0
$\begingroup$

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 ?

$\endgroup$
2
$\begingroup$

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 $M=A^n$.

$\endgroup$
0
$\begingroup$

Consider the following relations (unfamiliar notations are explained):

$$\begin{array} HHom_{A}(M,N)\ \underset{A}{\otimes} B & \stackrel{\rho}{\longrightarrow} & Hom_{B}(B\ \underset{A}{\otimes} M, B\ \underset{A}{\otimes} N ) \\ \updownarrow{\cong} & & \updownarrow{\cong} \\ Hom_{A}(\coprod A,N)\ \underset{A}{\otimes} B & & Hom_{B}(B\ \underset{A}{\otimes} \coprod A, B\ \underset{A}{\otimes} N ) \\ \updownarrow{\cong} & & \updownarrow{\cong} \\ \coprod Hom_{A}(A,N) \underset{A}{\otimes} B & & \coprod Hom_{B}(B\ \underset{A}{\otimes} A, B\ \underset{A}{\otimes} N ) \\ \updownarrow{\cong} & & \updownarrow{\cong} \\ \coprod N \underset{A}{\otimes} B & & \coprod Hom_{B}(B, B\ \underset{A}{\otimes} N ) \\ \updownarrow{\cong} & & \updownarrow{\cong} \\ \coprod N \underset{A}{\otimes} B & \stackrel{\cong}{\longleftrightarrow} & \coprod N \underset{A}{\otimes} B \\ \end{array}$$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.