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

The splitting field is the smallest subfield of $\mathbb{C}$ containing $\mathbb{Q}$ and all the roots of a polynomial under consideration. Equivalently it is generated over $\mathbb{Q}$ by those roots, and the polynomial splits there into linear factors.

Put ${\zeta }_6=e^{\pi i/3}$. All sixth roots of unity are powers of ${\zeta }_6$, so the required splitting field isIndeed ${\zeta }_6=(1+\sqrt{-3})/2$, and conversely $\sqrt{-3}=2{\zeta }_6-1$. Its field extension degree is two.

The polynomial identityshows that the roots are the ninth roots of unity which are not third roots. These are precisely the primitive ninth roots of unity, all powers of ${\zeta }_9=e^{2\pi i/9}$. Thus the splitting field is $\mathbb{Q}({\zeta }_9)$. The supplied irreducibility statement makes the displayed degree-six polynomial the minimal polynomial of ${\zeta }_9$, giving

Write $g(x)=x^6+x^3+1$. The second polynomial is $g(-x)$. Negating the roots of a polynomial does not change the field generated by them: each original root is minus a new root, and conversely. Hence this splitting field is again $\mathbb{Q}({\zeta }_9)$ and its field extension degree is six. Equivalently, it is the cyclotomic field $\mathbb{Q}({\zeta }_{18})=\mathbb{Q}({\zeta }_9)$, since ${\zeta }_{18}=-{\zeta }_9^5$.