Torsion Points - Elliptic Curves: Number Theory and Cryptography - page 85

Since n is odd, both ψ n and ψ n +2 δ are polynomials in x .Moreover,

( ψ n ψ n +2 δ ) 2 = ψ n ψ n +2 δ

vanishes at r . Therefore ψ n ψ n +2 δ vanishes at r .Since

φ n + δ = xψ n + δ − ψ n ψ n +2 δ ,

we find that φ n + δ ( r ) = 0. Therefore, φ n + δ and ψ n + δ have a common root.

Note that n + δ is even.
When considering the case that n is even, we showed that if φ 2 m and ψ 2 m
have a common root, then φ m and ψ 2 m have a common root. In the present
case, we apply this to 2 m = n + δ .Since n is assumed to be the smallest
index for which there is a common root, we have

n + δ
2 ≥ n.

This implies that n = 1. But clearly φ 1 = x and ψ 1 = 1 have no common

roots, so we have a contradiction.
This proves that φ n and ψ n have no common roots in all cases. Therefore,
as pointed out at the beginning of the proof, the multiplication by n map has
degree n 2 . This completes the proof of Corollary 3.7.

Recall from Section 2.9 that if α ( x, y )=( R ( x ) ,yS ( x )) is an endomorphism
of an elliptic curve E ,then α is separable if R ( x ) is not identically 0. Assume
n is not a multiple of the characteristic p of the field. From Theorem 3.6 we
see that the multiplication by n map has

x n 2 + ···
n 2 x n 2 − 1 +

R ( x )= .

···

The numerator of the derivative is n 2 x 2 n 2

− 2 +

=0,so R ( x )

= 0. Therefore,
multiplication by n is separable. From Corollary 3.7 and Proposition 2.21,
E [ n ], the kernel of multiplication by n , has order n 2 . The structure theorem
for finite abelian groups (see Appendix B) says that E [ n ]isisomorphicto

···

Z n 1 ⊕ Z n 2 ⊕···⊕ Z n k ,

for some integers n 1 ,n 2 ,...,n k with n i |n i +1 for all i .Let be a prime dividing
n 1 .Then |n i for all i . This means that E [ ] ⊆ E [ n ] has order k .Sincewe
have just proved that E [ ] has order 2 ,wemusthave k = 2. Multiplication by
n annihilates E [ n ] Z n 1 ⊕ Z n 2 ,sowemusthave n 2 |n .Since n 2 =# E [ n ]=
n 1 n 2 , it follows that n 1 = n 2 = n . Therefore,

E [ n ] Z n ⊕ Z n

when the characteristic p of the field does not divide n .