By Jürgen Müller

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Extra resources for Algebra

Example text

An ∈ K such that f = lc(f )· i=1 (X − ai ) ∈ K[X]. Hence n ϕ ϕ ′ f ϕ = lc(f )ϕ · i=1 (X − ai ) ∈ K ′ [X], thus L′ = K ′ (aϕ 1 , . . , an ) = K , and we let ϕ = ϕ. Let d ≥ 2, and let g ∈ K[X] be irreducible such that g | f , where we may assume that deg(g) ≥ 2. Then g ϕ ∈ K ′ [X] is irreducible such that g ϕ | f ϕ . Let a ∈ L be a root of g, and let a′ ∈ L′ be a root of g ϕ , hence there is an isomorphism ϕ : K(a) → K ′ (a′ ) extending ϕ. Then L/K(a) is a splitting field [L : K] for f , and L′ /K ′ (a′ ) is a splitting field for f ϕ .

Since Gi+1 Gi for i ∈ {1, . . , m−1}, the field extension Ki+1 /Ki is Galois such that Aut(Ki+1 /Ki ) ∼ = Aut(Km /Ki )/Aut(Km /Ki+1 ) ∼ = Z/pi Z is cyclic of prime order. = Gi /Gi+1 ∼ Hence we have n ∈ lcm{p1 , . . , pm−1 }, and let ζ ∈ M be a primitive n-th root of unity. Letting Ki′ := Ki (ζ) for i ∈ {1, . . , m} yields K ⊆ K(ζ) = K1′ ⊆ K2′ ⊆ ′ · · · ⊆ Km = L(ζ) = M , where since L/K is Galois, L(ζ)/K is Galois. Since ′ ′ Ki+1 /Ki is Galois, Ki+1 /Ki and Ki+1 /Ki′ are Galois for i ∈ {1, . .

Let K be a field and let f = i≥0 ai X i ∈ K[X]. Then ∂f i−1 ∈ K[X]. Hence we := the derivative of f is defined as ∂X i≥1 iai X ∂f ∂f have ∂X = 0 or deg( ∂X ) < deg(f ), and if K ⊆ L is a field extension then the derivative of f ∈ K[X] and the derivative of f ∈ L[X] coincide. The map ∂(f g) ∂f ∂g ∂ ∂X : K[X] → K[X] is K-linear, and we have ∂X = ∂X · g + f · ∂X for all i j f, g ∈ K[X]: We may assume that f = X and g = X where i, j ∈ N0 , and i+j ) ∂1 since ∂X = 0 we may assume i, j ≥ 1, then we have ∂(X = (i + j)X i+j−1 = ∂X iX i−1 X j + jX i X j−1 = ∂(X i ) ∂X · Xj + Xi · ∂(X j ) ∂X .

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Algebra by Jürgen Müller

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