8
$\begingroup$

Let $R$ be a ring, and let $S$ be an $R$-algebra. Let $M$ be an $R$-module. Under what conditions on $R$, $S,$ and $M$ is it true that we have the following isomorphism of $S$-modules? $$S \otimes_R \operatorname{Hom}_R(M, R) \simeq \operatorname{Hom}_S(S \otimes_R M, S)$$

I know there is an obvious map from the left-hand side above to the right-hand side (namely, send a map $\phi \colon M \to R$ to the map $\operatorname{id} \otimes_R\, \phi$), but I do not know what conditions are required for there to be an inverse map. Does this have anything to do with $M$ being torsion-free as an $R$-module?

$\endgroup$
1
7
$\begingroup$

Note that the RHS can be written $\text{Hom}_R(M, S)$. More generally, we might ask when we have

$$\text{Hom}_R(M, N) \otimes_R S \cong \text{Hom}_R(M, N \otimes_R S).$$

for three $R$-modules $M, N, S$. (Special cases of this general question occur frequently here and on MO: see, for example, here, here, and here.) The answer is no in general but yes if any of the following conditions holds.

  1. $M$ is finitely presented and $S$ is flat.
  2. $S$ is finitely presented and $M$ is projective.
  3. $M$ is finitely presented projective.
  4. $S$ is finitely presented projective.

A simple counterexample to the desired statement at the maximum level of generality is given by $R = \mathbb{Z}, M = S = \mathbb{Q}$. Note that here neither $M$ nor $S$ are finitely presented, and that $M$ is torsion-free.

$\endgroup$
2
  • $\begingroup$ Thanks a bunch! I asked the question hoping that dual might commute with arbitrary base-change for coherent sheaves, but I guess that's not true? Does this occur only for locally free sheaves? $\endgroup$ Mar 1 '17 at 23:56
  • 2
    $\begingroup$ @AshvinSwaminathan his criterion 1 (also to be found in the bourbaki books mentioned by georges) says that it holds for flat base change. you have no hope if you don't assume flatness (see georges' counterexample). it holds for locally free sheaves though, and this is criterion 3. $\endgroup$
    – Rüdiger
    Mar 2 '17 at 0:57
7
$\begingroup$

a) Your morphism $f:S \otimes_R \operatorname{Hom}_R(M, R) \to \operatorname{Hom}_S(S \otimes_R M, S)$ is in general not bijective, as shown by the example ($R=\mathbb Z,S=\mathbb Z/2,M=\mathbb Z/2)$ $$\mathbb Z/2 \otimes_\mathbb Z \operatorname{Hom}_\mathbb Z(\mathbb Z/2,\mathbb Z )=0 \to \operatorname{Hom}_{\mathbb Z/2}(\mathbb Z/2 \otimes_\mathbb Z \mathbb Z/2, \mathbb Z/2)=\mathbb Z/2$$

b) Your morphism is an isomorphism whenever at least one of the two modules $A$-modules $B$ or $M$ is projective of finite type: Bourbaki, Algebra, Chapter II, §5.4, Prop.8, page 283.

$\endgroup$
1
  • 1
    $\begingroup$ +1 for mentioning bourbaki. their books are the best source for base change results. $\endgroup$
    – Rüdiger
    Mar 2 '17 at 0:55

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.