Elliptic Curves over C - Elliptic Curves: Number Theory and Cryptography - page 303

the difference is a constant; call it C . The roots of the cubic polynomial

4 T 3 g 3 are the x -coordinates of the points of order 2, namely ℘ ( 2 ω 1 ),

℘ ( 2 ω 2 ), and ℘ ( 2 ( ω 1 + ω 2 )). Since there is no T 2 term, the sum of these three
roots is 0. Therefore,

− g 2 T −

f ( 1

2 ω 1 )+ f ( 1 2 ω 2 )+ f ( 1 2 ( ω 1 + ω 2 )) = 3 C.

The following lemma shows that C = 0, which yields the proposition.

LEMMA 9.36

f ( 1

2 ω 1 )+ f ( 1 2 ω 2 )+ f ( 1

2 ( ω 1 + ω 2 )) = 0 .

PROOF Thevaluesof u corresponding to the three values of z are

u = q 1 / 2 , u = q − 1 / 2 . u = − 1 ,

We may divide by the factor (2 πi/ω 2 ) 2 , hence we ignore it. The sum of
the three terms 1/12 yields 1/4, which cancels the value of u/ (1 − u ) 2 when
u = − 1. The final term inside the sum defining f ( z ) is independent of u and
thus yields

6 ∞

q n
(1 − q n ) 2 .

− (9.29)

n =1

We now consider the remaining terms.
The value u = − 1 (substituted into the sum in f ) yields

− 2 ∞

q n
(1 + q n ) 2 .

(9.30)

n =1

Combining (9.29) and (9.30) yields

− 8 ∞

q n + q 2 n + q 3 n
(1

. (9.31)

− q 2 n ) 2

n =1

The value u = q 1 / 2 (substituted into the sum in f ) yields (the value of

u ) 2 at u = q 1 / 2 is included in the first summation) u/ (1 −

∞

(1 − q n + 2 ) 2 + ∞

q n + 2

1
2

q n−

1
2

( q n− − 1) 2

n =0 n =1

(9.32)

=2 ∞

1
2

q n−

2 ) 2 .

1

(1 − q n−

n =1