# Noetherian Modules

Through this article, *A* is a fixed ring. For the first two sections, all modules are over *A*.

Recall that a submodule of a finitely generated module is not finitely generated in general. This will not happen if we constrain ourselves to a better behaved class of modules.

Definition.Let M be an A-module. Consider the set of submodules of M, ordered by reverse inclusion, i.e. if and only if . We say M is

noetherianif is a noetherian poset.

By proposition 1 here, *M* is noetherian if either of the following equivalent conditions holds.

- Every non-empty collection of submodules of
*M*has a maximal element. - If is a sequence of submodules of
*M*, then for some .

Philosophically, noetherian is a type of “finiteness” condition on modules as the results below show.

Proposition 1.Given an exact sequence of A-modules:

,

M is noetherian if and only if N and P are.

**Proof**

Without loss of generality, we assume and .

(⇒) Suppose *M** *is noetherian. *N* is noetherian since submodules of *N* are submodules of *M*. Similarly, *P* is noetherian since submodules of *P* correspond bijectively to submodules of *M* containing *N*.

(⇐) Let be a sequence of submodules of *M*. We get sequences of submodules

Since *N* and *M*/*N* are noetherian, there is a *k* such that and . Now apply the following to finish to job. ♦

**Exercise A**

Prove that if and *N* are submodules of *M* such that and , then .

Corollary 1.

- If M and N are noetherian modules, so is .
- If N and N’ are noetherian submodules of M, so is N + N’ (because it is a quotient of ).

# Classification of Noetherian Modules

Lemma 1.A noetherian module M is finitely generated.

**Proof**

Take the set of all finitely generated submodules of *M*. Since *M* is noetherian it has a maximal element . If pick any . Then is a finitely generated submodule of *M* which properly contains , which contradicts maximality of . ♦

Proposition 2.A module M is noetherian if and only if every submodule is finitely generated.

**Proof**

(⇒) Any submodule of a noetherian module is noetherian, and hence finitely generated.

(⇐) Pick any sequence of submodules of *M*. Let . Show that this is a submodule of *M*. By the given condition it is finitely generated by, say, . Then for some *n* we have . It follows that . ♦

Definition.The ring A is

noetherianif it is noetherian as a module over itself. By proposition 2, A is noetherian if and only if all its ideals are finitely generated.

In particular, a noetherian ring *A* satisfies a.c.c. on the set of its principal ideals. Hence in a noetherian integral domain, every non-zero element can be factored as a product of irreducibles.

Proposition 3.If the ring A is noetherian, an A-module M is noetherian if and only if it is finitely generated.

**Proof**

(⇒) Apply lemma 1. (⇐) Since *A* is noetherian as an *A*-module, so is any finite direct sum . And since is finitely generated, it is a quotient of some ; hence *M* is noetherian. ♦

*In summary, the class of finitely generated modules over noetherian rings is well-behaved: it is closed under taking submodules, quotient modules, finite direct sums and finite sums of submodules.*

**Exercise B**

Let *M*, *N* be modules; prove the following.

- If
*M*and*N*are finitely generated, so is . - If
*M*is noetherian and*N*is finitely generated, then is noetherian.

# Constructing Noetherian Rings

Thus the question now stands: which rings are noetherian? Obviously fields are noetherian, having only two ideals. Next we have the following easy constructions.

Lemma 2.Any principal ideal domain is noetherian.

**Proof**

Indeed we showed earlier that *A* satisfies a.c.c. on principal ideals. But all ideals of *A* are principal. ♦

Lemma 3.If A is a noetherian ring, so is any localization .

**Proof**

We showed that any ideal of is of the form for some ideal . If is generated by then is generated by . ♦

Lemma 4.If A and B are noetherian rings, so is .

**Proof**

We saw that any ideal of is of the form where are ideals of *A*, *B* respectively. ♦

Lemma 5.If A is noetherian, so is any quotient .

**Proof**

Any ideal of is of the form where is an ideal, so it is finitely generated. ♦

# Hilbert Basis Theorem

Hilbert Basis Theorem.If A is a noetherian ring, so is .

**Proof**

Suppose is not finitely generated; we pick of minimum degree. Recursively, pick of minimum degree. Let so that . Suppose the leading term of is , where .

The sequence of ideals of *A* eventually terminates so for some *n*. We write for some . Now

Let *g* be the sum of all these polynomials; then has degree and leading coefficient . Hence has degree and lies outside , which contradicts minimality of . ♦

Immediately we have:

Corollary 2.If A is a noetherian ring, so is any finitely generated A-algebra.

**Proof**

Such an algebra must be a (ring) quotient of for some *n*. ♦

**Examples**

Any finitely generated algebra over a field or is noetherian; more generally any localization of such a ring is noetherian.

**Exercise C**

Prove the following for *A*-algebras *B* and *C*.

- If
*B*and*C*are of finite type over*A*, so is . - If
*B*is of finite type over*A*, and*C*is noetherian, then is noetherian.

**Note**

In most cases of interest (e.g. algebraic number theory and algebraic geometry), our rings and modules will be noetherian.

Most *A*-algebras we look at are of finite type over *A* but *not finite over A*. In other words, they are finitely generated as *A*-algebras but not as *A*-modules. Thus, if we look at such an algebra as an *A*-module, it is no longer noetherian and we have to be careful in applying the above results.

Next, by proposition 1 here, if *M* is a finitely generated module over a noetherian ring, it is flat if and only if it is projective. But that does not mean we can safely ignore their difference when dealing with concrete examples, for in the case of flatness, we are usually more interested in flat algebras rather than modules.

In summary:

- flat
*A*-algebras correspond to families of varieties which “vary nicely”; - projective
*A*-modules correspond to locally free vector bundles.