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What is $\operatorname{rank}_{\mathbb{Z}} \mathbb{Q}$?

I think it is $\aleph_0$, but cannot figure out how the basis would look like.

Thank you in advance.

(I faced this when proving $(\mathbb{Q},+)$ is not a finitely generated group. If it is finitely generated, then $\mathbb{Q}$ is group-isomorphic to $\mathbb{Z}^r \times G$ where $G$ is some finite group. However, since $\mathbb{Q}$ is torsion-free, it is isomorphic to $\mathbb{Z}^r$. So the rank of $\mathbb{Q}$ over $\mathbb{Z}$ is finite, which I want to make a contradiction.)

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    $\begingroup$ If you define this "rank" to be the size of a maximal $\mathbb{Z}$-linearly independent subset of $\mathbb{Q}$, then it is $1$. In the usual definition, the rank of a unitary module $M$ over a unital ring $R$ is defined when (1) $M$ is a unitary free $R$-module and (2) $R$ has the invariant basis number property. Under this usual definition, the rank of $\mathbb{Q}$ is undefined as an abelian group, since it is not a free abelian group. $\endgroup$ – Batominovski Aug 9 '15 at 12:52
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    $\begingroup$ If you define the rank of an abelian group $G$ as $\dim_{\Bbb{Q}} G \otimes_{\Bbb{Z}} \Bbb{Q}$, then the rank of $\Bbb{Q}$ is $1$. $\endgroup$ – Crostul Aug 9 '15 at 12:54
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You're wrong: $\mathbf Q$ is not a free $\mathbf Z$-module, hence it has no rank.

To see this tae any two fractions, $\dfrac ab$ and $\dfrac cd$. A linear relation between them is: $$bc\cdot\frac ab-ad\cdot\frac cd=0.$$

More generally, no ring of fractions $S^{-1}A$ can be free over $A$, unless $A=S^{-1}A$.

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$\mathbb{Q}$ is not finitely generated because it is a division group. It is generated by countable elements because $\mathbb{Q}$ itself is countable.

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