11   Decomposition groups  Definition 11 1   Ramification    Let 0     be a prime ideal of  O K  and        O L   P 1 e 1   P r e r     with P i distinct prime ideals in  O L  and e i   0    i  e i is the ramification index  of P i over      ii  We say   ramifies  in L if some e i   1   Example  O K       t     O L       T     O K   O L sends t   T n  Then   t O L   T n O L  so the ramification index of   T   over   t   is n   Corresponds geometrically to the degree n of covering of Riemann surfaces        x   x n   Definition 11 2   Residue class degree    f i       O L   P i   O K       is the residue class degree  of P i over     Theorem 11 3    i   1 r e i f i     L   K     Proof  Let  S   O K          Exercise  properties of localisation      1    S   1 O L is the integral closure of   S   1 O K in L    2    msup  msup   S   1 O L   S   1 P 1 e 1   S   1 P r e r    3    S   1 O L   S   1 P i   O L   P i and   S   1 O K   S   1     O K       In particular   2  and  3  imply e i and f i don t change when we replace  O K and  O L by  S   1 O K and  S   1 O L   Thus we may assume that  O K is a discrete valuation ring  hence a PID   By Chinese remainder theorem  we have      O L     O L     i   1 r O L   P i e i       We count dimension as  k     O K     vector spaces   RHS  for each i  there exists a decreasing sequence of k suibspaces     0   P i e i   1   P i e i       P i   P i e i   O L   P i e i       Thus  dim k O L   P i e i     j   0 e i   1 dim k   P i j   P i j   1    Note that P i j   P i j   1 is an  O L   P i module and x   P i j   P i j   1 is a generator  for example can  prove this after localisation at P i    Then dim k P i j   P i j   1   f i and we have      dim k O L   P i e i   e i f i       and hence      dim k   i   1 r O L   P i e i     i   1 r e i f i       LHS  Structure theorem for finitely generated modules over PIDs tells us that  O L is a free module over  O K of rank n   Thus   O L     O L     O K       n as k vector spaces  hence   dim k O L     O L   n     Geometric analogue   f   X   Y a degree n cover of compact Riemann surfaces  For y   Y      n     x   f   1   y   e   x     where e x is the ramification index of x  Now assume L   K is Galois  Then for any     Gal   L   K         P i     O K     and hence     P i       P 1       P r     Proposition 11 4  The action of Gal   L   K   on   P 1       P r   is transitive    Proof  Suppose not  so that there exists i   j such that     P i     P j for all     Gal   L   K     By Chinese remainder theorem  we may choose  x   O L such that x   0 m o d P i  x   1 m o d     P i   for all     Gal   L   K    Then     N L   K   x           Gal   L   K       x     O K   P i       P j       Since P j prime  there exists     Gal   L   K   such that     x     P j  Hence x       1   P j    i e  x   0 m o d     1   P j    contradiction     Corollary 11 5  Suppose L   K is Galois  Then e 1       e r   e  f 1       f r   f  and we have n   e f r   Proof  For any     Gal   L   K   we have   i    O L           O L       P 1   e 1       P r   e r  hence e 1       e r    ii  O L   P i   O L       P i   via    Hence f 1       f r     If L   Kis an extension of complete discretely valued fields with normalised valuations  v L  v K and uniformisers    L     K  then the ramification index is e   e L   K   v L     K    The residue class degree is f     f L   K     k L   k     Corollary 11 6  Let L   K be a finite separable extension  Then   L   K     e f   O K a Dedekind domain   Definition 11 7   Decomposition    Let L   K be a finite Galois extension  The decomposition at a prime P of  O L is the subgroup of Gal   L   K   defined by      G P         Gal   L   K         P     P         Proposition 11 8  Assuming that   O K a Dedekind domain  L   K a finite Galois extension  0   P   O L a prime ideal  P       O K  Then    i  L P   K   is Galois    ii  There is a natural map      res   Gal   L P   K       Gal   L   K       which is injective and has image  G P   Proof   i  L   K Galois implies that L is a splitting field of a separable polynomial f   X     K   X    Hence  L P is the splitting field of  f   X     K   X    hence   L P   K   is Galois    ii  Let       Gal   L P   K      then     L     L since L   K is normal  hence we have a map   res   Gal   L P   K       Gal   L   K            L  Since L is dense in  L P  res is injective  By Lemma 8 2  we have            x     P     x   P     for all       Gal   L P   K     and  x   L P  Hence     P     P for all       Gal   L P   K     and hence  res         G P for all       Gal   L P   K       To show surjectivity  it suffices to show that        G P     e f     L P   K           Write    O L   P 1 e 1   P r e r   f     O L   P   O K        Then     G P       Gal   L   K     r   e f r r   e f  using Corollary 11 5       L P   K       e f  Apply Corollary 11 6 to   L P   K    noting that e   f don t change when we take completions