Let $\alpha = \zeta_7 + \zeta_7^{-1}$. Find the splitting field of $\alpha$ over $\mathbb{Q}$.

Question

Let $\alpha = \zeta_7 + \zeta_7^{-1}$. I want to find the splitting field of $\alpha$ over $\mathbb{Q}$. I know that $\operatorname{minpoly}_Q (\alpha)$ is $x^3+x^2-2x-1$. So $\mathbb{Q}(\alpha)$ is degree $3$ extension of $\mathbb{Q}$. I can factor $x^3+x^2-2x-1 = \left(x - (\zeta_7 + \frac{1}{\zeta_7})\right) \left(x^2 + (1+\zeta_7 + \frac{1}{\zeta_7})x + \frac{1}{\zeta_7 + \frac{1}{\zeta_7}}\right)$ but then im unable to further factorize the RHS term. What can I do from here?

Answer 1

For an odd prime $p$, let $\alpha_p =\zeta_p + \zeta_p^{-1}$. It is known that the cyclotomic field $K=\mathbf Q(\zeta_p)$ is a normal extension of $\mathbf Q$, with cyclic Galois group $G\cong (\mathbf Z/p)^*$. The subgroup of $G$ generated by the automorphism $\zeta_p \to \zeta_p^{-1}$ ("complex conjugation") is the unique subgroup of order $2$, and its fixed field $K$ is the unique subfield $L$ s.t. $K/L$ is quadratic. By definition, $\alpha_p \in L$. Moreover, $\zeta_p$ is obviously a root of the polynomial $X^2 -\alpha_p X+1$, so $K/\mathbf Q(\alpha_p)$ is quadratic, and $L=\mathbf Q(\alpha_p)$ and is the splitting field of $\alpha_p$. This is a very classical result from Galois theory, so I wonder whether there isn't some additional question.

Answer 2

The conjugated roots of $\alpha=2\cos\frac{2\pi}{7}$ are $2\cos\frac{4\pi}{7}$ and $2\cos\frac{6\pi}{7}$, so $\mathbb{Q}(\alpha)$ is the splitting field of $\alpha$ over $\mathbb{Q}$, by the cosine duplication and triplication formulas.