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

2
are x 1 , x 2 , and x 3 . The relation x 1 x 2 , x 3 = ν , a square, shows that α(x 1 ,y 1 )α(x 2 ,y 2 )·
α(x 3 , y 3 ) = 1. This and the calculation

b −bx b ∗ 2 ∗ 2
α ((0, 0) + (x, y)) = α , = mod(k ) = bx mod(k )
x x 2 x
= α(0, 0)α(x, y)

shows that α is a group homomorphism.
(5.7) Proposition. The sequence

ϕ 2 α k ∗
E[a, b] → E[−2a, a − 4b] →
∗ 2
(k )
is exact.
2
2
2
∗ 2
Proof. First α(ϕ(x, y)) = α(y /x , ∗) = (y.x) mod(k ) = 1 so that the compos-
2
ite is trivial. Next, if α(x, y) = 1, i.e., if α = t, then we choose two points (x + , y + )
2
and (x − , y − ), where x ± = 1/2(t − a ± y/t) and y ± = x ± t. We wish to show that
2
(x ± , y ± ) is on E[a, b]and ϕ(x ± , y ± ) = (x, y), where (x, y) is on E[−2a, a − 4b].
First we see that x + x − = b by the direct calculation
2 3 2 2 2
1 2 y x − 2ax + a x − y
x + x − = (x − a) − = = b
4 x 4x
2
2
3
since y = x − 2ax + (a − 4b)x.
Now a point (x ± , y ± ) is on E[a, b]ifand onlyif
2

y ± b
= x ± + a + = x ± + a + x ∓ .
x ± x ±
2
2
This is just the relation t = x + + a + x − , where x + + x − = t − a immediately
from the definition of x ± .
Finally we must show that ϕ(x ± , y ± ) = (x, y). For this, we calculate

2
y ± b 2 b
ϕ(x ± , y ± ) = , y ± 1 − = t , x ± t 1 −
x 2 x 2
y ±
± ±

b
= x, t x ± − = (x, t (x ± − x ∓ ))
x ±
y

= x, t ± = (x, ±y) .
t
This proves the proposition.
In 8, §2, see 8(2.3), we give a second interpretation of the homomorphism α in
terms of Galois cohomology.