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The line through P 3 =( x 3 ,y 3 )and
P 3 is x
x 3 = 0. The divisor of the
function x
x 3 is
div( x − x 3 )=[ P 3 ]+[ −P 3 ] 2[ ] .
(11.1)
Therefore,
div ax + by + c
x
=div( ax + by + c ) div( x−x 3 )=[ P 1 ]+[ P 2 ] [ −P 3 ] [ ] .
x 3
Since P 1 + P 2 = −P 3 on E , this may be rewritten as
]+div ax + by + c
x − x 3
.
[ P 1 ]+[ P 2 ]=[ P 1 + P 2 ]+[
The following important result is the analogue of Theorem 9.6.
THEOREM 11.2
Let E be an elliptic curve. Let D be a divisor on E with deg( D )=0 .Then
there isafunction f on E with
div( f )= D
ifand onlyif
sum( D )= ∞.
PROOF We have just shown that a sum [ P 1 ]+[ P 2 ] can be replaced by
[ P 1 + P 2 ]+[
] plus the divisor of a function, call it g . Note also that
sum(div( g )) = P 1 + P 2 ( P 1 + P 2 ) −∞ = ∞.
Equation (11.1) shows that [ P 1 ]+[ P 2 ]equals2[
] plus the divisor of a function
when P 1 + P 2 =
. Therefore, the sum of all the terms in D with positive
coe cients equals a single symbol [ P ] plus a multiple of [
] plus the divisor
of a function. A similar result holds for the sum of the terms with negative
coecients. Therefore, there are points P and Q on E , a function g 1 ,andan
integer n such that
D =[ P ]
[ Q ]+ n [
]+div( g 1 ) .
Also, since g 1 is the quotient of products of functions g with sum(div( g )) = ,
we have
sum(div( g 1 )) = ∞.
Since deg(div( g 1 )) = 0 by Proposition 11.1, we have
0=deg( D )=1 1+ n +0= n.
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