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

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

• Use the FTGT: it's the fixed field of complex conjugation, which has order 2. – Adam Hughes Aug 13 '16 at 15:28

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.
• But is it clear that there is an automorphism given by $\zeta\mapsto \zeta^{-1}$? Is there a way to show it without using that the $n$-th cyclotomic polynomial is irreducible? Because showing its irreducibility is quite involved and not trivial at all. – Claudius Aug 12 '16 at 3:13
• Why are $\zeta$ and $\zeta^{-1}$ roots of the same minimal polynomial? Your last statement is obviously false, since $\zeta+1$ is another primitive element of $\mathbb Q(\zeta)$, but $\zeta\mapsto \zeta+1$ does not define an automorphism. – Claudius Aug 12 '16 at 14:17