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

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 ?

• – Watson Nov 22 '18 at 13:45

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 $M=A^n$.
$$\begin{array} HHom_{A}(M,N)\ \underset{A}{\otimes} B & \stackrel{\rho}{\longrightarrow} & Hom_{B}(B\ \underset{A}{\otimes} M, B\ \underset{A}{\otimes} N ) \\ \updownarrow{\cong} & & \updownarrow{\cong} \\ Hom_{A}(\coprod A,N)\ \underset{A}{\otimes} B & & Hom_{B}(B\ \underset{A}{\otimes} \coprod A, B\ \underset{A}{\otimes} N ) \\ \updownarrow{\cong} & & \updownarrow{\cong} \\ \coprod Hom_{A}(A,N) \underset{A}{\otimes} B & & \coprod Hom_{B}(B\ \underset{A}{\otimes} A, B\ \underset{A}{\otimes} N ) \\ \updownarrow{\cong} & & \updownarrow{\cong} \\ \coprod N \underset{A}{\otimes} B & & \coprod Hom_{B}(B, B\ \underset{A}{\otimes} N ) \\ \updownarrow{\cong} & & \updownarrow{\cong} \\ \coprod N \underset{A}{\otimes} B & \stackrel{\cong}{\longleftrightarrow} & \coprod N \underset{A}{\otimes} B \\ \end{array}$$ As it can be seen we have $A$-isomorphisms in each column and in the last row. The fact that $M$ is a finitely generated free $A$-module allowed us to write $M\cong\coprod A$ (which maybe a little bit unfamiliar notation). All the other isomorphisms are derived from the properties of $\otimes$ and $Hom$.\ Finally according to the diagram it is apparent that $\rho$ is an $A$-isomorphism.