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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.

kpax
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  • Your trying it seems to use only the finiteness (artinian or noetherian) properties of $R(G)$. This makes me think that you are not on the right track. –  Dec 25 '13 at 13:30
  • actually I don't know what should I do,what is your Idea? – kpax Dec 25 '13 at 13:50
  • Yeah, I think user's feeling is correct. "R[G] artinian implies G finite" is a nontrivial proof appearing in Lambek's algebra book and Connell's thesis. – rschwieb Dec 25 '13 at 14:59
  • Are you trying to show "semisimple" in the sense that the Jacobson radical is zero, or semisimple in the sense that it is a direct sum of simple rings? –  Dec 25 '13 at 15:15
  • @ Jason Polak ,I must show that it is not semisimple,so I suppose it is,and I wanted to show it comes to a contradiction. – kpax Dec 25 '13 at 15:26
  • @ rschwieb,thank you for the reference,I will check it. – kpax Dec 25 '13 at 15:27
  • Lambek's algebra book,you mean that Lectures on Rings and Modules by Lambek? – kpax Dec 25 '13 at 15:38
  • @rschwieb, got it, thanks, deleting. – Andreas Caranti Dec 25 '13 at 18:49

1 Answers1

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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.
rschwieb
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  • Dear @user :Thanks, that was good inspiration for some additional comments about the partial progress. While it's possible to pursue the "suppose $G$ is infinite, now I'll show $R[G]$ isn't Artinian" idea, it would be excessively long for a solution I think. The "suppose $G$ is infinite, now I'll show $R[G]$ isn't Noetherian" idea just doesn't work (not without backtracking and picking up more conditions from semisimplicity.) – rschwieb Dec 25 '13 at 19:33
  • thank you for your complete explanation,and also happy new year,I wish the best for all. – kpax Dec 25 '13 at 20:41
  • @kpax Yeah, Happy Holidays to you too! – rschwieb Dec 25 '13 at 22:47
  • Dear @rschwieb : In working through your suggested approach, I became confused about the existence of the idempotent $e \in R$ as opposed to $e \in R[G]$. My understanding is that semisimplicty only guarantees the latter in this case? Everything else seems to work the same regardless of which is the case, so I am wondering if that is a typo, or there is something I am missing. Thanks – user1348 Sep 08 '16 at 18:33
  • @user1348 Hi, yes, I think you're right that I left out the $[G]$ by accident. Thanks for letting me know! – rschwieb Sep 08 '16 at 18:53
  • @MarianoSuárez-Álvarez For non-commutative ring is it true that ring $A$ is Artinian if and only if $A$ is Noetherian and $krdim A = 0$? – Lalbahadur Sahu Dec 28 '23 at 09:47
  • @LalbahadurSahu No. There are many rings which are Noetherian on both sides, simple (hence have krull dimension 0, if you are defining it by chains of prime ideals) but not Artinian. One such example is the first Weyl algebra over a field of characteristic $0$. – rschwieb Dec 28 '23 at 16:19
  • @MarianoSuárez-Álvarez That "probably" seems like a shot in the dark, but of course the input is welcome and I'm glad for that reference pointer, too. – rschwieb Dec 28 '23 at 16:53