Fundamental Theorem of Galois Theory

Theorem
Let $L/K$ be a finite Galois extension.

Let $H$ denote a subgroup of $\operatorname{Gal}(L/K)$ and $F$ denote an intermediate field.

Then the functions


 * $H \mapsto L_H$, and


 * $F \mapsto \operatorname{Gal}(L/F)$

are inclusion-reversing and inverses.

Moreover, these maps induce a bijection between the normal subgroups of $\operatorname{Gal}(L/K)$ and the normal, intermediate extensions of $L/K$.

Note that $\operatorname{Gal}(L/K)$ denotes the Galois group of the extension $L/K$.

Proof
First, we show that the maps are inclusion-reversing.

Let $K \subset F_1 \subset F_2 \subset L$, and let $G_i = \operatorname{Gal}(L/F_i)$.

If $\sigma\in G_2$, then $\sigma$ is an automorphism of $L$ which fixes $F_2$.

Since $F_1\subset F_2$, it follows that $\sigma$ fixes $F_1$ and consequently $\sigma \in G_1$.

Let $H_1\subset H_2 \subset \operatorname{Gal}(L/K)$, and let $F_i = L_{H_i}$.

If $x\in F_2$, then $\sigma(x) = x$ for all $\sigma\in H_2$.

Since $H_1\subset H_2$, the same equality holds for each element of $H_1$ and thus $x\in F_1$.

For the remainder of the proof, we let $G$ denote $\operatorname{Gal}(L/K)$ and for any field $K\subset F \subset L$ we let $G_F$ denote $\operatorname{Gal}(L/F)$.

Next, we demonstrate that the two functions described are inverses; in other words,


 * For any intermediate field $K \subset F \subset L$,


 * $F = L_{G_F}$.


 * For any subgroup $H\subset G$,


 * $H = G_{L_H}$.

For the first equality, we obviously have $F\subset L_{G_F}$.

Suppose $\alpha\in L_{G_F}\setminus F$, then $[F(\alpha):F]>1$.

We can express the minimal polynomial of $\alpha$ in terms of $G_F$ as:


 * $m_\alpha(x) = \displaystyle \prod_{\sigma\in G_F}(x-\sigma(\alpha))^\frac{1}{[L:F(\alpha)]}$

However, by our assumption, $\sigma(\alpha) = \alpha$ for each $\sigma$.

Thus:


 * $m_\alpha(x) = (x-\alpha)^\frac{[L:F]}{[L:F(\alpha)]} = (x-\alpha)^{[F(\alpha):F]}$

Since $[F(\alpha):F]>1$, this contradicts the separability of $L/F$.

Therefore, the first equality holds.

For the second equality, it is immediate that $H\subset G_{L_H}$.

Suppose $H$ were a proper subset of $G_{L_H}$.

By the Primitive Element Theorem, there exists an $\alpha\in L$ such that $L = L_H(\alpha)$.

Consider the polynomial


 * $f = \displaystyle \prod_{\sigma\in H}{(x-\sigma(\alpha))}$.

The coefficients of $f$ are evidently elements of $L_H$ and $f$ is monic by construction. However,


 * $[L:L_H] = \operatorname{deg}(f) = |H| < |G_{L_H}| = [L:L_H]$,

by Order of Galois Group Equals Degree of Extension. This is a contradiction and it follows that $H=G_{L_H}$.

Finally, we demonstrate the correspondence between normal subgroups of $G$ and the intermediate normal extensions of $K$.

Suppose $K\subset F \subset L$ is an intermediate field and $F/K$ is a normal extension.

We let $H = \operatorname{Gal}(F/K)$ denote the Galois group of interest.

Let $\sigma\in G$ and $\tau\in H$. We want to show that $\sigma^{-1}\tau\sigma\in H$ to conclude that $H$ is normal.

Since $F\subset L$, $\sigma$ restricts to an embedding of $F$ in $\overline{K}$.

However, since $F/K$ is a normal extension, the image of every embedding of $F$ is again $F$.

Thus, $\sigma$ restricts to an automorphism of $F$.

If $x\in F$, then $\sigma(x)\in F$. Since $\tau$ fixes $F$, $\tau(\sigma(x)) = \sigma(x)$.

Therefore, $\sigma^{-1}(\tau(\sigma(x)))=x$ and we conclude that $\sigma^{-1}\tau\sigma\in H$.

Next, suppose $H$ is a normal subgroup of $G$ and $F = L_H$.

Let $\tau\in H$ and $\sigma:F\mapsto \overline{K}$ be an embedding of $F$.

By Extension of Isomorphisms, we extend $\sigma$ to $\overline{\sigma}$, an automorphism of $L$.

Consider the composition $\hat{\sigma}^{-1}\tau\hat{\sigma} = \hat{\tau}\in H$ by our assumption of normality.

Then,


 * $\hat{\sigma}^{-1}\tau\hat{\sigma}(x) = \hat{\tau}(x) = x$,

which implies that $\tau(\hat{\sigma}(x)) = \hat{\sigma}(x)\in F$.

Since $x\in F$, we have $\hat{\sigma}(x) = \sigma(x)\in F$, exactly as we needed to show.