4   Hensel s Lemma  Theorem 4 1   Hensel s Lemma version 1    Assuming that     K           is a complete discretely valued field  f   X     O K   X    assume    a   O K such that   f   a         f     a     2  Then  there exists a unique  x   O K such that f   x     0 and   x   a       f     a       Proof  Let      O K be a uniformiser and let r   v   f     a      with v the normalised valuation  v         1   We construct a sequence   x n   n   1   in  O K such that     i  f   x n     0   m o d   n   2 r     ii  x n   1   x n   m o d   n   r    Take x 1   a  then f   x 1     0   m o d   1   2 r     Now we suppose we have constructed x 1       x n satisfying  i  and  ii   Define      x n   1   x n   f   x n   f     x n         Since x n   x 1   m o d   r   1    we have      v   f     x n       v   f     x i       r       and hence      f   x n   f     x n     0   m o d   n   r       by  i    It follows that x n   1   x n   m o d   n   r    so  ii  holds  Note that letting X   Y be indeterminates  we have      f   X   Y     f 0   X     f 1   X   Y   f 2   X   Y 2           where  f i   X     O K   X   and f 0   X     f   X    f 1   X     f     X    Thus     f   x n   1     f   x n     c f     x n     c 2 f 2   x N     c 2 f 2   x n             n   2 r   1     where c     f   x n   f     x n     Since c   0   m o d   n   r   and v   f i   x n       0 we have      f   x n   1     f   x n     f     x n   c   0   m o d   n   2 r   1         so  i  holds   Property  ii  implies that   x n   n   1   is Cauchy  so let  x   O K such that x n   x  Then f   x     lim n     f   x n     0 by  i    Moreover   ii  impies that     a   x 1   x n   m o d   r   1     n   a   x   m o d   r   1       x   a       f     a         This proves existence   Uniqueness  suppose x   also satisfies f     x     0    x     a       f     a      Set     x     x   0  Then       x     a       f     a       x   a         f     a           and the ultrametric inequality implies                x   x         f     a         f     x           But      0   f   x       f   x         f   x       0   f     x                         2       Hence   f     x               2  so   f     x              a contradiction     Corollary 4 2  Let    K           be a complete discretely valued field  Let  f   X     O K   X   and  c     k     O K   m a simple root of f     X       f   X     m o d m     k   X    Then there exists a unique  x   O K such that f   x     0  x   c     m o d m     Proof  Apply Theorem 4 1 to a lift  c   O K of c    Then   f   c       1     f     c     2 since c   is a simple root     Example  f   X     X 2   2 has a simple root modulo 7  Thus 2     7     7   Corollary 4 3       p         p     2           2     2 if p   2       2     3 if p   2     Proof  Case  p   2   Let  b     p    Applying  to f   X     X 2   b  we find that  b       p     2 if and only if b       p     2  Thus     p         p     2     p         p     2       2       p           p   1        We have an isomorphism        p                   p       given by   u   n     u p n  Thus        p         p     2         2     2       Case  p   2   Let b     2    Consider f   X     X 2   b  Note f     X     2 X   0   m o d 2    Let b   1   m o d 8    Then        f   1       2   3   2   2     f     1     2       Hensel s Lemma version 1 gives      b       2     2   b   1   m o d 8         Then        2         2     2         8               2     j       Again using   2       2        we find that   2           2     3     Remark   Proof uses the iteration      x n   1   x n   f   x n   f     x n         which is the non archimedean analogue of the unewton Raphson method    Theorem 4 4   Hensel s Lemma version 2    Assuming that     K           is a complete discretely valued field  f   X     O K   X    f     X       f   X     m o d m     k   X   factorises as f     X     g     X   h     X   in k   X    g     X   and h     X   coprime   Then  there is a factorisation      f   X     g   X   h   X       in  O K   X    with g     X     g   X     m o d m    h     X     h   X     m o d m   and deg g     deg g   Proof  Example Sheet 1     Corollary 4 5  Let    K           be a complete discretely valued field  Let      f   X     a n X n       a n   K   X       with a 0   a n   0  If f   X   is irreducible  then   a i     max     a 0       a n     for all i   Proof  Upon scaling  we may assume  f   X     O K   X   with max i     a i       1  Thus we need to show that max     a 0       a n       1  If not  let r minimal such that   a r     1  then 0   r   n  Thus we have      f     X     X r   a r       a n X n   r     m o d m         Then Theorem 4 4 implies f   X     g   X   h   X   with 0   deg   n