Spring 2016 - Problem 7

Solution.

By Dirichlet's theorem, we may find a prime $p$ such that $p \equiv 1 \mod 2n$. Let $\zeta_p$ be a primitive $p$-th root of unity so that $\Q\p{\zeta_p}$ has Galois group $C_{p-1}$. Note that complex conjugation is an automorphism of $\Q\p{\zeta_p}$ fixing $\Q$, so let $C_2$ be the subgroup generated by this.

Since $C_{p-1}$ is abelian, $C_2$ is a normal subgroup, so the field $E$ fixed by $C_2$ is Galois over $\Q$ with Galois group $C_{p-1}/C_2 \simeq C_{\p{p-1}/2}$. Moreover, $E \subseteq \R$ since it is fixed by complex conjugation. Finally, by choice of $p$, we know $n$ divides $\frac{p-1}{2}$, so let $k = \frac{p-1}{2n}$. As before, $C_k$ is a normal subgroup of $C_{\p{p-1}/2}$ whose fixed field $F$ is Galois over $\Q$ with Galois group $C_{\p{p-1}/2}/C_k \simeq C_n$. Since $F \subseteq E \subseteq \R$, we are done.