# Classify commutative rings $R$ such that $A \otimes_{\Bbb Z} B = A \otimes _{R} B$

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?

• This may be related to the concept of "solid ring". – Fernando Muro Dec 3 '18 at 11:10
• 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 3 '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 3 '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 3 '18 at 14:24
• 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. – Pierre-Yves Gaillard Dec 3 '18 at 19:06

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 $$A = B = R$$.)