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 ,
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