Singular

[quote="Wolfram Decker"]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[/quote]

Thanks ! I would prefer to avoid elimintation because this is a kind of detour. But well, if there is no other solutions...

Qing Liu

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

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 [b]sub-algebra[/b] (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