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The arithmetic genus of nonsingular curve C of degree d in PP^3 over an algebraically closed field is less than or equal to 1/2(d-1)(d-2). I must show it by comparing C with a suitable projection from a point into PP^2. How can I prove it?

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How did this problem come up? –  S. Carnahan Aug 23 '11 at 12:46
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Actually, it looks a lot like homework. –  Francesco Polizzi Aug 23 '11 at 13:30
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It is also done in Hartshorne chapter IV. –  J.C. Ottem Aug 23 '11 at 15:10
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1 Answer

$\frac{1}{2}(d-1)(d-2)$ is the genus of a smooth plane curve of degree $d$. If you project from $P^3$ to $P^2$ off a point not contained in $C$ you can always get a plane curve of the same degree with at most nodes as singularities, which is birational to $C$ (hence has the same genus). Each node lowers the genus of the image curve by one w.r.t the formula you give. Hence you have the desired inequality.

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