If $\mathbb Q \otimes_\mathbb Z \mathbb Q \cong \mathbb Q^\mathbb N$, why is $\mathbb Q \otimes_\mathbb Z \mathbb Q$ a $1$-dim $\mathbb Q$-v.s.

Question

In Dummit & Foote, it is an exercise to show that $\mathbb Q \otimes_\mathbb Z \mathbb Q$ is a $1$-dimensional $\mathbb Q$-vector space.

This is fairly easy: a $\mathbb Q$-basis for $\mathbb Q \otimes_\mathbb Z \mathbb Q$ is $\{1 \otimes 1\}$ since any simple tensor can be rewritten as $a/b \otimes c/d = 1 \otimes (a/b)(c/d)$. Therefore, any tensor can be rewritten as $1 \otimes x$. Then, the map $\mathbb Q \otimes_\mathbb Z \mathbb Q \to \mathbb Q$ where $1 \otimes x \mapsto x$ is a $\mathbb Q$-linear isomorphism.

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However, the $\mathbb Z$-module $\mathbb Q$ is isomorphic to the free $\mathbb Z$-module $\mathbb Z^\mathbb N$, and we know $M \otimes_\mathbb Z \mathbb Z^A \cong M^A$ for any $\mathbb Z$-module $M$. So,
$$\mathbb Q \otimes_\mathbb Z \mathbb Q \cong \mathbb Q \otimes_\mathbb Z \mathbb Z^\mathbb N \cong \mathbb Q^\mathbb N$$
which is an infinite dimensional $\mathbb Q$-vector space.

What is wrong here?

Answer 1

$\mathbb Q$ is a not a free $\mathbb Z$-module:

• $\mathbb Q \ne \frac{a}{b}\mathbb Z$. So, $\mathbb Q$ is not generated by a single element over $\mathbb Z$.
• If $\frac{a}{b}, \frac{c}{d} \in \mathbb Q$, then $bc\frac{a}{b} - ad\frac{c}{d}=0$. So, no set with more than two elements can be linearly independent.