What is the Galois group of \(x^n-1\) over \(\QQ\)?
Solution
Let \(\zeta_n\) be a primitive \(n\)th root of unity. The splitting field is \[ \QQ(\zeta_n). \] Every \(\QQ\)-automorphism is determined by the image of \(\zeta_n\), and it must send \(\zeta_n\) to another primitive \(n\)th root, \[ \zeta_n\longmapsto \zeta_n^a, \qquad a\in(\ZZ/n\ZZ)^\times. \] Conversely, each such \(a\) defines an automorphism. Therefore \[ \operatorname{Gal}(x^n-1/\QQ) \cong \operatorname{Gal}(\QQ(\zeta_n)/\QQ) \cong (\ZZ/n\ZZ)^\times. \] In particular the Galois group is abelian of order \(\varphi(n)\).