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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.

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    $\begingroup$ The submodule is again Noetherian, see also here, by definition! "Consider an ascending chain in the submodule $N\subseteq M$, and remember it is a chain in $M$ too!" $\endgroup$ – Dietrich Burde Oct 6 '16 at 7:51
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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.

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