Commutative Algebra/Generators and chain conditions

From Wikibooks, open books for an open world
Jump to navigation Jump to search

Generators[edit]

Definition 6.1 (generators of modules):

Let be a module over the ring . A generating set of is a subset such that

.

Example 6.2:

For every module , the whole module itself is a generating set.

Definition 6.3:

Let be a module. is called finitely generated if there exists a generating set of which has a finite cardinality.

Example 6.4: Every ring is a finitely generated -module over itself, and a generating set is given by .

Definition 6.5 (generated submodules):

Exercises[edit]

Noetherian and Artinian modules[edit]

Definition 6.6 (Noetherian modules):

Let be a module over the ring . is called a Noetherian module iff for every ascending chain of submodules

of , there exists an such that

.

We also say that ascending chains of submodules eventually become stationary.

Definition 6.7 (Artinian modules):

A module over a ring is called Artinian module iff for every descending chain of submodules

of , there exists an such that

.

We also say that descending chains of submodules eventually become stationary.

We see that those definitions are similar, although they define a bit different objects.

Using the axiom of choice, we have the following characterisation of Noetherian modules:

Theorem 6.8:

Let be a module over . The following are equivalent:

  1. is Noetherian.
  2. All the submodules of are finitely generated.
  3. Every nonempty set of submodules of has a maximal element.

Proof 1:

We prove 1. 2. 3. 1.

1. 2.: Assume there is a submodule of which is not finitely generated. Using the axiom of dependent choice, we choose a sequence in such that

;

it is possible to find such a sequence since we may just always choose , since is not finitely generated. Thus we have an ascending sequence of submodules

which does not stabilize.

2. 3.: Let be a nonempty set of submodules of . Due to Zorn's lemma, it suffices to prove that every chain within has an upper bound (of course, our partial order is set inclusion, i.e. ). Hence, let be a chain within . We write

.

Since every submodule is finitely generated, so is

.

We write , where only finitely many of the are nonzero. Hence, we have

for suitably chosen . Now each is eventually contained in some . Since the are an ascending sequence with respect to inclusion, we may just choose large enough such that all are contained within . Hence, is the desired upper bound.

3. 1.: Let

be an ascending chain of submodules of . The set has a maximal element and thus this ascending chain becomes stationary at .

Proof 2:

We prove 1. 3. 2. 1.

1. 3.: Let be a set of submodules of which does not have a maximal element. Then by the axiom of dependent choice, for each we may choose such that (as otherwise, would be maximal). Hence, using the axiom of dependent choice and starting with a completely arbitrary , we find an ascending sequence

which does not stabilize.

3. 2.: Let be not finitely generated. Using the axiom of dependent choice, we choose first an arbitrary and given we choose in . Then the set of submodules

does not have a maximal element, although it is nonempty.

2. 1.: Let

be an ascending chain of submodules of . Since these are finitely generated, we have

for suitable and . Since every submodule is finitely generated, so is

.

We write , where only finitely many of the are nonzero. Hence, we have

for suitably chosen . Now each is eventually contained in some . Hence, the chain stabilizes at , if is chosen as the maximum of those .

The second proof might be advantageous since it does not use Zorn's lemma, which needs the full axiom of choice.

We can characterize Noetherian and Artinian modules in the following way:

Theorem 6.9:

Let be a module over a ring , and let . Then the following are equivalent:

  1. is Noetherian.
  2. and are Noetherian.

Proof 1:

We prove the theorem directly.

1. 2.: is Noetherian since any ascending sequence of submodules of

is also a sequence of submodules of (check the submodule properties), and hence eventually becomes stationary.

is Noetherian, since if

is a sequence of submodules of , we may write

,

where . Indeed, "" follows from and "" follows from

.

Furthermore, is a submodule of as follows:

  • since and ,
  • since and .

Now further for each , as can be read from the definition of the by observing that . Thus the sequence

becomes stationary at some . But If , then also , since

.

Hence,

becomes stationary as well.

2. 1.: Let

be an ascending sequence of submodules of . Then

is an ascending sequence of submodules of , and since is Noetherian, this sequence stabilizes at an . Furthermore, the sequence

is an ascending sequence of submodules of , which also stabilizes (at , say). Set , and let . Let . Then and thus , that is for an and an . Now , hence . Hence . Thus,

is stable after .


Proof 2:

We prove the statement using the projection morphism to the factor module.

1. 2.: is Noetherian as in the first proof. Let

be a sequence of submodules of . If is the projection morphism, then

defines an ascending sequence of submodules of , as preserves inclusion (since is a function). Now since is Noetherian, this sequence stabilizes. Hence, since also preserves inclusion, the sequence

also stabilizes ( since is surjective).

2. 1.: Let

be an ascending sequence of submodules of . Then the sequences

and

both stabilize, since and are Noetherian. Now , since . Thus,

stabilizes. But since , the theorem follows.

Proof 3:

We use the characterisation of Noetherian modules as those with finitely generated submodules.

1. 2.: Let . Then and hence is finitely generated. Let . Then the module is finitely generated, with generators , say. Then the set generates since is surjective and linear.

2. 1.: Let now . Then is finitely generated, since it is also a submodule of . Furthermore,

is finitely generated, since it is a submodule of . Let be a generating set of . Let further be a finite generating set of , and set . Let be arbitrary. Then , hence (with suitable ) and thus , where ; we even have due to , which is why we may write it as a linear combination of elements of .

Proof 4:

We use the characterisation of Noetherian modules as those with maximal elements for sets of submodules.

1. 2.: If is a family of submodules of , it is also a family of submodules of and hence contains a maximal element.

If is a family of submodules of , then is a family of submodules of , which has a maximal element . Since is inclusion-preserving and for all , is maximal among .

2. 1.: Let be a nonempty family of submodules of . According to the hypothesis, the family , where is defined such that the corresponding are maximal elements of the family , is nonempty. Hence, the family , where

,

has a maximal element . We claim that is maximal among . Indeed, let . Then since . Hence, . Furthermore, let . Then , since . Thus for a suitable , which must be contained within and thus also in .

We also could have first maximized the and then the .

These proofs show that if the axiom of choice turns out to be contradictory to evident principles, then the different types of Noetherian modules still have some properties in common.

The analogous statement also holds for Artinian modules:

Theorem 6.10:

Let be a module over a ring , and let . Then the following are equivalent:

  1. is Artinian.
  2. and are Artinian.

That statement is proven as in proofs 1 or 2 of the previous theorem.

Lemma 6.11:

Let be modules, and let be a module isomorphism. Then

.

Proof:

Since is also a module isomorphism, suffices.

Let be Noetherian. Using that is an inclusion-preserving bijection of submodules which maps generating sets to generating sets (due to linearity), we can use either characterisation of Noetherian modules to prove that is Noetherian.

Theorem 6.12:

Let be modules and let be a surjective module homomorphism. If is Noetherian, then so is .

Proof:

Let be a submodule of . By the first isomorphism theorem, we have . By theorem 6.9, is Noetherian. Hence, by lemma 6.11, is Noetherian.

Exercises[edit]

  • Exercise 6.2.1: Is every Noetherian module finitely generated?
  • Exercise 6.2.2: We define the ring as the real polynomials in infinitely many variables, i.e. . Prove that is a finitely generated -module over itself which is not Noetherian.