Divisors - Elliptic Curves: Number Theory and Cryptography - page 365

must be at least n . If it is larger, then g ∈L ( D 1 ). Similarly for h .) Let u be

a uniformizer at P .Write

g = u n g 1 , h = u n h 1

with g 1 ( P )= c =0 ,∞ and h 1 ( P )= d =0 ,∞ .Then

dg − ch = u n ( dg 1 − ch 1 ) ,

and ( dg 1 − ch 1 )( P ) = 0. Therefore, dg − ch has order greater than n at P ,so

dg − ch ∈L ( D 1 ) .

Therefore any two such elements g, h ∈L ( D 2 ) are linearly dependent mod
L ( D 1 ). It follows that

( D 1 ) ≤ ( D 2 ) ≤ ( D 1 )+1 .

As pointed out above, this implies (4).
To prove (5), assume deg( D )=0. If L ( D ) = 0, we're done. Otherwise,
there exists 0 = f ∈L ( D ). Then

div( f )+ D ≥ 0 and deg(div( f )+ D )=0+0=0 .

Therefore,

div( f )+ D =0 .

Since D ∼ div( f )+ D =0,wehave

L ( D ) L (0) = K,

by (2) and (3). Therefore, ( D ) = 1. This proves (5).

A very fundamental result concerning divisors is the following.

THEOREM 11.15 (Riemann-Roch)
Given an algebraiccurve C ,there existsaninteger g (called the genus of C )
and a divisor K (called a canonical divisor )suchthat

( D ) − ( K−D )=deg( D ) − g +1

for alldivisors D .

For a proof, see [42] or [49]. The divisor K is the divisor of a differential on

C .

COROLLARY 11.16

deg( K )=2 g − 2 .