Artin's theorem on fixed fields

A finite group of field automorphisms yields a finite Galois extension with degree equal to the group order.
Artin’s theorem on fixed fields

Let LL be a field and let GG be a finite subgroup of the group of of LL. Define the

LG={xL:σ(x)=x for all σG}. L^G=\{x\in L:\sigma(x)=x\ \text{for all}\ \sigma\in G\}.

Theorem (Artin). If GAut(L)G\le \mathrm{Aut}(L) is finite and K=LGK=L^G, then:

  1. L/KL/K is a finite ;
  2. the restriction map identifies GG with the full : Gal(L/K)=G; \mathrm{Gal}(L/K)=G;
  3. in particular, [L:K]=G. [L:K]=|G|. This is a foundational input to the and explains why fixed fields and automorphism groups match so tightly.

Examples

  1. L=CL=\mathbb{C}, G={1,conj}G=\{1,\mathrm{conj}\} where conj(z)=z\mathrm{conj}(z)=\overline{z}.
    Then LG=RL^G=\mathbb{R}, so C/R\mathbb{C}/\mathbb{R} is Galois with [C:R]=2=G[\mathbb{C}:\mathbb{R}]=2=|G|.

  2. L=Q(2,3)L=\mathbb{Q}(\sqrt2,\sqrt3), and let GG be the subgroup of Aut(L)\mathrm{Aut}(L) generated by 22\sqrt2\mapsto-\sqrt2 and 33\sqrt3\mapsto-\sqrt3.
    Then GG has order 44, its fixed field is Q\mathbb{Q}, and Artin’s theorem gives [L:Q]=4[L:\mathbb{Q}]=4.

  3. L=FpnL=\mathbb{F}_{p^n} and let G=φG=\langle \varphi\rangle where φ(x)=xp\varphi(x)=x^p is Frobenius (see ).
    Then G=n|G|=n, LG=FpL^G=\mathbb{F}_p, and Artin’s theorem recovers that Fpn/Fp\mathbb{F}_{p^n}/\mathbb{F}_p is Galois of degree nn.