If $G$ is an infinite group, then the group ring $R(G)$ is not semisimple.

Question

Let $R$ be a ring and $G$ an infinite group. Prove that $R(G)$ (group ring) is not semisimple.

My idea was to suppose it is semisimple, then $R(G)$ is left artinian and $J(R(G))=0$. I was trying to make a ascending chain of ideals that won't stop, then it is not left noetherian, by Hopkins theorem it is not left artinian, a contradiction. I also tried to make a descending chain that won't stop, so it is not left artinian, but I wasn't successful. So please help me.

Answer 1

Comments on the partial progress.

Unfortunately, this is a case of something that sometimes happens with students: a strong condition was given (semisimple Artinian) and then you decided to quickly drop down to something weaker (Artinian/Noetherian) possibly because you feel more comfortable with that condition.

It is possible to prove that if $G$ is infinite then $R[G]$ can't be Artinian, but as I alluded to in the comments it's not easy enough to be a homework problem.

If you start with assuming $G$ is infinite and then try to show that $R[G]$ isn't Noetherian, you are unfortunately doomed to failure, because (I think I am recalling correctly) there are examples of infinite groups $G$ such that $R[G]$ is Noetherian.

This highlights some of the dangers of "simplifying" givens too quickly. We can prove it in the following way taking full advantage of supposed semisimplicity of $R[G]$. Specifically, we'll use the fact that ideals are summands.

Hints:

1. Let $A$ be the augmentation ideal of $R[G]$. That is, it is the kernel of the projection $R[G]\to R$ which sends $g\mapsto 1$ for all $g\in G$. Show that $1-g\in A$ for all $g\in G$.
2. Suppose $R[G]$ is semisimple. Then $A$ is a summand of $R[G]$, and in particular we can find an idempotent $e\in R$ such that $R[G]e=A$ and $R[G]e\oplus R[G]f=R[G]$ where $f=1-e$ is another idempotent. Prove that $(1-g)f=0$ for every $g\in G$ using the previous point. Consequently, $f=gf$ for all $g\in G$.
3. $f$ has to be nonzero, since the augmentation ideal is proper. Let $h\in G$ be a term of $f$ that has a nonzero coefficient. Show using the previous point that $gh$ appears with a nonzero coefficient in the expression of $f$.
4. Since every element of $G$ can be expressed as $gh$ for some $g$, this means that all elements of $G$ have nonzero coefficients in the expression of $f$... but it isn't possible for $f$ to have infinitely many nonzero terms. Thus $G$ isn't infinite.