Elliptic Curves over C - Elliptic Curves: Number Theory and Cryptography - page 281

PROOF First, we need to calculate j ( ρ )and j ( i ). Recall that τ corresponds
to the lattice L τ = Z τ + Z .Since ρ 2 = − 1 − ρ , it follows easily that ρL ρ ⊆ L ρ .

Therefore,

L ρ = ρ 3 L ρ ⊆ ρ 2 L ρ ⊆ ρL ρ ⊆ L ρ ,

so ρL ρ = L ρ . It follows from (9.14) that

g 2 ( L ρ )= g 2 ( ρL ρ )= ρ − 4 g 2 ( L ρ )= ρ − 1 g 2 ( L ρ ) .

Since ρ =1,wehave g 2 ( ρ )= g 2 ( L ρ ) = 0. Therefore,

g 2 ( L ρ ) 3
g 2 ( L ρ ) 3

j ( ρ ) = 1728 =0

− 27 g 3 ( L ρ ) 2

(note that the denominator is nonzero, by Proposition 9.9).
Similarly, τ = i corresponds to the lattice L i = Z i + Z ,and iL i = L i .
Therefore,

g 3 ( L i )= g 3 ( iL i )= i − 6 g 3 ( L i )= − g 3 ( L i ) ,

so g 3 ( i )= g 3 ( L i ) = 0. Therefore,

g 2 ( L i ) 3
g 2 ( L i ) 3

j ( i ) = 1728 = 1728 .

− 27 g 3 ( L i ) 2

We now look at the other values of τ . Consider the function h ( τ )= j ( τ ) − z .

Then h has a pole of order 1 at i ∞ and no other poles. By Proposition 9.16,

we have

2 ord i ( h )+
z

1

3 ord ρ ( h )+ 1

ord z ( h )=1 .

= i,ρ, ∞

If z =0 , 1728, then h has order 0 at ρ and at i . Therefore, h has a unique zero
in F ,so j ( τ )= z has a unique solution in F .If z = 1728, then (1 / 2)ord i ( h ) >
0. Since the order of h at a point is an integer, the order must be 0 when z =
i, ρ ; otherwise, the sum would be larger than 1. Also, there is no combination
of m/ 2+ n/ 3 that equals 1 except when either m =0or n = 0. Therefore,
j ( τ ) − 1728 has a double zero at i and no other zero in F . Similarly, j ( τ )has

atriplezeroat ρ and no other zero in F .

COROLLARY 9.19

Let τ 1 ,τ 2 ∈H .Then j ( τ 1 )= j ( τ 2 ) ifand onlyifthere exists ab

∈ SL 2 ( Z )

cd

su ch that

aτ 1 + b
cτ 1 + d

= τ 2 .

PROOF Proposition 9.13 gives one direction of the statement. Assume
conversely that j ( τ 1 )= j ( τ 2 ). Let τ 1 ,τ 2 ∈F map to τ 1 ,τ 2 via the action of