Fix(Gal($\bar{F}/F)) = F$ when $\bar{F}$ is separable over $F$

Question

Here $\bar{F}$ is an algebraic closure of $F$. I really don't know how to approach this problem since we can't use that being Galois is equivalent to being normal and separable since the extension is not given to be finite. I'm guessing this means I have to directly use automorphisms, but I'm not sure if that is right or even how to go about doing such a thing. Do I need to show that $F \subset \bar{F}$ is Galois or can I assume it since the question asks for Fix(Gal$(\bar{F}/F$))?

Answer 1

You know that $F \subseteq Fix(Gal(\bar{F}/F))$ by definition. So, show that for any $x\in \bar{F}\backslash F$, there is an automorphism fixing $F$ that does not fix $x$.