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