorem 8.11, A/R contains that is not equal to D.

of R.

every simple divisible submodule

of K

But if V is the pseudo valuation ring cor-

responding to A, then D c S I R c V/R, and hence D c h(V/R) = A/R. Thus A/R contains every simple divisible submodule of K. this contradicts

However,

the fact that K is a semi-simple divisible module,

and thus we see that S is a finitely generated R-module. Theorem i0.~. equivalent

I f R i__ssanalytically unramified,

too E.

Proof.

Let S be the integral closure of R.

S/R is finitely generated. 5.$.

then K i__ss

Hence K is equivalent

By Theorem 10.2 to Q/S by Lemma

If VI,...,V n are the pseudo valuation rings of R, then by

Theorem 7.8 we see that Q/S & Q/V I • ... • Q/V n. equivalent

K is

to Q/V l • ... • Q / V n-

By Theorem 10.2, E is a semi-simple E is equivalent R-modules

Therefore,

divisible module, and hence

to D I • ... • Dr, where the Di's are simple divisible

by Theorem 9.1.

By Theorem 9.5, D i is not equivalent

for i ~ J, and every equivalence

to Dj

class of simple divisible modules

has a representative among the Di's.

Thus t = n and (after renum-

93

bering)

we have D i ~ - ~ Q / V i by Corollary

8.4.

Therefore

E is equival-

ent to K. Remarks. that

R is said to be a Gorenstein

is, if K is an injective

13.1,

R is a Gorenstein

stein rings property false.

For there

This

exist

as we shall

find necessary and we shall

analytically divisible,

shows

10.4.

that the converse

Gorenstein

rings

see in Theorem

R = I;

in Theorem

Thus both Gorenin common

the

of Theorem

10.3 is

that are not analytically

14.16.

conditions

In Chapter X V we shall

for K to be equivalent

to E,

to the conditions

ring.

Assume

unramified.

Artinian

rings have

closely these are related

that R is a Gorenstein Theorem

unramified

and sufficient

see how

As we shall prove

ring if and only if K ~ E.

and analytically

that K-~'E.

unramified

R-module.

rin~ if inJ. dim.

that R i_ss~ Gorenstein Then K contains

R-module

rins,

o__[rthat R i~s

one and only on_~e uniserial,

fro___~meach equivalence

class

of such mod-

u!es. Proof.

Theorem

i0.~ and Theorem

9.5 combine

to yield an immed-

late proof. Remarks. theorem

We can now present

that we proved

Theorem

10.5.

followin G statements (I)

simplified

in [13, Th. 7.4] only with great

Let R b_~e ~ Noetherian

intesral

proof

of a

difficulty.

domain.

Then the

are equivalent:

R is a Noetherian

its completion

an extremely

local domain

H has ' onl____~y0he 2 r i m e

(2)

The integral

closure

(3)

Q/C is an inde¢omposable

(4)

R is a Noetherian

of Krull

ideal

dimension

I, a n d

of rank 0.

of R is a discrete R-module

valuation

rin[.

for every R-submodule

C

of q.

is an indeeomposable

local domain of Krull

R-module

for every

dimension

ideal I o f R.

i and Q/I

94

Proof.

(i) <m~> (2).

crete valuation domain.

ring,

establishes

rings

= A/R.

there

is a one-to-one

of R and the prime

the equivalence

(2) ~,,> (3)-

ideals

of (I) and

ring,

submodule

Noetherian

local

correspondence

closure

component

D of K is contained

component of R.

proper proper

divisible

submodule

of K, we have D c A/R.

Q/C = BI/C @ B2/C , where

sume that R c C. D 2 = h(B2/R);

of Q, and suppose

BI, B 2 are R-submodules

B 1 + B 2 = Q and B 1 n B 9 = C.

Without

Thus we have B I / R +

then by Corollary

5.2,

+ T2, where T 1 and T 9 are finitely we have K = D 1 + D 2 by Theorem

K.

Then we can as-

Let D 1 = h(BI/R)

submodules

of K.

of K, then we have K = D 1 + D 2 c A / R by the preceding

graph.

This

shows

I~t then B 1 = Q, and this shows

that we that

and

Hence

If both D 1 and D 2 are proper

modules

contradiction

that

BI/R ~ D 1 + T 1 and B2/R ~ D 2

generated

5.3.

of Q.

loss of generality,

B2/R =

every

in a maximal

of K, and since A/R is the only maximal

R-submodule

of R by

Since

submodule

nonzero

This

of R, and h(V/R)

divisible

Let C be a proper,

between

of rank 0 in H.

A is a closed

and it is the only closed

proper divisible

of R is a dis-

(2) immediately.

Let V be the integral

Since V is a valuation

Theorem 7.5;

closure

then R is a 1-dimensional

By Theorem 7.4,

the valuation

If the integral

can assume

sub-

para-

that D 1 = K.

Q/C is an indecomposable

R-mod-

ule. (3) ~ = > for every

(4).

ideal

Of course

I of R.

it is trivial

Suppose

ring.

Then R has two distinct

put

= R - (I 1 U

~

ensional Theorem

local,

rank 1 prime

domain with two distinct

decomposition.

(4) =,,,> (2).

ideals

Then C is a Noetherian

maximal

ideals

shows

local

I 1 and 12, and we 1-dim-

IIC and I2C.

Q/C ~ Q/ CI1 @ Q/CI2 , a nontrivlal

This contradiction

1-dimensional

Q/I is indecomposable

that R is not a 1-dimenslonal

1 2) and C = Rj .

4.1 we have:

that

direct

By sum

that R is a Noetherian,

domain.

Let S be a ring extension

of R in Q that

is

95

finitely

generated

as a n R - m o d u l e .

Q/S has a n o n - t r l v i a l S is i s o m o r p h i c

direct

If S is not a local ring,

sum d e c o m p o s t i o n

to an i d e a l of R, this

then

by T h e o r e m 4.1.

contradiction

shows

Since

that S is

a local ring. Let V be the i n t e g r a l discrete

valuation

ring.

c l o s u r e of R, and s u p p o s e T h e n t h e r e exist

s u c h that x + y ~ !.

Let S ~ R[x,y];

R-module,

s i n c e x and

y are i n t e g r a l

and hence

e i t h e r x or y is a unit

unit

in V, and this

t i o n ring.

contradiction

non-unlts

that V is not a x and y in V

t h e n S is a f i n i t e l y g e n e r a t e d o v e r R.

in S. shows

But

Thus S is a local

ring,

t h e n e i t h e r x or y is a

that V is a d i s c r e t e

valua-

C H A P T E R XI THE PRIMARY DECOMPOSITION

Throughout Macaulay

t h i s c h a p t e r R w i l l be a 1 - d i m e n s i o n a l ,

Cohen-

ring.

Definition. prime

local,

Let D be an A r t i n i a n ,

i d e a l of r a n k 0 in H.

ule if A n n H D is a P - p r i m a r y

T h e o r e m II.I.

We shall ideal,

divisible

or if D = O.

divisible

R - m o d u l e and P a

T h e m D is a P - p r i m a r y m o d u l e

if all of the s i m p l e d i v i s i b l e

and P a

say that D is a P - p r i m a r y m o d -

Let D be an A r t i n i a n ,

p r i m e i d e a l of rank 0 in H.

R-module

factor modules

if and o n l y

in a c o m p o s i t i o n

series

for D c o r r e s p o n d to P. Proof.

Suppose

0 = DO c DI c modules

of D.

that D is a P - p r i m a r y m o d u l e

... c D n = D be a c o m p o s i t i o n Then AnnHD c AnnH(Di+i/Di)

and let

s e r i e s of d i v i s i b l e

and A n n H ( D i + i / D i )

sub-

is a

p r i m e i d e a l of rank 0 in H by C o r o l l a r y 8.3. S i n c e A n n H D is a P - p r i m a r y AnnH(Di+lJDi) ponds

ideal,

it f o l l o w s

= P for all i = 0 , . . . , n - l ;

that

and h e n c e D i + I / D i c o r r e s -

to P. Conversely,

a composition

series

Let I = AnnHD; Since

suppose

that

every simple

of d i v i s i b l e

divisible

submodules

factor module

of D c o r r e s p o n d s

to P.

and we s h a l l p r o v e t h a t I is a P - p r i m a r y ideal.

PDi+ I c D i for all i, we h a v e pnD = O, and t h u s pn c I.

c a u s e I D i + I c Di, we h a v e that fg ~ I and f ~ P. t h e r e is a c o m p o s i t i o n

I c p.

We a s s e r t series

t h r o u g h fD b y T h e o r e m 5.10.

Let

f and g be e l e m e n t s

that fD = D.

of d i v i s i b l e

of H s u c h

F o r if fD ~ D, t h e n

submodules

of D p a s s i n g

Hence there exists a divisible

submod-

ule B of D s u c h that f D c B and C = B / f D is a s i m p l e d i v i s i b l e module

corresponding

tradiction

shows

h e n c e g E I.

to P.

But fC = 0, and h e n c e f ~ P.

that fD = D.

This proves

ideal.

R-

This con-

We t h e n h a v e 0 = (gf)D = gD,

that I is a P - p r i m a r y

Be-

and

of

97

Corollary R-module,

11.5.

Let D be a nonzero

where P is a prime

P-primary

divisible

Artinian

ideal of rank 0 in H, and let f ¢ H.

Then fD = D if and only if f # P. Proof. fn E AnnHD,

If f ¢ P, then there and hence fD ~ D.

Artinian, H.

ll.3.

divisible

an integer n > 0 such that

Conversely,

shown that fD = D in the course Corollary

exists

if f # P, then we

of the proof

Let O - ~ B - ~ C - ~ D - ~ O R-module,

have

of Theorem ll.1. be an exact

sequence

of

and let P be a prime ideal of rank 0 in

Then C is a P-primary m o d u l e

if and only if B and D are P-primary

modules. Proof.

By Theorem

ible submodules

5.10,

of C passing

there is a composition through

B.

Hence

series

of divis-

the set of simple

divisible

factor modules

for C is the union of the sets of simple

divisible

factor modules

for B and for D.

directly

from Theorem

CorollarY modules

11.4.

(2)

now follows

ll.1. (I)

and P is a prime

P-primary module

The corollary

I f DI,...,D n are Artinian,

divisible

ideal of rank 0 in H, then D I @

... @ D n is

if and only if every D i is a P-primary

module.

if D i~s an Artinian,

divisible

P-primary

divisible

submodules

of D, then B + C is also a P-primary

divisible

submodule

of D.

(3)

If D is an Artinian,

unique largest

P-primary

divisible

P-primary

divisible

submodu!es.

(4)

divisible

submodule

respectively,

Proof. from

Corollary

(1) follows

of D;

As for

C;

and B and C a~e

then D contains

C contains

and D/C has no nonzero

and P'-primary

immediately ll.3.

from Corollary

P-primary

Artinian

divisible

ll.3 and

(3) now follows

(4), let f ~ HomR(B,B'),

from

a

every

w i t h P ~ P', then HomR(B,B' ) = 0.

(1) and Corollary

ll.3.

R-module,

submodule

I f B and B' are P-primary

R-modules

follows

divisible

R-module,

R-

(2)

(2) and

and suppose

that f

98

is not zero. ll.3.

Then f(B) is both P-primary and P'-primary by Corollary

This contradiction Definition.

shows that HomR(B,B' ) = 0.

Let D be an Artinian,

prime ideal of rank 0 in H. m a r y divisible

divisible R-module and P a

We shall call the unique largest P-pri-

submodule of D that is guaranteed to exist by Corollary

ll.4 the P-primary component of D.

We shall denote this component

by Dp. Theorem ll.5.

Let D be an Artinian,

divisible R-module,

P1,...,Pn be the prime ideals of rank 0 i_~n H. (I)

D

(2)

D is equivalent

Proof.

= DPl

(1)

+ ..

.

Let C

+

DPn to

= DPl

Then there is a divisible

DPI;

Let A = DP2 + . ..

Then:

.

DP1 @ .. • @ DPn . + .

"-

+

and suppose that C W D.

DPn

submodule B of D such that C c B and B/C is

a simple divisible R-module. PI"

and let

We can assume that B/C corresponds

+ DPn,"

t h e n C/A i s

a homomorphic

to

image o f

and thus by Corollary 11.3, C/A is a Pl-primary module.

We

have an exact sequence: 0 -+ C/A -~ B/A -~ B/C -~ 0 where the extremes are Pl-primary modules.

Hence by Corollary ll.3,

B/A is a Pl-primary module. Let I = AnnHA; primary ideal, that I ~ Pl"

then

Q AnnHDPi c I. Since AnnHDPi is a Pii>2 (or is equal--to H in the case that DPi = O) we see

Choose h ~ I such that h ~ Pl"

Because B is an H-mod-

ule by Theorem 2.7, h induces an R-homomorphism f : B / A - ~ B by f(x + A) = hx for all x ¢ B.

defined

We assert that Ker f is a reduced

R-module. Suppose that Ker f is not reduced; divisible R-module F/A.

then Ker f contains a simple

Since B/A is Pl-primary we have A n n H F / A =

Pl"

If x ¢ F, then hx = f(x + A) = 0, and thus h ¢ A n n H F c AnnHF/A = Pl'

99

But h ~ PI' and this contradiction Let G = f(B/A) c B;

then G is a Pl-primary module by Corollary

ll.3, and hence G c DP1 c C. thus f(B/A) = f((G + A)/A). + Ker f.

shows that Ker f is reduced.

By Corollary ll.2 we have G = hG, and Therefore,

we have B/A = (G + A ) / A

Since Ker f is reduced, we have B = G + A by Theorem 5.3.

Thus B c C, and this is a contradiction. D

+ .

= DP1 (2)

-.

+

Hence we have

DPn-

Now AnnH(DP2 + ... + Dp ) D A AnnHDPi, n i~2

and hence

DP2 + ... + DPn has no nonzero Pl-primary component. DP1 0 (DP2 + ... + DPn) is a reduced R-module.

Thus

By statement

(1) we

have an exact sequence:

O-~OP1 0 (Op2 + ... + nPn) -~nP1 ~ (Op2 + ... + nPn) ~n-~o. Hence by Lemma 5.8, D is equivalent to DP1 ~ (DP2 + ... $ DPn).

By

induction on the number of nonzero primary components, DP2

+ .

.-

+

is equivalent to

DPn

DP 2

~ .

.-

~

DPn

.

Thus D is equival-

ent to DPl @ (DP2 ~ .,. ~ DPn ). Corollary ll.6.

A uniserial,

Artinian,

divisible R-module is

P-primary for some prime ideal P of rank 0 in H. Proof. Remarks.

This is an immediate consequence of Theorem ll.5. It is clear from Theorem ll.5 that if D is an Artinian

divisible R-module,

then a composition

series of divisible

submodules

of D can be found so that all of the simple divisible factor modules corresponding to a given prime ideal P of rank 0 in H can be arranged in sequence in a single block,

and that the position of these blocks

in the chain is arbitrary. Corollary ll.7.

If A is an Artinian divisible R-module,

given simple divisible R-module D is isomorphic in a composition

series of divisible

then a

to a factor module

submodules of A if and only if

100

D is equivalent Proof.

to a submodule of A.

This is clear from the preceding

remarks about Theorem

11.5. Theorem 11.8.

Let 0 = N I O ... n N n be a normal decomposition

o_~f 0 i_nn H, where N i is a Pi-primary prime ideals of rank 0 i_n H. U =

... A B n.

BI N

ideal of H and PI,...,Pn are the

Let K D R N i = Bi/R c K, and let

Then:

(i)

B i i_~sa strongly unramified

(2)

Bi +

n

ring extension of R i_nnQ.

Bj = Q, and thus Q/U % Q/B I • ... ~ Q/B n.

j~i (3)

l_~f A i is the closed component

of R corresponding

t__ooP i,

then B i c A i and B i ~ Aj for j ~ i. (4)

U is a finitely generated R-module,

ensional, main,

semi-local,

Cohen Macaulay

ring.

and thus U is a l-di m-

If R is an integral

then the ideals M i = BiM 0 U are all distinct

do-

and are the only

maximal ideals of U, and B i = UMi. (5)

K i_ss equivalent

(6)

Q/B i i_~s isomorphic

Q/B i i_~s equivalent (7)

to Q/B I ~ ... ~ Q/B n. to the Pi-primary

to the Pi-primary

K D R (j~iN Nj) = h(jji n Bj/R)

component

of Q/U an__~d

component of K. is the Pi-primary

component o_~f

K. Proof.

(i)

B i is a strongly unramified

extension of R in Q by

Theorem 6.6. (2)

between 1 and n; and k let A be a closed component of R such that A/R contains h( n B~ /R). j:l ~t k k By Corollary 6.7, h( n B~ /R) = K ®R ( n N~ ); and by Theorem 7.3, j=l ut j=l ut A/R = K @R P' where P is the prime ideal of rank 0 in H corresponding k to A. Then by Theorem 6.6, t=lONjt c p. Hence P = PJt for some integer

Let Jl'''''Jk be any set of integers

Jt and thus A = Ajt.

Therefore,

if j is an integer,

1 _< J 0

we shall give a proof of this prove that ~ ( A / A M ) ~(A/R)

= p.

= e.

Furthermore,

en f o r a l l n > o. Let ~ ( A / A M )

= a and

S(A/R)

= ~.

By Theorem 12.2 there

107

is a regular

element x of A such that Ax = AM.

Thus ^Mi/AM i+l

A/Ax for every i, and hence S ( A / A M n) = no for every n ~ O. all large k we have by Theorem - p.

Thus a k =

+ (ek - p).

AM =

t O i=l

~ ( A / A M k) = J~(A/M k) =

~(A/R)

~R/M

k) = ek

+ ~ ( R / M k) = 8

Letting k -*~, we see that a = e and 8 = P.

Remarks. A i = ANi.

12.1 that AM k = M k and

For

Let N 1,.. .,N t be the set of maximal

Then A O AiM is an Ni-primary (A 0 AiM ) is a normal

ideals

of A and let

ideal of A and

decomposition

of AM.

Since

t E @ Ai/AiM by the Chinese i=l t Remainder Theorem. Hence we see that ~ ( A / A M ) = Z ~ ( A i / A i M) • If i=l we let ~i be the length of Ai/AiM as a Ai-module and A/(A 0 AiM ) --- Ai/AiM , we have A/AM -=

k i = [Ai/AiN i : R/M],

and note that by Theorem 12.4 ~ ( A / A M ) = e, t then the preceding equation becomes e = E ki~ i, a formula due to i=l Northcott [21, Th. 2]. The next proposition

is a strengthened

version

of [21, Propo-

sition I]. Theorem

12.5.

Let a ~ M s and let I be a finitely

submodule

of Q such that I contains

~(I/aI)

~ se and ~ ( I / a I )

degree

element

of R.

R-

Then

= se if and only if a is superficial

o_~f

s.

Proof.

If a is a zero divisor

and thus without element

a regular

generated

of R.

loss of generality

Hence by Theorem

in R, then ~ ( I / a I )

we may assume that a is a regular

12.3 we m a y assume

Aa c AM s, we have S ( A / A a )

> se and S ( A / A a )

Aa = AM s by Theorem

By Theorem

a is superficial Corollary of degree

12.4.

of degree

12.6.

is infinite,

12.1,

that I = A.

Since

= se if and only if Aa = AM s if and only if

s.

If a ( M s and b ~ M t, then ab i_~s superficial

s + t if and only if a is superficial

of degree

a and b is

108

superficial

of d e g r e e

Proof. degree

If a is s u p e r f i c i a l

ficial

of d e g r e e

of d e g r e e

b is s u p e r f i c i a l

Ch. IV., define

Prop.

= M'I*.

5], M[X]

ideal

of f i n i t e denote

ideal

We have a n a t u r a l I* = R*I.

1-dimensional

ideal

length,

ab is

= S(A/Aa)

12.5.

Hence + S(A/Ab).

of d e g r e e

s and

t.

of R, then by

I* is an M * - p r i m a r y

of

that ab is super-

a is s u p e r f i c i a l

is a p r i m e

R* is a local,

and h e n c e

= (s + t)e by T h e o r e m

+ S(Aa/Aab)

12.5 that

of R, we d e f i n e

an M - p r i m a r y

suppose

Let X be an i n d e t e r m i n a t e

R* = R [ X ] M [ X ] .

is an i d e a l

= M n+s+t,

of R and thus A a / A a b ~ A/Ab.

of d e g r e e

Definition.

~(A/Aab)

= ~(A/Aa)

from Theorem

s and b is s u p e r f i c i a l

Conversely,

Then

element

se + te = ~ ( A / A a b ) It f o l l o w s

s + t.

s + t.

N o w b is a r e g u l a r

shall

of d e g r e e

t, then for large n, M n a b = M n + S b

superficial

module

t.

[18,

over R;

of rank i in R[X] imbedding

Cohen-Macaulay Ch.

shall d e n o t e

the m u l t i p l i c i t y

R c R*;

T h e n we have MX* =

IV,

Cor.

of R* and I* O R = I. we

then by

ring.

[18,

and we can and if I (MI)* If I is

1 and Prop.

2],

If A* is an R*-

its l e n g t h by ~ * ( A * ) .

and r e d u c t i o n

number

We

of R* by e* and

0", r e s p e c t i v e l y . The f o l l o w i n g

Theorem R,

t h e o r e m was c o m m u n i c a t e d

12.7

(D. Rees).

then ~ * ( J * / I * )

= ~(J/I).

(2)

e* = e and

(3)

R * / M * is infinite,

of d e g r e e

(1)

to me by D. Rees.

If" I c J are M - p r i m a r y

ideals

of

p* = 0. and thus R* has a s u p e r f i c i a l

element

1.

Proof.

If ~ ( J / I )

= n, t h e r e

exists

a chain

of M - p r l m a r y

I = I0 c I 1 c

... c I n = J such that I k / I k _ 1 "-- R / M for all

k = 1,...,n.

Thus for each k t h e r e is an e l e m e n t

I k = Ik_ 1 + Ra k and M a k c Ik_ 1. = J* is a c h a i n of M * - p r i m a r y

Therefore,

ideals

of R*;

ideals

a k ~ I k such that

I* = I~ c I~ c

... c I*n

and since I~ O R = I k,

109

the I~ are all distinct. M*a k = (Mak)* c I~_ 1. composition

series

We have I~ = I~_ 1 + R*a k and

Thus I~/I~. 1 ~ R*/M*,

from I* to J*.

If j > 0 is any integer,

and the I~ gives us a

Therefore,

2~*(J*/I*)

then • * ( ( M * ) J / ( M * )

j÷l)

= ~ * ( ( M J ) * / ( M J + l ) *) = ~ ( M J / M j+l) by the preceding this it follows

immediately

It is obvious

I am indebted the strengthening special

element

is infinite,

and thus by [24, p. 287],

to D. Rees for suggesting of the statement

the method

of the next theorem.

(unpublished)

1 in a C o h e n - M a c a u l a y

Theorem 12.8.

From

of degree 1.

case of a m o r e general

ideals of height

paragraph.

that e* = e and p* = 0.

that R*/M*

R* has a superficial

= ~(J/I).

Every M - p r i m a r y

result

of proof and It is a

of his concerning

ring.

ideal of R can be generated

by

e elements. Proof.

if I is an M - p r i m a r y

ber of generators

ideal of R, then the minimal

of I is equal to ~ ( I / M I ) ,

that ~ ( I / M I )

< e.

of generality

that R has a superficial

then a I c

and hence

MI,

Because

of Theorem

~(I/MI)

then R has a superficial

stance in which the existence given by the following Theorem element

of degree

Proof. AM = Ax. there

12.9.

element

< J~(i/aI)

= e by Theorem

1.

But

12.5.

12.7 that if R/M is in-

of degree

of such an element

If A is a local

1.

Another

is guaranteed

inis

ring,

then R has a superficial

1.

By Theorem

elements

other hand there

element

a of degree

loss

theorem.

12.2 there is an element x in A such that

Now M is generated

exist

and we wish to prove

12.7 we can assur~e without

We have noted in the proof of Theorem finite,

num-

by elements

al,...,a n over R, and hence n u~,~ ...,u n in A such that x = Z u~a-.~ ~ On the i:l

exist elements

Vl,...,v n in A such that a i = vix.

110

n n Thus x =( Z uivi)x, and hence 1 = Z uiv i. Becanse A is a local ring i=l i=l there is an index j, 1 ~ and ~ ( M n / M n+l) < e if n < v. Proof.

Because of Theorem 12.7 we can assume without

generality that R has a superficial

element a of degree 1.

loss of Let ~ be

the smallest integer such that ~ ( M n / M n+l) = e for all n > v. k ~ 0 be any integer; =

Z(Mk/Mka)

-

Let

then by Theorem 12.5 ~ ( M k / M k+l)

(Mk+I/Mka)

= e -

(Mk+I/Mka).

Therefore,

~ ( M k / M k+l) < e and equality holds if and only if M k+l = Mka.

But

if M k+l = Mka, then M n+l = Mna for all n > k, and hence by Theorem 12.5 ~ ( M n / M n+l) = e for all n ~ k.

In this case k ~ v, and this

proves the proposition. Definition.

Let k I be the smallest integer

L(R/M kl) = ek I - p; AM k 2 = M k2 .

Theorem 12.11.

such that

and let k 2 be the smallest integer such that

We have k I = k2 = ~, where ~ is the integer of

Theorem 12.10. Proof. = ekl; M kl c

We have ~h~(A/M kl) = ~ ( A / R )

and we have S ( A / A M k l ) k I ek I by Theorem 12.4. Since kl AM , we conclude that M = AM kl. Therefore, k 2 < k 1.

On the other hand, we have -

+ S ( R / M kl) = p + (ek I - p)

~(A/R)

~ ( R / M k2) = Z ( R / A M ~ )

= ~ ( A / A M k2)

= ek 2 - p by Theorem 12.4, and hence k I < k 2.

By Theorem 12.2 there is a regular element x in A such that Ax = AM. Therefore,

If n > k2, then M n = AM n = Ax n, and hence Mn/M n+l --- A/Ax. ~ ( M n / M n+l) = e by Theorem 12.4, and thus k2 _> v.

111

If kl > ~ , t h e n

~(21-1/Mkl)

= e, and h e n c e S ( R / 2

I-I)

k = ~(R/M

I)

dicting

. ~(Mkl-I/21)

the m i n i m a l i t y

Definition.

of k I.

Thus k I < ~ and hence k I = k 2 = ~.

If A is an R - s u b m o d u l e

of A by A -I = [q ¢ Q ideal,

= (ekl _ p) _ e = e(k I - i) - p, c o n t r a -

I qA c R].

Since R has a p r i n c i p a l

seen that R c~ M-I.

it is e a s i l y

of Q, we d e f i n e

Because

we have A A -I = R if and o n l y if A is i s o m o r p h i c that M M -I = R if and o n l y if R is a d i s c r e t e is not a d i s c r e t e Noetherian, perly

valuation

semi-local,

contains

R is a local to R.

ring,

Thus w e

valuation

then M M -I = M;

1-dimensional

M-prlmary

ring.

see If R

and thus M -I is a

Cohen-Macaulay

ring that pro-

R.

Theorem

12.12.

Proof.

By T h e o r e m

Ax = AM.

ring,

the i n v e r s e

We have Mn(Mn) -I = A -I for all n > v. 12.2,

there

is an e l e m e n t

If n k v and a = x n, we have Aa = M n.

x in A such that Hence

A -I = a(Mn) -I

and thus A -I = AA -I = Aa(Mn) -! = Mn(Mn) -I.

Theorem

12.13.

R is not a d i s c r e t e

valuation

Suppose

that R is not a d i s c r e t e

ring if and o n l y

if M -1 c A. Proof.

we have M ' I M = M. degree

By T h e o r e m

12.1,

s s u c h that A = M S a -1.

valuation

R has a s u p e r f i c i a l

ring.

Then

element

a of

Thus M-1A = M - 1 M S a -1 = MSa -1 = A, and

h e n c e M -1 c A. On the o t h e r hand uation

ring,

12.1,

In T h e o r e m

But t h i s c a n n o t

happen

p > e - I.

val-

b -1 ~ M -1 c A,

since A is a f i n i t e R - m o d u l e

1 2 . 1 5 we shall find n e c e s s a r y

12.14.

If R is a d i s c r e t e

Therefore,

and thus R is not a d i s c r e t e

tions for M "l to equal A.

Theorem

that M -1 c A.

then M = Rb for some b ~ M.

and thus AM = A. by Theorem

suppose

valuation and

ring.

sufficient

condi-

112

Proof. 12.4.

We have

~(A/M)

: f(A/R)

Since M c AM and ~ ( A / A M )

+ ~(R/M)

= 0 + I by T h e o r e m

= e by T h e o r e m

12.4,

we have

~ +l>e. The next p r o p o s i t i o n for

p to equal

Theorem

and

sufficient

e-

The f o l l o w i n g

statements

are equivalent:

(I)

p=

(2)

A = M -I or R is a d i s c r e t e

(3)

There

exists

a ( R such that M - l a = M.

(4)

There

exists

a c R such that ~

i.

(5)

S(M/M

(6)

~ ( M n / M n+l)

ring.

(i) ~ >

12.4.

=

Therefore, (3).

ring.

= Ma. element

Suppose

+ (4).

e

and thus AM = M - i M = M.

ring and M = Rb for some b ~ R.

(4) ~ > n > I.

12.13

that M - I M ~ M, and h e n c e M - I M : R.

valuation

-

a = b 2 we have

AM = Ax = M, and h e n c e x = a ~ M,

(3) ~---> (4). assume

ring by T h e o r e m

valuation

= 2f(A/M)

ring,

Letting

valuation

I.

A c M -I.

some b ~ R and h e n c e M "! = R(I/b).

discrete

of d e g r e e

: e for e v e r y n > I.

(2).

,,,~(A/R)

(2) ~ >

valuation

2) : e and R has a s u p e r f i c i a l

T h e n M -I c A by T h e o r e m

,,~'(A/AM)

conditions

e - I.

12.15.

Proof.

-

gives necessary

12.5.

element

But Ma c M 2 and

Thus we have Ma : M 2.

of d e g r e e

I.

Then

~ ( M / M 2) = e by as-

113

(6) = >

(i).

For large n we have ~ ( R / M n) = en - p.

~ ( R / M n) = ~ ( R / M )

+

... + s ( M n - I / M n) = 1 + (n-l)e by

Therefore en - p = 1 + (n-l)e = en - (e-l),

assumption. p = e-

+ ~(M/~)

But

and hence

I.

Northcott Theorem

has proved

12.16

the following

(Northcott).

theorem

The following

(see

[20] and

statements

[22]).

are equival-

ent: (I) uation

Every ideal of R is principal

P(n)

(3)

o =

o.

(4)

e =

1.

Proof.

= n;

where P(n)

The implications

(3) ~--~> (4) follows

a principal

is the Hilbert

(I) = >

immediately

that is, e = I.

discrete

It is an immediate

tion where

there is no similar

Theorem (i)

12.17.

theorem

P(n)

(3)

0 = l.

(4)

e = 2.

of this that

Therefore,

R is a

to characterize the situa-

theorem

The following

valuation

by three

statements

How-

p = 2, or the sitelements.

are equivalent:

by two elements,

but R

ring.

= 2n - i, where

If any of these conditions

by two elements.

to characterize

Every ideal of R can be generated

(2)

and

Hence assume

consequence

ideal of R.

every ideal can be generated

is not a discrete

I.

12.14.

every ideal of R can be generated

uation where

of R.

ring.

We shall prove an analogous

degree

val-

Then there is an integer n > 0 such that M n is

ideal of R.

valuation

polynomial

(2) ~---> (3) are trivial,

from Theorem

M M -I = R, and hence M is a principal

ever,

R is a discrete

ring ) .

(2)

(4);

(i.e.,

P(n)

hold,

is the Hilbert

polynomial

then R has a superficial

of R.

element

of

114

Proof.

(I) = = >

(2).

If M n is p r i n c i p a l

M - I M = R and h e n c e R is a d i s c r e t e tion.

Thus ~ ( M n / M n+l)

minimal S(MU/M

number

Thus P(n)

of g e n e r a t o r s

(3) ~ >

By T h e o r e m

(4).

12.16,

(4) ~ > ~(M~-l/M

we have e > 1.

(1).

~) = 1.

is true of

contradiction n+l)

conditions

we have

12.!4,

M n.

is the

By a s s u m p t i o n

= 2 f o r all n > 1.

~ = e - I = 2 - 1 = i.

we h a v e e < p + 1 = 2;

Therefore, 12.10,

(~-l)n÷l)

J ~ ( M ~ - l / M v) < e = 2, and thus

But e = 2, and t h i s

H e n c e by T h e o r e m

a of d e g r e e ~l s u c h that M 2 = Ma. can be g e n e r a t e d

Therefore

by T h e o r e m

P ~ Ra.

a and b are not l i n e a r l y

ideal

element

of R.

two e l e m e n t s Noetherian

is a d i s c r e t e

ring w h o s e m a x i m a l

valuation

ring.

Rb c p and rank P = 0, we rank 0 p r i m e Suppose

modulo

and the fact

shows that M = (a,b).

local

of R, and h e n c e

~,

that M can be g e n e r a t e d

Therefore,

This

P = Rb.

If

by

R / R b is a 1 - d i m e n s i o n a l

i d e a l is p r i n c i p a l

ideal of R is a p r i n c i p a l that I is a P - p r l m a r y

P = 0.

t h e n b ~ Ra +

H e n c e Rb is a p r i m e

see that

If

C h o o s e b ~ P such that b ~ Ra.

independent

This c o n t r a d i c t i o n ,

12.8

b y e = 2 elements.

But t h e n P = Pa n c M n+l f o r all n > l, and thus shows that

equivalent

that R has a s u p e r f i c i a l

rank 0 p r i m e

t h e n a # P since a is a r e g u l a r

1 2 . 1 0 we have

This is one of the

and we c o n c l u d e

that P is a n o n z e r o

of R and the same

If v - 1 > 0, then

that ~ - 1 = 0.

ideal

ideal

= e f o r some l a r g e n.

12.15,

and b y

e = 2.

H e n c e M ~-l is a p r i n c i p a l

of T h e o r e m

contradiction

= Ra.

12.15,

= e = 2 for all n > ~ = 1.

Suppose

P = Pa.

By T h e o r e m

shows

every M-primary

p c Ra,

to g e n e r a t e

H e n c e ~ ( M n / M u÷l)

(M~-l) n for all n > 1.

1 = S(M(~-l)n/M

element

to a s s u m p -

= 2n - I. Trivial.

~(Mn/M

required

e = 2, and b y T h e o r e m

(2) ---~> (3).

Theorem

ring c o n t r a r y

> 2 for all n > 1 b e c a u s e f ( M n / M n+l)

n÷l) < 2 f o r all n > 1.

Therefore,

valuation

f o r a n y n > l, then

We have

and t h u s R / R b

ideal;

and since

shown that e v e r y

ideal.

i d e a l of R.

Since pn c I for some

115

n > I, there is a smallest I = Rb t.

F o r if not,

integer t > i such that b t ¢ I.

choose y ¢ I such that y # Rb t.

w h e r e k < t and r ~ Rb = P. that b k ~ I.

But then

Then y = rb k

But I is P - p r i m a r y and r ~ P implies

This c o n t r a d i c t i o n

shows that I = Rb t.

Therefore

every

P-primary ideal is a p o w e r of Rb. It is easily seen that the i n t e r s e c t i o n

of a finite n u m b e r of

ideals that are powers of p r i n c i p a l prime ideals is again a p r i n c i p a l ideal.

Thus every unmixed

rank 0 ideal of R is a p r i n c i p a l

ideal.

N o w let J be a nonzero ideal of R that is not M - p r i m a r y not unmixed of rank 0. and I is unmixed

Then J = A A I, w h e r e A is an M - p r i m a r y ideal

of rank 0.

Then I = Rc for some c ~ M and hence

J = Bc, where B is an ideal of R. rank 0, then B can be generated spectively. elements. of rank O.

and is

If B is M-primary,

by two elements,

or unmixed

or 1 element,

But then in all cases J = Bc can be generated

of

re-

by two

Thus we can assume that B is n e i t h e r M - p r i m a r y n o r unmixed As in the case of J we see that B = BlC l, where c I ~ M

and B 1 is an ideal of R.

Therefore,

were to continus indefinitely, and hence J = 0.

J = BlClC c M 2.

we would

This c o n t r a d i c t i o n

If this process

have J c M n for all n > 0,

shows that the process must

stop

somewhere,

and when it does we see that J can be g e n e r a t e d by 1 or 2

elements.

Therefore,

ments.

every ideal of R can be generated

by two ele-

CHAPTER XIII

GORENSTEIN RINGS Throughout this chapter R will be a 1-dimensional local CohenMacaulay ring.

We shall freely use the notation of Chapter XII with-

out further definition. Definition.

R is said to be a Gorenstein ring if the injective

dimension of R as an R-module is I. The following theorem may be found in [i] and [14] in a more general setting. Theorem 13.1.

The following statements are equivalent:

(I)

R is a Gorenstein ring.

(2)

Q and K are injective R-modules.

(3)

K i_~san injective R-module.

(4)

K ~ E(R/M), the injective envelope of R/M.

(5)

M-I/R % R/M.

(6)

M -I can be generated by two elements.

(7)

I = I -I-I for every regular ideal of R.

(8) (9)

~(I/J)

= ~(J-i/l-l)

for all regular ideals J c I of R.

R has a proper, irreducible,

Proof.

regular, principal ideal.

If pl,...,pt are the prime ideals of rank 0 in R, then

Q = Rpl @ "'" @ Rpt" and hence Q is a direct sum of Artinian local rings.

Let E = E(R/M) and Epi

of R/M and R/Pi, indecomposable

respectively.

E(R/Pi ) be the injective envelopes By Theorem 4.5 these are the only

injective R-modules;

posable injective Q-modules; a direct sum of indecomposable inj. dimRQ ~ inj. dimQQ.

the ~ i ' s

are the only indecom-

and every injective R (or Q) module is injeetives.

It follows that

On the other hand we have Q ®R E = 0 and

Ep Thus if C is an injective R-module, then Q ~R C is Q ~R ~ i i" an injective Q-module. Hence if we apply the exact functor Q @R

117

to an injective

resolution for either R or Q over R we obtain an in-

jective resolution for Q as a Q-module.

Thus we have proved the

following: inj dimRQ : inj.dimQQ ~ inj.dimRR. (I) ~---> (2).

By the preceding

remarks we have inj.dimQQ ~ i.

It is sufficient to prove that Q is an injeetive Q-module.

For then

by the preceding equation Q would be an injective R-module.

Since Q

is a direct sum of Artinian local rings, it is sufficient to prove that if S is an Artinian local ring such that inj.dim.sS ~ I, then S is an injeetive S-module. Let N be the maximal ideal of S.

Then there is a nonzero ele-

ment x in S such that Nx = O, and hence SX = S/N. % Ext~(Sx,S) Ext~(A,S)

~ Ext~(S/Sx,S)

= O.

We have Ext~(S/N,S)

By induction on length we see that

= 0 for every S-module A of finite length.

ideal of S has finite length,

Because every

we have shown that S is an injective

S-module. (2) --> (3).

This is a trivial assertion.

(3) ---> (4).

It is easy to see that M-I/R ~ 0 and that K is an

essential extension of M-I/R.

Since M-I/R is isomorphic

to a finite

direct sum of copies of R/M, it follows that K is isomorphic finite direct sum of copies of E(R/M).

But K is an indecomposable

R-module by Theorem 4.7, and hence K is isomorphic (4) ~ > R-module.

(I).

to a

to E(R/M).

It is sufficient to prove that Q is an injective

From the given exact sequence O - ~ R - ~ Q - ~

E-+O,

we ob-

tain the exact sequence: (*)

0 -~ HomR(E,E ) ~ HomR(Q,E ) ~ E - ~ O.

Now HomR(E,E)

~ H, the completion of R, by Theorem 4.5, and hence

HomR(E,E ) is a flat R-module.

Using the abbreviation w.dim,

for weak

dimension, we observe that w.dimRE = 1 since E % K, and Q is a flat R-module.

Thus from exact sequence

(*) we see that

118 w. dimRHomR(Q, E )

~ 1.

I f I i s an i d e a l

o f R, t h e n by [3,

Ch. VI, P r o p .

5.3],

we have:

HomR(EXt~(R/I,Q) ,E) -= TOrnR(R/I,HOmR(Q, E)) for all n > O.

Hence we have ExtR(R/I,Q)

inj.dimRQ 2.

Thus

As in the proof of (1) ==> (2), we see that Q is

an injective R-module.

Therefore,

inJ.dimRR = l, and R is a Goren-

stein ring. (4) ~ > [x ~ E

(5).

(6).

(6) ~---> (5). M - I / R ~ R/M.

-

S(R/M)

I Mx = O) and R/M

I Mx = 0], we see that M-I/R % R/M.

(5) ~ >

~(M-I/M)

Since M-1/R = {x ~ K

This is an obvious assertion. If M-IM = R, then M -I is cyclic and hence

Hence we can assume that M-IM = M.

= S ( M - ! / M M -I) = 2.

But then

Therefore ~ ( M - ! / R )

= ~(M-I~)

= I, and we see that M-I/R ~ R/M.

(5) ~-~-> (7).

Let I be a regular ideal of R and let a be a

regular element of I.

Then m u l t i p l i c a t i o n

the other hand multiplication Ext~(R/I,Q)

= 0 for all n _~ O.

by a annihilates R/I.

by a is an isomorphism on Q.

On

Hence

As a consequence we have

ExtlR(R/I,R) ~ HomR(R/I,K ) ~ AnnKl = I-I/R.

Another consequence is

that every R-homomorphism from I into Q can be extended to an R-homomorphism of R into Q.

This latter statement implies directly that

there is a canonical isomorphism of I -I and HomR(i,R). Since Ra is an M-primary ideal we have a composition series of ideals:

Ra = I 0 c i I c ... c I n = ! such that lj+i/l j ~ R/M. I 0 = I0 I-I.

Because I 0 is a principal ideal, we have

We will assume that lj = I] I-I and prove that

-!-1 l j+ I = l j÷ I o

By induction this will establish

(7).

119

We have an exact sequence: 0 -~ H o m R ( I j + I , R ) -~ HomR(Ij,R ) -+ ExtI(R/M,R) . By our preceding

remarks we have ExtI(R/M,R)

tion M-I/R "- R/M.

--- M-l/R,

and by assump-

Thus the exact sequence can be written in the form: 0

~

I-lj+l~ I]1 ! R/M

where ~ is the inclusion map. Ij+ 1 c i-l-I j+l = I] !-I = Ij. 8 ~ O, and so 8 is onto.

If 8 = O, then I -I j+l = I ~I, and hence

But this is a contradiction,

and thus

We can now produce from this sequence an-

other exact sequence: 0 ~ I~ I-I X~ l-l'lj+l6_+ R/M

where ¥ is the inclusion map. a contradiction.

If 8 = O, then Ij

= I -! j+l and we have

Hence 6 is onto.

that I]~?I/Ij~__- % R/M.

Since I[ I-I = I we have proved 0 J' Now Ij+i/I j is a nonzero submodule of

-I-I : i-l-I Ij+ I /ij, and hence lj+ I j+l " (7) ~---> (8).

Let J c I be regular ideals of R.

m a r y ideals of R and hence we can find a composition

They are M-priseries of

ideals of R: J = i 0 c I ! c ... c I n = I such that lj+I/I j --- R/M.

i-I : and I j+-i 1 ~

!]i.

Then we have a series:

Inl m Inll c ... c Iol = j-l,

Thus ~ ( I / J )

e

-

I < v.

(I).

Since AM v : ~ ,

we have M v c A -I : M e'l , and thus

On the other hand AM e-I : AA "I = A -I = M e-I

v < e - I by the minimality Thus by Theorem 13.4, Lemma 13.~.

Proof.

Therefore v = e - I, and A -I = M v.

0 : ~I ve : ~I (e-l)e.

Assume that the maximal

er__ate_~d by two elements. < ~ ( M m / M m+l)

of v.

and thus

ideal M of R can be gen-

I f ~ ( M q / M q÷i) < q + I, then ~ ( M n / M n+l)

for every n and m such that q < m < n. Because of an obvious

recursive argument

cient to prove that Z ( M q + I / M q+2) < ~ ( M q / M q + i ) .

it is suffi-

Let a and b be ele-

ments of R that generate M, and define F i to be the subspace of Mq/M q+l generated by the images of b q, abq'l,a2bq-2,...,aib q-i in Mq/M q+l.

Similarly

define Gj to be the subspace of Mq+i/M q+2 gen-

erated by the images of b q+l, ab q, a2bq-l,...,aJb q+l-j in Mq+I/M q+2. We will prove first that if F k = Fk_ I (for k > 0), then Gk÷ I = Gk_ I. (!)

For if F k = Fk_ I, then we have:

akb q-k E (bq,abq-l,...,ak-lbq-k+l)

Multiplying

+ M q+l.

(I) by a and then by b we obtain:

(2)

ak+Ib q-k E (abq,a2bq-l,...,akb q'k+l) + M q+2

(3)

akb q - k ÷ l E ( b q + l ~ a b q , . . . , a k - l b q ' k + 2 )

and

Substituting

+ Mq÷2.

(3) in (2) we see that both ak+Ib q-k and akb q-k+l are

in (bq+l,ab q,...,ak-lb q-k+2) + M q+2.

Therefore,

we have shown that

Gk+ 1 = Gk_ 1. By hypothesis ~(F~)

there is an integer p such that j~(Fp)

= s + 1 if 0 O, then Fp = Fp_ I, and hence by the preceding para-

graph Gp+ I = % _ I. fore,

Therefore

Thus

S(Gp+I)

< ~(Fp)

in this case also.

there exists a largest integer t such that / ( G t + l )

There-

~f(Ft).

We will suppose that t < q and arrive at a contradiction. Case I:

S(Ft+I)

: ~(Ft)

We then have ~ ( G t + 2 ) This contradicts Case II.

~ ~(Gt+l)

the maximality /(Ft+l)

+ I. + I < ~f(F t) + i = S ( F t + I ) .

of t and hence this case cannot arise.

= ~(Ft).

In this case we have Ft+ 1 = Ft, and hence by the preceding paragraph Gt+ 2 = G t. : ~(Ft+l).

Thus 2~(Gt+2)

= ~f(Gt) ! ~ ( O t + l )

This also contradicts

the maximality

are the only two cases that can arise,

~ S(Ft) of t.

Since these

we have t = q and the lemma is

proved. Theorem 13.7.

(Step Theorem).

of R can be generated

by two elements.

L(Mn/~

+l)

=

In + i e

Proof.

Assume that the maximal Then if n < e - i~

--

_

if n >

e

Since there are only n + 1 monomials

two generators

of degree n in the

of M, we have / ( M n / M n+l) _< n + 1 for all n _> 0.

~ ( M n / M n+l) is bounded we conclude integer k > 0 such that

from Lemma 13.6 that there is an

function for n > k.

we have v = k and hence e = Z ( M V / M v+l) = k + I. and the theorem follows immediately Corollary 13.8.

~ = e - I.

(2)

p = ~1 (e-l) e

Then:

By Theorem 12.10, Thus v = e - 1

from Theorem 12.10.

Assume that the maximal

by two elements.

(I)

Since

~ ( M n / M n+l) = n + 1 for n _< k and

~ ( M n / M n+l) is a nonincreasing

generated

ideal M

ideal M of R can be

124

(3)

i -! = M e-I and A : (Me-l) -I.

Proof.

It is an i m m e d i a t e

that v = e - I. = ~(R/M)

= ~I

Using Theorem

+ ~ ( M / ~ 2) +

v(v+l) .

consequence

of T h e o r e m 1 2 . 1 0 and 13.7

13.7 we h a v e ev - p =

... + 2 ~ ( M v - I / M v) = I + 2 +

Therefore,

0 = ev - F vI

~ ( R / M v)

... +

(v + i) = e(e-l)

-y I

(e-l)e

= ~ (e-l)e. 2 S i n c e the m a x i m a l R is a G o r e n s t e i n Theorem

r i n g by T h e o r e m 13.2.

= ~ and A = A - I - ! = i~.~

Remarks.

Theorem

proved by Northcott i~ge

i d e a l of R can be g e n e r a t e d

A -I = M e-I by

(2)

and

(3)

of C o r o l l a r y

13.8 w e r e

in the s p e c i a l case w h e r e R is a h o m o m o r ~ i c

One c o n s e q u e n c e

is that the p r o o f s

Therefore,

elements,

(Me-l)-i

13.7 and

of a 2 - d i m e n s i o n a l

(2.3)].

by two

regular local

ring

[23, Th.

of t h e g r e ~ t e r g e n e r a l i t y

are c o r r e s p o n d i n g l y

I and Th.

2 and

of our t h e o r e m s

s i m p l e r and m o r e

elementary

in nature. The n e x t two p r o p o s i t i o n s the m u l t i p l i c i t y ment

is s u p e r f i c i a l

of d e g r e e I.

Theorem

Assume

13.9.

s u c h t h a t M n c Ra. f i c i a l of d e g r e e Proof. M n c Ra, ~

M

-

If a ¢ M - ~ ,

or not

a g i v e n ele-

i d e a l M of R can be g e n -

let n be the s m a l l e s t

integer

I.

We can a s s u m e

that

there is an i n t e g e r n s u c h that

t h e r e is n o t h i n g

to prove.

w e can c h o o s e b ~ M - M 2 s u c h t h a t M =

the l a r g e s t (i)

that the m a x i m a l

for d e t e r ~ i n i n g

T h e n n ~ e and n = e if and o n l y if a is s u p e r -

for otherwise ~,

criteria

of R and for d e c i d i n g w h e t h e r

e r a t e d by two e l e m e n t s .

a

give u s e f u l

integer

s u c h t h a t b n ~ Mqa,

Because (a,b).

If q is

then:

E i t h e r q > n - i or n > e.

Suppose

that

q + i < n < e.

(i) is false.

T h e n q < n - i and n < ~, and thus

N o w b n is a l i n e a r c o m b i n a t i o n

q + i b e c a u s e b n is in Mqa.

Since bn

~ Mq+la,

of m o n o m i a i s

of d e g r e e

the c o e f f i c i e n t

of

125

one of t h e s e m o n o m i a l s

is a u n i t

t i o n for this m o n o m i a l

and

other monomials that

~(MV/M~a)

~(MV/M

v+l)

By C o r o l l a r y 13.8,

(i) that q > n - I.

hence M n = Mn'la.

We c o n c l u d e

shows

q + I > e _ i.

(I).

of d e g r e e

see t h a t M V a = M ~+I.

h a v e n < ~ + !. follows from

Since

of the

S i n c e n > q + i, this

~ 7 H e n c e b y T h e o r e m 11.,,

a is s u p e r f i c i a l = e.

we

it as a l i n e a r c o m b i n a t i o n

q + 1 < e and e s t a b l i s h e s

that

M V a c M v+l,

H e n c e w e can solve the e q u a -

of d e g r e e q + I and b n.

This contradicts

have

express

S ( M q + I / M q+2 ) < q + 2.

Assume

in R.

I.

By T h e o r e m 1 2 . 5 we

= e b y T h e o r e m 1 2 . 1 0 and

H e n c e by t h e m l n i m a l i t y

of n we

v + 1 : e and t h u s n < e.

It

T h e r e f o r e b n ~ M q a c M n - l a and

from Theorem

1 2 . 5 that

and h e n c e b y T h e o r e m 1 3 . 7 w e h a v e n - 1 > e - !;

~ ( M n - I / M n) = e

that is n > e.

Thus

we h a v e n = e in t h i s case. N o w a s s u m e that a is not t h e n q > n - I by M n = Mn-la. sumption.

(I).

superficial

of d e g r e e

H e n c e b n ¢ M q a c Mn-la,

But t h e n a is

superficial

I.

If n < e,

s h o w i n g that

of d e g r e e 1 c o n t r a r y to as-

T h u s n > e in this case.

Theorem e r a t e ~ by two

13.10.

Assume

elements,

that the m a x i m a l

and let a ¢ M.

g r e e I if and o n l y if M e - l a = M e .

i d e a l M of R can be g e n -

T h e n a is s u p e r f i c i a l

Furtheremore,

of de-

e is the s m a l l e s t

i n t e g e r w i t h this p r o p e r t y . Proof.

Of c o u r s e if M e - l a = M e , t h e n a is s u p e r f i c i a l

I by d e f i n i t i o n . I.

Conversely,

assume

If M k - l a : M k, t h e n ~ ( M k - I / M k)

that a is s u p e r f i c i a l

of d e g r e e

= e by Theorem

Thus

k - I > v by T h e o r e m 12.10.

On the o t h e r hand

Theorem 12.10

: e by T h e o r e m

we

and

~(MV/MVa)

see that M V a = M v+i.

M e - l a : M e.

of d e g r e e

12.9.

~ ( M V / M v+l)

]2.5.

But v = e - i b y C o r o l l a r y

: e by

S i n c e M V a c M v+l, 1 3 . 8 and thus

CHAPTER XIV MULTI PLI CI TIES Throughout Macaulay

this chapter

ring.

As in earlier

and H is its completion. denotes sical

its

"divisible"

sense,

R will be a 1-dimensional,

then L(A)

chapters

M is the maximal

If A is an Artinian length.

= O, and we denote

We shall use all of the notation

further

explanation.

P-primary

ideals

Let P be a prim______eeeideal of H such that there

pr___operly bet wee_nn I and J.

ideal

Proof.

local

ideals properly

Thus HpJ/HpI

Now multiplications

between

ideal PHp, ~ Hp/PHp,

by elements

J/I is a torsion-free generated ideal

But this follows

H-module,

them.

ideals

the P-primary

and there are

Since Hp is a Noetherian that HpJ/HpI field

is a simple

of H/P.

J/I c HpJ/HpI.

H/P-module we conclude

This demon-

of rank 1.

Since J/I

that J/I is isomorphic

of H/P.

it is merely necessary immediately

N O J with a normal Definition.

of rank 0 i__n_nH

(N A J)/(N A I) is isomorphic

(N A J)/(N N I) % ((N N J) + I)/I c J/I,

the proof

ideal

of H - P on J/Z are monomor-

that

clude

ideal

the quotient

strates

Now

c of rank 0 i__q H and I ~ J

this means

and thus we have an imbedding

to a nonzero

without

does not exist a P-primary

Then

phisms,

is a finitely

length by

of H/P.

ring with maximal

Hp-module.

then L(A)

of earlier chapters

Clearly HpI c HpJ are PHp-primary

no PHp-primary

ideal of R

length in the clas-

its classical

Let N be an unmixed

such that P does not belong to N. to a nonzero

R-module,

If A has finite

J~(A).

Lemma 14.1.

local Cohen-

to show that N O J ~ N D I.

from comparing

decomposition

and thus to con-

a normal

decomposition

of

of N A I.

Let P be a prime ideal of rank 0 in H and let N be

component

of 0 in H.

Then it is well known

[18, Ch.

127

III,

Th. 5] that there

exists

a so-called

that is, a chain of P-primary

"composition

series"

for N;

ideals:

C

C

C

N = PI ~ P2 ~ "'" ,J Pk = m such that there are no P-primary for all i. the number Ch. m(P)

III,

ideals properly

Then the latent multiplicity of terms in the chain.

It is also well known

is independent

of the chain chosen.

Theorem

Le_~t PI,...,Pn

0 in H.

of P is defined

Th. 5]) that by a Jordan-Holder

1%.2.

between

Pi and Pi+l by L~(P) : k, (see

type of argument

[18,

the integer

be the set of prime ideals

of rank

Then:

L(K)

: E ~,(Pi ), i

and ~(Pi ) is the number

of times

simple divisible

module

corresponding

in a composition

series

of divisible

Proof.

Let 0 : N I N

(i.e.,

t h_ee multiplicity)

to Pi occurs submodules

... N N n be a normal

H, where N i is Pi-primary.

By patching

for each N i we obtain a descending

as a factor module

of K. decomposition

together

chain

that a

"composition

of unmixed

ideals

of 0 in series"

of rank 0

in H given by: ~

(l)

D

~

D

D

D

H ¢ PI ¢ PI2 ~ "'" ¢ N1 J N1 n P2 ¢ "'" ~ N1 n N 2 D

j... j ~ N i

If we tensor

this chain with K, we obtain by Theorem 6.6 a descending

chain of divisible

(2)

~ ~K

: O-

submodules

~R PI ¢ "'" ¢ ~ %

of K:

N1 ¢ "'" ¢ ~ %

(N1 n N2)

D

¢... ¢ ~ % O N i : o . i

Letting

N N Pi,t ~ N n Pi,t+l be a typical

step in chain

(I),

128

we observe

that

(N n Pi,t)/(N N Pi,t+l ) is isomorphic

ideal of H/P i by Lemma 14.1.

Thus

(K ®R

(N N Pi,t))/(K

Z K ~R (N A Pi,t)/(N A Pi,t+l)

is a simple

ponding

Thus chain

to Pi by Theorem 8.5.

of divisible

submodules

Corollary

1%.3.

of K.

Clearly

to a nonzero

divisible

~R (NDPI,t+I))

R-module

corres-

(2) is a composition

this proves

I f S is the integral

series

the theorem.

closure

of R

I n Q, then

n

L(S/R) Proof. n -

n

:

Z

:

z i:l

(.(Pi)

By Theorem I0.i and Theorem n

~(Pi)

-

n

=

Z

i:l

(~(Pi)

-

Corollary

14.3

~(Pi ) = i for all i = l,...,n. Definition.

unrmnified

its m u l t i p l i c i t y

regular

I is a finitely

ideal of B.

Cohen-Macaulay

by e T and its reduction

length

num-

that there is an integer

generated

torsion R-module

ideal

of B, then

and ideal of R.

n > 0 such that

ideal of B. length

torsion B-module

of R in Q.

of B and BM con-

(hence M-primary)

over B (and the same finite

a finitely

ideal

ideal of B, I = B(I N R),

I A R is a regular

readily

maximal

extension

If I is a regular

(BM) n c I, and thus I is a BM-pr[mary

erated

10.2).

local,

unramified

BM is a regular,

generated

B/I Z R/(I A R).

eral,

of the theorem

if and only if

(See also Theorem

Let B be a strongly

Then by Theorem 3.6,

finite

proof

If T = R, we shall drop the subscripts.

Remarks.

It follows

an immediate

If T is a 1-dimensional,

ring, we shall denote

every

: L(K)

i).

gives

[19, Th. 8] that R is analytically

tains

1%.2 we have L(S/R)

i=l

Remarks.

ber by PT"

- 1).

Therefore over R).

B/i has

In gen-

is also a finitely

and has the same finite

gen-

length over B as

over R. We shall recapitulate other theorems.

Let ~

some of the results

be the divisible

of Theorem 2.11 and

submodule

of B, and let

129

= B/~ , Q : Q/~,

and R : R / ( ~

O R).

Now

~

is an ideal o f Q, and

is the full ring of quotients of both ~ and B.

B is a strongly

unramified ring extension of R in Q, and B is a reduced ~-module. is a quasi-local

ring by Theorem 3.6.

Clearly we have Q / B ~ Q/B.

Let H be the completion of B in the B-topology. = HomB(Q/B,Q/B)

~ Hom~(Q/B,Q/~),

of B in the B-topology. J = HomR(K,B/R), Macaulay ring. B is a local,

Then

and hence H is also the completion

By Theorem 6.6, H ~ H/J, where

and thus H is a complete,

local,

l-dimensional

Cohen-

If B is a closed component of R, then by Theorem 7.2, 1-dimensional,

analytically

irreducible,

Cohen-Macaulay

ring. ~is

contained in every regular ideal of B, and in f a c t ~ I s

intersection of the regular ideals of B.

the

If I is a regular ideal of

B, and [ = I/ , then T is a regular ideal of B and every regular ideal of B arises in this way. completion of T over ~.

The completion of I over B is equal to the Thus by Theorem 2.8, the completion of I is

equal to H ® ~ ~ = HI, and hence is a regular ideal of H. Let M = M/(R N ~ ) ;

then M is the maximal ideal of R, and

BM = BM is the maximal ideal of the quasi-local

ring B.

The maximal

ideal of H is HM, and since HI is a regular ideal of H, it follows that HI is an HM-primary ideal of H.

In particular,

we have B/(BM) n ~ ~ / ( ~ ) n

Thus for large n we have

~ ( B / ( B M ) n) = e~n - p~.

~ ~/(~)n.

for any n > 0

We therefore define the multiplicity and re-

duction number of B (even though B is not a Noetherian local ring) by:

e B = e~ and PB = PH" We similarly define e~ = e~ and p~ = 0F. unramified

Now H ®R B is a strongly

ring extension of H in H @R Q' and (H @R B)/H % B/R.

J = HomR(K,B/R)

~ Hor~H(K,(H ® B)/H),

H ~R B is equal to H/J = H.

Thus

and hence the completion of

Therefore,

we have e H @R B = e~ = e B and

130

PH ~R B = 0~ = PB" We shall

say that

B if y ¢ (BM) s and then d e f i n e elements ficial

an e l e m e n t

(BM)ny =

y ¢ B is s u p e r f i c i a l

(BM) n+s for all large

the first n e i g h b o r h o o d

of B of d e g r e e

W i t h the r e m a r k s show that T h e o r e m by a s t r o n g l y

remains

unramified

12.1 is m o d i f i e d

of d e g r e e

see that

a is a s u p e r f i c i a l

BA c F, the f i r s t

Because Theorem

F(BM) ns = (BM) ns.

ring e x t e n s i o n

ring of R.

of

Then

element

element

a in R

of B of d e g r e e

neighborhood

s.

We

of BA is of the form x/a,

in B of d e g r e e

to B and F, t h e r e

element

(BM)na = (BM) n+s for all

s.

Therefore

by

ring of B.

(BA)a : (BM) s, we have Fa : F ( B A ) a

12.1 a p p l i e d

= (BA)a n.

unramified

ring of B.

Since

w h e r e x ~ (BM) s and a is s u p e r f i c i a l definition,

ring).

12.1 there is a s u p e r f i c i a l

have BA = (BM) Sa -I and thus e v e r y

of T h e o r e m

to its first n e i g h b o r h o o d

neighborhood

s such that A = MSa -I.

large n we

B of R in Q ((4)

valuation

Let B be a s t r o n g l y

By T h e o r e m

it is not hard to

true if we r e p l a c e R in that t h e o r e m

BA = B + A is the first n e i g h b o r h o o d Proof.

we h a v e m a d e

ring extension

R in Q, and let A be the first

We can

w ¢ (BE) s and y is a super-

to read that B is equal

14.4.

of n.

s (s a r b i t r a r y ) .

ring if and o n l y if B is a p s e u d o

Theorem

where

and d e f i n i t i o n s

12.1

values

s in

ring F of B to be the set of all

of Q of the form y = w/y,

element

of d e g r e e

= F(BM) s.

is an i n t e g e r

By

n > 0 such that

Thus we have Fa n = (F(BM)S) n = F(BM) ns = (BM) ns

Cancelling

a n we o b t a i n F = BA.

By T h e o r e m 6.1 we have

BA = B + A.

Remarks. Theorem

12.2

R in Q. cause

It is an i m m e d i a t e is true

It is also

Theorems

and 12.3,

12.4,

consequence

for any s t r o n g l y

unramified

easy to see that T h e o r e m 12.5 and 12.6

they are also true for B.

of T h e o r e m

depend

12.3

14.4 that

ring e x t e n s i o n

B of

is true for B.

o n l y on T h e o r e m s

Thus in p r o v i n g

12.1,

Be12.2

the f o l l o w i n g

131

theorems we can freely apply Theorems unramified

12.i through 12.6 to strongly

ring extensions of R in Q.

Theorem 14.5.

Let A be a closed component of R and let B/R be a

simple divisible R-submodule

of K corresponding to A.

If a is a reg-

ular element of R, then: ~(R/Ra) Proof.

= ~ I.

nonzero,

D B)) < ~ ( A / R ) .

by Theorem Thus we need

Hence it is sufficient

to

prove that (A A B)/R ~ 0. Since R is not a discrete valuation Theorem 12.13.

Now AnnB/RM c M-I/R;

an Artinian R-module.

ring, we have M -I c A b y

and AnnB/RM ~ O, since B/R is

Thus AnnB/RM c (M -I A B)/R c (A N B)/R, and

hence (A N B)/R ~ O. Theorem 14.15:

Assume that the maximal ideal of R can be gen-

crated by two elements.

Let A be a closed component

of R and B/R a

137

simple divisible

R-submodule

of K corresponding

t_~oA.

Then:

P = OB + PA + eBeA"

Proof.

If B = Q, then 0B = 0 = e B and A = R.

trivial in this case.

Thus we may assume that B ~ Q.

Since the completions homomorphic

elements.

H(B) and H(A) of B and A, respectively,

images of H by Theorem 6.6,

Cohen-Macaulay

they are local,

Theorem 14.4,

and reduction numbers of B and A are unchanged Thus b y C o r o l l a r y

1

by two

the multiplicities

upon passage to H(B)

1 3 . 8 we h a v e p = !

e(e-1) ,

1

PB = ~ eB(eB-l)'

and PA = ~ eA(eA-l)" By Corollary 14.6 we have 1 Therefore we have O = ~ (e B + eA)(e B + e A - l)

e = e B + e A. = 21 eB(eB-l)

are

l-dimensional,

rings whose maximal ideals can be generated

By the Remarks preceding

and H ( A ) .

The equation is

+ ~1 eA( eA_l ) + eBeA = PB + OA + eBeA-

Corollary

14.16.

Let AI,...,A n be the closed components

and ul,...,u n their associated

latent multiplicities.

the maximal ideal of R can be generated

of R

Assume that

by two elements.

Then:

I P = 7. i UiPAi + ~- iZ ui(ui-l)eAl. + i~jT. UiUjeAi eAj" Proof.

As we observed in the proof of Theorem 14.15 we have the

following equations by Corollary 13.8: i e(e-l) p = ~

and

PA i

= ~1 eAi(e

_i ) for i = I, ..,n.

1 2 1 1 Z Thus by Theorem 14.9 we have p - ~ e = - ~ e = - ~ i uieAi 1 2 = .Z Ui(PAi - ~ eAi ). Therefore, l i e2 i 2 p = ~ + 7, Ui - ~- 7 uie A i PAi i i = I(Z u i e ~ ) 2 i

1 Z 2 - ~ i uieAi + 7. I UiPAi

= 7. 7. UiUjeAieAj i UiPAi + i~j

+ ~I 7. 2 2 i (ui-ui) eAi"

138

Definition. analytically Theorem

ramified

10.2,

lytically finitely

We shall say that a Noetherian

a Noetherian

ramified

module.

ramified

Nagata

Noetherian

theorem proves

ramified

local

Theorem

14.16.

domains

ramified

[17] has given

local Gorenstein

several

ring T of Krull

1 is ana-

examples

dimension

of 1.

of analytically

of Krull dimension

1.

ramified

l, then there

By

is not a

of Krull

the existence

If R is an analytically

local domain of Krull dimension

dimension

closure

local domains

Hence the following Gorenstein

integral

is

unramified.

local domain of Krull

if and only if'its

generated

analytically

if it is not analytically

local domain

exists

dimension

Noetherian an analytically

1 such that

R c T c Q. Proof.

Let 0 = N 1 A

... O N n be a normal

H where N i is a Pi-primary ideals

of rank 0 in H.

ponding

i ~ J.

Let V i be the valuation

unramified

extension

Hence by Corollary

closure 14.3,

It is clear from the proof Thus L(Vi/Bi) analytically assume

= L(Q/Bi) ramified.

that R = B i.

ring V and L(V/R) Now there L(Q/B)

= 2.

ring of R. closed

are the prime

By Theorem

ll.8,

B i is a

and B i ~ Vj for

of B i.

of R, then L(S/R)

> 0 by Theorem

there is an index i such that u(Pi)>l.

of Theorem

14.2 that L(Q/Bi)

_ L(Q/Vi ) = u(Pi ) _ i > O. Therefore,

without

Hence the integral

= u(Pi).

Hence B i is

loss of generality,

closure

we may

of R is a valuation

> i.

exists

a divisible

By Theorem 6.1,

A.

submodule

B is a strongly

Since the integral

component

closure

of 0 in

ring of R corres-

of R in Q, B i c Vi,

Thus V i is the integral

If S is the integral 5.1.

ideal of H and P1,...,Pn

to Pi' and let K @R Ni = Bi/R c K.

strongly

decomposition

closure

By Theorems

B/R of K such that unramified

extension

of B is V, B has only one

7.5 and 5.1, V is a finitely

gen-

139 erated A-module. analytically L(Q/T)

If we adjoin these generators

to B we obtain an

ramified local domain T (of Krull dimension I) such that

= 2 and V is the only closed component of T.

The completion of T has only one rank 0 prime ideal P by Theorem 7.3.

If u(P) is the latent multiplicity

we have a(P) = L(Q/T) = 2. that e T = u(P) = 2.

of P, then by Theorem 14.2

Since eV = I, we see by Theorem 14.9

Hence by Theorem 12.17, every ideal of T can be

generated by 2 elements.

Therefore,

by Theorem 13.2, T is a Goren-

stein ring. Definition:

If S is an Artinian local ring, then we shall de-

note the length of S as an S-module by k(S). local, Cohen-Macaulay

If R is a 1-dimensional,

ring and p is a prime ideal of rank 0 in R, we

shall let ~(p) denote the length of a composition series of p-primary ideals between p and the p-primary component of 0 in R.

It is then

easy to see that U(P) = k ( ~ ) . Lemma 14.17: and I = Hp.

Let P be a prime ideal of rank 0 i n_nH , p = P A R,

Then H/I i__ssth__eecompletion of R/p, and u(P)

= U (P)U ( P / I ) • Proof.

It is easily seen that H/I is the completion of R/p.

Since (H/I)p/i ~ H J I p , = k(~)k(Hp/Ip).

what we have to prove is that k(Hp)

Let us take a composition

series for Rp:

= Jo o J1 ~ "'' o Jn = 0 where the Ji's are ideals of Rp and Ji/Ji+l ~ k, the quotient field of R/p.

Then we have a chain

Hp

®R P~ = Hp ®R J0 D Hp ®R J1 D

Since H p ® R Let R = R/p;

~

--- ~ ' then

...

D Hp ®R J n = 0 .

it is sufficient to prove that ~

®R k --- ~ / I p .

140 H ®R k ---- H ®R(R @~ k) --- (H ®R ~) ®R k -~ (H/i) @~ k.

Hence H ®R k is the full ring of quotients of H/I by Theorem 3.6. But if P = P1,...,Pm are the rank 0 prime ideals of H that contain I, m then Z @ HPi/IPi is also the full ring of quotients of H/I, and j=l m hence Hp ®R k ~ Hp @H (H ®R k) ~ Hp @H (i=IZ @ HPi/IPi) ~ Hp/Ip. Theorem 14.18: R and let Rj = R/pj. Proof.

Let Pl'''''~t be the prime ideals of rank 0 of Then e =

Z ~(pj)e~j. j=l Let Pjl,...,Pjk be the rank 0 prime ideals of H having

the property that PJi" N R = pj;

and let Ajl,...,Ajk be the closed

• 's. components of R that correspond to the PJi i = 1,...,k and we let Aji = Aji/Qpj.

Now Qpj c Aji for all

Then the Aji's are the closed

components of Rj, and hence by Theorem 14.9 k e~.j = i=iE u(Pjl/Hpj)e~j i . . and Theorem 14.9 that e = t : j=lZ U(pj)e~j.

Since e~i = eAi, it follows from Lemma 14.~ Z u(Pji)eA. = Z ~(pj)u(Pji/Hpj)e~j i i,j Ji i,j

CHAPTER X V THE C A N O N I C A L IDEAL OF R

Throughout Macaulay

ring.

this chapter R will be a local, We shall explore further

the relationship

for the existence

to e s t a b l i s h n e c e s s a r y

of a canonical

T h e o r e m 15.1.

Cohen-

than we have already done

between the modules K and E.

these relationships

1-dimenslonal,

We shall then use

and sufficient

conditions

ideal of R.

There is a one-to-one,

pondence between the set of n o n z e r o

order-reversing

divisible

submodules

corres-

D of E and

the set of unmixed ideals I of rank 0 in H given by:

AnnHD = I and AnnEl = D. Proof. I = AnnHE.

Let D be a nonzero

I is an unmixed Conversely,

R-module.

let I be an unmixed

Then D ~ HomH(H/I,E),

elements

Therefore,

This

by L e m m a 6.5,

ideal of rank 0 in H. ideal of rank 0 in H and let

and hence D is the injective

ope over H/I of the residue field of H/I. regular

of E and let

Then 0 = hbD = hD, and hence h ( I.

that H/I is a t o r s i o n - f r e e

D = AnnEl.

submodule

Suppose that h is an element of H and b a regular element

of R such that bh ( I. proves

divisible

of H/I.

Thus D is divisible by the

If b is a regular element of R, then b is a

regular element of H, and we let S = b + I, an element of H/I. Hb is an H M - p r i m a r y H, it follows

envel-

Since

ideal of H and I is an unmixed ideal of rank 0 in

that b is a regular element of H/I.

ible by b, and hence by b.

Therefore,

Thus D is divis-

D is a nonzero,

divisible R-

module. The c o r r e s p o n d e n c e we have established reversing c o r r e s p o n d e n c e H and the submodules C o r o l l a r y 15.2.

is a one-to-one,

order-

because of the duality between the ideals of

of E that is given in Theorem 4.5. There is a one-to-one,

order-reversing

cortes-

142

pondence

between

divisible

the set of divisible

submodules

of E.

submodules

Thus L(K)

of K and the set of

= L(E).

I

Proof.

By Theorem 6.6 there

correspondence

between

the set of proper

and the set of unmixed w i t h Theorem

15.3.

between

and B correspond, where

unramified

I = AnnHD.

extensions

15.1,

divisible

in Q, and H/I is the completion

we have

bedding

D m EB(B/MB ).

D c t(EB(B/MB)). 2.7,

Because

However,

H/I-module.

it follows

of E, and let

ideal of rank 0 in H. ring extension

of R

The residue

and in the proof of Theorem

15.1

envelope

Thus

D is a torsion

summand

between

over H/I of B/MB.

B-module,

is an H/I-module

of t(EB(B/MB))

we have by Theorem

since D is an extension

of

D and B that we have found is clearly

correspondence

elements,

other

Q is a semi-simple

full ring of quotients t(EB(B/MB))

submodule

unramified

t(EB(B/MB))

order-reversing

4.4,

of

that D = t(E B (B/MB)).

If R has no nilpotent in Corollary

submodule

over B o_~f B/MB.

Since EB(B/MB ) is an essential

This correspondence a one-to-one

D of E and

over B of B/MB and hence we have an im-

and thus D is a direct

injective B/MB,

to B/MB;

extension

envelope

of B by Theorem 6.6.

seen that D is the injective

D is an essential

submodules

corres-

B of R in Q such that if D

I is an unmixed

If B/R = K @R I, then B is a strongly

field of H/I is isomorphic

of K

other than O, then D ~ EB(B/MB ) .

Let D be a nonzero By Theorem

order-reversing

to the torsion

is the injective

elements

submodules

This fact combined

divisible

then D is isomorphic

~(B/MB)

order-preserving

proof of the corollary.

There is a one-to-one,

I~f R has no nilpotent Proof.

divisible

of rank 0 in H.

the set of nonzero,

the set of strongly

EB(B/MB),

ideals

15.1 gives an immediate

Corollary pondence

is a one-to-one,

is a direct

6.6 and 15.1.

than 0, then as we have seen ring.

of B, it follows summand

by Theorems

Because

from Theorem

of EB(B/MB ) .

Q is also the 1.4 that

Since EB(B/MB ) is an

143 essential

extension

= t(EB(B/MB)).

of B/MB this implies

Therefore,

that EB(B/MB)

by the preceding

paragraph

we have

O ~ EB(B/~2B). Theorem 15.4. divisible

composition

series

of

R-modules.

Proof.

Let P be a prime ideal

be the P-primary ll.10,

K and E h a v e e q u i v a l e n t

components

it is sufficient

latent multiplicity

of E and K, respectively.

to prove

to prove that L(Ep)

Let N be the P-primary series

that L(Ep)

of P, then by Theorem

Hence it is sufficient

composition

of rank 0 in H, and let ~

component

of P-primary

By Theorem

= L(Ep).

14.2,

and

If U is the

we have L(Kp)

= u.

= U also.

of 0 in H.

Then we have a

ideals:

P = N 1 { N2 ~ "'" ~ N u Putting D i = A n n U l , we have by Theorem submodules

15.1 a chain

of divisible

of E: C

C

C

C

0 d D 1 ~ 0 ! ~ D 2 d "'" ~

OU"

By Theorem 4.5,

AnnHD i = N i, and thus every D i is a P-primary

ible R-module.

Hence we have D u c ~ .

On the other hand, tion,

AnnHE P is a P-primary

and thus N c AnnHE P.

Ep c Du, and hence D u = Ep. Theorem

15.1 that L(D

theorem

is proved.

Definition. an R-module.

ideal of H by defini-

from Theorem

15.1 that

Since it is an immediate

) = U, we see that L ( ~ )

consequence

= U = L(Kp),

of

and the

Let I be a fixed regular ideal of R, and let A be

Then we shall denote the R-module

There is a canonical = f(x)

It follows

divis-

R-homomorphism

~ : A-*A**

for all x ~ A*.

If A is an R-submodule

of Q, let us define

HomR(A,I)

by A*.

given by ~(x)(f)

144

A I = {q ~ Q

I qA c I).

If A contains a regular element of R, we shall

show that there is a natural isomorphism of A* and A I.

For we have

an exact sequence: (#)

HomR(Q/A,Q) -~HOmR(Q,Q) ~ H O m R ( A , Q ) -*Ext~(Q/A,Q).

Since Q is a flat R-module, we have by [3, Ch. V, Prop. 4.1.3] that HomR(Q/A,Q ) ~ HomQ(Q ®R Q/A,Q) and Ext~(Q/A,Q)

~ Ext~(Q @R Q/A,Q).

But Q/A is a torsion R-module because A contains a regular element of R, and hence Q @R Q/A = 0 by Theorem i.I. HomR(Q/A,Q ) = 0 = Ext~(Q/A,Q). that HomR(Q,Q ) ~ HomR(A,Q ) .

It follows from exact sequence

Therefore,

into Q is achieved by the multiplication Q.

Thus we have (#)

every R-homomorphism from A on A by a unique element of

It follows from this that there is a natural isomorphism between

A* and A I . If J is a regular ideal of R, then clearly jI contains a regular element of R and is isomorphic to a regular ideal of R. is naturally isomorphic to ( j I ) I : J-~J**

with the containment

Thus (jI)*

and we can identify the mapping J c (jI)I.

If L is an ideal of R

such that J c L, then obviously we have L I c j I The ideal I is said to be a canonical ideal for R if ~ : J - ~ J * * is an isomorphism for every regular ideal J of R; j = (jl)l. = R.

In particular,

that is, if

we then have HOmR(l,l ) = (RI)* ~ (RI) I

It has been shown in Theorem 13.1 that R is a Gorenstein ring

if and only if R is a canonical ideal for R.

The next theorem is a

generalization of this result. Theorem 15.5.

Let I be a regular ideal of R.

lowing statements are equivalent: (1)

I is a canonical ideal for R.

(2)

M!/I ~ R/M.

(3)

Ext~(R/M,I) ~ R/M.

Then the fol-

145

(4)

Q/I ~ E

(5)

Q/I is an injective R-module.

(6)

Q and Q/I are inJective R-modules.

(7)

inj.dim.Rl = I.

Proof.

(I) ~---> (2).

Let L be an R-module such that ! c L c M i.

Since I I = (RI) I = R and (MI) ! = M, we see that M c L ! c R. L I = M, or L I = R; Therefore,

and hence L = (LI) I is equal to either M I or I.

MI/! is a simple R-module,

(2) < = >

(3).

Thus

and hence MI/I ~ R/M.

We have an exact sequence:

HomR(R/M,Q ) - + H O m R < R / M , Q / I ) - + E x t ~ ( R ~ , l )

-~Ext~(R/M,Q).

The end terms of this sequence are annihilated by M, and yet are torsion-free and divisible.

Thus the end terms of this sequence are O,

showing that we have an isomorphism:

Ext,(R/M,!)

% HomR(R/M,Q/I).

The socle of an R-module is defined to be the sum of all of its simple submodules.

It is clear that the socle of Q/I is equal to

MI/I, and is isomorphic phism:

Ext~(R/M,I)

to HomR(R/M,Q/I ) .

~ MI/I.

Thus we have an isomor-

The equivalence of (2) and (3) is now

apparent. (2) ~ > R-module.

(4).

Because I is a regular ideal,

R/I is an Artinian

Hence from the exact sequence:

O-~R/I - - Q / I - ~ K - ~ O we see that Q/! is an Artinian R-module. an essential extension of its socle. MI/I, and hence by (2) is isomorphic

Every A r t i n ~ n R-module is

Clearly the socle of Q/i is to R/M.

Every essential exten-

sion of R/M can be imbedded in E, and thus we can assume that Q/I c E. = L(K);

But from the preceding exact sequence we see that L(Q/I) and L(K) = L(E) by Corollary 15.2.

Therefore,

L(Q/I) = L(E),

and hence Q/I ~ E. (4) = >

(2).

Since the socle of Q/I is equal to MI/I,

and the

146

socle of E is equal to R/M, it follows

that

if Q/I ~ E, then

that

(RI) I = R.

MI/! ~ R/H. (4) ~---> (i). q ~ (RI)I;

then qI c I.

f : Q/I-~Q/I

and every R - s u b m o d u l e

Thus b e c a u s e

Therefore,

c R/I.

that

= r + I.

(HI) I = R.

If b is any regular element Since

= H

by Theorem

It follows

r ~ R such that q + I = f(l+I)

q ~ R, showing that

pose that q ¢ ((Rb)I) I.

Now HomR(E,E)

of E is an H - m o d u l e

f E H, we have f(R/I)

there exists an element

Let

Hence we can define an R - h o m o m o r p h i s m

by f(x + I) = qx + I for all x ~ Q.

by Theorem 4.5; 2.7.

We shall show first

of R, then Rb = ((Rb)I) I.

(Rb) I = b'lI,

For sup-

we have qb-lI m I.

Thus

qb -1 ~ I I = (RI) I = R, and hence q ~ Rb. Suppose

that J is a regular ideal of R such that J = (jI)I,

let L be an ideal of R such that L m J and L/J ~ R/M. that L = (L!) I.

and

We shall show

We have an exact sequence:

HomR(R/M,I ) - - H O m R ( L , I ) - ~ H O m R ( J , I ) - * E x t ~ ( R / M , I ) . The first term of this sequence and the last term is i s o m o r p h i c ence of (4) and (3).

is 0 b e c a u s e

to R/M because

I is torsion-free;

of the p r o v e n

equival-

Since HomR(L,I ) % L I and HOmR(J,I ) z j I

this

sequence becomes:

o ~ L i -- ji ~ R/M. If a = O, then L I = jI, and hence J = (jl)I = (LI)I D L. tradiction preceding

shows that a ~ 0, and thus a is onto. type of argument

This con-

A repetition

of the

shows that we can derive from this se-

quence another:

0 ~ (ji)i - (LI)I A ~/M. If 8 = 0, then J = (jI)I = (LI)I o L. 8 is onto.

Now L/J c (LI)I/j,

This c o n t r a d i c t i o n

and the last sequence

shows that

shows that

147

(LI)I/J is a simple R-module.

Therefore,

we have L = (LI) I.

Now let J be any regular ideal of R, and let b be a regular element of J.

Then we have a composition

series:

Jo = Rb c Jl c ... c Jn = J where Ji+i/Ji % R/M for all i < n.

It follows from the preceding

paragraphs

that we can climb up the ladder and obtain

Therefore,

I is a canonical ideal of R.

(4) < ~ >

(5).

It is trivial

that (4) ~-~-> (5), and hence assume

that Q/I is an injective R-module. it follows from Theorem 4.6, of copies of E.

sum can have only one term, (4) = >

(7).

Since Q/I is an Artinian R-module,

that Q/I is isomorphic

Since L(Q/I)

and thus Q / 1 %

Theorem I.I that E Z Q/I z Tor~(K,Q/I)

K D R H % K.

Ch. VI, Prop.

for all n ~ O.

image of HomR(R,E),

Thus by Corollary

and

1.2, the

Since HomR(I,E ) % K, we have by [3,

~ Tor~(HOmR(I,E),R/J ) ~ HomR(Ext~(R/J,I),E )

Since Tor~(K,R/J)

= 0 for all n ~ 2, it follows

= 0 for all n ~ 2, and hence inj.dimRl (6).

jective R-module.

It is clearly sufficient

= I.

As we observed in the proof of Theorem 13.1,

we obtain an injective

resolution

and hence Q ~R I ~ Q.

proof of Theorem 13.1,

if we

resolution over R of I,

over Q of Q ~R I. Thus inj.dimQQ < I.

this implies

that

to prove that Q is an in-

apply the exact functor Q D R • to an injective

lar ideal,

Thus

5.3]:

Tor~(K,R/J)

(7) ~ >

Z K ®R I.

we have by

show that HomR(I,E ) % K.

Let J be an ideal of R.

Ext~(R/J,l)

E.

Now HomR(I,E ) is a homomorphic

isomorphisms

sum

Z K ~R H°mR(K ~R I,E) Z K ~R H°mR(E'E)

thus is a torsion divisible R-module. preceding

to a direct

= L(K) = L(E), we see that this direct

Since Q/I is a torsion R-module,

K ~R H°mR(K'H°mR(I'E))

(jl)l = j.

But I is a reguAs in the

that Q is an injective

R-module.

148

(6) = >

(5).

This is a trivial assertion.

This concludes the proof of the theorem. Definition.

A commutative Noetherian ring of Krull dimension 0

is called a Gorenstein ring of dimension 0 if it is self injective. In the proof of Theorem 13.1 we showed that Q is self-injective if and only if it is an inJective R-module.

Thus we have the following

corollary. Corollary 15.6.

I f R has a canonical ideal,

then Q is a Goren-

stein ring of dimension O. Remarks.

Q being a Gorenstein ring of dimension 0 is a neces-

sary condition for R to have a canonical ideal, but it is not a sufficient condition.

For there is an example of a Noetherian local do-

main of Krull dimension I that does n ~

have a canonical ideal

(see

[5]). On the other hand, we shall see in the next theorem that H D R Q being a Gorenstein

ring of dimension 0 is a necessary and sufficient

condition for R to have a canonical ideal. answers

several other questions

Theorem %~.7.

The next theorem also

raised earlier in these notes.

The following statements are equiva!ent:

(I)

R has a canonical ideal.

(2)

H ®R Q' the full ring o~f quotients of H, is a Gorenstein

rin~ of dimension O. (3)

I_~f P is a prim e ideal of rank 0 i__n_nH, then the P-primary

component o_~f 0 i_n_nH is an irreducible

ideal.

(4)

There is a regular ideal I o_~fQ such that E Z Q/I.

(5)

K i_ss equivalent t__ooE.

(6)

There is an R-homomorphism o_~fK onto E.

(7)

K contains only one simpl 9 divisible R-module from each

equivalence class of such modules.

149

Proof. (4) < ~ > E ~ Q/I. Q/Rb.

(i) < ~ > (6).

(4).

This has been proved in Theorem 15.5.

Suppose that I is a regular ideal of R such that

If b is a regular element of I, then K is isomorphic

to

Since Rb c I we have an R-homomorphism of Q/Rb onto Q/I, and

hence an R-homomorphism of K onto E. an R-homomorphism Q/C.

of K onto E.

We have L(C/R)

lary 15.2.

= L(K)

Conversely,

suppose that f is

If C/R = Ker f, then E ~ K/Ker f - L(E),

and hence L(C/R)

= 0 by Corol-

Thus C is a finitely generated R-module by Theorem 5.1.

It follows that there exists a regular element b ¢ R such that bC = I is a regular ideal of R. (6) (5). proof of (4) < ~ >

We have E ~ Q/C ~ Q/bC = Q/I.

Let f be an R-homomorphism (6) we see that K e r f

equivalent to E by Lemma 5.8.

of K onto E.

is reduced.

That (5) implies

As in the

Hence K is

(6) follows immedi-

ately from the definition. (5) < ~ >

(7).

The assertion that (5) implies

quence of Theorem 9.5.

Conversely,

(7) is a conse-

suppose that K contains only one

simple divisible R-module from each equivalence class of such modules.

Let P be a prime ideal of rank 0 in H, and let D be the unique

simple divisible

submodule of Kp, the P-primary component of K.

is a nonzero R - h o m o m o r p h i s m g of D into E.

Because E is injective,

g can be extended to an R-homomorphism f of Kp into E. ll.3, Im f c Ep, the P-primary component of E. for otherwise it would contain D.

By Corollary

Now Ker f is reduced,

Thus we have L(Kp) = L(Im f).

Theorem 15.4, K and E have equivalent composition submodules,

and thus L(Kp) = L(Ep).

Ep = Im f.

Therefore,

Kp and ~

There

By

series of divisible

Hence L(Im f) = L(Ep),

and so

are equivalent by Corollary 5.13.

It follows that if P1,...,Pn are the prime ideals of rank O in H, then gPl ~ "'" ~ KPn is equivalent to EPl ~ ... • ~ n " are equivalent to theee two direct sums,

respectively,

ll.5, it follows that K and E are equivalent (7) < ~

(3).

Let P1 .... 'Pn

Since K and E by Theorem

to each other.

be the prime ideals of rank 0 in

150

H, and let NI,...,N n be the Pi-primary components of 0 in H.

Let D

and D' be simple divisible submodules of K corresponding to PI"

Then

by Theorem 14.8 we have D = K D R i and D' = K D R I', where I and I' are minimal ideals of H for PI"

By definition I = Jl A N2A...AN n

and I' = J~ D N 2 ~ ... D Nn, where Jl = H = J~ if PI = NI' and otherwise Jl and J~ are

Pl-primary

ideals of H that properly contain N 1

and are minimal with respect to that property.

Now h(D N D')

= h((K D R I) n (K D R I')) = K D R (I A I') = K D R (Jl A J~ n N 2 N ... N Nn) by Corollary 6.7.

Thus we see that

D ~ D' if and only if h(D n D') = 0 if and only if (Jl A J{) A N 2 N ... N N n = 0 if and only if Jl A J{ = N I.

It fol-

lows that K has only one simple divisible submodule corresponding PI if and only if N I is an irreducible (3) < = > let g

(2).

Hd

ideal.

With the notation of the preceding paragraph,

: H - (PI U ... U Pn ).

of H, and hence }~

to

Then H ~

Z H D R Q.

z Hp I @ ... @ HPn.

It is easy to see that

Thus H ~

every HPi is self-injeetive.

is the full ring of quotients

is self-injective if and only if

Furthermore,

of H if and only if 0 is an irreducible

N i is an irreducible

ideal of HPi.

HPi is an Artinian local ring, it is sufficient

Thus,

ideal

since

to prove that an

Artinian local ring S is self-injeetive if and only if 0 is an irreducible ideal of S. If S is self-injective, module.

from

then S is an indecomposable

injeetive S-

Because the annihilator of ! in S is the 0 ideal,

Theorem 4.5 that 0 is an irreducible ideal of S.

suppose that 0 is an irreducible ideal of S. ideal of S, then 0 is an irreducible E(S) = E(S/P) by Theorem %.5.

it follows

Conversely,

If P is the maximal

P-primary ideal of S, and hence

Since HOms(S/P,E(S/P))

Z S/P, it fol-

lows easily by induction on length that if A is any S-module of finite length,

then S ( A )

over S, we see that

= S(HOms(A,E(S/P)) ) . ~(S)

Since S has finite length

.....~ ( H o m s ( S , E ( S / P ) ) )

= ~(E(S/P)).

Because

151

S c E(S)

= E(S/P),

we see that S = E(S/P),

and hence

S is self-ln-

jective. Remarks. Theorems

(1)

The equivalence

of several

of the statements

15.5 and 15.7 has b e e n proved by a different

(2)

We have seen in Theorem

R if and only if R is a Gorenstein is isomorphic canonical

to E.

Theorem

15.7

if there is a m o n o m o r p h i s m R has a canonical

in [7].

13.1 that R is a canonical ring;

ideal for

that is, if and only if K

shows that in general

ideal if and only if K is equivalent

Theorem 13.1 we established

method

in

to E.

that R is a Gorenstein

R has a

In the proof of

ring if and only

of K into E, while Theorem

15.7

shows that

ideal if and only if there is an epimorphism

of K

onto E. (3)

It follows

from Theorem

15.7 that if S is an R-submodule

Q such that R c S and Q/S ~ E, then S is isomorphic ideal

of

to a regular

of R. (4)

Theorem

is a finitely by Theorem

10.3

states

generated

that if the integral

R-module,

then K is equivalent

15.7 we see that if R is analytically

R has a canonical

Theorem

closure

of R in Q

to E.

Hence

unramified,

then

ideal.

15.8.

Any two canonical

ideals

of R (if any exist)

are

isomorphic. Proof.

Suppose

J is a regular

that I and J are canonical

ideal of R, there is a regular

bI c J, and bI is a canonical loss of generality because

Q/1%

element

ideal isomorphic

we can assume that I c j.

E and Q/J ~ E we have an exact

ideals

to I.

of R.

Since

b in J.

Now

Hence without

Let B = J,/I;

then

sequence:

0-~B-~E-+E-.O Let B # = HomR(B,E); sequence:

then since HomR(E,E ) % H, we can derive

an exact

152

0 - + H ~ - ~ H - ~ B # --~0. By Theorem 2.7, ~ is an H-homomorphism, by a regular

element

Let L = Hg O R; H/Hg ~ R/L. Corollary ideals

by Theorem 2.8 we have H ®R L % HL = Hg and

(2) that L is a projective

in a local

~ng

are principal,

a ~ R such that L = Ra.

Since B # % H/Hg % R/Ra, B ~ B ##,

of Theorem =

~{(J/I)

13.1,

~(R/Ra). = ~(B)

Therefore,

By Theorem

= ~(J/Ja).

ideal of R.

and hence

length. that

S(B)

12.3, Because

I and J are isomorphic

from

Projective

there

exists

a regu-

Thus H/Hg % R/Ra.

that aB = O.

of finite

this implies

ideal of H, it follows

we have aB # = 0.

and hence it follows

N o w B is an R-module

~(B)

Hence we have B # % H/Hg.

Since H @R L is a projective

2.6

lar element

g of H.

and thus a is m u l t i p l i c a t i o n

By Theorem 4.6 we have

Thus we see that aJ c I.

As we observed = ~(B#).

S(R/Ra)

Therefore,

= S(J/Ja),

Ja c I it follows

ideals

of R.

in the proof

and hence that Ja = I.