I was studing some algebra by my own and I was wondering if this propiety is true:
Let $M$ be a module over a ring $A$ and $S\subseteq M$ a submodule, if $M$ is Noetherian, then $S$ is Noetherian.
I think I could use this prop:
Let $0\rightarrow M^\prime \rightarrow M \rightarrow M^{\prime\prime}\rightarrow 0$ be an exact sequence of $A$-modules. Then:
$M$ is Noetherian if and only if, $M^\prime$ and $M^{\prime\prime}$ are Noetherian.
and clearly I have $S\rightarrow M\rightarrow 0$ if I consider inclusion. But I can't find the other side homomorphism.
Thanks for your help.
Remember that being noetherian means it satisfies the ascending chain condition. $M$ is noetherian so any chain $S_i\subseteq S$ in $S$ will form a chain in $M$ such that $$S_1\subseteq S_2\subseteq S_3\subseteq S_4\subseteq S_5\subseteq\ldots$$, however by the fact that $M$ is neotherian and $S_i\subseteq M$ means that it will stop at some point, such that for $i>N$ that $S_i=S_{i+1}=\ldots$ and as such is noetherian.