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 $L/K$.

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 $H\subset G_{L_H}$.