Suppose we have an algebra $A$ (unital, associative), with an ideal $I \leq A$ and a finitely generated module $M$ over $A$.


It is possible to obtain both $\mathrm{End}\_A(M)$ and $\mathrm{End}\_A(M/IM)$ as subquotients of $M\_n(A)$ ($n\times n$ matrices with values in $A$), as given for example [here](https://mathoverflow.net/questions/30290/algebra-of-endomorphisms-of-f-g-modules-as-subquotients-of-matrix-algebras). I am interested in the following:



> 
> When can one get $\mathrm{End}\_A(M/IM)$ as a subquotient of $\mathrm{End}\_A(M)$? 
>  Are there simple counter examples, or should one hope that such a claim holds in general under some mild assumptions?
> 
> 
> 


In my particular settings, $A$ is a $\mathbb{C}$-algebra, finitely generated over its center, $Z(A)$, and the ideal is of the form $IA$ where $I$ is a maximal ideal of $Z(A)$, making $A/IA$ a finite dimensional algebra.
I'm attempting to show that commutativity of $\mathrm{End}\_A(M)$ implies commutativity of $\mathrm{End}\_A(M/IM)$, and an argument of the above form implies this statement.


Help is much appreciated.



Under what assumptions can endomorphisms of $M/IM$ be realized as a subquotient of endomorphisms of $M$?
Let $A=\mathbb{C}[x,y]$, $I=(x,y)$, and $M$ the $3$-dimensional $A$-module $(x,y)/(x^2,y^2)$ (with basis $\{x,y,xy\}$).


Then $M/IM$ is isomorphic to a direct sum of two copies of $\mathbb{C}=A/I$, so $\operatorname{End}\_A(M)\cong\mathbb{C}$, but $\operatorname{End}\_A(M/IM)\cong M\_2(\mathbb{C})$.


