Is $\mathbb{R}\otimes_{\mathbb{Z}}\mathbb{Q}\cong\mathbb{Q}$? Is $\mathbb{R}\otimes_{\mathbb{Z}}\mathbb{Q}\cong\mathbb{Q}$? I have tried several attempts to prove this using universial property, but eventually found that $\mathbb{R}\otimes_{\mathbb{Z}}\mathbb{Q}\cong \mathbb{R}$.
 A: If $A$ is an abelian group, then $A \otimes_{\mathbb{Z}} \mathbb{Q} = A \otimes_{\mathbb{Z}} (\mathbb{Z} \setminus \{0\})^{-1} \mathbb{Z}$ is the localization $(\mathbb{Z} \setminus \{0\})^{-1} A$. Thus, if all elements of  $\mathbb{Z} \setminus \{0\}$ already act as automorphisms of $A$ (equivalently, $A$ is the underlying group of a $\mathbb{Q}$-vector space), this simplifies to $A$. It applies in particular to $A=\mathbb{R}$.
A: I claim $\mathbb R$ and the bilinear map $p\colon\mathbb R\times\mathbb Q\to\mathbb R:(x,y)\mapsto xy$ satisfies the universal property of the tensor product $\mathbb R\otimes_\mathbb Z\mathbb Q$.
Let $M$ be a $\mathbb Z$-module (that is, an abelian group)  and let $f\colon\mathbb R\times\mathbb Q\to M$ be a bilinear map. Then $g\colon \mathbb R\to M$ defined by $g(x)=f(x,1)$ is a linear map such that $f=g\circ p$. To check this, let $(x,\frac nm)\in\mathbb R\times\mathbb Q$ be arbitrary, so that $g\circ p(x,\frac nm)=g(\frac nmx)$. Now, $m\cdot f(x,\frac nm)=f(x,n)=nf(x,1)=ng(x)=mg(\frac nmx)$, so we obtain the desired equality by dividing by $m$.
The uniqueness of $g$ can also be checked easily: if there is another map $g'$ with $f=g'\circ p$, then $(g-g')\circ p=0$, so $g=g'$ since $p$ is surjective.
P.S. Note that the only property of $\mathbb R$ I used here is: for any $x\in\mathbb R$ and nonzero integer $n$, there is a unique $y\in\mathbb R$ with $x=ny$. This property is sometimes called "uniquely divisible." That is, if $M$ is uniquely divisible, then $M\otimes_\mathbb Z\mathbb Q\cong M$.
