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It's stated here that for a commutative ring $R$, every simple module over $R$ is isomorphic as an $R$-module to a quotient ring of $R$ by a maximal ideal.

Intuitively this seems likely, since a quotient ring is a field if any only if we quotient by a maximal ideal, and a field has no nonzero proper ideals. This makes me suspect that an $R$-module $M$ is simple if and only if $M\cong R/I$ (as $R$-modules) for a maximal ideal $I$ of $R$.

Is there a proof of whether this is true or not?

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up vote 1 down vote accepted

Let $M$ be a simple module and let $m \in M$ be nonzero. By simplicity, $\{ rm : r \in R \} = R$, so the map $r \mapsto rm$ defines a natural surjective homomorphism $R \to M$ of $R$-modules giving an isomorphism $M \cong R/I$ where $I$ is the kernel. If $I$ is not maximal, then letting $m$ be any maximal ideal containing it we have another surjection $R/I \to R/m$ with nontrivial kernel, and this contradicts simplicity.

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If $M$ is a simple module for each $x\in M$ with $x\neq 0$, $Rx=M$, so take a $x\in M$ with $x\neq 0$, and consider the next morfism $f:R\longrightarrow M$ given by $f(a)=ax$, by the last observation is an epimorphism. Now apply the first isomorphism theorem and you get $M/\ker f\cong M$. Now the $\ker f=I$ is an ideal of $R$. To see that $I$ is maximal use the theorem of the bijective correspondence. Also the same theorem gives you a proof of the return.

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