- Identify all of the elements of the Galois group of \(x^p-2\) for \(p\) an odd prime (note: this has a complicated presentation).
Solution
Let \(\alpha=2^{1/p}\) and \(\zeta=\zeta_p\). The roots of \(x^p-2\) are \(\alpha\zeta^a\) for \(a\in\mathbb F_p\), and the splitting field is \[ K=\mathbb Q(\alpha,\zeta). \] Eisenstein at \(2\) gives \([\mathbb Q(\alpha):\mathbb Q]=p\), while \([\mathbb Q(\zeta):\mathbb Q]=p-1\). Their intersection has degree dividing both \(p\) and \(p-1\), hence is \(\mathbb Q\). Therefore \[ [K:\mathbb Q]=p(p-1). \]
Every \(\sigma\in G=\operatorname{Gal}(K/\mathbb Q)\) is determined by \[ \sigma(\alpha)=\alpha\zeta^a,\qquad \sigma(\zeta)=\zeta^b, \] with \(a\in\mathbb F_p\) and \(b\in\mathbb F_p^\times\). There are exactly \(p(p-1)=|G|\) such pairs, hence every pair occurs. Writing the corresponding automorphism as \(\sigma_{a,b}\), composition is \[ \sigma_{a_1,b_1}\sigma_{a_2,b_2} =\sigma_{a_1+b_1a_2,\,b_1b_2}. \] Thus \[ G\cong \mathbb F_p\rtimes\mathbb F_p^\times =\operatorname{AGL}_1(\mathbb F_p). \]
If \(g\) generates \(\mathbb F_p^\times\), set \(\tau=\sigma_{1,1}\) and \(\omega=\sigma_{0,g}\). Then \[ \tau^p=1,\qquad \omega^{p-1}=1,\qquad \omega\tau\omega^{-1}=\tau^g, \] so \[ G=\langle \tau,\omega\mid \tau^p=\omega^{p-1}=1,\ \omega\tau\omega^{-1}=\tau^g\rangle. \] These \(p(p-1)\) automorphisms are all elements of the Galois group.