# criterion for a module to be 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. ∎

A weaker form of the converse is the following:

###### 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 $a\in A_{k}$. ∎

Title criterion for a module to be noetherian CriterionForAModuleToBeNoetherian 2013-03-22 15:28:46 2013-03-22 15:28:46 mps (409) mps (409) 9 mps (409) Theorem msc 13E05