Example 2

\* Let f = X³ + X² − 2X − 1 ∈ Q[X]. The polynomial f is irreducible, for if not, then it would have a root x ∈ Q; on writing x = a/b with a, b ∈ Z, a and b relatively prime, we should find a(a² + ab − 2b²) = b³ and a³ = b(b² + 2ab − a²); but this implies that a divides b and b divides a, hence x = ±1, which is impossible. Put ξ = e^{2\pi i / 7} ∈ C; then the polynomial f has the roots α = ξ + ξ⁻¹, β = ξ² + ξ⁻², γ = ξ³ + ξ⁻³. We have β = a² − 2 and γ = a³ − 3α, hence the extension Q(α) is Galois over Q. The Galois group of Q(α) over Q is cyclic of order 3 and is generated by an element σ such that σ(α) = β, σ(β) = γ, σ(γ) = α. \*