By Wilkins D.R.

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Now i does not belong to Q(ξ), since Q(ξ) ⊂ R. Therefore the polynomial x2 + 1 is the minimum polynomial of i over Q(ξ). 21 now shows that [L: Q(ξ)] = [Q(ξ, i): Q(ξ)] = 2. 18) that [L: Q] = [L: Q(ξ)][Q(ξ): Q] = 8. 46). Another application of the Tower Law now shows that [L: Q(i)] = 4, since [L: Q] = [L: Q(i)][Q(i): Q] and [Q(i): Q] = 2. 21). But ξ is a root of x4 −2. Therefore x4 −2 is irreducible over Q(i), and is the minimum polynomial of ξ over Q(i). 31 then ensures the existence of an automorphism σ of L that sends ξ ∈ L to iξ and fixes each element of Q(i).

50), and therefore [M : K] = |Γ(L: K)/H| = p. 57 that M = K(α) for some element α ∈ M satisfying αp ∈ K. 55). The induction hypothesis ensures that f is solvable by radicals when considered as a polynomial with coefficients in M , and therefore the roots of f lie in some extension field of M obtained by successively adjoining radicals. But M is obtained from K by adjoining the radical α. Therefore f is solvable by radicals, when considered as a polynomial with coefficients in K, as required. 58, we see that a polynomial with coefficients in a field K of characteristic zero is solvable by radicals if and only if its Galois group ΓK (f ) over K is a solvable group.

Now, given any root αi of f , there exists some σ ∈ G such that αi = σ(α). Thus if g ∈ K[x] is a polynomial with coefficients in K which satisfies g(α) = 0 then g(αi ) = σ(g(α)) = 0, since the coefficients of g are fixed by σ. But then f divides g. Thus f is the minimum polynomial of α over K, as required. Definition A field extension is said to be a Galois extension if it is finite, normal and separable. 45 Let L be a field, let G be a finite subgroup of the group of automorphisms of L, and let K be the fixed field of G.