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 $\mathbb{C}$.

  • $\begingroup$ But I checked both cases: $g$ is continuous and $g$ is not? $\endgroup$ – Hulkster Aug 13 '17 at 2:35
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    $\begingroup$ If $g$ is not continuous, then the denominator has a zero. But that's a new polynomial, not the original one. $\endgroup$ – quasi Aug 13 '17 at 2:36
  • $\begingroup$ 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? $\endgroup$ – Hulkster Aug 13 '17 at 2:52
  • $\begingroup$ 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. $\endgroup$ – quasi Aug 13 '17 at 2:54
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    $\begingroup$ 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 $\mathbb{C}$. $\endgroup$ – quasi Aug 13 '17 at 3:00

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