Find an irreducible polynomial with zero $\cos(2\pi/7)$

Question

I have a test that asks to find $\operatorname{Irr}(\cos(2\pi/7),\mathbb Q)$.

So I need $f \in \mathbb Q[x]$, $f(\cos(2\pi/7))=0$ and $f$ irreducible.

How I can solve that, some hint?

Answer 1

Hint: Combine the following facts that you should know (if you don't, then you have more reviewing to do). Here $\zeta=e^{2\pi i/7}$.

1. The number $\zeta$ is a zero of the irreducible polynomial $$ p(x)=\frac{x^7-1}{x-1}=x^6+x^5+x^4+x^3+x^2+x+1. $$ Irreducibility can be shown by applying Eisenstein to $p(x\pm1)$ - don't remember which sign works. Also, observe that $$ \zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}=\zeta^{-3}p(\zeta). $$
2. We have $$ \zeta+\zeta^{-1}=2\cos\frac{2\pi}7. $$ Spend some quality time with the cube and square of this!