I'm trying to prove the following:
Let $L$ be an algebraically closed field and let $\sigma \in \mathrm{Aut}(L)$. I want to show that for any separable $\alpha \in L$ over $K = L^\sigma$ ($K$ is the fixed field under $\sigma$), we have that $K(\alpha)/K$ is a normal extension.
I've been stuck on this for quite some time now and started to wonder whether it is even true?
Basically what I managed to figure out is the following: If $n \in \mathbb N$ is chosen maximally such that $\alpha, \sigma(\alpha), \dots, \sigma^{n-1}(\alpha)$ are pairwise distinct, the minimal polynomial $f$ of $\alpha$ in $K[X]$ must divide
$P = \prod_{i=0}^{n-1} (X - \sigma^i(\alpha)) \in K[X]$
($P$ is in $K[X]$ because it is invariant under $\sigma$)
This implies that if $K(\alpha)$ is normal, we must have $K(\alpha) = K(\alpha, \sigma(\alpha), \dots, \sigma^{n-1}(\alpha))$, which is the case iff $K(\alpha)$ is invariant under $\sigma$.
I think that it might be possible to conlude $\sigma^i(\alpha) \in K(\alpha)$ by investigating the coefficients of $P$, but it seems a bit complicated: One could start out by observing
\begin{align*} \sigma(\alpha) \cdots \sigma^{n-1}(\alpha) \in K(\alpha) &\implies \sum_{i=1}^{n-1} \sigma(\alpha) \cdots \widehat{\sigma^i(\alpha)}\cdots \sigma^{n-1}(\alpha) \in K(\alpha)\\ \sum_{i=0}^{n-1} \sigma^i (\alpha ) \in K(\alpha) &\implies \sum_{} \sigma^i(\alpha) \sigma^j(\alpha)\in K(\alpha) \end{align*}
But I doubt that there is no other way. I would very much appreciate good hints rather than a complete solution.
Thank you =)