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In a textbook of representation theory I have encountered the following statement without proof:

Let $R$ be a commutative ring and $G$ a finite group. If $M$ is a simple $RG$-module then the annihilator of $M$ as $R$-module is a maximal ideal of $R$. Thus $M$ can be considered as a representation of $kG$ where $k$ is the quotient field $R/I$.

Could someone please help me to prove this or at least give me a reference where can I found a proof of it?

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    $\begingroup$ Every $RG$-module is the union of its $RG$-submodules that happen to be finitely generated as $R$-modules. Since $M$ is simple, it follows that $M$ is a finitely generated $R$-module. For every ideal $J$ of $R$, $JM$ is an $RG$-submodule of $M$. If $M$ is simple, then either $JM$ equals $M$ or $JM$ equals $\{0\}$. In particular, for every ideal $J$ that properly contains $\text{ann}(M)$, then $JM$ equals $M$. By Nakayama, there exists $r\in R$ with $1-r\in J$ and $r\in \text{ann}(M)$. Since $1-r$ and $r$ are in $J$, $1\in J$. Thus $\text{ann}(M)$ is a maximal ideal. $\endgroup$
    – Jason Starr
    Commented Jan 21, 2017 at 14:56
  • $\begingroup$ Thank you very much! I was missing the Nakayama lemma step; looked up in wikipedia and have seen the version for commutative rings. So the proof works for any R-algebra that is free of finite rank as R-module. $\endgroup$
    – user103474
    Commented Jan 21, 2017 at 16:02
  • $\begingroup$ Yes, the argument works for simple $A$-modules where $A$ is an associative $R$-algebra that is finitely generated as an $R$-module. $\endgroup$
    – Jason Starr
    Commented Jan 21, 2017 at 18:13

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