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 BETH KITCHEN B.Sc. , U n i v e r s i t y of B r i t i s h Columbia, 1969 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE i n the Department of Ma thema t i c s We accept t h i s t h e s i s as conforming to the r e q u i r e d standard THE UNIVERSITY OF BRITISH COLUMBIA May, 1972 In presenting t h i s thesis i n p a r t i a l f u l f i l m e n t of the requirements for an advanced degree at the University of B r i t i s h Columbia, I agree that the Library s h a l l make i t f r e e l y available for reference and study. I further agree that permission fo r extensive copying of t h i s thesis for scholarly purposes may be granted by the Head of my Department or by h i s representatives. It i s understood that copying or publication of t h i s thesis for f i n a n c i a l gain s h a l l not be allowed without my written permission. Department of The University of B r i t i s h Columbia Vancouver 8, Canada Date O^t^C /3 j /9?2~ S u p e r v i s o r : Dr. K. Hoechsmann ABSTRACT In t h i s t h e s i s we c o n s i d e r the group of f r a c t i o n a l l i n e a r t r a n s f o r m a t i o n s of a v a r i a b l e x over an a l g e b r a i c a l l y c l o s e d f i e l d k. The purpose of the t h e s i s i s to determine a l l f i n i t e subgroups of t h i s group whose orders are not d i v i s i b l e by the c h a r a c t e r i s t i c of k. ACKNOWLEDGMENTS I would l i k e to express my g r a t i t u d e to Dr. Hoechsmann f o r h i s guidance and encouragement throughout the p r e p a r a t i o n of t h i s t h e s i s . Dr. Hoechsmann's enthusiasm f o r the s u b j e c t made the task e n j o y a b l e . I would a l s o l i k e to thank Dr. Gamst f o r reading the t h e s i s and the N a t i o n a l Research C o u n c i l and the U n i v e r s i t y of B r i t i s h Columbia f o r f i n a n c i a l ass i s tance . TABLE OF CONTENTS Chapter I I n t r o d u c t i o n 1 Chapter I I Alg e b r a P r e l i m i n a r i e s 9 Chapter I I I D i f f e r e n t i a l A l g e b r a 13 Chapter IV The Schwarzian D e r i v a t i v e 22 Chapter V R a m i f i c a t i o n Theory 26 Chapter VI The Hurwitz Formula 45 Chapter VII Consequences of the Hurwitz Formula 50 Chapter VIII Proof of the Main Theorem 60 B i b l i o g r a p h y 73 CHAPTER I INTRODUCTION The i n t e n t i o n of t h i s t h e s i s i s to give an up to date v e r s i o n of part of Klein's treatise [3], and to extend Klein's r e s u l t from the complexes to any a l g e b r a i c a l l y c l o s e d f i e l d . We c o n s i d e r the group $J of f r a c t i o n a l l i n e a r t r a n s f o r m -a t i o n s of a v a r i a b l e x ax + b x 1—> — J cx + d where the elements a, b, c, d belong to an a l g e b r a i c a l l y c l o s e d f i e l d k. That i s , P S L 2 ( k ) , the i n v e r t i b l e 2 X 2 ma t r i c e s over k modulo t h e i r c e n t r e . Our aim i s to d e t e r -mine a l l f i n i t e subgroups of & having order not d i v i s i b l e by the c h a r a c t e r i s t i c of k. The f o l l o w i n g o b s e r v a t i o n i n the case where k = <C serves to motivate our r e s u l t i n the ge n e r a l case: For k = C, ^ i s isomorphic to the group of a n a l y t i c automorphisms of the Riemann sphere. We know that t h i s group has as f i n i t e subgroups the c y c l i c groups, the d i h e d r a l groups, the t e t r a h e d r a l group, the o c t a h e d r a l group and the i c o s a h e d r a l group. I t turns out that these groups a l s o e x i s t as f i n i t e subgroups of xb f o r general k, i f t h e i r order i s r e l a t i v e l y prime to the c h a r a c t e r i s t i c of k. In f a c t , we s h a l l show that these are e s s e n t i a l l y a l l such f i n i t e subgroups of . 2 . As evidence of the e x i s t e n c e of these c a n o n i c a l subgroups of is we give t h e i r g e n e r a t o r s . I f the order of a subgroup i s N and the c h a r a c t e r i s t i c of k i s p, assume that P X N. For the c y c l i c and d i h e d r a l groups i t i s c l e a r what the generators are. I The C y c l i c Group of order N i s generated by where e i s a p r i m i t i v e N root of u n i t y i n k. II The D i h e d r a l Group of order N = 2n i s generated by ( o i ) a n d ( l o) where e i s a p r i m i t i v e n*"*1 root of u n i t y i n k. For the three remaining c a n o n i c a l groups we d e r i v e the generators by c o n s i d e r i n g k = C . However, using w e l l -known p r o p e r t i e s of the groups i t can e a s i l y be shown that these are a l s o the generators f o r k any a l g e b r a i c a l l y c l o s e d f i e l d . Let G denote one of the three groups. View the f i g u r e F corresponding to G i . e . the t e t r a h e d r o n , o c t a -hedron or icosahedron, as embedded i n the Riemann sphere. Then elements of G are r o t a t i o n s of the sphere which t r a n s -form F i n t o i t s e l f . Every element of G must then be one of 3. the f o l l o w i n g : a r o t a t i o n R of order i = 3, 4 or 5 about a v e r t e x of F ( i corresponding to the number of faces meeting at any v e r t e x i n the t e t r a h e d r o n , octahedron or i c o s a h e d r o n ) , or a r o t a t i o n S of order 2 about the a x i s through the midpoint of an edge, or a r o t a t i o n T of order 3 about the a x i s through the midpoint of a f a c e . Since elements of each.of these three types are conjugate, to gen-e r a t e G i t s u f f i c e s to give one element of each type. ^ ^ We now g i v e the generators R, S and T f o r each group. A rough 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 r e l a t e d f i g u r e i s shown as an a i d to the reader i n t e r e s t e d i n checking the ge n e r a t o r s . I l l The T e t r a h e d r a l Group Let OJ be a p r i m i t i v e 3 rd root of uni t y . S t e r e o g r a p h i c P roj e ct i o n o f Te t rahedron R 1 1 (1) G can i n f a c t always be generated by two elements. See, f o r example §§71-74[5]. V The I c o s a h e d r a l Group Let e be a p r i m i t i v e 5th root of u n i t y - -, -3 + /5 Note: e + e - 1 = ~ S t e r e o g r a p h i c P r o j e ct ion of Icosahedron -3+/5~ -1 -1 With e x i s t e n c e of these subgroups of & accepted, we can now s t a t e our p r o p o s a l . TV) I t turns out that on the s t e r e o g r a p h i c p r o j e c t i o n of the icosahedron, r = ~ { z J ^ ) ^ t h e d i s t a n c e from the o r i g i n of the f i v e outer icosahedron v e r t i c e s . MAIN THEOREM Let k be an a l g e b r a i c a l l y c l o s e d f i e l d of c h a r a c t e r -i s t i c p. Any f i n i t e subgroup G of PSL2(k) f o r which p )( |G| i s isomorphic to one of the c a n o n i c a l groups I-V. In f a c t , i f G i s isomorphic to one of I-IV then G i s a c t u a l l y con-j u g a t e to that group. F u r t h e r , i n c h a r a c t e r i s t i c 0 a l l f i n i t e subgroups are conjugate to one of I-V. This r e s u l t immediately gives the f o l l o w i n g s p e c i a l -i z a t i o n . Let F be a f i n i t e s u b f i e l d of k c o n t a i n i n g p n elements Any subgroup of PSL^CF) whose order i s not d i v i s i b l e by p i s e i t h e r conjugate over k to one of I-IV or i s isomorphic to V. This r e s u l t i s s t a t e d i n a s t r o n g e r form i n Dickson [1] ( §256). Dickson's treatment of t h i s matter deals with a l l f i n i t e subgroups of PSL^CF), not only those with order prime to p. Our reasons f o r a v o i d i n g the most general case w i l l become c l e a r as our technique f o r proof develops. Note here that our methods w i l l be designed so as to handle any a l g e b r a i c a l l y c l o s e d f i e l d k, not f i n i t e f i e l d s . We now o u t l i n e our p l a n f o r a c c o m p l i s h i n g the p r o o f . Our f i r s t step i s to observe that the group of f r a c t i o n a l l i n e a r t r a n s f o r m a t i o n s of x over k i s isomorphic to A u t ^ k ( x ) , 7 . the group of automorphisms of k(x) over k. We e s t a b l i s h t h i s f a c t now and then choose to work with Aut^k(x) t h e r e a f t e r . Theorem k(x) = k(y) <=> y = *x t d f o r some elements a, b, c, d E k such that ad-bc 4 0. Proof ( <f ) k(y) c k(x) i s c l e a r . To show k(x) c k(y) we 3 X "fr* b must determine x i n terms of y. Rewrite y = ;—T i n ' cx + d /y\ /a b \ / x V matrix form as = I I I I * 1 \c d / 1/ /x\ la b \ - l From t h i s we have = -1 \ l j U d| \ l / a d ~ b c \ c -a la b\ where t h i s i s p o s s i b l e s i n c e det 1 = ad - be f 0. Hence -d + b \c d/ x = ^ , and t h e r e f o r e k(x) k ( y ) . cy - a 4 ) k(x) = k(y) i m p l i e s , i n p a r t i c u l a r , that y e k ( x ) . A fx) So y = -z-j—r f o r some A ( x ) , B(x) e k [ x ] , which we may assume to be r e l a t i v e l y prime. Let I(X) = B(X)y - A(X). I i s a non-zero polynomial s a t i s f i e d by x over k ( y ) . F u r t h e r , I i s i r r e d u c i b l e over k ( y ) : otherwise, i t would be p o s s i b l e to f a c t o r an i r r e d u c i b l e polynomial from each of the r e l a t i v e l y prime polynomials 8 . A(X) and B(X). Thus I(X) = I r r ( x , k ( y ) , X). Hence, by f i e l d theory, [k(x) : k ( y ) ] = deg I ( X ) . But [k(x) : k ( y ) ] = 1, and deg I(X) = max (deg A(X), deg B( X ) ) . Thus A(X) = aX + b and B(X) = cX + d f o r some elements a, b, c, d e k f o r which ad - be j> 0. ( I f ad - be = 0, (a b) and (c d) would be k - m u l t i p l e s of each other and y would t h e r e f o r e be i n k.) Hence, we have A(x) _ ax + b  y ~ B(x) cx + d' with a, b, c, d as s t a t e d . That t h i s e s t a b l i s h e s the d e s i r e d r e s u l t i s c l e a r . For, any element at Aut^k(x) i s such that o ( k ( x ) ) = k( a ( x ) ) = k ( y ) , say; and any element of the group of f r a c t i o n a l l i n e a r t r a n s f o r m a t i o n s of x over k ac t s by mapping x i - • " j " — j - = y, say, f o r some elements a, b, c, d E k such that ad - be 4 0 . In c o n s i d e r i n g f i n i t e subgroups of Aut^k(x) one n a t u r a l l y thinks of using G a l o i s theory and f i e l d theory. With these a l g e b r a i c techniques and the Hurwitz Formula (a r e s u l t dependent on d i f f e r e n t i a l a l g e b r a and r a m i f i c a t i o n theory f o r i t s d e r i v a t i o n ) , we deduce the f i r s t two con-c l u s i o n s of the Main Theorem, For the f i n a l c o n c l u s i o n i n c h a r a c t e r i s t i c 0, we invoke as w e l l the s o - c a l l e d Schwarzian (3) d e r i v a t i v e , an operator i n the theory of d i f f e r e n t i a l a l g e b r a . (3) Our use of the Schwarzian d e r i v a t i v e i n t h i s context f o l l -ows K l e i n [ 3 ] . Here he uses the Schwarzian i n f i n d i n g the f i n i t e subgroups of P S L ^ ^ ) . CHAPTER I I ALGEBRA PRELIMINARIES T e r m i n o l o g y : By an a l g e b r a i c f u n c t i o n f i e l d o v e r k we s h a l l mean a f i n i t e l y g e n e r a t e d e x t e n s i o n f i e l d o f k o f t r a n s c e n d e n c e d e g r e e 1. A r a t i o n a l f u n c t i o n f i e l d i s t h e n an a l g e b r a i c f u n c t i o n f i e l d h a v i n g o n l y one g e n e r a t o r . F o r use i n l a t e r s e c t i o n s we s h a l l need t h r e e s p e c i a l r e s u l t s f r o m the g e n e r a l t h e o r y o f a l g e b r a . Theorem 1 L e t x be a t r a n s c e n d e n t a l e l e m e n t o v e r a f i e l d k. L e t G be a f i n i t e s ubgroup o f A u t ^ k ( x ) . Then t h e r e e x i s t s an ele m e n t y t r a n s c e n d e n t a l o v e r k s u c h t h a t F i x G = k ( y ) N o t e : T h i s i s a s p e c i a l c a s e o f L U r o t h ' s Theorem w h i c h makes the same c l a i m about any s u b f i e l d o f k ( x ) . P r o o f L e t G = { o , , a _ , . . . , a } ^ . A u t . k ( x ) where n = I GI . I z n K D e f i n e y = a ^ ( x ) a ^  (x) • • • o"n (x) . C l e a r l y y i s t r a n s c e n d e n t a l o v e r k. We w i l l show t h a t y e F i x G and t h a t , i n f a c t , k ( y ) = F i x G. To show t h a t y e F i x G, we must show t h a t c ^ ( y ) = y f o r a l l i = l , . . . , n . F i x some i , 1 < i < n. Then 10 . (1) a .Cy) = a . ( a 1 ( x ) . . -a^x) ) = ( a ± a 1 ) ( x ) . . . ( a i a r ) ( X ) C l e a r l y , a . •a . e G f o r i = l , . . . , n . F u r t h e r , i f a. ^ a, 1 j j k then a.°a. ^  a.oa, . For, a . ° a . = a.«a. i m p l i e s that l j x k l j I k r a.  Lc a .» a . = a. \ a . • a, , so that a. = a, . l i j x i k j k Hence G = {a . •a,, . . • ,a . o a }• Thus from (1) we see that l 1 x n o^(y) = y and y i s indeed i n F i x G. We have now k(y) £ F i x G £ k ( x ) . Hence (2) [ k ( x ) : F i x G ] [ F i x G:k(y)] = [ k ( x ) : k ( y ) ] By G a l o i s theory we know that [ k ( x ) : F i x GJ = |G| = n. From (2) then we have ( 3 ) [ k ( x ) : k ( y ) ] > n We complete the proof by showing that [ k ( x ) : k ( y ) j = n. This f i n i s h e s the proof s i n c e from (2) i t then f o l l o w s that |Fix G:k(y)| = 1, and so, that F i x G = k ( y ) . In the I n t r o d u c t i o n we e s t a b l i s h e d that Aut^k(x) i s isomorphic to the group of f r a c t i o n a l l i n e a r t r a n s f o r -mations. Thus each a.(x) i s of the form a i X + ^ i , c x : — j — f o r c . x + d . x x some a., b., c., d . e k such that a.d. - b.c. 4 0. 1 ' 1 ' 1 1 X I 1 1 n a ^ x + b ^ fCx) Then y = II ; ; — = —T^-V' f o r some r e l a t i v e l y prime i = l C i X i g ( x ) 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 k(y) and F(x) = 0. i . e . F(X) = I r r ( x , k ( y ) , X). Hence [k ( x ) : k ( y ) ] = deg F(X) = max(deg f, deg g) < n. Together with (3) t h i s g i v e s the r e q u i r e d r e s u l t . Theorem 2 E x i s t e n c e of a Separating T r a n s c e n d e n t a l Let 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 over k, where k i s of c h a r a c t e r i s t i c p. Then there e x i s t s a t r a n s c e n d e n t a l x e K such 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. Thus we have K 3 k(x) r> k and K/k(x) a l g e b r a i c . Suppose K = k(x, x„ , . . . , x ) = k (x, , x O J...,x ), where the x 0,...,x ' 2 n 1 2 n 2 n are a l g e b r a i c over k, and x = x^. Use i n d u c t i o n on n to show ex i s t e n c e 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 n = 1 then K = k ( x ) , and t h i s i s c l e a r l y s eparable over k(x ). Assume that k(x^»•••> x n_^) z s separable over k ( x ^ ) . Since i s a l g e b r a i c over k ( x ^ ) , there e x i s t s an i r r e d u c i b l e polynomial f such that f ( x , , x ) = 0, where the c o e f f i c i e n t s I n of f are i n k. If x i s separable over k ( x j take the sep-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 that x i s not ° I n separable over k ( x ^ ) . Then f must be a polynomial i n x n ^ • If x, i s i n s e p a r a b l e over k(x ) then f i s a l s o a polynomial 1 n i n x A Thus 12 . f ( x 1 , x n ) = g ( x 1 P , x n P ) = [ g ( x 1 , x n ) ] P , where g i s some polynomial with c o e f f i c i e n t s i n k. But t h i s i m p l i e s that f i s r e d u c i b l e . C o n t r a d i c t i o n . Hence x, must be separable over k(x ). 1 n Thus we have ^ separable over k(x^) and x, separable over k(x ). So x. x , are a l l s e p a r a b l e 1 r n 1 n-1 over k(x ). T h e r e f o r e , take x to be the s e p a r a t i n g t r a n s -n n cendental . The f o l l o w i n g i s a standard theorem of a l g e b r a , a proof of which can be found on p. 171 [4] . Theorem 3 Let A be an a l g e b r a i c a l l y c l o s e d e x t e n s i o n of a f i e l d k, K an a l g e b r a i c extension of k. I f p : k «—»- A i s a monomorphism then p can be extended to a monomorphism of K i n t o A. 13. CHAPTER I I I DIFFERENTIAL ALGEBRA In the chapter on the Schwarzian d e r i v a t i v e we w i l l d e a l with d e r i v a t i o n s and w i l l need one g e n e r a l r e s u l t concerning them. This r e s u l t i s found i n the lemma of t h i s s e c t i o n . The theorem f o l l o w i n g the lemma s t a t e s that under s u i t a b l e c o n d i t i o n s d e r i v a t i o n s can be extended. This i s used i n the proof of the Main Theorem. The l a t t e r p a r t of t h i s chapter deals with d i f f e r e n t i a l s and r e l a t e d r e s u l t s which are necessary f o r the development of the Hurwitz Formula. (1) D e r i v a t i o n s Let k be a commutative r i n g , K a commutative k - a l g e b r a and V a K-module. Then a k - l i n e a r map D:K —»• V such that D(fg) = fDg + gDf f o r any elements f, g e K i s c a l l e d a k - d e r i v a t i o n . Note: This d e f i n i t i o n w i l l be used i n the case where k and K are both f i e l d s . Assume i n the f o l l o w i n g then that we have: k a f i e l d , K an e x t e n s i o n f i e l d of k, D:K —*• V a k - d e r i v a t i o n of K i n t o a K-module V. We d e f i n e the f i e l d o f cons tants f o r D to be K D = {f e K|Df = 0}. N o t i c e that k £ K^. For, i f a e k, D(a) = D(a'l) = a«D(l) = a«0 = 0. But i t i s not n e c e s s a r i l y true that K Q £ k. For, i f char k = p and f e K then f P e K D s i n c e D ( f P ) = p f p - 1 D f = 0. But f P need not be i n k. Also n o t i c e that D i s l i n e a r over K Q. For, i f f e K^ and g e K then D(fg) = fDg + gDf = fDg. Note: For f e K ( x ) , Df = D f*Dx where D denotes "formal 14 . d i f f e r e n t i a t i o n with re s p e c t to x". Denote by D X ( f ) , f o r f e K, the element D ( D ( . . . ( D / ( f ) ) . . . ) ) . Then we have the f o l l o w i n g i Lemma Let K be a f i e l d e x t e n s i o n of k, and l e t D:K —• K be a k - d e r i v a t i o n with f i e l d of constants . The elements ^1' ^2'**"'^n E ^ a r e  L i n e a T ± y dependent over i f and only i f det ( D 1 f j ) = 0, where i = 0, 1,..., n - l and j = 1, 2,..., n. Proof Assume ^ ' ' " ' ' ^ n a r e l i n e a * " l y dependent over . Then, without l o s s of g e n e r a l i t y , there e x i s t • • • e such that f, = X „ f „ +... + X f . 1 2 2 n n Since D i s l i n e a r over i t f o l l o w s that D 1 f , = X„D if„ + ... + X D X f 1 Z z n n f o r a l l i = l , . . . , n - l . The columns of the matrix ( D 1 f j ) are t h e r e f o r e l i n e a r l y dependent. Hence det ( D 1 f j ) = 0. Now assume that det ( D 1 f . ) = 0 and use i n d u c t i o n to J show that the f. must be l i n e a r l y dependent over K„. J D For n = 2, we have f ^ , f2 £ K. Assume both to be nonzero s i n c e otherwise they are l i n e a r l y dependent. Hence we have f l f2 D f l ° f2 = 0 This i m p l i e s that the columns are l i n e a r l y dependent over K, 15. Thus there e x i s t s A e K such that (1) f± = xf2 (2) D f x = A Df2 C l e a r l y we w i l l be done i f we can show A e . A p p l y i n g D to (1) we have D f 1 = A Df 2 + f 2 n A . Together with (2) t h i s y i e l d s f^DA = 0. But 4 0, so that DA must be 0. Hence A e . Assume now that det ( D X f j ) = 0 f o r j < n-1 i m p l i e s that the f are l i n e a r l y dependent. We w i l l show t h i s r e s u l t true f o r j = n. Let f , . . . , f n e K and assume none i s 0. Then assuming det ( D 1 f . ) = 0, there e x i s t A „,..., A £ K such that j ' 2 n (1) fx = A 2 f 2 + ... + A n f n (2) »f± = A 2 D f 2 + ... + A n D f n (n) D n 1 f 1 = A „ D n 1 f . + ... + A D n _ 1 f 1 2. I n n As f o r the case n = 2 we show that the A^ e . A p p l y i n g D to (1) gives Df, = (DA - f . + ... + DA -f ) + (A„Df +...+ A Df ) 1 2 1 n n 1 2 n n S u b t r a c t i n g (2) from t h i s gives (1 ' ) DA • f + ... + DA • f = 0 2 2 n n S i m i l a r l y , a p p l y i n g D to (2) we have D 2 f , = (DA 0•Df„ +...+ DA «Df ) + (A„D 2f +...+ A D 2 f ) 1 2 2 n n 2 2 n n S u b t r a c t i n g (3) y i e l d s (2') DA •Df +...+ DA -Df = 0 2. 2 n n 16 Repeating t h i s o p e r a t i o n n - l times i n a l l we o b t a i n (1') DX 0-f„ +...+ DX -f = 0 2 2 n n (2 1) DA 0-Df 0 +...+ DX 'Df = 0 2 2 n n ( n - l ' ) DA„-D n 2 f „ +...+ DX •Dn 2 f = 0 2 2 n n T h i s system can be w r i t t e n as f. ... (DX . DX ) n Df. Df n-2 * \ D f. n / 0 \0 I f det (D f j ) = 0 f o r j = 2,...,n then by the i n d u c t i o n assumption we are done. Assuming t h e r e f o r e that det ( D 1 f j ) f 0 f o r j = 2,...,n, we f i n d from the above system that (DA 2,..•,DA n) = 0. But then DA ± = 0 f o r i = 2,...,n, so that each A. e . Thus f . , . . . , f are l i n e a r l y dependent l D I n J r over , as d e s i r e d . Theorem Ex t e n s i o n of a D e r i v a t i o n I f a f i n i t e l y generated e x t e n s i o n K(x) over K i s s e p a r a b l y a l g e b r a i c then any d e r i v a t i o n D on K can be u n i q u e l y extended to K ( x ) . Proof Since D i s d e f i n e d on K, to extend D to K(x) we need only d e f i n e D(x). Let f(X) = I r r ( x , K, X) = a n X n +...+ + a Q , f o r a. e K. Then f ( x ) = a x° +...+ a,x + a_ = 0, and i n 1 0 a p p l y i n g any d e r i v a t i o n d to t h i s gives 17 . [ d ( a ) x n + d(a ^ x 1 " " 1 +...+ d ( a n ) ] + [ (na x 1 1" 1 +...+ a n ) d ( x ) ] = 0 v n H z i ^ 0/ . n _ ° , = f d ( x ) = f ' ( x ) Thus we must d e f i n e Dx by f d ( x ) Thi s d e f i n e s the only p o s s i b l e e x t e n s i o n of D to K ( x ) . Note: d i v i s i o n by f 1 ( x ) i n the d e f i n i t i o n of Dx makes sense s i n c e f'(x) ^  0 f o r x separable over K. ( 2 ) Dif f e r e n t i a l s Let 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 over k. Define D:K — > • ^ j r/k. t 0 ^ e t n e 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 sense: For any k - d e r i v a t i o n D:K >W, W a K-module, there e x i s t s a unique K-module map u: ^K/k *^ such that the f o l l o w i n g diagram commutes K y | JJ, i i 4-W Once we have e x i s t e n c e of such a u n i v e r s a l p a i r (d, being d e f i n e d by the u n i v e r s a l mapping p r o p e r t y , i t must be unique up to isomorphism. E x i s t e n c e of (d, ^J^/J^) : Take x to 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/k. (Such x e x i s t s by Theorem 2 , Chapter I I ) . Thus K = k(x, 9 ) f o r some separable element 9 e K. Now i n k ( x ) , formal d i f f e r e n t i a t i o n with r e s p e c t to x d e f i n e s a d e r i v a t i o n d 18. on k ( x ) . T h e n , K b e i n g s e p a r a b l e o v e r k ( x ) , c a n b e e x t e n d e d t o g i v e a d e r i v a t i o n d o n K. T a k e Q = Kde. K/k G i v e n D : K — y W a s a b o v e , d e f i n e u : ^ ^ / k = K < ^ — * * w by y ( f « d e ) = f • D O , f o r f e K. D e n o t i n g b y D Q f o r m a l 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 t o 6, we t h e n h a v e f o r f e K, ( p o d ) ( f ) = u ( d f ) = u ( D „ f « d e ) = D f • DG , w h i l e D f = D o f • D8 . H e n c e u » d = D, a s d e s i r e d . T h e K - m o d u l e ^ j ^ ^ i s k n o w n a s t h e m o d u l e o f d i f f e r e n t i a l s f o r t h e e x t e n s i o n K / k . We now d e f i n e w h a t we s h a l l mean b y a c o n t i n u o u s d i f f e r e n t i a l . I n p a r t i c u l a r , we s h a l l b e i n t e r e s t e d i n c o n t i n u o u s d i f f e r e n t i a t i o n i n a f i e l d o f f o r m a l p o w e r s e r i e s L e t K = k ( ( t ) ) b e a f i e l d o f f o r m a l p o w e r s e r i e s . K h a s a v a l u a t i o n " o r d " d e f i n e d o n i t b y o r d ( f ) = o r d e r o f f a s a f u n c t i o n o f t , f o r a n y f e K. T h i s v a l u a t i o n o r d d e f i n e s a m e t r i c o n K i n t h e u s u a l way. A t o p o l o g i c a l K - s p a c e i s a v e c t o r s p a c e W w i t h t o p -o l o g y i n w h i c h a d d i t i o n ( i ) W X W —*• W a n d s c a l a r m u l t i p l i c a t i o n ( i i ) K X W —>• W a r e c o n t i n u o u s . c D e f i n e d:K —>• fi„to be t h e u n i v e r s a l c o n t i n u o u s K/k 19 . d e r i v a t i o n i n the f o l l o w i n g sense: For any continuous k - d e r i v a t i o n D:K —»• W, W a t o p o l o g i c a l K-space, there e x i s t s a unique K-module map u such that the f o l l o w i n g diagram i s commutative d For any f i e l d of formal power s e r i e s K = k ( ( t ) ) c we w i l l p r e s e n t l y show that such a u n i v e r s a l p a i r (d, e x i s t s ; again, once we have e x i s t e n c e , uniqueness i s automatic E x i s t e n c e of (d, : Set ^^/y, ~ ^ ^ t ' w ^ e r e dt ^ s the generator of t h i s 1-dimensional module. Define d:K = k ( ( t ) ) —>• ^ K y k by d : t t—* dt. N o t i c e that Df = f'*Dt f o r any f e K where f' denotes the formal d e r i v a t i v e of f with r e s p e c t to t, and D:K —• W i s as above. Proof: Given f e K, by c o n t i n u i t y we can approximate f by f u n c t i o n s f e k f t , t X ] . Then, s i n c e f o r the f we J n n know that Df = f ''Dt, n n Df = l i m Df = l i m (f 1•Dt) n n n->-oo n-*-"1 But by ( i i ) of the t o p o l o g i c a l K-space d e f i n i t i o n l i m (f '-Dt) = ( l i m f 1 ) ' D t n n n-*-°° n->°° 20 . Hence Df = ( l i m f ')«Dt = f ' D t . ^ n n - > o o Then the u n i v e r s a l mapping pro p e r t y f o l l o w s e a s i l y . Define y: ^ £ / k = Kdt —> W by y: f ' d t —> f'Dt, f o r f e K. Then Df = f ' D t and ( y . d ) ( f ) = y ( d f ) = y ( f ' d t ) = f ' D t . (We have df = f '«dt, as f o r D, s i n c e d i s continuous.) That i s , D = y 0 d , as r e q u i r e d . We c a l l ^he module of continuous d i f f e r e n t i a l s of K/k. Next we d e f i n e the order of a continuous d i f f e r e n t i a l , c c Let co e Since ~ k ( ( t ) ) d t , we can w r i t e to as f«dt f o r some element f e k ( ( t ) ) . Then d e f i n e the order of co to be the order of the f u n c t i o n f . Write ord(co) = o r d ( f ) Remark: Let t ' be a power s e r i e s i n t of order 1. Then any f e K = k ( ( t ) ) can be r e w r i t t e n as a power s e r i e s i n t ' . So K = k ( ( t ' ) ) . For ord(co) to be w e l l - d e f i n e d , i t must be independent of the generator chosen f o r K. We must t h e r e f o r e show that the f o l l o w i n g h o l d s . Let t , t ' be generators of K oyer k. For w e ^K/k we have to = f'dt and to = g'dt' where f e k ( ( t ) ) and g e k ( ( t ' ) ) . Then o r d ( f ) = ord(g) Pro o f : t' e K = k ( ( t ) ) . Hence we can w r i t e t ' = u ( t ) ' t f o r u ( t) a u n i t i n k ( ( t ) ) . Then, l e t t i n g D t denote formal 21. d i f f e r e n t i a t i o n with r e s p e c t to t, we have dt' = d(u(t))«t + u ( t ) - d t = D ( u ( t ) ) ' d t ' t + u ( t ) ' d t = ( D t ( u ( t ) ) - t + u ( t ) ) - d t = u Q ( t ) - d t , say. I t i s easy to see that U g ( t ) i s a u n i t i n k ( ( t ) ) . For, u(t) a u n i t i m p l i e s ord ( u ( t ) ) = 0 and ord ( D t u ( t ) ) ) > 0. Thus ord (D(u(t))«t) > 1, so that ord (u ( t ) ) = ord ( D ( u ( t ) ) - t + u ( t ) ) = 0. We now have w = g'dt' - (g • ( t ) ) •dt. But a l s o co = f - d t . Hence f = g « U Q ( t ) . T h e r e f o r e ord ( f ) = ord (g'uCt)) = ord (g) + ord ( u 0Ct)) = ord (g) + 0 => ord (g) 22 . CHAPTER IV THE SCHWARZIAN DERIVATIVE D e f i n i t i o n of the Schwarzian D e r i v a t i v e : Let D:E —*• E be a d e r i v a t i o n on a f i e l d E, with f i e l d of constants k. For h e E-k d e f i n e the S chwarzian  D e r i v a t i v e of h with r e s p e c t to D to be where h' = Dh, h 1 ' = D 2h and h ' 1 ' = D 3h. Note: Since h e E-k, h' i s non-zero and t h e r e f o r e [h] makes sense. Theorem Let E be a f i e l d on which there i s d e f i n e d a d e r i v a t i o n D:E —*• E. Let k be the f i e l d of constants f o r D. Then f o r f , g e E-k there e x i s t a, b, c, d e k with ;id-bc ^ 0 such that f = tf-H<=> I f ] D - I*] D Proof Assuming f = a ^ "j" ^  f o r some a, b, c, d e k such eg + a that ad-bc f 0 , we see that t h i s i s e q u i v a l e n t to cfg - ag + df - b = 0 (1) A p p l y i n g D to t h i s g i v e s the equation c ( f g + fg') - ag' + d f = 0 (2) T h i s says that = f g + f g ' , h 2 = -g ' and h 3 = f are l i n e a r l y dependent over k. By the lemma on the Wronskian (1) ad-bc ^ 0 s i n c e otherwise f would be i n k, and so [ f ] P would not be d e f i n e d . 23 ( c f . §1, Chapter I I I ) we know that t h i s i s e q u i v a l e n t to det (D xh,) being 0. Hence f'g + f g ' -g f*'g + 2 f V + f g ' * -g f • ' » g + 3f ' V + 3 f ' g " + f g ' * ' -g Adding c o l 1 plus f ( c o l 2) plus - g ( c o l 3) gives 0 -g* f i ™ > t i f ' ' det (D h ) = f ' f • ' f ' * * 2f 'g •g "g = 0 3f " g * + 3f ' g " Or, expanding the determinant, -2f V (-g'f ' " + f ' g " ') + (3f ' 'g'+3f 'g' ') (-f » *g' + f ' g") = 0 M u l t i p l y i n g t h i s out we have 2 f ' f " ' ( g ' ) 2 - 2 ( f * ) 2 g ' g ' ' ' - 3 ( f ' ' ) 2 ( g ' ) 2 - 3 f ' f " g ' g " + 3 f ' f " g ' g " + 3 ( f ' ) 2 ( g " ) 2 = 0 S i m p l i f y i n g t h i s g i v e s (f ' ) 2 [ - 2 g ' g ' • • + 3 ( g " ) 2 ] + ( g ' ) 2 [ 2 f ' f " '-3(f ' ' ) 2 ] = 0 2 D i v i d i n g through by ( f ' g ' ) and t r a n s p o s i n g one of the two e x p r e s s i o n s , 2 f f " - 3 ( f ' ' ) 2 ( g ' ) ^ Or, f ' ' ' f ' ' 2 i f ' f ' - 3 That i s , [ f ] D - [ g ] D Remark: Our proof i s a s e r i e s of e q u i v a l ences except at sentences (1) and ( 2 ) . Here, assuming [ f ] D = [g]p and f o l l o w i n g the proof back 24 . to sentence (2) we know that there e x i s t a, c, d e k such that c ( f ' g + fg') - ag' + d f = 0 R e c a l l that t h i s simply means that D(cfg - ag + df) = 0 So, by d e f i n i t i o n of k, c f g - ag + df = b f o r some element b e k. Thus £ _ ag + b f o r some a, b, c, d e k eg + d such that ad-bc f 0. The c o n c l u s i o n of t h i s theorem can be r e s t a t e d i n a more u s e f u l form as C o r o l l a r y 1 k( f ) = k(g) [ f ] D = [ g ] n Proof This statement f o l l o w s d i r e c t l y from the above theorem and the theorem near the end of Chapter I. Remark: R e c a l l that we are going to need the Schwarzian f o r p r o v i n g the Main Theorem i n c h a r a c t e r i s t i c 0. What enables us to make use of the pre c e d i n g theorem on the Schwarzian 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 f a c t : I f the f i e l d E of the theorem i s an a l g e b r a i c f u n c t i o n f i e l d over an a l g e b r a i c a l l y c l o s e d f i e l d kg of c h a r a c t e r i s t i c 0, then the f i e l d of D-constants i s k^, Th i s holds s i n c e i n c h a r a c t e r i s t i c 0 any t r a n s c e n -d e n t a l i s s e p a r a t i n g ; i f D i s zero on a s e p a r a t i n g t r a n s -cendental then D can only be the t r i v i a l d e r i v a t i o n . 25 . For non-zero c h a r a c t e r i s t i c however, the D-constants may form a n o n - t r i v i a l e x t e n s i o n of kg. T h i s i s the f a c t which makes the Schwarzian of no use to us i n the gen e r a l case. The t r i v i a l p a r t of the preceding theorem y i e l d s the next r e s u l t . C o r o l l a r y 2 Let H be a f i n i t e group of automorphisms of k(x) where x i s an in d e t e r m i n a t e over k. Let F be the f i x e d f i e l d of H i n k ( x ) . Suppose D i s a d e r i v a t i o n on F with f i e l d of constants k. We know that D extends u n i q u e l y to k(x) ( s i n c e k ( x ) / F i s separable a l g e b r a i c ) . Then the Schwarzian of x l i e s i n F. That i s , [ x ] D e F Proof Let u e H = Gal ( k ( x ) / F ) . To show [ x ] D e F we need only show that y ( [ x ] p ) = [x] n« Since y i s an auto-morphism of k ( x ) , i t f o l l o w s that y ( [ x ] D ) v > ( * > J Y D Now uD:k(x) —»- k(x) i s a d e r i v a t i o n on k ( x ) , and uD| = D. r Thus, the ext e n s i o n of D to k(x) being unique, yD = D. The r e f o r e p ( [ x ] D ) = [ u ( x ) ] D Now y e Gal (k(x)/F) i m p l i e s that k(x) = k ( y ( x ) ) . But by C o r o l l a r y 1 t h i s i s e q u i v a l e n t to [ x ] D = [ y ( x ) ] D Hence M ( [ x ] ) - t x ^ j ) a s r e q u i r e d . 26. CHAPTER V RAMIFICATION THEORY Throughout t h i s chapter assume k to be an a l g e b r a i c -a l l y c l o s e d f i e l d , K to be an a l g e b r a i c f u n c t i o n f i e l d over k By Theorem 2, Chapter II we have e x i s t e n c e 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 i n K/k. Thus we can assume K to be a sep-a r a b l e e x t e n s i o n of a . r a t i o n a l f u n c t i o n f i e l d k ( x ) , f o r some t r a n s c e n d e n t a l x e K. For our purposes the f i e l d k(x) and i t s extensions are the main source of i n t e r e s t . However, where p o s s i b l e without a d d i t i o n a l work we s t a t e the f o l l o w i n g theory i n the g e n e r a l case. (1) P l a c e s The o r i g i n of t h i s theory i s i n the theory of meromorphic f u n c t i o n s on a Riemann s u r f a c e . P l a c e s are i n t r o d u c e d here as the concept which supersedes the n o t i o n of p o i n t on the Riemann s u r f a c e . I t w i l l be convenient i n the main tex t of our argument to be able to view a p l a c e i n a f i e l d as e i t h e r of two n o t i o n s . We d e f i n e each of these here and then note that they are i n f a c t e q u i v a l e n t . One concept of a p l a c e w i l l be as a v a l u a t i o n r i n g . Let & be a domain which i s not a f i e l d . & i s c a l l e d a v a l u a t i o n r i n g i f there i s an i r r e d u c i b l e element T T e O such that every z e & - {0} can be w r i t t e n u n i q u e l y i n the form z = U T T 1 1 , f o r some un i t u e & and some non-negative i n t e g e r n. An element T I as above i s c a l l e d a l o c a l 27 . u n i f o r m i z i n g parameter f o r &; any other l o c a l u n i f o r m i z i n g parameter i s of the form UIT , u a u n i t i n cr. Let L be the q u o t i e n t f i e l d of U. Then, f o r f i x e d l o c a l u n i f o r m i z i n g parameter i r , any z e L - {0} has a unique e x p r e s s i o n z = UT: 1 1, u a u n i t i n & and n e Z . The exponent n i s e a s i l y seen to be independent of the p a r t i c u l a r u n i f o r m i z i n g parameter i r . We c a l l the exponent n the order of z and w r i t e n = ord ( z ) . Define ord (0) = ». Then 0= {z e L|ord (z) * 0} and f = {z e L|ord (z) > 0} i s the unique maximal i d e a l of Our second n o t i o n of a p l a c e w i l l be as a v a l u a t i o n . A v a l u a t i o n on a f i e l d L i s a f u n c t i o n v:L —*• Z U {°°} s a t i s f y i n g ( i ) v(a) = «> i f f a = 0 ( i i ) v(ab) = v(a) + v ( b ) ( i i i ) v(a+b) > Min ( v ( a ) , v ( b ) ) Any such v a l u a t i o n d e f i n e s a v a l u a t i o n r i n g i n L. For, & - iz c L | V ( Z ) > 0} i s a v a l u a t i o n r i n g with maximal i d e a l ^ = {z e L|v(z) > 0} and q u o t i e n t f i e l d L. C o n v e r s e l y , a v a l u a t i o n r i n g i n L d e f i n e s a v a l u a t i o n on L. Namely, i f 0" i s a v a l u a t i o n r i n g with q u o t i e n t f i e l d L, then the f u n c t i o n ord:L —*• Z U {">) i s a v a l u a t i o n on L. Thus g i v i n g a v a l u a t i o n r i n g with q u o t i e n t f i e l d L i s the same as d e f i n i n g a v a l u a t i o n on L. Since these are e q u i v a l e n t n o t i o n s , we w i l l r e f e r to them both as p l a c e s . In f a c t , s i n c e s p e c i f y i n g the maximal i d e a l of a v a l u a t i o n r i n g 28. determines the r i n g , we w i l l normally use the f o l l o w i n g n o t a t i o n : by a p l a c e ~jQ of a f i e l d L we denote the maximal i d e a l of the v a l u a t i o n r i n g a s s o c i a t e d with fl. Denote the v a l u a t i o n r i n g i t s e l f by 6" (or 0^ i f n e c e s s a r y ) , and the a s s o c i a t e d v a l u a t i o n by o r d ^ . For u n i f o r m i t y c a l l Q the p l a c e r i n g of Given two f i e l d s and , L a f i n i t e e x t e n s i o n of , we w i l l want to be able to extend a given p l a c e on to a p l a c e on L^. This process can a c t u a l l y be c a r r i e d out f o r any e x t e n s i o n of L^; the reader i s r e f e r r e d f o r proof of t h i s f a c t to p. 299 [ 4 ] . To denote that a p l a c e 2^ i s an e x t e n s i o n of a placexp, we w r i t e ^|^» read p l i e s over ^ . Consider now the case L = k ( x ) . To see what the p l a c e s are i n t h i s f i e l d we need the f o l l o w i n g d e f i n i t i o n . Say a f u n c t i o n f e k(x) i s r e g u l a r at a p o i n t a e k i f f can be w r i t t e n as ^ f o r some g, h e k[x] such that h(a) ^ 0 . For each a e k, d e f i n e & = {f e k ( x ) | f r e g u l a r at a}. 3. Then i t i s f a i r l y easy to see that &^  i s a v a l u a t i o n r i n g with l o c a l u n i f o r m i z i n g parameter IT = x-a and q u o t i e n t f i e l d k ( x ) . A l s o , & = {— e k(x)|deg g > deg f} i s a v a l u a t i o n r i n g with l o c a l u n i f o r m i z i n g parameter — and q u o t i e n t f i e l d k ( x ) . In f a c t , these r i n g s are the only v a l u a t i o n r i n g s which c o n t a i n k and have q u o t i e n t f i e l d k ( x ) . 29 . So we have the f o l l o w i n g correspondence. {a e k} \J » < > {Q I a e k} U (3 < ) a 1 °° ( V a l u a t i o n r ings & | & O k and q u o t i e n t f i e l d of £> i s k(x)} Any v a l u a t i o n on k(x) i s thus a s s o c i a t e d to a unique r i n g Cf i n k(x) (a i n k U 0 0) . The p l a c e s i n k(x) t h e r e f o r e ct correspond to l o c a l u n i f o r m i z i n g parameters TT = x-a, a e k and TT = We s h a l l 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 22 i f i t i s a parameter f o r the v a l u a t i o n r i n g & d e f i n e d by . The f o l l o w i n g r e s u l t w i l l be u s e f u l l a t e r i n connec t i o n with p l a c e s on k ( x ) . Theorem Any element of k(x) has an equal number of zeros and poles . Proof f (x) Let h(x) = be an a r b i t r a r y element of k(x) where f , g e k [ x ] . Since k i s a l g e b r a i c a l l y c l o s e d we can f a c t o r f and g i n t o l i n e a r f a c t o r s . The zeros of h i n k w i l l then be given by the f a c t o r s of f . The number of zeros i n k, with m u l t i p l i c i t i e s , w i l l t h e r e f o r e be deg f = n, say. S i m i l a r l y , the number of poles of h i n k, with m u l t i p l i c i t i e s , w i l l be deg g = m, say. The behaviour of h at 0 0 balances out the number of zeros and p o l e s . For, i f n = m, h has no pole or zero at » and the proof i s complete. I f n < m, h has a zero of order m-n at 0 0, so that the number of zeros i s n + (m-n) = m. S i m i l a r l y , 30. i f n > m, we f i n d that the number of poles i s m + (n-m) = n. Now we make a sho r t remark about the connection of p l a c e s with i n t e g r a l i t y . Let L be an e x t e n s i o n of K. Suppose ^ i s a p l a c e of K with p l a c e r i n g / C , and "P i s a p l a c e of L l y i n g over ^ . Let & be the p l a c e r i n g of jp . I f a i s an element of L s a t i s f y i n g an i n t e g r a l equation over Q, i . e . an equation of the form m m-1 a + a m_^a +...+ a^ = 0, a^ e XT, then a must be i n &. Proof: Suppose & has l o c a l u n i f o r m i z i n g parameter TT , and assume a has a pole at n of order n. Then ord ^ (a) = -n. m m — l Thus, ord^(a ) = -mn while o r d ? l ( - a m _ 1 a a Q ) > -(m-l)n m m-1 = -mn + n. But s i n c e a s a t i s f i e s a = - a m - . . . - a ^ , t h i s cannot be t r u e . Hence a must have ord ^ (a) £ 0. a i s t h e r e f o r e i n &. We show now how p l a c e s give a r e l a t i o n between the e x t e n s i o n f i e l d K of k and a f i e l d of formal power s e r i e s . We w i l l see that each p l a c e of K gives r i s e to an i n j e c t i o n of K i n t o a s u i t a b l e power s e r i e s f i e l d . Let 2^ be a p l a c e i n K. As u s u a l , denote by & the place r i n g de f ine d . by p. , and by TT a l o c a l u n i f o r m i z i n g parameter of "p- . We want to show that to any element z e & there i s a s s o c i a t e d a unique power s e r i e s i n the l o c a l u n i f o r m i z i n g parameter TI . 3 1 . Let z £ &. Then z i s congruent modulo ^ to a unique element AQ e k. (Obtain AQ through the isomorphism k —>• &—y Thus Z - A Q = Z^TT f o r some z^ £ &. Repeating t h i s argument f o r z^ and using i n d u c t i o n on n, we o b t a i n unique A Q , A - ^ , A 2 > - - ' £ k such that 2 z = AQ + A ^TT + X + ... 2 Thus we have a map z >—>• A Q + A^TT + A ^ ^ +... which d e f i n e s an i n j e c t i o n of & i n t o k [ [ IT ] ] , the r i n g of formal power s e r i e s over k. T h i s homomorphism extends i n the obvious way to an i n j e c t i o n of K i n t o k ( ( i r ) ) , as d e s i r e d . Note: I f ^ i s a p l a c e of K, t h e n ^ d e f i n e s a m e t r i c on K and we can then form the completion of K with r e s p e c t to t h i s m e t r i c . Denote t h i s completion by K^. I f ir i s a l o c a l u n i f o r m i z i n g parameter f o r ^3 then i t i s not d i f f i c u l t to see that = k ( ( i T ) ) . Hence the above map g i v e s an i n j e c t i o n of K i n t o i t s completion with r e s p e c t top.. ( 2) Rami f i cat ion In t h i s s e c t i o n we d i s c u s s the i n t e r c o n n e c t i o n between p l a c e s and t h e i r e x tensions or r e s t r i c t i o n s . Assume L and K are both a l g e b r a i c f u n c t i o n f i e l d s over k, and L i s a f i n i t e e x t e n s i o n of K. Take a p l a c e ^3 i n L. Denote by o r d^ J ^ the r e s t r i c t i o n of ord^j to K. The values of ordjjlj, form some subgroup of Z- not n e c e s s a r i l y a l l of Z, of course. In any case, there i s a s m a l l e s t p o s i t i v e i n t e g e r e o c c u r r i n g i n t h i s subgroup, and a l l values of 32 . ordjj on K w i l l be m u l t i p l e s of e. C a l l e the r a m i f i c a t i o n  index of p . I f e / 1, say "fl, i s r a m i f i e d . In f u t u r e i n r e f e r r i n g to the r e s t r i c t i o n of ordp to K, we w i l l mean the normalized r e s t r i c t i o n i . e . the v a l u a t i o n 0 r (*?^K. e This n o r m a l i z a t i o n ensures that the r e s t r i c t i o n has s m a l l e s t p o s i t i v e value 1. By the r e s t r i c t i o n of $L to K, we w i l l o r d I mean the p l a c e of K d e f i n e d by ft'K. e From t h i s d e f i n i t i o n we deduce e a s i l y the f o l l o w i n g two r e s u l t s . (1) Suppose p. i s a pl a c e i n L with r a m i f i c a t i o n index e and i s the r e s t r i c t i o n of ~§X. to K. Let II, TT be l o c a l u n i f o r m i z i n g parameters f o r p. , ^ r e s p e c t i v e l y . Then TT = u n e , where U i s some u n i t i n Oi . r For, II a u n i f o r m i z i n g parameter f o r ^2 i m p l i e s that e e ord^(II) = 1, and hence that ord^II ) = e-1 = e. So ord^f.11 ) = us II i s an element of L having the same e TT order as TT with r e s p e c t to ^  . T h i s i m p l i e s that must be a u n i t i n {X, . ( 2 ) R a m i f i c a t i o n i n d i c e s m u l t i p l y i n a tower of f i e l d s . Suppose L^/L^/K i s a tower of f i e l d s and ^2 i s a pla c e of L 2> Let ^ , ^  be the r e s t r i c t i o n s of ^  to L^ and K r e s p e c t i v e l y . Suppose x£ has r a m i f i c a t i o n index e 1 over ^ , j"0_ r a m i f i c a t i o n index over ^ and r a m i f i c a t i o n index e over ^ . Then e = e ± ' e 2 ' T n e P r o ° f °f t h i s Is obvious 33 . We now st a t e some of the b a s i c f a c t s about r a m i f i c a t i o n needed. R e s u l t I giv e s the major f a c t . The r e s u l t s f o l l -owing t h i s are c o r o l l a r i e s of I i n the s p e c i a l case where the f i e l d extension i s G a l o i s . In most texts on r a m i f i c a t i o n theory, the r e s u l t c o r r e s p o n d i n g to our r e s u l t I takes on a more complicated form. The s i m p l i c i t y i n our case i s due to the f a c t that k i s a l g e b r a i c a l l y c l o s e d . I Let L/K be a f i n i t e s e p a r a b l e f i e l d e x t e n s i o n . Suppose j p Q i s a place on L, and ^ i s the r e s t r i c t i o n of g)0 to K. Consider the set of a l l p l a c e s }^ on L l y i n g o v e r ^ . Let e^ denote the r a m i f i c a t i o n index of ^  . Then Ee« « [L:K] Instead of g i v i n g a complete proof of t h i s r e s u l t , we s h a l l merely i n d i c a t e how i t comes about that the r a m i f -i c a t i o n i n d i c e s are connected to the degree of the e x t e n s i o n . For a complete proof the reader i s r e f e r r e d to C o r o l l a r y 2, p. 308 [ A ] . Consider the p l a c e on K. We are i n t e r e s t e d i n i t s extensions to L. We can view t h i s as a homomorphism^ ex t e n s i o n problem i n the f o l l o w i n g manner. Consider the homomorphism >|i : stf—*• k, where /f i s the p l a c e r i n g o f ^ . ( A c t u a l l y , iii : /C —>/?IJS^ —>• k) I f we extend ij; to a s u b r i n g of L which i s as l a r g e as p o s s i b l e , t h i s s u b r i n g w i l l be a p l a c e r i n g f o r an e x t e n s i o n )^ x n L. 34 . In f a c t , there i s a 1-1 correspondence between ^jj/^and such maximal extensions of i|i . We view our problem t h e r e f o r e i n terms of these extensions of . Since L i s separable over K, we can choose a gener-a t o r 6 f o r L over K such that 6 i s i n t e g r a l over Z^. Say 9 s a t i s f i e s the equation N N - l F(X) = X N + a N_ xX "•+...+ a Q = 0 where the a . e . x Consider C^ Q = xS [ 0 ] =^T(X)/F(X). The q u o t i e n t f i e l d of &Q i s L. Thus, i f we have extended <J to & Q , i t i s f a i r l y easy to see that t h i s determines a unique maximal e x t e n s i o n of I|J . Our problem t h e r e f o r e i s to extend to &^ . i s a l r e a d y d e f i n e d on/C. To extend to^<T[0], we must f i n d the p o s s i b l e v a l u e s f o r i p(6). 6 s a t i s f i e s F(X) = 0; hence f ( 6 ) must s a t i s f y the equation .F(X) = X N + ^ ( a N _ 1 ) X N _ 1 +.,.+ <Ka 0) = 0 where the c o e f f i c i e n t s <Ma_^ ) e k. Since k i s a l g e b r a i c a l l y c l o s e d , F(X) can be f a c t o r e d i n t o l i n e a r p o l y n o m i a l s , say as F(X) = ( X - x p 6 ! . . . ( X - A r ) 6 r , where the X. e k. But then we see that the p o s s i b l e v a l u e s f o r ^ ( 6 ) are p r e c i s e l y the r o o t s of F, namely, \^,...,\ . This gives us the f o l l o w i n g i n f o r m a t i o n : (1) there always w i l l be extensions of '•!> (there are always roo t s of F(X) = 0), and (2) there are at most N extensions of ii; (r < deg F = N) . 35 . C l e a r l y there are r i d e a l s , . . . ,"p^  of CTQ l y i n g o v e r ^ , namely = (9 - ^^)0Q- The r a m i f i c a t i o n index of i s the l a r g e s t power of (9 - ^ ) $ Q c o n t a i n i n g ^ - . This turns out to be the same as the l a r g e s t power of ( X - X^) d i v i d i n g F ( X ) , namely e^. A l l i n a l l then, we have I e . = deg F ( X ) = [L:K]. Note: As remarked i n the p r o o f , we o b t a i n 2 important f a c t s about the extensions of . Namely, that always has extensions i n L and that these extensions are f i n i t e i n number where t h i s number i s l e s s than or equal to N. I f there i s p r e c i s e l y one e x t e n s i o n of ^ i n L, we say that i s t o t a l l y r a m i f i e d . In t h i s case, I t e l l s us that the r a m i f i c a t i o n index of the e x t e n s i o n of x^. i s N, the degree of L over K. A l s o , we f i n d that the l o c a l u n i f o r m i z i n g parameter of the e x t e n s i o n i n L s a t i s f i e s a p a r t i c u l a r l y n i c e e q u a t i o n . Lemma E i s e n s t e i n Let L/K be a f i e l d e x t e n s i o n of degree N. Suppose the p l a c e on K i s t o t a l l y r a m i f i e d . Let p be the p l a c e L l y i n g over and take II to be a l o c a l u n i f o r m i z i n g para-on me t e r f o r ^ ? . Then there e x i s t z K such that N N - l n = c t T 1 i r +... + a., n + a _ N-l 1 0 where the a . z j£ f o r 1 < i < N-l and ord (o.) = 1. ^ t o r 1 < l < N - l a n d o r d ^ o Q , 3 6 . P r o o f II i s an e l e m e n t o f L and hence s a t i s f i e s an e q u a t i o n o f d e g r e e n o v e r K, f o r some n|N. T h u s , w i t h o u t l o s s o f g e n e r a l i t y , t h e r e e x i s t E K, i = 0 , 1,..., n-1, s u c h t h a t (*) I I n = a ^ n 1 1 " 1 +...+ a . n + a . n-1 1 0 From t h i s e q u a t i o n we see t h a t o r d ^ ( n n ) = n, and hence t h a t o r d ~ ( a ^ II n ^ +.;.+ a n) must a l s o be n. f- n- i u Now c o n s i d e r each term on t h e r i g h t o f t h e e q u a t i o n . F o r e a c h e l e m e n t s K, o r d ^ ( a ^ ) = N o r d ^ ( a ^ ) , s i n c e i s t o t a l l y r a m i f i e d . Thus o r d ^ ( a ^ ) = 0 (mod N ) , and t h e r e f o r e o r d ^ a ^ = 0 (mod n) . But t h e n o r d ^ o ^ I I 1 ) = o r d ^ ( a . ) + o r d ^ ) ( n 1 ) = i mod (n) . Hence (1) f o r each i , j from 0 t o n-1, i f i f j t h e n o r d ^ ( a ^ n 1 ) ordgj(a H J ) , and (2) o r d K ( a . I 1 ) i 0 (mod n) f o r 1 i i < n-1. fl l From (1) we see t h a t t n e terms on the r i g h t o f (*) a r e d i s t i n c t . T h e r e f o r e o r d ^ 0 n - l l i n " 1 + ' " + V = 0 l i < n - l ( o r d ^ a . n 1 ) ) We know t h i s o r d e r must be n. By (2) t h e o n l y term whose o r d e r can p o s s i b l y be n i s a ^ . Thus o r d ^ ( a g ) must be n, and o r d ^ ( a J 1 ) must be >n. In f a c t , s i n c e o r d ^ a ^ ) = 0 (mod N) and n|N, we f i n d n = N. Summing up, we have o r d ^ t a ) = N and o r d s ( a , ) = m.N, f o r some i n t e g e r m. > 1 T 0 fr i i i and 1 < i < n-1. T h i s i s the d e s i r e d c o n c l u s i o n . 37 . Our second b a s i c r e s u l t gives the r e l a t i o n s h i p among a l l extensions of a given p l a c e i n K, assuming that L/K i s G a l o i s . II Let G be the G a l o i s group of L/K and l e t ^ b e a p l a c e on K. Then G a c t s t r a n s i t i v e l y on {~p_!*Ei a p l a c e on L and That i s , given p l a c e s P]_ ,c^2 l y i n g over , there e x i s t s a e G such t h a t p ^ = For the proof of t h i s standard r e s u l t the reader i s r e f e r r e d to P r o p o s i t i o n 1 1 , p. 244 [ 4 ] . Assume L/K to be G a l o i s i n the f o l l o w i n g . We devel op a c o n n e c t i o n between r a m i f i c a t i o n i n d i c e s and c e r t a i n subgroups of the G a l o i s group of L/K. Let G be the G a l o i s group of L/K. Let be a p l a c e on L with p l a c e r i n g &. Define the i n e r t i a group of 2^ to 1 ^ = {o e G|a(a) = a(modp) f o r a l l a e & } Each a z I ^ leaves a l l elements of the p l a c e r i n g of ^2 f i x e d modulo JJ). Define the decomp o s i t i o n group of |}, Dp, to be composed of a l l elements of G l e a v i n g p. f i x e d . That i s , take D F = (a £ G|a£=#l } Then, i t turns out that i n our set-up, s i n c e k i s a l g e -b r a i c a l l y c l o s e d , these two groups are i d e n t i c a l . Lemma For any v a l u a t i o n 7) on L, 38 . Proof I t i s c l e a r that I ^  <S D^. Assume a e D^, and l e t a be an a r b i t r a r y element of 0". Then there i s a unique A E k, the f i e l d of c o n s t a n t s , such that a = A(mod|2.). (Obtain X through the isomorphism k —> & —• &/]p.) • A p p l y i n g a to t h i s we have a(a) = a(X)(modap) . But aX = X s i n c e X e k, and a^ Q = j~Q s i n c e a e D^. Hence a(a) H A(mod^), so that a does belong to I^j. We can now e a s i l y d e r i v e from I and II the f o l l o w i n g important r e l a t i o n : namely, that the i n e r t i a group of a p l a c e has order p r e c i s e l y equal to the r a m i f i c a t i o n index of the p l a c e . I l l Let j p be a p l a c e on L, and l e t e^j be i t s r a m i f i c a t i o n index. Then Proof Let be a p l a c e on K such that $)\^' Take S =; a p l a c e on L and "Pl^ -} • We know S i s f i n i t e by r e s u l t I j l e t |s| denote the number of elements i n S. Then our s i t u a t i o n i s as f o l l o w s : we have the G a l o i s group G of L/K a c t i n g t r a n s i t i v e l y on S (by I I ) , and the subgroup Dp of G i s Stabg(^J), the s t a b i l i z e r of jp i n G. Hence by the fundamental theorem on t r a n s f o r m a t i o n groups, | G: D-, | = |s| , IT 3 9 . or , • If! A l s o from I I , we deduce that a l l the e^j must be the same; say they are a l l e. Then E e<j = Z e = e | S | . From I we f i n d that e|s| = [ L : k ] ; by G a l o i s theory t h i s i s j u s t |G| . Thus e =|g"|> s o that = l1^}! = e» a s d e s i r e d T h i s r e s u l t g i v e s us two more f a c t s which w i l l prove very u s e f u l l a t e r on. C o r o l l a r y 1 Suppose I^/L^/K 1 S a G a l o i s tower of f i e l d s . Let p be a pl a c e on L^. I f l p i s contained i n Gal ( L ^ / L ^ ) , then i s not r a m i f i e d i n L^/K - a l l r a m i f i c a t i o n occurs i n L 2 / L ^ . This i s c l e a r using the f a c t that r a m i f i c a t i o n i n d i c e s are m u l t i p l i c a t i v e i n a tower of f i e l d s . C o r o l l a r y 2 The number of r a m i f i e d p l a c e s i n any sepa r a b l e e x t e n s i o n L/K i s f i n i t e . N o t i c e that here L/K need not be G a l o i s . Proof Suppose k i s the ground f i e l d f o r L and K, as usual Take x to 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/k, and l e t 6 be the generator of L over K. Thus we have L = K(3) and K(6) separable over k ( x ) . 40. Before g e t t i n g i n t o the proof we make two s i m p l i -f i c a t i o n s of the s i t u a t i o n . To show that there are f i n i t e l y many r a m i f i e d places on L/K, i t i s c l e a r l y s u f f i c i e n t to show that only f i n i t e l y many p l a c e s of L/k(x) are r a m i f i e d . S i m i l a r l y , i f we can show only f i n i t e l y many p l a c e s r a m i f i e d i n L/k(x) f o r some e x t e n s i o n L of L, then we are a l s o done. In p a r t i c u l a r , we take L to be the f i e l d o b t a ined by a d j o i n i n g to L a l l conjugates of 8 . For sim-p l i c i t y , denote L again by L. We now are c o n s i d e r i n g the G a l o i s e x t e n s i o n L / k ( x ) . Denote i t s G a l o i s group by G. By modifying the generator 6 of L s l i g h t l y , i f necessary, we can assume that 8 s a t i s f i e s an equation of the form F(T) = T n + f j " " 1 +...+ f„ n - l U where the f ^ e k [ x ] . Then, L being normal over k [ x ] , we can f a c t o r F(T) i n L as F(T) = n (T - 0-6) oeG Taking the formal d e r i v a t i v e of F with r e s p e c t to T, and e v a l u a t i n g at 8 we f i n d F1C@) = n ( 8 - a 8 ) oeG a ^ i d Now we come to counting the number of r a m i f i e d p l a c e s of L / k ( x ) . Suppose i s a r a m i f i e d p l a c e on L, not l y i n g over the place corresponding to 0 0 i n k ( x ) . (This excludes only f i n t e l y many p l a c e s by I.) Then 6 l i e s i n the pl a c e r i n g of i n L. For, i f -t|. i s the place on k(x) induced by ^ } and /f i s i t s place r i n g , then XT 2 k[x] . 41 . Thus 6 s a t i s f i e s an equation with c o e f f i c i e n t s in^tT". By the remark on i n t e g r a l i t y of s e c t i o n 1, 9 must t h e r e f o r e l i e i n £ . By I I I , ^2 r a m i f i e d i m p l i e s that i t s i n e r t i a group I £ i s n o n - t r i v i a l . Hence there i s some automorphism a- 4 i d i n G such that 6 - o6 = 0 ( m o d p ) . But then F ' ( 9 ) II (8 - o6) = 0 ( m odS). This means i n p a r t i c u l a r 0eG r a ^ i d that F ' ( 8 ) i s d i v i s i b l e by the u n i f o r m i z i n g parameter f o r Th i s statement holds f o r any r a m i f i e d p l a c e not over °° i n L/k(x) Since F ' ( 9 ) has only f i n i t e l y many zer o s , the number of r a m i f i e d p l a c e s must be f i n i t e . As an example of our r a m i f i c a t i o n theory we d e r i v e the r a m i f i c a t i o n i n d i c e s of the f i e l d e xtensions c o r r e s -ponding to the c a n o n i c a l groups. Assume that k = C and l e t G denote one of the c a n o n i c a l groups. Say |G| = N. As i n the I n t r o d u c t i o n , we view the f i g u r e F corresponding to G i . e . the plane N-gon, the dihedron, the t e t r a h e d r o n , the octahedron or the i c o s a h e d r o n , as embedded i n the Riemann sphere. R e c a l l that elements of G are then r o t a t i o n s of the sphere which leave F f i x e d . Now suppose that G i s one of the t e t r a h e d r a l , o c t a h e d r a l or i c o s a h e d r a l groups. Each element of G moves (1) Refer to §§1-9 Chp . 1[3] example. K l e i n looks at geometric p o i n t of view. f o r f u r t h e r d e t a i l s t h i s s i t u a t i o n from on t h i s a 42 . a p o i n t Pg on the sphere to another p o i n t on the sphere. C a l l the set of p o i n t s to which PQ i s moved by G i t s o r b i t . Host p o i n t s have o r b i t c o n t a i n i n g N p o i n t s . However, there are three types of p o i n t s which have s m a l l e r o r b i t s , namely: midpoints of edges of the f i g u r e F, c e n t r e p o i n t s of faces of F and v e r t i c e s of F. I t can e a s i l y be seen that the midpoint P of an edge i s l e f t f i x e d by p r e c i s e l y two elements of G (the h a l f turn about the a x i s through P and the midpoint of the o p p o s i t e edge, and the i d e n t i t y ) ; the c e n t r e p o i n t of any face i s l e f t f i x e d by three elements of G; any v e r t e x i s f i x e d by i elements of G where i = 3, 4 or 5, the number of faces a d j o i n i n g at a v e r t e x of F. A l s o , i t i s c l e a r that the p o i n t s i n any one of these three sets form a conjugacy c l a s s under G. Each such c l a s s of p o i n t s corresponds to the p l a c e s i n k(x) l y i n g over a given p l a c e i n F i x G. (Refer to pp. 28-29 f o r the correspondence between p o i n t s i n k and p l a c e s i n k ( x ) , ) The elements of G l e a v i n g a p o i n t i n any c l a s s f i x e d form the i n e r t i a group at that p o i n t . Since we know that the order of the i n e r t i a group i s equal to the r a m i f i c a t i o n index at a p o i n t , 2, 3 and i are the only non-unit r a m i f i c a t i o n i n d i c e s f o r G. T h i s t r i p l e , (2, 3, i ) , of r a m i f i c a t i o n i n d i c e s f o r the three n o n - t r i v i a l conjugacy c l a s s e s w i l l be c a l l e d the r a m i f i c a t i o n p a t t e r n of G. Suppose now that G i s a c y c l i c group of order N. Viewed as a group of automorphisms of the sphere, the 43. elements of G are obtained by r e p e t i t i o n of a s i n g l e r o t a t i o n . As such we see immediately that there are only two p o i n t s on the sphere which have o r b i t s c o n t a i n i n g l e s s than N p o i n t s - the two p o i n t s at e i t h e r end of the a x i s of r o t a t i o n . These pofnts are f i x e d by a l l N elements and are t h e r e f o r e non-conjugate. The r a m i f i c a t i o n p a t t e r n of G must then be (1, N. N) . For the d i h e d r a l group of order N = 2n the p o i n t s on the dihedron having an o r b i t of l e s s than 2n p o i n t s are: the two p o i n t s at the end of the a x i s of the c y c l i c order n r o t a t i o n (these are conjugate and f i x e d f o r n r o t a t i o n s ) ; the n v e r t i c e s (these are conjugate and f i x e d by a h a l f - t u r n about the a x i s through the v e r t e x and the i d e n t i t y ) ; the n mid-edge p o i n t s (these are a l s o conjugate and f i x e d by 2 r o t a t i o n s ) . Thus there are three d i s t i n c t conjugacy c l a s s e s with o r b i t s s m a l l e r than normal. We see by the above c o n s i d e r a t i o n s that the r a m i f i c a t i o n N p a t t e r n of G must t h e r e f o r e b (2, 2, -j) . For f u t u r e r e f e r e n c e , a l i s t of the r a m i f i c a t i o n p a t t e r n s i s given here. I C y c l i c group, order N (1, N, N) II D i h e d r a l group, order N (2, 2, -) I I I T e t r a h e d r a l group (2, 3, 3) IV O c t a h e d r a l group (2, 3, 4) V I c o s a h e d r a l group (2, 3, 5) 44 . For the general s i t u a t i o n , k any a l g e b r a i c a l l y c l o s e d f i e l d , i n determining the r a m i f i c a t i o n p a t t e r n s of the c a n o n i c a l groups, the geometry of our s i t u a t i o n g i v e s way to combinatorics of G-action on p l a c e s of k ( x ) . We w i l l see l a t e r that t h i s does r e s u l t i n the same p a t t e r n s . In f a c t , we w i l l determine that any f i n i t e subgroup of Aut.k(x) has to have one of the above r a m i f i c a t i o n p a t t e r n s . 45 CHAPTER VI THE HURWITZ FORMULA (1) Order of a D i f f e r e n t i a l ft^yk be the u n i v e r s a l Let 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 over k, with k a l g e b r a i c a l l y c l o s e d . Let d:K k - d e r i v a t i o n . (Cf. §2, Chapter I I I ) F i x a p l a c e |2 on K, and denote by the completion of K with r e s p e c t to p. . Then, f o r any l o c a l u n i f o r m i z i n g parameter TT f o r ^  , = k((ir c Let d- : K a—• ft be the u n i v e r s a l continuous k - d e r i v a t i o n y / k of Kp/k. Take i:K —* Kj, to be the i n j e c t i o n of K i n t o Kg). Then u n i v e r s a l i t y of d guarantees e x i s t e n c e of a . c unique K-module map i ^ : ^ / j . Q~ .. completing the K B / k. f o l l o w i n g diagram. K K/k Let a £ !1 f o l l o w s : K/k Define ord^(to), the order of to at?Q , as Map to to i^(to) = to^, say. Define ord^(to) = ord ( w^) ( r e c a l l that the order of a continuous d i f f e r e n t i a l was d e f i n e d i n §2, Chapter I I I ) For any d i f f e r e n t i a l to e ^ / ^ J ord^(to) = 0 f o r almost 46 . a l l p l a c e s ^ on K. Let 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/k. Then tt K ^ = Kdx. Thus f o r any to e ^ K ^ k we have w = fdx, f o r some f e K. Claim: ord^(o)) = 0 f o r almost a l l p l a c e s jp on K. Proof: ord-p(to) = o r d p ( f ) + ord-^(dx) We know that o r d p ( f ) = 0 almost everywhere s i n c e f e K. (f has order 0 at a l l p o i n t s except at i t s zeros and p o l e s , and these are f i n i t e i n number.) So we must show that ord^dx = 0 almost everywhere. Let jp be any p l a c e on K which i s u n r a m i f i e d and which does not l i e over the p l a c e - * ^ : x •—»• 0 0 i n k ( x ) . (This excludes "J2 from being only a f i n i t e number of places.) Suppose p l i e s over the p l a c e ^. of k ( x ) . , not being •^•Q : x 1—°°> must be of the form <^  : x i—> a, f o r some a e k - 0 0. Thus /} has l o c a l u n i f o r m i z i n g parameter TT = x - a But then "p. has l o c a l u n i f o r m i z i n g parameter II = ir = x - a. Then x = II + a, and dx = 1 • d IT. T h e r e f o r e , ord^(dx) = ord ( 1 ) = 0. (3) D e f i n i t i o n of 5(K/k) Let co e ^Yi/k' Define div(co) = n^ r < ^^^ U ^ where the product i s taken over a l l p l a c e s ~p. on K. T h i s i s a f i n i t e product by s e c t i o n (2); the f o l l o w i n g d e f i n i t i o n t h e r e f o r e makes sense. We d e f i n e deg ( d i v co) = E ord-n(co) 47 . Claim: deg ( d i v co ) i s independent of co . Proof: Let 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. Then ^K/k = K ( * x ' s o t n a t a n y t w ° d i f f e r e n t i a l s co , co ' d i f f e r at most by an element of K. (For, co = fdx and co ' = gdx f o r some f, g e K v co = ( f g "^co', where fg 1 e K.) Say co = hco 1 then. We can now compare deg ( d i v co) and deg ( d i v deg ( d i v co) = ord^(co) = E^  ord^Chto 1) = E ( o r d B h + ord^co') = (E ord^h) + (E ord^co') ( 1 ) = E ord_co' = deg ( d i v co ' ) . Since deg ( d i v co) i s an i n v a r i a n t depending only on the f i e l d e x t e n s i o n K over k, we denote i t with r e f e r e n c e to K/k alone, by 6(K/k). Note: I f K i s a r a t i o n a l f u n c t i o n f i e l d , say K = k(x) where x E K i s some in d e t e r m i n a t e over k, then 6(K/k) = -2 Proof : Assume K = k ( x ) , where x £ K i s some in d e t e r m i n a t e over k. Let co = dx e ft„/, . Then 6 (K/k) = E o r d K ( d x ) . K/K p p Now, a l l p l a c e s on K are of the form ( i ) fl : x »—*• a, a E k or (i±)p: x »—> °° Case ( i ) Take n = x - a to be the l o c a l u n i f o r m i z i n g parameter f o r p . Then dx = l ' d n , so that ord^(dx) = 0. (1) E o r d c h = 0 by the theorem of p. 29. 48 . Case ( i i ) Take TT = ^ to be the l o c a l u n i f o r m i z i n g parameter f o r ^ l . Then we have x = ^, so that dx = - ^ - 2 d i T . Thus ord^(dx) = -2. Hence deg ( d i v dx) = _! ord^(dx) = 0 + (-2) = -2 (4) The Hurwitz Formula Let L and K be f i e l d e x t ensions of k of transcendence degree 1, L a l g e b r a i c of degree N over K, and k of c h a r a c -t e r i s t i c p. Let e^j denote the r a m i f i c a t i o n index of the plac e p. on L. Write }^ e L to denote that |! i s a p l a c e on L. Then ?™MS" 6 < L / k ) - W > - « + * «p J 1 E L ( 2 ) where = e j, - 1 i f p | e^ and > e ^  - 1 i f p | e^. Proof: Let u) e ^Yi/\n ( a n c* s o e ^L/k^" Idea of p r o o f : Consider a s i n g l e p l a c e X£ on K. Compare ^(K) - ord^(co) to 6„(L) = a ^ _ o r d c ( c o ) . Then to r e l a t e <5(K/k) and 6 (L/k) , n o t i c e that 6 (K/k) « * 5-(K) and 5 (L/k) = „ ^ K ^ ( L ) Suppose |2 i s a p l a c e on L with r a m i f i c a t i o n index e. Let J|, be the p l a c e i n K l y i n g under p . Let TT, II be l o c a l u n i f o r m i z i n g parameters f o r ^ i n K and p i n L r e s p e c t i v e l y . Q Then TT = UH f o r some u n i t U i n L. (2) The o r i g i n a l § 2 Chapter XXI deduc t i o n of - V [2] t h i s formula can be found i n 49 Thus d TT = e u n e - 1 d 4 , n + (D u)n"d^n 1 ? TT "J = [ e u n 6 - 1 + ( D n u ) n e ] d ^ n Now we can r e l a t e d^(K) and d ^ ( L ) . 03 E ft..,. . Thus i „ ( c o ) = t o . , = f d^i r f o r some f E K ( ( T T ) ) , K/k ft ' ' J TI -"J- TT v v / / » and o r d ^ ( c o ) = o r d ^ ( f ^ ) = 6 ^ , ( K ) . What i s 6 ^ ( L ) ? co - = f d * TT = f [ e u n 6 - 1 + (D u ) n e ] d_n T TT f TT TT ^ Therefore o r d B ( c o ) = ord-»(f ) + ord J e U n 6 " 1 + (D U ) n e ) = ord^Cf^) + E , say. N o t i c e : e = e-1 i f p )( e, and E > e-1 i f p|e. With p now any pl a c e above AS^ and e^ i t s r a m i f i c a t i o n index, we have 5 Y . ( L ) = E ord^Cco) = Z o r d - ( f ) + E fi^j. IT* TT f\^t r = E e-ord (f ) + E e-= <5^(k) = 6 ^ ( K ) • E e + E e B = 6 f ( K ) . ( N ) + £ e , = 6 f ( K ) . N + ^ £, We t h e r e f o r e have the f o l l o w i n g r e l a t i o n between 5 ( L / k ) and 6 ( K / k ) . 6 (L/k) = E 6^,( L ) xfeK ^ = E 6 «.(K) • N + E E e_ = 6 (K/k)-N + E £ p  ? E L 50 . CHAPTER VII CONSEQUENCES OF THE HURWITZ FORMULA Assume throughout t h i s chapter that K i s a f i x e d f i e l d of r a t i o n a l f u n c t i o n s , p u r e l y t r a n s c e n d e n t a l of degree 1 over an a l g e b r a i c a l l y c l o s e d f i e l d k. We study G a l o i s extensions of K which are a l s o r a t i o n a l f u n c t i o n f i e l d s . A l though t h i s i s not the n a t u r a l p o i n t of view a r i s i n g from our problem, (the n a t u r a l one being to study s u b f i e l d s of a f i x e d extension) i t w i l l be seen l a t e r that the problem does reduce to t h i s s i t u a t i o n . (1) In the theorem on the Hurwitz Formula, i f both L and K are p u r e l y t r a n s c e n d e n t a l of degree 1 over k, then the formula takes on a p a r t i c u l a r l y simple form. Namely, i t y i e l d s a diophantine e q u a t i o n , the s o l u t i o n s of which give enough i n f o r m a t i o n to determine L to i s o -morphism over K i n most cases. We now determine the diophantine equation and i t s s o l u t i o n s . With L and K as above, from §3, Chapter VI we know that 6 ( L / k ) = 6(K/k) = -2. I f furthermore, the c h a r a c t e r i s t i c p of k does not d i v i d e e f o r any r a m i f i c a t i o n index e i n L/K, then the Hurwitz Formula becomes 2N - 2 = Z = E (e^- 1) 51 . indexing t h e ^ . i n K from 1 to r . Or, d i v i d i n g by N , we have 2 r 1 (*) 2 - — = i c i - — K } N • , e . ) i = l l Regarding t h i s as a d i o p h a n t i n e equation i n N and e^, i t i s f a i r l y easy to see that i t has r e l a t i v e l y few s o l u t i o n s . Assuming N > 1, then r = 2 or 3. For, r = 1 i m p l i e s R.S. (*) < 1, while L.S. (*) > 1; and r > 4 i m p l i e s R.S. (*) > 2, while L.S. (*) < 2. ( i ) L e t t i n g r=2 (*) becomes 2 = 1 1 N " e l e2 — , — > — . It f o l l o w s from our 6 1 x 6 2 be —. Hence e n = e„ = N. N 1 2 i + i e2 e 3 Since e ^  , e^ < N, we have equation that — , — e e 1 2 must ( i i ) Le t t i n g r = 3 (*) = i 4-6 1 Since not a l l e. can l be > So 1 - 1 = 2 If anothe r e^, say e 2 , i s e l = e 2= 2, e 3= n If n e i t h e r e 2 nor e^ i s 2 Ass ume e ^  = 3. Then 1 6 Thus e ^  < 6, and e must e 2 e 3 N e, 52 . have e± = 2, e 2 = 3, e 3 = 3, N = 12 e ; L = 2, e 2 = 3, e 3 = 4, N = 24 e l = 2» e 2 = 3, e 3 = 5, N = 60 So the Hurwitz Formula i n t h i s case has only f i v e systems of s o l u t i o n s . Notice that these systems (N and r a m i f -i c a t i o n i n d i c e s ) correspond to the group orders and r a m i f -i c a t i o n i n d i c e s f o r the f i v e c a n o n i c a l groups: c y c l i c d i h e d r a l , t e t r a h e d r a l , o c t a h e d r a l and i c o s a h e d r a l . T h i s i s the major f a c t donated by the Hurwitz Formula towards the proof of the Main Theorem. (2) The Hurwitz Formula shows that with L, K both as d e s c r i b e d i n ( 1 ) , [L:K] and the r a m i f i c a t i o n i n d i c e s of L/K can be one of only f i v e types. Since the set of r a m i f i c a t i o n i n d i c e s i s d i f f e r e n t f o r each type, i t i s enough to s p e c i f y t h i s to determine the type. We set up a n o t a t i o n f o r easy r e f e r e n c e to each type. Let -<J^ , 3 ^ e the t n r e e p l a c e s i n K at which r a m i f i c a t i o n o c c u r s . ( I t may occur at only two p l a c e s , i n which case i t does not matter what the t h i r d p l a c e i s . ) Let e^, e 2 > e^ be the r e s p e c t i v e r a m i f i c a t i o n i n d i c e s at these p l a c e s . Then we use the ordered t r i p l e (e^, e 2 > e^) to show the r a m i f i c a t i o n a t x^ , ^  , x^ 2 , x^ ^ i n L/K. C a l l t h i s the r a m i f i c a t i o n p a t t e r n R(L/K) of L/K. Hurwitz says that R(L/K) i s one of types 53 I (1, N, N) II (2, 2, 2J I I I C2, 3, 3) IV (2, 3, 4) 'V (2, 3, 5) In what f o l l o w s , we s h a l l show that knowing the r a m i f i c a t i o n p a t t e r n of an ex t e n s i o n of K i s enough to determine the ext e n s i o n i n cases I-IV, and i n case V i t determines the group of the ex t e n s i o n (assuming always that the ex t e n s i o n i s G a l o i s ) . For the lemma and theorem f o l l o w i n g , we f i x an ex t e n s i o n L of K. Say L = k ( x ) , f o r some x e L t r a n s c e n -d e n t a l over k. Also assume [L:K] = N. Lemma In L/K we have (*) I r r (x, K, T) = T N + a ^ T ^ " 1 +...+ a Q , where e i t h e r the a. e k or K = k ( a . ) , and a l l nonconstant have the same p o l e . Proof Choose come element y e K such that K = k ( y ) . Let I r r ( x , k ( y ) , T) = f N ( y ) T N + f N - ]_ (y) T N _ 1 + . . .+ f Q ( y ) . We know y = ^ f o r some h, g £ kfx] , so that g (x) I r r (x, k ( y ) , T) = h(T) - y g ( T ) . Thus y occurs to the f i r s t power only i n the minimal polynomial f o r x. Since we have unique f a c t o r i z a t i o n i n k ( y ) [ T ] , we see that a l l 54 . f i ( y ) must be of the form a^y + b^ where a^, b^ e k. Then we have , v _ N A a N - i y + b N - l TN-1 ^ A v + b o I r r (x, K, T) = T + —.r • T + ...+ , . a N y + b N a N y + b N N N - l = T + a N - l T + * ' , + a 0 ' S a y " By d e f i n i t i o n then, a l l which are nonconstant have the same pole (given by a^y + b ^ ) ; and a l l are of the form a . y + b . i i ; — r — • T h i s i s e i t h e r i n k ( i f a.b„ - a„b. = 0) or av.y + b„ 1 N N x N N generates K (by the theorem of Chapter I ) . Remark: I f R(L/K) i s type I with r a m i f i c a t i o n at (x) and (•—) , then i n the above lemma we can conclude that K = k(a.) f o r a l l i . For, l by the lemma f o l l o w i n g I of Chapter V, we know x s a t i s f i e s an equation of form (*) where each o, ^  has order > 1 at 0. Th i s together with the above lemma i m p l i e s that each must have order p r e c i s e l y 1, as d e s i r e d . Theorem 1 Suppose R(L/K) i s type I with r a m i f i c a t i o n at the pl a c e s (x) and (^). Then K = k ( x N ) . Proof By the preceding remark we know that I r r (x, K, T) = T N + a N _ 1 + . . . + a Q where each a. has order 1. Hence we can f i x y = a., i n x J 0 p a r t i c u l a r , as generator of K. Then a l l other a are of 55 . a.y + b. the form ;—-5— f o r some a. , b. , c. , d. e k such that c . v + d . x * x x* x x' x a.d. - b.c. 4 0. Now the preceding lemma s t a t e s that a l l x i I i • 1 have the same p o l e . Since has a pole at 0 0 (meaning —) a l l other ct ^  must have a pole at 0 0 . That i s , they must be of the form a. = a.y f o r some a. e k. Thus i x - 7 x (**) I r r (x, K, T) = T N + a N _ 1 y T N " 1 +...+ y. Now look at the equation s a t i s f i e d by —. Take T = 77 ^ J x U and d i v i d e by y i n (*') to f i n d that ^ s a t i s f i e s 1 — — 2 — N-1 N - + a„ ,U + a H „U +...+ a.U + U = 0 y N-1 N-2 1 Again a p p l y i n g the preceding lemma we f i n d that a l l a^ must be m u l t i p l e s of ^. That i s , they must a l l be 0. From (*') then we have y = - x N , or K, = k(y) = k ( x N ) . Remark: In Theorem 1 we assumed that r a m i f i c a t i o n o c c u r r e d at (x) and (—) merely f o r convenience i n computation. The theorem could have been s t a t e d as: i f R(L/K) i s type I and the p l a c e s at which r a m i f i c a t i o n occurs are known, then the e x t e n s i o n i s determined as c y c l i c of degree N. Thus, i f R(L/K) and R(L'/K) are both type I with r a m i f i c a t i o n at the same p l a c e s , then the extensions L and L' are isomor-phic over K. This i s the b a s i c f a c t upon which the proof of the f o l l o w i n g theorem r e l i e s . We a l s o make use of the f o l l o w i n g f a c t s from Chapter V. (1) r a m i f i c a t i o n index = order of i n e r t i a group f o r a f i x e d p l a c e ( r e s u l t I I I ) 5 6 . (2) r a m i f i c a t i o n i n d i c e s m u l t i p l y i n a tower of f i e l d s (3) degree of f i e l d e x t e n s i o n = sum of r a m i f i c a t i o n i n d i c e s ( r e s u l t I) Theorem 2 Let L/K be a G a l o i s e x t e n s i o n and L a r a t i o n a l f u n c t i o n f i e l d . Suppose the group of L/K i s s o l v a b l e . Then R ( L / K ) determine's L up to an isomorphism over K. Proof We c o n s i d e r the three cases where R(L/K) i s type I I , I I I or IV, I having been d e a l t with i n Theorem 1. II (2, 2, f ) N Assume f i r s t that -zj ^  2. Let H be the i n e r t i a N group of the p l a c e with r a m i f i c a t i o n index —. Let L Q = F i x H. Then Gal ( L Q / K ) i s a f i n i t e subgroup of Aut^k(x) and so must have r a m i f i c a t i o n p a t t e r n type I-V. Since |Gal ( L Q / K ) I = 2, R ( L Q / K ) can only be type I with N = 2. That i s , R ( L Q / K ) = (2, 2, 1). By Theorem 1 t h i s means that L Q i s determined to isomorphism over K. Let , 'P^ be the two p l a c e s i n L Q above • These p l a c e s must N N N ramify of index i n L / L Q . Hence R ( L / L Q ) = (—, —, 1). Again using Theorem 1 we f i n d that L i s determined to i s o -morphism over L Q , and so over K. If |- = 2, then R ( L / K ) = (2, 2, 2). We know that the only groups of order 4 are the c y c l i c group or the K l e i n 4-group; and the r a m i f i c a t i o n p a t t e r n of the c y c l i c 57 . group i s type I. Hence Gal (L/K) must be the K l e i n 4-group. Let H be the i n e r t i a group of any p o i n t ^ . ^ with r a m i f i c a t i o n index 2. Take L Q = F i x H. Again we f i n d that R ( L Q / K ) i s type I with N = 2, but now the p l a c e u n r a m i f i e d i n L Q / K i s not u n i q u e l y determined unless the above ^ i s s p e c i f i e d . We f i n d that R ( L Q / K ) i s e i t h e r (2, 2, 1), (2, 1, 2) or (1, 2, 2). Theorem 1.then y i e l d s that to isomorphism there are p r e c i s e l y three e x t e n s i o n f i e l d s L Q of K, Choose one of these, say L Q such that R ( L Q / K ) = (2, 2, 1 ) . As i n the argument f o r N ^ 4 we o b t a i n R ( L / L Q ) = (2, 2, 1 ) , and L i s determined. I I I (2, 3, 3) Since [L:K] = 12, Gal (L/K) has a subgroup H of index 3. Take L Q = F i x H. Then |Gal ( L Q / K ) | = 3, and R ( L Q / K ) i s t h e r e f o r e (1, 3, 3). L Q i s then determined over K. Le t , 'p 2 , ^ ? 3 be the three p l a c e s i n L Q l y i n g o v e r ^ , Each of these must ramify of index 2 i n L / L Q . Hence R ( L / L Q ) = (2, 2, 2). By the argument f o r II we know that L i s determined to isomorphism f o r such a r a m i f i c a t i o n p a t t e r n . That i s , L i s determined to i s o -morphism over K, as d e s i r e d . IV (2, 3, 4) In a group of order 24 there i s always a subgroup of index 2. Let H be such a subgroup i n Gal (L/K), and take L Q = F i x H, as u s u a l . We f i n d that R ( L Q / K ) must be (2, 1, 2), so that L Q i s determined. L e t Z / ^ , ^ ^ be the 5 8 . two p l a c e s i n L Q above ^ ! ^ E T ( ^ 3 ^ e t n e P x a c e i n ^ 0 a b o v e ^ . Then c l e a r l y R ( L / L Q) = ( 3 , 3 , 2 ) . From the p r e v i o u s case, L i s determined to isomorphism over L Q , and so over K , as s t a t e d . Theorem 3 Let L / K be a G a l o i s e x t e n s i o n and l e t L be a r a t i o n a l f u n c t i o n f i e l d . Let G be Gal ( L / K ) . I f R ( L / K ) i s type V then G i s simple. Proof Suppose H i s a normal subgroup i n G of order m > 1 . Let L Q = F i x H. Then |Gal ( L Q / K ) | = m. We show that such an e x t e n s i o n L Q / K cannot e x i s t . R e c a l l that R(L / K ) = ( 2 , 3 , 5 ) . 1 f ^ 3 l s r a m i f i e d i n L Q / K i t must be t o t a l l y r a m i f i e d here, so that R ( L Q / K ) must be e i t h e r ( 1 , 5 , 5 ) or ( 2 , 2 , 5 ) . But t h i s i s i m p o s s i b l e s i n c e two p l a c e s i n L / K would then have to r a m i f y , both with i n d i c e s a m u l t i p l e of 5 or 2 . Thus-^-^ i s u n r a m i f i e d i n L Q / K . Hence there e x i s t at l e a s t two p l a c e s , Z* 2 ^0 a ^ o v e ^ 3 * These ramify t o t a l l y i n L / L Q so that R ( L / L Q ) must be ( 5 , 5 , 1 ) But t h i s i m p l i e s [ L Q : K ] = 1 2 . being u n r a m i f i e d i n L Q / K determines R ( L Q / K ) as ( 1 2 , 1 2 , 1 ) . C o n t r a d i c t i o n . Remark: Theorems 1-3 a c t u a l l y e s t a b l i s h the f i r s t c l a i m of the Main Theorem, namely, that a f i n i t e subgroup G of Aut.k(x) f o r which p Jf J G| must be isomorphic to a c a n o n i c a l 59 • group. For, G determines K = F i x G, a r a t i o n a l f u n c t i o n f i e l d c L = k ( x ) , and by (1) we know that R(L/K) i s type I-V. Theorems 1 and 2 then show that i f R(L/K) i s type I-IV, the f i e l d e x t e n s i o n L/K i s determined to i s o -morphism- but t h i s c l e a r l y i m p l i e s that the group of the e x t e n s i o n , G, must be determined to isomorphism. Since we know that the c a n o n i c a l groups I<-IV have e x t e n s i o n s determined by R(L / K), G must be isomorphic to one of these c a n o n i c a l groups. Theorem 3 shows that i f R(L/K) i s type V then G = Gal (L/K) i s simple of order 60. But the i c o s a h e d r a l group i s the only simple group of order 60. Hence G, i n t h i s case too, i s isomorphic to a c a n o n i c a l group (1) Theorem 1, Chapter II guarantees that F i x G i s r a t i o n a l f u n c t i o n f i e l d . 6 0 . CHAPTER VIII PROOF OF THE MAIN THEOREM R e c a l l the hypotheses of the theorem; we have a f i n i t e subgroup G of Aut k ( x ) , where k i s an a l g e b r a i c a l l y c l o s e d f i e l d of c h a r a c t e r i s t i c p with p A\ |G| . The aim i s to show that G must be conjugate to one of the f i v e c a n o n i c a l groups i f the r a m i f i c a t i o n p a t t e r n of G i s type I-IV or i f p = 0 . The f i r s t c o nnection we o b t a i n between G and a c a n o n i c a l group i s d e r i v e d from the Hurwitz Formula, as i n Chapter VII . We found there that G must be isomorphic to a c a n o n i c a l group, say Gg. Our aim i s to show e x i s t e n c e of a e Aut^ k(x) such that G a = a Go  1 = G Q . We can reduce t h i s to a f i e l d isomorphism problem by the f o l l o w i n g argument. Let F = F i x G and F Q = F i x G Q. Then f o r o e A u t ^ k O x ) , OF = FQ G° = GQ P r o o f : G a <== GQ : Let ago" 1 e G° . Then, f o r f Q e F Q , (ago ^ ) ( f g ) = (ago ^ ) ( o f ) f o r some f e F such that of = f g . But ( o g o _ 1 ) (of) = ( o g ) ( f ) = of = f g . Thus ago" 1 i s i n G Q. GQ Q. G ° : Let gg e Gg and fg e Fg. Then, w r i t i n g f g as of f o r some f e F, we have gg(gf) = of. But then (a ^ g g C f ) ( f ) = a "'"(of) = f . This i m p l i e s a "*"ggO- = g, f o r some g e G, so that gg = ago 1 e G°. Thus, i f F i x G = k(y) and F i x Gg = k(yg) f o r some 61 elements y, y n e k(x) t r a n s c e n d e n t a l over k, then the p r o b l i s to f i n d a e Aut^k(x) such that (1) o C k C y ) ) = k C y - 0 ) Define o" to be the n a t u r a l isomorphism from k ( y ) to k C y ^ ) . That i s , d e f i n e a: k ( y ) — > • k ( y ^ ) by a : y •—*• y.g and a: a i—• a f o r a l l a e k . Then a w i l l be the r e q u i r e d map i f i t can be extended to an automorphism of k ( x ) . Extend a as f o l l o w s : Let K be an a l g e b r a i c c l o s u r e of k ( x ) . Then a can be extended to give an automorphism of K (by Theorem 3, Chapter I I ) . Thus we have the set-up K ^ • K k(x) k ( y ) > k ( a ( x ) ) = k(x) - k ( y n ) where now our job i s to show that k(x) = k ( a ( x ) ) . If the r a m i f i c a t i o n p a t t e r n of G and G Q i s type I-IV, then Theorems 1 and 2 of Chapter VII show that the e x t e n s i o n s k(x) and k ( a ( x ) ) are u n i q u e l y determined i n the a l g e b r a i c c l o s u r e . i . e . k(x) = k ( a ( x ) ) f o r these cases. Hence G i s conjugate to GQ here. The remaining case we have to c o n s i d e r i s p = 0 and (1) Assume, without l o s s of g e n e r a l i t y , y, yQ chosen such that r a m i f i c a t i o n occurs at 1, 0 and °°. 62 . the r a m i f i c a t i o n p a t t e r n of G being type V. For t h i s we need s e v e r a l t e c h n i c a l r e s u l t s concerning the Schwarzian of x with r e s p e c t to c e r t a i n d e r i v a t i o n s . These r e s u l t s are l i s t e d i n the f o l l o w i n g lemmas. Through the 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 k. I f D i s a d e r i v a t i o n on k(z) we know that i t can be extended to k(x) s i n c e k ( x ) / k ( z ) i s s e p a r a b l e . Thus d i f f e r e n t i a t i o n with r e s p e c t to some f u n c t i o n of z, f ( z ) , on k(z) can be extended to k ( x ) . Denote the Schwarzian of x with r e s p e c t to t h i s d e r i v a t i o n by t x ] f ( z ) . In the f i r s t lemma we r e l a t e the Schwarzian of x (3 ) with r e s p e c t to z to the Schwarzian of x with r e s p e c t to — and with r e s p e c t to z - c, f o r some c e k. z Lemma 1 ( i ) [ x ] z = \n [ x ] i z ( i i ) [ x ] z = f x ] z _ c f o r c e k. Proof Let D denote d i f f e r e n t i a t i o n with r e s p e c t to z on k(x) and l e t DQ denote d i f f e r e n t i a t i o n with r e s p e c t to — on k ( x ) . z 3 2 ( i ) R e c a l l that [x] = 2 ^ - 3 ( £ - X ^ 2 z Dx V Dx (2) P o s s i b l e by Theorem 1, Chapter I I , (3) A c t u a l l y we should w r i t e "The Schwarzian of x with r e s p e c t to the d e r i v a t i o n on k(x) induced by d i f f e r -e n t i a t i o n with r e s p e c t to z on k ( z ) " . 63 A r 1 D n x / D n x \ 2 JL = 2 - 3 I — i . To compare these z U 0 X V 0 X Schwarzians we f i r s t of a l l compare " l i k e " terms. N o t i c e that e x t e n s i v e use of the Chain Rule i s made. D Qx = D X - D Q Z and D 2x = D Q(D Qx) = D C D Q X ) - D Q Z D 2 X D C D 0 X ) . D Q Z D ( ; D 0 X I D ( D x . p 0 z j D Qx ~ Dx • D Qz ~ Dx " Dx' 2 D X . D Q Z + D X . D ( D Q Z ) Thus Dx 2 1 But D~z = -z . For, l e t y = —. Then D„ i s d i f f e r e n t i a t i o n 0 J z 0 1 1 2 with r e s p e c t to y, and D~z = D-. (—) = - —? = -z . T h e r e f o r e r J 0 0 y y ^  U ) D 0 X D ' X . (-z Z) + (-2z) D Q x D X s i m i l a r l y we have D Q X D ( D Q X ) . D Q Z D ( D Q X ) D Q X D X - D Q Z D X D[D x.z + Dx•2z ] Dx D 3 x - z 4 + D 2 x - 4 z 3 + D 2 x . 2 z 3 + Dx » 6 z 2 Dx Thus ( M D 0 X L ° 2 * , 0 0 4 . , 3 , , 2 z + — • 6 z + 6 z D Q X Dx Dx From (a) and (b) we have 64 . r i o r ° 3 x 4 j. D 2x /- ^  , /- 2 . r.D 2x,2 4 . D 2x 3 . . 2 z 0 D 3x 4 „,D 2x.2 4 " 2 r ^ T " z - 3 ( b T - ) , z r l 4 = [ x ] z - z ( i i ) This i s c l e a r s i n c e d i f f e r e n t i a t i o n with r e s p e c t to z i s the same as d i f f e r e n t i a t i o n with r e s p e c t to z - c f o r c e k. Define the p r i n c i p a l t e rm of C x ] z a t z ~ c f ° r c e k , to be that part of the Laurent s e r i e s f o r [x] at z - < r z which co n t a i n s n e g a t i v e powers of z - c. Lemma 2 Let e be th.e r a m i f i c a t i o n index of k ( x ) / k ( z ) at z - c, f o r c e k. Then the p r i n c i p a l term of f x J z a t z - c i s 2 ( i ) (z - c ) " 2 + A Q ( z r- c ) " 1 i f e > 1, where i s some constant element. ( i i ) 0 i f e = 1. Proof ( i ) Since the r a m i f i c a t i o n index at z - c i s e, we can w r i t e U(x - C) = z - c f o r U a u n i t i n k(x) and C a constant i n k(x) . N o t a t i o n : Notice that [x] = [x - C] . Thus we z z-c want the p r i n c i p a l term of [x - C ] ^ _ c where U(x - C) = z - c For s i m p l i c i t y of n o t a t i o n , w r i t e 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 d i f f e r e n t i a t i o n with r e s p e c t to z by D, with r e s p e c t to x by d. From Ux = z we have D U ' X 6 + U'Dx 6 =; 1 or, dU'Dx ' X 6 + U e x e - 1 D x = 1, Thus Dx(dU«x e + eUx 6" 1] = 1 Let W - 1 = 1 + ^ 7 7»x eU Then (a) Dx = - J - x 1 - 6 N o t i c e that W i s a 1-unit i . e . a u n i t whose i n i t i a l term i s 1. For, U a u n i t i m p l i e s dU has order S 0, and ~ r i s a l s o a u n i t . Hence ord (^-jr'x) > 0, so that the eU eU constant term of W ^ must be 1. D2x = D(Dx) = d(Dx)'Dx From (a) then we have (b) D x = (-^- - x + - d(-^) x )Dx The re f ore , v D 2x 1-e W -e , 1 „ ,WN 1-e ( C ) Dx~ = ~e~ U X + e d ( U > X w w — i s a u n i t , so ord (d(yj)) i 0. Hence l W l - e D 2x ord (— d(—)x ) > 1-e > -e. The i n i t i a l term of — e U Dx (4 ) i s t h e r e f o r e at worst that of (4) One i n i t i a l term i s worse than another i f i t has a l a r g e r negative exponent. N o t i c e that we cannot say above that the i n i t i a l term i s . . . because 1-e may be 0 66 1-e W -e 1-e • / T T e . - l 1-e T-1 — x = (Ux ) W = z W. e U e e 1-e -1 T h i s has i n i t i a l term z s i n c e W i s a 1 - u n i t . e From (a) and (b) we have ... 2 1-e ,W,2 l-2e 1 W , ,W\ 2 ( l - e ) ( d ) CT? (p) x , + ? u d ( U ) X order > l-2e order > 2-2e 3 2 Then, s i n c e D x = d(D x)»Dx, we have D 3x D x = d(D^x) = — — c y ^— f :^"(u^ x + ( t e r m s °f h i g h e r degree) e - 2 2 (Ux ) W +... (1- e) (1-2e) e^ (1- e) (1- 2e) e^ (1- e) (1- 2e) e2 where, again, we have used the f a c t that W i s a 1 - u n i t . Thus the term with l a r g e s t p o s s i b l e n e g a t i v e exponent i n [ x ] z i s !2<1->i1-2«> - 3 ( i ^ ) 2 ] 2 - 2 .2 - 6e + 4e 2 - 3 + 6e - 3e 2 -2 = ( ~e~2 ) z e 2 - 1 -2 = n Z e z 2 e - 1 -2 -1 Hence the p r i n c i p a l term of [xl i s * z + A^z r v z e^ 0 f o r some constant A Q . R e p l a c i n g z by z-c give s the r e s u l t as s t a t e d . ( i i ) In the proof of ( i ) we d i d not ever use the f a c t that e > 1. So f o r e = 1, proof ( i ) i s v a l i d . But now we want to show that f o r e = 1, the e n t i r e p r i n c i p a l term i s 0 67 . For e = 1 , l i n e (c) of ( i ) reads DT " d ( u } -T h i s we know has p r i n c i p a l term 0 . L i n e ( d ) , f o r e = 1 , b e come s TJ d ( u > Thus _ 3 0 DT" = D ( ° X ) - r ^ W M 2 4. W H 2 ^ and t h i s too c l e a r l y has p r i n c i p a l term 0 . Hence the p r i n c i p a l 3 2 D x D x 2 term of 2 - - 3 ( — ) must be 0 , as d e s i r e d . Dx Dx Lemma 3 Let e be the r a m i f i c a t i o n index of kQx} /k(z) at °s>, and assume e > 1. Then the expansion of I x l z a t °° 1 S °f the form e 2 - l 1 , . 1 . A 1 . A 1 2— —7 + A „ — r + A , — 7 - + A „ —r-e^ z^ 0 z J l z ^ 2 z-5 f o r some constants A Q , A ^ , ,••• Proof From Lemma l ( i ) , [ x l z = " " T r f x ] ^ - Using t h i s f a c t together with Lemma 2 ( i ) we f i n d that the terms of l e a s t p o s s i b l e degree i n the expansion of [ X 3 Z a t 0 0 a r e 1 , e 2 - l , 1 , - 2 . . , 1 , - 1 , 2 e - 1 , 1 * . . ,1 •n—(—T) + A (—rr) , f o r some constant A „ e ^ z ^  Q z J 0 68 . Note: Lemma 3 t e l l s us that i f there i s r a m i f i c a t i o n at oo then [x] has no pole at °° : i t has a zero at <» of z order at l e a s t 2. The next lemma, deduced using the pr e v i o u s two, give s the b a s i c r e s u l t needed to conclude the Main Theorem. Namely, that the Schwarzian of x with r e s p e c t to z i s a r a t i o n a l f u n c t i o n whose c o e f f i c i e n t s depend only on the r a m i f i c a t i o n i n k ( x ) / k ( z ) . Main Lemma There i s a r a t i o n a l f u n c t i o n r ( z ) £ k(z) such that [ x ] z = r ( z ) where the c o e f f i c i e n t s of r ( z ) depend only on the r a m i f i e d p l a c e s and r a m i f i c a t i o n i n d i c e s of k ( x ) / k ( z ) . Proof R a t i o n a l i t y of " x ] z f o l l o w s from C o r o l l a r y 2 of Chapter IV. Suppose [ x ] z = r ( z ) e k ( z ) . Assume without l o s s of g e n e r a l i t y the generator z of F i x G chosen such that r a m i f i c a t i o n occurs at 1, 0 and °° (t h a t i s , at the pl a c e s ( z - 1 ) , (z) and (—) ) with r a m i f -z i c a t i o n i n d i c e s e^, e^ and e^ r e s p e c t i v e l y . From Lemmas 2 and 3 we f i n d that r ( z ) has poles at most at 0 and 1: not at 0 0 or at any u n r a m i f i e d p l a c e . Thus r ( z ) has the f ° r m e 2 - 1 e 2 -1 <*> r ( 2 ) = e T ( ^ i T 2 + ~ l + e-T2- + z + C 1 e 2 z f o r some constants A, B, C, (C i s a con s t a n t , s i n c e i f 69 . i t were a nonconstant p o l y n o m i a l , r ( z ) would have a pole at co . C o n t r a d i c t i o n . ) From (*) we can c a l c u l a t e the i n i t i a l terms of the expansion of r ( z ) at <» i n terms of the constants A, B and C. But Lemma 3 al r e a d y t e l l s us what these i n i t i a l terms are. Equating l i k e c o e f f i c i e n t s i n the expansions w i l l a l low us to s o l v e f o r A, B and C, and so to w r i t e down an e x p l i c i t formula f o r r ( z ) . In (*) the i n i t i a l terms of r ( z ) at 0 0 are found by r e w r i t i n g r ( z ) i n terms of u = —. Then z = — and (*) y i e l d s r< z> • r(u-> = \ i ' 2 + r r +  1 h i 4 + I + c e l ( u - 1 } u " 1 e 2 ( u> v 2 2 1 u u 2 2 = —2 2 + A j ^ + ~ 2 u + Bu + C 61 1 _ U e2 e 2 - 1 = '-^—2 ("2 + 2 u 3 + 3u 4 +...) + A(u + u 2 + u 3 +...) 61 62 " 1 2 + u + Bu + C 62 2 - 2 e - 1 e - 1 = C + (A + B)u + ( 2 + A + 2 )u +... e l ' 62 Lemma 3 s t a t e s that the i n i t i a l terms of t h i s expansion must be 2 , 7 0 -e 3 " 2 3 4 = - u + A Q u + A 1 u e 3 0 2 Equating c o e f f i c i e n t s of u , u, u give s C = 0 A + B = 0 2 i 2 i 2 i e l e2 ~ 6 3 2~ + A + 2 " 2~ e l 62 6 3 1 1 1 T h e r e f o r e A = ~ + ~2 ^ ~2 " ^ a n d B = ^A e l e2 e 3 S u b s t i t u t i n g these v a l u e s i n t o equation (*) g i v e s 2 _ - 2 . e1 -2 e 2 ^ x - 2 1 1 1 r ( z ) = 1 2 (z-1) + — 2~z + (;-| + - | - - | . l ) K z ) C z e l e2 e l e2 6 3 The c o e f f i c i e n t s of r ( z ) do, t h e r e f o r e , depend only on e l ' e 2 ' e 3 ' Now we are i n shape to prove the remaining case i n the Main Theorem. Let D , DQ denote the extensions to k(x) of d i f f e r -e n t i a t i o n with r e s p e c t to y, i n k ( y ) , k(yQ> r e s p e c t i v e l y Assuming, without l o s s of g e n e r a l i t y , that the generators y, yQ have been chosen such that r a m i f i c a t i o n occurs at 1, 0 and 0 0, then k ( x ) / k ( y ) and k(x)/k(yQ) have the same r a m i f i e d p l a c e s and r a m i f i c a t i o n i n d i c e s . By the Main Lemma then, there i s a r a t i o n a l f u n c t i o n r with c o e f f i c i e n t s i n k depending only on the r a m i f i c a t i o n 71 i n d i c e s of 1, 0 and <» such that [x]D = r ( y ) and [x]D = r ( y Q ) . But then i t i s easy to see that a([x] ) = Ix] For, a([x]D) = a ( r ( y ) ) = r ( o ( y ) ) = r ( y Q ) = [x]D . A l s o , s i n c e a i s an isomorphism we f i n d that a([x]n) = [a(x)]n " u0 Then, from these l a s t two equations, we have [x]_ = [o(x)1 By C o r o l l a r i e s 1 and 2 of Chapter IV i t f o l l o w s that k(x) = k(a(x)), as d e s i r e d . Remarks: (1) In the hypotheses of the Main Theorem we s t a t e p )( | GI . In the proof ( d e r i v a t i o n of the Hurwitz Formula i n p a r t i c u l a r ) , we used only the f a c t that p J( e f o r any r a m i f i c a t i o n index e. So our o r i g i n a l h y p o t h e s i s might seem too s t r o n g , i n that i t i s p l a u s i b l e that the case p | |G| but p e f o r any r a m i f i c a t i o n index e a r i s e . But the r e s u l t of the theorem guarantees that t h i s cannot happen. For, the theorem says that i f p J( e f o r a l l e, then |G| and the r a m i f i c a t i o n i n d i c e s e must be one of the f i v e c a n o n i c a l types. We know f o r these that i f p | |G| (p a prime) then p = e, f o r some r a m i f i c a t i o n index e. 72. (2) It turns out to be true that for any algebraically 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 in fact hold for the icosahedral case i n characteristic p > 0 , although i t does not yield to proof using the techniques here developed for the case of characteristic 0 . That the theorem is true for a l l cases in characteristic p > 0 follows from Dickson [1]. From §259 we have: i f k^ is the algebraic closure of the prime f i e l d Z/pZ and i f a subgroup of Aut k n(x) is isomorphic to the icosahedral group then that subgroup is k0 conjugate to the icosahedral group. By standard techniques of group representations and group extensions this result holds for any algebraically closed f i e l d k . Dickson's techniques, however, do not work for both characteristic 0 and characteristic p . Dickson uses f i n i t e permutation groups to deduce his result for characteristic p — i n characteristic 0 the corresponding devices are simply not f i n i t e . Thus, in characteristic 0 , some other method such as the one used here is necessary. 73 . B i b l i o g r a p h y [1] L.E. Dickson, " L i n e a r Groups", Dover P u b l i c a t i o n s , Inc., New York, 1958. [2] A. Hurwitz, "Mathematische Werke", Bir k h a u s e r and Co., B a s e l , 1932. [3] F. K l e i n , "Lectures on the Icosahedron", Dover P u b l i c a t i o n s , Inc., New York, 1956. [4] S. Lang, "Algebra", Addison-Wesley P u b l i s h i n g Co., 1967 . [5] H. Weber, "Lehrbuch der A l g e b r a " , Vol.11, Chelsea P u b l i s h i n g Co., New York, 1961. 7 4 . 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 in any characteristic 2, 3, 5. It hinges on the easy arithmetic of what we shall c a l l affine extensions L/K : L = k(x) , K = k(y) and, k[y] k[x] . In other words, a pair L S K of rational function fields i s affine, i f generators x,y can be found such that y i s a polynomial in x . As above, k w i l l be algebraically closed. The following lemma i s a synthesis of the lemmas on pages 35 and 53. Lemma: L/K affine <=> some place of K is totally ramified in L Proof: (<=) Let p , ^  be the places in question. As in the lemmas on pages 35 and 53, we have /•*\ N N-l N-2 . (*) w + a ^ - j U w + a N_ 2u w + ... + u = 0 where w e L , u e K are local uniformizing parameters f o r p , ^ , respectively Put x = — , y = — . (=>) From an equation F(x) = y with a polynomial w u F of degree N , we get an Eisenstein equation (*) for w = — , u, = ~ '•> x y hence total ramification. 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) is reflected completely i n the factorization of i t s uniformizing parameter y - c = F(x) - c in k[x] For instance, i f there i s another totally 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 line 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 hence a f i e l d L containing fields K^, K 2 such that [L : K^] =5 , ( i = 1,2) . It i s easy to see that ramification in each L/K; must necessarily be as follows: r We choose the generator x of L so that the uniformizing parameters of iR' ^ f l ' ^ 1' 2^ 3 r e x' X ' X ~ a > X ~ ^ ' r e s P e c t * v e x y » w n e r e a + 3 = -4 By our lemma and its proof, each K^  i s generated by a y = F(x) . Let us i n s i s t that F(x) be monic and that y be a local parameter for the 76. place JO* . We must show that F(x) is uniquely determined by these data. Ramif ica t i o n of and gives us 3 2 (1) y = x (x + ax + b) and 2 2 (2) y - 6 = (x-a) (x-g) (x-y) respectively. 5 4 3 Accordingly, F(x) = x + ax + 6x . We shall 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 (x-a)(x-6) i s determined. Thus (2') y - c = (x 2 + -jax + -|b)2(x-y) Comparing the linear and quadratic terms of (2') to those of (1) yields (i) 3b - 8ya = 0 , 2 ( i i ) 12ab - 15yb - 8ya = 0 , whence 3a = 5y 8 2 Therefore (i) becomes: b = —a . Since we had normalized -4 = a + 3 4 - —a , we have a = 5 and b = 40 . Quod erat demonstrandum. 

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