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Let $K$ be a field of characteristic $0$ contained in its algebraic closure $A$. Let $s$ be an automorphism of $A$ over $K$, and let $F$ be the fixed field. How does one prove then that every finite extension of $F$ is cyclic?

First, I know this question has been asked once or twice before but none of the answers are actually satisfactory and most are either incomplete or just incorrect as pointed out on the page. Having not been able to find a hint/answer so far anywhere on the internet, I ask you stackexchange for some help. Unfortunately, I am completely clueless where to even start with this problem so all help/solutions are very much appreciated.

Thank you very much.

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    $\begingroup$ Trying to compose a list of questions where this has been discussed [1removed after a close examination] ,2,3,4 $\endgroup$ – Jyrki Lahtonen Jul 14 '17 at 20:44
  • $\begingroup$ What is $Gal(\overline{F}/F)$ ? (if $\sigma \in Gal(\overline{F}/F), \sigma \not \in \langle s \rangle$ then $F = \overline{F}^{\langle \sigma, s \rangle}$ which has to be a strict subfield of $\overline{F}^{\langle s \rangle}$ ?) $\endgroup$ – reuns Jul 14 '17 at 20:51
  • $\begingroup$ @user1952009 If $s$ has infinite order and $\sigma\in Gal(A/F)$ you can only proceed that way if $\sigma\notin\overline{\langle s\rangle}$. When $\langle s\rangle$ is an infinite cyclic group it need not be closed. $\endgroup$ – Jyrki Lahtonen Jul 14 '17 at 21:03
  • $\begingroup$ Daniele, which old answers do you find unsatisfactory? It would help the answerers if you listed some. Quite possibly the earlier answers (mine included) glossed over some point. $\endgroup$ – Jyrki Lahtonen Jul 14 '17 at 21:06
  • $\begingroup$ @JyrkiLahtonen, I thought to $s \in \hat{\mathbb{Z}}^\times$ acting on $\mathbb{Q}(\zeta_\infty)$. Do you have an example of a missing element $\sigma \in \overline{\langle s \rangle} \setminus \langle s \rangle$ ? $\endgroup$ – reuns Jul 14 '17 at 21:08
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Given a finite extension $L$ of $F$ we have that $L/F$ is finite and separable because $\text{char }F = \text{char }K = 0$.

We then have that $\bar{L}/F$ is Galois (and finite) where $\bar{L}$ is the normal closure of $L/F$.

The automorphism $s$ of $A/F$ restricts to an automorphism $\bar{s}$ of $\bar{L}/F$, and so $Gal(\bar{L}/F) \supseteq \left<\bar{s}\right>$.

The fixed field of $\left<\bar{s}\right>$ in $\bar{L}$ is $F$, so by the Galois correspondence we have $Gal(\bar{L}/F) = \left<\bar{s}\right>$.

Writing $L = F(\alpha)$ and $\sigma \in Aut(L/F)$ we can extend $\sigma$ to $\bar{\sigma} \in Gal(\bar{L}/F)$ by choosing a $\bar{\sigma} \in Gal(\bar{L}/F)$ such that $\bar{\sigma}(\alpha) = \sigma(\alpha)$.

This is valid because $\sigma(\alpha)$ is a conjugate of $\alpha$ over $F$, and a normal extension has automorphism group that is transitive over the set of conjugates of $\alpha$ over $F$.

We then have $\bar{\sigma} = \bar{s}^n$ for some $n \in \mathbb{Z}$.

Choose the smallest $n \gt 0$ such that the restriction $\sigma$ of $\bar{\sigma} = \bar{s}^n$ on $L$ is in $Aut(L/F)$.

Then $Aut(L/F) = \left<\sigma\right>$ because otherwise we have a $\tau$ that extends to some $\bar{s}^m$ with $n \nmid m$. But then $\tau\sigma^{-q}$ extends to some $\bar{s}^r$ with $0 \lt r \lt n$ where $m = qn + r$, contradicting the choice of $n$.

Thus the extension $L/F$ is cyclic.

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  • $\begingroup$ I am not completely sure that I am allowed to say that $\bar{L}/F$ is finite. But I think it is true because writing $L = F(\alpha)$ and $f(x) \in F[x]$ the minimal polynomial of $\alpha$ over $F$, then $\bar{L}$ should be the splitting field of $f$. $\endgroup$ – Tob Ernack Jul 14 '17 at 22:26
  • $\begingroup$ This seems like a nice an simple proof. Two questions though: a) What do you mean by 'transitive' in "automorphism group...over F". b) How do you conclude in the next line that sigma (with line over it)=s(line)^n for some n integer. $\endgroup$ – Daniele1234 Jul 14 '17 at 22:29
  • $\begingroup$ Sorry I understand b) now $\endgroup$ – Daniele1234 Jul 14 '17 at 22:33
  • $\begingroup$ For a), a group $G$ acts transitively on a set $S$ if for any $x, y \in S$ there is a $g \in G$ such that $gx = y$. To be fair, I haven't actually proven that claim. I didn't remember how to prove that, but this answer seems to do so. $\endgroup$ – Tob Ernack Jul 14 '17 at 23:00
  • $\begingroup$ And I would be interested in a proof check, I cannot consider myself particularly well trained in this subject. $\endgroup$ – Tob Ernack Jul 14 '17 at 23:04
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Let $L$ be a finite extension of $F$, since the characteristic is zero, $L:F$ is separable as well as $A:L$, so you can extend any automorphism $f$ of $L:F$ to $g$ of $A:F$ we have $g=s^n$. Let $l$ is $inf\{n: g=s^n\}$ where $g$ is an extension of an automorphism of $L:F$, $f^l$ preserves $L$ and is a generator of the Galois group of $L:K$.

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  • $\begingroup$ Why would $g$ be a power of $s$? True, we can come arbitrarily close to $g$ (w.r.t. the Krull topology) by a power of $s$ (which actually suffices, but is rather the whole point I think). In other words, a few words about $Gal(A/F)$ only being the closure of $\langle s\rangle$ may be needed? May be your thinking was organized differently, and I just couldn't follow? $\endgroup$ – Jyrki Lahtonen Jul 14 '17 at 20:58
  • $\begingroup$ I agree why would g be a power of s? (note I don't have any background in topology). Also, what exactly do you mean by 'inf' in: "Let l is inf(n:g=s^n). "Let l is" doesn't even make grammatical sense... $\endgroup$ – Daniele1234 Jul 14 '17 at 21:41

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