Solutions to Problem Set

Problem 3. The element 1 + 2 ∈ k[S3] is equal to (1 + ρ+ ρ2)/3. Thus M is the same as the image of (1 + ρ+ ρ2)/3, which I’ll denote by φ. Notice that if β ∈ L, then ρ(φβ) = (ρφ)β = φβ. This means that M is contained in the fixed field of ρ, i.e. M ⊆ L. Let’s show that in fact M = L. We need to prove the other inclusion, that L ⊆M . If β ∈ L, then ρ(β) = β, so φ(β) = (β + ρ(β) + ρ2(β))/3 = (β + β + β)/3 = β.