7   Elliptic Curves over Finite Fields  Lemma 7 1  Assuming that   A an abelian group  q   A     a positive definit quadratic form  Then        q   x   y     q   x     q   y         x   y       2 q   x   q   y     x   y   A       Proof  We may assume x   0  otherwise the result is clear  So q   x     0   Let m   n      Then     0   q   m x   n y     1 2   m x   n y   m x   n y     m 2 q   x     m n   x   y     n 2 q   y     q   x     m     x   y   2 q   x   n   2     q   y       x   y   2 4 q   x     n 2     Take m       x   y     n   2 q   x     0 to deduce      q   y       x   y   2 4 q   x     0     hence   x   y   2   4 q   x   q   y   and hence          x   y       2 q   x   q   y            Theorem 7 2   Hasse s Theorem    Assuming that   E     q an elliptic curve  Then          E     q       q   1       2 q       Proof  Recall Gal     q r     q   is cyclic of order r and generated by Frobenius x   x q   Let E have Weierstrass equation with coefficients a 1       a 6     q  so a i q   a i   Define the Frobenius endomorphism         E   E   x   y       x q   y q       This is an isogeny of degree q   Then      E     q       P   E       P     P     ker   1                              d x y     d   x q   y q   q x q   1 d x y q   0   q   0   m o d p         Lemma 6 3 tells us that        1                                 0       Hence 1     is separable   By Theorem 2 8 and the fact that 1     is a group homomorphism  we argue as in the proof of Theorem 6 5 that        ker   1             E     q     deg   1             deg   Hom   E   E       is a positive definite quadratic form  Theorem 5 7  and positive definiteness is obvious since non constant morphisms have positive degree    Lemma 7 1 gives        deg   1         1   deg       2 deg         Hence          E     q     1   q     2 q         Definition  For         End   E     Hom   E   E    we put             deg             deg     deg   and tr               1     Corollary 7 3  Assuming that   E     q an elliptic curve      End   E   the q power Frobenius  Then     E     q     q   1   tr       and    tr           2 q   7 1   Zeta functions  For K a number field  let        K   s           O K 1   N     s         O K prime   1   1   N     s     1       For K a function field  i e  K     q   C   where C     q is a smooth projective curve         K   s       x     C     1   1   N x   s     1       where        C       closed points on C        closed points are orbits for action of Gal     q       q   on C     q      and N x   q deg x  deg x is the size of orbit   We have   K   s     F   q   s   for some F         T          F   T       x     C     1   T deg x     1 log F   T       x     C     m   1   1 m T m deg x   log   1   x       m   1   x m m T d d T   log F   T         x     C     m   1   deg x T m deg x     n   1       x     C   deg x   n deg x   T n n   m deg x     n   1     C     q n   T n     Therefore      F   T     exp     n   1     C     q n   n T n         Definition   Zeta function   The zeta function Z C   T   of a smooth projective curve C     q is defined by      Z C   T     exp     n   1     C     q n   n T n         Theorem 7 4  Assuming that   E     q an elliptic curve with   E     q     q   1   a  Then      Z E   T     1   a T   q T 2   1   T     1   q T         Proof  Let     E   E be the q power Frobenius map  By Corollary 7 3       E     q     q   1   tr             Hence  tr         a  deg         q   Example Sheet 2  Q6 iii  implies   2   a     q   0  hence   n   2   a   n   1   q   n   0  so      tr     n   2     a tr     n   1     q tr     n     0       This second order difference equation with initial conditions  tr   1     2   tr         a has solution      tr     n       n     n       where       are roots of X 2   a X   q   0   Again by Corollary 7 3        E     q n     q n   1   tr     n     1   q n     n     n     Therefore     Z E   T     exp   n   1     T n n     q T   n n       T   n n       T   n n       1     T     1     T     1   T     1   q T     1   a T   q T 2   1   T     1   q T         Remark   Hasse s Theorem tells us that   a     2 q           and so                 q         Let K     q   E      K   s     0  so  Z E   q   s     0  so q s     or    Then q Re   s           or        so by        Re   s     1 2    This is an analog of the Riemann hypothesis