Elliptic Curves over C - Elliptic Curves: Number Theory and Cryptography - page 267

PROOF When | z | < | ω | ,

= ω − 2

(1 − ( z/ω )) 2 − 1
= ω − 2 ∞

1
( z − ω ) 2 −

1
ω 2

1

.

( n +1) z n
ω n

n =1

Therefore,

z 2 +
ω

∞

z n
ω n +2 .

℘ ( z )= 1 ( n +1)

=0 n =1

Summing over ω first, then over n , yields the result.

THEOREM 9.8
Let ℘ ( z ) be the W eierstra ss ℘ -function for a lattice L .Then

℘ ( z ) 2 =4 ℘ ( z ) 3

− 60 G 4 ℘ ( z ) − 140 G 6 .

PROOF

From Proposition 9.7,

℘ ( z )= z − 2 +3 G 4 z 2 +5 G 6 z 4 + ···
℘ ( z )= − 2 z − 3 +6 G 4 z +20 G 6 z 3 + ··· .

Cubing and squaring these two relations yields

℘ ( z ) 3 = z − 6 +9 G 4 z − 2 +15 G 6 + ···
℘ ( z ) 2 =4 z − 6 − 24 G 4 z − 2 − 80 G 6 + ··· .

Therefore,

f ( z )= ℘ ( z ) 2 4 ℘ ( z ) 3 +60 G 4 ℘ ( z ) + 140 G 6 = c 1 z + c 2 z 2 + − ···

is a power series with no constant term and with no negative powers of z .
But the only possible poles of f ( z ) are at the poles of ℘ ( z )and ℘ ( z ), namely,
the elements of L .Since f ( z ) is doubly periodic and, as we have just shown,
hasnopoleat0, f ( z ) has no poles. By Theorem 9.1, f ( z ) is constant. Since
the power series for f ( z ) has no constant term, f (0) = 0. Therefore, f ( z )is
identically 0.

It is customary to set

g 2 =60 G 4
g 3 = 140 G 6 .