# Do the algebraic integers form a free abelian group?

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.

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Pontryagin's criterion says that, for a countable, torsion-free, abelian group to be free, it suffices that every finitely many elements lie in a finitely generated pure subgroup. The rings $\mathcal O_K$ for finite extensions $K$ of $\mathbb Q$ show, in view of the result you quoted, that this criterion is satisfied. The axiom of choice is not needed for any of this, because the question is absolute between the full universe $V$ and Gödel's constructible universe $L$.
@David: Among torsion-free abelian groups, a subgroup $H$ of a group $G$ is said to be pure iff the quotient $G/H$ is torsion-free; equivalently, whenever $H$ contains a non-zero multiple of an element $x$ of $G$, it also contains $x$ itself. In particular, the only pure subgroups of $\mathbb Q$ are $\mathbb Q$ itself and 0. Also, any finite-rank subgroup is included in the purification of a finite subset $F$ (i.e., the smallest pure subgroup that includes $F$). – Andreas Blass Dec 22 '11 at 20:05