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Therefore, either λL or iλL the desired lattice.
If g 2 ( L )=0,then j = j ( L )=0,so A =0. Since A 3
27 B 2
=0by
assumption and since g 2 ( L ) 3
27 g 3 ( L ) 2
= 0 by Proposition 9.9, we have
B =0and g 3 ( L ) = 0. Choose μ ∈ C × such that
g 3 ( μL )= μ 6 g 3 ( L )= B.
Then g 2 ( μL )= μ 4 g 2 ( L )=0= A ,so μL is the desired lattice.
By Theorem 9.10, the map
C /L −→ E ( C )
is an isomorphism.
The elements of L are called the periods of L .
Theorem 9.21 gives us a good way to work with elliptic curves over C .For
example, let n be a positive integer and let E be an elliptic curve over C .
By Theorem 9.21, there exists a lattice L = Z ω 1 + Z ω 2 such that C /L is
isomorphic to E ( C ). It is easy to see that the n -torsion on C /L is given by
the points
n ω 1 + k
j
n ω 2 ,
0
j, k
n
1 .
It follows that
E [ n ]
Z n
Z n .
In fact, we can use this observation to give a proof of Theorem 3.2 for all fields
of characteristic 0.
COROLLARY 9.22
Let K be a fi eld of characteristic0,and et E be an elliptic curve over K .
Then
E [ n ]=
{
P
E ( K )
|
nP =
∞} Z n Z n .
PROOF Let L be the field generated by Q and the coecients of the
equation of E .Then L has finite transcendence degree over Q , hence can be
embedded into C (see Appendix C). Therefore, we can regard E as an elliptic
curve over C . Therefore, the n -torsion is Z n Z n .
There is a technical point to worry about. The definition of E [ n ]that
we have used requires the coordinates of the n -torsion to lie in the algebraic
closure of the base field. How can we be sure that the field K isn't so large
that it allows more torsion points than C ? Suppose that E [ n ] ⊂ E ( K )has
order larger than n 2 .Thenwecanchoose n 2 + 1 of these points and adjoin
their coordinates to L .Then L still has finite transcendence degree over Q ,
hence can be embedded into C . The coordinates of the n 2 +1 points will yield
 
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