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32     1. Elementary Properties of the Chord-Tangent Group Law on a Cubic Curve

        (u,v,w) = (0, 0, 0), (1, 0, 1), (−1, 0, −1), (0, 1, 1),and (0, −1, −1). This is not
        even a cubic equation, but under the transformation

                               2yw 2                 1
                           2
                          u =         and v = w 1 −     ,
                                x 2                  x
        we obtain
                                     4 2
                                  4w y            1    4
                              4             4
                             w =        + w   1 −     .
                                    x 4           x
                                      4
                    4
        Dividing by w and multiplying by x , we derive the cubic equation
                                         3  2     1
                                 2
                                     3
                                y = x − x + x − .
                                         2        4
        The equation factors giving one point (1/2, 0) of order 2

                                       1           1
                              2             2
                             y = x −       x − x +    .
                                       2           2
        Consider the lines y = λ(x−1/2) through (1/2, 0) and their other intersection points
        with the cubic. They have x-coordinates satisfying
                                      1     2      1

                                2
                               λ  1 −    = x − x +
                                      2            2
        or equivalently
                                               1     2
                                 2
                                         2
                            0 = x − (1 + λ )x + (1 + λ ).
                                               2
        The two other intersection points will coincide, that is, the line will be tangent to
        curve if and only if
                                      2 2
                                                 2
                                (1 + λ ) = 2(1 + λ ).
              2
                                                         2
        Since λ + 1 is nonzero, we can divide by it, and we obtain λ = 1or λ =+1, −1.
        Substituting back for
                                       2
                                                        2
          1. λ = 1, y = x − 1/2sothat0 = x − 2x + 1 = (x − 1) and (1, 1/2) is a point
            on the curve with 2(1, 1/2) = (1/2, 0),and
          2. λ =−1, gives by the same argument (1, −1/2) with 2(1, −1/2) = (1/2, 0),
            but this could have been deduced immediately from (1) by 2(1, −1/2) =
            −2(1, 1/2) =−(1/2, 0) = (1/2, 0). It is an application of the group structure.
           For this curve E with equation in normal form
                                         3  2     1
                                     3
                                2
                                y = x − x + x − ,
                                         2        4
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