# proof of fundamental theorem of algebra

If $f(x)\in \u2102[x]$ let $a$ be a root of $f(x)$ in some
extension^{} of $\u2102$. Let $K$ be a Galois closure of
$\u2102(a)$ over $\mathbb{R}$ and set $G=\mathrm{Gal}(K/\mathbb{R})$.
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 ${G}_{1}=\mathrm{Gal}(K/\u2102)$ is also a 2-group. If ${G}_{1}\ne 1$ choose ${G}_{2}\le {G}_{1}$ such that $[{G}_{1}:{G}_{2}]=2$, and set $M={K}^{{G}_{2}}$,
so that $[M:\u2102]=[{G}_{1}:{G}_{2}]=2$. But any polynomial^{} of
degree 2 over $\u2102$ has roots in $\u2102$ by the
quadratic formula, so such a field $M$ cannot exist. This
contradiction^{} shows that ${G}_{1}=1$. Hence $K=\u2102$ and $a\in \u2102$, 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 |