Answer to Question #17110 in Abstract Algebra for Tsit Lam

Question

Let R be a simple ring that is finite-dimensional over its center k. Let M be a finitely generated left R-module and let E = End(RM). Show that (dimkM)2 = (dimkR)(dimkE).

Answer 1

Say R = Mn(D), where D is a divisionring, with Z(D) = Z(R) = k. Let V be the unique simple left R-module, so M ∼ mV for some m. For d = dimkD, we have dimkM = m · dimkV = mnd. Since
E ∼ M_m(EndRV ) ∼ M_m(D),
it follows that (dimkR)(dimkE) = (n²d)(m²d)= (mnd)² = (dimkM)².