My question rises from the question: Let $\sigma \in Aut(K)$ have infinite order and $F = \mathcal{F}(\sigma)$. Show that if $K/F$ is algebraic, then $K$ is normal over $F$.
The answer by Tsemo Aristide is wonderful and he used the fact that $Q^{\sigma}=Q$ implies $Q\in F[X]$. However, what if $\chi(F)=p$ and $f$ is inseparable? Since if a polynomial $f$ is irreducible and inseparable on such $F$, then $f(X)=\overline{f}(X^{p^e})$ for some $e\in\mathbb N$ and $\overline f$ is irreducible but separable. Let $f(X)=(X^{p^e}-a_1)\cdots (X^{p^e}-a_n)=(X-\alpha_1)^{p^e}\cdots (X-\alpha_n)^{p^e}$ and $Q=(X-\alpha_1)\cdots (X-\alpha_n)$, then $Q^{\sigma}=Q$ but $Q\notin F[X]$ since $f$ is irreducible but $Q$ divides $f$.
It would be appreciated if someone could point out where I made any mistakes.