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Before prove the statement, we recall the following facts given , for example, in "Foundations of Galois Theory", Publishers of Physics and Mathematics Literature, Moscow, 1963.
(a) Let F⊂L be a normal extension. There exists a normal radical extension F⊂K such that K contains L;
(b) Let F⊂L be a normal extension, and let F⊂K be a normal radical extension. Then the Galois group G(L,F) is solvable;
(c) Let F⊂L and L⊂K be normal extensions. Then F⊂K is normal if and only if there exists a polynomial p(x) ∈ F[x], such that it's splitting field over L is K.
(d) Let F⊂L1 be a normal extension, and F⊂L2 be any extension. Let K be a composite of L1 and L2 (i.e. the smallest subfield containing both L1 and L2). Assume that F⊂K is normal. Then the Galois group of G(K, L2) is isomorphic of some subgroup of G(L1,F).
(e) Let F⊂L be a normal extension of the degree n, and let the Galois group G(L,F) be cyclic. Let the field F contains a primitive root of 1 of the degree n. Then F⊂L is a radical extension.
Let us now prove that the following conditions are equivalent.
(i) The roots of f are solvable;
(ii) The splitting field K of f is contained in some normal radical extension of F;
(iii) The Galois group of f is solvable.
(i)⇒(ii): The field K obviously is a normal extension of F. (a) implies that K is contained in some normal radical extension of F.
(ii)⇒(i) is obvious.
(ii)⇒(iii) follows immediately from (b).
(iii)⇒(ii): Let us first consider the case, where the Galois group G(K,F) is cyclic. Let [K:F]=m, and let ε be a primitive root of 1 of the order m. Consider the field K(ε). Obviously K(ε) is the splitting field of xm−1 over K, and hence the extension K⊂K(ε) is normal. From (c) it follows that the extension F⊂K(ε) is ...
The solution shows that the roots are solvable and discusses a solvable group.