Galois Correspondence

Let $L/K$ be a finite Galois extension with Galois group $G = \mathrm{Gal}(L/K)$ .

Let $\mathcal{I}(L/K)$ denote the set of intermediate fields $E$ with $K\subseteq E\subseteq L$ , and let $\mathrm{Sub}(G)$ be the set of subgroups of $G$ .

For $E\in\mathcal{I}(L/K)$ define

\Phi(E) = \mathrm{Gal}(L/E)=\{\sigma\in G:\sigma|_E=\mathrm{id}_E\},

and for a subgroup $H\le G$ define its fixed field

\Psi(H)=L^H=\{x\in L:\sigma(x)=x\ \text{for all }\sigma\in H\}.

Theorem (Galois correspondence). The assignments $\Phi$ and $\Psi$ are inverse bijections

\mathcal{I}(L/K)\ \longleftrightarrow\ \mathrm{Sub}(G),

and they reverse inclusions: if $E_1\subseteq E_2$ then $\mathrm{Gal}(L/E_2)\le \mathrm{Gal}(L/E_1)$ , and if $H_1\le H_2$ then $L^{H_2}\subseteq L^{H_1}$ .

Moreover, for $H\le G$ one has the degree/index formulas

[L:L^H]=|H|,\qquad [L^H:K]=[G:H],

which combine degree = group order for finite Galois extensions with the tower law.

Finally, for $E\in\mathcal{I}(L/K)$ with corresponding subgroup $H=\mathrm{Gal}(L/E)$ , the subextension $E/K$ is normal (equivalently, Galois) if and only if $H$ is a normal subgroup of $G$ , and then restriction induces an isomorphism

\mathrm{Gal}(E/K)\ \cong\ G/H.

Remarks

This correspondence is one standard formulation of the Fundamental Theorem of Galois Theory.

Examples

A biquadratic extension with Klein four group. Let $L=\mathbb{Q}(\sqrt2,\sqrt3)$ and $K=\mathbb{Q}$ . This is the splitting field of $(x^2-2)(x^2-3)$ , hence a finite Galois extension. The group $G=\mathrm{Gal}(L/\mathbb{Q})$ has four elements, determined by independent sign changes of $\sqrt2$ and $\sqrt3$ :

$\tau_2(\sqrt2)=-\sqrt2,\ \tau_2(\sqrt3)=\sqrt3$ ,
$\tau_3(\sqrt2)=\sqrt2,\ \tau_3(\sqrt3)=-\sqrt3$ ,
$\tau_6=\tau_2\tau_3$ .

The three subgroups of order $2$ correspond to the three quadratic intermediate fields:

L^{\langle\tau_2\rangle}=\mathbb{Q}(\sqrt3),\quad L^{\langle\tau_3\rangle}=\mathbb{Q}(\sqrt2),\quad L^{\langle\tau_6\rangle}=\mathbb{Q}(\sqrt6).

The full group fixes $\mathbb{Q}$ , and the trivial subgroup fixes $L$ .

A cyclotomic example. Let $L=\mathbb{Q}(\zeta_5)$ (a cyclotomic extension) and $K=\mathbb{Q}$ . Then $G\cong(\mathbb{Z}/5\mathbb{Z})^\times$ is cyclic of order $4$ . There is exactly one subgroup of order $2$ , hence exactly one intermediate field $E$ with $[E:\mathbb{Q}]=2$ , namely the maximal real subfield $E=\mathbb{Q}(\zeta_5+\zeta_5^{-1})=\mathbb{Q}(\sqrt5).$

Finite fields. For $L=\mathbb{F}_{p^n}$ and $K=\mathbb{F}_p$ , the extension is Galois with cyclic group (see finite-field Galois cyclicity). Subgroups of a cyclic group correspond to divisors of $n$ , so the intermediate fields are exactly $\mathbb{F}_{p^d}\quad (d\mid n),$ with $\mathbb{F}_{p^d}$ corresponding to the unique subgroup of $G$ of index $d$ .