2
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.