Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / A / 1 / i / Solution

The splitting field in $\mathbb{C}$ is the smallest subfield containing $\mathbb{Q}$ and every root of the polynomial. The sixth roots of unity are powers of ${\zeta }_6=e^{i\pi /3}$, so the splitting field of $x^6-1$ is $\mathbb{Q}({\zeta }_6)=\mathbb{Q}(\sqrt{-3})$.

For $x^6+x^3+1=(x^9-1)/(x^3-1)$, the roots are exactly the primitive ninth roots of unity, all powers of ${\zeta }_9=e^{2\pi i/9}$. Thus its splitting field is $\mathbb{Q}({\zeta }_9)$. The supplied irreducibility makes the displayed degree-six polynomial the minimal polynomial of ${\zeta }_9$, giving