# Problem 24 of section 2 of *Noncommutative Algebra* by Farb & Dennis

Problem 24 of section 2 of Noncommutative Algebra by Farb & Dennis states:

Let $R$ be an artinian ring and let $G$ be a finite group. Show that $R[G]$ is semisimple if and only if $R$ is semisimple and $|G|$ is invertible in $R$.

I tried to make some progress in assuming that $R$ is semisimple and $G$ is invertible in $R$ and proving that $R[G]$ is semisimple. I'm trying to prove it by showing that in $R[G]$, any submodule is a direct summand. So, given a submodule $M$ of $R[G]$, I'd first like to find a complement for it as a direct summand as $R$-modules (not sure how to do that) and then define a projection into $M$ and turn its kernel into a complement as $R[G]$ module. This is an attempt to immitate techniques learnt in class, but I'm very confused because $R$ is not a field.

This exercise is from the chapter discussing the Jacobson radical so it may be related.

I would appreciate a hint.

EDIT: I managed this direction. Now I'm trying the other one. $R[G]$ is semisimple. I can show that $R$ is then semisimple, but how do I show that $|G|$ is invertible in $R$? I tried to assume that $|G|$ is not invertible and find a nil ideal in $R[G]$ to get a contradiction, but couldn't do it. I considered 2 ideals so far:

1. $R\sum_{g \in G}1 \cdot g$
2. $\{\sum_{g \in G} a_g g \mid \sum_{g \in G}a_g = 0\}$.
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If you have a short exact sequence of $R[G]$-modules then it splits over $R$, right? Can't you do the same averaging process as in Maschke's theorem with this splitting? I might be missing something. –  Dylan Moreland Dec 15 '11 at 0:17
@DylanMoreland: Thanks. I understand. Now I'm trying the other direction (see edit). –  Jen Dec 17 '11 at 16:26
@Jen: Dear Jen, You might first want to try a simple case to get a feel for what is going on: take $G$ a cyclic group of order $2$ and $R$ to be the field $\mathbb F_2$; then take $G$ to be a cyclic group of prime order $p$ and $R$ to be the field $\mathbb F_p$. In each of these cases you should be able to compute by hand that $R[G]$ is not semi-simple. Once you understand those cases, you can try extending your intuition to the general case. Regards, –  Matt E Dec 17 '11 at 16:39
@MattE: In those examples, the only noninvertible element of $R$ is zero. So I take the first ideal I considered, square it, and get 0. What example should I take which is not a field? –  Jen Dec 17 '11 at 16:44
Got it. The first ideal I considered really was the key. –  Jen Dec 18 '11 at 1:06