D+F 13.4.1

Determine the splitting field and its degree over $\mathbb{Q}$ for $x^4-2$.

$\textit{Proof}$ : The four roots of $x^4-2$ can found by factoring: $$x^4-2=(x^2)^2-(\sqrt{2})^2=(x^2-\sqrt{2})(x^2+\sqrt{2})$$ $$=(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-\sqrt[4]{2}i)(x+\sqrt[4]{2}i)$$ These four roots can be thought of as the 4th roots of unity scaled by $\sqrt[4]{2}$.
In this factorization, we need to adjoin $\sqrt[4]{2}$ and $i$. Now, $$\mathbb{Q}[\sqrt[4]{2}]=\{a_0+a_1\sqrt[4]{2}+a_2\sqrt{2}+a_3\sqrt[4]{2}^3|a_i\in{\mathbb{Q}}\}$$ so, $$|\mathbb{Q}[\sqrt[4]{2}]:\mathbb{Q}|=4$$ But the 2 complex roots $\sqrt[4]{2}i$ and $-\sqrt[4]{2}i$ are not in $\mathbb{Q}[\sqrt[4]{2}]$, so we need to also adjoin $i$. Now, $$|\mathbb{Q}[\sqrt[4]{2}][i]:\mathbb{Q}[\sqrt[4]{2}]|=2$$ So, $$|\mathbb{Q}[\sqrt[4]{2}][i]:\mathbb{Q}|=2 \cdot 4 = 8$$ Thus, the splitting field of $x^4-2$ is a degree 8 extension.