C as algebraicly closed field

Suppose \(P \in \mathbb{C}[X]\) with \(\mathrm{deg}(P) = n \geq 0\) Suppose \(P\) has no root. w.l.o.g. we may assume, that \(P\) is monic, as division by the leading coefficient results in a rootless polynomial

Therefore, the map

\begin{align*} x \mapsto& \frac{P(x)}{\| P(x) \|} \end{align*}

is a continuous map from

\begin{align*} \mathbb{R}^2 \cong \mathbb{C} \rightarrow S^1 \end{align*}

NOte that the restriction to \(S^1\)

w.l.o.g. we may assume, that \(f\) is monic, otherwise division by the leading coefficient does not change the amount of roots.

Suppose \(f\) does not have a root. Then the map

\begin{align*} \varphi_{f}: \mathbb{C} \rightarrow& S^1 \subseteq \mathbb{C} \\ x \mapsto \frac{f(x)}{\| f(x) \|} \end{align*}

is continuous.

Note that \(\varphi_{f}\) is not surjective on \(\mathbb{C} \cong D^2\), hence the restriction has degree \(0\) (cf.nonzero homological degree and surjectivity on the disk) Therefore it is also nullhomotopic

Consider

\begin{align*} P_t = \sum_{i=0}^{n} \alpha_i t^{n-i} X^i \end{align*}

with

\begin{align*} P_1 = f \end{align*}

Then

\begin{align*} H: [0,1] \times S^1 \rightarrow& S^1 \\ (t,x) \mapsto \frac{P_t(x)}{\| P_t(x) \|} \end{align*}

is a homotopy from \(\varphi_f\) to

\begin{align*} P_0 = x^n \end{align*}

But

\begin{align*} \mathrm{deg}(x^n) = n \neq 0 \end{align*}

(cf. singular homological degree as group epimorphism)