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

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

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

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

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

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

Examples

1. 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 $L$.

2. 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

   $$E=\mathbb{Q}(\zeta_5+\zeta_5^{-1})=\mathbb{Q}(\sqrt5).$$

3. 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 $d$.