An Introduction to Galois Theory, Edition: version 23 Jan by Andrew Baker

By Andrew Baker

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Find expressions for ζ5 +ζ5−1 and ζ52 +ζ5−2 in terms of cos(2π/5). Hence find a rational polynomial which has cos(2π/5) as a root. 1-25. Let p > 0 be a prime and K be a field with char K = p. (a) Show that if ζ ∈ K is a p-th root of 1 then ζ = 1. Deduce that if m, n > 0 and p n, then every npm -th root of 1 in K is an n-th root of 1. (b) If a ∈ K, show that the polynomial X p − a ∈ K[X] has either no roots or exactly one root in K. 1. 1. Definition. Let K and L be fields and suppose that K ⊆ L is a subring.

These mappings have images √ √ 3 3 α0 Q( 2) = Q( 2), √ √ 3 3 α1 Q( 2) = Q( 2 ζ3 ), 43 √ √ 3 3 α2 Q( 2) = Q( 2 ζ32 ). 37. Proposition. Let F/K and L/K be extensions. (i) For p(X) ∈ K[X], each monomorphism α ∈ MonoK (L, F ) restricts to a function αp : Roots(p, L) −→ Roots(p, F ) which is an injection. (ii) If α ∈ MonoK (L, L), then αp : Roots(p, L) −→ Roots(p, L) is a bijection. Proof. (i) For u ∈ Roots(p, L) we have p(α(u)) = α(p(u)) = α(0) = 0, so α maps Roots(p, L) into Roots(p, F ). Since α is an injection its restriction to Roots(p, L) ⊆ L is also an injection.

70, [M : L][L : K] = (M : L)(L : K). This can only happen if [M : L] = (M : L) and [L : K] = (L : K), since (M : L) [M : L] and (L : K) [L : K]. 72 this implies that L/K and M/L are separable. Conversely, if L/K and M/L are separable then [M : L] = (M : L) and [L : K] = (L : K), hence [M : K] = [M : L][L : K] = (M : L)(L : K) = (M : K). Therefore M/K is separable. 6. 10 that a finite extension L/K is simple if there is an element u ∈ L for which L = K(u), and such an element is called a primitive element.

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