Cryptography Reference
In-Depth Information
lie on
E
, except possibly for
P
33
. We show in Theorem 2.6 that having the
eight points
P
ij
=
P
33
on
E
forces
P
33
to be on
E
.Since
3
intersects
E
at
the points
R, P
+
Q,
−
((
P
+
Q
)+
R
), we must have
−
((
P
+
Q
)+
R
)=
P
33
.
Similarly,
−
(
P
+(
Q
+
R
)) =
P
33
,so
−
((
P
+
Q
)+
R
)=
−
(
P
+(
Q
+
R
))
,
which implies the desired associativity.
There are three main technicalities that must be treated. First, some of
the points
P
ij
could be at infinity, so we need to use projective coordinates.
Second, a line could be tangent to
E
, which means that two
P
ij
could be
equal. Therefore, we need a careful definition of the order to which a line
intersects a curve. Third, two of the lines could be equal. Dealing with these
technicalities takes up most of our attention during the proof.
First, we need to discuss lines in
P
2
K
. The standard way to describe a line
is by a linear equation:
ax
+
by
+
cz
= 0. Sometimes it is useful to give a
parametric description:
x
=
a
1
u
+
b
1
v
y
=
a
2
u
+
b
2
v
(2.2)
z
=
a
3
u
+
b
3
v
where
u, v
run through
K
, and at least one of
u, v
is nonzero. For example, if
a
= 0, the line
ax
+
by
+
cz
=0
can be described by
x
=
−
(
b/a
)
u
−
(
c/a
)
v, y
=
u, z
=
v.
Suppose all the vectors (
a
i
,b
i
) are multiples of each other, say (
a
i
,b
i
)=
λ
i
(
a
1
,b
1
). Then (
x, y, z
)=
x
(1
,λ
2
,λ
3
) for all
u, v
such that
x
=0. Soweget
a point, rather than a line, in projective space. Therefore, we need a condition
on the coecients
a
1
,...,b
3
that ensure that we actually get a line. It is not
hard to see that we must require the matrix
⎛
⎞
a
1
b
1
a
2
b
2
a
3
b
3
⎝
⎠
to have rank 2 (cf. Exercise 2.12).
If (
u
1
,v
1
)=
λ
(
u
2
,v
2
)forsome
λ ∈ K
×
,then(
u
1
,v
1
)and(
u
2
,v
2
) yield
equivalent triples (
x, y, z
). Therefore, we can regard (
u, v
) as running through
points (
u
:
v
) in 1-dimensional projective space
P
1
K
. Consequently, a line
corresponds to a copy of the projective line
P
1
K
embedded in the projective
plane.
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