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I'm wondering if anyone has classified all the fields $K$ such that $Gal(\bar{K}/K)$ is abelian?

The only examples I'm aware of are: finite fields, the real numbers $\mathbb{R}$ and $k((T))$ where $k$ is any algebraically closed field of characteristic 0. (One might also add any algebraically closed field as an example not listed above...)

Is there any other example of such fields?

An possibly easier question would be: can one apriori prove that $Gal(\bar{K}/K)$ is procyclic as suggested by the examples above?

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    $\begingroup$ The real numbers can be replaced by real closed fields. Finite fields can be replaced by algebraic extensions of finite fields. Do you know all the closed subgroups of the profinite integers $\widehat{\mathbf Z}$? $\endgroup$
    – KConrad
    Feb 13, 2017 at 15:40
  • $\begingroup$ The finite fields can be replaced by quasi-finite fields. $\endgroup$
    – Watson
    Feb 13, 2017 at 15:44
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    $\begingroup$ Exercise 3, Chapter 8, Lang's "Algebra": $K$ = a maximal subfield of $\overline{\mathbb Q}$ not containing $\sqrt{2}$. $\endgroup$ Feb 13, 2017 at 16:33
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    $\begingroup$ Consider the iterated formal power series field $K=k((T_1))((T_2))\cdots((T_n))$ with $k$ an algebraically closed field of characteristic zero, and you'll get a field $K$ with $Gal(\overline{K}/K)=\widehat{\mathbb Z}^n$. $\endgroup$ Feb 13, 2017 at 20:52
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    $\begingroup$ @LeonidPositselski $\mathbb{Z}_p\times\mathbb{Z}/2$ cannot be the absolute Galois group of a field, for any prime $p$. More generally, the normalizer of an involution in an absolute Galois group of a field is the group of order 2 generated by this involution. See Proposition 19.4.3 in: I. Efrat, Valuations, Orderings, and Milnor K-Theory, AMS 2006. $\endgroup$
    – user05811
    Jun 15, 2017 at 18:38

1 Answer 1

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Geyer in Unendliche algebraische Zahlkörper, über denen jede Gleichung auflösbar von beschränkter Stufe ist, Satz 1.13 and the paragraph after that, gives a full characterization of which abelian profinite groups occur as absolute Galois groups: They are either $\mathbb{Z}/2\mathbb{Z}$ or $\prod_p\mathbb{Z}_p^{c(p)}$ for some cardinal numbers $c(p)$.

Side remark: For algebraic extensions of $\mathbb{Q}$, abelian absolute Galois groups are in fact procyclic, see Satz 2.3 in the same paper.

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    $\begingroup$ Note: these are exactly $\mathbf{Z}/2\mathbf{Z}$ and all torsion-free profinite abelian groups. $\endgroup$
    – YCor
    Sep 15, 2021 at 11:33

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