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I have asked this in MSE but there was no reply. Feel free to close if inappropriate.

Let $R$ be commutative ring, what can we say about the rings $R$ such that $A \otimes_{\Bbb Z} B \cong A \otimes_R B$ as abelian groups for all $A$ right $R$ module and $B$ left $R$ module?

Bryan Shih
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    This may be related to the concept of "solid ring". – Fernando Muro Dec 03 '18 at 11:10
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    Maybe you mean in the assumption that the canonical surjective homomorphism $A\otimes_ZB\to A\otimes_B R$ is an isomorphism for all $A,B$, which sounds more natural (and which I guess is implicit in your mind). – YCor Dec 03 '18 at 11:17
  • Not sure it helps but (if I'm not mistaken) the condition is clearly equivalent to $\text{Hom}R(A,B)=\text{Hom}{\mathbb Z}(A,B)$ for all $R$-modules $A$ and $B$. – Pierre-Yves Gaillard Dec 03 '18 at 14:12
  • It would be helpful if you gave some sample computations/results. For example, it appears that if $R$ is torsion-free as an abelian group, then it must be a subring of the rationals ... – David Handelman Dec 03 '18 at 14:24
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    I'll use Fernando Muro's hint: https://mathoverflow.net/q/95160/461 shows (I think) that $R$ has your property $\iff R$ is solid. Indeed $\implies$ is clear. It only remains to check that the solid rings given by the classification have your property, which (it seems to me) is not hard. – Pierre-Yves Gaillard Dec 03 '18 at 19:06

1 Answers1

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Fernando and Pierre-Yves in the comments are right; $R$ has this property (the version where the canonical map is an isomorphism, as YCor says in the comments) iff it is a solid ring, meaning the multiplication map $m : R \otimes_{\mathbb{Z}} R \to R$ is an isomorphism, and no commutativity assumption is needed. To see this really clearly note that we can rewrite the canonical map as

$$\text{id}_A \otimes m \otimes \text{id}_B : A \otimes_R (R \otimes_{\mathbb{Z}} R) \otimes_R B \to A \otimes_R R \otimes_R B.$$

Now the implication $\Rightarrow$ follows by letting $A = B = R$, and the implication $\Leftarrow$ follows by noting that if $m$ is an isomorphism then it's an isomorphism of $(R, R)$-bimodules.

(The idea behind rewriting things this way is that by a variation of the Eilenberg-Watts theorem, the category of functors taking as input a left $R$-module and a right $R$-module, cocontinuous in each variable, and returning as output an abelian group is equivalent to the category of $(R, R)$-bimodules, hence any natural transformation of such functors as in the OP can be analyzed as a bimodule homomorphism, namely the bimodule obtained by substituting $A = B = R$.)

Qiaochu Yuan
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