Let F be a field let R MnF and let M be the unique irreduci

Let F be a field, let R = M_n(F) and let M be the unique irreducible R-module. Prove that Hom_R(M,M) is isomorphic to F (as a ring).

Solution

HomR(M, M) is a ring has been proved earlier.

Let 0 6= F HomR(M, M). Then f : M M is an R-homomorphism. Now Ker(f) and Im(f) are submodules of M.

Since f 6= 0, Ker(f) 6= M and Im(f) 6= (0). Hence Ker(f) = (0) and Im(f) = Mfollows from the fact that M is simple. Thus f is an isomorphism and hence invertible.