# A semisimple ring has only finitely many isomorphism classes of simple modules

Consider the ring $$R= M_{n_1}(D_1)\oplus \dots \oplus M_{n_k}(D_k)$$ where $$D_1,\dots , D_k$$ are skew fields.

Is there an easy way to see that there are, up to isomorphism of left $$R$$-modules, only finitely many simple left $$R$$-modules?

Attempt: I managed to prove this for one factor, but my proof does not generalise and I don't think I can get an induction started.

You can prove quite generally that if $$R \times S$$ is a product of rings, every $$R \times S$$-module $$M$$ canonically decomposes as a product $$M_R \times M_S$$ of an $$R$$-module and an $$S$$-module, where $$M_R = (1, 0) M$$ and $$M_S = (0, 1) M$$. It follows that a simple $$R \times S$$-module is either a simple $$R$$-module or a simple $$S$$-module (and the same for indecomposable modules), hence that the number of simple modules over $$R \times S$$ is the number of simple modules over $$R$$ plus the number of simple modules over $$S$$. Then you're done by induction.