The Basic Theory - Elliptic Curves: Number Theory and Cryptography - page 62

THEOREM 2.31
Let E be the curve y 2

= x 2 ( x + a ) with 0 K .Let E ns ( K ) be the

nonsingular pointson E withcoord nates in K .Let α 2

= a ∈

= a .Cons der the

map

y + αx
y

ψ :( x, y ) →

αx ,

∞ → 1 .

−

1. If α ∈ K ,then ψ gives an isom orphism fro m E ns ( K ) to K × ,considered
as a m ultiplicative group.

2. If α ∈ K ,then ψ gives an isom orphism

K, u 2 av 2 =1 E ns ( K ) { u + αv | u, v ∈ − } ,

wherethe right hand side is a group under m ultiplication.

PROOF Let

t = y + αx

y − αx .
This may be solved for y/x to obtain

y

x = α t +1

t − 1 .

Since x + a =( y/x ) 2 ,weobtain

4 α 2 t
( t − 1) 2

y = 4 α 3 t ( t +1)
( t − 1) 3

x = and

(the second is obtained from the first using y = x ( y/x )). Therefore, ( x, y )
determines t and t determines ( x, y ), so the map ψ is injective, and is a
bijection in case (1).
In case (2), rationalize the denominator by multiplying the numerator and
denominator of ( y + αx ) / ( y − αx )by y + αx to obtain an expression of the
form u + αv :

( y + αx )
( y − αx ) = u + αv.
We can change the sign of α throughout this equation and preserve the equal-
ity. Now multiply the resulting expression by the original to obtain

αv )= ( y + αx )
( y − αx )

( y − αx )
( y + αx ) =1 .

u 2 av 2 =( u + αv )( u − −

Conversely, suppose u 2

− av 2 =1. Let
x = u +1
v

2

y = u +1
v

x.

− a,