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Homeproof of fundamental theorem of algebra

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# 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.

## Mathematics Subject Classification

30A99*no label found*12D99

*no label found*

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