Elliptic curves - Mathematics of Public Key Cryptography - page 141

Clearly, τ Q is a rational map that is defined everywhere on E and so is a morphism.
Since τ Q has inverse map τ − Q it follows that τ Q is an isomorphism of the curve E to itself
(though be warned that in the next section we will define isomorphism for pointed curves
and τ Q will not be an isomorphism in this sense).

Corollary 9.2.3 Let E 1 and E 2 be elliptic curves over E 2 be a rational

map. Then φ is the composition of a group homomorphism and a translation map.

k and let φ : E 1 →

Proof First, by Lemma 7.3.6 a rational map to a projective curve is a morphism. Now
let φ (

O E 1 ) = Q ∈ E 2 ( k ). The composition ψ = τ − Q ◦ φ is therefore a morphism. As in

Theorem 9.2.1 it is a group homomorphism.

Hence, every rational map between elliptic curves corresponds naturally to a map of
groups. Theorem 9.6.19 gives a partial converse.

Example 9.2.4 Let E : y 2 x 3

= + x and Q = (0 , 0). We determine the map τ Q on E .

Let P = ( x,y ) ∈ E ( k ) be a point such that P is neither Q nor O E . To add P and Q to

obtain ( x 3 ,y 3 ) we compute λ = ( y − 0) / ( x − 0) = y/x . It follows that

y 2
x 2 −

y 2 x 3 − 1

x

λ 2 x 3 = − x − 0 = x = =

x 2

and

= −

y
x 2 .

y 3 =− λ ( x 3 − 0) − 0

y/x 2 )awayfrom

Hence, τ Q ( x,y ) . It is clear that τ Q is a rational map

of degree 1 and hence an isomorphism of curves by Lemma 8.1.15 . Indeed, it is easy to see
that the inverse of τ Q is itself (this is because Q has order 2).
One might wish to write τ Q projectively (we write the rational map in the form mentioned
in Exercise 5.5.2 ). Replacing x by x/z and y by y/z gives τ Q ( x/z,y/z )

= (1 /x, − { O E ,Q }

yz/x 2 )

= ( z/x, −

from which we deduce

yz : x 2 ) . τ Q ( x : y : z ) = ( xz : − (9.7)

Note that this map is not defined at either O E = (0 : 1 : 0) or Q = (0:0:1),inthesense

that evaluating at either point gives (0 : 0 : 0).
To get a map defined at Q one can multiply the right-hand side of equation ( 9.7 ) through
by y to get

y 2 z : x 2 y ) x 3 xz 2 : x 2 y )

( xyz : − = ( xyz : − −

x 2 z 2

and dividing by x gives τ Q ( x : y : z ) = ( yz : − − : xy ). One can check that τ Q (0 :

0:1) = (0 : − 1:0) = (0 : 1 : 0) as desired. Similarly, to get a map defined at O E one can

multiply ( 9.7 )by x , rearrange, and divide by z to get

( x 2 xy : y 2

τ Q ( x : y : z ) = : − − xz ) ,

which gives τ Q (0 : 1 : 0) = (0 : 0 : 1) as desired.