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Answer to Question #17110 in Abstract Algebra for Tsit Lam

Question #17110
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).
Expert's answer
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 ∼ Mm(EndRV ) ∼ Mm(D),
it follows that (dimkR)(dimkE) = (n2d)(m2d)= (mnd)2 = (dimkM)2.

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