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I have seen a handful of proofs that any finitely generated module over a Noetherian ring is again Noetherian. I'm specifically trying to understand the following proof idea. It goes as this:

Observe that if $T\subseteq S$ are submodules of and $R$-module $M$ with $R$ Noetherian, and if $T$ is finitely generated (f.g.) and $S/T$ is f.g., then so is $S$. I get that.

Proceed by induction. If $M$ is $1$-generated, say $M=Rv$, then if $\phi\colon R\to M$ defined by $r\mapsto rv$ is an epimorphism, and so $M\cong R/\ker(\phi)$, and is Noetherian as the quotient of a Noetherian ring. Suppose it holds for all $k$-generated $R$-modules, for $k\leq n$. Let $M=\langle v_1,\dots,v_{n+1}\rangle$. Let $M'=\langle v_1,\dots, v_n\rangle$. Then let $T=S\cap M'$.

On second thought, I think $S$ is supposed to be an arbitrary submodule of $M$. Let $T=S\cap M'$. So $T$ is a submodule of $M'$, hence finitely generated. Then $$ S/T=S/(S\cap M')\cong SM'/M'. $$ How is $SM'/M'$ finitely generated? I suppose would give the conclusion.

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$SM'$ has no meaning for modules; this must be $S+M'$. –  user26857 Feb 17 '13 at 21:57
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up vote 4 down vote accepted

I assume that $S$ is a submodule of $M$ and you want to prove that $S$ is finitely generated.

$T=S\cap M'$ is finitely generated as a submodule of a noetherian module ($M'$ is noetherian by the induction hypothesis.)

$S/T=S/S\cap M'\simeq (S+M')/M'$ and the last one is a submodule of $M/M'$. But $M/M'$ is cyclic (generated by $\bar v_{n+1}$), hence noetherian, so $S/T$ is finitely generated.

That's all.

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Thanks YACP, I wrote down the isomorphisms badly and got confused. This clarified it. –  Tiffany Hwang Feb 17 '13 at 21:56
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