## Proof of Fundamental Theorem of Algebra

I am trying to make a proof of  Fundamental Theorem of Algebra which is seem to be understandable by secondary school student. Although I use some topological argument, but I think I could  cheat them except the point I have mentioned.

What I use about complex is open mapping theorem and $\dfrac{1}{f}$ is a holomorphic function.

Let $f:{\mathbb{C}}\rightarrow {\mathbb{C}}$ be a polynomial which is non-constant. It can extend to a continuous function, also denoted as $f$, by  $f:{\mathbb{S}}^2 \rightarrow {\mathbb{S}}^2$ such that $f$ maps $\infty$ to $f(\infty)=\infty$. It is well-defined and continuous by using limit. By open mapping theorem (can be replaced by elemantary argument?)(1),  $f({\mathbb{S}}^2 )$ is open ( for the neighborhood of $\infty$, $U_{\infty}$, observe that (1)$f(U_{\infty}-\infty)$ is open and (2) $\dfrac{1}{f}$ contain the set $\{z|0<|z| for some small $r$ (or by continuity of $f$), $f(U_{\infty})$ is open) . Also, $f({\mathbb{S}}^2 )$ is closed. By connectness, $f({\mathbb{S}}^2 )={\mathbb{S}}^2$, hence there exists $z$ such that $f(z)=0$.

(1) can be reduced to show that $\{f(z)||z| < r \}$ is open.

But still, I use open mapping theorem. Can I use more elementary way to show (1)?

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### 3 Responses to Proof of Fundamental Theorem of Algebra

1. kingleunglee says:

It seems it is not need to that 1/f is holomorphic.

2. KKK says:

We need it. The crucial point is that $f:\mathbb{S}^2\rightarrow \mathbb{S}^2$ is open, even at the point of infinity. How can we show this?
The continuity of $f$ (or 1/f) is not enough, what we need is the open mapping theorem (applied to 1/f), which is a nontrivial result.

3. KKK says:

By the way, this is KKK. I am now in Shanghai.