Let $A$ be commutative $k$-algebra over a field $k$, and $M$ be a finite dimensional indecomposable* $A$-module. In general, $\operatorname{End}_A(M)/J(\operatorname{End}_A(M))$ will be a division ring $D$. I struggle to find a convincing example where $D$ (or $\operatorname{End}_A(M)$) is not commutative itself. At best, I would like to find such an example for $M$ an $n$-graded module over $A=k[x_1,\dotsc,x_n]$. This yields to the following
Question: Can a finite dimensional $n$-graded indecomposable $k[x_1,\dotsc,x_n]$-module have a non-commutative endomorphism ring? Can its quotient by its radical be non-commutative? If not, can it have any endomorphism ring other than $k$?
Edit: Originally, I had asked for a simple module, which cannot be, as pointed out by Eric Wolfsey.
Edit: See this more structured question which also incorporates this one.
A simple module is in particular cyclic (generated by any nonzero element), which means every endomorphism of it is given by multiplying by some element of $A$. So since $A$ is commutative, the endomorphism ring is commutative (specifically, it is the quotient of $A$ by the ideal that annihilates $M$).
In the case that $M$ is additionally $n$-graded over $k[x_1,\dots,x_n]$, the story is even simpler. Let $I$ be the idea $(x_1,\dots,x_n)$. If you fix any nonzero homogeneous element $m\in M$, then $Im$ is a proper submodule of $M$ (it cannot contain $m$, since all its elements have higher degree than $m$). So by simplicity, $Im=0$. That means every $x_i$ acts by $0$ on $M$, and so any $k$-vector subspace of $M$ is also a submodule, so by simplicity $M$ is $1$-dimensional (and its endomorphism ring is $k[x_1,\dots,x_n]/I\cong k$).