proof of fundamental theorem of algebra

If $f(x)\in\mathbb{C}[x]$ let $a$ be a root of $f(x)$ in some extension of $\mathbb{C}$ . Let $K$ be a Galois closure of $\mathbb{C}(a)$ over $\mathbb{R}$ and set $G=\operatorname{Gal}(K/\mathbb{R})$ . Let $H$ be a Sylow 2-subgroup of $G$ and let $L=K^{H}$ (the fixed field of $H$ in $K$ ). By the Fundamental Theorem of Galois Theory we have $[L:\mathbb{R}]=[G:H]$ , an odd number. We may write $L=\mathbb{R}(b)$ for some $b\in L$ , so the minimal polynomial $m_{b,\mathbb{R}}(x)$ is irreducible over $\mathbb{R}$ and of odd degree. That degree must be 1, and hence $L=\mathbb{R}$ , which means that $G=H$ , a 2-group. Thus $G_{1}=\operatorname{Gal}(K/\mathbb{C})$ is also a 2-group. If $G_{1}\neq 1$ choose $G_{2}\leq G_{1}$ such that $[G_{1}:G_{2}]=2$ , and set $M=K^{G_{2}}$ , so that $[M:\mathbb{C}]=[G_{1}:G_{2}]=2$ . But any polynomial of degree 2 over $\mathbb{C}$ has roots in $\mathbb{C}$ by the quadratic formula, so such a field $M$ cannot exist. This contradiction shows that $G_{1}=1$ . Hence $K=\mathbb{C}$ and $a\in\mathbb{C}$ , 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