Localization
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IV
LOCALIZATIONS Definitions:
A commutative
n < ~ if e v e r y chain of p r i m e and there is at l e a s t
ring is said to have K r u l l
ideals
Noetherian
ring of K r u l l
R contains
a regular
mary component
ring of K r u l l
I x = O, or AnnR(x)
AnnR(x ) = [r¢
let T be a t o r s i o n
R
I rx = 0].
w h e r e M S ranges
R-module.
Mi-primary.
decomposition
can w r i t e
1 = rI +
x = rlx +
... + rnX.
hence
ideal
from
closed.
I, and let T be a
of R, d e f i n e
the M a - p r i -
is an M a - p r i m a r y
Clearly
Since
Therefore,
sum of the Ta's.
T a is a s u b m o d u l e
Cohen-Macaulay
of T.
ring and
Ta
element
of I~
(k ~ i),
... + rn, w h e r e
is an M i - p r i m a r y
ideal]
ideals
of R.
Furthermore,
is an RMjmOdUle
If L i = A Jk'
i = 1,...,n.
differs
Then
Let x be a n o n z e r o
Take a p r i m a r y
ideal of
of K r u l l d i m e n s i o n
Such a d o m a i n
dimension
over all of the m a x i m a l
and hence
Proof.
domain
Let R be a 1 - d i m e n s l o n a l
T=Z@
TMj
ring if it is a
that it m a y not be i n t e g r a l l y
If M a is a m a x i m a l
T h e o r e m 4.1.
%
ring.
say
of T by:
T a = [x C T where
We will
i, and if e v e r y m a x i m a l
A Noetherian
ring o n l y in the fact
R-module.
n + I terms.
Cohen-Macaulay
Cohen-Macaulay
Let R be a N o e t h e r i a n torsion
exactly
dimensional
element.
1 is a 1 - d i m e n s i o n a l a Dedekind
in R has at m o s t n + i terms,
one chain w i t h
that a ring R is a 1 - d i m e n s i o n a l
dimension
of T, and let I = AnnR(x).
I = Jl ~
then R = L 1 + r i ~ L i.
ideal,
H e n c e we
Jiri x = 0 for
AnnR(rix ) contains
or rix = O.
... + L n.
Ji is
Thus we h a v e
Jiri c I, we have
either
"'' ~ Jn w h e r e
a power
This p r o v e s
of M i and
that T is the
35
We prove next that the sum is direct. maximal
ideals
of R, and suppose
that x ( T 1 N
integers
M ~k2
kI k2 Since R = M 1 + (M 1
M knx n
= 0.
(j E> 1 Tj).
Then
k1 kl,...,k n such that M 1 x = 0 and
there are positive
...
Let MI,...,M n be distinct
shown that T = Z @ T~. a It is obvious that T a is uniquely
k .. Mnn)
-
we see that x = 0.
,
Hence we have
R - Ma, and hence
(T~)M~ = o.
(Ta)Ma Z T a.
Thus T M ~
Z~
by the elements
On the other hand,
of
if 8 ~ a, we have
(T~)M : ( T ~ ) M ~ T~.
8
Theorem 4.2.
divisible
a
Let R be a 1-dimensional
Cohen-Macaulay
ring.
Then hdRQ = I. Proof.
Case I.
R is a local
Let b be a regular J
= [bn].
Noetherian
rank 0 and is a m a x i m a l
ideals
ideal of R j .
Since R c R j
Let F be a Tree R-module define an R-eplmorphism
Case II.
f : F-*Q
has
every regular
General
that R j
free basis
= i/b n.
= Q. [x n] and
It is then easy
case. h-divisible
R-module.
by Theorem
Ext~(K,B)
an exact
Let us consider
ele-
of f, then P is a free R-module w i t h
that hdRQ = l, it is sufficient
(i)
Hence R~
is equal to its own full
c Q, it follows
by f(Xn)
the
Hence we have hdRQ = I.
Let B be a torsion
= 0.
of R j .
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