# What are the commutative rings $R$ for which $A \otimes _{\Bbb Z} B = A \otimes _R B$ as abelian groups?

This is a follow up.

What are the commutative rings $$R$$, for which given $$R$$-modules $$A$$ and $$B$$, $$A \otimes _{\Bbb Z} B = A \otimes _R B$$ as abelian groups?

This is true when $$R= \Bbb Q$$, or $$\Bbb Z_m$$. We can give an $$R$$-module structure $$A \otimes _{\Bbb Z} B$$ satisfying $$r (a \otimes b) = ra \otimes b$$.

When $$R= \Bbb Q$$, we get the additional fact that $$a \otimes rb=ra \otimes b$$. To see this, note that when $$r \in \Bbb N$$, we have

$$r a \otimes b = \sum a \otimes b = a \otimes rb$$ by bilinearity - which extends $$r$$ to $$\Bbb Z$$ too. When $$r=1/m$$, $$m \in \Bbb Z$$, $$\frac{1}{m}a \otimes b = \frac{1}{m} (a \otimes b) = \frac{1}{m} ( \sum (a \otimes \frac{1}{m} b)) = \frac{1}{m} (ma \otimes \frac{1}{m} b ) = a \otimes \frac{1}{m} b$$ Thus, we have equality for all $$r \in \Bbb Q$$.

I think generalizing to $$\Bbb Q$$ is as far as we can get for this naive strategy. I wonder if there exists better method for the classification.

• nice question. to make it self-contained, you might edit into it that $A$ and $B$ start life as $R$-modules (it's clear from reading your other post) Nov 28, 2018 at 10:41
• A guess: you might have this when $\Bbb Z\to R$ is an epimorphism (as it is in the cases $R = \Bbb Q$ and $R = \Bbb Z/n$). However, I do not have a proof of this (nor am I convinced that even if this is true, that these are all such rings). Nov 29, 2018 at 4:11

First, there is an ambiguity in your question. The usage of $$=$$ is ambiguous, but I'll interpret it as meaning that the natural map $$\newcommand\tens\otimes\newcommand\ZZ{\mathbb{Z}}A\tens_\ZZ B\to A\tens_R B$$ is an isomorphism.

Given this interpretation, we can give a simple criterion which rings with this property must satisfy. Namely the natural map $$R\tens_\ZZ R\to R$$ must be an isomorphism. This is also sufficient though, since if this is true, then $$A\tens_\ZZ B \simeq (A\tens_R R)\tens_\ZZ (R\tens_R B)\simeq A\tens_R (R\tens_\ZZ R)\tens_R B \simeq A\tens_R R \tens_R B \simeq A\tens_R B,$$ where I'm using $$\simeq$$ for natural isomorphism.

Thus the question is reduced to the question of for which rings is the natural map $$R\tens_\ZZ R \to R$$ an isomorphism. This is equivalent to asking for which rings is the diagram $$\newcommand\id{\operatorname{id}} \require{AMScd} \begin{CD} \ZZ @>\iota>> R \\ @V\iota VV @VV\id V \\ R @>\id >> R \end{CD}$$ a pushout diagram (where $$\iota$$ is the unique map $$\ZZ\to R$$.

Well, if it is, then for any pair of morphisms $$f,g : R\to S$$ with $$f\circ \iota = g\circ \iota$$, then there is a unique map $$h : R\to S$$ with $$f=h\circ \id = g$$. Thus $$\iota$$ is an epimorphism.

Conversely if $$\iota$$ is an epimorphism, then for any pair of morphisms $$f,g : R\to S$$ with $$f\circ \iota = g\circ \iota$$, then $$f=g$$, so the map $$h=f=g : R\to S$$ satisfies $$f=h\circ\id$$ and $$g=h\circ\id$$. Thus if $$\iota$$ is epic, this diagram is a pushout.

Hence a commutative ring $$R$$ has the property that $$A\tens_\ZZ B\simeq A\tens_R B$$ for all pairs $$A$$ and $$B$$ of $$R$$-modules if and only if the natural map $$\ZZ\to R$$ is an epimorphism.

Edit

As for what rings $$R$$ for which the natural map $$\ZZ\to R$$ is an epimorphism look like, I'm not sure. In general epis in the category of rings are complicated. That said, if I had to guess the answer in this case, my guess would be that these rings would be the subrings of $$\Bbb{Q}$$ and the rings $$\ZZ/n\ZZ$$, but that should probably be another question.

• I thought this might be the case, but I didn't come up with an argument - very nice! There's a mathoverflow post I saw a bit ago which described or linked to a paper classifying rings such that the map from $\Bbb Z$ is an epi, and its very similar to what you say, but you also need to allow some products or sums of such rings as well. Dec 15, 2018 at 23:26
• @Stahl Thanks, I'll have to go search for that, I'd be very interested in learning more about that.
– jgon
Dec 15, 2018 at 23:29
• Actually it appears to be one of the answers to the question I linked xP
– jgon
Dec 15, 2018 at 23:31