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

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 $$\mathbb{Q}$$.

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.
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 $$\Bbb Q(\zeta+\overline\zeta)\subseteq\Bbb R$$.