proof of fundamental theorem of algebra

If f(x)[x] let a be a root of f(x) in some extensionPlanetmathPlanetmathPlanetmathPlanetmath of . Let K be a Galois closure of (a) over and set G=Gal(K/). Let H be a Sylow 2-subgroup of G and let L=KH (the fixed field of H in K). By the Fundamental Theorem of Galois TheoryMathworldPlanetmath we have [L:]=[G:H], an odd numberMathworldPlanetmathPlanetmath. We may write L=(b) for some bL, so the minimal polynomial mb,(x) is irreduciblePlanetmathPlanetmath over and of odd degree. That degree must be 1, and hence L=, which means that G=H, a 2-group. Thus G1=Gal(K/) is also a 2-group. If G11 choose G2G1 such that [G1:G2]=2, and set M=KG2, so that [M:]=[G1:G2]=2. But any polynomialPlanetmathPlanetmath of degree 2 over has roots in by the quadratic formula, so such a field M cannot exist. This contradictionMathworldPlanetmathPlanetmath shows that G1=1. Hence K= and a, completing the proof.

Title proof of fundamental theorem of algebra
Canonical name ProofOfFundamentalTheoremOfAlgebra
Date of creation 2013-03-22 13:09:39
Last modified on 2013-03-22 13:09:39
Owner scanez (1021)
Last modified by scanez (1021)
Numerical id 5
Author scanez (1021)
Entry type Proof
Classification msc 30A99
Classification msc 12D99