## You are here

Homeproof of fundamental theorem of algebra

## Primary tabs

# 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*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff

## Recent Activity

new question: how to contest an entry? by zorba

new question: simple question by parag

Sep 26

new question: Latent variable by adam_reith

Sep 17

new question: Harshad Number by pspss

Sep 14

new problem: Geometry by parag