Why is $\operatorname{Hom}(M,N)$ not necessarily an $R$ module? 
Let $R$ be a ring, and $M,N$ be left $R$-modules. Then is it not true that $\operatorname{Hom}_R(M,N)$ has the structure of an $R$-module? 

I was reading the preface of the Homological Algebra book by Rotman and was quite surprised to learn that this is not the case. I think all the axioms for being a module are satisfied by $\operatorname{Hom}_R(M,N)$, but Rotman is very unlikely to make a mistake. What is it that I am missing?
Under what circumstances is this true?
 A: Any natural definition of $R$-action works only, when $R$ is commutative. For example, if you try to define
$$
(rf)(m)=r(f(m))
$$
for all $r\in R$, $f\in Hom_R(M,N)$, $m\in M$, then the mapping $rf$ fails to be homomorphism of $R$-modules in general. If $s\in R$ is such that $sr\neq rs$, then
$$
(rf)(sm)=r(f(sm))=r(sf(m))=(rs)(f(m))\neq s((rf)(m))
$$
in general.
OTOH, if one of the modules, $M$ or $N$, is an $(R,R)$-bimodule, then you do get a module structure.
A: What action of the ring $R$ do you expect on $\operatorname{Hom}_R(M, N)$?  The left action of $R$ on $M$ and $N$ is "used up" when you consider $R$-linear homomorphisms
$$
\operatorname{Hom}_R(M, N) \subseteq \operatorname{Hom}_{\Bbb{Z}}(M, N).
$$
If $M$ is in fact an $(R, R)$-bimodule (this is automatic if $R$ is commutative), then the right action of $R$ on $M$ yields a left action of $R$ on $\operatorname{Hom}_R(M, N)$ via
$$
(rf)(m) = f(mr).
$$
On the other hand, if $N$ is an $(R, R)$-bimodule, then the right action of $R$ on $N$ yields a right action of $R$ on $\operatorname{Hom}_R(M, N)$ via
$$
(fr)(m) = f(m)r.
$$
Of course, if both $M$ and $N$ are bimodules, then $\operatorname{Hom}_R(M, N) $ is a bimodule, as well.
A: This is true if $R$ is commutative.
Otherwise, say that you are dealing with left $R$-modules, for instance. If you attempt to define multiplication by $r$ by $(rh)(m) = rh(m)$ for any homomorphism $h \colon M \to N$, then you run into the problem that the mapping $rh$ may not be $R$-linear. 
For example, let $h \colon R \to R$ be the identity map. Then $k(s) = (rh)(s)$ would be $rs$. But in general, $k(s1) = rs \neq sr = sk(1)$.
A: Presumably, you'll want to define $\psi=r.\phi$ by
$$\psi(m)=r.\phi(m)$$
(where $r\in R, \phi\in\mathrm{Hom}_R(M,N)$ and $m\in M$.) However, this map, which is a morphism of abelian groups, need not be $R$-linear when $R$ isn't commutative : $R$-linearity would imply that for all $r'\in R$ (and all $m\in M$)
$\psi(r'.m)=r'.\psi(m)$, i.e., by $R$-linearity of $\phi$,
$$rr'.\phi(m)=r'r.\phi(m)$$
Since $rr'\neq r'r$ in general, there is no reason the above identity should hold.
