Cryptography Reference
In-Depth Information
x
y = x + y ,so y = 0. Similarly, y 1 =
36 yields x = 0. The point with
( x 1 ,y 1 )=
y , which means that z = 0. Therefore,
there are no solutions to x 3 + y 3 + z 3 =0when xyz
corresponds to x =
=0.
2.5.3
Quartic Equations
Occasionally, we will meet curves defined by equations of the form
v 2 = au 4 + bu 3 + cu 2 + du + e,
(2.6)
with a
K ,then
the equation (when it is nonsingular) can be transformed into a Weierstrass
equation by an invertible change of variables that uses rational functions with
coe cients in the field K . Note that an elliptic curve E defined over a field K
always has a point in E ( K ), namely
=0. Ifwehaveapoint( p, q ) lying on the curve with p, q
(whose projective coordinates (0 : 1 : 0)
certainly lie in K ). Therefore, if we are going to transform a curve C into
Weierstrass form in such a way that all coecients of the rational functions
describing the transformation lie in K , then we need to start with a point on
C that has coordinates in K .
There are curves of the form (2.6) that do not have points with coordinates
in K . This phenomenon will be discussed in more detail in Chapter 8.
Suppose we have a curve defined by an equation (2.6) and suppose we have
apoint( p, q ) lying on the curve. By changing u to u + p , we may assume
p = 0, so the point has the form (0 ,q ).
First, suppose q =0. If d = 0, then the curve has a singularity at ( u, v )=
(0 , 0). Therefore, assume d
=0. Then
( v
u 2 ) 2 = d ( 1
u ) 3 + c ( 1
u ) 2 + b ( 1
u )+ a.
This can be easily transformed into a Weierstrass equation in d/u and dv/u 2 .
The harder case is when q = 0. We have the following result.
THEOREM 2.17
Let K be a fi eld of characteristic not 2. C onsider the equation
v 2 = au 4 + bu 3 + cu 2 + du + q 2
with a, b, c, d, q ∈ K .Let
y = 4 q 2 ( v + q )+2 q ( du + cu 2 )
( d 2 u 2 / 2 q )
x = 2 q ( v + q )+ du
u 2
,
.
u 3
D efine
a 2 = c − ( d 2 / 4 q 2 ) ,
a 4 = 4 q 2 a,
a 1 = d/q,
3 =2 qb,
a 6 = a 2 a 4 .
Search WWH ::




Custom Search