Is it true that $A \otimes_{\Bbb Z} B= A \otimes _{ R} B$ for every commutative ring $R$?

Question

I have been obtaining a weird statement:

Let $R$ be a commutative ring $A,B$ both $R$-modules. Then $A \otimes_{\Bbb Z} B= A \otimes _{ R} B$ as abelian groups.

Proof: We may regard $A \otimes _{\Bbb Z} B$ as an $R$-module by acting $r(a \otimes b) = ra \otimes b$. There is thus an $R$-billinear map inducing $$A \otimes _R B \rightarrow A \otimes B$$ where $ a \otimes b \mapsto a \otimes b$. Conversely, there is also a billiner map inducing $$A \otimes B \rightarrow A \otimes _R B$$ where $a \otimes b \mapsto a \otimes b$. They form inverse to one another.

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What has gone wrong in the proof?

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EDIT: I believe I have found my main problem. So I took the $R$-module structure for $A \otimes B$ to be that induced from the map $$ R \times A \otimes B \rightarrow A \otimes B, (r,a \otimes b) \mapsto ra \otimes b $$ But this does not imply $ra \otimes b = a \otimes rb $. Which is required to induce the map $$A \otimes _R B \rightarrow A \otimes B $$

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This is seen in the example given by egreg, that with $\Bbb Z$ given the induced qutoient structure of $R= \Bbb Z[x]$, in $\Bbb Z \otimes _{\Bbb Z} R \cong \Bbb Z^{(N)}$ $$ 0 = x 1 \otimes 1 \not= 1 \otimes x $$ Since under the isomorphism, the latter item is $(1) $ in the 1st level of $\Bbb Z^{(N)}$.

Answer 1

You're doing too many steps implicitly.

In your second step, you want to say that the natural map $A\times B \to A \otimes_R B$ is $\mathbb{Z}$-bilinear. That's true, and so it gives a map $A\otimes_\mathbb{Z} B \to A\otimes_R B$.

In your first step, you want to say that the natural map $A\times B \to A \otimes_\mathbb{Z} B$ is $R$-bilinear. That's false; you've defined $r(a \otimes b) = ra\otimes b$, but that might not equal $a\otimes rb$. So there's no universal property to apply that gives us a map $A\otimes_R B \to A \otimes_\mathbb{Z} B$.

Answer 2

The statement is not true in general.

Consider $R=\mathbb{Z}[x]$. If $A$ is any $R$-module, then $$ A\otimes_R R\cong A $$ but $$ A\otimes_{\mathbb{Z}}R\cong A^{(\mathbb{N})} $$ (direct sum of a countable number of copies of $A$). Now take $A=\mathbb{Z}$, which is an $R$-module because it is $\mathbb{Z}[x]/(x)$.