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Let $R$ be a ring, and let $M$ be an $R$-module. Is it true that $M$ is a simple $R$-module if and only if $M\cong R/\mathfrak{p}$ for some prime ideal $\mathfrak{p}$ of $R$?

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No. $\mathbb Z /(0) \cong \mathbb Z$ is not $\mathbb Z $- simple although $(0)$ is prime. –  Georges Elencwajg Mar 25 '12 at 14:19

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No, $M$ is a simple $R$-module iff $M\cong R/\mathfrak{m}$ for some maximal ideal $\mathfrak{m}$. Clearly such modules are simple $R$-modules. Conversely, if $M$ is a simple $R$-module then for any $0\neq x\in M$ we must have $xR=M$ as it is clearly a submodule of $M$. But this means that $M\cong R/I$ where $I$ is the kernel of the map $r\mapsto xr$, and $I$ must be maximal as otherwise we could take some $J\supset I$ and have $0\neq R/J\subset M$.

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