ag.algebraic geometry - Sections of a divisor on elliptic curve

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.