Cryptography Reference
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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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