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?
3
-
$\begingroup$ How did this problem come up? $\endgroup$– S. Carnahan ♦Commented Aug 23, 2011 at 12:46
-
1$\begingroup$ Actually, it looks a lot like homework. $\endgroup$– Francesco PolizziCommented Aug 23, 2011 at 13:30
-
1$\begingroup$ It is also done in Hartshorne chapter IV. $\endgroup$– J.C. OttemCommented Aug 23, 2011 at 15:10
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
$\frac{1}{2}(d-1)(d-2)$ 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.
