Isogenies - Elliptic Curves: Number Theory and Cryptography - page 393

X, Y . Therefore the function Y 2 + A 1 XY + A 3 Y X 3 A 2 X 2 − − − A 4 X − A 6

can have poles only at the points of C .Itvanishesat , since it is O ( t ). We

want to show that it also vanishes at the nontrivial points of C . A calculation
(see Exercise 12.6) shows that
X ( P )= x ( P )+
∞

∞

[ x ( P + Q ) − x ( Q )] (12.2)

= Q ∈ C

Y ( P )= y ( P )+
∞

[ y ( P + Q ) − y ( Q )] . (12.3)

= Q ∈ C

In particular, X and Y are invariant under translation by elements of C .
Therefore, Y 2 + A 1 XY + A 3 Y − X 3 − A 2 X 2 − A 4 X − A 6 is invariant under

translation by elements of C . Since it vanishes at ∞ , it vanishes at all points of
C . Hence it has no poles. This means that it is constant (see Proposition 11.1).
Since it vanishes at ∞ ,itis0. Thisprovesthat X and Y satisfy the desired
generalized Weierstrass equation. The following shows that this equation gives
a nonsingular curve.

LEMMA 12.17
E 2 isnonsingular.

PROOF For simplicity, assume that the characteristic of K is not 2. By
completing the square, we may reduce to the case where A 1 = A 3 =0,sothe
equation of E 2 is

Y 2 = X 3 + A 2 X 2 + A 4 X + A 6 =( X

− e 1 )( X − e 2 )( X − e 3 ) .

We need to show that e 1 ,e 2 ,e 3 are distinct. Suppose that e 1 = e 2 .Then

e 3 = Y
X − e 1

2

X − .

Let F = Y/ ( X − e 1 ), which is a function on E .
The function X −e 3 on E has double poles at the points of C and no other
poles. Therefore, its square root, namely F , has simple poles at the points of
C and no other poles. Note that F is inv ar iant under translation by elements
of C ,sinceboth X and Y are. Let a a has N poles, where

N =# C ,ithas N zeros. If P is one of these zeros, then P + Q is also a zero
for each Q

∈ K .Since F −

C . This gives all of the N zeros, so we conclude that F = a
occurs for exactly N distinct points of E .
We now need a special case of what is known as the Riemann-Hurwitz
formula. Consider an algebraic curve C defined by a polynomial equation
G ( x, y ) = 0 over an algebraically closed field K .Let F ( x, y ) be a rational
function on C .Let n be the number of poles of F , counted with multiplicity.
If a ∈ K ,then F −a has n poles, hence n zeros. It can be shown that if F is not

∈