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?

  • 7
    $\begingroup$ This may be related to the concept of "solid ring". $\endgroup$ Dec 3, 2018 at 11:10
  • 6
    $\begingroup$ 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). $\endgroup$
    – YCor
    Dec 3, 2018 at 11:17
  • $\begingroup$ 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 $B$. $\endgroup$ Dec 3, 2018 at 14:12
  • $\begingroup$ 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 ... $\endgroup$ Dec 3, 2018 at 14:24
  • 1
    $\begingroup$ I'll use Fernando Muro's hint: 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. $\endgroup$ Dec 3, 2018 at 19:06

1 Answer 1


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 $A = B = R$.)

  • $\begingroup$ Welcome back, Qiaochu! $\endgroup$ Dec 3, 2018 at 19:52
  • $\begingroup$ It's nifty to know these solid rings are exactly those (commutative) rings $R$ for which the canonical map $\mathbb Z \rightarrow R$ is an epimorphism. Compare my very first MO post mathoverflow.net/a/110443/27465. $\endgroup$ Jun 27, 2019 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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