Is my example correct? $A \otimes_R B \ne A \otimes_\mathbb{Z} B$. Exercise IV.5.3.(a) (in Algebra by Hungerford) asks for an example to show that the following may actually occur for a ring $R$ and $R$-modules $A$, $B$(left and right modules, respectively).
$$ A \otimes_R B \ne A \otimes_\mathbb Z B. $$
So I bring $A=B=R=\mathbb R$. Check my argument and be critic if somethings wrong. No matter it is wrong or not, new examples are welcomed.
My answer. Observe that the set $X_1 = \{1,\ \pi,\ \pi^2 \}$ is linearly independent in $\mathbb R$ over $\mathbb Q$. We easily see that if $a_0 + a_1 \pi + a_2 \pi^2 =0$ with rational $a_i$, then all $a_i$ are zero since $\pi$ is not algebraic over $\mathbb Q$. Then we may extend $X_1$ to get a basis $X$ of $\mathbb R$ over $\mathbb Q$.
Define a set map $T$ of $X$ into $\mathbb R$ by
$$T(1)=1,\ \ \ \  T(\pi)=\pi^2,\ \ \ \  T(\pi^2)=\pi, $$
and $T(x)=0$ for any other $x \in X$. Then we may uniquely extend $T$ linearly to get a $\mathbb Q$-linear map $T:\mathbb R \to \mathbb R$.
Now define $f:\mathbb R \times \mathbb R \to \mathbb R$ by $f(x,\ y) = T(x)T(y)$. Then $f$ is $\mathbb Q$-bilinear. The universal property of tensor product says that there is a unique $\mathbb Z$-linear map $\bar f:\mathbb R \otimes_\mathbb Z \mathbb R \to \mathbb R$ induced by $f$. Then we have
$$ \bar f (\pi \otimes \pi) = T(\pi)T(\pi) = \pi^4 \ne \pi = T(\pi^2)T(1) = \bar f (\pi^2 \otimes 1). $$
Therefore $\pi \otimes \pi \ne \pi^2 \otimes 1$ in $\mathbb R \otimes_\mathbb Z \mathbb R$. But it is obvious that $\pi \otimes \pi = \pi^2 \otimes 1$ in $\mathbb R \otimes_\mathbb R \mathbb R$, hence the result.
 A: Your proof looks good to me, and you have successfully shown that the natural map $\mathbb{R}\otimes_\mathbb{Z}\mathbb{R}\to\mathbb{R}\otimes_\mathbb{R}\mathbb{R}$ is not an isomorphism. However, $\mathbb{R}\otimes_\mathbb{Z}\mathbb{R}$ and $\mathbb{R}\otimes_\mathbb{R}\mathbb{R}$ are isomorphic as abelian groups, since they are both isomorphic to $\mathbb{Q}^{\oplus2^{\aleph_0}}$. (Let me know if you would like a proof of this; it follows essentially immediately from the fact that both are $\mathbb{Q}$-vector spaces of size $2^{\aleph_0}$.) If you are interested, here is an example where the corresponding groups are in fact not even isomorphic as abelian groups:
Let $R=\mathbb{Z}[x]$, and take $A=\mathbb{Z}$, where $x$ acts on $A$ as $0$, and $B=\mathbb{Z}[x,x^{-1}]$ as a subring of $\operatorname{Frac}R\cong\mathbb{Q}(x)$. Then $A\otimes_R B=0$, since we have, for any $a\in A$ and $f\in B$: $$a\otimes_R f=a\otimes_R x(x^{-1}f)=ax\otimes_R x^{-1}f=0\otimes_R fx^{-1}=0\otimes_R 0.$$ However, $A\otimes_\mathbb{Z} B$ is non-zero; indeed, it is isomorphic to $B$ as a $\mathbb{Z}$-module, since $A\cong\mathbb{Z}$ as a $\mathbb{Z}$-module.
