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Let $G$ be a finite group. Consider the group algebra $\Bbb{C}[G]$ as a right module over itself. By Maschke's Theorem, $\Bbb{C}[G]$ is semisimple. How can we identify all the minimal right ideals of $\Bbb{C}[G]$? For example, from general module theory I know that if $I$ is a minimal right ideal in a ring $R$ (not necessarily commutative) then we have that $I^2 = 0$ or $I =eR$ for some idempotent $R$.

However my ring $R$ ( that is $\Bbb{C}[G]$ now) is semisimple, is there anything special now in this case?

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2 Answers 2

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Since $R$ is semisimple every right ideal is of the form $eR$ and every left ideal is of the form $Rf$ for idempotents $e,f$.

The minimal ones will be the ones generated by idempotents that cannot be written as a sum of two nonzero idempotents.


added a few more notes On the off-chance it helps your computations, the minimal left ideals are exactly the direct complements of maximal left ideals.

If you go so far as to decompose $R$ into a product of matrix rings over $\mathbb{C}$, then we can determine the minimal ideals completely in terms of that decomposition.

The subset $L_j$ of matrices which are zero off of column $j$ are left ideals, and in fact are minimal left ideals. If memory serves, every minimal left ideal is of the form $L_j u$, for some $j$ and for some $u$ a unit of $R$.

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I think you mean column $j$. –  user38268 Oct 4 '12 at 13:10
    
@BenjaLim Right, indeed I do :) Fixed - thanks! –  rschwieb Oct 4 '12 at 14:01

The minimal right ideals of $\mathbb C[G]$ are precisely the irreducible (right) subrepresentations of the right regular representation of $G$. It is well known that any irreducible representation of $G$ is present in the regular representation with multiplicitly equal to its dimension $d$. That means that, unless $d=1$, there are infinitely many copies of the irreducible representation that can be found inside the regular representation, which taken together fill a space of dimension $d^2$. Concretely, the left action of $\mathbb C[G]$ permutes its minimal right ideals.

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Thanks for your answer. However, I am looking for a general form for all minimal right ideals in my ring $R$. For example, we know that the group algebra as a left module over itself has all minimal left ideals of the form $Re$, $e$ a primitive idempotent. Is there a similar thing when I look at the group algebra now as a right module over itself? –  user38268 Oct 4 '12 at 7:21
    
Yes of course, left and right are completely equivalent (any right structure for an algebra $A$ is becomes a left structure for the opposite algebra $A^{\textrm{op}}$, whose multiplication is that of $A$ after swiching operands). But my answer shows that in general there are infinitely many primitive idempotents in your algebra. –  Marc van Leeuwen Oct 4 '12 at 8:53

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