Let $R$ be a finite commutative ring with identity and $G$ an finite abelian group. Is there any more conditions (on $R$ or on $G $) under which we can characterize maximal ideals of group ring $RG $, with maximal ideal of $R$ (the same as the relation between maximal ideals of $R$ and $R [[X]]$)?
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4$\begingroup$ Could you please clarify what kind of result you are hoping for? Obviously the solution is going to be quite different for $\mathbb{C}\cdot (\mathbb{Z}/6\mathbb{Z})$ and $\mathbb{Z}\cdot (\mathbb{C}^\times)$. $\endgroup$– Jason StarrCommented Sep 14, 2015 at 16:08
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UPDATE: the question has been changed to say that both $R$ and $G$ are finite, so I have modified my answer accordingly.
Any maximal ideal in $R[G]$ contains the ideal $\text{Nil}(R)$ of nilpotent elements in $R$, and it is standard in this context that $R/\text{Nil}(R)$ is a finite product of finite fields, so we can assume that $R$ itself is a finite field, and of course these are well-understood.
Given a maximal ideal $M<R[G]$, put $H_M=\ker(G\to(R[G]/M)^\times)$. Recall that any finite subgroup of the multiplicative group of a field is cyclic, so the quotient $C_M=G/H_M$ is cyclic, of order $d$ say. Now $M$ corresponds to a maximal ideal in $$ R[C_M]\simeq R[u]/(u^d-1) = R[u]/\prod_{e|d}\varphi_e(u). $$ However, by construction $C_M$ embeds in $R[G]/M$, so $u^e-1$ cannot go to zero for any $e<d$, so we really have a maximal ideal in $R[u]/\varphi_d(u)$. As $\varphi_d(x)$ and $\varphi'_d(x)$ are coprime, the ring $R[u]/\varphi_d(u)$ is reduced, and so is a finite product of finite field extensions of $R$.
answered Sep 14, 2015 at 17:27