My previous question about the theorem (apparently due to Dedekind -- thanks, Arturo Magidin!) that the ring $\overline{\mathbb{Z}}$ of all algebraic integers is a Bezout domain got me thinking about rings of integers in infinite algebraic extensions of $\mathbb{Q}$.

As I mentioned here, I was thinking about Kaplansky's formulation/generalization of Dedekind's theorem: if $R$ is a Dedekind domain with fraction field $K$ such that for all finite degree field extensions $L/K$ the ideal class group of the integral closure $R_L$ of $R$ in $L$ is a torsion abelian group, then the integral closure $S$ of $R$ in $\overline{K}$ is a Bezout domain.

Since we know (by nefarious arithmetic arguments...) that the Picard group of the ring of integers of a number field is finite, Kaplansky's theorem applies with $R = \mathbb{Z}$ and gives Dedekind's theorem.

Suppose we now look at Kaplansky's proof and see if it gives more: do we need to go all the way up to $\overline{\mathbb{Q}}$ in order to show that the ring of integers is a Bezout domain?

Indeed not. The key property of the ring $S$ used in the proof is that it is root closed, i.e., closed under extraction of $n$th roots, for all $n \in \mathbb{Z}^+$. There are lots of smaller -- but still infinite -- algebraic extensions of $\mathbb{Q}$ which have this property: the smallest is $\mathbb{Q}^{\operatorname{solv}}$, the maximal solvable extension of $\mathbb{Q}$.

But what about the ring of integers of $\mathbb{Q}^{\operatorname{ab}}$, the maximal abelian extension of $\mathbb{Q}$? Is this a Bezout domain? If not, what is its Picard group?

In general, if $R$ is a Dedekind domain with fraction field $K$, $L/K$ is any algebraic field extension and $S$ is the integral closure of $R$ in $L$, then if $[L:K]$ is infinite $S$ need not be a Dedekind domain but is always a Prufer domain: every finitely generated ideal is invertible. In particular $S$ is a Bezout domain iff $\operatorname{Pic}(S) = 0$. So one may ask:

What is known about the Picard groups of rings of integers of infinite algebraic extensions of $\mathbb{Q}$? What, if anything, is the connection to more classical algebraic number theory?

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    $\begingroup$ The Picard group of the ring of integers of ${\mathbb Q}^{ab}$ is a a countable direct sum of copies of ${\mathbb Q}/{\mathbb Z}$. See ams.org/mathscinet-getitem?mr=601678 $\endgroup$ – dke Dec 28 '10 at 19:18
  • $\begingroup$ Nice question Pete. $\endgroup$ – Adam Hughes Dec 28 '10 at 19:22
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    $\begingroup$ @dke: Thanks a lot. I think this is worth leaving as an answer. $\endgroup$ – Pete L. Clark Dec 28 '10 at 19:43
  • $\begingroup$ @Pete, I've sent you the paper. Another relevant result is by Kurihara: if you take the maximal abelian subfield of ${\mathbb Q}^{ab}$, the Picard group is trivial. Unfortunately I don't have that paper at hand. $\endgroup$ – dke Dec 28 '10 at 19:53
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    $\begingroup$ Oops, 'maximal abelian' should have been 'maximal real'. Damn uneditable comments... $\endgroup$ – dke Dec 28 '10 at 19:59

Let $K$ be a cyclotomic extension, and $a, b \in O_K$. If the ideal $A = (a,b)$ becomes principal in some cyclotomic extension $L/K$, then clearly you can solve the Bezout problem, and the converse holds, too: if the Bezout problem is solvable in some cyclotomic extension, then the ideal class of $A$ must capitulate there. Thus the question you are asking is whether the class group of a cyclotomic field capitulates in a cyclotomic extension. For certain pieces of the class group this is believed to be true (Greenberg), but in general I don't know the answer. Georges Gras and Bosca & Jaulent have studied the phenomenon of "abelian capitulation". From a first glance at their article it seems that the answer to your question is yes for the maximal real cyclotomic subfield of all cyclotomic extensions.


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