10   Elliptic Curves over Number Fields  The torsion subgroup    K            E   K an elliptic curve   Notation    a prime of K  i e  a prime  ideal in O K    K   is the   adic completion of K  valuation ring O     k     O K     residue field    Definition   Good reduction  prime      is a prime of good reduction  for E   K if E   K   has good reduction   Lemma 10 1  E   K has only finitely many primes of bad reduction   Proof  Take a Weierstrass equation for E with a 1       a 6   O K  E non singular implies that 0       O K   Write             1     r   r  factorisation into prime ideals    Let S       1         r    If     S then v           0  Hence E   K   has good reduction   Therefore   bad primes for E     S  hence is finite     Remark   If K has class number 1  e g  K      then we can always find a Weierstrass equation for E with a 1       a 6   O K which is minimal  at all primes     Basic group theory   If A is a finitely generatead abelian group then A     finite group       r  We call r the  rank   and   finite subgroup   is the torsion subgroup   Lemma 10 2  E   K   tors is finite   Proof  Take any prime    We saw that E   K     has a finite index subgroup A  say  with A     O           In particular  A is torsion free      E   K   tors   E   K     tors   E   K     A   finite         Lemma 10 3  Assuming that     is a prime of good reduction      n  Then  reduction modulo   gives an injective group homomorphism        E   K     n         k           Proof  Proposition 9 5 gives that   E   K           k     is a group homomorphism  with kernel E 1   K       Corollary 8 5 and     n gives that E 1   K     has no n torsion     Example  E      y 2   y   x 3   x        1 1   E has good reduction at all p   1 1   p   2   3   5   7   1 1   1 3             p    5  5  5  5     1 0   Lemma 10 3 gives                 tors   5   2 a for some a   0   E       tors   5   3 b for some b   0     Hence         tors   5   Let T     0   0     E        Calculation gives 5 T   0  Therefore E       tors       5     Example  E      y 2   y   x 3   x 2        4 3  p   2   3   5   7   1 1   1 3             p    5  6  1 0  8  9  1 9   Lemma 10 3 gives           E       tors   5   2 a for some a   0   E       tors   9   1 1 b for some b   0     Therefore E       tors     0     Therefore P     0   0   is a point of infinite order   In particular  E       is infinite    Example  E D  y 2   x 3   D 2 x   f   x    D     square free      2 6 D 6  If p   2 D  then            D     p     1     x     p     f   x   p     1         If p   3   m o d 4   then since f is odd         f     x   p         f   x   p         1 p     f   x   p         f   x   p         Hence             D     p     p   1          E D       tors     0     0   0         D   0             2     2       Let m     E D       tors   We have 4   m   p   1 forr all sufficiently large  p   2 m D  primes p with p   3   m o d 4    Hence m   4  otherwise get contradiction to Dirichlet s theorem on primes on arithmetic progressions    So     rank E D         1     x   y       y   0   y 2   x 3   D 2 x   Lecture 1 D is a congruent number     Lemma 10 4  Assuming that   E     is given by a Weierstrass equation with a 1       a 6      0   T     x   y     E       tors  Then    i  4 x   8 y       ii  If 2   a 1 or 2 T   0 then x   y      Proof   i  The Weierstrass equation defines a formal group    over    For r   1             p r   p         x   y     E     p     v p   x       2 r   v p   y       3 r       0         Then Theorem 9 2 gives      p r   p         p       if r   1 p   1  Hence      4   2   and      p   p   for p are odd are torsion free   So if 0   T     x   y     E       tors then     v 2   x       2 v 2   y       3 v p   x     0 v p   y     0     for all odd primes p    ii  Suppose  T       2   2    i e  v 2   x       2  v 2   y       3  Since              2   2       4   2         2           and      4   2   is torsion free  we get 2 T   0   Also   x   y     T     T     x     y   a 1 x   a 3    So     2 y   a 1 x   a 3   0 8 y   odd   a 1   4 x     odd   4 a 3   even   0     Hence a 1 is odd   So if 2 T   0 or a 1 is even then  T       2   2    so x   y         Example  y 2   x y   x 3   4 x   1       1 4   1 8     E         2     Theorem 10 5   Lutz Nagell    Assuming that   E      y 2   x 3   a x   b  a   b      0   T     x   y     E       tors  Then  x   y     and either y   0 or y 2     4 a 3   2 7 b 2     Proof  Lemma 10 4 gives that x   y       If 2 T   0 then y   0  Otherwise 0   2 T     x 2   y 2     E       tors   Lemma 10 4 gives x 2   y 2      But      x 2     f     x   2 y   2   2 x     hence y   f     x     E non singular gives that f   X   adn f     X   are coprime  so f   X   and f     X   2 are coprime  Therefore there exists g   h       X   satisfying      g   X   f   X     h   X   f     X   2   1       Doing this and clearing denominators gives        3 X 2   4 a   f     X   2   2 7   X 3   a X   b   f   X     4 a 3   2 7 b 2       Since y   f     x   and y 2   f   x    we get y 2     4 a 3   2 7 b 2       Remark   Mazur showed that if E     is an elliptic curve then      E       tors         n   1   n   1 2   n   1 1     2         2 n   1   n   4     Moreover  all 15 possibilities occur