Noetherian modules

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 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 $k\ge m$.