Cryptography Reference
In-Depth Information
We have
F
x
=
−
3
x
2
− Az
2
,
F
z
=
y
2
−
2
Axz −
3
Bz
2
.
F
y
=2
yz,
Suppose
P
=(
x
:
y
:
z
) is a singular point. If
z
=0,then
F
x
= 0 implies
x
=0and
F
z
= 0 implies
y
=0,so
P
= (0 : 0 : 0), which is impossible.
Therefore
z
=0,sowemaytake
z
= 1 (and therefore ignore it). If
F
y
=0,
then
y
=0. Since(
x
:
y
: 1) lies on the curve,
x
must satisfy
x
3
+
Ax
+
B
=0.
If
F
x
=
−
(3
x
2
+
A
)=0,then
x
is a root of a polynomial and a root of its
derivative, hence a double root. Since we assumed that the cubic polynomial
has no multiple roots, we have a contradiction. Therefore an elliptic curve has
no singular poi
nts
. Note that this is true even if we are considering points with
coordinates in
K
(= algebraic closure of
K
). In g
en
eral, by a
nonsingular
curve
we mean a curve with no singular points in
K
.
If we allow the cubic polynomial to have a multiple root
x
,thenitiseasyto
see that the curve has a singularity at (
x
: 0 : 1). This case will be discussed
in Section 2.10.
If
P
is a nonsingular point of a curve
F
(
x, y, z
) = 0, then the tangent line
at
P
is
F
x
(
P
)
x
+
F
y
(
P
)
y
+
F
z
(
P
)
z
=0
.
For example, if
F
(
x, y, z
)=
y
2
z − x
3
− Axz
2
− Bz
3
= 0, then the
tangent
line
at (
x
0
:
y
0
:
z
0
)is
(
−
3
x
0
− Az
0
)
x
+2
y
0
z
0
y
+(
y
0
−
2
Ax
0
z
0
−
3
Bz
0
)
z
=0
.
If we set
z
0
=
z
= 1, then we obtain
(
−
3
x
0
− A
)
x
+2
y
0
y
+(
y
0
−
2
Ax
0
−
3
B
)=0
.
Using the fact that
y
0
=
x
0
+
Ax
0
+
B
,wecanrewritethisas
3
x
0
−
(
−
A
)(
x
−
x
0
)+2
y
0
(
y
−
y
0
)=0
.
This is the tangent line in a
ne coordinates that we used in obtaining the
formulas for adding a point to itself on an elliptic curve. Now let's look at
the point at infinity on this curve. We have (
x
0
:
y
0
:
z
0
) = (0 : 1 : 0). The
tangent line is given by 0
x
+0
y
+
z
= 0, which is the “line at infinity” in
P
2
K
.
It intersects the elliptic curve only in the point (0 : 1 : 0). This corresponds
to the fact that
∞
+
∞
=
∞
on an elliptic curve.
LEMMA 2.5
Let
F
(
x, y, z
)=0
define a curve
C
.If
P
is a nonsingular point of
C
,then
there isexactlyone ine in
P
2
K
that intersects
C
toorderat east 2, and itis
the tangent to
C
at
P
.
PROOF
Let
L
be a line intersecting
C
to order
k ≥
1. Parameterize
L
by (2.2) and substitute into
F
. This yields
F
(
u, v
). Let (
u
0
:
v
0
) correspond
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