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
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