Page 185 -
P. 185

162    8. Descent and Galois Cohomology

        Conversely, any solution of this quartic equation yields a rational point on E[a, b].
           This quartic equation came from a down to earth factorization of the coordinates
        of a rational point on E[a, b], but in §1 we also derived quartic equations used to de-
        scribe principal homogeneous spaces representing elements in a Galois cohomology
        group. The Galois cohomology theory is contained in the following exact sequence:

                         E (Q)     1            1

                    0 →        → H (Q, T ) → ϕ H  Q, E Q    → 0
                        ϕE(Q)
                                                         2

        considered in (2.2), where E = E[a, b]and E = E[−2a, a − 4b]. Elements t in
                                                                  1
          1
        H (Q, T ) correspond to quadratic extensions of Q and their image in H (Q, E(Q))
        is described by a homogeneous space P t , a curve of genus 1, defined by the quartic
        given in (1.7)
                                       4
                                               2
                                2


                               N = d M − 2aM + d ,
                     2
                                                          2

        where d d = a − 4b. This curve associated with E[−2a, a − 4b] is exactly the
        same curve as the quartic above
                                             2 2
                                       4
                               2
                              N = b 1 M + aM e + b 2 e 4
                                            1

        associated with E[a, b]. The element t in H (Q, T ) comes from a point in E (Q)/
        ϕE(Q) if and only if this quartic in (1.7) has a rational point, i.e., the principal
        homogeneous space is trivial.
           The methods described above do allow one to calculate the rank of certain elliptic
        curves over Q.
                                                                 3
                                                            2
        (2.4) Example. Consider the curve E = E[0, p] with equation y = x − px for p
                                                  2
                                                       3
        a prime number. Then E is E[0, 4p] with equation y = x +4px. For the first curve


        the divisors of b =−p are b =±1, ±p, and the corresponding quartic equations
        are
            2     4    4   2      4     4   2      4   4   2       4   4
           N = M − pe , N =−M + pe , N = pM − e , N =−pM + e .
        For p ≡ 3(mod4), −1 is not a square mod p so that the second and third equations
        have no solution mod p, and, hence, no solution in the integers. This means that for
        p ≡ 3(mod4)

                                                      ∗ 2
                           im(α on E) ={1, −p} mod Q    .
        For the second curve the divisors of b = 4p are b =±1, ±2, ±4, ±p, ±2p, ±4p,

        and the corresponding quartic equations are
                                                            4
                                                                  4
                              4
                       4
                  2
                                   2
                                                      2
                                                4
                                          4
                N = M + 4pe ,     N = 2M + 2pe ,    N = 4M + pe ,
                                                                        2
        where we excluded the cases where both factors of 4p are negative since N is
        positive. For p ≡±3 (mod 8), 2 is not a square mod p so that the second equation
   180   181   182   183   184   185   186   187   188   189   190