## Show that two tensor products are isomorphic as modules

Show that $\mathbb{Q} \otimes_\mathbb{Q} \mathbb{Q}$ and $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q}$ are isomorphic as left $\mathbb{Q}$-modules.

Let $\frac{a}{b} \otimes \frac{c}{d}$ be a simple tensor in $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q}$. We have $\frac{a}{b} \otimes \frac{c}{d} = \frac{ad}{bd} \otimes \frac{c}{d}$ $= \frac{a}{bd} \cdot d \otimes \frac{c}{d}$ $= \frac{a}{bd} \otimes d \cdot \frac{c}{d}$ $= \frac{a}{bd} \otimes \frac{c}{1}$ $= \frac{ac}{bd} \otimes \frac{1}{1}$. In particular, every simple tensor (hence every element of $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q}$) can be written as $q \otimes 1 = q \cdot (1 \otimes 1)$ where $q \in \mathbb{Q}$. This gives (by the universal property of free modules) a unique $\mathbb{Q}$-module homomorphism $\Phi : \mathbb{Q} \rightarrow \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q}$ such that $q \mapsto q \otimes 1$. Certainly $\Phi$ is surjective.

Certainly every simple tensor (and every element) in $\mathbb{Q} \otimes_\mathbb{Q} \mathbb{Q}$ can be written in the form $q \cdot (1 \otimes 1)$. By the universal property of free modules, this yields a surjective $\mathbb{Q}$-module homomorphism $\Theta : \mathbb{Q} \rightarrow \mathbb{Q} \otimes_\mathbb{Q} \mathbb{Q}$.

Now define $\psi : \mathbb{Q} \times \mathbb{Q} \rightarrow \mathbb{Q}$ by $(a,b) \mapsto ab$. Certainly $\psi$ is $\mathbb{Q}$-bilinear (and also $\mathbb{Z}$-bilinear), and so induces a $\mathbb{Q}$-module homomorphism $\Psi : \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q} \rightarrow Q$ such that $\Psi(a \otimes b) = ab$. Certainly $\Psi$ is surjective, and so $\Phi$ and $\Theta$ are isomorphisms. Thus $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Q} \cong_\mathbb{Q} \mathbb{Q} \cong_\mathbb{Q} \mathbb{Q} \otimes_\mathbb{Q} \mathbb{Q}$.