UBC Theses and Dissertations

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UBC Theses and Dissertations

Finite groups of fractional linear transformations Kitchen, Vivien Beth 1972

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FINITE GROUPS OF FRACTIONAL LINEAR TRANSFORMATIONS by VIVIEN B.Sc.  , University  A THESIS THE  BETH  KITCHEN  of B r i t i s h  Columbia,  SUBMITTED IN PARTIAL  REQUIREMENTS FOR MASTER OF  in  THE  1969  FULFILMENT  OF  DEGREE OF  SCIENCE  the Department of Ma thema t i c s  We  accept  required  THE  this  thesis  as c o n f o r m i n g  to the  standard  UNIVERSITY  OF BRITISH COLUMBIA  May,  1972  In p r e s e n t i n g  t h i s t h e s i s i n p a r t i a l f u l f i l m e n t of the requirements f o r  an advanced degree at the U n i v e r s i t y of B r i t i s h Columbia, I agree t h a t the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r r e f e r e n c e  and  study.  I f u r t h e r agree t h a t p e r m i s s i o n f o r e x t e n s i v e copying o f t h i s t h e s i s f o r s c h o l a r l y purposes may by h i s r e p r e s e n t a t i v e s .  be granted by  permission.  Department o f The U n i v e r s i t y o f B r i t i s h Vancouver 8, Canada  Date  O^t^C  /3  Department or  I t i s understood t h a t copying or  of t h i s t h e s i s f o r f i n a n c i a l g a i n written  the Head of my  j  Columbia  /9?2~  s h a l l not be  publication  allowed without  my  Dr.  Supervisor :  K. Hoechsmann  ABSTRACT  In  this  thesis  we  linear  transformations  closed  field  all  finite  divisible  k.  consider  of a v a r i a b l e  The p u r p o s e  subgroups  t h e group  of t h i s  x o v e r an  of the t h e s i s group  by the c h a r a c t e r i s t i c  whose  o f k.  of f r a c t i o n a l algebraically  i s to d e t e r m i n e  orders  a r e not  ACKNOWLEDGMENTS  I would for of  like  h i s guidance this  thesis.  made t h e t a s k  t o e x p r e s s my g r a t i t u d e and e n c o u r a g e m e n t D r . Hoechsmann's  enjoyable.  I would  for  reading  and  the U n i v e r s i t y of B r i t i s h  ass i s tance .  t o D r . Hoechsmann  throughout enthusiasm also  t h e t h e s i s and t h e N a t i o n a l Columbia  like  the p r e p a r a t i o n f o r the subject  t o thank  Research  D r . Gamst  Council  for financial  TABLE  OF  CONTENTS  Chapter  I  Introduction  1  Chapter  II  Algebra  9  Chapter  III  Differential  Chapter  IV  The  Chapter  V  Ramification  Chapter  VI  The  Chapter  VII  Consequences  Chapter  VIII  Proof  Bibliography  Preliminaries Algebra  Schwarzian  Hurwitz  of  13  Derivative  22  Theory  26  Formula  45  of  the H u r w i t z  t h e M a i n Theorem  Formula  50 60 73  CHAPTER I  INTRODUCTION  The version result We  intention  thesis  i s to g i v e  an up t o d a t e  o f part of Klein's t r e a t i s e [3], and to extend Klein's from  the complexes  t o any a l g e b r a i c a l l y  t h e g r o u p $J o f f r a c t i o n a l  consider  ations  of t h i s  of a v a r i a b l e  linear  closed  field.  transform-  x ax  + b — J cx + d  x —> 1  where  the elements  closed  field  matrices  k.  over  That  o f k.  k = C, ^  automorphisms  having  order  to m o t i v a t e  i s isomorphic  not d i v i s i b l e observation i n  our r e s u l t  It subgroups  group,  i n the  We  the c y c l i c  of  analytic  know t h a t groups,  the o c t a h e d r a l  this  the d i h e d r a l  group  and t h e  group.  turns  out that  these  o f xb f o r g e n e r a l  to t h e c h a r a c t e r i s t i c these  t o t h e group  sphere.  subgroups  the t e t r a h e d r a l  icosahedral  prime  2 X 2  Our aim i s t o d e t e r -  The f o l l o w i n g  o f t h e Riemann  has as f i n i t e  groups,  algebraically  case:  For  group  centre.  of &  c a s e where k = <C s e r v e s  general  that  their  to an  the i n v e r t i b l e  2  subgroups  the c h a r a c t e r i s t i c  the  i s ,PSL (k),  k modulo  mine a l l f i n i t e by  a, b, c, d b e l o n g  are e s s e n t i a l l y  groups  k, i f t h e i r  also  exist  order  o f k.  In f a c t ,  a l l such  finite  as  finite  is relatively we  shall  subgroups  show of  .  2. As subgroups  evidence o f is  a subgroup P  X  the  give  i s N and  e x i s t e n c e of  their  the  the  generators I  where  The  cyclic  generators.  characteristic  and  canonical  I f the  order  o f k i s p,  of  assume  that  The  d i h e d r a l groups  Cyclic  Group  N  D i h e d r a l Group  where e i s a p r i m i t i v e  the  generators  three  are  also  of u n i t y i n  of o r d e r  i)  n*"*  1  a  root  remaining  N i s generated  n  what  ( l  of u n i t y  the  the  groups  generators  in  by  k.  N = 2n  d  i s generated  k.  derive  However, u s i n g  i t can  easily  f o r k any  be  well-  shown  that  algebraically  closed  field. Let  G denote  one  of  the  three  figure  F corresponding  to G i . e . the  hedron  or  as  icosahedron,  Then  elements  form  F into  of  G are  itself.  embedded  rotations  Every  of  element  groups.  View  tetrahedron,  i n the the of  Riemann  sphere  G must  by  o)  c a n o n i c a l g r o u p s we  c o n s i d e r i n g k = C.  by  known p r o p e r t i e s o f these  of o r d e r  root  (o  For  i t is clear  are.  e is a primitive II  the  these  N.  For the  we  of  the octasphere.  which  then  be  transone  of  3. the  following:  a vertex meeting  a t any v e r t e x  through  the a x i s  erate We  G i t suffices  now  rough  give  t o t h e number  S of order  of f a c e s  octahedron  2 about  the m i d p o i n t  these  three  to g i v e  the g e n e r a t o r s  o f each  R, S and T f o r e a c h  s t e r e o g r a p h i c p r o j e c t i o n of each  related  interested  3  Since  are conjugate,  one e l e m e n t  shown as an a i d t o t h e r e a d e r  T of order  of a f a c e .  types  or  the a x i s  o f an edge, o r a r o t a t i o n  through  of each.of  i = 3, 4 o r 5 a b o u t  i n the t e t r a h e d r o n ,  or a r o t a t i o n  the m i d p o i n t  elements  R of order  of F ( i corresponding  icosahedron),  about  a rotation  type. group.  to gen^^ A  figure i s  i n checking  the  generators. Ill  The  Tetrahedral  Group  Let  OJ be a p r i m i t i v e  3  rd  root  of uni ty.  S tereographic P roj e ct i o n o f Te t r a h e d r o n  1 R  (1)  G can i n f a c t a l w a y s be g e n e r a t e d See, f o r example §§71-74[5].  1 by two  elements.  V  The  Icosahedral  Let  e be  Note:  -  Group  a primitive -,  e + e - 1 =  -3 +  ~  /5  5th  root  of  unity  S tereographic P roj e ct i o n of Icosahedron  -1 -3+/5~  With can TV)  now  s t a t e our  It turns the  the  existence  out  of  the  these  subgroups  of &  accepted,  we  proposal. that  icosahedron,  origin  of  -1  five  on  the  r = ~ { outer  stereographic z  J  ^ )  ^  icosahedron  t  h  e  p r o j e c t i o n of distance  vertices.  from  MAIN THEOREM Let istic  p.  k be  an  algebraically  Any  finite  subgroup  is  isomorphic  to one  of  if  G i s isomorphic  jugate  to t h a t  finite  subgroups  This  to one  group. are  result  the  closed  field  G of PSL2(k)  f o r which  c a n o n i c a l groups  of  I-IV  then  of c h a r a c t e r -  I-V.  In  G is actually  Further, in characteristic conjugate  immediately  to one  gives  of  the  p )(  |G|  fact, con-  0 all  I-V.  following  special-  ization.  Let Any  F be  subgroup  either  ( §256).  all  finite  Dickson's  Our  will  become c l e a r  here  that  our  algebraically  Our  first  linear  k to one  i s stated  subgroups  to p.  We  over  result  [1]  subfield  of k c o n t a i n i n g p  o f PSL^CF) whose o r d e r  conjugate  This  prime  a finite  now step  I-IV  of PSL^CF),  our  matter  not  only  those with  f o r a v o i d i n g the most for proof  be  designed  closed  field  k,  not  our  plan  i s to o b s e r v e  so  finite  as  t r a n s f o r m a t i o n s of x over  the  deals  to  V.  with  order  general  develops. to h a n d l e  case Note any  fields.  for accomplishing  that  p is  i n Dickson  this  technique  by  i s isomorphic  of  methods w i l l  outline  or  divisible  i n a s t r o n g e r form  treatment  reasons as  of  i s not  elements  n  group  of  the  proof.  fractional  k i s isomorphic  to A u t ^ k ( x ) ,  7 . the group  of automorphisms  this  now  fact  and  o f k ( x ) o v e r k.  then choose  We  to work w i t h  establish  Aut^k(x)  thereafter.  Theorem k(x) = k(y) < = > some e l e m e n t s  a, b,  y = *x  c, d E k s u c h  that  t  d  for  ad-bc  0.  4  Proof (  <f  must  )  k(y) c  determine  k(x)  i s clear.  x i n terms /y\  matrix  form  as  o f y. '  this  we  \c  la  have  =  d /  we  "fr* b ;—T i n cx + d  3 X  1/  b\-l  U  \lj  k(y)  b\/xV III*  /a  1  show k ( x ) c  Rewrite y =  I  =  /x\ From  To  l / ~ \ c -a la b\ where t h i s i s p o s s i b l e s i n c e d e t 1 = ad - be f 0. -d + b \c d/ x = ^ , and t h e r e f o r e k ( x ) k(y). cy - a 4  )  d|  -1  k(x) = k(y) i m p l i e s ,  \  a  d  b  in particular,  c  that  Hence  y e k(x).  A fx) So  y = -z-j—r  to be  relatively Let  satisfied k(y):  f o r some A ( x ) , B ( x )  we  may  assume  prime.  I(X) = B(X)y  - A(X).  by x o v e r k ( y ) .  o t h e r w i s e , i t would  polynomial  e k [ x ] , which  from each  I i s a non-zero p o l y n o m i a l  Further, be  I is irreducible  possible  o f the r e l a t i v e l y  to f a c t o r  an  over irreducible  prime p o l y n o m i a l s  8. A(X)  and B ( X ) .  field  theory,  Thus  I(X) = I r r ( x ,  k(y), X).  [ k ( x ) : k ( y ) ] = deg I ( X ) .  and  deg I ( X ) = max  and  B(X) = cX + d f o r some e l e m e n t s  ad  - be j> 0.  k-multiples Hence, we  o f each o t h e r  That  this  any e l e m e n t  fractional  c,  x  i  In  at  the d e s i r e d  Aut^k(x)  i s such  transformations  -•  "j"—j-  that  thinks  these  result  theory  be i n k.)  =  y,  that  i s clear.  o(k(x))  =  of x over  k a c t s by  s a y , f o r some e l e m e n t s  a , b,  ad - be 4 0 .  finite  of using  subgroups  Galois  a l g e b r a i c techniques dependent  0, we  invoke  o f A u t ^ k ( x ) one  theory  and f i e l d  and t h e H u r w i t z  on d i f f e r e n t i a l  o f the M a i n Theorem,  characteristic  result  o f t h e group o f  algebra  f o r i t s d e r i v a t i o n ) , we deduce  clusions  therefore  a, b, c, d as s t a t e d .  considering  naturally  (a  (a b) and ( c d) would be  and y w o u l d  establishes  linear  d E k such  With  a , b, c, d e k f o r w h i c h  = k ( y ) , s a y ; and any e l e m e n t  mapping  Thus A(X) = aX + b  A ( x ) _ ax + b ~ B(x) cx + d'  with  k(a(x))  = 1,  have y  For,  But [ k ( x ) : k ( y ) ]  (deg A ( X ) , deg B ( X ) ) .  ( I f ad - be = 0,  Hence, by  as w e l l  Formula  and r a m i f i c a t i o n  the f i r s t  F o r the f i n a l  theory.  two  con-  conclusion i n  the s o - c a l l e d  Schwarzian  (3) derivative  , an o p e r a t o r  i n the theory  of  differential  algebra. (3) Our use o f t h e S c h w a r z i a n d e r i v a t i v e i n t h i s c o n t e x t follows K l e i n [ 3 ] . Here he uses t h e S c h w a r z i a n i n f i n d i n g the f i n i t e s u b g r o u p s o f P S L ^ ^ ) .  CHAPTER  II  ALGEBRA  PRELIMINARIES  Terminology: By shall  mean  a  finitely  transcendence  degree A  function  field  For  from  Theorem  1 Let  element  y  having  only  one  subgroup  the  same  field  k  we  of k of  theory  over  about  i s then  of  shall  an a l g e b r a i c  k  need  three  special  algebra.  element  of Aut^k(x).  i s a special  claim  field  we  transcendental  transcendental  This  over  generator.  sections  Fix Note:  extension  function  the g e n e r a l  a finite  field  1.  rational  x be a  func t i o n  generated  use i n l a t e r  results  G be  an a l g e b r a i c  such  over  Then  a  field  there  k.  exists  Let  an  that  G = k(y)  case  of LUroth's  any s u b f i e l d  of  Theorem  which  makes  k(x).  Proof Let Define over  k.  k(y)  =  {o,,a_,...,a } ^ . A u t . k ( x ) I z n K  y = a ^ ( x ) a ^ ( x ) • • • o" ( x ) .  Clearly  n  We  will  show  that  y  e F i x G and  where n = y  I GI .  i s transcendental  that,  in  fact,  F i x G.  To for  G =  show  that  a l l i = l,...,n.  y  e F i x G,  we  must  show  F i x some  i , 1 < i < n.  that  c^(y) = y  Then  10 . a . C y ) = a . ( a ( x ) . . -a^x) )  (1)  1  = Clearly, then a. l  (a  ±  a )(x)...(a 1  a )(X)  i  r  a . •a . e G f o r i = l , . . . , n . 1 j  a . ° a . ^ a.oa, . l j x k  a . = a, . j k  Hence  G = {a . • a , , . . • ,a . o a l 1 x n  o^(y)  = y and y i s i n d e e d i n F i x G.  By  From  Thus  G:k(y)]  t h e o r y we know  that  (1) we  =  see t h a t  Hence  [k(x):k(y)]  [k(x):Fix  [k(x):k(y)]  complete  the p r o o f by showing  This  finishes  |Fix  G : k ( y ) | = 1, and s o , t h a t In  is  from  GJ =  |G| = n.  (2) t h e n we have  ( 3 ) We  }•  have now k ( y ) £ F i x G £ k ( x ) .  [k(x):Fix G][Fix Galois  that  r  c a .» a . = a. \ a . • a, , so t h a t i j x i k  (2)  i f a. ^ a, j k  F o r , a . ° a . = a.«a. i m p l i e s l j I k  L  We  Further,  the p r o o f s i n c e  the I n t r o d u c t i o n  we  i s o m o r p h i c to t h e group  mations.  Thus  each  1 '  1  n Then y =  a.(x)  i=l  C  i  X  (2) i t t h e n  i  follows  established  that  Aut^k(x)  linear a  i  X  +  transfor-  ^ i ,  c  : — j —  that  for  a.d. - b . c . 4 0. 1 1  fCx) —T^-V'  = g  (  x  )  that  Fix G = k(y).  X I  ; —  = n.  c.x + d . x x  1  ;  from  [k(x):k(y)j  i s o f t h e form  x  a^x+b^  II  that  of f r a c t i o n a l  some a . , b . , c . , d . e k s u c h 1 '  > n  f o r some  relatively  prime  11. polynomials  f ( x ) , g ( x ) e k [ x ] such  that  deg  f , deg g < n.  Let  F(X) = g(X)y - f ( X ) . F(X) i s i r r e d u c i b l e  over  and  F ( x ) = 0.  Hence  [k(x):k(y)] Together  Theorem  i.e.  F(X) = I r r ( x ,  = deg F ( X ) = max(deg  with  2  (3) t h i s  gives  k ( y ) , X).  f , deg g) < n.  the r e q u i r e d  E x i s t e n c e of a S e p a r a t i n g  Let  result.  Transcendental  K be an a l g e b r a i c f u n c t i o n f i e l d  k i s of c h a r a c t e r i s t i c x e K such  p.  Then  k(y)  there  over  exists  k,  where  a transcendental  that K/k(x) i s s e p a r a b l e .  Proof Let  x e K be  t r a n s c e n d e n t a l over  K 3 k ( x ) r> k and K / k ( x ) a l g e b r a i c . O J  are  n  a l g e b r a i c over  existence  1 2  ) , where  the  n  k, and x = x^.  n = 1 then  Use  x ,...,x 0  2  n  i n d u c t i o n on n to show  K = k ( x ) , and  this  is clearly  separable  k(x ). Assume  Since  that k ( x ^ » • • • > _ ^ ) x  f such  f a r e i n k.  z  s  n  i s a l g e b r a i c over  polynomial of  have  of a s e p a r a t i n g t r a n s c e n d e n t a l .  If over  Thus we  Suppose  K = k ( x , x„ , . . . , x ) = k (x, , x . . . , x ' 2  k.  that  If x  separable  k ( x ^ ) , there  exists  over an  k(x^). irreducible  f ( x , , x ) = 0, where the c o e f f i c i e n t s I n i s s e p a r a b l e o v e r k ( x j t a k e the s e p -  n 1 a r a t i n e t r a n s c e n d e n t a l to be x.. Assume t h a t x i s not ° I n s e p a r a b l e over k ( x ^ ) . Then f must be a p o l y n o m i a l i n ^ • x  n  If  x, i s i n s e p a r a b l e o v e r 1  in  x A  Thus  k ( x ) then n  f i s also  a  polynomial  12 . f(x ,x ) 1  where this  = g(x  n  P 1  ,  x  g i s some p o l y n o m i a l i m p l i e s that  x, s e p a r a b l e 1 over k(x ). n r  )  =  with  [g(x , 1  over  x )] , P  n  coefficients  f is reducible.  x, must be s e p a r a b l e 1 Thus we  P n  i n k.  Contradiction.  But Hence  k(x ). n  have  ^ separable  over  k ( x ^ ) and  over  k(x ). So x. x , are a l l separable n 1 n-1 T h e r e f o r e , take x to be t h e s e p a r a t i n g t r a n s n  cendental .  The a proof  f o l l o w i n g i s a standard  of which  can be  found  theorem  on p . 171  of a l g e b r a ,  [4] .  Theorem 3 Let field  A be an a l g e b r a i c a l l y  k, K an a l g e b r a i c e x t e n s i o n  is  a monomorphism  of  K into  A.  then  closed extension o f k.  p can be e x t e n d e d  of a  I f p : k «—»- A to a  monomorphism  13. CHAPTER I I I  In deal  with  DIFFERENTIAL  the  chapter  derivations  concerning  them.  section.  The  that  under  This  i s used  i n the  part  of  chapter  the  and  This  this  this which  Hurwitz  Formula.  will  result  theorem  and  are  proof  of  deals  any  k be  This  and  K are  have:  for  lemma  lemma  the  of  states  can  differentials  be  extended.  The and  development  both  =  {f  fDg  latter  related  of  the  the  Also e K  Note:  D  since  notice  k. D(f )  that  then D(fg) For  For,  f e K  P  i f char  = pf  p - 1  Df  = fDg  =  +  ( x ) , Df  gDf  of  k,  for D  k £ K^. But  = 0. over  case  where  following  But K. Q  to  For,  i t i s not  k = p and  D is linear  that  then  that  —*• V  D:K  V.  o f cons t a n t s  that  such  k-derivation.  i n the  a K-module  true  £  a  field  = 0.  —»• V  k-algebra  gDf  extension  field  = 0}.  D:K  the  = D ( a ' l ) = a « D ( l ) = a«0 Q  +  used  D(a)  K  K a commutative  Assume i n  K an  K into  be  that  e K  e K|Df  =  will  fields.  of  define  ring,  Notice  g  the  will  result  M a i n Theorem.  f, g e K i s c a l l e d  k a field,  We  P  the  Then a k - l i n e a r map  definition  a k-derivation  f  in  derivations  the  a commutative  elements  Note:  D  general  i s found  with  necessary  V a K-module.  for  K  one  following  D(fg)  we  need  d e r i v a t i v e we  Derivations Let  k  Schwarzian  suitable conditions  results  (1)  on  ALGEBRA  f  f e K P  For,  need  be if a e  k,  necessarily then not  be  i f f e K^  in and  fDg.  = D f*Dx where D  denotes  "formal  k.  14 . differentiation Denote  with  respect  by D ( f ) ,  to  x".  f o r f e K,  X  D(D(...(D (f))...)). i  the element  Then we  /  have  the  following  Lemma Let be  extension  a k - d e r i v a t i o n with f i e l d  ^1' if  K be a f i e l d  ^2'**"'^n det  (D fj) 1  E  ^  a  r  e  L  i  n  e  a  T  = 0, where  ±  o f k,  and  of c o n s t a n t s y  dependent  i = 0,  l e t D:K .  The  over  —• K  elements  i f and  only  1,..., n - l and j = 1,  2,..., n.  Proof Assume ^ ' ' " ' ' ^ n Then, w i t h o u t such  a  r  li  e  n e  a*"ly  l o s s of g e n e r a l i t y ,  dependent  there  • • •  e  D i s linear  f , = X „ f „ +... + X f . 1 2 2 n n i t follows that  over  D f, = X„D f„ + 1 Z z 1  a l li=l,...,n-l.  therefore  linearly  Now show t h a t For nonzero  assume  The  columns  dependent.  that  of the m a t r i x  n (D fj)  are  1  (D fj) 1  =  0.  ( D f . ) = 0 and use i n d u c t i o n J l i n e a r l y dependent over K„. D  be  n = 2, we  have  otherwise  D f X  n  Hence  det  the f . must J  since  ... + X  i  det  we  exist  .  that  Since  for  over  1  f^, f  2  they  £ K.  Assume b o t h  are l i n e a r l y  to  to be  dependent.  Hence  have f  l  f  2 = 0  D  This  implies  that  f  l  ° 2  the columns  f  are l i n e a r l y  dependent  over  K,  15. Thus  there  Clearly  exists  we  will  D to  (1)  we  this  yields  A e K  be  Hence A e  the  result  f  (2)  Df  i f we  f  are  true  that  linearly  for j =  = A Df2  x  n 2  show A e  A.  .  Together so  that  Applying  with  DA  (2)  must  be  0.  f  , ..., f  ( D f j ) = 0 f o r j < n-1  dependent.  e K and  n  exist  (n) the  (1)  case n =  implies  X  We  will  show  this  n.  1  to  2  4 0,  det  2 we  assume none i s 0. A „,..., A  f  = A f  ±  =  2  »f  D  n  n  2  x  (2)  D  f  But  ( D f . ) = 0, t h e r e j '  for  xf  can  +  2  (1)  As  =  ±  .  Let det  that  (1)  = A Df  1  = 0.  Assume now that  done  have D f f^DA  such  1  f  1  1  £ K  such  ...  +  +  2  A Df 2  +  2  = A„D  n  1  f . +  the  A^  assuming  that  A f n  +  n  A Df n  ...  I  2.  show t h a t  ...  Then  e  +  n  A  .  n  D  n _ 1  f  n  Applying  gives  Subtracting  Df, = (DA - f . + ... 1 2 1 (2) f r o m t h i s g i v e s (1 ' )  Similarly,  applying  D  to  -f ) + n n  • f + ... 2 2 (2) we have DA  2  0  (3)  DA  DA  D f , = (DA • D f „ +...+ 1 2 2 Subtracting  +  «Df n n  +  ) +  DA n  (A„Df  •f n  DA  •Df 2.  +...+ 2  (A„D f 2 2  +...+  2  DA  -Df n  = n  A Df ) n n  =0  yields (2')  +...+  1 2  0  A D f ) n n 2  16 Repeating  this  operation (1')  D X - f „ +...+ 2 2 D A - D f +...+ 2 2  1  0  (n-l') system  can  be  n  2  f„  written  2  =  0  n  DX  •D n  n  2  f  n  =  0  ...  Df. . DX  =0  n 'Df n  +...+  obtain  as  f.  (DX  -f  n DX  0  DA„-D 2  i n a l l we  DX  0  (2 )  This  n - l times  Df  0  ) n \D  If  det  (D  are  done.  j = 2,...,n, we  find  ( D A , . . • , D A ) = 0. 2  so  But  n  that  e a c h A.l  over  f.  n  f j ) = 0 for j =  a s s u m p t i o n we for  n-2 *  , as  e  D.  2,...,n t h e n by  uniquely  induction  therefore  from  above s y s t e m  then  the DA  = 0 for  ±  f .I , . . n. , f  that  i =  are  det  (D fj) 1  that  2,...,n,  l i n e a r l yJ  d e rp e n d e n t  desired. Extension  a finitely  separably  the  Assuming  Thus  Theorem If  \0  /  algebraic extended  to  generated then  any  of  a  Derivation  extension  K(x)  d e r i v a t i o n D on  over K K can  is  be  K(x).  Proof Since need  only  define  D(x).  f(X)  = Irr(x,  Let for  a. i  e K.  applying  D is defined  any  Then  f(x)  derivation  on  K,  K,  X)  = a x° n d to  to  extend  = a X n  +...+ this  n  +...+  a,x gives  D  to K(x)  we  +  a ,  + a_ = 0, 1 0  and  Q  f  0  17 . [d(a v  n  )  + d(a ^x "" Hzi^  n  1  x  +...+  1  d(a )]  +  n  0/  [ (na x " 1 1  .  _  n  = f (x) must  +...+  a )d(x)]  °,  n  = f'(x)  d  Thus we  1  d e f i n e Dx  by f (x) d  This  defines  the o n l y  possible extension  Note:  division  sense  since f'(x) ^ 0 f o r x separable  (2)  by f ( x ) i n the d e f i n i t i o n 1  o f Dx  over  makes  K.  Dif f e r e n t i a l s Let  D:K  o f D to K ( x ) .  K be an a l g e b r a i c f u n c t i o n f i e l d  — > • ^jr/k.  t  ^  0  e  t  n  over  k.  Define  u n i v e r s a l k - d e r i v a t i o n i n the f o l l o w i n g  e  sense: For exists  any k - d e r i v a t i o n D:K  a u n i q u e K-module  following  diagram  map  >W,  u:  W a K-module,  ^K/k  * ^ such  there  that the  commutes K  y  | JJ, i i  4W Once  we  being  have  existence  of such  a universal pair  d e f i n e d by t h e u n i v e r s a l mapping  be u n i q u e  up  to i s o m o r p h i s m .  Existence  of  (d,  ^J^/J^)  for  x exists  by Theorem  some s e p a r a b l e  differentiation  element  with  i t must  :  Take x to be a s e p a r a t i n g (Such  property,  (d,  transcendental  2 , Chapter 9 e K.  respect  i n K/k.  I I ) . Thus K = k ( x , 9 )  Now  i n k ( x ) , formal  to x d e f i n e s a d e r i v a t i o n d  = 0  18. K  on  k ( x ) .  Then,  to  give  derivation  a  Take  y(f«de)  =  — y  f•DO,  d i f f e r e n t i a t i o n  u»d  =  The for  the  D,  as  continuous  Let a  In  in  the  f  ology  e  K.  ^ j ^ ^ i s  be  extended  u:  Denoting to  6,  = D  we  ^^/k  by then  D  =  ^—**  w  formal  Q  have  f • DG , w h i l e  K <  f o r  Df =  D  f o  K,  e  f • D8 .  known  as  the  module  of  d i f f e r e n t i a l s  K/k.  what  we  shall  p a r t i c u l a r ,  we  be  a  a  (f) = This  order  shall  of on  of  valuation  mean  f i e l d  "ord" defined  f  ord  by be  of  by  as  a  continuous  interested formal  formal i t  a  power  power  function  defines  a  i n series  series.  of  t,  metric  on  K  way.  topological  i n which  scalar  can  desired.  K-space  i s  a  vector  space  W  with  top-  addition W  X  W  —*•  W  m u l t i p l i c a t i o n (ii)  are  K.  respect  k((t))  (i) and  e  f i e l d  usual A  f  define  K  valuation  any  above,  i n  ord for  as  d i f f e r e n t i a t i o n  =  k ( x ) ,  K.  on  f o r  define  d i f f e r e n t i a l .  has  W  K-module  now  over  Kde.  with  extension  We  K  separable  = u ( d f ) = u(D„f«de)  (pod)(f) Hence  d  =  K/k  D:K  Given by  Q  being  K  X  W  —>•  W  continuous.  Define  d:K  —>•  c  fi„to K/k  be  the  universal  continuous  19 . derivation  in  For  any  topological such  that  the  following  continuous  K-space,  the  sense:  k-derivation  there  following  D:K  e x i s t s a unique  diagram  is  —»• W,  W  a  K-module map  u  commutative  d  For  any  field  of  formal  power s e r i e s K =  k((t)) c  we  will  exists;  presently again,  Existence  of  Set  show t h a t  once we  (d,  ^^/y,  1-dimensional  such  have  a universal  existence,  pair  (d,  uniqueness  is  automatic  :  ~ ^ ^  t  ' ^ w  module.  e  r  e  dt  Define  ^  d:K  the  s  generator  of  = k ( ( t ) ) —>• ^ y K  this by  k  d : t t—* d t . Notice the is  formal as  that  Df  = f'*Dt  d e r i v a t i v e of  f o r any  f with  f e K where  respect  to  t , and  f'  denotes  D:K  —•  W  above.  Proof: Given f by J  functions  know t h a t  Df  n  f  Df  e kft, t  n  = f  f e K,  n  by X  ] .  by  ( i i ) of  = lim  the  Then,  since  can for  approximate the  ''Dt, Df  n  = lim n-*-"  1  n->-oo  But  c o n t i n u i t y we  (f  1  n  •Dt)  t o p o l o g i c a l K-space  l i m ( f '-Dt) n n-*-°°  =  definition  (lim f )'Dt n n->°° 1  f  n  we  20 . Hence  Df =  ( l i m f ')«Dt = f ' D t . ^ n n  Then Define Then (We  = Kdt —>  k  Df = f ' D t and  follows  K.  = f'Dt.  d i s continuous.)  required.  0  call  since  easily.  f'Dt, f o r f e  (y.d)(f) = y(df) = y ( f ' d t )  i s , D = y d , as  ^he module  of continuous  differentials  K/k.  Next we  define  c Let  the o r d e r  Since  ~ k((t))dt,  f o r some e l e m e n t f e k ( ( t ) ) .  co to be  the o r d e r  of the f u n c t i o n ord(co) =  Remark: any  L e t t ' be  independent  therefore Let we  a power  have  differential,  f.  can w r i t e define  ord(f)  o f the g e n e r a t o r  as a power  chosen  the f o l l o w i n g  1.  =  f o r K.  Then  series in t'. i t must We  must  holds.  of K oyer  to = f ' d t and to = g ' d t ' where Then o r d ( f )  of  Write  s e r i e s i n t of order  t , t ' be g e n e r a t o r s  to as  the o r d e r  F o r ord(co) to be w e l l - d e f i n e d ,  show t h a t  e k((t')).  we  Then  f e K = k ( ( t ) ) c a n be r e w r i t t e n  So K = k ( ( t ' ) ) . be  of a continuous  c  co e  f«dt  g  property  W by y: f ' d t —>  d f = f '«dt, as f o r D,  We of  t h e u n i v e r s a l mapping  y: ^ £ /  have  That  - > o o  k.  For w e  ^K/k  f e k ( ( t ) ) and  ord(g)  Proof: t' for  e K = k((t)).  u(t) a unit  in k((t)).  Hence Then,  we  can w r i t e  letting  D  t  t' =  denote  u(t)'t  formal  21. differentiation  with  dt'  It For,  Thus o r d  a unit  We co =  f-dt.  + now  t , we  (u(t))'dt't +  =  (D (u(t))-t t  =  u (t)-dt, see  ord  u(t))  =  (u(t))  so  that  = 0 and  ord  (u  in  ord  k((t)). (D u(t)))  (t)) =  t  ord  0.  have w = g ' d t '  -  Hence f = g « U Q ( t ) . ord  u(t))-dt  U g ( t ) i s a unit  that  implies  +  u(t)'dt  say.  Q  to  have  u(t)-dt  = D  ( D ( u ( t ) ) « t ) > 1,  (D(u(t))-t  to  = d(u(t))«t +  i s easy  u(t)  respect  ( f ) = ord =  (g •  ( t ) ) •dt.  Therefore  (g'uCt))  ord  (g)  + ord  = ord  (g)  +  => o r d  (g)  0  (u Ct)) 0  But  also  >  0.  22 . CHAPTER IV  THE SCHWARZIAN  Definition  o f the S c h w a r z i a n  Let field  DERIVATIVE  Derivative:  D:E —*• E be a d e r i v a t i o n on a f i e l d  of constants  Derivative  k.  of h with  F o r h e E-k d e f i n e respect  where h ' = Dh, h ' = D h 1  Note:  Since  makes  sense.  2  E, w i t h  the S chwarzian  to D to be  and h ' ' = D h . 1  3  h e E-k, h' i s n o n - z e r o  and t h e r e f o r e [h]  Theorem Let D:E —*• E. for  E be a f i e l d  L e t k be t h e f i e l d  f , g e E-k t h e r e  such  on w h i c h  exist  there  i s defined  of constants  a derivation  f o r D.  a, b , c, d e k w i t h  Then  ;id-bc ^ 0  that  f = tf-H<=> D - I*] If]  D  Proof Assuming that  ad-bc f  f = ^ "j" ^ f o r some a , b, c, d e k s u c h eg + a a  0  , we  cfg (1)  Applying  see t h a t  this  are (1)  This  D to t h i s  says  linearly  to  - ag + d f - b = 0 gives  the e q u a t i o n  c ( f g + f g ' ) - ag' + d f (2)  i s equivalent  that  dependent  = 0  = f g + fg', h over  k.  2  = -g ' and h  3  =  f  By t h e lemma on the W r o n s k i a n  ad-bc ^ 0 s i n c e o t h e r w i s e f would [ f ] P would n o t be d e f i n e d .  be i n k, and so  23 (cf.  §1,  det  Chapter  (D h,) being x  I I I ) we 0.  f'g det  (D h  ) =  c o l 1 plus  2 f V  +  fg'*  3f ' V  +  g  +  f(col  2)  + 3f'g"  plus  >  •g  3f " g *  -2f V  the  +  3f ' g "  f'  -g  f • '  this  gives f f''  t i  =  determinant,  out  2  we  2  2  2  = 0  gives  ( f ' ) [ - 2 ' ' • •+ 3 ( g " ) ] + ( g ' ) [ 2 f ' f " ' - 3 ( f ' ' ) ] 2  2  2  2  g  =  2 Dividing two  through  by  (f'g')  and  transposing  one  of  the  expressions, 2 f  f  -  "  3(f ' ' )  2  (g')^ Or,  f ' ' ' f'  f' ' 2 f'  That i s , Remark: sentences Here,  [f] Our  proof  (1)  assuming  and  0  2  -3f'f " g ' g " + 3f'f " g ' g " + 3 ( f ' ) ( g " )  g  =  have  2  this  0  "g  2f'f " ' ( g ' ) - 2 ( f * ) g ' g ' ''-3(f' ' ) ( g ' )  Simplifying  f ' * *  -g  (-g'f ' " + f ' g " ') + ( 3 f ' 'g'+3f 'g' ') (-f » *g' + f ' g " )  Multiplying  to  -g  fg' * '  -g*  2f i'g™  expanding  +  - g ( c o l 3)  0  Or,  is equivalent  fg'  f • ' » Adding  this  Hence  +  f*'g  know t h a t  D  -  is a  i  [g]  -  3  D  series  of  equivalences  except  at  (2).  [f]  D  =  [g]p  and  f o l l o w i n g the  proof  back  0  24 . to  sentence  (2) we know  that  there  e x i s t a, c, d e k such  that c(f'g Recall  that  this  + f g ' ) - ag' + d f  simply  D(cfg So,  by d e f i n i t i o n  b e k.  means  that  - ag + d f ) = 0  o f k, c f g - ag + d f = b  conclusion  useful  f o r some  element  Thus £ _ ag + b eg + d  The  = 0  f o r some a, b, c, d e k s u c h t h a t ad-bc f 0.  of t h i s  theorem  c a n be r e s t a t e d  i n a more  form as  Corollary  1 k(f)  = k(g)  [f]  D  =[g]  n  Proof This theorem  Recall  proving  enables  that  If function  from  t h e above  t h e end o f C h a p t e r I . to need  the Schwarzian  i n c h a r a c t e r i s t i c 0.  use o f t h e p r e c e d i n g  theorem  the f i e l d field  This  over  E of the theorem  holds  the f i e l d  since  i s separating;  field  of D-constants  i n c h a r a c t e r i s t i c 0 any  i f D i s zero  t h e n D can o n l y  fact:  i s an a l g e b r a i c  an a l g e b r a i c a l l y c l o s e d  0, t h e n  What  on t h e  i n c h a r a c t e r i s t i c 0 i s the f o l l o w i n g  characteristic  cendental  directly  we a r e g o i n g  t h e Main Theorem  us t o make  Schwarzian  dental  follows  and t h e t h e o r e m n e a r  Remark: for  statement  kg o f  i s k^, transcen-  on a s e p a r a t i n g  be t h e t r i v i a l  trans-  derivation.  25 . For  non-zero  form  characteristic  a non-trivial  makes  next  extension  the Schwarzian  The  however,  trivial  the D-constants  of kg.  This  may  i s the f a c t  o f no use t o us i n t h e g e n e r a l  part  o f the p r e c e d i n g  which  case.  theorem y i e l d s the  result.  Corollary  2  Let  H be a f i n i t e  group  where x i s an i n d e t e r m i n a t e field  of H i n k ( x ) .  field  of constants  k(x)  (since  Schwarzian  k(x)/F  over  Suppose k.  We  of automorphisms  know  i n F.  L e t F be t h e f i x e d  D i s a derivation that  i s separable  of x l i e s  k.  of k(x)  D extends  algebraic).  on F w i t h u n i q u e l y to  Then t h e  That i s ,  [x]  e F  D  Proof Let need  only  morphism  u e H = Gal (k(x)/F). show  that  y([x]p)  of k(x), i t follows y([x] ) D  Now  =  To show  [x] «  Since  n  [ x ] e F we D  y i s an  auto-  that v>(*>J  u D : k ( x ) —»- k ( x ) i s a d e r i v a t i o n  Y  D  on k ( x ) , and uD|  = D. r  Thus,  the e x t e n s i o n  of D to k ( x ) b e i n g  unique,  yD = D.  Therefore p([x] ) D  Now by  y e Gal (k(x)/F) Corollary  1 this  implies  ) -  t ^j) x  [u(x)]  a  s  D  =  D  that k(x) = k ( y ( x ) ) .  i s e q u i v a l e n t to [x]  Hence M ( [ x ]  =  [y(x)]  required.  D  But  26. CHAPTER V  RAMIFICATION  Throughout ally By  closed  field, 2,  Theorem  K  extension  and  to be  of  an  I I we  i n K/k.  assume k algebraic  have  Thus we  x e K.  i t s extensions  are  can  our  the  main  the  following  (1)  Places The  origin  meromorphic introduced of  point  the  h e r e as the  text  i n a f i e l d as here  and  & be  of  every  that  z = n.  that of  which  i s not i s an  1 1  T I as  we  be u  above  state  We  be  and  notion  convenient  define  a  in  place  each  of  these  equivalent. a valuation &  is called  written e &  are  the  to v i e w  as  of  Places  i r r e d u c i b l e element  f o r some u n i t  element  able  a field.  can  for  field  theory  be  in fact  will  sep-  interest.  supersedes  It w i l l  are  a  k(x), the  surface.  to be  a place  separating  to be  of  over  case.  notions.  they  z e & - {0}  source  field  a  field  i s i n the  a Riemann  two  i f there  UTT ,  An  theory  argument  of  purposes  general  concept  a domain w h i c h  such  integer  the  our  concept  ring  form  on  e i t h e r of  valua tion  the  this  algebraic-  a d d i t i o n a l work  Riemann s u r f a c e .  then note  One Let  of  i n the  functions  on  main  theory  an  function  assume K  For  However, where p o s s i b l e w i t h o u t  to be  existence  a.rational function  some t r a n s c e n d e n t a l k(x)  chapter  Chapter  transcendental arable  this  THEORY  is called  a  a  TT e O  uniquely some  ring.  in  non-negative local  k  27 . uniformizing parameter quotient  parameter  i s of field  the  z =  UT: , 11  seen  write  in &  We  n = ord  call  (z).  e L|ord  {z  the  u n i q u e maximal  Our  second  A valuation  on  for  of  has  the  the  f  ideal  of  notion  of  L  (0) =  Let  the  order  easily  of  z  and  Then  will  is a function  n is  uniformizing  e L|ord  a place  the  uniformizing  exponent  = ».  {z  L be  expression  particular  ord  and  a field  The  n  uniformizing  i n cr.  a unique  exponent  Define  local  fixed local  n e Z.  and  (z) * 0}  0=  other  UIT , u a u n i t  Then,  independent ir.  parameter  any  z e L - {0}  u a unit  to be  form  of U.  p a r a m e t e r i r , any  f o r &;  > 0}  (z)  be  v:L  as  a  is  valuation.  —*• Z U {°°}  satisfying (i)  v(a)  (ii)  v(ab)  (iii)  v ( a + b ) > Min  Any  such v a l u a t i o n  & -  iz  ideal  c L|V(Z) ^ =  {z  a valuation if  defines  > 0}  a valuation  notions,  we  will  specifying  (b)  (v(a),  on  L. to  r i n g i n L.  r i n g with  quotient  quotient  field  Since  ideal  L  field  as of  L.  on  L.  field  L,  on  are  For,  maximal  L.  i s the  these  them b o t h  maximal  v(b))  a valuation  quotient  refer the  v  {">) i s a v a l u a t i o n  r i n g with  defining  since  and  r i n g with  o r d : L —*• Z U  a valuation  +  a valuation  ring in L defines  0" i s a v a l u a t i o n  function  = v(a)  is a valuation  e L | v ( z ) > 0}  0  = «> i f f a =  Conversely, Namely, then  the  Thus  giving  same  as  equivalent  places.  In  a valuation  fact, ring  28. determines  the  ring,  we  will  normally  use  the  following  notation: by ideal  of  the  valuation  a place  valuation  ring  by 6" ( o r 0^  itself  v a l u a t i o n by  place  of  Given of  , we to  two  will  a place  out  for  for  proof  of  p l a c e ^2 i s an plies  extension  this  can  the  299  and  [4].  a placexp,  we  the  extension  actually  To  the  the  Q  call  a given  reader  maximal  Denote  a finite  to e x t e n d  the  place be  on  carried  is referred denote  that  w r i t e ^|^»  a  read  over ^ .  in this  a  function  be  written  For  , L  o f L^;  of  denote  uniformity  process  to p.  extension  Consider are  fact  For  able  This  L we  i f necessary),  and  to be  L^.  on  any  ord^.  fields  want  a field  a s s o c i a t e d w i t h fl.  ring  associated ring  ~jQ of  now  field  we  f e k(x) as ^  e a c h a e k,  the  case L = k ( x ) .  need  the  see  what  the  following definition.  is regular  f o r some g,  h  define &  {f  =  To  at  a point  e k[x]  such  Say  a e k if f that  can  h(a)  ^  at  a}.  e k(x)|f regular  places  0.  3.  Then  i t is fairly  ring  with  field  local  k(x).  valuation quotient valuation  easy  field  see  uniformizing  Also, &  ring  to  with  =  {—  local  k(x).  r i n g s which  In  that  &^  is a valuation  parameter  e k(x)|deg  g > deg  uniformizing fact,  contain  these k and  IT = x-a  and f}  is a  parameter —  r i n g s are  the  have q u o t i e n t  quotient  and only field  k(x).  29 . So  we  have {a  the f o l l o w i n g e k}  correspondence.  \J » < > {Q  I a  Cf  < )  U (3 °°  r i n g s & | & O k and q u o t i e n t  (Valuation Any  e k}  a  1  v a l u a t i o n on k ( x ) i s thus i n k ( x ) (a i n k U  00  ) .  field  o f £> i s k ( x ) }  a s s o c i a t e d to a unique  The p l a c e s  ring  i n k(x) therefore  ct  correspond  to l o c a l We  u n i f o r m i z i n g parameters  and  TT =  for  22 i f i t i s a p a r a m e t e r  defined  by  say TT i s a l o c a l  u n i f o r m i z i n g parameter  f o r the v a l u a t i o n r i n g  &  .  The with  shall  TT = x - a , a e k  following result  places  will  be u s e f u l l a t e r  i n connection  on k ( x ) .  Theorem Any and  element  o f k ( x ) has an e q u a l  number  of  zeros  poles .  Proof f (x) Let where can  factor k will  of  zeros  in  be g i v e n  i n k, w i t h  behaviour poles.  so  then  k, w i t h  Since  f and g i n t o  f = n, s a y .  proof  be an a r b i t r a r y  f, g e k[x].  in  deg  h(x) =  linear by  Similarly,  00  i s complete. t h e number  factors.  balances  will  of h  say.  of zeros  or zero  h has a z e r o  is n +  of h  The number  of p o l e s  o u t t h e number  I f n < m,  we  t h e r e f o r e be  be deg g = m,  h has no p o l e  of zeros  The z e r o s  will  t h e number  of k(x)  closed  the factors of f .  multiplicities,  F o r , i f n = m,  that  k i s algebraically  multiplicities, of h at  element  and  a t » and t h e  of order  (m-n) = m.  The  m-n  at  Similarly,  0 0  ,  30. if  n > m,  we  Now of  we  places  K with  Let  & be  of  the  an  place  satisfying  the  poles  remark a b o u t  extension  ring/C,  place  an  number o f  is m +  the  (n-m)  =  n.  connection  integrality.  L be  the  that  make a s h o r t  with  Let of  find  of  o f jp .  integral  Suppose ^  "P i s a p l a c e  and  ring  K.  If a  equation  place  of L l y i n g  i s an  o v e r Q,  is a  element  i . e . an  over ^ .  of  L  equation  form  a t h e n a must  m  be  + a _^a  m-1  +...+  m  in  = 0,  a^  a^  e XT,  &.  Proof: Suppose & assume a has Thus, o r d ^ ( a  has  a pole m  local at  ) = -mn  u n i f o r m i z i n g p a r a m e t e r TT ,  n of  while  order ord  ? l  n.  Then o r d ^ (a) =  (-a _ a m  m —l  a )  1  this a  i s  + n.  But  c a n n o t be therefore We  extension We  will  of  K  field  ^2 be  ring  parameter  K of  satisfies a  places  k and  of  a place  We  = -a  -...-a^,  m  want  i n K.  to  z e &  there  local  u n i f o r m i z i n g parameter  a relation of  K gives  power s e r i e s  de f i n e d . by p. , and o f"p-.  give  a field  each p l a c e  a suitable  Let place  that  how  formal rise  usual,  between power  to an  TT  denote  a  local  show t h a t  to any  i s a s s o c i a t e d a unique TI .  0.  the  series. injection  field.  As by  -(m-l)n  m-1  Hence a must have o r d ^ (a) £  true. i n &.  show now  see  into  since a  -n.  >  Q  m = -mn  and  by  &  the  uniformizing element  power s e r i e s  in  the  31.  Let  z £ &.  A Q e k.  element  AQ,  unique  f o r z^ and  A-^,  modulo ^  the  using induction  £ k such  A2>--'  z = AQ +  to a  unique  isomorphism  Z - A Q = Z^TT f o r some z^  Thus  argument  z i s congruent  ( O b t a i n AQ through  k —>• &—y this  Then  on  £ &.  Repeating  n, we  obtain  that  2 A ^TT +  X  +  ...  2 Thus we an  injection  series way  to an  If ^  and  can  to  injection ( 2)  then  form  the  =  k ( ( i T ) )  of K i n t o  ring  , as  k((ir))  .  which d e f i n e s  o f f o r m a l power  extends  i n the  obvious  desired.  t h e n ^ d e f i n e s a m e t r i c on  completion  this  +...  A^^  of K w i t h  completion  by  f o r ^3 t h e n  Hence  the  i t s completion  K^.  r e s p e c t to I f ir i s a  i t i s not  above map with  K  difficult  gives  respect  an  top..  Rami f i c a t i o n In  this  section  between p l a c e s and Assume L and and  of K i n t o  Denote  that  +  T h i s homomorphism  u n i f o r m i z i n g parameter  see  A^TT  k [ [ IT ] ] , the  i s a p l a c e o f K,  metric.  local  into  injection  Note:  this  of &  o v e r k.  we  z >—>• A Q +  have a map  by  of  ordjjlj,  of  course.  integer  their  K are both  L is a finite  Denote  we  ord^J^ form  d i s c u s s the  interconnection  e x t e n s i o n s or  restrictions.  algebraic  e x t e n s i o n o f K. the  e occurring  Take  f i e l d s over  case,  o f Z-  not  subgroup,  The  necessarily  there i s a smallest  in this  k,  a p l a c e ^3 i n L.  r e s t r i c t i o n o f ord^j to K.  some s u b g r o u p  In any  function  and  values a l l of  positive  a l l values  of  Z,  32 . o r d j j on K w i l l index  of p .  referring the  be m u l t i p l e s  to the r e s t r i c t i o n  of ordp  restriction  normalization  positive  Call  e the r a m i f i c a t i o n  I f e / 1, say"fl,i s r a m i f i e d .  normalized  This  o f e.  that  by  mean  *?^K. e  has  smallest  o f $L t o K, we  By t h e r e s t r i c t i o n  of K d e f i n e d  0 r (  the r e s t r i c t i o n  will  I  ord  mean t h e p l a c e  t o K, we w i l l  i . e . the v a l u a t i o n  ensures  1.  value  In f u t u r e i n  ft'K. e  From two  this  definition  we deduce e a s i l y t h e f o l l o w i n g  results.  (1)  Suppose p. i s a p l a c e  i n L with  and  i s the r e s t r i c t i o n  o f~§X.to K.  uniformizing  f o r p. , ^  parameters  TT  where U i s some u n i t  =  r a m i f i c a t i o n index e L e t II, TT be  respectively.  ord^(II)  = 1,  un ,  i n Oi .  II a u n i f o r m i z i n g and hence us  Then  e  r  For,  local  II  that  parameter e  ord^II  f o r ^2 i m p l i e s  ) = e-1  i s an e l e m e n t  that  e So ord^f.11 ) =  = e.  of L having  t h e same  e TT  order be  as  a unit  (2)  TT w i t h  respect  to ^ .  This  implies  indices multiply  i n a tower o f f i e l d s .  Suppose L^/L^/K i s a tower o f f i e l d s  and  of L > 2  Let ^ , ^  K respectively.  be t h e r e s t r i c t i o n s  e over ^ .  and ^2 i s a of ^  to L^  Suppose x£ has r a m i f i c a t i o n i n d e x  o v e r ^ , j"0_ r a m i f i c a t i o n i n d e x index  must  i n {X, .  Ramification  place  that  Then e =  e  ± '  e  1  o v e r ^ and r a m i f i c a t i o n e  2 '  T  n  e  P  r o  °f  °f t h i s  Is obvious  33 . We needed. owing the  now  state  Result  I gives  field  extension  corresponding form.  texts  on r a m i f i c a t i o n  to o u r r e s u l t I t a k e s  The s i m p l i c i t y i n o u r c a s e  I  g)0 to K.  e^ denote  merely  ication  p.  indicate  a complete 308  extension  proof  how  field  extension.  index  ^} on L l y i n g o v e r ^ .  of ^ .  proof  i t comes a b o u t  Then  of t h i s that  to t h e d e g r e e  the r e a d e r  the p l a c e to L .  problem  Consider  ij;  that  i s referred  r e s u l t , we  the r a m i f -  of the e x t e n s i o n . to C o r o l l a r y  2,  [A].  extensions  place  complicated  i s the r e s t r i c t i o n o f  a complete  are connected  Consider its  where  « [L:K]  of g i v i n g  indices  case  foll-  the r e s u l t  on a more  the s e t of a l l p l a c e s  the r a m i f i c a t i o n  Instead  theory,  separable  on L, and ^  Ee«  shall  The r e s u l t s  closed.  i s a place Consider  ramification  i s due t o t h e f a c t  L e t L/K be a f i n i t e  Suppose j p Q  For  fact.  about  i s Galois.  is algebraically  Let  the major  facts  t h i s a r e c o r o l l a r i e s of I i n the s p e c i a l  I n most  k  some o f t h e b a s i c  ring  subring  will  We  can v i e w  (Actually, of L w h i c h  be a p l a c e  ring  are i n t e r e s t e d i n  this  i n the f o l l o w i n g  t h e homomorphism  o f ^ .  to a s u b r i n g  We  on K.  >|i :  as a homomorphism^  manner. stf—*•  k, where  iii : /C —>/?IJS^ i s as l a r g e  —>• k)  /f i s t h e I f we  as p o s s i b l e ,  f o r an e x t e n s i o n  ^)  extend  this x  n  L.  34 . In  fact,  such in  there  maximal  terms  of  extensions these  Since ator 9  6  i s a 1-1  the  o f i|i .  our  problem  can  choose  L i s separable  the  K  such  We  view  of  .  over  K,  that  we  6 is integral  ^jj/^and therefore  a  gener-  o v e r Z^.  Say  equation  F(X) where  between  extensions  f o r L over  satisfies  correspondence  a. e  = X  N  +  N  a _ X N  x  N-l "•+...+ a  Q  0  =  .  x  C o n s i d e r C^Q = xS [ 0 ] = ^ T ( X ) / F ( X ) . of &Q  field it  i s L.  is fairly  maximal  to  must  the  have  that Our  the  defined  this  <JJ to & Q ,  extended determines  problem  = X  N  +  a  unique  t h e r e f o r e i s to  N  can  be  To  the  +.,.+  N _ 1  1  extend  for ip(6).  satisfy  ^(a _ )X  coefficients  F(X)  on/C.  possible values  h e n c e f ( 6 ) must  .F(X)  closed,  see  o f I|J .  already  find  = 0;  where  to  i f we  quotient  &^.  is  F(X)  easy  extension  extend  we  Thus,  The  to^<T[0], 6  satisfies  equation 0  <Ka ) = 0  <Ma_^) e k.  Since  factored into  linear  k is  algebraically  polynomials,  say  as  F(X) = ( X - x p ! . . . ( X - A ) r , 6  6  r  where for  gives there  roots (2)  X.  ^ ( 6 ) are  This (1)  the  of  there  us  e k.  But  precisely the  are  the  we  see  roots  of  that F,  the  possible  values  namely, \^,...,\  .  following information:  always w i l l F(X)  then  = 0),  be  extensions  of  '•!> ( t h e r e a r e  always  and  a t most  N extensions  of  ii;  ( r < deg  F = N) .  35 . Clearly over^, of  namely  i s the  This  there are =  - X^)  we  have I e . = deg  facts has in  about  the  this  the  degree  that  of L over  K.  parameter  Lemma  Eisenstein  the p l a c e  on  on  L  me t e r  lying  N  =ct  and  T 1  ir  N-l  ^  where  power  of  one  important  that  extensions than  or  always are  equal  e x t e n s i o n of ^ In  of  this  case,  finite to  N.  i n L, I  we  tells  e x t e n s i o n o f x^. i s N,  the find  that  the  extension i n L  local  satisfies  equation.  e x t e n s i o n of  K is totally  Then  largest  obtain 2  i s less  the  a field  over  f o r ^? .  n  be  containing^-.  A l l i n a l l then,  these  index  of  nice  L/K  index  Namely,  A l s o , we  a particularly  Let  .  ramified.  ramification  uniformizing  the  e^.  of  number  is totally  ramification  ^)$Q  i n the p r o o f , we  i n L and  lying  [L:K].  there i s p r e c i s e l y  that that  =  extensions  number where If  us  the  The  same as namely  F(X)  remarked  extensions  say  F(X),  dividing  As  the  o f CTQ  , . . . ,"p^  (9 -  power of  to be  (X  Note:  (9 - ^^)0Q-  largest  t u r n s out  r ideals  take  there  N-l  tor  ramified. II to be  exist  a  degree  N.  Suppose  Let p  be  local  u n i f o r m i z i n g para-  z K such  the  that  +... + a., n + a _ 1  <  l  <  1  N-l  0  and  the a . z j£ f o r 1 < i < N - l and  o r d ^ o  ord  Q  ,  (o.)  =  1.  place  36. Proof II i s a n e l e m e n t of  degree  n over  generality,  there  (*)  equation  that  ord~(a  ^ II  f- n - i  Now  totally  consider  ramified. = 0  1 1  K,  an  without  i = 0 , 1,...,  equation  loss  of  n-1, such  that  a) u  = n,  n  must  n  0  ord^(n )  each  term  s K,  ord^(a^)  (mod n ) .  Thus,  1  see that  Thus  satisfies  +...+ a . n + a .  1  n-1  ^ +.;.+  n  element  o r d ^ a ^  ^n "  we  n|N.  E  exist  n  this  each  f o r some  II = a  From  For  K,  o f L and hence  on  be  hence  n.  the r i g h t  ord^(a^) But then  also  and  of the equation.  = N ord^(a^) , since = 0  i s  (mod N ) , a n d t h e r e f o r e  ord^o^II )  = ord^(a.)  1  +  o r d ^ ) ( n ) = i mod ( n ) . 1  Hence (1)  f o r each  i , j  ord ^(a^n ) 1  K  ( 1 ) we  see that  distinct. o  know  order and (mod  d  ^  0  n - l  this  ord^(a J N)  (mod  ) , and  n)  f o r 1 i i < n-1.  tne terms  on  the r i g h t  of  (*) a r e  Therefore l  i  n  "  order  1  )  +  ' "  +  0li<n-l  =  be n.  we  be >n. find  = N and o r d ( a , ) fr i  1 < i < n-1.  V  By  be n i s a ^ .  must  a n d n|N,  1  must  can p o s s i b l y  ord^ta) T 0 and  r  0  J  then  l  fl  We  i  ord (a.I )  From  0 t o n-1, i f i f j  ordgj(a H  1  (2)  from  s  This  = m.N, i  1  (2) t h e o n l y  Thus  In f a c t ,  n = N.  (ord^a.n ))  ord^(ag) since  Summing f o r some  i s the d e s i r e d  term must  whose be n,  ord^a^)  u p , we  = 0  have  integer  conclusion.  m. > 1 i  37 . Our  second b a s i c  among a l l e x t e n s i o n s L/K  Let  ^be  of a g i v e n  G be  a place  on  K.  {~p_!*Ei a p l a c e  on  L and  lying  over  the  the  Galois  relationship  i n K,  assuming  of  a connection  subgroups  of  Let L with  Each  the  place  That  that  Define  on  p l a c e s P]_ c^2 ,  thatp^ result  = the  reader  [4]. following.  of L/K. the  a l l elements  Define  composed  the  let  We  devel  certain  L/K.  e G|a(a) = a ( m o d p ) of  Let  inertia  be  group  a l l elements  (a  £ G|a£=#l }  place  o f ^2 to  for a l l a e & } the  place  ring  decomp o s i t i o n group  of  a  of  o f ^2  o f |},  G l e a v i n g p. f i x e d .  i s , take D  Then,  of  G a l o i s group  r i n g &.  modulo JJ).  to be  G a l o i s i n the  G a l o i s group  a z I ^ leaves  fixed Dp,  the  {o  244  p.  and  between r a m i f i c a t i o n i n d i c e s and  G be  1^ =  to be  such  standard  r e f e r r e d to P r o p o s i t i o n 1 1 ,  o f L/K  i s , given  a e G  this  group  transitively  That  exists  proof  Assume L/K  on  the  place  Then G a c t s  there  ,  For  op  gives  is Galois.  II  is  result  i t turns  braically  F  =  out  closed,  that these  i n our two  set-up,  groups  Lemma For  any  v a l u a t i o n 7) on  L,  are  since k i s algeidentical.  38 .  Proof It a be  I ^ <S D^.  i s c l e a r that  an a r b i t r a r y e l e m e n t  A E k,  the f i e l d  of 0".  of c o n s t a n t s ,  we  have  and a^Q = j~Q a does  a e D^.  can now  easily  important  relation:  place  order  has  the  —•  .  there  that  i s a unique  a = A(mod|2.).  &/]p.) •  But aX  and l e t  (Obtain  Applying  = X since  Hence a ( a ) H A ( m o d ^ ) ,  a to  X e k, so  that  b e l o n g to I^j. We  of  &  a ( a ) = a(X)(modap) since  Then such  X t h r o u g h the i s o m o r p h i s m k —> this  Assume a e D^,  derive  namely,  f r o m I and  that  p r e c i s e l y equal  I I the  the i n e r t i a  following  group  of a  to the r a m i f i c a t i o n i n d e x  place.  L e t j p be  Ill ramification  index.  a place  on L, and  l e t e^j be i t s  Then  Proof  a  let  Let  be a p l a c e  place  on L and "Pl^-} •  |s| d e n o t e  situation L/K  acting  i s as f o l l o w s : transitively  fundamental  We  t h a t $)\^'  we on  have  on  i n S.  the G a l o i s  S (by I I ) , and  the s t a b i l i z e r  theorem  Take  know S i s f i n i t e  the number o f e l e m e n t s  Dp o f G i s Stabg(^J), the  on K s u c h  group  of jp i n G.  transformation  by  Then  the  S =; result  Ij  our G of  subgroup Hence  groups,  by  | G: D-, | = IT  |s| ,  39.  • If!  or ,  Also the  from  I I , we  same; say t h e y  From  I we  just  |G| .  find  very  that  e|s| =  =  Corollary  later  [ L : k ] ; by G a l o i s =  us two more  l ^}! 1  =  theory e  »  a  s  this i s  desired  f a c t s which w i l l  prove  1  p be a p l a c e  in  Then E e<j = Z e = e | S | .  that  o  a l l t h e e^j must be  on.  Suppose I^/L^/K  then  s  r e s u l t gives  useful  that  a r e a l l e.  Thus e |g"|>  This  deduce  on L ^ .  1  S  a  Galois  tower o f f i e l d s .  I f l p i s contained  i s not r a m i f i e d  i n L^/K -  Let  i n Gal ( L ^ / L ^ ) ,  a l l ramification  occurs  L /L^. 2  This indices  the f a c t  that  a r e m u l t i p l i c a t i v e i n a tower o f  Corollary  extension  ramification fields.  2  The  Notice  i s clear using  number o f r a m i f i e d  places  i n any  separable  L/K i s f i n i t e .  that  h e r e L/K need  n o t be G a l o i s .  Proof Suppose k i s t h e g r o u n d Take x to be a s e p a r a t i n g  of L over  and  over k ( x ) .  separable  f o r L and K, as  transcendental  6 be t h e g e n e r a t o r K(6)  field  K.  usual  i n K/k, and l e t  Thus we have L =  K(3)  40. Before fications  getting into  o f the s i t u a t i o n .  many  ramified places  show  that  only  Similarly,  are  done.  In p a r t i c u l a r ,  by a d j o i n i n g  plicity,  denote  Galois By  necessary, the  many p l a c e s  there  by L.  to  many  places  L o f L, t h e n we  take  Denote  L t o be t h e f i e l d of 8 .  For sim-  i t s Galois  8 satisfies  that  group  by G.  6 of L s l i g h t l y , i f  the g e n e r a t o r  can assume  are f i n i t e l y  We now a r e c o n s i d e r i n g  L/k(x).  modifying  simpli-  sufficient  to L a l l conjugates  L again  extension  we  finitely  we  two  of L/k(x) are r a m i f i e d .  f o r some e x t e n s i o n  obtained  the  make  To show t h a t  i f we c a n show o n l y  i n L/k(x)  we  on L/K, i t i s c l e a r l y  finitely  ramified also  the p r o o f  an e q u a t i o n  of  form F(T)  where can  = T  the f ^ e k [ x ] .  + f j " " n-l  n  1  +...+  Then, L b e i n g  f„ U  normal over  k[x],  we  f a c t o r F ( T ) i n L as F(T)  n (T -  =  0-6)  oeG Taking  the formal  evaluating  d e r i v a t i v e of F with  a t 8 we  respect  t o T, and  find  = n  F C@) 1  (8  -  a 8 )  oeG a^id  Now we come to c o u n t i n g places not  lying  (This in  of L / k ( x ) . over  excludes  the p l a c e  k(x)  induced  Suppose  the p l a c e only  ring  of  the number o f r a m i f i e d  i s a ramified place  corresponding  fintely  to  many p l a c e s  i n L.  00  on L,  i n k(x).  by I.)  Then  F o r , i f -t|. i s the p l a c e  by ^} and /f i s i t s p l a c e  ring,  6 lies on  t h e n XT 2 k [ x ] .  41 . Thus 6 the  satisfies  remark on  an  equation  integrality  with  of  coefficients  section  1,  in^tT".  9 must  By  therefore l i e  in£. I I I , ^2  By  ramified  I £ is non-trivial. in  G such  F'(9)  that (8  II  Hence  6 - o6 - o6)  0eG  implies  its inertia  group  i s some a u t o m o r p h i s m a- 4 i d  there  (modp).  = 0 = 0  that  (modS).  But This  then means i n p a r t i c u l a r  r  a^id that  F'(8)  This  statement  is divisible  F'(9)  Since  ramified  has  places  As the  holds  an  only  to  the  canonical view  N-gon,  the the  of  our  which  the  leave  F fixed.  octahedral  field  suppose or  that  number  we  extensions  l e t G denote As  to G  G are  i n the then  G i s one  icosahedral  groups.  of  one  i n the  tetrahedron,  embedded of  the  over  for °° i n of  derive  corres-  groups.  |G| = N.  elements  Now  the  Say  that  not  r a m i f i c a t i o n theory  C and  as  parameter  finite.  k =  dihedron,  place  many z e r o s ,  figure F corresponding  icosahedron,  Recall  (1)  groups.  uniformizing  ramified  finitely  canonical  Assume t h a t  the  f o r any  must be  example  the  r a m i f i c a t i o n i n d i c e s of  ponding  we  by  of  Introduction,  i . e . the  the  or  sphere.  r o t a t i o n s of  Each  plane  octahedron  Riemann  the  the  the  sphere  tetrahedral, element  of  G moves  R e f e r to §§1-9 Chp . 1 [ 3 ] f o r f u r t h e r d e t a i l s on example. K l e i n l o o k s a t t h i s s i t u a t i o n from a g e o m e t r i c p o i n t of v i e w .  this  L/k(x)  42 . a point  Pg  on  Call  the  Host  points  are  set  three  the  sphere  of p o i n t s  of  namely:  midpoints  of  of  faces  the  midpoint  elements of the  4  any or  sets  the  an  edge  of of  the  form  place  any  number o f  a conjugacy  corresponds  between  points  (Refer  at  inertia point,  that group  2,  point.  3 and  i are  f o r G.  This  indices  f o r the  three  the  as  to the  the  a group  of  by  G.  identity);  the  elements  G where i =  at  a vertex one  Each  i n k(x)  28-29  of  such  lying  f o r the  in k(x),  )  the  (2,  3,  non-trivial  F.  these  three  class  of  a  given  correspondence elements  the  inertia  order  of  at  the  a  i ) , of r a m i f i c a t i o n  of  G is a cyclic  automorphisms  of  ramification  conjugacy  of  classes  will  G. group the  of  3,  over  The  form  know t h a t  non-unit  two and  three  of  that  P  r a m i f i c a t i o n index  only  triple,  that  through  class fixed we  seen  axis  i n any  ramification pattern  Suppose now Viewed  Since  i s equal  indices  called  i n any  be  the  adjoining  places  centrepoints  easily  the  there  orbits,  precisely  fixed  places  G its orbit.  by  edge, and  points  sphere.  fixed  about  to pp.  and  G leaving a point  group  the  the  f i g u r e F, I t can  c l a s s under  to  in k  the F.  the  However,  smaller  i elements  faces  on  i s moved by  have  is left by  point  N points.  is left  turn  face  is fixed  i n F i x G.  of  opposite  i t i s clear that  points  be  edges  (the h a l f  vertex  5,  Also,  of  G  which  v e r t i c e s of  P of  midpoint  centrepoint G;  F and  PQ  containing  points of  another  to w h i c h  have o r b i t  types  to  of  order  sphere,  the  N.  43. elements  of  G are  obtained  rotation.  As  s u c h we  two  on  the  points  than N p o i n t s -  axis  of  rotation.  are  therefore  of  G must  then  For on  the  two  the  two  ( 1 , N.  having  points are  The  group an  of  orbit  less  of  the  order  n r o t a t i o n (these  are  conjugate  and  fixed  by  conjugacy see  by  n v e r t i c e s (these the  axis  2 rotations).  above G must  therefore  For  future  reference,  b  are  (2,  2,  a list  2n of  points the  fixed  the  there  that  than  vertex  are are  than  points  and  Cyclic  group,  II  Dihedral  III  Tetrahedral  IV  Octahedral  V  Icosahedral  order  group,  N  order  group group group  N  fixed  and  also three  cyclic  for n  the  conjugate distinct  normal.  We  the r a m i f i c a t i o n N -j) . of  the r a m i f i c a t i o n  here.  I  the  elements  the  conjugate  smaller  considerations  i s given  of  all N  axis  (these  Thus  orbits  of  patterns  by  and  through  n mid-edge p o i n t s  c l a s s e s with  the  pattern  the  only  containing  N = 2n  of  end  identity);  are  ramification pattern  order  the  a h a l f - t u r n about  there  N) .  at  by  a single  e i t h e r end  fixed  the  the  that  at  are:  rotations);  points  of  have o r b i t s  non-conjugate.  dihedral  dihedron  immediately  T h e s e pofnts  be  the  repetition  sphere which  less  and  see  by  N)  (1,  N,  (2,  2, -)  (2,  3,  3)  (2,  3,  4)  (2,  3,  5)  44 . For  the  general  closed  field,  of  canonical  the  gives  way  We  will  In  fact,  to  see we  A u t . k ( x ) has  situation,  i n determining groups,  the  combinatorics later will  that  this  determine  to have one  the  of  k any  ramification patterns  geometry  of  of  G-action  does that the  algebraically  result any  our  on  situation  places  i n the  finite  of  same  subgroup  k(x). patterns. of  above r a m i f i c a t i o n p a t t e r n s .  45 THE  CHAPTER VI  Order  (1) Let  HURWITZ  of a D i f f e r e n t i a l  K be an a l g e b r a i c  algebraically  closed.  k-derivation.  FORMULA  function  L e t d:K  ft^yk  ( C f . §2, Chapter  III)  K, and d e n o t e by  the completion  Then,  uniformizing  Let of Kg).  f o r any l o c a l  d- : y  c K —•  ft  /k  a  Kp/k.  Take  be  of K with  respect  to p. .  TT f o r ^ ,  continuous  = k((ir  k-derivation  i : K — * Kj, t o be t h e i n j e c t i o n o f K i n t o  K-module map i ^ :  following  k, w i t h k  the u n i v e r s a l  parameter  be t h e u n i v e r s a l  over  F i x a p l a c e |2 on  Then u n i v e r s a l i t y o f d g u a r a n t e e s  unique  existence  of a  .c Q~ .. c o m p l e t i n g t h e K B / k.  ^ / j .  diagram. K  Let  field  a £ !1K/k  K/k  Define  ord^(to),  the order  o f to at?Q , as  follows: Map to t o i^(to) = to^, s a y . Define  ord^(to)  (recall  that  defined  i n §2, C h a p t e r I I I )  For  the order  any d i f f e r e n t i a l  = ord ( ^) w  of a continuous  to e ^ / ^ J ord^(to)  differential  was  = 0 f o r almost  46 . all  places ^  Let tt ^  on  x be  = Kdx.  K  some f  K.  a separating  Thus  f o r any  transcendental  to e ^ ^ K  k  we  have  i n K/k. w =  fdx,  Then for  e K.  Claim:  ord^(o))  = 0  f o r almost  a l l places  jp  on  K.  Proof: ord-p(to) We f  e K.  and  know t h a t ( f has  poles,  show  that  and  jp  not  "J2 from p lies  Suppose  a  x  1  —°°>  e k -  But  00  (1)  II +  any  =  finite  the  being  the of  and  K which  a finite  So  we  i s unramified 00  in k(x).  number  of  k(x).  must  being  f o r some  u n i f o r m i z i n g parameter  co e ^Yi/k'  product  i s taken  product  by  makes  Therefore,  II =  TT = x - a  ir = x -  ord^(dx) =  5(K/k)  Define over  section  sense.  deg  of  a l l places (2);  We  ( d i v co) =  n^ ^^^ ^  div(co) =  the  r<  ~p. on  U  K.  following  define E ord-n(co)  which  (This  u n i f o r m i z i n g parameter  = 1 • d IT.  and  places.)  , not  f o r m <^ : x i — > a,  local  dx  i t s zeros  0.  Let  therefore  i n number.)  p l a c e ^. of  the  at  since  everywhere.  on  only  everywhere  except  p l a c e - * ^ : x •—»•  Definition  finite  a l l points  place  local  a,  = 0 almost  = 0 almost  Thus /} has  (3)  the  are  over  "p. has  then  Then x = ord  these  must be  .  ord-^(dx)  ordp(f) 0 at  l i e over  excludes  •^•Q :  be  +  order  ord^dx  Let does  = ordp(f)  where This  is a  definition  a.  47 . Claim:  deg ( d i v co ) i s i n d e p e n d e n t  o f co .  Proof: Let ^K/k  =  most  K (  * ' x  x be a s e p a r a t i n g t r a n s c e n d e n t a l i n K. s  o  t  n  a  t  a  n  y  by an e l e m e n t  some f , g e K co = hco  then.  1  deg  t  w  ° differentials  o f K.  ( F o r , co = f d x and co ' = gdx f o r  can now  ( d i v co) =  compare  +  deg ( d i v co) i s an i n v a r i a n t  field  e x t e n s i o n K over  Note:  (div  ord^co') (  1  )  = deg ( d i v co ' ) .  Since  by  deg ( d i v co) and deg  (E o r d ^ h ) + (E ord^co')  = E ord_co'  K/k a l o n e ,  Say  1  B  =  e K.)  1  ord^(co) = E^ o r d ^ C h t o )  = E (ord h  to  co , co ' d i f f e r a t  v co = ( f g "^co', where f g We  Then  depending  k, we d e n o t e  only  i t with  on t h e  reference  6(K/k).  If K i s a rational  function  where x E K i s some i n d e t e r m i n a t e  field, over  say K = k ( x )  k, t h e n  6 ( K / k ) = -2  Proof : Assume K = k ( x ) , where x £ K i s some over  k.  L e t co = dx e ft„/, . K/K Now,  a l l places (i)  fl  (i±)p: Case  (i)  parameter (1)  E  Then  6 (K/k)  = E ord (dx) . K  p  ord h c  p  on K a r e o f t h e f o r m  : x »—*• a, a E k  or  x »—> °°  Take n = x - a to be t h e l o c a l for p .  indeterminate  uniformizing  Then dx = l ' d n , so t h a t o r d ^ ( d x )  = 0 by t h e theorem o f p. 29.  = 0.  48 . Case  Take TT = ^  (ii)  parameter  for^l.  Thus o r d ^ ( d x ) Hence  to be  Then we  =  the  have x = ^,  deg  ( d i v dx)  = _!  = The Let degree  L  1,  teristic  p l a c e p. on on  L.  Hurwitz K  and  dx  = -  ^-2diT.  ord^(dx) (-2)  -2  field  of degree  L e t e^j d e n o t e L.  that  Formula be  L algebraic p.  so  -2.  = 0 +  (4)  local uniformizing  the  e x t e n s i o n s o f k of  transcendence  N over  charac-  K,  and  ramification  W r i t e ^} e L to d e n o t e  that  k of  index  of  the  |! i s a p l a c e  Then  ?™MS"  6  <  L  /  k  - W > - «  )  +  *  «p  J 1 E L  (2) = e j, -  where  1 if p |  e^  and  > e ^ - 1 i f p | e^.  Proof: u) e ^Yi/\n  Let Consider to  6„(L)  notice  a single =  a  (  a n c  c  6 (K/k)  s  o  ^L/k^"  e  p l a c e X£ on  ^_ord (co).  that  *  « *  K.  Then  J|, be  5-(K)  the p l a c e i n K l y i n g  uniformizing Then TT = UH (2)  The  parameters  Compare  to r e l a t e  Suppose |2 i s a p l a c e on Let  Idea  for ^  and  ^(K)  - ord^(co)  =  6 (L/k) ,  „^ ^(L) K  ramification  under p . i n K and  proof:  <5(K/k) and  5 (L/k)  L with  of  Let p  TT, II be  in L  index  e.  local  respectively.  Q  f o r some u n i t  o r i g i n a l deduc t i o n  § 2 Chapter XXI - V [2]  U i n L. of  this  formula  can  be  found  in  49 Thus d TT = e u n  e - 1  d ,n  we  and What  [eun  ft..,. . K/k  E  ord^(co)  is  "J  TT  +  6 - 1  (D u)n ]d^n e  n  d^(K) and d ^ ( L ) .  can r e l a t e 03  (D u)n"d^n  ?  = Now  +  4  1  Thus  i „ ( c o )  =  ft  = ord^(f^)  = f  to.,  ''J  f o r some  d^ir TI -"J-  f  TT  E  = 6^,(K).  6 ^ ( L ) ?  co T  =  Therefore  f d * TT = f [ e u n TT  f  +  6 - 1  ( D u ) n ] d_n e  TT  TT  Notice:  6  B  have  Y  (D  U)n ) e  > e-1 i f p | e .  above AS^ and e ^ i t s r a m i f i c a t i o n  any p l a c e 5 .(L)  +  1  , say.  = e-1 i f p )( e, and E  e  With p now  ^  = ord-»(f ) + o r d J e U n "  ord (co)  = ord^Cf^) + E  = =  E Z  ord^Cco)  E  e-ord  fi^j.  =  ord-(f IT*  ) + E f\^t  TT  (f  index,  r  ) + E e-  = <5^(k) 6^(K) •  =  = 6 (K).( f  = We 5(L/k)  K((TT)), v v / / »  and  therefore  E N  6 (K).N + f  have  e + E e ) +  £  ^  £,  B  ,  e  the f o l l o w i n g  relation  6(K/k). 6 (L/k) =  E 6^,(L) xfeK ^ = E 6 «.(K) • N + E  E  =  p  6 (K/k)-N +  E £ ?  E L  e_  between  we  50 . CHAPTER V I I  CONSEQUENCES  Assume field  of r a t i o n a l  1 over  though  of  of K which  this  does  chapter that purely  closed  are also  (the natural  to t h i s  (1)  k.  We  rational point  o f view  be seen  later  on t h e H u r w i t z  L and K a r e p u r e l y  then  the f o r m u l a t a k e s on a p a r t i c u l a r l y  of  which  morphism  give  enough  we know t h a t  With  fields.  Al  from  subfields that  the problem  Formula, i f  of degree simple  1 o v e r k, form.  t h e d i o p h a n t i n e e q u a t i o n and i t s  L and K as a b o v e ,  e i n L/K, t h e n  Z  from  =  e f o r any  Formula E  §3, C h a p t e r VI  I f f u r t h e r m o r e , the  not d i v i d e  the Hurwitz  - 2 =  L to i s o -  cases.  p o f k does  2N  Galois  arising  to d e t e r m i n e  6 ( L / k ) = 6 ( K / k ) = -2.  characteristic index  transcendental  information  now d e t e r m i n e  solutions.  o f degree  a d i o p h a n t i n e e q u a t i o n , the s o l u t i o n s  o v e r K i n most  We  study  function  both  i t yields  fixed  situation.  In t h e theorem  Namely,  K is a  one b e i n g t o s t u d y  extension) i t w i l l  reduce  FORMULA  transcendental  field  i s not the n a t u r a l  problem, a fixed  this  functions,  an a l g e b r a i c a l l y  extensions  our  throughout  OF THE HURWITZ  becomes  ( e ^ - 1)  ramification  51 . indexing  the^. in K  from  1 to r .  2 = i N • ,  2 -  }  c i  —  -  i=l  e^,  this  as  i t is fairly  easy  to see  Assuming  For, and (i)  a diophantine  N > 1,  that then  r = 1 i m p l i e s R.S.  r > 4 i m p l i e s R.S. Letting  by  N , we  (*)  i t has  < 1,  while  in N  relatively  r = 2 or  (*)  > 2,  equation  few  3.  while  L.S.  and  (*)  L.S. <  (*)  1  =  1  e  e  6  6  x  n  =i 6  41  can be  e. l  i + i e  2  e  3  >  So 1 2 If  anothe r e  If  l  = e=  neithe r  Ass ume  e^, 2  e  2  say  -1 = e  2,  e=  n  nor  e^ i s  2  e ^ = 3.  e^ <  6,  and  e  2  e  3  , is  2  3  Then 1 6  Thus  1;  2.  " l 2 S i n c e e ^ , e^ < N, we have — , — > — . I t f o l l o w s from 1 2 Hence e = e„ = N. equation that — , — must be — . N 1 2 e e 1 2 (ii) Le t t i n g r = 3 (*) N  not a l l  >  r=2 (*) becomes 2  Since  have  l  Regarding  solutions.  dividing  1 — e. )  r  (*) K  Or,  e  must  N  e,  our  52 . have e  = 2,  e  2  =  3,  e  3  =  3,  N =  12  e ; L  = 2,  e  2  =  3,  e  3  =  4,  N =  24  ±  e  So  the  of  solutions.  l  = » 2  Hurwitz  Notice  indices)  ication  indices  is  the  the  major  for  the  (2)  the  Main  described  [L:K]  (1),  ramification  up  be  L e t -<J^ , ramification which Let  e^,  2 >  at  these p l a c e s .  to  show  the  the  for 3 ^  be  e  R(L/K) i s one  of  only  systems  ramif-  groups:  and  ramif-  cyclic  icosahedral.  the  Formula  This towards  pattern  five  t  to  r  e  occur  to  a t x^, ^ , x^  the  R(L/K) of  the  two  third  type.  type.  i n K at  only  ordered 2  each  each  ramification  the  K  Since  determine  at  m a t t e r what  use  types.  places  e  L,  ramification  reference n  with  is different for  respective  Then we  types  the  the  this  easy  ( I t may  ramification  ramification  and  shows t h a t  and  indices  i t does not e^  of  specify  occurs.  case e  one  to  a notation  five  orders  Hurwitz  in  of  set  the  (N  group  and  Formula  the  We  the  canonical  by  only  systems  Hurwitz  L/K  i t i s enough  to  60  Theorem.  of  type,  N =  c a s e has  these  five  5,  octahedral  indices set  can  =  3  this  that  donated  The as  e  correspond  fact  of  3,  in  tetrahedral,  proof  both  2  Formula  ication  dihedral,  =  e  which  places, place  in  is.)  indices  triple  (e^,  e  , x^ ^ i n L/K.  Call  this  L/K.  says  that  Hurwitz  2 >  e^)  53  In what ramification determine  II  (2,  2,  2  III  C2,  3,  3)  IV  (2,  3,  4)  'V  (2,  3,  5)  pattern  t h e group  For  dental  N, N)  f o l l o w s , we  the e x t e n s i o n  extension  (1,  shall  show t h a t knowing t h e  i n cases  over  k.  o f K i s enough to V i t  I - I V , and i n c a s e  of the e x t e n s i o n  (assuming  always  i s Galois).  t h e lemma and t h e o r e m L o f K.  J  o f an e x t e n s i o n  the e x t e n s i o n  determines that  I  f o l l o w i n g , we  Say L = k ( x ) ,  Also  assume  f i x an  f o r some x e L  transcen-  [ L : K ] = N.  Lemma In L/K we (*)  have  I r r ( x , K, T) = T  where e i t h e r have  N  + a^T^"  1  a ,  +...+  Q  t h e a . e k o r K = k ( a . ) , and a l l n o n c o n s t a n t  t h e same  pole.  Proof Choose Let We  come e l e m e n t  y e K such  I r r ( x , k ( y ) , T) = f ( y ) T N  N  + f  N - ]  that  _ (y) T  N _ 1  K = k(y). + . . .+  f (y). Q  know y = ^ f o r some h, g £ k f x ] , so t h a t g (x)  Irr  ( x , k ( y ) , T) = h ( T ) - y g ( T ) .  power o n l y have  unique  i n the m i n i m a l factorization  Thus y o c c u r s  polynomial  f o r x.  i n k ( y ) [ T ] , we  to the f i r s t  Since  we  see t h a t a l l  54 . f (y)  must  i  we  be o f t h e form  a^y + b^ where  a ^ , b^ e k.  Then  have , _ N N-iy N - l N-1 ^ (x, K, T) = T + —.r • T a y + b a  Irr  v  +  T  N  By d e f i n i t i o n same p o l e  N  then,  N-l  a  T  +  a l l  *'  ,  +  0'  a  S  a  y  "  which are nonconstant  This i s either  have t h e  i n k ( i f a . b „ - a „ b . = 0) o r 1 N N x  v  generates  K (by t h e t h e o r e m o f C h a p t e r I ) .  Remark:  I f R(L/K)  (•—) , t h e n  lemma we  N  ( g i v e n by a^y + b ^ ) ; and a l l a r e o f t h e form  a .y + b . i i ;—r—• a .y + b„ N N  and  o , . a y + b b  N-l +  T  +  A  N  N =  v + ...+  b  A  i s type  I with  ramification  a t (x)  i n t h e above  can c o n c l u d e  that K = k(a.) f o r a l l i . For,  l by  t h e lemma  an  equation  f o l l o w i n g I of Chapter o f form  This  together  must  have  Theorem  with  order  (*) where t h e above  precisely  satisfies  e a c h o, ^ has o r d e r lemma  implies that  > 1 a t 0.  each  1, as d e s i r e d .  1 Suppose R(L/K)  places  V, we know x  i s type  I with  ramification  at the  (x) and ( ^ ) . Then K = k ( x ) . N  Proof By  the p r e c e d i n g Irr  where  each  particular,  remark we know  ( x , K, T) = T  a. has o r d e r x as g e n e r a t o r  1.  + a  N  Hence  o f K.  we  Then  N  _  that 1  + . .. + a  Q  can f i x y = a., i n 0 J  a l l other  a  are of  55 . a.y the  + b. ;—-5— f o r some a. , b. , c. , d. e k s u c h c.v+d. x* x x* x x' x  form  a.d. - b . c . 4 0. x i I i have all  Now  the p r e c e d i n g  t h e same p o l e .  Since  •  lemma  states  00  that a l l 1  has a p o l e  o t h e r ct ^ must have a p o l e a t  that  .  That  at  (meaning  00  i s , they  —)  must  be  o f t h e form a . = a.y f o r some a. e k. Thus i x x (**) I r r ( x , K, T) = T + a _ y T " +...+ 7  N  N  N  Now and  l o o k a t the e q u a t i o n ^ by y i n (*') to f i n d  divide  1 — — - + a„ ,U + a y Again must  N-1  a p p l y i n g the p r e c e d i n g of ^.  That  find  i s , t h e y must  t h e n we have y = - x , o r K, = k ( y ) =  theorem  1 we  c o u l d have been  I and t h e p l a c e s a t w h i c h then Thus,  assumed  stated  ramification  K.  We (1)  also  theorem  make  for  a fixed  From  ramification i n computation.  i f R(L/K) i s t y p e occurs  as c y c l i c type  fact  a r e known,  of degree I with  upon w h i c h  N.  ramification  the p r o o f  relies.  use o f t h e f o l l o w i n g  ramification  a l l be 0.  t h e e x t e n s i o n s L and L ' a r e i s o m o r -  T h i s i s the b a s i c  the f o l l o w i n g  a l l a^  N  that  as:  i f R(L/K) and R ( L ' / K ) a r e b o t h  over  that  f o r convenience  the e x t e n s i o n i s d e t e r m i n e d  t h e same p l a c e s , t h e n  =0  k(x ).  N  a t (x) and (—) m e r e l y  of  — N-1 N a.U + U  lemma we  occurred  phic  ^  1  I n Theorem  at  that  +...+  Remark:  The  J  N-2  be m u l t i p l e s  (*')  „U  by — . Take T = 77 x U satisfies  satisfied  2 H  y.  1  1  index place  = order  facts  from  of i n e r t i a  (result III)  Chapter  group  V.  56 .  (2)  ramification  (3)  degree  of  indices  Theorem  field  multiply  i n a tower  e x t e n s i o n = sum  (result  of  of  fields  ramification  I)  2 L/K  Let function  be  field.  R(L/K)  Then  indices  a Galois Suppose  the  L  determine's  L  e x t e n s i o n and  up  of L/K  group to an  a  rational  is  solvable.  isomorphism  over  K.  Proof We  c o n s i d e r the  II,  I I I or  IV,  II  (2, 2,  f)  Assume group LQ  of  that  so must  Then have  ramify Again  of  morphism If the  L Q , and  |- = 2,  o n l y groups  Klein  4-group;  then  so  I with N =  that  over  K.  of o r d e r 4 a r e and  the  Aut^k(x)  type  By  =  of  o n l y be  find  R(L/K)  Let  I-V.  p l a c e s i n L Q above N index in L/LQ. Hence  over  inertia N  type  Theorem  to i s o m o r p h i s m  1 we  1.  pattern  two  u s i n g Theorem  the  subgroup  1).  L Q i s determined  L e t H be  (LQ/K) i s a f i n i t e  ( 2 , 2,  that  i n Theorem  index —.  Q  =  with  i s type  ramification  can  is,  the  N -zj ^ 2.  R(L /K)  That  be  2,  R(LQ/K)  dealt  ramification  (L /K)I Q  =  Gal  |Gal  'P^  been  the p l a c e w i t h  = F i x H.  and  I having  first  R(L/K)  t h r e e c a s e s where  over  • L  =  2).  cyclic  Let  ,  p l a c e s must N N (—, —, 1).  We  know  group  pattern  2.  means  i s determined  ( 2 , 2,  ramification  this  K.  These  R(L/LQ)  the  1  Since  of  or the  to  iso-  that the cyclic  57 . group Let  i s type  H be  index  2.  is  type  is  not  We  find  (1,  Take L  Choose  III  Le t  morphism  L  Gal  In  of =  be  3  these  Klein  with  the  R(L /K)  that  Q  above ^ 1),  that  in L Q / K  obtain  specified. 2)  or  isomorphism  fields  that  is  (2, 1,  to  4-group.  ramification  place unramified  (2, 2,  pattern. K,  as  (L/K)  |Gal  3).  LQ  the  2).  has  Then  must  i s determined  over  (2, 3,  1,  such  = F i x H.  Q  2  ramification  (2,  find  (2, 2,  ^ 4 we  = 12,  , 'p , ^?  Each  that  index  the  extension LQ  N  we  yields  i s t h e r e f o r e ( 1 , 3,  K.  LQ  for  [L:K]  Q  take  say  the  L Q of =  R(LQ/K)  K, (2,  =  R(L/LQ)  2,  1).  (2, 2,  3)  Hence R ( L / L )  of  three  be  point^.^  unless  1.then  these,  Take L  over^,  IV  now  must  determined.  3.  R(LQ/K)  know  but  determined  argument  Since  over  2,  Q  of  (2, 3,  index  Again  Theorem  one  L is  = F i x H.  precisely  i n the  and  o f any  R(L /K) i s either  that  are  Q  (L/K)  group  N =  I with  2).  Hence G a l  inertia  uniquely  2,  there  As  the  I.  (L /K)| Q  i s then  three places  ramify By  a subgroup  the  of  That  is, L  3,  of and  determined in L  argument  lying  Q  2 in L / L Q .  index  to i s o m o r p h i s m  =  H  f o r II  f o r such  i s determined  we a  to  iso-  desired.  4) a group  2.  order  L e t H be  = F i x H, 2 ) , so  of  as  that  such  usual.  24  there  i s always  a subgroup We  find  L Q i s determined.  i n Gal  that  a  subgroup  (L/K),  R(LQ/K)  LetZ/^,^^  and  must be  be  the  1),  58.  two  places  above^.  Then  previous  K,  ^  above ^ ! clearly  L  case,  and so o v e r  as  E  (^3  T  R(L/L ) Q  i s determined  = to  ^  e  t  n  (3,  P  e  3,  x  a  c  e  i  2).  isomorphism  ^  n  0  From  the  over  LQ,  stated.  3  Theorem  L / K be  Let function V then  LQ  in  field.  G is  a Galois extension Let  G be  and  (L/K).  Gal  l e t L be  If R(L/K)  a  rational  is  type  simple.  Proof Suppose H i s a n o r m a l LQ  Let an  = F i x H.  LQ/K  must  be  since  either  two  indices  there  These  ramify  But  5,  this  determines  exist  at  implies  5)  5 or  in  [LQ:K]  R(LQ/K)  as  We  order  1.  m >  show t h a t  (12,  5).  1  f  ^  2,  then  l  s  3  such  so  ramified R(LQ/K)  that  5).  But  have  to r a m i f y ,  this  is  impossible both  2.  Thus-^-^ i s u n r a m i f i e d  two  places  L/LQ =  3,  (2,  or  least  totally  = m.  r a m i f i e d here,  i n L / K would  a m u l t i p l e of  Hence  = (2,  totally  (1,  places  Q  i n G of  exist.  that R(L/K)  i t must be  (L /K)|  |Gal  L Q / K cannot  extension Recall  in  Then  subgroup  so  that  12.  R(L/LQ)  being  12,  1).  ^0  , Z* 2  must  in LQ/K. ^  a  be  unramified  with  o  v  e  ^3 *  (5,  in  5,  LQ/K  Contradiction.  Remark: Theorems 1-3  actually  the M a i n Theorem, n a m e l y , Aut.k(x)  f o r which  establish  that a f i n i t e  p Jf J G| must  be  the  first  subgroup  isomorphic  claim  of  G of  to a c a n o n i c a l  1)  59 group.  F o r , G determines c  field  L = k ( x ) , and by (1)  type  I-V.  Theorems  type  I-IV,  the f i e l d  morphism-  but t h i s  extension,  G, must  we know  that  V then  we know  Hence G, i n t h i s  function  clearly  i m p l i e s that  be d e t e r m i n e d  i f R(L/K) i s  Theorem  i s simple  i s the o n l y case  have  be i s o m o r p h i c  3 shows  that  of order simple  Theorem 1, C h a p t e r II g u a r a n t e e s rational function f i e l d .  60.  of the Since  extensions to one o f t h e s e  i f R(L/K)  group  too, i s isomorphic  to i s o -  the group  to i s o m o r p h i s m .  I<-IV  •  R(L/K) i s  e x t e n s i o n L/K i s d e t e r m i n e d  G = G a l (L/K) group  that  show t h a t  the c a n o n i c a l g r o u p s  groups.  icosahedral  (1)  1 and 2 t h e n  by R ( L / K ) , G must  determined canonical  K = F i x G, a r a t i o n a l  i s type  But t h e of order  60.  to a c a n o n i c a l group  that  Fix G is  60. CHAPTER V I I I  PROOF  Recall  the hypotheses  finite  subgroup  closed  field  is  o f t h e theorem;  we have a  of c h a r a c t e r i s t i c G must  groups  algebraically  p w i t h p A\ |G| .  The aim  be c o n j u g a t e to one o f t h e f i v e  i f the r a m i f i c a t i o n  pattern  of G i s  I-IV or i f p = 0.  type  The  first  group  i s derived  VII.  We  group, such  THEOREM  G o f Aut k ( x ) , where k i s an  to show t h a t  canonical  OF THE MAIN  found  G  Let  the Hurwitz that  G must  between G and a c a n o n i c a l  Formula,  as i n C h a p t e r  be i s o m o r p h i c t o a c a n o n i c a l  Our aim i s to show e x i s t e n c e  = a Go  a  isomorphism  from  there  s a y Gg.  that  c o n n e c t i o n we o b t a i n  problem  = GQ.  1  We  can r e d u c e  by t h e f o l l o w i n g  F = F i x G and F OF =  Q  to a  field  Then f o r o e A u t ^ k O x ) ,  Q  G°  this  argument.  = Fix G .  FQ  of a e A u t ^ k ( x )  = GQ  Proof: G (ago But  ^)(fg) (ogo  GQ Q. G ° : of  <== GQ :  a  _ 1  Let ago"  = (ago ^ ) ( o f )  ) (of)  = (og)(f)  1  e G° .  Then,  f o r some f e F s u c h = of  = fg.  L e t gg e Gg and f g e F g .  that  gg = ago  Thus,  This 1  implies  a  "*"ggO-  = g,  Q  e F , Q  that  Thus a g o "  Then,  f o r some f e F, we have g g ( g f ) = o f .  a "'"(of) = f .  for f  1  writing But t h e n  of = f g .  is in G . Q  f g as (a  ^ggCf)(f)  f o r some g e G, so  e G°.  i f F i x G = k ( y ) and F i x Gg = k ( y g ) f o r some  =  61 elements is  y, y  to f i n d  e k(x)  n  t r a n s c e n d e n t a l over  a e Aut^k(x)  such  oCkCy))  D e f i n e o" to be to k C y ^ ) .  That  a:  map  i f i t can  Extend  a as Let  be  =  extended  an  t o g i v e an  II).  0  to an  Thus we  closure  automorphism have  the ^  job  If  the  ramification  (1)  k(y)  the  required  i s to show t h a t  t h e n Theorems k(x)  algebraic  and  1 and  of K  pattern  (by Theorem  Hence G i s c o n j u g a t e c a s e we  can  3,  k(x)  =  k(x)  n  k(x)  =  k(a(x)).  o f G and  2 of Chapter  i.e.  Then a  k(y )  VII  k ( a ( x ) ) are u n i q u e l y  closure.  remaining  of k ( x ) .  • K  -  our  The  be  > k(a(x))  where now  cases.  from  set-up  k(y)  the  isomorphism  automorphism o f k ( x ) .  k(x)  extensions  (1)  kCy- )  Then a w i l l  algebraic  K  I-IV,  probl  follows: K be  extended  Chapter  be  the  a : k ( y ) — > • k ( y ^ ) by a : y •—*• y.g  a i—• a f o r a l l a e k .  and  then  that  the n a t u r a l  i s , define  k,  G Q i s type show t h a t  determined  the in  = k ( a ( x ) ) f o r these  to G Q h e r e . have  to c o n s i d e r i s p = 0  Assume, w i t h o u t l o s s o f g e n e r a l i t y , y , yQ t h a t r a m i f i c a t i o n o c c u r s a t 1, 0 and °°.  chosen  and such  62 . the  ramification  need of  several technical  x with  are  p a t t e r n of G being  respect  listed  results  to c e r t a i n  type  V.  concerning  For this  the  derivations.  we  Schwarzian  These  results  i n t h e f o l l o w i n g lemmas.  Through  t h e lemmas l e t k ( z ) = F i x G f o r some z z  k(x),  (2 ) z  t r a n s c e n d e n t a l over  know is  that  to k(x) s i n c e  Thus d i f f e r e n t i a t i o n  with  the Schwarzian  by  f  (  z  )  In  of x with  respect  on k ( z ) we  k(x)/k(z)  respect  o f z, f ( z ) , on k ( z ) c a n be e x t e n d e d  Denote tx]  If D i s a derivation  i t c a n be e x t e n d e d  separable.  function  k.  to some  to k ( x ) .  to t h i s  derivation  . the f i r s t  lemma we  relate  the Schwarzian  of x  (3 ) w i t h r e s p e c t to z — and w i t h r e s p e c t z  to the Schwarzian of x w i t h to z - c, f o r some c e k.  r e s p e c t to  Lemma 1 (i)  [x]  z  =  \n [ x ] i z  (ii)  [x]  z  =  fx] _ z  f o r c e k.  c  Proof Let on  D denote  differentiation  k ( x ) and l e t DQ d e n o t e  with  differentiation  respect with  to z  respect  to — on k ( x ) . z  (i)  (2) (3)  Recall  that  [x] = z  3 2 ^ Dx  -  3  2 (£- ^ V Dx X  2  P o s s i b l e by Theorem 1, C h a p t e r I I , A c t u a l l y we s h o u l d w r i t e "The S c h w a r z i a n o f x w i t h r e s p e c t to t h e d e r i v a t i o n on k ( x ) i n d u c e d by d i f f e r e n t i a t i o n w i t h r e s p e c t to z on k ( z ) " .  63 A  r 1 =  /  x  2  -  z  0  U  Schwarzians that  n  D  JL  we  V  X  first  extensive  Thus  2  D x  =  2  X  D C D  D x  ~  Q  2  i . 0  Q  =  Q  X ) . D  Q  Z  "like"  Rule  terms.  Notice  i s made.  DCDQX)-D  D ( ; D  Dx • D Q z  these  DX-DQZ  =  D (D x)  0  To compare  X  use o f t h e C h a i n Q  D  \  x  of a l l compare  D x  and  n  D  3 I—  ~  0  X I  Q  Z  D ( D  Dx  "  x.p zj 0  Dx'  2 X . D Q Z  D  +  D X . D ( D Q Z )  Dx But with  U  D~z 0  = -z  respect r  2  1 For, l e t y = —. z  .  J  to y, and D~z 0 J  0 D x  )  D  X  D  we  X  . (-z ) Z  DX  Q  similarly  '  Then D„ i s d i f f e r e n t i a t i o n 0 1 1 2 = D-. (—) = - —? = -z . Therefore 0 y y^  +  (-2z)  have  DQX  D ( D Q X ) . D Q Z  D ( D Q X )  DQX  DX-DQZ  DX  D[D x.z + Dx•2z ] Dx D x-z 3  4  + D x-4z Dx 2  3  + D x.2z 2  Thus  (  M  From  D  0  L  X  0X DQ  Dx  (a) and  (b) we  4 z  °  2  *  . Dx + —  have  ,  0  , 3 , , 2 • 6z + 6z  3  +  Dx»6z  2  64 . r i  or°  3 x  j. D x  4  /- ^ , /- 2  2  . .D x,2  4 . D x  2  3 . . 2  2  r  z D x 4 r ^ T " 3  0  "  2  r[ x l ] -z  = (ii)  z  „,D x.2 4 b T 2  3  (  )  ,  z  4  z  This  i s clear  since d i f f e r e n t i a t i o n  z i s t h e same as d i f f e r e n t i a t i o n for  with  with  respect  r e s p e c t to to z - c  c e k.  Define  to be t h a t which  t e rm o f C ]  the p r i n c i p a l part r  contains  x  a  t  ~  z  z  of the Laurent negative  series  c  f°  r c  k,  e  f o r [x] a t z - < z  powers o f z - c.  Lemma 2 Let  e be th.e r a m i f i c a t i o n  z - c, f o r c e k. z - c is  Then t h e p r i n c i p a l  of k ( x ) / k ( z ) at term  of f J  z  c)"  i f e > 1,  x  a  t  2  (i)  (z - c ) " + A ( z 2  Q  where (ii)  index  r-  i s some c o n s t a n t  1  element.  0 i f e = 1.  Proof (i)  Since  write in  the r a m i f i c a t i o n  U(x - C)  a t z - c i s e, we can  = z - c f o r U a unit  i n k ( x ) and C a  constant  k(x) .  Notation: want For  index  Notice  the p r i n c i p a l simplicity  that  term  [x] = z  [x - C] . z-c  o f [x - C ] ^ _ where  of n o t a t i o n , w r i t e  c  Thus we  U(x - C)  = z - c  x f o r x - C, z f o r z - c .  65 . The  problem  then  Is to f i n d  the p r i n c i p a l  term  of * " x  z  where  e Ux  = z.  Denote to  differentiation  with  respect  t o z by D, w i t h  respect  x by d. From Ux  = z we have DU'X  or,  dU'Dx'X  Thus  Dx(dU«x  Let  W  =  -1  1  +  e  6  + U'Dx  6  + Uex  e - 1  =; 1  6  Dx  = 1,  + eUx " ] = 1 6  1  ^77»x  eU  Then (a)  Dx = Notice  term  ~ r eU  i s 1.  -  term  From  i . e . a unit  implies  whose  initial  dU has o r d e r S 0, and  Hence o r d (^-jr'x) > 0, so t h a t t h e eU  o f W ^ must 2  6  W i s a 1-unit  a unit.  Dx  (b)  1  For, U a unit  i s also  constant  that  -J-x  be 1.  = D(Dx) = d(Dx)'Dx  ( a ) t h e n we have  D x = (-^- - x  + - d(-^) x  )Dx  The r e f o r e , (  v C  )  D x Dx~ 2  =  1-e W ~e~ U X  -e , 1 „ ,W e U>  N  +  d (  1-e  X  w w — i s a u n i t , so o r d (d(yj)) i 0. Hence l W l - e ord (— d ( — ) x ) > 1-e > - e . The i n i t i a l e U  D x of — Dx 2  term  (4) is (4)  therefore  at worst  that of  One i n i t i a l term i s worse t h a n a n o t h e r i f i t has a l a r g e r n e g a t i v e exponent. N o t i c e t h a t we c a n n o t s a y above t h a t t h e i n i t i a l t e r m i s . . . b e c a u s e 1-e may be 0  66 1-e W -e 1-e • e.-l 1-e T-1 — x = (Ux ) W = z W. e U e e / T T  This  has i n i t i a l From  ... (  d  (a) and (b) we  2  1-e  ,W,2  (p)  CT?  )  order  l-2e , +  1  ? u  3  W d (  , ,W\ U  X  > 2-2e  2 we  have  + = d(D^x) = —(1— e) ^— ( 1 - 2e)^"(u^ e -2 2 e^ (Ux ) W c  x  2(l-e)  )  order  s i n c e D x = d(D x ) » D x , (1- e) ( 1 - 2e) e^  D x D  x  have  > l-2e  3 Then,  1-e -1 z since W i s a 1-unit. e  term  f:  y  x  (  °f  t e r m s  higher  degree)  +...  (1- e) ( 1 - 2e) 2 e  where, a g a i n , Thus [x]  z  we have  t h e term w i t h  used  largest  1  12  -  .2  =  (  e  =  Hence  possible negative  e  some  3(i^) ] 2  1-unit.  exponent i n  2  n  z  Z  term o f [ x l z  AQ.  -2  2 )  z  -2  v  constant  2  2  6e + 4 e - 3 + 6e - 3 e ~e~2  - 1  2  the p r i n c i p a l r  as  that W i s a  is  !2<->i-«> -  for  the f a c t  Replacing  is  2 e - 1 *  z  -2  e^  + A^z  -1  0  z by z-c g i v e s  the r e s u l t  stated.  (ii) that  In the proof e > 1.  we want  o f ( i ) we  d i d not ever  So f o r e = 1, p r o o f  to show t h a t  use t h e f a c t  (i)is valid.  f o r e = 1, the e n t i r e  But now  principal  term i s 0  67 . For  e = 1, line  (c) of ( i ) reads  "  DT  This  d  u  (  we know has p r i n c i p a l  -  }  0.  term  (d),fore = 1 ,  Line  b e come s  TJ _3  Thus  DT"  0  =  D (  -  and  u>  d (  ° r  X  ^  W  )  M  4.  2  H  W  ^  2  t o o c l e a r l y has p r i n c i p a l term 0 . Hence 3 2 D x D x 2 of 2 - 3 ( — ) must be 0 , as d e s i r e d . Dx Dx  this  term  the p r i n c i p a l  Lemma 3 Let and  assume  the  form  e be t h e r a m i f i c a t i o n e > 1.  Then  index  the expansion  o f kQx} /k(z) a t °s>, of I l x  a  °°  t  1  z  S  °f  e -l 1 , . 1 . 1 . 1 2— —7 + A „ — r + A , — 7 - + A „ —re^ z^ 0 z l z ^ 2 z2  A  A  J  for  some c o n s t a n t s A Q , A ^ ,  5  ,•••  Proof From Lemma l ( i ) ,  together  with  Lemma  possible  degree  [  x  l  2 ( i ) we f i n d  i n the expansion 1  ,e -l ,1,-2 2  2 •n—(—T)  e e- 1^  ,1 z ^*  Using  ""Trfx]^-  = z  .  that  t h e terms  of [ 3 X  .  this  a  t  00  a  r  fact  of l e a s t e  Z  ,1,-1,  + A (—rr) . . Q ,1 z J  ,  f o r some c o n s t a n t  A„  0  68 . Note:  Lemma  at oo then order  i f there  a t °° :  i s ramification  i t has a zero  a t <» o f  a t l e a s t 2. next  basic  Namely,  lemma, d e d u c e d  result  that  rational  needed  using  the p r e v i o u s  to conclude  the Schwarzian  function  ramification Main  us t h a t  [x] has no p o l e z  The the  3 tells  whose  the Main  of x with  Theorem.  respect  c o e f f i c i e n t s depend  two, g i v e s  to z i s a  only  on t h e  i n k(x)/k(z).  Lemma There  i s a rational [x]  where  function  and r a m i f i c a t i o n  £ k ( z ) such  that  = r(z)  z  the c o e f f i c i e n t s of r ( z )  places  r(z)  depend  indices  only  on t h e r a m i f i e d  of k ( x ) / k ( z ) .  Proof Rationality Chapter  IV.  of "  Suppose  [x]  Assume w i t h o u t of  F i x G chosen  °° ( t h a t  such  ]  x  z  r(z)  = z  loss that  i s , at the places  follows  from  Corollary  2 of  e k(z).  of g e n e r a l i t y ramification (z-1),  the g e n e r a t o r z  occurs  a t 1, 0 and  ( z ) and (—) ) w i t h  ramif-  z ication  indices  e^, e^ and e^ r e s p e c t i v e l y .  2 and 3 we f i n d not f  °  r  <*>  at  that  has p o l e s  o r a t any u n r a m i f i e d  00  e  m  r  r(z)  (  2  )  2  = eT(^iT2  e  ~l  1 for  place.  - 1 +  some c o n s t a n t s  a t most  2  a t 0 and 1:  Thus r ( z )  has t h e  -1  + -T2e  From Lemmas  +  z  +  C  e z 2  A, B, C,  (C i s a c o n s t a n t ,  since i f  69 . it  were a n o n c o n s t a n t  at  co .  A,  (*) we c a n c a l c u l a t e  expansion B and C.  terms will  But Lemma  are.  the i n i t i a l  o f r ( z ) a t <» i n terms  Equating  a pole  of the c o n s t a n t s  3 already t e l l s  like  terms o f  us what  coefficients  these  initial  i n the e x p a n s i o n s  a l l o w us to s o l v e f o r A, B and C, and so t o w r i t e  down an e x p l i c i t In by  have  Contradiction.) From  the  p o l y n o m i a l , r ( z ) would  formula  for r(z).  (*) t h e i n i t i a l  rewriting  terms  r ( z ) i n terms  of r ( z ) at  of u = —.  a r e found  00  Then  z = — and  (*) y i e l d s r  < > z  • u-> = \ r(  e  i' 2 l u(  1  +  r  }  r u"  +  h i 4 2 u>  1  e  1  2 u  2  6  1  e  u + A j ^  2  + ~  1 _ U  -  2  2  2  + Bu + C  2  ("  2  +  2  u  3  + 3u  4  1  +...) 6  +  e = C + (A + B ) u + (  2  2  e  Lemma  3 states  must be  that  l  + A(u + u  2 " 2u  2  + u  3  +...)  1  6  expansion  v  u  2  c  1  = '-^—2 6  e  +  (  2  = —1  I  +  + Bu + C  2  1  2  e  + A +  2  '  6  the i n i t i a l  -  1  )u  +...  2 terms  of this  2  ,  3  e  =  " e  7  u  2  + A  u  Q  3  + A  u  1  3  e  i  l  C  =  0  A + B  =  0  e +  A  2  +  ~  "  6  A  ~ l  =  +  e  Substituting  these  r(z) =  1 1 ~2 ^ ~2 " ^ 2 3 e  values  e  l '  n  d  e  = ^  B  into  + —  equation  A  (*)  gives  .  2 ^  x  - 2  1  2~z  e  1  1  + (;-| + - | - - | .  2  l  e  e  c o e f f i c i e n t s o f r ( z ) do, t h e r e f o r e ,  2  6  l)Kz)Cz  3  depend  o n l y on  2' 3' e  Now the Main  we a r e i n shape  entiation  D , DQ denote with  respect  Assuming, without yQ have been  1, 0 and  00  ramified By r with  to p r o v e  the remaining  case i n  Theorem.  Let  y,  e  l  e  The  -2  a  e  2  (z-1)  2  2~ 3  6  2 _ 1  1  3  gives  i  2  6  2 2  1 Therefore  i  2  2~ l  e  e  2  c o e f f i c i e n t s o f u , u, u  2  -  4  0 Equating  0  , then  loss  chosen  the extensions to y,  i n k ( y ) , k(yQ>  of g e n e r a l i t y , such  k(x)/k(y)  that  that  then,  respectively  the g e n e r a t o r s  ramification  and k ( x ) / k ( y Q )  p l a c e s and r a m i f i c a t i o n t h e Main Lemma  to k ( x ) o f d i f f e r -  have  occurs at t h e same  indices.  there  c o e f f i c i e n t s i n k depending  is a rational  only  function  on t h e r a m i f i c a t i o n  71 indices  o f 1, 0 and <» s u c h [x]  But  then  =  D  i t i s easy  that  r ( y ) and  [x]  =  D  r(y ). Q  to see t h a t  a([x] ) = Ix] a([x] )  For,  D  since  = a(r(y))  = r(o(y))  a i s an i s o m o r p h i s m  we f i n d  = r(y ) = Q  n  "  from  these  last  By =  =  Also,  n  0  have  [o(x)1  C o r o l l a r i e s 1 and 2 o f C h a p t e r  k(a(x)),  .  u  two e q u a t i o n s , we  [x]_  D  that  a([x] ) = [a(x)] Then,  [x]  IV i t f o l l o w s  that  k(x)  as d e s i r e d .  Remarks: (1)  In the h y p o t h e s e s  p )( | GI . in  In the p r o o f  particular),  ramification too  strong,  but  p  of  index  theorem  only  the f a c t  i t i s plausible  types.  that  indices  that  So o u r o r i g i n a l  guarantees  says  state  ( d e r i v a t i o n of the Hurwitz  e f o r any r a m i f i c a t i o n i n d e x  ramification  then  e.  i n that  the theorem  the  we used  o f t h e M a i n Theorem we  that  this  p J( e f o r any  h y p o t h e s i s might  that  the case  e arise. cannot  We know f o r t h e s e  be one o f t h e f i v e  that  p | |G|  For,  |G| and t h e canonical  i f p | |G| (p a p r i m e )  p = e, f o r some r a m i f i c a t i o n i n d e x e.  seem  But t h e r e s u l t  happen.  i f p J( e f o r a l l e, t h e n e must  Formula  72. (2)  I t turns out to be true that f o r any a l g e b r a i c a l l y closed f i e l d  k , a f i n i t e subgroup of Aut^k(x)  must be conjugate to a canonical group.  That i s , the major claim of our Main Theorem does i n fact hold for the icosahedral case i n c h a r a c t e r i s t i c  p > 0 , although i t does not y i e l d  to proof using the techniques here developed f o r the case of c h a r a c t e r i s t i c 0 . That the theorem i s true for a l l cases i n c h a r a c t e r i s t i c follows from Dickson [1].  From §259 we have:  algebraic closure of the prime f i e l d Aut  k (x) n  k  Z/pZ  i f k^  p > 0  i s the  and i f a subgroup of  i s isomorphic to the icosahedral group then that subgroup i s  0  conjugate to the icosahedral group.  By standard techniques of group  representations and group extensions t h i s r e s u l t holds f o r any a l g e b r a i c a l l y closed f i e l d  k .  Dickson's techniques, however, do not work f o r both c h a r a c t e r i s t i c 0  and c h a r a c t e r i s t i c  p .  Dickson uses f i n i t e permutation groups to  deduce h i s r e s u l t f o r c h a r a c t e r i s t i c  p —  i n characteristic  corresponding devices are simply not f i n i t e .  0  the  Thus, i n c h a r a c t e r i s t i c  some other method such as the one used here i s necessary.  0 ,  73 . Bibliography  [1]  L . E . D i c k s o n , " L i n e a r G r o u p s " , Dover New Y o r k , 1958.  Publications, Inc.,  [2]  A. H u r w i t z , " M a t h e m a t i s c h e Werke", B i r k h a u s e r B a s e l , 1932.  [3]  F. K l e i n ,  [4]  S. L a n g ,  [5]  H. Weber,  and Co.,  " L e c t u r e s on the I c o s a h e d r o n " , Dover P u b l i c a t i o n s , I n c . , New Y o r k , 1956. "Algebra", Addison-Wesley 1967 .  Publishing  Co.,  "Lehrbuch der A l g e b r a " , V o l . 1 1 , C h e l s e a P u b l i s h i n g Co., New Y o r k , 1961.  74.  Postscript to Beth Kitchen's Thesis  by Klaus Hoechsmann  After the completion of this thesis and shortly before i t was  due to be handed i n , we stumbled on a simple proof of the icosahedral  case, which works i n any c h a r a c t e r i s t i c  2, 3, 5.  I t hinges on the easy  arithmetic of what we s h a l l c a l l a f f i n e extensions K = k(y)  and,  k[y]  k[x]  .  In other words, a p a i r  function f i e l d s i s a f f i n e , i f generators y  i s a polynomial i n  x  .  L/K  As above,  x,y k  : L = k(x) , L S K  of r a t i o n a l  can be found such that  w i l l be a l g e b r a i c a l l y closed.  The following lemma i s a synthesis of the lemmas on pages 35 and 53.  Lemma:  L/K  Proof:  (<=)  a f f i n e <=>  some place of  K  i s t o t a l l y ramified i n  Let p , ^ be the places i n question.  L  As i n the lemmas on  pages 35 and 53, we have  /•*\ (*)  N w  where Put F  + a^-jU w  w e L , u e K  x = — , y = — w u of degree  N  N-l + a _ u w N  2  N-2 . + ... + u = 0  are l o c a l uniformizing parameters f o r p , ^ , respectively .  (=>)  From an equation  F(x) = y  , we get an Eisenstein equation (*) f o r  hence t o t a l ramification.  with a polynomial w = — , u, x  =  ~ y  '•>  75.  Remark:  In an affine extension, the s p l i t t i n g of any place of  K  (other than the x^. singled out above) i s r e f l e c t e d completely i n the f a c t o r i z a t i o n of i t s uniformizing parameter  y - c = F(x) - c  in  k[x]  For instance, i f there i s another t o t a l l y ramified place, we obtain a pure equation  y - c = (x - b)N  Our derivation of the icosahedral equation w i l l follow the same l i n e s . To prove the conjugacy of two icosahedral groups, we choose a subgroup of index 5 i n each of them.  These are conjugate; making them  equal by a suitable inner automorphism, we now have two icosahedral groups which intersect i n a tetrahedral group fields  K^,  K  2  such that  [L : K^] = 5  see that ramification i n each  L/K;  hence a f i e l d  , ( i = 1,2)  .  L  containing  I t i s easy to  must necessarily be as follows:  r  We choose the generator x iR' ^ f l ' ^1' ^2  3 r e  x' ' X  X  of ~  a >  X  By our lemma and i t s proof, each us i n s i s t that  F(x)  L  so that the uniformizing parameters of  ~ ^ '  r e s  P  e c t  *  v e x  y»  w  n  e  r  K^ i s generated by a  be monic and that  y  e  a + 3 = -4 y = F(x)  . Let  be a l o c a l parameter f o r the  76.  place  JO* . We must show that  F(x)  i s uniquely determined by these  data. Ramif i f i c a t i o n of  and  gives us  (1)  3 2 y = x (x + ax + b)  (2)  2 2 y - 6 = (x-a) (x-g) (x-y)  and respectively.  5 4 3 Accordingly, F(x) = x + ax + 6x . We s h a l l see that a = 5 , b = 40 2 2 Obviously, x (x-a)(x-3) must divide the derivative F'(x) = (5x + 4ax + 2 3b)x  , whence  (2')  (x-a)(x-6)  i s determined.  Thus  y - c = ( x + -jax + -|b) (x-y) 2  2  Comparing the l i n e a r and quadratic terms of (2') to those of (1) y i e l d s  (i)  (ii)  3b - 8ya = 0 2 12ab - 15yb - 8ya = 0  ,  , whence  3a = 5y  8 2 Therefore ( i ) becomes:  b = —a  4 - —a  and  , we have  a = 5  .  b = 40  Since we had normalized .  -4 = a + 3  Quod erat demonstrandum.  

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