degree of extension re: roots of unity to the $2^n$ power and their conjugates

Question

Let $\zeta_{2^{n+2}}$ be a $2^{n+2}$th root of unity, and let $\overline\zeta_{2^{n+2}}$ be its complex conjugate.

I am looking for help in showing that $[\mathbb{Q}(\zeta_{2^{n+2}}): \mathbb{Q}(\zeta_{2^{n+2}} + \overline\zeta_{2^{n+2}})] = 2$.

Since I know that $\overline\zeta_{2^{n+2}} = \zeta_{2^{n+2}}^{-1}$, I was thinking that $\zeta_{2^{n+2}}$ satisfies the polynomial $x^2 - x(\zeta_{2^{n+2}} + \overline\zeta_{2^{n+2}}) + 1$, but I don't know how to show this polynomial is irreducible over $\mathbb{Q}(\zeta_{2^{n+2}} + \overline\zeta_{2^{n+2}})$ to complete the proof. I am familiar with tools such as Eisenstein's Criterion (with transformations), mod $p$ irreducibility, etc., just not how to adapt them to extension fields of $\mathbb{Q}$ as opposed to just $\mathbb{Q}$.

Answer 1

Hint Since $\zeta_{2^{n+2}}$ satisfies the polynomial $x^2 - x(\zeta_{2^{n+2}} + \overline\zeta_{2^{n+2}}) + 1$ you have $$[\mathbb{Q}(\zeta_{2^{n+2}}): \mathbb{Q}(\zeta_{2^{n+2}} + \overline\zeta_{2^{n+2}})] \leq 2$$

To complete your proof you need to show that $[\mathbb{Q}(\zeta_{2^{n+2}}): \mathbb{Q}(\zeta_{2^{n+2}} + \overline\zeta_{2^{n+2}})] \neq 1$.

If you assume by contradiction that $[\mathbb{Q}(\zeta_{2^{n+2}}): \mathbb{Q}(\zeta_{2^{n+2}} + \overline\zeta_{2^{n+2}})] = 1$ then you get $$\mathbb{Q}(\zeta_{2^{n+2}}) = \mathbb{Q}(\zeta_{2^{n+2}} + \overline\zeta_{2^{n+2}})] $$

But this implies $ \zeta_{2^{n+2} } \in\mathbb{Q}(\zeta_{2^{n+2}} + \overline\zeta_{2^{n+2}})] \subseteq \mathbb R$ which is a contradiction.

Answer 2

I'll write $\zeta$ for your $\zeta_{2^{n+2}}$. Then $\zeta$ is not real, but $\zeta+\overline\zeta$ is. So $X^2-(\zeta+\overline\zeta)X+1$ is irreducible over $\Bbb R$ and a fortiori over $\Bbb Q(\zeta+\overline\zeta)\subseteq\Bbb R$.