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Let $A$ be an associative algebra. I am wondering under what conditions we can get all irreducible representations of $A$ as left ideals $A\cdot e$ with $e\in A$ an idempotent.

This is certainly the case for the group ring $\mathbb C[S_n]$ of the symmetric group, but I couldn't even find out whether it holds in general for group rings $\mathbb C[G]$ of finite groups $G$ - in fact, some references seem to suggest the opposite.

I would be very glad if someone could shed some light on this, i.e. name a condition for all irreps of $A$ to arise as left ideals generated by an idempotent.

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Your question is about semisimple algebras. The main fact that you have to use is that $A\cdot e$ is a projective module for any idempotent $e\in A$: Indeed for an idempotent $e\in A$ we have $A=Ae\oplus A(1-e)$. Hence $Ae$ is a direct summand of $A$ and hence projective (see for example [Lam: Lectures on Modules and Rings, Example 2.12A]).

Now suppose that every irreducible representation of $A$ is of the form $Ae$. In particular every irreducible representation is projective. Hence every module is a direct sum of irreducibles by Jordan-Hölder theorem. So this means that the algebra is semisimple.

So what group algebras are semisimple: This is Maschke's theorem. A group algebra is semisimple iff the characteristic of the field does not divide the group order. So in particular this holds for all finite groups over the complex numbers.

For an example where this does not hold one just has to find a non-semisimple algebra with a non-projective simple. For example the irreducible module $\mathbb{F}_2$ over the group algebra $\mathbb{F}_2[\mathbb{Z}/2\mathbb{Z}]$ is an example.

EDIT (to answer the comment): For the converse: For every finite dimensional algebra we have that $A=\bigoplus_{i=1}^n Ae_i$, where $e_i$ are a complete set of primitive orthogonal idempotents and the $Ae_i$ give all indecomposable projective modules. Now general theory (as for example explained in this question) tells that all simple modules over a general finite dimensional algebra (not necessarily semisimple) are given by $Ae_i/\operatorname{rad} Ae_i$, where $\operatorname{rad} Ae_i$ is its unique maximal submodule. Now in case of a semisimple algebra the indecomposable projective modules are simple and hence the unique maximal submodule vanishes. Hence all the simples are of the form $Ae_i$.

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You started from the assumption that every irreducible representation of $A$ has the form $Ae$ and deduced that $A$ is semisimple, which is fine. Assuming $A$ is semisimple, why is every irreducible representation of $A$ isomorphic to some $Ae$? –  Jesko Hüttenhain Jan 9 '13 at 12:05
    
@JeskoHüttenhain Answered the comment in the EDIT. –  Julian Kuelshammer Jan 9 '13 at 12:13
    
Sweet. Thanks a bunch! –  Jesko Hüttenhain Jan 9 '13 at 12:21
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