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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?

Ashvin Swaminathan
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    Possibly related: https://math.stackexchange.com/questions/1728363/is-it-true-that-operatornamehom-rm-n-otimes-r-p-operatornamehom-rm-n – Watson May 21 '18 at 20:02

2 Answers2

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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.

Community
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Qiaochu Yuan
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  • 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? – Ashvin Swaminathan Mar 01 '17 at 23:56
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    @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. – Rüdiger Mar 02 '17 at 00:57
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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.

Georges Elencwajg
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