Divisors - Elliptic Curves: Number Theory and Cryptography - page 368

PROPOSITION 11.20
ψ isabijection.

PROOF Suppose Q 1 = Q 2 are such that ψ ( Q 1 )= ψ ( Q 2 ), hence

f ( Q 1 )= f ( Q 2 )= a and g ( Q 1 )= g ( Q 2 )= b

for some a, b .Since f − a has a double pole at P and g − b has a triple pole
at P ,

div( f − a )=[ Q 1 ]+[ Q 2 ] − 2[ P ]
div( g − b )=[ Q 1 ]+[ Q 2 ]+[ R ] − 3[ P ]

for some R . Subtracting yields

[ R ] − [ P ] = div(( g − b ) / ( f − a )) ∼ 0 .

By Corollary 11.18, this means that R = P . Therefore,

div( g − b )=[ Q 1 ]+[ Q 2 ] − 2[ P ] ,

so g has only a double pole at P . This contradiction proves that ψ is an
injection.
To prove surjectivity, let ( a, b ) ∈ E ( K ). We want to find P with ψ ( P )=

( a, b ). Since f a has a double pole at P and since the divisor of a fu nction

has degree 0, there are (not necessarily distinct) points Q 1 ,Q 2 ∈

−

C ( K )such

that

div( f − a )=[ Q 1 ]+[ Q 2 ] − 2[ P ] .
For a given x -coordinate a , there are two possible y -coordinates b and b for
points on E .If g ( Q i )= b for some i =1 , 2, we have ψ ( Q i )=( a, b )andwe're
done. Therefore, suppose

g ( Q 1 )= g ( Q 2 )= b .

Then ψ ( Q 1 )= ψ ( Q 2 )=( a, b ). Since ψ is injective, Q 1 = Q 2 ,so

div( f − a )=2[ Q 1 ] − 2[ P ] .

Let u be a uniformizing parameter at Q 1 .Then

a = u 2 f 1 , b = ug 1

f − g −

with f 1 ( Q 1 ) =0 ,∞ and g 1 ( Q 1 ) = ∞ (possibly g 1 ( Q 1 ) = 0). Substituting into
(11.10) and using the fact that ( a, b ) ∈ E yields

( ug 1 )(2 b + a 1 a + a 3 )= u 2 h