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Let $R$ Noetherian and $M$ finitely generated over $R$, prove $M$ is Noetherian.

Proof

We can suppose WLOG that $M\cong R^{\oplus s}$. The proof goes by induction. For $s=1$ it's obvious. Suppose $R^{\oplus s-1}$ is finitely generated. Consider $\varphi:R^{\oplus s}\to R^{\oplus s-1}$ the projection on the $s-1$ first coordinate. Let $N$ a submodule of $M$. We know that $K:=\ker \varphi|_N$ is a submodule of $R^{\oplus s}$ and since $\varphi|_N\neq 0$, we have that $K\neq R^{\oplus s}$. Now as I asked here, a submodule of a finitely generated module is not necessarily finitely generated. But in my couse it's written that $K$ is finitely generated, and since a submodule of a finitely generated module has no reason to be finitely generated, I don't understand why $K$ is finitely generated.

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    $\begingroup$ "We can suppose WLOG that $M\cong R^{\oplus s}$." No, a f.g. module doesn't have to be free. Look at $R=\Bbb Z, M=\Bbb Z/2\Bbb Z$ (but this becomes true if $R$ is a PID and $M$ is torsion-free, for instance). $\endgroup$ – Watson Jan 29 '17 at 12:57
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    $\begingroup$ A submodule of a finitely generated $R$-module is guaranteed to be finitely generated if $R$ is Noetherian. That's a big reason why Noetherian is such an important property. $\endgroup$ – Arthur Jan 29 '17 at 12:58
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    $\begingroup$ @Watson: I agree, but since $M$ is finitely generated, $M\cong R^{\oplus s}/\ker \varepsilon$ of $\varepsilon: R^{\oplus s}\to M$ surjective. By the way, if $R^{\oplus}$ is Notherian, then so is $R^{\oplus}/\ker \varepsilon$. Therefore, we just have to prove that $R^{\oplus s}$ is notherian, that's why we can suppose WLOG that $M=R^{\oplus s}$ $\endgroup$ – user386627 Jan 29 '17 at 13:03
  • $\begingroup$ @Arthur: I know, but that's what we want to prove, therefore we can't use this. $\endgroup$ – user386627 Jan 29 '17 at 13:05
  • $\begingroup$ It is sufficient to prove that the direct sum of two noetherian $R$-modules is noetherian ; this follows from $N$ and $M/N$ noetherian $\implies M$ noetherian. $\tag^*{}$ Here, $\mathrm{ker}(\phi\vert_N) = (0^{\oplus(s-1)} \oplus R) \cap N$ can be identified with a submodule of $R$, which is noetherian, so that $\mathrm{ker}(\phi\vert_N)$ is f.g. $\endgroup$ – Watson Jan 29 '17 at 13:20
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It is easy to see the equality $$\mathrm{ker}(\phi\vert_N) = (0^{\oplus(s-1)} \oplus R) \cap N.$$

Therefore, $\mathrm{ker}(\phi\vert_N)$ can be identified with a submodule of $R$, which is noetherian, so that $\mathrm{ker}(\phi\vert_N)$ is finitely generated as $R$-module.

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    $\begingroup$ The is a very well answer. Thanks a lot ! (and for all your other answer in my previous post). $\endgroup$ – user386627 Jan 29 '17 at 13:32
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What is $\ker\varphi$? It is isomorphic to $R$. Therefore $\ker\varphi|_N$ is an $R$-submodule of $R$, which is finitely generated because $R$ is a Noetherian ring. Note this is precisely the point in the proof where we need the fact that $R$ is Noetherian.

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