How is $e^{i2\pi /7}$ a solution to a quadratic over $\mathbb{Q}(\cos (2\pi/7))$? I'm doing some exercises in the cyclotomic extension $\mathbb{Q}_7$. I have $\omega=e^{i2\pi /7}$, and I know $\omega+\omega^6=\omega+\omega^{-1}=2\cos(2\pi/7)$. My book says that then $\omega$ is a solution to a quadratic over $\mathbb{Q}(\cos 2\pi /7)$, but I don't see why. I only see a polynomial of degree 6. Where's the quadratic?
 A: Andrea's answer tells you the quadratic equation that $\omega$ satisfies over $\mathbb{Q}(\omega+\omega^{-1})$. I will attempt to explain why this quadratic exists, via some hints (in the form of questions). These assume you're familiar with Galois theory.


*

*What is the degree $[\mathbb{Q}(\omega):\mathbb{Q}]$?

*What is the subgroup $H\subseteq \text{Gal}(\mathbb{Q}(\omega)/\mathbb{Q})$ that fixes $\omega+\omega^{-1}=\omega+\overline{\omega}$?

*What is the fixed field of $H$, i.e. $F=\mathbb{Q}(\omega)^H$?

*What is the degree $[\mathbb{Q}(\omega):F]$? 

*What does that mean $\deg(\text{Irr}(\omega,F))$ is?
(note that $\text{Irr}(\alpha,K)$ is one notation for the minimal polynomial of some $\alpha$ over a field $K$).
A: Let 
$$y=(1/2)\left(x+\frac{1}{x}\right).$$
If $x=\omega$, then $y=\cos\left(2\pi/7\right)$.
Thus $\omega$ is a solution of the equation
$$(1/2)\left(x+\frac{1}{x}\right)= \cos\left(2\pi/7\right).$$
With some manipulation (multiply through by $x$, rearrange), this equation can be written equivalently as a quadratic equation in $x$, with coefficients in $\mathbb{Q}(\cos(2\pi/7))$.
Comment: There is nothing particularly special about $7$, insofar as this calculation goes.  But there is a connection to a problem that (probably) goes back to Archimedes, namely the problem of doing a straightedge and compass construction of the regular heptagon.  The problem comes down to constructing $\cos(2\pi/7)$.  It is easy to see that $\omega$ is a solution of the equation
$$x^6+x^5+x^4+x^3+x^2+x+1=0.$$
This may be rewritten as
$$x^3+x^2+x+1+x^{-1}+x^{-2}+x^{-3}=0.$$
Let $y=(1/2)(x+x^{-1})$.
So $x+x^{-1}=2y$.  Square. We get $x^2+2+x^{-2}=4y^2$, and therefore
$x^2+x^{-2}=4y^2-2$.
Cube $x+x^{-1}$. We get $x^3+3x+3x^{-1}+x^{-3}$,
so $x^3+x^{-3}=8y^3-6y$.
So $\cos(2\pi/7)$ is a root of the equation $8y^3 +4y^2-4y-1=0$.
Now an algebraic argument much like the proof that the $60^\circ$ angle is not trisectable shows that $\cos(2\pi/7)$ is not constructible.
A: Here it is:
$$\omega^2=\omega(\omega+\omega^{-1})-1.$$
