Elliptic curves - Mathematics of Public Key Cryptography - page 151

→ E be another Exercise 9.6.20 Let the notation be as in Theorem 9.6.19 .Let ψ : E

isogeny over k such that ker( ψ ) = G . Show that ψ = λ ◦ φ where λ is an automorphism

of E (or, if

k is finite, the composition of an isogeny and a Frobenius map). Similarly,

→ k = = ◦

if ψ : E E 2 is an isogeny over with ker( ψ ) G then show that ψ λ φ where

λ : E → k

E 2 is an isomorphism over of elliptic curves.

→ E be an isogeny over We now present the dual isogeny. Let φ : E k . Then the r e is

( E ) a group homomorphism φ ∗ :Pic 0

k

Pic 0
k

( E ). Since Pic 0
k

→ ( E ) is identified w it h E ( k ) i n

a standard way (and similarly for E ) one gets a group homomorphism from E ( ).

Indeed, the next result shows that this is an isogeny of elliptic curves; this is not trivial as φ ∗
is defined set-theoretically and it is not possible to interpret it as a rational map in general.

k )to E ( k

Theorem 9.6.21 Let E and E be elliptic curves over

→ E be a non-zero k

. Let φ : E

of degree m. Then there is a non-zero isogeny φ : E isogeny over k → E over k such that

φ

◦ φ = [ m ]: E → E.

Indeed, φ is unique (see Exercise 9.6.22 ).

Pic 0
k

Proof Let α 1 : E ( k ) → ( E ) be the canonical map P → ( P ) − ( O E ) and similarly for

α 2 : E α − 2 as above. We refer to Theorem III.6.1 of

[ 505 ] (or Section 21.1 of [ 560 ] for elliptic curves over

( E ). We have φ

Pic 0
k

φ ∗ ◦

→ = α 1 ◦

C ) for the details.

→ E is a non-zero isogeny over

Exercise 9.6.22 Suppose as in Theorem 9.6.21 that φ : E

of degree m . Show that if ψ : E = φ . k → E is any isogeny such that ψ ◦ φ = [ m ] then ψ

Definition 9.6.23 Let E and E be elliptic curves over → E be a non-zero

k and let φ : E

. The isogeny φ : E

k →

isogeny over E of Theorem 9.6.21 is called the dual isogeny .

Example 9.6.24 Let E be an elliptic curve over F q and π q : E → E the q -power Frobenius

map. The dual isogeny π q is called the Verschiebung . Since π q ◦ π q = [ q ] it follows that

[ q ]( x 1 /q ,y 1 /q ). Example 9.10.2 gives another way to write the Verschiebung.

π q ( x,y ) =

Exercise 9.6.25 Let E : y 2 x 3

= + a 6 over k with char( k ) = 2 , 3. Let ζ 3 ∈ k be such that

ζ 3 + ζ 3 + 1 = 0 and define the isogeny ρ ( O E ) = O E and ρ ( x,y ) = ( ζ 3 x,y ). Show that

ρ 2 (where ρ 2 ρ = means ρ ◦ ρ ).

Exercise 9.6.26 Recall E , E and φ from Example 9.6.9 . Show that φ : E → E is given by

Y 2

X 2 )

4 X 2 , Y ( D

−

φ ( X,Y )

=

8 X 2

and that φ ◦ φ ( x,y ) = [2]( x,y ).

We list some properties of the dual isogeny.