I want to prove the following, $$\mathbb{Q}\cong \mathbb{Q}\otimes_\mathbb{Z}\mathbb{Q}.$$
I know that the map $f:\mathbb{Q}\rightarrow \mathbb{Q}\otimes_\mathbb{Z}\mathbb{Q}$ defined by $f(q)=q\otimes 1$ is a homomorphism so I just need to show that it is bijective. It is obviously onto, but for one to one I take $\frac{n}{d}$ from the kernel of $f$. So $0=f(n/d) = n/d\otimes 1$, but I don't know how this implies $n/d=0$. Any suggestions?
It is often the case that it is more convenient to prove something is an isomorphism directly by writing down its inverse, rather than by the indirect route of proving it is bijective.
This case is no different: the evident inverse is the function
$$ g : \mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} : x \otimes y \mapsto xy $$
so all you need to do is show that $g$ is well defined, and that $f \circ g$ and $g \circ f$ are the identity maps.