Annihilator as a maximal ideal

Question

Let $R[G]$ be the group ring of a finite group $G$ over a commutative ring $R$ with $1$. Let $S$ be a simple $R[G]$-module and let $I$ be the annihilator in $R$ of $S$. Show that $I$ is a maximal ideal in $R$.

What I have so far: Suppose $I\subseteq M\subseteq R$ for some ideal $M$ and $M\neq I$, then pick $m\in M-I$. For a non-zero $x\in S$, $m\cdot x=y$ for some $0\neq y\in S$. Since $S$ is a simple $R[G]$-module, $x=(\sum_{i=1}^nr_ig_i)y$ for some $r_1,...,r_n\in R$ and $g_1,...,g_n\in G$. Then $m\sum_{i=1}^nr_ig_i-1$ annihilates $y$. I don't know how I can conclude that $m$ is a unit in $R$

Answer 1

Let $P$ be the annihilator of $S$ in $R[G]$, so $P\cap R=I$. Since $R[G]$ is finitely generated as an $R$-module, it follows that $S$ is finitely generated as an $R$-module. Thus $S$ has a maximal (proper) $R$-submodule $N$. For each $g\in G$, the set $gN$ is an $R$-submodule of $S$. It is easy to see that $gN\cong N$ and $S/gN\cong S/N$ as $R$-modules. If we let $J$ denote the annihilator of $S/N$ (and hence of each $S/gN$) in $R$, then $J$ is a maximal ideal since $S/N$ is simple.

The intersection $\cap_{g\in G}\, gN$ is a proper $R[G]$-submodule of $S$, since it's closed under multiplication by $R$ and $G$. Thus $\cap_{g\in G}\, gN=0$. As usual, this implies that $S$ embeds in $\oplus_{g\in G}\, S/gN$ as an $R$-module. As noted above, the direct sum is annihilated by $J$ and so $S$ is annihilated by $J$. Hence $J\subseteq P\cap R$. But $J$ is maximal, so we must have $J=P\cap R$, proving that $I=P\cap R$ is a maximal ideal of $R$.

Edited because my notation didn't agree with the OP.

Answer 2

If you're willing to use PI theory, here's a proof. Let $P$ be the annihilator of $S$ in $R[G]$. The ideal $P$ is primitive by definition and the ring $R[G]$ is a PI (polynomial identity) ring since it's a finitely generated module over a commutative ring. Kaplansky's Theorem implies primitive ideals in PI rings are maximal.

Edit: This answer is the wrong conclusion because I misread the problem. This shows $P$ is maximal, but the OP wants to show $P\cap R$ is maximal. However, Kaplansky's Theorem in this situation also says that $P\cap R$ is maximal.