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
1Dimensional CohenMacaulay Rings
SpringerVerlag Berlin. Heidelberg • New York 1973
A M S Subject Classifications (1970): 1302, 13Cxx, 13E05, 1 3 E l 0 , 13F05, 1 3 H x x
I S B N 3540063277 SpringerVerlag 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 0387063277 SpringerVerlag 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, reuse 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 SpringerVerlag Berlin  Heidelberg 1973. Library of Congress Catalog Card Number 7380869. Printed in Germany. Offsetdruck: Julius Bettz, Hemsbach/Bergstr.
TABLE
Chapter
I
hDivisible 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
SemiSimple 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 1dimensional,
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
CohenMacaulay
of such a ring may be
In fact, we shall be able
about
1dimensional
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 1dimenslonal, 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
CohenMacaulay
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
1dimenslonal,
the elementary
about
torsionfree,
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
1dimensional
a local problem,
Rmodules
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
CohenMacaulay
to attain the p r e c i s i o n
nonzero
of an A r t i n i a n mod
CohenMacaulay
of finding the structure
1dimensional
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 Rmodule of finite length and of a d i v i s i b l e A r t i n i a n Rmodule. A r t i n i a n divisible submodules
Rmodule
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
Rmod
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
Rmodules
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
Rmodules
over a Noetherian
ring,
we actually go to the extreme of "throwing them away". The JordanH~lder
theorem for Artinian divisible Rmodules
gives
us the concept of the divisible length of an Artinian Rmodule, function that proves to be most useful.
a
Since modulo its divisible
submodule an Artinian Rmodule has finite length in the classical sense, we see that we have two numerical
invariants
to describe an
Artinian Rmodule. We prove that there are only a finite number of equivalence classes of simple divisible correspondence
Rmodules;
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
Rmodules. The set of isomorphism given simple,
divisible
classes
Rmodule is in onetoone
with the ideal class semigroup 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 Rmodules
divisible Rmodules 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 semisimple equivalent
to a finite direct
divisible
Rmodule to be one that is
sum of simple,
We find that every Artinian divisible
divisible Rmodules.
Rmodule is semis~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 Pprimary correspond
the completion module
of R.
Rmodules
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 conRmodules. divisible
ring extension
ring extensions
components
of R.
of the cor
Rmodule
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
Rmod
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
Pprimary
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.
Rmodule;
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 Rmodule
code" for all Artinian,
as a factor
Rmodules
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
Rmodule
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 semislmple
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 Madic
topology
basis
in [16].
for the treatment
in the notes.
Chapter
divisible modules
While
elementary
of strongly
gives
ring may be reduced
We also
Rmodules
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 1dimensional,
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 Rtop
Madic
of the two topologies
tool of the duality
Noetherian
Madic
and thus H is the ordinary
reworked
hdivisible
of H, the completion
is based
torsionfree
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 Rtopology
and homo
theorems
through
but has been
commutative
Much of the analysis
to
that have to be used.
hdivisible,
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
CohenMacaulay
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 1dimensional
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 hdivisible,
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 Rmodule
is
of an Rmodule B,
then B is also divisible.
of a module
an Rmodule 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 Rmodule
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 Rmodule 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 torsionfree. if the m u l t i p l i c a t i o n s
of R are all monomorphisms. is torsionfree.
no nonzero
on the modA submodule
If A is a submodule
an Rmodule B, and if both A and B / A are torsionfree, torsionfree.
submodules.
of
then B is also
An Rmodule is said to be a torsion Rmodule
torsionfree
An R
if it has
An Rmodule B has a unique
largest
torsion
submodule
of B, and B/t(B)
t(B)
that contains
is torsionfree.
order if there is a regular
submodule
B is said to be torsion of bounded
element
easy to see that an Rmodule
every torsion
r of R such that rB = 0.
is both torsionfree
It is
and divisible
if
and only if it is a Qmodule. A more useful
concept
than divisible
plying homological
methods
is that of hdivisible.
said to be hdivisible free and divisible
if it is a homomorphic
Rmodule;
image of a Qmodule. R is an integral
if it is a homomorphic
A direct Consequently module
h(B)
sum of hdivisible
said to be hreduced, that is, if h(B) B/h(B)
mental
exact
I.i.
Rmodules
B contains
a unique
It follows
is divisible.
If B is an Rmodule,
largest
hdivisible
submodule
of B.
hdivisible
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__sshreduced
Rmodule
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 hdivisible.
every hdivisible
Unfortunately,
conditions
Theorem
Rmodule
if it has no nonzero
= 0.
is hreduced.
sufficient
Rmodule.
to say about the converse
that contains
is
(in particular,
is hdivisible
image of an injective
an Rmodule
An Rmodule
image of a torsion
ring
then an Rmodule
that an hdivisible
shall have something
of ap
that is, if it is an Rhomomorphic
If Q is a semisimple
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 Qmodule,
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 torsionfree,
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 Qmodule, and thus Im B c h(B). mapping onto h(B). maps onto x.
it is torsionfree
On the other hand,
and divisible
there is a Qmodule V
Let x E h(B) and choose an element y ¢ V that
Since Qy c V, there is an Rhomomorphism 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 hreduced if and only if HomR(Q,B) Corollary 1.2.
Let B be a torsion Rmodule.
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
Rmodule,
= 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 Rhomo
If S = ker ~,
such that y ~ f(K).
is h
a in S N R.
Then v ( a I) = y
an induced
Hence we
Rhomo
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 hdivisible,
~ : 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
Rhomomorphism.
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
Rmodule.
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__sshreduced i f an__dd only i f H o m R ( K , B ) = O.
(3)
I__ffA and C are
hdivisible
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
hdlvi______sibl____eeRmodule,
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
torslonfree
and prove that B is a torsion
Rmodule.
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 torsionfree
of B, and hence is equal to O.
Rmodule,
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
torslonfree
t(B) O C = O, we may assume without
erality that B has no nonzero
an Rhomomorphism
torsionfree
and hence x is a torsion
shown that B is a torsion
Rmodule,
Therefore
element
of B.
and this completes
the proof of the theorem. Definition.
An Rmodule
if HomR(Q,C ) = 0 and Ext~(Q,C) sion module order,
is hreduced.
then Ext~(Q,B)
EXt~(Q,B)
establishing Theorem
(1)
Rmodule
Rmodule
a cotor
of bounded
and torsionfree,
Thus a torsion
Rmodule,
and hence of bounded
and A is any Qmodule,
of F.
then
For we have HomR(A,C ) = 0 because
Now we can write A = F/P,
P is a Qsubmodule
in particular,
Rmodule
Rmodule.
HomR(A,C ) = 0 = Ext~(A,C).
modules.
Thus,
is both torsion
If C is a eotorsion
hreduced.
= 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 Qmodule
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 hreduced. 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 Rmodules.
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 hreduced.
Suppose that B is a torsion Rmodule, module containing C.
Thus
and that E is an injective R
Then we have an exact sequence of hreduced
Rmodules: O*HOmR(B,C) We have Ext~(Q,HOmR(B,E))
[3, Ch. Vl, Prop. 5.1]. torsion Rmodule.
*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 Rmodule,
be a free Rmodule mapping onto B with kernel A.
and let F
Then we have an
exact sequence of hreduced Rmodules. 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 Rmodule.
Hence HomR(B,C ) is cotorsion by Theorem
1.5. Definition.
The homological dimension of an Rmodule 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 Rmodules 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 Rmodule B that is a
homomorphic image of an inJective Rmodule. Theorem 1.7.
The following statements are equivalent:
(1)
hdRQ : 1
(2)
If 0  ~ A  ~
B~ C~ 0 is an exact sequence of Rmodules and
A and B are cotorsion, then C is cotorsion. (3)
Ext~(K,B) is a cotorsion Rmodule for all Rmodules B.
(4)
Ext~(K,B) = 0 for all hdivisible Rmodules B.
(5)
Ext~(K,B) = 0 for all torsion, hdlvisible Rmodules B
(6)
hdRK = I.
Proof. (1) ~ > reduced Rmodule.
(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 Rmodule and hdRQ = I, the end terms of this sequence are zero.
Thus Ext~(Q,C) = O, and hence C is a cotorsion
Rmodule. (2) ~> (3). containing B.
Let B be an Rmodule and E an injective Rmodule
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 Rmodule. sion Rmodule.
Hence by (2), Im a is a cotor
But then by (2) again, Ext~(K,B) is also a cotorsion
Rmodule. (3) = >
(4).
If B is an hdivislble Rmodule, 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) Rmodule is hdivisible,
= 0.
It is easy
and hence every
image of an inJective Rmodule is hdivislble.
We have
an exact sequence: HomR(K,B/t(B))
~Ext~(K,t(B))
Clearly HomR(K,B/t(B)) Qmodule,
= 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 hdivisible. Let x ¢ t(B); f : Q~B
then by Theorem I.I there is an Rhomomorphism
such that f(1) = x.
If S = Ker f, then S contains
ular element of R and hence Q/S is a torsion Rmodule. Im f c t(B), and from this it follows (6) ==> (i).
This assertion
is obvious.
The gollowing
(I)
B/h(B)
is hreduced for every Rmodule B.
(2)
If 0*A
two statements
(1) ~=> (2).
then B is also hdivisible.
Let A be a submodule of an Rmodule B,
and suppose that A and B/A are hdlvlsible. is a homomorphic
Since B/h(B)
are equivalent:
~ B  ~ C ~ 0 is an exact sequence of Rmodules
such that A and C are hdivisible,
B/h(B)
Therefore
that t(B) is hdlvisible.
Lemma 1.8.
Proof.
a reg
image of B/A.
is hreduced
Then A c h(B),
Thus B/h(B)
and hence
is hdivisible.
by (1), we see that B/h(B) = 0.
Therefore
B is hdivlsible. (2) ~> (1). taining h(B). (2).
Let B be an Rmodule and A a submodule of B con
If A/h(B)
But then A = h(B), Theorem 1.9.
is hdlvisible,
then A is hdivislble
showing that B/h(B)
If hdRQ = I, then B/h(B)
by
is hreduced. is hreduced
for every
Rmodule B.
On the other hand, if Q is a semisimple
ring, then the
two statements are equivalent. Proof.
If hdRQ = l, then by Theorem 1.7, Ext~(K,B)
torsion Rmodule. 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 hreduced.
On the other hand let us assume that Q is a semisimple ring and that B/h(B)
is hreduced for every Rmodule B.
sion hdlvisible Rmodule.
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 Rmodules
and,
O. By Lemma 1.8, C is an h
since Q is a semlsimple
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 Rmodule is hdlvislble. 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 semisimple 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 Rmodule and C a cotorsion Rmodule
containing A such that C/A is torsionfree
and divisible.
Then the
following statements are true: (I)
If D is a cotorsion Rmodule,
then every Rhomomorphism
from A into D can be lifted uniquely to an Rhomomorphism of C into D.
10
(2) sionfree
I f D is a cotorsion and divisible,
Proof.
Rmodule 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 Rmodule.
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 Rmodule,
the Rtopology
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 Rmodule point
limit of the Rmodules
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 Rmodules 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.
~ : AA
~(x) where
We have a natural
= (x + rA) for every x ~ A. r ranges
over the regular
Theorem 2.1. bounded
order.
Rhomomorphism
given by
The kernel elements
Let A be an Rmodule
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 torslonfree.
Rmodule.
and thus if A is torslonfree,
X is also tor
sionfree.
(4) A/~(A) is Proof. all regular divisible
(i)
torsionfree
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 torsionfree.
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 
Rmodule.
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
Rmodule.
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 Rmodule,
@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 ® (arb)
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*.
Rmodule.
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
Rhomomorphism.
: 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(rl+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 = (cl+R)
@R A);
t h e n if r is a r e g u l a r
a r ~ A such that
f(rl+R)
element
= (rl+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 . = (clrl+R)
® ca r = (clrl+R)
for any regular element =
f(rl+R)
~ rb = (c'l÷R) ® b = f(rl+R).
s ~ R we have:
 sf(slrl+R)
= O.
Thus (rl+R) ® a r
(r'l+R)
® (arars)
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 Rmodule.
(2)
A is complete in the Rtopology.
(3)
A ~ HomR(K,K @R A).
(4)
A ~ HomR(K,B)
torsionfree
Rmodule.
Then the fol
are equivalent:
(1)
Proof.
1.5.
Let A be a torsionfree
for some Rmodule B.
(i) ~> (2).
By Theorem 2.1, A is cotorsion,
and divisible.
But A/A is a reduced Rmodule by Theorem
Hence A = A, and thus A is complete
in the Rtopology.
(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
torsionfree,
cotorsion Rmodules
onto the category ~ of torsion hdivlsible 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. Rtopology. 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 torsionfree
R i H ~ Ext~(Q,R) submodule
Rmodule•
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
Rmodule
Rmodule
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 torsionfree
sequence
is a direct
of H, then
= svy = f(vy) ring.
Rmodule.
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 Rtopology is equal to the completion of R in the Rtopology;
that is, H is complete in the Rtopology.
Corollary 2.6.
(l)
Assume that R is a reduced Rmodule.
I f Q is a semisimple ring,
Then:
then H is a faithfully flat
Rmodule. (2)
If H is a faithfully flat Rmodule and A is any Rmodule,
then H @R A is a finitely generated ~roJectiv ~ Hmodule if and only if A is a finitely generated projective Rmodule. (3)
I f R is a Noetherian local ring and hdRQ = l, then H is a
faithfully flat Rmodule and H @R A is thee completion o f A for every finitely generated Rmodule A. Proof. Ext~(Q,R)
(1)
Since Q is a semlsimple
is a projective Qmodule.
follows that Ext~(Q,R)
ring, the Qmodule
Because Q is flat over R, it
is a flat Rmodule.
(2)
Hence exact sequence
of Corollary 2.5 shows that H is a faithfully flat Rmodule
(see
[2]). (2)
This is a property of faithfully flat ring extensions of
R (see [2]). (3) sequence:
If A is a finitely generated Rmodule, Fl~F0
free Rmodules.
~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 Hmodules. fore Im a is an hreduced Rmodule,
There
and hence Ker a is a cotorsion
Rmodule by Theorem 1.5.
But since hdRQ = l, Im a is a cotorsion
Rmodule 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 Rmodule. 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 Rmodule. because H/R is a Qmodule. 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 torsionfree and divisible.
If A is the completion of
A, then ~ is a cotorsion Rmodule and A / A is torsionfree 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 torsionfree Tor~(H,R/I)
= 0
It follows that H and H/R are flat Rmodules.
Thus,
as is well known (see [2]), H is a faithfully flat Rmodule. Theorem 2.7.
(I)
i_~f T is a torsion Rmodule,
Thus T has the s~ructure of an Hmodule, unique.
and this structure is
Every Rsubmodule of T is also an Hsubmodule of T.
is any Hmodule, (2)
then T ~ H D R T.
then HomR(T,A ) = HomH(T,A ) .
I f C is a eotorsion Rmodule,
has the structure of an Hmodule, A is any Hmodule, 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 Hmodule structure extending that of R
defined by fx = 81(f ® x) for all f c H and x ~ T. Suppose that there is another Hmodule 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
Rmodule,
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. Hmodule
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 quasilocal quasilocal
ring w i t h o n l y a f i n i t e
a semilocal
ring.
ring ring is number
i9 Theorem 2.8.
There is a onetoone correspondence between the
set of regular ideals I of R and the set of regular ideals J of H such that kl(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 torsionfree and divisible.
Further
= O.
If J is an ideal of H such that kl(J) is a regular ideal of
R then J = H(kl(J)) (3)
HI is a regular ideal of H;
and H/J ~ R/(XI(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 quasilocal
ring with maximal ideal M, then H i__ss
quasilocal 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 torsionfree Rmodule, 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 Rmodule.
H is
20 also a cotorsion Theorem
Rmodule,
Rmodule
by
1.5.
Let i' = kl(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' = XI(HI). Because kl(HI)
= I, we have HI n X(R)
H/k(R)
is a divisible
H/X(R)
=
Rmodule,
(Hi+k(R))/k(R)
is a torsionfree
and divisible
If ~ is the divisible a regular
ideal of R.
2.1,
then K e r
~
=
Therefore
~ / ~ is torsionfree 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
Rmodule.
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
Rmodule.
~
= ~.
hence ~(I) ~ I / ~
0 ~H Since
of R, then
(HI)/~ is also torsionfree
HI is a cotorsion is isomorphic
, and
~
= HI/X(1),
2.5,
Rmodule.
submodule
pletion of I in the Rtopology
By Corollary
and thus H = HI + k(R).
HI/(HI n k(R))
~
= k(I).
= O.
ideal of H such that I = kl(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/kl(J).
= HI.
Rmodule,
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 quasilocal
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 torsionfree
Rmodule
of R in Q,
o f S in the Stopology.
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 hdivisiblue,
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 Rhomomorphism x ~ Q/S. Rmodule. 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 torsionfree
becomes:
H ~ H(S) *0.
= HomR(Q/S,Q/S).
Then s = a/b where
the sequence
proving
Rmodule,
the desired
equality.
since Q/S is a divisible
Rmod
ule. By Theorem 2.7, = HomH(Q/S,Q/S).
Q/S is an Hmodule
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
Rhomomorphism 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 Rhomomorphism 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, hdivisible,
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 torsionfree.
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 onetoone
between the se ~ of rings B between R and Q and the set
of rings 0 between H and H ®R Q ~yen blf 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 torsionfree
Proof. Tor~(H,Q/B)
torsion modules. torsion modules.
then B c H @R B, and H ®R B = HB = H + B. and divisible Rmodule.
Since Q/B and B/R are torsion Rmodules, = 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 Qmodule, 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 torsionfree
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
Rmodule.
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
Rmodule
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. torsionfree,
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 Rmodule,
and hence there
h = bg + ~. Therefore,
Thus b(kg) 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
sionfree,
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.
Rmodule,
(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
Rmodules
of R;
hdivisible
and H, the
of R.
Rmodules
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 torsionfree
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.
Rmodule.
If we let ~ element
the image of b
, then ba = bd for some element
R is torsionfree.
that b is a regular
of R, and b = b + ~ ,
But then b(ad)
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
hdivisible 1.2,
hdivisible
Rmodule
R, and hence ~ is a regular ideal of R.
cotorsion
Rmodule
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 Rtopology
where
of R, there is an ele
q = a/b where a E R and b is a regular
H°mR(K'K
hdivisible
Rmodule.
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 torsionfree ootorsion
A is a torsion
element b of R that is contained
= I//~.
completions.
On the other hand if A is an Rmodule,
fashion,
Let I be a regular
regular
HomR(K,K )
is of the form K @R A for some Rmodule
~module.
it is a torsicnfree
regular
Hence K is
and clearly K ~R A ~ K D R A % K @ ~ (R ~R A) is a
In similar
that an Rmodule
+ R)/R = O.
Furthermore,
then R D R A % A, and hence K D R A ~ K ~ Rmodule.
and thus Q is the full
: (Q~
and of course K ~ K.
A by Corollary
Q is equal to
of R.
an Rmodule,
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
Rtopology. T h e o r e m 2.12.
H is complete
in the Htopology.
Furthermore,
if
27
an Rmodule cotorsion
is a cotorsion
(or torsion)
Proof. Hmodule.
(or torsion)
Let k : R  ~ H be t h e ring homomorphism
also a torsion
element
of R, then k(b)
Hmodule.
Hmodule,
Htopology.
Rmodule.
By Corollary
Hmodule.
Let V be the full
ring of quotients
Rmodule,
HomH(V,C ) c HomR(V,C ) = O.
element
= 0.
Therefore
Let us consider
where A is an Hmodule,
Rmodule,
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 Hhomomorphisms.
Since
s : A  ~ C such that
s
But HomR(A,C ) = HomH(A,C ) by Theorem 2.7,
s is an Hhomomorphism.
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
Rmodule
and hence V is a Qmodule.
0C
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)
Rmodule
Thus a torsion
ficient
ExtH(V,C)
then it is also a
Hmodule.
H is a torsionfree
cotorsion
Rmodule
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.
torsionfree
III
EXTENSIONS
Let A be a c o m m u t a t i v e
as an Rmodule.
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.
torsionfree 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 Rmodule,
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
Rmodule,
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 Atopology on any Amodule coincide. has one of the following properties
Furthermore,
an Amodule
as an Amodule if and only if it
has the same property as an Rmodule: (I) (5)
torsion,
divisible, Proof.
(6)
(2)
hreduced,
(3)
cotorsion,
(4)
complete,
hdivisible.
It is obvious from Theorem 3.1 that the Rtopology and
the Atopology on an Amodule coincide
and that (1), (4), ~und (5) are
t rue. Let B be an Amodule. 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 hdivisible (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 torslonfree as an Rmodule 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.
Rmodule.
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,
Rmodule.
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).
Rmodule
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
Rmodule,
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
Rsubmodule
of H by C o r o l l a r y
element
proves
R
to H @R Q"
hdivisible
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 hreduced.
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.
Rsubmodule
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
Rmodule.
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
Rmodule. 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
Rmodule.
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)
Rmodule.
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 Rsubmodule
is a proper
over A.
ideal of R.
% R/M.
ideal M, then AM i s
AM contains
Rmodule,
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 quasilocal
(7)
S.
(resp.
(resp.
If R is a quasilocal
r_egular maximal
ule.
is also a finitely
If every regular ideal of R is finitely
then every regular
maximal
Amodule
Then in Rmodule.
If S
R, then AS/S is a divisible
Thus if AS ~ A ~R Q' then AS is a compatible
Smod
ring extension
of
If A c S, then S is also an Amodule. (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
Amodule
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 quasilocal
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
Rmodule
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
quasilocal
ideal AM.
(7).
Let S be an R  s u b m o d u l e
Rmodule,
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 Rmodule. 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
Smodule.
It fol
Rmodule 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 Amodule. (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 Vmodule
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 Vmodule,
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
Rmodule.
Miprimary.
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
CohenMacaulay
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
[email protected] 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
Rmodule.
n + I terms.
CohenMacaulay
CohenMacaulay
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 1dimensional
CohenMacaulay
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 Rmodule define an Replmorphism
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. hdivisible
Rmodule.
by Theorem
Ext~(K,B)
an exact
Let us consider
ele
of f, then P is a free Rmodule 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+lbXn].
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 semilocal
basis
of the m a x i m a l
Then there is a onetoone
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,
hdivisible
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 semilocal
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 Rmodule 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 1dimensional is divisible
Of course
of R.
CohenMacaulay
ring.
if and only if it is hdivisible.
an hdivisible
Rmodule
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.
Rmodule,
and let x ~ D.
let Xn+ 1 be an element
an Rhomomorphism
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 hdivisible
Case II:
General
we see that D/h(D)
the torsion M a ranges
I.
Rmodule.
is hreduced
that D is hreduced,
sum of
Rmodule.
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
Rhomomorphism
Case.
Let D be a divisible
sion,
that f is a welldefined
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~amodule 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.
Rmodule.
image
Thus TMa = 0 for all M a, and hence D is torsion
But then D is hdivisible, Corollary
4.4.
w i t h n o nilpotent module
Q ~ RP1 @
elements
... @ Rp,.~
other than zero. Rmodule
Let P1,...,Pt
have 0 = P1 0
and therefore
Let R be a 1dimensional
of every divisible
Proof.
of Q, and thus TMa is an h
and thus Q is a semisimple
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 1dimensional
The theorem
is now
1.4.
CohenMacaulay
it is sufficient
For if D is a divisible is a Qmodule,
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 torsionsub
ideals
Since R has no nonzero
... 0 Pt"
CohenMacaulay
is a direct
be the prime
D = O.
by the dimensions
Rmodule,
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
RMamodules
ideals M S of R.
of the vector
Hence
space components
If R is a 1dimensional 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.
CohenMacaulay
ring with nonzero
define Artinian
that they are torsion Rmodules,
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
1dimensional
R
our attention
CohenMacaulay
some results
of a general
nature. Definitions. reducible
An ideal of a commutative
if it is not the intersection
An Rmodule
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
Rmodule
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
Rmodule
is a direct
injective
Rmodule.
sum of indecomposable
Rmodules. Let P be a prime ideal of R and Rp the Padlc
Then E(R/P)
is also the injective as an ~pmodule
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 Rmodule. (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_~sPjprimary 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 Rsubmodules
A1 c ~
ring, A an Rmodule,
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
Rmodule.
then A is said to
A Noetherian Rmodule 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 Rmodule.
An
Artinian module over an integral domain or a 1dimensional Noetherian CohenMacaulay
ring is e torsion Rmodule.
If an Rmodule 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 Madic
topology,
ideal M, let R be the completion
and let E = E(R/M)
Let A be an Rmodule.
Noetherian
b_~e the injective
Then A is an Artinian
H o m R ( A , E ) is a Noetherian
Rmodule.
of R in the
envelope
Rmodule
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
Rmodule
E n is a direct
Conversely,
sum of n > 0 copies
let B be an Rmodule.
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
Rmodule.
~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 nonunit y ~ B/R.
quasilocal
c
R and A ~ Q and B ~ Q. Then
But then regular
ring,
loss of generality
(c 1 + R) = rlx ~ 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 quasilocal 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 Rsubmod
and t in R such that x = rc 1 + R and y = tc 1 + R. 1
and the
Rmodule.
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 cld 1 ¢ A U B, and hence we can assume that
cld 1 ~ A.
Thus d 1 = c(cld 1) ~ A A B = R, contradicting
that d is not a unit of R.
Therefore,
the fact
K is an indecomposable
Rmod
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 Rhomomorphism
~ : 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 Meprimary
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, 1dimensional, Theorem 5.0.
(1)
CohenMacaulay ring with maximal ideal M.
The following statements are true.
The Rtp~01ogy and the Madic topology o_nnannRmodule are
the same.
(2)
H is complete and is the completion of R in the Madic
topology.
(3)
H is a complete, Noetherian, local, 1dimenslonal Cohen
Macaulay ring with maximal ideal HM. (4)
H is a faithfully flat Rmodule.
(5)
I_~f A is a finitely generated Rmodule, then H D R A is the
completion of A in the R (o__~rMadic) topology. Proof.
(1)
A regular ideal of R is an Mprimary ideal of R,
and hence contains a power of M.
It follows easily from this that
the Rtopology and the Madic topology on an Rmodule 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 Rtopology is equal to
the completion of R in the Rtopology; Rtopology.
that is, H is complete in the
Therefore H is complete in the Madic topology, and is
the Madic 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 1dimensional 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, quasilocal ring with maximal ideal HM.
Clearly the (HM)adic topology on H is
equal to the Madlc 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 HMprimary 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,
1dimensional,
CohenMacaulay
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.
1dlmensional~
Cohen
As before, Q is the full ring of quotients of R
We have hdRQ = 1 and hdRK ~ 1.
R is the Rtopology,
and H = HomR(K,K ) .
divisible Rmodule isomorphic
H/R is a torsionfree and
to ExtlCQ,R) i ~
H is the completion of
~
6
44
Every divisible nilpotent
elements,
ule is a direct divisible ule,
Rmodule
is hdivisible.
then the torsion
summand.
submodule,
and B/h(B)
then its completion
to HomR(K,K
is reduced.
then h(B)
is a cotorsion
Rmod
is its
If B is a torsion
If A is a torsionfree
the maximal
Rmodule
ideal of R by M.
if and only if it is an Mprimary
the Madic
topology
Noetherian,
local,
on an Rmodule 1dimensional,
ideal HM and H/HM ~ R/M. a finitely
of a divisible
Rmod
R
and is i s o m o ~ h i e
D R A).
We shall denote regular
submodule
If B is an Rmodule,
then h(B) ~ K ~R H°mR(K'B)"
module,
If R has no nonzero
generated
ideal.
The Rtopology
and
are the same.
H is a complete,
CohenMacaulay
ring w i t h maximal
H is a faithfully
Rmodule,
An ideal of R is
flat Rmodule.
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
Hmodules An Artinian
is alse true.
Definitions. define
Theorem
and Artinian Rmodule
Artinian
reduced.
By the Nakayama
Conversely,
generated.
the duality of Theorem 4.6 beRmodules
Rmodules
= [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 funcHmodule,
suppose
generated
and I is an ideal of R we shall
Rmodule. (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 Rmodules.
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 Rmodule
the annihilator
annihilator
of H/HM over H.
We also have available
tween Noetherian tor H O m R (  , E ) .
envelope
envelope
length). Rmodule
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
Rmodule
by Theorem 4.5.
and this contradiction
Therefore,
proves
that A is
generated.
Corollary finitely sense.
Rr is an Mprimary
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,
Rmodule,
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 Rmodules:
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 torsionfree
= O, and by the preceding
that HomR(Q,C ) = 0.
exact
Hence C has no nonzero
is a torsion
and divisible.
sequence
this implies
hdivisible
submodules,
and thus C is reduced. Definitions. module
We will say that an Rmodule
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
Rmodule
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
Rmodule
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 MI/R and thus K c E(MI/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).
MI/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 MI/R is a finitely
It is easy to see that K is an essential
Since MI/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(MI/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) Hsubmodules Rmodule 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) ~ > Hmodule
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
Rmodules.
submodules,
Suppose
divisible
We assert
divisible
submodule
implication. of (3) = >
Rmodule
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 +
Rmodules
(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
Rmodule
Rmodule
divisible
Rmodule.
if and only if every proper
generated.
Then D Rsubmod
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
Rmodule,
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 Rmodules are equivalent if
each is a homomorphic
image of the other.
This is clearly an equiv
alence relation on the class of Rmodules. Rmodules,
we shall denote this by A  ~ B ,
If A and B are equivalent and we shall denote the
class of Rmodules equivalent to A by [A]. Lemma 5.8.
If A is an Artinian divisible Rmodule and B is a
reduced submodule of A, then A/B is equivalent simple divisible Rmodule,
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 Rmodule,
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 Rmodule and B c C submodules of A such
that C/B is a simple divisible Rmodule.
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 Rmodule 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 Rmodule,
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 onetoone
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 Rmodules.
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.
Rmodule.
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 Rmodule,
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 Rmodule,
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 Rmodules.
 ~ 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 Rmodule.
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 Rmodule. 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
Rmodules,
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
Rmodules,
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 Rmodules,
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 59.
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 Rsubmodule 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 Rsubmodule
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
Rmod
Rmodules
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 Rsubmodule
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,
Rmodule
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)
Rmodule.
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
Rmodule.
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. Rmodule,
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 Rmodules
•(A/B).
~(A/B)
But
that A cannot be isomorphic
c be regular element B = Rc1/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 Rmodule.
a sum of Artinian divisible Proof. Corollary
Rmodules.
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,
CohenMacaulay
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 onetoone
(i.e. Mprlmary)
(i.e. HMprimary)
ideals
ideals
We
however,
is an unramified
2.5 that H is a t o r s i o n  f r e e
Rmodule
CohenMacaulay
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 Rmodule,
1dimensional
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
Rmodule.
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
1dimen
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
Rmodules.
Let A be an Rmodule
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
Rmodule. 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 Rmodule
such that A c S c Q, then S is an A
module. (b)
A finitely
as an Rmodule, (c)
generated
torsion
and the same finite
If J is a regular
Amodule
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
Rmodule.
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 1dlmensional
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 KrullAklzuki.
(For a different
proof,
see [17,
Theorem 33.2].
C o r o l l a r y 6.2. Noetherian
(Theorem of KrullAkizuki).
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 Rmodule
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 1dimenslonal 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).
semilocal 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 SMlmOdule 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 Stopology.
Then we have an exact sequence: 0 ~ HomR(K,B/R ) ~ H ~ H(S)
where ~ is a ring homomorphism. as an Rmodule.
Proof.
Furthermore,
semilocal,
0
S/B has finite length
Noetherian,
1dimensional
ring.
It is clear that HomR(K,B/R)
exact sequence:
S/B ~
Therefore H(S) is a finitely generated Hmodule,
and hence is a complete, CohenMacaulay
~
~ 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 Rmodule
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 torsionfree
Rmodule 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
Rmodule,
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
Rmodule.
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
Rmodule.
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
Rmodule 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 onetoone, 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
sionfree Rmodule, 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 Rmodule.
Hence we have Q/B = K D R H(B)  K D R H/I.
H/I is complete in the Btopology, by Theorem 3.2.
it is complete in the Rtopology
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 semilocal 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 Btopology.
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. sionfree Bmodule
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 Bmodule 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 Rmodule.
Hence
(HM)(J/I)
= J/i.
But J is an
ideal of H, and thus J/I is a finitely generated Hmodule. 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 Hmodule
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 onetoone
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
torsionfree and divisible. Theorem 6.8.
(I)
There is a onetoone 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 onetoone 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 onetoone 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 1dimensional,
local, CohenMacaulay 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 onetoone 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 ~ Hmodule,
between the set of ideals
= d(0).
Then ~
ring extension of H in
is a torsionfree
and hence is an H ®R Qmodule; 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, CohenMacaulay Let 01 ~ q / ~ ;
N H) is a complete,
1dlmenslonal
ring and Q1 is its full ring of quotients.
then H 1 c 01 c Q1 and D 1 is a reduced Himodule.
see that O1/H 1 ~ O/(H +
dQ ) is a homomorphic
is a nonzero divisible Himodule.
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 Hmodule and (H + /~)/i~O
~ H/(H n Mg) is reduced.
is the divisible submodule of H + /_0. is a divisible Hmodule,
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 torsionfree 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 1dimensional,
local,
CohenMacaulay
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(lrb) = 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 Rmodule.
If B'/R = h(B/R),
of R.
Let ~
be the divisible
and divisible,
and hence
~
submodule
Since Q is a semlslmple
e E ~
such that Qe = ~
is a p r i n c i p a l
=
of B.
is a Qmodule;
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 Rmodule.
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 quasilocal ments
Since ~
Rmodule.
We p r o c e e d by induction divisible
of R.
as R
every
Rmodule.
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
Rmodule.
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
Rmodule.
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
Rmodule
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) ~
Rmodule,
A R), and this is a contradiction
0 and R / ( ~
closure
Since
Therefore,
valuation
(6) ~ >
every localization generated
of V with
Rmodule,
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).
Rendomorphism divisors.
Rsubmodule
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
Rmodule. Since E is a simple
of E is an epimorphism.
But H Z HomR(E,E)
divisible
Rmodule,
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
Rmodulein
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 Amodule.
(3)
H(A), the completion of A in the Atopology,
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 Amodule is a torsion,
divisible Amodule if and only if it is a torsion, ule.
Furthermore,
by Theorem 3.6
Q is also an Amodule.
divisible Rmod
(7), every Rmodule between A and
Thus A is a closed component of R if and
only if Q / A is a simple divisible Amodule. (2) ~ >
(3)
Since Q/A is a simple divisible Amodule,
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 Qmodule 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 Rmodule. 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 Amodule.
Since
Q / A ~ Q/A, we see that Q/A is a simple divisible Amodule. 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 Amodule.
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,
Btopology
we recall
There is a onetoone
set of closed components
its completion
correspondence
in the
and hence there is a onetoone
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
orderpreserving
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
onetoone 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.
CohenMacaulay
ring,
R and Q will be called
then the the pseudo
to R.
There is a onetoone
components
rings V belonging
A valuation
ring V is a pseudo valuation
of a pseudo valuation
If R is a ldimensional,
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/
[email protected] 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 Rmodule, it is a
finitely generated torsion Rmodule 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 Rmodule 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 Rmodule of finite length,
n
and hence S is a
finitely generated Tmodule. 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 + Ami ~ 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 hreduced. But Ck/T = (C k + Ak)/Ak,
and since Q/A k is a simple divisible
Thus
Rmodule,
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 Rmodule
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 Rmodule 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 semilocal 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 79.
[19, Th. 5]. There is a onetoone 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 Hmodule 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/Pmodule,
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 onetoone 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 1dimensional,
local, Cohen
ring.
Lemma 8.1.
l_~f T I and T 2 are equivalent
torsion Rmodules,
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 Hmodules
Hhomomorphisms. 
we have Repimorphisms
It follows
immediately
and f and g are
from this that AnnHT 1
AnnHT 2. Definition.
class.
Let T be a torsion Rmodule,
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
Rmodul~,
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 Rhom_~oThen:
determined
by
class of D.
D is Isomorphic t__~oQ/C, and C is isomorphic
to a regular
ideal of A. (3) Rmodule
If I is a regula r ideal of A, then Q/I is a simple divisible equivalent
Rmodule 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 Amodule; sion Rmodule.
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 Amodule,
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 onetoone
classes
the set o__ffclosed components
correspondence
[D] of simple divisible
between
Rmodules,
A o_~f R i_.nnQ given by Q / A ' v D
and
and
A/R = h(Ker f), where f i_~s any Rhomomorphism o f K onto D. (2)
There is also a one rtoone 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,
Rmodule.
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
Rmodules
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 Rmodule we shall let
G([D]) denote the set of isomorphism classes of Rmodules D' equivalent to D. If A is any ring we shall let G(A) denote the semigroup
of
isomorphism classes of regular ideals of A. Theorem 8,~. P ~ AnnH(D), o_~f R.
Let D be a simple divisible Rmodule,
let
and let H = H/P, the corresponding analytic cpmponent
Then there is a onetoone 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 Rmodule equivalent t_2oD. Proof.
Let D' be an Rmodule 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 Rtopology,
and
hence by Theorem 3.2 to the completion of I in the Atopology.
Now
is the completion of A by Theorem 73, 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 Rmodule,
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 Rmodule
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 Rmodule should be something special.
That this is indeed the case is shown by the following
theorem. Theorem 8,~.
Let D be a simple divisible Rmodule 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 Rmodule 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 Rmodule, 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 Hmodule, 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 Rmodule 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 Rmodules.
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 Rmodul_.~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 Rmodule 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 Rmodule, 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 Rmodule, we have
HomR(D,E) ~ O.
Let f be a nonzero element of HomR(D,E).
Then Im f
84 iS an Artinian divisible Rmodule and HOmR(Im f, K) c HomR(D,K ). Thus we can assume that D is an Artinian divisible Rmodule.
There
is a submodule C of D such that D/C is a simple divisible Rmodule. Since HomR(D/C,K) c HomR(D,K ) we can assume that D is a simple divisible Rmodule.
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 SEMISIMPLE
Throughout Macaulay
this
semisimple
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
Rmodule
Rmodules.
see in the next
Rmodule
semlslmple
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
Rmodule.
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
Rmodule. sum of s i m p l e
divisible
Rmod
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
Rmodule.
... @ B n. (2) m >
Since
n F~ L ( B i)  L(B) ,, n  n = 0, and il 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
Rmodules, 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 = fl(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
Rmodules. Remarks. divisible
Rmodule
the integral remarks
We shall prove
the converse
Corollary
9.2.
5.14).
simple divisible are nonzero
Rmodule.
of divisible Lemma
A is a submodule Artinian
divisible
if and only
Rmodule.
if
(See the
is true for
images
Artinian
submodule
R
of B.
Rmodules.
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
Rmodule
The same a s s e r t i o n
r o ~
every nonzero
homomorphic
Definition.
every Artinian
Let B b__ee~ semislmple , 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 semisimple
closure
following
in a later theorem
from semlsimplicity
Rmodule
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
Rmodule.
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 Rmodule
one and only one uniserial,
equivalent
divisible,
that
an Rhomo~*.
the theorem. Thus the
is redundant. to E.
Artinian
Then B con
Rmodule
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,
Rmodule
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.
Rmodule,
D2, and E are Hmodules.
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 Replmorphlsm
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), 828. (2)
N. Bourbaki,
"Elements de
[email protected], 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.
[email protected] and A. Grothendieck,
[email protected] I", SpringerVerlag,
"Elements de
[email protected] Berlin,Heidelberg,
New York
(1971) (5)
D. Ferrand and M. Raynaud,
[email protected]",
"Fibres Formelles d'un Local
Ann. Scient. Ec. Norm. Sup. 4 e Serie, t.3,
(1970), 295311. (6)
R. Hamsher,
"On the structure of a one dimensional quotient
field", J. of Algebra, 19 (1971), 416425. (7)
J. Herzog and E. Kunz, MacaulayRings",
"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), 649685. (9)
E. Matlis,
"Injective modules over Noetherian rings", Pacific
J. Math., 8 (1958), 511528. (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), 569586. (12) E. Matlis,
"Cotorsion modules", Memoirs Amer. Math. Soc., No. 49,
(1964). (13) E. Matlis,
"Decomposable modules",
(1966), 147179.
Trans. Amer. Math. Soc., 125
154
(14) E. Matlis,
"Reflexive Domains", J. Algebra, 8 (1968), 133.
(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 Qrings",
(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 onedimensional local rings"
Proc. Glasgow Math. Assoc., 2 (1956), 159169. (20) D. G. Northcott,
"On the notion of a first neighborhood ring",
Proc. Camb. Phil. Soc., 53 (1957), 4356. (21) D. G. Northcott,
"The theory of onedimensional rings", Proc.
London Math. Soc., 8 (1958), 388415. (22) D. G. Northcott,
"The reduction number of a onedimensional
local ring", Mathematic&, (23) D. G. Northcott,
6 (1959), 8790.
"An algebraic relation connected with the
theory of curves'[, J. London Math. Soc., 34 (1959), 195204. (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)
hdivisible
8
hreduced
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)
1dimensional
4O
(42)
Pprimary component
104
(43)
Pprimary divisible module
102
(44)
pseudo valuation ring
(45)
Q, the full ring of quotients
(46)
quasilocal
(47)
Rtopology
(48)
reduced module
(49)
reduction number
(50)
regular element
(51)
regular ideal
17
(52)
semilocal ring
24
(53)
semisimple 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)
torsionfree module
7
7
the classical length of A
CohenMacaulay. 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 1dimensional,
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 CoroliI lary 8.4 Q / V i is a simple divisible Rmodule. 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
Rmodule.
(3)
L(K) = n, where n i__ssthe number
of distinct
t.ion rin~s o~f R. (4)
K is a semisimple
divisible
Rmodule.
(5)
E i_~s ! semisimple
divisible
Rmodule.
pseudo
valua
91
(6)
Every nonzero Artinian divisible
(7)
I_~f C and D are Artinlan divisible
are equivalent Proof.
R_~modul__~ei_~ssemislmpl e. Rmodules,
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 Rmodule. (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 semisimple finitely generated 5.8;
Rmodule,
divisible
by Corollary 8.3, Rmodule.
then K is equivalent
and hence K is semisimple (4) =.=> (5).
Rmodule
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 semisimple (5) ="> (6).
divisible
A nonzero Artinian divisible
image of K m for
Rmodule.
Rmodule
is isomor
phic to a submodule of E m for some m > O, by Theorem 5.5; it is a semisimple (6) =t> (7).
divisible
5.14.
Rmodule
by Corollary
and hence
9.2.
Let C and D be Artinian divisible Rmodules.
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 =
[email protected]~Dm, and hence K and D
Therefore K is a semisimple divisible
module by Theorem 9.1. (4) ==> (2). ule. D.
Suppose that S is not a finitely generated Rmod
Then by Theorem 5.1, S/R contains a simple divisible Rmodule 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 semisimple divisible module,
and thus we see that S is a finitely generated Rmodule. 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 semisimple E is equivalent Rmodules
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
Rmodule.
rin~ if inJ. dim.
that R i_ss~ Gorenstein Then K contains
Rmodule
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 Rmodule
valuation
rin[.
for every Rsubmodule
C
of q.
is an indeeomposable
local domain of Krull
Rmodule
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 onetoone
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 Rsubmodules
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
Rsubmodule
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 1dimensional
By Theorem 7.4,
the valuation
If the integral
can assume
sub
para
that D 1 = K.
Q/C is an indecomposable
Rmod
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 1dim
IIC and I2C.
Q/C ~ Q/ CI1 @ Q/CI2 , a nontrivlal
This contradiction
1dimensional
Q/I is indecomposable
that R is not a 1dimenslonal
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];
Rmodule,
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
nonunlts
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.
Rmodule
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 Rmodule,
11.5.
Let D be a nonzero
where P is a prime
Pprimary
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 Rmodule,
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 Pprimary m o d u l e
if and only if B and D are Pprimary
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
Pprimary 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 Pprimary
module.
if D i~s an Artinian,
divisible
Pprimary
divisible
submodules
of D, then B + C is also a Pprimary
divisible
submodule
of D.
(3)
If D is an Artinian,
unique largest
Pprimary
divisible
Pprimary
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
Pprimary
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.
Rmodule,
submodule
I f B and B' are Pprimary
Rmodules
follows
divisible
Rmodule,
R
(2)
(2) and
and suppose
that f
98
is not zero. ll.3.
Then f(B) is both Pprimary 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 Rmodule and P a
We shall call the unique largest Ppri
submodule of D that is guaranteed to exist by Corollary
ll.4 the Pprimary component of D.
We shall denote this component
by Dp. Theorem ll.5.
Let D be an Artinian,
divisible Rmodule,
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 Rmodule. 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 Plprimary module.
We
have an exact sequence: 0 + C/A ~ B/A ~ B/C ~ 0 where the extremes are Plprimary modules.
Hence by Corollary ll.3,
B/A is a Plprimary module. Let I = AnnHA; primary ideal, that I ~ Pl"
then
Q AnnHDPi c I. Since AnnHDPi is a Pii>2 (or is equalto H in the case that DPi = O) we see
Choose h ~ I such that h ~ Pl"
Because B is an Hmod
ule by Theorem 2.7, h induces an Rhomomorphism f : B / A  ~ B by f(x + A) = hx for all x ¢ B.
defined
We assert that Ker f is a reduced
Rmodule. Suppose that Ker f is not reduced; divisible Rmodule F/A.
then Ker f contains a simple
Since B/A is Plprimary 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 Plprimary 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 Plprimary component. DP1 0 (DP2 + ... + DPn) is a reduced Rmodule.
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 Rmodule is
Pprimary 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 Rmodule,
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 Rmodule,
given simple divisible Rmodule 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 Piprimary 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 Rmodule,
ensional, main,
semilocal,
Cohen Macaulay
ring.
and thus U is a ldi 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 Piprimary
to the Piprimary
K D R (j~iN Nj) = h(jji n Bj/R)
component
of Q/U an__~d
component of K. is the Piprimary
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 Niprimary (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 Aimodule 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.
1dimensional
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
CohenMacaulay 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
~(211/Mkl)
= e, and h e n c e S ( R / 2
II)
k = ~(R/M
I)
dicting
. ~(MklI/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~ MI.
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
semilocal,
contains
R is a local to R.
ring,
Thus w e
valuation
then M M I = M;
1dimensional
Mprlmary
ring.
see If R
and thus M I is a
CohenMacaulay
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 M1A = 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 + (nl)e by
Therefore en  p = 1 + (nl)e = en  (el),
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 Mprimary
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
Pprimary 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 Mprimary,
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 1dimensional 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 Rmodule 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 Rmodules.
(3)
K i_~san injective Rmodule.
(4)
K ~ E(R/M), the injective envelope of R/M.
(5)
MI/R % R/M.
(6)
M I can be generated by two elements.
(7)
I = I II for every regular ideal of R.
(8) (9)
~(I/J)
= ~(Ji/ll)
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 Rmodules;
posable injective Qmodules; 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 Rmodule, then Q ~R C is Q ~R ~ i i" an injective Qmodule. 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 Qmodule.
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 Qmodule.
For then
by the preceding equation Q would be an injective Rmodule.
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 Smodule. 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 Smodule A of finite length.
ideal of S has finite length,
Because every
we have shown that S is an injective
Smodule. (2) > (3).
This is a trivial assertion.
(3) > (4).
It is easy to see that MI/R ~ 0 and that K is an
essential extension of MI/R.
Since MI/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
Rmodule by Theorem 4.7, and hence K is isomorphic (4) ~ > Rmodule.
(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 Rmodule.
Using the abbreviation w.dim,
for weak
dimension, we observe that w.dimRE = 1 since E % K, and Q is a flat Rmodule.
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 Rmodule.
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 MI/R % R/M.
(5) ~ >
~(MI/M)
Since M1/R = {x ~ K
This is an obvious assertion. If MIM = R, then M I is cyclic and hence
Hence we can assume that MIM = M.
= S ( M  ! / M M I) = 2.
But then
Therefore ~ ( M  ! / R )
= ~(MI~)
= I, and we see that MI/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 = II/R.
Another consequence is
that every Rhomomorphism from I into Q can be extended to an Rhomomorphism 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 Mprimary 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 II.
Because I 0 is a principal ideal, we have
We will assume that lj = I] II 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 MI/R " R/M.
 Ml/R,
and by assump
Thus the exact sequence can be written in the form: 0
~
Ilj+l~ I]1 ! R/M
where ~ is the inclusion map. Ij+ 1 c ilI 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~ II X~ ll'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[ II = I we have proved 0 J' Now Ij+i/I j is a nonzero submodule of
II : ilI 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 Mpriseries of
ideals of R: J = i 0 c I ! c ... c I n = I such that lj+I/I j  R/M.
iI : and I j+i 1 ~
!]i.
Then we have a series:
Inl m Inll c ... c Iol = jl,
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 eI : AA "I = A I = M eI
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 (el)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,a2bq2,...,aib qi 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, a2bql,...,aJb q+lj 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 qk E (bq,abql,...,aklbqk+l)
Multiplying
+ M q+l.
(I) by a and then by b we obtain:
(2)
ak+Ib qk E (abq,a2bql,...,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 qk and akb qk+l are
in (bq+l,ab q,...,aklb qk+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 (el) 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 eI and A : (Mel) 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(el)
y I
(el)e
= ~ (el)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 eI 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,
(Mel)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 = Mnla. sumption.
(I).
superficial
of d e g r e e
H e n c e b n ¢ M q a c Mnla,
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 1dimensional,
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.
Pprimary
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 torsionfree generated ideal
But this follows
Hmodule,
them.
ideals
the Pprimary
and there are
Since Hp is a Noetherian that HpJ/HpI field
is a simple
of H/P.
J/I c HpJ/HpI.
H/Pmodule 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 Pprimary
Then
phisms,
is a finitely
length by
of H/P.
ring with maximal
Hpmodule.
then L(A)
of earlier chapters
Clearly HpI c HpJ are PHpprimary
no PHpprimary
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
Rmodule,
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 socalled
that is, a chain of Pprimary
"composition
series"
for N;
ideals:
C
C
C
N = PI ~ P2 ~ "'" ,J Pk = m such that there are no Pprimary 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 JordanHolder
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 Piprimary.
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))
Rmodule
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.
CohenMacaulay
by e T and its reduction
length
num
that there is an integer
generated
torsion Rmodule
ideal
of B, then
and ideal of R.
n > 0 such that
ideal of B. length
torsion Bmodule
of R in Q.
of B and BM con
(hence Mprimary)
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 BMpr[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 1dimensional,
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 quasilocal
ring by Theorem 3.6.
Clearly we have Q / B ~ Q/B.
Let H be the completion of B in the Btopology. = HomB(Q/B,Q/B)
~ Hom~(Q/B,Q/~),
of B in the Btopology. 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,
ldimensional
Cohen
If B is a closed component of R, then by Theorem 7.2, 1dimensional,
analytically
irreducible,
CohenMacaulay
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 quasilocal
ring B.
The maximal
ideal of H is HM, and since HI is a regular ideal of H, it follows that HI is an HMprimary 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 Rsubmodule
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 MI/R;
an Artinian Rmodule.
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
Rsubmodule
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,
CohenMacaulay
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(e1) ,
1
PB = ~ eB(eBl)'
and PA = ~ eA(eAl)" 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(eBl)
are
ldimensional,
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(uil)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(el) 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 (uiui) 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 Piprimary 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 Amodule. 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 Smodule by k(S). local, CohenMacaulay
If R is a 1dimensional,
ring and p is a prime ideal of rank 0 in R, we
shall let ~(p) denote the length of a composition series of pprimary ideals between p and the pprimary 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
1dimenslonal,
We shall then use
and sufficient
conditions
ideal of R.
There is a onetoone,
pondence between the set of n o n z e r o
orderreversing
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,
Rmodule.
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 onetoone,
order
because of the duality between the ideals of
of E that is given in Theorem 4.5. There is a onetoone,
orderreversing
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/Imodule.
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.
Bmodule,
is an H/Imodule
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 semisimple
full ring of quotients t(EB(B/MB))
submodule
unramified
t(EB(B/MB))
orderreversing
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 onetoone
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
orderreversing
to the torsion
is the injective
elements
submodules
This fact combined
divisible
then D is isomorphic
~(B/MB)
orderpreserving
proof of the corollary.
There is a onetoone,
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 onetoone,
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
Rmodules.
Proof.
Let P be a prime ideal
be the Pprimary 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 Pprimary series
that L(Ep)
of P, then by Theorem
Hence it is sufficient
composition
of rank 0 in H, and let ~
component
of Pprimary
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 Pprimary
ible Rmodule.
Hence we have D u c ~ .
On the other hand, tion,
AnnHE P is a Pprimary
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 Rmodule.
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 Rmodule
There is a canonical = f(x)
It follows
divis
Rhomomorphism
~ : A*A**
for all x ~ A*.
If A is an Rsubmodule
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 Rmodule, 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 Rmodule 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 Rhomomorphism 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 Rmodule.
(6)
Q and Q/I are inJective Rmodules.
(7)
inj.dim.Rl = I.
Proof.
(I) ~> (2).
Let L be an Rmodule 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 Rmodule,
(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 torsionfree 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 Rmodule 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) ~ > Rmodule.
(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 Rmodule. an essential extension of its socle. MI/I, and hence by (2) is isomorphic
Every A r t i n ~ n Rmodule 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 qblI 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 torsionfree;
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 Rmodule.
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 Rmodule. it follows from Theorem 4.6, of copies of E.
sum can have only one term, (4) = >
(7).
Since Q/I is an Artinian Rmodule,
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 Rmodule.
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 Rmodule. preceding
to a direct
= L(K) = L(E), we see that this direct
Since Q/I is a torsion Rmodule,
K ~R H°mR(K'H°mR(I'E))
(jl)l = j.
But I is a reguAs in the
that Q is an injective
Rmodule.
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 selfinjective if and only if it is an inJective Rmodule.
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 Pprimary
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 Rhomomorphism o_~fK onto E.
(7)
K contains only one simpl 9 divisible Rmodule 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 Rhomomorphism of Q/Rb onto Q/I, and
hence an Rhomomorphism of K onto E. an Rhomomorphism 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 Rmodule 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 Rhomomorphism (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 Rmodule 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 Pprimary 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 Rhomomorphism f of Kp into E. ll.3, Im f c Ep, the Pprimary 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 Piprimary 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
Plprimary
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 selfinjeetive.
is the full ring of quotients
is selfinjective 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 selfinjeetive if and only if 0 is an irreducible ideal of S. If S is selfinjective, 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
Pprimary 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 Smodule 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 selfln
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 Rsubmodule
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
Rmodule,
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 Hhomomorphism, 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 Rmodule
~(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.