Is \(\mathbb{Q}(\sqrt[3]{21})\) normal? What is its splitting field? What is its Galois group? Describe the lattice of subfields.
Solution
Let \(\alpha=\sqrt[3]{21}\) and let \(\omega\) be a primitive cube root of unity. The polynomial \[ f(x)=x^3-21 \] is irreducible over \(\mathbb Q\) by Eisenstein at \(3\) (or \(7\)). Its roots are \(\alpha,\omega\alpha,\omega^2\alpha\).
Since \(\mathbb Q(\alpha)\subset\mathbb R\), it contains only the real root, so it is not normal over \(\mathbb Q\). The splitting field is \[ K=\mathbb Q(\alpha,\omega)=\mathbb Q(\alpha,\sqrt{-3}). \] Now \([\mathbb Q(\alpha):\mathbb Q]=3\) and \(\omega\notin\mathbb R\), so \([K:\mathbb Q(\alpha)]=2\); hence \([K:\mathbb Q]=6\).
The discriminant of \(x^3-21\) is \[ -27\cdot 21^2, \] which is not a square in \(\mathbb Q\). Therefore the Galois group of the irreducible cubic is \[ \operatorname{Gal}(K/\mathbb Q)\cong S_3. \] One may take generators \[ \sigma(\alpha)=\omega\alpha,\quad \sigma(\omega)=\omega, \qquad \tau(\alpha)=\alpha,\quad \tau(\omega)=\omega^2, \] with \(\sigma^3=\tau^2=1\) and \(\tau\sigma\tau=\sigma^{-1}\).
By Galois correspondence, the intermediate fields are: \[ \mathbb Q, \quad \mathbb Q(\omega), \quad \mathbb Q(\alpha),\quad \mathbb Q(\omega\alpha),\quad \mathbb Q(\omega^2\alpha), \quad K. \] The quadratic field \(\mathbb Q(\omega)=\mathbb Q(\sqrt{-3})\) corresponds to \(A_3\), and the three cubic fields correspond to the three subgroups of order \(2\) in \(S_3\).