The Basic Theory - Elliptic Curves: Number Theory and Cryptography - page 63

Then ( x, y )isonthecurve E and

ψ ( x, y )= ( y/x )+ α
( y/x ) − α

= u +1+ αv
u +1 − αv

= u + αv

(the last equality uses the fact that u 2 − av 2 = 1). Therefore, ψ is surjective,

hence is a bijection in case (2), too.
It remains to show that ψ is a homomorphism. Suppose ( x 1 ,y 1 )+( x 2 ,y 2 )=
( x 3 ,y 3 ). Let

t i = y i + αx i

y i − αx i .

We must show that t 1 t 2 = t 3 .
When ( x 1 ,y 1 ) =( x 2 ,y 2 ), we have
x 3 = y 2 −

2

y 1

− a − x 1 − x 2 .

x 2 − x 1

4 α 2 t i
( t i − 1) 2

and y i = 4 α 3 t i ( t i +1)

Substituting x i = and simplifying yields

( t i − 1) 3

4 t 3
( t 3 − 1) 2

4 t 1 t 2

=

( t 1 t 2 − 1) 2 .

(2.11)

Similarly,

−y 3 = y 2 −

( x 3 − x 1 )+ y 1

y 1

x 2 − x 1

yields

4 α 3 t 3 ( t 3 +1)
( t 3 − 1) 3

= 4 α 3 t 1 t 2 ( t 1 t 2 +1)
( t 1 t 2 − 1) 3

.

The ratio of this equation and (2.11) yields

t 3 − 1

t 3 +1 = t 1 t 2 − 1 t 1 t 2 +1 .

This simplifies to yield

t 1 t 2 = t 3 ,

as desired.
The case where ( x 1 ,y 1 )=( x 2 ,y 2 ) is similar, and the cases where one or
more of the points is ∞ are trivial. This completes the proof.

One situation where the above singular curves arise naturally is when we
are working with curves with integral coe cients and reduce modulo various
primes. For example, let E be y 2 = x ( x + 35)( x − 55). Then we have

E mod 5 : y 2 x 3 , ≡

E mod 7 : y 2 ≡ x 2 ( x +1) ,

E mod 11 : y 2 ≡ x 2 ( x +2) .