It is a well-known fact, proved in every introductory textbook on algebraic number theory, that if $K$ is an algebraic number field, i.e. a *finite* extension of $\mathbb{Q}$, then its ring $\mathcal{O}_K$ of integers is a free abelian group.

Does this statement still hold for arbitrary algebraic extensions of $\mathbb{Q}$? In particular, is the underlying abelian group of the ring $\mathcal{O}_{\overline{\mathbb{Q}}}$ of all algebraic integers free abelian?

Should this be true, I am also interested whether anything is known about the dependence of this statement on the axiom of choice, and similar logical questions.