# stably free modules over semisimple rings are free

This is exercise 1.1 of Chapter 1 in the K-book:

If $$R$$ is a semisimple ring, then $$R$$ is a direct sum of a finite number of simple modules. Furthermore, every stably free module over $$R$$ is free.

For the first part, I guess this is a direct consequence of Artin-Wedderburn Theorem. For the second part, let $$M$$ be a stably free module over $$R$$. Since $$R$$ is semisimple, then $$M$$ is a semisimple module, and by Jordan-Holder Theorem we know the length of $$M$$ is an invariant, then the rank of $$M$$ is invariant for all free modules, so $$R$$ satisfies IBP. I am not sure how to proceed from here. My guess would be if I can write down the corresponding short exact sequence of $$R^n\cong R^m\oplus M$$, then I can pick a basis for $$R^m$$ and extend it to a basis in $$R^n$$, so that $$M$$ becomes isomorphic to the image which is $$R^{n-m}$$. Would that be correct?

• @MarianoSuárez-Álvarez thanks for the comment. Would you mind elaborating on it a bit? I am not sure what you meant by "multiplicity" in this case. Sep 3, 2023 at 2:29

## 1 Answer

I think I have resolved my issue after a few days of thinking. Here is my revised solution:

The first part can be revised, as it should follow from the proof idea of Artin-Wedderburn Theorem. Since $$R$$ is semisimple, then $$R$$ is isomorphic to a finite direct sum of simple modules, so we can write $$R\cong \bigoplus\limits_{i=1}^m I_i^{\oplus n_i}$$, and therefore induces an isomorphism of endomorphism rings $$R\cong \operatorname{End}(R)\cong \bigoplus\limits_{i=1}^m \operatorname{End}(I_i^{\oplus n_i})\cong \bigoplus\limits_{i=1}^m M_{n_i}(\operatorname{End}(I_i))$$. Note that each matrix ring is simple as an $$R$$-module.

Now suppose $$M$$ is a stably free $$R$$-module, then we have $$M\oplus R^m\cong R^n$$, and $$M$$ is also a semisimple $$R$$-module, so we can write down a decomposition into finite number of simple $$R$$-modules, and by the uniqueness of simple factorization of semisimple modules, we count the dimensions and conclude that $$M$$ is a free $$R$$-module.