# Any submodule of any finitely generated $R$-module is finitely generated if $R$ is Noetherian

I am trying to prove that if $R$ is a Noetherian ring then any submodule of any finitely-generated $R$-module is also finitely-generated.

What I have tried: I know that any finitely generated $R$-module is isomorphic to $R^n/N$ for some $n$ and $N$. Since submodules of $R^n/N$ correspond to submodules of $R^n$ containing $N$, and $R/M$ is finitely generated if $R$ is finitely generated, it suffices to show that any submodule of $R^n$ is finitely generated. The condition that every submodule be finitely generated is equivalent to every increasing chain of submodules terminating. Now since $R$ is Noetherian every increasing chain of ideals terminates. I need to use this to show that every increasing chain of submodules in $R^n$ terminates. I am not sure how to show this though.

I would appreciate any different solutions too.

You need to prove that if $M$ and $N$ are Noetherian modules, then so is $M\oplus N$. Then from the Noetherianness of $R$, inductively one gets $R^n$ Noetherian.
Let $\pi:M\oplus N\to N$ to be the projection. Given an increasing chains $(A_n)$ of submodules of $M\oplus N$, then $(M\cap A_n)$ and $\pi(B_n)$ are increasing chains of submodules of $M$ and $N$ respectively, so they both stabilise. One now checks that $P_n\subseteq P_{n+1}$, $P_n\cap M=P_{n+1}\cap M$ and $\pi(P_n)=\pi(P_{n+1})$ implies $P_n=P_{n+1}$. Therefore $(P_n)$ stabilises.
• Thank you. I assume that should be $\pi(A_n)$ not $\pi(B_n)$? – user544680 Jun 5 '18 at 22:05