# $R$ semisimple $\implies M_n(R)$ is semsimple

I'm working out of TY Lam's first course in noncommutative rings. I'd like to show that if $$R$$ is a semismiple ring, $$M_n(R)$$ is as well. The obvious answer to me seems to be that since $$R$$ is a (left) semisimple ring and $$M_n(R)$$ is a left $$R$$-module, we have $$M_n(R)$$ is semisimple. This follows from the fact that all left $$R$$-modules are semisimple if $$R$$ is semisimple as a ring.

What is confusing me is the section the question is occurring in. It isn't appearing with the basic properties of semisimple rings, but with the structure of semisimple rings which deals with the Wedderburn decomposition. Am I missing something and my logic is incorrect? Or is there something else going on that I'm missing? Or is it just an odd placement?

Your argument is wrong: it proves that $$M_n(R)$$ is semisimple as an $$R$$-module, but you need instead prove that $$M_n(R)$$ is semisimple as an $$M_n(R)$$-module. Your argument would more generally claim that any algebra over a field is semisimple because a filed is semisimple, which is obviously wrong.
Probably it's possible to prove this statement correctly by analyzing the structure of $$M_n(R)$$-submodules of $$M_n(R)$$, but a very easy proof follows from the Artin-Wedderburn classification (if $$R$$ is a product of matrix rings $$M_{k_i}(D_i)$$ then $$M_n(R)$$ is a product of matrix rings $$M_{n\cdot k_i}(D_i)$$, by viewing an $$(n\cdot k_i)\times(n\cdot k_i)$$ matrix as a block matrix with $$n\times n$$ blocks)