6   The Invariant Differential  Let C be an algebraic curve over K   K     Definition   Space of differentials   The space of differentials    C is the K   C   vector space generated by d f for f   K   C    subject to the relations   i  d   f   g     d f   d g   ii  d   f g     f d g   g d f   iii  d a   0   a   K  Fact     C is a 1 dimensional K   C   vector space   Let 0         C  Let P   C be a smooth point  and t   K   C   a uniformiser at P   Then     f d t for some f   K   C      We define ord p         ord p   f    independent of choice of t    We assume C is a smooth projective curve   Definition  div           P   C ord P       P   Div   C     Note   This is a divisor  i e  ord p         0 for all but finitely many P   C   Definition   Regular differential   A differential       C is regular  if  div         0  i e  it has no poles      g   C     dim K         C   div         0         As a consequence of Riemann Roch we have   If 0         C  then  deg   div       2 g   2   Fact   Suppose f   K   C      ord p   f     n   0  If char K   n  then ord p   d f     n   1   Lemma 6 1  Assuming that   char K   2  E  y 2     x   e 1     x   e 2     x   e 3    e 1   e 2   e 3   K distinct  Then      d x y is a differential on E with no zeroes or poles   g   E     1  In particular  the K vector space of regular differentials on E is 1 dimensional  spanned by     Proof  Let T i     e i   0     E   2       0   T 1   T 2   T 3        div   y       T 1       T 2       T 3     3   0      1      For P   E     0        div   x   x p       P         P     2   0         If  P   E   E   2    then ord p   x   x p     1   ord p   d x     0   If  P   E   E   2   then ord p   x   x p     1   ord p   d x     0   If P   T i then ord p   x   x p     2   ord p   d x     1   If P   0 then ord p   x       2   ord p   d x       3   Therefore      div   d x       T 1       T 2       T 3     3   0      2       1   and  2   implies  div   d x y     0     Definition  For     C 1   C 2 a non constant morphism we define             C 2     C 1 f d g         f   d       g       Lemma 6 2  Assuming that   P   E        P   E   E X   P   X         d x y as above  Then    P          we say   is the invariant differential     Proof    P     is a regular differential on E  So   P         p     The map E     1  P     P is a morphism of smooth projective curves but not  surjective  misses 0 and     Therefore it is constant  by Theorem 2 8    i e  there exists     K   such that   P           for all P   E   Taking P   0 E shows     1     Remark   If K                E        z         z           z          d x y         z   d z       z     d z        invariant under z   z     const      Lemma 6 3  Assuming that           Hom   E 1   E 2      an invariant differential on E 2  Then                                  Proof  Write E   E 2     E   E   E       P   Q     P   Q p r 1     P   Q     P p r 2     P   Q     Q     Fact     E   E is a 2 dimensional K   E   E   vector space with basis p r 1     and p r 2       Therefore              f p r 1       g p r 2      1      for some f   g   K   E   E     For fixed Q   E  let       Q   E   E   E P     P   Q       Applying   Q   to  1    gives           Q             Q   f     p r 1   Q             Q   g     p r 2   Q         Q                 Q   f     by Lemma 6 2     0     Therefore   Q   f   1 for all Q   E  so f   P   Q     1 for all P   Q   E   Similarly g   P   Q     1 for all P   Q   E        1             p r 1       p r 2           Now pull back by     E 1   E   E P         P         P         to get                                            Lemma 6 4  Assuming that       C 1   C 2 a non constant morphism  Then    is separable  if and only if         C 1     C 2 is non zero   Proof  Omitted     Example    m     1     0    multiplicative group             m     m x   x n      n   2 integer          d x     d   x n     n x n   1 d x  So if char K   n then   is separable   Theorem 2 8 implies       1   Q     deg   for all but finitely many Q     m   But   is a group homomorphism  so            1   Q       ker   Q       for all Q     m  Thus   ker     deg     n and hence K     K     contains exactly n n th roots of unity    Theorem 6 5  Assuming that   char K   n  Then   E   n           n     2   Proof  Lemma 6 3   induction gives   n         n     char K   n implies   n   is separable  So     n     1   Q     deg   n   for all but finitely many Q   E  But   n   is a group homomorphism  so    n     1   Q       E   n   for all Q   E   Putting these two statements together gives        E   n     deg   n     Corollary 5 9 n 2       Group theory  structure theorem  gives that      E   n         d 1             d t       for some 1   d 1   d 2       d t   n   Let p be a prime with p   d 1  Then  E   p           p     t  But    E   p     p 2  so t   2   But    E   p     p 2  so t   2  i e   E   n         d 1         d 2    Since d 1   d 2   n and d 1 d 2   n 2  we get d 1   d 2   n   Thus  E   n           n     2     Remark   If char K   p then   p   is inseparable  It can be shown that      either E   p r         p r     r   1  ordinary  or E   p     0  supersingular      Do not use this remark on Example Sheet 2