Real subfield of cyclotomic field is generated by $\zeta+\zeta^{-1}$

Question

Let $p\neq 2$ a prime number, $\zeta=e^{\frac{2i\pi}{p}}$ and $\alpha=2\cos\left(\frac{2\pi}{p}\right)$. We consider the field extensions $F=\mathbb Q(\zeta)$ and $E=F\cap \mathbb R$ of $\mathbb Q$. I have shown that $[F:\mathbb Q]=p-1$, that $\zeta+\zeta^{-1}\in E$ and that $t^2-(\zeta+\zeta^{-1})t+1$ is the minimal polynomial of $\zeta$ on $E$. Now I'm trying to show that $E=\mathbb Q(\alpha)$. I have shown that $E\supset \mathbb Q(\alpha)$ but I do not arrive to show the inclusion $E\subset \mathbb Q(\alpha)$.

Answer 1

$E = F \cap \mathbb R$ is precisely the fixed field of complex conjugation acting on $F$ in the only possible way. By Galois theory, it follows that $F$ is a quadratic extension of $E$. Since you have shown that $F$ is a quadratic extension of $\mathbb Q(\alpha)$, and that $E \supset \mathbb Q(\alpha)$, it follows that $E=\mathbb Q(\alpha)$.

Answer 2

Basically you have $[\mathbb Q(\xi):\mathbb Q(\xi+\xi^{-1})]=2$ because of the minimal polynomial you've found. Then notice that $[\mathbb Q(\xi):E] >1$, then since $\mathbb Q(\xi+\xi^{-1}) \subset E$, we are done.