Let $k$ be a algebraically closed field and let $R$ be a semi-simple finite dimensional $k$-algebra. Show that the number of primitive central idempotents of $R$ is the same as the dim$_k Z(R)$.
I suspect that the center will be spanned by these idempotents, but I'm not sure how to relate them. Please guide me.
By the Artin-Wedderburn theorem, the ring is a unique direct sum of $n$ matrix rings over $k$ (since $k$ is algebraically closed).
Each one has a one has center isomorphic to $k$, and the center of the whole ring is the direct sum of centers of the parts, so the center has dimension $n$ over $k$.
Each centrally primitive idempotent produces an ideal summand which is one of the n simple factors. So the number of centrally primitive idempotents is equal to n as well.