Question

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.

Thanks.

Answer 1

Let $e_1, \dots, e_n$ be a basis for $A^{\oplus n}$ and let $m_1, \dots, m_n$ be generators for $M$ so that your map $f: A^{\oplus n} \twoheadrightarrow M$ has $f(e_i) = m_i$. Now, suppose $\varphi \in \mathrm{End}_A(M)$. For each $i$, choose $a_{i,j} \in A$ such that $$\varphi(m_i) = \sum_j a_{i,j} m_j.$$ (Obviously, this choice is not necessarily unique.) Now, define $\tilde{\varphi} \in \mathrm{End}_A(A^{\oplus n})$ by setting $$\tilde{\varphi}(e_i) = \sum_j a_{i,j} e_j.$$ You can show that $f \circ \tilde{\varphi} = \varphi \circ f$, which means that $\tilde{\varphi} \in B$ and $g(\tilde{\varphi}) = \varphi$.