Question

Show that if R is semisimple, so is Mn(R).

Answer 1

Let $R \sim \prod_{i=1}^{r} M_{n_i}(D_i)$, where the $D_i$'s are suitable division rings. Then $\mathbf{M}_n(R) \sim \prod_{i} M_n\left(M_{n_i}(D_i)\right) \cong \prod_{i} M_{nn_i}(D_i)$ which is a semisimple ring.