Isogenies - Elliptic Curves: Number Theory and Cryptography - page 389

Assume for simplicity that the elliptic curves are in Weierstrass form: E i :
y i = x i + A i x i + B i . The isogeny α 2 can be described by ( x 2 ,y 2 )= α 2 ( x 1 ,y 1 )=
( r 1 ( x 1 ) ,y 1 r 2 ( x 1 )), where r 1 and r 2 are rat ional functions with coe c ients in
the field K . This allows us to regard K ( x 2 ,y 2 ) as a subfield of K ( x 1 ,y 1 ).
Write r 1 ( x 1 )= p ( x 1 ) /q ( x 1 ), wh ere p and q are polynomials with no common
factors. Then p ( T ) − x 2 q ( T ) ∈ K ( x 2 )[ T ] is irre du cible of degree N =deg α 2
(see Exercise 12.7). Ther efore, the exte nsio n K ( x 1 ) /K ( x 2 ) has de gree N .

By Lemma 11.5, y i = x i + A i x i + B i ∈ K ( x i ). Therefore, [ K ( x i ,y i ):

K ( x i )] = 2. It follows that

2[ K ( x 1 ,y 1 ): K ( x 2 ,y 2 )] = [ K ( x 1 ,y 1 ): K ( x 2 ,y 2 )][ K ( x 2 ,y 2 ): K ( x 2 )]
=[ K ( x 1 ,y 1 ): K ( x 1 )][ K ( x 1 ): K ( x 2 )] = 2 N,

so [ K ( x 1 ,y 1 ): K ( x 2 ,y 2 )] = N .
Let Q be in the kernel of α 2 .Translationby Q gives a map

σ Q :( x 1 ,y 1 ) → ( x 1 ,y 1 )+ Q =( f ( x 1 ,y 1 ) ,g ( x 1 ,y 1 )) .

This is an automorphism of K ( x 1 ,y 1 ) (see Exercise 12.9). Since

σ Q ( x 2 ,y 2 )= σ Q ( α 2 ( x 1 ,y 1 )) = α 2 (( x 1 ,y 1 )+ Q )= α 2 ( x 1 ,y 1 )=( x 2 ,y 2 ) ,

this automorphism acts as the identity on the field K ( x 2 ,y 2 ). A result from
field theory says that if G is a finite group of automorphisms of a field L ,
then the subfield of elements fixed by G is of degree # G below L (see, for
example, [71]). If α 2 is separable, there are N (= deg α 2 ) automorphisms given
by translation by ele me nts of the ker nel of α 2 , so the fixed field of this group
is of deg ree N be low K ( x 1 ,y 1 ). Since K ( x 2 ,y 2 ) is containe d in this fixed field,
and [ K ( x 1 ,y 1 ): K ( x 2 ,y 2 )] = N , the fixed field is exactly K ( x 2 ,y 2 ).
The same an alysis applie s t o α 3 .If α 2 and α 3 are separable with the same
kernel, then K ( x 2 ,y 2 )and K ( x 3 ,y 3 )arethefixedfieldofthesamegroupof
automorphisms, hence

K ( x 2 ,y 2 )= K ( x 3 ,y 3 ) .

Therefore, x 2 ,y 2 are rational functions of x 3 ,y 3 ,and x 3 ,y 3 are rational func-
tions of x 2 ,y 2 .Write

x 2 = R 1 ( x 3 ,y 3 ) , 2 = R 2 ( x 3 ,y 3 )

for rational functions R 1 ,R 2 .Then

γ :( x 3 ,y 3 ) → ( x 2 ,y 2 )=( R 1 ( x 3 ,y 3 ) ,R 2 ( x 3 ,y 3 ))

gives a map E 3 → E 2 . Similarly, there exists β : E 2 → E 3 ,and γ◦β =idon E 2
and β ◦ γ =idon E 3 . By translating the images of β and γ (that is, change
β to β − β ( ∞ ), and similarly for γ ), we may assume that β ( ∞ )= ∞ and