NOETHERIAN PROPERTIES OF N0NN0ETHERIAN MODULES
B.A., University of Colorado, Denver, 1992
A thesis submitted to the
Faculty of the Graduate School of the
University of Colorado at Denver
in partial fulfillment
of the requirements for the degree of
Master of Science
Applied Mathematics
by
Michael David Snell
1994
This thesis for the Master of Arts
degree by
Michael David Snell
has been approved for the
Department of
Mathematics
by
Date
Snell, Michael David (M.S., Applied Mathematics)
Noetherian Properties of NonNoetherian Modules
Thesis directed by Professor Sylvia Lu
ABSTRACT
It is well known that submodules of Noetherian modules possess a primary
decomposition. The class of Noetherian modules is contained in the class of Laskerian
modules modules satisfying the property that every proper submodule has a
primary decomposition. The ascending chain condition for residual submodules (accr)
states that for every submodule N and every finitely generated ideal B, the ascending
chain of submodules N: B C N: B2 C terminates. The class of modules satisfying
(accr) contains the class of Laskerian modules, we investigate properties of exact
sequences, localization, the weak ArtinRees property and the Krull intersection
theorem, that can be extended from Noetherian modules to Laskerian modules and
(accr) modules.
This abstract accurately represents the content of the
recommend its publication.
Signed
candidates thesis. I
m
Contents
1. Preliminaries 1
1.1. Radicals and Residuals of Submodules 1
1.2. Modules of Fractions 2
1.3. Primary Submodules 4
1.4. Primary Decomposition 9
2. Noetherian Modules 12
2.1. Preliminaries 12
2.2. Modules of Fractions 16
2.3. Primary Decomposition 17
2.4. Krull Intersection Theorem 18
3. Laskerian Modules 20
3.1. Exact Sequences 20
3.2. Modules of Fractions 22
3.3. Ascending Chains of Residuals 23
4. ACCR Modules 28
4.1. Preliminaries 2 8
4.2. Localization 30
4.3. Krull Intersection 30
4.4. Polynomial Rings 31
References 33
IV
1. Preliminaries
1.1. Radicals and Residuals of Submodules.
Definition 1. For a submodule N of the Amodule M, the radical of N in M is the
ideal of A
vm(N) = {a G A  akM C jV,for some nonnegative integer 7}.
It follows that when we view A as an Amodule and I is an ideal of A,
vA(I) = {a E A \ an I for some nonnegative integer n}. When refering to the
radical of an ideal I, we will use the notation yfl instead of t(7).
Proposition 1.1. If Ny, ...,Nk are submodules of the Amodule M, then
tM(n*=1iVi) = n?=1t M(Ni).
Proof. For 7 = 2, let N\ and N2 be submodules of M. Suppose a is in the radical of
Ni and the radical of N2 and let x be an element of M. Then there is a nonnegative
integer j for which a?x G Ni and a*x G N2, so a is in the radical of the intersection of
Ni and N2. Conversely, if a is in the radical of the intersection, then given an x in M,
a^x is in Nx and N2 for some nonnegative integer j. Thus a is in the radical of Ni
and the radical of N2. The general case follows by induction.
The proposition fails to hold for infinite intersections as the following example shows.
Example. Consider the ring of integers Z, and let p be a prime ideal. For any positive
integer 7, we have = p, so that f)t jVp^ = p. But the infinite intersection of the
p* is (0), and the radical of (0) is (0) because Z is an integral domain.
Given two submodules N and K we can form the ideal
N: K = {a G A  aK C N}.
Similarly, given a submodule N of M and an ideal I of A we can define the residual
submodule
N: I = {m Â£ M \ ml C N}.
1
Since N: M is an ideal of A we can look at its radical. An element a 6 A is in
VNTM if and only if there is a nonnegative integer k such that akM C N. In other
words,
VN: M = tM(N).
The annihilates of an element x in the Amodule M is the set of all a Â£ A for
which ax = 0. Given a submodule N of M, the annihilator of N is the set of a Â£ A
for which ax = 0 for all x Â£ N. The annihilator of an element or of a submodule is an
ideal of A denoted by Ann x or Ann N respectively. Given an ideal I of A, a prime
ideal p containing I is a minimal prime of I if p contains no other prime ideal
containing I.
Definition 2. A prime ideal p of A is associated with the Amodule M, written
p Â£ Assa(M), if it is a minimal prime of Anna; for some x Â£ M.
1.2. Modules of Fractions.
The rational numbers are formed from the integers by an equivalence relation on
fractions. This construction can be carried out in an arbitrary ring or module by a
slight generalization.
A multiplicatively closed subset 5 of a ring A is a set containing 1 that is
closed under multiplication. We will assume that S contains no zero divisors so that
zero is not in S.
Given an Amodule M we can form the set of fractions 51M as the set of all
with m Â£ M and s Â£ S. Let ~ be a relation on S~lM defined by
771 71
~ <#> there exists s' Â£ S such that s'{mt ns) 0.
s i
Then ~ is an equivalence relation and the equivalence classes form an Amodule when
we define addition by
m n mt + ns
s t st
and multiplication by elements of A as
to am
s s
2
This module S1M is called the module of fractions.
If we let A M in the above construction we get a ring S'1 A called the ring of
fractions. Note that A is mapped cannonically into 51A by a i> j.
We can make S~1M into an .S'1 A module by defining
a m am
s t st
An element m E M is mapped cannonically into S~1M by m k . If we extend the
submodule A under the cannonical homomorphism M S~1M, we get the S1M
submodule S~1N. Then the preimage of S~1N, which contains A, is a submodule of
M called the saturation of A in M with respect to S and is denoted 65(A). The
extension of the saturation of A is again S~1N. Another way to describe the
saturation of A in if is
65(A) = {* E M  sx E A for some s in 5}.
Thus the kernel of the cannonical homomorphism is the saturation of (0) in if.
Commonly used multiplicatively closed subsets are A p for a prime ideal p and
the set {a*}Â£i0 for some a not a divisor of zero. Also, the intersection of two
multiplicatively closed subsets is again a multiplicatively closed subset.
Example. In the ring Z of integers, let S = {2fc}Â£L0. Then 51Z is the set of fractions
n/2k. For the ideal (3223) = (72), the saturation of (72) is the ideal (32) = (9).
Proposition 1.2. If S is a multiplicatively closed subset of A that does not meet the
ideal a, there exists a prime ideal containing 0 and not meeting S.
Proof. See [2, page 71, corollary 2]
Corollary 1.1. Every proper ideal of a ring is contained in a maximal ideal.
Proof. Let A be a ring and I a proper ideal. Then I contains no units of A and the
set S of units of A is a multiplicatively closed subset of A. By proposition 1.2 there is
a prime ideal containing I and not meeting S. Let J be a maximal element of the
prime ideals containing I, then J is maximal in A and J contains I.
3
1.3. Primary Submodules.
Definition 3. A proper submodule N of M is primary if for a Â£ A and x Â£ M,
ax G N implies that either x G N or a Â£ xM(N) y/N: M.
Example. The ideal (pn) in Z, generated by pn for prime p, is primary. First note
that \/(pn) = (p). Then rx Â£ (pn) means rx = pnt for some integer t. Then if
x (J: (pn), x has fewer than n factors of p, so that r must have at least one factor of p.
Thus r Â£ (p) = y/(pn).
Proposition 1.3. If the submodule N of the Amodule M is primary, then rm(N) is
a prime ideal of A.
Proof. Suppose ab Â£ tm(N) while a ^ tm(N)> for a,b Â£ A. Then ab Â£ tm(N) implies
that (ab)kM C N for some positive integer k. Since a Â£ rm(N), there is some x Â£ M
for which akx N. Then we have bk{akx) Â£ N which implies that bk Â£ rM{N) as N is
primary. But bk Â£ tm(N) implies b Â£ Zm(N), so Xm(N) is prime.
When N is primary in M and p = Xm(N) is a prime ideal of A, we say that N is
pprimary in M, or that the prime ideal p belongs to N. Note that the prime ideal p
can belong to more than one primary module. In the example above, the ideal (p)
belongs to every ideal generated by a power of p.
Example. Consider the integers modulo the ideal generated by pn, Z/(p"), as a
Zmodule. The nonzero element pn~l + (p") of Z/(p") is annihilated by everything
that is a multiple of p, so (p) = Annp"1 + (pn) is in Assz(Z/(p")). On the other
hand, the annihilator of every nonzero element of Z/(p") is some power of p, and (p)
is the only minimal prime ideal containing the powers of p, hence (p) is the only
element in Assz(Z/(pn)).
In this last example, the set of primes associated with Z/(pn) consists of only one
ideal. The reason for this is the fact that (p") is (p)primary in Z, as we will now
show.
4
Proposition 1.4. A submodule A of the Amodule M is pprimary in M if and only
if Assa(M/N) = {p}.
Proof. (=>) Suppose N is pprimary in M. Let x be an nonzero element of M/N and
let a: be an element of M which maps to x under the cannonical homomorphism.
Then a Â£ Anns implies ax is in A. Now x is not in A as x is nonzero in M/N, so the
fact that N is primary in M says that a is in tM(A) = p. Thus p contains the
annihilator of each nonzero element of M/N.
Now suppose that q is any prime ideal in Ass {M/N) and let y be a nonzero
element of M/N for which q is a minimal prime containing Ann y. From the above
result, q C p. Let a be in p = rm(N). Then if y maps to y, we have any Â£ N for some
n > 0. It follows that an E Ann y which is contained in q. So a E q as this ideal is
prime. Thus p = q is the only element of Ass (M/N).
(t=) Suppose that Ass (M/N) = {p}. Let ax Â£ N, a E A,x Â£ N. Then a is an
element of Anna: and hence an element of p. Now for any nonzero y in M/N, p is the
only minimal prime containing Ann?/, so a is nilpotent in A/ Anny. Thus any is in A
for some n > 0. As y was arbitrary it follows that a E tM(lV) and that N is primary
in M.
Corollary 1.2. Let : M > M' be an epimorphism of Amodules and p a prime
ideal of A. If N is a submodule of M containing the kernel of and N' the
corresponding submodule of M', then N is pprimary in M if and only if N' is
pprimary in M'.
Proof. We have that M/N is isomorphic to M'/N' and so Ass (M/N) = {p} if and
only if Ass (M'/N') = {p}.
Proposition 1.5. If N\,..., N/. are pprimary submodules of M, then their
intersection nf=1 A, is pprimary in M.
Proof. By proposition 1.1, tjw(nf=1 At) = p. Suppose that, for r Â£ A and x E M,
rx E nf=1Ni while x (/ flf=1 A,. Then x is not in Nj for some j. But Nj is pprimary
and rx E Nj, so we must have r Â£ p = tM(ntL1A,).
5
When considering the cannonical homomorphism from a module into its module
of fractions for some multiplicatively closed subset we cannot use corollary 1.2
because the map is not a epimorphism. We will show however that if the image of a
submodule is primary in the module of fractions, then the saturation of the original
module is primary. This gives us a way to enlarge the original module to a primary
module containing it which will play a role in expressing a module as an intersection
of primary modules.
Proposition 1.6. Let S be a multiplicatively closed subset of A. If N' is an
S,1pprimary submodule of the 51A module S~1M, then 4>~1(N') = N is a
pprim ary submodule of M, where : M * S~1M is the cannonical homomorphism.
Proof. We will show first that Xs1m(N') = S
a x
Suppose is in S~1vM(N) and let be an element of S~1M which is not in N'.
s t
Then x Â£ N, but akx E N for some nonnegative k as a E tThus
= impUes 7 G 's'm(N').
Conversely, suppose is in Vsim(N') and let x be an element of M which is not
s
in N. Then
/a\k tx takx . k *r
U) T = WgN ,mi,lies axeN
Thus a is in vM(N) and finally is in 51rAf(lV). This proves equality.
&
SCLX
Now to show that N is primary in M, let ax E N, x ^ N. Thenis in N'
s
sx a a
while is not, so is in Xsim(N') because N' is primary. By the first result, is
s 1 1
in S~1Vm(IV) and therefore a is in tm(N) and N is primary in M.
Proposition 1.7. Let S be a multiplicatively closed subset of the ring A and let M
be an ^.module. Then Ass5m(5'1M) = {5,_1p: p E Assa(M) andp fl S = 0}.
Proof. If p is a prime ideal of A not meeting S and if p contains some ideal I of A,
then p contains the saturation of I with respect to S. This follows from the primality
of p, for if sa E I for some s E S, then sa E p while s Â£ p, so a must be in p.
6
Let p E Assa(M) not meet S. Suppose p D Anna; minimally, then
sx
p D s(Anna;) minimally and therefore 5,_1p is a minimal prime containing Ann .
So 51p is in Asssi a(S~1M).
X
Now suppose iS:p is a minimal prime containing Ann Then by taking inverse
s
images under the cannonical mapping we see that p D s(Annai) D Anna;. Suppose
we have another prime ideal q satisfying p D q D Anna;. Then 51q contains Ann f
and the minimality of 51p shows that q = p. Hence p is a minimal prime of Anna;
and p is in Assa(M).
Corollary 1.3. Given a multiplicatively closed subset S of A, there is a one to one
correspondence between the pprimary submodules N of M where p does not meet S,
and the S1 pprimary submodules S_1N of 51A.
Proof. By proposition 1.6 we have that the preimage of any 5,1pprimary submodule
of S~XM is pprimary in M. Now by proposition 1.7 we see that if N is pprimary in
M and p does not meet S, then Assa(M/N) = {p} and
Ass5iA(51M/1V) = {51pp G Assa(M/N) and p n S = 0} = {S^p}.
So S_1N. is 51pprimary in S~1M.
This corollary gives us a way to focus attention on some subset of the prime
ideals of A by using the multiplicatively closed subset of elements not in any of the
prime ideals under consideration. In particular we will be interested in primes
associated with some submodule, so the next few results show; how these primes are
related and what happens to them in the process of localization.
Lemma 1.1. Let N be a submodule of the Amodule M. Then
Assa(N) C Assa(M) C Assa(M/N) U Assa(N).
Proof. The first inclusion is obvious. Let p E Ass^(M) and suppose p is not in
Assa(N). Let x be an element of M such that p is a minimal prime containing Anna;.
Then as p contains Annaa; D Anna; for all a not in p, it is clear that ax Â£ N for all a
not in p, for otherwise p would be in Ass^A). But this means that p contains
7
TV: x = Ann a; in M/N. The series of inclusions Ann x C TV: x C p shows that p is in
Ass a(M/N).
Lemma 1.2. Let p be a minimal prime of the ideal a and let S = A p. Then for
any a Â£ p, there exists an s Â£ S and an integer k > 0 such that sak Â£ a.
Proof. Suppose otherwise. Then for all k > 0 and all s Â£ S, sak ^ a. Form the
multiplicative set S' = S U {a*} which meets p but not a. By proposition 1.2, there
exists a prime ideal containing o which does not meet S'. But then this prime ideal is
strictly contained in p which is a contradiction.
It was pointed out before that the kernel of the localization map M S~1M is
the saturation of (0) in M with respect to S. The following proposition shows how
the primes associated with a module M can be split into two sets, those associated
with the kernel and those associated with the factor module of M mod the kernel.
Proposition 1.8. Let TV be the kernel of the cannonical homomorphism
M S~1M for some multiplicatively closed subset S of A.
(1) Assa(N) = {p G Assa(M): p n 5/0}
(2) Assa(M/N) = {p Â£ Assa(M): p fl S = 0}
Hence Assa{M) = AssA(M/N) U Ass^(lV).
Proof. If p Â£ Assa(M) does not meet S, then for any x Â£ N we have Ann x fl S ^ 0
and p Anna;. Thus p is not in Ass^(iV) and must therfore be in AssA(M/N) by
lemma 1.1.
If p does meet S and if t Â£ p fl S, then tksx = 0 for some k > 0 by lemma 1.2,
where p is a minimal prime containg Anna;. It follows that a; is in TV and p is in
Ass^(TV).
Suppose Now that p is in AssJi(TV) and let p be a minimal prime of Ann y for
some y Â£ TV. As sy = 0 for some s Â£ S, p meets S. The fact that p is in Assa(M)
follows from lemma 1.1.
Finally suppose p is in AssA(M/N) and let x be an element of M/N for which p
is minimal containing Annx. It is sufficient to show that p is in Assa(M). Let x be
8
an element of M which maps to x and suppose p is not a minimal prime containing
Anna;. Let q be a prime ideal satisfying Anna: C q C p. Then also q does not meet S
and so q contains Anna:. But p is a minimal prime containing Anns, so we must have
p = q and we are done.
Corollary 1.4. If p is a minimal element of the set of primes in Assa(M) for some
Amodule M, the saturation of (0) C M with respect to p in M is pprimary in M.
Proof. If p is minimal in Assa(M), every other prime ideal of this set meets
S = A p. Thus by proposition 1.8, AssA(M/N) = {p} and by proposition 1.4 we see
that N is pprimary in M.
1.4. Primary Decomposition.
Definition 4. A submodule N of the Amodule M is said to possesses a primary
decomposition in M if it can be represented as the intersection of finitely many
primary submodules.
Example. In the ring of polynomials in one indeterminant over the rationals, Q[x], let
g be any polynomial. Write g as a product of irreducible polynomials
g = /i1/!2... frr The ideal generated by g is the intersection of the ideals generated
by the each of which is primary.
Given a primary decomposition of a submodule N, say N = Ni fl fl Nr, we
can remove any A, that is contained in some other Nj. After removing all such
superfluous submodules, the remaining decomposition is called irredundant. Now if
we have an irredundant decomposition and a prime ideal belongs to more than one
submodule in the decomposition we may, by proposition 1.5, replace the A< that
belong to the prime ideal with their intersection. After doing this, a prime ideal
belongs to at most one component of the decomposition. An irredundant
decompositon for which the primes belonging to the components are distinct is called
a normal decomposition. Every module with a primary decomposition has a
normal decomposition.
9
Proposition 1.9. Let JV = JVX fl D JV* be a primary decompositon of JV C M,
where each JV, is p,primary. Let S be a multiplicatively closed subset of A and
suppose the submodules are ordered so that pi,... ,pm do not meet S while the rest
do. Then JVX D fl Nm = s(JV), the saturation of JV.
h'
Proof. Let x G JVi fl fl JVm. For j = m + 1, let Sj3 be an element of pj n S
satisfying s*3x G JV. Put s = IIi=m+i s]3 Then sx G JV and so x G s(JV).
Now suppose y G s(JV). Then there exists an s' in S such that s'y G JV. For
i = 1,..., m we have that s'y G JV, while s' ^ p,. Since is p2primary, y is in Nt for
all i between 1 and m. This proves that &S(N) = Ni D fl Nm.
Proposition 1.10. Let M be an Amodule and let JV be a submodule of M. If N has
a primary decomposition in M, then for an arbitrary multiplicatively closed subset S
of A, the saturation of N in M with respect to S is of the form JV: a for some a G A.
Proof. Let JV = JVa D fl JVr be a normal decomposition of JV where each JV, is
prprimary and let S' be a multiplicatively closed subset of A. As in proposition 1.9,
let the saturation of JV with respect to S' be JVi fl fl Nm. For j = m + 1,..., r, let
sj Â£ pj n S and put s = IIj=m+i sj Then T = {s}0 is a multiplicatively closed
subset of A which meets pm+i,..., pr and does not meet pi,... ,pm. So by proposition
1.9 we have
es(N) = eT(N) = NinnNm.
Recall that we have pj = tM(Nj) = y/Nj: M. As s G p; for j = m + 1,..., r it follows
that skjM C Nj for some kj > 0. If we put k = max{&_,} we have
skM C JVm+i fl D JVr and we see that ^(JV) = JV: sk, for if
x G s(JV) = JVf fl D JVm we have skx G JVm+1 D fl JVr, so x G JV. On the other
hand, if x G JV: sk then skx G JV implies x G JVi fl fl JVm = 5(JV). The
proposition follows from putting a = sk.
Proposition 1.11. Let M be an Amodule and JV a submodule of M. If p is a prime
ideal of A, a (f p such that Q = N: a is pprimary in M, then JV = Q fl (JV + aM).
Furthermore, JV is pprimary in (JV + aM).
10
Proof. The inclusion N C Q n (N + aM) is obvious. Let x = n + am be an element of
the intersection, with n Â£ N and m Â£ M. Then am = x n is in Q as both x and n
are. As Q is pprimary and a Â£ p, we have m Â£ Q which implies that am Â£ N.
Therefore x = n + am is in A.
To show that N is pprimary in (N + aM), suppose rx Â£ N for some r Â£ A and
x Â£ (N + aM) N. Then x has the form n + am for some n Â£ N and m Â£ M, and
rx = rn + ram. By assumption rx is in N, so rx rn is in N implying that
ram Â£ N. Then ram being in N implies that rm is in Q = N: a which is pprimary.
If m were in Q, then x = n + am would be in N, contradicting our assumption, hence
m is not in Q and we must have r Â£ p. This completes the proof.
Proposition 1.12. Let M be an Amodule and N a submodule of M. Suppose there
exists a b Â£ A and an a: Â£ M for which bnx N for all n > 0. If there is an n such
that N:bn = N:bn+1, then N = (N + Abnx) n (N: bn).
Proof. Let y = n + rbnx be an element of N: bn, where n Â£ N and r Â£ A. Then
y n = rbnx is an element of N :bn and therefore rb2nx is in N. So rx is in
N: b2n = N: bn from which we conclude that y Â£ N. The other inclusion is
obvious.
11
2. Noetherian Modules
2.1. Preliminaries.
Given a module, we can partially order the submodules by inclusion. A module
satisfies the ascending chain condition (ACC) on submodules if every ascending chain
of submodules terminates, and it satisfies the maximal condition if every nonempty
subset of submodules has a maximal element.
Proposition 2.1. For a module M, the following conditions are equivalent:
(1) Every submodule of M is finitely generated
(2) M satisfies the ascending chain condition on submodules
(3) M satisfies the maximal condition.
Proof. (1 => 2) Suppose every submodule of M is finitely generated and let
Ni C A2 C ...
be an ascending chain of submodules. The set union A = U^A, is a submodule of
M and must therefore be finitely generated. If {ei,..., ek} generate A, then for each
1 < j < & the generator e, is in some Aq.. Let J = max{*j}, then each e, is in the
submodule Nj. It follows that N = Nj and therefore that the chain terminates.
(2 => 3) Let ft be a nonempty set of submodules of M and suppose ft has no
maximal element. Let Ai be an element of ft. As ft has no maximal element, there
must be a submodule A2 in ft which strictly contains Ai. By repeatedly applying this
argument we can construct an ascending chain of submodules which never terminates,
contradicting condition (2).
(3 => 1) Let A be a submodule of M and let ft be the set of submodules of A
which are finitely generated. The set ft is nonempty as it contains the zero
submodule, so let A' be a maximal element of ft. Let {ex,... ,ek} generate A' and let
e be an element of A. The submodule of A generated by {ei,..., ek, e} contains A'
and is finitely generated, so by the maximality of N1 we see that e is an element of
A'. Hence A' = A and every submodule of M is finitely generated.
12
Definition 5. A module which satisfies any of the equivalent conditions of
proposition 2.1 is called a Noetherian module.
Example. The integers is a Noetherian ring. In fact Z is a principal ideal domain, so
every ideal of a Dedekind domain can be generated by at most two elements, so
Dedekind domains are Noetherian.
Example. The additive group of polynomials of degree less than or equal to n over
the rationals is a module over generated by {1, x,x2,..., xn}. This is a Noetherian
module. Furthermore, Q[X] is a Noetherian ring, but Q[X1?X2,...] is not a
Noetherian ring as the ideal (X1,X2,...) is not finitely generated.
Example. A vector space over a field is a module; the submodules being the vector
subspaces. Any finite dimensional vector space is an example of a Noetherian module,
while infinite dimensional vector spaces are not Noetherian.
Theorem 2.1. (Hilbert Basis Theorem) A is a Noetherian ring if and only if A[X] is
a Noetherian ring.
Proof. (=>) By contradiction, suppose A[X] is not Noetherian, and let I be an ideal of
A[X] which cannot be finitely generated. Choose fi in I of least degree and continue
as follows: if fk has been chosen, select fk+i of least degree in / (/i,... ,/*), where
(/x,...,fk) is the ideal of A[X] generated by the /1?..., fk. Denote by nk and ak the
degree of fk and the leading term of fk respectively. By method of choice,
ni
(ax) C (a1,a2) C..C (au...,ak) C
form a strictly ascending chain of ideals. To verify this, suppose
(ax, ...,ak) = (ox,..., ct*+x) for some k. Then a*+1 can be written as X)i=i riaii where
every ideal is generated by only one element, satisfying condition one. More generally,
Ti (E A. Then
13
is a polynomial of smaller degree than fk which is also in I (/i,... ,/*), contrary to
our assumption. Therefore A is also not Noetherian.
(4=) Conversely, suppose A[X] is Noetherian and let Ik C I2 C I3 C be an
ascending sequence of ideals of A. By hypothesis, the chain E A[X] C I2A[X] C
terminates, so there is an integer n for which InA[X] = In+iA[X], Then
InA[X] fl A = In+iA[X] fl A. Suppose b is in I+1, then b is also in
In+\A[X] fl A InA[X\ fl A and hence also in In. Since we already have In C In+
the chain of ideals terminates and A must be Noetherian.
Lemma 2.1. Let N, N' and E be submodules of an Amodule M. If N C Nr,
N n E = N' n E and N + E = N' + E, then N = N'.
Proof. Let n' be an element of N'. Then n' is in N + E and we can write n' = n + e
for some n G N and e E E. Because n E N C N', n' n is in both N' and E and
therefore in IV D E which is in IV. Thus n' Â£ N and N = N'.
Theorem 2.2. Given an exact sequence of Amodules
M is Noetherian if and only if M' and M" are Noetherian.
Proof. We will verify the ascending chain property for submodules in the following.
(=^) Suppose M satisfies the ascending chain condition for submodules. By the
injectivity of a, any ascending chain of submodules of M' is isomorphic to an
ascending chain of submodules of M and therefore terminates. For M", there is a one
to one inclusion preserving correspondence between the submodules of M" and the
submodules of M which contain a(M'). So any ascending chain of submodules of M"
must also terminate.
(<=) Suppose now that M' and M" satisfy the ascending chain condition for
submodules and let
JVi c n2 c ...
14
be an ascending chain of submodules of M. Then
ot^Ni) C a~x(N2) C and C /?(iV2) C
are ascending chains of submodules of Ikf' and M" respectively. By hypothesis there
is an integer u such that if m > u, a~1(Nm) = a~1(Nu) and 0(Nm) f3(Nu). Using
the fact that
a(a 1(E)) = E fl Im(a) and (5 1(/3(E)) E + Ker(/3)
for any submodule E of M, it follows that a(a1(iVm)) = Nm n a(M') and
a(a1(iVu)) = Nu fl a(M'), so Nm fl a(M') = Nu n a(M'). Furthermore,
t3\P{Nm)) = Nm + a(M') and p\/3(Nu)) = Nu + a(M'), so
Nm + a(M') = Nu + a(M'). Because Nu C Nm, it follows from the previous lemma
that Nm Nu and M is Noetherian.
Corollary 2.1. The Amodules Mt (1 < i < k) are Noetherian if and only if the
direct sum *=1 Mi is a Noetherian Amodule.
Proof. (By induction on k) If k 1 there is nothing to prove. Suppose the corollary
is true for k = m 1. The result follows from theorem 2.2 and the exact sequence
m m
*=1 2=1
for each j.
Theorem 2.3. If A is a Noetherian ring and M is a finitely generated Amodule,
then M is Noetherian.
Proof. Let {ei,..., e*} generate M as an Amodule. Then there exists an
epimorphism
eiA^eiA.
i=l i=l
The Amodule A is Noetherian and the mapping A e^A is an epimorphism of
Amodules for each generator. By theorem 2.2 each e, A is Noetherian and it follows
15
from corollary 2.1 that the direct sum f=1 e,A is Noetherian. Then again by
theorem 2.2, the epimorphism
k k
@eiA^'Â£eiA = M
i1 i=l
shows that M is Noetherian.
In the following theorem we use the fact that M is an Amodule if and only if M
is an A/ Ann(M)module. A proof of this can be found in [10, page 65, proposition 24]
Theorem 2.4. Let M be a finitely generated Amodule. M is Noetherian if and only
if A/ Ann(M) is a Noetherian ring.
Proof. (<Â£=) This follows from theorem 2.3.
(=>) Suppose M is Noetherian. Let {ei,..., ek} generate M and let a* = Ann(e,).
The epimorphism of Amodules A e,A has kernel a,, so A/a, is isomorphic to e,A
which, being a submodule of a Noetherian module is itself Noetherian. So each A/o,
is Noetherian which implies that the direct sum f=1 A/a, is Noetherian. Now the
homomorphism
A * A/ai
i= 1
has kernel H*=1 a, = Ann(M), so A/ Ann(M) is isomorphic to a submodule of the
direct sum and hence is itself Noetherian.
2.2. Modules of Fractions.
In this section we will show that for any multiplicatively closed subset S of the
ring A, if M is a Noetherian Amodule, then 51M is a Noetherian S'1 AModule.
To do this we need to first look at the result of taking a submodule of the module of
fractions, contracting it to a submodule of M, and extending back into the module of
fractions.
Lemma 2.2. If N' is a submodule of S~XM, then
WW) = N1,
where is the cannonical homomorphism.
16
Proof. See [10, page 137, proposition 2]
Theorem 2.5. Let S be a multiplicatively closed subset of A and M an Amodule.
If M is Noetherian, then S~XM is Noetherian.
Proof. Let N\ C N'2 C be an ascending chain of submodules of S~lM. Then the
ascending chain
of submodules of M terminates. Then the chain of their images under (p does too, but
this chain is our original chain by the previous lemma. So every ascending chain of
submodules of S~XM terminates, and it is therefore Noetherian.
2.3. Primary Decomposition.
A submodule of M is irreducible in M if it cannot be written as the
intersection of two submodules, each of which properly contains it.
Proposition 2.2. Suppose N is irreducible in M. Then the following two conditions
are equivalent:
(1) For all b G A, the sequence N: bn is stationary,
(2) N is primary in M.
Proof. (1 => 2) Let r Â£ tM(N). Then there exists an x G M for which rkx $ N for all
k > 0. However, by hypothesis, there is an n > 0 for which N: rn = N: r"+1. By
propositon 1.12, N = (N + Arnx) n(A: rn), but N is irreducible and therefore must
equal one of these submodules. If IV = (IV + Arnx), then rnx is in N, contrary to the
choice of x and r. So N = (N: rn). Now suppose ry 6 N, then rny G N implies y is
in N, so N is primary.
(2 =*> 1) Let b G A, then if b G Vm(N) there exists a k > 0 such that bkM C N, in
other words N: bk = M. Thus N: bk is stationary. If on the other hand b is not in
tjw (IV), then N: bk = N for all k > 0 and again N: bn is stationary.
Theorem 2.6. If M is a Noetherian Amodule, then every submodule of M has a
primary decompostion.
17
Proof. We will show that every submodule can be written as a finite intersection of
irreducible submodules, from which the lemma will follow by proposition 2.2
Let ft be the set of submodules of M which cannot be written as an intersection
of finitely many irreducible submodules and suppose ft is not empty. As M is
Noetherian, ft has a maximal element N. N is not irreducible, so N can be written as
the intersection of two submodules which strictly contain it, say N = N1 f] N". By
the maximality of N we see that N' and N" are not in ft and hence each can be
written as a finite intersection of irreducible submodules. But this means that N can
also, so we must have ft equal to the empty set and the lemma is proved.
2.4. Krull Intersection Theorem.
In this section we prove the Krull intersection theorem. To do this we use a form
of Nakayamas Lemma from [9, page 11].
Lemma 2.3. Let A be a ring, M a finitely generated Amodule and I an ideal of A.
Suppose IM = M, then there exists an element a Â£ A of the form a = 1 + x for some
x Â£ I such that aM = 0.
Proof. Let M = e\A +1 ekA. If k = 1, then putting x = 0 gives e^A = e\I which
implies M = e1A = 0. By induction on the number of generators, suppose the lemma
is true for k 1. Put M' = M/ekA. By the induction hypothesis there is an x Â£ I for
which (1 + x)M' 0. This means that (1 j x)M C ekA. Now IM M implies
(1 + x)IM C I(ekA) = ekI, so (1 + x)ek = yek for some y Â£ I. Then (1 + x y)ek = 0
and so (1 + x y){ 1 + x)M = 0. Furthermore, (1 + x j/)(l + x) = 1 + x2 y xy
where each of x2, y and xy is in I. The lemma follows by putting
a = 1 + (x2 y xy).
Theorem 2.7. Let M be a Noetherian Amodule and I an ideal of A. An element
m Â£ M is in fl^=i InM if and only if m = rm for some r in I.
Proof. (=$>) Put N = nInM, then N is a submodule of M and therefore IN, also a
submodule, has a primary decomposition by theorem 2.6. Let IN = Nx fl fl Nk be
a primary decomposition of IN, where each A, is p,primary. For each i = 1,..., k, if
18
/ C p, then there is an n > 0 for which InM C p?M C TV,. Then also N C TV,. If
however I % p,, then because IN C iVz and TV, is primary, again N C TV,. Therefore
JVC IVifl.flJV* = /JV
and N = IN. By lemma 2.3, there is an element r of / for which (1 + r)N = 0. In
other words, if m Â£ N = flInM then (1 + r)m = 0, or m = (r)m.
(<$=) If m = rm for some r in /, then m = rnm for all n > 0 and clearly m is in
nlnM.
The Jacobson radical of a ring is the intersection of all maximal ideals of the
ring and is characterized by the fact that x is in the Jacobson radical if and only if
1 + xy is a unit for all y in the ring.
Theorem 2.8. Let I be an ideal of a ring A. Then the following statements are
equivalent:
(1) / is contained in the Jacobson radical;
(2) For all Noetherian Amodules M, r\=1InM = (0).
Proof. Suppose I is contained in the Jacobson radical and let M be a Noetherian
i4module. By theorem 2.7 it follows that m 6 n^L1/"M if and only if m = rm for
some r G I. Since I is contained in the Jacobson radical, 1 + ra is a unit for all a Â£ A.
Then 0 = m rm = (1 r)m implies that m = 0asl risa unit in A. It follows
that n jfM = (0).
Now suppose r\%L1InM = (0) for all Noetherian Amodules M. Let m be a
maximal ideal of A. Then ^4/m is an ^module whose only submodule is (0), so it is
Noetherian. Then 7(A/m) is a submodule of A/m and so must be equal to either
A/m or (0). If /(A/m) = A/m then flÂ£L1/n(A/m) = A/m contradicting our
assumption. So it must be that /(A/m) = (0), in other words / C m. It follows that I
is contained in every maximal ideal of A and hence in the Jacobson radical.
19
3. Laskerian Modules
It was shown in the previous section that every submodule of a Noetherian
module has a primary decomposition. In this section we will investigate modules
whose defining property is that all proper submodules have primary decompositions.
In section 3.1 we give an example of a Laskerian ring that is not Noetherian, and
extend the results on exact sequences obtained for Noetherian modules to Laskerian
modules. In section 3.2 we show that the localization of a Laskerian module is
Laskerian. In section 3.3, a characterization of Laskerian modules as modules
satisfying the two conditions (1) for all submodules N of M, Ass(M/N) is finite, and
(2) for all submodules N and all elements a of A, the ascending chain of residual
submodules N: a C N: a2 C terminates.
3.1. Exact Sequences.
Definition 6. An Amodule M is Laskerian if every proper submodule of M has a
primary decomposition.
From theorem 2.6 it follows that every Noetherian module is Laskerian. We now
give an example of a Laskerian module which is not Noetherian to show that the class
of Laskerian modules strictly contains the class of Noetherian modules.
Example. Let A be a ring with one prime ideal m such that m" = (0) for some n > 1.
Let a be a proper ideal of A (necessarily contained in tn) and suppose ab 6 a, while
b Â£ a. Then as every element not contained in m is a unit, a must be contained in
m = t^(a). Hence a is mprimary and every ideal is primary, showing that A is
Laskerian.
Example. Let K be a field and V an infinite dimensional vector space. Form the set
K V and define addition termwise and multiplication by
(a, v){a' ,v') = (aa',av' + a'v). Then K V is a ring that is Laskerian but not
Noetherian. The ideals of K 0 V are the sets (0, W) = {(0,iu)u; 6 W}, where W is a
subspace of V, and the only prime ideal is (0, V). As the previous example showed,
this is a Laskerian ring, but it is not Noetherian because every infinite ascending
20
chain Wi C W2 C ... of subspaces of V leads to an infinite ascending chain of ideals
of jr v, (o,Wi) c (o,w2) c ....
Lemma 3.1. Let M be an Amodule and M' a submodule. Let N be a submodule of
M containing M'. Then N has a primary decomposition in M if and only if N/M'
has a primary decomposition in M/M1.
Proof. This follows from the fact that
N = N1nnNr N/M' = Ni/M' D fl JVr/M'
and corollary 1.2.
Theorem 3.1. Given an exact sequence of Amodules
0 H M' A M A M" > 0,
M is Laskerian if and only if M' and M" are Laskerian.
Proof. Without loss of generality, we may assume that M' is a submodule of M and
that M" is M/M'.
(=>) Suppose M is a Laskerian Amodule and M' a submodule. Since every
submodule of M is also a submodule of M, M is Laskerian. Let N/M' be a
submodule of M/M', where N is a submodule of M containing M'. Then N has a
primary decomposition in M and so by the lemma 3.1 N/M' also has a primary
decomposition. This shows that M' and M/M' are Laskerian if M is Laskerian.
(<$=) Conversely, suppose M' and M/M' are Laskerian and let N be a submodule
of M. By lemma 3.1, N has a primary decomposition in M if and only if N/(N fl M')
has a primary decomposition in M/(N fl M'). The image of N under the mapping
M M/M' is (N + M')/M'. Since M/M' is Laskerian, (N + M')/M' has a primary
decomposition. But (N + M')/M' is isomorphic to N/(N fl M'), so it too has a
primary decomposition and therefore, by the previous remark, N has a primary
decomposition in M.
Corollary 3.1. The Amodules Mi (1 < i < k) are Laskerian if and only if their
direct sum is Laskerian.
21
Proof. This follows from theorem 3.1 and Induction on k in the following exact
sequence,
k k
0 > Ms 0Mi .0M; > 0.
*=1 i=1
i?j
Corollary 3.2. If A is a Laskerian ring and M a finitely generated Amodule, then
M is Laskerian.
Proof. First suppose M is generated by one element e. Then eA is isomorphic to
A/(Ann e) and so eA is Laskerian by proposition 3.1. Then by corollary 3.1 we see
that the direct sum "=1 e,A is Laskerian. The corollary follows from the
epimorphism
0e,A > ^ejA
2=1 2 = 1
and proposition 3.1.
3.2. Modules of Fractions.
Proposition 3.1. Let M be an Amodule and S any multiplicatively closed subset of
A. If M is Laskerian, then S~1M is Laskerian.
Proof. Let N be a submodule of M with primary decomposition N = Ni fl fl Nr,
where each IV, is pjprimary. By 1.3, for each index i satisfying p, fl S = 0, S~xNi is
piprimary in S~1M, while for each index j satisfying pj D S ^ 0, S~xNj = S~lM.
Thus
S~lN = S\nrk=1Nk) = n S'Ni,
where i runs over all indicies satisfying p, D S = 0. As each submodule of S~lM is of
the form S1N for some submodule N of M, the proposition follows.
22
3.3. Ascending Chains of Residuals.
Lemma 3.2. If N is a pprimary submodule of a module M and a Â£ A, then the
ascending chain of submodules
N: a C N: a2 C C N: ak C
is eventually stationary.
Proof. If a p = y/N: M, then N: aj = N for all j, and the chain terminates. If,
however, a Â£ p, then a* Â£ N: M for some positive integer j. Then N: aj = M and
the chain again terminates.
Proposition 3.2. If N is a submodule of an Amodule M which admits a primary
decomposition, and a is an element of A, then the ascending chain of submodules
N:aCN:a2CCN:ak ex
terminates.
Proof. Let IV = lV1nnIVrbea primary decomposition of N. Then
iV:o' = (nr=1JVi):o# = nU(^.:fl*)
As each of the terms (iV*: a*) is eventually stationary, so is the intersection of these
terms.
Corollary 3.3. Every irreducible submodule of a Laskerian module is primary.
Proof. This follows from proposition P:IrreduciblePrimary.
Corollary 3.4. Let M be a Laskerian Amodule, N a submodule and a Â£ A. For
sufficiently large k,
N = (N: ak)C\(N + akM).
Proof. We clearly have the inclusion N C (N: ak) fl (N + akM). Let a; be an element
of (N: ak) fl (N + akM) and write x = n + akm, where n Â£ N and m Â£ M. To show
that x Â£ N, multiply both sides by ak. Since x Â£ (N: ak), akx Â£ N. This implies that
a2km Â£ N. If & is large enough that the ascending chain in the previous proposition is
23
stationary for all integers larger than or equal to k, then N: a2k = N: ak and we have
m G N: a2k = N: ak. Thus akm 6 N and so x = n + akm is also in N.
Lemma 3.3. If an Amodule M is the direct sum of a finite number of submodules
Mi, (1 < i < t), then AssA(M) = U*_1Ass>1(Mi).
Proof. By induction it suffices to prove the case t = 2. So let M = Mx M2. By
lemma 1.1 we see that Assj4(Mi) U Assa(M2) C Assa(M). Conversely, by the same
lemma, Assa(M) C AssA(MX) U Assa(M/Mx), but Assa(M/Mi) = AssA(M2) and so
equality holds.
Lemma 3.4. If the submodule N of M has a normal decomposition
N = Ni fl D Nt, where each Nx is p,primary, then Ass(M/N) = {p1;..., pt}.
Proof. To show that pj is in Ass (M/N), note that if x is in each Nj except for Ni,
then pkx N for some k if and only if p E p,. This is because N, is piprimary. Thus
p, is in Ass (M/N).
Consider the homomorphism
This is injective, so M/N is isomorphic to a submodule of @\=XM/Ni and therefore
the primes associated to M/N is a subset of the primes associated to the direct sum.
But by lemma 3.3, Ass(@M/N{) = UAss(M/N{) = {pi,... ,p4}. Hence
Ass(M/N) C {pl5... ,pi) which, combined with the above result shows that equality
holds.
Theorem 3.2. An Amodule M is Laskerian if and only if the following two
properties hold:
(1) For every submodule N of M and every element a G A, the ascending chain of
submodules
(N: a) C (N: a2) C (N: a3) C
is eventually stationary.
24
(2) For all submodules N of M, the set Assa(M/N) of primes associated to M/N
is finite.
Proof. (=>) Suppose M is a Laskerian Amodule. Condition 1 is just proposition 3.2,
and condition 2 is lemma 3.4.
(<$=) Suppose now that conditions 1 and 2 hold for the Amodule M, and let N
be a submodule of M. Let AssA(M/N) = {pi,... pt} be the finite set of primes
assoctiated to M/N. Without loss of generality, assume pi is minimal in this set. For
each of the other primes p*, (2 < i < t), choose an a, 6 p, that is not in pi and let
a = a2a3 at. Let S = {a*}Â£i0, a multiplicatively closed subset of A that does not
meet pi. By assumption 2, the chain
(N: a) C (N: a2) C
terminates, which implies that &s(N) = (IV: ak) for some sufficiently large integer k.
We now want to show that N: ak is piprimary. Since &s(N)/N is the kernel of the
cannonical homomorphism M/N i* S~1(M/N), proposition 1.8 says that
Absa(M/6s(N)) = = {p.p( n S = 0} = {Pi},
so it follows from proposition 1.4 that 6s(lV) = (N: ak) is piprimary. Now by
proposition 1.11 we can write
N = (N: ak)n(N + akM).
It also follows from proposition 1.11 that N is piprimary in N + akM, in other
words, Assa((N + akM)/N) = {pi}. Now the primes associated with a union of
submodules is equal to the union of the primes associated with each submodule (see
[11, page 20, 3.3]), which makes the set of submodules E/N of M/N whose only
associated prime is pi an inductive set. Every totally ordered set has an upper bound,
that upper bound being the union of the submodules in the chain.
Let G be a submodule of M such that G/N is maximal among the submodules
containing (N + akM)/N such that AssA(G/N) = {pi} We will show that
25
N = (N: ak) (1 G, that G is maximal among submodules whose intersection with
N: ak is N and that AssA(M/G) = {p2,p3,... ,pt}.
Clearly we have N C (N: ak) fl G, so suppose x is in (N: ak) fl G. Then akx is in
N. But N is pi primary in G, so x G G and akx G N implies that ak is in pi unless x
is already in N. Hence equality holds.
To show that G is maximal among submodules whose intersection with N: ak is
N, suppose G' D G. Then there is some x in G' such that, without loss of generality,
p2 is the only prime minimal over N: x. If p2 2$ Pi5 then there is a p G pi that is not
in p2 for which px G N: ak. But then px is in N: ak and in G', but px is not in N. If,
on the other hand, p2 D pi, then it must be that p2 = N: x, in which case akx G N
because ak G p2 by construction. In this case, x is in G' and in N: ak, but x is not in
N. So G is maximal among the submodules whose intersection with N: ak is N.
Now we can show that AssA(M/G) {p2,... ,pt}. We do this by showing that
for any y G M G), if p' is a minimal prime of G: y then p' is a minimal prime of
N: z for some z in M, and therefore p' is in Assa(M[N).
Suppose that y G M G and that p' 2 G: y minimally. Then we have the
containments
P' 2G:yDN:y.
If p' is a minimal prime of N: y then p' is in AssA(M/N). Otherwise there must be
some p, associated to M/N such that
p'2G:y2piDN:y.
In this case we must have p' = G: y. Enlarge the submodule G by adding y to it,
G C G + Ay,
where the containment is strict by the choice of y. Then by the maximality of G as a
submodule whose intersection with N: ak is N, there must be some g G G and
rGlp' such that z = g + ry is in N: ak but not in G. Consider N: z. If s G p',
then sz = sg + sry is in N: ak by assumption, and in G because s is in G: y = p'.
Then sz G N and we have p' C N: z. Conversely, suppose s G N: z. Then sz is in N
26
which implies that sry is in G, so that sr Â£ p'. But r Â£ p' implies that s is in p', so we
have p' C N: z and hence equality. This means that p' is in Assa(M/N) and so we
finally conclude that AssA{M/N) {p2,... ,p}.
Thus we have taken N and represented it as N = (N: ak) fl G, where N: ak is
piprimary in M and assAM/G = {p2,... ,pi}. By repeating this process t 1 more
times we arrive at a primary decomposition of N where each p, associated to M/N
belongs to some factor of the decompositon.
27
4. ACCR Modules
From theorem 3.2, an Amodule M is Laskerian if and only if; for every
submodule N and every a Â£ A, the ascending chain
N: a C IV: a2 C N: a3 C
terminates, and for every submodule N, the set of prime ideals Assa(M/N) is finite.
In this section we will investigate modules which are required only to satisfy the first
condition, called the ascending chain condition on residual submodules (accr). The
study of modules satisfying (accr) is pioneered by Dr. Sylvia Lu, and a thourough
treatment of the subject can be found in [6], [7] and [8].
4.1. Preliminaries.
Definition 7. An Amodule M is said to satisfy the ascending chain condition on
residual submodules (accr) if for every submodule N and every finitely generated
ideal B of A, the ascending chain
N: B C N: B2 C N: B3 C
terminates.
In [6, theorem 1, p. 305] it is shown that the property (accr) is equivalent to
property (1) of theorem 3.2 which is stated only for principle ideals. Laskerian
modules, and therefore Noetherian modules, satisfy (accr). One of the properties of
Laskerian modules is that for every submodule N and every element a Â£ A, we can
write N = (N: ak) fl (N + akM). In general, a module satisfies (accr) if and only if
for every submodule N and every finitely generated ideal B, we can write
N = (N: Bk) fl(lV + BkM) for some sufficiently large integer k. For a proof see [6,
corollary, page 306].
Theorem 4.1. An Amodule M satisfies (accr) if and only if every factor module
M/N satisfies (accr) for nonzero submodules N.
28
Proof. Let M be an .Amodule that satisfies (accr) and let N be a submodule of M.
For an arbitrary submodule E/N of M/N and finitely generated ideal B of A, we
have E/N: Bk = (E: Bk)/N. Since E: Bk becomes stationary, so must E/N: Bk.
Conversely, suppose every factor module M/N satisfies (accr). If
E: B C E: B2 C is a chain of residual submodules of M that does not terminate,
we may assume that E is not the zero submodule. For otherwise there is an integer n
for which E: Bn strictly contains E and we may consider the chain starting at
E: Bn. But then the nonterminating chain E: B C E: B2 C leads to a
nonterminating chain of submodules of M/E, (0): B C (0): B2 C , which
contradicts the assumption that M/E satisfies (accr).
Theorem 4.2. Given the exact sequence of submodules
0  M' > M  M" * 0,
M satisfies (accr) if and only if M' and M" satisfy (accr).
Proof. There is no loss of generality by assuming that M' is a submodule of M and
that M" = M/M'. Suppose M satisfies (accr). Then M' as a submodule clearly
satisfies (accr), and by theorem 4.1 we see that M" does too. Conversely, suppose M'
and M/M' satisfy (accr). For any N, a submodule of M, and any element a Â£ A, we
want to show that N: a C N: a2 C terminates. In the factor module we have
(IV + M')/M': a C (IV + M')/M': a2 C terminates, so there is some k for which
(N + M')/M': ak = (N + Af')/M': ak+1. But
(IV + M')/M': ak = ((IV + M'): ak)/M', so that (IV + M'): ak = (N + M'): ak+1.
Similarly, we have (N D M'): a1 = (N fl M'): at+1 for some integer t. We will show
that N: ak+t = N: ak+t+1.
Suppose x is in N: ak+t+1. Then ak+t+1x G N and also in
(IV + M'): ak+t+1 (N + M') \ ak. So akx = n + m for some n G IV and m G M'.
Then a*+t+1a: = at+1(n + m) = at+1n + a<+1m is in N, hence at+1m G IV n M;, so that
m G (IV fl M'): at+1 = (N fl M'): at. This implies that afm is in N, so that finally,
ak+ix = afin + m) = afn + ofm G N. In other words a; is in N: ak+t. Hence M
29
satisfies (accr).
Corollary 4.1. The 4modules Mi, (1 < i < t), satisfy (accr) if and only if the direct
sum satisfies (accr).
Proof. The proof is similar to that of corollary 2.1.
Corollary 4.2. If 4 satisfies (accr), then any finitely generated 4module satisfies
(accr).
Proof. For a set of generators e1}... ,et of M, each module 4e, satisfies (accr), so
then does the direct sum 4es. Then M is a factor module of the direct sum and
hence M satisfies (accr).
4.2. Localization.
In this section we show that the process of forming the module of fractions
preserves the property of satisfying (accr).
Theorem 4.3. If the 4module M satisfies (accr) and S is any multiplicatively
closed subset of 4, then 51M satisfies (accr).
Proof. Every submodule of S~XM is of the form S1N for some submodule N of M.
Furthermore we have S1N: (b/s) = S~1(N: b). So any sequence of submodules
S~1N: (b/s) C S~1N: (b/s)2 C must terminate as the corresponding sequence
N: b C N: b2 C in M terminates.
4.3. Krull Intersection.
Note that in the proof Nakayamas lemma, used to prove the Krull intersection
theorem for Noetherian modules, [see 2.7] explicit use of the Noetherian property of
the module was used. Furthermore, embedded in the proof is the use of the property
that if N is pprimary in M, then pkM C N. This property does not hold in general
even for Laskerian modules, so a new approach is needed.
30
Definition 8. The ArtinRees property is said to hold for an ideal B of A and a
submodule N of M if there exists a positive integer n such that N n BhM C BN for
all integers h> n.
Lemma 4.1. If an Amodule M satisfies (accr), then the ArtinRees property holds
for every finitely generated ideal B of A and every submodule N of M.
Proof. Since M satisfies (accr), there is an n such that, for all h > n,
BN = (BN: Bh) fl (BN + BhM). On the right hand side we have (BN: Bh) D N
and (BN + BhM) D BhM, so the right side contains N fl BhM from which the
lemma follows.
Lemma 4.2. If the ArtinRees property holds for an ideal B of a ring A, and every
cyclic submodule of M, then an element m 6 M is in fl^L1BnM if and only if
m = rm for some r in B.
Proof. Suppose m is in 1BnM, then the submodule Am is also contained in the
intersection and hence in BkM for each positive integer k. Since the ArtinRees
property holds for B and Am, for all sufficiently large integers h we have
Am fl BhM C Bm. But Am fl BhM = Am by the previous remarks. It follows that
Am C Bm and therefore that equality holds. Then for some r G B, rm = m. The
other direction is trivial, so the lemma is proved.
Theorem 4.4. Let M be an Amodule satisfying (accr) and B a finitely generated
ideal of A. An element m G M is in fl^Li BnM if and only if m = rm for some r in B.
Proof. Since M satisfies (accr), and B is finitely generated, the ArtinRees property
holds for B and every submodule of M. In particular it holds for B and every cyclic
submodule. The theorem then follows from lemmas 4.1 and 4.2.
4.4. Polynomial Rings.
In section 2.1 we showed that a ring A is Noetherian if and only if the ring of
polynomials A[X] is Noetherian. In this section we show that the ring of polynomials
A[X] satisfies (accr) if and only if it is Noetherian.
31
Theorem 4.5. A[X] satisfies (accr) if and only if A is Noetherian.
Proof. If A is Noetherian, then A[X] is Noetherian and therfore satisfies (accr). So
suppose A[X\ satisfies (accr) and let I0 C /i c h C be an ascending chain of
ideals in A. If this sequence does not terminate, consider the ideal I' of polynomials
whose coefficient on a;7 is in I). Since the chain of ideals does not terminate, there is
some element a in In+1 that is not in In, and therefore axn+1 is in I1 while axn is not
This shows that the sequence /': C I': x2 C does not terminate, contradicting
the hypothesis.
32
References
1. M. Atiyah and I. Macdonald, Introduction to Commutative Algebra, AddisonWesley,
Reading, 1969.
2. N. Bourbaki, Commutative Algebra, Hermann, Paris, 1972.
3. R. Gilmer and W. Heinzer, The Laskerian Property, Power Series Rings and Noetherian
Spectra, Proc. Amer. Math. Soc. 79, 1316.
4. W. Heinzer and J. Ohm, On the Noetherianlike Rings of E. G. Evans, Proc. Amer.
Math. Soc. 34, 7374.
5. E. Kunz, Introduction to Commutative Algebra and Algebraic Geometry, Birkhauser,
Boston, 1985.
6. C. P. Lu, Modules Satisfying ACC on a Certain Type of Colons, Pacific Journal of
Mathematics, Vol. 131, No. 2, 303318.
7. C. P. Lu, Modules and Rings Satisfying (ACCR), Proc. Amer. Math. Soc. Vol. 117, No. 1,
510.
8. C. P. Lu, Two Noetherian Properties of Rings Satisfying (accr), To Appear in Nova
Journal of Algebra and Geometry.
9. H. Matsumura, Commutative Algebra, 2nd ed., Mathematics Lecture Note Series 56,
BenjaminCummings, Reading 1980.
10. D. Northcott, Lessons on Rings, Modules and Multiplicities, Cambridge University Press,
London,1968.
11. N. Radu, Lec\ii De Algebra III, Universitatea Din BucureÂ§ti, BucureÂ§ti, 1981.
33

PAGE 1
NOETHERIAN PROPERTIES OF NONNOETHERIAN MODULES by Michael David Snell B.A., University of Colorado, Denver, 1992 A thesis submitted to the Facuity of the Graduate School of the University of Colorado at Denver in partial fulfillment of the requirements for the degree of Master of Science Applied Mathematics 1994
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! I I This thesis for the Master of Arts degree by Michael David Snell has been approved for the Department of Mathematics by Date
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Snell, Michael David (M.S., Applied Mathematics) Noetherian Properties of NonNoetherian Modules Thesis directed by Professor Sylvia Lu ABSTRACT It is well known that submodules of Noetherian modules possess a primary decomposition. The class of Noetherian modules is contained in the class of Laskerian modules modules satisfying the property that every proper submodule has a primary decomposition. The ascending chain condition for residual submodules (accr) states that for every sub module N and every finitely generated ideal B, the ascending chain of sub modules N: B <;:; N: B2 <;:; terminates. The class of modules satisfying ( accr) contains the class of Laskerian modules. we investigate properties of exact sequences, localization, the weak ArtinRees property and the Krull intersection theorem, that can be extended from Noetherian modules to Laskerian modules and ( accr) modules. This abstract accurately represents the content of the candidate's thesis. I recommend its publication. Signed iii
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CONTENTS 1. Preliminaries 1 1.1. Radicals and Residuals of Submodules 1 1.2. Modules of Fractions 2 1.3. Primary Submodules 4 1.4. Primary Decomposition 9 2. Noetherian Modules 12 2.1. Preliminaries 12 2.2. Modules of Fractions 16 2.3. Primary Decomposition 17 2.4. Krull Intersection Theorem 18 3. Laskerian Modules 20 3.1. Exact Sequences 20 3.2. Modules of Fractions 22 3.3. Ascending Chains of Residuals 23 4. ACCR Modules 28 4.1. Preliminaries 28 4.2. Localization 30 4.3. Krull Intersection 30 4.4. Polynomial Rings 31 References 33 iv
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1. PRELIMINARIES 1.1. Radicals and Residuals of Submodules. Definition 1. For a submodule N of the Amodule M, the radical of N in M is the ideal of A tM(N) ={a E A I ak M N,for some nonnegative integer k}. It follows that when we view A as an Amodule and I is an ideal of A, tA(I) ={a E A I an E I for some nonnegative integer n}. When refering to the radical of an ideal I, we will use the notation VJ instead of t(I). Proposition 1.1. If ... Nk are submodules of the Amodule M, then Proof. Fork= 2, let N1 and N2 be submodules of M. Suppose a is in the radical of N1 and the radical of N2 and let x be an element of M. Then there is a nonnegative integer j for which ai x E N1 and ai x E N2 so a is in the radical of the intersection of N1 and N2 Conversely, if a is in the radical of the intersection, then given an x in M, ai xis in N1 and N2 for some nonnegative integer j. Thus a is in the radical of N1 and the radical of N2 The general case follows by induction. 0 The proposition falls to hold for infinite intersections as the following example shows. Example. Consider the ring of integers Z, and let p be a prime ideal. For any positive integer k, we have Jiik = p, so that = p. But the infinite intersection of the p' is (0), and the radical of (0) is (0) because Z is an integral domain. Given two submodules N and K we can form the ideal N: K ={a E A I aK N}. Similarly, given a submodule N of M and an ideal I of A we can define the residual sub module N: I= {mE M I ml N}. 1
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Since N: M is an ideal of A we can look at its radical. An element a E A is in vN: M if and only if there is a nonnegative integer k such that ak M <;; N. In other words, The annihilator of an element x in the Amodule M is the set of all a E A for which ax = 0. Given a submodule N of M, the annihilator of N is the set of a E A for which ax = 0 for all x E N. The annihilator of an element or of a submodule is an ideal of A denoted by Ann x or Ann N respectively. Given an ideal I of A, a prime ideal p containing I is a minimal prime of I if p contains no other prime ideal containing I. Definition 2. A prime ideal p of A is associated with the Amodule M, written p E AssA(M), if it is a minimal prime of Annx for some x EM. 1.2. Modules of Fractions. The rational numbers are formed from the integers by an equivalence relation on fractions. This construction can be carried out in an arbitrary ring or module by a slight generalization. A multiplicatively closed subset Sofa ring A is a set containing 1 that is closed under multiplication. We will assume that S contains no zero divisors so that zero is not in S. Given an Amodule M we can form the set of fractions S1 M as the set of all !!! with mE M and s E S. Let a relation on s1 M defined by m n , exists s E S such that s (mtns) = 0. s t Then is an equivalence relation and the equivalence classes form an Amodule when we define addition by m n mt + ns s + t = s'=tand multiplication by elements of A as m am a=. 8 s 2
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This modules1M is called the module of fractions. If we let A = M in the above construction we get a ring s1 A called the ring of fractions. Note that A is mapped cannonically into S1 A by a f+ '!We can make s1 M into an S1 A module by defining am am = s t st An element mE M is mapped cannonically into s1M by m f+ T If we extend the submodule N under the cannonical homomorphism M _, s1 M, we get the s1 M submodule s1 N. Then the preimage of S1 N, which contains N, is a submodule of M called the saturation of N in M with respect to Sand is denoted 65(N). The extension of the saturation of N is again S1 N. Another way to describe the saturation of N in M is 6s(N) = {x EM I sx EN for somes inS}. Thus the kernel of the cannonical homomorphism is the saturation of (0) in M. Commonly used multiplicatively closed subsets are Ajl for a prime ideal p and the set {a'}k'=o for some a not a divisor of zero. Also, the intersection of two multiplicatively closed subsets is again a multiplicatively closed subset. Example. In the ring Z of integers, letS= {2'}r;,0 Then s'z is the set of fractions n/2. For the ideal (3223 ) = (72), the saturation of (72) is the ideal (32 ) = (9). Proposition 1.2. If Sis a multiplicatively closed subset of A that does not meet the ideal a, there exists a prime ideal containing a and not meeting S. Proof. See [2, page 71, corollary 2] 0 Corollary 1.1. Every proper ideal of a ring is contained in a maximal idea]. Proof. Let A be a ring and I a proper ideal. Then I contains no units of A and the set S of units of A is a multiplicatively closed subset of A. By proposition 1.2 there is a prime ideal containing I and not meeting S. Let J be a maximal element of the prime ideals containing I, then J is maximal in A and J contains I. 0 3
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1.3. Primary Submodules. Definition 3. A proper submodule N of M is primary if for a E A and x EM, ax EN implies that either x EN or a E tM(N) = vN: M. Example. The ideal (p") in Z, generated by p" for prime p, is primary. First note that .;r:pn) = (p). Then rx E (p") means rx = p"t for some integer t. Then if x rt (p" ), x has fewer than n factors of p, so that r must have at least one factor of p. Thus r E (p) = .j(jj"). Proposition 1.3. If the submodule N of the Amodule M is primary, then tM(N) is a prime ideal of A. Proof. Suppose abE tM(N) while art tM(N), for a, bE A. Then abE tM(N) implies that (ab)' M N for some positive integer k. Since art tM(N), there is some x EM for which ax rt N. Then we have b(a'x) EN which implies that b E tM(N) as N is primary. But b' E rM(N) implies bE tM(N), so tM(N) is prime. D When N is primary in M and p = rM(N) is a prime ideal of A, we say that N is pprimary in M, or that the prime ideal p belongs toN. Note that the prime ideal p can belong to more than one primary module. In the example above, the ideal (p) belongs to every ideal generated by a power of p. Example. Consider the integers modulo the ideal generated by p", Zj(p"), as a Zmodule. The nonzero element p"1 + (p") of Zj(p") is annihilated by everything that is a multiple of p, so (p) = Annp"1 + (p") is in Assz(Z/(p")). On the other hand, the annihilator of every nonzero element of Z/(p") is some power of p, and (p) is the only minimal prime ideal containing the powers of p, hence (p) is the only element in Assz(Z/(p")). In this last example, the set of primes associated with Z/(p") con.sists of only one ideal. The reason for this is the fact that (p") is (p)primary in Z, as we will now show. 4
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Proposition 1.4. A submodule N of the Amodule Jvi is pprimary in Jvi if and only if AssA(MjN) = {p}. Proof. (=>)Suppose N is pprimary in M. Let x be an nonzero element of M/N and let x be an element of Jvi which maps to x under the cannonical homomorphism. Then a E Ann x implies ax is in N. Now x is not in N as x is nonzero in M j N, so the fact that N is primary in M says that a is in tM(N) = p. Thus p contains the annihilator of each nonzero element of Jvi j N. Now suppose that q is any prime ideal in Ass(M/N) and let y be a nonzero element of Jvi j N for which q is a minimal prime containing Ann fi. From the above result, q t;;; p. Let a be in p = tM(N). Then if y maps toy, we have a"y EN for some n > 0. It follows that a" E Ann y which is contained in q. So a E q as this ideal is prime. Thus p = q is the only element of Ass(M/N). (o) Suppose that Ass(M/N) = {p}. Let axE N, a E A,x N. Then a is an element of Annx and hence an element of p. Now for any nonzero yin MjN, pis the only minimal prime containing Ann y, so a is nilpotent in A/ Ann fi. Thus a"y is in N for some n > 0. As y was arbitrary it follows that a E tM(N) and that N is primary in M. 0 Corollary 1.2. Let >: M+ M' be an epimorphism of Amodules and p a prime ideal of A. If N is a submodule of Jvi containing the kernel of > and N' the corresponding submodule of M', then N is pprimary in Jvi if and only if N' is pprimary in M'. Proof. We have that M/N is isomorphic to M'/N' and so Ass(M/N) = {p} if and only if Ass(M'/N') = {p}. 0 Proposition 1.5. If N, ... Nk are pprimary submodules of Jvi, then their intersection nf=1 N, is pprimary in M. Proof. By proposition 1.1, tM(nf=1N;) = p. Suppose that, for rEA and x EM, rx E nf=1N; while x nf=1N;. Then xis not inN; for some j. But N; is pprimary and rx EN;, so we must haver E p = tM(nf=1N;). 0 5
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When considering the cannonical homomorphism from a module into its module of fractions for some multiplicatively closed subset, we cannot use corollary 1.2 because the map is not a epimorphism. We will show however that if the image of a submodule is primary in the module of fractions, then the saturation of the original module is primary. This gives us a way to enlarge the original module to a primary module containing it which will play a role in expressing a module as an intersection of primary modules. Proposition 1.6. Let S be a multiplicatively closed subset of A. If N' is an S1pprimary submodule of the S1 A module S1 M, then s1 M is the cannonical homomorphism. Proof. We will show first that tsM(N') = S1tM(N). Suppose'!:_ is in s1rM(N) and let :_be an element of S1 M which is not inN'. s t Then x N, but ax EN for some nonnegative k as a E tM(N). Thus ( ) k k a x ax . = E N Implies s t st Conversely, suppose '!:is in tsM(N') and let x be an element of M which is not s inN. Then (a) tx tax =.EN' implies ax EN. s t ts Thus a is in tM(N) and finally '!:_is in S1rM(N). This proves equality. s Now to show that N is primary in M, let axE N, x N. Then sax is inN' s while sx is not, so '!:is in tsM(N') because N' is primary. By the first result, '!:is s 1 1 in S1tM(N) and therefore a is in tM(N) and N is primary in M. D Proposition 1. 7. Let S be a multiplicatively closed subset of the ring A and let M be an Amodule. Then AsssA(S1 M) = {S1p: p E AssA(M) andp n S = 0}. Proof. If p is a prime ideal of A not meeting S and if p contains some ideal I of A, then p contains the saturation of I with respect to S. This follows from the primality of !J, for if sa E I for some s E S, then sa E IJ while s p, so a must bein p. 6
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Let p E AssA(M) not meetS. Suppose p :J Annx minimally, then 1 sx p :J 65(Annx) minimally and theTefore spis a minimal prime containing Ann. s So S1p is in Asss'A(S1 M). Now suppose s1p is a minimal prime containing Ann:.. Then by taking inverse s images under the cannonical mapping we see that p :J 65(Annx) :J Annx. Suppose we have another prime ideal q satisfying p :J q :J Annx. Then s1q contains and the minimality of s1p shows that q = p. Hence pis a minimal prime of Annx and pis in AssA(M). 0 Corollary 1.3. Given a multiplicatively closed subset S of A, there is a one to one correspondence between the pprimary submodules N of M where p does not meet S, and the sIpprimary submodules sl N of sl A. Proof. By proposition 1.6 we have that the preimage of any S1pprimary submodule of s1 M is pprimary in M. Now by proposition 1.7 we see that if N is pprimary in M and p does not meetS, then AssA(M/N) = {p} and So S1 N is S1pprimary in S1 M. 0 This corollary gives us a way to focus attention on some subset of the prime ideals of A by using the multiplicatively closed subset of elements not in any of the prime ideals under consideration. In particular we will be interested in primes associated with some submodule, so the next few results show how these primes are related and what happens to them in the process of localization. Lemma 1.1. Let N be a submodule of the Amodule M. Then Proof. The first inclusion is obvious. Let p E AssA(M) and suppose pis not in AssA(N). Let x be an element of M such that pis a minimal prime containing Annx. Then as p contains Ann ax :J Ann x for all a not in p, it is clear that ax N for all a not in jl, for otherwise p would be in AssA(N). Bnt this means that p cc:mta.ins 7
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N: x = Ann x in M / N. The series of inclusions Ann x <;;; N: x <;;; p shows that p is in AssA(M/N). D Lemma 1.2. Let p be a minimal prime of the ideal a and let S = Ap. Then for any a E p, there exists an s E S and an integer k > 0 such that sa' E a. Proof. Suppose otherwise. Then for all k > 0 and all s E S, sa' a. Form the multiplicative setS'= S U {a'} which meets p but not a. By proposition 1.2, there exists a prime ideal containing a which does not meet S'. But then this prime ideal is strictly contained in p which is a contradiction. D It was pointed out before that the kernel of the localization map M >+ s1 M is the saturation of (0) in M with respect to S. The following proposition shows how the primes associated with a module M can be split into two sets, those associated with the kernel and those associated with the factor module of M mod the kernel. Proposition 1.8. Let N be the kernel of the cannonical homomorphism M _, s1 M for some multiplicatively closed subset S of A. (1) AssA(N) = {p E AssA(M): p n S i 0} (2) AssA(MjN) = {p E AssA(M): p n S = 0} Hence AssA(M) = AssA(M/N) U AssA(N). Proof. If p E AssA(M) does not meet S, then for any x EN we have Ann x n S "I 0 and p 1J Ann x. Thus p is not in Ass A ( N) and must therfore be in Ass A ( M / N) by lemma 1.1. If p does meet S and if t E p n S, then t' sx = 0 for some k > 0 by lemma 1.2, where pis a minimal prime containg Annx. It follows that xis inN and pis in AssA(N). Suppose Now that pis in AssA(N) and let p be a minimal prime of Ann y for some yEN. As sy = 0 for somes E S, p meets S. The fact that pis in AssA(M) follows from lemma 1.1. Finally suppose pis in AssA(M/N) and let x be an element of M/N for which p is minimal containing Annx. It is sufficient to show that pis in AssA(M). Let x be 8
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an element of M which maps to x and suppose p is not a minimal prime containing Ann x. Let q be a prime ideal satisfying Ann x c q C p. Then also q does not meet 8 and so q contains Annx. But pis a minimal prime containing Annx, so we must have p = q and we are done. 0 Corollary 1.4. If pis a minimal element of the set of primes in AssA(M) for some Amodule M, the saturation of (0) C M with respect top in M is pprimary in M. Proof. If pis minimal in AssA(M), every other prime ideal of this set meets 8 =Ap. Thus by proposition 1.8, AssA(M/N) = {p} and by proposition 1.4 we see that N is pprimary in M. 0 1.4. Primary Decomposition. Definition 4. A submodule N of the Amodule M is said to possesses a primary decomposition in M if it can be represented as the intersection of finitely many primary submodules. Example. In the ring of polynomials in one indeterminant over the rationals,
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Proposition 1.9. Let N = N1 n n N, be a primary decompositon of N c M, where each N; is p;primary. Let S be a multiplicatively closed subset of A and suppose the submodules are ordered so that p1 . Pm do not meet S while the rest do. Then N1 n n Nm = 65(N), the saturation of N. Proof. Let x E N1 n n Nm. For j = m + 1, ... k, let sJi be an element of P; n S satisfying sJix EN. Puts= TIJ;m+l sJi. Then sx EN and sox E 6s(N). Now suppose y E 65(N). Then there exists an s' in S such that s'y EN. For i = 1, ... m we have that s'y EN, whiles' rt p,. Since N; is p;primary, y is inN, for all i between 1 and m. This proves that 65(N) = N1 n n Nm. 0 Proposition 1.10. Let M be an Amodule and let N be a submodule of M. If N has a primary decomposition in M, then for an arbitrary multiplicatively closed subset S of A, the saturation of N in M with respect to Sis of the form N: a for some a E A. Proof. Let N = N1 n n N, be a normal decomposition of N where each N, is p;primary and let S be a multiplicatively closed subset of A. As in proposition 1.9, let the saturation of N with respect to S be N 1 n n N mFor j = m + 1, ... r, let s; E P; n S and put s = TI};m+l s;. Then T = { s'};:;0 is a multiplicatively closed subset of A which meets Pm+JoPr and does not meet p, ... ,Pm So by proposition 1.9 we have 6s(N) = 6T(N) = N; n n Nm. Recall that we have P; = rM(N;) = .j N;: M. As s E P; for j = m + 1, ... r it follows that s''M N; for some k; > 0. If we put k = max{k;} we have s' M Nm+l n n N, and we see that 65(N) = N: s', for if x E 65(N) = N; n n Nm we have s'x E Nm+l n n N, sox EN. On the other hand, if x EN: s' then s'x EN implies x E N1 n n Nm = 6s(N). The proposition follows from putting a = s'. 0 Proposition 1.11. Let M be an Amodule and N a submodule of M. If pis a prime ideal of A, a rt p such that Q = N: a is pprimary in M, then N = Q n (N +aM). Furthermore, N is pprimary in (N +aM). 10
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Proof. The inclusion N Q n (N +aM) is obvious. Let x = n +am be an element of the intersection. with n E N and m E M. Then am = x n is in Q as both x and n are. As Q is pprimary and p, we have mE Q which implies that amEN. Therefore x = n + am is in N. To show that N is pprimary in (N +aM), suppose rx E N for some r E A and x E (N +aM)N. Then x has the form n +am for some n EN and mE M, and rx = rn + ram. By assumption rx is in N, so rx rn is in N implying that ram E N. Then ram being in N implies that rm is in Q = N: a which is pprimary. If m were in Q, then x = n + am would be in N, contradicting our assumption, hence m is not in Q and we must have r E j). This completes the proof. D Proposition 1.12. Let M be an Amodule and N a submodule of M. Suppose there exists a b E A and an x E M for which b" x N for all n > 0. If there is an n such that N: b" = N: b"+1 then N = ( N + Abnx) n ( N: bn ). Proof. Let y = n + rb"x be an element of N: bn, where n E N and r E A. Then yn = rbnx is an element of N: bn and therefore rb2nx is in N. So rx is in N: b2n = N: b" from which we conclude that y E N. The other inclusion is obvious. D 11
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2. NOETHERIAN MODULES 2.1. Preliminaries. Given a module, we can partially order the submodules by inclusion. A module satisfies the ascending chain condition (ACC) on submodules if every ascending chain of submodules terminates, and it satisfies the maximal condition if every non empty subset of submodules has a maximal element. Proposition 2.1. For a module M, the following conditions are equivalent: (1) Every submodule of M is finitely generated (2) M satisfies the ascending chain condition on submodules (3) M satisfies the maximal condition. Proof. (1 '* 2) Suppose every submodule of M is finitely generated and let be an ascending chain of submodules. The set union N = is a submodule of M and must therefore be finitely generated. If { e 1 ... ek} generate N, then for each 1::; j::; k the generator e; is in some N,,. Let J = max{i;}, then each e; is in the sub module NJ. It follows that N = NJ and therefore that the chain terminates. (2 '* 3) Let U be a nonempty set of submodules of M and suppose fl has no maximal element. Let N1 be an element of U. As fl has no maximal element, there must be a submodule N2 in U which strictly contains N1 By repeatedly applying this argument we can construct an ascending chain of submodules which never terminates, contradicting condition (2). (3 '* 1) Let N be a submodule of M and let fl be the set of submodules of N which are finitely generated. The set fl is nonempty as it contains the zero sub module, so let N' be a maximal element of fl. Let {e1 e.} generate N' and let e be an element of N. The submodule of N generated by { e, ... e., e} contains N' and is finitely generated, so by the maximality of N' we see that e is an element of N'. Hence N' = N and every submodule of M is finitely generated. D 12
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Definition 5. A module which satisfies any of the equivalent conditions of proposition 2.1 is called a Noetherian module. Example. The integers is a Noetherian ring. In fact Z is a principal ideal domain, so every ideal is generated by only one element, satisfying condition one. More generally, every ideal of a Dedekind domain can be generated by at most two elements, so Dedekind domains are Noetherian. Example. The additive group of polynomials of degree less than or equal ton over the rationals is a module over IQl generated by {1, x, x 2 ... xn }. This is a Noetherian module. Furthermore, IQ[X] is a Noetherian ring, but IQ[XJ.X2 J is not a Noetherian ring as the ideal (X1 X2 ) is not finitely generated. Example. A vector space over a field is a module; the submodules being the vector subspaces. Any finite dimensional vector space is an example of a Noetherian module, while infinite dimensional vector spaces are not Noetherian. Theorem 2.1. (Hilbert Basis Theorem) A is a Noetherian ring if and only if A[X] is a Noetherian ring. Proof. ( =*') By contradiction, suppose A[X] is not Noetherian, and let I be an ideal of A[ X] which cannot be finitely generated. Choose /1 in I of least degree and continue as follows: if !k has been chosen, select /k+l of least degree in IU1o ... fk), where (!,, ... !k) is the ideal of A[X] generated by the /1o ... fk Denote by nk and ak the degree of !k and the leading term of fk respectively. By method of choice, n1 ::; n 2 ::; ::; nk ::; .. and the ideals of A form a strictly ascending chain of ideals. To verify this, suppose (a 1 ... ,ak) =(a" ... ,ak+l) for some k. Then ak+l can be written as 2::;=, r,a,, where r; EA. Then 13
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is a polynomial of smaller degree than j, which is also in I(!1 .. h), contrary to our assumption. Therefore A is also not Noetherian. ( <:=) Conversely, suppose A[X] is Noetherian and let I1 I2 I3 be an ascending sequence of ideals of A. By hypothesis, the chain I1A[X] I2A[X] terminates, so there is an integer n for which I,A[X] = I,+1A[X]. Then I, A[ X] n A= I,+lA[X] n A. Suppose b is in I,+1 then b is also in I,+1A[X] n A= InA[ X] n A and hence also in InSince we already have In In+1 the chain of ideals terminates and A must be Noetherian. D Lemma 2.1. Let N, N' and E be submodules of an Amodule M. If N N', NnE= N' n E and N + E = N' + E, then N = N'. Proof. Let n' be an element of N'. Then n' is in N + E and we can write n' = n + e for some n EN and e E E. Because n N', n'nisin both N' and E and therefore inN n E which is inN. Thus n' EN and N = N'. D Theorem 2.2. Given an exact sequence of Amodules M is Noetherian if and only if M' and M" are Noetherian. Proof. We will verify the ascending chain property for submodules in the following. ( =;.) Suppose M satisfies the ascending chain condition for submodules. By the injectivity of a, any ascending chain of submodules of M' is isomorphic to an ascending chain of submodules of M and therefore terminates. For M", there is a one to one inclusion preserving correspondence between the submodules of M" and the submodules of M whlch contain a(M'). So any ascending chain of submodules of M" must also terminate. ( <:=) Suppose now that M' and M" satisfy the ascending chain condition for submodules and let 14
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be an ascending chain of submodules of M. Then are ascending chains of submodules of 1\!l' and M" respectively. By hypothesis there is an integer u such that if m > u, a1(Nm) = a1(Nu) and f3(Nm) = f3(Nu) Using the fact that a(a1()) =En Im(a) and {31({3()) = E + Ker(/3) for any submodule E of M, it follows that a(a1(Nm)) = Nm n a(M') and a(a1(Nu)) =Nun a(M'), so Nm n a(M') =Nun a(M'). Furthermore, {31({3(Nm)) = Nm + a(M') and {31({3(Nu)) = Nu + a(M'), SO Nm + a(M') = Nu + a(M'). Because Nu Nm, it follows from the previous lemma that Nm = Nu and M is Noetherian. 0 Corollary 2.1. The Amodules M; (1 :<:; i :<:; k) are Noetherian if and only if the direct sum EB:=1 M; is a Noetherian Amodule. Proof. (By induction on k) If k = 1 there is nothing to prove. Suppose the corollary is true for k = m 1. The result follows from theorem 2.2 and the exact sequence m m 0 __, M;> @M;+ @M; __, 0 i:::::l i:::::l '#j for each j. 0 Theorem 2.3. If A is a Noetherian ring and M is a finitely generated Amodule, then M is Noetherian. Proof. Let { e1 ... ek} generate M as an Amodule. Then there exists an epimorphism k k @e;A> L;e;A. i:l i=l The Amodule A is Noetherian and the mapping A+ e;A is an epimorphism of Amodules for each generator. By theorem 2.2 each e;A is Noetherian apd it follows 15
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from corollary 2.1 that the direct sum EB:=l e;A is Noetherian. Then again by theorem 2.2, the epimorphism k k ElJe;A+ I;e;A = M i:::::l i:::::l shows that M is Noetherian. 0 In the following theorem we use the fact that M is an Amodule if and only if M is an A/ Ann(M)module. A proof of this can be found in [10, page 65, proposition 24] Theorem 2.4. Let M be a finitely generated Amodule. M is Noetherian if and only if A/ Ann(M) is a Noetherian ring. Proof. (<=) This follows from theorem 2.3. ( =;.) Suppose M is Noetherian. Let { e1 ... ek} generate M and let U; = Ann( e; ). The epimorphism of Amodules A + e;A has kernel U;, so AjU; is isomorphic to e;A which, being a submodule of a Noetherian module is itself Noetherian. So each AjU; is Noetherian which implies that the direct sum Aja; is Noetherian. Now the homomorphism k A+ ElJAja; i=l has kernel n:=1 a; = Ann( M), so Aj Ann( M) is isomorphic to a submodule of the direct sum and hence is itself Noetherian. 0 2.2. Modules of Fractions. In this section we will show that for any multiplicatively closed subset S of the ring A, if M is a Noetherian Amodule, then s1 M is a Noetherian s1 AModule. To do this we need to first look at the result of taking a submodule of the module of fractions, contracting it to a submodule of M, and extending back into the module of fractions. Lemma 2.2. If N' is a submodule of S'M, then (,p1(N')) = N', where is the cannonical homomorphism. 16
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Proof. See [10, page 137, proposition 2] 0 Theorem 2.5. Let S be a multiplicatively closed subset of A and M an Amodule. If M is Noetherian, then s1 M is Noetherian. Proof. Let N'1 <;; N'2 <;; be an ascending chain of submodules of S1 M. Then the ascending chain of sub modules of M terminates. Then the chain of their images under does too, but this chain is our original chain by the previous lemma. So every ascending chain of sub modules of s1 M terminates, and it is therefore Noetherian. 0 2.3. Primary Decomposition. A submodule of M is irreducible in M if it cannot be written as the intersection of two submodules, each of which properly contains it. Proposition 2.2. Suppose N is irreducible in M. Then the following two conditions are equivalent: (1) For all bE A, the sequence N: b" is stationary, (2) N is primary in M. Proof. (1 =? 2) Let r tM(N). Then there exists an x EM for which r'x N for all k > 0. However, by hypothesis, there is an n > 0 for which N: r" = N: rn+1 By propositon 1.12, N = (N + Ar"x)n(N: r"), but N is irreducible and therefore must equal one of these submodules. If N = (N + Ar"x), then r"x is inN, contrary to the choice of x and r. SoN= (N: r"). Now suppose ryE N, then r"y EN implies y is in N, so N is primary. (2 =? 1) Let bE A, then if bE rM(N) there exists a k > 0 such that b' M <;; N, in other words N: b' = M. Thus N: b' is stationary. If on the other hand b is not in tM(N), then N: b' = N for all k > 0 and again N: b" is stationary. 0 Theorem 2.6. If M is a Noetherian Amodule, then every submodule of M has a primary decompostion. 17
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Proof. We will show that every submodule can be written as a finite intersection of irreducible submodules, from which the lemma will follow by proposition 2.2 Let l1 be the set of submodules of .M which cannot be written as an intersection of finitely many irreducible submodules and suppose l1 is not empty. As M is Noetherian, fl has a maximal element N. N is not irreducible, soN can be written as the intersection of two submodules which strictly contain it, say N = N' n N". By the max:imality of N we see that N' and N" are not in l1 and hence each can be written as a finite intersection of irreducible submodules. But this means that N can also, so we must have l1 equal to the empty set and the lemma is proved. D 2.4. Krull Intersection Theorem. In this section we prove the Krull intersection theorem. To do this we use a form of Nakayama's Lemma from [9, page 11]. Lemma 2.3. Let A be a ring, M a finitely generated Amodule and I an ideal of A. Suppose I1'vf = M, then there exists an element a E A of the form a = 1 + x for some x E I such that aM = 0. Proof. Let M = e1A + + e,A. If k = 1, then putting x = 0 gives e1A = e1I which implies M = e1A = 0. By induction on the number of generators, suppose the lemma is true for k1. Put M' = M / e,A. By the induction hypothesis there is an x E I for which (1 + x)M' = 0. This means that (1 + x)M <;; e,A. Now IM = M implies (1 + x)IM <;; I(e,A) = e,I, so (1 + x)e, = ye, for some y E I. Then (1 + xy)e, = 0 and so (1 + xy)(1 + x)M = 0. Furthermore, (1 + xy)(1 + x) = 1 + x2 yxy where each of x2 y and xy is in I. The lemma follows by putting a= 1 + (x2 yxy). D Theorem 2.7. Let M be a Noetherian Amodule and I an ideal of A. An element m E M is in n::'=1 I" M if and only if m = rm for some r in I. Proof. (=>)Put N = nr M, then N is a submodule of M and therefore IN, also a submodule, has a primary decomposition by theorem 2.6. Let IN= N1 n n N, be a primary decomposition of IN, where each N; is )l;primary. For each i ;= 1, ... ,k, if 18
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I<;; p, then there is ann> 0 for which I" M <;; P7 M <;; N,. Then also N <;; N,. If however I )1: p,, then because IN<;; N; and N; is primary, again N <;; N,. Therefore and N =IN. By lemma 2.3, there is an element r of I for which (1 + r)N = 0. In other words, if mEN= ni" M then (1 + r)m = 0, or m = ( r)m. (=)If m = rm for some r in I, then m = r"m for all n > 0 and clearly m is in ni"M. o The Jacobson radical of a ring is the intersection of all maximal ideals of the ring and is characterized by the fact that x is in the Jacobson radical if and only if 1 + xy is a unit for all y in the ring. Theorem 2.8. Let I be an ideal of a ring A. Then the following statements are equivalent: (1) I is contained in the Jacobson radical; (2) For all Noetherian Amodules M, n':(;1 I" M = (0). Proof. Suppose I is contained in the Jacobson radical and let M be a Noetherian Amodule. By theorem 2.7 it follows that mE n';;J" M if and only if m = rm for some r E I. Since I is contained in the Jacobson radical, 1 + ra is a unit for all a E A. Then 0 = mrm = (1r)m implies that m = 0 as 1r is a unit in A. It follows that nc;;1I" M = (0). Now suppose n';;1I"M = (0) for all Noetherian Amodules M. Let m be a maximal ideal of A. Then A/m is an Amodule whose only submodule is (0), so it is Noetherian. Then I(A/m) is a submodule of A/m and so must be equal to either A/m or (0). If I(A/m) = A/m then nc;;1I"(A/m) = Ajm contradicting our assumption. So it must be that I(A/m) = (0), in other words I C m. It follows that 1 is contained in every maximal ideal of A and hence in the Jacobson radical. 0 19
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3. LASKERIAN MODULES It was shown in the previous section that every submodule of a Noetherian module has a primary decomposition. In this section we will investigate modules whose defining property is that all proper submodules have primary decompositions. In section 3.1 we give an example of a Laskerian ring that is not Noetherian, and extend the results on exact sequences obtained for Noetherian modules to Laskerian modules. In section 3.2 we show that the localization of a Laskerian module is Laskerian. In section 3.3, a characterization of Laskerian modules as modules satisfying the two conditions (1) for all submodules N of M, Ass(M/N) is finite, and (2) for all submodules N and all elements a of A, the ascending chain of residual submodules N: a <:;; N: a' <:;; terminates. 3.1. Exact Sequences. Definition 6. An Amodule M is Laskerian if every proper submodule of M has a primary decomposition. From theorem 2.6 it follows that every Noetherian module is Laskerian. We now give an example of a Laskerian module which is not Noetherian to show that the class of Laskerian modules strictly contains the class of Noetherian modules. Example. Let A be a ring with one prime ideal m such that m" = (0) for some n > 1. Let a be a proper ideal of A (necessarily contained in m) and suppose abE a, while b f/: n. Then as every element not contained in m is a unit, a must be contained in m = tA(a). Hence n ismprimary and every ideal is primary, showing that A is Laskerian. Example. Let K be a field and V an infinite dimensional vector space. Form the set J( Ell V and define addition termwise and multiplication by (a,v)(a',v') = (aa',av' + a'v). Then J( Ell Vis a ring that is Laskerian but not Noetherian. The ideals of K Ell V are the sets (0, W) = {(O,w)lw E W}, where W is a subspace of V, and the only prime ideal is (0, V). As the previous example showed, this is a Laskerian ring, but it is not Noetherian because every infinite ascending 20
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chain W1 C W2 C ... of subspaces of V leads to an infinite ascending chain of ideals of K EB V, (0, W1 ) c (0, W2 ) c .... Lemma 3.1. Let M be an Amodule and M' a submodule. Let N be a submodule of M containing M'. Then N has a primary decomposition in M if and only ifNI M' has a primary decomposition in M I M'. Proof. This follows from the fact that N=N1n .. nNr > NIM'=NJ/M'n ... nNriM' and corollary 1.2. D Theorem 3.1. Given an exact sequence of Amodules 0 M' M L M" + 0, M is Laskerian if and only if M' and M" are Laskerian. Proof. Without loss of generality, we may assume that M' is a submodule of M and that M" is MIM'. ( =;.) Suppose M is a Laskerian Amodule and M' a submodule. Since every sub module of M' is also a sub module of M, M' is Laskerian. Let N IM' be a submodule of MIM', where N is a submodule of M containing M'. Then N has a primary decomposition in M and so by the lemma 3.1 N I M' also has a primary decomposition. This shows that M' and MIM' are Laskerian if M is Laskerian. suppose M' and MIM' are Laskerian and let N be a submodule of M. By lemma 3.1, N has a primary decomposition in M if and only if NI(N n M') has a primary decomposition in MI(N n M'). The image of N under the mapping M ._.. MIM' is (N + M')IM'. Since MIM' is Laskerian, (N + M')IM' has a primary decomposition. But (N + M')IM' is isomorphic to NI(N n M'), so it too has a primary decomposition and therefore, by the previous remark, N has a primary decomposition in M. D Corollary 3.1. The Amodules M, (1 :::; i :::; k) are Laskerian if and only if their direct sum EBf=1 M, is LaskeriaJJ.. 21
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Proof. This follows from theorem 3.1 and induction on kin the following exact sequence, k k 0+ Mj+ EJjM; +Ef:jM,+ 0. i::;: 1 i::;:l i#.j 0 Corollary 3.2. If A is a Laskerian ring and M a finitely generated Amodule, then M is Laskerian. Proof. First suppose M is generated by one element e. Then eA is isomorphic to A/(Anne) and so eA is Laskerian by proposition 3.1. Then by corollary 3.1 we see that the direct sum EB7=1 e,A is Laskerian. The corollary follows from the epimorphism n n Ef:je,A+ _L:C;A i==l i::::l and proposition 3.1. 0 3.2. Modules of Fractions. Proposition 3.1. Let M be an Amodule and S any multiplicatively closed subset of A. If M is Laskerian, then s1 M is Laskerian. Proof. Let N be a submodule of M with primary decomposition N = N1 n n N, where each N, is !\primary. By 1.3, for each index i satisfying p; n S = 0, s1 N, is p;primary in s1 M, while for each index j satisfying pj n S 1 0, s1 Nj = s1 M. Thus where i runs over all indicies satisfying p, n S = 0. As each submodule of s1 M is of the form s'N for some submodule N of M, the proposition follows. 0 22
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3.3. Ascending Chains of Residuals. Lemma 3.2. If N is a pprimary submodule of a module M and a E A, then the ascending chain of submodules N: a C N: a2 C C N: ak C is eventually stationary. Proof. If a f/:_ p = v N: M, then N: aj = N for all j, and the chain terminates. If, however, a E p, then aj E N: M for some positive integer j. Then N: aj = M and the chain again terminates. 0 Proposition 3.2. If N is a submodule of an Amodule M which admits a primary decomposition, and a is an element of A, then the ascending chain of submodules N: a C N: a 2 C C N: ak C terminates. Proof. Let N = N1 n n N, be a primary decomposition of N. Then As each of the terms ( N,: ak) is eventually stationary, so is the intersection of these terms. 0 Corollary 3.3. Every irreducible submodule of a Laskerian module is primary. Proof. This follows from proposition P:IrreduciblePrimary. 0 Corollary 3.4. Let M be a Laskerian Amodule, N a submodule and a E A. For sufficiently large k, N = (N: ak)n (N + akM). Proof. We clearly have the inclusion N C (N: ak) n (N + ak M). Let x be an element of (N: ak) n (N + ak M) and write x = n + akm, wheren EN and mE M. To show that x EN, multiply both sides by ak. Since x E (N: ak), akx EN. This implies that a2km EN. If k is large enough that the ascending chain in the previous proposition is 23
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stationary for all integers larger than or equal to k, then N: a''= N: a' and we have rn E N: a2k = N: a'. Thus a' m E N and so x = n + a' m is also in N. D Lemma 3.3. If an Amodule M is the direct sum of a finite number of submodules M,, (1::; i::; t), then AssA(M) = u;=,AssA(M,). Proof. By induction it suffices to prove the case t = 2. So let M = M1 Ell M2 By lemma 1.1 we see that AssA(MJ) U AssA(M2 ) <;; AssA(M). Conversely, by the same lemma, AssA(M) <;; AssA(MJ) U AssA(MIM,), but AssA(MIM,) = AssA(M,) and so equality holds. D Lemma 3.4. If the sub module N of M has a normal decomposition N = N1 n nN,, where each N1 is p;primary, thenAss(MIN) = {p, ... ,pt}. Proof. To show that p, is in Ass( MIN), note that if xis in each N; except for N,, then p1x EN for some kif and only if pEp,. This is because N, is p,primary. Thus p; is in Ass(M IN). Consider the homomorphism (MIN) N;/N This is injective, so MIN is isomorphic to a submodule of Elll=1MIN; and therefore the primes associated to MIN is a subset of the primes associated to the direct sum. But by lemma 3.3, Ass(EBMIN;) = UAss(MIN;) = {p, ... ,p,}. Hence Ass( MIN)<;; {p1 ,p,} which, combined with the above result shows that equality holds. D Theorem 3.2. An Amodule M is Laskerian if and only if the following two properties hold: (1) For every submodule N of M and every element a E A, the ascending chain of submodules (N: a) <;; (N: a2 ) <;; (N: a3 ) <;; is eventually stationary. 24
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(2) For all submodules N of M, the set AssA(M/N) of primes associated to M/N is finite. Proof. ( =;.) Suppose M is a Laskerian Amodule. Condition 1 is just proposition 3.2, and condition 2 is lemma 3.4. (=) Suppose now that conditions 1 and 2 hold for the Amodule M, and let N be a submodule of M. Let AssA(M/N) = {p1 . ,p,} be the finite set of primes assoctiated to M / N. Without loss of generality, assume p1 is minimal in this set. For each of the other primes p,, (2 :0: i:::; t), choose an a; E p; that is not in p1 and let a= a2a3 a,. LetS= {a'}k'=o' a multiplicatively closed subset of A that does not meet p1 By assumption 2, the chain terminates, which implies that 6s(N) = (N: a') for some sufficiently large integer k. We now want to show that N: a' is p1primary. Since 65(N)/N is the kernel of the cannonical homomorphism M/N >+ s1(M/N), proposition 1.8 says that so it follows from proposition 1.4 that 65(N) = (N: a') is p,primary. Now by proposition 1.11 we can write N = (N: a')n(N + a'M). It also follows from proposition 1.11 that N is p1primary inN+ a' M, in other words, AssA((N +a' M)/N) ={pi}. Now the primes associated with a union of submodules is equal to the union of the primes associated with each submodule (see [11, page 20, 3.3]), which makes the set of submodules E/N of M/N whose only associated prime is p1 an inductive set. Every totally ordered set has an upper bound, that upper bound being the union of the submodules in the chain. Let G be a submodule of M such that G/N is maximal among the submodules containing (N +a' M)/N such that AssA(GjN) = {p1}. We will show thq,t 25
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N = (N: ak) n G, that G is maximal among submodules whose intersection with N: ak is Nand that AssA(M/G) = {p,,Pa, ... ,pt}. Clearly we (N: ak) n G, so suppose xis in (N: ak) n G. Then akx is in N. But N is p1 primary in G, sox E G and akx EN implies that ak is in p1 unless x is already in N. Hence equality holds. To show that G is maximal among submodules whose intersection with N: ak is N, suppose G' ::J G. Then there is some x in G' such that, without loss of generality, p2 is the only prime minimal over N: x. If p21; p1 then there is apE P1 that is not in p2 for which px E N: ak. But then px is in N: ak and in G', but px is not in N. If, on the other hand, p2 ::J Ph then it must be that p2 = N: x, in which case ak x E N because ak E p2 by construction. In this case, x is in G' and in N: ak, but x is not in N. So G is maximal among the submodules whose intersection with N: ak is N. Now we can show that AssA(M/G) = {p2 ,p1}. We do this by showing that for any y E M G), if p' is a minimal prime of G: y then p' is a minimal prime of N: z for some z in M, and therefore p' is in AssA(M/N). Suppose that y E M G and that p' ;;? G: y minimally. Then we have the containments p'2G:y2N:y. If p' is a minimal prime of N: y then p' is in AssA(M/N). Otherwise there must be some p; associated to M / N such that p' 2 G: y 2 lJ; 2 N: y. In this case we must have p' = G: y. Enlarge the submodule G by adding y to it, G c G+Ay, where the containment is strict by the choice of y. Then by the maximality of G as a submodule whose intersection with N: a is N, there must be some g E G and r E Ap' such that z = g + ry is in N: ak but not in G. Consider N: z. If s E p', then sz = sg + sry is in N: a by assumption, and in G because s is in G: y = p'. Then sz E N and we have p' !:;; N: z. Conversely, suppose s E N: z. Then sz is in N 26
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which implies that sry isinG, so that sr E p'. But r p' implies that sis in p', so we have p' <;;; N: z aud hence equality. This means that p' is in AssA(Af/N) and so we finally conclude that AssA(M/N) = {p2 . ,pt}. Thus we have taken Nand represented it as N = (N: ak) n G, where N: ak is p1primary in M and assAM/G = {p2 . ,p,}. By repeating this process t 1 more times we arrive at a primary decomposition of N where each Pi associated to Nf / N belongs to some factor of the decompositon. 0 27
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4. ACCR lv10DULES From theorem 3.2, an Amodule M is Laskerian if and only if; for every submodule N and every a E A, the ascending chain terminates, and for every submodule N, the set of prime ideals Ass A ( M / N) is finite. In this section we will investigate modules which are required only to satisfy the first condition, called the ascending chain condition on residual submodules (accr). The study of modules satisfying ( accr) is pioneered by Dr. Sylvia Lu, and a thourough treatment of the subject can be found in [6], [7] and [8]. 4.1. Preliminaries. Definition 7. An Amodule M is said to satisfy the ascending chain condition on residual submodules (accr) if for every sub module N and every finitely generated ideal B of A, the ascending chain N: B N: B2 N: B3 terminates. In [6, theorem 1, p. 305] it is shown that the property (accr) is equivalent to property (1) of theorem 3.2 which is stated only for principle ideals. Laskerian modules, and therefore Noetherian modules, satisfy (accr). One of the properties of Laskerian modules is that for every submodule N and every element a E A, we can write N = (N: ak) n (N + ak M). In general, a module satisfies (accr) if and only if for every submodule Nand every finitely generated ideal B, we can write N = (N: Bk) n (N + Bk M) for some sufficiently large integer k. For a proof see [6, corollary, page 306]. Theorem 4.1. An Amodule M satisfies (accr) if and only if every factor module M/N satisfies (accr) for nonzero submodules N. 28
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Pmof. Let M be an Amodule that satisfies (accr) and let N be a submodule of M. For an arbitrary submodule E/N of M/N and finitely generated ideal B of A, we have E/N: B' == (E: B')fN. Since E: B' becomes stationary, so must E/N: B'. Conversely, suppose every factor module M/N satisfies (accr). If E: B E: B2 is a chain of residual submodules of M that does not terminate, we may assume that E is not the zero submodule. For otherwise there is an integer n for which E: B" strictly contains E and we may consider the chain starting at E: Bn. But then the non terminating chain E: B E: B2 leads to a nonterminating chain of submodules of M/E, (0): B (0): B' contradicts the assumption that M / E satisfies ( accr). D Theorem 4.2. Given the exact sequence of submodules 0 __, A1' __, A1 __, M" __, 0, M satisfies ( accr) if and only if M' and M" satisfy ( accr ). Proof. There is no loss of generality by assuming that M' is a submodule of M and that M" == M/M'. Suppose M satisfies (accr). Then M' as a submodule clearly satisfies (accr), and by theorem 4.1 we see that M" does too. Conversely, suppose M' and M/M' satisfy (accr). For any JY, a submodule of M, and any element a E A, we want to show that N: a N: a2 terminates. In the factor module we have (N + M')/M': (N + M')/M': a2 terminates, so there is some k for which (N + M')/M': a'== (N + M')/M': a'+'. But (N + M')/M': a' == ((N + M'): a')/M', so that (N + M'): a' == (N + M'): a'+1 Similarly, we have (N n M'): a'== (N n M'): a'+1 for some integer t. We will show that N: a'+' == N: ak+'+1 Suppose x is in N: ak+t+l. Then a'+t+l x E N and also in (N + M'): a'+t+l == (N + M'): a'. So a'x == n + m for some n EN and mE M'. Then a'+'+lx == a'+'(n + m) == a'+'n + at+1m is inN, hence at+'m EN n M', so that mE (N n M'): a'+'== (N n M'): a'. This implies that a'm is inN, so that finally, a'+'x == a'(n + m) == a'n +a 1m EN. In other words xis inN: ak+t. Hence M 29
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satisfies (accr). D Corollary 4.1. The Amodules M;, (1:::; i:::; t), satisfy (accr) if and only if the direct sum Elli;1M; satisfies (accr). Proof. The proof is similar to that of corollary 2.1. D Corollary 4.2. If A satisfies (accr), then any finitely generated Amodule satisfies ( accr ). Proof. For a set of generators e1 .. e, of 1;1, each module Ae; satisfies (accr), so then does the direct sum E!lAe;. Then M is a factor module of the direct sum and hence M satisfies (accr). D 4.2. Localization. In this section we show that the process of forming the module of fractions preserves the property of satisfying ( accr ). Theorem 4.3. If the Amodule M satisfies (accr) and Sis any multiplicatively closed subset of A, then s1M satisfies (accr). Proof. Every submodule of s1 M is of the form s1 N for some submodule N of M. Furthermore we have s1 N: (b/ s) = s'(N: b). So any sequence of submodules s1 N: (bjs) s1 N: (bjs)2 must terminate as the corresponding sequence N: b N: b2 in M terminates. D 4.3. Krull Intersection. Note that in the proof Nakayama's lemma, used to prove the Krull intersection theorem for Noetherian modules, [see 2.7] expliCit use of the Noetherian property of the module was used. Furthermore, embedded in the proof is the use of the property that if N is Jlprimary in M, then p M N. This property does not hold in general even for La.skerian modules, so a new approach is needed. 30
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Definition 8. The ArtinRees property is said to hold for an ideal B of A and a sub module N of M if there exists a positive integer n such that N n Bh M EN for all integers h 2': n. Lemma 4.1. If an Amodule M satisfies ( accr ), then the ArtinRees property holds for every finitely generated ideal B of A and every submodule N of M. Proof. Since M satisfies ( accr ), there is an n such that, for all h 2': n, EN= (EN: Bh) n (EN+ Bh Af). On the right hand side we have (EN: Bh) ;:;> N and (EN+ Bh M) ;:;> Bh M, so the right side contains N n Bh M from which the lemma follows. D Lemma 4.2. If the ArtinRees property holds for an ideal B of a ring A, and every cyclic submodule of M, then an element mE M is in M if and only if m = rm for some r in B. Proof. Suppose m is in M, then the submodule Am is also contained in the intersection and hence in Bk M for each positive integer k. Since the ArtinRees property holds for B and Am, for all sufficiently large integers h we have Am n Bh M Em. But Am n Bh M = Am by the previous remarks. It follows that Am Em and therefore that equality holds. Then for some r E B, rm = m. The other direction is trivial, so the lemma is proved. D Theorem 4.4. Let M be an Amodule satisfying (accr) and B a finitely generated ideal of A. An element mE M is in n;:'=1 B" M if and only if m = rm for some r in B. Proof. Since M satisfies (accr), and B is finitely generated, the ArtinRees property holds for B and every submodule of M. In particular it holds for B and every cyclic submodule. The theorem then follows from lemmas 4.1 and 4.2. 0 4.4. Polynomial Rings. In section 2.1 we showed that a ring A is Noetherian if and only if the ring of polynomials A[ X] is Noetherian. In this section we show that the ring of polynomials A[X] satisfies (accr) if and only if it is Noetherian. 31
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Theorem 4.5. A[ X] satisfies (accr) if and only if A is Noetherian. Proof. If A is Noetherian, then A[X] is Noetherian and therfore satisfies (accr). So suppose A[X] satisfies ( accr) and let I0 <;; I1 c I2 <;; be an ascending chain of ideals in A. If this sequence does not terminate, consider the ideal I' of polynomials whose coefficient on x; is in I;. Since the chain of ideals does not terminate, there is some element a in In+! that is not in In, and therefore ax"+l is in I' while ax" is not. This shows that the sequence I': x <;; I': x2 <;; ... does not terminate, contradicting the hypothesis. 0 32
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REFERENCES 1. M. Atiyah and I. Macdonald, Introduction to Commutative Algebra, AddisonWesley, Reading, 1969. 2. N. Bourbaki, Commutative Algebra, Hermann, Paris, 1972. 3. R. Gilmer and W. Heinzer, The Laskerian Property, Power Smes Rings and Noethenan Spectra, Proc. Amer. Math. Soc. 79, 1316. 4. W. Heinzer and J. Ohm, On the N oethenanlike Rings of E. G. Evans, Pro c. A mer. Math. Soc. 34, 7374. 5. E. Kunz. Introduction to Commutative Algebra and Algebraic Geometry, Birkhauser, Boston, 1985. 6. C. P. Lu, Modules Satisfying ACC on a Certam Type of Colons, Pacific Journal of Mathematics, Vol. 131, No.2, 303318. 7. C. P. Lu, Modules and Rings Satisfying (ACCR), Proc. Amer. Math. Soc. Vol. 117, No.1, 510. 8. C. P. Lu, Two Noetherian Properties of Rings Satisfyzng (accr), To Appear in Nova Journal of Algebra and Geometry. 9. H. Matsumura, Commutative Algebra, 2nd ed., Mathematics Lecture Note Series 56, BenjaminCummings, Reading 1980. 10. D. Northcott, Lessons on Rings, Modules and Multiplicities, Cambridge University Press, London, 1968. 11. N. Radu, Lectii De Algebra III, Universitatea Din Bucureti, Bucureti, 1981. 33
