$ρ:Hom_A(M,N)\otimes_A B \rightarrow Hom_B(B\otimes_A M,B\otimes_A N)$, if $M$ is free of finite rank over $A$, then $ρ$ is an isomorphism?

Question

Let $B$ be a faithfully flat $A$-algebra, M be a finitely generated $A$-module and $N$ be an $A$-module.

Then consider the natural homomorphism

$ρ:Hom_A(M,N)\otimes_A B \rightarrow Hom_B(B\otimes_A M,B\otimes_A N)$

how to prove that if $M$ is free of finite rank over $A$, then $ρ$ is an isomorphism ?

Answer 1

First check it for $M=A$: $$\def\HH{\operatorname{Hom}}\def\tens{\otimes_AB} \rho\colon\HH_A(A,N)\tens\to\HH_B(A\tens,N\tens) $$ In this case both the domain and the codomain are naturally isomorphic to $N\tens$ (as $B$-modules).

Now check that if $\rho$ is an isomorphism for $M_1$ and $M_2$, then it is also for $M=M_1\oplus M_2$.

This provides the induction step, as you can suppose $M=A^n$.