Fundamental theorem of Algebra from the Brouwer fixed point theorem.

Here is my attempt to prove the Fundamental theorem of algebra from the Brouwer fixed point theorem.

Lemma (Brouwer fixed point theorem). If $f:D_r\rightarrow D_r$ is a continous function, then there is a point $z_0 \in D_r$ such that $f(z_0)=z_0$.

Theorem (Fundamental theorem of algebra). Every non-constant complex polynomial $$p(z)=a_n z^n+a_{n-1}z^{n-1}+...+a_1 z+a_0,$$ where $a_0 \neq 0$, vanish somewhere in $\mathbb{C}$.

Proof: If the polynomial $p$ doesn't vanish in $\mathbb{C}$, then for every $z\in \mathbb{C}$ we have the following

$$g(z)=\frac{-a_0}{a_n z^{n-1}+a_{n-1}z^{n-2}+...+a_1}\neq z.$$

If $g$ is not continuous, then the polynimial $a_n z^{n-1}+a_{n-1}z^{n-2}+...+a_1$ must vanish somewhere in $\mathbb{C}$, and we are done since $n \in \mathbb{N}$ is arbitrary.

Suppose that $g$ is continuous. Now $g$ has the growth condition $|g(z)|\rightarrow 0$ as $|z|\rightarrow \infty$, so there is a constant $R>0$ such that $g(z)\in D_R$ when $z\in D_R$. Due to the Brouwer fixed point theorem $g$ can't be continuous since $g(z)\neq z$ in $D_R$. Again we are done.

Your error relates to your claim $$g(z)=\frac{-a_0}{a_n z^{n-1}+a_{n-1}z^{n-2}+\cdots +a_1}\neq z.$$

"If $g$ is not continuous, then the polynomial $a_n z^{n-1}+a_{n-1}z^{n-2}+\cdots +a_1$ must vanish somewhere in $\mathbb{C}$, and we are done since $n \in \mathbb{N}$ is arbitrary."

More specifically, your error is the fragment

" . . . and we are done since $n \in \mathbb{N}$ is arbitrary."

What you actually proved is that at least one of the polynomials $$a_n z^{n-1}+a_{n-1}z^{n-2}+\cdots +a_1$$ $$a_n z^n+a_{n-1}z^{n-1}+\cdots +a_1 z+a_0$$ has a root in $\mathbb{C}$.

• But I checked both cases: $g$ is continuous and $g$ is not? – Hulkster Aug 13 '17 at 2:35
• If $g$ is not continuous, then the denominator has a zero. But that's a new polynomial, not the original one. – quasi Aug 13 '17 at 2:36
• My aim was to show that if a general n. degree polynomial doesn't vanish, then a general (n-1). degree polynomial do vanish. So apparently that's not enough? – Hulkster Aug 13 '17 at 2:52
• The goal is to show that every $n$-th degree polynomial $p(z)$ has a root in $\mathbb{C}$, Once you choose it, it's no longer free. Let $q(z)$ be the denominator of $g(z)$, The polynomial $q(z)$ depends on $p(z)$, so it's also not free. – quasi Aug 13 '17 at 2:54
• Thus, what you proved is: Given an $n$-th degree polynomial $p(z)$, the associated polynomial $q(z)$ is such that at least one of $p(z),q(z)$ has a root in $\mathbb{C}$. – quasi Aug 13 '17 at 3:00