Divisors - Elliptic Curves: Number Theory and Cryptography - page 364

1. If deg D< 0 ,then L

( D )=0 .

2. If D 1 ∼

D 2 then L ( D 1 ) L ( D 2 ) .

3. L

(0) = K .

4. ( D ) <

∞ .

5. If deg( D )=0 then ( D )=0 or 1 .

PROOF Proposition 11.1 holds for all curves (see [38]), not just elliptic
curves, and we'll use it in this more general context throughout the present
proof. For example, we need that deg(div( f )) = 0 for functions f that are
not identically 0.
If L = 0, then there exists f =0with

div( f )+ D ≥ 0 ,

which implies that

deg( D ) = deg(div( f )+ D ) ≥ 0 .

This proves (1).
If D 1 ∼ D 2 ,then D 1 = D 2 + div( g )forsome g .Themap

L ( D 1 ) →L ( D 2 )

f → fg

is easily seen to be an isomorphism. This proves (2).
If 0 0. Since deg(div( f )) = 0, we must have

div( f ) = 0, which means that f has no zeros or poles.

= f ∈L (0), then div( f ) ≥

The analogue of

Proposition 11.1 says that f must be a constant. Therefore,

L (0) = K

and (0) = 1. This proves (3) and also proves (4) for D =0.
We can get from 0 to an arbitrary divisor by adding or subtracting one point
at a time. We'll show that each such modification changes the dimension by
at most one. Therefore, the end result will be a finite dimensional vector

space.
Suppose that D 1 ,D 2 are two divisors with D 2 = D 1 +[ P ]forsomepoint
P .Then

L ( D 1 ) ⊆L ( D 2 ) .

Suppose there exist g, h ∈L ( D 2 )with g, h ∈L ( D 1 ). Let −n be the coecient
of [ P ]in D 2 .Thenboth g and h must have order n at P .(Theorderof g