0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ @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. $\endgroup$ Sep 3, 2023 at 2:29

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .