Elliptic Curves over Q - Elliptic Curves: Number Theory and Cryptography - page 202

The slope of the tangent line to the curve can be found by implicit differ-
entiation:

ds
dt

=3 t 2 + As 2 +2 Ast ds

dt +3 Bs 2 ds dt ,

so

3 t 2 + As 2 ds

dt 3 Bs 2 .

If the line t = c is tangent to the curve at ( s, t ), then 1 − 2 Ast − 3 Bs 2 =0.
But s ≡ t ≡ 0(mod p ) implies that

=

1 − 2 Ast −

3 Bs 2 1 − 2 Ast − ≡ 1 ≡ 0(mod p ) .

Therefore, t = c is not tangent to the curve.

If d = 0, then our line is of the form in the lemma. But it passes through
the points P 1 and P 2 ,sowemusthave P 1 = P 2 , and the line is tangent to the
curve. Changing back to x, y coordinates, we obtain P 1 = P 2 . The definition
of the group law says that since the points P 1 and P 2 are equal, the line
ax + by + d =0istangentat( x, y ). As pointed out above, this means that
at + b + ds =0istangentat( s, t ). The lemma says that this cannot happen.
Therefore, d =0.
Dividing by d ,weobtain

s = αt + β

for some α, β ∈ Q .Then P 1 ,P 2 ,P 3 lie on the line s = αt + β .

LEMMA 8.5

t 2 + t 1 t 2 + t 1 + As 2

α =

1 − A ( s 1 + s 2 ) t 1 − B ( s 2 + s 1 s 2 + s 1 ) .

t 1 ). Since s i = t i + As i t i + Bs i , PROOF If t 1 = t 2 ,then α =( s 2 − s 1 ) / ( t 2 −

we have

( s 2 − s 1 ) 1 − A ( s 1 + s 2 ) t 1 − B ( s 2 + s 1 s 2 + s 1 )
=( s 2 − A ( s 2 − s 1 ) t 1 − B ( s 2 − s 1 ) s 1 ) −

As 2 t 2 − Bs 2 ) As 1 t 1 − Bs 1 )+ As 2 ( t 2 −

=( s 2 − − ( s 1 − t 1 )

= t 2 − t 1 + As 2 ( t 2 − t 1 )
=( t 2 − t 1 )( t 2 + t 1 t 2 + t 1 + As 2 ) .

This proves that ( s 2 − s 1 ) / ( t 2 − t 1 ) equals the expression in the lemma.
Now suppose that t 1 = t 2 . Since a line t = c with c ≡ 0(mod p ) intersects
the curve s = t 3 + As 2 t + Bs 3 in only one point with s ≡ 0(mod p )by
Lemma 8.4, the points ( s 1 ,t 1 )and( s 2 ,t 2 ) must be equal. The line s = αt + β