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| Arithmetic genus https://www.singular.uni-kl.de/forum/viewtopic.php?f=10&t=1884 |
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| Author: | QL [ Fri Nov 05, 2010 6:27 pm ] |
| Post subject: | Arithmetic genus |
Hi, I know that Singular can compute the arithmetic genus of a curve given by equations. Now I have a curve C (in P^5) given by paramatrization, namely as image of an explicit birational morphism C' -> P^5 (in fact C' is just a disjoint union of some copies of P^1, so C' is the normalization of C). Is it possible to compute the arithmetic genus of C ? If we want to do it by local computations, this amongs to compute the linear dimension of k[t_1] \oplus k[t_2] \oplus ... \oplus k[t_5] mod the sub-algebra (and not the ideal) generated by some explicit polynomials F_1(t_1,...,t_5), ..., F_N(t_1,...,t_5). For irreducible curves, e.g. y^2=x^{2m+1}, the local contribution of the singular point is the dimension of the quotient vector space k[t]/k[t^2, t^{2m+1}] which is generated by the classes of t, t^3, ..., t^{2m-1}. But in my situation, there are several irreducible components, and I just can not handle the direct computations. I think there is a way to do it with Singular. Thanks for advics ! Qing Liu Université Bordeaux 1 |
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| Author: | Wolfram Decker [ Sat Nov 06, 2010 4:39 pm ] |
| Post subject: | Re: Arithmetic genus |
You may use elimination to compute the explicit equations of your curve. Then compute the Hilbert polynomial to get the arithmetic genus. Regards, Wolfram Decker |
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| Author: | QL [ Sat Nov 06, 2010 10:07 pm ] |
| Post subject: | Re: Arithmetic genus |
Wolfram Decker wrote: You may use elimination to compute the explicit equations of your curve. Then compute the Hilbert polynomial to get the arithmetic genus. Regards, Wolfram Decker Thanks ! I would prefer to avoid elimintation because this is a kind of detour. But well, if there is no other solutions... Qing Liu |
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