Elliptic Curves over C - Elliptic Curves: Number Theory and Cryptography - page 295

so g also has a double pole at 0. Therefore, g has a total of 2 n 2 poles, counting
multiplicities.
The function g has a zero at z = w when nw ≡±w ≡ 0(mod L ). For such

w ,

dz g ( z ) z = w

d

= n℘ ( nw ) − ℘ ( w )= ±n℘ ( w ) − ℘ ( w )=( ±n − 1) ℘ ( w ) .

Since the zeros of ℘ ( z ) occur when z = ω j / 2, we have g ( w )

=0when

w = ω j / 2, so such w are simple zeros of g . Moreover, when n is odd, n ( ω j / 2) ≡

ω j / 2, so the points ω j / 2 are at least double zeros of g in this case.
If nw ≡ w (mod L ), then ( n − 1) w ≡ 0. Let δ =0if n is even and δ =1
if n is odd. There are ( n − 1) 2 − 1 − 3 δ points w with ( n − 1) w =0and

=0 ,ω j / 2. Similarly, there are ( n +1) 2 w − 1 − 3 δ points w with ( n +1) w =0

and w =0 ,ω j / 2. There are at least 6 δ zeros (counting multiplicities) at the

points ω j / 2. Therefore, we have accounted for at least

( n − 1) 2

− 1 − 3 δ +( n +1) 2 − 1 − 3 δ +6 δ =2 n 2

zeros. Since g ( z ) has exactly 2 n 2 poles, we have found all the zeros and their
multiplicities.
The function

f n − 1 f n +1
f n
has a double pole at each of the zeros of f n .If w ≡ 0and( n ± 1) w ≡ 0then
f n± 1 has a simple zero at w .Ifboth( n +1) w ≡ 0and( n − 1) w ≡ 0, then
2 w ≡ 0, so w ≡ ω j / 2forsome j . Therefore, f n− 1 f n +1 has a simple zero at
each w with ( n ± 1) w ≡ 0, except for those where w ≡ ω j / 2, at which points
it has a double zero. At z = 0, the expansions of the functions yield

f n − 1 f n +1

f n =
( − 1) n ( n − 1)
z ( n− 1) 2 − 1

( − 1) n +2 ( n +1)
z ( n +1) 2 − 1

( − 1) n +1 n
z n 2 − 1

2

+ ··· + ··· + ···

= 1 −

n 2 z − 2 + ··· ,

1

f n− 1 f n +1 /f n so there is a double zero at z = 0. Therefore, − has the same

divisor as ℘ ( nz ) ℘ ( z ), so the two functions are constant multiples of each

other. Since their expansions at 0 have the same leading coe cient, they must
be equal. This proves the lemma.

−

LEMMA 9.29
f 2 n +1 = f n +2 f n − f n +1 f n− 1 .