Elliptic Curves over C - Elliptic Curves: Number Theory and Cryptography - page 269

THEOREM 9.10
Let L be a lattice an d let E be the ellipticcurve y 2 =4 x 3

− g 2 x − g 3 .The

map

Φ : C /L −→ E ( C )
z −→ ( ℘ ( z ) ,℘ ( z ))
0 −→ ∞

isanisom orphism of groups.

PROOF The surjectivity is easy. Let ( x, y ) ∈ E ( C ). Since the function

℘ ( z ) x has a double pole, Theorem 9.1 implies that it has zeros, so there

exists z

−

∈ C such that ℘ ( z )= x . Theorem 9.8 implies that

℘ ( z ) 2 = y 2 ,

so ℘ ( z )= ±y .If ℘ ( z )= y , we're done. If ℘ ( z )= −y ,then ℘ ( −z )= y and
℘ ( −z )= x ,so −z → ( x, y ).
Suppose ℘ ( z 1 )= ℘ ( z 2 )and ℘ ( z 1 )= ℘ ( z 2 ), and z 1 ≡ z 2 mod L .Theonly
poles of ℘ ( z )arefor z ∈ L . Therefore, if z 1 is a pole of ℘ ,then z 1 ∈ L and
z 2 ∈ L ,so z 1 ≡ z 2 mod L . Now assume z 1 is not a pole of ℘ ,so z 1 is not in
L . The function

h ( z )= ℘ ( z ) − ℘ ( z 1 )

has a double pole at z = 0 and no other poles in F . By Theorem 9.1, it has
exactly two zeros. Suppose z 1 = ω i / 2forsome i . From Equation 9.6, we
know that ℘ ( ω i / 2) = 0, so z 1 is a double root of h ( z ), and hence is the only
root. Therefore z 2 = z 1 . Finally, suppose z 1 is not of the form ω i / 2. Since
h ( −z 1 )= h ( z 1 ) = 0, and since z 1 ≡−z 1 mod L , the two zeros of h are z 1

and − z 1 mod L . Therefore, z 2 ≡− z 1 mod L .But

y = ℘ ( z 2 )= ℘ ( −z 1 )= −℘ ( z 1 )= −y.

This means that ℘ ( z 1 )= y =0. But ℘ ( z ) has only a triple pole, so has only
three zeros in F . From Equation 9.6, we know that these zeros occur at ω i / 2.
This is a contradiction, since z

= ω i / 2. Therefore, z 1 ≡ z 2 mod L ,soΦis

injective.
Finally, we need to show that Φ is a group homomorphism. Let z 1 ,z 2 ∈ C
and let

Φ( z i )= P i =( x i ,y i ) .

Assume that both P 1 ,P 2 are finite and that the line through P 1 ,P 2 intersects
E in three distinct finite points (this means that P 1 = ±P 2 ,that2 P 1 + P 2 = ∞ ,
and that P 1 +2 P 2 = ∞ ). For a fixed z 1 , this excludes finitely many values of
z 2 . There are two reasons for these exclusions. The first is that the addition
law on E has a different formula when the points are equal. The second is