Cryptography Reference
In-Depth Information
to
P
.Then
F
=(
v
0
u
u
0
v
)
k
H
(
u, v
)forsome
H
(
u, v
) with
H
(
u
0
,v
0
)
−
=0.
Therefore,
F
u
(
u, v
)=
kv
0
(
v
0
u − u
0
v
)
k−
1
H
(
u, v
)+(
v
0
u − u
0
v
)
k
H
u
(
u, v
)
and
F
v
(
u, v
)=
u
0
v
)
k−
1
H
(
u, v
)+(
v
0
u
u
0
v
)
k
H
v
(
u, v
)
.
−
ku
0
(
v
0
u
−
−
It follows that
k ≥
2 if and only if
F
u
(
u
0
,v
0
)=
F
v
(
u
0
,v
0
)=0.
Suppose
k ≥
2. The chain rule yields
F
u
=
a
1
F
x
+
a
2
F
y
+
a
3
F
z
=0
,
F
v
=
b
1
F
x
+
b
2
F
y
+
b
3
F
z
= 0
(2.3)
at
P
. Recall that since the parameterization (2.2) yields a line, the vectors
(
a
1
,a
2
,a
3
)and(
b
1
,b
2
,b
3
) must be linearly independent.
Suppose
L
is another line that intersects
C
to order at least 2. Then we
obtain another set of equations
a
1
F
x
+
a
2
F
y
+
a
3
F
z
=0
,
b
1
F
x
+
b
2
F
y
+
b
3
F
z
=0
at
P
.
If the vectors
a
=(
a
1
,a
2
,a
3
)and
b
=(
b
1
,b
2
,b
3
) span the same plane in
K
3
as
a
=(
a
1
,a
2
,a
3
)and
b
=(
b
1
,b
2
,b
3
), then
a
=
α
a
+
β
b
,
b
=
γ
a
+
δ
b
for some invertible matrix
αβ
γδ
. Therefore,
u
a
+
v
b
=(
uα
+
vγ
)
a
+(
uβ
+
vδ
)
b
=
u
1
a
+
v
1
b
for a new choice of parameters
u
1
,v
1
. This means that
L
and
L
are the same
line.
If
L
and
L
are different lines, then
a
,
b
and
a
,
b
span different planes, so
the vectors
a
,
b
,
a
,
b
must span all of
K
3
.Since(
F
x
,F
y
,F
z
) has dot product
0 with these vectors, it must be the 0 vector. This means that
P
is a singular
point, contrary to our assumption.
Finally, we need to show that the tangent line intersects the curve to order
at least 2. Suppose, for example, that
F
x
=0at
P
. The cases where
F
y
=0
and
F
z
= 0 are similar. The tangent line can be given the parameterization
x
=
−
(
F
y
/F
x
)
u −
(
F
z
/F
x
)
v,
y
=
u,
z
=
v,
so
a
1
=
−F
y
/F
x
,b
1
=
−F
z
/F
x
,a
2
=1
,b
2
=0
,a
3
=0
,b
3
=1
in the notation of (2.2). Substitute into (2.3) to obtain
F
u
=(
−F
y
/F
x
)
F
x
+
F
y
=0
,
F
v
=(
−F
z
/F
x
)
F
x
+
F
z
=0
.
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