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Why is the complex projective planes the most natural place to look to consider solutions of polynomial equation?

Why is the complex plane $\mathbb{C}$ adequate for polynomial equations of one variable? Is this because the Riemann sphere is topologically similar to $\mathbb{C}$?

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Why have you asked several questions, and not accepted answers to any of them? If you don't like the answers, why do you keep asking questions? If you don't know about accepting answers, why haven't you read the faq? –  Gerry Myerson Nov 5 '11 at 9:19
Oh, I didn't know about that. I have accepted the previous good answers now. :) –  alext87 Nov 5 '11 at 9:48

1 Answer 1

The complex plane is adequate because there is a theorem that says so. It's called The Fundamental Theorem Of Algebra, and you shouldn't have any trouble finding lots of information about it in books and online. Warning: the proof takes a bit of heavy lifting.

EDIT: For a polynomial equation in two variables, even something as simple as $x-y=0$, there will generally be "points at infinity", so you need to go to projective space for a full understanding of the solutions. In one variable, only finitely many solutions, no points at infinity, no need to projectivize.

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So why for bivariate polynomials is it not good enough to look in $\mathbb{C}^2$. Why is everything in the $\mathbb{C}\mathbb{P}^2$? –  alext87 Nov 5 '11 at 10:31
$\mathbb{CP}^2$ has nice geometric properties: For instance, two lines always intersect in a point. In $\mathbb{C}^2$, you can have parallel lines that do not satisfy this condition. –  Jesko Hüttenhain Nov 5 '11 at 14:27

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