Homomorphism of modules and Tensor Product. Let $\phi: A \rightarrow B$ be a ring homomorphism. Let $M$ be an $A$-module. We can think $B$ as $A$-module via the map $\phi$ defined by  $\phi:A\times B \rightarrow B$, $(a,b)\mapsto\phi(a)\cdot b$.
So we can construct the $A$-module $M \otimes_A B$. Furthermore, $M \otimes_AB$ can be thought of as a $B$-module via $B\times (B \otimes _AM)\rightarrow (B\otimes _A M)$ defined by $(b',(b\otimes m))\mapsto b'b\otimes m$. 
Question: Why is $\operatorname{Hom}_A(M,N)=\operatorname{Hom}_B(B\otimes_A M,N)$ where $N$ is an $B$-module and in the left hand side $N$ is considered as a $A$-module as described above?

Hint: Given an $A$-linear map $f : M \to N$, define $B \times M \to N$ by $(b,m) \mapsto b \,  f(m)$. Check that this is $A$-bilinear, hence lifts to an $A$-linear map $h : B \otimes_A M \to N$ characterized by $h(b \otimes m)= b \, f(m)$. Check that it is actually $B$-linear. Conversely, given a $B$-linear map $h : B \otimes_A M \to N$, check that $f : M \to N$, $m \mapsto h(1 \otimes m)$ is $A$-linear. Finally, prove that these constructions are inverse to each other.
