Abstract Algebra Theory and Applications - Computer Science ...

380 CHAPTER 21 GALOIS THEORYis a normal series. Since G(E/F (ω)) andG(E/F )/G(E/F (ω)) ∼ = G(F (ω)/F )are both abelian, G(E/F ) is solvable.□Lemma 21.17 Let F be a field of characteristic zero and let E be a radicalextension of F . Then there exists a normal radical extension K of F thatcontains E.Proof. Since E is a radical extension of F , there exist elements α 1 , . . . , α r ∈K and positive integers n 1 , . . . , n r such thatwhere α n 11 ∈ F andE = F (α 1 , . . . , α r ),α n ii∈ F (α 1 , . . . , α i−1 )for i = 2, . . . , r. Let f(x) = f 1 (x) · · · f r (x), where f i is the minimal polynomialof α i over F , and let K be the splitting field of K over F . Everyroot of f(x) in K is of the form σ(α i ), where σ ∈ G(K/F ). Therefore, forany σ ∈ G(K/F ), we have [σ(α 1 )] n 1∈ F and [σ(α i )] n i∈ F (α 1 , . . . , α i−1 ) fori = 2, . . . , r. Hence, if G(K/F ) = {σ 1 = id, σ 2 , . . . , σ k }, then K = F (σ 1 (α j ))is a radical extension of F .□We will now prove the main theorem about solvability by radicals.Theorem 21.18 Let f(x) be in F [x], where charF = 0. If f(x) is solvableby radicals, then the Galois group of f(x) over F is solvable.Proof. Let K be a splitting field of f(x) over F . Since f(x) is solvable,there exists an extension E of radicals F = F 0 ⊂ F 1 ⊂ · · · F n = E. Since F iis normal over F i−1 , we know by Lemma 21.17 that E is a normal extensionof each F i . By the Fundamental Theorem of Galois Theory, G(E/F i ) is anormal subgroup of G(E/F i−1 ). Therefore, we have a subnormal series ofsubgroups of G(E/F ):{id} ⊂ G(E/F n−1 ) ⊂ · · · ⊂ G(E/F 1 ) ⊂ G(E/F ).Again by the Fundamental Theorem of Galois Theory, we know thatG(E/F i−1 )/G(E/F i ) ∼ = G(F i /F i−1 ).By Lemma 21.16, G(F i /F i−1 ) is solvable; hence, G(E/F ) is also solvable.□The converse of Theorem 21.18 is also true. For a proof, see any of thereferences at the end of this chapter.