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 $R$.