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.


  • 2
    $\begingroup$ This is Lemma 8.2.8 in McConnell and Robson (Noncommmutative Noetherian Rings, AMS Graduate Studies in Mathematics, Vol. 30). They don't give a proof that your $g$ is surjective, but leave it for the reader to check. $\endgroup$ – PersonX Jul 2 '10 at 11:15
  • $\begingroup$ Dear @Sergiy Kozerenko: You have made a significant number of very minor edits (specifically, seven) to old questions in the past two hours. That is considered too much. Moreover, your edits arguably do not actually improve the readability of the questions. $\endgroup$ – Ricardo Andrade Nov 22 '13 at 12:47
  • $\begingroup$ Dear @Ricardo Andrade: If we have TeX, we should use it. $\endgroup$ – Sergiy Kozerenko Nov 22 '13 at 17:17
  • $\begingroup$ Dear @Sergiy Kozerenko: Mathjax is indeed a useful tool for MathOverflow. However, just because it's there does not mean we have to use it. For one, it significantly slows down rendering in the browser (for example, see the discussion at meta.mathoverflow.net/questions/462/…), and not all browsers or people allow mathjax to run. If mathjax is not running, a post heavy with TeX may be essentially unreadable. $\endgroup$ – Ricardo Andrade Nov 22 '13 at 17:26
  • $\begingroup$ Perhaps more importantly, introducing mathjax into a post may actually make it less readable (even when mathjax is running), if TeX is not properly wielded: as a simple example, compare the spacing in "Spec K" with "$Spec \mathbb{K}$" (by the way, I took this from one of your edits today). While I have your attention, TeX/mathjax should also not be used to format spacing, such as now appears in the above question: $End(M)\ ?$; spacing should not be controlled within mathjax unless absolutely necessary, and the question mark should also be outside of the dollar signs. $\endgroup$ – Ricardo Andrade Nov 22 '13 at 17:32

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 $g(\tilde{\varphi}) = \varphi$.

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