Intuition for equivalent definitions of Galois extension Assume $K/F$ is a finite field extension. Then the following are equivalent:


*

*$|\operatorname{Gal}(K/F)| = [K:F]$.

*The fixed field of $\operatorname{Gal}(K/F)$ is $F$.

*$K$ is the splitting field of some $f(x) \in F[x]$.

*Every $f(x) \in F[x]$ has either all or none of its roots in $K$.


(Notation is fairly standard, I think: $\operatorname{Gal}(K/F)$ is the group of $K$-automorphisms fixing $F$; $[K: F]$ is the dimension of $K$ as a vector space over $F$.)
I can follow the "small picture" logical arguments for why they're equivalent, but I have very little intuitive sense for why they should be. These conditions all describe a very "nice"/"clean" extension, and I find myself like a linear algebra student who thinks all the equivalent conditions in the invertible matrix theorem are "hard to memorize" because he or she doesn't yet see the geometric picture of a linear transformation that makes everything "obvious".
Is there some grand unifying big picture that makes all these equivalent definitions fall out "for free"? (Obviously, obviousness is subjective, but if you have some mental picture of field extensions that makes this obvious to you, I'd like to know what that picture is.)
 A: Not really, but there is a rephrasing from which the whole proof takes a few lines. Given $K/F$

*

*Write it as a tower of simple extensions $E_0=F,E_m  = K, E_j=E_{j-1}[x_j]/(f_j(x_j))$


*From it  construct all the $E_0$-algebra homomorphisms $\sigma:E_m\to \overline{E_m}$
(one root of $f_1$ determinates an homomorphism $E_1\to \overline{E_m}$, apply it to the coefficients of $f_2$, then repeat iteratively)


*$E_m/E_0$ separable is the same as each $f_j$ separable which is the same as $\prod_j \deg_s(f_j)=|Hom_{E_0}(E_m,\overline{E_m})|=[E_m:E_0]=\prod_j \deg(f_j)$


*The normal / splitting field condition is the same as $\sigma(E_m)\subset E_m$ ie. $\sigma(E_m)=E_m$ ie. $Hom_{E_0}(E_m,\overline{E_m})= Aut(E_m/E_0)$


*The fixed field $E_0=E_m^{Aut(E_m/E_0)}$ condition is the same as every $a\in E_m,\not \in E_0$ has a distinct conjugate $b$, the isomorphism $\sigma:E_0(a)\to E_0(b)$ extends to some $\sigma\in Hom_{E_0}(E_m,\overline{E_m})=Aut(E_m/E_0)$, ie. $E_m/E_0$ is normal and separable.
