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Is the following proof that $\mathbb{Q}$ is not finitely generated correct?

If $\mathbb{Q}$ were finitely generated so would the quotient module $\mathbb{Q}/\mathbb{Z}$ (generated by the cosets of the generators of $\mathbb{Q}$). But the latter is a torsion module, so by the classification theorem it would have the form $\bigoplus_{i=1}^n \mathbb{Z} / p_i^{r_i}$ (torsion module -> no free part). But this is finite, whereas $\mathbb{Q} / \mathbb{Z}$ is clearly infinite.

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    Finitely generated as a module over what?2012-03-28
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    $\mathbb Z$, I assume.2012-03-28
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    Sounds OK. But if $N$ is the lcm of the denominators of the supposed generators, that forces $\frac{1}{2N}$ to be not in $\mathbb{Q}$, contradiction. Less machinery.2012-03-28
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    It's good to have multiple proofs of things. And this technique is more general.2012-03-28
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    Multiple proofs are indeed good, a twice-proved theorem is twice as true. This is particularly the case if the proofs are really *different*. But in your proof and the one I suggest, it all comes down to divisibility, so the arguments are in essence the same.2012-03-29
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    @Andre: I think the notion of "sameness" you need to say that is far too inclusive to be meaningful. Bean's proof really has a different flavor and exercises a different tool set than the one you cite. And to be glib, one could say that it all comes down to modules anyways; after all, isn't your proof a calculation using invertible ideals?2012-03-29
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    The proof I gave would work for any $\mathbb{Z}$-module which admits an infinite torsion quotient. It doesn't rely on knowing the elements explicitly, as the lcm one does.2012-03-29

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