Cryptography Reference
In-Depth Information
9.3 Isomorphisms of elliptic curves
We have already defined isomorphisms of algebraic varieties. It is natural to ask when two
Weierstrass equations are isomorphic. Since one can compose any isomorphism with a
translation map it is sufficient to restrict attention to isomorphisms
φ
:
E
→
E
such that
φ
(
O
E
)
=
O
E
.
Formally, one defines a
pointed curve
to be a curve
C
over a field
k
together with a
(
C,P
0
)isan
fixed
k
-rational point
P
0
.An
is
o
morphism of pointed curves
φ
:(
C,P
0
)
→
→
C
over
=
P
0
. When one refers to an
isomorphism
φ
:
C
k
of varieties such that
φ
(
P
0
)
elliptic curve one usually means the pointed curve (
E,
O
E
).
O
E
) and (
E,
Definition 9.3.1
Let (
E,
O
E
) be elliptic
c
urves over
k
.An
isomorphism
→
E
is an isomorphism over
of elliptic curves
φ
:
E
k
of algebraic varieties such that
=
O
E
. If there is an isomorphism from
E
to
E
then we write
E
=
E
.
φ
(
O
E
)
By Theorem
9.2.1
, an isomorphism of elliptic curves is a group homomorphism over
k
.
Exercise 9.3.2
Let
E
1
and
E
2
be elliptic curves over
k
. Show that if
E
1
is isomorphic over
k
to
E
2
then
E
1
(
k
) is isomorphic as a group to
E
2
(
k
). In particular, if
k = F
q
is a finite
field then #
E
1
(
F
q
)
=
#
E
2
(
F
q
).
Note that the translation map
τ
Q
is not considered to be an isomorphism of the pointed
curve (
E,
O
E
) to itself, unless
Q
=
O
E
in which case
τ
Q
is the identity map.
Exercise 9.3.3
Exercises
7.2.6
and
7.2.7
give simplified Weierstrass models for elliptic
curves when char(
3. Verify that there are isomorphisms, from a general Weierstrass
equation to these models that fix
k
)
=
O
E
.
Theorem 9.3.4
Let
k
be
a field and E
1
, E
2
elliptic curves over
k
. Every isomorphism from
E
1
to E
2
defined over
k
restricts to an affine isomorphism of the form
(
u
2
x
r,u
3
y
su
2
x
φ
(
x,y
)
=
+
+
+
t
)
(9.8)
where u,r,s,t
∈ k
. The isomorphism is defined over
k
if and only if u,r,s,t
∈ k
.
Proof
See Proposition III.3.1(b) of [
505
].
k
=
2
,
3 and let
a
4
,a
6
∈ k
be such that 4
a
4
+
27
a
6
=
Definition 9.3.5
Suppose char(
)
0.
For the short Weierstrass equation
y
2
z
=
x
3
+
a
4
xz
2
+
a
6
z
3
, define the
j
-invariant
4
a
4
4
a
4
+
j
(
E
)
=
1728
.
27
a
6
Suppose char(
k
)
=
2 and
a
2
,a
6
∈ k
with
a
6
=
0. For the short Weierstrass equation
y
2
z
x
3
a
2
x
2
z
a
6
z
3
define the
j
-invariant
+
xyz
=
+
+
j
(
E
)
=
1
/a
6
and for
E
:
y
2
z
yz
2
x
3
a
4
xz
2
a
6
z
3
(we now allow
a
6
=
+
=
+
+
0) define
j
(
E
)
=
0.
