Show that the splitting field of \(f(x) = x^3-2\) over \(\mathbb{Q}\) is \(\mathbb{Q}(\sqrt[3]{2}, \zeta_3)\) (where \(\zeta_3 = e^{2\pi i/3}\)).
Solution
Let \(\alpha=\sqrt[3]{2}\) and let \(\zeta_3\) be a primitive cube root of unity. The roots of \[ x^3-2 \] are \[ \alpha,\qquad \zeta_3\alpha,\qquad \zeta_3^2\alpha. \] Hence \[ K=\mathbb Q(\alpha,\zeta_3) \] contains all roots, so the splitting field \(L\) is contained in \(K\).
Conversely, \(L\) contains \(\alpha\) and \(\zeta_3\alpha\), hence also \[ \zeta_3=\frac{\zeta_3\alpha}{\alpha}. \] Therefore \(K\subseteq L\), and so \[ L=\mathbb Q(\sqrt[3]{2},\zeta_3). \]
Moreover, \([\mathbb Q(\alpha):\mathbb Q]=3\) by Eisenstein at \(2\), and \(\zeta_3\notin\mathbb R\) while \(\mathbb Q(\alpha)\subset\mathbb R\). Thus adjoining \(\zeta_3\) has degree \(2\), giving \[ [L:\mathbb Q]=6. \]