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By Jürgen Müller

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Since FixL (τ ) ∩ FixL (σ) = M ∩ K = Q and L = Q(M, K), we get τ, σ = G and τ ∩ σ = {1}, respectively. We have τ σ : ζ → ζ, ρ → ζ −1 ρ, hence τ σ = τ −1 . Thus we have G ∼ = D8 . From the subgroup lattice of G we determine all intermediate fields of L/Q: We have FixL ( σ, τ 2 ) = K ∩ Q(ζ, ρ2 ) = Q(ρ2 ) and FixL (στ 2 ) = K ′ , as well as FixL ( στ, τ 2 ) = Q(ζρ2 ). To find FixL (στ ) and FixL (στ 3 ) we use the trace map Tr στ = id + στ : L → FixL (στ ): We have Tr στ (ζ) = ζ + ζ −1 = 0 as well as Tr στ (ρ) = ρ + ζρ = 2 (1 + ζ)ρ =: ω.

There is a separable radical extension M/K such that f splits in M [X], then Aut(L/K) is soluble. b) If char(K) | [L : K] and Aut(L/K) is soluble then f is solvable by radicals. Proof. a) There is a Galois radical extension M/K such that f splits in M [X], where Aut(M/K) is soluble. There is a splitting field L/K for f such that L ⊆ M , hence Aut(L/K) ∼ = Aut(M/K)/Aut(M/L) is soluble. b) There is a Galois radical extension M/K such that L ⊆ M . 10) Symmetric polynomials. a) Let K be a field, let n ∈ N, and let n X = {X1 , .

Since ∂(X∂X−1) = nX n−1 = 0 ∈ Fp [X], thus 1 ∈ gcd(X n − 1, nX n−1 ), this is a contradiction. ♯ We thus have µζn = Φn , hence [Q(ζn ) : Q] = ϕ(n). Since Φn splits in Q(ζn )[X], we conclude that Q(ζn ) ⊆ C is the splitting field for Φn in C, hence Q(ζn )/Q is Galois. Thus we have |Aut(Q(ζn )/Q)| = ϕ(n) = |(Z/nZ)∗ |, and for any k ∈ (Z/nZ)∗ there is an automorphism ϕk : Q(ζn ) → Q(ζn ) : ζn → ζnk extending idQ ; in particular ϕ−1 is the restriction of complex conjugation to Q(ζn ). Hence (Z/nZ)∗ → Aut(Q(ζn )/Q) : k → ϕk is an isomorphism.

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