Does this property generalize to modules over noncommutative rings?

Question

Let $M$ be a finite simple module over a ring $R$. For commutative $R$, it is easy to show that for every $r \in R$, the function $M \rightarrow M; x \mapsto rx$ is either bijective or constantly $0$. I'm wondering if this is also the case for noncommutative $R$ (I'm only interested in the case when $R$ is finite if that makes a difference). I tried proving it, but to no avail. It's probably false in the noncommutative case, but I wasn't able to find a counterexample, as I'm not very experienced with modules.

Answer 1

Let $F$ be a finite field and consider $F^2$ as an $M_2(F)$ module. It is simple (no nonzero submodules) since any non-zero vector can be moved to any other one by some element of $M_2(F)$ (proof: either one is a scalar multiple of the other and you can take the diagonal matrix with that scalar, or they form a basis and you can take the transformation flip-flopping them). But you can write down a matrix with a kernel, e.g. $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$.