By Paul J. McCarthy

ISBN-10: 0486666514

ISBN-13: 9780486666518

*Math.*experiences. Over 2 hundred workouts. Bibliography.

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If m = [K1 : L] then m divides n and therefore L contains a primitive mth root of unity. Then by the first part of the proof there is a radical tower L = F0 ~ F1 ~ • • • £ Fr = K1 . But then k s; k(') = L = F0 £ F 1 ~ • • • £ Fr = K 1 is a radical tower over k and since K £ K 1 it follows that f(x) = 0 is solvable by radicals. Proof The condition is sufficient by the definition of solvable group and necessary by virtue of the lemma. II 2. Let G be a finite solvable group and H a subgroup of G. Then His solvable.

We have A' = {a E A aa(a)-1 = 1 for all a E G} = 14. Let k be a field that contains a primitive nth root of unity. Then there is a one-one correspondence between the subgroups H of k* which contain k*n and for which the index (H: k*n) is finite and the finite Abelian extensions of exponent n of k in C. The correspondence is Sl Multiplicative Kummer theory I {a E A I a(a) = a for all a E G} =A f"\k*=k*. Also I G' = {a E G a(a) = a for all a E A}. Thus a E G' if and only if x(a) = 1 for all x EGA.

J 17. Let f(x) be a nonconstant polynomial in k[x]. Then f(x) = 0 is solvable by radicals if and only if the Galois group off(x) is solvable. THEOREM Proof Let K be a splitting field over k of f(x) and assume that G = G(K/k) is solvable. We shall show that f(x) = 0 is solvable by radicals. Let n = [K: k] = #G(K/k). Assume first that k contains a primitive nth root of unity. Then k contains a primitive mth root of unity for all positive integers m that divide n. Let G = G0 2 G1 2 · · · 2 Gr = 1 be a chain of subgroups of G such that for i = 1, ...

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