criterion for a module to be noetherian

Theorem 1.

A module is noetherian if and only if all of its submodules andquotients (http://planetmath.org/QuotientModule) are noetherian.

Proof.

Suppose $M$ is Noetherian (over a ring $R$), and $N\subseteq M$ a submodule. Since any submodule of $M$ is finitely generated, any submodule of $N$, being a submodule of $M$, is finitely generated as well. Next, if $A/N$ is a submodule of $M/N$, and if $a_1,\mathrm{\dots },a_n$ is a generating set for $A\subseteq M$, then $a_1+N,\mathrm{\dots },a_n+N$ is a generating set for $A/N$. Conversely, if every submodule of $M$ is Noetherian, then $M$, being a submodule itself, must be Noetherian.∎

A weaker form of the converse is the following:

Theorem 2.

If $N\mathrm{\subseteq }M$ is a submodule of $M$ such that $N$ and $M\mathrm{/}N$ are Noetherian, then $M$ is Noetherian.

Proof.

Suppose $A_1\subseteq A_2\subseteq \mathrm{\cdots }$ is an ascending chain of submodules of $M$. Let $B_i=A_i\cap N$, then $B_1\subseteq B_2\subseteq \mathrm{\cdots }$ is an ascending chain of submodules of $N$. Since $N$ is Noetherian, the chain terminates at, say $B_n$. Let $C_i=(A_i+N)/N$, then $C_1\subseteq C_2\subseteq \mathrm{\cdots }$ is an ascending chain of submodules of $M/N$. Since $M/N$ is Noetherian, the chain stops at, say $C_m$. Let $k=\max (m,n)$. Then we have $B_k=B_{k+1}$ and $C_k=C_{k+1}$. We want to show that $A_k=A_{k+1}$. Since $A_k\subseteq A_{k+1}$, we need the other inclusion. Pick $a\in A_{k+1}$. Then $a+N=b+N$, where $b\in A_k$. This means that $a-b\in N$. But $b\in A_{k+1}$ as well, so $a-b\in N\cap A_{k+1}$. Since $N\cap A_k=N\cap A_{k+1}$, this means that $a-b\in A_k$ or $a\in A_k$.∎