A degree-three Galois extension of $\mathbb{Q}$ inside $\mathbb{Q}({\zeta }_7)$

By [L1] the Galois group of $\mathbb{Q}(\zeta )\mathrm{/}\mathbb{Q}$ is isomorphic to the finite subgroup $(\mathbb{Z}\mathrm{/}7{)}^{\times }$ of the unit group of the field $\mathbb{Z}\mathrm{/}7$, so [L4] makes it cyclic; its order is $6$. Thus [L2] gives a unique subgroup $H$ of order $2$, and [L3] makes its fixed field $F$ have degree $[F:\mathbb{Q}]=6\mathrm{/}2=3$.

The subgroup $H$ is generated by the class $[-1]$, so it acts by complex conjugation. Therefore $\zeta +{\zeta }^{-1}$ is fixed by $H$ and lies in $F$.

Put $x:=\zeta +{\zeta }^{-1}$. Then ${\zeta }^2+{\zeta }^{-2}=x^2-2,{\zeta }^3+{\zeta }^{-3}=x^3-3x\mathrm{.}$ Since $1+\zeta +{\zeta }^2+{\zeta }^3+{\zeta }^4+{\zeta }^5+{\zeta }^6=0$, dividing by ${\zeta }^3$ gives $1+(\zeta +{\zeta }^{-1})+({\zeta }^2+{\zeta }^{-2})+({\zeta }^3+{\zeta }^{-3})=0.$ Substituting the expressions above yields $x^3+x^2-2x-1=0.$

The element $x$ is not rational: if it were, then $\zeta$ would satisfy the quadratic polynomial $t^2-xt+1\in \mathbb{Q}[t]$, which would force $[\mathbb{Q}(\zeta ):\mathbb{Q}]\le 2$, contradicting [L1]. Since $x\in F$, the prime degree $[F:\mathbb{Q}]=3$ from step 1.1 leaves only the subfields $\mathbb{Q}$ and $F$, so $\mathbb{Q}(x)=F$. Therefore the minimal polynomial of $x$ has degree $3$, and the cubic from step 3.1 is that minimal polynomial.

The ambient Galois group is cyclic and hence abelian, so $H$ is normal. By [L5], the fixed field $F\mathrm{/}\mathbb{Q}$ is Galois and $\mathrm{Gal}(F\mathrm{/}\mathbb{Q})$ is isomorphic to the quotient by $H$, which has order $3$. The group is cyclic by [L6], so $F\mathrm{/}\mathbb{Q}$ is a cyclic cubic Galois extension.