Degree of finite extension of $\mathbb{Q}$

Question

Let $\zeta = e^{2\pi i / 7}$. I know the minimal polynomial of $\zeta$ over $\mathbb{Q}$ is $\sum_{i=0}^{6} x^{i}$. But what is $[ \mathbb{Q}(\zeta) : \mathbb{Q}(\zeta) \cap \mathbb{R}]$. I saw that it is $2$ but I can't find the minimal polynomial of degree $2$. We have that $a \zeta ^{2} + b \zeta + c = 0$, where $a,b,c \in \mathbb{Q}(\zeta) \cap \mathbb{R}$. I know that $\cos(2 \pi n / 7) \in \mathbb{Q}(\zeta) \cap \mathbb{R}$, and I want that $a \cdot \sin(4 \pi / 7) + b \cdot \sin ( 2 \pi /7 ) = 0 : a,b \in \mathbb{Q}(\zeta) \cap \mathbb{R}$. But I can't seem to find it.

Answer 1

You want a quadratic polynomial with real coefficients where $\zeta$ is a root. That means that its complex conjugate $\zeta^6$ must be the other root. Thus Vieta's formulas tells you that the polynomial you're after is $$ x^2-(\zeta+\zeta^6)x+\zeta\cdot\zeta^6\\ =x^2-2\cos(2\pi/7)x+1 $$

Answer 2

For an odd prime, the cyclotomic field $K=\mathbf Q(\zeta_p)$ has cyclic Galois group of order $\frac {p-1}2$, hence admits a unique subextension $L$ s.t. $[K:L]=2$, which is the subfield fixed by "complex conjugation", which is $L=K\cap \mathbf R=\mathbf Q(\zeta_p + {\zeta_p}^{-1})$. Then the minimal polynomial of $\zeta_p$ over $L$ is obviously $X^2 -(\zeta_p + {\zeta_p}^{-1})X+\zeta_p {\zeta_p}^{-1}=X^2 - 2cos (2\pi/p)+1$.