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
Previous Page Next Page