Let $A$ be an abelian group, let $B$ be a PID and $C$ be an unitary commutative ring which is also a $B$-module. Then all abelian group homomorphisms between $A$ and $B$ consist an abelian group $\text{Hom}(A,B)$ and it has a natural $B$-module structure. In the same way $\text{Hom}(A,C)$ is a $C$-module.
$\textbf{Question}$: Then do we have the $C$-module isomorphism?$\text{Hom}(A,B)\bigotimes_BC\cong_C\text{Hom}(A,C)$
By universal property, we have a obvious $C$-module homomorphism $f\bigotimes c\rightarrow (a\rightarrow f(a)\circ c)$, but I can't construct the inverse, and it seems to me the two $C$-modles can't be isomorphic.
Thanks!
For $A = \mathbb{Z}_p, B = \mathbb{Z}$, and $C = \mathbb{Z}_p$, the module $\operatorname{Hom}(A, B)\otimes_B C = 0$, but $\operatorname{Hom}(A, C)$ is nonzero. This is similar to asking if every map $A\to C$ factors through $B$, which is clearly false in general.