Finding a unique subfield of $\mathbb{Q}(\zeta)$ of degree $2$? 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.
 A: This is just a remark to the effect that Arturo's answer, although mostly correct, contains a few inaccuracies. Specifically, the field $\mathbf{Q}(\zeta)$ is of degree $6$, and so one needs to be careful writing elements in terms of the (seven) elements $\zeta^i$ for $i = 0$ to $6$. It would probably be better to write elements of $\mathbf{Q}(\zeta)$ as $a + b \zeta + c \zeta^2 + d \zeta^3 + e \zeta^4 + f \zeta^5$ and replace all occurrences of $\zeta^6$ by
$$\zeta^6 = -1 - \zeta - \zeta^2  -\zeta^3 - \zeta^4 - \zeta^5.$$
A: 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 $\mathbb{Q}(\sqrt{-7})$.
A: Since $\zeta$ satisfies $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$, which is irreducible over $\mathbb{Q}$ (e.g., do Eisenstein to $p(x+1)$), then $\mathbb{Q}(\zeta)$ is of degree $6$ over $\mathbb{Q}$. Moreover, since the "other" roots of this polynomial are $\zeta^i$, $i=2,\ldots,6$, then $\mathbb{Q}(\zeta)$ is the splitting field of this polynomial, and the Galois group is completely determined by what it does to $\zeta$. Since $[\mathbb{Q}(\zeta)\colon\mathbb{Q}]=6$, the Galois group is either isomorphic to $S_3$ or to $\mathbb{Z}_6$ (cyclic of order $6$). Either way, there is a unique subgroup of order $3$, whose fixed field is a subextension of degree $2$ over $\mathbb{Q}$. This is what you are looking for.
Some corrections below to using a basis for the extension.
Now, the elements of $\mathbb{Q}(\zeta)$ are of the form
$$a + b\zeta + c\zeta^2 + d\zeta^3 + e\zeta^4 + f\zeta^5 +g\zeta^6,\qquad a,b,c,d,e,f,g\in\mathbb{Q},$$
which can be rewritten to be in the form
$$a + b\zeta + c\zeta^2 + d\zeta^3 + e\zeta^4 + f\zeta^5$$
(for other rationals) using the relation $\zeta^6 = -1-\zeta-\zeta^2-\zeta^3-\zeta^4-\zeta^5$; the expression is then unique.
In fact, the Galois group is isomorphic to $\mathbb{Z}_6$: the map that sends $\zeta$ to $\zeta^3$ has order $6$ and determines an automorphism. The subgroup of order $3$ is generated by the square of this element, which sends $\zeta$ to $\zeta^2$. The subextension of degree $2$ is the fixed field of this automorphism. This automorphism maps
$$\begin{align*}
a+b\zeta+c\zeta^2 + d\zeta^3 + e\zeta^4 + f\zeta^5 &\longmapsto a + b\zeta^2 + c\zeta^4 + d\zeta^6+ e\zeta + f\zeta^3\\
&= a + b\zeta^2 + c\zeta^4 + d(-1-\zeta-\zeta^2-\zeta^3-\zeta^4-\zeta^5) + e\zeta + f\zeta^3\\
&= (a-d)+ (e-d)\zeta + (b-d)\zeta^2 + (f-d)\zeta^3 + (c-d)\zeta^4  -d\zeta^5.
\end{align*}$$
If the element is fixed by this map, we must have $d=0$ (so $a=a-d$), hence $f=0$, and $b=c=e$. 
So an element is fixed by this map if and only if it is of the form
$$a + b(\zeta + \zeta^2 + \zeta^4),\qquad a,b\in\mathbb{Q}.$$
Any such element which is not in $\mathbb{Q}$ will generate the extension. So taking $$\zeta+\zeta^2+\zeta^4$$ will do.
Now you can do the same sort of thing you did in part (a). If you square $\zeta+\zeta^2+\zeta^4$, you get
$$\begin{align*}
(\zeta+\zeta^2+\zeta^4)^2 &= \zeta^2 + \zeta^4 + \zeta + 2\zeta^3 + 2\zeta^5 + 2\zeta^6\\
&= -1 + \zeta^3+\zeta^5+\zeta^6\\
 &= -2-(\zeta+\zeta^2+\zeta^4);
\end{align*}$$
thus you have that $\beta=\zeta+\zeta^2+\zeta^4$ satisfies the polynomial
$$x^2 + x + 2.$$
This polynomial is irreducible over $\mathbb{Q}$, and its discriminant is $-7$. So the splitting field is given by $\mathbb{Q}(\sqrt{-7})$. This is the field you want.
