How to prove that any finite extension field for this field would be cyclic This question was asked in mid term exam of Field Theory quite earlier this year and I was unable to completely solve and so I am looking for help here.

Let $K$ be a field, $\bar K$ an algebraic closure of $K$, and $\sigma \in \operatorname{Aut}_K(\bar{K})$. Let
$F = \{u \in \bar{K} : \sigma(u)=u\}$.
Then $F$ is a field and every finite dimensional extension of $F$ is cyclic.

I have proved $F$ to be a field but unable to prove that any finite dimensional extension would be cyclic. I took the extension to be $K = F(u_1,\dots,u_k)$ but I am unable to get any ideas why it should be cyclic and I would require some help.
 A: Take any finite Galois extension $E/F$. Verify that its Galois group must be generated by $\sigma|_E$. Then given any finite separable extension $E/F$, apply the previous sentence to its Galois closure $\hat{E}/F$, and use that cyclic groups are abelian to conclude that $E=\hat{E}$ was already Galois over $F$. Finally, note that $F$ must be perfect, since if $\sigma(x^p)=x^p$, then $\sigma(x)=x$.
A: OK. Just to add a few details.
(1) If $F$ has characteristic $p$, and $x\in F$, pick a $p$-th root $y$ of $x$ in $\overline{F}$, then $y^p = x = \sigma(x) = \sigma(y^p) = \sigma(y)^p\Rightarrow (y-\sigma(y))^p = 0\Rightarrow y = \sigma(y)\Rightarrow y\in F$, therefore every element of $F$ has a $p$-th root in it. If $F$ has characteristic $0$, it's already perfect. Either way, $F$ is perfect, and any extension of $F$ is separable.
(2) For a finite extension $E$ of $F$, pick its normal closure $\hat E$ of $E$, so that $\sigma|_{\hat E}\in G:=\text{Gal}(\hat E/F)$ is well-defined. By $E/F$ is separable, $\hat E/F$ is Galois, so we can apply the fundamental theorem of Galois theory. Let $H=\langle \sigma|_{\hat E}\rangle\le G$ be the subgroup generated by $\sigma|_{\hat E}$. By Galois correspondence $$[G:H]=[(\hat E)^H:F]$$ where $(\hat E)^H:=\{x\in \hat E: \tau(x)=x, \forall \tau\in H\}$. If $x\in (\hat E)^H$, then $x$ is fixed by $\sigma|_{\hat E}$ hence $\sigma$, therefore by the definition of $F$, $x\in F$. This shows that $(\hat E)^H=F$. Therefore $[G:H]=1$, $G=H$ is a cyclic group.
(3) As $G$ is cyclic in particular abelian, all its subgroups are normal. In particular, $\{\tau\in G: \tau(x)=x, \forall x\in E\}$ is normal. Thus as part of the fundamental theorem, $E/F$ is already Galois, hence $\hat E=E$, and it's been shown that $G=\text{Gal}(E/F)$ is cyclic.
