Cryptography Reference
In-Depth Information
[
x
] be such that
y
2
Let
H
(
x
)
,F
(
x
)
∈ k
+
H
(
x
)
y
=
F
(
x
) is a non-singular affine curve.
2
is given by
Define
D
=
max
{
deg(
F
(
x
))
,
deg(
H
(
x
))
+
1
}
. The projective closure of
C
in
P
y
2
z
D
−
2
z
D
−
1
H
(
x/z
)
y
z
D
F
(
x/z
)
.
+
=
(10.1)
Exercise 10.1.3
Show that if
D>
2 then there are at most two points at infinity on the
curve of equation (
10.1
). Show further that if
D>
3 and deg(
F
)
>
deg(
H
)
+
1 then there
is a unique point (0 : 1 : 0) at infinity, which is a singular point.
In Definition
10.1.10
we will define the genus of a hyperelliptic curve in terms of the
degree of the hyperelliptic equation. To do this, it will be necessary to have conditions
that ensure that this degree is minimal. Example
10.1.4
and Exercise
10.1.5
show how a
hyperelliptic equation that is a variety can be isomorphic to an equation of significantly
lower degree (remember that isomorphism is only defined for varieties).
Example 10.1.4
The curve
y
2
x
200
x
101
x
3
+
xy
=
+
+
+
1 over
F
2
is isomorphic over
F
2
to the curve
Y
2
x
3
x
100
).
+
xY
=
+
1 via the map (
x,y
)
→
(
x,Y
+
be any field. Show that the affine algebraic variety
y
2
2
x
3
)
y
Exercise 10.1.5
Let
k
+
(1
−
=
x
6
x
3
−
1 is isomorphic to a variety having an equation of total degree 2. Show
that the resulting curve has genus 0.
+
+
x
+
Lemma 10.1.6
Let
k
be a perfect field of characteristic 2 and h
(
x
)
,f
(
x
)
∈ k
[
x
]
. Suppose
the hyperelliptic equation C
:
y
2
+
h
(
x
)
y
=
f
(
x
)
is a variety. Then it is isomorphic over
to Y
2
k
+
H
(
x
)
Y
=
F
(
x
)
where one of the following conditions hold:
1.
deg(
F
(
x
))
>
2deg(
H
(
x
))
and
deg(
F
(
x
))
is odd;
2.
deg(
F
(
x
))
2
d and the equation u
2
=
2deg(
H
(
x
))
=
+
H
d
u
+
F
2
d
has no solution in
k
H
d
x
d
H
d
−
1
x
d
−
1
F
2
d
x
2
d
(where H
(
x
)
=
+
+···+
H
0
and F
(
x
)
=
+···+
F
0
);
3.
deg(
F
(
x
))
<
deg(
H
(
x
))
.
cx
i
Proof
Let
d
H
=
deg(
H
(
x
)) and
d
F
=
deg(
F
(
x
)). The change of variables
y
=
Y
+
transforms
y
2
H
(
x
)
cx
i
. Hence, if
deg(
F
(
x
))
>
2deg(
H
(
x
)) and deg(
F
(
x
)) is eve
n th
en one can remove the leading coef-
ficient by choosing
i
F
(
x
)to
Y
2
c
2
x
2
i
+
H
(
x
)
y
=
+
H
(
x
)
Y
=
F
(
x
)
+
+
=
√
F
2
i
(remember that char(
=
deg(
F
(
x
))
/
2 and
c
k
)
=
2 and
k
is
perfect so
c
deg(
F
(
x
))
<
2deg(
H
(
x
)) then one can
remove the leading coefficient
F
j
x
i
from
F
by taking
i
∈ k
). Similarly, if deg(
H
(
x
))
≤
j
=
F
j
/H
d
H
.
Repeating these processes yields the first and third claims. The second case follows
easily.
=
j
−
deg(
H
(
x
)) and
c
=
Note that in the second case in Lemma
1
0
.1.6
one can lower the degree using a
k
-
isomorphism. Hence, geometrically (i.e., over
k
) one can assume that a hyperelliptic equa-
tion is of the form of case 1 or 3.
