Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / A (Introduction to Fields)

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

Replacing $x$ by $-x$ in the irreducible polynomial from part (i) gives $x^6-x^3+1$. This substitution is an invertible automorphism of $\mathbb{Q}[x]$, so it preserves irreducibility. The roots are the primitive eighteenth roots, and the splitting field is $\mathbb{Q}({\zeta }_{18})$. They are all powers of one such root, so its irreducible degree-six polynomial givesIn fact this is the same field as in part (i): ${\zeta }_9={\zeta }_{18}^2$ and ${\zeta }_{18}=-{\zeta }_9^5$.