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Proof. (Gf solvable =⇒ f solvable). Let f ∈ F [X] have solvable Galois group. Let F = F [ζ] where ζ is a primitive nth root of 1 for some large n—for example, n = (deg f)! will do. The lemma shows that the Galois group G of f as an element of F [X] is a subgroup of Gf , and hence is solvable. This means that there is a sequence of subgroups G = Gm ⊃ Gm−1 ⊃ · · · ⊃ G1 ⊃ G0 = {1} such that each Gi is normal in Gi+1 and Gi+1 /Gi is cyclic (even of prime order, but we don’t need this). Let E be a splitting field of f(X) over F , and let Fi = E Gi .

Write the minimum polynomial of α as f(X) = X n + a1 X n−1 + · · · + an = (X − αi ). Then cα (X) = (f(X))m = X mn + ma1X mn−1 + · · · + am n, and so TrE/F (α) = −ma1 = m αi , FIELDS AND GALOIS THEORY 51 and NmE/F (α) = (−1)mn am n = ( αi )m . 35. (a) Consider the extension C ⊃ R. If α ∈ C \ R, then cα (X) = f(X) = X 2 − 2 (α)X + |α|2 . If α ∈ R, then cα (X) = (X − a)2. (b) Let E = Q[α, i] be the splitting field of X 8 − 2 (see Exercise 16). The minimum √ polynomial of α = 8 2 is X 8 − 2, and so TrQ[α]/Q α = 0; NmQ[α]/Q α = −2; TrE/Q α = 0.

S. MILNE 5. Applications of Galois Theory In this section, we apply the Fundamental Theorem of Galois Theory to obtain other results about polynomials and extensions of fields. 1. Primitive element theorem. Recall that a finite extension of fields E/F is simple if E = F [α] for some element α of E. Such an α is called a primitive element of E. We shall show that (at least) all separable extensions have primitive elements. √ √ Consider for example Q[ 2, 3]/Q. We know (see Exercise 13) that its Galois group over Q is a 4-group < σ, τ >, where √ √ √ √ σ √2 = − 2 2 = 2 τ √ √ √ , σ 3 = τ 3 = − 3.

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