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.
Drop "non-zero" and add "non-$M$" and you are on the clear: you just need to use the bijective order-preserving correspondence between submodules of $M/N$ and submodules $K\subseteq M$ such that $K\supseteq N$. If there was a submodule $H\subsetneq M/N$ such that $H\ne N/N$, then $N\subsetneq H^c\subsetneq M$.