11   Kummer Theory  Let K be a field and char K   n  Assume   n   K   Lemma 11 1  Assuming that       K       K     n a finite subgroup  L   K     n    Then  L   K is Galois and Gal   L   K     Hom         n     Proof    n   K gives L   K normal  and char K   n gives that L   K is separable  So L   K is Galois   Define the Kummer pairing             Gal   L   K           n       x         x n   x n     Well defined   Suppose         L with   n     n   x  Then         n   1  so         n   K  Then                 for all     Gal   L   K    hence                        for all     Gal   L   K     Bilinear              x           x n     x n   x n x n         x         x         x y       x y n x y n     x n x n   y n y n         x         y       Non degenerate   Let     Gal   L   K    If       x     1 for all x      then        x n   x n   x         adn hence   fixes L pointwise  i e      1   Let x   K     n      If       x     1 for all     Gal   L   K    then       x n   x n       Gal   L   K         So x n   K  so x     K     n  i e  x   K     n     is the identity element   We get injective group homomorphisms   i  Gal   L   K     Hom         n      ii      Hom   Gal   L   K       n      i  implies Gal   L   K   is abelian and of exponent dividing n   Fact   If G is a finite abelian group of exponent dividing n then Hom   G     n     G  non canonically    So        Gal   L   K        i           ii    Gal   L   K           Therefore  i  and  ii  are isomorphisms     Example  Gal       2   3   5                 2     3   Definition   Abelian extension   We say L   K is abelian  if it is Galois  and has abelian Galois group   Similarly for other group terminology  e g  we can say that L   K has exponent dividing n to mean that it is Galois  with Galois group having exponent dividing n    Theorem 11 2  There is a bijection          finite subgroups     K       K     n       finite abelian extension L   K of exponent dividing n       K     n     L     n   K     K     n     L     Proof   i  Let     K       K     n be a finite subgroup  Let L   K     n   and         L     n   K     K     n  We must show           Clearly          So      L   K     n     K       n     L       So K     n     K       n    So Lemma 11 1 gives                   Since          it follows that            ii  Let L   K be a finite abelian extension of exponent dividing n  Let       L     n   K     K     n  Then K     n     L and we aim to prove this inclusion is an equality  Let G   Gal   L   K    Thu Kummer pairing gives an injection     Hom   G     n     Claim   This map is surjective   Granted the claim        K     n     K     Lemma 11 1         by claim   G       L   K       Since K     n     L it follows that L   K     n     Proof of claim   Let     G     n be a group homomorphism  Distinct automorphisms are linearly independent  So there exists a   L such that            G           1     a       y   0       Let     G  Then         y           G           1       a           G         1       1     a                   G           1     a               y           Therefore     y n     y n for all     G  Hence y n   K   Let x   y n  Then x   K       L     n  Then x   K     n       Also  by                   y   y     x n x n  So the map     Hom   G     n   sends x      This proves the claim     Proposition 11 3  Assuming    K a number field     n   K   S a finite set of promes of K Then there are only finitely many extensions L   K such that   i  L   K is a finite abelian extension of exponent dividing n   ii  L   K is unramified at all     S  Proof  Theorem 11 2 gives L   K     n   for some     K       K     n a finite subgroup   Let   be a prime of K        O L   P 1 e 1   P r e r     for P 1       P r some distinct primes in O L   If x   K   represents an element of   then      n v P i   x n     v P i   x     e i v     x         If     S then all e i   1  so v     x     0   m o d n    Therefore     K   S   n   where     K   S   n       x   K       K     n   v     x     0   m o d n         S         The proof is completed by the next lemma     Lemma 11 4  K   S   n   is finite   Proof  The map     K   S   n           n       S   x     v     x     m o d n         S     is a group homomorphism with kernel K       n     Since   S        it suffices to prove the lemma with S       If x   K   represents an element of K       n   then   x       n for some fractional ideal    There is a short exact sequence     0   O K       O K     n   K       n     Cl K   0 x   K     n               Cl K       and O K   being a finitely generated abelian group  Dirichlet s unit theorem  gives us that K       n   is finite