Set of elements fixed by Galois group of $E$ over $F$ if $F$ (requesting alternate proof)

Question

Proposition: Let $E$ be the splitting field for some separable polynomial over $F$. Then the set of elements $E_G$ fixed by all automorphisms in $G(E \setminus F)$ is equal to $F$.

My book gives a nice, simple argument based the degree of the extension in terms of the intermediate field and the equivalence of $|G(E\setminus F)|$ and $[E:F].$ I understand this proof well enough.

Is there a proof that proceeds in a more direct manner?

It's clear that $F \subset E_G.$ Now, for any $a \in E_G$, every automorphism $\sigma \in G(E\setminus F)$ has $\sigma(a) = a$. It seems "reasonable" in some sense that the collection of all automorphisms that fix $F$ will never all haphazardly fix some additional element of $E$, but I couldn't formalize this in any way.

Answer 1

Let $a$ be an element of $E$ not in $F$, then $a$ is the zero of some polynomial over $F$ which necessarily has degree 2 or higher (if degree=1 then $a$ must be in $F$ - work it out). That polynomial must have ANOTHER root $a'$ (also not in $F$). Then there must be an automorphism in $G(E / F)$ which maps $a$ to $a'$, so $a$ cannot be in $E_G$.

Answer 2

Let $ a \in E_G$ and $\sigma : F(a) \to \overline{E}$ be a morphism inducing the identity on $F$. Extend $\sigma$ to $\tau : E \to \overline{E}$, where $\overline{E}$ is the algebraic closure of $E$. By normality $\tau$ is an automorphism of $E$ inducing the identity on $F$. Then $\tau \in G$. But then $\tau(a)=a$, and there is only one extension of $\sigma$ to an embedding $E \hookrightarrow \overline{E}$. This number is, by definition, the separable degree of $F \subset F(a)$. But $a$ is separable over $F$, so $[F(a) : F]=$ separabale degree$=1$ and $a \in F$.

(This is more or less Lang's Algebra proof).