The Basic Theory - Elliptic Curves: Number Theory and Cryptography - page 53

PROOF Since φ q ( x, y )=( x q ,y q ), the map is given by rational functions
(in f act, by p ol ynomials) and the degree is q . The main point is th at φ q :
E ( F q ) → E ( F q ) is a homomorphism. Let ( x 1 ,y 1 ) , ( x 2 ,y 2 ) ∈ E ( F q ) with

x 1 = x 2 .Thesumis( x 3 ,y 3 ), with

y 1
x 2 − x 1

y 2 −

x 3 = m 2

− x 1 − x 2 , y 3 = m ( x 1 − x 3 ) − y 1 , where m =

(we are working with the Weierstrass form here; the proof for the generalized
Weierstrass form is essentially the same). Raise everything to the q th power
to obtain

y 2 − y 1
x 2 − x 1

x 3 = m 2 x 1 − x 2 , y 3 = m ( x 1 − x 3 ) y 1 ,

where m = − − .

This says that

φ q ( x 3 ,y 3 )= φ q ( x 1 ,y 1 )+ φ q ( x 2 ,y 2 ) .
The cases where x 1 = x 2 or where one of the points is ∞ are checked similarly.
However, there is one subtlety that arises when adding a point to itself. The
formula says that 2( x 1 ,y 1 )=( x 3 ,y 3 ), with

where m = 3 x 1 + A
2 y 1

x 3 = m 2 − 2 x 1 , y 3 = m ( x 1 − x 3 ) − y 1 , .

When this is raised to the q th power, we obtain

where m = 3 q ( x 1 ) 2 + A q x 3 = m 2 − 2 x 1 , y 3 = m ( x 1 − x 3 ) − y 1 , .

2 q y 1

Since 2 , 3 ,A ∈ F q ,wehave2 q =2 , 3 q =3 ,A q = A . This means that we
obtain the formula for doubling the point ( x 1 ,y 1 )on E (if A q didn't equal A ,
we would be working on a new elliptic curve with A q in place of A ).
Since φ q is a homomorphism given by rational functions, it is an endo-
morphism of E .Since q =0in F q , the derivative of x q

is identically zero.

Therefore, φ q is not separable.

The following result will be crucial in the proof of Hasse's theorem in Chap-
ter 4 and in the proof of Theorem 3.2.

PROPOSITION 2.21
Let α =0 be a separable endom orphism of an ellipticcurve E .Then

deg α =# Ker ( α ) ,

w here K er ( α ) isthe kernel of the hom om orphism α : E ( K )

→ E ( K ) .

If α

=0 is not separable, then

deg α> # Ker ( α ) .