8   Formal Groups  Definition   I adic topology   Let R be a ring and I   R an ideal  The I adic  topology  on R has basis   r   I n   r   R   n   1     Definition   Cauchy sequence   A sequence   x n   in R is Cauchy  if   k     N such that   m   n   N  x m   x n   I k   Definition   Complete   R is complete  if   i    n   0 I n     0     ii  every Cauchy sequence converges  Useful remark   if x   I then      1 1   x   1   x   x 2         so 1   x   R     Example  R     p  I   p   p   R         t      I     t     Lemma 8 1   Hensel s Lemma    Assuming that   R is complete with respect to an ideal I  F   R   X    s   1  a   R satisfies F   a     0   m o d I s    F     a     R    Then  there exists a unique b   R such that F   b     0 and b   0   m o d I s     Proof  Let u   R   with F   a     u   m o d I    e g  we could take u   F     a     Replacing F   X   by F   X   a   u  we may assume a   0  F     0     1   m o d I     We put x 0   0      x n   1   x n   F   x n    1      Easy induction gives      x n   0   m o d I s     n  2      Let      F   X     F   Y       X   Y     F     0     X G   X   Y     Y H   X   Y      3      for some G   H   R   X   Y     Claim   x n   1   x n   m o d I n   s   for all n   0   Proof  By induction on n  Case n   0 is true   Suppose x n   x n   1   m o d I n   s   1    Then      F   x n     F   x n   1       x n   x n   1     1   c     m o d I n   s       for some c   I  Hence      F   x n     F   x n   1     x n   x n   1   m o d I n   s       Hence      x n   F   x n     x n   1   F   x n   1     m o d I n   s       and hence x n   1   x n   m o d I n   s     This proves the claim   Therefore   x n   n   0 is Cauchy  Since R is complete  we have x n   b as n     for some b   R   Taking limit n     in  1   gives b   b   F   b    hence F   b     0   Taking limit n     in  2   gives b   0   m o d I s     Uniqueness is proved using  3   and the  useful remark   if x   I then 1   x   R           E   Y 2 Z   a 1 X Y Z   a 3 Y Z 2   X 3   a 2 X 2 Z   a 4 X Z 2   a 6 Z 3       Affine piece Y   0  t     X Y  w     Z Y      w   t 3   a 1 t w   a 2 t 2 w   a 3 w 2   a 4 t w 2   a 6 w 3     f   t   w         We apply Lemma 8 1 with     R       a 1       a 6       t     I     t   F   X     X   f   t   X     R   X       s   3  a   0   Check      F   0       f   t   0       t 3   0   m o d I 3   F     0     1   a 1 t   a 2 t 2   R       Hence there exists a unique w   t     R       a 1       a 6       t     such that     w   t     f   t   w   t     w   t     0   m o d t 3       Remark   Taking u   1 in the proof of Lemma 8 1  w   t     lim n     w n   t   where w 0   t     0  w n   1   t     f   t   w n   t       In fact       w   t     t 3   1   A 1 t   A 2 t 2           where A 1   a 1  A 2   a 1 2   a 2  A 3   a 1 3   2 a 1 a 2   2 a 3     Lemma 8 2  Assuming that   R an integral domain which is complete with respect to an ideal I  a 1       a 6   R  K   Frac   R    Then             I           t   w     E   K     t   w   I       is a subgroup of E   K     Note   By uniqueness in Hensel s lemma              I         t   w   t       E   K     t   I         Proof  Taking   t   w       0   0   show  0 E       I     So it suffices to show that  P 1   P 2       I   then           P 1   P 2     P 3       I          P 1   P 2       I   implies t 1   t 2   I  w 1   w   t 1     I  w 2   w   t 2     I      w   t       n   2   A n   2 t n   1   A 0   1         w   t 2     w   t 1   t 2   t 1 t 1   t 2 w     t 1   t 1   t 2     n   2   A n   2   t 1 n   t 1 n   1 t 2       t 2 n     I     w 1     t 1   I     Substituting w     t     into w   f   t   w   gives       t       t 3   a 1 t     t         a 2 t 2     t         a 3     t       2   a 4 t     t       2   a 6     t       3 A     coefficient of t 3     1   a 2     a 4   2   a 6   3 B     coefficient of t 2     a 1     a 2     a 3   2   2 a 4       3 a 6   2       We have A   R   and B   I  Hence t 3     B A   t 1   t 2   I  w 3     t 3       I     Taking R       a 1       a 6       t      I     t    then Lemma 8 2  gives that there exists         a 1       a 6       t     with     0     0  and          1     t   w   t             t     w       t             Taking R       a 1       a 6       t 1   t 2      I     t 1   t 2    Lemma 8 2 gives that there exists F       a 1       a 6       t 1   t 2     with F   0   0     0 and        t 1   w   t 1         t 2   w   t 2         F   t 1   t 2     w   F   t 1   t 2           and      F   X   Y     X   Y   a 1 X Y   a 2   X 2 Y   X Y 2           By properties of the group law  we deduce   i  F   X   Y     F   Y   X     ii  F   X   0     X and F   0   Y     Y   iii  F   X   F   Y   Z       F   F   X   Y     Z     iv  F   X       X       0  Definition   Formal group   Let R be a ring  A formal group  over R is a power series F   X   Y     R     X   Y     satisfying   i  F   X   Y     F   Y   X     ii  F   X   0     X and F   0   Y     Y   iii  F   X   F   Y   Z       F   F   X   Y     Z    This looks like it would only define a monoid  but in fact inverses are guaranteed to exist in this context   Exercise   Show that for any formal group  there exists a unique          X       X       R     X         such that F   X       X       0   Example   i  F   X   Y     X   Y  called     a    ii  F   X   Y     X   Y   X Y     1   X     1   Y     1  called     m    iii  F   X   Y      as above   called      Definition   Morphism   isomorphic  formal groups    Let F and G be formal groups  over R given by power series F and G    i  A morphism  f   F   G is a power series f   R     T     such that f   0     0 satisfying      f   F   X   Y       G   f   x     f   Y            ii  F   G if there exists F   f G and G   g F morphisms such that f   g   X       g   f   X       X   Theorem 8 3   All formal groups are isomorphic    Assuming that   char R   0  Then  any formal group F over R is isomorphic to     a over R      More precisely   i  There is a unique power series      log   T     T   a 2 2 T 2   a 3 3 T 3         with a i   R such that      log   F   X   Y       log   X     log   Y              ii  There is a unique power series      exp   T     T   b 2 2     b 3 3   T 3         with b i   R such that      exp   log   T       log   exp   T       T       Proof   i  Notation F 1   X   Y       F   x   X   Y    Uniqueness   Let      p   T     d d T log   T     1   a   2 T   a 3 T 2           Differentiating      with respect to X gives      p   F   X   Y     F 1   X   Y     p   X     0       Putting X   0 gives p   Y   F 1   0   Y     1  hence p   Y     F 1   0   Y     1  so p is uniquely determined by F and hence log is too   Existence   Let      p   T     F 1   0   T     1   1   a 2 T   a 3 T 2          say   Let      log   T     T   a 2 2 T 2   a 3 3 T 3           Calculate     F   F   X   Y     Z     F   X   F   Y   Z     F 1   F   X   Y     Z   F 1   X   Y     F 1   X   F   Y   Z     d d x F 1   Y   Z   F 1   0   Y     F 1   0   F   Y   Z     put X   0   F 1   Y   Z   p   Y     1   p   F   Y   Z       1   F 1   Y   Z   p   F   Y   Z       p   Y   log   F   Y   Z       log   Y     h   Z   integrate w r t  Y     for some power series h   Symmetry Y   Z gives h   Z     log   Z     This proves existence    ii  We prove a lemma first     Lemma 8 4  Assuming that   f   T     a T       R     T     with a   R    Then  there exists a unique g   T     a   1 T       R     T     such that f   g   T       g   f   T       T   Proof  We construct polynomials g n   T     R   T   such that     f   g n   T       T   m o d T n   1   g n   1   T     g n   T     m o d T n   1       Then g   T     lim n     g n   T   satisfies f   g   T       T   To start the induction  we set g 1   T     a   1 T  Now suppose n   2 and g n   1   T   exists  Then      f   g n   1   T       T   b T n   m o d T n   1         We put g n   T     g n   1   T       T n for some     R to be chosen later   Then     f   g n   T       f   g n   1   T       T n     f   g n   1   T         a T n   m o d T n   1     T     b     a   T n   m o d T n   1       We take       b a  a   R     b   R       R    This completes the induction step   We get g   T     a   1 T       R     T     such that f   g   T       T  Applying the same construction to g gives h   T     a T       R     T     such that g   h   T       T   Now note f   T     f   g   h   T         h   T       Theorem 8 3  ii  now follows by Lemma 8 4 and Q12 from Example Sheet 2   Notation  Let F  e g      a      m      be a formal group given by a power series F   R     X   Y       Suppose R is a ring complete  with respect to ideal I  For x   y   I  put      x   F y   F   x   y     I       Then F   I         I     F   is an abelian group   Examples        a   I       I            m   I       1   I             I     subgroup of E   K   in Lemma 8 2  Corollary 8 5  Assuming that   F a formal group over R  n     such that n   R    Then    i    n     F   F is an isomorphism of formal groups   ii  If R is complete with respect to ideal I then F   I       n F   I   is an isomorphism of groups  In particular  F   I   has no n torsion   Proof  We have       1     T     T   n     T     F     n   1     T     T        for n   0 use     1     T         T      Since      F   X   Y     X   Y   X Y           we get        2     T     F   T   T   2 T       R     T         and by induction we get        n     T     n T       R     T           Lemma 8 4 shows that if n   R   then   n   is an isomorphism  This proves  i   and  ii  follows