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On p832 of Coombes, Harbater - Hurwitz familes and arithmetic Galois groups, the following is claimed:

Let $K$ be a number field, take $1 \neq \omega \in \mathrm{Gal}(\bar{\mathbb{Q}}/K)$, and let $L_{\omega}$ be the fixed field of $\omega$. By Artin-Schreier the case when $\omega$ has finite order is boring, so suppose it has infinite order. Then $\Gamma = \mathrm{Gal}(\bar{\mathbb{Q}}/L_{\omega})$ is a free profinite group.

Question: Why is $\Gamma$ free profinite?

$\Gamma$ must be the smallest closed subgroup containing $\omega$, or the closure of the cyclic subgroup $<\omega>$. So it will be a procyclic group, a quotient of $\hat{\mathbb{Z}}$. So in order to be free, we need $\Gamma \cong \hat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$. How do we know this will be the case?

This doesn't mesh with another result I thought to be true: Suppose $K$ is a field, $\alpha \in K^{sep} \backslash K$ an algebraic element, and $F$ a field maximal with respect to not containing $\alpha$ such that $K^{sep}/F/K$. Then $\mathrm{Gal}(K^{sep}/F)$ is either isomorphism to $\mathbb{Z}/2\mathbb{Z}$ or to some $\mathbb{Z}_p$.

So what's the deal?

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  • $\begingroup$ Since $\hat{\mathbf{Z}}$ contains a subgroup isomorphic to $\mathbf{Z}_p$, $\mathrm{Gal}(\bar{\mathbb{Q}}/K)$ contains a copy of $\mathbf{Z}_p$ too (for any $p$) if you know that it contains $\hat{\mathbf{Z}}$. So the claims sounds absurd. $\endgroup$
    – YCor
    Jun 26, 2017 at 22:40
  • $\begingroup$ Ok, I was thinking along the same lines. Perhaps Coombes-Harbater meant projective rather than free, because that was really all they needed for the argument. $\endgroup$ Jun 26, 2017 at 23:06
  • $\begingroup$ btw an infinite procyclic group need not be isomorphic to a subgroup of $\widehat{\mathbf{Z}}$. For instance $\prod_p\mathbf{Z}/p\mathbf{Z}$. Why can't $\omega$ generate such a group? (Indeed a torsion-free procyclic group is isomorphic to a subgroup of $\widehat{\mathbf{Z}}$). $\endgroup$
    – YCor
    Jun 26, 2017 at 23:19
  • $\begingroup$ Oh, thanks, you're right. It doesn't not need to be a subgroup, but it will be a quotient, right? I will fix that. $\endgroup$ Jun 26, 2017 at 23:24
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    $\begingroup$ Yes, you're right. I had not come to a concrete realization that it was false, only that I did not understand. Perhaps you could say I trust Coombes and Harbater more than myself. In any case, I will try to rework the question. $\endgroup$ Jun 27, 2017 at 12:07

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