D+F 13.4.4

Determine the splitting field and its degree over $\mathbb{Q}$ for $x^6-4$.

Splitting fields are the smallest field extensions which allow $f(x)$ to factor/split completely. $$x^6-4=(x^3)^2-(2)^2=(x^3+2)(x^3-2)$$ $$=(x+\sqrt[3]{2})(x-\sqrt[3]{2}\zeta_6)(x-\sqrt[3]{2}\zeta_6^5)(x-\sqrt[3]{2})(x-\sqrt[3]{2}\zeta_6^2)(x-\sqrt[3]{2}\zeta_6^4)$$ Where $\zeta_6=\frac{1+\sqrt{-3}}{2}$. These are the 6th roots of unity scaled by $\sqrt[3]{2}$. So, in order for all 6 of these roots to be contained in an extension of $\mathbb{Q}$, we need to adjoin $\sqrt{-3}$ and $\sqrt[3]{2}$. So, the splitting field of $x^6-4$ is $\mathbb{Q}[\sqrt[3]{2},\sqrt{-3}]$. Now, $|\mathbb{Q}[\sqrt{-3}]:\mathbb{Q}|=2$ and $|\mathbb{Q}[\sqrt[3]{2}]:\mathbb{Q}|=3$. By corollary 22 p. 529, $$|\mathbb{Q}[\sqrt[3]{2},\sqrt{-3}]:\mathbb{Q}|=2 \cdot 3 = 6$$