13   Unramified and totally ramified extensions of local fields  Let L   K be a finite separable extension of non archimedean local fields  Corollary 11 6 implies        L   K     e L   K f L   K           Lemma 13 1  Assuming that   M   L   K finite separable extensions of local fields  Then    i  f M   K   f L   K f M   L   ii  e M   K   e L   K f M   L  Proof   i  f M   K     k M   k       k M   k L     k L   k     f M   L f L   K    ii   i  and          Definition 13 2   Unramified   ramified   totally ramified    The extension L   K is said to be    unramified  if e L   K   1  equivalently f L   K     L   K      ramified  if e L   K   1  equivalently f L   K     L   K      totally ramified  if e L   K     L   K    equivalently f L   K   1    From now on in this course  if unspecified L   K is a finite separable extension of  non archimedean   local fields  Also  all local fields that we consider from now on will be non archimedean   Theorem 13 3  Assuming that   L   K a finite separable extension of non archimedean local fields  Then  there exists a field K 0  K   K 0   L and such that   i  K 0 is unramified   ii  L   K 0 is totally ramified  Moreover   L   K 0     e L   K    K 0   K     f L   K and K 0   K is Galois   Proof  Let k     q  so that k L     q f  f L   K   f  Set m   q f   1            q f   L the Teichm ller map for L   Let   m           for   a generator of   q f      m a primitive m th root of unity  Set K 0   K     m     L  then K 0   K is Galois and has residue field k 0     q         k L   Hence f L   K 0   1  i e  L   K 0 is totally ramified   Let res   Gal   K 0   K     Gal   k 0   k   be the natural map  For     Gal   K 0   K    We have       m       m if       m       m mod m  since   m   K 0       m   k 0   by Hensel s Lemma version 1    Hence res is injective  Thus   Gal   K 0   K         Gal   k 0   k       f K 0   K  so   K 0   K     f K 0   K   Hence res is an isomorphism  and K 0   K is unramified     Theorem 13 4  Assuming that   k     q  n   1  Then  there exists a unique unramified L   K of degree n  Moreover  L   K is Galois and the natural  Gal   L   K     Gal   k L   k   is an isomorphism  In particular  Gal   L   K       Frob L   K   is cyclic  where Frob L   K   x     x q mod m L for all  x   O L   Proof  For n   1  take L   K     m   where m   q n   1   As in Theorem 13 3       Gal   L   K       Gal   k L   K     Gal     q n     q         Hence Gal   L   K   is cyclic  generated by a lift of x   x q   Uniqueness  L   K of degree n unramified  Then Teichm ller gives   m   L  so L   K     m       Corollary 13 5  L   K a finite Galois extension  Then res   Gal   L   K     Gal   k L   k   is surjective    Proof  res factorises as      Gal   L   K     Gal   K 0   K       Gal   k L   k           Definition 13 6   Inertial subgroup    The inertial subgroup is      I L   K   ker   Gal   L   K     Gal   k L   k           Since e L   K f L   K     L   K    we have    I L   K     e L   K   I L   K   Gal   L   K 0     K 0 as in Theorem 13 3   Definition 13 7   Eisenstein polynomial    f   x     x n   a n   1 x n   1       a 0   O K   x   is Eisenstein  if v K   a i     1 for all i  and v K   a 0     1   Fact   f   x   Eisenstein  implies f   x   irreducible   Theorem 13 8   i  Let L   K finite totally ramified    L   O L a uniformiser  Then the minimal polynomial of   L is Eisenstein and   O L   O K     L    hence L   K     L      ii  Conversely  if  f   x     O K   x   is Eisenstein and a root of if f  then L     K         K is totally ramified and   is a uniformiser of L   Proof   i    L   K     e   e L   K  Let      f   x     x m   a m   1 x m   1       a 0   O K   x       the minimal polynomial for   L  Then m   e  Since v L   K       e    we have v L   a i   i     e mod e  for i   m  Hence these terms have distinct valuations  As        L m       i   0 m   1 a i   L i       we have      m   v L     L m     min 0   i   m   1   i   e v K   a i         hence v K   a i     1 for all i   Hence v K   a 0     1 and m   e  Thus f   x   is Eisenstein  and L   K     L    For y   L  we write y     i   0 e   1   L i b i  b i   K  Then      v L   y     min 0   i   e   1   i   e v K   b i           Thus     y   O L   v L   y     0   v K   b i     0   i   y   O K     L        ii  Let f   x     x n   a n   1 x n   1       a 0 is Eisenstein and e   e L   K  Thus v L   a i     e and v L   a 0     e  If v L         0  we have      v L     n     v L     i   0 n   1 a i   i       hence v L         0  For i   0  v L   a i   i     e   v L   a 0    Therefore      v L     n     v L       i   0 n   1 a i   i     v L   a 0     e       Hence n v L         e  But n     L   K     e  so n   e and v L         1      13 1   Structure of Units  Let   K            e     e K     p    a uniformiser  in K   Proposition 13 9  Assuming that   r   e p   1  Then  exp   x       n   0   x n n   converges on    r O K and induces an isomorphism          r O K             1     r O K             Proof     v K   n       e v p   n       e   n   s p   n     p   1 Example Sheet 1   e   n   1 p   1       For  x     r O K and n   1      v K   x n n       n r   e   n   1   p   1   r     n   1     r   e p   1       0     Hence v K   x n n         as n      Thus exp   x   converges   Since v K   x n n       r for all n   1   exp   x     1     r O K   Consider log   1     r O K     r O K      log   1   x       n   1       1   n   1 n x n     which converges as before   Recall identities in       X   Y          exp   X   Y     exp   X   exp   Y   exp   log   1   X       1   X log   exp   X       X     Thus   exp       r O K             1     r O K       is an isomorphism     K any  local field   U K     O K         O K uniformiser   Definition 13 10   s th unit group    For s      the s th unit group U K   s   is defined by      U K   s       1     s O K             Set U K   0     U K  Then we have          U K   s     U K   s   1         U K   0     U K       Proposition 13 11   i  U K   0     U K   i       k           k   O K        ii  U K   s     U K   s   1       k       for s   1  Proof   i  Reduction modulo     O K     k   is surfective with kernel  1     O K   U K   1      ii  f   U K   s     k  1     s x   x m o d          1     s x     1     s y     1     s   x   y     s x y         x   y     s x y   x   y mod    hence f is a group homomorphism  surjective with kernel U K   s   1        Remark   Let    K     p        Proposition 13 9     implies that there exists finite index subgroup of  O K   isomorphism to    O K         Example    p  p   2  e   1  take r   1  Then       p             p           1   p   p           p   1         p x     x mod p   x   x mod p         p   2  take  r   2        2             4           1   p 2   p         2       2 x     x mod 4   x     x         where          x         1 x   1   m o d 4     1 x     1   m o d 4       So         p         p     2         2   if p   2       2     2 if p   2