I have to show:
An $A$-module M is isomorphic to an A-module of the form $A/m$ for some maximal ideal m of A $\Longleftrightarrow$ M is a simple $A$-module
I have already proved this direction $'\Longleftarrow '$, but now I have trouble showing the other side. I tried doing it with a contraposition, saying $m$ is not maximal and hence there exists an ideal $m \subsetneq I \subsetneq M$ and there is an isomorphism $f: A/m \to M$. Then $0 \subsetneq f(A/m) \subsetneq M$ is a submodule of $M$, so $M$ is not simple. This feels wrong, was it wrong to state that $f$ exists in my contraposition? Can somebody help me?
$A/m$ is simple because the $A$-submodules of $A/m$ correspond exactly to the ideals of $A$ containing $m$. But since $m$ is maximal the only such ideal is $m$ itself. Hence, the only submodules of $A/m$ are $0,A/m$ which by definition means that $A/m$ is simple. It may be worth pointing out that $A/m\neq 0$ because $m\subset A\setminus A^\times$.