Cryptography Reference
In-Depth Information
This can be proved as follows. Factor off the highest possible power of
v
,say
v
k
.Then
S
(
u, v
)vanishestoorder
k
at (1 : 0), and
S
(
u, v
)=
v
k
S
0
(
u, v
) with
S
0
(1
,
0)
=0. Since
S
0
(
u,
1) is a polynomial of degree 3
−
k
, the polynomial
S
0
(
u,
1) can have at most 3
−
k
zeros, counting multiplicities (it has exactly
3
= (1 : 0) can be written
in the form (
u
:1),so
S
0
(
u, v
) has at most 3
− k
zeros. Therefore,
S
(
u, v
)has
at most
k
+(3
− k
) = 3 zeros in
P
1
K
.
It follows easily that the condition that
S
(
u, v
) vanish to order at least
k
could be replaced by the condition that
S
(
u, v
) vanish to order exactly
k
.
However, it is easier to check “at least” than “exactly.” Since we are allowing
the possibility that
R
(
u, v
) is identically 0, this remark does not apply to
R
.
Let (
u
0
,
:
v
0
)beanypointin
P
1
K
not equal to any of the (
u
i
:
v
i
). (
Technical
point
:If
K
has only two elements, then
P
1
K
has only three elements. In this
case, enlarge
K
to
GF
(4). The
α
we obtain is forced to be in
K
since it is the
ratio of a coecient of
R
and a coecient of
S
, both of which are in
K
.) Since
S
can have at most three zeros,
S
(
u
0
,v
0
)
=0. Let
α
=
R
(
u
0
,v
0
)
/S
(
u
0
,v
0
).
Then
R
(
u, v
)
− αS
(
u, v
) is a cubic homogeneous polynomial that vanishes at
the four points (
u
i
:
v
i
),
i
=0
,
1
,
2
,
3. Therefore
R − αS
must be identically
zero.
−
k
if
K
is algebraically closed). All points (
u
:
v
)
Returning to the proof of the theorem, we note that
C
and
m
1
m
2
m
3
vanish
at the points (
u
i
:
v
i
),
i
=1
,
2
,
3. Moreover, if
k
of the points
P
1
j
are the
same point, then
k
of the linear functions vanish at this point, so the product
m
1
(
u, v
)
m
2
(
u, v
)
m
3
(
u, v
) vanishes to order at least
k
. By assumption,
C
vanishes to order at least
k
in this situation. By the lemma, there exists a
constant
α
such that
C
=
α m
1
m
2
m
3
.
Let
C
1
(
x, y, z
)=
C
(
x, y, z
)
−
αm
1
(
x, y, z
)
m
2
(
x, y, z
)
m
3
(
x, y, z
)
.
The line
1
can be described by a linear equation
1
(
x, y, z
)=
ax
+
by
+
cz
=
0. At least one coecient is nonzero, so let's assume
a
= 0. The other cases
are similar. The parameterization of the line
1
can be taken to be
x
=
−
(
b/a
)
u −
(
c/a
)
v,
y
=
u,
z
=
v.
(2.4)
C
1
(
u, v
)=
C
1
(
Then
(
c/a
)
v, u, v
). Write
C
1
(
x, y, z
) as a polynomial
in
x
with polynomials in
y, z
as coecients. Writing
−
(
b/a
)
u
−
x
n
=(1
/a
n
)((
ax
+
by
+
cz
)
−
(
by
+
cz
))
n
=(1
/a
n
)((
ax
+
by
+
cz
)
n
+
···
)
,
we can rearrange
C
1
(
x, y, z
) to be a polynomial in
ax
+
by
+
cz
whose coe
-
cients are polynomials in
y, z
:
C
1
(
x, y, z
)=
a
3
(
y, z
)(
ax
+
by
+
cz
)
3
+
···
+
a
0
(
y, z
)
.
(2.5)
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