Galois theory question

Question

Let α be a complex primitive 43rd root of 1. Prove that there is an extension field F of the rational numbers such that $[F(\alpha): F] = 14$. I don't know how to start, can someone help me out?

Answer 1

Note that $43$ is prime. For any prime $p$ and $p$-th root of unity $\zeta$ we have $$\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) \cong (\mathbb{Z}_p)* \cong \mathbb{Z}_{p-1},$$ the first isomorphism is given by $\varphi \mapsto \overline{a}$ if $\varphi(\zeta)=\zeta^a$, the second one is a known fact. So for $p=43$ we have that the corresponding galois group is isomorphic to the cyclic group of order $42$. Now conveniently, $14$ divides $42$, and since a cyclic group of order $n$ has a subgroup for each of its divisors of the corresponding order, (take in this example the subgroup generated by $\overline{3}$ in $\mathbb{Z}_{42}$), we can consider this subgroup, call it $H$. By the galois correspondence we have a corresponding field, $F:=\mathbb{Q}(\alpha)^H$, with the property

$$[\mathbb{Q}(\alpha) : F]=|H|=14$$ Note that $F(\alpha)=\mathbb{Q}(\alpha)$, which establishes the result together with the above equality.