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 $n\times n$ 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 $x$ from 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.