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?