# 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 ∼ M

it follows that (dimkR)(dimkE) = (n

E ∼ M

_{m}(EndRV ) ∼ M_{m}(D),it follows that (dimkR)(dimkE) = (n

^{2}d)(m^{2}d)= (mnd)^{2}= (dimkM)^{2}.Need a fast expert's response?

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