Cryptography Reference
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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
iμ
(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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