Clarification on module structure over matrices Let $R$ be a ring and $M$ an $R-$module. I know that to make $\operatorname{Hom}_R(R^n,R^m)$ an $R-$module I need $R$ to be commutative. However, if we consider $R^{n\times m }$, we can give it $R-$module structure using the usual sum of matrices and scalar multiplication, independently of $R$ being commutative.
But, I find it strange that we can define a module structure in one space and not in the other as there is a natural way to think of a matrix as a morphism. So, my question is if there is a reason why the module structure on matrices needs the ring to be commutative.
 A: It isn't necessary for $R$ to be commutative to make the hom set a module. There are a couple structures on it that are very natural provided the modules you're working with are bimodules -- as is the case with $R^n$.
If $R,S,T$ are rings and $A$ is an $S,R$ bimodule and $B$ is an $T,R$ bimodule, then
$\mathrm{Hom}(A_R, B_R)$ has a natural right $S$ module structure using the rule $(f\cdot s)(x):=f(sx)$ and natural left $T$ module structure using the rule $(t\cdot f)(x):=tf(x)$. In fact, the hom set is a $T, S$ bimodule.
In particular, you can have $R=S=T$.
Since $R^n$ is an $R,R$ bimodule for any $n$, you can apply this always to $\mathrm{Hom}(R^n_R, R^m_R)$, which is the same thing as matrices over $R$.
But it is true that given arbitrary one-sided modules $A_R$, $B_R$ with no further information, there is not enough to build these two actions and the best one can say is that the hom set is an abelian group.

If you are thinking of left $R$ modules instead, everything can be mirrored: if $A$ is an $R,S$ bimodule and $B$ is an $R,T$ bimodule then $\mathrm{Hom}(_RA, _RB)$ gains a natural $S,T$ bimodule structure via $(s\cdot f\cdot t)(x):=f(xt)s$. This is left $R$ linear because $(s\cdot f\cdot t)(rx)=f(rxt)s=rf(xt)s=r(s\cdot f\cdot t)(x)$
