Let L/KL/K be a finite , and write

G=Gal(L/K)G = \mathrm{Gal}(L/K)

for its .

For a subgroup HGH\le G, the of HH is

LH={xL:σ(x)=x for all σH}.L^H=\{x\in L:\sigma(x)=x\text{ for all }\sigma\in H\}.

For an KELK\subseteq E\subseteq L, write

Gal(L/E)={σG:σE=idE}.\mathrm{Gal}(L/E)=\{\sigma\in G:\sigma|_E=\mathrm{id}_E\}.

Theorem (Fundamental theorem of Galois theory). The assignments

HLHandEGal(L/E)H \longmapsto L^H \qquad\text{and}\qquad E \longmapsto \mathrm{Gal}(L/E)

give an inclusion-reversing bijection between subgroups HGH\le G and intermediate fields EE with KELK\subseteq E\subseteq L. Under this correspondence:

  1. [L:E]=Gal(L/E)[L:E]=|\,\mathrm{Gal}(L/E)\,| and [E:K]=[G:Gal(L/E)][E:K]=[G:\mathrm{Gal}(L/E)].
  2. E/KE/K is Galois if and only if Gal(L/E)G\mathrm{Gal}(L/E)\trianglelefteq G is a normal subgroup; in that case there is a natural isomorphism
    Gal(E/K)G/Gal(L/E).\mathrm{Gal}(E/K)\cong G/\mathrm{Gal}(L/E).
Remarks

This is the conceptual content packaged in the explicit .

Examples

  1. L=Q(2)L=\mathbb{Q}(\sqrt2) over K=QK=\mathbb{Q}. Then GC2G\cong C_2 has exactly two subgroups, so the only intermediate fields are Q\mathbb{Q} and Q(2)\mathbb{Q}(\sqrt2).
  1. L=Q(ζ8)L=\mathbb{Q}(\zeta_8) over Q\mathbb{Q}, where ζ8\zeta_8 is a primitive 8th root of unity (see ). Here G(Z/8Z)×C2×C2G\cong (\mathbb{Z}/8\mathbb{Z})^\times\cong C_2\times C_2, which has three distinct index-2 subgroups. Correspondingly, L/QL/\mathbb{Q} has three distinct quadratic intermediate fields.
  1. Finite fields: L=FpnL=\mathbb{F}_{p^n} over K=FpK=\mathbb{F}_p. The extension is Galois, and GG is cyclic of order nn generated by ; see . Subgroups of a cyclic group correspond to divisors of nn, so the intermediate fields are exactly Fpd\mathbb{F}_{p^d} for dnd\mid n.