# Splitting field of separable polynomial and the fixed field of the Galois group

Working my way through this resource on abstract algebra, and a few questions pop up from time to time.

Here is a statement regarding fields and Galois groups I find puzzling:

Proposition 23.17: Let be $$E$$ be a splitting field over $$F$$ of a separable polynomial. Then $$E_{G(E/F)} = F$$.

Here, $$G(E/F) : = \{\sigma \in \mathrm{Aut}(E) : \sigma(\alpha) = \alpha, \; \forall \alpha \in F\}$$ is the Galois group of $$E$$ over $$F$$, and $$E_{G(E/F)} = \{ \alpha \in E : \sigma(\alpha) = \alpha, \; \forall \sigma \in G(E/F)\}$$ is the fixed field of $$E$$ w.r.t $$G(E/F)$$.

To me, this intuitively feels like a tautology - i.e. using brackets for comprehension, it is the statement that "the sub-field of $$E$$ consisting of elements fixed by [the group whose members leave $$F$$ invariant] is $$F$$".

There is a short proof provided, which does not make this clear to me:

Let $$G=G(E/F)$$. Clearly, $$F \subset E_G \subset E$$. Also, $$E$$ must be a splitting field of $$E_G$$ and $$G(E/F) = G(E/E_G)$$. By Theorem 23.7 [which is about a splitting field $$E$$ for a polynomial in $$F[x]$$ which is separable]

$$|G| = [E:E_G] = [E:F]$$.

Therefore $$[E_G:F]=1$$. Consequently $$E_G =F$$.

What makes the Proposition a (relatively) non-trivial statement? The key piece seems to be that we have a splitting field whose minimal polynomial over $$F$$ is separable in $$E$$, but I can't see a way through.

(This is slightly before the fundamental theorem on Galois theory is proven, so it would be helpful to see this without using it.)

What is clear is that every element in $$F$$ is fixed by every automorphism of $$E$$ that fixes $$F$$.
However, to go the opposite way, i.e. to claim that every element of $$E$$ that is fixed by [every automorphism of $$E$$ that fixes $$F$$] must belong to $$F$$, is not trivial.
This non-trivial direction is equivalent to (by taking contrapositive) the assertion that if I give you a random element $$\alpha$$ of $$E$$ that is not in $$F$$, you have to give me an automorphism of $$E$$ that fixes $$F$$ but not $$\alpha$$.
An example where this is not true is $$\Bbb Q(\sqrt2)$$ as an extension of $$\Bbb Q$$; here we do not have enough automorphisms, since any automorphism must send $$\sqrt2$$ to a cube root of $$2$$, but the only cube root of $$2$$ in $$\Bbb Q(\sqrt2)$$ is $$\sqrt2$$ itself; i.e. every automorphism is the identity.
So in this case, "the group of automorphisms that fix $$\Bbb Q$$" consists just of the identity, and so the sub-field of $$\Bbb Q(\sqrt2)$$ fixed by "the group of automorphisms that fix $$\Bbb Q$$" is unfortunately just $$\Bbb Q(\sqrt2)$$ itself, instead of $$\Bbb Q$$.