Galois Theory: Prove that every finite extension over $F$ is cyclic (more details in question)

Question

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.

Answer 1

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 $L:K$.

Answer 2

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.