Fundamental Theorem of Galois Theory

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

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

Let $\operatorname{Gal} \left({L / K}\right)$ denote the Galois group of the extension $L / K$.

The mappings:


 * $H \mapsto L_H$, and


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

are inclusion-reversing and inverses.

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

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

Let $K \subset F_1 \subset F_2 \subset L$.

Let $G_i = \operatorname{Gal} \left({L / F_i}\right)$.

Let $\sigma \in G_2$.

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 $H_1 \subset H_2 \subset \operatorname{Gal} \left({L / K}\right)$.

Let $F_i = L_{H_i}$.

Let $x \in F_2$.

Then $\sigma \left({x}\right) = x$ 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:
 * let $G$ denote $\operatorname{Gal} \left({L / K}\right)$
 * for any field $K \subset F \subset L$ let $G_F$ denote $\operatorname{Gal} \left({L / F}\right)$.

Next, we demonstrate that the two functions described are inverses.

That is:


 * 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 $F \subset L_{G_F}$.

Let $\alpha \in L_{G_F} \setminus F$.

Then:
 * $\left[{F \left({\alpha}\right) : F}\right] > 1$

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


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

However, by our assumption, $\sigma \left({\alpha}\right) = \alpha$ for each $\sigma$.

Thus:

Since $\left[{F \left({\alpha}\right) : F}\right] > 1$, 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 $L = L_H \left({\alpha}\right)$.

Consider the polynomial:
 * $\displaystyle f = \prod_{\sigma \mathop \in H} \left({x - \sigma \left({\alpha}\right)}\right)$

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

However:


 * $\left[{L : L_H}\right] = \operatorname{deg} \left({f}\right) = \left\vert{H}\right\vert < \left\vert{G_{L_H}}\right\vert = \left[{L : L_H}\right]$

by definition of Galois 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 $H = \operatorname{Gal} \left({F / K}\right)$ 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.

We have that $F \subset L$.

Thus $\sigma$ 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 $F$.

Let $x \in F$.

Then $\sigma \left({x}\right) \in F$.

We have that $\tau$ fixes $F$.

Thus:
 * $\tau \left({\sigma \left({x}\right)}\right) = \sigma \left({x}\right)$

Therefore:
 * $\sigma^{-1} \left({\tau\left({\sigma \left({x}\right)}\right)}\right) = x$

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:


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

which implies that:
 * $\tau \left({\hat \sigma \left({x}\right)}\right) = \hat \sigma \left({x}\right) \in F$

Since $x \in F$:
 * $\hat \sigma \left({x}\right) = \sigma \left({x}\right) \in F$

which was required to be shown.