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M − 1} such that n = km + l; hence we have k gcd(n, m) = gcd(m, l) ⊆ Z. Then X n − 1 = (X m − 1) · i=1 X m(k−i)+l + (X l − 1) ∈ Z[X] implies gcd(q n − 1, q m − 1) = gcd(q m − 1, q l − 1) ⊆ Z and thus gcd(q n − 1, q m − 1) = q d − 1, where 0 < d ∈ gcd(n, m). Writing K ∗ as the disjoint union of its conjugacy classes, where T ⊆ K ∗ is a set of representatives, we from Z(K)∗ = {a ∈ T ; na = n} get |K ∗ | = ∗ n | n |Z(K)∗ | + a∈T ,na

Since ω τ √ = ζ 2 ω = ω we have ω ∈ Q(ζρ2 ), hence FixL (στ ) = 2π −1 Q(ω); since ζ8 := exp( 8 ) = √12 · (1 + ζ) ∈ C we have ω = ζ8 ρ3 ∈ C, hence ω 4 = −8, thus µω = X 4 + 8 ∈ Q[X]. Since στ 3 = (στ )σ , we have FixL (στ 3 ) = FixL (στ )σ = Q(ω σ ) = Q((1 − ζ)ρ). 9) Finite fields. Let p ∈ N be a prime and n ∈ N. Since Fpn is a splitn ting field for X p − X ∈ Fp [X], the field extension Fpn /Fp is Galois, and we have |Aut(Fpn /Fp )| = [Fpn : Fp ] = n. Let the Frobenius automorphism k ϕp ∈ Aut(Fpn /Fp ) have order k | n.

M − 1}; radical extensions are finite, and Kummer extensions are radical extensions. If L/K is a separable radical extension then the ni can be chosen such that char(K) | ni , for all i ∈ {1, . . , m − 1}: By choosing intermediate fields appropriately we may assume that Ki+1 = Ki and that ni is a prime, hence assuming ni = char(K) implies X ni − ai = X ni − bni i = (X − bi )ni , contradicting the separability of Ki+1 /Ki . If L/K is a radical extension such that char(K) | ni for i ∈ {1, . .