Past exam of the number theory course of the University of Oxford / 2011 / A / 2

The fixed field isIt is a subfield because every field automorphism preserves sums, products and inverses.

Set $F=K^G$. If $[K:F]$ is infinite the desired finite lower bound is immediate. Otherwise let $b_1,\dots ,b_d$ be an $F$-basis of $K$. If $|G|>d$, the homogeneous system over $K$has a nonzero solution. For $a={\sum }_j{a}_j{b}_j$, with $a_j\in F$, every $\sigma$ fixes $a_j$, so the same coefficients give ${\sum }_{\sigma }{c}_{\sigma }\sigma (a)=0$. This contradicts linear independence of distinct field embeddings. Consequently

The Fundamental theorem of Galois theory says that for a finite Galois extension $L/F$ with group $G$, the maps $H\mapsto L^H$ and $E\mapsto \mathrm{Gal}(L/E)$ are mutually inverse inclusion-reversing correspondences between subgroups and intermediate fields. They satisfy $[L:L^H]=|H|$ and $[L^H:F]=[G:H]$. An intermediate extension $E/F$ is Galois exactly when its subgroup is normal, in which case its group is the quotient by that subgroup.

Let $a=\sqrt{2}$, $b=\sqrt{3}$, $c=i$. The real field $\mathbb{Q}(a,b)$ has degree $4$: if $\sqrt{3}=u+v\sqrt{2}$ with $u,v\in \mathbb{Q}$, squaring forces $uv=0$ and either $u^2=3$ or $v^2=3/2$, both impossible. Adjoining the nonreal $i$ doubles the degree. All three square-root signs can be changed independently, givingWrite ${\sigma }_{uvw}$ for the automorphism sending $(a,b,c)$ to $((-1{)}^{u}a,(-1{)}^{v}b,(-1{)}^{w}c)$, where $(u,v,w)\in {\mathbb{F}}_2^3$. Every nonzero vector generates an order-two subgroup, so there are $7$ such subgroups. Their fixed fields are:

Each listed field has degree $4$ and is fixed by the indicated sign change, which proves equality with the fixed field.

There are also $7$ order-four subgroups: they are the kernels of the seven nonzero linear functionals on ${\mathbb{F}}_2^3$. For $e=(e_1,e_2,e_3)\ne 0$, define $H_e=\{{\sigma }_{uvw}:e_{1}u+e_{2}v+e_{3}w=0\}$. ThenThus the seven quadratic fields have radicands $2,3,-1,6,-2,-3,-6$. Their square classes are distinct, so all are different. Finally the trivial subgroup fixes $L$, and all of $G$ fixes $\mathbb{Q}$. This accounts for every subgroup and intermediate field.