16   Descent by Cyclic Isogeny  Let E   E   be elliptic curves  over a number field K  and     E   E   an isogeny  of degree n   Suppose E                 n   generated by T   E     K    Then E           n as a Gal   K     K   module     S   e   S   T         Have a short exact sequence of Gal   K     K   modules     0     n   E     E     0       Get long exact sequence    the   is by Hilbert 90     Theorem 16 1  Assuming that   f   K   E     and g   K   E    div   f     n   T     n   0        f   g n  Then      P     f   P     m o d   K     n   for all P   E     K       0   T     Proof  Let Q       1 P  Then      P     H 1   K     n   is represented by       Q   Q   E           n      e     Q   Q   T     g     Q   Q   X   g   X   for any X   E avoiding zeroes and poles of g   g     Q   g   Q   pick X   Q       g   Q     g   Q       f   P   n f n   P            f   g n  f   P     g   Q   n    But        H 1   K     n     K       K     n       x n x n       x     Hence     P     f   P     m o d   K     n       16 1   Descent by 2 isogeny     E   y 2   x   x 2   a x   b   E     y 2   x   x 2   a   x   b         b   a 2   4 b     0  a       2 a  b     a 2   4 b          E   E     x   y         y x   2   y   x 2   b   x 2         E     E   x   y       1 4   y x   2   y   x 2   b     8 x 2       E           0   T    T     0   0     E   K     E               0   T      T       0   0     E     K     Proposition 16 2  There is a group homomorphism     E     K     K       K     2   x   y       x mod   K     2 if x   0 b   mod   K     2 if x   0     with kernel   E   K     Proof  Either  Apply Theorem 16 1 with f   x   K   E      g   y x   K   E     Or  direct calculation   see Example Sheet 4          E   E   K       E     K     K       K     2   E     E     K     E   K     K       K     2     Lemma 16 3  2 rank E   K       Im   E       Im   E     4   Proof  If A   f B   g C are homomorphisms of abelian groups  then there is an exact sequence   Since           2   E  we get an exact sequence      0   E   K                 2     E   K     2       E     K                   2     E     K     E   K       Im     E           E   K   2 E   K     E   K       E     K       Im     E     0       Therefore        E   L     2 E   K       E   K     2         Im   E       Im   E     4    1      Mordell Weil  E   K           r  where   is a finite group and r   rank E   K     E   K   2 E   K       2         2     r   E   K     2         2     Since   is finite  we have that   2   and      2   have the same order  and therefore        E   K     2 E   K       E   K     2       2 r    2      So we are done  by using  1   and  2       Lemma 16 4  Assuming that   K is a number field  a   b   O K  Then  Im     E     K   S   2    where S         b     Proof  We must show that x   y   K  y 2   x   x 2   a x   b   and v     b     0 then v     x     0   m o d 2     Case  v     x     0   Lemma 9 1 gives that for some r   1  v     x       2 r and v     y       3 r  so done   Case  v     x     0   Then v     x 2   a x   b     0  So v     x     v     y 2     2 v     y       Lemma 16 5  Assuming that   b 1 b 2   b  Then  b 1   K       Im     E    if and only if      w 2   b 1 u 4   a u 2 v 2   b 2 v 4         is soluble for u   v   w   K not all zero    Proof  If b 1     K     2 or b 2     K     2 then both conditions are satisfied  So we may assume b 1   b 2     K     2   Now note b 1   K     2   Im     E   if and only if there exists   x   y     E   K   such that x   b 1 t 2 for some t   K    This implies      y 2     b 1 t 2     b 1 2 t 2   a b 1 t 2   b       hence        y b 1 t   2   b 1 t 4   a t 2   b b 1     b 2       So      has solution   u   v   w       t   1   y b 1 t     Conversely  if   u   v   w   is a solution to       then u v   0 and   b 1   u v   2   b 1 u w v 3     E   K       Now take K       Example  E  y 2   x 3   x  a   0  b     1    Im     E         1                   2   E    y 2   x 3   4 x   Im     E           1   2                   2      b 1     1 w 2     u 4   4 v 4 b 1   2 w 2   2 u 4   2 v 4 b 1     2 w 2     2 u 4   2 v 4     The first and last lines are insoluble over    squares are non negative   The middle line does have a solution    u   v   w       1   1   2    Therefore Im     E         2                   2   Hence 2 rank E         2   2 4   1  so rank E         0   So 1 is not  a congruent number    Example  E  y 2   x 3   p x  p a prime which is 5 modulo 8      b 1     1 w 2     u 4   p v 4     This is insoluble over    hence Im     E       p                   2   E    y 2   x 3   4 p x   Im     E           1   2   p                   2   Note    E     T           4 p           2       p           2      b 1   2 w 2   2 u 4   2 p v 4 b 1   2 w 2     2 u 4   2 p v 4 b 1   p w 2   p u 4   4 v 4     Suppose the first line is soluble  Then without loss of generality u   v   w     with gcd   u   v     1  If p   u  then p   w and then p   v  contradiction  Therefore w 2   2 u 4     0   m o d p    So   2 p       1  which contradicts p   5   m o d 8     Likewise the second line has no solution since     2 p       1   TODO   Example   Lindemann   E  y 2   x 3   1 7 x      Im     E       1 7                   2       E    y 2   x 3   6 8 x      Im     E           1   2   1 7                   2       b 1   2  w 2   2 u 4   3 4 v 4   Replacing w by 2 w and dividing by 2 gives      C   2 w 2   u 4   1 7 v 4       Notation  C   K         u   v   w     K 3     0   satisfying equation above        where   u   v   w         u     v     2 w   for all     K     C     2       since 1 7       2     4   C     1 7       since 2       1 7     2   C           since 2       Therefore C     v       for all places v of     Suppose   u   v   w     C        Without loss of generality say u   v   w      gcd   u     v     1 and w   0   If 1 7   w then 1 7   u and then 1 7   v  contradiction  So if p   w then p   1 7 and    1 7 p       1  Therefore   p 1 7       1 7 p     1  using quadratic reciprocity    If p is odd  also   2 1 7       1    Therefore   w 1 7       1  But 2 w 2   u 4   m o d 1 7    so 2       1 7     4       1     4    contradiction   So C            i e  C is a counterexample to the Hasse Principle  It represents a non trivial element of  T S   E         16 2   Birch Swinnerton Dyer Conjecture  Let E     be an elliptic curve   Definition   L   E   s     Define      L   E   s       p L p   E   s       where      L p   E   s         1   a p p   s   p 1   2 s     1 if good reduction   1   p   s     1 if split multiplicative reduction   1   p   s     1 if nonsplit multiplicative reduction 1 if additive reduction     where          p     1   p   a p   Hasse s Theorem implies   a p     2 p  which shows that L   E   s   converges for Re   s     3 2   Theorem   Wiles  Breuil  Conrad  Diamon  Taylor   L   E   s   is the L function of a weight 2 modular form  and hence has an analytic continuation to all of    and a function equation relating L   E   s     L   E   2   s      Conjecture   Weak Birch Swinnerton Dyer Conjecture   ord s   1 L   E   s     rank E            r say    Conjecture   Strong Birch Swinnerton Dyer Conjecture   ord s   1 L   E   s     rank E        which we shall call r  and      lim s   1 1   s   1   r L   E   s       E Reg E         T S   E           p c p   E     E       tors   2       where   c p is given by      c p   E     Tamagawa number of E     p     E     p     E 0     p         If E         E       tors     P 1   P 2       P r    then Reg E         det     P i   P j     i   j   1       r  where               P   Q         P   Q         P         Q           E is given by         E     E         d x 2 y   a 1 x   a 3         where a 1       a 6     are coefficients of a globally minimal Weierstrass equation for E       Theorem   Kolyvagin   If ord s   1 L   E   s   is 0 or 1  then Weak Birch Swinnerton Dyer holds  and also    T S   E