10   Dedekind domains and extensions  Let L   K be a finite extension  For x   L  we write Tr L   K   x     K for the trace of the K linear map L   L  y   x y   If L   K is separable of degree n and   1         n   L   K  denotes the set of embeddings of L into an algebraic closure K    then Tr L   K   x       i   1 n   i   x     K   Lemma 10 1  Assuming that   L   K a finite separable extension of fields  Then  the symmetric bilinear pairing                 K   x   y     Tr L   K   x y       is non degenerate   Proof  L   K separable tells us that L   K       for some     L  Consider the matrix A for           in the K basis for L given by 1           n   1   Then A i j   Tr L   K     i   j       B B     i j where      B     1 1   1   1         2           n                 1     n   1     2     n   1       n     n   1         So      det A   det   B   2       1   i   j   n     i           j           2      Vandermonde determinant   which is non zero since   i           j       for i   j by separability     Exercise   On Example Sheet 3 we will show that a finite extension L   K is separable if and only if the trace form is non degenerate   Theorem 10 2  Assuming that   O K a Dedekind domain  L a finite separable extension of  K   Frac   O K    Then  the integral closure  O L of  O K in L is a Dedekind domain   Proof  O L a subring of L  hence  O L is an integral domain   Need to show    i  O L is Noetherian    ii  O L is integrally closed in L    iii  Every   0 prime ideal P in  O L is maximal   Proofs    i  Let e 1       e n   L be a K basis for L  Upon scaling by K  we may assume  e i   O L for all i  Let f i   L be the dual basis with respect to the trace form            Let  x   O L  and write x     i   1 n   i f i     i   K  Then    i   Tr L   K   x e i     O K    For any  z   O L  Tr L   K   z   is a sum of elements in K   which are integral over  O K  Hence Tr L   K   z     K is integral over  O K  hence  Tr L   K   z     O K    Thus   O L   O K f 1       O K f n   L  Since  O K is Noetherian   O L is finitely generated as an  O K module  hence  O L is Noetherian    ii  Example Sheet 2    iii  Let P be a non zero prime ideal of  O L  and  p     P   O K be a prime ideal of  O K   Let 0   x   P  Then x satisfies an equation      x n   a n   1 x n   1       a 0   0   a i   O K       with a 0   0  Then  a 0   P   O K is a non zero element of p  hence p is non zero  hence p is maximal   We have   O K   p   O L   P  and  O L   P is a finite dimensional vector space over  O K   p  Since  O L   P is an integral domain and finite  it is a field     Remark   Theorem 10 2 holds without the assumption that L   K is separable    Corollary 10 3  The ring of integers of a number field is a Dedekind domain    Convention    O K is the ring of integers of a number field    p   O K a non zero prime ideal  We normalise       p  absolute value associated to  v p  as defined in Definition 9 6   by    x   p     N p     v p   x    where  N p     O K   p     In the following theorems and lemmas we will have    O K a Dedekind domain  K   Frac   O K    L   K finite separable  O L the integral closure of  O K in L  which is a Dedekind domain by Theorem 10 2     Lemma 10 4  Assuming that   0   x   O   K  Then        x       p   0 prime ideal p v p   x         Proof  x O K     p       p O K     p     v p   x   by definition of  v p   x     Lemma follows from property of localisation     I   J   I O K     p     J O K     p       for all prime ideals p     Notation  P   O L   p   O K non zero prime ideals  We write P   p if  p O L   P 1 e 1   P r e r and P     P 1       P r    e i   0  P distinct     Theorem 10 5  Assuming that   O K   O L  K  L as usual  for p a non zero prime ideal of  O K  we write  p O L P 1 e 1   P r e r  Then  the absolute values on L extending       p  up to equivalence  are precisely       P 1             P L   Proof  By Lemma 10 4 for any  0   x   O K and i   1       r we have v P i   x     e i v p   x    Hence  up to equivalence        P i extends       p   Now suppose       is an absolute value on L extending       p  Then       is bounded on    hence is non archimedean  Let R     x   L     x     1     L be the valuation ring for L with respect to        Then  O K   R  and since R is integrally closed  in L  Lemma 6 8     we have  O L   R  Set     P       x   O L     x     1     m R   O L      where m R is the maximal ideal of R    Hence P a prime ideal in  O L  It is non zero since p   P  Then  O L     p     R  since  s   O L   P     s     1   But  O L     p   is a discrete valuation ring  hence a maximal subring of L  so  O L     p     R  Hence       is equivalent  to       p  Since       extends       p   P   O K   p so P 1 e 1   P r e r   P  so P   P i for some i     Let K be a number field  If     K         is a real or complex embedding  then x         x       defines  an absolute value on K  Example Sheet 2  denoted           Corollary 10 6  Let K be a number field with ring of integers  O K  Then any absolute value on K is equivalent to either   i        p for some non zero prime ideal of  O K    ii          for some     K           Proof  Case 1         non archimedean  Then           is equivalent to       p for some prime p by Ostrowski s Theorem  Theorem 10 5 gives that       is equivalent to       p for some      O K a prime ideal with      p   Case 2         archimedean  See Example Sheet 2     10 1   Completions  O K a Dedekind domain  L   K a finite separable extension   Let      O K   P   O L be non zero prime ideals with  P       We write K   and L P for the completions of K and L with respect to the absolute values         and       P respectively   Lemma 10 7   i  The natural   P   L   K K     L P is surjective    ii    L P   K P       L   K     Proof  Let M   L K     Im     P     L P   Write L   K       then M   K          Hence M is a finite extension of K   and   M   K         L   K    Moreover M is complete  Theorem 6 1   and since L   M   L P  we have M   L P     Lemma 10 8   Chinese remainder theorem    Assuming that   R a ring  I 1       I n   R ideals  I i   I j   R for all i   j  Then    i    i   1 n     i   1 n I i    I say     ii  R   I     I   1 n R   I i   Proof  Example Sheet 2     Theorem 10 9  The natural map      L   K K       P     L P     is an isomorphism    Proof  Write L   K       and let f   X     K   X   be the minimal polynomial of    Then we have      f   X     f 1   X     f r   X     K     X       where  f i   X     K     X   are distinct irreducible  separable   Since L   K   X     f   X        L   K K     K     X     f i   X       i   1 r K     X     f i   X         Set  L i   K     X     f i   X   a finite extension of  K    Then L i contains both  K   and L  use  K   X     f   x     K     X     f i   X   injective since morphism of fields   Moreover L is dense inside L i  approximate coefficients of K     X     f i   X   with an element of K   X     f i   X      The theorem follows from the following three claims    1    L i   L P for some prime P of  O L dividing      2    Each P appears at most once    3    Each P appears at least once   Proof of claims    1    Since    L i   K          there is a unique absolute value on L i extending          Theorem 10 5 gives us that         L  is equivalent to       P for some  P      Since L is dense in L and L i is complete  we have L i   L P    2    Suppose     L i   L j is an isomorphism preserving L and  K    then          K     X     f i   X     K     X     f i   X       takes x to x and hence f i   f i    3    By Lemma 10 7  the natural map    P   L   K K     L P is surjective for any prime  P      Since L P is a field    P factors through L i for some i  and hence L i   L P by surjectivity of   P     Example  K      L       i    f   X     X 2   1  Hensel s Lemma version 1 gives us that   1     5  Hence   5   splies in     i    i e   5 O L     1   2   Corollary 10 10  Let  0       O K a prime ideal  For x   L we have      N L   K   x       P     N L P   L     x         Proof  Let B 1       B r be bases for   L P 1       L P r as  K   vector spaces  Then B     i B i is a basis for  L   K K   over  K    Let   mult   x     B  respectively   mult   x     B i  denote the matrix for   mult   x     L   K K     L   K K    respectively  L P i   L P i  with respect to the basis B  respectively B i   Then        mult   x     B       mult   x     B 1     mult   x     B r       hence     N L   K   x     det     mult   x     B       i   1 r det   mult   x     B i     i   1 r N L P i   K     x