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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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    Oh, I didn't know about that. I have accepted the previous good answers now. :)2011-11-05

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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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    $\mathbb{CP}^2$ 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.2011-11-05