Let A be a C algebra, where C is a commutative ring with 1, and M be a finitely generated left A-module.
Question: Is it true that we can always find a positive integer n, a C subalgebra B of M_n(A) and an ideal J of B such that B/J is isomorphic to End(M)? If not, what other conditions are needed to make the statement true?
By isomorphism I mean a C algebra isomorphism. M_n(A), as always, is the algebra of n-by-n matrices with entries from A. By End(M) I mean the algebra of A- homomorphisms from M to M.
So I'm looking for a homomorphism from some C subalgebra B of M_n(A) onto End(M). Well, I know that there exists a natural surjection f from A^n to M for some positive integer n because M is finitely generated over A. One way to define a map g from M_n(A) to End(M) is to define g(a)(m)=f(ax), for all a in A and m in M, where x is any element of A^n with f(x) = m. Ok, this map has obviously the well-definedness issue and that prevents g to be defined on the whole M_n(A). So, we choose B to be the set of those elements a in M_n(A) such that f(ax)=0, for all x in the kernel of f. Now g is well-defined on B and B is a C subalgebra of M_n(A). What I'm having trouble with is to show that g is surjective!
PS. I need the above to show that if S is a subalgebra of R and R is finitely generated as S module, then the Gelfand Kirillov dimension of R and S are equal. I didn't know how to prove it directly using the definition. So if anybody knows a direct proof, that would also be great.