Xrfgtjtk

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.

First, there is an ambiguity in your question. The usage of $=$ is ambiguous, but I'll interpret it as meaning that the natural map $newcommandtensotimesnewcommandZZ{mathbb{Z}}Atens_ZZ Bto Atens_R B$ is an isomorphism.

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

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

Well, if it is, then for any pair of morphisms $f,g : Rto S$ with $fcirc iota = gcirc iota$, then there is a unique map $h : Rto S$ with $f=hcirc id = g$. Thus $iota$ is an epimorphism.

Conversely if $iota$ is an epimorphism, then for any pair of morphisms $f,g : Rto S$ with $fcirc iota = gcirc iota$, then $f=g$, so the map $h=f=g : Rto S$ satisfies $f=hcircid$ and $g=hcircid$. Thus if $iota$ is epic, this diagram is a pushout.

Hence a commutative ring $R$ has the property that $Atens_ZZ Bsimeq Atens_R B$ for all pairs $A$ and $B$ of $R$-modules if and only if the natural map $ZZto R$ is an epimorphism.

Edit

As for what rings $R$ for which the natural map $ZZto R$ 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 $ZZ/nZZ$, but that should probably be another question.