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By C. Menini, F. van Oystaeyen

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The roots of the polynomial in this extension field are α, β and γ, where α=ξ+ p , 3ξ β = ωξ + ω 2 p , 3ξ γ = ω2ξ + ω3 p . 3ξ Now let us consider the possibilities for the Galois group Γ(L: K), where L is a splitting field for f over K. Now L = K(α, β, γ), where α, β and γ are the roots of f . Also a K-automorphism of L must permute the roots of f amongst themselves, and it is determined by its action on these roots. , the group of permutations of a set of 3 objects), and thus the possibilities for the order of Γ(L: K) are 1, 2, 3 and 6.

Suppose c = 0. The roots of the polynomial xp − c are distinct, and each pth root of unity is the ratio of two roots of xp − c. Therefore M = K(α, ω), where αp = c and ω is some primitive pth root of unity. 32). 50), we see that Γ(M : K(ω)) is a normal subgroup of Γ(M : K), and Γ(M : K)/Γ(M : K(ω)) is isomorphic to Γ(K(ω): K). 53). It therefore suffices to show that Γ(M : K(ω)) is also Abelian. Now the field M is obtained from K(ω) by adjoining an element α satisfying αp = c. Therefore each automorphism σ in Γ(M : K(ω)) is uniquely determined by the value of σ(α).

Example Consider the quintic polynomial f where f (x) = x5 − 6x + 3. 17) can be used to show that f is irreducible over Q. Now f (−2) = −17, f (−1) = 8, f (1) = −2 and f (2) = 23. The Intermediate Value Theorem ensures that f has at least 3 distinct real roots. If f had at least 4 distinct real roots then Rolle’s Theorem would ensure that the number of distinct real roots of f and f would be at least 3 and 2 respectively. But zero is the only root of f since f (x) = 20x3 . Therefore f must have exactly 3 distinct real roots.

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Abstract Algebra - A Comprehensive Trtmt by C. Menini, F. van Oystaeyen


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