Cryptography Reference
In-Depth Information
Exercise 8.5.19
Let
C
be a curve over
k
.Let
x
1
,x
2
∈ k
(
C
) be separating elements and
h
1
,h
2
∈ k
(
C
). Show that
h
1
dx
1
is equivalent to
h
2
dx
2
if and only if
h
1
∂x
1
h
2
=
∂x
2
.
1
). Since
1
)
Example 8.5.20
We determine
k
(
P
k
(
P
= k
(
x
) the differentials are
d
(
f
(
x
))
=
(
∂f/∂x
)
dx
for
f
(
x
)
∈ k
(
x
).
The following theorem, that all differentials on a curve are multiples of
dx
where
x
is a
separating element, is a direct consequence of the definition.
Theorem 8.5.21
Let C be a curve over
k
and let x be a separating element. Let ω
∈
k
(
C
)
.
Then ω
=
hdx for some h
∈ k
(
C
)
.
Exercise 8.5.22
Prove Theorem
8.5.21
.
This result shows that
k
(
C
)isa
k
(
C
)-vector space of dimension 1 (we know that
k
(
C
)
={
0
}
since
dx
=
0if
x
is a separating element). Therefore, for any
ω
1
,ω
2
∈
k
(
C
)
with
ω
2
=
0 there is a unique function
f
∈ k
(
C
) such that
ω
1
=
fω
2
. We define
ω
1
/ω
2
to
be
f
(see Proposition II.4.3 of Silverman [
505
]).
We now define the divisor of a differential by using uniformisers. Recall from
Lemma
8.5.8
that a uniformiser
t
P
is a separating element and so
dt
P
=
0.
Definition 8.5.23
Let
C
be a curve over
k
.Let
ω
∈
k
(
C
),
ω
=
0 and let
P
∈
C
(
k
)have
uniformiser
t
P
∈ k
(
C
). Then the
order
of
ω
at
P
is
v
P
(
ω
):
=
v
P
(
ω/dt
P
). The
divisor of a
differential
is
div(
ω
)
=
v
P
(
ω
)(
P
)
.
P
∈
C
(
k
)
Lemma 8.5.24
Let C be a c
ur
ve over
k
and let ω be a differential on C. Then v
P
(
ω
)
=
0
for only finitely many P
∈
C
(
k
)
and so
div(
ω
)
is a divisor.
Proof
See Proposition II.4.3(e) of Silverman [
505
].
Exercise 8.5.25
Show that
v
P
(
hdx
)
=
v
P
(
h
)
+
v
P
(
dx
) and
v
P
(
df
)
=
v
P
(
∂f/∂t
P
).
Lemma 8.5.26
The functions v
P
(
ω
)
and
div(
ω
)
in Definition
8.5.23
are well-defined (both
with respect to the choice of representative for ω and choice of t
P
).
Exercise 8.5.27
Prove Lemma
8.5.26
.
and ω,ω
∈
Lemma 8.5.28
Let C be a curve over
k
k
(
C
)
. Then:
deg(div(
ω
))
.
2.
div(
ω
)
is well-defined up to principal divisors.
1.
deg(div(
ω
))
=
Exercise 8.5.29
Prove Lemma
8.5.28
.
