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 $R$.
Best Answer
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.