- Prove that the Galois group of the polynomial \(x^{n}-1\) over \(Q\) is abelian.
Solution
1. Let \(\zeta_n\) be a primitive \(n\)-th root of unity and let\[ K=\mathbb Q(\zeta_n). \]Then \(K\) is the splitting field of \(x^n-1\) over \(\mathbb Q\).
Every root of \(x^n-1\) is a power of \(\zeta_n\), hence belongs to \(\mathbb Q(\zeta_n)\). Conversely, every splitting field contains a primitive \(n\)-th root of unity, so it contains \(\mathbb Q(\zeta_n)\).
2. For every \(\sigma\in\operatorname{Gal}(K/\mathbb Q)\), there is a unique residue class \(a_\sigma\in(\mathbb Z/n\mathbb Z)^\times\) such that\[ \sigma(\zeta_n)=\zeta_n^{a_\sigma}. \]
An automorphism preserves multiplicative order, so \(\sigma(\zeta_n)\) is again a primitive \(n\)-th root of unity. The primitive \(n\)-th roots are exactly \(\zeta_n^a\) with \(\gcd(a,n)=1\), and the exponent is unique modulo \(n\).
3. The map\[ \Theta:\operatorname{Gal}(K/\mathbb Q)\longrightarrow (\mathbb Z/n\mathbb Z)^\times, \qquad \Theta(\sigma)=a_\sigma, \]is an injective group homomorphism.
If \(\sigma(\zeta_n)=\zeta_n^a\) and \(\tau(\zeta_n)=\zeta_n^b\), then \[ (\sigma\tau)(\zeta_n) =\sigma(\zeta_n^b) =\sigma(\zeta_n)^b =\zeta_n^{ab}, \] so \(\Theta(\sigma\tau)=\Theta(\sigma)\Theta(\tau)\). If \(\Theta(\sigma)=1\), then \(\sigma(\zeta_n)=\zeta_n\). Since \(K=\mathbb Q(\zeta_n)\), this forces \(\sigma=\operatorname{id}_K\). Hence \(\Theta\) is injective.
4. Therefore \(\operatorname{Gal}(x^n-1/\mathbb Q)\) is abelian.
By 3, the Galois group is isomorphic to a subgroup of \((\mathbb Z/n\mathbb Z)^\times\), and this latter group is abelian. Every subgroup of an abelian group is abelian.