Cryptography Reference
In-Depth Information
We need to quantify the order to which a line intersects a curve at a point.
The following gets us started.
LEMMA 2.2
Let
G
(
u, v
)
be a nonzero hom ogeneous polynom ial and let
(
u
0
:
v
0
)
∈
P
1
K
.
Thenthere existsaninteger
k ≥
0
and a polynom ial
H
(
u, v
)
with
H
(
u
0
,v
0
)
=
0
su ch that
u
0
v
)
k
H
(
u, v
)
.
G
(
u, v
)=(
v
0
u
−
PROOF
Suppose
v
0
=0. Let
m
be the degree of
G
.Let
g
(
u
)=
G
(
u, v
0
).
By factoring out as large a power of
u − u
0
as possible, we can write
g
(
u
)=
(
u − u
0
)
k
h
(
u
)forsome
k
and for some polynomial
h
of degree
m − k
with
h
(
u
0
)
=0. Let
H
(
u, v
)=(
v
m−k
/v
0
)
h
(
uv
0
/v
), so
H
(
u, v
) is homogeneous of
degree
m − k
.Then
G
(
u, v
)=
v
v
0
m
g
uv
0
v
=
v
m−k
v
0
(
v
0
u − u
0
v
)
k
h
uv
0
v
u
0
v
)
k
H
(
u, v
)
,
=(
v
0
u
−
as desired.
If
v
0
=0,then
u
0
= 0. Reversing the roles of
u
and
v
yields the proof in
this case.
Let
f
(
x, y
)=0(where
f
is a polynomial) describe a curve
C
in the ane
plane and let
x
=
a
1
t
+
b
1
,y
=
a
2
t
+
b
2
be a line
L
written in terms of the parameter
t
.Let
f
(
t
)=
f
(
a
1
t
+
b
1
,a
2
t
+
b
2
)
.
Then
L
intersects
C
when
t
=
t
0
if
f
(
t
0
)=0. If(
t − t
0
)
2
divides
f
(
t
),
then
L
is tangent to
C
(if the point corresponding to
t
0
is nonsingular. See
Lemma 2.5). More generally, we say that
L
intersects
C
to order
n
at the
point (
x, y
) corresponding to
t
=
t
0
if (
t − t
0
)
n
is the highest power of (
t − t
0
)
that divides
f
(
t
).
The homogeneous version of the above is the following. Let
F
(
x, y, z
)bea
homogeneous polynomial, so
F
= 0 describes a curve
C
in
P
2
K
.Let
L
be a
line given parametrically by (2.2) and let
F
(
u, v
)=
F
(
a
1
u
+
b
1
v, a
2
u
+
b
2
v, a
3
u
+
b
3
v
)
.
We say that
L
intersects
C
to order
n
at the point
P
=(
x
0
:
y
0
:
z
0
)
corresponding to (
u
:
v
)=(
u
0
:
v
0
)if(
v
0
u − u
0
v
)
n
is the highest power of
F
(
u, v
). We denote this by
(
v
0
u − u
0
v
) dividing
ord
L,P
(
F
)=
n.
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