4
$\begingroup$

Let $R$ be a commutative ring. Let $M$ be an $R$-module with the following property:

For every commutative $R$-algebra $A$ and every $R$-module $N$ the canonical map $\mathrm{Hom}_R(M,N) \otimes_R A \to \mathrm{Hom}_R(M,N \otimes_R A)$ is an isomorphism.

Does this imply that $M$ is finitely generated projective? If not, what happens when we even assume the following property?

For every homomorphism of commutative $R$-algebras $A \to B$ and every $A$-module $N$ the canonical map $\mathrm{Hom}_R(M,N) \otimes_A B \to \mathrm{Hom}_R(M,N \otimes_A B)$ is an isomorphism.

If this is too hard in general, what about explicit examples for $R$?

  • 0
    In your 1st question, the fact that $A$ is an algebra does not seem to play any role.2012-12-24

1 Answers 1

4

If the canonical map is an iso for all modules $A$, then taking $N=R$ and $A=M$, we get an iso $\hom_R(M,R)\otimes_RM\to\hom_R(M,M)$. The preimage of $1_M$ gives a finite dual basis for $M$.

If you want the hypothesis on algebras only, you can probably reduce to the above by considering the $R$-algebra $A=R\oplus M$ with $M\cdot M=0$. One has to check this does give the hypothesis on all modules —I haven't looked :-)

  • 0
    Since we only need to take $N=R$ to get what we want, which is a bimodule for free, there is no loss, really! And then the canonical map ends up being an iso for all bimodules $N$. In a way, the question as stated *is* considering $N$ to be an $R$-bimodule, with the same action on both sides :-)2012-12-26