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.

  • $\begingroup$ 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) $\endgroup$
    – hunter
    Nov 28, 2018 at 10:41
  • $\begingroup$ 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). $\endgroup$
    – Stahl
    Nov 29, 2018 at 4:11

1 Answer 1


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.


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.

  • $\begingroup$ 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. $\endgroup$
    – Stahl
    Dec 15, 2018 at 23:26
  • $\begingroup$ @Stahl Thanks, I'll have to go search for that, I'd be very interested in learning more about that. $\endgroup$
    – jgon
    Dec 15, 2018 at 23:29
  • 1
    $\begingroup$ Actually it appears to be one of the answers to the question I linked xP $\endgroup$
    – jgon
    Dec 15, 2018 at 23:31

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