14   Dual isogenies and the Weil pairing  Let K be a perfect field and E   K an elliptic curve   Proposition 14 1  Assuming that       E   K     be a finite Gal   K     K   stable subgroup  Then  there exists an elliptic curve E     K and a separable isogeny     E   E   defined over K  with kernel    such that every isogeny     E   E   with       factors uniquely via     Proof  Omitted  see Silverman  Chapter III  Proposition 4 12      Proposition 14 2  Assuming that       E   E   an isogeny of degree n  Then  there exists a unique isogeny       E     E such that           n     Proof  Case    is separable   We have   ker       n  hence  ker     E   n    Apply Proposition 14 1 with       n     Case    is inseparable   omitted   Uniqueness   Suppose   1       2       n    Then     1     2       0  so deg     1     2   deg     0  Then deg     1     2     0  hence   1     2     Remark    i  Write E 1   E 2 to mean  E 1 and E 2 are isogenous    Then   is an equivalence relation    ii  deg   n     n 2 gives that deg       deg   and   n         n      iii  Note                       n   E     n   E           Hence           n   E     In particular               iv  If E     E       E   then                    v  If     End   E   then        2     Tr         deg     0       So          Tr                         deg           Hence   Tr                 Lemma 14 3  Assuming that           Hom   E   E      Then                        Proof   i  If E   E    then this follows from Tr             Tr         Tr          ii  In general  let     E     E be any isogeny  e g        Then                                                                              Hence the result follows     Remark   In Silverman s book he proves Lemma 14 3 first  and uses this to show deg   Hom   E   E         is a quadratic form    Notation     sum   Div   E     E   n p   P     formal sum     n P   P     a d d u s i n g g r o u p l a w     Recall     E     Pic 0   E   P       P       0         Therefore sum D     D   for all D   Div 0   E     We deduce   Lemma 14 4  Assuming that   D   Div   E    Then  D   0  if and only if deg D   0 and sum D   0 E   We will now discuss Weil pairing   Let     E   E   be an isogeny  of degree n  with Dual isogeny       E     E  Assume char K   n  hence        separable    We define the Weil pairing      e     E         E               n       Let T   E          Then n T   0  so there exists f   K     E       such that      div   f     n   T     n   0         Pick T 0   E   K     with     T 0     T  Then            T           0       P   E         P   T 0       P   E         P       has sum      n T 0         T 0       T   0       So there exists g   K     E     such that      div   g           T           0         Now     div       f           div f     n         T           0       div   g n       Therefore     f   c g n for some c   K       Rescaling f  we can say without loss of generality c   1  i e      f   g n   For S   E       we get     S     div g       div g   so   S   g     g for some     K      i e      g   X   S   g   X   is independent of choice of X   E   K       Now        n   g   X   S   n g   X   n   f       X   S     f       X       1     since S   E        Hence       n   We define      e     S   T     g   X   S   g   X         Proposition 14 5  e is bilinear and non degenerate   Proof   i  Linearity in first argument      e   S 1   S 2   T     g   X   S 1   S 2   g   X   S 2   g   X   S 2   g   X     e   S   T   e   S 2   T        ii  Linearity in second argument  Let T 1   T 2   E                div   f 1     n   T 1     n   0       f   g 1 n div   f 2     n   T 2     n   0       f 2   g 2 n     There exists h   K     E     such that      div   h       T 1       T 2       T 1   T 2       0         Then put f   f 1 f 2 h n and g   g 1 g 2     h   Check   div   f     n   T 1   T 2     n   0    Yes           f       f 1     f 2       h   n     g 1 g 2       h     n   g n       Therefore     e   S   T 1   T 2     g   X   S   g   X     g 1   X   S   g 1   X   g 2   X   S   g 2   X   h       X     h       X   S         1 since S   E         e   S   T 1   e   S   T 2        iii  e is non degenerate  Fix T   E            Suppose  e   S   T     1 for all S   E        Then   S   g   g for all S   E        Have K     E         K     E     is a Galois extension with Galois group E        S   E       acts on K     E   via   S      Therefore g       h for some h   K     E      So     f   g n         h n    So f   h n  and hence  div   h       T       0    So T   0     We ve shown E           Hom   E           n    It is an isomorphism since   E           E             n   Remark    i  If E   E       are defined over K then  e is Galois equivariant  i e        Gal   K     K      S   E          T   E              e     S     T         e   S   T            ii  Taking       n     E   E  so         n    gives      e n   E   n     E   n       n 2   n       Corollary 14 6  Assuming that   E   n     E   K    Then    n   K   Proof  Let  T   E   n   have order n  Non degeneracy of e n  implies that there exists  S   E   n   such that e n   S   T   is a primitive n th root of unity  say   n   Then            n         e n   S   T       e n     S     T       n     for all     Gal   K     K    So   n   K     Example  There does not exist E     with E       tors         3     2   Remark   In fact  e n is alternating  i e  e n   T   T     1 for all  T   E   n      This implies e n   S   T     e n   T   S     1