This post is the third of three parts, culminating in a proof of the (classical) fundamental theorem of Galois Theory:

1) Finite Separable Algebras
2) Galois Connections
3) The Fundamental Theorem of Galois Theory

We use the same notation as we used in (1) and (2), so read those if you don’t know any of the terms here. For the reader who is looking for the minimal route to a proof of the fundamental theorem, I have put a star (*) next to the essential parts of this post.

Definition:* Let $K/k$ be a finite field extension. We say $K /k$ is Galois if $K$ is a $K$-separable $k$-algebra.

Lemma: Let $K/k$ be a finite field extension, and let $\overline{K}$ be the algebraic closure of $K$. $K/k$ is Galois if and only if $K$ is separable and the canonical map $[K, K]_k \rightarrow [K, \overline{K}]_k$ is surjective.

Proof: If the canonical map above is surjective, then $\# [K, K]_k = \# [ K, \overline{K} ]_k$, so $\# [K, K]_k = \# [K, \overline{K}]_k = \text{dim}_k (K)$. Hence $K$ is $K$-separable.

Conversely, suppose $K$ is $K$-separable. Then $\text{dim}_k (K) = \# [K, K]_k$, so $\# [ K, \overline{K} ]_k \geq \# [K, K]_k = \text{dim}_k (K)$, so that $K$ is separable. Since $\# [K, K ]_k = [K, \overline{K} ]_k$, the map $[K, K]_k \rightarrow [K, \overline{k}]_k$ is surjective, as desired.

The Fundamental theorem of Galois theory asserts that there is a Galois correspondence between intermediate fields of a Galois field extension $K /k$ and subgroups of $\text{Aut}_k (K)$. Going along with more recent approaches to the theorem, we choose here to write in terms of group actions instead of subgroups. Here we show that there is a Galois correspondence between intermediate fields of a finite Galois field extension $K/k$ and quotient $G$-sets of $G$. This exercise shows that the two approaches are equivalent:

Exercise: there is a preorder isomorphism between subgroups of a group $G$ and quotient $G$-sets of $G$. The correspondence sends a subgroup $H$ of $G$ to the quotient $G$-set $G/H$ and a quotient $G$-set $X$ with quotient map $\pi : G \rightarrow X$ to $\text{Stab}_{G} (\pi (e))$, where $e$ is the identity element in $G$.

Setup: * Let $\text{Sub}(K)$ be the preorder of subobjects of $K$ in the category of $k$-algebras. Let $\text{Quot}(G)$ be the preorder of quotient objects of $G$ in the category of $G$-sets.

For a $k$-algebra $F$, the set $[F, K]_k$ of $k$-algebra maps from $F$ to $K$ has the structure of a $G$-set. $\sigma \in G$ acts on $[F, K]_k$ by sending $\iota : F \rightarrow K$ to $\sigma \circ \iota : F \rightarrow K$. Define a preorder map $[-, K]_k : \text{Sub}(K) \rightarrow \text{Quot}(G)$ sending a $k$-algebra $F$ with monomorphism $\iota : F \rightarrow K$ to the $G$-set $[F, K]_k$ with epimorphism $G \rightarrow [F, K]_k$ sending $\sigma$ to $\sigma \circ \iota$.

For a $G$-set $X$, the set $[X, K]_G$ of $G$-set maps from $X$ to $K$ has the structure of a $k$-algebra. For two $G$-sets $f : X \rightarrow K$ and $g : X \rightarrow K$, we set $f + g : X \rightarrow K$ to send $x$ to $f(x) + g(x)$ and $fg : X \rightarrow Y$ to send $x$ to $f(x)g(x)$. For an element $a \in k$ and $f \in [ X, K]_G$, we set $af : X \rightarrow K$ to send $x$ to $af(x) = f(ax)$. Define a preorder map $[-, K]_G : \text{Quot}(G) \rightarrow \text{Sub}(K)$ sending a $G$-set $X$ with epimorphism $\pi : G \rightarrow X$ to the $k$-algebra $[X, K]_k$ with monomorphism $[X, K]_k \rightarrow K$ sending $\phi$ to $\phi \circ \pi (1)$.

$[-, K]_k$ and $[-, K]_G$ form a Galois connection. Indeed, taking a subobject $F$ of $K$ and a quotient object $X$ of $G$, any map $\phi : F \rightarrow [X, K]_G$ of $k$-algebras gives a map $\psi : X \rightarrow [F, K]_k$ of $G$-sets sending $x \in X$ to the map $\psi(x) : F \rightarrow K$ of $k$-algebras sending $a$ to $\phi(a)(x)$. Conversely, any map $\psi : X \rightarrow [F, K]_k$ of $G$-sets gives a map $\phi : F \rightarrow [X, K]_G$ of $k$-algebras sending $a \in F$ to the map $\phi(a) : X \rightarrow K$ of $G$-sets sending $x$ to $\psi(x)(a)$.

At last, we have the fundamental theorem of Galois theory:

Theorem: (The Fundamental Theorem of Galois Theory) Let $K/k$ be a finite Galois field extension. Put $G = [K, K]_k$. There is a preorder isomorphism between $\text{Quot}(G)$ and $\text{Sub}(K)$ given by $[-, K]_k$ and $[-, K]_G$.

Proof: Take a $k$-algebra $F \in \text{Sub}(K)$. Put $\hat{F} = [[F, K]_k, K]_G$. Then $[\hat{F}, K]_k = [[[F, K]_k, K]_G, K]_k \cong [F, K]_k$ by the main proposition in my post on Galois connections. Hence $\# [\hat{F}, K]_k = \# [F, K]_k$. $K$ is $K$-separable since it is Galois (we showed this above). $F$ and $\hat{F}$ are both subobjects of $K$, and so they are both $K$-separable, by the stability results in my post on finite separable algebras. So $\text{dim}_k (F) = \# [ F, K]_k$ and $\text{dim}_k (\hat{F}) = \# [ \hat{F}, K]_k$. It follows that $\text{dim}_k (F) = \text{dim}_k (\hat{F})$. Now the inclusion $F \rightarrow \hat{F}$ (see my post on Galois connections) is an injection of $k$-algebras of the same finite dimension, and so must be an isomorphism.

Conversely, take a $G$-set $X \in \text{Quot}(G)$. Put $F = [X, K]_G$, and $\hat{X} = [F, K]_k$. The quotient map $\pi : G \rightarrow X$ gives an inclusion map $[\pi, K]_G : F \rightarrow K$. This inclusion tells us that $F$ is $K$-separable, one of the stability properties in my post on finite separable algebras. So $\text{dim}(F) = \# [F, K ] = \# \hat{X}$. Also, $\# X \leq \text{dim}(F)$ by part (ii) of the main theorem in my post on finite separable algebras. Now the canonical map $X \rightarrow \hat{X}$ is a surjection and $\# X \leq \# \hat{X}$, so that it must be an isomorphism of $G$-sets.