Show that $\operatorname{Hom}_R(E, F)$ is an $R$-module in a natural way. 
Let $R$ be a conmutative ring, and $E,F$ are modules, show that $\operatorname{Hom}_R(E, F)$ is an $R$-module in a natural way. Is this still true if $R$ is not conmutative?

I think in the to following operations: $f,g\in \operatorname{Hom}_R(E, F)$ then $(f+g)(x)=f(x)+g(x)$, if $f\in \operatorname{Hom}_R(E,F)$ and $r\in R$ then $(rf)(x)=rf(x)$. I don't see why it does not work for no conmutative rings. Why?
Thanks!
 A: The reason is, basically, that  $rf$ defined this way would usually fail to be a homomorphism of $R$-modules.
Assume that for every $r \in R$, $g_r=rf$ is a homomorphism of (left) $R$-modules as well. 
Then for every $r,s \in R$ and every $x \in M$, we have
$$(rs)\cdot f(x)=r\cdot(s\cdot f(x))=r\cdot f(sx)=[rf](sx)=g_r(sx)=s\cdot g_r(x)=\\=s \cdot [rf](x)=s\cdot(r \cdot f(x))=(sr)\cdot f(x) \;\;,$$
that is, $(rs-sr)f(x)=0$ for all $r,s \in R$ and all $x \in E$.
This implies that $\text{Im }f$ has actually a structure of a $R/[R,R]$-module, where $[R,R]$ denotes the ideal generated by all the possible elements of the form $rs-sr, r,s \in R$. Thus,  $\text{Im }f$ has actually a structure of a module over a commutative ring. 
Whenever this fails (that is, there exists $y \in F$ such that $y=f(x)$ for some homomorphism $f:E\rightarrow F$ such that $y$ is not annihilated by some element of the form $rs-sr$), then the multiplication of homomorphism defined as above will not be even correctly defined of $\mathrm{Hom}_R(E,F)$.
