Originally Posted by

**NonCommAlg** this is a good question! the answer is No! because, for example, if $\displaystyle x$ is transcendental over $\displaystyle F$ and $\displaystyle E=F(x)$, then

it can be proved (not very easily though!) that $\displaystyle E^{\text{Gal}(E/F)}=F.$ the main part of the proof is to show that $\displaystyle \text{Gal}(E/F)$

consists of all maps $\displaystyle \sigma$ defined by $\displaystyle \sigma(f(x))=f(\frac{ax+b}{cx+d}), \ \forall f \in E,$ where $\displaystyle a,b,c,d \in F$ and $\displaystyle ad - bc \neq 0.$