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.)

1 Answer

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 $$\Bbb Q(\sqrt[3]2)$$ as an extension of $$\Bbb Q$$; here we do not have enough automorphisms, since any automorphism must send $$\sqrt[3]2$$ to a cube root of $$2$$, but the only cube root of $$2$$ in $$\Bbb Q(\sqrt[3]2)$$ is $$\sqrt[3]2$$ 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 $$\Bbb Q(\sqrt[3]2)$$ fixed by "the group of automorphisms that fix $$\Bbb Q$$" is unfortunately just $$\Bbb Q(\sqrt[3]2)$$ itself, instead of $$\Bbb Q$$.