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98 4. Families of Elliptic Curves and Geometric Properties of Torsion Points

2
2x 1 + x 2 = σ − a.
We just calculated σ above so that
2
x 2 = σ − a − 2x 1
2
2 !
y 2 s − b
1
= + − a − 2x 1 .
x 1 2y 1
Dropping the subscript 1 on x 1 and y 1 ,wehave
2 2 3 2 2 3
x − b x − bx + y − ax − 2x
x 2 = +
4y 2 x 2
2 2
x − b
=
4y 2
3
2
2
using y = x + ax + bx. This is just the x-coordinate in the above calculation
of ˆϕ(ϕ(x, y)). For the y-coordinate we use y = σ(x − x 1 ) + y 1 and hence y =
2
(y 1 /x 1 + (x − b)/2y 1 )(x 2 − x 1 ) + y 1 . Again dropping the subscript 1 on x 1 and y 1
1
and substituting the above expression for x 2 , we obtain
2 2 2
y x − b x − b
y 2 = + − x + y
x 2y 4y 2
2 3 2 2 2 2 2 2
x x − b + 2y x − b − 4x y x − b
=
8xy 3
2 2 2 2 2 2 2
x x − b + 2x y − 2by − 4x y

2
= x − b
8xy 3
x x − b − 2y y − ax
2 2 2 2 2 2
2
= x − b .
2 3
8x y
3
2
2
Now using x − bx = y − ax − 2bx,wehave
2
3 2 2 2
x − bx − 2y y − ax

4 2 2 2 4 3 2 2 4 2 2
= y − 2y ax + 2bx + a x + 4abx + 4b x − 2y + 2ax y

4 3 2 2 4 3 2 2
=−y − 4bx x + ax + bx + a x + 4abx + 4b x
" #
4 2 4
=− y − a − 4b x .
Hencewededucethat2(x 1 , y 1 ) =ˆϕϕ(x 1 , y 1 ).
Although we have not checked directly that ϕ is a group homomorphism, the
previous relation ˆϕ(ϕ(x, y)) = 2(x, y) and the following useful results show that the
composite