# Every non-zero noetherian module has a simple quotient module.

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 $N\subsetneq H^c\subsetneq M$.
• Oops yes I was trying to emulate a proof for a similar statement for artinian modules. I've always thought there was a correspondence between submodules of $M/N$ and submodules $K$ such that $N\subseteq K \subseteq M$. Is this somehow via the map $M \rightarrow M/N$? – lego323 Dec 19 '16 at 11:44
• Yes, if you call the quotient map $\pi$, the correspondence is $K\to\pi(K)$ (this is well-defined because $\pi$ is surjective), with inverse $\overline K\mapsto \pi^{-1}(\overline K)$ (this is always defined for a general homomorphism). They are one the inverse of the other because $N=\ker\pi$. – user228113 Dec 19 '16 at 11:49
If $M$ itself is simple we're done. Otherwise, take a nontrivial proper submodule $N_0\subset M$, and let $L_0=M/N_0$. If $L_0$ is simple we're done. Otherwise, there's some $P_0\subset L_0$. Take $N_1:=\text{ker}(M\to L_0/P_0)$. Clearly $N_0\subset N_1$; moreover this is a proper inclusion since $P_0\neq 0$. We can repeat this process to get $L_1\supset P_1$ and $N_1\subset N_2$ and so on. We get a chain of proper inclusions
$0\neq N_0 \subset N_1 \subset N_2 \subset \cdots \subset M$