# A proof of the fundamental theorem of algebra using homotopy

fundamental theorem of algebra는 보통 Liouville’s theorem로 증명하는 것이 가장 일반적이지만 그래도 쫌 다르게 (폼나게?) 증명하는 증명법을 소개한다.

이 내용은 모두 C.T.C Wall책[1]에 있는 내용임.

1. notation >

$X,Y$ are topological spaces.
$I$ is the unit interval on $\mathbb{R}$ ( = [0,1] ).
$S^1$ is the unit circle on complex plane.
$H^0 (X)$ is the set of continuous map $X \to \mathbb{Z}$. (every subset of $\mathbb{Z}$ is open.)
$\pi_0(X)$ is path components of $X$.
$F(X)$ is free abelian group of $X$.
$H_0(X) := F(\pi_0(X))$
$[Y,X]$ is set of homotopy classes of maps $f : Y \to X$
$H^1(X) := [X,S^1]$
$e : \mathbb{R} \to S^1$, $e(t) := \exp(2\pi it)$

2. definition > lifting maps from $S^1$ up to $\mathbb{R}$

If $f : X \to S^1$ is continuous map, does there exist $\hat{f}$ ?

If there is, $\hat{f}$ is lifting map of $f$

3. theorem >

Any continuous map $f : I \to S^1$ has a lift $\hat{f} : I \to \mathbb{R}$ which is unique up to translation by integer.
증명 생략

4. definition > degree of map

for any continuous map $f : S^1 \to S^1$,

there exist $g : I \to \mathbb{R}$
$e(g(1)) = f(e(1)) = f(e(0)) = e(g(0))$ (∵ $e(0) = e(1) = 1$)
$g(1)-g(0)$ is an integer
$\deg f := g(1)-g(0)$
사실 degree는 $S^1$ 에서 한 바퀴 돌 동안에 $f$ 의 image에서 돈 횟수이다.

5. a horde of theorems >

1. $\deg(f\circ g) = \deg f + \deg g$
2. degree 자체가 function이므로 $\deg : H^1(S^1) \to \mathbb{Z}$
The degree map is group isomorphism.
3. The two homotopic maps have same degree.
4. the following conditions are equivalent
i) $f$ is nullhomotopic
ii) $\deg f = 0$
iii) $f$ has continuous lift $\hat{f} : S^1 \to \mathbb{R}$
5. If the map $f$ is not surjective, $\deg f = 0$. (i.e f is nullhomotopic)

3번만 빼면 쉽다.

6. theorem > Fundamental theorem of algebra

proof >
We will suppose not, and find a contradiction.
Write the polynomial as

$\displaystyle P(z) = \sum_{i=0}^n a_i z^i$

Without loss of generality, we may suppose $a_n=1$
We define a map $F : S^1 \times \mathbb{R}^+ \to S^1$ by $F(z,r) = P(rz)/|P(rz)|$
If we write $f_r(z) = F(z,r)$, then $F$ provides a homotopy between the maps $f_r$. Now $f_0$ is constant, so has zero degree. We will show that for $r$ large enough, $f_r$ has degree $n$. This will give the required contradiction.

Choose

$\displaystyle R > \max\left( \sum_{i=0}^{n-1} |a_i| , 1\right)$

Then for $|z| = 1$,

$\displaystyle \left| \sum_{i=0}^{n-1} a_i (Rz)^i \right| \le \sum_{i=0}^{n-1} |a_i|R^i \le R^{n-1} \sum_{i=0}^{n-1} |a_i| < R^n = |(Rz)^n|$

So

$\displaystyle \left| \frac{\sum_{i=0}^{n-1}a_i(Rz)^i}{(Rz)^n}\right| < 1$

It follows that $P(Rz)/(Rz)^n$ has a positive real part. Hence so has

$\displaystyle \frac{P(Rz)}{(Rz)^n} \left|\frac{(Rz)^n}{P(Rz)}\right| = \frac{f_R(z)}{z^n}$

So, the map $f_R(z)/z^n$ has zero degree. Hence $\deg f_R$ equals the degree of the map $z^n$.

The theorem now follows

2014.7.10

2017.1.20
The Fundamental Theorem of Algebra (with the Fundamental Group) in Math ∩ Programming

[1] C. T. C. Wall, A geometric introduction to topology. Addison-Wesley 1972, Dover 1993.