22 2. Elliptic curves

Example 2.2.3. Let d ∈ Z, d = 0 and let E be the elliptic curve

given by the cubic equation

X3

+ Y

3

=

dZ3

with O = [1, −1, 0]. The reader should verify that E is a smooth

curve. We wish to find a Weierstrass equation for E and, indeed, one

can find a change of variables ψ : E → E given by

ψ([X, Y, Z]) = [12dZ, 36d(X − Y ), X + Y ] = [x, y, z]

such that

zy2

=

x3

−

432d2z3.

The map ψ is invertible; the inverse

map

ψ−1

: E → E is

ψ−1([x,

y, z]) =

36dz + y

72d

,

36dz − y

72d

,

x

12d

.

In aﬃne coordinates, the change of variables is going from X3 +Y 3 =

d to the curve y2 = x3 − 432d2:

ψ(X, Y ) =

12d

X + Y

,

36d(X − Y )

X + Y

,

ψ−1(x,

y) =

36d + y

6x

,

36d − y

6x

.

Definition 2.2.4. Let E : f(x, y) = 0 be an elliptic curve with origin

O, and let E : g(X, Y ) = 0 be an elliptic curve with origin O . We

say that E and E are isomorphic over Q if there is an invertible

change of variables ψ : E → E , defined by rational functions with

coeﬃcients in Q, such that ψ(O) = O .

Example 2.2.5. Sometimes, a curve given by a quartic polynomial

can be isomorphic over Q to another curve given by a cubic polyno-

mial. For instance, consider the curves

C/Q : V

2

= U

4

+ 1 and E/Q :

y2

=

x3

− 4x.

The map ψ : C → E given by

ψ(U, V ) =

2(V + 1)

U 2

,

4(V + 1)

U 3