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

$${N}_{1}\subseteq {N}_{2}\subseteq \cdots $$

eventually stabilizes; i.e., there is some ${n}_{0}$ such that ${N}_{k}={N}_{{n}_{0}}$ for every $k\ge {n}_{0}$.

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 $R$.

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 ${N}_{1}\u228a{N}_{2}$. But then ${N}_{2}$ cannot be maximal either, so there is some ${N}_{3}$ in $S$ with ${N}_{2}\u228a{N}_{3}$. Continuing as such, using the Axiom of Choice we can then 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 ${N}^{\prime}$, which by definition of $S$ is finitely generated. We claim ${N}^{\prime}=N$. Suppose not, so that there is some $x\in N\mathrm{\setminus}{N}^{\prime}$. Then the submodule generated by ${N}^{\prime}$ and $x$ is a finitely generated submodule of $N$ that is strictly larger than ${N}^{\prime}$, violating the maximality of ${N}^{\prime}$. Thus, we must have ${N}^{\prime}=N$ and hence $N$ is finitely generated. So (2) implies (3).

Finally, suppose (3) holds and let ${N}_{1}\subseteq {N}_{2}\subseteq \cdots $ 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 ${n}_{1},{n}_{2},\dots ,{n}_{a}$. Since $N$ is a union, each ${n}_{i}$ is contained in some submodule ${N}_{{j}_{i}}$. Choose any integer $m$ greater than ${j}_{1},\dots ,{j}_{a}$. Then every ${n}_{i}$ is contained in ${N}_{m}$, hence $N\subseteq {N}_{m}$, and hence $N={N}_{m}$. This implies ${N}_{k}={N}_{m}=N$ for all $k\ge m$.

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.