TITLE: Finding a unique subfield of $\mathbb{Q}(\zeta)$ of degree $2$?
QUESTION [11 upvotes]: I'm trying to pick up a little bit of Galois theory to see its applications to classical geometry and constructions of regular $n$-gons, the regular $17$-gon to be exact. This problem comes from Hartshorne's Classical Geometry and appears in Section 29 on Gauss's construction of a regular $17$-gon.

Let $\zeta=\cos(2\pi/7)+i\sin(2\pi/7)$ and let $\alpha=\zeta+\zeta^{-1}$.
(a) Find the minimal polynomial for $\alpha$ over $\mathbb{Q}$.
(b) Show that $\mathbb{Q}(\zeta)$ contains a unique subfield of $E$ of degree $2$ over $\mathbb{Q}$. Find an integer $d$ for which $E=\mathbb{Q}(\sqrt{d})$.

For $(a)$, I noticed that $\alpha=\zeta+\zeta^{-1}=\zeta+\zeta^6=2\cos(2\pi/7)$. 
Also,
$$
\alpha^2=\zeta^2+2+\zeta^5
$$
$$
\alpha^3=\zeta^3+\zeta^4+3\alpha
$$
and so
$$
\alpha^3+\alpha^2+\alpha=\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+2+3\alpha
$$
and then $\alpha^3+\alpha^2-2\alpha=1$ since $\zeta$ is a root of $\Phi_7$, the seventh cyclotomic polynomial. So I believe the minimal polynomial of $\alpha$ is $x^3+x^2-2x-1$, which is irreducible by the rational roots test.
However for part (b), I'm coming up blank. I couldn't find an expression for $\zeta$ in terms of square roots, so I'm not really sure what $\mathbb{Q}(\zeta)$ looks like, nor can I hazard a guess as to what $d$ may be. I figured if $\zeta=\sqrt{10+2\sqrt{5}}$ or something, I could guess $d=5$ and attempt to work with that. Unfortunately, I'm not too keen with algebra, so my question is, how would one go about part (b)? Thanks.

REPLY [2 votes]: In this case $d = -7$. In general, if $p \equiv 1 \mod{4}$ then $\mathbb{Q}(\sqrt{p}) \subseteq \mathbb{Q}(\zeta_p)$ and if $p \equiv 3 \mod{4}$ then $\mathbb{Q}(\sqrt{-p}) \subseteq \mathbb{Q}(\zeta_p)$. There are several proofs of this fact, but the only elementary one that comes to mind involves Gauss sums: Let $g = \sum_{a=1}^6 \chi(a) \zeta_7^a$ where $\chi(a) = \pm 1$ depending on whether or not $a$ is a quadratic residue mod $7$. You can show that $g\bar{g} = 7$ but that $\bar{g} = -g$. So $g^2 = -7$. So $g = \pm \sqrt{-7}$. So $\mathbb{Q}(\zeta_7)$ contains $\mathbb{Q}(\sqrt{-7})$.