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Suppose $M$ is a finitely generated module over $R$ where $R$ is a Noetherian ring. Prove that $M$ is a noetherian module.

Proof: Since $M$ is finitely generated module, there exists a free module $N$ such that $M \cong N/\ker(f)$ where $f:N \rightarrow M$. Since $N$ is free, it can be expressed as direct sum of isomorphic copy of underlying ring, i.e. $N \cong\bigoplus R$. Since $R$ is noetherian ring, $\bigoplus R$ is also noetherian. Since $\ker(f)$ is a submodule of $\bigoplus R$, $\ker(f)$ is also noetherian, which implies that $M$ is noetherian.

Is my proof correct? Is it true that direct sum of noetherian ring is noetherian?

Remark: Since $N$ and $\ker(f)$ are both noetherian, then $N/\ker(f)$ is also noetherian, and hence $M$ is also noetherian.

Since $M$ is finitely generated, the free module $N$ is of finite rank, and hence is isomorphic to finitely many copies of $R$.

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    $\begingroup$ Dear Idonknow, Every module is a quotient of a free module. Where are you using finite generation? And why does it matter whether or not $ker(f)$ is Noetherian? In other words, your proof is on the one hand incomplete, and on the other hand includes irrelevant facts. You would do well to write out the implicit arguments that you are using (e.g. to deduce that $\oplus R$ is Noetherian, and to deduce from this that $M$ is Noetherian) in detail. Then you might have a better idea of how the fully correct argument should go. Regards, $\endgroup$
    – Matt E
    Sep 13, 2013 at 18:51
  • $\begingroup$ The added remark is true, but if $N'<N$, with $N$ Noetherian, it's superfluous to check that $N'$ is Noetherian before concluding $N/N'$ is Noetherian. The real issue here is figuring out why $N$ is Noetherian, as MattE mentions. $\endgroup$
    – rschwieb
    Sep 13, 2013 at 19:01
  • $\begingroup$ Since $N \cong \bigoplus R$ and $\bigoplus R$ is noetherian, isn't $N$ will also be noetherian? $\endgroup$
    – Idonknow
    Sep 13, 2013 at 19:06
  • $\begingroup$ @Idonknow Of course, but we do not know if it's expected of you to prove a finite free rank module is Noetherian. $\endgroup$
    – rschwieb
    Sep 13, 2013 at 19:51

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Yes, it's true that finite direct sums of Noetherian modules are Noetherian (so any finite rank free module of a Noetherian ring is Noetherian.) This will be a key point, as will the fact that quotients of Noetherian modules are Noetherian.

But there are a few problems. Firstly, you said $N$ was free, but you didn't somehow exclude the case where it used infinitely many copies of $R$. It won't be Noetherian unless you do this.

Secondly, you said "$\ker(f)$ is Noetherian, which implies $M$ is Noetherian." Why would this be? Find a non-Noetherian module with a simple submodule and look at the quotient by the simple submodule. The quotient is certainly not Noetherian, even though the submodule is a finitely generated kernel of a projection.

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9 years later... I'll have to prove it myself.

Proof. Since $R$ is Noetherian, then the direct sum $R^n$ is Noetherian (finite $n$). And since every $R$-module $M$ is isomorphic to a quotient of $R^n$, i.e., $M\cong R^n/Ker(f)$, where $f: R^n \to M$ onto. Then we can consider this exact sequence: $$ 0\to Ker(f) \to R^n \to R^n/Ker(f) \to 0 $$ Since $R^n$ is Noetherian, then both $Ker(f)$ and $R^n/Ker(f)$ are Noetherian. Hence $M(\cong R^n/Ker(f))$ is Noetherian.

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  • $\begingroup$ class! Absolute class! $\endgroup$
    – ben huni
    Mar 2, 2023 at 21:01

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