I am trying to prove that $\Bbb Q\otimes_{\Bbb Z}\Bbb Q\cong\Bbb Q$. I know that $\frac a b \otimes \frac c d = 1\otimes \frac{ac}{bd}$ and therefore it makes sense to define the map $\varphi:\Bbb Q\to \Bbb Q\otimes_{\Bbb Z}\Bbb Q$ by $$\varphi(q) = 1\otimes q$$ and we need to confirm that this is an isomorphism.
The part I struggle with here is injectivity, since if we assume $1\otimes q=1\otimes r$ then we need to prove $q=r$. But from the definition of the tensor product we only know that if $H\subseteq \Bbb Q\times \Bbb Q$ is generated by familiar relations, such that $\Bbb Q\otimes_{\Bbb Z}\Bbb Q = \Bbb Q\times\Bbb Q/H$ then we have that there is some $(s,t)\in H$ such that $(1,q)=(1,r)+(s,t)$, so $(0,q-r)=(s,t)$ and therefore $s=0$ and $q-r=t$. This implies $(0,q-r)\in H$, but from here, I'm not sure what to do with it.
I could argue that there is some $\Bbb Z$-linear combination of the generators of $H$ which equals $(0,q-r)$, but I don't see any way to make use of this fact. Since I don't know much else about the tensor product, then I'm somewhat out of ideas at this point.
I have seen this post: Showing $\mathbb{Q}\cong \mathbb{Q}\otimes_\mathbb{Z}\mathbb{Q}$.
and tried to prove that this inverse map is well-defined, and it seems to me that this runs into exactly the same problem.
$V=\Bbb Q\otimes_{\Bbb Z}\Bbb Q$ is an extension of scalars from $\Bbb{Z}$ to $\Bbb{Q}$ with multiplication defined on the left coordinate $a(u\otimes v) = (au)\otimes v$ for $a\in \Bbb{Q}, u\otimes v\in V$. You already know however that $(au)\otimes v = u\otimes (av)$. The map you defined $$\varphi(q) = 1\otimes q$$ is surjective, and it is $\Bbb{Q}$-linear because $$\varphi(aq) = 1\otimes (aq) = a\otimes q = a(1\otimes q) = a\varphi(q)$$ So $\varphi$ is a surjective $\Bbb{Q}$-linear map from a $1$-dimensional vector space over $\Bbb{Q}$ to a vector space over $\Bbb{Q}$. It follows that either $V=0$ or $V=\Bbb{Q}$. So it's enough to prove for example that $1\otimes 1 \ne 0$ in $V$ and for that you can define any explicit binlinear map $\Bbb{Q} \times \Bbb{Q}\to M$ for some vector space $M$ over $\Bbb{Q}$ such that $(1,1)$ does not map to zero.