Base-free computation: Center of $M_n(R)$ Let $R$ be a (possibly noncommutative) ring and $M$ a free module. I think it is true that $Z(\operatorname{End}(M))\cong Z(R)$ and $Z(\operatorname{Aut}(M))\cong Z(R^\times)$, by the usual Matrix argument. I am having a hard time finding a base free proof. Thank you.
 A: At least if $M$ is a finite free module, the first statement holds. One abstract way to see this is that


*

*$M_n(R)$ is Morita equivalent to $R$, and

*Taking the center is Morita invariant: the center of a ring $R$ can be defined as the center in the categorical sense of the category $\text{Mod}(R)$.


The second statement is then equivalent to the claim that $Z(GL_n(R)) \cong Z(M_n(R))^{\times}$, which will hold whenever the condition that an element $g \in GL_n(R)$ is central in $GL_n(R)$ is equivalent to the condition that it's central in $M_n(R)$. An easy way to guarantee this is if elements of $GL_n(R)$ generate $M_n(R)$ as an abelian group. This might always be true but I don't know if there will be a clean basis-free proof at this level of generality. Over, say, an infinite field there's a straightforward proof using determinants, but already over a finite field I'm not sure what to do other than a computation with matrices.  
If $M$ isn't a finite free module then I don't know what happens. It's certainly no longer true that $\text{End}(M)$ is Morita equivalent to $R$. 
