Let F - minimal subfield in $\mathbb{C}$, such that $\mathbb{Q} \subseteq F$ and $\{x \in \mathbb{C} | x^4 - x^2 +1 = 0 \} \subseteq F$. Find the degree of the extension $[F : \mathbb{Q}]$.
I understand that $p(x):=x^4-x^2+1\implies x^2_{1,2}=\frac{1\pm\sqrt 3\,i}2=e^{\pm\pi i/3}\implies x_{1,2,3,4}=\pm w_1:=\pm e^{\pi i/6},\,\pm w_2:=\pm e^{-\pi i/6}$ and the answer is 4. But I don't quite understand how to prove that
Assuming $x^4-x^2+1$ is irreducible, take one of its roots, like $w_1$. Note that $F$ is the smallest subfield of $\Bbb C$ that contains $w_1$ and $\Bbb Q$, as any subfield of $\Bbb C$ that contains $w_1$ also automatically contains $-w_1$ and $\pm w_2=\pm w_1^5$.
Since the minimal polynomial $x^4-x^2+1$ of $w_1$ over $\Bbb Q$ has degree $4$ (we know it's minimal because it's irreducible), this means that $F$, as a vector space over $\Bbb Q$, is spanned by $1,w_1,w_1^2,w_1^3$ (any higher power of $w_1$ can be reduced to a lower power using the minimal polynomial). By definition, this gives $[F:\Bbb Q]=4$.