Isogenies - Elliptic Curves: Number Theory and Cryptography - page 390

γ ( . By Theorem 12.10, these maps are homomorphisms. Therefore,

β is an isomorphism, so E 2 and E 3 are isomorphic, as claimed. Moreover,

∞ )= ∞

β ◦ α 2 ( x 1 ,y 1 )= β ( x 2 ,y 2 )=( x 3 ,y 3 )= α 3 ( x 1 ,y 1 ) ,

so β ◦ α 2 = α 3 .

REMARK 12.13 If α 2 and α 3 are defined over K , then it is possible to
show that E 2 and E 3 are isomorphic over K . See [109, Exercise 3.13].

A very important property of isogenies is the existence of dual isogenies.
We already proved this in the case of elliptic curves over C . In the following,
we treat elliptic curves over arbitrary fields.

THEOREM 12.14
Let α : E 1 →

E 2 be an isogeny of elliptic curves. T hen there existsa dual

isogeny α : E 2 → E 1 su ch that α ◦ α ismu tiplication by deg α on E 1 .

PROOF We give the proof only in the case that deg α is not divisible
by the characteristic of the field K . The proof in the general case involves
working with inseparable extensions of fields. See [109].
Let N =deg α .ThenKer( α ) E 1 [ N ], and α ( E 1 [ N ]) is a subgroup of

E 1 of order N . We show in Theorem 12.16 that there exists an isogeny
α 2 : E 2 → E 3 ,forsome E 3 , such that Ker( α 3 )= α ( E 1 [ N ]). Then α 2 ◦ α has
kernel equal to E 1 [ N ]. The map E 1 → E 1 given by multiplication by N has
the same kernel. By Proposition 12.12, there is an isomorphism β : E 3 → E 1
such that β ◦ α 2 ◦ α is multiplication by N .Let α = β ◦ α 2 .

⊂

The map α is unique, its degree is deg α ,and α ◦ α equals multiplication
by deg( α )on E 2 . See Exercise 12.10.
If α and β are isogenies from E 1 to E 2 ,then α + β is defined by ( α + β )( P )=
α ( P )+ β ( P ). If α = −β , this is an isogeny. It can be shown that α + β = α + β .
See [109].

REMARK 12.15 There is an inseparable isogeny for which the dual
isogeny can be constructed easily. If E is an elliptic curve over the finite
field F q , then the q th power Frobenius endomorphism can be regarded as an
isogeny of degree q from E to itself. We know that φ 2 − aφ + q = 0 for some

integer a . Therefore,

( a − φ ) ◦ φ = q =deg φ,

so φ = a − φ is the dual isogeny for φ .