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I'm reading Clausen, Fast Fourier Transforms, and he states (p. 34):

Let A be an associative algebra.
A is semisimple $\Leftrightarrow$ A is a direct sum of minimal left ideals.

Can anyone explain why?

Followup question: since A is semisimple, its left A-modules are completely reducible, i.e. every left A-module is the direct sum of simple left A-modules. Is there an easy correspondence between the simple A-modules and the left ideals of A?

share|cite|improve this question – user26857 Nov 23 '12 at 23:46

The equivalence between these conditions in an associative ring $R$ are covered in almost every book with noncommutative rings:

  1. Every right ideal of $R$ is a direct summand. (Same for left ideals)

  2. $R$ is a direct sum of simple right ideals. (Same for left ideals)

  3. Every $R$ module is a direct sum of simple $R$ modules.

As for the followup question (usually it's better to post these separately, but this one is easy enough to cover here) the simple right $R$ modules will all be represented by the simple right ideals. There may be duplicate copies though!

Here is how to see it. If $M$ is a simple right $R$ module, then there is a homomorphism $R\rightarrow M$ that is onto. Thus $M\cong R/T$ where $T$ is a maximal right ideal. But since $R$ is semisimple $R\cong T\oplus N$, and you can verify that $N$ is a minimal right ideal. Finally, $M\cong N$.

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