Let \(p \in \mathbb{Z}\) be a prime number. Then describe the elements of the Galois group of the polynomial \(x^{p}-2\).
Solution
Let \(\alpha=\sqrt[p]{2}\) and \(\zeta=\zeta_p\), and work over \(\QQ\).
1. The splitting field is \(L=\QQ(\alpha,\zeta)\), and \([L:\QQ]=p(p-1)\).
The roots of \(x^p-2\) are \(\alpha\zeta^a\), \(0\le a<p\), so \(L=\QQ(\alpha,\zeta)\). By Eisenstein at \(2\), \([\QQ(\alpha):\QQ]=p\), and \([\QQ(\zeta):\QQ]=p-1\). Both divide \([L:\QQ]\) and are coprime, so \(p(p-1)\mid[L:\QQ]\); and \([L:\QQ]\le[\QQ(\alpha):\QQ][\QQ(\zeta):\QQ]=p(p-1)\).
2. The elements of \(\Gal(L/\QQ)\) are the \(p(p-1)\) automorphisms\[\sigma_{a,b}\colon\alpha\mapsto\zeta^a\alpha,\qquad\zeta\mapsto\zeta^b,\qquad a\in\ZZ/p\ZZ,\ b\in(\ZZ/p\ZZ)^\times ,\]and \(\sigma_{a,b}\mapsto\begin{pmatrix}b&a\\0&1\end{pmatrix}\) is an isomorphism \(\Gal(L/\QQ)\cong\ZZ/p\ZZ\rtimes(\ZZ/p\ZZ)^\times\).
An automorphism sends \(\alpha\) to a root of \(x^p-2\) and \(\zeta\) to a primitive \(p\)th root of unity, so it is some \(\sigma_{a,b}\), and it is determined by \((a,b)\) since \(L=\QQ(\alpha,\zeta)\). By step 1 there are \(p(p-1)\) automorphisms, so every pair \((a,b)\) occurs. From \(\sigma_{a,b}\sigma_{a',b'}(\alpha)=\sigma_{a,b}(\zeta^{a'}\alpha)=\zeta^{ba'+a}\alpha\) and \(\sigma_{a,b}\sigma_{a',b'}(\zeta)=\zeta^{bb'}\), composition matches the product of the matrices.