Sections of a divisor on elliptic curve

Question

I'm interested in producing explicit bases for the sections of a line bundle on an embedded genus 1 curve. Let me restrict to the first case that I don't know how to do, so that I can be as concrete as possible. Take a smooth genus 1 curve E defined over QQ by an explicit cubic equation C₀ in QQ[x,y,z]. Let D be a divisor of degree 6 on E, and I_D be defined by four cubic equations (C₀,C₁,C₂,C₃). Note that D does not lie on any conic. Riemann-Roch says that h^0(E,O_E(D))=6, and I'd like to find an explicit basis of rational functions for this vector space. How do I find such a basis?

Note that if D sat on a conic defined by Q, then finding a basis is relatively easy: we could simply choose the functions x^2/Q, xy/Q, ..., z^2/Q as our rational functions.

Answer 1

For an explicit example I believe Magma can do this: check out the this part of the documentation.

Answer 2

I think I know how to answer this now. The main point is that O_E(D) is the dual of I_D. Namely: O_E(D)=sheafHom(I_D, O_E). Thus, H^0(E,O_E(D))=Hom(I_D, O_E).

This can be computed explicitly in any computer algebra package. Or you can see how to compute it as follows. Take a free presentation of I_D as an O_E-module. In the case I asked about, this yields:

O_E³(-4)-->O_E³(-3)-->I_D.

Label the first map F. Then Hom(I_D, O_E) is just the kernel of the map of free modules:

Hom(O_E³(-3), O_E)--> Hom(O_E³(-4), O_E)

induced by composition with F. Thus, computing a free presentation of the ideal sheaf I_D yields a presentation of H^0(E,O_E(D)) as the kernel of a map of free modules.

Answer 3

I would try to play with rational functions of the form xⁿ/Q₁Q₂, perhaps they will form a big part of a basis?

Answer 4

Let H be a hyperplane section of your cubic and let x be one of the three non trivial halves of D-2H in Pic^0(X). If you embed E in the complete linear system |H+x|, then D now sits an a conic; all you needed is a cubic field extension.