Elliptic Curves over C - Elliptic Curves: Number Theory and Cryptography - page 298

PROOF The function ℘ ( nz ) has triple poles at all points of ( C /L )[ n ].
Therefore, there are 3 n 2 poles. Since the zeros of ℘ are at the points in
( C /L )[2] other than 0, the zeros of ℘ ( nz ) are at the points that are in
( C /L )[2 n ] but not in ( C /L )[ n ]. There are 3 n 2 such points. Since the num-
ber of zeros equals the number of poles, all of these zeros are simple. The
expansion of ℘ ( nz )at z =0is

℘ ( nz )= − 2

n 3 z 3 +

··· .

The function f 2 n /f n is easily seen to have the same divisor as ℘ ( nz )and
their expansions at z = 0 have the same leading coecients. Therefore, the
functions are equal.

Finally, we can prove the main result of this section.

THEOREM 9.33
Let E be an elliptic curve over a field of characteristic not 2, let n be a
positive integer, and let ( x, y ) be a pointon E .Then
n ( x, y )= φ n

,

ω n
ψ n

ψ n ,

where φ n , ψ n ,and ω n are defined inSection 3.2.

PROOF First, assume E is defined over a field of characteristic 0. As
above, we regard E as being defined over C .Wehave
( x, y )= ℘ ( z ) ,

2 ℘ ( z ) ,

n ( x, y )= ℘ ( nz ) ,

2 ℘ ( nz )

1 1

for some z . Therefore,

f n − 1 f n +1
f n

℘ ( nz )= ℘ ( z ) −

℘f n − f n − 1 f n +1
f n

=

xψ n − ψ n − 1 ψ n +1
ψ n

= (by Lemma 9.31)

φ n

ψ n .
This proves the formula for the x -coordinate.
For the y -coordinate, observe that the definition of ω n can be rewritten as

=

ω n = 1
2

ψ 2 n

ψ n .