P. Fitzpatrick's Advanced Calculus [Int'l Student edn] PDF

By P. Fitzpatrick

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Extra resources for Advanced Calculus [Int'l Student edn]

Example text

K, and nk = 1. We shall prove that a polynomial with coefficients in a field K of characteristic zero is solvable by radicals if and only if its Galois group ΓK (f ) over K is a solvable group. Let L be a field, and let p be a prime number that is not equal to the characteristic of L. Suppose that the polynomial xp − 1 splits over L. Then the polynomial xp − 1 has distinct roots, since its formal derivative pxp−1 is non-zero at each root of xp − 1. An element ω of L is said to be a primitive pth root of unity if ω p = 1 and ω = 1.

Moreover σ(α) is also a root of xp − c, and therefore σ(α) = αω j for some integer j. Thus if σ and τ are automorphisms of M belonging to Γ(M : K(ω)), and if σ(α) = αω j and τ (α) = αω k , then σ(τ (α)) = τ (σ(α)) = αω j+k , since σ(ω) = τ (ω) = ω. Therefore σ ◦ τ = τ ◦ σ. We deduce that Γ(M : K(ω)) is Abelian, and thus Γ(M : K) is solvable, as required. 53 Let f be a polynomial with coefficients in a field K of characteristic zero, and let K = K(α), where α ∈ K satisfies αp ∈ K for some prime number p.

K are distinct and are the elements of the orbit of α under the action of G on L. 31 Proof Let f (x) = (x − α1 )(x − α2 ) · · · (x − αm ). Then the polynomial f is invariant under the action of G, since each automorphism in the group G permutes the elements α1 , α2 , . . , αk and therefore permutes the factors of f amongst themselves. It follows that the coefficients of the polynomial f belong to the fixed field K of G. Thus α is algebraic over K, as it is a root of the polynomial f . Now, given any root αi of f , there exists some σ ∈ G such that αi = σ(α).

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Advanced Calculus [Int'l Student edn] by P. Fitzpatrick

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