Commutative Algebra (VI): Chain Conditions
14 Feb 2019We have previously considred finite generated and finite presented modules. In this post, we look at other finiteness properties of modules based on chain conditions.
Chain conditions
Proposition: Let be a partially ordered set. Then the following conditions are equivalent:
- Any nonempty has a maximal element (ie. an element such that for all );
- Any ascending chain of elements of must stop after finitely many steps (ie. there does not exist an infinite ascending chain);
- Any nondecreasing chain of elements in must be eventually constant (ie. there exists such that for all ).
Proof: () The contrapositive is clear: if is an infinite ascending chain, then does not have a maximal element.
() By removing repeated terms from a nondecreasing chain, we get an ascending chain, which only has finitely many terms by hypothesis; hence all terms after the first appearance of the largest term are equal to the largest term.
() We prove the contrapositive. Take nonempty such that has no maximal element. By the axiom of choice, there exists a function such that for all . Pick any , and inductively define . Then is an infinite ascending chain, which in particular is not eventually constant.
The equivalent conditions above are called the ascending chain condition (a.c.c.) for a partially ordered set . By reversing the order, we define the descending chain condition (d.c.c.) analogously.
Proposition: If satisfies a.c.c. (resp. d.c.c.) and , then also satisfies a.c.c. (resp. d.c.c.).
Proof: Every nonempty has a maximal (resp. minimal) element; in particular, so does every nonempty .
Noetherian and Artinian modules
Let be a (commutative unital) ring, and an -module. Note that the set of submodules of form a partially ordered set under inclusion. We say that is a Noetherian module (resp. Artinian module) if its set of submodules satisfies a.c.c. (resp. d.c.c.).
For instance, consider as a -module. Its submodules are and for integers . Hence ascending chains of (nonzero) submodules correspond to decreasing sequences of positive integers, which must terminate in finitely many steps by the well-ordering property of . Thus is Noetherian. However, we have the infinite descending sequence
so is not Artinian.
For another example, let be the abelian group of dyadic rationals,
Now consider the -module . For any submodule , note that if and is odd, then is invertible mod implies that . Hence we must have one of the following:
Hence descending chains of submodules of correspond to decreasing sequences of , which must terminate after finitely many steps. Thus is Artinian. However, we have the infinite ascending sequence
so is not Noetherian.
The Noetherian and Artinian properties carry over to submodules and quotient modules:
Proposition: Let be a Noetherian (resp. Artinian) -module, and let be a submodule. Then and are also Noetherian (resp. Artinian).
Proof: Note that the set of submodules of is a subset of the set of submodules of . Also, by the lattice isomorphism theorem, the set of submodules of is order-isomorphic to a subset of the set of submodules of (namely, those which contain ). The result now follows from the previous proposition. .
In the other direction, we have the following result:
Proposition: Let be an -module, and let be a submodule. If and are both Noetherian (resp. Artinian), then is also Noetherian (resp. Artinian).
Proof: We prove the statement for Noetherian modules; the proof for Artinian modules is analogous.
Let be a nondecreasing chain of submodules of . By hypothesis, the nondecreasing chains
are both eventually constant. Hence there exists such that for all , we have
But , so for all .
Hence the nondecreasing chain is eventually constant, so is Noetherian.
Recall that a composition series of is a chain of submodules
such that each is simple. We have proven in the previous post that any two composition series of have the same length, known as the (composition) length .
Proposition: Let be an -module. Then has a composition series if and only if it is both Noetherian and Artinian.
Proof: () We proved that for any submodules of , we have . Hence any ascending or descending chain of submodules of has at most elements, so is Noetherian and Artinian.
() Let , and inductively define to be a minimal submodule strictly containing . Since is Artinian, this always exists unless . Note that by minimality, has no proper submodules except , ie. is simple.
Now we have an ascending chain of submodules of . Since is Noetherian, this chain must stop after finitely many steps, so for some . Then
is a composition series for .
Hence being both Noetherian and Artinian is equivalent to the finiteness of composition length.
Noetherian and Artinian rings
We say that is a Noetherian ring (resp. Artinian ring) if it is Noetherian (resp. Artinian) as an -module; equivalently, if its set of ideals satisfies a.c.c. (resp. d.c.c.). The above results for Noetherian and Artinian modules yield the following:
Proposition: Let be a Noetherian (resp. Artinian) ring, and let be an ideal. Then is also Noetherian (resp. Artinian).
Proof: We proved that is a Noetherian (resp. Artinian) -module. Now note that the -submodules of are precisely the -submodules, namely the ideals of .
Hence the Noetherian and Artinian properties carry over to quotient rings, but not subrings in general (since subrings of are not -modules).
Proposition: Let be a Noetherian (resp. Artinian) ring. Then any finite -module is also Noetherian (resp. Artinian).
Proof: First we induct on to show that the free module is Noetherian (resp. Artinian). This is given for . If this is true for , then the -modules and are Noetherian (resp. Artinian), hence by the previous result, so is .
Now any finite -module is a quotient of some free module , and hence is also Noetherian (resp. Artinian).
Corollary: If is a Noetherian (resp. Artinian) ring and is a ring which is finitely generated as an -module, then is also a Noetherian (resp. Artinian) ring.
Proof: By the previous proposition, is a Noetherian (resp. Artinian) -module. Since every -submodule of is also an -submodule, is also a Noetherian (resp. Artinian) -module.
A warning to the reader: note that, for instance, is not finitely generated as an -module, even though it is finitely generated as an -algebra (by ). For Noetherian rings, the extension of the above result to finitely generated algebras uses the Hilbert basis theorem, which we will cover in the next post.
Artinian rings
We have seen that the Noetherian and Artinian conditions for modules are independent. However, for rings we have the following surprising result:
Akizuki-Hopkins-Levitzki Theorem: Any Artinian ring is Noetherian.
We proceed in several steps.
Proposition: Let be an Artinian ring. Then has finitely many maximal ideals.
Proof: Suppose otherwise; then let be pairwise distinct maximal ideals. For any , pick for . Since is prime, we have
Then we have the infinite descending chain of ideals
which is the desired contradiction.
Proposition: Let be an Artinian ring. Then the Jacobson radical is nilpotent.
Proof: Write , and consider the nonincreasing chain
which must be eventually constant; hence there exists with .
Assume for sake of contradiction that . Consider the family of ideals such that ; since , this family is nonempty, so it has a minimal element .
Now there exists such that , so and implies by minimality of ; in particular, is finitely generated as an -module.
Also, note that
so implies by minimality of . By Nakayama’s lemma, we have , so , which is the desired contradiction.
Corollary: Let be an Artinian ring. Then every prime ideal of is maximal.
Proof: Let the maximal ideals of be , and let be any prime ideal of . Since the Jacobson radical is nilpotent, we have
for some . Hence we must have for some , so .
Proof of Akizuki-Hopkins-Levitzki Theorem: Let be an Artinian ring, with maximal ideals , and Jacobson radical satisfying . Consider the chain of ideals
For any two consecutive terms in this sequence, say and , note that is a -vector space, whose -vector subspaces are precisely its -submodules. Hence
so is finite-dimensional over . Thus . By additivity of composition length, we get , so has a composition series implies is a Noetherian ring.
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