Let K be a field, K[x] an irreducible polynomial of degree n 5 and
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Let K be a field, ∫ ϵ K[x] an irreducible polynomial of degree n ≥ 5 and F a splitting field of ∫ over K. Assume that AutKF ≅ Sn. Let u be a root of ∫ in F. Then
(a) K(u) is not Galois over K; [K(u) : KJ = n and AutKK(u) = I (and hence is solvable).
(b) Every normal closure over K that contains u also contains an isomorphic copy of F.
(c) There is no radical extension field E of K such that E ⊃ K(u) ⊃ K.
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ANSWER a Ku is not Galois over K Since is irreducible it follows that Ku is isomorphic to Kx which is a finite extension of K of degree n Since AutKF ...View the full answer
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Related Book For
Algebra Graduate Texts In Mathematics 73
ISBN: 9780387905181
8th Edition
Authors: Thomas W. Hungerford
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