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with
A 1 = μ 4 A, B 1 = μ 6 B.
(In the generalized Weierstrass equation y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x +
a 6 , this change of variables yields new coecients μ i a i . This explains the
numbering of the coecients.)
Define the j-invariant of E to be
4 A 3
4 A 3 +27 B 2 .
j = j ( E ) = 1728
Note that the denominator is the negative of the discriminant of the cubic,
hence is nonzero by assumption. The change of variables (2.8) leaves j un-
changed. The converse is true, too.
THEOREM 2.19
Let y 1 = x 1 + A 1 x 1 + B 1 and y 2 = x 2 + A 2 x 2 + B 2 be twoelliptic curves w ith
j -invariants j 1 and j 2 , respectively. If j 1 = j 2 ,then there exists μ
=0 in K
(= algebraicc osure of K )suchthat
A 2 = μ 4 A 1 ,
B 2 = μ 6 B 1 .
Thetra n sform ation
x 2 = μ 2 x 1 ,
2 = μ 3 y 1
takes one equation tothe other.
PROOF
First, assume that A 1
= 0. Since this is equivalent to j 1
=0,we
=0. Choose μ such that A 2 = μ 4 A 1 .Then
also have A 2
4 A 2
4 A 2 +27 B 2
4 A 1
4 A 1 +27 B 1
4 μ 12 A 2
4 μ 12 A 2 +27 B 1
4 A 2
4 A 2 +27 μ 12 B 1
=
=
=
,
which implies that
B 2 =( μ 6 B 1 ) 2 .
μ 6 B 1 .If B 2 = μ 6 B 1 , we're done. If B 2 =
μ 6 B 1 ,then
Therefore B 2 =
±
change μ to (where i 2 =
1). This preserves the relation A 2 = μ 4 A 1 and
also yields B 2 = μ 6 B 1 .
If A 1 =0,then A 2 =0. Since4 A i +27 B i
=0,wehave B 1 ,B 2
=0. Choose
μ such that B 2 = μ 6 B 1 .
There are two special values of j that arise quite often:
1. j = 0: In this case, the elliptic curve E has the form y 2 = x 3 + B .
2. j = 1728: In this case, the elliptic curve has the form y 2 = x 3 + Ax .
 
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