Let $A, B$ be commutative rings with one and let $M$ be an $A$-module, $f: A \rightarrow B$ a ring homomorphism. Consider the (right) $B$-module $M \otimes_A B$.
What can we say about $\operatorname{Hom}_B(M \otimes_A B, N)$ for a $B$-module (and hence $A$-module) $N$? I was hoping that this can somehow be expressed as a Hom-set of $A$-modules, but I'm not sure how to do this. Many thanks in advance.
Denote by $N_A$ the $A$ module obtained from $N$ by restriction of scalars (i.e. $a\ast n=f(a)\cdot n$).
Then we have a map $$b: \text {Hom}\: _ {A-\text {lin}}(M,N_A) \to \text {Hom}\: _ {B-\text {lin}}(M\otimes _A B,N) $$ sending $u:M\to N_A$ to $b(u)= u^+: M\otimes _A B \to N$, which last map is characterized by $ u^+(m\otimes b)=f(m).b$ .
This map $b$ is bijective and its inverse is $$c: \text {Hom}\: _ {B-\text {lin}}(M\otimes _A B,N) \to \text {Hom} \: _ {A-\text {lin}}(M,N_A) $$ sending $v:M\otimes _A B\to N$ to $c(v)=v_-:M\to N_A$, which last map is defined by $v_-(m)=v(m\otimes 1_B)$
So this realizes your hope that "$\operatorname{Hom}_B(M \otimes_A B, N)$ [...] can somehow be expressed as a Hom-set of $A$-modules".
Informally but more vividly we might sum this up by saying that $B$-linear maps $M\otimes B\to N$ amount to the same as $A$-linear maps $M\to N$ .