How can I prove that $[\mathbb{Q}(\omega + \omega^{-1}) :\mathbb{Q}]=\frac{\varphi (n)}{2}$?

Question

If $n>2$, $\omega \in \mathbb{C}$ an $n$-th primitive root of unity then $[\mathbb{Q}(\omega + \omega^{-1}) :\mathbb{Q}]=\frac{\varphi (n)}{2}$. ($\varphi (n)$ is the Euler totient function.)

How can I prove it?

Answer 1

We have $$ \phi(n)=[\mathbb{Q}(\omega):\mathbb{Q}]=[\mathbb{Q}(\omega):\mathbb{Q}(\omega +\omega^{-1})][\mathbb{Q}(\omega +\omega^{-1}):\mathbb{Q}], $$ where $[\mathbb{Q}(\omega):\mathbb{Q}(\omega +\omega^{-1})]=2$, since $t^2-(\omega+\omega^{-1})t+1$ is the minimal polynomial of $\omega$ over $\mathbb{Q}(\omega+\omega^{-1})$. It follows that $$[\mathbb{Q}(\omega +\omega^{-1}):\mathbb{Q}]=\frac{\phi(n)}{2}.$$