Degree of $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}(\zeta_n+\zeta_{n}^{-1})$

Question

Let $\zeta_n$ be a $n$-th primitive root of unity. How to prove that $[\mathbb{Q}(\zeta_n):\mathbb{Q}(\zeta_n+\zeta_{n}^{-1})]=2$ ?

Answer 1

This is only true for $n>2$. Since $\zeta_n$ is a root of the polynomial $$X^2 -(\zeta_n+\zeta_n^{-1})X + 1 = (X-\zeta_n)(X-\zeta_n^{-1})\in \mathbb Q(\zeta_n+\zeta_n^{-1})[X], $$ it follows that $[\mathbb Q(\zeta_n): \mathbb Q(\zeta_n+\zeta_n^{-1})]\le 2$. Now, we have $\mathbb Q(\zeta_n+\zeta_n^{-1})\subseteq \mathbb R$ and $\zeta_n\notin \mathbb R$. Together it follows that $\mathbb Q(\zeta_n)\neq \mathbb Q(\zeta_n+\zeta_n^{-1})$ and hence the degree must be 2.