Theorem 1   A module is noetherian if and only if all of its submodules and quotients 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 $a_1,\ldots, a_n$ is a generating set for $A\subseteq M$ , then $a_1+N,\ldots,a_n+N$ is a generating set for $A/N$ . Conversely, if every submodule of $M$ is Noetherian, then $M$ , being a submodule itself, must be Noetherian.

Theorem 2   If $N\subseteq M$ is a submodule of $M$ such that $N$ and $M/N$ are Noetherian, then $M$ is Noetherian.

Proof. Suppose $A_1\subseteq A_2\subseteq \cdots $ is an ascending chain of submodules of $M$ . Let $B_i=A_i\cap N$ , then $B_1\subseteq B_2\subseteq \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 \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 $a\in A_k$ .