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This is just a question wondering whether a concrete (enough) result that can be proved using the axiom of choice can be proved without it. The result being that the group of principal ideals $P_K$ of a number field $K$ is free abelian.

One can show using the axiom of choice that a sub-group of a free abelian group is again gree abelian so that, as $P_K$ is a sub-group of the free abelian group $Id_K$ of all ideals of $K$, $P_K$ is itself free abelian.

Does anyone know of an axiom of choice free proof of this result?

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    $\begingroup$ Aren't number fields countable? $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 17, 2015 at 6:23

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To elaborate on Asaf's comment: the usual proof does not actually need the axiom of choice. The proof that a subgroup of a free abelian group is free actually shows (without using AC) that a subgroup of a well-orderable free abelian group is free. If $K$ is a number field, then $K$ is countable. The set $Id_K$ is also countable, being a quotient of the set of finite subsets of $K$ (every fractional ideal is finitely generated). So we can conclude that any subgroup of $Id_K$ is free without needing AC.

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  • $\begingroup$ Ha, I did not realise that! Cheers! $\endgroup$
    – lrg
    Commented Jul 18, 2015 at 8:36

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