Let $R$ be an arbitrary ring and $M$ a non-zero noetherian R-module. Show it has a simple quotient module.
My idea: Let $S$ be the set of non-zero submodules of $M$. Then $M$ noetherian implies that there is a maximal element, $N\subseteq M$ say, which is an $R$-submodule. Then claim $M/N$ is a simple quotient. I'm pretty sure that's the right answer but I'm not sure how to prove that it is simple. I'm guessing we take a submodule of $M/N$ and somehow contradict the maximality of $N\subseteq M$? Any help would be appreciated, thank you.