9   Elliptic Curves over Local Fields  Let K be a field  complete with respect to discrete valuation v   K         Notation  Valuation ring    ring of integers  will be denoted by      O K     x   K     v   x     0       0         Unit group will be denoted by      O K       x   K     v   x     0         The maximal ideal will be denoted by   O K where v         1   The residue field will be denoted by k   O K     O K   We assume char K   0 and char k   p   0  For example  K     p  O K     p  k     p   Let E   K be an elliptic curve   Definition   Integral   minimal Weierstrass equation   A Weierstrass equation for E with coefficients a 1       a 6   K is integral  if a 1       a 6   O K and minimal  if v       is minimal among all integral Weierstrass equations for E   Remark    i  Putting x   u 2 x    y   u 3 y   gives a i   u i a i    Therefore integral Weierstrass equations exist    ii  a 1       a 6   O K       O K   v         0  Therefore minimal Weierstrass equations exist    iii  If char k   2   3 then there exists minimal Weierstrass equation of the form y 2   x 3   a x   b   Lemma 9 1  Assuming that   E   K have integral Weierstrass equation     y 2   a 1 x y   a 3 y   x 3   a 2 x 2   a 4 x   a 6     0   P     x   y     E   K    Then  either x   y   O K or        v   x       2 s v   y       3 s     for some s   1    Compare with Q5 from Example Sheet 1    Proof  Throughout this proof  LHS and RHS refer to the Weierstrass equation of the curve   Case  v   x     0   If v   y     0 then v   LHS     0 and v   RHS     0  Therefore x   y   O K   Case  v   x     0   v   LHS     min   2 v   y     v   x     v   y     v   y     and v   RHS     3 v   x    We get 3 possible inequalities from this  and each of them gives v   y     v   x     Now     v   LHS     2 v   y   v   RHS     3 v   x       so        v   x       2 s v   y       3 s     for some s   1     If K is complete  then O K is complete  with respect to   r O K  for any r   1    We fix a minimal Weierstrass equation for E   K  Get formal group    over O K   Taking I     r O K  with r   1  in Lemma 8 2 gives            I         x   y     E   K       x y     1 y     r O K       0         x   y     E   K     v   x y     r   v   1 y     r       0         x   y     E   K     v   x       2 s   v   y       3 s for s   r       0    using Lemma 9 1         x   y     E   K     v   x       2 r   v   y       3 r       0    using Lemma 9 1       By Lemma 8 2 this is a subgroup of E   K    say E r   K            E 2   K     E 1   K     E   K         More generally  for F a formal group over O K         F     2 O K     F     O K         We claim that   F     r O K       O K       for r sufficiently large   F     r O K   F     r   1 O K       k       for r   1   Reminder  char K   0  char k   p   0   Theorem 9 2  Assuming that   K a local field with char K   0  char k   p   0  F a formal group over O K  e   v   p    r   e p   1  Then      log   F     r O K           a     r O K       is an isomorphism of groups with inverse      exp       a     r O K       F     r O K         Remark       a     r O K         r O K           O K        x       r x   Proof  For x     r O K we must show the power series log   x   and exp   x   converge to elements in   r O K   Recall      exp   T     T   b 2 2   T 2   b 3 3   T 3         for some b i   O K   Claim  v p   n       n   1 p   1   Proof of claim       v p   n         r   1     n p r       r   1   n p r   n   1 p 1   1 p   n p   1       Therefore       p   1   v p   n       n   p   1   v p   n       n   1      we go from  to       1 by noting that the LHS is in      This proves the claim   Now     v   b n x n n       n r   e   n   1 p   1       n   1     r   e p   1       0   r     This is always   r and     as n      Therefore exp   x   converges to an element in   r O K   Same method works for log     Lemma 9 3  F     r O K   F     r   1 O K       k       for all r   1   Proof  Definition of formal group gives      F   X   Y     X   Y   X Y             So if x   y   O K      F     r x     r y       r   x   y     m o d   r   1         Therefore     F     r O K       k         r x   x mod       is a surjective group homomorphism with kernel F     r   1 O K       Corollary  Assuming that     k        Then  F     O K   has a subgroup of finite index isomorphic to   O K         Notation  Reduction modulo       O K   O K   O K   k x   x       Proposition 9 4  Assuming that   E   K be an elliptic curve  Then  the reduction mod   of any two minimal Weierstrass equations for E define isomorphic curves over k   Proof  Say Weierstrass equations are related by   u   r   s   t    u   K    r   s   t   K  Then   1   u 1 2   2  Both equations minimal gives us that v     1     v     2    hence u   O K     Transformation formula for the a i and b i   O K is integrally closed  hence r   s   t   O K   The Weierstrass equations obtained by reducing mod   are now related by        r     s     t           k    r     s     t     k     Definition  The reduction      k of E   K is defined by the reduction of a minimal Weierstrass equation   E has good reduction  if    is non singular  and so an elliptic curve   otherwise has bad reduction   For an integral Weierstrass equation    If v         0 then good reduction   If 0   v         1 2 then bad reduction   If v         1 2 then beware that the equation might not be minimal   There is a well defined map          2   K       2   k     x   y   z       x         z          choose a representative with min   v   x     v   y     v   z       0    We restrict to give        E   K         K   P   P       If P     x   y     E   K   then by Lemma 9 1 either   x   y   O K in which case  P       x           or v   x       2 s  v   y       3 s for some s   1  in which case P     x   y   1         3 s x     3 s y     3 s   and P       0   1   0     Therefore      E 1   K     I d i d n   t m a n a g e t o w r i t e t h i s b e f o r e i t w a s r u b b e d o f f      kernel of reduction    Notation              if E has good reduction       singular point   if E has bad reduction     The chord and tangent process still defines a group law on     ns   In cases of bad reduction      ns     m  over k or possibly a quadratic extension of k  or     ns     a  over k    For simplicity we suppose char k   2   Then       y 2   f   x    deg f   3              ns       a   x   y     x y   t   2   t   3       t   point at         0     Let P 1   P 2   P 3 lie on the line a x   b y   1  Write P i     x i   y i    t i   x i y i  Then x i 3   y i 2   y i 2   a x i   b y i    So t i 3   a t i   b   0  So t 1   t 2   t 3 are the roots of T 3   a T   b   0  Looking at coefficient of T 2 gives t 1   t 2   t 3   0   Definition   E 0   K       E 0   K       P   E   K     P       ns   k       Proposition 9 5  E 0   K   is a subgroup of E   K   and reduction modulo   is a surjective group homomorphism   E 0   K       ns   k     Note   If E   K has good reduction  then this is a surjective group homomorphism   E   K         k     Proof  Group homomorphism   A line l in   2 defined over K has equation      l   a X   b Y   c Z   0 a   b   c   K       We may assume min   v   a     v   b     v   c       0  Reduction modulo   gives a line         l       X   b   Y   c   Z   0       If P 1   P 2   P 3   E   K   with P 1   P 2   P 3   0 then these points lie on a line l  So P   1   P   2   P   3 lie on the line l    If  P   1   P   2       ns k   then  P   3       ns k     So if P 1   P 2   E 0   K   then P 3   E 0   K   and P   1   P   2   P   3   0    Exercise  check this still works if     P   1   P   2   P   3         Surjective   Let f   x   y     y 2   a 1 x y   a 3 y     x 3        Let   P       ns   k       0    say  P       x   0     0   for some x 0   y 0   O K   Since P   non singular  either    i    f   x   x 0   y 0       0   m o d        ii    f   y   x 0   y 0       0   m o d       If  i  then put g   t     f   t   y 0     O K   t    Then        g   x 0     0   m o d     g     x 0     O K       Hensel s lemma gives us that there exists b   O K such that        g   b     0 b   x 0   m o d         Then   b   y 0     E   K   has erduction P    asdfadsf    Recall that for r   1 we put      E r   K         x   y     v   x       2 r   v   y       3 r       0         If r   e p   1  these give           E r   K         O K             E 2   K             2 O K     E 1   K             O K     E 0   K     E   K         where for i   1  each E i   1   K     E i   K   gives a quotient isomorphic to   k         We have   E 0   K   E 1   K       ns   k     What about E   K   E 0   K     Lemma 9 6  Assuming that     k        Then  E 0   K     E   K   has finite index   Proof    k       implies that O K   r O K is finite for all r   1   Hence      O K   lim r       O K   r O K     is a profinite group  hence compact   Then   n   K   is the union of sets          a 0       a i   1   1   a i   1       a n     a j   O K       and hence compact  for the   adic topology    Now note E   K       2   K   is a closed subset  hence compact   So E   K   is a compact topological group   If     has a singular point    x   0     0   then      E   K     E 0   K         x   y     E   K     v   x   x 0     1   v   y   y 0     1       is a closed subset of E   K   hence E 0   K   is an open subgroup of E   K     The cosets of E 0   K   are an open cover of E   K    Hence   E   K     E 0   K             Definition   Tamagawa number   c K   E       E   K     E 0   K     is called the Tamagawa number   Remark    i  If we have good reduction then  c K   E     1  converse is false     ii  It can be shown that  either  c K   E     v        or  c K   E     4  for the above statements  it is essential that we work with minimal Weierstrass equations   We deduce   Theorem 9 7  Assuming that     K     p        Then  E   K   contains a subgroup of finite index isomorphic to   O K         Let   K     p       and L   K a finite extension  Let the residue fields be k   and k  and let f     k     k     Facts    i    L   K     e f    ii  If L   K is Galois then the natural map Gal   L   K     Gal   k     k   is surjective with kernel of order e   Definition   Unramified   L   K is unramified  if e   1   Fact   For each m   1   i  k has a unique extension of degree m  say k m     ii  K has a unique unramified extension of degree m  say K m    These extensions are Galois  with cyclic Galois groups   Definition   Maximal unramified extension   K nr     m   1 K m  inside K      maximal unramified extension   Theorem 9 8  Assuming that     K     p        E   K has good reduction  p   n  P   E   K    Then  K     n     1 P     K is unramified   Notation       n     1   P       Q   E   K       n Q   P            K     Q 1       Q r       K   x 1       x r   y 1       y r         where Q i     x i   y i     Proof  For each m   1 there is a short exact sequence         0   E 1   K m     E   K m         K m     0       Taking   m   1 gives a commutative diagram with exact rows    An isomorphism by Corollary 8 5 applied over each K m  using p   n here    surjective by Theorem 2 8  kernel         n     2 by Theorem 6 5  again using p   n    Snake lemma gives        E   K nr     n           n     2 E   K nr   n E   K nr     0     So if P   E   K   then there exists  Q   E   K nr   such that n Q   P and        n     1 P     Q   T   T   E   n       E   K nr         Hence  K     n     1 P     K nr and so K     n     1 P     K is unramified