Show that the splitting field of \(f(x) = x^3-2\) over \(\mathbb{Q}\) is \(\mathbb{Q}(\sqrt[3]{2}, \zeta_3)\).
Solution
Let \[ \alpha=\sqrt[3]{2}, \qquad \zeta_3=e^{2\pi i/3}. \] The three roots of \(x^3-2\) are \[ \alpha,\qquad \alpha\zeta_3,\qquad \alpha\zeta_3^2. \] Hence all roots lie in \(\mathbb Q(\alpha,\zeta_3)\), so the splitting field \(L\) satisfies \[ L\subseteq\mathbb Q(\alpha,\zeta_3). \] Conversely, \(L\) contains \(\alpha\) and \(\alpha\zeta_3\), so \[ \zeta_3=\frac{\alpha\zeta_3}{\alpha}\in L. \] Thus \(\alpha,\zeta_3\in L\), giving the reverse inclusion. Therefore \[ \boxed{L=\mathbb Q(\sqrt[3]{2},\zeta_3).} \]
The polynomial \(x^3-2\) is Eisenstein at \(2\), so \([\mathbb Q(\alpha):\mathbb Q]=3\). Since \(\mathbb Q(\alpha)\subset\mathbb R\) while \(\zeta_3\notin\mathbb R\), adjoining \(\zeta_3\) has degree \(2\). Hence \[ [L:\mathbb Q]=6. \]