Commutative Algebra/Generators and chain conditions

Generators

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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

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Noetherian and Artinian modules

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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

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  • 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.