FINITE GROUPS OF FRACTIONAL LINEAR TRANSFORMATIONS by VIVIEN B.Sc. , University A THESIS THE BETH KITCHEN of B r i t i s h Columbia, SUBMITTED IN PARTIAL REQUIREMENTS FOR MASTER OF in THE 1969 FULFILMENT OF DEGREE OF SCIENCE the Department of Ma thema t i c s We accept required THE this thesis as c o n f o r m i n g to the standard UNIVERSITY OF BRITISH COLUMBIA May, 1972 In p r e s e n t i n g t h i s t h e s i s i n p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at the U n i v e r s i t y of B r i t i s h Columbia, I agree t h a t the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r r e f e r e n c e and study. I f u r t h e r agree t h a t p e r m i s s i o n f o r e x t e n s i v e copying o f t h i s t h e s i s f o r s c h o l a r l y purposes may by h i s r e p r e s e n t a t i v e s . be granted by permission. Department o f The U n i v e r s i t y o f B r i t i s h Vancouver 8, Canada Date O^t^C /3 Department or I t i s understood t h a t copying or of t h i s t h e s i s f o r f i n a n c i a l g a i n written the Head of my j Columbia /9?2~ s h a l l not be publication allowed without my Dr. Supervisor : K. Hoechsmann ABSTRACT In this thesis we linear transformations closed field all finite divisible k. consider of a v a r i a b l e The p u r p o s e subgroups t h e group of t h i s x o v e r an of the t h e s i s group by the c h a r a c t e r i s t i c whose o f k. of f r a c t i o n a l algebraically i s to d e t e r m i n e orders a r e not ACKNOWLEDGMENTS I would for of like h i s guidance this thesis. made t h e t a s k t o e x p r e s s my g r a t i t u d e and e n c o u r a g e m e n t D r . Hoechsmann's enjoyable. I would for reading and the U n i v e r s i t y of B r i t i s h ass i s tance . t o D r . Hoechsmann throughout enthusiasm also t h e t h e s i s and t h e N a t i o n a l Columbia like the p r e p a r a t i o n f o r the subject t o thank Research D r . Gamst Council for financial TABLE OF CONTENTS Chapter I Introduction 1 Chapter II Algebra 9 Chapter III Differential Chapter IV The Chapter V Ramification Chapter VI The Chapter VII Consequences Chapter VIII Proof Bibliography Preliminaries Algebra Schwarzian Hurwitz of 13 Derivative 22 Theory 26 Formula 45 of the H u r w i t z t h e M a i n Theorem Formula 50 60 73 CHAPTER I INTRODUCTION The version result We intention thesis i s to g i v e an up t o d a t e o f part of Klein's t r e a t i s e [3], and to extend Klein's from the complexes t o any a l g e b r a i c a l l y t h e g r o u p $J o f f r a c t i o n a l consider ations of t h i s of a v a r i a b l e linear closed field. transform- x ax + b — J cx + d x —> 1 where the elements closed field matrices k. over That o f k. k = C, ^ automorphisms having order to m o t i v a t e i s isomorphic not d i v i s i b l e observation i n our r e s u l t It subgroups group, i n the We the c y c l i c of analytic know t h a t groups, the o c t a h e d r a l this the d i h e d r a l group and t h e group. turns out that these o f xb f o r g e n e r a l to t h e c h a r a c t e r i s t i c these t o t h e group sphere. subgroups the t e t r a h e d r a l icosahedral prime 2 X 2 Our aim i s t o d e t e r - The f o l l o w i n g o f t h e Riemann has as f i n i t e groups, algebraically case: For group centre. of & c a s e where k = <C s e r v e s general that their to an the i n v e r t i b l e 2 subgroups the c h a r a c t e r i s t i c the i s ,PSL (k), k modulo mine a l l f i n i t e by a, b, c, d b e l o n g are e s s e n t i a l l y groups k, i f t h e i r also exist order o f k. In f a c t , a l l such finite as finite is relatively we shall subgroups show of . 2. As subgroups evidence o f is a subgroup P X the give i s N and e x i s t e n c e of their the the generators I where The cyclic generators. characteristic and canonical I f the order o f k i s p, of assume that The d i h e d r a l groups Cyclic Group N D i h e d r a l Group where e i s a p r i m i t i v e the generators three are also of u n i t y i n of o r d e r i) n*"* 1 a root remaining N i s generated n what ( l of u n i t y the the groups generators in by k. N = 2n d i s generated k. derive However, u s i n g i t can easily f o r k any be well- shown that algebraically closed field. Let G denote one of the three figure F corresponding to G i . e . the hedron or as icosahedron, Then elements form F into of G are itself. embedded rotations Every of element groups. View tetrahedron, i n the the of Riemann sphere G must by o) c a n o n i c a l g r o u p s we c o n s i d e r i n g k = C. by known p r o p e r t i e s o f these of o r d e r root (o For i t is clear are. e is a primitive II the these N. For the we of the octasphere. which then be transone of 3. the following: a vertex meeting a t any v e r t e x through the a x i s erate We G i t suffices now rough give t o t h e number S of order of f a c e s octahedron 2 about the m i d p o i n t these three to g i v e the g e n e r a t o r s o f each R, S and T f o r e a c h s t e r e o g r a p h i c p r o j e c t i o n of each related interested 3 Since are conjugate, one e l e m e n t shown as an a i d t o t h e r e a d e r T of order of a f a c e . types or the a x i s o f an edge, o r a r o t a t i o n through of each.of i = 3, 4 o r 5 a b o u t i n the t e t r a h e d r o n , or a r o t a t i o n the m i d p o i n t elements R of order of F ( i corresponding icosahedron), about a rotation type. group. to gen^^ A figure i s i n checking the generators. Ill The Tetrahedral Group Let OJ be a p r i m i t i v e 3 rd root of uni ty. S tereographic P roj e ct i o n o f Te t r a h e d r o n 1 R (1) G can i n f a c t a l w a y s be g e n e r a t e d See, f o r example §§71-74[5]. 1 by two elements. V The Icosahedral Let e be Note: - Group a primitive -, e + e - 1 = -3 + ~ /5 5th root of unity S tereographic P roj e ct i o n of Icosahedron -1 -3+/5~ With can TV) now s t a t e our It turns the the existence out of the these subgroups of & accepted, we proposal. that icosahedron, origin of -1 five on the r = ~ { outer stereographic z J ^ ) ^ icosahedron t h e p r o j e c t i o n of distance vertices. from MAIN THEOREM Let istic p. k be an algebraically Any finite subgroup is isomorphic to one of if G i s isomorphic jugate to t h a t finite subgroups This to one group. are result the closed field G of PSL2(k) f o r which c a n o n i c a l groups of I-IV then of c h a r a c t e r - I-V. In G is actually Further, in characteristic conjugate immediately to one gives of the p )( |G| fact, con- 0 all I-V. following special- ization. Let Any F be subgroup either ( §256). all finite Dickson's Our will become c l e a r here that our algebraically Our first linear k to one i s stated subgroups to p. We over result [1] subfield of k c o n t a i n i n g p o f PSL^CF) whose o r d e r conjugate This prime a finite now step I-IV of PSL^CF), our matter not only those with f o r a v o i d i n g the most for proof be designed closed field k, not our plan i s to o b s e r v e so finite as t r a n s f o r m a t i o n s of x over the deals to V. with order general develops. to h a n d l e case Note any fields. for accomplishing that p is i n Dickson this technique by i s isomorphic of methods w i l l outline or divisible i n a s t r o n g e r form treatment reasons as of i s not elements n group of the proof. fractional k i s isomorphic to A u t ^ k ( x ) , 7 . the group of automorphisms this now fact and o f k ( x ) o v e r k. then choose We to work w i t h establish Aut^k(x) thereafter. Theorem k(x) = k(y) < = > some e l e m e n t s a, b, y = *x c, d E k s u c h that t d for ad-bc 0. 4 Proof ( <f must ) k(y) c determine k(x) i s clear. x i n terms /y\ matrix form as o f y. ' this we \c la have = d / we "fr* b ;—T i n cx + d 3 X 1/ b\-l U \lj k(y) b\/xV III* /a 1 show k ( x ) c Rewrite y = I = /x\ From To l / ~ \ c -a la b\ where t h i s i s p o s s i b l e s i n c e d e t 1 = ad - be f 0. -d + b \c d/ x = ^ , and t h e r e f o r e k ( x ) k(y). cy - a 4 ) d| -1 k(x) = k(y) i m p l i e s , \ a d b in particular, c that Hence y e k(x). A fx) So y = -z-j—r to be relatively Let satisfied k(y): f o r some A ( x ) , B ( x ) we may assume prime. I(X) = B(X)y - A(X). by x o v e r k ( y ) . o t h e r w i s e , i t would polynomial e k [ x ] , which from each I i s a non-zero p o l y n o m i a l Further, be I is irreducible possible o f the r e l a t i v e l y to f a c t o r an over irreducible prime p o l y n o m i a l s 8. A(X) and B ( X ) . field theory, Thus I(X) = I r r ( x , k(y), X). [ k ( x ) : k ( y ) ] = deg I ( X ) . and deg I ( X ) = max and B(X) = cX + d f o r some e l e m e n t s ad - be j> 0. k-multiples Hence, we o f each o t h e r That this any e l e m e n t fractional c, x i In at the d e s i r e d Aut^k(x) i s such transformations -• "j"—j- that thinks these result theory be i n k.) = y, that i s clear. o(k(x)) = of x over k a c t s by s a y , f o r some e l e m e n t s a , b, ad - be 4 0 . finite of using subgroups Galois a l g e b r a i c techniques dependent 0, we invoke o f A u t ^ k ( x ) one theory and f i e l d and t h e H u r w i t z on d i f f e r e n t i a l o f the M a i n Theorem, characteristic result o f t h e group o f algebra f o r i t s d e r i v a t i o n ) , we deduce clusions therefore a, b, c, d as s t a t e d . considering naturally (a (a b) and ( c d) would be and y w o u l d establishes linear d E k such With a , b, c, d e k f o r w h i c h = k ( y ) , s a y ; and any e l e m e n t mapping Thus A(X) = aX + b A ( x ) _ ax + b ~ B(x) cx + d' with k(a(x)) = 1, have y For, But [ k ( x ) : k ( y ) ] (deg A ( X ) , deg B ( X ) ) . ( I f ad - be = 0, Hence, by as w e l l Formula and r a m i f i c a t i o n the f i r s t F o r the f i n a l theory. two con- conclusion i n the s o - c a l l e d Schwarzian (3) derivative , an o p e r a t o r i n the theory of differential algebra. (3) Our use o f t h e S c h w a r z i a n d e r i v a t i v e i n t h i s c o n t e x t follows K l e i n [ 3 ] . Here he uses t h e S c h w a r z i a n i n f i n d i n g the f i n i t e s u b g r o u p s o f P S L ^ ^ ) . CHAPTER II ALGEBRA PRELIMINARIES Terminology: By shall mean a finitely transcendence degree A function field For from Theorem 1 Let element y having only one subgroup the same field k we of k of theory over about i s then of shall an a l g e b r a i c k need three special algebra. element of Aut^k(x). i s a special claim field we transcendental transcendental This over generator. sections Fix Note: extension function the g e n e r a l a finite field 1. rational x be a func t i o n generated use i n l a t e r results G be an a l g e b r a i c such over Then a field there k. exists Let an that G = k(y) case of LUroth's any s u b f i e l d of Theorem which makes k(x). Proof Let Define over k. k(y) = {o,,a_,...,a } ^ . A u t . k ( x ) I z n K y = a ^ ( x ) a ^ ( x ) • • • o" ( x ) . Clearly n We will show that y e F i x G and where n = y I GI . i s transcendental that, in fact, F i x G. To for G = show that a l l i = l,...,n. y e F i x G, we must show F i x some i , 1 < i < n. that c^(y) = y Then 10 . a . C y ) = a . ( a ( x ) . . -a^x) ) (1) 1 = Clearly, then a. l (a ± a )(x)...(a 1 a )(X) i r a . •a . e G f o r i = l , . . . , n . 1 j a . ° a . ^ a.oa, . l j x k a . = a, . j k Hence G = {a . • a , , . . • ,a . o a l 1 x n o^(y) = y and y i s i n d e e d i n F i x G. By From Thus G:k(y)] t h e o r y we know that (1) we = see t h a t Hence [k(x):k(y)] [k(x):Fix [k(x):k(y)] complete the p r o o f by showing This finishes |Fix G : k ( y ) | = 1, and s o , t h a t In is from GJ = |G| = n. (2) t h e n we have ( 3 ) We }• have now k ( y ) £ F i x G £ k ( x ) . [k(x):Fix G][Fix Galois that r c a .» a . = a. \ a . • a, , so t h a t i j x i k (2) i f a. ^ a, j k F o r , a . ° a . = a.«a. i m p l i e s l j I k L We Further, the p r o o f s i n c e the I n t r o d u c t i o n we i s o m o r p h i c to t h e group mations. Thus each 1 ' 1 n Then y = a.(x) i=l C i X (2) i t t h e n i follows established that Aut^k(x) linear a i X + transfor- ^ i , c : — j — that for a.d. - b . c . 4 0. 1 1 fCx) —T^-V' = g ( x ) that Fix G = k(y). X I ; — = n. c.x + d . x x 1 ; from [k(x):k(y)j i s o f t h e form x a^x+b^ II that of f r a c t i o n a l some a . , b . , c . , d . e k s u c h 1 ' > n f o r some relatively prime 11. polynomials f ( x ) , g ( x ) e k [ x ] such that deg f , deg g < n. Let F(X) = g(X)y - f ( X ) . F(X) i s i r r e d u c i b l e over and F ( x ) = 0. Hence [k(x):k(y)] Together Theorem i.e. F(X) = I r r ( x , = deg F ( X ) = max(deg with 2 (3) t h i s gives k ( y ) , X). f , deg g) < n. the r e q u i r e d E x i s t e n c e of a S e p a r a t i n g Let result. Transcendental K be an a l g e b r a i c f u n c t i o n f i e l d k i s of c h a r a c t e r i s t i c x e K such p. Then k(y) there over exists k, where a transcendental that K/k(x) i s s e p a r a b l e . Proof Let x e K be t r a n s c e n d e n t a l over K 3 k ( x ) r> k and K / k ( x ) a l g e b r a i c . O J are n a l g e b r a i c over existence 1 2 ) , where the n k, and x = x^. n = 1 then Use x ,...,x 0 2 n i n d u c t i o n on n to show K = k ( x ) , and this is clearly separable k(x ). Assume Since that k ( x ^ » • • • > _ ^ ) x f such f a r e i n k. z s n i s a l g e b r a i c over polynomial of have of a s e p a r a t i n g t r a n s c e n d e n t a l . If over Thus we Suppose K = k ( x , x„ , . . . , x ) = k (x, , x . . . , x ' 2 k. that If x separable k ( x ^ ) , there exists over an k(x^). irreducible f ( x , , x ) = 0, where the c o e f f i c i e n t s I n i s s e p a r a b l e o v e r k ( x j t a k e the s e p - n 1 a r a t i n e t r a n s c e n d e n t a l to be x.. Assume t h a t x i s not ° I n s e p a r a b l e over k ( x ^ ) . Then f must be a p o l y n o m i a l i n ^ • x n If x, i s i n s e p a r a b l e o v e r 1 in x A Thus k ( x ) then n f i s also a polynomial 12 . f(x ,x ) 1 where this = g(x n P 1 , x g i s some p o l y n o m i a l i m p l i e s that x, s e p a r a b l e 1 over k(x ). n r ) = with [g(x , 1 over x )] , P n coefficients f is reducible. x, must be s e p a r a b l e 1 Thus we P n i n k. Contradiction. But Hence k(x ). n have ^ separable over k ( x ^ ) and over k(x ). So x. x , are a l l separable n 1 n-1 T h e r e f o r e , take x to be t h e s e p a r a t i n g t r a n s n cendental . The a proof f o l l o w i n g i s a standard of which can be found theorem on p . 171 of a l g e b r a , [4] . Theorem 3 Let field A be an a l g e b r a i c a l l y k, K an a l g e b r a i c e x t e n s i o n is a monomorphism of K into A. then closed extension o f k. p can be e x t e n d e d of a I f p : k «—»- A to a monomorphism 13. CHAPTER I I I In deal with DIFFERENTIAL the chapter derivations concerning them. section. The that under This i s used i n the part of chapter the and This this this which Hurwitz Formula. will result theorem and are proof of deals any k be This and K are have: for lemma lemma the of states can differentials be extended. The and development both = {f fDg latter related of the the Also e K Note: D since notice k. D(f ) that then D(fg) For For, f e K P i f char = pf p - 1 Df = fDg = + ( x ) , Df gDf of k, for D k £ K^. But = 0. over case where following But K. Q to For, i t i s not k = p and D is linear that then that —*• V D:K V. o f cons t a n t s that such k-derivation. i n the a K-module true £ a field = 0. —»• V k-algebra gDf extension field = 0}. D:K the = D ( a ' l ) = a « D ( l ) = a«0 Q + used D(a) K K a commutative Assume i n K an K into be that e K e K|Df = will fields. of define ring, Notice g the will result M a i n Theorem. f, g e K i s c a l l e d k a field, We P the Then a k - l i n e a r map definition a k-derivation f in derivations the a commutative elements Note: D general i s found with necessary V a K-module. for K one following D(fg) we need d e r i v a t i v e we Derivations Let k Schwarzian suitable conditions results (1) on ALGEBRA f f e K P For, need be if a e k, necessarily then not be i f f e K^ in and fDg. = D f*Dx where D denotes "formal k. 14 . differentiation Denote with respect by D ( f ) , to x". f o r f e K, X D(D(...(D (f))...)). i the element Then we / have the following Lemma Let be extension a k - d e r i v a t i o n with f i e l d ^1' if K be a f i e l d ^2'**"'^n det (D fj) 1 E ^ a r e L i n e a T = 0, where ± o f k, and of c o n s t a n t s y dependent i = 0, l e t D:K . The over —• K elements i f and only 1,..., n - l and j = 1, 2,..., n. Proof Assume ^ ' ' " ' ' ^ n Then, w i t h o u t such a r li e n e a*"ly l o s s of g e n e r a l i t y , dependent there • • • e D i s linear f , = X „ f „ +... + X f . 1 2 2 n n i t follows that over D f, = X„D f„ + 1 Z z 1 a l li=l,...,n-l. therefore linearly Now show t h a t For nonzero assume The columns dependent. that of the m a t r i x n (D fj) are 1 (D fj) 1 = 0. ( D f . ) = 0 and use i n d u c t i o n J l i n e a r l y dependent over K„. D be n = 2, we have otherwise D f X n Hence det the f . must J since ... + X i det we exist . that Since for over 1 f^, f 2 they £ K. Assume b o t h are l i n e a r l y to to be dependent. Hence have f l f 2 = 0 D This implies that f l ° 2 the columns f are l i n e a r l y dependent over K, 15. Thus there Clearly exists we will D to (1) we this yields A e K be Hence A e the result f (2) Df i f we f are true that linearly for j = = A Df2 x n 2 show A e A. . Together so that Applying with DA (2) must be 0. f , ..., f ( D f j ) = 0 f o r j < n-1 dependent. e K and n exist (n) the (1) case n = implies X We will show this n. 1 to 2 4 0, det 2 we assume none i s 0. A „,..., A f = A f ± = 2 »f D n n 2 x (2) D f But ( D f . ) = 0, t h e r e j ' for xf can + 2 (1) As = ± . Let det that (1) = A Df 1 = 0. Assume now that done have D f f^DA such 1 f 1 1 £ K such ... + + 2 A Df 2 + 2 = A„D n 1 f . + the A^ assuming that A f n + n A Df n ... I 2. show t h a t ... Then e + n A . n D n _ 1 f n Applying gives Subtracting Df, = (DA - f . + ... 1 2 1 (2) f r o m t h i s g i v e s (1 ' ) Similarly, applying D to -f ) + n n • f + ... 2 2 (2) we have DA 2 0 (3) DA DA D f , = (DA • D f „ +...+ 1 2 2 Subtracting + «Df n n + ) + DA n (A„Df •f n DA •Df 2. +...+ 2 (A„D f 2 2 +...+ 2 DA -Df n = n A Df ) n n =0 yields (2') +...+ 1 2 0 A D f ) n n 2 16 Repeating this operation (1') D X - f „ +...+ 2 2 D A - D f +...+ 2 2 1 0 (n-l') system can be n 2 f„ written 2 = 0 n DX •D n n 2 f n = 0 ... Df. . DX =0 n 'Df n +...+ obtain as f. (DX -f n DX 0 DA„-D 2 i n a l l we DX 0 (2 ) This n - l times Df 0 ) n \D If det (D are done. j = 2,...,n, we find ( D A , . . • , D A ) = 0. 2 so But n that e a c h A.l over f. n f j ) = 0 for j = a s s u m p t i o n we for n-2 * , as e D. 2,...,n t h e n by uniquely induction therefore from above s y s t e m then the DA = 0 for ± f .I , . . n. , f that i = are det (D fj) 1 that 2,...,n, l i n e a r l yJ d e rp e n d e n t desired. Extension a finitely separably the Assuming Thus Theorem If \0 / algebraic extended to generated then any of a Derivation extension K(x) d e r i v a t i o n D on over K K can is be K(x). Proof Since need only define D(x). f(X) = Irr(x, Let for a. i e K. applying D is defined any Then f(x) derivation on K, K, X) = a x° n d to to extend = a X n +...+ this n +...+ a,x gives D to K(x) we + a , + a_ = 0, 1 0 and Q f 0 17 . [d(a v n ) + d(a ^x "" Hzi^ n 1 x +...+ 1 d(a )] + n 0/ [ (na x " 1 1 . _ n = f (x) must +...+ a )d(x)] °, n = f'(x) d Thus we 1 d e f i n e Dx by f (x) d This defines the o n l y possible extension Note: division sense since f'(x) ^ 0 f o r x separable (2) by f ( x ) i n the d e f i n i t i o n 1 o f Dx over makes K. Dif f e r e n t i a l s Let D:K o f D to K ( x ) . K be an a l g e b r a i c f u n c t i o n f i e l d — > • ^jr/k. t ^ 0 e t n over k. Define u n i v e r s a l k - d e r i v a t i o n i n the f o l l o w i n g e sense: For exists any k - d e r i v a t i o n D:K a u n i q u e K-module following diagram map >W, u: W a K-module, ^K/k * ^ such there that the commutes K y | JJ, i i 4W Once we being have existence of such a universal pair d e f i n e d by t h e u n i v e r s a l mapping be u n i q u e up to i s o m o r p h i s m . Existence of (d, ^J^/J^) for x exists by Theorem some s e p a r a b l e differentiation element with i t must : Take x to be a s e p a r a t i n g (Such property, (d, transcendental 2 , Chapter 9 e K. respect i n K/k. I I ) . Thus K = k ( x , 9 ) Now i n k ( x ) , formal to x d e f i n e s a d e r i v a t i o n d = 0 18. K on k ( x ) . Then, to give derivation a Take y(f«de) = — y f•DO, d i f f e r e n t i a t i o n u»d = The for the D, as continuous Let a In in the f ology e K. ^ j ^ ^ i s be extended u: Denoting to 6, = D we ^^/k by then D = ^—** w formal Q have f • DG , w h i l e K < f o r Df = D f o K, e f • D8 . known as the module of d i f f e r e n t i a l s K/k. what we shall p a r t i c u l a r , we be a a (f) = This order shall of on of valuation mean f i e l d "ord" defined f ord by be of by as a continuous interested formal formal i t a power power function defines a i n series series. of t, metric on K way. topological i n which scalar can desired. K-space i s a vector space W with top- addition W X W —*• W m u l t i p l i c a t i o n (ii) are K. respect k((t)) (i) and e f i e l d usual A f define K valuation any above, i n ord for as d i f f e r e n t i a t i o n = k ( x ) , K. on f o r define d i f f e r e n t i a l . has W K-module now over Kde. with extension We K separable = u ( d f ) = u(D„f«de) (pod)(f) Hence d = K/k D:K Given by Q being K X W —>• W continuous. Define d:K —>• c fi„to K/k be the universal continuous 19 . derivation in For any topological such that the following continuous K-space, the sense: k-derivation there following D:K e x i s t s a unique diagram is —»• W, W a K-module map u commutative d For any field of formal power s e r i e s K = k((t)) c we will exists; presently again, Existence of Set show t h a t once we (d, ^^/y, 1-dimensional such have a universal existence, pair (d, uniqueness is automatic : ~ ^ ^ t ' ^ w module. e r e dt Define ^ d:K the s generator of = k ( ( t ) ) —>• ^ y K this by k d : t t—* d t . Notice the is formal as that Df = f'*Dt d e r i v a t i v e of f o r any f with f e K where respect to t , and f' denotes D:K —• W above. Proof: Given f by J functions know t h a t Df n f Df e kft, t n = f f e K, n by X ] . by ( i i ) of = lim the Then, since can for approximate the ''Dt, Df n = lim n-*-" 1 n->-oo But c o n t i n u i t y we (f 1 n •Dt) t o p o l o g i c a l K-space l i m ( f '-Dt) n n-*-°° = definition (lim f )'Dt n n->°° 1 f n we 20 . Hence Df = ( l i m f ')«Dt = f ' D t . ^ n n Then Define Then (We = Kdt —> k Df = f ' D t and follows K. = f'Dt. d i s continuous.) required. 0 call since easily. f'Dt, f o r f e (y.d)(f) = y(df) = y ( f ' d t ) i s , D = y d , as ^he module of continuous differentials K/k. Next we define c Let the o r d e r Since ~ k((t))dt, f o r some e l e m e n t f e k ( ( t ) ) . co to be the o r d e r of the f u n c t i o n ord(co) = Remark: any L e t t ' be independent therefore Let we a power have differential, f. can w r i t e define ord(f) o f the g e n e r a t o r as a power chosen the f o l l o w i n g 1. = f o r K. Then series in t'. i t must We must holds. of K oyer to = f ' d t and to = g ' d t ' where Then o r d ( f ) of Write s e r i e s i n t of order t , t ' be g e n e r a t o r s to as the o r d e r F o r ord(co) to be w e l l - d e f i n e d , show t h a t e k((t')). we Then f e K = k ( ( t ) ) c a n be r e w r i t t e n So K = k ( ( t ' ) ) . be of a continuous c co e f«dt g property W by y: f ' d t —> d f = f '«dt, as f o r D, We of t h e u n i v e r s a l mapping y: ^ £ / have That - > o o k. For w e ^K/k f e k ( ( t ) ) and ord(g) Proof: t' for e K = k((t)). u(t) a unit in k((t)). Hence Then, we can w r i t e letting D t t' = denote u(t)'t formal 21. differentiation with dt' It For, Thus o r d a unit We co = f-dt. + now t , we (u(t))'dt't + = (D (u(t))-t t = u (t)-dt, see ord u(t)) = (u(t)) so that = 0 and ord (u in ord k((t)). (D u(t))) (t)) = t ord 0. have w = g ' d t ' - Hence f = g « U Q ( t ) . ord u(t))-dt U g ( t ) i s a unit that implies + u(t)'dt say. Q to have u(t)-dt = D ( D ( u ( t ) ) « t ) > 1, (D(u(t))-t to = d(u(t))«t + i s easy u(t) respect ( f ) = ord = (g • ( t ) ) •dt. Therefore (g'uCt)) ord (g) + ord = ord (g) + => o r d (g) 0 (u Ct)) 0 But also > 0. 22 . CHAPTER IV THE SCHWARZIAN Definition o f the S c h w a r z i a n Let field DERIVATIVE Derivative: D:E —*• E be a d e r i v a t i o n on a f i e l d of constants Derivative k. of h with F o r h e E-k d e f i n e respect where h ' = Dh, h ' = D h 1 Note: Since makes sense. 2 E, w i t h the S chwarzian to D to be and h ' ' = D h . 1 3 h e E-k, h' i s n o n - z e r o and t h e r e f o r e [h] Theorem Let D:E —*• E. for E be a f i e l d L e t k be t h e f i e l d f , g e E-k t h e r e such on w h i c h exist there i s defined of constants a derivation f o r D. a, b , c, d e k w i t h Then ;id-bc ^ 0 that f = tf-H<=> D - I*] If] D Proof Assuming that ad-bc f f = ^ "j" ^ f o r some a , b, c, d e k s u c h eg + a a 0 , we cfg (1) Applying see t h a t this are (1) This D to t h i s says linearly to - ag + d f - b = 0 gives the e q u a t i o n c ( f g + f g ' ) - ag' + d f (2) i s equivalent that dependent = 0 = f g + fg', h over k. 2 = -g ' and h 3 = f By t h e lemma on the W r o n s k i a n ad-bc ^ 0 s i n c e o t h e r w i s e f would [ f ] P would n o t be d e f i n e d . be i n k, and so 23 (cf. §1, det Chapter (D h,) being x I I I ) we 0. f'g det (D h ) = c o l 1 plus 2 f V + fg'* 3f ' V + g + f(col 2) + 3f'g" plus > •g 3f " g * -2f V the + 3f ' g " f' -g f • ' this gives f f'' t i = determinant, out 2 we 2 2 2 = 0 gives ( f ' ) [ - 2 ' ' • •+ 3 ( g " ) ] + ( g ' ) [ 2 f ' f " ' - 3 ( f ' ' ) ] 2 2 2 2 g = 2 Dividing two through by (f'g') and transposing one of the expressions, 2 f f - " 3(f ' ' ) 2 (g')^ Or, f ' ' ' f' f' ' 2 f' That i s , Remark: sentences Here, [f] Our proof (1) assuming and 0 2 -3f'f " g ' g " + 3f'f " g ' g " + 3 ( f ' ) ( g " ) g = have 2 this 0 "g 2f'f " ' ( g ' ) - 2 ( f * ) g ' g ' ''-3(f' ' ) ( g ' ) Simplifying f ' * * -g (-g'f ' " + f ' g " ') + ( 3 f ' 'g'+3f 'g' ') (-f » *g' + f ' g " ) Multiplying to -g fg' * ' -g* 2f i'g™ expanding + - g ( c o l 3) 0 Or, is equivalent fg' f • ' » Adding this Hence + f*'g know t h a t D - is a i [g] - 3 D series of equivalences except at (2). [f] D = [g]p and f o l l o w i n g the proof back 0 24 . to sentence (2) we know that there e x i s t a, c, d e k such that c(f'g Recall that this + f g ' ) - ag' + d f simply D(cfg So, by d e f i n i t i o n b e k. means that - ag + d f ) = 0 o f k, c f g - ag + d f = b conclusion useful f o r some element Thus £ _ ag + b eg + d The = 0 f o r some a, b, c, d e k s u c h t h a t ad-bc f 0. of t h i s theorem c a n be r e s t a t e d i n a more form as Corollary 1 k(f) = k(g) [f] D =[g] n Proof This theorem Recall proving enables that If function from t h e above t h e end o f C h a p t e r I . to need the Schwarzian i n c h a r a c t e r i s t i c 0. use o f t h e p r e c e d i n g theorem the f i e l d field This over E of the theorem holds the f i e l d since i s separating; field of D-constants i n c h a r a c t e r i s t i c 0 any i f D i s zero t h e n D can o n l y fact: i s an a l g e b r a i c an a l g e b r a i c a l l y c l o s e d 0, t h e n What on t h e i n c h a r a c t e r i s t i c 0 i s the f o l l o w i n g characteristic cendental directly we a r e g o i n g t h e Main Theorem us t o make Schwarzian dental follows and t h e t h e o r e m n e a r Remark: for statement kg o f i s k^, transcen- on a s e p a r a t i n g be t h e t r i v i a l trans- derivation. 25 . For non-zero form characteristic a non-trivial makes next extension the Schwarzian The however, trivial the D-constants of kg. This may i s the f a c t o f no use t o us i n t h e g e n e r a l part o f the p r e c e d i n g which case. theorem y i e l d s the result. Corollary 2 Let H be a f i n i t e group where x i s an i n d e t e r m i n a t e field of H i n k ( x ) . field of constants k(x) (since Schwarzian k(x)/F over Suppose k. We of automorphisms know i n F. L e t F be t h e f i x e d D i s a derivation that i s separable of x l i e s k. of k(x) D extends algebraic). on F w i t h u n i q u e l y to Then t h e That i s , [x] e F D Proof Let need only morphism u e H = Gal (k(x)/F). show that y([x]p) of k(x), i t follows y([x] ) D Now = To show [x] « Since n [ x ] e F we D y i s an auto- that v>(*>J u D : k ( x ) —»- k ( x ) i s a d e r i v a t i o n Y D on k ( x ) , and uD| = D. r Thus, the e x t e n s i o n of D to k ( x ) b e i n g unique, yD = D. Therefore p([x] ) D Now by y e Gal (k(x)/F) Corollary 1 this implies ) - t ^j) x [u(x)] a s D = D that k(x) = k ( y ( x ) ) . i s e q u i v a l e n t to [x] Hence M ( [ x ] = [y(x)] required. D But 26. CHAPTER V RAMIFICATION Throughout ally By closed field, 2, Theorem K extension and to be of an I I we i n K/k. assume k algebraic have Thus we x e K. i t s extensions are can our the main the following (1) Places The origin meromorphic introduced of point the h e r e as the text i n a f i e l d as here and & be of every that z = n. that of which i s not i s an 1 1 T I as we be u above state We be and notion convenient define a in place each of these equivalent. a valuation & is called written e & are the to v i e w as of Places i r r e d u c i b l e element f o r some u n i t element able a field. can for field theory be in fact will sep- interest. supersedes It w i l l are a k(x), the surface. to be a place separating to be of over case. notions. they z e & - {0} source field a field i s i n the a Riemann two i f there UTT , An theory argument of purposes general concept a domain w h i c h such integer the our concept ring form on e i t h e r of valua tion the this algebraic- a d d i t i o n a l work Riemann s u r f a c e . then note One Let of i n the functions on main theory an function assume K For However, where p o s s i b l e w i t h o u t to be existence a.rational function some t r a n s c e n d e n t a l k(x) chapter Chapter transcendental arable this THEORY is called a a TT e O uniquely some ring. in non-negative local k 27 . uniformizing parameter quotient parameter i s of field the z = UT: , 11 seen write in & We n = ord call (z). e L|ord {z the u n i q u e maximal Our second A valuation on for of has the the f ideal of notion of L (0) = Let the order easily of z and Then will is a function n is uniformizing e L|ord a place the uniformizing exponent = ». {z L be expression particular ord and a field The n uniformizing i n cr. a unique exponent Define local fixed local n e Z. and (z) * 0} 0= other UIT , u a u n i t Then, independent ir. parameter any z e L - {0} u a unit to be form of U. p a r a m e t e r i r , any f o r &; > 0} (z) be v:L as a is valuation. —*• Z U {°°} satisfying (i) v(a) (ii) v(ab) (iii) v ( a + b ) > Min Any such v a l u a t i o n & - iz ideal c L|V(Z) ^ = {z a valuation if defines > 0} a valuation notions, we will specifying (b) (v(a), on L. to r i n g i n L. r i n g with quotient quotient field Since ideal L field as of L. on L. field L, on are For, maximal L. i s the these them b o t h maximal v(b)) a valuation quotient refer the v {">) i s a v a l u a t i o n r i n g with defining since and r i n g with o r d : L —*• Z U a valuation + a valuation ring in L defines 0" i s a v a l u a t i o n function = v(a) is a valuation e L | v ( z ) > 0} 0 = «> i f f a = Conversely, Namely, then the Thus giving same as equivalent places. In a valuation fact, ring 28. determines the ring, we will normally use the following notation: by ideal of the valuation a place valuation ring by 6" ( o r 0^ itself v a l u a t i o n by place of Given of , we to two will a place out for for proof of p l a c e ^2 i s an plies extension this can the 299 and [4]. a placexp, we the extension actually To the the Q call a given reader maximal Denote a finite to e x t e n d the place be on carried is referred denote that w r i t e ^|^» a read over ^ . in this a function be written For , L o f L^; of denote uniformity process to p. extension Consider are fact For able This L we i f necessary), and to be L^. on any ord^. fields want a field a s s o c i a t e d w i t h fl. ring associated ring ~jQ of now field we f e k(x) as ^ e a c h a e k, the case L = k ( x ) . need the see what the following definition. is regular f o r some g, h define & {f = To at a point e k[x] such Say a e k if f that can h(a) ^ at a}. e k(x)|f regular places 0. 3. Then i t is fairly ring with field local k(x). valuation quotient valuation easy field see uniformizing Also, & ring to with = {— local k(x). r i n g s which In that &^ is a valuation parameter e k(x)|deg g > deg uniformizing fact, contain these k and IT = x-a and f} is a parameter — r i n g s are the have q u o t i e n t quotient and only field k(x). 29 . So we have {a the f o l l o w i n g e k} correspondence. \J » < > {Q I a Cf < ) U (3 °° r i n g s & | & O k and q u o t i e n t (Valuation Any e k} a 1 v a l u a t i o n on k ( x ) i s thus i n k ( x ) (a i n k U 00 ) . field o f £> i s k ( x ) } a s s o c i a t e d to a unique The p l a c e s ring i n k(x) therefore ct correspond to l o c a l We u n i f o r m i z i n g parameters and TT = for 22 i f i t i s a p a r a m e t e r defined by say TT i s a l o c a l u n i f o r m i z i n g parameter f o r the v a l u a t i o n r i n g & . The with shall TT = x - a , a e k following result places will be u s e f u l l a t e r i n connection on k ( x ) . Theorem Any and element o f k ( x ) has an e q u a l number of zeros poles . Proof f (x) Let where can factor k will of zeros in be g i v e n i n k, w i t h behaviour poles. so then k, w i t h Since f and g i n t o f = n, s a y . proof be an a r b i t r a r y f, g e k[x]. in deg h(x) = linear by Similarly, 00 i s complete. t h e number factors. balances will of h say. of zeros or zero h has a z e r o is n + of h The number of p o l e s o u t t h e number I f n < m, we t h e r e f o r e be be deg g = m, h has no p o l e of zeros The z e r o s will t h e number of k(x) closed the factors of f . multiplicities, F o r , i f n = m, that k i s algebraically multiplicities, of h at element and a t » and t h e of order (m-n) = m. The m-n at Similarly, 0 0 , 30. if n > m, we Now of we places K with Let & be of the an place satisfying the poles remark a b o u t extension ring/C, place an number o f is m + the (n-m) = n. connection integrality. L be the that make a s h o r t with Let of find of o f jp . integral Suppose ^ "P i s a p l a c e and ring K. If a equation place of L l y i n g i s an o v e r Q, is a element i . e . an over ^ . of L equation form a t h e n a must m be + a _^a m-1 +...+ m in = 0, a^ a^ e XT, &. Proof: Suppose & assume a has Thus, o r d ^ ( a has a pole m local at ) = -mn u n i f o r m i z i n g p a r a m e t e r TT , n of while order ord ? l n. Then o r d ^ (a) = (-a _ a m m —l a ) 1 this a i s + n. But c a n n o t be therefore We extension We will of K field ^2 be ring parameter K of satisfies a places k and of a place We = -a -...-a^, m want i n K. to z e & there local u n i f o r m i z i n g parameter a relation of K gives power s e r i e s de f i n e d . by p. , and o f"p-. give a field each p l a c e a suitable Let place that how formal rise usual, between power to an TT denote a local show t h a t to any i s a s s o c i a t e d a unique TI . 0. the series. injection field. As by -(m-l)n m-1 Hence a must have o r d ^ (a) £ true. i n &. show now see into since a -n. > Q m = -mn and by & the uniformizing element power s e r i e s in the 31. Let z £ &. A Q e k. element AQ, unique f o r z^ and A-^, modulo ^ the using induction £ k such A2>--' z = AQ + to a unique isomorphism Z - A Q = Z^TT f o r some z^ Thus argument z i s congruent ( O b t a i n AQ through k —>• &—y this Then on £ &. Repeating n, we obtain that 2 A ^TT + X + ... 2 Thus we an injection series way to an If ^ and can to injection ( 2) then form the = k ( ( i T ) ) of K i n t o ring , as k((ir)) . which d e f i n e s o f f o r m a l power extends i n the obvious desired. t h e n ^ d e f i n e s a m e t r i c on completion this +... A^^ of K w i t h completion by f o r ^3 t h e n Hence the i t s completion K^. r e s p e c t to I f ir i s a i t i s not above map with K difficult gives respect an top.. Rami f i c a t i o n In this section between p l a c e s and Assume L and and of K i n t o Denote that + T h i s homomorphism u n i f o r m i z i n g parameter see A^TT k [ [ IT ] ] , the i s a p l a c e o f K, metric. local into injection Note: this of & o v e r k. we z >—>• A Q + have a map by of ordjjlj, of course. integer their K are both L is a finite Denote we ord^J^ form d i s c u s s the interconnection e x t e n s i o n s or restrictions. algebraic e x t e n s i o n o f K. the e occurring Take f i e l d s over case, o f Z- not subgroup, The necessarily there i s a smallest in this k, a p l a c e ^3 i n L. r e s t r i c t i o n o f ord^j to K. some s u b g r o u p In any function and values a l l of positive a l l values of Z, 32 . o r d j j on K w i l l index of p . referring the be m u l t i p l e s to the r e s t r i c t i o n of ordp restriction normalization positive Call e the r a m i f i c a t i o n I f e / 1, say"fl,i s r a m i f i e d . normalized This o f e. that by mean *?^K. e has smallest o f $L t o K, we By t h e r e s t r i c t i o n of K d e f i n e d 0 r ( the r e s t r i c t i o n will I ord mean t h e p l a c e t o K, we w i l l i . e . the v a l u a t i o n ensures 1. value In f u t u r e i n ft'K. e From two this definition we deduce e a s i l y t h e f o l l o w i n g results. (1) Suppose p. i s a p l a c e i n L with and i s the r e s t r i c t i o n o f~§X.to K. uniformizing f o r p. , ^ parameters TT where U i s some u n i t = r a m i f i c a t i o n index e L e t II, TT be respectively. ord^(II) = 1, un , i n Oi . II a u n i f o r m i z i n g and hence us Then e r For, local II that parameter e ord^II f o r ^2 i m p l i e s ) = e-1 i s an e l e m e n t that e So ord^f.11 ) = = e. of L having t h e same e TT order be as a unit (2) TT w i t h respect to ^ . This implies indices multiply i n a tower o f f i e l d s . Suppose L^/L^/K i s a tower o f f i e l d s and of L > 2 Let ^ , ^ K respectively. be t h e r e s t r i c t i o n s e over ^ . and ^2 i s a of ^ to L^ Suppose x£ has r a m i f i c a t i o n i n d e x o v e r ^ , j"0_ r a m i f i c a t i o n i n d e x index must i n {X, . Ramification place that Then e = e ± ' e 1 o v e r ^ and r a m i f i c a t i o n e 2 ' T n e P r o °f °f t h i s Is obvious 33 . We needed. owing the now state Result I gives field extension corresponding form. texts on r a m i f i c a t i o n to o u r r e s u l t I t a k e s The s i m p l i c i t y i n o u r c a s e I g)0 to K. e^ denote merely ication p. indicate a complete 308 extension proof how field extension. index ^} on L l y i n g o v e r ^ . of ^ . proof i t comes a b o u t Then of t h i s that to t h e d e g r e e the r e a d e r the p l a c e to L . problem Consider ij; that i s referred r e s u l t , we the r a m i f - of the e x t e n s i o n . to C o r o l l a r y 2, [A]. extensions place complicated i s the r e s t r i c t i o n o f a complete are connected Consider its where « [L:K] of g i v i n g indices case foll- the r e s u l t on a more the s e t of a l l p l a c e s the r a m i f i c a t i o n Instead theory, separable on L, and ^ Ee« shall The r e s u l t s closed. i s a place Consider ramification i s due t o t h e f a c t L e t L/K be a f i n i t e Suppose j p Q For fact. about i s Galois. is algebraically Let the major facts t h i s a r e c o r o l l a r i e s of I i n the s p e c i a l I n most k some o f t h e b a s i c ring subring will We can v i e w (Actually, of L w h i c h be a p l a c e ring are i n t e r e s t e d i n this i n the f o l l o w i n g t h e homomorphism o f ^ . to a s u b r i n g We on K. >|i : as a homomorphism^ manner. stf—*• k, where iii : /C —>/?IJS^ i s as l a r g e —>• k) /f i s t h e I f we as p o s s i b l e , f o r an e x t e n s i o n ^) extend this x n L. 34 . In fact, such in there maximal terms of extensions these Since ator 9 6 i s a 1-1 the o f i|i . our problem can choose L i s separable the K such We view of . over K, that we 6 is integral ^jj/^and therefore a gener- o v e r Z^. Say equation F(X) where between extensions f o r L over satisfies correspondence a. e = X N + N a _ X N x N-l "•+...+ a Q 0 = . x C o n s i d e r C^Q = xS [ 0 ] = ^ T ( X ) / F ( X ) . of &Q field it i s L. is fairly maximal to must the have that Our the defined this <JJ to & Q , extended determines problem = X N + a unique t h e r e f o r e i s to N can be To the +.,.+ N _ 1 1 extend for ip(6). satisfy ^(a _ )X coefficients F(X) on/C. possible values h e n c e f ( 6 ) must .F(X) closed, see o f I|J . already find = 0; where to i f we quotient &^. is F(X) easy extension extend we Thus, The to^<T[0], 6 satisfies equation 0 <Ka ) = 0 <Ma_^) e k. Since factored into linear k is algebraically polynomials, say as F(X) = ( X - x p ! . . . ( X - A ) r , 6 6 r where for gives there roots (2) X. ^ ( 6 ) are This (1) the of there us e k. But precisely the are the we see roots of that F, the possible values namely, \^,...,\ . following information: always w i l l F(X) then = 0), be extensions of '•!> ( t h e r e a r e always and a t most N extensions of ii; ( r < deg F = N) . 35 . Clearly over^, of namely i s the This there are = - X^) we have I e . = deg facts has in about the this the degree that of L over K. parameter Lemma Eisenstein the p l a c e on on L me t e r lying N =ct and T 1 ir N-l ^ where power of one important that extensions than or always are equal e x t e n s i o n of ^ In of this case, finite to N. i n L, I we tells e x t e n s i o n o f x^. i s N, the find that the extension i n L local satisfies equation. e x t e n s i o n of K is totally Then largest obtain 2 i s less the a field over f o r ^? . n be containing^-. A l l i n a l l then, these index of nice L/K index Namely, A l s o , we a particularly Let . ramified. ramification uniformizing the e^. of number is totally ramification ^)$Q i n the p r o o f , we i n L and lying [L:K]. there i s p r e c i s e l y that that = extensions number where If us the The same as namely F(X) remarked extensions say F(X), dividing As the o f CTQ , . . . ,"p^ (9 - power of to be (X Note: (9 - ^^)0Q- largest t u r n s out r ideals take there N-l tor ramified. II to be exist a degree N. Suppose Let p be local u n i f o r m i z i n g para- z K such the that +... + a., n + a _ 1 < l < 1 N-l 0 and the a . z j£ f o r 1 < i < N - l and o r d ^ o ord Q , (o.) = 1. place 36. Proof II i s a n e l e m e n t of degree n over generality, there (*) equation that ord~(a ^ II f- n - i Now totally consider ramified. = 0 1 1 K, an without i = 0 , 1,..., equation loss of n-1, such that a) u = n, n must n 0 ord^(n ) each term s K, ord^(a^) (mod n ) . Thus, 1 see that Thus satisfies +...+ a . n + a . 1 n-1 ^ +.;.+ n element o r d ^ a ^ ^n " we n|N. E exist n this each f o r some II = a From For K, o f L and hence on be hence n. the r i g h t ord^(a^) But then also and of the equation. = N ord^(a^) , since = 0 i s (mod N ) , a n d t h e r e f o r e ord^o^II ) = ord^(a.) 1 + o r d ^ ) ( n ) = i mod ( n ) . 1 Hence (1) f o r each i , j ord ^(a^n ) 1 K ( 1 ) we see that distinct. o know order and (mod d ^ 0 n - l this ord^(a J N) (mod ) , and n) f o r 1 i i < n-1. tne terms on the r i g h t of (*) a r e Therefore l i n " order 1 ) + ' " + 0li<n-l = be n. we be >n. find = N and o r d ( a , ) fr i 1 < i < n-1. V By be n i s a ^ . must a n d n|N, 1 must can p o s s i b l y ord^ta) T 0 and r 0 J then l fl We i ord (a.I ) From 0 t o n-1, i f i f j ordgj(a H 1 (2) from s This = m.N, i 1 (2) t h e o n l y Thus In f a c t , n = N. (ord^a.n )) ord^(ag) since Summing f o r some i s the d e s i r e d term must whose be n, ord^a^) u p , we = 0 have integer conclusion. m. > 1 i 37 . Our second b a s i c among a l l e x t e n s i o n s L/K Let ^be of a g i v e n G be a place on K. {~p_!*Ei a p l a c e on L and lying over the the Galois relationship i n K, assuming of a connection subgroups of Let L with Each the place That that Define on p l a c e s P]_ c^2 , thatp^ result = the reader [4]. following. of L/K. the a l l elements Define composed the let We devel certain L/K. e G|a(a) = a ( m o d p ) of Let inertia be group a l l elements (a £ G|a£=#l } place o f ^2 to for a l l a e & } the place ring decomp o s i t i o n group of a of o f ^2 o f |}, G l e a v i n g p. f i x e d . i s , take D Then, of G a l o i s group r i n g &. modulo JJ). to be G a l o i s i n the G a l o i s group a z I ^ leaves fixed Dp, the {o 244 p. and between r a m i f i c a t i o n i n d i c e s and G be 1^ = to be such standard r e f e r r e d to P r o p o s i t i o n 1 1 , o f L/K i s , given a e G this group transitively That exists proof Assume L/K on the place Then G a c t s there , For op gives is Galois. II is result i t turns braically F = out closed, that these i n our two set-up, groups Lemma For any v a l u a t i o n 7) on L, are since k i s algeidentical. 38 . Proof It a be I ^ <S D^. i s c l e a r that an a r b i t r a r y e l e m e n t A E k, the f i e l d of 0". of c o n s t a n t s , we have and a^Q = j~Q a does a e D^. can now easily important relation: place order has the —• . there that i s a unique a = A(mod|2.). &/]p.) • But aX and l e t (Obtain Applying = X since Hence a ( a ) H A ( m o d ^ ) , a to X e k, so that b e l o n g to I^j. We of & a ( a ) = a(X)(modap) since Then such X t h r o u g h the i s o m o r p h i s m k —> this Assume a e D^, derive namely, f r o m I and that p r e c i s e l y equal I I the the i n e r t i a following group of a to the r a m i f i c a t i o n i n d e x place. L e t j p be Ill ramification index. a place on L, and l e t e^j be i t s Then Proof a let Let be a p l a c e place on L and "Pl^-} • |s| d e n o t e situation L/K acting i s as f o l l o w s : transitively fundamental We t h a t $)\^' we on have on i n S. the G a l o i s S (by I I ) , and the s t a b i l i z e r theorem Take know S i s f i n i t e the number o f e l e m e n t s Dp o f G i s Stabg(^J), the on K s u c h group of jp i n G. transformation by Then the S =; result Ij our G of subgroup Hence groups, by | G: D-, | = IT |s| , 39. • If! or , Also the from I I , we same; say t h e y From I we just |G| . find very that e|s| = = Corollary later [ L : k ] ; by G a l o i s = us two more l ^}! 1 = theory e » a s this i s desired f a c t s which w i l l prove 1 p be a p l a c e in Then E e<j = Z e = e | S | . that o a l l t h e e^j must be on. Suppose I^/L^/K then s r e s u l t gives useful that a r e a l l e. Thus e |g"|> This deduce on L ^ . 1 S a Galois tower o f f i e l d s . I f l p i s contained i s not r a m i f i e d i n L^/K - Let i n Gal ( L ^ / L ^ ) , a l l ramification occurs L /L^. 2 This indices the f a c t that a r e m u l t i p l i c a t i v e i n a tower o f Corollary extension ramification fields. 2 The Notice i s clear using number o f r a m i f i e d places i n any separable L/K i s f i n i t e . that h e r e L/K need n o t be G a l o i s . Proof Suppose k i s t h e g r o u n d Take x to be a s e p a r a t i n g of L over and over k ( x ) . separable f o r L and K, as transcendental 6 be t h e g e n e r a t o r K(6) field K. usual i n K/k, and l e t Thus we have L = K(3) 40. Before fications getting into o f the s i t u a t i o n . many ramified places show that only Similarly, are done. In p a r t i c u l a r , by a d j o i n i n g plicity, denote Galois By necessary, the many p l a c e s there by L. to many places L o f L, t h e n we take Denote L t o be t h e f i e l d of 8 . For sim- i t s Galois 8 satisfies that group by G. 6 of L s l i g h t l y , i f the g e n e r a t o r can assume are f i n i t e l y We now a r e c o n s i d e r i n g L/k(x). modifying simpli- sufficient to L a l l conjugates L again extension we finitely we two of L/k(x) are r a m i f i e d . f o r some e x t e n s i o n obtained the make To show t h a t i f we c a n show o n l y i n L/k(x) we on L/K, i t i s c l e a r l y finitely ramified also the p r o o f an e q u a t i o n of form F(T) where can = T the f ^ e k [ x ] . + f j " " n-l n 1 +...+ Then, L b e i n g f„ U normal over k[x], we f a c t o r F ( T ) i n L as F(T) n (T - = 0-6) oeG Taking the formal evaluating d e r i v a t i v e of F with a t 8 we respect t o T, and find = n F C@) 1 (8 - a 8 ) oeG a^id Now we come to c o u n t i n g places not lying (This in of L / k ( x ) . over excludes the p l a c e k(x) induced Suppose the p l a c e only ring of the number o f r a m i f i e d i s a ramified place corresponding fintely to many p l a c e s i n L. 00 on L, i n k(x). by I.) Then F o r , i f -t|. i s the p l a c e by ^} and /f i s i t s p l a c e ring, 6 lies on t h e n XT 2 k [ x ] . 41 . Thus 6 the satisfies remark on an equation integrality with of coefficients section 1, in^tT". 9 must By therefore l i e in£. I I I , ^2 By ramified I £ is non-trivial. in G such F'(9) that (8 II Hence 6 - o6 - o6) 0eG implies its inertia group i s some a u t o m o r p h i s m a- 4 i d there (modp). = 0 = 0 that (modS). But This then means i n p a r t i c u l a r r a^id that F'(8) This statement is divisible F'(9) Since ramified has places As the holds an only to the canonical view N-gon, the the of our which the leave F fixed. octahedral field suppose or that number we extensions l e t G denote As to G G are i n the then G i s one icosahedral groups. of one i n the tetrahedron, embedded of the over for °° i n of derive corres- groups. |G| = N. elements Now the Say that not r a m i f i c a t i o n theory C and as parameter finite. k = dihedron, place many z e r o s , figure F corresponding icosahedron, Recall (1) groups. uniformizing ramified finitely canonical Assume t h a t the f o r any must be example the r a m i f i c a t i o n i n d i c e s of ponding we by of Introduction, i . e . the the or sphere. r o t a t i o n s of Each plane octahedron Riemann the the the sphere tetrahedral, element of G moves R e f e r to §§1-9 Chp . 1 [ 3 ] f o r f u r t h e r d e t a i l s on example. K l e i n l o o k s a t t h i s s i t u a t i o n from a g e o m e t r i c p o i n t of v i e w . this L/k(x) 42 . a point Pg on Call the Host points are set three the sphere of p o i n t s of namely: midpoints of of faces the midpoint elements of the 4 any or sets the an edge of of the form place any number o f a conjugacy corresponds between points (Refer at inertia point, that group 2, point. 3 and i are f o r G. This indices f o r the three the as to the the a group of by G. identity); the elements G where i = at a vertex one Each i n k(x) 28-29 of such lying f o r the in k(x), ) the (2, 3, non-trivial F. these three class of a given correspondence elements the inertia order of at the a i ) , of r a m i f i c a t i o n of G is a cyclic automorphisms of ramification conjugacy of classes will G. group the of 3, over The form know t h a t non-unit two and three of that P r a m i f i c a t i o n index only triple, that through class fixed we seen axis i n any ramification pattern Suppose now Viewed Since i s equal indices called i n any be the adjoining places centrepoints easily the there orbits, precisely fixed places G its orbit. by edge, and points sphere. fixed about to pp. and G leaving a point group the the f i g u r e F, I t can c l a s s under to in k the F. the However, smaller i elements faces on i s moved by have is left by point N points. is left turn face is fixed i n F i x G. of opposite i t i s clear that points be edges (the h a l f vertex 5, Also, of G which v e r t i c e s of P of midpoint centrepoint G; F and PQ containing points of another to w h i c h have o r b i t types to of order sphere, the N. 43. elements of G are obtained rotation. As s u c h we two on the points than N p o i n t s - axis of rotation. are therefore of G must then For on the two the two ( 1 , N. having points are The group an of orbit less of the order n r o t a t i o n (these are conjugate and fixed by conjugacy see by n v e r t i c e s (these the axis 2 rotations). above G must therefore For future reference, b are (2, 2, a list 2n of points the fixed the there that than vertex are are than points and Cyclic group, II Dihedral III Tetrahedral IV Octahedral V Icosahedral order group, N order group group group N fixed and also three cyclic for n the conjugate distinct normal. We the r a m i f i c a t i o n N -j) . of the r a m i f i c a t i o n here. I the elements the conjugate smaller considerations i s given of all N axis (these Thus orbits of patterns by and through n mid-edge p o i n t s c l a s s e s with the pattern the only containing N = 2n of end identity); are ramification pattern order the a h a l f - t u r n about there N) . at by a single e i t h e r end fixed the the that at are: rotations); points of have o r b i t s non-conjugate. dihedral dihedron immediately T h e s e pofnts be the repetition sphere which less and see by N) (1, N, (2, 2, -) (2, 3, 3) (2, 3, 4) (2, 3, 5) 44 . For the general closed field, of canonical the gives way We will In fact, to see we A u t . k ( x ) has situation, i n determining groups, the combinatorics later will that this determine to have one the of k any ramification patterns geometry of of G-action does that the algebraically result any our on situation places i n the finite of same subgroup k(x). patterns. of above r a m i f i c a t i o n p a t t e r n s . 45 THE CHAPTER VI Order (1) Let HURWITZ of a D i f f e r e n t i a l K be an a l g e b r a i c algebraically closed. k-derivation. FORMULA function L e t d:K ft^yk ( C f . §2, Chapter III) K, and d e n o t e by the completion Then, uniformizing Let of Kg). f o r any l o c a l d- : y c K —• ft /k a Kp/k. Take be of K with respect to p. . TT f o r ^ , continuous = k((ir k-derivation i : K — * Kj, t o be t h e i n j e c t i o n o f K i n t o K-module map i ^ : following k, w i t h k the u n i v e r s a l parameter be t h e u n i v e r s a l over F i x a p l a c e |2 on Then u n i v e r s a l i t y o f d g u a r a n t e e s unique existence of a .c Q~ .. c o m p l e t i n g t h e K B / k. ^ / j . diagram. K Let field a £ !1K/k K/k Define ord^(to), the order o f to at?Q , as follows: Map to t o i^(to) = to^, s a y . Define ord^(to) (recall that defined i n §2, C h a p t e r I I I ) For the order any d i f f e r e n t i a l = ord ( ^) w of a continuous to e ^ / ^ J ord^(to) differential was = 0 f o r almost 46 . all places ^ Let tt ^ on x be = Kdx. K some f K. a separating Thus f o r any transcendental to e ^ ^ K k we have i n K/k. w = fdx, Then for e K. Claim: ord^(o)) = 0 f o r almost a l l places jp on K. Proof: ord-p(to) We f e K. and know t h a t ( f has poles, show that and jp not "J2 from p lies Suppose a x 1 —°°> e k - But 00 (1) II + any = finite the being the of and K which a finite So we i s unramified 00 in k(x). number of k(x). must being f o r some u n i f o r m i z i n g parameter co e ^Yi/k' product i s taken product by makes Therefore, II = TT = x - a ir = x - ord^(dx) = 5(K/k) Define over section sense. deg of a l l places (2); We ( d i v co) = n^ ^^^ ^ div(co) = the r< ~p. on U K. following define E ord-n(co) which (This u n i f o r m i z i n g parameter = 1 • d IT. and places.) , not f o r m <^ : x i — > a, local dx i t s zeros 0. Let therefore i n number.) p l a c e ^. of the at since everywhere. on only everywhere except p l a c e - * ^ : x •—»• Definition finite a l l points place local a, = 0 almost = 0 almost Thus /} has (3) the are over "p. has then Then x = ord these must be . ord-^(dx) ordp(f) 0 at l i e over excludes •^•Q : be + order ord^dx Let does = ordp(f) where This is a definition a. 47 . Claim: deg ( d i v co ) i s i n d e p e n d e n t o f co . Proof: Let ^K/k = most K ( * ' x x be a s e p a r a t i n g t r a n s c e n d e n t a l i n K. s o t n a t a n y by an e l e m e n t some f , g e K co = hco then. 1 deg t w ° differentials o f K. ( F o r , co = f d x and co ' = gdx f o r can now ( d i v co) = compare + deg ( d i v co) i s an i n v a r i a n t field e x t e n s i o n K over Note: (div ord^co') ( 1 ) = deg ( d i v co ' ) . Since by deg ( d i v co) and deg (E o r d ^ h ) + (E ord^co') = E ord_co' K/k a l o n e , Say 1 B = e K.) 1 ord^(co) = E^ o r d ^ C h t o ) = E (ord h to co , co ' d i f f e r a t v co = ( f g "^co', where f g We Then depending k, we d e n o t e only i t with on t h e reference 6(K/k). If K i s a rational function where x E K i s some i n d e t e r m i n a t e field, over say K = k ( x ) k, t h e n 6 ( K / k ) = -2 Proof : Assume K = k ( x ) , where x £ K i s some over k. L e t co = dx e ft„/, . K/K Now, a l l places (i) fl (i±)p: Case (i) parameter (1) E Then 6 (K/k) = E ord (dx) . K p ord h c p on K a r e o f t h e f o r m : x »—*• a, a E k or x »—> °° Take n = x - a to be t h e l o c a l for p . indeterminate uniformizing Then dx = l ' d n , so t h a t o r d ^ ( d x ) = 0 by t h e theorem o f p. 29. = 0. 48 . Case Take TT = ^ (ii) parameter for^l. Thus o r d ^ ( d x ) Hence to be Then we = the have x = ^, deg ( d i v dx) = _! = The Let degree L 1, teristic p l a c e p. on on L. Hurwitz K and dx = - ^-2diT. ord^(dx) (-2) -2 field of degree L e t e^j d e n o t e L. that Formula be L algebraic p. so -2. = 0 + (4) local uniformizing the e x t e n s i o n s o f k of transcendence N over charac- K, and ramification W r i t e ^} e L to d e n o t e that k of index of the |! i s a p l a c e Then ?™MS" 6 < L / k - W > - « ) + * «p J 1 E L (2) = e j, - where 1 if p | e^ and > e ^ - 1 i f p | e^. Proof: u) e ^Yi/\n Let Consider to 6„(L) notice a single = a ( a n c c 6 (K/k) s o ^L/k^" e p l a c e X£ on ^_ord (co). that * « * K. Then J|, be 5-(K) the p l a c e i n K l y i n g uniformizing Then TT = UH (2) The parameters Compare to r e l a t e Suppose |2 i s a p l a c e on Let Idea for ^ and ^(K) - ord^(co) = 6 (L/k) , „^ ^(L) K ramification under p . i n K and proof: <5(K/k) and 5 (L/k) L with of Let p TT, II be in L index e. local respectively. Q f o r some u n i t o r i g i n a l deduc t i o n § 2 Chapter XXI - V [2] U i n L. of this formula can be found in 49 Thus d TT = e u n e - 1 d ,n we and What [eun ft..,. . K/k E ord^(co) is "J TT + 6 - 1 (D u)n ]d^n e n d^(K) and d ^ ( L ) . can r e l a t e 03 (D u)n"d^n ? = Now + 4 1 Thus i „ ( c o ) = ft = ord^(f^) = f to., ''J f o r some d^ir TI -"J- f TT E = 6^,(K). 6 ^ ( L ) ? co T = Therefore f d * TT = f [ e u n TT f + 6 - 1 ( D u ) n ] d_n e TT TT Notice: 6 B have Y (D U)n ) e > e-1 i f p | e . above AS^ and e ^ i t s r a m i f i c a t i o n any p l a c e 5 .(L) + 1 , say. = e-1 i f p )( e, and E e With p now ^ = ord-»(f ) + o r d J e U n " ord (co) = ord^Cf^) + E = = E Z ord^Cco) E e-ord fi^j. = ord-(f IT* ) + E f\^t TT (f index, r ) + E e- = <5^(k) 6^(K) • = = 6 (K).( f = We 5(L/k) K((TT)), v v / / » and therefore E N 6 (K).N + f have e + E e ) + £ ^ £, B , e the f o l l o w i n g relation 6(K/k). 6 (L/k) = E 6^,(L) xfeK ^ = E 6 «.(K) • N + E E = p 6 (K/k)-N + E £ ? E L e_ between we 50 . CHAPTER V I I CONSEQUENCES Assume field of r a t i o n a l 1 over though of of K which this does chapter that purely closed are also (the natural to t h i s (1) k. We rational point o f view be seen later on t h e H u r w i t z L and K a r e p u r e l y then the f o r m u l a t a k e s on a p a r t i c u l a r l y of which morphism give enough we know t h a t With fields. Al from subfields that the problem Formula, i f of degree simple 1 o v e r k, form. t h e d i o p h a n t i n e e q u a t i o n and i t s L and K as a b o v e , e i n L/K, t h e n Z from = e f o r any Formula E §3, C h a p t e r VI I f f u r t h e r m o r e , the not d i v i d e the Hurwitz - 2 = L to i s o - cases. p o f k does 2N Galois arising to d e t e r m i n e 6 ( L / k ) = 6 ( K / k ) = -2. characteristic index transcendental information now d e t e r m i n e solutions. o f degree a d i o p h a n t i n e e q u a t i o n , the s o l u t i o n s o v e r K i n most We study function both i t yields fixed situation. In t h e theorem Namely, K is a one b e i n g t o s t u d y extension) i t w i l l reduce FORMULA transcendental field i s not the n a t u r a l problem, a fixed this functions, an a l g e b r a i c a l l y extensions our throughout OF THE HURWITZ becomes ( e ^ - 1) ramification 51 . indexing the^. in K from 1 to r . 2 = i N • , 2 - } c i — - i=l e^, this as i t is fairly easy to see Assuming For, and (i) a diophantine N > 1, that then r = 1 i m p l i e s R.S. r > 4 i m p l i e s R.S. Letting by N , we (*) i t has < 1, while in N relatively r = 2 or (*) > 2, equation few 3. while L.S. and (*) L.S. < (*) 1 = 1 e e 6 6 x n =i 6 41 can be e. l i + i e 2 e 3 > So 1 2 If anothe r e If l = e= neithe r Ass ume e^, 2 e 2 say -1 = e 2, e= n nor e^ i s 2 e ^ = 3. e^ < 6, and e 2 e 3 , is 2 3 Then 1 6 Thus 1; 2. " l 2 S i n c e e ^ , e^ < N, we have — , — > — . I t f o l l o w s from 1 2 Hence e = e„ = N. equation that — , — must be — . N 1 2 e e 1 2 (ii) Le t t i n g r = 3 (*) N not a l l > r=2 (*) becomes 2 Since have l Regarding solutions. dividing 1 — e. ) r (*) K Or, e must N e, our 52 . have e = 2, e 2 = 3, e 3 = 3, N = 12 e ; L = 2, e 2 = 3, e 3 = 4, N = 24 ± e So the of solutions. l = » 2 Hurwitz Notice indices) ication indices is the the major for the (2) the Main described [L:K] (1), ramification up be L e t -<J^ , ramification which Let e^, 2 > at these p l a c e s . to show the the for 3 ^ be e R(L/K) i s one of only systems ramif- groups: and ramif- cyclic icosahedral. the Formula This towards pattern five t to r e occur to a t x^, ^ , x^ the R(L/K) of the two third type. type. i n K at only ordered 2 each each ramification the K Since determine at m a t t e r what use types. places e L, ramification reference n with is different for respective Then we types the the this easy ( I t may ramification ramification and shows t h a t and indices i t does not e^ of specify occurs. case e one to a notation five orders Hurwitz in of set the (N group and Formula the We the canonical by only systems Hurwitz L/K i t i s enough to 60 Theorem. of type, N = c a s e has these five 5, octahedral indices set can = 3 this that donated The as e correspond fact of 3, in tetrahedral, proof both 2 Formula ication dihedral, = e which places, place in is.) indices triple (e^, e , x^ ^ i n L/K. Call this L/K. says that Hurwitz 2 > e^) 53 In what ramification determine II (2, 2, 2 III C2, 3, 3) IV (2, 3, 4) 'V (2, 3, 5) pattern t h e group For dental N, N) f o l l o w s , we the e x t e n s i o n extension (1, shall show t h a t knowing t h e i n cases over k. o f K i s enough to V i t I - I V , and i n c a s e of the e x t e n s i o n (assuming always i s Galois). t h e lemma and t h e o r e m L o f K. J o f an e x t e n s i o n the e x t e n s i o n determines that I f o l l o w i n g , we Say L = k ( x ) , Also assume f i x an f o r some x e L transcen- [ L : K ] = N. Lemma In L/K we (*) have I r r ( x , K, T) = T where e i t h e r have N + a^T^" 1 a , +...+ Q t h e a . e k o r K = k ( a . ) , and a l l n o n c o n s t a n t t h e same pole. Proof Choose Let We come e l e m e n t y e K such I r r ( x , k ( y ) , T) = f ( y ) T N N + f N - ] that _ (y) T N _ 1 K = k(y). + . . .+ f (y). Q know y = ^ f o r some h, g £ k f x ] , so t h a t g (x) Irr ( x , k ( y ) , T) = h ( T ) - y g ( T ) . power o n l y have unique i n the m i n i m a l factorization Thus y o c c u r s polynomial f o r x. i n k ( y ) [ T ] , we to the f i r s t Since we see t h a t a l l 54 . f (y) must i we be o f t h e form a^y + b^ where a ^ , b^ e k. Then have , _ N N-iy N - l N-1 ^ (x, K, T) = T + —.r • T a y + b a Irr v + T N By d e f i n i t i o n same p o l e N then, N-l a T + a l l *' , + 0' a S a y " which are nonconstant This i s either have t h e i n k ( i f a . b „ - a „ b . = 0) o r 1 N N x v generates K (by t h e t h e o r e m o f C h a p t e r I ) . Remark: I f R(L/K) (•—) , t h e n lemma we N ( g i v e n by a^y + b ^ ) ; and a l l a r e o f t h e form a .y + b . i i ;—r—• a .y + b„ N N and o , . a y + b b N-l + T + A N N = v + ...+ b A i s type I with ramification a t (x) i n t h e above can c o n c l u d e that K = k(a.) f o r a l l i . For, l by t h e lemma an equation f o l l o w i n g I of Chapter o f form This together must have Theorem with order (*) where t h e above precisely satisfies e a c h o, ^ has o r d e r lemma implies that > 1 a t 0. each 1, as d e s i r e d . 1 Suppose R(L/K) places V, we know x i s type I with ramification at the (x) and ( ^ ) . Then K = k ( x ) . N Proof By the p r e c e d i n g Irr where each particular, remark we know ( x , K, T) = T a. has o r d e r x as g e n e r a t o r 1. + a N Hence o f K. we Then N _ that 1 + . .. + a Q can f i x y = a., i n 0 J a l l other a are of 55 . a.y the + b. ;—-5— f o r some a. , b. , c. , d. e k s u c h c.v+d. x* x x* x x' x form a.d. - b . c . 4 0. x i I i have all Now the p r e c e d i n g t h e same p o l e . Since • lemma states 00 that a l l 1 has a p o l e o t h e r ct ^ must have a p o l e a t that . That at (meaning 00 i s , they —) must be o f t h e form a . = a.y f o r some a. e k. Thus i x x (**) I r r ( x , K, T) = T + a _ y T " +...+ 7 N N N Now and l o o k a t the e q u a t i o n ^ by y i n (*') to f i n d divide 1 — — - + a„ ,U + a y Again must N-1 a p p l y i n g the p r e c e d i n g of ^. That find i s , t h e y must t h e n we have y = - x , o r K, = k ( y ) = theorem 1 we c o u l d have been I and t h e p l a c e s a t w h i c h then Thus, assumed stated ramification K. We (1) also theorem make for a fixed From ramification i n computation. i f R(L/K) i s t y p e occurs as c y c l i c type fact a r e known, of degree I with upon w h i c h N. ramification the p r o o f relies. use o f t h e f o l l o w i n g ramification a l l be 0. t h e e x t e n s i o n s L and L ' a r e i s o m o r - T h i s i s the b a s i c the f o l l o w i n g a l l a^ N that as: i f R(L/K) and R ( L ' / K ) a r e b o t h over that f o r convenience the e x t e n s i o n i s d e t e r m i n e d t h e same p l a c e s , t h e n =0 k(x ). N a t (x) and (—) m e r e l y of — N-1 N a.U + U lemma we occurred phic ^ 1 I n Theorem at that +...+ Remark: The J N-2 be m u l t i p l e s (*') „U by — . Take T = 77 x U satisfies satisfied 2 H y. 1 1 index place = order facts from of i n e r t i a (result III) Chapter group V. 56 . (2) ramification (3) degree of indices Theorem field multiply i n a tower e x t e n s i o n = sum (result of of fields ramification I) 2 L/K Let function be field. R(L/K) Then indices a Galois Suppose the L determine's L e x t e n s i o n and up of L/K group to an a rational is solvable. isomorphism over K. Proof We c o n s i d e r the II, I I I or IV, II (2, 2, f) Assume group LQ of that so must Then have ramify Again of morphism If the L Q , and |- = 2, o n l y groups Klein 4-group; then so I with N = that over K. of o r d e r 4 a r e and the Aut^k(x) type By = of o n l y be find R(L/K) Let I-V. p l a c e s i n L Q above N index in L/LQ. Hence over inertia N type Theorem to i s o m o r p h i s m 1 we 1. pattern two u s i n g Theorem the subgroup 1). L Q i s determined L e t H be (LQ/K) i s a f i n i t e ( 2 , 2, that i n Theorem index —. Q = with i s type ramification can is, the N -zj ^ 2. R(L /K) That be 2, R(LQ/K) dealt ramification (L /K)I Q = Gal |Gal 'P^ been the p l a c e w i t h = F i x H. and I having first R(L/K) t h r e e c a s e s where over • L = 2). cyclic Let , p l a c e s must N N (—, —, 1). We know group pattern 2. means i s determined ( 2 , 2, ramification this K. These R(L/LQ) the 1 Since of or the to iso- that the cyclic 57 . group Let i s type H be index 2. is type is not We find (1, Take L Choose III Le t morphism L Gal In of = be 3 these Klein with the R(L /K) that Q above ^ 1), that in L Q / K obtain specified. 2) or isomorphism fields that is (2, 1, to 4-group. ramification place unramified (2, 2, pattern. K, as (L/K) |Gal 3). LQ the 2). has Then must i s determined over (2, 3, 1, such = F i x H. Q 2 ramification (2, find (2, 2, ^ 4 we = 12, , 'p , ^? Each that index the extension LQ N we yields i s t h e r e f o r e ( 1 , 3, K. LQ for [L:K] Q take say the L Q of = R(LQ/K) K, (2, = R(L/LQ) 2, 1). (2, 2, 3) Hence R ( L / L ) of three be point^.^ unless 1.then these, Take L over^, IV now must determined. 3. R(LQ/K) know but determined argument Since over 2, Q of (2, 3, index Again Theorem one L is = F i x H. precisely i n the and o f any R(L /K) i s either that are Q (L/K) group N = I with 2). Hence G a l inertia uniquely 2, there As the I. (L /K)| Q i s then three places ramify By a subgroup the of That is, L 3, of and determined in L argument lying Q 2 in L / L Q . index to i s o m o r p h i s m = H f o r II f o r such i s determined we a to iso- desired. 4) a group 2. order L e t H be = F i x H, 2 ) , so of as that such usual. 24 there i s always a subgroup We find L Q i s determined. i n Gal that a subgroup (L/K), R(LQ/K) LetZ/^,^^ and must be be the 1), 58. two places above^. Then previous K, ^ above ^ ! clearly L case, and so o v e r as E (^3 T R(L/L ) Q i s determined = to ^ e t n (3, P e 3, x a c e i 2). isomorphism ^ n 0 From the over LQ, stated. 3 Theorem L / K be Let function V then LQ in field. G is a Galois extension Let G be and (L/K). Gal l e t L be If R(L/K) a rational is type simple. Proof Suppose H i s a n o r m a l LQ Let an = F i x H. LQ/K must be since either two indices there These ramify But 5, this determines exist at implies 5) 5 or in [LQ:K] R(LQ/K) as We order 1. m > show t h a t (12, 5). 1 f ^ 2, then l s 3 such so ramified R(LQ/K) that 5). But have to r a m i f y , this is impossible both 2. Thus-^-^ i s u n r a m i f i e d two places L/LQ = 3, (2, or least totally = m. r a m i f i e d here, i n L / K would a m u l t i p l e of Hence = (2, totally (1, places Q i n G of exist. that R(L/K) i t must be (L /K)| |Gal L Q / K cannot extension Recall in Then subgroup so that 12. R(L/LQ) being 12, 1). ^0 , Z* 2 must in LQ/K. ^ a be unramified with o v e ^3 * (5, in 5, LQ/K Contradiction. Remark: Theorems 1-3 actually the M a i n Theorem, n a m e l y , Aut.k(x) f o r which establish that a f i n i t e p Jf J G| must be the first subgroup isomorphic claim of G of to a c a n o n i c a l 1) 59 group. F o r , G determines c field L = k ( x ) , and by (1) type I-V. Theorems type I-IV, the f i e l d morphism- but t h i s extension, G, must we know that V then we know Hence G, i n t h i s function clearly i m p l i e s that be d e t e r m i n e d i f R(L/K) i s Theorem i s simple i s the o n l y case have be i s o m o r p h i c 3 shows that of order simple Theorem 1, C h a p t e r II g u a r a n t e e s rational function f i e l d . 60. of the Since extensions to one o f t h e s e i f R(L/K) group too, i s isomorphic to i s o - the group to i s o m o r p h i s m . I<-IV • R(L/K) i s e x t e n s i o n L/K i s d e t e r m i n e d G = G a l (L/K) group that show t h a t the c a n o n i c a l g r o u p s groups. icosahedral (1) 1 and 2 t h e n by R ( L / K ) , G must determined canonical K = F i x G, a r a t i o n a l i s type But t h e of order 60. to a c a n o n i c a l group that Fix G is 60. CHAPTER V I I I PROOF Recall the hypotheses finite subgroup closed field is o f t h e theorem; we have a of c h a r a c t e r i s t i c G must groups algebraically p w i t h p A\ |G| . The aim be c o n j u g a t e to one o f t h e f i v e i f the r a m i f i c a t i o n pattern of G i s I-IV or i f p = 0. type The first group i s derived VII. We group, such THEOREM G o f Aut k ( x ) , where k i s an to show t h a t canonical OF THE MAIN found G Let the Hurwitz that G must between G and a c a n o n i c a l Formula, as i n C h a p t e r be i s o m o r p h i c t o a c a n o n i c a l Our aim i s to show e x i s t e n c e = a Go a isomorphism from there s a y Gg. that c o n n e c t i o n we o b t a i n problem = GQ. 1 We can r e d u c e by t h e f o l l o w i n g F = F i x G and F OF = Q to a field Then f o r o e A u t ^ k O x ) , Q G° this argument. = Fix G . FQ of a e A u t ^ k ( x ) = GQ Proof: G (ago But ^)(fg) (ogo GQ Q. G ° : of <== GQ : a _ 1 Let ago" = (ago ^ ) ( o f ) ) (of) = (og)(f) 1 e G° . Then, f o r some f e F s u c h = of = fg. L e t gg e Gg and f g e F g . that gg = ago Thus, This 1 implies a "*"ggO- = g, Q e F , Q that Thus a g o " Then, f o r some f e F, we have g g ( g f ) = o f . a "'"(of) = f . for f 1 writing But t h e n of = f g . is in G . Q f g as (a ^ggCf)(f) f o r some g e G, so e G°. i f F i x G = k ( y ) and F i x Gg = k ( y g ) f o r some = 61 elements is y, y to f i n d e k(x) n t r a n s c e n d e n t a l over a e Aut^k(x) such oCkCy)) D e f i n e o" to be to k C y ^ ) . That a: map i f i t can Extend a as Let be = extended an t o g i v e an II). 0 to an Thus we closure automorphism have the ^ job If the ramification (1) k(y) the required i s to show t h a t t h e n Theorems k(x) algebraic and 1 and of K pattern (by Theorem Hence G i s c o n j u g a t e c a s e we can 3, k(x) = k(x) n k(x) = k(a(x)). o f G and 2 of Chapter i.e. Then a k(y ) VII k ( a ( x ) ) are u n i q u e l y closure. remaining of k ( x ) . • K - our The be > k(a(x)) where now cases. from set-up k(y) the isomorphism automorphism o f k ( x ) . k(x) extensions (1) kCy- ) Then a w i l l algebraic K I-IV, probl follows: K be extended Chapter be the a : k ( y ) — > • k ( y ^ ) by a : y •—*• y.g a i—• a f o r a l l a e k . and then that the n a t u r a l i s , define k, G Q i s type show t h a t determined the in = k ( a ( x ) ) f o r these to G Q h e r e . have to c o n s i d e r i s p = 0 Assume, w i t h o u t l o s s o f g e n e r a l i t y , y , yQ t h a t r a m i f i c a t i o n o c c u r s a t 1, 0 and °°. chosen and such 62 . the ramification need of several technical x with are p a t t e r n of G being respect listed results to c e r t a i n type V. concerning For this the derivations. we Schwarzian These results i n t h e f o l l o w i n g lemmas. Through t h e lemmas l e t k ( z ) = F i x G f o r some z z k(x), (2 ) z t r a n s c e n d e n t a l over know is that to k(x) s i n c e Thus d i f f e r e n t i a t i o n with the Schwarzian by f ( z ) In of x with respect on k ( z ) we k(x)/k(z) respect o f z, f ( z ) , on k ( z ) c a n be e x t e n d e d Denote tx] If D i s a derivation i t c a n be e x t e n d e d separable. function k. to some to k ( x ) . to t h i s derivation . the f i r s t lemma we relate the Schwarzian of x (3 ) w i t h r e s p e c t to z — and w i t h r e s p e c t z to the Schwarzian of x w i t h to z - c, f o r some c e k. r e s p e c t to Lemma 1 (i) [x] z = \n [ x ] i z (ii) [x] z = fx] _ z f o r c e k. c Proof Let on D denote differentiation k ( x ) and l e t DQ d e n o t e with differentiation respect with to z respect to — on k ( x ) . z (i) (2) (3) Recall that [x] = z 3 2 ^ Dx - 3 2 (£- ^ V Dx X 2 P o s s i b l e by Theorem 1, C h a p t e r I I , A c t u a l l y we s h o u l d w r i t e "The S c h w a r z i a n o f x w i t h r e s p e c t to t h e d e r i v a t i o n on k ( x ) i n d u c e d by d i f f e r e n t i a t i o n w i t h r e s p e c t to z on k ( z ) " . 63 A r 1 = / x 2 - z 0 U Schwarzians that n D JL we V X first extensive Thus 2 D x = 2 X D C D D x ~ Q 2 i . 0 Q = Q X ) . D Q Z "like" Rule terms. Notice i s made. DCDQX)-D D ( ; D Dx • D Q z these DX-DQZ = D (D x) 0 To compare X use o f t h e C h a i n Q D \ x of a l l compare D x and n D 3 I— ~ 0 X I Q Z D ( D Dx " x.p zj 0 Dx' 2 X . D Q Z D + D X . D ( D Q Z ) Dx But with U D~z 0 = -z respect r 2 1 For, l e t y = —. z . J to y, and D~z 0 J 0 D x ) D X D we X . (-z ) Z DX Q similarly ' Then D„ i s d i f f e r e n t i a t i o n 0 1 1 2 = D-. (—) = - —? = -z . Therefore 0 y y^ + (-2z) have DQX D ( D Q X ) . D Q Z D ( D Q X ) DQX DX-DQZ DX D[D x.z + Dx•2z ] Dx D x-z 3 4 + D x-4z Dx 2 3 + D x.2z 2 Thus ( M From D 0 L X 0X DQ Dx (a) and (b) we 4 z ° 2 * . Dx + — have , 0 , 3 , , 2 • 6z + 6z 3 + Dx»6z 2 64 . r i or° 3 x j. D x 4 /- ^ , /- 2 2 . .D x,2 4 . D x 2 3 . . 2 2 r z D x 4 r ^ T " 3 0 " 2 r[ x l ] -z = (ii) z „,D x.2 4 b T 2 3 ( ) , z 4 z This i s clear since d i f f e r e n t i a t i o n z i s t h e same as d i f f e r e n t i a t i o n for with with respect r e s p e c t to to z - c c e k. Define to be t h a t which t e rm o f C ] the p r i n c i p a l part r contains x a t ~ z z of the Laurent negative series c f° r c k, e f o r [x] a t z - < z powers o f z - c. Lemma 2 Let e be th.e r a m i f i c a t i o n z - c, f o r c e k. z - c is Then t h e p r i n c i p a l of k ( x ) / k ( z ) at term of f J z c)" i f e > 1, x a t 2 (i) (z - c ) " + A ( z 2 Q where (ii) index r- i s some c o n s t a n t 1 element. 0 i f e = 1. Proof (i) Since write in the r a m i f i c a t i o n U(x - C) a t z - c i s e, we can = z - c f o r U a unit i n k ( x ) and C a constant k(x) . Notation: want For index Notice the p r i n c i p a l simplicity that term [x] = z [x - C] . z-c o f [x - C ] ^ _ where of n o t a t i o n , w r i t e c Thus we U(x - C) = z - c x f o r x - C, z f o r z - c . 65 . The problem then Is to f i n d the p r i n c i p a l term of * " x z where e Ux = z. Denote to differentiation with respect t o z by D, w i t h respect x by d. From Ux = z we have DU'X or, dU'Dx'X Thus Dx(dU«x Let W = -1 1 + e 6 + U'Dx 6 + Uex e - 1 =; 1 6 Dx = 1, + eUx " ] = 1 6 1 ^77»x eU Then (a) Dx = Notice term ~ r eU i s 1. - term From i . e . a unit implies whose initial dU has o r d e r S 0, and Hence o r d (^-jr'x) > 0, so t h a t t h e eU o f W ^ must 2 6 W i s a 1-unit a unit. Dx (b) 1 For, U a unit i s also constant that -J-x be 1. = D(Dx) = d(Dx)'Dx ( a ) t h e n we have D x = (-^- - x + - d(-^) x )Dx The r e f o r e , ( v C ) D x Dx~ 2 = 1-e W ~e~ U X -e , 1 „ ,W e U> N + d ( 1-e X w w — i s a u n i t , so o r d (d(yj)) i 0. Hence l W l - e ord (— d ( — ) x ) > 1-e > - e . The i n i t i a l e U D x of — Dx 2 term (4) is (4) therefore at worst that of One i n i t i a l term i s worse t h a n a n o t h e r i f i t has a l a r g e r n e g a t i v e exponent. N o t i c e t h a t we c a n n o t s a y above t h a t t h e i n i t i a l t e r m i s . . . b e c a u s e 1-e may be 0 66 1-e W -e 1-e • e.-l 1-e T-1 — x = (Ux ) W = z W. e U e e / T T This has i n i t i a l From ... ( d (a) and (b) we 2 1-e ,W,2 (p) CT? ) order l-2e , + 1 ? u 3 W d ( , ,W\ U X > 2-2e 2 we have + = d(D^x) = —(1— e) ^— ( 1 - 2e)^"(u^ e -2 2 e^ (Ux ) W c x 2(l-e) ) order s i n c e D x = d(D x ) » D x , (1- e) ( 1 - 2e) e^ D x D x have > l-2e 3 Then, 1-e -1 z since W i s a 1-unit. e term f: y x ( °f t e r m s higher degree) +... (1- e) ( 1 - 2e) 2 e where, a g a i n , Thus [x] z we have t h e term w i t h used largest 1 12 - .2 = ( e = Hence possible negative e some 3(i^) ] 2 1-unit. exponent i n 2 n z Z term o f [ x l z AQ. -2 2 ) z -2 v constant 2 2 6e + 4 e - 3 + 6e - 3 e ~e~2 - 1 2 the p r i n c i p a l r as that W i s a is !2<->i-«> - for the f a c t Replacing is 2 e - 1 * z -2 e^ + A^z -1 0 z by z-c g i v e s the r e s u l t stated. (ii) that In the proof e > 1. we want o f ( i ) we d i d not ever So f o r e = 1, p r o o f to show t h a t use t h e f a c t (i)is valid. f o r e = 1, the e n t i r e But now principal term i s 0 67 . For e = 1, line (c) of ( i ) reads " DT This d u ( we know has p r i n c i p a l - } 0. term (d),fore = 1 , Line b e come s TJ _3 Thus DT" 0 = D ( - and u> d ( ° r X ^ W ) M 4. 2 H W ^ 2 t o o c l e a r l y has p r i n c i p a l term 0 . Hence 3 2 D x D x 2 of 2 - 3 ( — ) must be 0 , as d e s i r e d . Dx Dx this term the p r i n c i p a l Lemma 3 Let and assume the form e be t h e r a m i f i c a t i o n e > 1. Then index the expansion o f kQx} /k(z) a t °s>, of I l x a °° t 1 z S °f e -l 1 , . 1 . 1 . 1 2— —7 + A „ — r + A , — 7 - + A „ —re^ z^ 0 z l z ^ 2 z2 A A J for some c o n s t a n t s A Q , A ^ , 5 ,••• Proof From Lemma l ( i ) , together with Lemma possible degree [ x l 2 ( i ) we f i n d i n the expansion 1 ,e -l ,1,-2 2 2 •n—(—T) e e- 1^ ,1 z ^* Using ""Trfx]^- = z . that t h e terms of [ 3 X . this a t 00 a r fact of l e a s t e Z ,1,-1, + A (—rr) . . Q ,1 z J , f o r some c o n s t a n t A„ 0 68 . Note: Lemma at oo then order i f there a t °° : i s ramification i t has a zero a t <» o f a t l e a s t 2. next basic Namely, lemma, d e d u c e d result that rational needed using the p r e v i o u s to conclude the Schwarzian function ramification Main us t h a t [x] has no p o l e z The the 3 tells whose the Main of x with Theorem. respect c o e f f i c i e n t s depend two, g i v e s to z i s a only on t h e i n k(x)/k(z). Lemma There i s a rational [x] where function and r a m i f i c a t i o n £ k ( z ) such that = r(z) z the c o e f f i c i e n t s of r ( z ) places r(z) depend indices only on t h e r a m i f i e d of k ( x ) / k ( z ) . Proof Rationality Chapter IV. of " Suppose [x] Assume w i t h o u t of F i x G chosen °° ( t h a t such ] x z r(z) = z loss that i s , at the places follows from Corollary 2 of e k(z). of g e n e r a l i t y ramification (z-1), the g e n e r a t o r z occurs a t 1, 0 and ( z ) and (—) ) w i t h ramif- z ication indices e^, e^ and e^ r e s p e c t i v e l y . 2 and 3 we f i n d not f ° r <*> at that has p o l e s o r a t any u n r a m i f i e d 00 e m r r(z) ( 2 ) 2 = eT(^iT2 e ~l 1 for place. - 1 + some c o n s t a n t s a t most 2 a t 0 and 1: Thus r ( z ) has t h e -1 + -T2e From Lemmas + z + C e z 2 A, B, C, (C i s a c o n s t a n t , since i f 69 . it were a n o n c o n s t a n t at co . A, (*) we c a n c a l c u l a t e expansion B and C. terms will But Lemma are. the i n i t i a l o f r ( z ) a t <» i n terms Equating a pole of the c o n s t a n t s 3 already t e l l s like terms o f us what coefficients these initial i n the e x p a n s i o n s a l l o w us to s o l v e f o r A, B and C, and so t o w r i t e down an e x p l i c i t In by have Contradiction.) From the p o l y n o m i a l , r ( z ) would formula for r(z). (*) t h e i n i t i a l rewriting terms r ( z ) i n terms of r ( z ) at of u = —. a r e found 00 Then z = — and (*) y i e l d s r < > z • u-> = \ r( e i' 2 l u( 1 + r } r u" + h i 4 2 u> 1 e 1 2 u 2 6 1 e u + A j ^ 2 + ~ 1 _ U - 2 2 2 + Bu + C 2 (" 2 + 2 u 3 + 3u 4 1 +...) 6 + e = C + (A + B ) u + ( 2 2 e Lemma 3 states must be that l + A(u + u 2 " 2u 2 + u 3 +...) 1 6 expansion v u 2 c 1 = '-^—2 6 e + ( 2 = —1 I + + Bu + C 2 1 2 e + A + 2 ' 6 the i n i t i a l - 1 )u +... 2 terms of this 2 , 3 e = " e 7 u 2 + A u Q 3 + A u 1 3 e i l C = 0 A + B = 0 e + A 2 + ~ " 6 A ~ l = + e Substituting these r(z) = 1 1 ~2 ^ ~2 " ^ 2 3 e values e l ' n d e = ^ B into + — equation A (*) gives . 2 ^ x - 2 1 2~z e 1 1 + (;-| + - | - - | . 2 l e e c o e f f i c i e n t s o f r ( z ) do, t h e r e f o r e , 2 6 l)Kz)Cz 3 depend o n l y on 2' 3' e Now the Main we a r e i n shape entiation D , DQ denote with respect Assuming, without yQ have been 1, 0 and 00 ramified By r with to p r o v e the remaining case i n Theorem. Let y, e l e The -2 a e 2 (z-1) 2 2~ 3 6 2 _ 1 1 3 gives i 2 6 2 2 1 Therefore i 2 2~ l e e 2 c o e f f i c i e n t s o f u , u, u 2 - 4 0 Equating 0 , then loss chosen the extensions to y, i n k ( y ) , k(yQ> of g e n e r a l i t y , such k(x)/k(y) that that then, respectively the g e n e r a t o r s ramification and k ( x ) / k ( y Q ) p l a c e s and r a m i f i c a t i o n t h e Main Lemma to k ( x ) o f d i f f e r - have occurs at t h e same indices. there c o e f f i c i e n t s i n k depending is a rational only function on t h e r a m i f i c a t i o n 71 indices o f 1, 0 and <» s u c h [x] But then = D i t i s easy that r ( y ) and [x] = D r(y ). Q to see t h a t a([x] ) = Ix] a([x] ) For, D since = a(r(y)) = r(o(y)) a i s an i s o m o r p h i s m we f i n d = r(y ) = Q n " from these last By = = Also, n 0 have [o(x)1 C o r o l l a r i e s 1 and 2 o f C h a p t e r k(a(x)), . u two e q u a t i o n s , we [x]_ D that a([x] ) = [a(x)] Then, [x] IV i t f o l l o w s that k(x) as d e s i r e d . Remarks: (1) In the h y p o t h e s e s p )( | GI . in In the p r o o f particular), ramification too strong, but p of index theorem only the f a c t i t i s plausible types. that indices that So o u r o r i g i n a l guarantees says state ( d e r i v a t i o n of the Hurwitz e f o r any r a m i f i c a t i o n i n d e x ramification then e. i n that the theorem the we used o f t h e M a i n Theorem we that this p J( e f o r any h y p o t h e s i s might that the case e arise. cannot We know f o r t h e s e be one o f t h e f i v e that p | |G| For, |G| and t h e canonical i f p | |G| (p a p r i m e ) p = e, f o r some r a m i f i c a t i o n i n d e x e. seem But t h e r e s u l t happen. i f p J( e f o r a l l e, t h e n e must Formula 72. (2) I t turns out to be true that f o r any a l g e b r a i c a l l y closed f i e l d k , a f i n i t e subgroup of Aut^k(x) must be conjugate to a canonical group. That i s , the major claim of our Main Theorem does i n fact hold for the icosahedral case i n c h a r a c t e r i s t i c p > 0 , although i t does not y i e l d to proof using the techniques here developed f o r the case of c h a r a c t e r i s t i c 0 . That the theorem i s true for a l l cases i n c h a r a c t e r i s t i c follows from Dickson [1]. From §259 we have: algebraic closure of the prime f i e l d Aut k (x) n k Z/pZ i f k^ p > 0 i s the and i f a subgroup of i s isomorphic to the icosahedral group then that subgroup i s 0 conjugate to the icosahedral group. By standard techniques of group representations and group extensions t h i s r e s u l t holds f o r any a l g e b r a i c a l l y closed f i e l d k . Dickson's techniques, however, do not work f o r both c h a r a c t e r i s t i c 0 and c h a r a c t e r i s t i c p . Dickson uses f i n i t e permutation groups to deduce h i s r e s u l t f o r c h a r a c t e r i s t i c p — i n characteristic corresponding devices are simply not f i n i t e . 0 the Thus, i n c h a r a c t e r i s t i c some other method such as the one used here i s necessary. 0 , 73 . Bibliography [1] L . E . D i c k s o n , " L i n e a r G r o u p s " , Dover New Y o r k , 1958. Publications, Inc., [2] A. H u r w i t z , " M a t h e m a t i s c h e Werke", B i r k h a u s e r B a s e l , 1932. [3] F. K l e i n , [4] S. L a n g , [5] H. Weber, and Co., " L e c t u r e s on the I c o s a h e d r o n " , Dover P u b l i c a t i o n s , I n c . , New Y o r k , 1956. "Algebra", Addison-Wesley 1967 . Publishing Co., "Lehrbuch der A l g e b r a " , V o l . 1 1 , C h e l s e a P u b l i s h i n g Co., New Y o r k , 1961. 74. Postscript to Beth Kitchen's Thesis by Klaus Hoechsmann After the completion of this thesis and shortly before i t was due to be handed i n , we stumbled on a simple proof of the icosahedral case, which works i n any c h a r a c t e r i s t i c 2, 3, 5. I t hinges on the easy arithmetic of what we s h a l l c a l l a f f i n e extensions K = k(y) and, k[y] k[x] . In other words, a p a i r function f i e l d s i s a f f i n e , i f generators y i s a polynomial i n x . L/K As above, x,y k : L = k(x) , L S K of r a t i o n a l can be found such that w i l l be a l g e b r a i c a l l y closed. The following lemma i s a synthesis of the lemmas on pages 35 and 53. Lemma: L/K Proof: (<=) a f f i n e <=> some place of K i s t o t a l l y ramified i n Let p , ^ be the places i n question. L As i n the lemmas on pages 35 and 53, we have /•*\ (*) N w where Put F + a^-jU w w e L , u e K x = — , y = — w u of degree N N-l + a _ u w N 2 N-2 . + ... + u = 0 are l o c a l uniformizing parameters f o r p , ^ , respectively . (=>) From an equation F(x) = y , we get an Eisenstein equation (*) f o r hence t o t a l ramification. with a polynomial w = — , u, x = ~ y '•> 75. Remark: In an affine extension, the s p l i t t i n g of any place of K (other than the x^. singled out above) i s r e f l e c t e d completely i n the f a c t o r i z a t i o n of i t s uniformizing parameter y - c = F(x) - c in k[x] For instance, i f there i s another t o t a l l y ramified place, we obtain a pure equation y - c = (x - b)N Our derivation of the icosahedral equation w i l l follow the same l i n e s . To prove the conjugacy of two icosahedral groups, we choose a subgroup of index 5 i n each of them. These are conjugate; making them equal by a suitable inner automorphism, we now have two icosahedral groups which intersect i n a tetrahedral group fields K^, K 2 such that [L : K^] = 5 see that ramification i n each L/K; hence a f i e l d , ( i = 1,2) . L containing I t i s easy to must necessarily be as follows: r We choose the generator x iR' ^ f l ' ^1' ^2 3 r e x' ' X X of ~ a > X By our lemma and i t s proof, each us i n s i s t that F(x) L so that the uniformizing parameters of ~ ^ ' r e s P e c t * v e x y» w n e r K^ i s generated by a be monic and that y e a + 3 = -4 y = F(x) . Let be a l o c a l parameter f o r the 76. place JO* . We must show that F(x) i s uniquely determined by these data. Ramif i f i c a t i o n of and gives us (1) 3 2 y = x (x + ax + b) (2) 2 2 y - 6 = (x-a) (x-g) (x-y) and respectively. 5 4 3 Accordingly, F(x) = x + ax + 6x . We s h a l l see that a = 5 , b = 40 2 2 Obviously, x (x-a)(x-3) must divide the derivative F'(x) = (5x + 4ax + 2 3b)x , whence (2') (x-a)(x-6) i s determined. Thus y - c = ( x + -jax + -|b) (x-y) 2 2 Comparing the l i n e a r and quadratic terms of (2') to those of (1) y i e l d s (i) (ii) 3b - 8ya = 0 2 12ab - 15yb - 8ya = 0 , , whence 3a = 5y 8 2 Therefore ( i ) becomes: b = —a 4 - —a and , we have a = 5 . b = 40 Since we had normalized . -4 = a + 3 Quod erat demonstrandum.
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Finite groups of fractional linear transformations Kitchen, Vivien Beth 1972
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Title | Finite groups of fractional linear transformations |
Creator |
Kitchen, Vivien Beth |
Publisher | University of British Columbia |
Date Issued | 1972 |
Description | In this thesis we consider the group of fractional linear transformations of a variable x over an algebraically closed field k. The purpose of the thesis is to determine all finite subgroups of this group whose orders are not divisible by the characteristic of k. |
Subject |
Transformation groups. Algebras, Linear. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2011-04-05 |
Provider | Vancouver : University of British Columbia Library |
Rights | For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use. |
DOI | 10.14288/1.0302202 |
URI | http://hdl.handle.net/2429/33278 |
Degree |
Master of Science - MSc |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
Campus |
UBCV |
Scholarly Level | Graduate |
Aggregated Source Repository | DSpace |
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