Cryptography Reference
In-Depth Information
Lemma 9.6.6
Let E
1
and E
2
be elliptic curves over
k
. Then
Hom
k
(
E
1
,E
2
)
is a
group with addition defined by
(
φ
1
+
=
Hom
k
(
E
1
,E
1
)
is a ring with addition defined in the same way and with multiplication
defined by composition.
φ
2
)(
P
)
=
φ
1
(
P
)
+
φ
2
(
P
)
. Furthermore,
End
k
(
E
1
)
Proof
The main task is to show that if
φ
1
,φ
2
:
E
1
→
E
2
are morphisms then so is (
φ
1
+
φ
2
).
The case
φ
2
=−
φ
1
is trivial, so assume
φ
2
=−
φ
1
.Let
U
be an open set such that:
φ
1
and
φ
2
are regular on
U
;
P
φ
2
(
P
).
That such an open set exists is immediate for all but the final requirement, but one can also
show that the points such that
φ
1
(
x,y
)
∈
U
implies
φ
1
(
P
)
=
O
E
2
and
φ
2
(
P
)
=
O
E
2
;
φ
1
(
P
)
=−
=−
φ
2
(
x,y
) form a closed subset of
E
1
as long
as
φ
1
=−
E
2
.
Finally, since composition of morphisms is a morphism it is easy to check that End
k
(
E
1
)is
a ring.
φ
2
. Then using equation (
9.3
) one obtains a rational map (
φ
1
+
φ
2
):
E
1
→
By Exercise
8.1.12
,if
φ
1
:
E
1
→
E
2
and
φ
2
:
E
2
→
E
3
are non-constant isogenies then
deg(
φ
2
◦
deg(
φ
2
)deg(
φ
1
). This fact will often be used.
An important example of an isogeny is the multiplication by
n
map.
φ
1
)
=
Exercise 9.6.7
Show that [
n
] is an isogeny.
Example 9.6.8
Let
E
:
y
2
x
3
=
+
x
. Then the map [2] :
E
→
E
is given by the rational
function
(
x
2
.
1)
2
x
)
,
y
(
x
6
5
x
4
5
x
2
−
+
−
−
1)
[2](
x,y
)
=
4(
x
3
+
8(
x
3
+
x
)
2
O
E
together with the three points (
x
P
,
0) such that
x
P
+
The kernel of [2] is
x
P
=
0. In
other words, the kernel is the set of four points of order dividing 2.
∈ N
We now give a simple example of an isogeny that is not [
n
]forsome
n
.The
derivation of a special case of this example is given in Example
25.1.5
.
A
2
Example 9.6.9
Let
A,B
∈ k
be such that
B
=
0 and
D
=
−
4
B
=
0. Consider the
elliptic curve
E
:
y
2
x
(
x
2
=
+
Ax
+
B
) over
k
, which has the point (0
,
0) of order 2. There
is an elliptic curve
E
and an isogeny
φ
:
E
→
E
such that ker(
φ
)
={
O
E
,
(0
,
0)
}
. One can
verify that
y
2
x
2
−
x
2
)
+
+
−
x
2
x
2
,
y
(
B
Ax
B
,y
B
=
=
φ
(
x,y
)
x
2
x
has the desired kernel, and the image curve is
E
:
Y
2
=
X
(
X
2
−
+
2
AX
D
).
Before proving the next result we need one exercise (which will also be used later).
Exercise 9.6.10
Let
y
2
x
3
a
2
x
2
+
a
1
xy
+
a
3
y
=
+
+
a
4
x
+
a
6
be an elliptic curve over
k
. Show that if char(
k
)
=
2 and
a
1
=
0 then there are no points (
x,y
) of order 2. Show
that if char(
k
)
=
2 and
a
1
=
0 then (
x,y
) has order 2 if and only if
x
=
a
3
/a
1
. Hence, if
char(
k
)
=
2 then #
E
[2]
∈{
1
,
2
}
.
