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Let $\mathbb C G$ be the group ring of a finite group $G$ and let $V$ be an irreducible $\mathbb C G$ module. I'm having trouble showing that $V$ is projective. Any help?


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What do you know about group rings? –  Mariano Suárez-Alvarez Jul 20 '11 at 22:05
This really depends where you're starting from. Maschke's theorem is relevant if you know that result. –  Geoff Robinson Jul 20 '11 at 22:05
I think Alex B. gave a great answer here in which he explains why $V$ is a direct summand of $\mathbb{C}G$ as a $\mathbb{C}G$ module. –  t.b. Jul 20 '11 at 22:08
@Mariano: I know that $\mathbb C G$ is semisimple ($rad(\mathbb C G) = 0$). –  Eric O. Korman Jul 20 '11 at 22:10
@Eric Is a module over a semisimple ring also semisimple? –  Dylan Moreland Jul 20 '11 at 22:15
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2 Answers

up vote 3 down vote accepted

I have decided to record an answer which makes reference to the standard theorems in this area. It is not that different from either Mariano's answer or Akhil Mathew's deleted answer (although the way I am expressing it does not use categorical language so may be more widely accessible).

First a basic proposition / definition. For a left module $M$ over a not necessarily commutative ring $R$, the following are equivalent:
(i) $M$ is a direct sum of simple submodules.
(ii) Every submodule of $M$ is a direct summand.
(iii) The least submodule of $M$ containing every simple submodule of $M$ is $M$ itself.
A module satisfying these equivalent conditions is called semisimple.

A ring $R$ is called semisimple if every left $R$-module is semisimple. There are several equivalent conditions, for instance that $R$ be semisimple as a left $R$-module. Importantly for this question, $R$ is semisimple iff every short exact sequence of $R$-modules splits iff every $R$-module is projective iff every $R$-module is injective.

Reference for the above facts: $\S 2.1$ of my noncommutative algebra notes.

Thus if you are trying to show that an $R$-module is projective, you can go home early if you know that the ring $R$ is semisimple: then, and only then, it doesn't matter what the $R$-module is. But in fact one of the first nontrivial results in representation theory tells you that this holds in your situation.

Theorem (Maschke): For a group $G$ and a field $K$, the group ring $K[G]$ is semisimple iff $G$ is finite and the characteristic of $K$ does not divide $G$. In particular, for every finite group $G$, $\mathbb{C}[G]$ is semisimple.

[Comment: I think the second sentence is actually what was proved by Maschke, but the other parts of it are easy to prove and it is traditional to credit the full result to him, at least in the case when $G$ is finite.]

Reference: $\S 2.5$ of my noncommutative algebra notes. Note that the argument that Mariano gives in his answer is essentially the proof of Maschke's Theorem: given a $K[G]$-submodule $W$ of $V$, first you choose a complement as a $K$-vector space. This need not be $G$-stable, so you make it $G$-stable by an averaging construction, which works if(f) you can divide by $\#G$ in $K$.

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Thanks for this answer! It's very useful for me since I was having issues reconciling the different notions of semi-simplicity and using the right notion makes my question very simple to answer. –  Eric O. Korman Jul 21 '11 at 15:44
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A simple way is the following:

Let $V$ be a simple left $\mathbb CG$-module. Let $v_0\in V$ be a non-zero element, and let $\phi:\mathbb CG\to V$ be the unique $\mathbb CG$-linear map such that $\phi(1)=v_0$. Since the image of $\phi$ is a submodule of $V$ which is not zero, it must be all of $V$: in other words, $\phi$ is surjective.

Now let $s:V\to\mathbb CG$ be any $\mathbb C$-linear map such that $\phi\circ s=\mathrm{id}_V$. Such map exists because $V$ is a projective $\mathbb C$-module! We define a new map, $\tilde s:V\to\mathbb CG$ as follows: if $v\in V$, let $$\tilde s(v)=\frac1{|G|}\sum_{g\in G}gs(g^{-1}v).$$ It is easy to see that $\tilde s:V\to\mathbb CG$ is a morphism of $\mathbb CG$-modules and that $\phi\circ\tilde s=\mathrm{id}_V$, by a direct computation. Using these two facts, it follows that the image of $\tilde s$ is a submodule of $\mathbb CG$ isomorphic to $V$, and that $\mathbb CG\cong\operatorname{im}\tilde s\oplus\ker \phi$. In particular, $V$ is projective.

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