Isogenies - Elliptic Curves: Number Theory and Cryptography - page 381

Regard each ω ( k i as a two-dimensional vector over R . Then the area of the
fundamental parallelogram of L k is | det( ω ( k )

,ω ( k )

) | .Since

1 2

det αω (1 1 ,αω (1)

=det( a ij )det ω (2 1 ,ω (2) ,

2 2

the index of αL 1 in L 2 , which is the ratio of the areas of the fundamental
parallelograms, equals | det( a ij ) | .

REMARK 12.2 A potential source of confusion is the following. Suppose
a lattice L 1 is contained in L 2 ,so L 2 is a larger lattice than L 1 .Let F 1 and F 2
be fundamental parallelograms for these lattices. Then F 2 is sm aller than F 1 .
For example, let L 1 =2 Z +2 i Z and L 2 = Z + i Z .Then L 1 ⊂ L 2 . The unit

square is a fundamental parallelogram for L 2 , while the square with corners
at 0 , 2 , 2 i, 2+2 i is a fundamental parallelogram for L 1 .

Define the degree of [ α ] to be the index [ L 2 : αL 1 ]. If α = 0, define
thedegreetobe0. If N is the degree, we say that C /L 1 and C /L 2 are N -
isogenous. The existence of the dual isogeny, defined below, shows that if E 1
and E 2 are N -isogenous, then E 2 and E 1 are N -isogenous, so this relation is
symmetric.

PROPOSITION 12.3
If α

=0 ,then #Ker([ α ]) = deg([ α ]) .

PROOF Let z ∈ C .Then[ α ]( z )=0 ⇐⇒ αz ∈ L 2 ,so

Ker([ α ]) = α − 1 L 2 /L 1

L 2 /αL 1 ,

where the isomorphism is given by multiplication by α . Therefore, the order
of the kernel is the index, which is the degree.

If Ker([ α ]) = α − 1 L 2 /L 1 is cyclic, we say that [ α ]isa cyclic isogeny .
In general, Ker([ α ]) is a finite abelian group with at most two generators
(coming from the generators of L 2 ), so it has the form Z n 1 ⊕ Z n 2 with n 1 |n 2
(see Appendix B). Therefore, the isogeny equals multiplication by n 1 on E 1
composed with a cyclic isogeny whose kernel has order n 2 /n 1 (Exercise 12.2).
Let α =0andlet N =deg([ α ]). Define the dual isogeny

[ α ]: C /L 2 −→ C /L 1

to be the map given by multiplication by N/α . We need to show this is well
defined: Since N =[ L 2 : αL 1 ], we have NL 2 ⊆ αL 1 . Therefore, ( N/α ) L 2 ⊆
L 1 , as desired.