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

Question

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 $R$.

Answer 1

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.

Answer 2

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.