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|<r\} 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.

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