proof of fundamental theorem of algebra

If $f(x)\\in\\mathbb{C}[x]$ let $a$ be a root of $f(x)$ in someextension 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 odddegree. That degree must be 1, and hence $L=\\mathbb{R}$ , whichmeans that $G=H$ , a 2-group. Thus $G_{1}=\\operatorname{Gal}(K/\\mathbb{C})$ is also a 2-group. If $G_{1}\eq 1$ 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 ofdegree 2 over $\\mathbb{C}$ has roots in $\\mathbb{C}$ by thequadratic formula, so such a field $M$ cannot exist. Thiscontradiction shows that $G_{1}=1$ . Hence $K=\\mathbb{C}$ and $a\\in\\mathbb{C}$ , completing the proof.

单词	ProofOfFundamentalTheoremOfAlgebra
释义	proof of fundamental theorem of algebra If $f(x)\\in\\mathbb{C}[x]$ let $a$ be a root of $f(x)$ in someextension 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 odddegree. That degree must be 1, and hence $L=\\mathbb{R}$ , whichmeans that $G=H$ , a 2-group. Thus $G_{1}=\\operatorname{Gal}(K/\\mathbb{C})$ is also a 2-group. If $G_{1}\eq 1$ 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 ofdegree 2 over $\\mathbb{C}$ has roots in $\\mathbb{C}$ by thequadratic formula, so such a field $M$ cannot exist. Thiscontradiction shows that $G_{1}=1$ . Hence $K=\\mathbb{C}$ and $a\\in\\mathbb{C}$ , completing the proof.
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