Question regarding notation of $Hom$ of modules I was working on a problem and I have been struggling to understand the notation:

Let $f: A \to B$ be a ring homomorphism. Let $M, N$ be $B$-modules. Show that $Hom_{B} (M,N) \subset Hom_{A} (M,N)$ is a subgroup.

I understand that $Hom_{B} (M,N)$ is the set of $B$-module homomorphisms. But how do I view $Hom_{A} (M,N)$? Is there any natural way to view this set as a group with $f: A \to B$ being given?

PS: I am not looking for any solutions.
 A: First, you need to understand how $M$ and $N$ are A-modules, the action of $a\in A$ is given by $a*m=f(a)m$, note $f(a)\in B$ and so the $f(a)$ action is given by the $B$ action .
Now both $Hom_A(M, N)$ and $Hom_B(M, N)$ are sets of function from $M$ to $N$ satisfies certain conditions, both of them are abelian group (if you are not sure about it prove this!), so all you need is to prove the inclusion.
A: Once I struggled in understanding stuffes like this too, especially when learning the definition and properties of tensor product of $R$ modules.
Maybe the keypoint is to understand Bimodule, here the definition is taken from Rotman's An introduction to Homological Algebra:

Definition. Let $R$ and $S$ be rings and let $M$ be an abelian group. Then $M$
is an $(R, S)$-bimodule, denoted by $_R M_S$, if $M$ is a left $R$-module and a right
$S$-module, and the two scalar multiplications are related by an associative law:
$r(ms) = (rm)s$
for all $r ∈ R$, $m ∈ M$, and $s ∈ S$.

With this definition and some basic facts, like:

*

*Any $R$ module $M$ is a $(R, \Bbb{Z})$-bimodule.

*if $R$ is a commutative ring and $M$ is a $R$ module, then $M$ is a $(R, R)$ bimodule.

The problem may be formalized as:

Show that $\operatorname{Hom}_B(M, N)$ is a $(B, \Bbb{Z})$-bimodule and $\operatorname{Hom}_A(M, N)$ is a $(A, \Bbb{Z})$-bimodule. And as $\Bbb{Z}$ module, $\operatorname{Hom}_B(M, N) \subset \operatorname{Hom}_A(M, N)$. This viewpoint is equivalent to forget the multiplication of the Hom set.

Or:

Show that $\operatorname{Hom}_B(M, N)$ is a $(B, B)$-bimodule and $\operatorname{Hom}_A(M, N)$ is a $(A, B)$-bimodule. And as $B$ module, $\operatorname{Hom}_B(M, N) \subset \operatorname{Hom}_A(M, N)$

