Divisors - Elliptic Curves: Number Theory and Cryptography - page 344

Then D has degree 0 and an easy calculation shows that sum( D )= .There-

fore, D is the divisor of a function. Let's find the function. The line through
(0 , 0) and (2 , 4) is y

∞

− 2 x = 0. It is tangent to E at (2 , 4), so

div( y − 2 x )=[(0 , 0)] + 2[(2 , 4)] − 3[ ∞ ] .

The vertical line through (2 , 4) is x − 2 = 0, and

div( x − 2) = [(2 , 4)] + [(2 , − 4)] − 2[ ∞ ] .

Therefore,

D =[(2 , − 4)] + div y

+[(4 , 5)] + [(6 , 3)] − 3[ ∞ ] .

− 2 x

x − 2

Similarly, we have
[(4 , 5)] + [(6 , 3)] = [(2 , 4)] + [ ∞ ]+div y + x +2
x − 2

,

which yields
D =[(2 , − 4)] + div y − 2 x
x − 2

+[(2 , 4)] + div y + x +2
x − 2

− 2[ ∞ ] .

Since we have already calculated div( x − 2), we use this to conclude that

D = div( x − 2) + div y

+div y + x +2
x

− 2 x

x − 2 − 2

=div ( y

.

− 2 x )( y + x +2)

x − 2

This function can be simplified. The numerator is

2 x )( y + x +2)= y 2 2 x 2 +2 y ( y − − xy − − 4 x

= x 3 − xy − 2 x 2 +2 y (since y 2 = x 3 +4 x )

=( x − 2)( x 2

− y ) .

Therefore,

D =div( x 2

− y ) .

P roof of Lem m a 11.3:
Suppose P = Q and div( h )=[ P ] − [ Q ]. Then, for any constant c ,the
function h − c has a simple pole at Q and therefore, by Proposition 11.1, it