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I'm reading Donaldson's book on Riemann surfaces and came across the following problem.

Let $p$ be a quadratic homogeneous polynomial in 3 variables. Using classification of forms or otherwise, show that the Riemann surface it defines (its solution set in projective space) is equivalent to the Riemann sphere.

I am a bit confused what classification of quadratic forms in $\mathbb{C}$ looks like. I know that over $\mathbb{R}$, we can write the quadratic form as a matrix and diagonalize it by the spectral theorem and then scale to get something of the form $\sum x_i^2 -\sum x_j^2$. But we can't always diagonalize a complex symmetric matrix.

Even assuming that we can write it in the form $x^2+y^2+z^2$, I am having trouble seeing how this is the Riemann sphere. Naively, if we set $z = 0$ we get the point at infinity, and otherwise if we fix $z=1$ it looks like we get two copies of $\mathbb{P}^1$. This is probably not the right approach to begin with but I'm struggling to proceed.

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    $\begingroup$ First, you certainly can always diagonalize a quadratic form over an algebraically closed field not of characteristic $2$... Second, think of the parametrization of Pythagorean triples... some equivalent form gives $t \to ({t\over t^2+1},{t^2-1\over t^2+1}$ parametrizing $\{x,y:x^2+y^2=1\}$... $\endgroup$ – paul garrett Jul 27 at 17:35

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