6   Extensions of complete valued fields  Theorem 6 1  Assuming that     K           is a complete discretely valued field  L   K a finite extension of degree n  Then    i        extends uniquely to an absolute value        L on L defined by        y   L     N L   L   y     1 n   y   L        ii  L is complete with respect to        L   Recall   If L   K is finite  N L   K   L   K is defined by N L   K   y     det K   m u l t   y   where m u l t   y     L   L is the K linear map induced by multiplication by y   Facts   N L   K is multiplicative   Let X n   a n   1 X n   1       a 0   K   X   be the minimal polynomial of y   L  Then N L   K   y       a 0 m for some m   1  in fact  m is the degree of L   K   y      N L   K   y     0   y   0   Definition 6 2   Norm    Let    K           be a non archimedean valued field  V a vector space over K  A norm on V is a function         V       0 satisfying    i    x     0   x   0    ii      x             x   for all     K  x   V    iii    x   y     max     x       y     for all x   y   V   Example  If V is finite  dimensional and e 1       e n is a basis of V  The supremum        sup on V is defined by        x   sup   max i   x i         where x     i   1 n x i e i   Exercise          sup is a norm    Definition 6 3   Equivalent norms    Two norms        1 and        2 on V are equivalent if there exists C   D       0 such that      C   x   1     x   2   D   x   1   x   V       Fact   A norm defines a topology on V  and equivalent norms induce the same topology   Proposition 6 4  Assuming that     K           is a complete non archimedean valued field  V a finite dimensional vector space over K  Then  V is complete with respect to        sup   Proof  Let   v i   i   1   be a Cauchy sequence in V  and let e 1       e n be a basis for V   Write v i     j   1 n x j i e j  Then   x j i   i   1   is a Cauchy sequence in K  Let x j i   x j   K  then v i   v       j   1 n x j e j     Theorem 6 5  Assuming that     K           is a complete non archimedean valued field  V a finite dimensional vector space over K  Then  any two norms on K are  equivalent  In particular  V is complete with respect to any norm  using Proposition 6 4     Proof  Since equivalence defines an equivalence relation on the set of norms  it suffices to show that any norm        is equivalent to        sup   Let e 1       e n be a basis for V  and set  D     max i   e i     0  Then for x     i   1 n x i e i  we have        x     max i   x i e i     max i   x i     e i     D max i   x i     D   x   sup       To find C such that   C       sup          we induct on n   dim V   For n   1     x       x 1 e 1       x 1     e 1    so take  C     e 1     For n   1  set V i   span   e 1       e i   1   e i   1       e n     By induction  V i is complete with respect to         hence closed   Then e i   V i is closed for all i  and hence      S       i   1 n e i   V i     is a closed subset not containing 0  Thus there exists c   0 such that B   0   C     S     where  B   0   C       x   V     x     C     Let 0   x     i   1 n x i e i and suppose     x j     max i   x i    Then    x   sup     x j    and 1 x j   S  Thus   x i x j     C  and hence        x     C   x j     C   x   sup       V is complete since it is complete with respect to        sup  see Proposition 6 4       Definition 6 6   Integral closure    Let R be a subring of S  We say s   S is integral  over R if there exists a monic polynomial f   X     R   X   such that f   s     0   The integral closure  R i n t   S   of R inside S is defined to be      R i n t   S       s   S   s integral over R         We say R is integrally closed  in S if R i n t   S     R   Proposition 6 7  R int   S   is a subring of S  Moreover   R int   S   is integrally closed in S   Proof  Example Sheet 2     Lemma 6 8  Assuming that     K           is non archimedean valued field  Then   O K is integrally closed in K   Proof  Let x   K be integral  over  O K  Without loss of generality  x   0  Let  f   X     X n   a n   1 X n   1       a 0   O K   X   such that f   x     0  Then      x     a n   1 1 x       a 0 1 x n   1       If    x     1  we have     a n   1 1 x       a 0 1 x n   1     1  Thus     x     1   x   O K     Lemma 6 9  O L is the integral closure of  O K inside L   Proof  Let 0   y   L and let      f   X     X d   a d   1 X d   1       a 0   K   X       be the minimal  monic  polynomial of y   Claim   y integral over  O K if and only if f   X       K   X         Clear       Let  g   X     O K   X   monic such that g   y     0  Then f   g  in K   X     and hence every root of f is a root of g  So every root of f in K   is integral over  O K  so a i are integral over  O K for i   0       d   1   Hence  a i   O k  by Lemma 6 8    By Corollary 4 5    a i     max     a 0     1   for i   0       d   1  By property of N L   K  we have N L   K   y       a 0 m for m   1   Hence     y   O L     N L   K   y       1     a 0     1   C o r o l l a r y 4   5   a i     1   i   i e  a i   O K     Thus msub  msub int   L     O L and proves the Lemma     Proof of Theorem  6 1     We first show         L     N L   K         1 n satisfies the three axioms in the definition of absolute value    i       y   L   0     N L   K   y     1 n   0   N L   K   y     0   y   0      ii       y 1 y 2   L     N L   K   y 1   y 2     1 n     N L   K   y 1   N L   K   y 2     1 n     N L   K   y 1     1 n   N L   K   y 2     1 n     y 1   L   y 2   L      iii  Set   O L     y   L     y   L   1    Claim    O L is the integral closure of  O K inside L   Assuming this  we prove  iii   Let x   y   L  and without loss of generality assume     x   L     y   L  Then   x y   L hence  x y   O L  Since  1   O L and  O Lis a ring  we have  1   x y   O L and hence   1   x y   L   1  Hence     x   y   L     y   L   max     x   L     y   L   thus  iii  is satisfied   So we have proved that       L is an absolute value on L   Since N L   K   x     x n for x   K    x   L extends       on K   If        L   is another absolute value on L extending         then       L         L   are norms on L   Theorem 6 5 tells us that       L           L induce the same topology on L  Hence       L           L c for some c   0  by Proposition 1 4   since       L   extends        we have c   1   Now we show that L is complete with respect to       L  this is immediate by Theorem 6 5     Let   K           be a complete discretely valued field   Corollary 6 10  Let L   K be a finite extension  Then   i  L is discretely valued with respect to       L    ii  O L is the integral closure of  O K in L   Proof   i  v a valuation on K  v L valuation on L such that v L extends v  Let n     L   K     and let y   L    Then   y   L     N L   K   y     1 n hence v L   y     1 n v   N L   K   y      hence v L   L       1 n v   K      so v L is discrete    ii  Lemma 6 9      Corollary 6 11  Let K     K be an algebraic closure of K  Then       extends to a unique absolute value       K   on K     Proof  Let x   K    then x   L for some L   K finite  Define   x   K       x   L  Well defined  i e  independent of L by the uniqueness in Theorem 6 1   The axioms for        K   to be an absolute value can be checked over finite extensions   Uniqueness  clear     Remark         K   on K   is never discrete  For example  K     p   p n     p   for all n       0  Then     v p   p n     1 n v   p     1 n         p   is not complete with respect to          p   Example Sheet 2    p     completion of    p   with respect to          p    then   p is algebraically closed    Proposition 6 12  Assuming that   L   K finite extension of complete discretely valued fields    i    O K is compact    ii   The extension of residue fields k L   k is finite and separable   Then  there exists      O L such that   O L   O K         Later we ll prove that the  i  implies  ii    Proof  We ll choose      O L such that    there exists      O L       a uniformiser for  O L  O K         k L surjective  k L   k separable tells us that there exists       k L such that k L   k           Let      O L a lift of      and  g   X     O K   X   a monic lift of the minimal polynomial of       Fix    L   O L a uniformiser  Then g     X     k   X   irreducible and separable  hence g         0 mod   L and g             0 mod   L   If g         0 mod   L 2  then     g         L     g           L g         mod   L 2       Thus      v L   g         L       v L     L g             v L     L     1        v L normalised valuation on L    Thus either v L   g           1 or v L   g         L       1  Upon possibly replacing   by       L  we may assume v L   g           1   Set      g         O K       a uniformiser  Then  O K         L is the image of a continuous map      O K n   L   x 0       x n   1       i   0 n x i a i     where n     K         K    Since  O K is compact   O K         L is compact  hence closed  Since k L   k           O K       contians a set of coset representatives for   k L   O L     O L   Let  y   O L  Then Proposition 3 4 gives us      y     i   0     i   i     i   O K           Then  y m     i   0 m   i   i   O K        Hence  y   O K        since  O k       is closed