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}.

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