Cryptography Reference
In-Depth Information
We refer to Section III.1 of [
505
] for the definition of the
j
-invariant for general
Weierstrass equations.
Theorem 9.3.6
Let
k
be a field and
E
1
, E
2
elliptic curves over
k
. Then there is an
isomorphism from E
1
to E
2
defined over
k
if and only if j
(
E
1
)
=
j
(
E
2
)
.
Proof
See Proposition III.1.4(b) of [
505
] or Theorem 2.19 of [
560
].
Exercise 9.3.7
Let
E
:
y
2
x
3
=
+
a
4
x
+
a
6
be an elliptic curve. Show that
j
(
E
)
=
0 if and
only if
a
4
=
0 and
j
(
E
)
=
1728 if and only if
a
6
=
0. Suppose char(
k
)
=
2
,
3. Let
j
∈ k
,
j
=
0
,
1728. Show that the elliptic curve
3
j
1728
2
j
1728
E
:
y
2
x
3
=
+
j
x
+
−
−
j
has
j
(
E
)
=
j
.
Exercise 9.3.8
Let
E
1
:
y
2
x
3
,
E
2
:
y
2
x
3
1 and
E
3
:
y
2
x
3
+
y
=
+
y
=
+
+
y
=
+
x
be
F
2
. Since
j
(
E
1
)
=
=
=
elliptic curve
s
over
j
(
E
2
)
j
(
E
3
)
0 it follows that there are isomor-
F
2
from
E
1
to
E
2
and from
E
1
to
E
3
. Write down such isomorphisms.
phisms over
Exercise 9.3.9
Let
E
1
,E
2
be elliptic curves over
F
q
that are isomorphic over
F
q
. Show that
the discrete logarithm problem in
E
1
(
F
q
) is equivalent to the discrete logarithm problem
in
E
2
(
F
q
). In other words, the discrete logarithm problem on
F
q
-isomorphic curves has
exactly the same security.
9.4 Automorphisms
Definition 9.4.1
Let
E
be an elliptic curv
e
over
k
.An
automorphism
of
E
is an isomor-
phism from (
E,
O
E
) to itself defined over
k
. The set of all automorphisms of
E
is denoted
Aut(
E
).
O
E
. Under composition, Aut(
E
)formsa
group, The identity of the group is the identity map.
We stress that an automorphism maps
O
E
to
Example 9.4.2
The map
φ
(
P
)
=−
P
is an automorphism that is not the identit
y
map. On
y
2
x
3
=
+
1 over
k
,themap
ρ
(
x,y
)
=
(
ζ
3
x,y
) is an automorphism where
ζ
3
∈ k
satisfies
ζ
3
=
1.
E
xercise 9.4.3
Show that if
E
1
and
E
2
are elliptic curves over
k
that are isomorphic over
then Aut(
E
1
)
=
k
Aut(
E
2
).
k
|
Theorem 9.4.4
Let E be an elliptic curve over
. Then
#Aut(
E
)
is even and
#Aut(
E
)
24
.
More precisely:
#Aut(
E
)
=
2
if j
(
E
)
=
0
,
1728
,
#Aut(
E
)
=
4
if j
(
E
)
=
1728
and
char(
k
)
=
2
,
3
,
#Aut(
E
)
=
6
if j
(
E
)
=
0
and
char(
k
)
=
2
,
3
,