Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
math.uni-duesseldorf.de/~wisbauer/book.pdf –  user26857 Nov 23 '12 at 23:46
add comment

1 Answer 1

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, $M\cong N$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.