From what I have read, for a field extension $E/F$, the automorphism group $Aut(E/F)$ is a Galois group when $E/F$ is Galois, that is, the extension is normal and separable, or equivalently, when $E$ is a splitting field of a separable polynomial with coefficients in $F$.
My question is, is there any intuitive way to explain why this specific definition of a group results in the bijection in the fundamental theorem of Galois theory?
Moreover, I believe the tower of fields $F \subset F_{1} \subset F_{2} \cdots \subset F_{m} = E$, where each intermediate fields are galois extensions of previous ones, correspond to a tower of galois groups, where each subgroup extension are normal. I am confused by the use of word "normal", where in the tower of groups, they essentially say that the left coset is equal to the right coset, while in the tower of fields, the "normal" means every polynomial that is irreducible in $F$ either has no roots in $E$ or splits in $E$. Is there a one-to-one correspondence between the concept of "normal extension" or are they simply different concepts?
Any clarification will be much appreciated.
Best Answer
If $E/F$ is finite normal and separable then so is $E/K$ for any $E/K/F$. Then $K=E^{Aut(E/K)}$ (the subfield of $E$ fixed by $Aut(E/K)$): clearly $K\subset E^{Aut(E/K)}$. If $a\in E^{Aut(E/K)},\not \in K$, because $E/K$ is separable then $a$'s $K$-minimal polynomial has another root $b$. The natural isomorphism $K(a)\to K(b)$ extends to some $K$-homomorphism $\sigma: K(a,a_2)\to \overline{E},K(a,a_2,a_3)\to \overline{E}$ and so on until we get $\sigma:E\to \overline{E}$. Then $\sigma(\overline{E})\subset E$ because $E/K$ is normal. Whence $\sigma\in Aut(E/K),\sigma(a)\ne a$ so in fact $a$ doesn't exist.
$K/F$ is normal iff $g(K)=K$ for all $g\in Aut(E/F)$ iff $E^{Aut(E/K)}=E^{g^{-1}Aut(E/K)g}$ iff $Aut(E/K)$ is normal in $Aut(E/F)$.