Noetherian modules

A finiteness condition for modules

Many additional properties of modules arise if we assume some type of "finiteness condition" on the module. One such condition is the following:

Definition of Noetherian module

A left $R$ -module $M$ is Noetherian (or satisfies the ascending chain condition on submodules) if there are no infinite strictly increasing chains of submodules in $M$ . In other words, every increasing chain of submodules

N_{1} \subseteq N_{2} \subseteq \dots

eventually stabilizes; i.e., there is some $k_{0}$ such that $N_{k} = N_{k_{0}}$ for every $k \geq k_{0}$ .

A ring $R$ is Noetherian if it is Noetherian as a left module over itself; i.e., if there are no strictly increasing infinite chains of left ideals in $R$ .

There are certainly other "finiteness conditions" one can imagine (how about a descending chain condition?), but the above is one of the most popular, likely since it leads to many structural results.

It is useful to have some alternative characterizations of this Noetherian property. Fortunately, we have the following:

Noetherian and finite generation

The following are equivalent for a left $R$ -module $M$ :

$M$ is Noetherian.
Every nonempty collection of submodules of $M$ contains a maximal member (under inclusion).
Every submodule of $M$ is finitely generated.

Let's take a look at how the above proposition is proved. First assume $M$ is Noetherian. Let $S$ be any nonempty collection of submodules of $M$ but suppose, towards a contradiction, that $S$ does not contain a maximal element. Choose any $N_{1} \in S$ . By assumption $N_{1}$ is not maximal (as $S$ has no maximal member), so there is some $N_{2} \in S$ with $N_{1} ⊊ N_{2}$ . But then $N_{2}$ cannot be maximal either, so there is some $N_{3}$ in $S$ with $N_{2} ⊊ N_{3}$ . Continuing as such, using the Axiom of Choice we can construct an infinite strictly increasing sequence of submodules, violating our assumption that $M$ is Noetherian. So, (1) implies (2).

Now assume (2) holds and let $N$ be any submodule of $M$ . Let $S$ be the collection of all finitely generated submodules of $N$ . By assumption (2) this collection $S$ contains a maximal element $N^{'}$ , which by definition of $S$ is finitely generated. We claim $N^{'} = N$ . Suppose not, so that there is some $n \in N ∖ N^{'}$ . Then the submodule generated by $N^{'}$ and $n$ is a finitely generated submodule of $N$ that is strictly larger than $N^{'}$ , violating the maximality of $N^{'}$ . Thus, we must have $N^{'} = N$ and hence $N$ is finitely generated. So (2) implies (3).

Finally, suppose (3) holds and let $N_{1} \subseteq N_{2} \subseteq \dots$ be a chain of submodules of $M$ . Let $N$ be the union of this family of submodules (and so $N$ is itself a submodule of $M$ ). Then by assumption (3) $N$ is finitely generated, say by elements $n_{1}, n_{2}, \dots, n_{l}$ . Since $N$ is a union, each $n_{i}$ is contained in some submodule $N_{j_{i}}$ . Choose any integer $k_{0}$ greater than $j_{1}, \dots, j_{l}$ . Then every $n_{i}$ is contained in $N_{k_{0}}$ , hence $N \subseteq N_{k_{0}}$ , and hence $N = N_{k_{0}}$ . This implies $N_{k} = N_{k_{0}} = N$ for all $k \geq k_{0}$ .

Recall that a ring $R$ can be viewed as a left $R$ -module over itself, using its internal multiplication for the $R$ -action. Moreover, the ideals of $R$ are then exactly the submodules of $R$ when it is viewed as a left $R$ -module.^[1] Now, if $R$ is a PID then every ideal is principal (hence finitely generated), so by the above theorem we have the following:

Corollary

Every principal ideal domain is Noetherian (when viewed as a module over itself).

Before moving on, we need to:

Beware!

Even if a left $R$ -module $M$ is finitely generated, it can have submodules that are not finitely generated! For example, let $R = F [x_{1}, x_{2}, \dots]$ be a polynomial ring in infinitely many variables over a field $F$ . As an $R$ -module, $R$ is generated by the element $1$ (and hence is itself finitely generated). However, the submodules of the $R$ -module $R$ are exactly the ideals of the ring $R$ . In particular, consider the ideal/submodule $I$ generated by the set ${x_{1}, x_{2}, \dots}$ . You can show this ideal is not finitely generated.

So the condition that $M$ be Noetherian is stronger than the condition that $M$ be finitely generated.

Long story short: being Noetherian is stronger than being finitely generated.

Suggested next note

Linear independence, rank and the structure of free modules

We really should be defining a functor of some type, but it's a bit tricky to phrase correctly. Can you figure out how? ↩︎