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?