Cryptography Reference
In-Depth Information
∂F/∂T
=
0; as a function on
C
we have
F
(
x,f
)
=
0 and so
∂F
∂x
δ
(
x
)
∂F
∂T
δ
(
f
)
.
0
=
δ
(
F
(
x,f
))
=
+
(8.1)
This motivates the following definition.
Definition 8.5.9
Let
C
be a curve over
k
and let
x
∈ k
(
C
) be a separating element. Let
y
∈ k
(
C
). Let
F
(
x,T
) be a rational function such that
F
(
x,y
)
=
0. Define
∂y
∂x
=−
(
∂F/∂x
)
/
(
∂F/∂T
)
evaluated at
y
.
Lemma 8.5.10
The value ∂y/∂x in Definition
8.5.9
is well-defined. More precisely, if F
and F
are rational functions such that F
(
x,y
)
F
(
x,y
)
=
=
0
then
(
∂F/∂x
)
/
(
∂F/∂y
)
=
(
∂F
/∂x
)
/
(
∂F
/∂y
)
and if z
≡
y in
k
(
C
)
then ∂z/∂x
≡
∂y/∂x.
Proof
The first claim follows from equation (
8.1
). For the second claim, if
z
=
y
in
k
(
C
)
then they satisfy the same minimal polynomial.
It remains to show that this construction does give a derivation.
Lemma 8.5.11
Let C be a curve over
k
and x
∈ k
(
C
)
a separating element. The function
δ
:
k
(
C
)
→ k
(
C
)
defined by δ
(
y
)
=
∂y/∂x as in Definition
8.5.9
is
k
-linear and satisfies
the product rule.
Furthermore, if f
=
H
(
y
)
∈ k
(
C
)
is another function, where H
(
T
)
∈ k
(
x
)[
T
]
is a
polynomial, then
∂H
∂x
−
(
∂F/∂x
)
/
(
∂F/∂T
)
∂H
∂T
δ
(
f
)
=
(8.2)
evaluated at y, where F is as in Definition
8.5.9
.
Proof
See Proposition IV.1.4 and Lemma IV.1.3 of Stichtenoth [
529
].
Example 8.5.12
Let
C
:
y
2
x
3
=
+
x
+
1 over
Q
. Note that
x
is a separating element. To
y
2
(
x
3
compute
∂y/∂x
one uses the fact that
F
(
x,y
)
=
−
+
x
+
1)
=
0in
k
(
C
) and so
(3
x
2
∂y/∂x
1)
/
(2
y
).
Consider the function
f
(
x,y
)
=
+
=
xy
and let
δ
(
f
)
=
∂f/∂x
. Then
δ
(
f
)
=
xδ
(
y
)
+
y
=
x
(3
x
2
+
+
=
(3
x
3
+
+
2
y
2
)
/
(2
y
)
=
(5
x
3
+
+
1)
/
(2
y
)
y
x
3
x
2)
/
(2
y
).
(
x
3
Exercise 8.5.13
Let
k
(
C
) be as in Example
8.5.12
. Show that
δ
(
y/x
)
=
−
x
−
2)
/
(2
yx
2
).
Lemma 8.5.14
Let C be a curve over
k
and let x,y
∈ k
(
C
)
be separating elements. Then
the corresponding derivations on
k
(
C
)
satisfy the chain rule, namely
∂f
∂y
=
∂f
∂x
∂x
∂y
.
