Finite automata and algebraic extensions of function fields

\noindent Journal de Th\ ’{ e} o r i e des Nombres

\noindent de Bordeaux 18 ( 2006 ) , 3 79 − 420

\centerline{F i n i t e \quad automata \quad and \quad a l g e b r a i c \quad ex t en s i on s \quad o f }

\centerline{ f unc t i on \quad f i e l d s }

\centerline{par Kiran S . KEDLAYA }

\centerline{R\ ’{ e}sum\ ’{ e} . \quad On donne une d e s c r i p t i o n , dans l e langage des auto − }

\centerline{mates f i n i s , de l a c l $ \hat{o} $ ture a lg \ ’{ e} br ique du corps des f o n c t i o n s r a t i o n − }

\centerline{ n e l l e s $ F { q } ( t ) $ sur un corps f i n i $ F { q } . $ Cette d e s c r i p t i o n , qui g \ ’{ e}n\ ’{ e} ra − }

\centerline{ l i s e un r \ ’{ e} s u l t a t de C h r i s t o l , emploie l e corps de Hahn − Mal ’ cev − }

\centerline{Neumann des ‘ ‘ s \ ’{ e} r i e s f o r m e l l e s g \ ’{ e}n\ ’{ e} r a l i s \ ’{ e} es ’ ’ sur $ F { q }. $ En passant , }

\centerline{on obt i en t une ca rac t \ ’{ e} r i s a t i o n des ensembles bien ordonn \ ’{ e} s de }

\centerline{nombres r a t i o n n e l s dont l e s repr \ ’{ e} s e n t a t i o n s $ p − $ adiques sont g \ ’{ e}n\ ’{ e} − }

\centerline{ r \ ’{ e} es par un automate f i n i , \quad et on pr \ ’{ e} s ente des techn iques pour }

\centerline{ c a l c u l e r dans l a c l $ \hat{o} $ ture a lg \ ’{ e} br ique ; c e s t echn iques i n c l u e n t une }

\centerline{ v e r s i on en ca rac t \ ’{ e} r i s t i q u e non n u l l e de l ’ a lgor i thme de Newton − }

\centerline{Puiseux pour d\ ’{ e} terminer l e s d\ ’{ e}veloppements locaux des courbes }

\centerline{ planes . On c on j e c t u r e une g \ ’{ e}n\ ’{ e} r a l i s a t i o n de nos r \ ’{ e} s u l t a t s au cas }

\centerline{de p l u s i e u r s v a r i a b l e s . }

\centerline{Abstract . \quad We give an automata − t h e o r e t i c d e s c r i p t i o n o f the a l − }

\centerline{ gebra i c c l o s u r e o f the r a t i o n a l func t i on f i e l d $ F { q } ( t ) $over a f i n i t e f i e l d }

\centerline{ $ F { q } , $ g e n e r a l i z i n g a r e s u l t o f C h r i s t o l . The d e s c r i p t i o n o ccur s with in }

\centerline{ the Hahn − Mal ’ cev − Neumann f i e l d o f \quad ‘ ‘ g e n e r a l i z e d power s e r i e s ’ ’ }

\centerline{ over $ F { q } . $ In pas s ing , we obta in a c h a r a c t e r i z a t i o n o f we l l − ordered }

\centerline{ s e t s o f r a t i o n a l numbers whose base $ p $ expans ions are generated }

\centerline{by a f i n i t e automaton , and e x h i b i t some techn iques f o r computing }

\centerline{ in the a l g e b r a i c c l o s u r e ; the se in c lude an adaptat ion to p o s i t i v e }

\centerline{ c h a r a c t e r i s t i c o f Newton ’ s a lgor i thm f o r f i n d i n g l o c a l expans ions }

\centerline{ o f plane curves . We a l s o c on j e c tu r e a g e n e r a l i z a t i o n o f our r e s u l t s }

\centerline{ to s e v e r a l v a r i a b l e s . }

\centerline {1 . \quad In t roduc t i on }

\noindent 1 . 1 . \ h f i l l C h r i s t o l ’ s theorem , and i t s l im i t s . \ h f i l l Let $ F { q }$be a f i n i t e f i e l d o f char −

\noindent a c t e r i s t i c $ p , $ and l e t $ F { q } ( t ) $ and $ F { q } (( t ) ) $ denote the f i e l d s o f r a t i o n a l f u n c t i o n s

\noindent and o f formal ( Laurent ) power s e r i e s , r e s p e c t i v e l y , over $ F { q } . $C h r i s t o l [ 4 ] ( s e e

\noindent a l s o [ 5 ] ) proved that an element $ x = \sum ˆ{ \ infty } { i = 0 }x { i } t ˆ{ i }$ o f $ F { q } ( ( t ) ) $ i s a l g e b r a i c over

\begin { a l i g n ∗}\ r u l e {3em}{0 .4 pt}\end{ a l i g n ∗}

\centerline{Manuscrit re $ c c e d i l l a $ u l e 3 mai 2005 . }

Journal de Theacuteorie des Nombresde Bordeaux 18 open parenthesis 2006 closing parenthesis comma 3 79 hyphen 420Finite .. automata .. and .. algebraic .. extensions .. offunction .. fieldspar Kiran S period KEDLAYAReacutesumeacute period .. On donne une description comma dans le langage des auto hyphenmates finis comma de la cl ocircumflex ture algeacutebrique du corps des fonctions ration hyphennelles F sub q open parenthesis t closing parenthesis sur un corps fini F sub q period Cette description comma

qui geacuteneacutera hyphenlise un reacutesultat de Christol comma emploie le corps de Hahn hyphen Mal quoteright cev hyphenNeumann des quotedblleft seacuteries formelles geacuteneacuteraliseacutees quotedblright sur F sub q period En

passant commaon obtient une caracteacuterisation des ensembles bien ordonneacutes denombres rationnels dont les repreacutesentations p hyphen adiques sont geacuteneacute hyphenreacutees par un automate fini comma .. et on preacutesente des techniques pourcalculer dans la cl ocircumflex ture algeacutebrique semicolon ces techniques incluent uneversion en caracteacuteristique non nulle de l quoteright algorithme de Newton hyphenPuiseux pour deacuteterminer les deacuteveloppements locaux des courbesplanes period On conjecture une geacuteneacuteralisation de nos reacutesultats au casde plusieurs variables periodAbstract period .. We give an automata hyphen theoretic description of the al hyphengebraic closure of the rational function field F sub q open parenthesis t closing parenthesis over a finite fieldF sub q comma generalizing a result of Christol period The description o ccurs withinthe Hahn hyphen Mal quoteright cev hyphen Neumann field of .. quotedblleft generalized power series quoted-

blrightover F sub q period In passing comma we obtain a characterization of well hyphen orderedsets of rational numbers whose base p expansions are generatedby a finite automaton comma and exhibit some techniques for computingin the algebraic closure semicolon these include an adaptation to positivecharacteristic of Newton quoteright s algorithm for finding lo cal expansionsof plane curves period We also conjecture a generalization of our resultsto several variables period1 period .. Introduction1 period 1 period .... Christol quoteright s theorem comma and it s l imits period .... Let F sub q be a finite field

of char hyphenacteristic p comma and let F sub q open parenthesis t closing parenthesis and F sub q open parenthesis open

parenthesis t closing parenthesis closing parenthesis denote the fields of rational functionsand of formal open parenthesis Laurent closing parenthesis power series comma respectively comma over F sub

q period Christol open square bracket 4 closing square bracket open parenthesis seealso open square bracket 5 closing square bracket closing parenthesis proved that an element x = sum sub i = 0

to the power of infinity x sub i t to the power of i of F sub q open parenthesis open parenthesis t closing parenthesisclosing parenthesis i s algebraic over

hlineManuscrit re ccedilla u le 3 mai 2005 period

Journal de Theorie des Nombresde Bordeaux 18 ( 2006 ) , 3 79 - 420

Finite automata and algebraic extensions offunction fields

par Kiran S . KEDLAYAResume . On donne une description , dans le langage des auto -mates finis , de la cl o ture algebrique du corps des fonctions ration -

nelles Fq(t) sur un corps fini Fq. Cette description , qui genera -lise un resultat de Christol , emploie le corps de Hahn - Mal ’ cev -Neumann des “ series formelles generalisees ” sur Fq. En passant ,

on obtient une caracterisation des ensembles bien ordonnes denombres rationnels dont les representations p− adiques sont gene -rees par un automate fini , et on presente des techniques pourcalculer dans la cl o ture algebrique ; ces techniques incluent uneversion en caracteristique non nulle de l ’ algorithme de Newton -Puiseux pour determiner les developpements locaux des courbesplanes . On conjecture une generalisation de nos resultats au cas

de plusieurs variables .Abstract . We give an automata - theoretic description of the al -

gebraic closure of the rational function field Fq(t) over a finite fieldFq, generalizing a result of Christol . The description o ccurs within

the Hahn - Mal ’ cev - Neumann field of “ generalized power series ”over Fq. In passing , we obtain a characterization of well - ordered

sets of rational numbers whose base p expansions are generatedby a finite automaton , and exhibit some techniques for computing

in the algebraic closure ; these include an adaptation to positivecharacteristic of Newton ’ s algorithm for finding lo cal expansionsof plane curves . We also conjecture a generalization of our results

to several variables .1 . Introduction

1 . 1 . Christol ’ s theorem , and it s l imits . Let Fq be a finitefield of char -acteristic p, and let Fq(t) and Fq((t)) denote the fields of rational functionsand of formal ( Laurent ) power series , respectively , over Fq. Christol [ 4 ]( seealso [ 5 ] ) proved that an element x =

∑∞i=0 xit

i of Fq((t)) i s algebraic over

Manuscrit re ccedilla u le 3 mai 2005 .

\noindent 3 80 \quad Kiran S . Kedlaya

\noindent $ F { q } ( t ) ( $ that i s , i s the root o f a monic polynomial in one v a r i a b l e with c o e f −

\noindent f i c i e n t s in $ F { q } ( t ) ) $ i f and only i f f o r each $ c \ inF { q } , $ the s e t o f base $ p $ expans ions

o f the i n t e g e r s $ i $ f o r which $ x { i } = c $ i s generated by a f i n i t e automaton .

However , t h i s i s not the end o f the s to ry , f o r the re are monic polynomia l sover $ F { q } ( t ) $ which do not have any roo t s in $ F { q } ( ( t

) ) , $ even i f you en l a rge the

\noindent f i n i t e f i e l d and / or r e p l a c e $ t $ by a root . An example , due to Cheval ley [ 3 ] , i sthe polynomial

\ [ x ˆ{ p } − x − t ˆ{ − 1 } . \ ]

\noindent Note that t h i s i s a phenomenon r e s t r i c t e d to p o s i t i v e c h a r a c t e r i s t i c ( and

\noindent caused \quad by \quad wild \quad r a m i f i c a t i o n ) : \quad an \quad o ld \quad theorem \quad o f Puiseux \quad [ 18 , \quad Propos i −t ion I I . 8 ] i m p l i e s that i f $ K $ i s a f i e l d o f c h a r a c t e r i s t i c 0 , then any monic poly −

\noindent nomial o f degree $ n $ over $ K ( t ) $ f a c t o r s i n to l i n e a r po lynomia l s over$ L ( ( t ˆ{ 1 / n } ) ) $

f o r some f i n i t e ex tens i on f i e l d $ L $ o f $ K $ and some p o s i t i v e i n t e g e r $ n. $

\noindent 1 . 2 . \quad Beyond C h r i s t o l ’ s theorem : g e n e r a l i z e d power s e r i e s . \quad As sug −ges ted by Abhyankar \quad [ 1 ] , \quad the s i t u a t i o n de s c r ibed in the prev ious s e c t i o ncan be remedied by a l l ow ing c e r t a i n ‘ ‘ g e n e r a l i z e d power s e r i e s ’ ’ ; the se werein f a c t f i r s t introduced by Hahn [ 8 ] in 1 907 . We w i l l d e f i n e the se more pre −c i s e l y in Sec t i on 3 . 1 ; f o r now , th ink o f a g e n e r a l i z e d power s e r i e s as a s e r i e s$ \sum { i \ in I } x { i } t ˆ{ i }$ where the index s e t \quad $ I $ i s a we l l − ordered subset o f the r a t i o n a l s

\noindent ( i . e . , a subset conta in ing no i n f i n i t e dec r ea s ing sequence ) . For example , inthe r ing o f g e n e r a l i z e d power s e r i e s over $ F { p } , $ Cheval ley ’ s polynomial has theroo t s

\begin { a l i g n ∗}x = c + t ˆ{ − 1 / p } + t ˆ{ − 1 / p ˆ{ 2 }} + \cdot \cdot\cdot \\ f o r c = 0 , 1 , . . . , p − 1 .\end{ a l i g n ∗}

\hspace ∗{\ f i l l }Denote the f i e l d o f g e n e r a l i z e d power s e r i e s over $ F { q }$ by $ F { q }( ( t ˆ{ Q } ) ) . $ Then

\noindent i t turns out that \quad $ F { q } ( ( t ˆ{ Q } ) ) $ \quad i s \quad a l g e b r a i c a l l y c l o s e d , \quad and one can e x p l i c i t l yc h a r a c t e r i z e those o f i t s e lements which are the roo t s o f po lynomia l s over$ F { q } ( ( t ) ) [ 1 1 ] . $ One then may ask whether one can , in the ve in o f C h r i s t o l , g ive

\noindent an automata − t h e o r e t i c c h a r a c t e r i z a t i o n o f the e lements o f $ F { q } (( t ˆ{ Q } ) ) $ \ h f i l l which

\noindent are r oo t s o f monic polynomia l s over $ F { q } ( t ) . $

In t h i s paper , we g ive such an automata − t h e o r e t i c c h a r a c t e r i z a t i o n . ( Thec h a r a c t e r i z a t i o n appeared p r e v i o u s l y in the unpubl ished pr ep r i n t [ 13 ] ; t h i spaper i s an updated and expanded ve r s i o n o f that one . ) \quad In the proce s s ,we c h a r a c t e r i z e we l l − ordered s e t s o f nonnegat ive r a t i o n a l numbers with t e r −minating base $ b $ expans ions \quad $ ( b > 1 $ \quad an i n t e g e r ) \quad which are generated by af i n i t e automaton , and d e s c r i b e some techn iques that may be u s e f u l f o r com −puting in the a l g e b r a i c c l o s u r e o f $ F { q } ( t ) , $ such as an analogue o f Newton ’ s

\noindent a lgor i thm . \quad ( One th ing we do not do i s g ive an independent d e r i v a t i o n o fC h r i s t o l ’ s theorem ; the new r e s u l t s here are e s s e n t i a l l y orthogona l to thatr e s u l t . ) \quad Whether one can use automata in p r a c t i c e to perform some s o r to f ‘ ‘ i n t e r v a l a r i thmet i c ’ ’ i s an i n t r i g u i n g ques t i on about which we w i l l not

3 80 .. Kiran S period KedlayaF sub q open parenthesis t closing parenthesis open parenthesis that is comma i s the root of a monic polynomial

in one variable with coef hyphenficients in F sub q open parenthesis t closing parenthesis closing parenthesis if and only if for each c in F sub q

comma the set of base p expansionsof the integers i for which x sub i = c i s generated by a finite automaton periodHowever comma this i s not the end of the story comma for there are monic polynomialsover F sub q open parenthesis t closing parenthesis which do not have any roots in F sub q open parenthesis

open parenthesis t closing parenthesis closing parenthesis comma even if you enlarge thefinite field and slash or replace t by a root period An example comma due to Chevalley open square bracket 3

closing square bracket comma i sthe polynomialx to the power of p minus x minus t to the power of minus 1 periodNote that this i s a phenomenon restricted to positive characteristic open parenthesis andcaused .. by .. wild .. ramification closing parenthesis : .. an .. old .. theorem .. of Puiseux .. open square

bracket 18 comma .. Proposi hyphent ion II period 8 closing square bracket implies that if K is a field of characteristic 0 comma then any monic poly

hyphennomial of degree n over K open parenthesis t closing parenthesis factors into linear polynomials over L open

parenthesis open parenthesis t to the power of 1 slash n closing parenthesis closing parenthesisfor some finite extension field L of K and some positive integer n period1 period 2 period .. Beyond Christol quoteright s theorem : generalized power series period .. As sug hyphengested by Abhyankar .. open square bracket 1 closing square bracket comma .. the situation described in the

previous sectioncan be remedied by allowing certain quotedblleft generalized power series quotedblright semicolon these werein fact first introduced by Hahn open square bracket 8 closing square bracket in 1 907 period We will define these

more pre hyphencisely in Section 3 period 1 semicolon for now comma think of a generalized power series as a seriessum sub i in I x sub i t to the power of i where the index set .. I is a well hyphen ordered subset of the rationalsopen parenthesis i period e period comma a subset containing no infinite decreasing sequence closing parenthesis

period For example comma inthe ring of generalized power series over F sub p comma Chevalley quoteright s polynomial has therootsx = c plus t to the power of minus 1 slash p plus t to the power of minus 1 slash p to the power of 2 plus times

times times for c = 0 comma 1 comma period period period comma p minus 1 periodDenote the field of generalized power series over F sub q by F sub q open parenthesis open parenthesis t to the

power of Q closing parenthesis closing parenthesis period Thenit turns out that .. F sub q open parenthesis open parenthesis t to the power of Q closing parenthesis closing

parenthesis .. i s .. algebraically closed comma .. and one can explicitlycharacterize those of it s elements which are the roots of polynomials overF sub q open parenthesis open parenthesis t closing parenthesis closing parenthesis open square bracket 1 1

closing square bracket period One then may ask whether one can comma in the vein of Christol comma givean automata hyphen theoretic characterization of the elements of F sub q open parenthesis open parenthesis t

to the power of Q closing parenthesis closing parenthesis .... whichare roots of monic polynomials over F sub q open parenthesis t closing parenthesis periodIn this paper comma we give such an automata hyphen theoretic characterization period open parenthesis Thecharacterization appeared previously in the unpublished preprint open square bracket 13 closing square bracket

semicolon thispaper i s an updated and expanded version of that one period closing parenthesis .. In the process commawe characterize well hyphen ordered set s of nonnegative rational numbers with ter hyphenminating base b expansions .. open parenthesis b greater 1 .. an integer closing parenthesis .. which are generated

by afinite automaton comma and describe some techniques that may be useful for com hyphenputing in the algebraic closure of F sub q open parenthesis t closing parenthesis comma such as an analogue of

Newton quoteright salgorithm period .. open parenthesis One thing we do not do is give an independent derivation ofChristol quoteright s theorem semicolon the new results here are essentially orthogonal to thatresult period closing parenthesis .. Whether one can use automata in practice to perform some sortof quotedblleft interval arithmetic quotedblright i s an intriguing question about which we will not

3 80 Kiran S . Kedlaya

Fq(t)( that is , i s the root of a monic polynomial in one variable with coef -ficients in Fq(t)) if and only if for each c ∈ Fq, the set of base p expansionsof the integers i for which xi = c i s generated by a finite automaton .

However , this i s not the end of the story , for there are monic polynomialsover Fq(t) which do not have any roots in Fq((t)), even if you enlarge thefinite field and / or replace t by a root . An example , due to Chevalley [ 3] , i s the polynomial

xp − x− t−1.

Note that this i s a phenomenon restricted to positive characteristic ( andcaused by wild ramification ) : an old theorem of Puiseux[ 18 , Proposi - t ion II . 8 ] implies that if K is a field of characteristic 0, then any monic poly -nomial of degree n over K(t) factors into linear polynomials over L((t1/n))for some finite extension field L of K and some positive integer n.1 . 2 . Beyond Christol ’ s theorem : generalized power series. As sug - gested by Abhyankar [ 1 ] , the situation describedin the previous section can be remedied by allowing certain “ generalizedpower series ” ; these were in fact first introduced by Hahn [ 8 ] in 1 907 .We will define these more pre - cisely in Section 3 . 1 ; for now , think ofa generalized power series as a series

∑i∈I xit

i where the index set I is awell - ordered subset of the rationals( i . e . , a subset containing no infinite decreasing sequence ) . For example, in the ring of generalized power series over Fp, Chevalley ’ s polynomialhas the roots

x = c+ t−1/p + t−1/p2

+ · · ·forc = 0, 1, ..., p− 1.

Denote the field of generalized power series over Fq by Fq((tQ)). Thenit turns out that Fq((tQ)) i s algebraically closed , and one can ex-plicitly characterize those of it s elements which are the roots of polynomialsover Fq((t))[11]. One then may ask whether one can , in the vein of Christol, givean automata - theoretic characterization of the elements of Fq((tQ)) whichare roots of monic polynomials over Fq(t).

In this paper , we give such an automata - theoretic characterization . (The characterization appeared previously in the unpublished preprint [ 13] ; this paper i s an updated and expanded version of that one . ) Inthe process , we characterize well - ordered set s of nonnegative rationalnumbers with ter - minating base b expansions (b > 1 an integer )which are generated by a finite automaton , and describe some techniquesthat may be useful for com - puting in the algebraic closure of Fq(t), suchas an analogue of Newton ’ salgorithm . ( One thing we do not do is give an independent derivationof Christol ’ s theorem ; the new results here are essentially orthogonal tothat result . ) Whether one can use automata in practice to performsome sort of “ interval arithmetic ” i s an intriguing question about whichwe will not

\hspace ∗{\ f i l l }F i n i t e automata and a l g e b r a i c ex t en s i on s o f func t i on f i e l d s \quad 38 1

\noindent say anything c o n c l u s i v e , though we do make a few s p e c u l a t i v e comments

\noindent in Sec t i on 8 . 1 .

\noindent 1 . 3 . \ h f i l l Struc ture o f the paper . \ h f i l l To conclude t h i s i n t r o d u c t i o n , we d e s c r i b e

\noindent the contents o f the remaining chapter s o f the paper .

In Chapter 2 , we c o l l e c t some r e l e v a n t background mate r i a l on determin −i s t i c f i n i t e automata ( Sec t i on 2 . 1 ) , nonde t e rm in i s t i c f i n i t e automata ( Sec −t ion 2 . 2 ) , and the r e l a t i o n s h i p between automata and base $ b $ expans ions o fr a t i o n a l numbers ( Sec t i on 2 . 3 ) .

In Chapter 3 , we c o l l e c t some r e l e v a n t background mate r i a l on gene ra l −i z ed power s e r i e s ( Sec t i on 3 . 1 ) , a l g e b r a i c e lements o f f i e l d ex t en s i on s ( Sec −t ion 3 . 2 ) , and a d d i t i v e po lynomia l s ( Sec t i on 3 . 3 ) .

In \quad Chapter 4 , \quad we \quad s t a t e \quad our main theorem \quad r e l a t i n g \quad g e n e r a l i z e d \quad powers e r i e s a l g e b r a i c over $ F { q } ( t ) $ with automata , to be proved l a t e r in the paper .

\noindent We formulate the theorem and note some c o r o l l a r i e s \quad ( Sec t i on 4 . 1 ) , \quad thenr e f i n e the \quad statement \quad by check ing i t s c o m p a t i b i l i t y with decimation o f apower s e r i e s ( Sec t i on 4 . 2 ) .

In Chapter 5 , we g ive one complete proo f o f the main theorem , which inone d i r e c t i o n r e l i e s on a c e r t a i n amount o f s o p h i s t i c a t e d a l g e b r a i c machin −ery . We g ive a f a i r l y d i r e c t proo f that automatic g e n e r a l i z e d power s e r i e sare a l g e b r a i c ( Sec t i on 5 . 1 ) , then g ive a proo f o f the r e v e r s e i m p l i c a t i o n bys p e c i a l i z i n g the r e s u l t s o f [ 1 1 ] \quad ( Sec t i on 5 . 2 ) ; the dependence on [ 1 1 ] i s thesource o f the r e l i a n c e on a l g e b r a i c t o o l s , such as Artin − S c h r e i e r theory .

In Chapter 6 , we c o l l e c t more r e s u l t s about f i e l d s with a va lua t i on , s p e c i f −i c a l l y in the case o f p o s i t i v e c h a r a c t e r i s t i c . We r e c a l l ba s i c p r o p e r t i e s o ftw i s t ed polynomia l s ( Sec t i on 6 . 1 ) and Newton polygons ( Sec t i on 6 . 2 ) , g ivea ba s i c form o f Hensel ’ s lemma on s p l i t t i n g polynomia ls ( Sec t i on 6 . 3 ) , andadapt t h i s r e s u l t to tw i s t ed polynomia l s ( Sec t i on 6 . 4 ) .

In Chapter 7 , we g ive a second proo f o f the r e v e r s e i m p l i c a t i o n o f themain \quad theorem , \quad r e p l a c i n g the \quad a l g e b r a i c \quad methods \quad o f the prev ious \quad chapterwith more e x p l i c i t c o n s i d e r a t i o n s o f automata . To do t h i s , we analyze thet r a n s i t i o n graphs o f automata which g ive r i s e to g e n e r a l i z e d power s e r i e s( Sec t i on \quad 7 . 1 ) , \quad show that the c l a s s o f automatic g e n e r a l i z e d power s e r i e si s c l o s e d under add i t i on \quad and m u l t i p l i c a t i o n \quad ( Sec t i on \quad 7 . 2 ) , \quad and e x h i b i t \quad ap o s i t i v e − c h a r a c t e r i s t i c analogue o f Newton ’ s i t e r a t i o n ( Sec t i on 7 . 3 ) .

In Chapter 8 , we r a i s e some f u r t h e r que s t i on s about the a l g o r i t h m i c s o fautomatic g e n e r a l i z e d power s e r i e s ( Sec t i on 8 . 1 ) , and about a p o t e n t i a l gen −e r a l i z a t i o n o f the m u l t i v a r i a t e analogue o f C h r i s t o l ’ s theorem ( Sec t i on 8 . 2 ) .

\noindent Acknowledgments . \ h f i l l Thanks to Bjorn \ h f i l l Poonen f o r br ing ing the work o f

\noindent C h r i s t o l to the author ’ s a t t e n t i o n , to Er ic Rains f o r some i n t r i g u i n g sugges −t i on s concern ing e f f i c i e n t r e p r e s e n t a t i o n s , to Richard Stanley f o r po in t ingout the term ‘ ‘ cactus ’ ’ , and to George Bergman f o r h e l p f u l d i s c u s s i o n s . The

Finite automata and algebraic extensions of function fields .. 38 1say anything conclusive comma though we do make a few speculative commentsin Section 8 period 1 period1 period 3 period .... Structure of the paper period .... To conclude this introduction comma we describethe contents of the remaining chapters of the paper periodIn Chapter 2 comma we collect some relevant background material on determin hypheni stic finite automata open parenthesis Section 2 period 1 closing parenthesis comma nondeterministic finite

automata open parenthesis Sec hyphent ion 2 period 2 closing parenthesis comma and the relationship between automata and base b expansions ofrational numbers open parenthesis Section 2 period 3 closing parenthesis periodIn Chapter 3 comma we collect some relevant background material on general hyphenized power series open parenthesis Section 3 period 1 closing parenthesis comma algebraic elements of field

extensions open parenthesis Sec hyphent ion 3 period 2 closing parenthesis comma and additive polynomials open parenthesis Section 3 period 3 closing

parenthesis periodIn .. Chapter 4 comma .. we .. state .. our main theorem .. relating .. generalized .. powerseries algebraic over F sub q open parenthesis t closing parenthesis with automata comma to be proved later in

the paper periodWe formulate the theorem and note some corollaries .. open parenthesis Section 4 period 1 closing parenthesis

comma .. thenrefine the .. statement .. by checking it s compatibility with decimation of apower series open parenthesis Section 4 period 2 closing parenthesis periodIn Chapter 5 comma we give one complete proof of the main theorem comma which inone direction relies on a certain amount of sophisticated algebraic machin hyphenery period We give a fairly direct proof that automatic generalized power seriesare algebraic open parenthesis Section 5 period 1 closing parenthesis comma then give a proof of the reverse

implication byspecializing the results of open square bracket 1 1 closing square bracket .. open parenthesis Section 5 period 2

closing parenthesis semicolon the dependence on open square bracket 1 1 closing square bracket is thesource of the reliance on algebraic tools comma such as Artin hyphen Schreier theory periodIn Chapter 6 comma we collect more results about fields with a valuation comma specif hyphenically in the case of positive characteristic period We recall basic properties oftwisted polynomials open parenthesis Section 6 period 1 closing parenthesis and Newton polygons open paren-

thesis Section 6 period 2 closing parenthesis comma givea basic form of Hensel quoteright s lemma on splitting polynomials open parenthesis Section 6 period 3 closing

parenthesis comma andadapt this result to twisted polynomials open parenthesis Section 6 period 4 closing parenthesis periodIn Chapter 7 comma we give a second proof of the reverse implication of themain .. theorem comma .. replacing the .. algebraic .. methods .. of the previous .. chapterwith more explicit considerations of automata period To do this comma we analyze thetransition graphs of automata which give rise to generalized power seriesopen parenthesis Section .. 7 period 1 closing parenthesis comma .. show that the class of automatic generalized

power seriesi s closed under addition .. and multiplication .. open parenthesis Section .. 7 period 2 closing parenthesis comma

.. and exhibit .. apositive hyphen characteristic analogue of Newton quoteright s iteration open parenthesis Section 7 period 3

closing parenthesis periodIn Chapter 8 comma we raise some further questions about the algorithmics ofautomatic generalized power series open parenthesis Section 8 period 1 closing parenthesis comma and about a

potential gen hypheneralization of the multivariate analogue of Christol quoteright s theorem open parenthesis Section 8 period 2

closing parenthesis periodAcknowledgments period .... Thanks to Bjorn .... Poonen for bringing the work ofChristol to the author quoteright s attention comma to Eric Rains for some intriguing sugges hyphent ions concerning efficient representations comma to Richard Stanley for pointingout the term quotedblleft cactus quotedblright comma and to George Bergman for helpful discussions period The

Finite automata and algebraic extensions of function fields 38 1

say anything conclusive , though we do make a few speculative commentsin Section 8 . 1 .1 . 3 . Structure of the paper . To conclude this introduction , wedescribethe contents of the remaining chapters of the paper .

In Chapter 2 , we collect some relevant background material on determin- i stic finite automata ( Section 2 . 1 ) , nondeterministic finite automata( Sec - t ion 2 . 2 ) , and the relationship between automata and base bexpansions of rational numbers ( Section 2 . 3 ) .

In Chapter 3 , we collect some relevant background material on general -ized power series ( Section 3 . 1 ) , algebraic elements of field extensions (Sec - t ion 3 . 2 ) , and additive polynomials ( Section 3 . 3 ) .

In Chapter 4 , we state our main theorem relating gen-eralized power series algebraic over Fq(t) with automata , to be provedlater in the paper .We formulate the theorem and note some corollaries ( Section 4 . 1 ) ,then refine the statement by checking it s compatibility with decimationof a power series ( Section 4 . 2 ) .

In Chapter 5 , we give one complete proof of the main theorem , which inone direction relies on a certain amount of sophisticated algebraic machin -ery . We give a fairly direct proof that automatic generalized power seriesare algebraic ( Section 5 . 1 ) , then give a proof of the reverse implicationby specializing the results of [ 1 1 ] ( Section 5 . 2 ) ; the dependenceon [ 1 1 ] is the source of the reliance on algebraic tools , such as Artin -Schreier theory .

In Chapter 6 , we collect more results about fields with a valuation , specif- ically in the case of positive characteristic . We recall basic properties oftwisted polynomials ( Section 6 . 1 ) and Newton polygons ( Section 6 . 2 ), give a basic form of Hensel ’ s lemma on splitting polynomials ( Section 6. 3 ) , and adapt this result to twisted polynomials ( Section 6 . 4 ) .

In Chapter 7 , we give a second proof of the reverse implication of themain theorem , replacing the algebraic methods of the previouschapter with more explicit considerations of automata . To do this , weanalyze the transition graphs of automata which give rise to generalizedpower series ( Section 7 . 1 ) , show that the class of automaticgeneralized power series i s closed under addition and multiplication (Section 7 . 2 ) , and exhibit a positive - characteristic analogue ofNewton ’ s iteration ( Section 7 . 3 ) .

In Chapter 8 , we raise some further questions about the algorithmics ofautomatic generalized power series ( Section 8 . 1 ) , and about a potentialgen - eralization of the multivariate analogue of Christol ’ s theorem ( Section8 . 2 ) .Acknowledgments . Thanks to Bjorn Poonen for bringing the work ofChristol to the author ’ s attention , to Eric Rains for some intriguing sugges- t ions concerning efficient representations , to Richard Stanley for pointingout the term “ cactus ” , and to George Bergman for helpful discussions .The


\noindent author was supported p r e v i o u s l y by an NSF Postdocto ra l Fe l lowsh ip andc u r r e n t l y by NSF grant DMS − 400727 .

\centerline {2 . \quad Automata }

In t h i s chapter , we r e c a l l no t i ons and f i x notat ion and termino logy re −garding f i n i t e automata . We take as our r e f e r e n c e [ 2 , Chapter 4 ] . We notein pas s ing that a s u f f i c i e n t l y d i l i g e n t reader should be ab le to reproducethe p roo f s o f a l l c i t e d r e s u l t s in t h i s chapter .

\noindent 2 . 1 . \quad Dete rm in i s t i c automata .

\noindent D e f i n i t i o n 2 . 1 . 1 . \ h f i l l A d e t e r m i n i s t i c f i n i t e automaton , or DFA f o r shor t , i s

\noindent a tup l e $ M = ( Q , \Sigma , \delta , q 0 , F ), $ where

\centerline{ $ \bullet Q $ i s a f i n i t e s e t ( the s t a t e s ) ; }

\centerline{ $ \bullet \Sigma $ i s another s e t ( the input a lphabet ) ; }

\centerline{ $ \bullet \delta $ i s a func t i on from $ Q \times \Sigma $ to $ Q( $ the t r a n s i t i o n func t i on ) ; }

\centerline{ $ \bullet q 0 \ in Q $ i s a s t a t e ( the i n i t i a l s t a t e ) ; }

\centerline{ $ \bullet F $ i s a subset o f $ Q ( $ the accept ing s t a t e s ) . }

\noindent D e f i n i t i o n 2 . 1 . 2 . \quad Let $ \Sigma ˆ{ ∗ }$ denote the s e t o f f i n i t e sequences c o n s i s t i n g o fe lements o f $ \Sigma ; $ we w i l l r e f e r to e lements o f $ \Sigma $ as \quad c h a r a c t e r s and elementso f $ \Sigma ˆ{ ∗ }$ as s t r i n g s . We i d e n t i f y e lements o f $ \Sigma $ with one − element s t r i n g s , anddenote concatenat ion o f s t r i n g s by j u x t a p o s i t i o n : \quad that \quad i s , \quad i f

$ s $ \quad and $ t $ \quad ares t r i n g s , then $ s t $ i s the s t r i n g composed o f the e lements o f $ s $ f o l l owed by thee lements o f $ t . $ We d e f i n e a language ( over $ \Sigma ) $ to be any subset o f

$ \Sigma ˆ{ ∗ } . $

\hspace ∗{\ f i l l } I t i s sometimes convenient to r e p r e s e n t a DFA us ing a t r a n s i t i o n graph .

\noindent D e f i n i t i o n 2 . 1 . 3 . \ h f i l l Given a DFA $ M = ( Q , \Sigma ,\delta , q 0 , F ) , $ the t r a n s i t i o n graph

\noindent o f $ M $ i s the edge − l a b e l e d d i r e c t e d graph ( p o s s i b l y with loops ) on the ver tex

\noindent s e t $ Q , $ with an edge from $ q \ in Q $ to $ q ˆ{ \prime } \ inQ $ l a b e l e d by $ s \ in \Sigma $ i f $ \delta ( q , s ) = q ˆ{ \prime }. $The t r a n s i t i o n graph a l s o comes equipped with a d i s t i n g u i s h e d ver tex cor −responding to $ q 0 , $ and a d i s t i n g u i s h e d subset o f the ver tex s e t cor re spond ingto $ F ; $ from these data , one can recove r $ M $ from i t s t r a n s i t i o n graph .

One can a l s o imagine a DFA as a machine with a keyboard conta in ingthe e lements o f $ \Sigma , $ which can be at any t ime in any o f the s t a t e s . When onep r e s s e s a key , the machine t r a n s i t i o n s to a new s t a t e by apply ing $ \delta $ to thecur rent s t a t e and the key pre s sed . One can then extend the t r a n s i t i o n func −t ion to s t r i n g s by p r e s s i n g the correspond ing keys in sequence . Formally ,we extend $ \delta $ to a func t i on $ \delta ˆ{ ∗ } : Q \times \Sigma ˆ{ ∗ }\rightarrow Q $ by the r u l e s

\ [ \delta ˆ{ ∗ } ( q , \ varnothing ) = q , \delta ˆ{ ∗ } ( q ,xa ) = \delta ( \delta ˆ{ ∗ } ( q , x ) , a ) ( q \ inQ , x \ in \Sigma ˆ{ ∗ } , a \ in \Sigma ) . \ ]

\noindent D e f i n i t i o n 2 . 1 . 4 . \quad We say that $ M $ accept s a s t r i n g $ x \ in\Sigma ˆ{ ∗ }$ i f $ \delta ˆ{ ∗ } ( q 0 , x ) \ in F , $

and otherwi se say i t r e j e c t s $ x . $ The s e t o f s t r i n g s accepted by $ M $ i s c a l l e d

3 82 .. Kiran S period Kedlayaauthor was supported previously by an NSF Postdoctoral Fellowship andcurrently by NSF grant DMS hyphen 400727 period2 period .. AutomataIn this chapter comma we recall notions and fix notation and terminology re hyphengarding finite automata period We take as our reference open square bracket 2 comma Chapter 4 closing square

bracket period We notein passing that a sufficiently diligent reader should be able to reproducethe proofs of all cited results in this chapter period2 period 1 period .. Deterministic automata periodDefinition 2 period 1 period 1 period .... A deterministic finite automaton comma or DFA for short comma i sa tuple M = open parenthesis Q comma Capital Sigma comma delta comma q 0 comma F closing parenthesis

comma wherebullet Q is a finite set open parenthesis the states closing parenthesis semicolonbullet Capital Sigma is another set open parenthesis the input alphabet closing parenthesis semicolonbullet delta is a function from Q times Capital Sigma to Q open parenthesis the transition function closing

parenthesis semicolonbullet q 0 in Q i s a state open parenthesis the initial state closing parenthesis semicolonbullet F i s a subset of Q open parenthesis the accepting states closing parenthesis periodDefinition 2 period 1 period 2 period .. Let Capital Sigma to the power of * denote the set of finite sequences

consisting ofelements of Capital Sigma semicolon we will refer to elements of Capital Sigma as .. characters and elementsof Capital Sigma to the power of * as strings period We identify elements of Capital Sigma with one hyphen

element strings comma anddenote concatenation of strings by juxtaposition : .. that .. i s comma .. if s .. and t .. arestrings comma then st is the string composed of the elements of s followed by theelements of t period We define a language open parenthesis over Capital Sigma closing parenthesis to be any

subset of Capital Sigma to the power of * periodIt i s sometimes convenient to represent a DFA using a transition graph periodDefinition 2 period 1 period 3 period .... Given a DFA M = open parenthesis Q comma Capital Sigma comma

delta comma q 0 comma F closing parenthesis comma the transition graphof M is the edge hyphen labeled directed graph open parenthesis possibly with loops closing parenthesis on the

vertexset Q comma with an edge from q in Q to q to the power of prime in Q labeled by s in Capital Sigma if delta

open parenthesis q comma s closing parenthesis = q to the power of prime periodThe transition graph also comes equipped with a distinguished vertex cor hyphenresponding to q 0 comma and a distinguished subset of the vertex set correspondingto F semicolon from these data comma one can recover M from its transition graph periodOne can also imagine a DFA as a machine with a keyboard containingthe elements of Capital Sigma comma which can be at any t ime in any of the states period When onepresses a key comma the machine transitions to a new state by applying delta to thecurrent state and the key pressed period One can then extend the transition func hyphent ion to strings by pressing the corresponding keys in sequence period Formally commawe extend delta to a function delta to the power of * : Q times Capital Sigma to the power of * right arrow Q

by the rulesdelta to the power of * open parenthesis q comma varnothing closing parenthesis = q comma delta to the power of

* open parenthesis q comma xa closing parenthesis = delta open parenthesis delta to the power of * open parenthesisq comma x closing parenthesis comma a closing parenthesis open parenthesis q in Q comma x in Capital Sigma tothe power of * comma a in Capital Sigma closing parenthesis period

Definition 2 period 1 period 4 period .. We say that M accepts a string x in Capital Sigma to the power of * ifdelta to the power of * open parenthesis q 0 comma x closing parenthesis in F comma

and otherwise say it rejects x period The set of strings accepted by M i s called


author was supported previously by an NSF Postdoctoral Fellowship andcurrently by NSF grant DMS - 400727 .

2 . AutomataIn this chapter , we recall notions and fix notation and terminology re -

garding finite automata . We take as our reference [ 2 , Chapter 4 ] . We notein passing that a sufficiently diligent reader should be able to reproduce theproofs of all cited results in this chapter .2 . 1 . Deterministic automata .Definition 2 . 1 . 1 . A deterministic finite automaton , or DFA forshort , i sa tuple M = (Q,Σ, δ, q0, F ), where

• Q is a finite set ( the states ) ;• Σ is another set ( the input alphabet ) ;

• δ is a function from Q× Σ to Q( the transition function ) ;• q0 ∈ Q i s a state ( the initial state ) ;

• F i s a subset of Q( the accepting states ) .Definition 2 . 1 . 2 . Let Σ∗ denote the set of finite sequencesconsisting of elements of Σ; we will refer to elements of Σ as charactersand elements of Σ∗ as strings . We identify elements of Σ with one - elementstrings , and denote concatenation of strings by juxtaposition : that is , if s and t are strings , then st is the string composed of theelements of s followed by the elements of t. We define a language ( overΣ) to be any subset of Σ∗.

It i s sometimes convenient to represent a DFA using a transition graph.Definition 2 . 1 . 3 . Given a DFA M = (Q,Σ, δ, q0, F ), the transitiongraphof M is the edge - labeled directed graph ( possibly with loops ) on thevertexset Q, with an edge from q ∈ Q to q′ ∈ Q labeled by s ∈ Σ if δ(q, s) = q′.The transition graph also comes equipped with a distinguished vertex cor -responding to q0, and a distinguished subset of the vertex set correspondingto F ; from these data , one can recover M from its transition graph .

One can also imagine a DFA as a machine with a keyboard containingthe elements of Σ, which can be at any t ime in any of the states . When onepresses a key , the machine transitions to a new state by applying δ to thecurrent state and the key pressed . One can then extend the transition func- t ion to strings by pressing the corresponding keys in sequence . Formally, we extend δ to a function δ∗ : Q× Σ∗ → Q by the rules

δ∗(q,∅) = q, δ∗(q, xa) = δ(δ∗(q, x), a) (q ∈ Q, x ∈ Σ∗, a ∈ Σ).

Definition 2 . 1 . 4 . We say that M accepts a string x ∈ Σ∗ ifδ∗(q0, x) ∈ F, and otherwise say it rejects x. The set of strings accepted byM i s called

\hspace ∗{\ f i l l }F i n i t e automata and a l g e b r a i c ex t en s i on s o f func t i on f i e l d s \quad 383

\noindent the language accepted by $ M $ and denoted $ L ( M ) . $ A language i s s a id to be

\noindent r e g u l a r i f i t i s accepted by some DFA .

\noindent Lemma 2 . 1 . 5 . \ h f i l l ( a ) \ h f i l l The c o l l e c t i o n o f r e g u l a r languages i s c l o s ed under

complement , \quad f i n i t e \quad union , \quad and f i n i t e \quad i n t e r s e c t i o n . \quad Also , \quad any \quad l an −guage c o n s i s t i n g o f a s i n g l e s t r i n g i s r e g u l a r .

\hspace ∗{\ f i l l }( b ) \quad The c o l l e c t i o n o f r e g u l a r languages i s c l o s ed under r e v e r s a l ( the op −

\centerline{ e r a t i o n on s t r i n g s tak ing $ s { 1 } \cdot \cdot \cdot s { n }$to $ s { n } \cdot \cdot \cdot s { 1 } ) . $ }

\hspace ∗{\ f i l l }( c ) \quad A language i s r e g u l a r i f and only i f i t i s generated by s ome r e g u l a r

\centerline{ exp r e s s i on ( s ee [ 2 , 1 . 3 ] f o r a d e f i n i t i o n ) . }

\noindent Proof . \ h f i l l ( a ) i s s t r a i g h t f o r w a r d , ( b ) i s [ 2 , Coro l l a ry 4 . 3 . 5 ] , and ( c ) i s Kleene ’ s

\noindent theorem [ 2 , Theorem $ 4 . 1 . 5 ] . \ square $

The Myhi l l − Nerode theorem [ 2 , Theorem 4 . 1 . 8 ] g i v e s an i n t r i n s i c charac −t e r i z a t i o n o f r e g u l a r languages , without r e f e r e n c e to an a u x i l i a r y automa −

\noindent ton .

\noindent D e f i n i t i o n 2 . 1 . 6 . \quad Given a language $ L $ over $ \Sigma , $d e f i n e the equ iva l ence r e l a −

t ion $ \sim L $ on $ \Sigma ˆ{ ∗ }$ as f o l l o w s $ : x \sim L y $ i f and only i f f o r a l l$ z \ in \Sigma ˆ{ ∗ } , xz \ in L $ i f and

\begin { a l i g n ∗}only i f yz \ in L .\end{ a l i g n ∗}

\noindent Lemma 2 . 1 . 7 ( Myhi l l − Nerode theorem ) . \quad The language $ L $ i s r e g u l a r i f andonly i f $ \Sigma ˆ{ ∗ }$ has only f i n i t e l y many equ iva l ence c l a s s e s under $ \sim

L . $

\centerline{Moreover , i f $ L $ i s r e g u l a r , then the DFA in which : }

\centerline{ $ \bullet Q $ i s the s e t o f equ iva l ence c l a s s e s under $ \sim L ; $}

$ \bullet \delta , $ app l i ed to the c l a s s o f some $ x \ in \Sigma ˆ{ ∗ }$and some $ s \ in \Sigma , $ r e tu rn s the

c l a s s o f $ xs ; $

\centerline{ $ \bullet q 0 $ i s the c l a s s o f the empty s t r i n g ; }

\centerline{ $ \bullet F $ i s the s e t o f c l a s s e s o f e lements o f $ L ; $ }

\noindent gene ra t e s $ L $ and has fewer s t a t e s than any nonisomorphic DFA which a l s ogene ra t e s $ L [ 2 , $ Coro l l a ry 4 . 1 . 9 ] .

I t w i l l a l s o be convenient to permit automata to make non − binary dec i −s i o n s about s t r i n g s .

\noindent D e f i n i t i o n 2 . 1 . 8 . \quad A d e t e r m i n i s t i c f i n i t e automaton with output , or DFAOf o r shor t , i s a tup l e $ M = ( Q , \Sigma , \delta , q 0 ,\Delta , \tau ) , $ where





\centerline{ $ \bullet \Delta $ i s a f i n i t e s e t ( the output a lphabet ) ; }

\centerline{ $ \bullet \tau $ i s a func t i on from $ Q $ to $ \Delta ( $ the output func t i on ) . }

\noindent A DFAO $ M $ g i v e s r i s e to a func t i on $ f M : \Sigma ˆ{ ∗ } \rightarrow\Delta $ by s e t t i n g $ f M ( w ) = $

\noindent $ \tau ( \delta ˆ{ ∗ } ( q 0 , w ) ) . $ Any func t i on$ f : \Sigma ˆ{ ∗ } \rightarrow \Delta $ equal to $ f M $ f o r some DFAO$ M $ i s

Finite automata and algebraic extensions of function fields .. 383the language accepted by M and denoted L open parenthesis M closing parenthesis period A language is said to

beregular if it i s accepted by some DFA periodLemma 2 period 1 period 5 period .... open parenthesis a closing parenthesis .... The collection of regular

languages is c losed undercomplement comma .. finite .. union comma .. and finite .. intersection period .. Also comma .. any .. lan

hyphenguage consisting of a s ingle string is regular periodopen parenthesis b closing parenthesis .. The collection of regular languages is c los ed under reversal open

parenthesis the op hypheneration on strings taking s sub 1 times times times s sub n to s sub n times times times s sub 1 closing parenthesis

periodopen parenthesis c closing parenthesis .. A language is regular if and only if it is generated by s ome regularexpression open parenthesis s ee open square bracket 2 comma 1 period 3 closing square bracket for a definition

closing parenthesis periodProof period .... open parenthesis a closing parenthesis i s straightforward comma open parenthesis b closing

parenthesis i s open square bracket 2 comma Corollary 4 period 3 period 5 closing square bracket comma and openparenthesis c closing parenthesis is Kleene quoteright s

theorem open square bracket 2 comma Theorem 4 period 1 period 5 closing square bracket period squareThe Myhill hyphen Nerode theorem open square bracket 2 comma Theorem 4 period 1 period 8 closing square

bracket gives an intrinsic charac hyphenterization of regular languages comma without reference to an auxiliary automa hyphenton periodDefinition 2 period 1 period 6 period .. Given a language L over Capital Sigma comma define the equivalence

rela hyphent ion thicksim L on Capital Sigma to the power of * as follows : x thicksim L y if and only if for all z in Capital

Sigma to the power of * comma xz in L if andonly if yz in L periodLemma 2 period 1 period 7 open parenthesis Myhill hyphen Nerode theorem closing parenthesis period .. The

language L is regular if andonly if Capital Sigma to the power of * has only finitely many equivalence c lasses under thicksim L periodMoreover comma if L i s regular comma then the DFA in which :bullet Q is the set of equivalence classes under thicksim L semicolonbullet delta comma applied to the class of some x in Capital Sigma to the power of * and some s in Capital Sigma

comma returns theclass of xs semicolonbullet q 0 is the class of the empty string semicolonbullet F i s the set of classes of elements of L semicolongenerates L and has fewer states than any nonisomorphic DFA which alsogenerates L open square bracket 2 comma Corollary 4 period 1 period 9 closing square bracket periodIt will also be convenient to permit automata to make non hyphen binary deci hyphensions about strings periodDefinition 2 period 1 period 8 period .. A deterministic finite automaton with output comma or DFAOfor short comma i s a tuple M = open parenthesis Q comma Capital Sigma comma delta comma q 0 comma

Capital Delta comma tau closing parenthesis comma wherebullet Q is a finite set open parenthesis the states closing parenthesis semicolonbullet Capital Sigma is another set open parenthesis the input alphabet closing parenthesis semicolonbullet delta is a function from Q times Capital Sigma to Q open parenthesis the transition function closing

parenthesis semicolonbullet q 0 in Q i s a state open parenthesis the initial state closing parenthesis semicolonbullet Capital Delta is a finite set open parenthesis the output alphabet closing parenthesis semicolonbullet tau is a function from Q to Capital Delta open parenthesis the output function closing parenthesis periodA DFAO M gives rise to a function f M : Capital Sigma to the power of * right arrow Capital Delta by setting f

M open parenthesis w closing parenthesis =tau open parenthesis delta to the power of * open parenthesis q 0 comma w closing parenthesis closing parenthesis

period Any function f : Capital Sigma to the power of * right arrow Capital Delta equal to f M for some DFAO M is

Finite automata and algebraic extensions of function fields 383

the language accepted by M and denoted L(M). A language is said to beregular if it i s accepted by some DFA .Lemma 2 . 1 . 5 . ( a ) The collection of regular languages is c losedunder

complement , finite union , and finite intersection . Also, any lan - guage consisting of a s ingle string is regular .

( b ) The collection of regular languages is c los ed under reversal (the op -

eration on strings taking s1 · · · sn to sn · · · s1).( c ) A language is regular if and only if it is generated by s ome

regularexpression ( s ee [ 2 , 1 . 3 ] for a definition ) .

Proof . ( a ) i s straightforward , ( b ) i s [ 2 , Corollary 4 . 3 . 5 ] , and (c ) is Kleene ’ stheorem [ 2 , Theorem 4.1.5]. �

The Myhill - Nerode theorem [ 2 , Theorem 4 . 1 . 8 ] gives an intrinsiccharac - terization of regular languages , without reference to an auxiliaryautoma -ton .Definition 2 . 1 . 6 . Given a language L over Σ, define theequivalence rela - t ion ∼ L on Σ∗ as follows : x ∼ Ly if and only if for allz ∈ Σ∗, xz ∈ L if and

onlyifyz ∈ L.

Lemma 2 . 1 . 7 ( Myhill - Nerode theorem ) . The language L isregular if and only if Σ∗ has only finitely many equivalence c lasses under∼ L.

Moreover , if L i s regular , then the DFA in which :• Q is the set of equivalence classes under ∼ L;

• δ, applied to the class of some x ∈ Σ∗ and some s ∈ Σ, returns the classof xs;

• q0 is the class of the empty string ;• F i s the set of classes of elements of L;

generates L and has fewer states than any nonisomorphic DFA which alsogenerates L[2, Corollary 4 . 1 . 9 ] .

It will also be convenient to permit automata to make non - binary deci- sions about strings .Definition 2 . 1 . 8 . A deterministic finite automaton with output ,or DFAO for short , i s a tuple M = (Q,Σ, δ, q0,∆, τ), where


• δ is a function from Q× Σ to Q( the transition function ) ;• q0 ∈ Q i s a state ( the initial state ) ;• ∆ is a finite set ( the output alphabet ) ;

• τ is a function from Q to ∆( the output function ) .A DFAO M gives rise to a function fM : Σ∗ → ∆ by settingfM(w) =τ(δ∗(q0, w)). Any function f : Σ∗ → ∆ equal to fM for some DFAO M i s


\noindent c a l l e d a f i n i t e − s t a t e func t i on ; note that $ f $ i s a f i n i t e − s t a t e func t i on i f and

\noindent only i f $ f ˆ{ − 1 } ( d ) $ \ h f i l l i s \ h f i l l a r e g u l a r language f o r each$ d \ in \Delta [ 2 , $ \ h f i l l Theorems 4 . 3 . 1

\noindent and 4 . 3 . 2 ] .

We w i l l i d e n t i f y each DFA with the DFAO with output a lphabet \{ 0 , 1 \}which outputs 1 on a s t r i n g i f the o r i g i n a l DFA accept s the s t r i n g and 0otherwi se .

I t w i l l a l s o be u s e f u l to have d e v i c e s that can operate on the c l a s s o fr e g u l a r languages .

\noindent D e f i n i t i o n 2 . 1 . 9 . A f i n i t e − s t a t e t ransducer i s a tup l e $ T = (Q , \Sigma , \delta , q 0 , \Delta , \lambda ) , $

where





\centerline{ $ \bullet \Delta $ i s a f i n i t e s e t ( the output a lphabet ) ; }

\centerline{ $ \bullet \lambda $ i s a func t i on from $ Q \times \Sigma $ to$ \Delta ∗ ( $ the output func t i on ) . }

\noindent I f the output o f $ \lambda $ i s always a s t r i n g o f l ength $ k , $ we say the t ransducer$ T $ i s

\begin { a l i g n ∗}k − uniform .\end{ a l i g n ∗}

A transducer $ T $ g i v e s r i s e to a func t i on $ f T : \Sigma ˆ{ ∗ } \rightarrow\Delta ∗ $ as f o l l o w s : g iven

a s t r i n g $ w = s { 1 } \cdot \cdot \cdot s { r } \ in \Sigma ˆ{ ∗ }( $ with each $ s { i } \ in \Sigma ) , $ put $ q i = \delta ˆ{ ∗ }( s { 1 } \cdot \cdot \cdot s { i } ) $ f o r

\begin { a l i g n ∗}i = 1 , . . . , r , and de f i ne \\ f T ( w ) = \lambda

( q 0 , s { 1 } ) \lambda ( q 1 , s { 2 } ) \cdot \cdot\cdot \lambda ( q r − 1 , s { r } ) .\end{ a l i g n ∗}

\noindent That i s , f e ed $ w $ in to the t ransducer and at each step , use the cur rent s t a t eand the next t r a n s i t i o n to produce a p i e c e o f output , then s t r i n g toge the r

\noindent the outputs . For $ L \subseteq \Sigma ˆ{ ∗ }$ and $ L ˆ{ \prime }\subseteq \Delta ∗ $ languages , we wr i t e

\ [\ begin { a l i gned } f T ( L ) = \{ f T ( w ) : w \ in L\} \\

f ˆ{ − 1 } { T } ( L ˆ{ \prime } ) = \{ w \ in \Sigma ˆ{ ∗ } : fT ( w ) \ in L ˆ{ \prime } \} . \end{ a l i gned }\ ]

\noindent Then one has the f o l l o w i n g r e s u l t [ 2 , Theorems 4 . 3 . 6 and 4 . 3 . 8 ] .

\noindent Lemma 2 . 1 . 10 . \ h f i l l Let $ T $ be a f i n i t e − s t a t e t ransducer . I f$ L \subseteq \Sigma ˆ{ ∗ }$ \ h f i l l i s a r e g u l a r

\noindent language , then $ f T ( L ) $ i s a l s o r e g u l a r ; i f $ L ˆ{ \prime }\subseteq \Delta ˆ{ ∗ }$ i s a r e g u l a r language , then

\noindent $ f ˆ{ − 1 } { T } ( L ˆ{ \prime } ) $ \quad i s a l s o r e g u l a r .

\noindent 2 . 2 . \quad Nondetermin i s t i c automata and m u l t i p l i c i t i e s . \quad Although theydo not expand the boundar ies o f the theory , i t w i l l be u s e f u l in p r a c t i c e toa l low so − c a l l e d ‘ ‘ nonde t e rm in i s t i c automata ’ ’ .

\noindent D e f i n i t i o n 2 . 2 . 1 . \ h f i l l A nonde t e rm in i s t i c f i n i t e automaton , or NFA f o r shor t ,

\noindent i s a tup l e $ M = ( Q , \Sigma , \delta , q 0 , F) , $ where


3 84 .. Kiran S period Kedlayacalled a finite hyphen state function semicolon note that f i s a finite hyphen state function if andonly if f to the power of minus 1 open parenthesis d closing parenthesis .... i s .... a regular language for each d

in Capital Delta open square bracket 2 comma .... Theorems 4 period 3 period 1and 4 period 3 period 2 closing square bracket periodWe will identify each DFA with the DFAO with output alphabet open brace 0 comma 1 closing bracewhich outputs 1 on a string if the original DFA accepts the string and 0otherwise periodIt will also be useful to have devices that can operate on the class ofregular languages periodDefinition 2 period 1 period 9 period A finite hyphen state transducer i s a tuple T = open parenthesis Q comma

Capital Sigma comma delta comma q 0 comma Capital Delta comma lambda closing parenthesis commawherebullet Q is a finite set open parenthesis the states closing parenthesis semicolonbullet Capital Sigma is another set open parenthesis the input alphabet closing parenthesis semicolonbullet delta is a function from Q times Capital Sigma to Q open parenthesis the transition function closing

parenthesis semicolonbullet q 0 in Q i s a state open parenthesis the initial state closing parenthesis semicolonbullet Capital Delta is a finite set open parenthesis the output alphabet closing parenthesis semicolonbullet lambda i s a function from Q times Capital Sigma to Capital Delta * open parenthesis the output function

closing parenthesis periodIf the output of lambda is always a string of length k comma we say the transducer T i sk hyphen uniform periodA transducer T gives rise to a function f T : Capital Sigma to the power of * right arrow Capital Delta * as

follows : givena string w = s sub 1 times times times s sub r in Capital Sigma to the power of * open parenthesis with each s

sub i in Capital Sigma closing parenthesis comma put q i = delta to the power of * open parenthesis s sub 1 timestimes times s sub i closing parenthesis for

i = 1 comma period period period comma r comma and define f T open parenthesis w closing parenthesis =lambda open parenthesis q 0 comma s sub 1 closing parenthesis lambda open parenthesis q 1 comma s sub 2 closingparenthesis times times times lambda open parenthesis q r minus 1 comma s sub r closing parenthesis period

That is comma feed w into the transducer and at each step comma use the current stateand the next transition to produce a piece of output comma then string togetherthe outputs period For L subset equal Capital Sigma to the power of * and L to the power of prime subset equal

Capital Delta * languages comma we writeLine 1 f T open parenthesis L closing parenthesis = open brace f T open parenthesis w closing parenthesis : w in

L closing brace Line 2 f sub T to the power of minus 1 open parenthesis L to the power of prime closing parenthesis= open brace w in Capital Sigma to the power of * : f T open parenthesis w closing parenthesis in L to the powerof prime closing brace period

Then one has the following result open square bracket 2 comma Theorems 4 period 3 period 6 and 4 period 3period 8 closing square bracket period

Lemma 2 period 1 period 10 period .... Let T be a finite hyphen state transducer period If L subset equal CapitalSigma to the power of * .... is a regular

language comma then f T open parenthesis L closing parenthesis is also regular semicolon if L to the power ofprime subset equal Capital Delta to the power of * is a regular language comma then

f sub T to the power of minus 1 open parenthesis L to the power of prime closing parenthesis .. is also regularperiod

2 period 2 period .. Nondeterministic automata and multiplicities period .. Although theydo not expand the boundaries of the theory comma it will be useful in practice toallow so hyphen called quotedblleft nondeterministic automata quotedblright periodDefinition 2 period 2 period 1 period .... A nondeterministic finite automaton comma or NFA for short commai s a tuple M = open parenthesis Q comma Capital Sigma comma delta comma q 0 comma F closing parenthesis

comma wherebullet Q is a finite set open parenthesis the states closing parenthesis semicolon


called a finite - state function ; note that f i s a finite - state function ifandonly if f−1(d) i s a regular language for each d ∈ ∆ [2, Theorems 4 . 3. 1and 4 . 3 . 2 ] .

We will identify each DFA with the DFAO with output alphabet { 0 , 1} which outputs 1 on a string if the original DFA accepts the string and 0otherwise .

It will also be useful to have devices that can operate on the class ofregular languages .Definition 2 . 1 . 9 . A finite - state transducer i s a tuple T =(Q,Σ, δ, q0,∆, λ), where


• δ is a function from Q× Σ to Q( the transition function ) ;• q0 ∈ Q i s a state ( the initial state ) ;• ∆ is a finite set ( the output alphabet ) ;

• λ i s a function from Q× Σ to ∆ ∗ ( the output function ) .If the output of λ is always a string of length k, we say the transducer T is

k − uniform.

A transducer T gives rise to a function fT : Σ∗ → ∆∗ as follows : given astring w = s1 · · · sr ∈ Σ∗ ( with each si ∈ Σ), put qi = δ∗(s1 · · · si) for

i = 1, ..., r, anddefine

fT (w) = λ(q0, s1)λ(q1, s2) · · · λ(qr − 1, sr).

That is , feed w into the transducer and at each step , use the current stateand the next transition to produce a piece of output , then string togetherthe outputs . For L ⊆ Σ∗ and L′ ⊆ ∆∗ languages , we write

fT (L) = {fT (w) : w ∈ L}f−1T (L′) = {w ∈ Σ∗ : fT (w) ∈ L′}.

Then one has the following result [ 2 , Theorems 4 . 3 . 6 and 4 . 3 . 8 ] .Lemma 2 . 1 . 10 . Let T be a finite - state transducer . If L ⊆ Σ∗ isa regularlanguage , then fT (L) is also regular ; if L′ ⊆ ∆∗ is a regular language ,thenf−1T (L′) is also regular .2 . 2 . Nondeterministic automata and multiplicities . Al-though they do not expand the boundaries of the theory , it will be usefulin practice to allow so - called “ nondeterministic automata ” .Definition 2 . 2 . 1 . A nondeterministic finite automaton , or NFAfor short ,i s a tuple M = (Q,Σ, δ, q0, F ), where

• Q is a finite set ( the states ) ;

F i n i t e automata and a l g e b r a i c ex t en s i on s o f f unc t i on f i e l d s \quad 385$ \bullet \Sigma $ i s another s e t ( the input a lphabet ) ;

\hspace ∗{\ f i l l } $ \bullet \delta $ i s a func t i on from $ Q \times \Sigma $ to the power s e t o f$ Q ( $ the t r a n s i t i o n

\ [ f unc t i on ) ; \ ]


\centerline{ $ \bullet F $ i s a subset o f $ Q ( $ the accept ing s t a t e s ) . }

\noindent For $ M $ an NFA and $ w = s { 1 } . . . s { n } \ in \Sigma ˆ{ ∗ }( $ with $ s { i } \ in \Sigma $ f o r $ i = 1 , . . . , n ), $ d e f i n ean accept ing path f o r $ w $ to be a sequence o f s t a t e s $ q 1 , . . .

, q n \ in Q $ such that$ q i \ in \delta ( q i − 1 , s { i } ) $ f o r $ i = 1

, . . . , n $ and $ q n \ in F . $ Def ine the language accepted by$ M $ as the s e t o f s t r i n g s $ w \ in \Sigma ˆ{ ∗ }$ f o r which the re e x i s t s an accept ing path .

In f o rma l l y , \quad an NFA i s a machine which may make a cho i c e o f how tot r a n s i t i o n based on the cur rent s t a t e and key pre s sed , or may not be ab le

\noindent to make any t r a n s i t i o n at a l l . I t accept s a s t r i n g i f the re i s some way i t cant r a n s i t i o n from the i n i t i a l s t a t e in to an accept ing s t a t e v ia the correspond −

\noindent ing key p r e s s e s .

Every DFA can be viewed \quad as \quad an NFA , \quad and the language \quad accepted i sthe \quad same under both i n t e r p r e t a t i o n s ; \quad hence every language \quad accepted bysome DFA i s \quad a l s o accepted by some NFA . \quad The converse i s \quad a l s o t rue \quad [ 2 ,Theorem 4 . 1 . 3 ] , so f o r t h e o r e t i c a l purposes , i t i s t y p i c a l l y s u f f i c i e n t to workwith the conceptua l l y s imp le r DFAs . However , the conver s i on from an NFAof $ n $ s t a t e s may produce a DFA with as many as $ 2 ˆ{ n }$ s t a t e s , so in p r a c t i c et h i s i s not u s u a l l y a good idea .

\centerline{We w i l l need the f o l l o w i n g q u a n t i t a t i v e var i ant o f [ 2 , Theorem 4 . 1 . 3 ] . }

\noindent Lemma \quad 2 . 2 . 2 . \quad Fix a p o s i t i v e \quad i n t e g e r $ n . $ \quad Let$ M = ( Q , \Sigma , \delta , q 0 , F ) $ \quad be \quad anNFA , \quad and l e t $ f : \Sigma ˆ{ ∗ } \rightarrow Z / n Z $ \quad be the func t i on that a s s i g n s to

$ w \ in \Sigma ˆ{ ∗ }$ \quad thenumber o f accept ing paths f o r $ w $ \quad in $ M , $ \quad reduced modulo $ n

. $ \quad Then $ f $ \quad i s \quad a

\noindent f i n i t e − s t a t e func t i on .

\noindent Proof . \ h f i l l We cons t ruc t a DFAO $ M ˆ{ \prime } = ( Q ˆ{ \prime }, \Sigma ˆ{ \prime } , \delta ˆ{ \prime } , q ˆ{ \prime } { 0 ˆ{ , }} \Delta ˆ{ \prime }, \tau ˆ{ \prime } ) $ with the property

\noindent that $ f = f M ˆ{ \prime } , $ as f o l l o w s . Let $ Q ˆ{ \prime }$denote the s e t o f f u n c t i o n s from $ Q $ to $ Z / n Z , $

\noindent and put $ \Sigma ˆ{ \prime } = \Sigma . $ Def ine the func t i on $ \delta ˆ{ \prime }: Q ˆ{ \prime } \times \Sigma \rightarrow Q ˆ{ \prime }$ as f o l l o w s : g iven a

func t i on $ g : Q \rightarrow Z / n Z $ and an element $ s \ in \Sigma, $ l e t $ \delta ˆ{ \prime } ( g , s ) : Q \rightarrow Z / nZ $ be

the func t i on given by

\ [\ begin { a l i gned } \delta ˆ{ \prime } ( g , s ) ( q ) = \sum g( q 1 ) . \\

q 1 ˆ{ \ in } Q : \delta ( q 1 , s ) = q \end{ a l i gned }\ ]

\noindent Let $ q ˆ{ \prime } { 0 } : Q \rightarrow Z / n Z $ be the func t i on ca r ry ing$ q 0 $ to 1 and a l l other s t a t e s to

0 . Put $ \Delta ˆ{ \prime } = Z / n Z , $ and l e t $ \tau ˆ{ \prime }: Q ˆ{ \prime } \rightarrow Z / n Z $ be the func t i on given by

\ [\ begin { a l i gned } \tau ˆ{ \prime } ( g ) = \sum g ( q ) . \\q \ in F \end{ a l i gned }\ ]

\noindent Then $ M ˆ{ \prime }$ has the d e s i r e d p r o p e r t i e s $ . \ square $Note that lemma 2 . 2 . 2 s t i l l works i f we a l low the va lue s o f $ \delta $ to be m u l t i s e t s .

Finite automata and algebraic extensions of function fields .. 385bullet Capital Sigma is another set open parenthesis the input alphabet closing parenthesis semicolonbullet delta i s a function from Q times Capital Sigma to the power set of Q open parenthesis the transitionfunction closing parenthesis semicolonbullet q 0 in Q i s a state open parenthesis the initial state closing parenthesis semicolonbullet F i s a subset of Q open parenthesis the accepting states closing parenthesis periodFor M an NFA and w = s sub 1 period period period s sub n in Capital Sigma to the power of * open parenthesis

with s sub i in Capital Sigma for i = 1 comma period period period comma n closing parenthesis comma definean accepting path for w to be a sequence of states q 1 comma period period period comma q n in Q such thatq i in delta open parenthesis q i minus 1 comma s sub i closing parenthesis for i = 1 comma period period period

comma n and q n in F period Define the language accepted byM as the set of strings w in Capital Sigma to the power of * for which there exists an accepting path periodInformally comma .. an NFA i s a machine which may make a choice of how totransition based on the current state and key pressed comma or may not be ableto make any transition at all period It accepts a string if there i s some way it cantransition from the initial state into an accepting state via the correspond hyphening key presses periodEvery DFA can be viewed .. as .. an NFA comma .. and the language .. accepted i sthe .. same under both interpretations semicolon .. hence every language .. accepted bysome DFA is .. also accepted by some NFA period .. The converse is .. also true .. open square bracket 2 commaTheorem 4 period 1 period 3 closing square bracket comma so for theoretical purposes comma it i s typically

sufficient to workwith the conceptually simpler DFAs period However comma the conversion from an NFAof n states may produce a DFA with as many as 2 to the power of n states comma so in practicethis is not usually a good idea periodWe will need the following quantitative variant of open square bracket 2 comma Theorem 4 period 1 period 3

closing square bracket periodLemma .. 2 period 2 period 2 period .. Fix a positive .. integer n period .. Let M = open parenthesis Q comma

Capital Sigma comma delta comma q 0 comma F closing parenthesis .. be .. anNFA comma .. and let f : Capital Sigma to the power of * right arrow Z slash n Z .. be the function that assigns

to w in Capital Sigma to the power of * .. thenumber of accepting paths for w .. in M comma .. reduced modulo n period .. Then f .. is .. afinite hyphen state function periodProof period .... We construct a DFAO M to the power of prime = open parenthesis Q to the power of prime

comma Capital Sigma to the power of prime comma delta to the power of prime comma q sub 0 to the powerof comma to the power of prime Capital Delta to the power of prime comma tau to the power of prime closingparenthesis with the property

that f = f M to the power of prime comma as follows period Let Q to the power of prime denote the set offunctions from Q to Z slash n Z comma

and put Capital Sigma to the power of prime = Capital Sigma period Define the function delta to the power ofprime : Q to the power of prime times Capital Sigma right arrow Q to the power of prime as follows : given a

function g : Q right arrow Z slash n Z and an element s in Capital Sigma comma let delta to the power of primeopen parenthesis g comma s closing parenthesis : Q right arrow Z slash n Z be

the function given byLine 1 delta to the power of prime open parenthesis g comma s closing parenthesis open parenthesis q closing

parenthesis = sum g open parenthesis q 1 closing parenthesis period Line 2 q 1 to the power of in Q : delta openparenthesis q 1 comma s closing parenthesis = q

Let q sub 0 to the power of prime : Q right arrow Z slash n Z be the function carrying q 0 to 1 and all otherstates to

0 period Put Capital Delta to the power of prime = Z slash n Z comma and let tau to the power of prime : Qto the power of prime right arrow Z slash n Z be the function given by

Line 1 tau to the power of prime open parenthesis g closing parenthesis = sum g open parenthesis q closingparenthesis period Line 2 q in F

Then M to the power of prime has the desired properties period squareNote that lemma 2 period 2 period 2 still works if we allow the values of delta to be multisets period

Finite automata and algebraic extensions of function fields 385 • Σ is another set( the input alphabet ) ;

• δ i s a function from Q× Σ to the power set of Q ( the transition

function);

• q0 ∈ Q i s a state ( the initial state ) ;• F i s a subset of Q( the accepting states ) .

For M an NFA and w = s1...sn ∈ Σ∗( with si ∈ Σ for i = 1, ..., n), definean accepting path for w to be a sequence of states q1, ..., qn ∈ Q such thatqi ∈ δ(qi− 1, si) for i = 1, ..., n and qn ∈ F. Define the language accepted by Mas the set of strings w ∈ Σ∗ for which there exists an accepting path .

Informally , an NFA i s a machine which may make a choice of how totransition based on the current state and key pressed , or may not be ableto make any transition at all . It accepts a string if there i s some way it cantransition from the initial state into an accepting state via the correspond -ing key presses .

Every DFA can be viewed as an NFA , and the language ac-cepted i s the same under both interpretations ; hence every languageaccepted by some DFA is also accepted by some NFA . The converseis also true [ 2 , Theorem 4 . 1 . 3 ] , so for theoretical purposes , it i stypically sufficient to work with the conceptually simpler DFAs . However ,the conversion from an NFA of n states may produce a DFA with as manyas 2n states , so in practice this is not usually a good idea .

We will need the following quantitative variant of [ 2 , Theorem 4 . 1 . 3 ] .Lemma 2 . 2 . 2 . Fix a positive integer n. Let M =(Q,Σ, δ, q0, F ) be an NFA , and let f : Σ∗ → Z/nZ be the functionthat assigns to w ∈ Σ∗ the number of accepting paths for w in M,reduced modulo n. Then f is afinite - state function .Proof . We construct a DFAO M ′ = (Q′,Σ′, δ′, q′0,∆

′, τ ′) with the propertythat f = fM ′, as follows . Let Q′ denote the set of functions from Q toZ/nZ,and put Σ′ = Σ. Define the function δ′ : Q′ × Σ → Q′ as follows : given afunction g : Q → Z/nZ and an element s ∈ Σ, let δ′(g, s) : Q → Z/nZ be thefunction given by

δ′(g, s)(q) =∑

g(q1).

q1∈Q : δ(q1, s) = q

Let q′0 : Q→ Z/nZ be the function carrying q0 to 1 and all other states to 0. Put ∆′ = Z/nZ, and let τ ′ : Q′ → Z/nZ be the function given by

τ ′(g) =∑

g(q).

q ∈ F

Then M ′ has the desired properties . � Note that lemma 2 . 2 . 2 stillworks if we allow the values of δ to be multisets .


\noindent 2 . 3 . \quad Base expans ions \quad and \quad automatic \quad f u n c t i o n s . \quad In t h i s s e c t i o n , wemake p r e c i s e the not ion o f ‘ ‘ a func t i on on $ Q $ computable by a f i n i t e automa −ton ’ ’ , and u l t im a t e l y r e l a t e i t to the not ion o f an ‘ ‘ automatic sequence ’ ’ from[ 2 , Chapter 5 ] . To do t h i s , we need to f i x a way to input r a t i o n a l numbersin to an automaton , by choos ing some convent ions about base expans ions .

Let $ b > 1 $ be a f i x e d p o s i t i v e i n t e g e r . Al l automata in t h i s s e c t i o n w i l lhave input alphabet $ \Sigma = \Sigma { b } = \{ 0 , 1 , . .

. , b − 1 , . \} , $ which we i d e n t i f y withthe base $ b $ d i g i t s and rad ix po int .

\noindent D e f i n i t i o n 2 . 3 . 1 . \quad A s t r i n g $ s = s { 1 } . . . s { n }\ in \Sigma ˆ{ ∗ }$ i s s a id to be a v a l i d base $ b $

expansion i f $ s { 1 } \ne 0 , s { n } \ne 0 , $ and exac t l y one o f$ s { 1 } , . . . , s { n }$ i s equal to the

rad ix po int . I f $ s $ i s a v a l i d base $ b $ expansion and $ s { k }$ i s the rad ix po int , thenwe d e f i n e the value o f $ s $ to be

\ [\ begin { a l i gned } k − 1 n \\v ( s ) = \sum s { i } b ˆ{ k − 1 − i } + \sum s { i }

b ˆ{ k − i } . \\i = 1 i = k + 1 \end{ a l i gned }\ ]

\noindent I t \quad i s c l e a r that \quad no two v a l i d \quad s t r i n g s have the \quad same value ; \quad we may thusunambiguously d e f i n e $ s $ to be the base $ b $ expansion o f $ v ( s )

. $ Let $ S { b }$ be the

\noindent s e t o f nonnegat ive $ b − $ ad ic r a t i o n a l s , i . e . , numbers o f the form$ m / b ˆ{ n }$ f o r some

nonnegat ive i n t e g e r s $ m , n ; $ i t i s a l s o c l e a r that the s e t o f va lue s o f v a l i d base$ b $ expans ions i s p r e c i s e l y $ S { b } . $ For $ v \ in S { b } , $ wr i t e

$ s ( v ) $ f o r the base $ b $ expansiono f $ v . $

\noindent Lemma 2 . 3 . 2 . \quad The s et o f v a l i d base $ b $ expans ions i s a r e g u l a r language .

\noindent Proof . \quad The language $ L { 1 }$ o f s t r i n g s with no l ead ing zero i s r e g u l a r by v i r t u eo f Lemma 2 . 1 . 7 : \quad the equ iva l ence c l a s s e s under $ \sim L { 1 }$ \quad c o n s i s t o f the emptys t r i n g , a l l nonempty s t r i n g s in $ L { 1 } , $ and a l l nonempty s t r i n g s not in

$ L { 1 } . $ Thelanguage $ L { 2 }$ o f s t r i n g s with no t r a i l i n g zero i s a l s o r e g u l a r : the equ iva l encec l a s s e s under $ \sim L { 2 }$ \quad c o n s i s t o f a l l s t r i n g s in $ L { 2 } , $\quad and a l l s t r i n g s not in $ L { 2 } . $

\noindent ( One could a l s o apply Lemma 2 . 1 . 5 ( b ) to $ L { 1 }$ to show that$ L { 2 }$ i s r e g u l a r , or

\noindent v i c e ver sa . ) The language $ L { 3 }$ o f s t r i n g s with exac t l y one rad ix po int i s a l s or e g u l a r : the equ iva l ence c l a s s e s under $ \sim L { 3 }$ \quad c o n s i s t o f a l l s t r i n g s with zeropo in t s , a l l s t r i n g s with one po int , and a l l s t r i n g s with more than one po int .Hence $ L { 1 } \cap L { 2 } \cap L { 3 }$ i s r e g u l a r by Lemma 2 . 1 . 5 ( a ) , as d e s i r e d

$ . \ square $

The \quad ‘ ‘ r e a l world ’ ’ \quad convent ion f o r base $ b $ expans ions i s a b i t more compli −cated than what we are us ing : normally , one omits the rad ix po int whenthere are no d i g i t s a f t e r i t , one adds a l e ad ing zero in f r o n t o f the rad ixpo int when there are no d i g i t s be f o r e i t , and one r e p r e s e n t s 0 with a s i n g l ezero ra the r than a bare rad ix po int \quad ( or the empty s t r i n g ) . \quad This w i l l notchange anything e s s e n t i a l , thanks to the f o l l o w i n g lemma .

\noindent Lemma \quad 2 . 3 . 3 . \quad Let $ S $ \quad be \quad a s et o f nonnegat ive$ b − $ ad ic r a t i o n a l s . \quad Then the

s et o f expans ions o f $ S , $ \quad under our convent ion , \quad i s a r e g u l a r language i f andonly i f the s e t o f ‘ ‘ r e a l world ’ ’ expans ions o f $ S $ i s a r e g u l a r language .

3 86 .. Kiran S period Kedlaya2 period 3 period .. Base expansions .. and .. automatic .. functions period .. In this section comma wemake precise the notion of quotedblleft a function on Q computable by a finite automa hyphenton quotedblright comma and ultimately relate it to the notion of an quotedblleft automatic sequence quoted-

blright fromopen square bracket 2 comma Chapter 5 closing square bracket period To do this comma we need to fix a way

to input rational numbersinto an automaton comma by choosing some conventions about base expansions periodLet b greater 1 be a fixed positive integer period All automata in this section willhave input alphabet Capital Sigma = Capital Sigma sub b = open brace 0 comma 1 comma period period period

comma b minus 1 comma period closing brace comma which we identify withthe base b digits and radix point periodDefinition 2 period 3 period 1 period .. A string s = s sub 1 period period period s sub n in Capital Sigma to

the power of * i s said to be a valid base bexpansion if s sub 1 equal-negationslash 0 comma s sub n equal-negationslash 0 comma and exactly one of s sub

1 comma period period period comma s sub n i s equal to theradix point period If s i s a valid base b expansion and s sub k i s the radix point comma thenwe define the value of s to beLine 1 k minus 1 n Line 2 v open parenthesis s closing parenthesis = sum s sub i b to the power of k minus 1

minus i plus sum s sub i b to the power of k minus i period Line 3 i = 1 i = k plus 1It .. i s clear that .. no two valid .. strings have the .. same value semicolon .. we may thusunambiguously define s to be the base b expansion of v open parenthesis s closing parenthesis period Let S sub

b be theset of nonnegative b hyphen adic rationals comma i period e period comma numbers of the form m slash b to

the power of n for somenonnegative integers m comma n semicolon it i s also clear that the set of values of valid baseb expansions i s precisely S sub b period For v in S sub b comma write s open parenthesis v closing parenthesis

for the base b expansionof v periodLemma 2 period 3 period 2 period .. The s et of valid base b expansions is a regular language periodProof period .. The language L sub 1 of strings with no leading zero is regular by virtueof Lemma 2 period 1 period 7 : .. the equivalence classes under thicksim L sub 1 .. consist of the emptystring comma all nonempty strings in L sub 1 comma and all nonempty strings not in L sub 1 period Thelanguage L sub 2 of strings with no trailing zero is also regular : the equivalenceclasses under thicksim L sub 2 .. consist of all strings in L sub 2 comma .. and all strings not in L sub 2 periodopen parenthesis One could also apply Lemma 2 period 1 period 5 open parenthesis b closing parenthesis to L

sub 1 to show that L sub 2 i s regular comma orvice versa period closing parenthesis The language L sub 3 of strings with exactly one radix point is alsoregular : the equivalence classes under thicksim L sub 3 .. consist of all strings with zeropoints comma all strings with one point comma and all strings with more than one point periodHence L sub 1 cap L sub 2 cap L sub 3 i s regular by Lemma 2 period 1 period 5 open parenthesis a closing

parenthesis comma as desired period squareThe .. quotedblleft real world quotedblright .. convention for base b expansions is a bit more compli hyphencated than what we are using : normally comma one omits the radix point whenthere are no digits after it comma one adds a leading zero in front of the radixpoint when there are no digits before it comma and one represents 0 with a singlezero rather than a bare radix point .. open parenthesis or the empty string closing parenthesis period .. This will

notchange anything essential comma thanks to the following lemma periodLemma .. 2 period 3 period 3 period .. Let S .. be .. a s et of nonnegative b hyphen adic rationals period ..

Then thes et of expansions of S comma .. under our convention comma .. is a regular language if andonly if the set of quotedblleft real world quotedblright expansions of S is a regular language period


2 . 3 . Base expansions and automatic functions . Inthis section , we make precise the notion of “ a function on Q computableby a finite automa - ton ” , and ultimately relate it to the notion of an“ automatic sequence ” from [ 2 , Chapter 5 ] . To do this , we need tofix a way to input rational numbers into an automaton , by choosing someconventions about base expansions .

Let b > 1 be a fixed positive integer . All automata in this section willhave input alphabet Σ = Σb = {0, 1, ..., b − 1, .}, which we identify with thebase b digits and radix point .Definition 2 . 3 . 1 . A string s = s1...sn ∈ Σ∗ i s said to be a validbase b expansion if s1 6= 0, sn 6= 0, and exactly one of s1, ..., sn i s equal tothe radix point . If s i s a valid base b expansion and sk i s the radix point, then we define the value of s to be

k − 1 n

v(s) =∑

sibk−1−i +

∑sib

k−i.

i = 1 i = k + 1

It i s clear that no two valid strings have the same value ; wemay thus unambiguously define s to be the base b expansion of v(s). LetSb be theset of nonnegative b− adic rationals , i . e . , numbers of the form m/bn

for some nonnegative integers m,n; it i s also clear that the set of values ofvalid base b expansions i s precisely Sb. For v ∈ Sb, write s(v) for the base bexpansion of v.Lemma 2 . 3 . 2 . The s et of valid base b expansions is a regularlanguage .Proof . The language L1 of strings with no leading zero is regular byvirtue of Lemma 2 . 1 . 7 : the equivalence classes under ∼ L1 consistof the empty string , all nonempty strings in L1, and all nonempty stringsnot in L1. The language L2 of strings with no trailing zero is also regular :the equivalence classes under ∼ L2 consist of all strings in L2, and allstrings not in L2.( One could also apply Lemma 2 . 1 . 5 ( b ) to L1 to show that L2 i sregular , orvice versa . ) The language L3 of strings with exactly one radix point is alsoregular : the equivalence classes under ∼ L3 consist of all strings with zeropoints , all strings with one point , and all strings with more than one point. Hence L1 ∩ L2 ∩ L3 i s regular by Lemma 2 . 1 . 5 ( a ) , as desired . �

The “ real world ” convention for base b expansions is a bit morecompli - cated than what we are using : normally , one omits the radixpoint when there are no digits after it , one adds a leading zero in frontof the radix point when there are no digits before it , and one represents 0with a single zero rather than a bare radix point ( or the empty string ) .This will not change anything essential , thanks to the following lemma .Lemma 2 . 3 . 3 . Let S be a s et of nonnegative b− adicrationals . Then the s et of expansions of S, under our convention ,is a regular language if and only if the set of “ real world ” expansions ofS is a regular language .

F i n i t e automata and a l g e b r a i c ex t en s i on s o f f unc t i on f i e l d s \quad 387Proof . \quad We may as we l l assume f o r s i m p l i c i t y that $ 0 element−s l a s h S $

s i n c e any s i n g l e t o n

\noindent language i s r e g u l a r . Put

\ [ S { 1 } = S \cap ( 0 , 1 ) , S { 2 } = S \cap Z ,S { 3 } = S \setminus ( S { 1 } \cup S { 2 } ) . \ ]

\noindent Then the expans ions o f $ S , $ under e i t h e r convent ion , form a r e g u l a r languagei f and only i f the same i s t rue o f $ S { 1 } , S { 2 } , S { 3 } . $ Namely , under our convent ion ,the language o f s t r i n g s with no d i g i t s be f o r e the rad ix po int and the lan −guage o f s t r i n g s with no d i g i t s a f t e r the rad ix po int are r e g u l a r . Under the‘ ‘ r e a l world ’ ’ \quad convent ion , the language o f s t r i n g s with no rad ix po int andthe language o f s t r i n g s with a s i n g l e 0 be f o r e the rad ix po int are r e g u l a r .

The \quad expans ions \quad o f $ S { 3 }$ \quad are \quad the \quad same \quad in \quad both \quad ca s e s , \quad so \quad we \quad can \quad i gno r ethem . For $ S { 1 } , $ note that the language o f i t s r e a l world expans ions i s reg −u la r i f and only i f the language o f the r e v e r s e s o f those s t r i n g s i s r e g u l a r( Lemma 2 . 1 . 5 ( b ) ) , i f and only i f the language o f those r e v e r s e s with a rad ix

\noindent po int added in f r o n t i s r e g u l a r ( c l e a r ) , i f and only i f the language o f the

\noindent r e v e r s e s o f those ( which are the expans ions under our convent ion ) i s reg −u la r . For $ S { 2 } , $ note that the language o f r e a l world expans ions i s r e g u l a r i fand only i f the language o f those s t r i n g s with the i n i t i a l z e r o e s removed i sr e g u l a r $ . \ square $

\noindent D e f i n i t i o n \quad 2 . 3 . 4 . \quad Let \quad $ M $ \quad be a DFAO \quad with input \quad alphabet \quad$ \Sigma { b } . $ \quad We say

a s t a t e $ q \ in Q $ i s pre rad ix \quad ( re sp . \quad pos t rad ix ) \quad i f the re e x i s t s a v a l i d base$ b $

expansion $ s = s { 1 } \cdot \cdot \cdot s { n }$ with $ s { k }$equal to the rad ix po int such that , i f we

s e t \quad $ q i = \delta ( q i − 1 , s { i } ) , $ \quad then$ q = q i $ \quad f o r some $ i < k ( $ resp . \quad f o r some $ i \geqk ) . $

That i s , \quad when t r a c i n g through the t r a n s i t i o n s produced by $ s , q $ appearsbe f o r e ( re sp . a f t e r ) the t r a n s i t i o n producing the rad ix po int . Note that i fthe language accepted by $ M $ c o n s i s t s only o f v a l i d base $ b $ expans ions , thenno s t a t e can be both pre rad ix and pos t rad ix , or e l s e $ M $ would accept somes t r i n g conta in ing more than one rad ix po int .

\noindent D e f i n i t i o n \quad 2 . 3 . 5 . \quad Let \quad $ \Delta $ be a f i n i t e \quad s e t . \quad A func t i on \quad$ f : S { b } \rightarrow \Delta $ i s \quad $ b − $

automatic i f the r e i s a DFAO $ M $ with input a lphabet $ \Sigma $ and output alphabet$ \Delta $ \quad such \quad that \quad f o r \quad any \quad $ v \ in S { b }

, f ( v ) = f M ( s ( v ) ) . $ \quad By \quad Lemma \quad 2 . 3 . 2 , \quad i t \quad i s

\noindent equ iva l en t to r e q u i r e that f o r some symbol $ \ star element−s l a s h \Delta, $ the re i s a DFAO $ M $ with

\noindent input a lphabet $ \Sigma $ and output a lphabet $ \Delta \cup \{ \ star\} $ such that

\ [ f M ( s ) = \ l e f t \{\ begin { a l i gned } & f ( v ( s ) ) si s a v a l i d base b expansion \\

& \ star otherwi s e . \end{ a l i gned }\ right . \ ]

\noindent We say a subset $ S $ o f $ S { b }$ i s $ b − $ r e g u l a r i f i t s c h a r a c t e r i s t i c f unc t i on

\ [ \chi S ( s ) = \ l e f t \{\ begin { a l i gned } & 1 s \ in S \\& 0 s element−s l a s h S \end{ a l i gned }\ right . \ ]

\noindent i s $ b − $ automatic ; then a func t i on $ f : S { b } \rightarrow\Delta $ i s $ b − $ automatic i f and only i f

$ f ˆ{ − 1 } ( d ) $ i s $ b − $ r e g u l a r f o r each $ d \ in \Delta. $

Finite automata and algebraic extensions of function fields .. 387Proof period .. We may as well assume for simplicity that 0 element-slash S since any singletonlanguage is regular period PutS sub 1 = S cap open parenthesis 0 comma 1 closing parenthesis comma S sub 2 = S cap Z comma S sub 3 = S

backslash open parenthesis S sub 1 cup S sub 2 closing parenthesis periodThen the expansions of S comma under either convention comma form a regular languageif and only if the same i s true of S sub 1 comma S sub 2 comma S sub 3 period Namely comma under our

convention commathe language of strings with no digits before the radix point and the lan hyphenguage of strings with no digits after the radix point are regular period Under thequotedblleft real world quotedblright .. convention comma the language of strings with no radix point andthe language of strings with a single 0 before the radix point are regular periodThe .. expansions .. of S sub 3 .. are .. the .. same .. in .. both .. cases comma .. so .. we .. can .. ignorethem period For S sub 1 comma note that the language of its real world expansions is reg hyphenular if and only if the language of the reverses of those strings is regularopen parenthesis Lemma 2 period 1 period 5 open parenthesis b closing parenthesis closing parenthesis comma

if and only if the language of those reverses with a radixpoint added in front i s regular open parenthesis clear closing parenthesis comma if and only if the language of

thereverses of those open parenthesis which are the expansions under our convention closing parenthesis i s reg

hyphenular period For S sub 2 comma note that the language of real world expansions i s regular ifand only if the language of those strings with the initial zeroes removed i sregular period squareDefinition .. 2 period 3 period 4 period .. Let .. M .. be a DFAO .. with input .. alphabet .. Capital Sigma sub

b period .. We saya state q in Q i s preradix .. open parenthesis resp period .. postradix closing parenthesis .. if there exists a valid

base bexpansion s = s sub 1 times times times s sub n with s sub k equal to the radix point such that comma if weset .. q i = delta open parenthesis q i minus 1 comma s sub i closing parenthesis comma .. then q = q i .. for

some i less k open parenthesis resp period .. for some i greater equal k closing parenthesis periodThat is comma .. when tracing through the transitions produced by s comma q appearsbefore open parenthesis resp period after closing parenthesis the transition producing the radix point period Note

that ifthe language accepted by M consists only of valid base b expansions comma thenno state can be both preradix and postradix comma or else M would accept somestring containing more than one radix point periodDefinition .. 2 period 3 period 5 period .. Let .. Capital Delta be a finite .. set period .. A function .. f : S sub

b right arrow Capital Delta i s .. b hyphenautomatic if there i s a DFAO M with input alphabet Capital Sigma and output alphabetCapital Delta .. such .. that .. for .. any .. v in S sub b comma f open parenthesis v closing parenthesis = f M

open parenthesis s open parenthesis v closing parenthesis closing parenthesis period .. By .. Lemma .. 2 period 3period 2 comma .. it .. i s

equivalent to require that for some symbol big star element-slash Capital Delta comma there i s a DFAO M withinput alphabet Capital Sigma and output alphabet Capital Delta cup open brace big star closing brace such thatf M open parenthesis s closing parenthesis = Case 1 f open parenthesis v open parenthesis s closing parenthesis

closing parenthesis s is a valid base b expansion Case 2 big star otherwise periodWe say a subset S of S sub b i s b hyphen regular if it s characteristic functionchi S open parenthesis s closing parenthesis = Case 1 1 s in S Case 2 0 s element-slash Si s b hyphen automatic semicolon then a function f : S sub b right arrow Capital Delta i s b hyphen automatic

if and only iff to the power of minus 1 open parenthesis d closing parenthesis is b hyphen regular for each d in Capital Delta

period

Finite automata and algebraic extensions of function fields 387 Proof . We mayas well assume for simplicity that 0element− slashS since any singletonlanguage is regular . Put

S1 = S ∩ (0, 1), S2 = S ∩ Z, S3 = S \ (S1 ∪ S2).

Then the expansions of S, under either convention , form a regular languageif and only if the same i s true of S1, S2, S3. Namely , under our convention, the language of strings with no digits before the radix point and the lan -guage of strings with no digits after the radix point are regular . Under the“ real world ” convention , the language of strings with no radix point andthe language of strings with a single 0 before the radix point are regular .

The expansions of S3 are the same in both cases ,so we can ignore them . For S1, note that the language of its realworld expansions is reg - ular if and only if the language of the reverses ofthose strings is regular ( Lemma 2 . 1 . 5 ( b ) ) , if and only if the languageof those reverses with a radixpoint added in front i s regular ( clear ) , if and only if the language of thereverses of those ( which are the expansions under our convention ) i s reg -ular . For S2, note that the language of real world expansions i s regular ifand only if the language of those strings with the initial zeroes removed i sregular . �Definition 2 . 3 . 4 . Let M be a DFAO with inputalphabet Σb. We say a state q ∈ Q i s preradix ( resp . postradix) if there exists a valid base b expansion s = s1 · · · sn with sk equal tothe radix point such that , if we set qi = δ(qi− 1, si), then q = qifor some i < k ( resp . for some i ≥ k). That is , when tracingthrough the transitions produced by s, q appears before ( resp . after ) thetransition producing the radix point . Note that if the language acceptedby M consists only of valid base b expansions , then no state can be bothpreradix and postradix , or else M would accept some string containingmore than one radix point .Definition 2 . 3 . 5 . Let ∆ be a finite set . Afunction f : Sb → ∆ i s b− automatic if there i s a DFAO Mwith input alphabet Σ and output alphabet ∆ such that for anyv ∈ Sb, f(v) = fM(s(v)). By Lemma 2 . 3 . 2 , it i sequivalent to require that for some symbol ?element − slash∆, there i s aDFAO M withinput alphabet Σ and output alphabet ∆ ∪ {?} such that

fM(s) =

{f(v(s)) sisavalidbasebexpansion

? otherwise.

We say a subset S of Sb i s b− regular if it s characteristic function

χS(s) =

{1 s ∈ S0 selement− slashS

i s b− automatic ; then a function f : Sb → ∆ i s b− automatic if andonly if f−1(d) is b− regular for each d ∈ ∆.

\noindent 3 88 \quad Kiran S . KedlayaLemma 2 . 3 . 6 . Let $ S \subseteq S { b }$ be a subset . \quad Then f o r any

$ r \ in N $ and any $ s \ in S { b } , $

\noindent $ S $ i s $ b − $ r e g u l a r i f and only i f

\ [ rS + s = \{ rx + s : x \ in S \} \ ]

\noindent i s $ b − $ r e g u l a r .

\noindent Proof . \quad As in [ 2 , Lemmas 4 . 3 . 9 and 4 . 3 . 1 1 ] , one can cons t ruc t a f i n i t e − s t a t et ransducer that performs the operat i on $ x \mapsto rx + s $ on v a l i d base

$ b $ ex −pans ions read from r i g h t \quad to l e f t , \quad by \quad s imply t r a n s c r i b i n g the usua l handc a l c u l a t i o n . ( Remember that r e v e r s i n g the s t r i n g s o f a language p r e s e r v e sr e g u l a r i t y by Lemma 2 . 1 . 5 , so the re i s no harm in read ing base $ b $ expans ionsbackwards . ) Lemma 2 . 1 . 1 0 then y i e l d s the d e s i r e d r e s u l t $ . \ square $

We conclude t h i s s e c t i o n by not ing the r e l a t i o n s h i p with the not ion o f‘ ‘ automatic sequences ’ ’ \quad from \quad [ 2 , \quad Chapter 5 ] . In \quad [ 2 , D e f i n i t i o n 5 . 1 . 1 ] , a se −quence $ \{ a { l } \} ˆ{ \ infty } { l = 0 }$ over $ \Delta $ i s s a id to be

$ b − $ automatic i f the r e i s a DFAO $ M $ with

\noindent input a lphabet $ \{ 0 , . . . , b − 1 \} $ and output alphabet$ \Delta $ such that f o r any s t r i n g

\noindent $ s = s { 1 } \cdot \cdot \cdot s { n } , $ i f we put $ v( s ) = \sum ˆ{ n } { i = 1 } s { i } b ˆ{ n − 1 − i } , $ then$ a { v ( s ) } = f M ( s ) . $ Note that

\noindent t h i s means $ M $ must eva luate c o r r e c t l y even on s t r i n g s with l ead ing z e r o e s ,but by [ 2 , Theorem 5 . 2 . 1 ] , i t i s equ iva l en t to r e q u i r e that the re e x i s t s suchan $ M $ only having the property that $ a { v ( s ) } = f M ( s

) $ when $ s { 1 } \ne 0 . $ I t f o l l o w s

\noindent r e a d i l y that $ \{ a { l } \} ˆ{ \ infty } { l = 0 }$ i s $ b − $automatic i f and only i f f o r some symbol $ \ star element−s l a s h \Delta , $

\noindent the func t i on $ f : S { b } \rightarrow \Delta \cup \{ \ star\} $ de f ined by

\ [ f ( x ) = \ l e f t \{\ begin { a l i gned } & a { x } x \ in Z \\& \ star otherwi s e \end{ a l i gned }\ right . \ ]

\noindent i s $ b − $ automatic .

\centerline {3 . \quad Algebra i c p r e l i m i n a r i e s }

In t h i s chapter , we r e c a l l the a l g e b r a i c machinery that w i l l go in to thefo rmulat ion o f Theorem 4 . 1 . 3 .

\noindent 3 . 1 . \quad Genera l i zed \quad power s e r i e s . \quad Let $ R $ be an a r b i t r a r y r ing . Then ther ing o f ord inary power \quad s e r i e s \quad over \quad $ R $ can \quad be \quad i d e n t i f i e d \quad with the r ing o ff u n c t i o n s from $ Z { \geq 0 }$ to $ R , $ with add i t i on g iven termwise and m u l t i p l i c a t i o ng iven by convo lut ion

\ [\ begin { a l i gned } ( f g ) ( k ) = \sum f ( i ) g ( j ); \\

i + j = k \end{ a l i gned }\ ]

\noindent the l a t t e r makes sense because f o r any f i x e d $ k \ in Z { \geq 0 }, $ the re are only f i n i t e l y

\noindent many p a i r s \ h f i l l $ ( i , j ) \ in Z ˆ{ 2 } { \geq 0 }$ \ h f i l l such that$ i + j = k . $ \ h f i l l In order to g e n e r a l i z e t h i s

\noindent c on s t r u c t i on \quad to index s e t s \quad other than $ Z { \geq 0 } , $\quad we \quad w i l l have \quad to r e s t r i c t \quad the

nonzero va lue s o f the f u n c t i o n s so that computing $ fg $ i n v o l v e s adding onlyf i n i t e l y many nonzero e lements o f $ R . $ The r e c i p e f o r doing t h i s dates back

3 88 .. Kiran S period KedlayaLemma 2 period 3 period 6 period Let S subset equal S sub b be a subset period .. Then for any r in N and any

s in S sub b commaS is b hyphen regular if and only ifrS plus s = open brace rx plus s : x in S closing braceis b hyphen regular periodProof period .. As in open square bracket 2 comma Lemmas 4 period 3 period 9 and 4 period 3 period 1 1 closing

square bracket comma one can construct a finite hyphen statetransducer that performs the operation x arrowright-mapsto rx plus s on valid base b ex hyphenpansions read from right .. to left comma .. by .. simply transcribing the usual handcalculation period open parenthesis Remember that reversing the strings of a language preservesregularity by Lemma 2 period 1 period 5 comma so there i s no harm in reading base b expansionsbackwards period closing parenthesis Lemma 2 period 1 period 1 0 then yields the desired result period squareWe conclude this section by noting the relationship with the notion ofquotedblleft automatic sequences quotedblright .. from .. open square bracket 2 comma .. Chapter 5 closing

square bracket period In .. open square bracket 2 comma Definition 5 period 1 period 1 closing square bracketcomma a se hyphen

quence open brace a sub l closing brace sub l = 0 to the power of infinity over Capital Delta i s said to be bhyphen automatic if there i s a DFAO M with

input alphabet open brace 0 comma period period period comma b minus 1 closing brace and output alphabetCapital Delta such that for any string

s = s sub 1 times times times s sub n comma if we put v open parenthesis s closing parenthesis = sum sub i= 1 to the power of n s sub i b to the power of n minus 1 minus i comma then a sub v open parenthesis s closingparenthesis = f M open parenthesis s closing parenthesis period Note that

this means M must evaluate correctly even on strings with leading zeroes commabut by open square bracket 2 comma Theorem 5 period 2 period 1 closing square bracket comma it i s equivalent

to require that there exists suchan M only having the property that a sub v open parenthesis s closing parenthesis = f M open parenthesis s

closing parenthesis when s sub 1 equal-negationslash 0 period It followsreadily that open brace a sub l closing brace sub l = 0 to the power of infinity i s b hyphen automatic if and only

if for some symbol big star element-slash Capital Delta commathe function f : S sub b right arrow Capital Delta cup open brace big star closing brace defined byf open parenthesis x closing parenthesis = Case 1 a sub x x in Z Case 2 big star otherwisei s b hyphen automatic period3 period .. Algebraic preliminariesIn this chapter comma we recall the algebraic machinery that will go into theformulation of Theorem 4 period 1 period 3 period3 period 1 period .. Generalized .. power series period .. Let R be an arbitrary ring period Then thering of ordinary power .. series .. over .. R can .. be .. identified .. with the ring offunctions from Z sub greater equal 0 to R comma with addition given termwise and multiplicationgiven by convolutionLine 1 open parenthesis fg closing parenthesis open parenthesis k closing parenthesis = sum f open parenthesis i

closing parenthesis g open parenthesis j closing parenthesis semicolon Line 2 i plus j = kthe latter makes sense because for any fixed k in Z sub greater equal 0 comma there are only finitelymany pairs .... open parenthesis i comma j closing parenthesis in Z sub greater equal 0 to the power of 2 .... such

that i plus j = k period .... In order to generalize thisconstruction .. to index set s .. other than Z sub greater equal 0 comma .. we .. will have .. to restrict .. thenonzero values of the functions so that computing fg involves adding onlyfinitely many nonzero elements of R period The recipe for doing this dates back

3 88 Kiran S . Kedlaya Lemma 2 . 3 . 6 . Let S ⊆ Sb be a subset .Then for any r ∈ N and any s ∈ Sb,S is b− regular if and only if

rS + s = {rx+ s : x ∈ S}

is b− regular .Proof . As in [ 2 , Lemmas 4 . 3 . 9 and 4 . 3 . 1 1 ] , one can construct afinite - state transducer that performs the operation x 7→ rx+ s on validbase b ex - pansions read from right to left , by simply transcribingthe usual hand calculation . ( Remember that reversing the strings of alanguage preserves regularity by Lemma 2 . 1 . 5 , so there i s no harm inreading base b expansions backwards . ) Lemma 2 . 1 . 1 0 then yields thedesired result . �

We conclude this section by noting the relationship with the notion of “automatic sequences ” from [ 2 , Chapter 5 ] . In [ 2 , Definition 5. 1 . 1 ] , a se - quence {al}∞l=0 over ∆ i s said to be b− automatic if therei s a DFAO M withinput alphabet {0, ..., b− 1} and output alphabet ∆ such that for any strings = s1 · · · sn, if we put v(s) =

∑ni=1 sib

n−1−i, then av(s) = fM(s). Note thatthis means M must evaluate correctly even on strings with leading zeroes, but by [ 2 , Theorem 5 . 2 . 1 ] , it i s equivalent to require that thereexists such an M only having the property that av(s) = fM(s) when s1 6= 0.It followsreadily that {al}∞l=0 i s b− automatic if and only if for some symbol ?element−slash∆,the function f : Sb → ∆ ∪ {?} defined by

f(x) =

{ax x ∈ Z? otherwise

i s b− automatic .3 . Algebraic preliminaries

In this chapter , we recall the algebraic machinery that will go into theformulation of Theorem 4 . 1 . 3 .3 . 1 . Generalized power series . Let R be an arbitrary ring. Then the ring of ordinary power series over R can be identifiedwith the ring of functions from Z≥0 to R, with addition given termwise andmultiplication given by convolution

(fg)(k) =∑

f(i)g(j);

i+ j = k

the latter makes sense because for any fixed k ∈ Z≥0, there are only finitelymany pairs (i, j) ∈ Z2

≥0 such that i+ j = k. In order to generalize thisconstruction to index set s other than Z≥0, we will have torestrict the nonzero values of the functions so that computing fg involvesadding only finitely many nonzero elements of R. The recipe for doing thisdates back


\noindent to Hahn \quad [ 8 ] \quad ( although the term \quad ‘ ‘ Mal ’ cev − Neumann s e r i e s ’ ’ \quad f o r an ob j e c to f the type we d e s c r i b e i s p r eva l ent ) , \quad and we r e c a l l i t now ; \quad s ee a l s o \quad [ 1 5 ,Chapter 1 3 ] .

\noindent D e f i n i t i o n 3 . 1 . 1 . \quad Let $ G $ be a t o t a l l y ordered abe l i an group ( wr i t t en ad −d i t i v e l y ) with i d e n t i t y element 0 ; that i s $ , G $ i s an abe l i an group equippedwith a binary r e l a t i o n $ > $ such that f o r a l l $ a , b , c \ in G , $

\ [\ begin { a l i gned } a negat i ons l a sh−g r e a t e r a \\a greate r−n e g a t i o n s l a s h b , b greate r−n e g a t i o n s l a s h a \Rightarrow a

= b \\a > b , b > c \Rightarrow a > c \\a > b \Leftrightarrow a + c > b + c . \end{ a l i gned }\ ]

\noindent Let $ P $ be the s e t o f $ a \ in G $ f o r which $ a > 0 ; P $i s c a l l e d the p o s i t i v e cone o f

\begin { a l i g n ∗}G .\end{ a l i g n ∗}

\noindent Lemma 3 . 1 . 2 . \ h f i l l Let $ S $ be a subset o f $ G . $ \ h f i l l Then the f o l l o w i n g two c o n d i t i o n s

\noindent are equ iva l en t .

\centerline {( a ) \quad Every nonempty subset o f $ S $ has a minimal e lement . }

\centerline {( b ) \quad There i s no i n f i n i t e dec r ea s ing sequence $ s { 1 } > s { 2 }> \cdot \cdot \cdot $ \quad with in $ S . $ }

\noindent Proof . \quad I f ( a ) ho lds but ( b ) did not , then the s e t $ \{ s { 1 }, s { 2 } , . . . \} $ would not havea minimal element , \quad a c o n t r a d i c t i o n . Hence ( a ) \quad i m p l i e s \quad ( b ) . Converse ly , i f

$ T $ were a subset o f $ S $ with no s m a l l e s t element , then f o r any $ s { i }\ in T , $ we

could choose $ s { i + 1 } \ in T $ with $ s { i } > s { i + 1 }, $ thus forming an i n f i n i t e dec r ea s ingsequence . Hence ( b ) i m p l i e s ( a $ ) . \ square $

\noindent D e f i n i t i o n 3 . 1 . 3 . A subset $ S $ o f $ G $ i s we l l − ordered i f i t s a t i s f i e s e i t h e r o f the

\noindent equ iva l en t c o n d i t i o n s o f Lemma 3 . 1 . 2 . ( Those who p r e f e r to avoid assumingthe axiom of cho i c e should take ( a ) to be the d e f i n i t i o n , as the i m p l i c a t i o n( b $ ) \Longrightarrow ( $ a ) r e q u i r e s cho i c e . )

For $ S { 1 } , . . . , S { n } \subseteq G , $ wr i t e $ S { 1 }+ \cdot \cdot \cdot + S { n }$ f o r the s e t o f e lements o f $ G $ o f

the form $ s { 1 } + \cdot \cdot \cdot + s { n }$ f o r $ s { i } \ inS { i } ; $ in case $ S { 1 } = \cdot \cdot \cdot = S { n } , $ we abbrev ia te

\noindent t h i s notat ion to $ S ˆ{ + n } . ( $ In [ 1 5 ] the notat ion $ nS $i s used in s t ead , but we havea l r eady de f ined t h i s as the d i l a t i o n o f $ S $ by the f a c t o r $ n . ) $ \quad Then one cane a s i l y v e r i f y the f o l l o w i n g ( or s ee [ 1 5 , Lemmas 1 3 . 2 . 9 and 13 . 2 . 1 0 ] ) .

\noindent Lemma 3 . 1 . 4 . \quad ( i ) \quad I f $ S { 1 } , . . . , S { n }$\quad are \quad we l l − ordered subse t s \quad o f $ G , $ \quad then

$ S { 1 } + \cdot \cdot \cdot + S { n }$ i s we l l − ordered .

( i i ) \quad I f $ S { 1 } , \cdot \cdot \cdot , S { n }$ are we l l − ordered subse t s o f$ G , $ then f o r any $ x \ in G , $ the

number o f $ n − $ t u p l e s $ ( s { 1 } , . . . , s { n } ) \ inS { 1 } \times \cdot \cdot \cdot \times S { n }$ such that $ s { 1 }+ \cdot \cdot \cdot + $

\ [ s { n } = x i s f i n i t e . \ ]

\hspace ∗{\ f i l l }( i i i ) \quad I f $ S $ i s a we l l − ordered subset o f $ P , $ \quad then$ \ t i lde {S} = \cup ˆ{ \ infty } { n = 1 } S ˆ{ + n }$ \quad a l s o i s we l l −

\ [ ordered ; moreover , \cap ˆ{ \ infty } { n = 1 }ˆ{ \ t i lde {S} ˆ{ + n }}= \ varnothing . \ ]

Finite automata and algebraic extensions of function fields .. 389to Hahn .. open square bracket 8 closing square bracket .. open parenthesis although the term .. quotedblleft

Mal quoteright cev hyphen Neumann series quotedblright .. for an objectof the type we describe i s prevalent closing parenthesis comma .. and we recall it now semicolon .. see also ..

open square bracket 1 5 commaChapter 1 3 closing square bracket periodDefinition 3 period 1 period 1 period .. Let G be a totally ordered abelian group open parenthesis written ad

hyphenditively closing parenthesis with identity element 0 semicolon that is comma G i s an abelian group equippedwith a binary relation greater such that for all a comma b comma c in G commaLine 1 a negationslash-greater a Line 2 a greater-negationslash b comma b greater-negationslash a double stroke

right arrow a = b Line 3 a greater b comma b greater c double stroke right arrow a greater c Line 4 a greater bLeftrightarrow a plus c greater b plus c period

Let P be the set of a in G for which a greater 0 semicolon P i s called the positive cone ofG periodLemma 3 period 1 period 2 period .... Let S be a subset of G period .... Then the following two conditionsare equivalent periodopen parenthesis a closing parenthesis .. Every nonempty subset of S has a minimal e lement periodopen parenthesis b closing parenthesis .. There is no infinite decreasing sequence s sub 1 greater s sub 2 greater

times times times .. within S periodProof period .. If open parenthesis a closing parenthesis holds but open parenthesis b closing parenthesis did not

comma then the set open brace s sub 1 comma s sub 2 comma period period period closing brace would not havea minimal element comma .. a contradiction period Hence open parenthesis a closing parenthesis .. implies ..

open parenthesis b closing parenthesis period Conversely comma ifT were a subset of S with no smallest element comma then for any s sub i in T comma wecould choose s sub i plus 1 in T with s sub i greater s sub i plus 1 comma thus forming an infinite decreasingsequence period Hence open parenthesis b closing parenthesis implies open parenthesis a closing parenthesis

period squareDefinition 3 period 1 period 3 period A subset S of G i s well hyphen ordered if it satisfies either of theequivalent conditions of Lemma 3 period 1 period 2 period open parenthesis Those who prefer to avoid assumingthe axiom of choice should take open parenthesis a closing parenthesis to be the definition comma as the impli-

cationopen parenthesis b closing parenthesis equal-arrowdblright open parenthesis a closing parenthesis requires choice

period closing parenthesisFor S sub 1 comma period period period comma S sub n subset equal G comma write S sub 1 plus times times

times plus S sub n for the set of elements of G ofthe form s sub 1 plus times times times plus s sub n for s sub i in S sub i semicolon in case S sub 1 = times times

times = S sub n comma we abbreviatethis notation to S to the power of plus n period open parenthesis In open square bracket 1 5 closing square

bracket the notation nS is used instead comma but we havealready defined this as the dilation of S by the factor n period closing parenthesis .. Then one caneasily verify the following open parenthesis or see open square bracket 1 5 comma Lemmas 1 3 period 2 period

9 and 13 period 2 period 1 0 closing square bracket closing parenthesis periodLemma 3 period 1 period 4 period .. open parenthesis i closing parenthesis .. If S sub 1 comma period period

period comma S sub n .. are .. well hyphen ordered subsets .. of G comma .. thenS sub 1 plus times times times plus S sub n is well hyphen ordered periodopen parenthesis ii closing parenthesis .. If S sub 1 comma times times times comma S sub n are well hyphen

ordered subsets of G comma then for any x in G comma thenumber of n hyphen tuples open parenthesis s sub 1 comma period period period comma s sub n closing parenthesis

in S sub 1 times times times times times S sub n such that s sub 1 plus times times times pluss sub n = x is finite periodopen parenthesis iii closing parenthesis .. If S is a well hyphen ordered subset of P comma .. then S-tilde = cup

sub n = 1 to the power of infinity S to the power of plus n .. also is well hyphenordered semicolon moreover comma cap sub n = 1 to the power of infinity to the power of S-tilde to the power

of plus n = varnothing period


to Hahn [ 8 ] ( although the term “ Mal ’ cev - Neumann series ”for an object of the type we describe i s prevalent ) , and we recall it now; see also [ 1 5 , Chapter 1 3 ] .Definition 3 . 1 . 1 . Let G be a totally ordered abelian group (written ad - ditively ) with identity element 0 ; that is , G i s an abeliangroup equipped with a binary relation > such that for all a, b, c ∈ G,

anegationslash− greateraagreater − negationslashb, bgreater − negationslasha⇒ a = b

a > b, b > c⇒ a > c

a > b⇔ a+ c > b+ c.

Let P be the set of a ∈ G for which a > 0;P i s called the positive cone of

G.

Lemma 3 . 1 . 2 . Let S be a subset of G. Then the following twoconditionsare equivalent .

( a ) Every nonempty subset of S has a minimal e lement .( b ) There is no infinite decreasing sequence s1 > s2 > · · · within S.Proof . If ( a ) holds but ( b ) did not , then the set {s1, s2, ...} wouldnot have a minimal element , a contradiction . Hence ( a ) implies (b ) . Conversely , if T were a subset of S with no smallest element , thenfor any si ∈ T, we could choose si+1 ∈ T with si > si+1, thus forming aninfinite decreasing sequence . Hence ( b ) implies ( a ). �Definition 3 . 1 . 3 . A subset S of G i s well - ordered if it satisfieseither of theequivalent conditions of Lemma 3 . 1 . 2 . ( Those who prefer to avoidassuming the axiom of choice should take ( a ) to be the definition , as theimplication ( b ) =⇒ ( a ) requires choice . )

For S1, ..., Sn ⊆ G, write S1 + · · · + Sn for the set of elements of G of theform s1 + · · ·+ sn for si ∈ Si; in case S1 = · · · = Sn, we abbreviatethis notation to S+n.( In [ 1 5 ] the notation nS is used instead , but wehave already defined this as the dilation of S by the factor n.) Then onecan easily verify the following ( or see [ 1 5 , Lemmas 1 3 . 2 . 9 and 13 . 2. 1 0 ] ) .Lemma 3 . 1 . 4 . ( i ) If S1, ..., Sn are well - ordered subsetsof G, then S1 + · · ·+ Sn is well - ordered .

( ii ) If S1, · · ·, Sn are well - ordered subsets of G, then for any x ∈ G,the number of n− tuples (s1, ..., sn) ∈ S1 × · · · × Sn such that s1 + · · ·+

sn = xisfinite.

( iii ) If S is a well - ordered subset of P, then S = ∪∞n=1S+n

also is well -

ordered;moreover,∩∞n=1S+n

= ∅.

\noindent 390 \quad Kiran S . Kedlaya

\noindent D e f i n i t i o n 3 . 1 . 5 . \quad Given a func t i on $ f : G \rightarrowR , $ the support o f $ f $ i s the

s e t o f $ g \ in G $ such that $ f ( g ) \ne 0 . $ A g e n e r a l i z e d Laurent s e r i e s over$ R $ with

exponents in $ G $ i s a func t i on $ f : G \rightarrow R $ whose support i s we l l − ordered ; i fthe support i s conta ined in $ P \cup \{ 0 \} , $ we c a l l $ f $ a g e n e r a l i z e d power s e r i e s .We t y p i c a l l y r ep r e s e n t the g e n e r a l i z e d Laurent s e r i e s $ f $ in s e r i e s notat ion

\noindent $ \sum { i } f ( i ) t ˆ{ i } , $ \quad and wr i t e $ R \ l l b r a c k e tt ˆ{ G } \ r rb ra cke t $ \quad and $ R ( ( t ˆ{ G } ) ) $ \quad f o r the s e t s o f g e n e r a l i z e d power

s e r i e s and g e n e r a l i z e d Laurent s e r i e s , r e s p e c t i v e l y , over $ R $ with exponentsin $ G . $

\hspace ∗{\ f i l l }Thanks to Lemma 3 . 1 . 4 , the termwise sum and convo lut ion product are

\noindent we l l − de f ined \ h f i l l binary ope ra t i on s \ h f i l l on \ h f i l l $ R \ l l b r a c k e tt ˆ{ G } \ r rb ra cke t $ \ h f i l l and \ h f i l l $ R ( ( t ˆ{ G } ) ) , $ \ h f i l l which form r i n g s

\noindent under the ope ra t i on s . A nonzero element o f $ R ( ( t ˆ{ G } ) ) $i s a un i t i f and only i f

i t s f i r s t nonzero c o e f f i c i e n t i s a un i t [ 1 5 , Theorem 1 3 . 2 . 1 1 ] ; in p a r t i c u l a r , i f

\noindent $ R $ i s a f i e l d , then so i s $ R ( ( t ˆ{ G } ) ) . $

\noindent 3 . 2 . \quad Algebra i c \quad e lements \quad o f f i e l d s . \quad In t h i s s e c t i o n , we r e c a l l the d e f i −n i t i o n o f a l g e b r a i c i t y o f an element o f one f i e l d over a s u b f i e l d , and thenreview some c r i t e r i a f o r a l g e b r a i c i t y . \quad Nothing in t h i s \quad s e c t i o n i s \quad even re −motely o r i g i n a l , as can be conf irmed by any s u f f i c i e n t l y d e t a i l e d abs t r a c ta lgebra textbook .

\noindent D e f i n i t i o n 3 . 2 . 1 . Let $ K \subseteq L $ be f i e l d s . Then $ \alpha\ in L $ i s sa id to be a l g e b r a i c

over $ K $ i f the re e x i s t s a nonzero polynomial $ P ( x ) \ in K [ x] $ over $ K $ such that

$ P ( \alpha ) = 0 . $ We say $ L $ i s a l g e b r a i c over $ K $ i f every element o f$ L $ i s a l g e b r a i c

over $ K . $

\noindent Lemma 3 . 2 . 2 . \ h f i l l Let $ K \subseteq L $ be f i e l d s . \ h f i l l Then$ \alpha \ in L $ i s a l g e b r a i c i f and only i f

\noindent $ \alpha $ i s conta ined in a subr ing o f $ L $ conta in ing $ K $ which i s f i n i t e d imens iona l

\noindent as a $ K − $ vec to r space .

\noindent Proof . \quad We may as we l l assume $ \alpha \ne 0 , $ \quad as otherwi se both a s s e r t i o n s \quad arec l e a r . I f $ \alpha $ i s conta ined in a subr ing $ R $ o f $ L $ which has f i n i t e dimension

$ m $as a $ K − $ vec to r space , then $ 1 , \alpha , . . . , \alpha ˆ{ m }$

must be l i n e a r l y dependent over $ K , $y i e l d i n g a polynomial over $ K $ with $ \alpha $ as a root . Converse ly , i f $ P

( \alpha ) = 0 $ f o rsome polynomial $ P ( x ) \ in K [ x ] , $ we may take $ P (

x ) = c { 0 } + c { 1 } x + \cdot \cdot \cdot + c { n } x ˆ{ n }$with $ c { 0 } , c { n } \not= 0 . $ In that case ,

\ [ \{ a { 0 } + a { 1 } \alpha + \cdot \cdot \cdot + a { n −1 } \alpha ˆ{ n − 1 } : a { 0 } , . . . , a { n − 1 } \ inK \} \ ]

\noindent i s a subr ing o f $ L $ conta in ing $ K $ and $ \alpha , $ o f dimension at most$ n $ as a vec to r

space over $ K . \ square $

\noindent Coro l l a ry 3 . 2 . 3 . \quad Let $ K \subseteq L $ be f i e l d s . I f $ \dim { K }L < \ infty , $ then $ L $ i s a l g e b r a i c

over $ K . $

\noindent Lemma 3 . 2 . 4 . \quad Let $ K \subseteq L \subseteq M $ \quad be f i e l d s , \quad with$ L $ \quad a l g e b r a i c over $ K . $ \quad For

any $ \alpha \ in M , \alpha $ i s a l g e b r a i c over $ K $ i f and only i f i t i s a l g e b r a i c over$ L . $

390 .. Kiran S period KedlayaDefinition 3 period 1 period 5 period .. Given a function f : G right arrow R comma the support of f i s theset of g in G such that f open parenthesis g closing parenthesis equal-negationslash 0 period A generalized Laurent

s eries over R withexponents in G i s a function f : G right arrow R whose support is well hyphen ordered semicolon ifthe support is contained in P cup open brace 0 closing brace comma we call f a generalized power series periodWe typically represent the generalized Laurent series f in series notationsum sub i f open parenthesis i closing parenthesis t to the power of i comma .. and write R llbracket t to the

power of G rrbracket .. and R open parenthesis open parenthesis t to the power of G closing parenthesis closingparenthesis .. for the set s of generalized power

series and generalized Laurent series comma respectively comma over R with exponentsin G periodThanks to Lemma 3 period 1 period 4 comma the termwise sum and convolution product arewell hyphen defined .... binary operations .... on .... R llbracket t to the power of G rrbracket .... and .... R open

parenthesis open parenthesis t to the power of G closing parenthesis closing parenthesis comma .... which form ringsunder the operations period A nonzero element of R open parenthesis open parenthesis t to the power of G closing

parenthesis closing parenthesis i s a unit if and only ifit s first nonzero coefficient i s a unit open square bracket 1 5 comma Theorem 1 3 period 2 period 1 1 closing

square bracket semicolon in particular comma ifR i s a field comma then so i s R open parenthesis open parenthesis t to the power of G closing parenthesis closing

parenthesis period3 period 2 period .. Algebraic .. elements .. of fields period .. In this section comma we recall the defi hyphennition of algebraicity of an element of one field over a subfield comma and thenreview some criteria for algebraicity period .. Nothing in this .. section is .. even re hyphenmotely original comma as can be confirmed by any sufficiently detailed abstractalgebra textbook periodDefinition 3 period 2 period 1 period Let K subset equal L be fields period Then alpha in L is said to be algebraicover K if there exists a nonzero polynomial P open parenthesis x closing parenthesis in K open square bracket x

closing square bracket over K such thatP open parenthesis alpha closing parenthesis = 0 period We say L i s algebraic over K if every element of L is

algebraicover K periodLemma 3 period 2 period 2 period .... Let K subset equal L be fields period .... Then alpha in L is algebraic if

and only ifalpha is contained in a subring of L containing K which is finite dimensionalas a K hyphen vector space periodProof period .. We may as well assume alpha equal-negationslash 0 comma .. as otherwise both assertions .. areclear period If alpha is contained in a subring R of L which has finite dimension mas a K hyphen vector space comma then 1 comma alpha comma period period period comma alpha to the power

of m must be linearly dependent over K commayielding a polynomial over K with alpha as a root period Conversely comma if P open parenthesis alpha closing

parenthesis = 0 forsome polynomial P open parenthesis x closing parenthesis in K open square bracket x closing square bracket

comma we may take P open parenthesis x closing parenthesis = c sub 0 plus c sub 1 x plus times times times plusc sub n x to the power of n

with c sub 0 comma c sub n negationslash-equal 0 period In that case commaopen brace a sub 0 plus a sub 1 alpha plus times times times plus a sub n minus 1 alpha to the power of n minus

1 : a sub 0 comma period period period comma a sub n minus 1 in K closing bracei s a subring of L containing K and alpha comma of dimension at most n as a vectorspace over K period squareCorollary 3 period 2 period 3 period .. Let K subset equal L be fields period If dimension sub K L less infinity

comma then L is algebraicover K periodLemma 3 period 2 period 4 period .. Let K subset equal L subset equal M .. be fields comma .. with L ..

algebraic over K period .. Forany alpha in M comma alpha is algebraic over K if and only if it is algebraic over L period

390 Kiran S . Kedlaya

Definition 3 . 1 . 5 . Given a function f : G → R, the supportof f i s the set of g ∈ G such that f(g) 6= 0. A generalized Laurent s eriesover R with exponents in G i s a function f : G→ R whose support is well- ordered ; if the support is contained in P ∪ {0}, we call f a generalizedpower series . We typically represent the generalized Laurent series f inseries notation∑i f(i)ti, and write RtG and R((tG)) for the set s of generalized power

series and generalized Laurent series , respectively , over R with exponentsin G.

Thanks to Lemma 3 . 1 . 4 , the termwise sum and convolution productarewell - defined binary operations on RtG and R((tG)), which form ringsunder the operations . A nonzero element of R((tG)) i s a unit if and onlyif it s first nonzero coefficient i s a unit [ 1 5 , Theorem 1 3 . 2 . 1 1 ] ; inparticular , ifR i s a field , then so i s R((tG)).3 . 2 . Algebraic elements of fields . In this section, we recall the defi - nition of algebraicity of an element of one field overa subfield , and then review some criteria for algebraicity . Nothing inthis section is even re - motely original , as can be confirmed by anysufficiently detailed abstract algebra textbook .Definition 3 . 2 . 1 . Let K ⊆ L be fields . Then α ∈ L is said to bealgebraic over K if there exists a nonzero polynomial P (x) ∈ K[x] over Ksuch that P (α) = 0. We say L i s algebraic over K if every element of L isalgebraic over K.Lemma 3 . 2 . 2 . Let K ⊆ L be fields . Then α ∈ L is algebraic ifand only ifα is contained in a subring of L containing K which is finite dimensionalas a K− vector space .Proof . We may as well assume α 6= 0, as otherwise both assertionsare clear . If α is contained in a subring R of L which has finite dimensionm as a K− vector space , then 1, α, ..., αm must be linearly dependent overK, yielding a polynomial over K with α as a root . Conversely , if P (α) = 0for some polynomial P (x) ∈ K[x], we may take P (x) = c0 + c1x + · · · + cnx

n

with c0, cn 6= 0. In that case ,

{a0 + a1α+ · · ·+ an−1αn−1 : a0, ..., an−1 ∈ K}

i s a subring of L containing K and α, of dimension at most n as a vectorspace over K. �Corollary 3 . 2 . 3 . Let K ⊆ L be fields . If dimK L <∞, then Lis algebraic over K.Lemma 3 . 2 . 4 . Let K ⊆ L ⊆M be fields , with L algebraicover K. For any α ∈ M,α is algebraic over K if and only if it isalgebraic over L.


\noindent Proof . \quad Clea r l y i f $ \alpha $ i s the root o f a polynomial with c o e f f i c i e n t s in$ K , $ that

same polynomial has c o e f f i c i e n t s in $ L . $ Converse ly , suppose $ \alpha $ i s a l g e b r a i cover $ L ; $ i t i s then conta ined in a subr ing $ R $ o f $ M $ which i s f i n i t e d imens iona lover $ L . $ That subr ing i s generated over $ L $ by f i n i t e l y many elements , each o fwhich i s a l g e b r a i c over $ K $ and hence l i e s in a subr ing $ R { i }$ o f $ M $

which i s f i n i t ed imens iona l over $ K . $ Taking the r ing generated by the $ R { i }$ g i v e s a subr ingo f $ M $ which i s f i n i t e d imens iona l over $ K $ and which conta in s $ \alpha

. $ Hence $ \alpha $ i s

\noindent a l g e b r a i c over $ K . \ square $

\noindent Lemma 3 . 2 . 5 . \quad Let $ K \subseteq L $ be f i e l d s . I f $ \alpha, \beta \ in L $ are a l g e b r a i c over $ K , $ thens o are $ \alpha + \beta $ and $ \alpha \beta . $ I f $ \alpha \ne 0

, $ then moreover $ 1 / \alpha $ i s a l g e b r a i c over $ K . $

\noindent Proof . \ h f i l l Suppose that $ \alpha , \beta \ in L $ are a l g e b r a i c over$ K ; $ we may assume $ \alpha , \beta \not= $

\noindent 0 , e l s e everyth ing i s c l e a r . Choose po lynomia l s $ P ( x ) = c { 0 }+ c { 1 } x + \cdot \cdot \cdot + c { m } x ˆ{ m }$

\noindent and $ Q ( x ) = d { 0 } + d { 1 } x + \cdot \cdot\cdot + d { n } x ˆ{ n }$ with $ c { 0 } , c { m } , d { 0 } , d { n }\not= 0 $ such that $ P ( \alpha ) = $

\begin { a l i g n ∗}Q ( \beta ) = 0 . Then \\ R = brace l e f tmid−b r a c e l e f t b t \begin { array }{ cccc } m− 1 n − 1 \\ \sum & \sum a { i j } \alpha ˆ{ i } \beta ˆ{ j } & : a { i j } &\ in K \\ i = 0 & j = 0 \end{ array } brace r i ghtbt−bracer ightmid\end{ a l i g n ∗}

\noindent i s a subr ing o f $ L $ conta in ing $ K , $ o f dimension at most $ mn $as a vec to r space

over $ K , $ conta in ing $ \alpha + \beta $ and $ \alpha \beta . $ Hence both o f those are a l g e b r a i c over

\noindent $ K . $ Moreover $ , 1 / \alpha = − ( c { 1 } + c { 2 }\alpha + \cdot \cdot \cdot + c { m } \alpha ˆ{ m − 1 } ) / c { 0 }$i s conta ined in $ R , $ so

i t too i s a l g e b r a i c over $ K . \ square $

\noindent D e f i n i t i o n 3 . 2 . 6 . A f i e l d $ K $ i s a l g e b r a i c a l l y c l o sed i f every polynomial over

\noindent $ K $ has a root , or e q u i v a l e n t l y , i f every polynomial over $ K $ s p l i t s complete ly( f a c t o r s i n to l i n e a r po lynomia ls ) . \quad I t can be shown \quad ( us ing Zorn ’ s lemma )that every f i e l d $ K $ i s conta ined in 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 ; the e lementso f such a f i e l d which are a l g e b r a i c over $ K $ form a f i e l d $ L $ which i s botha l g e b r a i c a l l y c l o s e d and a l g e b r a i c over $ K . $ Such a f i e l d i s c a l l e d an a l g e b r a i cc l o s u r e o f $ K ; $ i t can be shown to be unique up to noncanonica l i somorphism ,but we won ’ t need t h i s .

In \quad p r a c t i c e , \quad we \quad w i l l \quad always \quad con s id e r f i e l d s \quad conta ined in \quad$ F { q } ( ( t ˆ{ Q } ) ) , $ \quad and

c on s t r u c t i ng a l g e b r a i c a l l y c l o s e d f i e l d s conta in ing them i s s t r a i g h t f o r w a r d .That i s because i f $ 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 and $ G $ i s a d i v i s i b l egroup ( i . e . , m u l t i p l i c a t i o n by any p o s i t i v e i n t e g e r i s a b $ i j $ e c t i o n on

$ G ) , $ then

\noindent the f i e l d $ K ( ( t ˆ{ G } ) ) $ i s a l g e b r a i c a l l y c l o s e d . ( The case$ G = Q , $ which i s the only

case we need , i s t r ea t ed e x p l i c i t l y in [ 1 2 , Propos i t i on 1 ] ; f o r the gene ra l caseand much more , s e e [ 10 , Theorem 5 ] . ) Moreover , i t i s easy ( and does not

\noindent r e q u i r e the axiom of cho i c e ) to cons t ruc t an a l g e b r a i c c l o s u r e $\overline {\ }{ F } { q }$o f $ F { q } : $ order

the e lements o f $ F { q }$ with 0 coming f i r s t , then l i s t the monic po lynomia l s over$ F { q }$ in l e x i c o g r a p h i c order and s u c c e s s i v e l y ad jo in roo t s o f them . Then the

\noindent f i e l d $\overline {\ }{ F } { q ˆ{ ( ( t }}ˆ{ Q } ) ) $ i s a l g e b r a i c a l l y c l o s e d and conta in s$ F { q } ( ( t ˆ{ Q } ) ) . $

Finite automata and algebraic extensions of function fields .. 39 1Proof period .. Clearly if alpha i s the root of a polynomial with coefficients in K comma thatsame polynomial has coefficients in L period Conversely comma suppose alpha i s algebraicover L semicolon it i s then contained in a subring R of M which i s finite dimensionalover L period That subring i s generated over L by finitely many elements comma each ofwhich i s algebraic over K and hence lies in a subring R sub i of M which is finitedimensional over K period Taking the ring generated by the R sub i gives a subringof M which i s finite dimensional over K and which contains alpha period Hence alpha i salgebraic over K period squareLemma 3 period 2 period 5 period .. Let K subset equal L be fields period If alpha comma beta in L are algebraic

over K comma thens o are alpha plus beta and alpha beta period If alpha equal-negationslash 0 comma then moreover 1 slash alpha

is algebraic over K periodProof period .... Suppose that alpha comma beta in L are algebraic over K semicolon we may assume alpha

comma beta negationslash-equal0 comma else everything is clear period Choose polynomials P open parenthesis x closing parenthesis = c sub 0

plus c sub 1 x plus times times times plus c sub m x to the power of mand Q open parenthesis x closing parenthesis = d sub 0 plus d sub 1 x plus times times times plus d sub n x to

the power of n with c sub 0 comma c sub m comma d sub 0 comma d sub n negationslash-equal 0 such that P openparenthesis alpha closing parenthesis =

Q open parenthesis beta closing parenthesis = 0 period Then R = Row 1 m minus 1 n minus 1 Row 2 sum suma sub ij alpha to the power of i beta to the power of j : a sub ij in K Row 3 i = 0 j = 0 .

i s a subring of L containing K comma of dimension at most mn as a vector spaceover K comma containing alpha plus beta and alpha beta period Hence both of those are algebraic overK period Moreover comma 1 slash alpha = minus open parenthesis c sub 1 plus c sub 2 alpha plus times times

times plus c sub m alpha to the power of m minus 1 closing parenthesis slash c sub 0 is contained in R comma soit too i s algebraic over K period squareDefinition 3 period 2 period 6 period A field K is algebraically c lo sed if every polynomial overK has a root comma or equivalently comma if every polynomial over K split s completelyopen parenthesis factors into linear polynomials closing parenthesis period .. It can be shown .. open parenthesis

using Zorn quoteright s lemma closing parenthesisthat every field K i s contained in an algebraically closed field semicolon the elementsof such a field which are algebraic over K form a field L which is bothalgebraically closed and algebraic over K period Such a field is called an algebraicc losure of K semicolon it can be shown to be unique up to noncanonical i somorphism commabut we won quoteright t need this periodIn .. practice comma .. we .. will .. always .. consider fields .. contained in .. F sub q open parenthesis open

parenthesis t to the power of Q closing parenthesis closing parenthesis comma .. andconstructing algebraically closed fields containing them i s straightforward periodThat i s because if K i s an algebraically closed field and G i s a divisiblegroup open parenthesis i period e period comma multiplication by any positive integer i s a b ij ection on G

closing parenthesis comma thenthe field K open parenthesis open parenthesis t to the power of G closing parenthesis closing parenthesis is

algebraically closed period open parenthesis The case G = Q comma which i s the onlycase we need comma i s treated explicitly in open square bracket 1 2 comma Proposition 1 closing square bracket

semicolon for the general caseand much more comma see open square bracket 10 comma Theorem 5 closing square bracket period closing

parenthesis Moreover comma it i s easy open parenthesis and does notrequire the axiom of choice closing parenthesis to construct an algebraic closure overbar F sub q of F sub q :

orderthe elements of F sub q with 0 coming first comma then li st the monic polynomials overF sub q in lexicographic order and successively adjoin roots of them period Then thefield overbar F sub q to the power of open parenthesis open parenthesis t to the power of Q closing parenthesis

closing parenthesis is algebraically closed and contains F sub q open parenthesis open parenthesis t to the power ofQ closing parenthesis closing parenthesis period


Proof . Clearly if α i s the root of a polynomial with coefficients inK, that same polynomial has coefficients in L. Conversely , suppose α i salgebraic over L; it i s then contained in a subring R of M which i s finitedimensional over L. That subring i s generated over L by finitely manyelements , each of which i s algebraic over K and hence lies in a subringRi of M which is finite dimensional over K. Taking the ring generated bythe Ri gives a subring of M which i s finite dimensional over K and whichcontains α. Hence α i salgebraic over K. �Lemma 3 . 2 . 5 . Let K ⊆ L be fields . If α, β ∈ L are algebraicover K, then s o are α + β and αβ. If α 6= 0, then moreover 1/α isalgebraic over K.Proof . Suppose that α, β ∈ L are algebraic over K; we may assume α, β 6=0 , else everything is clear . Choose polynomials P (x) = c0 + c1x+ · · ·+ cmx

m

and Q(x) = d0 + d1x+ · · ·+ dnxn with c0, cm, d0, dn 6= 0 such that P (α) =

Q(β) = 0.Then

R = braceleftmid− braceleftbtm− 1n− 1∑ ∑

aijαiβj : aij ∈ K

i = 0 j = 0bracerightbt− bracerightmid

i s a subring of L containing K, of dimension at most mn as a vector spaceover K, containing α+ β and αβ. Hence both of those are algebraic overK. Moreover , 1/α = −(c1 + c2α+ · · ·+ cmα

m−1)/c0 is contained in R, so it tooi s algebraic over K. �Definition 3 . 2 . 6 . A field K is algebraically c lo sed if everypolynomial overK has a root , or equivalently , if every polynomial over K split s completely( factors into linear polynomials ) . It can be shown ( using Zorn ’ slemma ) that every field K i s contained in an algebraically closed field ;the elements of such a field which are algebraic over K form a field L whichis both algebraically closed and algebraic over K. Such a field is called analgebraic c losure of K; it can be shown to be unique up to noncanonicali somorphism , but we won ’ t need this .

In practice , we will always consider fields contained inFq((tQ)), and constructing algebraically closed fields containing them i sstraightforward . That i s because if K i s an algebraically closed field andG i s a divisible group ( i . e . , multiplication by any positive integer i s ab ij ection on G), thenthe field K((tG)) is algebraically closed . ( The case G = Q, which i s theonly case we need , i s treated explicitly in [ 1 2 , Proposition 1 ] ; for thegeneral case and much more , see [ 10 , Theorem 5 ] . ) Moreover , it i seasy ( and does notrequire the axiom of choice ) to construct an algebraic closure Fq of Fq : orderthe elements of Fq with 0 coming first , then li st the monic polynomialsover Fq in lexicographic order and successively adjoin roots of them . Thenthefield FQ

q((t)) is algebraically closed and contains Fq((tQ)).


\noindent 3 . 3 . \quad Addit ive \quad polynomia l s . \quad In p o s i t i v e c h a r a c t e r i s t i c , \quad i t i s convenientto r e s t r i c t a t t e n t i o n to a s p e c i a l c l a s s o f po lynomia l s , the \quad ‘ ‘ a d d i t i v e ’ ’ \quad poly −

\noindent nomials . F i r s t , we r e c a l l a standard r e c i p e ( analogous to the co n s t r uc t i o n

\noindent o f Vandermonde determinants ) f o r producing such polynomia ls .

\noindent Lemma 3 . 3 . 1 . \quad Let $ K $ be a f i e l d o f c h a r a c t e r i s t i c $ p >0 . $ \quad Given $ r { 1 } , . . . , r { n } \ in $

$ K , $ the Moore determinant

\ [\ l e f t (\ begin { array }{ cccc } r { 1 } & r { 2 } & \cdot \cdot \cdot & r { n }\\ r { p } { 1 } &r−p { 2 } & \cdot \cdot \cdot & r ˆ{ p } { n }\\ . & . & . & . \\ .& . & . & . \\ . & . & . & . \\ p−r ˆ{ n − 1 } { 1 } & r { p }ˆ{ n− 1 } { 2 } & \cdot \cdot \cdot & r ˆ{ p ˆ{ n − 1 }} { n }\end{ array }\ right ) \ ]

\noindent van i shes i f and only i f $ r { 1 } , . . . , r { n }$ \quad are l i n e a r l y dependent over$ F { p } . $

\noindent Proof . \ h f i l l Viewed as a polynomial in $ r { 1 } , . . . , r { n }$over $ F { p } , $ the Moore determinant

\noindent i s d i v i s i b l e by each o f the l i n e a r forms $ c { 1 } r { 1 } + \cdot\cdot \cdot + c { n } r { n }$ f o r $ c { 1 } , . . . , c { n }\ in F { p }$

\noindent not a l l ze ro . Up to s c a l a r m u l t i p l e s , the r e are $ p ˆ{ n − 1 } +\cdot \cdot \cdot + p + 1 $ such forms ,

so the determinant \quad i s d i v i s i b l e by the product o f the se forms . \quad However ,the determinant v i s i b l y i s a homogeneous polynomial in the $ r { i }$ o f degree

\noindent $ p ˆ{ n − 1 } + \cdot \cdot \cdot + p + 1 , $ so i t must be equal to the product o f the l i n e a r f a c t o r st imes a constant . The d e s i r e d r e s u l t f o l l o w s $ . \ square $

\noindent D e f i n i t i o n 3 . 3 . 2 . A polynomial $ P ( z ) $ over a f i e l d $ K $o f c h a r a c t e r i s t i c $ p > 0 $

\noindent i s s a id to be a d d i t i v e ( or l i n e a r i z e d ) i f i t has the form

\begin { a l i g n ∗}P ( z ) = c { 0 } z + c { 1 } z ˆ{ p } + \cdot \cdot \cdot

+ c { n } z ˆ{ p ˆ{ n }}\\ f o r some c { 0 } , . . . , c { n } \ inK .\end{ a l i g n ∗}

\noindent Lemma 3 . 3 . 3 . \quad Let $ P ( z ) $ \quad be a nonzero polynomial over a f i e l d$ K $ o f char −

a c t e r i s t i c $ p > 0 , $ and l e t $ L $ be an a l g e b r a i c c l o sure o f $ K . $\quad Then the f o l l o w i n g

\noindent c o n d i t i o n s are equ iva l en t .

\centerline {( a ) \quad The polynomial $ P ( z ) $ \quad i s a d d i t i v e . }

( b ) \quad The equat ion $ P ( y + z ) = P ( y ) + P ( z) $ ho lds as a formal i d e n t i t y o f

po lynomia l s .

\centerline {( c ) \quad The equat ion $ P ( y + z ) = P ( y ) +P ( z ) $ ho lds f o r a l l $ y , z \ in L . $ }

( d ) \quad The roo t s o f $ P $ in $ L $ form an $ F { p } − $ vec to r space under add i t i on , a l l r oo t soccur to the same m u l t i p l i c i t y , and that m u l t i p l i c i t y i s a power o f $ p . $

\noindent Proof . \quad The i m p l i c a t i o n s ( a $ ) \Longrightarrow ( $ b $ ) \Longrightarrow( $ c ) are c l e a r , and ( c $ ) \Longrightarrow ( $ b ) ho lds

because the f i e l d $ L $ must be i n f i n i t e . We next check that ( d $ ) \Longrightarrow( $ a ) . Let

$ V \subset L $ be the s e t o f r oo t s , and l e t $ p ˆ{ e }$ be the common m u l t i p l i c i t y . Let$ Q ( z ) $

\noindent be the Moore determinant o f $ z ˆ{ p ˆ{ e }} , r { p }ˆ{ e } { 1 } , \cdot\cdot \cdot , r ˆ{ p ˆ{ e }} { m } . $ By Lemma 3 . 3 . 1 , the roo t s

o f $ Q $ are p r e c i s e l y the e lements o f $ V , $ and each occurs with m u l t i p l i c i t y atl e a s t $ p ˆ{ e } . $ However , deg $ ( Q ) = p ˆ{ e + m } = $ deg

$ ( P ) , $ so the m u l t i p l i c i t i e s must be

392 .. Kiran S period Kedlaya3 period 3 period .. Additive .. polynomials period .. In positive characteristic comma .. it is convenientto restrict attention to a special class of polynomials comma the .. quotedblleft additive quotedblright .. poly

hyphennomials period First comma we recall a standard recipe open parenthesis analogous to the constructionof Vandermonde determinants closing parenthesis for producing such polynomials periodLemma 3 period 3 period 1 period .. Let K be a field of characteristic p greater 0 period .. Given r sub 1 comma

period period period comma r sub n inK comma the Moore determinantRow 1 r sub 1 r sub 2 times times times r sub n Row 2 r p sub 1 r-p sub 2 times times times r sub n to the power

of p Row 3 period period period period Row 4 period period period period Row 5 period period period period Row6 p-r sub 1 to the power of n minus 1 r p sub 2 to the power of n minus 1 times times times r sub n to the power ofp to the power of n minus 1 .

vanishes if and only if r sub 1 comma period period period comma r sub n .. are linearly dependent over F subp period

Proof period .... Viewed as a polynomial in r sub 1 comma period period period comma r sub n over F sub pcomma the Moore determinant

i s divisible by each of the linear forms c sub 1 r sub 1 plus times times times plus c sub n r sub n for c sub 1comma period period period comma c sub n in F sub p

not all zero period Up to scalar multiples comma there are p to the power of n minus 1 plus times times timesplus p plus 1 such forms comma

so the determinant .. is divisible by the product of these forms period .. However commathe determinant visibly i s a homogeneous polynomial in the r sub i of degreep to the power of n minus 1 plus times times times plus p plus 1 comma so it must be equal to the product of

the linear factorst imes a constant period The desired result follows period squareDefinition 3 period 3 period 2 period A polynomial P open parenthesis z closing parenthesis over a field K of

characteristic p greater 0i s said to be additive open parenthesis or linearized closing parenthesis if it has the formP open parenthesis z closing parenthesis = c sub 0 z plus c sub 1 z to the power of p plus times times times plus

c sub n z to the power of p to the power of n for some c sub 0 comma period period period comma c sub n in Kperiod

Lemma 3 period 3 period 3 period .. Let P open parenthesis z closing parenthesis .. be a nonzero polynomialover a field K of char hyphen

acteristic p greater 0 comma and let L be an algebraic c lo sure of K period .. Then the followingconditions are equivalent periodopen parenthesis a closing parenthesis .. The polynomial P open parenthesis z closing parenthesis .. is additive

periodopen parenthesis b closing parenthesis .. The equation P open parenthesis y plus z closing parenthesis = P open

parenthesis y closing parenthesis plus P open parenthesis z closing parenthesis holds as a formal identity ofpolynomials periodopen parenthesis c closing parenthesis .. The equation P open parenthesis y plus z closing parenthesis = P open

parenthesis y closing parenthesis plus P open parenthesis z closing parenthesis holds for all y comma z in L periodopen parenthesis d closing parenthesis .. The roots of P in L form an F sub p hyphen vector space under addition

comma all rootsoccur to the same multiplicity comma and that multiplicity is a power of p periodProof period .. The implications open parenthesis a closing parenthesis equal-arrowdblright open parenthesis b

closing parenthesis arrowdblright-equal open parenthesis c closing parenthesis are clear comma and open parenthesisc closing parenthesis equal-arrowdblright open parenthesis b closing parenthesis holds

because the field L must be infinite period We next check that open parenthesis d closing parenthesis equal-arrowdblright open parenthesis a closing parenthesis period Let

V subset L be the set of roots comma and let p to the power of e be the common multiplicity period Let Q openparenthesis z closing parenthesis

be the Moore determinant of z to the power of p to the power of e comma r p sub 1 to the power of e commatimes times times comma r sub m to the power of p to the power of e period By Lemma 3 period 3 period 1 commathe roots

of Q are precisely the elements of V comma and each occurs with multiplicity atleast p to the power of e period However comma deg open parenthesis Q closing parenthesis = p to the power of

e plus m = deg open parenthesis P closing parenthesis comma so the multiplicities must be


3 . 3 . Additive polynomials . In positive characteristic ,it is convenient to restrict attention to a special class of polynomials , the“ additive ” poly -nomials . First , we recall a standard recipe ( analogous to the constructionof Vandermonde determinants ) for producing such polynomials .Lemma 3 . 3 . 1 . Let K be a field of characteristic p > 0. Givenr1, ..., rn ∈ K, the Moore determinant

r1 r2 · · · rnrp1 r − p2 · · · rpn. . . .. . . .. . . .

p− rn−11 rpn−12 · · · rpn−1

n

vanishes if and only if r1, ..., rn are linearly dependent over Fp.Proof . Viewed as a polynomial in r1, ..., rn over Fp, the Mooredeterminanti s divisible by each of the linear forms c1r1 + · · ·+ cnrn for c1, ..., cn ∈ Fpnot all zero . Up to scalar multiples , there are pn−1 + · · ·+ p+ 1 such forms ,so the determinant is divisible by the product of these forms . However, the determinant visibly i s a homogeneous polynomial in the ri of degreepn−1 + · · · + p + 1, so it must be equal to the product of the linear factors times a constant . The desired result follows . �Definition 3 . 3 . 2 . A polynomial P (z) over a field K of characteristicp > 0i s said to be additive ( or linearized ) if it has the form

P (z) = c0z + c1zp + · · ·+ cnz

pn

forsomec0, ..., cn ∈ K.

Lemma 3 . 3 . 3 . Let P (z) be a nonzero polynomial over a fieldK of char - acteristic p > 0, and let L be an algebraic c lo sure of K.Then the followingconditions are equivalent .

( a ) The polynomial P (z) is additive .( b ) The equation P (y + z) = P (y) + P (z) holds as a formal identity of

polynomials .( c ) The equation P (y + z) = P (y) + P (z) holds for all y, z ∈ L.

( d ) The roots of P in L form an Fp− vector space under addition, all roots occur to the same multiplicity , and that multiplicity is a powerof p.Proof . The implications ( a ) =⇒ ( b ) =⇒ ( c ) are clear , and ( c) =⇒ ( b ) holds because the field L must be infinite . We next check that( d ) =⇒ ( a ) . Let V ⊂ L be the set of roots , and let pe be the commonmultiplicity . Let Q(z)be the Moore determinant of zp

e

, rpe1, · · ·, rpe

m . By Lemma 3 . 3 . 1 , the rootsof Q are precisely the elements of V, and each occurs with multiplicity atleast pe. However , deg (Q) = pe+m = deg (P ), so the multiplicities must be


\noindent exac t l y $ p ˆ{ e } , $ and $ P $ must equal $ Q $ t imes a s c a l a r . S ince$ Q $ i s v i s i b l y a d d i t i v e ,

\noindent so i s $ P . $

I t remains to check that \quad ( c $ ) \Longrightarrow ( $ d ) . Given ( c ) , note that the roo t s o f$ P $ in $ L $ form an $ F { p } − $ vec to r space under add i t i on ; \quad a l s o , \quad i f

$ r \ in L $ i s a root

\noindent o f $ P , $ then $ P ( z + r ) = P ( z ) , $ so a l l r oo t s o f$ P $ have the same m u l t i p l i c i t y .

Let $ V $ be the roo t s o f $ P , $ choose gene ra to r s $ r { 1 } , . . ., r { m }$ o f $ V $ as an $ F { p } − $ vec to r

\noindent space , and l e t $ Q ( z ) $ be the Moore determinant o f $ z ,r { 1 } , . . . , r { m } . $ Then $ P ( z ) = $

$ cQ ( z ) ˆ{ n }$ f o r some constant $ c , $ where $ n $ i s the common m u l t i p l i c i t y o f the roo t so f $ P ( $ because $ Q $ has no repeated roo t s , \quad by the a n a l y s i s o f the prev iousparagraph ) . Suppose that $ n $ i s not a prime power ; then the polynomia l s $ (

y + $$ z ) ˆ{ n }$ and $ y ˆ{ n } + z ˆ{ n }$ are not i d e n t i c a l l y equal , because the binomial c o e f f i c i e n t

\noindent $ ( ˆ{ n } { p ˆ{ i }} ) , $ f o r $ i $ the l a r g e s t i n t e g e r such that$ p ˆ{ i }$ d i v i d e s $ n , $ i s not d i v i s i b l e by $ p . $

\noindent Thus there e x i s t va lue s o f $ y , z $ in $ L $ f o r which $ ( y+ z ) ˆ{ n } \not= y ˆ{ n } + z ˆ{ n } . $ \quad Since

$ L $ i s a l g e b r a i c a l l y c l o s e d $ , Q $ i s s u r j e c t i v e as a map from $ L $ to i t s e l f ; we canthus choose $ y , z \ in L $ such that $ ( Q ( y ) + Q ( z

) ) ˆ{ n } \not= Q ( y ) ˆ{ n } + Q ( z ) ˆ{ n } . $ S ince $ Q $i s

a d d i t i v e , t h i s means that $ P ( y + z ) \ne P ( y ) + P( z ) , $ contrary to hypothes i s .We conclude that $ n $ must be a prime power . Hence ( c $ ) \Longrightarrow

( $ d ) , and theproo f i s complete $ . \ square $

The f o l l o w i n g obse rvat i on i s sometimes known as \quad ‘ ‘ Ore ’ s lemma ’ ’ \quad ( as in[ 2 , Lemma 1 2 . 2 . 3 ] ) .

\noindent Lemma 3 . 3 . 4 . \ h f i l l For $ K \subseteq L $ f i e l d s o f c h a r a c t e r i s t i c$ p > 0 $ and $ \alpha \ in L , \alpha $ i s

\noindent a l g e b r a i c over $ K $ i f and only i f i t i s a root o f some a d d i t i v e polynomial over

\begin { a l i g n ∗}K .\end{ a l i g n ∗}

\noindent Proof . \quad Clea r l y i f $ \alpha $ i s a root o f an a d d i t i v e polynomial over$ K , $ then $ \alpha $ i s

a l g e b r a i c over $ K . $ \quad Converse ly , i f $ \alpha $ i s a l g e b r a i c , then$ \alpha , \alpha ˆ{ p } , . . . $ \quad cannot a l l

be l i n e a r l y independent , \quad so the re must \quad be \quad a l i n e a r r e l a t i o n \quad o f the form

\noindent $ c { 0 } \alpha + c { 1 } \alpha ˆ{ p } + \cdot \cdot \cdot+ c { n } \alpha ˆ{ p ˆ{ n }} = 0 $ with $ c { 0 } , . . . , c { n }\ in K $ not a l l ze ro $ . \ square $

Our next \quad lemma g e n e r a l i z e s Lemma 3 . 3 . 4 to \quad ‘ ‘ semi − l i n e a r ’ ’ \quad systems o fequat ions .

\noindent Lemma 3 . 3 . 5 . \quad Let $ K \subseteq L $ \quad be f i e l d s \quad o f c h a r a c t e r i s t i c$ p > 0 , $ \quad l e t $ A , B $ \quad be

$ n \times n $ matr i ce s with e n t r i e s in $ K , $ \quad at l e a s t one o f which i s i n v e r t i b l e , \quad andl e t w $ \ in K ˆ{ n }$ \quad be any ( co lumn ) vec tor . \quad Suppose v $ \ in

L ˆ{ n }$ \quad i s a vec to r such that

\noindent $ A v ˆ{ \sigma } + B $ v $ = $ w , where $ \sigma $ denotes the$ p − $ th power Frobenius map . \quad Then the

e n t r i e s o f v are a l g e b r a i c over $ K . $

\noindent Proof . \quad Suppose $ A $ i s i n v e r t i b l e . Then f o r $ i = 1 ,2 , . . . , $ we can wr i t e $ v ˆ{ \sigma ˆ{ i }} = $

$ U { i }$ v $ + w { i }$ f o r some $ n \times n $ matrix $ U { i }$over $ K $ and some $\ l e f t . w { i } \ in K\begin { array }{ c} n \\ . \end{ array }Such\right . $v e c t o r s

\noindent span a vec to r space over $ K $ o f dimension at most $ n ˆ{ 2 } + n; $ f o r some $ m , $ wecan thus f i n d $ c { 0 } , . . . , c { m }$ not a l l ze ro such that

\begin { a l i g n ∗}\ tag ∗{$ ( 3 . 3 . 6 ) $} c { 0 } v + c { 1 } v ˆ{ \sigma } + \cdot\cdot \cdot + c { m } v ˆ{ \sigma ˆ{ m }} = 0 .\end{ a l i g n ∗}

Finite automata and algebraic extensions of function fields .. 393exactly p to the power of e comma and P must equal Q t imes a scalar period Since Q is visibly additive commaso i s P periodIt remains to check that .. open parenthesis c closing parenthesis equal-arrowdblright open parenthesis d closing

parenthesis period Given open parenthesis c closing parenthesis comma note that the roots ofP in L form an F sub p hyphen vector space under addition semicolon .. also comma .. if r in L is a rootof P comma then P open parenthesis z plus r closing parenthesis = P open parenthesis z closing parenthesis

comma so all roots of P have the same multiplicity periodLet V be the roots of P comma choose generators r sub 1 comma period period period comma r sub m of V as

an F sub p hyphen vectorspace comma and let Q open parenthesis z closing parenthesis be the Moore determinant of z comma r sub 1

comma period period period comma r sub m period Then P open parenthesis z closing parenthesis =cQ open parenthesis z closing parenthesis to the power of n for some constant c comma where n i s the common

multiplicity of the rootsof P open parenthesis because Q has no repeated roots comma .. by the analysis of the previousparagraph closing parenthesis period Suppose that n is not a prime power semicolon then the polynomials open

parenthesis y plusz closing parenthesis to the power of n and y to the power of n plus z to the power of n are not identically equal

comma because the binomial coefficientparenleftbig sub p to the power of i to the power of n parenrightbig comma for i the largest integer such that p

to the power of i divides n comma is not divisible by p periodThus there exist values of y comma z in L for which open parenthesis y plus z closing parenthesis to the power

of n negationslash-equal y to the power of n plus z to the power of n period .. SinceL i s algebraically closed comma Q is surjective as a map from L to itself semicolon we canthus choose y comma z in L such that open parenthesis Q open parenthesis y closing parenthesis plus Q open

parenthesis z closing parenthesis closing parenthesis to the power of n negationslash-equal Q open parenthesis yclosing parenthesis to the power of n plus Q open parenthesis z closing parenthesis to the power of n period Since Qi s

additive comma this means that P open parenthesis y plus z closing parenthesis equal-negationslash P openparenthesis y closing parenthesis plus P open parenthesis z closing parenthesis comma contrary to hypothesis period

We conclude that n must be a prime power period Hence open parenthesis c closing parenthesis equal-arrowdblrightopen parenthesis d closing parenthesis comma and the

proof i s complete period squareThe following observation i s sometimes known as .. quotedblleft Ore quoteright s lemma quotedblright .. open

parenthesis as inopen square bracket 2 comma Lemma 1 2 period 2 period 3 closing square bracket closing parenthesis periodLemma 3 period 3 period 4 period .... For K subset equal L fields of characteristic p greater 0 and alpha in L

comma alpha isalgebraic over K if and only if it is a root of some additive polynomial overK periodProof period .. Clearly if alpha is a root of an additive polynomial over K comma then alpha i salgebraic over K period .. Conversely comma if alpha i s algebraic comma then alpha comma alpha to the power

of p comma period period period .. cannot allbe linearly independent comma .. so there must .. be .. a linear relation .. of the formc sub 0 alpha plus c sub 1 alpha to the power of p plus times times times plus c sub n alpha to the power of p

to the power of n = 0 with c sub 0 comma period period period comma c sub n in K not all zero period squareOur next .. lemma generalizes Lemma 3 period 3 period 4 to .. quotedblleft semi hyphen linear quotedblright ..

systems ofequations periodLemma 3 period 3 period 5 period .. Let K subset equal L .. be fields .. of characteristic p greater 0 comma ..

let A comma B .. ben times n matrices with entries in K comma .. at least one of which is invertible comma .. andlet w in K to the power of n .. be any open parenthesis co lumn closing parenthesis vector period .. Suppose v

in L to the power of n .. is a vector such thatA v to the power of sigma plus B v = w comma where sigma denotes the p hyphen th power Frobenius map

period .. Then theentries of v are algebraic over K periodProof period .. Suppose A i s invertible period Then for i = 1 comma 2 comma period period period comma we

can write v to the power of sigma to the power of i =U sub i v plus w sub i for some n times n matrix U sub i over K and some w sub i in Row 1 n Row 2 period .

vectorsspan a vector space over K of dimension at most n to the power of 2 plus n semicolon for some m comma wecan thus find c sub 0 comma period period period comma c sub m not all zero such thatEquation: open parenthesis 3 period 3 period 6 closing parenthesis .. c sub 0 v plus c sub 1 v to the power of

sigma plus times times times plus c sub m v to the power of sigma to the power of m = 0 period


exactly pe, and P must equal Q t imes a scalar . Since Q is visibly additive,so i s P.

It remains to check that ( c ) =⇒ ( d ) . Given ( c ) , note that theroots of P in L form an Fp− vector space under addition ; also , ifr ∈ L is a rootof P, then P (z+ r) = P (z), so all roots of P have the same multiplicity . LetV be the roots of P, choose generators r1, ..., rm of V as an Fp− vectorspace , and let Q(z) be the Moore determinant of z, r1, ..., rm. Then P (z) =cQ(z)n for some constant c, where n i s the common multiplicity of the rootsof P ( because Q has no repeated roots , by the analysis of the previousparagraph ) . Suppose that n is not a prime power ; then the polynomials(y+ z)n and yn+zn are not identically equal , because the binomial coefficient(npi), for i the largest integer such that pi divides n, is not divisible by p.

Thus there exist values of y, z in L for which (y+z)n 6= yn+zn. Since L i salgebraically closed , Q is surjective as a map from L to itself ; we can thuschoose y, z ∈ L such that (Q(y) +Q(z))n 6= Q(y)n+Q(z)n. Since Q i s additive ,this means that P (y+ z) 6= P (y) +P (z), contrary to hypothesis . We concludethat n must be a prime power . Hence ( c ) =⇒ ( d ) , and the proof is complete . �

The following observation i s sometimes known as “ Ore ’ s lemma ”( as in [ 2 , Lemma 1 2 . 2 . 3 ] ) .Lemma 3 . 3 . 4 . For K ⊆ L fields of characteristic p > 0 andα ∈ L,α isalgebraic over K if and only if it is a root of some additive polynomial over

K.

Proof . Clearly if α is a root of an additive polynomial over K, then α is algebraic over K. Conversely , if α i s algebraic , then α, αp, ... cannotall be linearly independent , so there must be a linear relation ofthe formc0α+ c1α

p + · · ·+ cnαpn = 0 with c0, ..., cn ∈ K not all zero . �

Our next lemma generalizes Lemma 3 . 3 . 4 to “ semi - linear ”systems of equations .Lemma 3 . 3 . 5 . Let K ⊆ L be fields of characteristicp > 0, let A,B be n × n matrices with entries in K, at least one ofwhich is invertible , and let w ∈ Kn be any ( co lumn ) vector .Suppose v ∈ Ln is a vector such thatAvσ + B v = w , where σ denotes the p− th power Frobenius map .Then the entries of v are algebraic over K.Proof . Suppose A i s invertible . Then for i = 1, 2, ..., we can write

vσi

= Ui v +wi for some n× n matrix Ui over K and some wi ∈ Kn.Such

vectorsspan a vector space over K of dimension at most n2 +n; for some m, we canthus find c0, ..., cm not all zero such that

c0v + c1vσ + · · ·+ cmvσ

m

= 0. (3.3.6)


\noindent Apply Lemma 3 . 3 . 4 to each component in ( 3 . 3 . 6 ) to deduce that the e n t r i e so f v are a l g e b r a i c over $ K . $

\hspace ∗{\ f i l l }Now suppose $ B $ i s i n v e r t i b l e . There i s no harm in e n l a r g i n g $ L, $ so we may

\noindent as we l l assume that $ L $ i s c l o s e d under tak ing $ p − $ th roo t s , i . e$ . , L $ i s p e r f e c t .

Then the map $ \sigma : L \rightarrow L $ i s a b $ i j $ e c t i o n . Let $ K ˆ{ \prime }$be the s e t o f $ x \ in L $ f o r which

the re e x i s t s a nonnegat ive i n t e g e r $ i $ such that $ x ˆ{ \sigma ˆ{ i }} \ in K; $ then $ \sigma : K ˆ{ \prime } \rightarrow K ˆ{ \prime }$

i s a l s o a b $ i j $ e c t i o n , and each element o f $ K ˆ{ \prime }$ i s a l g e b r a i c over$ K . $

For $ i = 1 , 2 , . . . , $ we can now wr i t e $ v ˆ{ \sigma ˆ{ −i }} = U { i }$ v $ + w { i }$ f o r some $ n \times n $ matrix

$ U { i }$ over $ K ˆ{ \prime }$ and some $ w { i } \ in ( K ˆ{ \prime }) ˆ{ n } . $ As above , we conclude that the e n t r i e s

o f v are a l g e b r a i c over $ K ˆ{ \prime } . $ However , any element $ \alpha \ inL $ a l g e b r a i c over $ K ˆ{ \prime }$

i s a l g e b r a i c over $ K : $ i f $ d { 0 } + d { 1 } \alpha + \cdot \cdot\cdot + d { m } \alpha ˆ{ m } = 0 $ f o r $ d { 0 } , . . . , d { m }\ in K ˆ{ \prime }$ not

a l l ze ro , then we can choose a nonnegat ive i n t e g e r $ i $ such that $ d ˆ{ \sigma ˆ{ i }} { 0 }, . . . , d ˆ{ \sigma ˆ{ i }} { m }$belong to $ K , $ and $ d ˆ{ \sigma ˆ{ i }} { 0 } + d ˆ{ \sigma ˆ{ i }} { 1 }\alpha ˆ{ p ˆ{ i }} + \cdot \cdot \cdot + d ˆ{ \sigma ˆ{ i }} { m } \alpha ˆ{ mp ˆ{ i }}= 0 . $ We conclude that the

e n t r i e s o f v are a l g e b r a i c over $ K , $ as d e s i r e d $ . \ square $

\centerline {4 . \quad Genera l i zed power s e r i e s and automata }

In t h i s chapter , we s t a t e the main theorem ( Theorem 4 . 1 . 3 ) \quad and somer e l a t e d r e s u l t s ; i t s proo f ( or ra the r p roo f s ) w i l l occupy much o f the r e s t o fthe paper .

\noindent 4 . 1 . \ h f i l l The main theorem : s t atement and p r e l i m i n a r i e s . \ h f i l l We are now

\noindent ready to s t a t e our g e n e r a l i z a t i o n o f C h r i s t o l ’ s theorem , the main t h e o r e t i c a l

\noindent r e s u l t o f t h i s paper . For context , we f i r s t s t a t e a form o f C h r i s t o l ’ s theorem( compare \quad [ 4 ] , \quad [ 5 ] , and a l s o [ 2 , Theorem 1 2 . 2 . 5 ] ) . \quad Reminder

$ : F { q } ( t ) $ denotesthe f i e l d o f r a t i o n a l f u n c t i o n s over $ F { q } , $ i . e . , the f i e l d o f f r a c t i o n s o f the r ing

\begin { a l i g n ∗}o f po lynomia ls F { q } [ t ] .\end{ a l i g n ∗}

\noindent Theorem 4 . 1 . 1 ( C h r i s t o l ) . Let $ q $ be a power o f the prime $ p, $ and l e t $ \{ a { i } \} ˆ{ \ infty } { i = 0 }$

\noindent be a sequence over $ F { q } . $ \quad Then the s e r i e s $ \sum ˆ{ \ infty } { i= 0 } a { i } t ˆ{ i } \ in F { q } \ l l b r a c k e t t \ r rb ra cke t $ \quad i s a l g e b r a i c over

$ F { q } ( t ) $ \quad i f and only i f the sequence $ \{ a { i } \} ˆ{ \ infty } { i= 0 }$ i s $ p − $ automatic .

We now formulate our g e n e r a l i z a t i o n o f C h r i s t o l ’ s theorem . Reca l l that$ S { p }$ i s the s e t o f numbers o f the form $ m / p ˆ{ n } , $ f o r $ m

, n $ nonnegat ive i n t e g e r s .

\noindent D e f i n i t i o n 4 . 1 . 2 . \quad Let $ q $ be a power o f the prime $ p , $and l e t $ f : Q \rightarrow F { q }$ be

a func t i on whose support $ S $ i s we l l − ordered . We say the g e n e r a l i z e d Laurents e r i e s $ \sum { i } f ( i ) t ˆ{ i }$ i s $ p − $ quas i − automatic i f the f o l l o w i n g c o n d i t i o n s hold .

( a ) \quad For some i n t e g e r s $ a $ and $ b $ with $ a > 0 , $ the s e t $ aS+ b = \{ a i + b : i \ in S \} $

i s conta ined in $ S { p } , $ i . e . , c o n s i s t s o f nonnegat ive $ p − $ ad ic r a t i o n a l s .

( b ) \quad For some $ a , b $ f o r which ( a ) ho lds , the func t i on $ f { a, } b : S { p } \rightarrow F { q }$ givenby $ f { a , } b ( x ) = f ( ( x − b ) / a ) $ i s

$ p − $ automatic .

\noindent Note that by Lemma 2 . 3 . 6 , i f ( b ) ho lds f o r a s i n g l e cho i c e o f $ a, b $ s a t i s f y i n g( a ) , then ( b ) ho lds a l s o f o r any cho i c e o f $ a , b $ s a t i s f y i n g ( a ) . In case ( a ) and( b ) hold with $ a = 1 , b = 0 , $ we say the s e r i e s i s $ p − $

automatic .

394 .. Kiran S period KedlayaApply Lemma 3 period 3 period 4 to each component in open parenthesis 3 period 3 period 6 closing parenthesis

to deduce that the entriesof v are algebraic over K periodNow suppose B i s invertible period There i s no harm in enlarging L comma so we mayas well assume that L i s closed under taking p hyphen th roots comma i period e period comma L is perfect

periodThen the map sigma : L right arrow L is a b ij ection period Let K to the power of prime be the set of x in L

for whichthere exists a nonnegative integer i such that x to the power of sigma to the power of i in K semicolon then sigma

: K to the power of prime right arrow K to the power of primei s also a b ij ection comma and each element of K to the power of prime i s algebraic over K periodFor i = 1 comma 2 comma period period period comma we can now write v to the power of sigma to the power

of minus i = U sub i v plus w sub i for some n times n matrixU sub i over K to the power of prime and some w sub i in open parenthesis K to the power of prime closing

parenthesis to the power of n period As above comma we conclude that the entriesof v are algebraic over K to the power of prime period However comma any element alpha in L algebraic over K

to the power of primei s algebraic over K : if d sub 0 plus d sub 1 alpha plus times times times plus d sub m alpha to the power of m

= 0 for d sub 0 comma period period period comma d sub m in K to the power of prime notall zero comma then we can choose a nonnegative integer i such that d sub 0 to the power of sigma to the power

of i comma period period period comma d sub m to the power of sigma to the power of ibelong to K comma and d sub 0 to the power of sigma to the power of i plus d sub 1 to the power of sigma to

the power of i alpha to the power of p to the power of i plus times times times plus d sub m to the power of sigmato the power of i alpha to the power of mp to the power of i = 0 period We conclude that the

entries of v are algebraic over K comma as desired period square4 period .. Generalized power series and automataIn this chapter comma we state the main theorem open parenthesis Theorem 4 period 1 period 3 closing paren-

thesis .. and somerelated results semicolon it s proof open parenthesis or rather proofs closing parenthesis will occupy much of the

rest ofthe paper period4 period 1 period .... The main theorem : st atement and preliminaries period .... We are nowready to state our generalization of Christol quoteright s theorem comma the main theoreticalresult of this paper period For context comma we first state a form of Christol quoteright s theoremopen parenthesis compare .. open square bracket 4 closing square bracket comma .. open square bracket 5 closing

square bracket comma and also open square bracket 2 comma Theorem 1 2 period 2 period 5 closing square bracketclosing parenthesis period .. Reminder : F sub q open parenthesis t closing parenthesis denotes

the field of rational functions over F sub q comma i period e period comma the field of fractions of the ringof polynomials F sub q open square bracket t closing square bracket periodTheorem 4 period 1 period 1 open parenthesis Christol closing parenthesis period Let q be a power of the prime

p comma and let open brace a sub i closing brace sub i = 0 to the power of infinitybe a sequence over F sub q period .. Then the s eries sum sub i = 0 to the power of infinity a sub i t to the

power of i in F sub q llbracket t rrbracket .. is algebraic overF sub q open parenthesis t closing parenthesis .. if and only if the sequence open brace a sub i closing brace sub

i = 0 to the power of infinity is p hyphen automatic periodWe now formulate our generalization of Christol quoteright s theorem period Recall thatS sub p i s the set of numbers of the form m slash p to the power of n comma for m comma n nonnegative integers

periodDefinition 4 period 1 period 2 period .. Let q be a power of the prime p comma and let f : Q right arrow F sub

q bea function whose support S i s well hyphen ordered period We say the generalized Laurentseries sum sub i f open parenthesis i closing parenthesis t to the power of i i s p hyphen quasi hyphen automatic

if the following conditions hold periodopen parenthesis a closing parenthesis .. For some integers a and b with a greater 0 comma the set aS plus b =

open brace ai plus b : i in S closing braceis contained in S sub p comma i period e period comma consists of nonnegative p hyphen adic rationals periodopen parenthesis b closing parenthesis .. For some a comma b for which open parenthesis a closing parenthesis

holds comma the function f sub a comma b : S sub p right arrow F sub q givenby f sub a comma b open parenthesis x closing parenthesis = f open parenthesis open parenthesis x minus b

closing parenthesis slash a closing parenthesis i s p hyphen automatic periodNote that by Lemma 2 period 3 period 6 comma if open parenthesis b closing parenthesis holds for a single choice

of a comma b satisfyingopen parenthesis a closing parenthesis comma then open parenthesis b closing parenthesis holds also for any

choice of a comma b satisfying open parenthesis a closing parenthesis period In case open parenthesis a closingparenthesis and

open parenthesis b closing parenthesis hold with a = 1 comma b = 0 comma we say the series i s p hyphenautomatic period


Apply Lemma 3 . 3 . 4 to each component in ( 3 . 3 . 6 ) to deduce thatthe entries of v are algebraic over K.

Now suppose B i s invertible . There i s no harm in enlarging L, so wemayas well assume that L i s closed under taking p− th roots , i . e ., L isperfect . Then the map σ : L → L is a b ij ection . Let K ′ be the set ofx ∈ L for which there exists a nonnegative integer i such that xσ

i ∈ K; thenσ : K ′ → K ′ i s also a b ij ection , and each element of K ′ i s algebraic overK.

For i = 1, 2, ..., we can now write vσ−i

= Ui v +wi for some n× n matrixUi over K ′ and some wi ∈ (K ′)n. As above , we conclude that the entries ofv are algebraic over K ′. However , any element α ∈ L algebraic over K ′ i salgebraic over K : if d0 + d1α + · · ·+ dmα

m = 0 for d0, ..., dm ∈ K ′ not all zero, then we can choose a nonnegative integer i such that dσ

i

0 , ..., dσi

m belong toK, and dσ

i

0 + dσi

1 αpi + · · ·+ dσ

i

mαmpi = 0. We conclude that the entries of v are

algebraic over K, as desired . �4 . Generalized power series and automata

In this chapter , we state the main theorem ( Theorem 4 . 1 . 3 ) andsome related results ; it s proof ( or rather proofs ) will occupy much of therest of the paper .4 . 1 . The main theorem : st atement and preliminaries . Weare nowready to state our generalization of Christol ’ s theorem , the main theoret-icalresult of this paper . For context , we first state a form of Christol ’ stheorem ( compare [ 4 ] , [ 5 ] , and also [ 2 , Theorem 1 2 . 2 . 5 ] ) .Reminder : Fq(t) denotes the field of rational functions over Fq, i . e . , thefield of fractions of the ring

ofpolynomialsFq[t].

Theorem 4 . 1 . 1 ( Christol ) . Let q be a power of the prime p, andlet {ai}∞i=0

be a sequence over Fq. Then the s eries∑∞i=0 ait

i ∈ Fqt is algebraic overFq(t) if and only if the sequence {ai}∞i=0 is p− automatic .

We now formulate our generalization of Christol ’ s theorem . Recall thatSp i s the set of numbers of the form m/pn, for m,n nonnegative integers .Definition 4 . 1 . 2 . Let q be a power of the prime p, and letf : Q → Fq be a function whose support S i s well - ordered . We say thegeneralized Laurent series

∑i f(i)ti i s p− quasi - automatic if the following

conditions hold .( a ) For some integers a and b with a > 0, the set aS+b = {ai+b : i ∈ S}

is contained in Sp, i . e . , consists of nonnegative p− adic rationals .( b ) For some a, b for which ( a ) holds , the function fa,b : Sp → Fq

given by fa,b(x) = f((x− b)/a) i s p− automatic .Note that by Lemma 2 . 3 . 6 , if ( b ) holds for a single choice of a, bsatisfying ( a ) , then ( b ) holds also for any choice of a, b satisfying ( a) . In case ( a ) and ( b ) hold with a = 1, b = 0, we say the series i s p−automatic .


\noindent Theorem 4 . 1 . 3 . \ h f i l l Let $ q $ be a power o f the prime $ p , $\ h f i l l and l e t $ f : Q \rightarrow F { q }$ be a

\noindent f unc t i on whose support i s we l l − ordered . \ h f i l l Then the corre spond ing g e n e r a l i z e d

\noindent Laurent s e r i e s $ \sum { i } f ( i ) t ˆ{ i } \ in F { q } (( t ˆ{ Q } ) ) $ \quad i s a l g e b r a i c over $ F { q } ( t ) $ \quad i f and only i f i t

i s $ p − $ quas i − automatic .

We w i l l g ive two proo f s o f Theorem 4 . 1 . 3 in due course . In both ca s e s ,we use Propos i t i on 5 . 1 . 2 to deduce the i m p l i c a t i o n \quad ‘ ‘ automatic i m p l i e s a l −gebra i c ’ ’ . For the r e v e r s e i m p l i c a t i o n ‘ ‘ a l g e b r a i c i m p l i e s automatic ’ ’ , we usePropos i t i on 5 . 2 . 7 f o r a conceptua l proo f and Propos i t i on 7 . 3 . 4 f o r a morea l go r i thmi c proo f . Note , however , that both o f the p roo f s in t h i s d i r e c t i o nr e l y on C h r i s t o l ’ s theorem , so we do not obta in an independent d e r i v a t i o no f that r e s u l t .

\noindent Coro l l a ry \ h f i l l 4 . 1 . 4 . \ h f i l l The g e n e r a l i z e d Laurent s e r i e s $ \sum { i }f ( i ) t ˆ{ i } \ in F { q } ( ( t ˆ{ Q } ) ) $ \ h f i l l i s

\noindent a l g e b r a i c over $ F { q } ( t ) $ \ h f i l l i f and only i f f o r each $ \alpha\ in F { q } , $ the g e n e r a l i z e d Laurent

\noindent s e r i e s

\ [\ begin { a l i gned } \sum t ˆ{ i }\\i \ in f ˆ{ − 1 } ( \alpha ) \end{ a l i gned }\ ]

\noindent i s a l g e b r a i c over $ F { q } ( t ) . $

\centerline{We mention another c o r o l l a r y f o l l o w i n g [ 2 , Theorem 1 2 . 2 . 6 ] . }

\noindent D e f i n i t i o n 4 . 1 . 5 . \ h f i l l Given two g e n e r a l i z e d Laurent s e r i e s $ x =\sum { i } x { i } t ˆ{ i }$ and

\noindent $ y = \sum { i } y i ˆ{ t ˆ{ i }}$ in $ F { q } ( ( t ˆ{ Q }) ) , $ then $ \sum { i } ( x { i } y i ) t ˆ{ i }$ i s a l s o a g e n e r a l i z e d Laurent s e r i e s ;

\noindent i t i s c a l l e d the Hadamard product and denoted $ x \odot y . $ \quad Then one has thef o l l o w i n g a s s e r t i o n , \quad which in the case o f ord inary power s e r i e s i s due toFurstenberg [ 7 ] .

\noindent Coro l l a ry \ h f i l l 4 . 1 . 6 . \ h f i l l I f $ x , y \ in F { q } ( (t ˆ{ Q } ) ) $ \ h f i l l are \ h f i l l a l g e b r a i c \ h f i l l over $ F { q } ( t ) , $\ h f i l l then s o \ h f i l l i s

\begin { a l i g n ∗}x \odot y .\end{ a l i g n ∗}

\noindent Proof . \quad Thanks to Theorem 4 . 1 . 3 , t h i s f o l l o w s from the f a c t that i f$ f : \Sigma ˆ{ ∗ } \rightarrow $

$ \Delta { 1 }$ \quad and $ g : \Sigma ˆ{ ∗ } \rightarrow \Delta { 2 }$\quad are f i n i t e − s t a t e f u n c t i o n s , then so i s $ f \times g : \Sigma ˆ{ ∗ }\rightarrow $

$ \Delta { 1 } \times \Delta { 2 } ; $ the proo f o f the l a t t e r i s s t r a i g h t f o r w a r d ( or compare [ 2 , Theo −rem $ 5 . 4 . 4 ] ) . \ square $

\noindent 4 . 2 . \quad Decimation \quad and \quad a l g e b r a i c i t y . \quad Before \quad we \quad attack \quad Theorem \quad 4 . 1 . 3proper , \quad i t w i l l be h e l p f u l to know that the p r e c i s e cho i c e o f $ a , b $

in The −orem 4 . 1 . 3 , which does not matter on the automatic s i d e ( D e f i n i t i o n 4 . 1 . 2 ) ,a l s o does not matter on the a l g e b r a i c s i d e .

\noindent D e f i n i t i o n 4 . 2 . 1 . \ h f i l l For $ \tau \ in $ Gal $ ( F { q } /F { p } ) , $ regard $ \tau $ as an automorphism o f

\noindent $ F { q } ( t ) $ and $ F { q } ( ( t ˆ{ Q } ) ) $ by a l l ow ing i t to act on c o e f f i c i e n t s . That i s ,

\ [\ l e f t (\ begin { array }{ c} \sum x { i } t ˆ{ i }\\ i \end{ array }\ right ) \tau =\sum { i } x ˆ{ \tau } { i } t ˆ{ i } . \ ]

Finite automata and algebraic extensions of function fields .. 395Theorem 4 period 1 period 3 period .... Let q be a power of the prime p comma .... and let f : Q right arrow F

sub q be afunction whose support is well hyphen ordered period .... Then the corresponding generalizedLaurent series sum sub i f open parenthesis i closing parenthesis t to the power of i in F sub q open parenthesis

open parenthesis t to the power of Q closing parenthesis closing parenthesis .. is algebraic over F sub q openparenthesis t closing parenthesis .. if and only if it

is p hyphen quasi hyphen automatic periodWe will give two proofs of Theorem 4 period 1 period 3 in due course period In both cases commawe use Proposition 5 period 1 period 2 to deduce the implication .. quotedblleft automatic implies al hyphengebraic quotedblright period For the reverse implication quotedblleft algebraic implies automatic quotedblright

comma we useProposition 5 period 2 period 7 for a conceptual proof and Proposition 7 period 3 period 4 for a morealgorithmic proof period Note comma however comma that both of the proofs in this directionrely on Christol quoteright s theorem comma so we do not obtain an independent derivationof that result periodCorollary .... 4 period 1 period 4 period .... The generalized Laurent series sum sub i f open parenthesis i closing

parenthesis t to the power of i in F sub q open parenthesis open parenthesis t to the power of Q closing parenthesisclosing parenthesis .... is

algebraic over F sub q open parenthesis t closing parenthesis .... if and only if for each alpha in F sub q commathe generalized Laurent

s eriesLine 1 sum t to the power of i Line 2 i in f to the power of minus 1 open parenthesis alpha closing parenthesisis algebraic over F sub q open parenthesis t closing parenthesis periodWe mention another corollary following open square bracket 2 comma Theorem 1 2 period 2 period 6 closing

square bracket periodDefinition 4 period 1 period 5 period .... Given two generalized Laurent series x = sum sub i x sub i t to the

power of i andy = sum sub i y i to the power of t to the power of i in F sub q open parenthesis open parenthesis t to the power

of Q closing parenthesis closing parenthesis comma then sum sub i open parenthesis x sub i y i closing parenthesis tto the power of i i s also a generalized Laurent series semicolon

it is called the Hadamard product and denoted x odot y period .. Then one has thefollowing assertion comma .. which in the case of ordinary power series i s due toFurstenberg open square bracket 7 closing square bracket periodCorollary .... 4 period 1 period 6 period .... If x comma y in F sub q open parenthesis open parenthesis t to the

power of Q closing parenthesis closing parenthesis .... are .... algebraic .... over F sub q open parenthesis t closingparenthesis comma .... then s o .... is

x odot y periodProof period .. Thanks to Theorem 4 period 1 period 3 comma this follows from the fact that if f : Capital Sigma

to the power of * right arrowCapital Delta sub 1 .. and g : Capital Sigma to the power of * right arrow Capital Delta sub 2 .. are finite

hyphen state functions comma then so i s f times g : Capital Sigma to the power of * right arrowCapital Delta sub 1 times Capital Delta sub 2 semicolon the proof of the latter i s straightforward open parenthesis

or compare open square bracket 2 comma Theo hyphenrem 5 period 4 period 4 closing square bracket closing parenthesis period square4 period 2 period .. Decimation .. and .. algebraicity period .. Before .. we .. attack .. Theorem .. 4 period 1

period 3proper comma .. it will be helpful to know that the precise choice of a comma b in The hyphenorem 4 period 1 period 3 comma which does not matter on the automatic side open parenthesis Definition 4

period 1 period 2 closing parenthesis commaalso does not matter on the algebraic side periodDefinition 4 period 2 period 1 period .... For tau in Gal open parenthesis F sub q slash F sub p closing parenthesis

comma regard tau as an automorphism ofF sub q open parenthesis t closing parenthesis and F sub q open parenthesis open parenthesis t to the power of

Q closing parenthesis closing parenthesis by allowing it to act on coefficients period That is commaRow 1 sum x sub i t to the power of i Row 2 i . tau = sum i x sub i to the power of tau t to the power of i period


Theorem 4 . 1 . 3 . Let q be a power of the prime p, and letf : Q→ Fq be afunction whose support is well - ordered . Then the correspondinggeneralizedLaurent series

∑i f(i)ti ∈ Fq((tQ)) is algebraic over Fq(t) if and only if

it is p− quasi - automatic .We will give two proofs of Theorem 4 . 1 . 3 in due course . In both cases ,

we use Proposition 5 . 1 . 2 to deduce the implication “ automatic impliesal - gebraic ” . For the reverse implication “ algebraic implies automatic” , we use Proposition 5 . 2 . 7 for a conceptual proof and Proposition 7. 3 . 4 for a more algorithmic proof . Note , however , that both of theproofs in this direction rely on Christol ’ s theorem , so we do not obtainan independent derivation of that result .Corollary 4 . 1 . 4 . The generalized Laurent series∑i f(i)ti ∈ Fq((tQ)) is

algebraic over Fq(t) if and only if for each α ∈ Fq, the generalizedLaurents eries ∑

ti

i ∈ f−1(α)

is algebraic over Fq(t).We mention another corollary following [ 2 , Theorem 1 2 . 2 . 6 ] .

Definition 4 . 1 . 5 . Given two generalized Laurent series x =∑i xit

i

andy =

∑i yi

ti in Fq((tQ)), then∑i(xiyi)t

i i s also a generalized Laurent series ;it is called the Hadamard product and denoted x � y. Then one has thefollowing assertion , which in the case of ordinary power series i s due toFurstenberg [ 7 ] .Corollary 4 . 1 . 6 . If x, y ∈ Fq((tQ)) are algebraic overFq(t), then s o is

x� y.

Proof . Thanks to Theorem 4 . 1 . 3 , this follows from the fact that iff : Σ∗ → ∆1 and g : Σ∗ → ∆2 are finite - state functions , then soi s f × g : Σ∗ → ∆1 ×∆2; the proof of the latter i s straightforward ( orcompare [ 2 , Theo - rem 5.4.4]). �4 . 2 . Decimation and algebraicity . Before weattack Theorem 4 . 1 . 3 proper , it will be helpful to know thatthe precise choice of a, b in The - orem 4 . 1 . 3 , which does not matteron the automatic side ( Definition 4 . 1 . 2 ) , also does not matter on thealgebraic side .Definition 4 . 2 . 1 . For τ ∈ Gal (Fq/Fp), regard τ as anautomorphism ofFq(t) and Fq((tQ)) by allowing it to act on coefficients . That is ,( ∑

xiti

i

)τ =

∑i

xτi ti.

\noindent 396 \quad Kiran S . KedlayaLet \quad $ \sigma \ in $ \quad Gal $ ( F { q } / F { p } ) $ \quad denote \quad the

$ p − $ power Frobenius \quad map ; \quad note that \quad the

\noindent convent ion we j u s t introduced means that $ x ˆ{ p } = x ˆ{ \sigma }$ i f$ x \ in F { q } , $ but not i f

\begin { a l i g n ∗}x \ in F { q } ( t ) or x \ in F { q } ( ( t ˆ{ Q } ) ) .\end{ a l i g n ∗}

\noindent Lemma 4 . 2 . 2 . \ h f i l l Let $ a , b $ be i n t e g e r s with $ a > 0. $ \ h f i l l Then $ \sum { i } x { i } t ˆ{ i } \ in F { q } ( ( t ˆ{ Q }) ) $ \ h f i l l i s

\noindent a l g e b r a i c over $ F { q } ( t ) $ \quad i f and only i f $ \sum { i }x { a i + b } t ˆ{ i }$ i s a l g e b r a i c over $ F { q } ( t ) . $

\noindent Proof . \ h f i l l I t s u f f i c e s to prove the r e s u l t in the case $ a = 1 $and in the case $ b = 0 , $

\noindent as the gene ra l case f o l l o w s by apply ing these two in s u c c e s s i o n . The case

\noindent $ a = 1 $ i s s t r a i g h t f o r w a r d : i f $ x = \sum { i } x { i }t ˆ{ i }$ i s a root o f the polynomial $ P ( z ) $ over

\noindent $ F { q } ( t ) , $ then $ x ˆ{ \prime } = \sum { i } x { i+ b } t ˆ{ i } = \sum { i } x { i } t ˆ{ i − b }$ i s a root o f the polynomial$ P ( zt ˆ{ b } ) , $

and v i c e ver sa .

\hspace ∗{\ f i l l }As f o r the case $ b = 0 , $ we can f u r t h e r break i t down in to two ca s e s , one

\noindent in which $ a = p , $ the other in which $ a $ i s coprime to $ p. $ We t r e a t the former

\noindent case f i r s t . I f $ x = \sum { i } x { i } t ˆ{ i }$ i s a root o f the polynomial$ P ( z ) = \sum c { j } z ˆ{ j }$ over

\noindent $ F { q } ( t ) , $ then $ x ˆ{ \prime } = \sum { i } x { \pi }t ˆ{ i } = \sum { i } x { i } t ˆ{ i / p }$ i s a root o f the polynomial

\ [ \sum c ˆ{ \sigma } { j } z ˆ{ pj }\ ]

\noindent over $ F { q } ( t ) . $ Converse ly , i f $ x ˆ{ \prime }$ i s a root o f the polynomial$ Q ( z ) = \sum d { j } z ˆ{ j }$ over

$ F { q } ( t ) , $ then $ x $ i s a root o f the polynomial

\begin { a l i g n ∗}\sum ( d ˆ{ p } { j } ) ˆ{ \sigma ˆ{ − 1 }} z ˆ{ j }\\ over F { q } (

t ) .\end{ a l i g n ∗}

Now suppose that $ b = 0 $ and $ a $ i s coprime to $ p . $ Let $ \tau: F { q } ( ( t ˆ{ Q } ) ) \rightarrow F { q } ( ( t ˆ{ Q } )) $

denote the automorphism $ \sum x { i } t ˆ{ i } \mapsto \sum x { i } t ˆ{ a i }; $ then $ \tau $ a l s o ac t s on $ F { q } ( t ) . $ I f

$ x = \sum { i } x { i } t ˆ{ i }$ i s a root o f the polynomial $ P (z ) = \sum c { j } z ˆ{ j }$ \quad over $ F { q } ( t ) , $ then

$ x ˆ{ \prime } = \sum { i } x { a i } t ˆ{ i } = \sum { i } x { i } t ˆ{ i/ a }$ i s a root o f the polynomial

\ [ \sum c ˆ{ \tau }ˆ{ j } { − }ˆ{ 1 } z ˆ{ j }\ ]

\noindent over $ F { q } ( t ˆ{ 1 / a } ) ; $ s i n c e $ F { q } ( t ˆ{ 1/ a } ) $ i s f i n i t e d imens iona l over $ F { q } ( t ) , x ˆ{ \prime }$i s a l g e b r a i cover $ F { q } ( t ) $ \quad by Lemma 3 . 2 . 4 . \quad Converse ly , i f $ x ˆ{ \prime }$

i s a root o f the polynomial$ Q ( z ) = \sum d { j } z ˆ{ j }$ over $ F { q } ( t ) , $

then $ x $ i s a root o f the polynomial

\begin { a l i g n ∗}\sum c ˆ{ \tau }{ j } z ˆ{ j }\\ over F { q } ( t ) .\end{ a l i g n ∗}

We have now proved the statement o f the lemma in case $ a = 1 $ and $ b $ i sa r b i t r a r y , in case $ a = p $ and $ b = 0 , $ and in case $ a $ i s coprime to

$ p $ and $ b = 0 . $As noted above , the se three ca s e s toge the r imply the d e s i r e d r e s u l t $ . \ square $

396 .. Kiran S period KedlayaLet .. sigma in .. Gal open parenthesis F sub q slash F sub p closing parenthesis .. denote .. the p hyphen power

Frobenius .. map semicolon .. note that .. theconvention we just introduced means that x to the power of p = x to the power of sigma if x in F sub q comma

but not ifx in F sub q open parenthesis t closing parenthesis or x in F sub q open parenthesis open parenthesis t to the

power of Q closing parenthesis closing parenthesis periodLemma 4 period 2 period 2 period .... Let a comma b be integers with a greater 0 period .... Then sum sub i x

sub i t to the power of i in F sub q open parenthesis open parenthesis t to the power of Q closing parenthesis closingparenthesis .... is

algebraic over F sub q open parenthesis t closing parenthesis .. if and only if sum sub i x sub ai plus b t to thepower of i is algebraic over F sub q open parenthesis t closing parenthesis period

Proof period .... It suffices to prove the result in the case a = 1 and in the case b = 0 commaas the general case follows by applying these two in succession period The casea = 1 i s straightforward : if x = sum sub i x sub i t to the power of i i s a root of the polynomial P open

parenthesis z closing parenthesis overF sub q open parenthesis t closing parenthesis comma then x to the power of prime = sum sub i x sub i plus b t

to the power of i = sum sub i x sub i t to the power of i minus b i s a root of the polynomial P open parenthesis ztto the power of b closing parenthesis comma

and vice versa periodAs for the case b = 0 comma we can further break it down into two cases comma onein which a = p comma the other in which a i s coprime to p period We treat the formercase first period If x = sum sub i x sub i t to the power of i is a root of the polynomial P open parenthesis z

closing parenthesis = sum c sub j z to the power of j overF sub q open parenthesis t closing parenthesis comma then x to the power of prime = sum sub i x sub pi t to

the power of i = sum sub i x sub i t to the power of i slash p i s a root of the polynomialsum c sub j to the power of sigma z to the power of pjover F sub q open parenthesis t closing parenthesis period Conversely comma if x to the power of prime is a root

of the polynomial Q open parenthesis z closing parenthesis = sum d sub j z to the power of j overF sub q open parenthesis t closing parenthesis comma then x i s a root of the polynomialsum open parenthesis d sub j to the power of p closing parenthesis to the power of sigma to the power of minus

1 z to the power of j over F sub q open parenthesis t closing parenthesis periodNow suppose that b = 0 and a is coprime to p period Let tau : F sub q open parenthesis open parenthesis t to

the power of Q closing parenthesis closing parenthesis right arrow F sub q open parenthesis open parenthesis t tothe power of Q closing parenthesis closing parenthesis

denote the automorphism sum x sub i t to the power of i mapsto-arrowright sum x sub i t to the power of aisemicolon then tau also acts on F sub q open parenthesis t closing parenthesis period If

x = sum sub i x sub i t to the power of i is a root of the polynomial P open parenthesis z closing parenthesis =sum c sub j z to the power of j .. over F sub q open parenthesis t closing parenthesis comma then

x to the power of prime = sum sub i x sub ai t to the power of i = sum sub i x sub i t to the power of i slash a is a root of the polynomial

sum c to the power of tau from j to minus to the power of 1 z to the power of jover F sub q open parenthesis t to the power of 1 slash a closing parenthesis semicolon since F sub q open

parenthesis t to the power of 1 slash a closing parenthesis is finite dimensional over F sub q open parenthesis tclosing parenthesis comma x to the power of prime is algebraic

over F sub q open parenthesis t closing parenthesis .. by Lemma 3 period 2 period 4 period .. Conversely commaif x to the power of prime i s a root of the polynomial

Q open parenthesis z closing parenthesis = sum d sub j z to the power of j over F sub q open parenthesis t closingparenthesis comma then x i s a root of the polynomial

sum c to the power of tau j z to the power of j over F sub q open parenthesis t closing parenthesis periodWe have now proved the statement of the lemma in case a = 1 and b i sarbitrary comma in case a = p and b = 0 comma and in case a i s coprime to p and b = 0 periodAs noted above comma these three cases together imply the desired result period square

396 Kiran S . Kedlaya Let σ ∈ Gal (Fq/Fp) denote the p− powerFrobenius map ; note that theconvention we just introduced means that xp = xσ if x ∈ Fq, but not if

x ∈ Fq(t)orx ∈ Fq((tQ)).

Lemma 4 . 2 . 2 . Let a, b be integers with a > 0. Then∑i xit

i ∈ Fq((tQ)) isalgebraic over Fq(t) if and only if

∑i xai+bt

i is algebraic over Fq(t).Proof . It suffices to prove the result in the case a = 1 and in the caseb = 0,as the general case follows by applying these two in succession . The casea = 1 i s straightforward : if x =

∑i xit

i i s a root of the polynomial P (z)overFq(t), then x′ =

∑i xi+bt

i =∑i xit

i−b i s a root of the polynomial P (ztb), andvice versa .

As for the case b = 0, we can further break it down into two cases , onein which a = p, the other in which a i s coprime to p. We treat the formercase first . If x =

∑i xit

i is a root of the polynomial P (z) =∑cjz

j overFq(t), then x′ =

∑i xπt

i =∑i xit

i/p i s a root of the polynomial∑cσj z

pj

over Fq(t). Conversely , if x′ is a root of the polynomial Q(z) =∑djz

j overFq(t), then x i s a root of the polynomial

∑(dpj )

σ−1

zj

overFq(t).

Now suppose that b = 0 and a is coprime to p. Let τ : Fq((tQ))→ Fq((tQ))denote the automorphism

∑xit

i 7→∑xit

ai; then τ also acts on Fq(t). Ifx =

∑i xit

i is a root of the polynomial P (z) =∑cjz

j over Fq(t), thenx′ =

∑i xait

i =∑i xit

i/a i s a root of the polynomial∑cτ j−

1zj

over Fq(t1/a); since Fq(t1/a) is finite dimensional over Fq(t), x′ is algebraic overFq(t) by Lemma 3 . 2 . 4 . Conversely , if x′ i s a root of the polynomialQ(z) =

∑djz

j over Fq(t), then x i s a root of the polynomial

∑cτ jzj

overFq(t).

We have now proved the statement of the lemma in case a = 1 and b i sarbitrary , in case a = p and b = 0, and in case a i s coprime to p and b = 0.As noted above , these three cases together imply the desired result . �


\centerline {5 . \quad Proof o f the main theorem : ab s t r a c t approach }

In t h i s chapter , we g ive a proo f o f Theorem 4 . 1 . 3 . While the proo f in the‘ ‘ automatic i m p l i e s a l g e b r a i c ’ ’ \quad d i r e c t i o n i s f a i r l y e x p l i c i t , the proo f in ther e v e r s e d i r e c t i o n r e l i e s on the r e s u l t s o f [ 1 1 ] , and hence i s f a i r l y conceptua l .We w i l l g ive a more e x p l i c i t approach to the r e v e r s e d i r e c t i o n in the next

\noindent chapter .

\noindent 5 . 1 . \quad Automatic i m p l i e s a l g e b r a i c . \quad In t h i s s e c t i o n , we e s t a b l i s h the ‘ ‘ au −tomatic i m p l i e s a l g e b r a i c ’ ’ d i r e c t i o n o f Theorem 4 . 1 . 3 . The proo f i s a s l i g h tmod i f i c a t i on o f the usua l argument used to prove the corre spond ing d i −r e c t i o n o f C h r i s t o l ’ s theorem ( as in [ 2 , Theorem 1 2 . 2 . 5 ] ) . \quad ( Note that t h i sd i r e c t i o n o f Theorem 4 . 1 . 3 w i l l be invoked in both proo f s o f the r e v e r s ed i r e c t i o n . )

\noindent Lemma 5 . 1 . 1 . \ h f i l l Let $ p $ be a prime number , \ h f i l l and l e t $ S $\ h f i l l be a $ p − $ r e g u l a r subset

\noindent o f $ S { p } . $ \quad Then $ \sum { i \ in S } t ˆ{ i } \ in F { p }\ l l b r a c k e t t ˆ{ Q } \ r rb ra cke t $ \quad i s a l g e b r a i c over $ F { p } ( t ). $

\noindent Proof . \quad Let $ L $ be the language o f s t r i n g s o f the form $ s (v ) $ f o r $ v \ in S , $ and l e t

$ M $ be a DFA which accept s $ L . $

For $ n $ a nonnegat ive i n t e g e r , l e t $ s ˆ{ \prime } ( n ) $ be the base $ p $expansion o f $ n $ minus

the f i n a l rad ix po int . For each pre rad ix s t a t e $ q \ in Q , $ l e t $ T { q }$be the s e t o f

\noindent nonnegat ive i n t e g e r s $ n $ such that $ \delta ˆ{ ∗ } ( q 0 , s ˆ{ \prime }( n ) ) = q , $ put $ f ( q ) = \sum { i \ in T { q }}t ˆ{ i } , $ and

\noindent l e t $ U { q }$ be the s e t o f p a i r s $ ( q ˆ{ \prime } , d ) \ inQ \times \{ 0 , . . . , p − 1 \} $ such that $ \delta (q ˆ{ \prime } , d ) = q . $

\noindent ( Note that t h i s f o r c e s $ q ˆ{ \prime }$ to be pre rad ix . ) Then i f $ q\not= q 0 , $ we have

\ [\ begin { a l i gned }\ l e f t . f ( q ) = \sum t ˆ{ d } f ( q ˆ{ \prime } )\ begin { a l i gned } &p \\

& , \end{ a l i gned }\ right .\\( q ˆ{ \prime } , d ) \ in U { q }\end{ a l i gned }\ ]

\noindent whereas i f $ q = q 0 , $ we have

\ [\ begin { a l i gned }\ l e f t . f ( q 0 ) = 1 + \sum t ˆ{ d } f ( q ˆ{ \prime } )\ begin { a l i gned } &p \\

& . \end{ a l i gned }\ right .\\( q ˆ{ \prime } , d ) \ in U { q }\end{ a l i gned }\ ]

\noindent By Lemma $ 3 . 3 . 5 , f ( q ) $ i s a l g e b r a i c over $ F { q }( t ) $ f o r each pre rad ix s t a t e $ q . $

\hspace ∗{\ f i l l }For $ x \ in S { p } \cap [ 0 , 1 ) , $ l e t $ s ˆ{ \prime\prime } ( x ) $ be the base $ p $ expansion o f $ x $ minus the i n i t i a l

\noindent rad ix po int . For each pos t rad ix s t a t e $ q \ in Q , $ l e t $ V { q }$be the s e t o f $ x \ in S { p } \cap [ 0 , 1 ) $

such that $ \delta ˆ{ ∗ } ( q , s ˆ{ \prime \prime } ( x ) ) $ i s a f i n a l s t a t e , and put$ g ( q ) = \sum { i \ in V { q }} t ˆ{ i } . $ Then i f $ q $ i s

non − f i n a l , we have

\ [\ begin { a l i gned } p − 1 \\g ( q ) ˆ{ p } = \sum t ˆ{ d } g ( \delta ( q , d ) )

, \\d = 0 \end{ a l i gned }\ ]

\noindent whereas i f $ q $ i s f i n a l , then

\ [\ begin { a l i gned } p − 1 \\g ( q ) ˆ{ p } = 1 + \sum t ˆ{ d } g ( \delta ( q , d

) ) . \\d = 0 \end{ a l i gned }\ ]

\noindent By Lemma $ 3 . 3 . 5 , g ( q ) $ i s a l g e b r a i c f o r each pos t rad ix s t a t e$ q . $

Finite automata and algebraic extensions of function fields .. 3975 period .. Proof of the main theorem : abstract approachIn this chapter comma we give a proof of Theorem 4 period 1 period 3 period While the proof in thequotedblleft automatic implies algebraic quotedblright .. direction i s fairly explicit comma the proof in thereverse direction relies on the results of open square bracket 1 1 closing square bracket comma and hence i s fairly

conceptual periodWe will give a more explicit approach to the reverse direction in the nextchapter period5 period 1 period .. Automatic implies algebraic period .. In this section comma we establish the quotedblleft

au hyphentomatic implies algebraic quotedblright direction of Theorem 4 period 1 period 3 period The proof is a slightmodification of the usual argument used to prove the corresponding di hyphenrection of Christol quoteright s theorem open parenthesis as in open square bracket 2 comma Theorem 1 2 period

2 period 5 closing square bracket closing parenthesis period .. open parenthesis Note that thisdirection of Theorem 4 period 1 period 3 will be invoked in both proofs of the reversedirection period closing parenthesisLemma 5 period 1 period 1 period .... Let p be a prime number comma .... and let S .... be a p hyphen regular

subsetof S sub p period .. Then sum sub i in S t to the power of i in F sub p llbracket t to the power of Q rrbracket ..

is algebraic over F sub p open parenthesis t closing parenthesis periodProof period .. Let L be the language of strings of the form s open parenthesis v closing parenthesis for v in S

comma and letM be a DFA which accepts L periodFor n a nonnegative integer comma let s to the power of prime open parenthesis n closing parenthesis be the base

p expansion of n minusthe final radix point period For each preradix state q in Q comma let T sub q be the set ofnonnegative integers n such that delta to the power of * open parenthesis q 0 comma s to the power of prime

open parenthesis n closing parenthesis closing parenthesis = q comma put f open parenthesis q closing parenthesis= sum sub i in T sub q t to the power of i comma and

let U sub q be the set of pairs open parenthesis q to the power of prime comma d closing parenthesis in Q timesopen brace 0 comma period period period comma p minus 1 closing brace such that delta open parenthesis q to thepower of prime comma d closing parenthesis = q period

open parenthesis Note that this forces q to the power of prime to be preradix period closing parenthesis Then ifq negationslash-equal q 0 comma we have

Line 1 f open parenthesis q closing parenthesis = sum t to the power of d f open parenthesis q to the power ofprime Case 1 p Case 2 comma Line 2 open parenthesis q to the power of prime comma d closing parenthesis in Usub q

whereas if q = q 0 comma we haveLine 1 f open parenthesis q 0 closing parenthesis = 1 plus sum t to the power of d f open parenthesis q to the

power of prime Case 1 p Case 2 period Line 2 open parenthesis q to the power of prime comma d closing parenthesisin U sub q

By Lemma 3 period 3 period 5 comma f open parenthesis q closing parenthesis is algebraic over F sub q openparenthesis t closing parenthesis for each preradix state q period

For x in S sub p cap open square bracket 0 comma 1 closing parenthesis comma let s to the power of prime primeopen parenthesis x closing parenthesis be the base p expansion of x minus the initial

radix point period For each postradix state q in Q comma let V sub q be the set of x in S sub p cap open squarebracket 0 comma 1 closing parenthesis

such that delta to the power of * open parenthesis q comma s to the power of prime prime open parenthesis xclosing parenthesis closing parenthesis is a final state comma and put g open parenthesis q closing parenthesis =sum sub i in V sub q t to the power of i period Then if q i s

non hyphen final comma we haveLine 1 p minus 1 Line 2 g open parenthesis q closing parenthesis to the power of p = sum t to the power of d g

open parenthesis delta open parenthesis q comma d closing parenthesis closing parenthesis comma Line 3 d = 0whereas if q i s final comma thenLine 1 p minus 1 Line 2 g open parenthesis q closing parenthesis to the power of p = 1 plus sum t to the power

of d g open parenthesis delta open parenthesis q comma d closing parenthesis closing parenthesis period Line 3 d =0

By Lemma 3 period 3 period 5 comma g open parenthesis q closing parenthesis is algebraic for each postradixstate q period


5 . Proof of the main theorem : abstract approachIn this chapter , we give a proof of Theorem 4 . 1 . 3 . While the proof in

the “ automatic implies algebraic ” direction i s fairly explicit , the proofin the reverse direction relies on the results of [ 1 1 ] , and hence i s fairlyconceptual . We will give a more explicit approach to the reverse directionin the nextchapter .5 . 1 . Automatic implies algebraic . In this section , weestablish the “ au - tomatic implies algebraic ” direction of Theorem 4 . 1 .3 . The proof is a slight modification of the usual argument used to provethe corresponding di - rection of Christol ’ s theorem ( as in [ 2 , Theorem1 2 . 2 . 5 ] ) . ( Note that this direction of Theorem 4 . 1 . 3 will beinvoked in both proofs of the reverse direction . )Lemma 5 . 1 . 1 . Let p be a prime number , and let S be a p−regular subsetof Sp. Then

∑i∈S t

i ∈ FptQ is algebraic over Fp(t).Proof . Let L be the language of strings of the form s(v) for v ∈ S, andlet M be a DFA which accepts L.

For n a nonnegative integer , let s′(n) be the base p expansion of n minusthe final radix point . For each preradix state q ∈ Q, let Tq be the set ofnonnegative integers n such that δ∗(q0, s′(n)) = q, put f(q) =

∑i∈Tq t

i, andlet Uq be the set of pairs (q′, d) ∈ Q× {0, ..., p− 1} such that δ(q′, d) = q.( Note that this forces q′ to be preradix . ) Then if q 6= q0, we have

f(q) =∑

tdf(q′)p

,

(q′, d) ∈ Uq

whereas if q = q0, we have

f(q0) = 1 +∑

tdf(q′)p

.

(q′, d) ∈ Uq

By Lemma 3.3.5, f(q) is algebraic over Fq(t) for each preradix state q.For x ∈ Sp ∩ [0, 1), let s′′(x) be the base p expansion of x minus the initial

radix point . For each postradix state q ∈ Q, let Vq be the set of x ∈ Sp∩ [0, 1)such that δ∗(q, s′′(x)) is a final state , and put g(q) =

∑i∈Vq t

i. Then if q i snon - final , we have

p− 1

g(q)p =∑

tdg(δ(q, d)),

d = 0

whereas if q i s final , then

p− 1

g(q)p = 1 +∑

tdg(δ(q, d)).

d = 0

By Lemma 3.3.5, g(q) is algebraic for each postradix state q.

\noindent 398 \quad Kiran S . KedlayaF i n a l l y , note that

\ [\ begin { a l i gned } \sum t ˆ{ i } = \sum f ( q ) g ( q ˆ{ \prime }) , \\

i \ in S q , q ˆ{ \prime }\end{ a l i gned }\ ]

\noindent the sum running over pre rad ix $ q $ and pos t rad ix $ q ˆ{ \prime } . $This sum i s a l g e b r a i c

over $ F { q } ( t ) $ by Lemma 3 . 2 . 5 , as d e s i r e d $ . \ square $

\noindent Propos i t i on 5 . 1 . 2 . \ h f i l l Let $ \sum { i } x { i } t ˆ{ i } \ inF { q } ( ( t ˆ{ Q } ) ) $ \ h f i l l be a $ p − $ quas i − automatic gener −

\noindent a l i z e d Laurent s e r i e s . \quad Then $ \sum { i } x { i } t ˆ{ i }$ i s a l g e b r a i c over$ F { q } ( t ) . $

\noindent Proof . \quad Choose i n t e g e r s $ a , b $ as in D e f i n i t i o n 4 . 1 . 2 . For each$ \alpha \ in F { q } , $ l e t $ S { \alpha }$

be the s e t o f $ j \ in Q $ such that $ x { ( j − b ) / a } = \alpha. $ Then each $ S { \alpha }$ i s $ p − $ r e g u l a r , so

\noindent Lemma 5 . 1 . 1 i m p l i e s that $ \sum { j \ in S { \alpha }} t ˆ{ j }$i s a l g e b r a i c over $ F { q } ( t ) . $ By Lemma 3 . 2 . 5 ,

\ [ \sum { i } x { a i + b } t ˆ{ i } = \sum { \alpha \ in F { q }} \alpha\ l e f t (\ begin { array }{ cc } \sum & t ˆ{ j }\\ j \ in S { \alpha }\end{ array }\ right ) \ ]

\noindent i s a l s o a l g e b r a i c over $ F { q } ( t ) ; $ by Lemma $ 4 . 2. 2 , \sum { i } x { i } t ˆ{ i }$ i s a l s o a l g e b r a i c over

\ [ F { q } ( t ) . \ square \ ]

\noindent 5 . 2 . \quad Algebra i c \quad i m p l i e s \quad automatic . \quad We next prove the \quad ‘ ‘ a l g e b r a i c im −p l i e s automatic ’ ’ d i r e c t i o n o f Theorem 4 . 1 . 3 . Unfortunate ly , the techn iqueso r i g i n a l l y used to prove C h r i s t o l ’ s theorem ( as in [ 2 , Chapter 1 2 ] ) do nots u f f i c e to g ive a proo f o f t h i s d i r e c t i o n . In t h i s s e c t i o n , we w i l l get aroundt h i s by us ing the c h a r a c t e r i z a t i o n o f the a l g e b r a i c c l o s u r e o f $ F { q } ( (

t ) ) $ with in

\noindent $ F { q } ( ( t ˆ{ Q } ) ) $ provided by [ 1 1 ] . This proo f thus i n h e r i t s the property o f [ 1 1 ] o f

\noindent being a b i t ab s t r a c t , as [ 1 1 ] uses some Galo i s theory and p r o p e r t i e s o f f i n i t eex t en s i on s o f f i e l d s in p o s i t i v e c h a r a c t e r i s t i c ( namely Artin − S c h r e i e r theory ,which comes from an argument in Galo i s cohomology ) . I t a l s o r e q u i r e s in −voking the \quad ‘ ‘ a l g e b r a i c i m p l i e s automatic ’ ’ \quad d i r e c t i o n o f C h r i s t o l ’ s theoremi t s e l f . We w i l l g ive a second , more computat iona l ly e x p l i c i t proo f o f t h i sd i r e c t i o n l a t e r ( Propos i t i on 7 . 3 . 4 ) .

\noindent D e f i n i t i o n 5 . 2 . 1 . \quad For $ c $ a nonnegat ive i n t e g e r , l e t $ T { c }$be the subset o f $ S { p }$

given by

\ [ T { c } = \{ n − b { 1 } p ˆ{ − 1 } − b { 2 } p ˆ{ − 2 } −\cdot \cdot \cdot : n \ in Z { > 0 } , b { i } \ in \{ 0 ,. . . , p − 1 \} , \sum b { i } \ leq c \} . \ ]

Then [ 1 1 , Theorem 1 5 ] g i v e s a c r i t e r i o n f o r a l g e b r a i c i t y o f a g e n e r a l i z e dpower s e r i e s not over the r a t i o n a l func t i on f i e l d $ F { q } ( t ) , $ but over the Laurents e r i e s f i e l d $ F { q } ( ( t ) ) . $ I t can be s ta t ed as f o l l o w s .

\noindent Propos i t i on 5 . 2 . 2 . \quad For $ x = \sum { i } x { i } t ˆ{ i }\ in F { q } ( ( t ˆ{ Q } ) ) , x $ i s a l g e b r a i c over $ F { q } (( t ) ) $

i f and only i f the f o l l o w i n g c o n d i t i o n s hold .

\hspace ∗{\ f i l l }( a ) \quad There e x i s t i n t e g e r s $ a , b , c \geq 0 $ such that the support o f$ \sum { i } x { ( i − b ) / a } t ˆ{ i }$

\centerline{ i s conta ined in $ T { c } . $ }

398 .. Kiran S period KedlayaFinally comma note thatLine 1 sum t to the power of i = sum f open parenthesis q closing parenthesis g open parenthesis q to the power

of prime closing parenthesis comma Line 2 i in S q comma q to the power of primethe sum running over preradix q and postradix q to the power of prime period This sum i s algebraicover F sub q open parenthesis t closing parenthesis by Lemma 3 period 2 period 5 comma as desired period

squareProposition 5 period 1 period 2 period .... Let sum sub i x sub i t to the power of i in F sub q open parenthesis

open parenthesis t to the power of Q closing parenthesis closing parenthesis .... be a p hyphen quasi hyphen automaticgener hyphen

alized Laurent s eries period .. Then sum sub i x sub i t to the power of i is algebraic over F sub q open parenthesist closing parenthesis period

Proof period .. Choose integers a comma b as in Definition 4 period 1 period 2 period For each alpha in F sub qcomma let S sub alpha

be the set of j in Q such that x sub open parenthesis j minus b closing parenthesis slash a = alpha period Theneach S sub alpha is p hyphen regular comma so

Lemma 5 period 1 period 1 implies that sum sub j in S sub alpha t to the power of j is algebraic over F sub qopen parenthesis t closing parenthesis period By Lemma 3 period 2 period 5 comma

sum i x sub ai plus b t to the power of i = sum alpha in F sub q alpha Row 1 sum t to the power of j Row 2 j inS sub alpha .

i s also algebraic over F sub q open parenthesis t closing parenthesis semicolon by Lemma 4 period 2 period 2comma sum sub i x sub i t to the power of i is also algebraic over

F sub q open parenthesis t closing parenthesis period square5 period 2 period .. Algebraic .. implies .. automatic period .. We next prove the .. quotedblleft algebraic im

hyphenplies automatic quotedblright direction of Theorem 4 period 1 period 3 period Unfortunately comma the tech-

niquesoriginally used to prove Christol quoteright s theorem open parenthesis as in open square bracket 2 comma

Chapter 1 2 closing square bracket closing parenthesis do notsuffice to give a proof of this direction period In this section comma we will get aroundthis by using the characterization of the algebraic closure of F sub q open parenthesis open parenthesis t closing

parenthesis closing parenthesis withinF sub q open parenthesis open parenthesis t to the power of Q closing parenthesis closing parenthesis provided by

open square bracket 1 1 closing square bracket period This proof thus inherits the property of open square bracket1 1 closing square bracket of

being a bit abstract comma as open square bracket 1 1 closing square bracket uses some Galois theory andproperties of finite

extensions of fields in positive characteristic open parenthesis namely Artin hyphen Schreier theory commawhich comes from an argument in Galois cohomology closing parenthesis period It also requires in hyphenvoking the .. quotedblleft algebraic implies automatic quotedblright .. direction of Christol quoteright s theoremit self period We will give a second comma more computationally explicit proof of thisdirection later open parenthesis Proposition 7 period 3 period 4 closing parenthesis periodDefinition 5 period 2 period 1 period .. For c a nonnegative integer comma let T sub c be the subset of S sub pgiven byT sub c = open brace n minus b sub 1 p to the power of minus 1 minus b sub 2 p to the power of minus 2 minus

times times times : n in Z sub greater 0 comma b sub i in open brace 0 comma period period period comma p minus1 closing brace comma sum b sub i less or equal c closing brace period

Then open square bracket 1 1 comma Theorem 1 5 closing square bracket gives a criterion for algebraicity of ageneralized

power series not over the rational function field F sub q open parenthesis t closing parenthesis comma but overthe Laurent

series field F sub q open parenthesis open parenthesis t closing parenthesis closing parenthesis period It can bestated as follows period

Proposition 5 period 2 period 2 period .. For x = sum sub i x sub i t to the power of i in F sub q open parenthesisopen parenthesis t to the power of Q closing parenthesis closing parenthesis comma x is algebraic over F sub q openparenthesis open parenthesis t closing parenthesis closing parenthesis

if and only if the following conditions hold periodopen parenthesis a closing parenthesis .. There exist integers a comma b comma c greater equal 0 such that the

support of sum sub i x sub open parenthesis i minus b closing parenthesis slash a t to the power of iis contained in T sub c period

398 Kiran S . Kedlaya Finally , note that∑ti =

∑f(q)g(q′),

i ∈ S q, q′

the sum running over preradix q and postradix q′. This sum i s algebraicover Fq(t) by Lemma 3 . 2 . 5 , as desired . �Proposition 5 . 1 . 2 . Let

∑i xit

i ∈ Fq((tQ)) be a p− quasi -automatic gener -alized Laurent s eries . Then

∑i xit

i is algebraic over Fq(t).Proof . Choose integers a, b as in Definition 4 . 1 . 2 . For each α ∈ Fq,let Sα be the set of j ∈ Q such that x(j−b)/a = α. Then each Sα is p− regular, soLemma 5 . 1 . 1 implies that

∑j∈Sα t

j is algebraic over Fq(t). By Lemma 3 .2 . 5 , ∑

i

xai+bti =

∑α∈Fq

α

( ∑tj

j ∈ Sα

)i s also algebraic over Fq(t); by Lemma 4.2.2,

∑i xit

i is also algebraic over

Fq(t). �

5 . 2 . Algebraic implies automatic . We next provethe “ algebraic im - plies automatic ” direction of Theorem 4 . 1 . 3 .Unfortunately , the techniques originally used to prove Christol ’ s theorem( as in [ 2 , Chapter 1 2 ] ) do not suffice to give a proof of this direction .In this section , we will get around this by using the characterization of thealgebraic closure of Fq((t)) withinFq((tQ)) provided by [ 1 1 ] . This proof thus inherits the property of [ 1 1 ]ofbeing a bit abstract , as [ 1 1 ] uses some Galois theory and properties offinite extensions of fields in positive characteristic ( namely Artin - Schreiertheory , which comes from an argument in Galois cohomology ) . It alsorequires in - voking the “ algebraic implies automatic ” direction ofChristol ’ s theorem it self . We will give a second , more computationallyexplicit proof of this direction later ( Proposition 7 . 3 . 4 ) .Definition 5 . 2 . 1 . For c a nonnegative integer , let Tc be thesubset of Sp given by

Tc = {n− b1p−1 − b2p−2 − · · · : n ∈ Z>0, bi ∈ {0, ..., p− 1},∑

bi ≤ c}.

Then [ 1 1 , Theorem 1 5 ] gives a criterion for algebraicity of a generalizedpower series not over the rational function field Fq(t), but over the Laurentseries field Fq((t)). It can be stated as follows .Proposition 5 . 2 . 2 . For x =

∑i xit

i ∈ Fq((tQ)), x is algebraic overFq((t)) if and only if the following conditions hold .

( a ) There exist integers a, b, c ≥ 0 such that the support of∑i x(i−b)/at

i

is contained in Tc.

\noindent F i n i t e automata and a l g e b r a i c ex t en s i on s o f f unc t i on f i e l d s \quad 399( b ) \quad For some $ a , b , c $ \quad as in \quad ( a ) , \quad the re \quad e x i s t p o s i t i v e i n t e g e r s

$ M $ \quad and $ N $

\centerline{ such that every s equence $ \{ c { n } \} ˆ{ \ infty } { n = 0 }$\quad o f the form }

\begin { a l i g n ∗}\ tag ∗{$ ( 5 . 2 . 3 ) $} c { n } = x { ( m − b − b { 1 }p ˆ{ − 1 } − \cdot \cdot \cdot − b { j − 1 } p ˆ{ − j + 1 }− p ˆ{ − n } ( b { j } p ˆ{ − j } + \cdot \cdot \cdot } ) ) /a ˆ{ , }\end{ a l i g n ∗}

with $ j $ a nonnegat ive i n t e g e r $ , m $ a p o s i t i v e i n t e g e r , and $ b { i }\ in \{ 0 , . . . , p − $

1 \} such that $ \sum b { i } \ leq c , $ becomes even tua l l y p e r i o d i c with per iod l ength

\centerline{ d i v i d i n g $ N $ a $ f−t $ er at most $ M $ terms . }

\noindent Moreover , in t h i s case , \quad ( b ) ho lds f o r any $ a , b , c $as in ( a ) .

\centerline{Beware that i t i s p o s s i b l e to choose $ a , b $ so that the support o f$ \sum x { ( i − b ) / a } t ˆ{ i }$ }

\noindent i s conta ined in $ S { p }$ and yet not have ( a ) \quad s a t i s f i e d f o r any cho i c e o f$ c . $ For

\noindent example , the support o f $ x = \sum ˆ{ \ infty } { i = 0 } t ˆ{ (1 − p ˆ{ − i } ) / ( p − 1 ) }$ i s conta ined in $ S { p }$ and in

\noindent $ p \ f r a c { 1 }{ − 1 } T { 1 } , $ but i s not conta ined in $ T { c }$f o r any $ c . $

\centerline{We f i r s t t r e a t a s p e c i a l case o f the \quad ‘ ‘ a l g e b r a i c i m p l i e s automatic ’ ’ \quad imp l i − }

\noindent ca t i on which i s orthogona l to C h r i s t o l ’ s theorem .

\noindent Lemma 5 . 2 . 4 . \quad Suppose that $ x = \sum { i } x { i } t ˆ{ i }\ in F { q } ( ( t ˆ{ Q } ) ) $ has support in $ ( 0 , 1 ] \cap $

$ T { c }$ f o r s ome \quad nonnegat ive \quad i n t e g e r $ c , $ \quad and that$ x $ \quad i s \quad a l g e b r a i c \quad over $ F { q } ( ( t ) ) . $

\begin { a l i g n ∗}Then :\end{ a l i g n ∗}

\centerline {( a $ ) x $ i s $ p − $ automatic ; }

\centerline {( b $ ) x $ i s a l g e b r a i c over $ F { q } ( t ) ; $ }

\centerline {( c $ ) x $ l i e s in a f i n i t e s e t determined by $ q $ and $ c . $}

\noindent Proof . \quad Note that ( b ) f o l l o w s from ( a ) by v i r t u e o f Propos i t i on 5 . 1 . 2 , so i t

\noindent s u f f i c e s to prove ( a ) and ( c ) . The c r i t e r i o n from Propos i t i on 5 . 2 . 2 a p p l i e swith $ a = 1 , b = 0 $ and the g iven value o f $ c , $ so we have that every sequence

$ \{ c { n } \} ˆ{ \ infty } { n = 0 }$ o f the form

\begin { a l i g n ∗}\ tag ∗{$ ( 5 . 2 . 5 ) $} c { n } = x { 1 − b { 1 } p ˆ{ − 1 }− \cdot \cdot \cdot − b { j − 1 } p ˆ{ − j + 1 } − p ˆ{ −n } ( b { j } p ˆ{ − j } + \cdot \cdot \cdot } ) ,\end{ a l i g n ∗}

\noindent with $ b { i } \ in \{ 0 , . . . , p − 1 \} $ such that$ \sum b { i } \ leq c , $ becomes even tua l l y p e r i o d i c

with per iod l ength d i v i d i n g $ N $ a f t e r at most $ M $ terms .

\centerline{Def ine an equ iva l ence r e l a t i o n on $ S { p }$ \quad as f o l l o w s . \quad Declare two elements }

\noindent o f $ S { p }$ to be equ iva l en t i f one can obta in the base $ b $ expansion o f one from

\noindent the base \quad $ b $ expansion o f the other by repea t ing the f o l l o w i n g opera t i on :r e p l a c e a conse cu t i v e s t r i n g o f $ M + u + vN $ z e r o e s by a conse cu t i v e s t r i n go f $ M + u + wN $ z e r o e s , where $ u , v , w $ may be any nonnegat ive i n t e g e r s . Thec r i t e r i o n o f Propos i t i on 5 . 2 . 2 then a s s e r t s that i f $ i , j \ in S { p }$

s a t i s f y $ i \sim j , $

\noindent then \quad $ x { 1 − i } = x { 1 − j } ; $ \quad a l s o , \quad the equ iva l ence r e l a t i o n \quad i s \quad c l e a r l y \quad s t a b l e \quad under

\noindent concatenat ion with a f i x e d p o s t s c r i p t .

\noindent Under \quad t h i s \quad equ iva l ence \quad r e l a t i o n , \quad each \quad equ iva l ence \quad c l a s s \quad has \quad a \quad uniques h o r t e s t element , \quad namely the one in which no nonzero d i g i t in the base$ b $ expansion i s preceded by $ M + N $ z e r o e s . On one hand , t h i s means that$ x $ i s determined by f i n i t e l y many c o e f f i c i e n t s , so ( c ) f o l l o w s . On the other

\noindent hand , \quad by the Myhi l l − Nerode theorem \quad ( Lemma 2 . 1 . 7 ) , \quad i t \quad f o l l o w s that the

Finite automata and algebraic extensions of function fields .. 399open parenthesis b closing parenthesis .. For some a comma b comma c .. as in .. open parenthesis a closing

parenthesis comma .. there .. exist positive integers M .. and Nsuch that every s equence open brace c sub n closing brace sub n = 0 to the power of infinity .. of the formEquation: open parenthesis 5 period 2 period 3 closing parenthesis .. c sub n = x sub open parenthesis m minus

b minus b sub 1 p to the power of minus 1 minus times times times minus b sub j minus 1 p to the power of minusj plus 1 minus p to the power of minus n open parenthesis b sub j p to the power of minus j plus times times timesclosing parenthesis closing parenthesis slash a to the power of comma

with j a nonnegative integer comma m a positive integer comma and b sub i in open brace 0 comma period periodperiod comma p minus

1 closing brace such that sum b sub i less or equal c comma becomes eventually periodic with period lengthdividing N a f-t er at most M terms periodMoreover comma in this case comma .. open parenthesis b closing parenthesis holds for any a comma b comma

c as in open parenthesis a closing parenthesis periodBeware that it i s possible to choose a comma b so that the support of sum x sub open parenthesis i minus b

closing parenthesis slash a t to the power of ii s contained in S sub p and yet not have open parenthesis a closing parenthesis .. satisfied for any choice of c

period Forexample comma the support of x = sum sub i = 0 to the power of infinity t to the power of open parenthesis 1

minus p to the power of minus i closing parenthesis slash open parenthesis p minus 1 closing parenthesis is containedin S sub p and in

p 1 divided by minus 1 T sub 1 comma but i s not contained in T sub c for any c periodWe first treat a special case of the .. quotedblleft algebraic implies automatic quotedblright .. impli hyphencation which is orthogonal to Christol quoteright s theorem periodLemma 5 period 2 period 4 period .. Suppose that x = sum sub i x sub i t to the power of i in F sub q

open parenthesis open parenthesis t to the power of Q closing parenthesis closing parenthesis has support in openparenthesis 0 comma 1 closing square bracket cap

T sub c for s ome .. nonnegative .. integer c comma .. and that x .. is .. algebraic .. over F sub q openparenthesis open parenthesis t closing parenthesis closing parenthesis period

Then :open parenthesis a closing parenthesis x is p hyphen automatic semicolonopen parenthesis b closing parenthesis x is algebraic over F sub q open parenthesis t closing parenthesis semicolonopen parenthesis c closing parenthesis x li es in a finite set determined by q and c periodProof period .. Note that open parenthesis b closing parenthesis follows from open parenthesis a closing paren-

thesis by virtue of Proposition 5 period 1 period 2 comma so itsuffices to prove open parenthesis a closing parenthesis and open parenthesis c closing parenthesis period The

criterion from Proposition 5 period 2 period 2 applieswith a = 1 comma b = 0 and the given value of c comma so we have that every sequenceopen brace c sub n closing brace sub n = 0 to the power of infinity of the formEquation: open parenthesis 5 period 2 period 5 closing parenthesis .. c sub n = x sub 1 minus b sub 1 p to

the power of minus 1 minus times times times minus b sub j minus 1 p to the power of minus j plus 1 minus p tothe power of minus n open parenthesis b sub j p to the power of minus j plus times times times closing parenthesiscomma

with b sub i in open brace 0 comma period period period comma p minus 1 closing brace such that sum b sub iless or equal c comma becomes eventually periodic

with period length dividing N after at most M terms periodDefine an equivalence relation on S sub p .. as follows period .. Declare two elementsof S sub p to be equivalent if one can obtain the base b expansion of one fromthe base .. b expansion of the other by repeating the following operation :replace a consecutive string of M plus u plus vN zeroes by a consecutive stringof M plus u plus wN zeroes comma where u comma v comma w may be any nonnegative integers period Thecriterion of Proposition 5 period 2 period 2 then asserts that if i comma j in S sub p satisfy i thicksim j commathen .. x sub 1 minus i = x sub 1 minus j semicolon .. also comma .. the equivalence relation .. is .. clearly ..

stable .. underconcatenation with a fixed postscript periodUnder .. this .. equivalence .. relation comma .. each .. equivalence .. class .. has .. a .. uniqueshortest element comma .. namely the one in which no nonzero digit in the baseb expansion i s preceded by M plus N zeroes period On one hand comma this means thatx i s determined by finitely many coefficients comma so open parenthesis c closing parenthesis follows period On

the otherhand comma .. by the Myhill hyphen Nerode theorem .. open parenthesis Lemma 2 period 1 period 7 closing

parenthesis comma .. it .. follows that the

Finite automata and algebraic extensions of function fields 399 ( b ) For some a, b, cas in ( a ) , there exist positive integers M and N

such that every s equence {cn}∞n=0 of the form

cn = x(m−b−b1p−1−···−bj−1p−j+1−p−n(bjp−j+···))/a, (5.2.3)

with j a nonnegative integer ,m a positive integer , and bi ∈ {0, ..., p−1 } such that

∑bi ≤ c, becomes eventually periodic with period length

dividing N a f − t er at most M terms .Moreover , in this case , ( b ) holds for any a, b, c as in ( a ) .Beware that it i s possible to choose a, b so that the support of

∑x(i−b)/at

i

i s contained in Sp and yet not have ( a ) satisfied for any choice of c. Forexample , the support of x =

∑∞i=0 t

(1−p−i)/(p−1) is contained in Sp and inp 1−1T1, but i s not contained in Tc for any c.

We first treat a special case of the “ algebraic implies automatic ” impli -cation which is orthogonal to Christol ’ s theorem .Lemma 5 . 2 . 4 . Suppose that x =

∑i xit

i ∈ Fq((tQ)) has supportin (0, 1]∩ Tc for s ome nonnegative integer c, and that x isalgebraic over Fq((t)).

Then :

( a ) x is p− automatic ;( b ) x is algebraic over Fq(t);

( c ) x li es in a finite set determined by q and c.Proof . Note that ( b ) follows from ( a ) by virtue of Proposition 5 . 1. 2 , so itsuffices to prove ( a ) and ( c ) . The criterion from Proposition 5 . 2 . 2applies with a = 1, b = 0 and the given value of c, so we have that everysequence {cn}∞n=0 of the form

cn = x1−b1p−1−···−bj−1p−j+1−p−n(bjp−j+···), (5.2.5)

with bi ∈ {0, ..., p − 1} such that∑bi ≤ c, becomes eventually periodic

with period length dividing N after at most M terms .Define an equivalence relation on Sp as follows . Declare two elementsof Sp to be equivalent if one can obtain the base b expansion of one fromthe base b expansion of the other by repeating the following operation :replace a consecutive string of M + u+ vN zeroes by a consecutive string ofM + u + wN zeroes , where u, v, w may be any nonnegative integers . Thecriterion of Proposition 5 . 2 . 2 then asserts that if i, j ∈ Sp satisfyi ∼ j,then x1−i = x1−j ; also , the equivalence relation is clearlystable underconcatenation with a fixed postscript .Under this equivalence relation , each equivalence class hasa unique shortest element , namely the one in which no nonzero digitin the base b expansion i s preceded by M +N zeroes . On one hand , thismeans that x i s determined by finitely many coefficients , so ( c ) follows .On the other

hand , by the Myhill - Nerode theorem ( Lemma 2 . 1 . 7 ) , itfollows that the


function f : Sp → Fq given by f(i) = x1−i is p− automatic . ( More precisely ,Lemma 2 . 1 . 7 implies that the inverse image of each element of Fq underfi s p− regular , and hence f i s p− automatic . ) Since there i s an obvioustrans - ducer that perfoms the operation i 7→ 1 − i on the valid base bexpansions of elements of Sp ∩ (0, 1]( namely , transcribe the usual handcomputation ) ,x i s p− automatic , and ( a ) follows . �Lemma 5 . 2 . 6 . Suppose that x1, ..., xm ∈ Fq((tQ)) all satisfythe hypothe - s is of Lemma 5 . 2 . 4 for the same value of c, and thatx1, ..., xm are linearly dependent over Fq((t)). Then x1, ..., xm are alsolinearly dependent over

Fq.

Proof . If x1, ..., xm are linearly dependent over Fq((t)), then by clearingdenominators , we can find a nonzero linear relation among them of the formc1x1 + · · ·+cmxm = 0, where each ci i s in Fqt. Write ci =

∑∞j=0 ci,jt

j for ci,j ∈ Fq;we then have

0 =

j=0∑∞

m∑ci,jxii = 1

tj

in Fq((tQ)).However , the support of the quantity in parentheses i s containedin (j, j+1]; in particular , these supports are disjoint for different j. Thus forthe sum to be zero , the summand must be zero for each j; thati s ,

∑mi=1 ci,jxi = 0 for each j. The ci,j cannot all be zero or else

c1, ..., cm would have all been zero , so we obtain a nontrivial linear relationamongx1, ..., xm over Fq, as desired . �

We now establish the “ algebraic implies automatic ” implication ofThe -orem 4 . 1 . 3 .Proposition 5 . 2 . 7 . Let x =

∑xit

i ∈ Fq((tQ)) be ageneralized powers eries which is algebraic over Fq(t). Then x is p− quasi - automatic .Proof . Choose a, b, c as in Proposition 5 . 2 . 2 , and putyi = x(i−b)/a andy =

∑i yi

ti

, so that y is algebraic over Fq(t) ( by Lemma 4 . 2 . 2 ) andhassupport in Tc. Note that for any positive integer m, yq

m

is also algebraic overFq(t) and also has support in Tc. By Lemma 3 . 3 . 4 , wecan find a polynomial P (z) =

∑mi=0 ciz

qi over Fq(t) such that P (y− y0) = 0.We may assume without loss of generality that cm 6= 0, and that cl = 1,where l i s the smallest nonnegative integer for which cl 6= 0.

Let V be the set of elements of Fq((tQ)) which satisfy the hypothesesof Lemma 5 . 2 . 4 ; then V is a finite set which i s a vector spaceover Fq,


\noindent f unc t i on $ f : S { p } \rightarrow F { q }$ given by $ f (i ) = x { 1 − i }$ i s $ p − $ automatic . ( More p r e c i s e l y ,Lemma 2 . 1 . 7 i m p l i e s that the i n v e r s e image o f each element o f $ F { q }$ under

$ f $

\noindent i s $ p − $ r e g u l a r , and hence $ f $ i s $ p − $ automatic . ) S ince the re i s an obvious t rans −ducer that perfoms the operat i on $ i \mapsto 1 − i $ on the v a l i d base $ b $

expans ionso f e lements o f $ S { p } \cap ( 0 , 1 ] ( $ namely , t r a n s c r i b e the usua l hand computation ) ,

\noindent $ x $ i s $ p − $ automatic , and ( a ) f o l l o w s $ . \ square $

\noindent Lemma 5 . 2 . 6 . \quad Suppose that $ x { 1 } , . . . , x { m }\ in F { q } ( ( t ˆ{ Q } ) ) $ \quad a l l s a t i s f y the hypothe −

s i s o f Lemma 5 . 2 . 4 f o r the same value o f $ c , $ and that $ x { 1 } ,. . . , x { m }$ are l i n e a r l ydependent over $ F { q } ( ( t ) ) . $ \quad Then $ x { 1 } , .

. . , x { m }$ \quad are a l s o l i n e a r l y dependent over

\begin { a l i g n ∗}F { q } .\end{ a l i g n ∗}

\noindent Proof . \quad I f $ x { 1 } , . . . , x { m }$ are l i n e a r l y dependent over$ F { q } ( ( t ) ) , $ then by c l e a r i n g

denominators , we can f i n d a nonzero l i n e a r r e l a t i o n among them of the form$ c { 1 } x { 1 } + \cdot \cdot \cdot + c { m } x { m } = 0 , $

where each $ c { i }$ i s in $ F { q } \ l l b r a c k e t t \ r rb ra cke t . $ Write$ c { i } = \sum ˆ{ \ infty } { j = 0 } c { i , j } t ˆ{ j }$ f o r

$ c { i , j } \ in F { q } ; $ we then have

\ [ 0 = \sum ˆ{ j = 0 } { \ infty } \ l e f t (\ begin { array }{ c} m \\ \sum c { i, j } x { i }\\ i = 1 \end{ array }\ right ) t ˆ{ j }\ ]

\noindent in $ F { q } ( ( t ˆ{ Q } ) ) . $ However , the support o f the quant i ty in parenthese s i s conta ined

\noindent in $ ( j , j + 1 ] ; $ in p a r t i c u l a r , the se supports are d i s j o i n t f o r d i f f e r e n t$ j . $ Thus

f o r the \quad sum to be zero , \quad the summand must \quad be zero f o r each $ j; $ \quad that i s ,

$ \sum ˆ{ m } { i = 1 } c { i , j } x { i } = 0 $ f o r each $ j . $\quad The $ c { i , j }$ \quad cannot \quad a l l be zero or e l s e $ c { 1 } ,. . . , c { m }$would have a l l been zero , so we obta in a n o n t r i v i a l l i n e a r r e l a t i o n among

\noindent $ x { 1 } , . . . , x { m }$ over $ F { q } , $ as d e s i r e d$ . \ square $

\hspace ∗{\ f i l l }We now e s t a b l i s h the \quad ‘ ‘ a l g e b r a i c i m p l i e s automatic ’ ’ \quad i m p l i c a t i o n o f The −

\noindent orem 4 . 1 . 3 .

\noindent Propos i t i on \ h f i l l 5 . 2 . 7 . \ h f i l l Let $ x = \sum x { i } t ˆ{ i }\ in F { q } ( ( t ˆ{ Q } ) ) $ \ h f i l l be \ h f i l l a g e n e r a l i z e d power

\noindent s e r i e s which i s a l g e b r a i c over $ F { q } ( t ) . $ \quad Then $ x $i s $ p − $ quas i − automatic .

\noindent Proof . \ h f i l l Choose \ h f i l l $ a , b , c $ \ h f i l l as \ h f i l l in \ h f i l l Propos i t i on \ h f i l l 5 . 2 . 2 , \ h f i l l and put \ h f i l l$ y i = x { ( i − b ) / a }$ \ h f i l l and

\noindent $ y = \sum { i } y i ˆ{ t ˆ{ i }} { , }$ \ h f i l l so that $ y $ i s a l g e b r a i c over$ F { q } ( t ) ( $ by Lemma 4 . 2 . 2 ) \ h f i l l and has

\noindent support in $ T { c } . $ Note that f o r any p o s i t i v e i n t e g e r $ m ,y ˆ{ q ˆ{ m }}$ i s a l s o a l g e b r a i c

over $ F { q } ( t ) $ \quad and \quad a l s o has \quad support \quad in $ T { c }. $ \quad By Lemma \quad 3 . 3 . 4 , \quad we can f i n d \quad apolynomial $ P ( z ) = \sum ˆ{ m } { i = 0 } c { i } z ˆ{ q ˆ{ i }}$

over $ F { q } ( t ) $ such that $ P ( y − y 0 ) = 0 . $We may

assume without l o s s o f g e n e r a l i t y that $ c { m } \not= 0 , $ and that $ c { l }= 1 , $ where $ l $ i s

the s m a l l e s t nonnegat ive i n t e g e r f o r which $ c { l } \ne 0 . $

Let \quad $ V $ be the s e t o f e lements o f $ F { q } ( ( t ˆ{ Q } ) ) $ \quad which s a t i s f y the hypotheseso f Lemma 5 . 2 . 4 ; \quad then \quad $ V $ i s \quad a f i n i t e s e t which i s \quad a vec to r space over

$ F { q } , $

\noindent each o f whose e lements i s $ p − $ automatic and a l s o a l g e b r a i c over $ F { q }( t ) . $ Let

$ v { 1 } , . . . , v { r }$ \quad be \quad a b a s i s \quad o f $ V $\quad over \quad $ F { q } ; $ \quad by \quad Lemma \quad $ 5 . 2 . 6, v { 1 } , . . . , v { r }$ \quad are \quad a l s o

\noindent l i n e a r l y independent over $ F { q } ( ( t ) ) . $

400 .. Kiran S period Kedlayafunction f : S sub p right arrow F sub q given by f open parenthesis i closing parenthesis = x sub 1 minus i is p

hyphen automatic period open parenthesis More precisely commaLemma 2 period 1 period 7 implies that the inverse image of each element of F sub q under fi s p hyphen regular comma and hence f i s p hyphen automatic period closing parenthesis Since there i s an

obvious trans hyphenducer that perfoms the operation i arrowright-mapsto 1 minus i on the valid base b expansionsof elements of S sub p cap open parenthesis 0 comma 1 closing square bracket open parenthesis namely comma

transcribe the usual hand computation closing parenthesis commax i s p hyphen automatic comma and open parenthesis a closing parenthesis follows period squareLemma 5 period 2 period 6 period .. Suppose that x sub 1 comma period period period comma x sub m in F

sub q open parenthesis open parenthesis t to the power of Q closing parenthesis closing parenthesis .. all satisfy thehypothe hyphen

s is of Lemma 5 period 2 period 4 for the same value of c comma and that x sub 1 comma period period periodcomma x sub m are linearly

dependent over F sub q open parenthesis open parenthesis t closing parenthesis closing parenthesis period .. Thenx sub 1 comma period period period comma x sub m .. are also linearly dependent over

F sub q periodProof period .. If x sub 1 comma period period period comma x sub m are linearly dependent over F sub q open

parenthesis open parenthesis t closing parenthesis closing parenthesis comma then by clearingdenominators comma we can find a nonzero linear relation among them of the formc sub 1 x sub 1 plus times times times plus c sub m x sub m = 0 comma where each c sub i i s in F sub q llbracket

t rrbracket period Write c sub i = sum sub j = 0 to the power of infinity c sub i comma j t to the power of j forc sub i comma j in F sub q semicolon we then have0 = sum from j = 0 to infinity Row 1 m Row 2 sum c sub i comma j x sub i Row 3 i = 1 . t to the power of jin F sub q open parenthesis open parenthesis t to the power of Q closing parenthesis closing parenthesis period

However comma the support of the quantity in parentheses i s containedin open parenthesis j comma j plus 1 closing square bracket semicolon in particular comma these supports are

disjoint for different j period Thusfor the .. sum to be zero comma .. the summand must .. be zero for each j semicolon .. that i s commasum sub i = 1 to the power of m c sub i comma j x sub i = 0 for each j period .. The c sub i comma j .. cannot

.. all be zero or else c sub 1 comma period period period comma c sub mwould have all been zero comma so we obtain a nontrivial linear relation amongx sub 1 comma period period period comma x sub m over F sub q comma as desired period squareWe now establish the .. quotedblleft algebraic implies automatic quotedblright .. implication of The hyphenorem 4 period 1 period 3 periodProposition .... 5 period 2 period 7 period .... Let x = sum x sub i t to the power of i in F sub q open parenthesis

open parenthesis t to the power of Q closing parenthesis closing parenthesis .... be .... a generalized powers eries which is algebraic over F sub q open parenthesis t closing parenthesis period .. Then x is p hyphen quasi

hyphen automatic periodProof period .... Choose .... a comma b comma c .... as .... in .... Proposition .... 5 period 2 period 2 comma ....

and put .... y i = x sub open parenthesis i minus b closing parenthesis slash a .... andy = sum sub i y i to the power of t to the power of i sub comma .... so that y is algebraic over F sub q open

parenthesis t closing parenthesis open parenthesis by Lemma 4 period 2 period 2 closing parenthesis .... and hassupport in T sub c period Note that for any positive integer m comma y to the power of q to the power of m is

also algebraicover F sub q open parenthesis t closing parenthesis .. and .. also has .. support .. in T sub c period .. By Lemma

.. 3 period 3 period 4 comma .. we can find .. apolynomial P open parenthesis z closing parenthesis = sum sub i = 0 to the power of m c sub i z to the power

of q to the power of i over F sub q open parenthesis t closing parenthesis such that P open parenthesis y minus y 0closing parenthesis = 0 period We may

assume without loss of generality that c sub m negationslash-equal 0 comma and that c sub l = 1 comma wherel i s

the smallest nonnegative integer for which c sub l equal-negationslash 0 periodLet .. V be the set of elements of F sub q open parenthesis open parenthesis t to the power of Q closing parenthesis

closing parenthesis .. which satisfy the hypothesesof Lemma 5 period 2 period 4 semicolon .. then .. V is .. a finite set which i s .. a vector space over F sub q

commaeach of whose elements is p hyphen automatic and also algebraic over F sub q open parenthesis t closing parenthesis

period Letv sub 1 comma period period period comma v sub r .. be .. a basis .. of V .. over .. F sub q semicolon .. by ..

Lemma .. 5 period 2 period 6 comma v sub 1 comma period period period comma v sub r .. are .. alsolinearly independent over F sub q open parenthesis open parenthesis t closing parenthesis closing parenthesis

period

each of whose elements is p− automatic and also algebraic over Fq(t). Letv1, ..., vr be a basis of V over Fq; by Lemma 5.2.6, v1, ..., vrare alsolinearly independent over Fq((t)).


By the criterion of Proposition 5 . 2 . 2 , we can write y − y0 =∑∞j=0 vjt

j

with each vj ∈ V. In other words , y is an Fq((t))− linear combinationofelements of V. Likewise , for each positive integer m, (y− y0)q

m

is an Fq((t))−linear combination of elements of V. For i = l, ...,m, write (y − y0)q

i

=∑rj=1 ai,jvj with ai,j ∈ Fq((t)); then the ai,j are uniquely determined by

Lemma 5 . 2 . 6 .By the same reasoning , for j = 1, ..., r, we can write

r

vqh =∑

bh,jvj

h = 1

for some bh,j ∈ Fq[t]. This means that

ai,j =∑

bh,jaqi−1,h (i = l + 1, ...,m; j = 1, ..., r). (5.2.8)

h

Moreover , the equation

(y − y0)ql

= −cl+1((y − y0)ql

)q − · · · − cm((y − y0)qm−1

)q

,

which holds because P (y−y0) = 0 and cl = 1 by hypothesis, can be rewritten as

r m− 1 r∑al,jvj =

∑−ci+1(

∑ai,hvh)q

j = 1 i = l h = 1

m− 1 r r

=∑−ci+1

∑∑aqi,hbh,jvj .

i = l h = 1j = 1

Equating coefficients of vj yields a system of equations of the form

m− 1 r

al,j =∑∑

dg,haqg,h (5.2.9)

g = lh = 1

witheachdg,h ∈ Fq(t).

Combining ( 5 . 2 . 8 ) and ( 5 . 2 . 9 ) yields a system of equationswhich trans - lates into a matrix equation of the form described by Lemma3 . 3 . 5 , in whichthe matrix B described therein is the identity . By Lemma 3 . 3 . 5 , theai,j are algebraic over Fq(t) for i = l, ...,m.


\hspace ∗{\ f i l l }By the c r i t e r i o n o f Propos i t i on 5 . 2 . 2 , we can wr i t e $ y − y0 = \sum ˆ{ \ infty } { j = 0 } v { j } t ˆ{ j }$

\noindent with each $ v { j } \ in V . $ In other words $ , y $ i s an $ F { q }( ( t ) ) − $ l i n e a r combination o f

\noindent e lements o f $ V . $ Likewise , f o r each p o s i t i v e i n t e g e r $ m , (y − y 0 ) ˆ{ q ˆ{ m }}$ i s an $ F { q } ( ( t ) ) − $

l i n e a r combination o f e lements o f $ V . $ For $ i = l , . . . ,m , $ wr i t e $ ( y − y 0 ) ˆ{ q ˆ{ i }} = $

$ \sum ˆ{ r } { j = 1 } a { i , j } v { j }$ \quad with $ a { i ,j } \ in F { q } ( ( t ) ) ; $ then the $ a { i , j }$ \quad are unique ly determined by

\noindent Lemma 5 . 2 . 6 .

\centerline{By the same reason ing , f o r $ j = 1 , . . . , r , $ we can wr i t e }

\ [\ begin { a l i gned } r \\v ˆ{ q }{ h } = \sum b { h , j } v { j }\\h = 1 \end{ a l i gned }\ ]

\noindent f o r some $ b { h , j } \ in F { q } [ t ] . $ This means that

\begin { a l i g n ∗}\ tag ∗{$ ( 5 . 2 . 8 ) $} a { i , j } = \sum b { h , j } a ˆ{ q }{ i } { −1 , h } ( i = l + 1 , . . . , m ; j = 1 , .. . , r ) . \\ h\end{ a l i g n ∗}

\noindent Moreover , the equat ion

\ [\ l e f t . ( y − y 0 ) ˆ{ q ˆ{ l }} = − c { l + 1 } ( ( y −y 0 ) ˆ{ q ˆ{ l }} ) ˆ{ q } − \cdot \cdot \cdot − c { m } ( ( y− y 0 ) ˆ{ q ˆ{ m − 1 }} )\ begin { a l i gned } & q \\

& , \end{ a l i gned }\ right . \ ]

\noindent which ho lds \quad because \quad $ P ( y − y 0 ) = 0 $ \quad and \quad$ c { l } = 1 $ \quad by hypothes i s , \quad can \quad be

r e w r i t t e n as

\ [\ begin { a l i gned } r m − 1 r \\\sum a { l , j } v { j } = \sum − c { i + 1 } ( \sum a { i

, h } v { h } ) ˆ{ q }\\j = 1 i = l h = 1 \\m − 1 r r \\= \sum − c { i + 1 } \sum \sum a ˆ{ q } { i , h } b { h ,

j } v { j } . \\i = l h = 1 j = 1 \end{ a l i gned }\ ]

\noindent Equating c o e f f i c i e n t s o f $ v { j }$ y i e l d s a system o f equat ions o f the form

\begin { a l i g n ∗}m − 1 r \\\ tag ∗{$ ( 5 . 2 . 9 ) $} a { l , j } = \sum \sum

d { g , h } a ˆ{ q } { g , h }\\ g = l h = 1 \\ with each d { g, h } \ in F { q } ( t ) .\end{ a l i g n ∗}

Combining ( 5 . 2 . 8 ) and ( 5 . 2 . 9 ) y i e l d s a system o f equat ions which t rans −l a t e s i n to a matrix equat ion o f the form desc r ibed by Lemma 3 . 3 . 5 , in which

\noindent the matrix $ B $ de sc r ibed t h e r e i n i s the i d e n t i t y . By Lemma 3 . 3 . 5 , the$ a { i , j }$ are

a l g e b r a i c over $ F { q } ( t ) $ f o r $ i = l , . . . , m. $

S ince each $ a { i , j }$ \quad be longs to $ F { q } ( ( t ) ) , $\quad C h r i s t o l ’ s theorem \quad ( Theorem 4 . 1 . 1 )

i m p l i e s \quad that \quad $ a { i , j }$ \quad i s $ p − $ automatic \quad f o r$ i = l , . . . , m − 1 $ \quad and $ j = 1 , . . ., r . $

\noindent This i m p l i e s that $ ( y − y 0 ) ˆ{ q ˆ{ l }} = \sum ˆ{ r } { j= 1 } a { l , j } v { j }$ i s $ p − $ automatic , as f o l l o w s . Write

\noindent $ ( y − y 0 ) ˆ{ q ˆ{ l }} = \sum ˆ{ \ infty } { k = 0 }w { k } t ˆ{ k }$ with $ w { k } \ in V . $ \quad Then the func t i on \quad$ k \mapsto w { k }$ \quad i s $ p − $

automatic because the $ a { i , j }$ are $ p − $ automatic . So we can bu i ld an automatonthat , g iven a base $ b $ expansion , s o r t s the pre rad ix s t r i n g $ k $ accord ing to the

Finite automata and algebraic extensions of function fields .. 40 1By the criterion of Proposition 5 period 2 period 2 comma we can write y minus y 0 = sum sub j = 0 to the

power of infinity v sub j t to the power of jwith each v sub j in V period In other words comma y is an F sub q open parenthesis open parenthesis t closing

parenthesis closing parenthesis hyphen linear combination ofelements of V period Likewise comma for each positive integer m comma open parenthesis y minus y 0 closing

parenthesis to the power of q to the power of m is an F sub q open parenthesis open parenthesis t closing parenthesisclosing parenthesis hyphen

linear combination of elements of V period For i = l comma period period period comma m comma write openparenthesis y minus y 0 closing parenthesis to the power of q to the power of i =

sum sub j = 1 to the power of r a sub i comma j v sub j .. with a sub i comma j in F sub q open parenthesis openparenthesis t closing parenthesis closing parenthesis semicolon then the a sub i comma j .. are uniquely determinedby

Lemma 5 period 2 period 6 periodBy the same reasoning comma for j = 1 comma period period period comma r comma we can writeLine 1 r Line 2 v to the power of q h = sum b sub h comma j v sub j Line 3 h = 1for some b sub h comma j in F sub q open square bracket t closing square bracket period This means thatEquation: open parenthesis 5 period 2 period 8 closing parenthesis .. a sub i comma j = sum b sub h comma

j a to the power of q i sub minus 1 comma h open parenthesis i = l plus 1 comma period period period comma msemicolon j = 1 comma period period period comma r closing parenthesis period h

Moreover comma the equationopen parenthesis y minus y 0 closing parenthesis to the power of q to the power of l = minus c sub l plus 1 open

parenthesis open parenthesis y minus y 0 closing parenthesis to the power of q to the power of l closing parenthesisto the power of q minus times times times minus c sub m open parenthesis open parenthesis y minus y 0 closingparenthesis to the power of q to the power of m minus 1 Case 1 q Case 2 comma

which holds .. because .. P open parenthesis y minus y 0 closing parenthesis = 0 .. and .. c sub l = 1 .. byhypothesis comma .. can .. be

rewritten asLine 1 r m minus 1 r Line 2 sum a sub l comma j v sub j = sum minus c sub i plus 1 open parenthesis sum a

sub i comma h v sub h closing parenthesis to the power of q Line 3 j = 1 i = l h = 1 Line 4 m minus 1 r r Line 5 =sum minus c sub i plus 1 sum sum a sub i comma h to the power of q b sub h comma j v sub j period Line 6 i = l h= 1 j = 1

Equating coefficients of v sub j yields a system of equations of the formm minus 1 r Equation: open parenthesis 5 period 2 period 9 closing parenthesis .. a sub l comma j = sum sum

d sub g comma h a sub g comma h to the power of q g = l h = 1 with each d sub g comma h in F sub q openparenthesis t closing parenthesis period

Combining open parenthesis 5 period 2 period 8 closing parenthesis and open parenthesis 5 period 2 period 9closing parenthesis yields a system of equations which trans hyphen

lates into a matrix equation of the form described by Lemma 3 period 3 period 5 comma in whichthe matrix B described therein is the identity period By Lemma 3 period 3 period 5 comma the a sub i comma

j arealgebraic over F sub q open parenthesis t closing parenthesis for i = l comma period period period comma m

periodSince each a sub i comma j .. belongs to F sub q open parenthesis open parenthesis t closing parenthesis

closing parenthesis comma .. Christol quoteright s theorem .. open parenthesis Theorem 4 period 1 period 1 closingparenthesis

implies .. that .. a sub i comma j .. i s p hyphen automatic .. for i = l comma period period period comma mminus 1 .. and j = 1 comma period period period comma r period

This implies that open parenthesis y minus y 0 closing parenthesis to the power of q to the power of l = sum subj = 1 to the power of r a sub l comma j v sub j i s p hyphen automatic comma as follows period Write

open parenthesis y minus y 0 closing parenthesis to the power of q to the power of l = sum sub k = 0 to thepower of infinity w sub k t to the power of k with w sub k in V period .. Then the function .. k arrowright-mapstow sub k .. is p hyphen

automatic because the a sub i comma j are p hyphen automatic period So we can build an automatonthat comma given a base b expansion comma sorts the preradix string k according to the

Since each ai,j belongs to Fq((t)), Christol ’ s theorem ( Theorem4 . 1 . 1 ) implies that ai,j i s p− automatic for i = l, ...,m− 1and j = 1, ..., r.

This implies that (y−y0)ql

=∑rj=1 al,jvj i s p− automatic , as follows . Write

(y − y0)ql

=∑∞k=0 wkt

k with wk ∈ V. Then the function k 7→ wkis p− automatic because the ai,j are p− automatic . So we can build anautomaton that , given a base b expansion , sorts the preradix string kaccording to the


\noindent value o f $ w { k } , $ then handles the pos t rad ix s t r i n g by i m i t a t i n g some automatonthat computes $ w { k } . $

S ince $ ( y − y 0 ) ˆ{ q ˆ{ l }}$ i s $ p − $ automatic $ , y −y 0 $ i s $ p − $ quas i − automatic , as then i s $ y , $

as then i s $ x . $ This y i e l d s the d e s i r e d r e s u l t $ . \ square $

Note that Propo s i t i on s 5 . 1 . 2 and 5 . 2 . 7 toge the r g ive a complete proo f o fTheorem 4 . 1 . 3 . We w i l l g ive a second proo f o f a statement equ iva l en t toPropos i t i on 5 . 2 . 7 l a t e r ( s e e Propos i t i on 7 . 3 . 4 ) .

\centerline {6 . \quad Polynomials over valued f i e l d s }

In t h i s chapter , we int roduce some a d d i t i o n a l a l g e b r a i c t o o l s that w i l lhe lp us g ive a more a l go r i thmi c proo f o f the ‘ ‘ a l g e b r a i c i m p l i e s automatic ’ ’i m p l i c a t i o n o f Theorem 4 . 1 . 3 . Nothing in t h i s chapter i s p a r t i c u l a r l y nove l ,but the mate r i a l may not be f a m i l i a r to n o n s p e c i a l i s t s , so we g ive a d e t a i l e dp r e s en t a t i on .

\noindent $ 6 . 1 . T−w $ i s t e d \quad polynomial \quad r i n g s . \quad Ore ’ s \quad lemma \quad ( Lemma \quad 3 . 3 . 4 ) \quad a s s e r t sthat every a l g e b r a i c element o f a f i e l d o f c h a r a c t e r i s t i c $ p $ over a s u b f i e l di s a root o f an a d d i t i v e polynomial . \quad I t i s sometimes more convenient toview a d d i t i v e polynomia l s as the r e s u l t o f apply ing ‘ ‘ tw i s t ed polynomia ls ’ ’in the Frobenius operator . These po lynomia l s a r i s e n a t u r a l l y in the theoryo f Dr in f e l d modules ; s e e f o r i n s t anc e [ 9 ] .

\centerline{Throughout t h i s s e c t i o n , the f i e l d $ K $ w i l l have c h a r a c t e r i s t i c $ p> 0 . $ }

\noindent D e f i n i t i o n 6 . 1 . 1 . \ h f i l l Let $ K \{ F \} $ denote the noncommutative r ing whose e l e −

\noindent ments are f i n i t e formal sums $ \sum ˆ{ m } { i = 0 } c { i } F ˆ{ i }, $ added componentwise and mult i −

\noindent p l i e d by the r u l e

\ [\ l e f t (\ begin { array }{ c} m \\ \sum c { i } F ˆ{ i }\\ i = 0 \end{ array }\ right ) \ l e f t (\ begin { array }{ c} n \\\sum d { j } F ˆ{ j }\\ j = 0 \end{ array }\ right ) = \sum ˆ{ k = 0 } { m+ n } \ l e f t (\ begin { array }{ cc } \sum & c { i } d ˆ{ p ˆ{ i }} { j }\\ i + j= k \end{ array }\ right ) F ˆ{ k } . \ ]

\noindent The r ing $ K \{ F \} $ i s c a l l e d the tw i s t ed polynomial r i ng over$ K . ( $ Note that

the same d e f i n i t i o n can be made r e p l a c i n g the $ p − $ power endomorphism byany endomorphism o f $ K , $ but we w i l l only use t h i s p a r t i c u l a r form o f thec on s t r u c t i on . )

\noindent D e f i n i t i o n \quad 6 . 1 . 2 . \quad As \quad in \quad the \quad polynomial \quad case , \quad the \quad degree \quad o f a \quad nonzerotw i s t ed polynomial $ \sum c { i } F ˆ{ i }$ i s the l a r g e s t $ i $ such that \quad

$ c { i } \ne 0 ; $ \quad we conven −t i o n a l l y take the degree o f the zero polynomial to be $ − \ infty . $ The degree o fthe product o f two nonzero tw i s t ed polynomia ls i s the sum of t h e i r i n d i v i d −ual degree s ; in p a r t i c u l a r , the r ing $ K \{ F \} $ i s an i n t e g r a l domain .

Twisted polynomial r i n g s admit the f o l l o w i n g r i g h t d i v i s i o n a lgor i thm ,j u s t as in the usua l polynomial case .

402 .. Kiran S period Kedlayavalue of w sub k comma then handles the postradix string by imitating some automatonthat computes w sub k periodSince open parenthesis y minus y 0 closing parenthesis to the power of q to the power of l i s p hyphen automatic

comma y minus y 0 i s p hyphen quasi hyphen automatic comma as then i s y commaas then is x period This yields the desired result period squareNote that Propositions 5 period 1 period 2 and 5 period 2 period 7 together give a complete proof ofTheorem 4 period 1 period 3 period We will give a second proof of a statement equivalent toProposition 5 period 2 period 7 later open parenthesis see Proposition 7 period 3 period 4 closing parenthesis

period6 period .. Polynomials over valued fieldsIn this chapter comma we introduce some additional algebraic tools that willhelp us give a more algorithmic proof of the quotedblleft algebraic implies automatic quotedblrightimplication of Theorem 4 period 1 period 3 period Nothing in this chapter i s particularly novel commabut the material may not be familiar to nonspecialists comma so we give a detailedpresentation period6 period 1 period T-w isted .. polynomial .. rings period .. Ore quoteright s .. lemma .. open parenthesis Lemma

.. 3 period 3 period 4 closing parenthesis .. assertsthat every algebraic element of a field of characteristic p over a subfieldi s a root of an additive polynomial period .. It i s sometimes more convenient toview additive polynomials as the result of applying quotedblleft twisted polynomials quotedblrightin the Frobenius operator period These polynomials arise naturally in the theoryof Drinfeld modules semicolon see for instance open square bracket 9 closing square bracket periodThroughout this section comma the field K will have characteristic p greater 0 periodDefinition 6 period 1 period 1 period .... Let K open brace F closing brace denote the noncommutative ring

whose ele hyphenments are finite formal sums sum sub i = 0 to the power of m c sub i F to the power of i comma added

componentwise and multi hyphenplied by the ruleRow 1 m Row 2 sum c sub i F to the power of i Row 3 i = 0 . Row 1 n Row 2 sum d sub j F to the power of j

Row 3 j = 0 . = sum from k = 0 to m plus n Row 1 sum c sub i d sub j to the power of p to the power of i Row 2 iplus j = k . F to the power of k period

The ring K open brace F closing brace i s called the twisted polynomial ring over K period open parenthesis Notethat

the same definition can be made replacing the p hyphen power endomorphism byany endomorphism of K comma but we will only use this particular form of theconstruction period closing parenthesisDefinition .. 6 period 1 period 2 period .. As .. in .. the .. polynomial .. case comma .. the .. degree .. of a ..

nonzerotwisted polynomial sum c sub i F to the power of i i s the largest i such that .. c sub i equal-negationslash 0

semicolon .. we conven hyphent ionally take the degree of the zero polynomial to be minus infinity period The degree ofthe product of two nonzero twisted polynomials i s the sum of their individ hyphenual degrees semicolon in particular comma the ring K open brace F closing brace i s an integral domain periodTwisted polynomial rings admit the following right division algorithm commajust as in the usual polynomial case period


value of wk, then handles the postradix string by imitating some automatonthat computes wk.

Since (y− y0)ql

i s p− automatic , y− y0 i s p− quasi - automatic , as theni s y, as then is x. This yields the desired result . �

Note that Propositions 5 . 1 . 2 and 5 . 2 . 7 together give a completeproof of Theorem 4 . 1 . 3 . We will give a second proof of a statementequivalent to Proposition 5 . 2 . 7 later ( see Proposition 7 . 3 . 4 ) .

6 . Polynomials over valued fieldsIn this chapter , we introduce some additional algebraic tools that will

help us give a more algorithmic proof of the “ algebraic implies automatic ”implication of Theorem 4 . 1 . 3 . Nothing in this chapter i s particularlynovel , but the material may not be familiar to nonspecialists , so we give adetailed presentation .6.1. T−w isted polynomial rings . Ore ’ s lemma ( Lemma3 . 3 . 4 ) asserts that every algebraic element of a field of characteristicp over a subfield i s a root of an additive polynomial . It i s sometimesmore convenient to view additive polynomials as the result of applying “twisted polynomials ” in the Frobenius operator . These polynomials arisenaturally in the theory of Drinfeld modules ; see for instance [ 9 ] .

Throughout this section , the field K will have characteristic p > 0.Definition 6 . 1 . 1 . Let K{F} denote the noncommutative ringwhose ele -ments are finite formal sums

∑mi=0 ciF

i, added componentwise and multi -plied by the rule m∑

ciFi

i = 0

n∑djF

j

j = 0

=

k=0∑m+n

( ∑cid

pi

j

i+ j = k

)F k.

The ring K{F} i s called the twisted polynomial ring over K. ( Note thatthe same definition can be made replacing the p− power endomorphism byany endomorphism of K, but we will only use this particular form of theconstruction . )Definition 6 . 1 . 2 . As in the polynomial case ,the degree of a nonzero twisted polynomial

∑ciF

i i s the largest isuch that ci 6= 0; we conven - t ionally take the degree of the zeropolynomial to be −∞. The degree of the product of two nonzero twistedpolynomials i s the sum of their individ - ual degrees ; in particular , thering K{F} i s an integral domain .

Twisted polynomial rings admit the following right division algorithm ,just as in the usual polynomial case .

\centerline{F i n i t e automata and a l g e b r a i c ex t en s i on s o f func t i on f i e l d s \quad 403 }

\noindent Lemma 6 . 1 . 3 . \quad Let $ S ( F ) $ \quad and $ T ( F ) $\quad be twi s t ed polynomia l s over $ K , $ \quad with

$ T ( F ) $ nonzero and deg $ ( T ) = d \geq 0 . $ Then there e x i s t s a unique pa i r$ Q ( F ) , R ( F ) $

o f tw i s t ed polynomia l s over $ K $ with deg $ ( R ) < d , $ such that$ S = QT + R . $

\noindent Proof . \quad Write $ T ( F ) = \sum ˆ{ d } { i = 0 } c { i }F ˆ{ i }$ with $ c { d } \ne 0 . $ Ex i s tence f o l l o w s by induct i on

on deg $ ( S ) $ \quad and the f a c t that we can cons t ruc t a l e f t mu l t ip l e o f$ T $ o f any

p r e s c r i b e d degree $ e \geq d $ with any p r e s c r i b e d l ead ing c o e f f i c i e n t $ a( $ namely

\noindent $ ( a / ˆ{ c−p }ˆ{ e − d } { d } ) F ˆ{ e − d } T ) . $Uniqueness f o l l o w s from the f a c t that i f $ S = QT + R = $

\noindent $ Q ˆ{ \prime } T + R ˆ{ \prime }$ are two decompos i t ions o f the d e s i r e d form , then$ R − R ˆ{ \prime } = ( Q ˆ{ \prime } − $

$ Q ) T $ i s a l e f t mu l t ip l e o f $ T $ but deg $ ( R − R ˆ{ \prime }) < $ deg $ ( T ) , $ so $ R − R ˆ{ \prime } = 0 . \ square $

\centerline{We may view twi s t ed polynomia l s over $ K $ as a d d i t i v e ope ra to r s on any }

\noindent f i e l d $ L $ conta in ing $ K $ by d e c l a r i n g that

\ [\ l e f t (\ begin { array }{ c} \sum c { i } F ˆ{ i }\\ i \end{ array }\ right ) ( z )= \sum { i } c { i } z ˆ{ p ˆ{ i }} . \ ]

\noindent Lemma \quad 6 . 1 . 4 . \quad Let \quad $ L $ \quad be \quad an \quad a l g e b r a i c \quad c l o s u r e \quad o f$ K , $ \quad and \quad l e t \quad $ T ( F ) = $

\noindent $ \sum ˆ{ d } { i = 0 } c { i } F ˆ{ i }$ \quad be \quad a nonzero tw i s t ed polynomial o f degree$ d $ \quad over $ K . $ \quad Then the

ke rne l $ \ker { L } ( T ) $ \quad o f $ T $ ac t ing on $ L $ i s an $ F { p }− $ vec to r space o f dimension $ \ leq d , $

with e q u a l i t y i f and only i f $ c { 0 } \ne 0 . $

\noindent Proof . \quad The ke rne l i s an $ F { p } − $ vec to r space because $ T( F ) $ i s an a d d i t i v e opera −

to r on $ L ( $ Lemma 3 . 3 . 3 ) , and the dimension bound holds because $ T( F ) ( z ) $ i s

\noindent a polynomial in $ z $ o f degree $ p ˆ{ d } . $ The e q u a l i t y case ho lds because the formald e r i v a t i v e in $ z $ o f $ T ( F ) ( z ) $ i s $ c { 0 } , $ and t h i s d e r i v a t i v e van i shes i f and only i f$ T ( F ) ( z ) $ has no repeated roo t s over $ L . \ square $

\noindent Lemma 6 . 1 . 5 . \quad Let $ S ( F ) $ \quad and $ T ( F ) $\quad be twi s t ed polynomia ls over $ K , $ \quad with

\noindent the constant c o e f f i c i e n t o f $ T $ nonzero , \quad and l e t $ L $ \quad be an a l g e b r a i c c l o s u r e o f$ K . $ \quad Then $ S $ i s a l e $ f−t $ mu l t ip l e o f $ T ( $ that i s $ ,

S = QT $ f o r s ome $ Q \ in K \{ F \} ) $i f and only i f $ \ker { L } ( T ) \subseteq \ker { L } ( S ) . $

\noindent Proof . \quad I f $ S = QT , $ then $ T ( F ) ( z ) =0 $ i m p l i e s $ S ( F ) ( z ) = 0 , $ so $ \ker { L } ( T )\subseteq $

$ \ker { L } ( S ) . $ Converse ly , suppose that $ \ker { L } ( T )\subseteq \ker { L } ( S ) , $ and wr i t e $ S = QT + $

$ R $ by the r i g h t d i v i s i o n a lgor i thm ; we then have $ \ker { L } ( T ) \subseteq\ker { L } ( R ) $ as we l l .By Lemma $ 6 . 1 . 4 , \ker { L } ( T ) $ i s an $ F { p } − $

vec to r space o f dimension deg $ ( T ) ; $ i f $ R $ i snonzero , then $ \ker { L } ( R ) $ i s an $ F { p } − $ vec to r space o f dimension at most deg

$ ( R ) < $deg $ ( T ) . $ But that would c o n t r a d i c t the i n c l u s i o n $ \ker { L } (

T ) \subseteq \ker { L } ( R ) , $ so wemust have $ R = 0 $ and $ S = QT , $ as d e s i r e d $ . \ square $

\noindent 6 . 2 . \quad Newton polygons . \quad The theory o f Newton polygons i s a c r i t i c a l in −

\noindent g r ed i en t in the computat ional and t h e o r e t i c a l study o f valued f i e l d s . We

\noindent r e c a l l a b i t o f t h i s theory here .

Finite automata and algebraic extensions of function fields .. 403Lemma 6 period 1 period 3 period .. Let S open parenthesis F closing parenthesis .. and T open parenthesis F

closing parenthesis .. be twisted polynomials over K comma .. withT open parenthesis F closing parenthesis nonzero and deg open parenthesis T closing parenthesis = d greater

equal 0 period Then there exists a unique pair Q open parenthesis F closing parenthesis comma R open parenthesisF closing parenthesis

of twisted polynomials over K with deg open parenthesis R closing parenthesis less d comma such that S = QTplus R period

Proof period .. Write T open parenthesis F closing parenthesis = sum sub i = 0 to the power of d c sub i F tothe power of i with c sub d equal-negationslash 0 period Existence follows by induction

on deg open parenthesis S closing parenthesis .. and the fact that we can construct a left multiple of T of anyprescribed degree e greater equal d with any prescribed leading coefficient a open parenthesis namelyopen parenthesis a slash to the power of c-p sub d to the power of e minus d closing parenthesis F to the power

of e minus d T closing parenthesis period Uniqueness follows from the fact that if S = QT plus R =Q to the power of prime T plus R to the power of prime are two decompositions of the desired form comma then

R minus R to the power of prime = open parenthesis Q to the power of prime minusQ closing parenthesis T is a left multiple of T but deg open parenthesis R minus R to the power of prime closing

parenthesis less deg open parenthesis T closing parenthesis comma so R minus R to the power of prime = 0 periodsquare

We may view twisted polynomials over K as additive operators on anyfield L containing K by declaring thatRow 1 sum c sub i F to the power of i Row 2 i . open parenthesis z closing parenthesis = sum i c sub i z to the

power of p to the power of i periodLemma .. 6 period 1 period 4 period .. Let .. L .. be .. an .. algebraic .. c losure .. of K comma .. and .. let ..

T open parenthesis F closing parenthesis =sum sub i = 0 to the power of d c sub i F to the power of i .. be .. a nonzero twisted polynomial of degree d ..

over K period .. Then thekernel kernel sub L open parenthesis T closing parenthesis .. of T acting on L is an F sub p hyphen vector space

of dimension less or equal d commawith equality if and only if c sub 0 equal-negationslash 0 periodProof period .. The kernel i s an F sub p hyphen vector space because T open parenthesis F closing parenthesis

is an additive opera hyphentor on L open parenthesis Lemma 3 period 3 period 3 closing parenthesis comma and the dimension bound holds

because T open parenthesis F closing parenthesis open parenthesis z closing parenthesis i sa polynomial in z of degree p to the power of d period The equality case holds because the formalderivative in z of T open parenthesis F closing parenthesis open parenthesis z closing parenthesis i s c sub 0

comma and this derivative vanishes if and only ifT open parenthesis F closing parenthesis open parenthesis z closing parenthesis has no repeated roots over L

period squareLemma 6 period 1 period 5 period .. Let S open parenthesis F closing parenthesis .. and T open parenthesis F

closing parenthesis .. be twisted polynomials over K comma .. withthe constant coefficient of T nonzero comma .. and let L .. be an algebraic c losure ofK period .. Then S is a le f-t multiple of T open parenthesis that is comma S = QT for s ome Q in K open brace

F closing brace closing parenthesisif and only if kernel sub L open parenthesis T closing parenthesis subset equal kernel sub L open parenthesis S

closing parenthesis periodProof period .. If S = QT comma then T open parenthesis F closing parenthesis open parenthesis z closing

parenthesis = 0 implies S open parenthesis F closing parenthesis open parenthesis z closing parenthesis = 0 commaso kernel sub L open parenthesis T closing parenthesis subset equal

kernel sub L open parenthesis S closing parenthesis period Conversely comma suppose that kernel sub L openparenthesis T closing parenthesis subset equal kernel sub L open parenthesis S closing parenthesis comma and writeS = QT plus

R by the right division algorithm semicolon we then have kernel sub L open parenthesis T closing parenthesissubset equal kernel sub L open parenthesis R closing parenthesis as well period

By Lemma 6 period 1 period 4 comma kernel sub L open parenthesis T closing parenthesis i s an F sub p hyphenvector space of dimension deg open parenthesis T closing parenthesis semicolon if R i s

nonzero comma then kernel sub L open parenthesis R closing parenthesis i s an F sub p hyphen vector space ofdimension at most deg open parenthesis R closing parenthesis less

deg open parenthesis T closing parenthesis period But that would contradict the inclusion kernel sub L openparenthesis T closing parenthesis subset equal kernel sub L open parenthesis R closing parenthesis comma so we

must have R = 0 and S = QT comma as desired period square6 period 2 period .. Newton polygons period .. The theory of Newton polygons is a critical in hyphengredient in the computational and theoretical study of valued fields period Werecall a bit of this theory here period


Lemma 6 . 1 . 3 . Let S(F ) and T (F ) be twisted polynomialsover K, with T (F ) nonzero and deg (T ) = d ≥ 0. Then there exists aunique pair Q(F ), R(F ) of twisted polynomials over K with deg (R) < d,such that S = QT +R.Proof . Write T (F ) =

∑di=0 ciF

i with cd 6= 0.Existence follows by inductionon deg (S) and the fact that we can construct a left multiple of T of anyprescribed degree e ≥ d with any prescribed leading coefficient a( namely(a/c−pe−dd )F e−dT ). Uniqueness follows from the fact that if S = QT +R =Q′T +R′ are two decompositions of the desired form , then R−R′ = (Q′− Q)Tis a left multiple of T but deg (R−R′) < deg (T ), so R−R′ = 0. �

We may view twisted polynomials over K as additive operators on anyfield L containing K by declaring that( ∑

ciFi

i

)(z) =

∑i

cizpi .

Lemma 6 . 1 . 4 . Let L be an algebraic c losureof K, and let T (F ) =∑di=0 ciF

i be a nonzero twisted polynomial of degree d over K.Then the kernel kerL(T ) of T acting on L is an Fp− vector space ofdimension ≤ d, with equality if and only if c0 6= 0.Proof . The kernel i s an Fp− vector space because T (F ) is an additiveopera - tor on L( Lemma 3 . 3 . 3 ) , and the dimension bound holdsbecause T (F )(z) i sa polynomial in z of degree pd. The equality case holds because the formalderivative in z of T (F )(z) i s c0, and this derivative vanishes if and only ifT (F )(z) has no repeated roots over L. �Lemma 6 . 1 . 5 . Let S(F ) and T (F ) be twisted polynomialsover K, withthe constant coefficient of T nonzero , and let L be an algebraic closure of K. Then S is a le f − t multiple of T ( that is , S = QT fors ome Q ∈ K{F}) if and only if kerL(T ) ⊆ kerL(S).Proof . If S = QT, then T (F )(z) = 0 implies S(F )(z) = 0, so kerL(T ) ⊆kerL(S). Conversely , suppose that kerL(T ) ⊆ kerL(S), and write S = QT+ R bythe right division algorithm ; we then have kerL(T ) ⊆ kerL(R) as well . ByLemma 6.1.4, kerL(T ) i s an Fp− vector space of dimension deg (T ); if R is nonzero , then kerL(R) i s an Fp− vector space of dimension at most deg(R) < deg (T ). But that would contradict the inclusion kerL(T ) ⊆ kerL(R), sowe must have R = 0 and S = QT, as desired . �6 . 2 . Newton polygons . The theory of Newton polygons is acritical in -gredient in the computational and theoretical study of valued fields . Werecall a bit of this theory here .


\noindent D e f i n i t i o n \ h f i l l 6 . 2 . 1 . \ h f i l l For $ x \ in F { q } ( ( t ˆ{ Q }) ) $ \ h f i l l nonzero , \ h f i l l l e t $ v ( x ) $ \ h f i l l denote the smal l −

\noindent e s t element o f the support o f $ x ; $ we c a l l $ v $ the va lua t i on on$ F { q } ( ( t ˆ{ Q } ) ) . $ We

a l s o f o rma l l y put $ v ( 0 ) = \ infty . $ The func t i on $ v $ has the usua l p r o p e r t i e s o f a

\begin { a l i g n ∗}va luat i on : \\ v ( x + y ) \geq \min \{ v ( x ) , v

( y ) \} \\ v ( xy ) = v ( x ) + v ( y ) .\end{ a l i g n ∗}

\noindent Given a nonzero polynomial $ P ( z ) = \sum { i } c { i } z ˆ{ i }$over $ F { q } ( ( t ˆ{ Q } ) ) , $ we d e f i n e theNewton polygon o f $ P $ to be the lower boundary o f the lower convex h u l l o fthe s e t o f po in t s $ ( − i , v ( c { i } ) ) . $ The s l o p e s o f t h i s polygon are c a l l e d the s l o p e so f $ P ; $ f o r $ r \ in Q , $ we d e f i n e the m u l t i p l i c i t y o f $ r $ as a s l ope o f

$ P $ to be thewidth ( d i f f e r e n c e in $ x − $ coo rd ina t e s between the endpoints ) o f the segmento f the Newton polygon o f $ P $ o f s l ope $ r , $ \quad or 0 i f no such segment \quad e x i s t s .We say $ P $ i s pure \quad ( o f s lope $ r ) $ i f a l l o f the s l o p e s o f $ P $

are equal to $ r . $ Weconven t i ona l l y d e c l a r e that $ \ infty $ may a l s o be a s l ope , and i t s m u l t i p l i c i t y asa s l ope o f a polynomial $ P $ i s the order o f van i sh ing o f $ P $ at $ z =

0 . $

\noindent Lemma 6 . 2 . 2 . \ h f i l l Let $ P $ and $ Q $ be nonzero polynomia l s over$ F { q } ( ( t ˆ{ Q } ) ) . $ \ h f i l l Then

\noindent f o r each $ r \ in Q \cup \{ \ infty \} , $ \quad the m u l t i p l i c i t y o f$ r $ as a s l ope o f $ P + Q $ i s the sum

of the m u l t i p l i c i t i e s o f $ r $ as a s l ope o f $ P $ and o f $ Q . $

\noindent Proof . \ h f i l l The case $ r = \ infty $ i s c l e a r , so we assume $ r\ in Q . $ Write $ P ( z ) = \sum { i } c { i } z ˆ{ i }$

\noindent and $ Q ( z ) = \sum { j } d { j } z ˆ{ j } . $ Let $ (− e , v ( c { e } ) ) $ and $ ( − f , v ( d { f } ) )( $ resp $ . ( − g , v ( d { g } ) ) $ and

\noindent $ ( − h , v ( d { h } ) ) ) $ be the l e f t and r i g h t endpoints , r e s p e c t i v e l y , o f the ( p o s s i b l ydegenerate ) segment in which the Newton polygon o f $ P ( $ resp . o f $ Q ) $

meetsi t s support l i n e o f s l ope $ r . $ Then

\ [\ begin { a l i gned } v ( c { i } ) + r i \geq v ( c { e } ) + re= v ( c { f } ) + r f \\

v ( d { j } ) + r j \geq v ( d { g } ) + rg = v ( d { h }) + rh \end{ a l i gned }\ ]

\noindent with s t r i c t i n e q u a l i t y i f $ i element−s l a s h [ e , f ] $ or $ js l a sh−element [ g , h ] . $ For each $ k , $

\ [ i + ˆ{ \min } j = k ˆ{ \{ v ( c { i } d { j } ) } + rk \}\geq v ( c { e } ) + re + v ( d { g } ) + rg \ ]

\noindent and the i n e q u a l i t y i s s t r i c t in each o f the f o l l o w i n g ca s e s :

\ [\ begin { a l i gned } \bullet k element−s l a s h [ f + h , e + g ] ; \\\bullet k = e + g and i \ne e ; \\\bullet k = f + h and i \not= h . \end{ a l i gned }\ ]

\noindent I f we wr i t e $ R = PQ $ and $ R ( z ) = \sum a { k } z ˆ{ k }, $ i t f o l l o w s that

\ [ \min { k } \{ v ( a { k } ) + rk \} \geq v ( c { e } )+ re + v ( d { g } ) + rg , \ ]

404 .. Kiran S period KedlayaDefinition .... 6 period 2 period 1 period .... For x in F sub q open parenthesis open parenthesis t to the power

of Q closing parenthesis closing parenthesis .... nonzero comma .... let v open parenthesis x closing parenthesis ....denote the small hyphen

est element of the support of x semicolon we call v the valuation on F sub q open parenthesis open parenthesis tto the power of Q closing parenthesis closing parenthesis period We

also formally put v open parenthesis 0 closing parenthesis = infinity period The function v has the usual propertiesof a

valuation : v open parenthesis x plus y closing parenthesis greater equal minimum open brace v open parenthesisx closing parenthesis comma v open parenthesis y closing parenthesis closing brace v open parenthesis xy closingparenthesis = v open parenthesis x closing parenthesis plus v open parenthesis y closing parenthesis period

Given a nonzero polynomial P open parenthesis z closing parenthesis = sum sub i c sub i z to the power of iover F sub q open parenthesis open parenthesis t to the power of Q closing parenthesis closing parenthesis commawe define the

Newton polygon of P to be the lower boundary of the lower convex hull ofthe set of points open parenthesis minus i comma v open parenthesis c sub i closing parenthesis closing parenthesis

period The slopes of this polygon are called the slopesof P semicolon for r in Q comma we define the multiplicity of r as a slope of P to be thewidth open parenthesis difference in x hyphen coordinates between the endpoints closing parenthesis of the

segmentof the Newton polygon of P of slope r comma .. or 0 if no such segment .. exists periodWe say P i s pure .. open parenthesis of s lope r closing parenthesis if all of the slopes of P are equal to r period

Weconventionally declare that infinity may also be a slope comma and its multiplicity asa slope of a polynomial P i s the order of vanishing of P at z = 0 periodLemma 6 period 2 period 2 period .... Let P and Q be nonzero polynomials over F sub q open parenthesis open

parenthesis t to the power of Q closing parenthesis closing parenthesis period .... Thenfor each r in Q cup open brace infinity closing brace comma .. the multiplicity of r as a slope of P plus Q is the

sumof the multiplicities of r as a slope of P and of Q periodProof period .... The case r = infinity is clear comma so we assume r in Q period Write P open parenthesis z

closing parenthesis = sum sub i c sub i z to the power of iand Q open parenthesis z closing parenthesis = sum sub j d sub j z to the power of j period Let open parenthesis

minus e comma v open parenthesis c sub e closing parenthesis closing parenthesis and open parenthesis minus f commav open parenthesis d sub f closing parenthesis closing parenthesis open parenthesis resp period open parenthesis minusg comma v open parenthesis d sub g closing parenthesis closing parenthesis and

open parenthesis minus h comma v open parenthesis d sub h closing parenthesis closing parenthesis closingparenthesis be the left and right endpoints comma respectively comma of the open parenthesis possibly

degenerate closing parenthesis segment in which the Newton polygon of P open parenthesis resp period of Qclosing parenthesis meets

it s support line of slope r period ThenLine 1 v open parenthesis c sub i closing parenthesis plus ri greater equal v open parenthesis c sub e closing

parenthesis plus re = v open parenthesis c sub f closing parenthesis plus rf Line 2 v open parenthesis d sub j closingparenthesis plus rj greater equal v open parenthesis d sub g closing parenthesis plus rg = v open parenthesis d subh closing parenthesis plus rh

with strict inequality if i element-slash open square bracket e comma f closing square bracket or j slash-elementopen square bracket g comma h closing square bracket period For each k comma

i plus to the power of minimum j = k to the power of open brace v open parenthesis c sub i d sub j closingparenthesis plus rk closing brace greater equal v open parenthesis c sub e closing parenthesis plus re plus v openparenthesis d sub g closing parenthesis plus rg

and the inequality is strict in each of the following cases :Line 1 bullet k element-slash open square bracket f plus h comma e plus g closing square bracket semicolon Line

2 bullet k = e plus g and i equal-negationslash e semicolon Line 3 bullet k = f plus h and i negationslash-equal hperiod

If we write R = PQ and R open parenthesis z closing parenthesis = sum a sub k z to the power of k comma itfollows that

minimum k open brace v open parenthesis a sub k closing parenthesis plus rk closing brace greater equal v openparenthesis c sub e closing parenthesis plus re plus v open parenthesis d sub g closing parenthesis plus rg comma


Definition 6 . 2 . 1 . For x ∈ Fq((tQ)) nonzero , let v(x) denotethe small -est element of the support of x; we call v the valuation on Fq((tQ)). We alsoformally put v(0) =∞. The function v has the usual properties of a

valuation :

v(x+ y) ≥ min{v(x), v(y)}v(xy) = v(x) + v(y).

Given a nonzero polynomial P (z) =∑i ciz

i over Fq((tQ)), we define theNewton polygon of P to be the lower boundary of the lower convex hull ofthe set of points (−i, v(ci)). The slopes of this polygon are called the slopesof P ; for r ∈ Q, we define the multiplicity of r as a slope of P to be thewidth ( difference in x− coordinates between the endpoints ) of the segmentof the Newton polygon of P of slope r, or 0 if no such segment exists .We say P i s pure ( of s lope r) if all of the slopes of P are equal to r.We conventionally declare that ∞ may also be a slope , and its multiplicityas a slope of a polynomial P i s the order of vanishing of P at z = 0.Lemma 6 . 2 . 2 . Let P and Q be nonzero polynomials overFq((tQ)). Thenfor each r ∈ Q ∪ {∞}, the multiplicity of r as a slope of P + Q is thesum of the multiplicities of r as a slope of P and of Q.Proof . The case r =∞ is clear , so we assume r ∈ Q. Write P (z) =

∑i ciz

i

and Q(z) =∑j djz

j . Let (−e, v(ce)) and (−f, v(df ))( resp .(−g, v(dg)) and(−h, v(dh))) be the left and right endpoints , respectively , of the ( possiblydegenerate ) segment in which the Newton polygon of P ( resp . of Q) meetsit s support line of slope r. Then

v(ci) + ri ≥ v(ce) + re = v(cf ) + rf

v(dj) + rj ≥ v(dg) + rg = v(dh) + rh

with strict inequality if ielement− slash[e, f ] or jslash− element[g, h]. For eachk,

i+min j = k{v(cidj) + rk} ≥ v(ce) + re+ v(dg) + rg

and the inequality is strict in each of the following cases :

• kelement− slash[f + h, e+ g];

• k = e+ gandi 6= e;

• k = f + handi 6= h.

If we write R = PQ and R(z) =∑akz

k, it follows that

mink{v(ak) + rk} ≥ v(ce) + re+ v(dg) + rg,


\noindent with e q u a l i t y f o r $ k = f + h $ and $ k = e + g $ but not f o r any$ k element−s l a s h [ f + h , e + g ] . $

I t f o l l o w s that the m u l t i p l i c i t y o f $ r $ as a s l ope o f $ R $ i s $ e + g− ( f + h ) = $

$ ( e − f ) + ( g − h ) , $ proving the d e s i r e d r e s u l t $ .\ square $

\noindent Coro l l a ry \ h f i l l 6 . 2 . 3 . \ h f i l l Let $ P ( z ) $ \ h f i l l be a nonzero polynomial over$ F { q } ( ( t ˆ{ Q } ) ) $ \ h f i l l which

\noindent f a c t o r s as a product $ Q 1 \cdot \cdot \cdot Q { n }$ o fpure po lynomia ls ( e . g . , l i n e a r po lynomia ls ) .Then f o r each $ r \ in Q \cup \{ \ infty \} , $ the sum of the degree s o f the

$ Q i , $ \quad over those $ i $f o r which $ Q i $ has s lope $ r , $ \quad i s equal to the m u l t i p l i c i t y o f

$ r $ as a s l ope o f $ P . $

\noindent 6 . 3 . \quad Slope s p l i t t i n g s . \quad We now r e c a l l a s p e c i a l form o f Hensel ’ s lemma ,which makes i t p o s s i b l e to s p l i t po lynomia l s by t h e i r s l o p e s .

\noindent D e f i n i t i o n 6 . 3 . 1 . \ h f i l l A s u b f i e l d $ K $ o f $ F { q } ( ( t ˆ{ Q }) ) $ i s c l o s ed under the va luat i on $ v $

\noindent i f f o r any sequence $ \{ z { n } \} ˆ{ \ infty } { n = 0 }$ such that$ v ( z { n } − z { n + 1 } ) \rightarrow \ infty $ as $ n \rightarrow\ infty , $ the re

\noindent e x i s t s $ z \ in K $ such that $ v ( z { n } − z ) \rightarrow\ infty $ as $ n \rightarrow \ infty . $ For in s t anc e $ , F { q } ( (t ) ) $ i s

\noindent c l o s e d , as i s any f i n i t e ex tens i on o f $ F { q } ( ( t ) ) . $

\noindent Lemma 6 . 3 . 2 . \ h f i l l Let $ K $ be a s u b f i e l d o f $ F { q } ( (t ˆ{ Q } ) ) $ \ h f i l l c l o s ed under the va lua t i on

\noindent $ v . $ Let $ P ( z ) $ \quad be a nonzero polynomial over $ K, $ \quad and choose $ u \ in Q $ which doesnot occur as a s lope o f $ P . $ \quad Then there e x i s t s a f a c t o r i z a t i o n $ P

= QR , $ where$ Q $ i s a polynomial over $ K $ with a l l s l ope s l e s s than $ u , $ and $ R $

i s a polynomial

\noindent over $ K $ with a l l s l o p e s g r e a t e r than $ u . $

\noindent Proof . \quad Pick $ r , s \ in Q $ \quad such that \quad $ r <s $ \quad and the i n t e r v a l \quad $ [ r , s ] $ \quad conta in s \quad $ u $

but does not conta in any s l o p e s o f $ P . $ Let $ S $ be the r ing o f formal sums

\noindent $ \sum { j \ in Z } a { j } z ˆ{ j }$ over $ K $ such that $ v( a { j } ) + r j \geq 0 $ f o r $ j \geq 0 $ and $ v ( a { j }) + s j \geq 0 $ f o r

\noindent $ j \ leq 0 . $ Def ine the va lua t i on $ v { u }$ on $ S $ by the formula

\ [ v { u } ( \sum a { j } z ˆ{ j } ) = \min { j } \{ v ( a { j }) + uj \} . \ ]

\noindent Given $ x = \sum a { j } z ˆ{ j } \ in S , $ wr i t e $ f (x ) = \sum { j > 0 } a { j } z ˆ{ j } , $ so that $ f ( f (x ) ) = f ( x ) . $

\hspace ∗{\ f i l l }Write $ P ( z ) = \sum { i } c { i } z ˆ{ i } , $ put$ d = $ deg $ P , $ l e t $ e $ be the unique i n t e g e r that

\noindent minimizes $ v ( c { e } ) + ue , $ and put $ x { 0 } = c ˆ{ −1 } { e } z ˆ{ − e } P ( z ) \ in S . $ Def ine the sequence

$ \{ x { h } \} ˆ{ \ infty } { h = 0 }$ by the r e cu r r ence $ x { h + 1 }= x { h } ( 1 − f ( x { h } ) ) $ f o r $ h = 0 , 1 , .. . . $

Put \quad $ \ e l l = v { u } ( x { 0 } − 1 ) , $ \quad so that $ \ e l l> 0 . $ \quad We now show by induct i on that

$ v { u } ( x { h } − 1 ) \geq \ e l l $ and $ v { u } ( f ( x { h }) ) \geq ( h + 1 ) \ e l l $ f o r a l l $ h . $ \quad The f i r s t i n e q u a l i t y

c l e a r l y i m p l i e s the second f o r $ h = 0 ; $ g iven both i n e q u a l i t i e s f o r some$ h , $ we

\noindent have

\ [\ begin { a l i gned } v { u } ( x { h + 1 } − 1 ) = v { u } ( x { h }− x { h } f ( x { h } ) − 1 ) \\\geq \min \{ v { u } ( x { h } − 1 ) , v { u } ( x { h }

) + v { u } ( f ( x { h } ) ) \} \\\geq \min \{ \ e l l , ( h + 1 ) \ e l l \} \\\geq \ e l l \end{ a l i gned }\ ]

Finite automata and algebraic extensions of function fields .. 405with equality for k = f plus h and k = e plus g but not for any k element-slash open square bracket f plus h

comma e plus g closing square bracket periodIt follows that the multiplicity of r as a slope of R i s e plus g minus open parenthesis f plus h closing parenthesis

=open parenthesis e minus f closing parenthesis plus open parenthesis g minus h closing parenthesis comma proving

the desired result period squareCorollary .... 6 period 2 period 3 period .... Let P open parenthesis z closing parenthesis .... be a nonzero

polynomial over F sub q open parenthesis open parenthesis t to the power of Q closing parenthesis closing parenthesis.... which

factors as a product Q 1 times times times Q sub n ofpure polynomials open parenthesis e period g period commalinear polynomials closing parenthesis period

Then for each r in Q cup open brace infinity closing brace comma the sum of the degrees of the Q i comma ..over those i

for which Q i has s lope r comma .. is equal to the multiplicity of r as a slope of P period6 period 3 period .. Slope splittings period .. We now recall a special form of Hensel quoteright s lemma commawhich makes it possible to split polynomials by their slopes periodDefinition 6 period 3 period 1 period .... A subfield K of F sub q open parenthesis open parenthesis t to the

power of Q closing parenthesis closing parenthesis i s c los ed under the valuation vif for any sequence open brace z sub n closing brace sub n = 0 to the power of infinity such that v open parenthesis

z sub n minus z sub n plus 1 closing parenthesis right arrow infinity as n right arrow infinity comma thereexists z in K such that v open parenthesis z sub n minus z closing parenthesis right arrow infinity as n right

arrow infinity period For instance comma F sub q open parenthesis open parenthesis t closing parenthesis closingparenthesis i s

closed comma as is any finite extension of F sub q open parenthesis open parenthesis t closing parenthesis closingparenthesis period

Lemma 6 period 3 period 2 period .... Let K be a subfield of F sub q open parenthesis open parenthesis t to thepower of Q closing parenthesis closing parenthesis .... c los ed under the valuation

v period Let P open parenthesis z closing parenthesis .. be a nonzero polynomial over K comma .. and choose uin Q which does

not occur as a s lope of P period .. Then there exists a factorization P = QR comma whereQ is a polynomial over K with all s lopes less than u comma and R is a polynomialover K with all slopes greater than u periodProof period .. Pick r comma s in Q .. such that .. r less s .. and the interval .. open square bracket r comma s

closing square bracket .. contains .. ubut does not contain any slopes of P period Let S be the ring of formal sumssum sub j in Z a sub j z to the power of j over K such that v open parenthesis a sub j closing parenthesis plus rj

greater equal 0 for j greater equal 0 and v open parenthesis a sub j closing parenthesis plus sj greater equal 0 forj less or equal 0 period Define the valuation v sub u on S by the formulav sub u open parenthesis sum a sub j z to the power of j closing parenthesis = minimum j open brace v open

parenthesis a sub j closing parenthesis plus uj closing brace periodGiven x = sum a sub j z to the power of j in S comma write f open parenthesis x closing parenthesis = sum sub

j greater 0 a sub j z to the power of j comma so that f open parenthesis f open parenthesis x closing parenthesisclosing parenthesis = f open parenthesis x closing parenthesis period

Write P open parenthesis z closing parenthesis = sum sub i c sub i z to the power of i comma put d = deg Pcomma let e be the unique integer that

minimizes v open parenthesis c sub e closing parenthesis plus ue comma and put x sub 0 = c sub e to the powerof minus 1 z to the power of minus e P open parenthesis z closing parenthesis in S period Define the sequence

open brace x sub h closing brace sub h = 0 to the power of infinity by the recurrence x sub h plus 1 = x subh open parenthesis 1 minus f open parenthesis x sub h closing parenthesis closing parenthesis for h = 0 comma 1comma period period period period

Put .. l = v sub u open parenthesis x sub 0 minus 1 closing parenthesis comma .. so that l greater 0 period ..We now show by induction that

v sub u open parenthesis x sub h minus 1 closing parenthesis greater equal l and v sub u open parenthesis f openparenthesis x sub h closing parenthesis closing parenthesis greater equal open parenthesis h plus 1 closing parenthesisl for all h period .. The first inequality

clearly implies the second for h = 0 semicolon given both inequalities for some h comma wehaveLine 1 v sub u open parenthesis x sub h plus 1 minus 1 closing parenthesis = v sub u open parenthesis x sub

h minus x sub h f open parenthesis x sub h closing parenthesis minus 1 closing parenthesis Line 2 greater equalminimum open brace v sub u open parenthesis x sub h minus 1 closing parenthesis comma v sub u open parenthesisx sub h closing parenthesis plus v sub u open parenthesis f open parenthesis x sub h closing parenthesis closingparenthesis closing brace Line 3 greater equal minimum open brace l comma open parenthesis h plus 1 closingparenthesis l closing brace Line 4 greater equal l


with equality for k = f + h and k = e+ g but not for any kelement− slash[f +h, e+g]. It follows that the multiplicity of r as a slope of R i s e+g−(f+h) =(e− f) + (g − h), proving the desired result . �Corollary 6 . 2 . 3 . Let P (z) be a nonzero polynomial over Fq((tQ))whichfactors as a product Q1···Qn ofpure polynomials ( e . g . , linear polynomials) . Then for each r ∈ Q ∪ {∞}, the sum of the degrees of the Qi, overthose i for which Qi has s lope r, is equal to the multiplicity of r as aslope of P.6 . 3 . Slope splittings . We now recall a special form of Hensel’ s lemma , which makes it possible to split polynomials by their slopes .Definition 6 . 3 . 1 . A subfield K of Fq((tQ)) i s c los ed under thevaluation vif for any sequence {zn}∞n=0 such that v(zn − zn+1)→∞ as n→∞, thereexists z ∈ K such that v(zn − z)→∞ as n→∞. For instance ,Fq((t)) i sclosed , as is any finite extension of Fq((t)).Lemma 6 . 3 . 2 . Let K be a subfield of Fq((tQ)) c los ed under thevaluationv. Let P (z) be a nonzero polynomial over K, and choose u ∈ Q whichdoes not occur as a s lope of P. Then there exists a factorization P = QR,where Q is a polynomial over K with all s lopes less than u, and R is apolynomialover K with all slopes greater than u.Proof . Pick r, s ∈ Q such that r < s and the interval [r, s]contains u but does not contain any slopes of P. Let S be the ring offormal sums∑j∈Z ajz

j over K such that v(aj) + rj ≥ 0 for j ≥ 0 and v(aj) + sj ≥ 0 forj ≤ 0. Define the valuation vu on S by the formula

vu(∑

ajzj) = min

j{v(aj) + uj}.

Given x =∑ajz

j ∈ S, write f(x) =∑j>0 ajz

j , so that f(f(x)) = f(x).

Write P (z) =∑i ciz

i, put d = deg P, let e be the unique integer thatminimizes v(ce)+ue, and put x0 = c−1e z−eP (z) ∈ S. Define the sequence {xh}∞h=0

by the recurrence xh+1 = xh(1− f(xh)) for h = 0, 1, ....Put ` = vu(x0 − 1), so that ` > 0. We now show by induction

that vu(xh − 1) ≥ ` and vu(f(xh)) ≥ (h + 1)` for all h. The firstinequality clearly implies the second for h = 0; given both inequalities forsome h, wehave

vu(xh+1 − 1) = vu(xh − xhf(xh)− 1)

≥ min{vu(xh − 1), vu(xh) + vu(f(xh))}≥ min{`, (h+ 1)`}

≥ `

\noindent 406 \quad Kiran S . Kedlayaand

\ [\ begin { a l i gned } v { u } ( f ( x { h + 1 } ) ) = v { u } ( f( x { h } − x { h } f ( x { h } ) ) ) \\

= v { u } ( f ( x { h } − f ( x { h } ) − ( x { h } −1 ) f ( x { h } ) ) ) \\

= v { u } ( f ( x { h } ) − f ( f ( x { h } ) ) − f( ( x { h } − 1 ) f ( x { h } ) ) ) \\

= v { u } ( f ( ( x { h } − 1 ) f ( x { h } ) ) ) \\\geq v { u } ( ( x { h } − 1 ) f ( x { h } ) ) \\\geq \ e l l + ( h + 1 ) \ e l l \\= ( h + 2 ) \ e l l . \end{ a l i gned }\ ]

\noindent Thus the sequence $ \{ x { h } \} $ converges to a l i m i t $ x \ inS . $

By co n s t r uc t i o n $ , f ( x ) = 0 , $ so $ x $ has no p o s i t i v e powers o f$ z . $ By induct i on

on $ h , z ˆ{ e } x { h }$ has no negat ive powers o f $ z $ f o r each $ h, $ so $ z ˆ{ e } x $ a l s o has no negat ivepowers o f $ z . $ Thus $ z ˆ{ e } x $ i s a polynomial in $ z $ o f degree exac t l y

$ e , $ and i t s s l o p e s

\noindent are a l l g r e a t e r than or equal to $ s . $

\hspace ∗{\ f i l l }On the other hand , we have $ x = x { 0 } ( 1 − f ( x { 0 }) ) ( 1 − f ( x { 1 } ) ) \cdot \cdot \cdot , $ so that

\noindent $ x { 0 } x ˆ{ − 1 }$ has only p o s i t i v e powers o f $ z . $ However$ , z ˆ{ e − d } x { 0 } x ˆ{ − 1 }$ has no p o s i t i v e

\noindent powers o f $ z , $ so $ x { 0 } x ˆ{ − 1 }$ i s a polynomial in $ z $o f degree exac t l y $ d − e , $ and i t s

\noindent s l o p e s are a l l l e s s than or equal to $ r . $

\hspace ∗{\ f i l l }We thus obta in the d e s i r e d f a c t o r i z a t i o n as $ P = QR $ with $ Q( z ) = c { e } x { 0 } x ˆ{ − 1 }$

\ [ and R ( z ) = z ˆ{ e } x . \ square \ ]

\noindent Coro l l a ry 6 . 3 . 3 . \ h f i l l Let $ K $ be a s u b f i e l d o f $ F { q } (( t ˆ{ Q } ) ) $ c l o s e d under the va luat i on

\noindent $ v . $ \quad Then every monic polynomial $ P $ \quad over $ K $ \quad admits a unique f a c t o r i z a t i o n$ Q 1 \cdot \cdot \cdot Q { n }$ in to monic polynomia l s , \quad such that each

$ Q i $ i s pure o f some s lope

\begin { a l i g n ∗}\ tag ∗{$ s { i } , $} and s { 1 } < s { 2 } < \cdot \cdot \cdot <s { n } .\end{ a l i g n ∗}

\noindent Proof . \quad The e x i s t e n c e o f the f a c t o r i z a t i o n f o l l o w s from Lemma 6 . 3 . 2 ; \quad theuniqueness f o l l o w s from unique f a c t o r i z a t i o n f o r po lynomia l s over $ K $ andthe a d d i t i v i t y o f m u l t i p l i c i t i e s ( Lemma $ 6 . 2 . 2 ) . \ square $

\noindent D e f i n i t i o n 6 . 3 . 4 . \quad We c a l l the f a c t o r i z a t i o n g iven by Coro l l a ry 6 . 3 . 3 thes lope f a c t o r i z a t i o n o f the polynomial $ P . $

\noindent 6 . 4 . \quad Slope s p l i t t i ng s f o r tw i s t ed polynomia ls .

\noindent Remark 6 . 4 . 1 . \ h f i l l Throughout t h i s s e c t i o n , we w i l l use the f a c t that f o r$ K $

\noindent any s u b f i e l d o f $ F { q } ( ( t ˆ{ Q } ) ) , $ the va lua t i on$ v $ extends unique ly to any a l g e b r a i c

\noindent c l o s u r e o f $ K [ 18 , $ Propos i t i on I I . 3 ] . We do t h i s f o r conceptua l c l a r i t y ; not

\noindent doing so ( at the expense o f making the arguments mess i e r and more com −

\noindent puta t i ona l ) would not shed any l i g h t on how to compute with automatic

\noindent s e r i e s .

\noindent Lemma 6 . 4 . 2 . \ h f i l l Let $ K $ be a s u b f i e l d o f $ F { q } ( (t ˆ{ Q } ) ) $ \ h f i l l c l o s ed under the va lua t i on

\noindent $ v . $ Let $ P $ be a monic a d d i t i v e polynomial over $ K , $ and l e t$ Q 1 \cdot \cdot \cdot Q { n } t ˆ{ p ˆ{ d }}$ \ h f i l l be i t s

\noindent s lope f a c t o r i z a t i o n . \quad Then $ Q { n } t ˆ{ p ˆ{ d }}$ \quad i s a l s o a d d i t i v e .

406 .. Kiran S period KedlayaandLine 1 v sub u open parenthesis f open parenthesis x sub h plus 1 closing parenthesis closing parenthesis = v sub

u open parenthesis f open parenthesis x sub h minus x sub h f open parenthesis x sub h closing parenthesis closingparenthesis closing parenthesis Line 2 = v sub u open parenthesis f open parenthesis x sub h minus f open parenthesisx sub h closing parenthesis minus open parenthesis x sub h minus 1 closing parenthesis f open parenthesis x sub hclosing parenthesis closing parenthesis closing parenthesis Line 3 = v sub u open parenthesis f open parenthesis x subh closing parenthesis minus f open parenthesis f open parenthesis x sub h closing parenthesis closing parenthesis minusf open parenthesis open parenthesis x sub h minus 1 closing parenthesis f open parenthesis x sub h closing parenthesisclosing parenthesis closing parenthesis Line 4 = v sub u open parenthesis f open parenthesis open parenthesis x sub hminus 1 closing parenthesis f open parenthesis x sub h closing parenthesis closing parenthesis closing parenthesis Line5 greater equal v sub u open parenthesis open parenthesis x sub h minus 1 closing parenthesis f open parenthesis xsub h closing parenthesis closing parenthesis Line 6 greater equal l plus open parenthesis h plus 1 closing parenthesisl Line 7 = open parenthesis h plus 2 closing parenthesis l period

Thus the sequence open brace x sub h closing brace converges to a limit x in S periodBy construction comma f open parenthesis x closing parenthesis = 0 comma so x has no positive powers of z

period By inductionon h comma z to the power of e x sub h has no negative powers of z for each h comma so z to the power of e x

also has no negativepowers of z period Thus z to the power of e x is a polynomial in z of degree exactly e comma and it s slopesare all greater than or equal to s periodOn the other hand comma we have x = x sub 0 open parenthesis 1 minus f open parenthesis x sub 0 closing

parenthesis closing parenthesis open parenthesis 1 minus f open parenthesis x sub 1 closing parenthesis closingparenthesis times times times comma so that

x sub 0 x to the power of minus 1 has only positive powers of z period However comma z to the power of e minusd x sub 0 x to the power of minus 1 has no positive

powers of z comma so x sub 0 x to the power of minus 1 i s a polynomial in z of degree exactly d minus e commaand it s

slopes are all less than or equal to r periodWe thus obtain the desired factorization as P = QR with Q open parenthesis z closing parenthesis = c sub e x

sub 0 x to the power of minus 1and R open parenthesis z closing parenthesis = z to the power of e x period squareCorollary 6 period 3 period 3 period .... Let K be a subfield of F sub q open parenthesis open parenthesis t to

the power of Q closing parenthesis closing parenthesis closed under the valuationv period .. Then every monic polynomial P .. over K .. admits a unique factorizationQ 1 times times times Q sub n into monic polynomials comma .. such that each Q i is pure of some s lopeEquation: s sub i comma .. and s sub 1 less s sub 2 less times times times less s sub n periodProof period .. The existence of the factorization follows from Lemma 6 period 3 period 2 semicolon .. theuniqueness follows from unique factorization for polynomials over K andthe additivity of multiplicities open parenthesis Lemma 6 period 2 period 2 closing parenthesis period squareDefinition 6 period 3 period 4 period .. We call the factorization given by Corollary 6 period 3 period 3 thes lope factorization of the polynomial P period6 period 4 period .. Slope splitt ings for twisted polynomials periodRemark 6 period 4 period 1 period .... Throughout this section comma we will use the fact that for Kany subfield of F sub q open parenthesis open parenthesis t to the power of Q closing parenthesis closing paren-

thesis comma the valuation v extends uniquely to any algebraicclosure of K open square bracket 18 comma Proposition II period 3 closing square bracket period We do this for

conceptual clarity semicolon notdoing so open parenthesis at the expense of making the arguments messier and more com hyphenputational closing parenthesis would not shed any light on how to compute with automaticseries periodLemma 6 period 4 period 2 period .... Let K be a subfield of F sub q open parenthesis open parenthesis t to the

power of Q closing parenthesis closing parenthesis .... c los ed under the valuationv period Let P be a monic additive polynomial over K comma and let Q 1 times times times Q sub n t to the

power of p to the power of d .... be itss lope factorization period .. Then Q sub n t to the power of p to the power of d .. is also additive period

406 Kiran S . Kedlaya and

vu(f(xh+1)) = vu(f(xh − xhf(xh)))

= vu(f(xh − f(xh)− (xh − 1)f(xh)))

= vu(f(xh)− f(f(xh))− f((xh − 1)f(xh)))

= vu(f((xh − 1)f(xh)))

≥ vu((xh − 1)f(xh))

≥ `+ (h+ 1)`

= (h+ 2)`.

Thus the sequence {xh} converges to a limit x ∈ S.By construction , f(x) = 0, so x has no positive powers of z. By induction

on h, zexh has no negative powers of z for each h, so zex also has no negativepowers of z. Thus zex is a polynomial in z of degree exactly e, and it sslopesare all greater than or equal to s.

On the other hand , we have x = x0(1− f(x0))(1− f(x1)) · ··, so thatx0x−1 has only positive powers of z. However , ze−dx0x

−1 has no positivepowers of z, so x0x

−1 i s a polynomial in z of degree exactly d− e, and it sslopes are all less than or equal to r.

We thus obtain the desired factorization as P = QR with Q(z) = cex0x−1

andR(z) = zex. �

Corollary 6 . 3 . 3 . Let K be a subfield of Fq((tQ)) closed under thevaluationv. Then every monic polynomial P over K admits a unique factor-ization Q1 · · ·Qn into monic polynomials , such that each Qi is pure ofsome s lope

ands1 < s2 < · · · < sn. si,

Proof . The existence of the factorization follows from Lemma 6 . 3 .2 ; the uniqueness follows from unique factorization for polynomials overK and the additivity of multiplicities ( Lemma 6.2.2). �Definition 6 . 3 . 4 . We call the factorization given by Corollary 6 .3 . 3 the s lope factorization of the polynomial P.6 . 4 . Slope splitt ings for twisted polynomials .Remark 6 . 4 . 1 . Throughout this section , we will use the fact thatfor Kany subfield of Fq((tQ)), the valuation v extends uniquely to any algebraicclosure of K[18, Proposition II . 3 ] . We do this for conceptual clarity ; notdoing so ( at the expense of making the arguments messier and more com -putational ) would not shed any light on how to compute with automaticseries .Lemma 6 . 4 . 2 . Let K be a subfield of Fq((tQ)) c los ed under thevaluationv. Let P be a monic additive polynomial over K, and let Q1 · · ·Qntp

d

beitss lope factorization . Then Qnt

pd is also additive .


Proof . Let L be an algebraic closure of K. Let s1, ..., sn be the slopes ofP,and let V be the roots of P in L; then the possible valuations of the nonzeroelements of V are precisely s1, ..., sn. Moreover , an element of V i s a rootof Qnt

pd if and only if v(x) ≥ sn, and this subset of V is an Fp− subspace ofV. By Lemma 3.3.3, Qnt

pd i s also additive . �In terms of twisted polynomials , we obtain the following analogue of

theslope factorization .Definition 6 . 4 . 3 . Let P (F ) be a twisted polynomial over Fq((tQ)).Forr ∈ Q, we say P (F ) i s pure of s lope r if P has nonzero constant term andthe ordinary polynomial P (F )(z)/z i s pure of slope r. We conventionally saythat any power of F i s pure of slope ∞.Proposition 6 . 4 . 4 . Let K be a subfield of Fq((tQ)) c lo sed underthe val -uation v, and let P (F ) be a monic twisted polynomial over K. Thenthere exists a factorization P = Q1 · · · Qn of P into monic twistedpolynomialsover K, in which each Qi is pure of some slope .Proof . Let R(z) be the highest finite slope factor in the slope factor-ization of P (F )(z), t imes the slope ∞ factor ( a power of z). By Lemma6.4.2, R(z) i s additive , so we have R(z) = Q(F )(z) for some twisted polynomialQ. By Lemma 6 . 1 . 5 , we can factor P = P1Q for some P1 of lower degreethan P ;repeating the argument yields the claim . �

We can split further at the expense of enlarging q.Proposition 6 . 4 . 5 . Let K be a subfield of Fq((tQ)) closed underthe valu -ation v, and let P (F ) be a monic twisted polynomial over K whichis pure of s lope 0 . Then the polynomial P (F )(z) factorscompletely ( into linear factors ) over K ⊗Fq Fq′, for s ome powerq′ of q.

Proof . Write P (F ) =∑di=0 ciF

i with cd = 1. Since P i s pure of slope 0 ,we have v(c0) = 0 and v(ci) ≥ 0 for 1 ≤ i ≤ d− 1.

Let ai ∈ Fq be the constant coefficient of ci; then for some power q′ ofq, the polynomial

∑aiz

pi has pd distinct roots in Fq′. Let r ∈ Fq′be a nonzero root of

∑aiz

pi ; then P (F )(z+ r) has one slope greater than0 and all others equal to 0 . The slope factorization of P (F )(z+ r) then hasa linear factor ; in other words , P (F )(z) has a unique root r′ ∈ K ⊗Fq Fq′with v(r − r′) > 0. By the same argument , P (F )(z) factors completely over

K ⊗Fq Fq′,asdesired. �

Corollary 6 . 4 . 6 . Let K be a subfield of Fq((tQ)), containing allfractional powers of t and c los ed under the valuation v, andlet P (F ) be a monic twisted polynomial over K. Then forsome power q′ of q, there exists afactorization P = Q1 · · ·Qn of P into monic linear twistedpolynomials


\noindent Proof . \ h f i l l Let $ L $ be an a l g e b r a i c c l o s u r e o f $ K . $ Let $ s { 1 }, . . . , s { n }$ be the s l o p e s o f $ P , $

\noindent and l e t $ V $ be the roo t s o f $ P $ in $ L ; $ then the p o s s i b l e va lua t i o n s o f the nonzero

\noindent e lements o f $ V $ are p r e c i s e l y $ s { 1 } , . . . , s { n }. $ Moreover , an element o f $ V $ i s a root

\noindent o f $ Q { n } t ˆ{ p ˆ{ d }}$ i f and only i f $ v ( x ) \geq s { n }, $ and t h i s subset o f $ V $ i s an $ F { p } − $ subspace o f

\noindent $ V . $ By Lemma $ 3 . 3 . 3 , Q { n } t ˆ{ p ˆ{ d }}$i s a l s o a d d i t i v e $ . \ square $

\hspace ∗{\ f i l l } In terms o f tw i s t ed polynomia l s , we obta in the f o l l o w i n g analogue o f the

\noindent s l ope f a c t o r i z a t i o n .

\noindent D e f i n i t i o n \ h f i l l 6 . 4 . 3 . \ h f i l l Let $ P ( F ) $ \ h f i l l be a twi s t ed polynomial over$ F { q } ( ( t ˆ{ Q } ) ) . $ For

\noindent $ r \ in Q , $ we say $ P ( F ) $ i s pure o f s lope $ r $i f $ P $ has nonzero constant term andthe ord inary polynomial $ P ( F ) ( z ) / z $ i s pure o f s l ope $ r

. $ We conven t i ona l l ysay that any power o f $ F $ i s pure o f s l ope $ \ infty . $

\noindent Propos i t i on 6 . 4 . 4 . \ h f i l l Let $ K $ \ h f i l l be a s u b f i e l d o f $ F { q }( ( t ˆ{ Q } ) ) $ \ h f i l l c l o sed under the va l −

\noindent uat ion $ v , $ \quad and l e t $ P ( F ) $ \quad be a monic tw i s t ed polynomial over$ K . $ \quad Then there

e x i s t s a f a c t o r i z a t i o n $ P = Q 1 \cdot \cdot \cdot Q { n }$ \quad o f$ P $ in to monic tw i s t ed polynomia ls

\noindent over $ K , $ in which each $ Q i $ i s pure o f some s l ope .

\noindent Proof . \quad Let $ R ( z ) $ be the h i ghe s t f i n i t e s l ope f a c t o r in the s l ope f a c t o r i z a t i o no f $ P ( F ) ( z ) , $ t imes the s l ope $ \ infty $ f a c t o r ( a power o f

$ z ) . $ By Lemma $ 6 . 4 . 2 , R ( z ) $i s a d d i t i v e , so we have $ R ( z ) = Q ( F ) ( z ) $ f o r some twi s t ed polynomial

$ Q . $ ByLemma 6 . 1 . 5 , we can f a c t o r $ P = P { 1 } Q $ f o r some $ P { 1 }$ o f lower degree than

$ P ; $

\noindent r epea t ing the argument y i e l d s the c la im $ . \ square $

\centerline{We can s p l i t f u r t h e r at the expense o f e n l a r g i n g $ q . $ }

\noindent Propos i t i on 6 . 4 . 5 . \ h f i l l Let $ K $ be a s u b f i e l d o f $ F { q } (( t ˆ{ Q } ) ) $ \ h f i l l c l o s e d under the valu −

\noindent at i on $ v , $ \quad and l e t $ P ( F ) $ \quad be a monic tw i s t ed polynomial over$ K $ which i s pure

o f s lope 0 . \quad Then \quad the polynomial $ P ( F ) ( z ) $ \quad f a c t o r s \quad complete ly \quad ( i n to \quad l i n e a rf a c t o r s ) over $ K \otimes { F { q }} F { q } \prime { , }$ f o r s ome power

$ q ˆ{ \prime }$ o f $ q . $

\noindent Proof . \quad Write $ P ( F ) = \sum ˆ{ d } { i = 0 } c { i }F ˆ{ i }$ with $ c { d } = 1 . $ S ince $ P $ i s pure o f s l ope 0 , we

have $ v ( c { 0 } ) = 0 $ and $ v ( c { i } ) \geq 0 $ f o r$ 1 \ leq i \ leq d − 1 . $

\hspace ∗{\ f i l l }Let $ a { i } \ in F { q }$ be the constant c o e f f i c i e n t o f $ c { i }; $ then f o r some power $ q ˆ{ \prime }$ o f

\noindent $ q , $ \quad the polynomial $ \sum a { i } z ˆ{ p ˆ{ i }}$ \quad has$ p ˆ{ d }$ d i s t i n c t r oo t s in $ F { q } \prime { . }$ \quad Let $ r \ in F { q }\prime $ \quad be a

nonzero root o f $ \sum a { i } z ˆ{ p ˆ{ i }} ; $ \quad then $ P ( F )( z + r ) $ \quad has one s l ope g r e a t e r than 0and a l l o the r s equal to 0 . The s l ope f a c t o r i z a t i o n o f $ P ( F ) ( z +

r ) $ then has

\noindent a l i n e a r f a c t o r ; in other words $ , P ( F ) ( z ) $ has a unique root$ r ˆ{ \prime } \ in K \otimes { F { q }} F { q } \prime $

\noindent with $ v ( r − r ˆ{ \prime } ) > 0 . $ By the same argument$ , P ( F ) ( z ) $ f a c t o r s complete ly over

\ [ K \otimes { F { q }} F { q } \prime { , } as d e s i r e d . \ square \ ]

\noindent Coro l l a ry 6 . 4 . 6 . \quad Let $ K $ be a s u b f i e l d o f $ F { q } ( (t ˆ{ Q } ) ) , $ conta in ing a l l f r a c t i o n a l

powers \quad o f $ t $ \quad and c l o s ed under the \quad va luat i on $ v , $ \quad and l e t$ P ( F ) $ \quad be \quad a \quad monic

tw i s t ed polynomial \quad over $ K . $ \quad Then f o r some power $ q ˆ{ \prime }$\quad o f $ q , $ \quad the re \quad e x i s t s \quad a

\noindent f a c t o r i z a t i o n $ P = Q 1 \cdot \cdot \cdot Q { n }$ \ h f i l l o f$ P $ \ h f i l l i n t o \ h f i l l monic \ h f i l l l i n e a r \ h f i l l tw i s t ed polynomia l s

\begin { a l i g n ∗}over K \otimes { F { q }} F { q } \prime .\end{ a l i g n ∗}

Finite automata and algebraic extensions of function fields .. 407Proof period .... Let L be an algebraic closure of K period Let s sub 1 comma period period period comma s sub

n be the slopes of P commaand let V be the roots of P in L semicolon then the possible valuations of the nonzeroelements of V are precisely s sub 1 comma period period period comma s sub n period Moreover comma an

element of V i s a rootof Q sub n t to the power of p to the power of d if and only if v open parenthesis x closing parenthesis greater

equal s sub n comma and this subset of V is an F sub p hyphen subspace ofV period By Lemma 3 period 3 period 3 comma Q sub n t to the power of p to the power of d i s also additive

period squareIn terms of twisted polynomials comma we obtain the following analogue of theslope factorization periodDefinition .... 6 period 4 period 3 period .... Let P open parenthesis F closing parenthesis .... be a twisted

polynomial over F sub q open parenthesis open parenthesis t to the power of Q closing parenthesis closing parenthesisperiod For

r in Q comma we say P open parenthesis F closing parenthesis i s pure of s lope r if P has nonzero constant termand

the ordinary polynomial P open parenthesis F closing parenthesis open parenthesis z closing parenthesis slash zi s pure of slope r period We conventionally

say that any power of F i s pure of slope infinity periodProposition 6 period 4 period 4 period .... Let K .... be a subfield of F sub q open parenthesis open parenthesis

t to the power of Q closing parenthesis closing parenthesis .... c lo sed under the val hyphenuation v comma .. and let P open parenthesis F closing parenthesis .. be a monic twisted polynomial over K

period .. Then thereexists a factorization P = Q 1 times times times Q sub n .. of P into monic twisted polynomialsover K comma in which each Q i is pure of some slope periodProof period .. Let R open parenthesis z closing parenthesis be the highest finite slope factor in the slope

factorizationof P open parenthesis F closing parenthesis open parenthesis z closing parenthesis comma t imes the slope

infinity factor open parenthesis a power of z closing parenthesis period By Lemma 6 period 4 period 2 comma Ropen parenthesis z closing parenthesis

i s additive comma so we have R open parenthesis z closing parenthesis = Q open parenthesis F closing parenthesisopen parenthesis z closing parenthesis for some twisted polynomial Q period By

Lemma 6 period 1 period 5 comma we can factor P = P sub 1 Q for some P sub 1 of lower degree than Psemicolon

repeating the argument yields the claim period squareWe can split further at the expense of enlarging q periodProposition 6 period 4 period 5 period .... Let K be a subfield of F sub q open parenthesis open parenthesis t to

the power of Q closing parenthesis closing parenthesis .... closed under the valu hyphenation v comma .. and let P open parenthesis F closing parenthesis .. be a monic twisted polynomial over K which

is pureof s lope 0 period .. Then .. the polynomial P open parenthesis F closing parenthesis open parenthesis z closing

parenthesis .. factors .. completely .. open parenthesis into .. linearfactors closing parenthesis over K oslash sub F sub q F sub q prime sub comma for s ome power q to the power

of prime of q periodProof period .. Write P open parenthesis F closing parenthesis = sum sub i = 0 to the power of d c sub i F to

the power of i with c sub d = 1 period Since P i s pure of slope 0 comma wehave v open parenthesis c sub 0 closing parenthesis = 0 and v open parenthesis c sub i closing parenthesis greater

equal 0 for 1 less or equal i less or equal d minus 1 periodLet a sub i in F sub q be the constant coefficient of c sub i semicolon then for some power q to the power of

prime ofq comma .. the polynomial sum a sub i z to the power of p to the power of i .. has p to the power of d distinct

roots in F sub q prime sub period .. Let r in F sub q prime .. be anonzero root of sum a sub i z to the power of p to the power of i semicolon .. then P open parenthesis F closing

parenthesis open parenthesis z plus r closing parenthesis .. has one slope greater than 0and all others equal to 0 period The slope factorization of P open parenthesis F closing parenthesis open paren-

thesis z plus r closing parenthesis then hasa linear factor semicolon in other words comma P open parenthesis F closing parenthesis open parenthesis z

closing parenthesis has a unique root r to the power of prime in K oslash sub F sub q F sub q primewith v open parenthesis r minus r to the power of prime closing parenthesis greater 0 period By the same argument

comma P open parenthesis F closing parenthesis open parenthesis z closing parenthesis factors completely overK oslash sub F sub q F sub q prime sub comma as desired period squareCorollary 6 period 4 period 6 period .. Let K be a subfield of F sub q open parenthesis open parenthesis t to the

power of Q closing parenthesis closing parenthesis comma containing all fractionalpowers .. of t .. and c los ed under the .. valuation v comma .. and let P open parenthesis F closing parenthesis

.. be .. a .. monictwisted polynomial .. over K period .. Then for some power q to the power of prime .. of q comma .. there ..

exists .. afactorization P = Q 1 times times times Q sub n .... of P .... into .... monic .... linear .... twisted polynomialsover K oslash sub F sub q F sub q prime period

overK ⊗Fq Fq′.


\noindent Proof . \quad We may proceed by induct i on on deg $ P ; $ i t s u f f i c e s to show that i f$ P $ i s not l i n e a r , then i t i s a l e f t mu l t ip l e o f some l i n e a r tw i s t ed polynomial

\noindent over $ K \otimes { F { q }} F { q } \prime $ f o r some $ q ˆ{ \prime }. $ By Propos i t i on 6 . 4 . 4 , we may reduce to the case

\noindent where $ P $ i s pure o f some s l ope ; by r e s c a l i n g the polynomial $ P (F ) ( z ) ( $ and

\noindent us ing the f a c t that $ K $ conta in s a l l f r a c t i o n a l powers o f $ t ) , $we may reduce

\noindent to the case where $ P $ i s pure o f s l ope 0 . By Propos i t i on $ 6 . 4. 5 , P ( F ) ( z ) $ then

\noindent s p l i t s complete ly over $ K \otimes { F { q }} F { q } \prime $ f o r some$ q ˆ{ \prime } . $ Choose any one − dimens iona l

\noindent $ F { p } − $ subspace o f the s e t o f r oo t s o f $ P ( F ) ( z) ; $ by Lemma 3 . 3 . 3 , the se form the

\noindent r oo t s o f $ Q ( F ) ( z ) $ f o r some monic l i n e a r tw i s t ed polynomial$ Q $ over $ K \otimes { F { q }} F { q } \prime { . }$

\noindent By Lemma 6 . 1 . 5 , we can wr i t e $ P = P { 0 } Q $ f o r some $ P { 0 }; $ t h i s completes the

induct i on and y i e l d s the d e s i r e d r e s u l t $ . \ square $

\centerline {7 . \quad Proof o f the main theorem : conc re t e approach }

In t h i s chapter , we study the p r o p e r t i e s o f automatic power s e r i e s morec l o s e l y , with the end o f g i v ing a more down − to − earth proo f ( f r e e o f any de −

\noindent pendence on Galo i s theory or the l i k e ) o f the ‘ ‘ a l g e b r a i c i m p l i e s automatic ’ ’i m p l i c a t i o n o f Theorem 4 . 1 . 3 . In so doing , we int roduce some techn iqueswhich may be o f use in e x p l i c i t l y computing in the a l g e b r a i c c l o s u r e o f

$ F { q } ( t ) ; $ however , we have not made any attempt to \quad ‘ ‘ p r a c t i c a l i z e ’ ’ \quad the se tech −

\noindent niques . Just how e f f i c i e n t l y one can do such computing i s a problem worthyo f f u r t h e r study ; we d i s c u s s t h i s ques t i on b r i e f l y in the next chapter .

\noindent $ 7 . 1 . T−r $ a n s i t i o n digraphs o f automata . \quad We begin the chapter with amore conc re t e study o f the automata that g ive r i s e to g e n e r a l i z e d power

\noindent s e r i e s .

\noindent D e f i n i t i o n 7 . 1 . 1 . \quad Let $ M $ be a DFAO . Given a s t a t e $ q 1\ in Q , $ a s t a t e $ q \ in Q $

i s \quad sa id to be \quad r eachab l e from $ q 1 $ \quad i f $ \delta ˆ{ ∗ } (q 1 , s ) = q $ f o r some s t r i n g $ s \ in \Sigma ˆ{ ∗ } , $

and \quad unreachable from $ q 1 $ \quad otherwi se ; \quad i f $ q 1 = q0 , $ \quad we \quad s imply \quad say that \quad $ q $ i s

r eachab l e or unreachable . Any s t a t e from which a s t a t e unreachable from$ q 0 $ \quad can be reached must i t s e l f be unreachable from $ q 0 . $ \quad A DFAO with no

unreachable s t a t e s i s s a id to be minimal ; g iven any DFAO , one can removea l l o f i t s unreachable s t a t e s to obta in a minimal DFAO that accept s thesame language .

\noindent D e f i n i t i o n \quad 7 . 1 . 2 . \quad A s t a t e $ q \ in Q $ i s sa id to be r e l e v a n t i f the re e x i s t s af i n a l s t a t e reachab l e from $ q , $ and i r r e l e v a n t otherwi se . Any s t a t e which i sr eachab l e from an i r r e l e v a n t s t a t e must i t s e l f be i r r e l e v a n t .

\noindent D e f i n i t i o n 7 . 1 . 3 . \ h f i l l Let $ M $ be a DFA with input alphabet $ \Sigma { b }= \{ 0 , . . . , b − $

\noindent 1 , . \} . We say $ M $ i s we l l − formed ( resp . we l l − ordered ) i f the language acceptedby $ M $ c o n s i s t s o f the v a l i d base $ b $ expans ions o f the e lements o f an a r b i t r a r y( resp . a we l l − ordered ) subset o f $ S { b } . $

408 .. Kiran S period KedlayaProof period .. We may proceed by induction on deg P semicolon it suffices to show that ifP is not linear comma then it i s a left multiple of some linear twisted polynomialover K oslash sub F sub q F sub q prime for some q to the power of prime period By Proposition 6 period 4

period 4 comma we may reduce to the casewhere P i s pure of some slope semicolon by rescaling the polynomial P open parenthesis F closing parenthesis

open parenthesis z closing parenthesis open parenthesis andusing the fact that K contains all fractional powers of t closing parenthesis comma we may reduceto the case where P i s pure of slope 0 period By Proposition 6 period 4 period 5 comma P open parenthesis F

closing parenthesis open parenthesis z closing parenthesis thensplits completely over K oslash sub F sub q F sub q prime for some q to the power of prime period Choose any

one hyphen dimensionalF sub p hyphen subspace of the set of roots of P open parenthesis F closing parenthesis open parenthesis z closing

parenthesis semicolon by Lemma 3 period 3 period 3 comma these form theroots of Q open parenthesis F closing parenthesis open parenthesis z closing parenthesis for some monic linear

twisted polynomial Q over K oslash sub F sub q F sub q prime sub periodBy Lemma 6 period 1 period 5 comma we can write P = P sub 0 Q for some P sub 0 semicolon this completes

theinduction and yields the desired result period square7 period .. Proof of the main theorem : concrete approachIn this chapter comma we study the properties of automatic power series moreclosely comma with the end of giving a more down hyphen to hyphen earth proof open parenthesis free of any

de hyphenpendence on Galois theory or the like closing parenthesis of the quotedblleft algebraic implies automatic quoted-

blrightimplication of Theorem 4 period 1 period 3 period In so doing comma we introduce some techniqueswhich may be of use in explicitly computing in the algebraic closure ofF sub q open parenthesis t closing parenthesis semicolon however comma we have not made any attempt to ..

quotedblleft practicalize quotedblright .. these tech hyphenniques period Just how efficiently one can do such computing is a problem worthyof further study semicolon we discuss this question briefly in the next chapter period7 period 1 period T-r ansition digraphs of automata period .. We begin the chapter with amore concrete study of the automata that give rise to generalized powerseries periodDefinition 7 period 1 period 1 period .. Let M be a DFAO period Given a state q 1 in Q comma a state q in Qi s .. said to be .. reachable from q 1 .. if delta to the power of * open parenthesis q 1 comma s closing parenthesis

= q for some string s in Capital Sigma to the power of * commaand .. unreachable from q 1 .. otherwise semicolon .. if q 1 = q 0 comma .. we .. simply .. say that .. q i sreachable or unreachable period Any state from which a state unreachable fromq 0 .. can be reached must itself be unreachable from q 0 period .. A DFAO with nounreachable states i s said to be minimal semicolon given any DFAO comma one can removeall of its unreachable states to obtain a minimal DFAO that accepts thesame language periodDefinition .. 7 period 1 period 2 period .. A state q in Q is said to be relevant if there exists afinal state reachable from q comma and irrelevant otherwise period Any state which i sreachable from an irrelevant state must it self be irrelevant periodDefinition 7 period 1 period 3 period .... Let M be a DFA with input alphabet Capital Sigma sub b = open brace

0 comma period period period comma b minus1 comma period closing brace period We say M i s well hyphen formed open parenthesis resp period well hyphen

ordered closing parenthesis if the language acceptedby M consists of the valid base b expansions of the elements of an arbitraryopen parenthesis resp period a well hyphen ordered closing parenthesis subset of S sub b period


Proof . We may proceed by induction on deg P ; it suffices to show that ifP is not linear , then it i s a left multiple of some linear twisted polynomialover K ⊗Fq Fq′ for some q′. By Proposition 6 . 4 . 4 , we may reduce to thecasewhere P i s pure of some slope ; by rescaling the polynomial P (F )(z) ( andusing the fact that K contains all fractional powers of t), we may reduceto the case where P i s pure of slope 0 . By Proposition 6.4.5, P (F )(z) thensplits completely over K ⊗Fq Fq′ for some q′. Choose any one - dimensionalFp− subspace of the set of roots of P (F )(z); by Lemma 3 . 3 . 3 , these formtheroots of Q(F )(z) for some monic linear twisted polynomial Q over K ⊗Fq Fq′.By Lemma 6 . 1 . 5 , we can write P = P0Q for some P0; this completes theinduction and yields the desired result . �

7 . Proof of the main theorem : concrete approachIn this chapter , we study the properties of automatic power series more

closely , with the end of giving a more down - to - earth proof ( free of anyde -pendence on Galois theory or the like ) of the “ algebraic implies automatic ”implication of Theorem 4 . 1 . 3 . In so doing , we introduce some techniqueswhich may be of use in explicitly computing in the algebraic closure of Fq(t);however , we have not made any attempt to “ practicalize ” these tech-niques . Just how efficiently one can do such computing is a problem worthyof further study ; we discuss this question briefly in the next chapter .7.1. T− r ansition digraphs of automata . We begin the chapterwith a more concrete study of the automata that give rise to generalizedpowerseries .Definition 7 . 1 . 1 . Let M be a DFAO . Given a state q1 ∈ Q,a state q ∈ Q i s said to be reachable from q1 if δ∗(q1, s) = qfor some string s ∈ Σ∗, and unreachable from q1 otherwise ; ifq1 = q0, we simply say that q i s reachable or unreachable .Any state from which a state unreachable from q0 can be reached mustitself be unreachable from q0. A DFAO with no unreachable states i s saidto be minimal ; given any DFAO , one can remove all of its unreachablestates to obtain a minimal DFAO that accepts the same language .Definition 7 . 1 . 2 . A state q ∈ Q is said to be relevant ifthere exists a final state reachable from q, and irrelevant otherwise . Anystate which i s reachable from an irrelevant state must it self be irrelevant.Definition 7 . 1 . 3 . Let M be a DFA with input alphabetΣb = {0, ..., b−1 , . } . We say M i s well - formed ( resp . well - ordered ) if the languageaccepted by M consists of the valid base b expansions of the elements of anarbitrary ( resp . a well - ordered ) subset of Sb.


The goa l o f t h i s s e c t i o n i s to c h a r a c t e r i z e minimal we l l − ordered DFAs v iat h e i r t r a n s i t i o n graphs ; to do so , we use the f o l l o w i n g d e f i n i t i o n s . ( By way o fmot ivat ion , a connected und i rec ted graph i s c a l l e d a ‘ ‘ cactus ’ ’ i f each ver texo f the graph l i e s on exac t l y one minimal c y c l e . In r e a l l i f e , the saguaro i sa s p e c i f i c type o f cactus ind igenous to the southwest United Sta t e s andnorthwest Mexico . )

\noindent D e f i n i t i o n \quad 7 . 1 . 4 . \quad A d i r e c t e d graph $ G = ( V ,E ) $ \quad equipped with a d i s t i n −

guished ver tex $ v \ in V $ i s c a l l e d a rooted saguaro i f i t s a t i s f i e s the f o l l o w i n gc o n d i t i o n s .

\centerline {( a ) \quad Each ver tex o f $ G $ l i e s on at most one minimal c y c l e . }

\centerline {( b ) \quad There e x i s t d i r e c t e d paths from $ v $ to each ver tex o f $ G. $ }

\noindent In t h i s case , we say that $ v $ i s a root o f the saguaro . ( Note that a saguaro mayhave more than one root , because any ver tex l y i n g on the same minimalc y c l e as a root i s a l s o a root . ) A minimal c y c l e o f a rooted saguaro i s c a l l e d

\noindent a l o be . An edge o f a rooted saguaro i s c y c l i c i f i t l i e s on a lobe and a c y c l i co therw i se .

\noindent D e f i n i t i o n 7 . 1 . 5 . Let $ G = ( V , E ) $ be a rooted saguaro . A proper$ b − $ l a b e l i n g

\noindent o f $ G $ i s a func t i on $ \ e l l : E \rightarrow \{ 0 , . .. , b − 1 \} $ with the f o l l o w i n g p r o p e r t i e s .

\centerline {( a ) \quad I f $ v , w , x \ in V $ and $ vw , vx \ inE , $ then $ \ e l l ( vw ) \not= \ e l l ( vx ) . $ }

\hspace ∗{\ f i l l }( b ) \quad I f $ v , w , x \ in V , vw \ in E $ l i e s on a lobe , and$ vx \ in E $ does not l i e on a

\ [ l obe , then \ e l l ( vw ) > \ e l l ( vx ) . \ ]

\noindent Theorem 7 . 1 . 6 . Let $ M $ be a DFA with input alphabet $ \Sigma { b }, $ which i s minimal

\noindent and we l l − formed . \quad Then $ M $ \quad i s we l l − ordered i f and only i f f o r each r e l e v a n tpos t rad ix s t a t e $ q , $ the subgraph $ G { q }$ o f the t r a n s i t i o n graph c o n s i s t i n g o f r e l −evant s t a t e s r eachab l e from $ q $ i s a rooted saguaro with root $ q , $ \quad equipped witha proper $ b − $ l a b e l i n g .

\noindent Proof . \quad F i r s t suppose that $ M $ i s we l l − ordered . Let $ q $ be a r e l e v a n t pos t rad ixs t a t e o f $ M ; $ \quad note that \quad a l l t r a n s i t i o n s \quad from \quad $ q $\quad are l a b e l e d \quad by elements \quad o f

$ \{ 0 , . . . , b − 1 \} , $ s i n c e a v a l i d base $ b $ expansion cannot have two rad ix po in t s .Suppose that $ q $ admits a c y c l i c t r a n s i t i o n by $ s \ in \{ 0 , . .

. , b − 1 \} $ and a l s o

\noindent admits a t r a n s i t i o n to a r e l e v a n t s t a t e by $ s ˆ{ \prime } \not= s . $We w i l l show that $ s > s ˆ{ \prime } . $

To see t h i s , choose $ w \ in \Sigma ˆ{ ∗ }$ o f minimal l ength such that $ \delta ˆ{ ∗ }( q , sw ) = q . $

S ince $ M $ i s minimal $ , q $ i s r eachab l e , so we can choose $ w { 0 } \ in\Sigma ˆ{ ∗ }$ \quad such that

$ \delta ˆ{ ∗ } ( q 0 , w { 0 } ) = q . $ S ince $ \delta ( q, s ˆ{ \prime } ) $ i s r e l e v a n t , we can choose $ w { 1 } \ in \Sigma ˆ{ ∗ }$such that

\noindent $ \delta ˆ{ ∗ } ( q , s ˆ{ \prime } w { 1 } ) \ in F . $ Then a l l o f the s t r i n g s

\ [ w { 0 } s ˆ{ \prime } w { 1 } , w { 0 } sws ˆ{ \prime } w { 1 } , w { 0 }swsws ˆ{ \prime } w { 1 } , . . . \ ]

\noindent are accepted by $ M ; $ however , i f $ s ˆ{ \prime } > s , $ the se form the v a l i d base$ b $ expans ions

o f an i n f i n i t e dec r ea s ing sequence , which would c o n t r a d i c t the f a c t that $ M $

\noindent i s we l l − formed . Hence $ s > s ˆ{ \prime } . $

Finite automata and algebraic extensions of function fields .. 409The goal of this section i s to characterize minimal well hyphen ordered DFAs viatheir transition graphs semicolon to do so comma we use the following definitions period open parenthesis By

way ofmotivation comma a connected undirected graph i s called a quotedblleft cactus quotedblright if each vertexof the graph lies on exactly one minimal cycle period In real life comma the saguaro i sa specific type of cactus indigenous to the southwest United States andnorthwest Mexico period closing parenthesisDefinition .. 7 period 1 period 4 period .. A directed graph G = open parenthesis V comma E closing parenthesis

.. equipped with a distin hyphenguished vertex v in V i s called a rooted saguaro if it satisfies the followingconditions periodopen parenthesis a closing parenthesis .. Each vertex of G lies on at most one minimal cycle periodopen parenthesis b closing parenthesis .. There exist directed paths from v to each vertex of G periodIn this case comma we say that v i s a root of the saguaro period open parenthesis Note that a saguaro mayhave more than one root comma because any vertex lying on the same minimalcycle as a root i s also a root period closing parenthesis A minimal cycle of a rooted saguaro is calleda lo be period An edge of a rooted saguaro i s cyclic if it lies on a lobe and acyclicotherwise periodDefinition 7 period 1 period 5 period Let G = open parenthesis V comma E closing parenthesis be a rooted

saguaro period A proper b hyphen labelingof G is a function l : E right arrow open brace 0 comma period period period comma b minus 1 closing brace

with the following properties periodopen parenthesis a closing parenthesis .. If v comma w comma x in V and vw comma vx in E comma then l open

parenthesis vw closing parenthesis negationslash-equal l open parenthesis vx closing parenthesis periodopen parenthesis b closing parenthesis .. If v comma w comma x in V comma vw in E lies on a lobe comma and

vx in E does not lie on alobe comma then l open parenthesis vw closing parenthesis greater l open parenthesis vx closing parenthesis

periodTheorem 7 period 1 period 6 period Let M be a DFA with input alphabet Capital Sigma sub b comma which is

minimaland well hyphen formed period .. Then M .. is well hyphen ordered if and only if for each relevantpostradix state q comma the subgraph G sub q of the transition graph consisting of rel hyphenevant states reachable from q is a rooted saguaro with root q comma .. equipped witha proper b hyphen labeling periodProof period .. First suppose that M is well hyphen ordered period Let q be a relevant postradixstate of M semicolon .. note that .. all transitions .. from .. q .. are labeled .. by elements .. ofopen brace 0 comma period period period comma b minus 1 closing brace comma since a valid base b expansion

cannot have two radix points periodSuppose that q admits a cyclic transition by s in open brace 0 comma period period period comma b minus 1

closing brace and alsoadmits a transition to a relevant state by s to the power of prime negationslash-equal s period We will show that

s greater s to the power of prime periodTo see this comma choose w in Capital Sigma to the power of * of minimal length such that delta to the power

of * open parenthesis q comma sw closing parenthesis = q periodSince M i s minimal comma q is reachable comma so we can choose w sub 0 in Capital Sigma to the power of *

.. such thatdelta to the power of * open parenthesis q 0 comma w sub 0 closing parenthesis = q period Since delta open

parenthesis q comma s to the power of prime closing parenthesis i s relevant comma we can choose w sub 1 in CapitalSigma to the power of * such that

delta to the power of * open parenthesis q comma s to the power of prime w sub 1 closing parenthesis in F periodThen all of the strings

w sub 0 s to the power of prime w sub 1 comma w sub 0 sws to the power of prime w sub 1 comma w sub 0swsws to the power of prime w sub 1 comma period period period

are accepted by M semicolon however comma if s to the power of prime greater s comma these form the validbase b expansions

of an infinite decreasing sequence comma which would contradict the fact that Mi s well hyphen formed period Hence s greater s to the power of prime period


The goal of this section i s to characterize minimal well - ordered DFAsvia their transition graphs ; to do so , we use the following definitions . (By way of motivation , a connected undirected graph i s called a “ cactus ”if each vertex of the graph lies on exactly one minimal cycle . In real life ,the saguaro i s a specific type of cactus indigenous to the southwest UnitedStates and northwest Mexico . )Definition 7 . 1 . 4 . A directed graph G = (V,E) equippedwith a distin - guished vertex v ∈ V i s called a rooted saguaro if it satisfiesthe following conditions .

( a ) Each vertex of G lies on at most one minimal cycle .( b ) There exist directed paths from v to each vertex of G.

In this case , we say that v i s a root of the saguaro . ( Note that asaguaro may have more than one root , because any vertex lying on thesame minimal cycle as a root i s also a root . ) A minimal cycle of a rootedsaguaro is calleda lo be . An edge of a rooted saguaro i s cyclic if it lies on a lobe andacyclic otherwise .Definition 7 . 1 . 5 . Let G = (V,E) be a rooted saguaro . A proper b−labelingof G is a function ` : E → {0, ..., b− 1} with the following properties .

( a ) If v, w, x ∈ V and vw, vx ∈ E, then `(vw) 6= `(vx).( b ) If v, w, x ∈ V, vw ∈ E lies on a lobe , and vx ∈ E does not lie on a

lobe, then`(vw) > `(vx).

Theorem 7 . 1 . 6 . Let M be a DFA with input alphabet Σb, which isminimaland well - formed . Then M is well - ordered if and only if foreach relevant postradix state q, the subgraph Gq of the transition graphconsisting of rel - evant states reachable from q is a rooted saguaro withroot q, equipped with a proper b− labeling .Proof . First suppose that M is well - ordered . Let q be a relevantpostradix state of M ; note that all transitions from q are labeledby elements of {0, ..., b−1}, since a valid base b expansion cannot have tworadix points . Suppose that q admits a cyclic transition by s ∈ {0, ..., b−1}and alsoadmits a transition to a relevant state by s′ 6= s. We will show that s > s′.To see this , choose w ∈ Σ∗ of minimal length such that δ∗(q, sw) = q. SinceM i s minimal , q is reachable , so we can choose w0 ∈ Σ∗ such thatδ∗(q0, w0) = q. Since δ(q, s′) i s relevant , we can choose w1 ∈ Σ∗ such thatδ∗(q, s′w1) ∈ F. Then all of the strings

w0s′w1, w0sws

′w1, w0swsws′w1, ...

are accepted by M ; however , if s′ > s, these form the valid base b expansionsof an infinite decreasing sequence , which would contradict the fact that Mi s well - formed . Hence s > s′.

\noindent 4 1 0 \quad Kiran S . Kedlaya

This i m p l i e s immediately that a r e l e v a n t pos t rad ix s t a t e cannot l i e onmore than one minimal c y c l e , so $ G { q }$ i s a rooted saguaro . Moreover , no two

\noindent edges from the same ver tex o f $ G { q }$ carry the same l a b e l , and what we j u s t

\noindent proved i m p l i e s that any c y c l i c edge must car ry a g r e a t e r l a b e l than anyother edge from the same ver tex . Hence $ G { q }$ c a r r i e s a proper $ b − $ l a b e l i n g .

\hspace ∗{\ f i l l }Suppose conve r s e l y that each $ G { q }$ i s a rooted saguaro ca r ry ing a proper$ b − $

\noindent l a b e l i n g , but assume by way o f c o n t r a d i c t i o n that the language accepted by$ M $ \quad i n c l u d e s \quad an \quad i n f i n i t e \quad dec r ea s ing \quad sequence \quad

$ x { 1 } , x { 2 } , . . . . $ \quad We \quad say \quad the$ m − $ th d i g i t o f the base $ b $ expansion o f $ x { i } ( $ count ing from the l e f t ) i s s t a t i c

i f $ x { i } , x { i + 1 } , . . . $ \quad a l l have the same $ m− $ th d i g i t . Then the expansion o f each $ x { i }$

beg ins with an i n i t i a l segment \quad ( p o s s i b l y empty ) \quad o f s t a t i c d i g i t s , \quad and thenumber o f s t a t i c d i g i t s tends to i n f i n i t y with $ i . $ There e x i s t s a unique i n f i −n i t e sequence $ s { 1 } , s { 2 } , . . . $ \quad o f e lements o f $ \Sigma $

such that each i n i t i a l segment o fs t a t i c d i g i t s o c cu r r ing in the expansion o f some $ x { i }$ has the form $ s { 1 }\cdot \cdot \cdot s { m }$

f o r some $ m ( $ depending on $ i ) . $

The number o f pre rad ix d i g i t s in the expansion o f $ x { i }$ \quad i s \quad at \quad most \quad thecorre spond ing number f o r $ x { 1 } , $ so the sequence $ s { 1 } , s { 2 }

, . . . $ \quad must inc lude one( and only one ) \quad rad ix po int . \quad Def ine $ q m = \delta ˆ{ ∗ } (

q 0 , s { 1 } \cdot \cdot \cdot s { m } ) ; $ \quad then each \quad$ q m $

i s r e l e v a n t , \quad and $ q m $ \quad i s po s t rad ix f o r $ m $ \quad s u f f i c i e n t l y l a r g e . \quad Moreover , \quad onlyf i n i t e l y many o f the pos t rad ix t r a n s i t i o n s from $ q m $ to $ q m + 1 $ can be a c y c l i c ,as o therwi se some such t r a n s i t i o n would repeat , but then would v i s i b l y bepart o f a c y c l e .

Choose $ x { i }$ in the dec r ea s ing sequence whose f i r s t $ m $ d i g i t s are a l l s t a t i c ,f o r $ m $ l a r g e enough that $ q m $ i s pos t rad ix and the t r a n s i t i o n from $ q

n $ to $ q n + 1 $i s c y c l i c f o r a l l $ n \geq m ( $ again , t h i s amounts to tak ing $ i $ s u f f i c i e n t l y l a r g e ) .Let $ t { 1 } t { 2 } \cdot \cdot \cdot \ in \Sigma ˆ{ ∗ }$ denote the base

$ b $ expansion o f $ x { i } . $ S ince $ x { i + 1 } < x { i } , $ the ree x i s t s \quad a s m a l l e s t \quad i n t e g e r $ n > m $ such that \quad $ s { n } \not=

t { n } , $ \quad and that \quad i n t e g e r $ n $s a t i s f i e s $ s { n } < t { n } . $ However , \quad the t r a n s i t i o n from $ q

n $ to $ q n + 1 $ \quad i s c y c l i c , \quad andso the i n e q u a l i t y \quad $ s { n } < t { n }$ \quad v i o l a t e s \quad the d e f i n i t i o n o f a proper \quad

$ b − $ l a b e l i n g .This \quad c o n t r a d i c t i o n \quad means \quad that \quad the re \quad cannot \quad e x i s t \quad an i n f i n i t e \quad dec r ea s ingsubsequence , so $ M $ i s we l l − ordered , as d e s i r e d $ . \ square $

\noindent 7 . 2 . \quad Arithmet ic f o r we l l − ordered automata . \quad We next v e r i f y d i r e c t l ythat the s e t o f automatic s e r i e s i s a r ing ; t h i s f o l l o w s from Theorem 4 . 1 . 3and Lemma 3 . 2 . 5 , but g i v ing a d i r e c t proo f w i l l on one hand lead to a secondproo f o f Theorem 4 . 1 . 3 , and on the other hand sugges t ways to implementcomputations on automatic s e r i e s in p r a c t i c e . ( However , we have not madeany attempt here to opt imize the e f f i c i e n c y o f the computations ; t h i s w i l lr e q u i r e f u r t h e r study . )

\centerline{We f i r s t note that adding automatic s e r i e s i s easy . }

\noindent Lemma 7 . 2 . 1 . \ h f i l l Let $ x , y \ in F { q } ( ( t ˆ{ Q }) ) $ \ h f i l l be $ p − $ automatic ( re sp $ . p − $ quas i − automa −

\noindent t i c ) . \quad Then $ x + y $ i s $ p − $ automatic ( re sp $ . p− $ quas i − automatic ) .

4 1 0 .. Kiran S period KedlayaThis implies immediately that a relevant postradix state cannot lie onmore than one minimal cycle comma so G sub q i s a rooted saguaro period Moreover comma no twoedges from the same vertex of G sub q carry the same label comma and what we justproved implies that any cyclic edge must carry a greater label than anyother edge from the same vertex period Hence G sub q carries a proper b hyphen labeling periodSuppose conversely that each G sub q i s a rooted saguaro carrying a proper b hyphenlabeling comma but assume by way of contradiction that the language accepted byM .. includes .. an .. infinite .. decreasing .. sequence .. x sub 1 comma x sub 2 comma period period period

period .. We .. say .. them hyphen th digit of the base b expansion of x sub i open parenthesis counting from the left closing parenthesis

is staticif x sub i comma x sub i plus 1 comma period period period .. all have the same m hyphen th digit period Then

the expansion of each x sub ibegins with an initial segment .. open parenthesis possibly empty closing parenthesis .. of static digits comma ..

and thenumber of static digits tends to infinity with i period There exists a unique infi hyphennite sequence s sub 1 comma s sub 2 comma period period period .. of elements of Capital Sigma such that each

initial segment ofstatic digits occurring in the expansion of some x sub i has the form s sub 1 times times times s sub mfor some m open parenthesis depending on i closing parenthesis periodThe number of preradix digits in the expansion of x sub i .. i s .. at .. most .. thecorresponding number for x sub 1 comma so the sequence s sub 1 comma s sub 2 comma period period period ..

must include oneopen parenthesis and only one closing parenthesis .. radix point period .. Define q m = delta to the power of *

open parenthesis q 0 comma s sub 1 times times times s sub m closing parenthesis semicolon .. then each .. q mi s relevant comma .. and q m .. i s postradix for m .. sufficiently large period .. Moreover comma .. onlyfinitely many of the postradix transitions from q m to q m plus 1 can be acyclic commaas otherwise some such transition would repeat comma but then would visibly bepart of a cycle periodChoose x sub i in the decreasing sequence whose first m digits are all static commafor m large enough that q m is postradix and the transition from q n to q n plus 1i s cyclic for all n greater equal m open parenthesis again comma this amounts to taking i sufficiently large closing

parenthesis periodLet t sub 1 t sub 2 times times times in Capital Sigma to the power of * denote the base b expansion of x sub i

period Since x sub i plus 1 less x sub i comma thereexists .. a smallest .. integer n greater m such that .. s sub n negationslash-equal t sub n comma .. and that ..

integer nsatisfies s sub n less t sub n period However comma .. the transition from q n to q n plus 1 .. i s cyclic comma ..

andso the inequality .. s sub n less t sub n .. violates .. the definition of a proper .. b hyphen labeling periodThis .. contradiction .. means .. that .. there .. cannot .. exist .. an infinite .. decreasingsubsequence comma so M is well hyphen ordered comma as desired period square7 period 2 period .. Arithmetic for well hyphen ordered automata period .. We next verify directlythat the set of automatic series i s a ring semicolon this follows from Theorem 4 period 1 period 3and Lemma 3 period 2 period 5 comma but giving a direct proof will on one hand lead to a secondproof of Theorem 4 period 1 period 3 comma and on the other hand suggest ways to implementcomputations on automatic series in practice period open parenthesis However comma we have not madeany attempt here to optimize the efficiency of the computations semicolon this willrequire further study period closing parenthesisWe first note that adding automatic series i s easy periodLemma 7 period 2 period 1 period .... Let x comma y in F sub q open parenthesis open parenthesis t to the

power of Q closing parenthesis closing parenthesis .... be p hyphen automatic open parenthesis resp period p hyphenquasi hyphen automa hyphen

tic closing parenthesis period .. Then x plus y is p hyphen automatic open parenthesis resp period p hyphenquasi hyphen automatic closing parenthesis period

4 1 0 Kiran S . Kedlaya

This implies immediately that a relevant postradix state cannot lie onmore than one minimal cycle , so Gq i s a rooted saguaro . Moreover , notwoedges from the same vertex of Gq carry the same label , and what we justproved implies that any cyclic edge must carry a greater label than any otheredge from the same vertex . Hence Gq carries a proper b− labeling .

Suppose conversely that each Gq i s a rooted saguaro carrying a properb−labeling , but assume by way of contradiction that the language acceptedby M includes an infinite decreasing sequence x1, x2, .... Wesay the m− th digit of the base b expansion of xi ( counting from theleft ) is static if xi, xi+1, ... all have the same m− th digit . Then theexpansion of each xi begins with an initial segment ( possibly empty )of static digits , and the number of static digits tends to infinity with i.There exists a unique infi - nite sequence s1, s2, ... of elements of Σ suchthat each initial segment of static digits occurring in the expansion of somexi has the form s1 · · · sm for some m( depending on i).

The number of preradix digits in the expansion of xi i s at mostthe corresponding number for x1, so the sequence s1, s2, ... must includeone ( and only one ) radix point . Define qm = δ∗(q0, s1 · · · sm);then each qm i s relevant , and qm i s postradix for m sufficientlylarge . Moreover , only finitely many of the postradix transitions fromqm to qm+1 can be acyclic , as otherwise some such transition would repeat, but then would visibly be part of a cycle .

Choose xi in the decreasing sequence whose first m digits are all static, for m large enough that qm is postradix and the transition from qn toqn + 1 i s cyclic for all n ≥ m( again , this amounts to taking i sufficientlylarge ) . Let t1t2 · ·· ∈ Σ∗ denote the base b expansion of xi. Since xi+1 < xi,there exists a smallest integer n > m such that sn 6= tn, andthat integer n satisfies sn < tn. However , the transition from qnto qn+ 1 i s cyclic , and so the inequality sn < tn violates thedefinition of a proper b− labeling . This contradiction means thatthere cannot exist an infinite decreasing subsequence , so M iswell - ordered , as desired . �7 . 2 . Arithmetic for well - ordered automata . We nextverify directly that the set of automatic series i s a ring ; this follows fromTheorem 4 . 1 . 3 and Lemma 3 . 2 . 5 , but giving a direct proof willon one hand lead to a second proof of Theorem 4 . 1 . 3 , and on theother hand suggest ways to implement computations on automatic series inpractice . ( However , we have not made any attempt here to optimize theefficiency of the computations ; this will require further study . )

We first note that adding automatic series i s easy .Lemma 7 . 2 . 1 . Let x, y ∈ Fq((tQ)) be p− automatic ( resp .p−quasi - automa -tic ) . Then x+ y is p− automatic ( resp .p− quasi - automatic ) .

\hspace ∗{\ f i l l }F i n i t e automata and a l g e b r a i c ex t en s i on s o f func t i on f i e l d s \quad 4 1 1

\noindent Proof . \quad The cla im f o r $ p − $ quas i − automatic s e r i e s f o l l o w s from the cla im f o r$ p − $ automatic s e r i e s by Lemma 2 . 3 . 6 , so we may as we l l assume that $ x

, y $ are$ p − $ automatic . Write $ x = \sum x { i } t ˆ{ i }$ and $ y =

\sum y i ˆ{ t ˆ{ i }} ; $ then by assumption , thef u n c t i o n s $ i \mapsto x { i }$ \quad and $ i \mapsto y i $ on $ S { p }$\quad are $ p − $ automatic . Hence the func t i on

\noindent $ i \mapsto ( x { i } , y i ) $ i s a l s o $ p − $ automatic , as then i s the func t i on$ i \mapsto x { i } + y i . $ This

y i e l d s the c la im $ . \ square $

\hspace ∗{\ f i l l }We next t a c k l e the s t i c k i e r s ub j e c t o f m u l t i p l i c a t i o n o f automatic s e r i e s .

\noindent Lemma 7 . 2 . 2 . \ h f i l l Let $ x , y \ in F { q } ( ( t ˆ{ Q }) ) $ \ h f i l l be $ p − $ automatic ( re sp $ . p − $ quas i − automa −

\noindent t i c ) . \quad Then $ xy $ i s $ p − $ automatic ( re sp $ . p − $quas i − automatic ) .

Beware that t h i s proo f works p r imar i l y with r eve r s ed base $ p $ expans ions ;the not ions o f l e ad ing and t r a i l i n g z e r o e s w i l l be in terms o f the r eve r s edexpans ions , so l e ad ing z e r o e s are in the l e a s t s i g n i f i c a n t p l a c e s .

\noindent Proof . \quad Again , i t s u f f i c e s to t r e a t the automatic case . Write each o f$ x $ and $ y $

as an $ F { q } − $ l i n e a r combination o f g e n e r a l i z e d power s e r i e s o f the form$ \sum { i \ in S } t ˆ{ i }$

\noindent f o r some $ S \subset S { p } ; $ by Lemma 7 . 2 . 1 , i f the product o f two s e r i e s o f t h i s formi s always $ p − $ automatic , then so i s $ xy . $

\hspace ∗{\ f i l l }That i s , we may assume without l o s s o f g e n e r a l i t y that $ x = \sum { i\ in A { 1 }} t ˆ{ i }$ and

\noindent $ y = \sum { i \ in A { 2 }} t ˆ{ i } . $ Let $ S $ be the subset o f$ \Sigma ˆ{ ∗ } { p } \times \Sigma ˆ{ ∗ } { p }$ c o n s i s t i n g o f p a i r s $ ( w { 1 }, w { 2 } ) $

\noindent with the f o l l o w i n g p r o p e r t i e s .

\centerline {( a $ ) w { 1 }$ and $ w { 2 }$ have the same length . }

\centerline {( b $ ) w { 1 }$ and $ w { 2 }$ each end with 0 . }

( c $ ) w { 1 }$ and $ w { 2 }$ each have a s i n g l e rad ix po int , and both are in the samep o s i t i o n .

( d ) \quad After removing l ead ing and t r a i l i n g z e r o e s $ , w { 1 }$ and $ w { 2 }$become the

r eve r s ed v a l i d base $ p $ expans ions o f some $ i , j \ in S { p } . $

\centerline {( e ) \quad The pa i r $ ( i , j ) $ be longs to $ A { 1 } \timesA { 2 } . $ }

\noindent By ( a ) , we may view $ S $ as a language over $ \Sigma { p } \times\Sigma { p } ; $ i t i s s t r a i g h t f o r w a r d

\noindent to v e r i f y that t h i s language i s in f a c t r e g u l a r . Let $ M = ( Q ,\Sigma , \delta , q 0 , F ) $ be aDFA accept ing $ S , $ with $ \Sigma = \Sigma { p } \times \Sigma { p }

. $

\hspace ∗{\ f i l l }Def ine an NFA $ M ˆ{ \prime } = ( Q ˆ{ \prime } , \Sigma ˆ{ \prime }, \delta ˆ{ \prime } , q ˆ{ \prime } { 0 ˆ{ , }} F ˆ{ \prime } ) , $ in which$ \delta ˆ{ \prime }$ takes mu l t i s e t va lue s ,

\noindent as f o l l o w s . \ h f i l l Put \ h f i l l $ Q ˆ{ \prime } = Q \times \{ 0, 1 \} $ \ h f i l l and \ h f i l l $ \Sigma ˆ{ \prime } = \Sigma { p } . $ \ h f i l l For \ h f i l l$ ( q , i ) \ in Q ˆ{ \prime }$ \ h f i l l and $ s \ in $

\noindent $ \{ 0 , . . . , p − 1 \} , $ we inc lude $ ( q ˆ{ \prime }, 0 ) ( $ resp $ . ( q ˆ{ \prime } , 1 ) ) $ in $ \delta ˆ{ \prime }( ( q , i ) , s ) $ once f o r each

pa i r $ ( t , u ) \ in \{ 0 , . . . , p − 1 \} \times\{ 0 , . . . , p − 1 \} $ with $ t + u + i < p ( $resp $ . t + u + i \geq p ) $

\noindent and $ t + u + i \equiv s ( $ mod $ p ) $ such that $ \delta( q , ( t , u ) ) = q ˆ{ \prime } ; $ f o r $ ( q , i )\ in Q ˆ{ \prime }$ and $ s $

\noindent equal to the rad ix po int , we inc lude $ ( q ˆ{ \prime } , i ) $ in$ \delta ( ( q , i ) , s ) $ i f $ \delta ( q , ( s ,s ) ) = q ˆ{ \prime }$

\noindent ( and we never in c lude $ ( q ˆ{ \prime } , 1 − i ) ) . $ Put$ q ˆ{ \prime } { 0 } = ( q 0 , 0 ) $ and put $ F ˆ{ \prime } = F\times \{ 0 \} . $

Suppose $ w $ i s a s t r i n g which , upon removal o f l e ad ing and t r a i l i n g z e r o e s ,becomes the r eve r s ed v a l i d base $ p $ expansion o f some $ z \ in S { p } . $\quad Then the

\noindent number o f accept ing paths o f $ w $ \quad in $ M ˆ{ \prime }$ \quad i s \quad equal to the number o f p a i r s$ ( w { 1 } , w { 2 } ) \ in S $ which sum to $ w $ \quad with \quad i t s l e ad ing \quad and t r a i l i n g z e r o e s \quad under

Finite automata and algebraic extensions of function fields .. 4 1 1Proof period .. The claim for p hyphen quasi hyphen automatic series follows from the claim forp hyphen automatic series by Lemma 2 period 3 period 6 comma so we may as well assume that x comma y arep hyphen automatic period Write x = sum x sub i t to the power of i and y = sum y i to the power of t to the

power of i semicolon then by assumption comma thefunctions i arrowright-mapsto x sub i .. and i arrowright-mapsto y i on S sub p .. are p hyphen automatic period

Hence the functioni mapsto-arrowright open parenthesis x sub i comma y i closing parenthesis i s also p hyphen automatic comma

as then i s the function i mapsto-arrowright x sub i plus y i period Thisyields the claim period squareWe next tackle the stickier subject of multiplication of automatic series periodLemma 7 period 2 period 2 period .... Let x comma y in F sub q open parenthesis open parenthesis t to the

power of Q closing parenthesis closing parenthesis .... be p hyphen automatic open parenthesis resp period p hyphenquasi hyphen automa hyphen

tic closing parenthesis period .. Then xy is p hyphen automatic open parenthesis resp period p hyphen quasihyphen automatic closing parenthesis period

Beware that this proof works primarily with reversed base p expansions semicolonthe notions of leading and trailing zeroes will be in terms of the reversedexpansions comma so leading zeroes are in the least significant places periodProof period .. Again comma it suffices to treat the automatic case period Write each of x and yas an F sub q hyphen linear combination of generalized power series of the form sum sub i in S t to the power of

ifor some S subset S sub p semicolon by Lemma 7 period 2 period 1 comma if the product of two series of this

formi s always p hyphen automatic comma then so i s xy periodThat is comma we may assume without loss of generality that x = sum sub i in A sub 1 t to the power of i andy = sum sub i in A sub 2 t to the power of i period Let S be the subset of Capital Sigma sub p to the power of

* times Capital Sigma sub p to the power of * consisting of pairs open parenthesis w sub 1 comma w sub 2 closingparenthesis

with the following properties periodopen parenthesis a closing parenthesis w sub 1 and w sub 2 have the same length periodopen parenthesis b closing parenthesis w sub 1 and w sub 2 each end with 0 periodopen parenthesis c closing parenthesis w sub 1 and w sub 2 each have a single radix point comma and both are

in the sameposition periodopen parenthesis d closing parenthesis .. After removing leading and trailing zeroes comma w sub 1 and w sub

2 become thereversed valid base p expansions of some i comma j in S sub p periodopen parenthesis e closing parenthesis .. The pair open parenthesis i comma j closing parenthesis belongs to A

sub 1 times A sub 2 periodBy open parenthesis a closing parenthesis comma we may view S as a language over Capital Sigma sub p times

Capital Sigma sub p semicolon it i s straightforwardto verify that this language i s in fact regular period Let M = open parenthesis Q comma Capital Sigma comma

delta comma q 0 comma F closing parenthesis be aDFA accepting S comma with Capital Sigma = Capital Sigma sub p times Capital Sigma sub p periodDefine an NFA M to the power of prime = open parenthesis Q to the power of prime comma Capital Sigma to the

power of prime comma delta to the power of prime comma q sub 0 to the power of comma to the power of prime Fto the power of prime closing parenthesis comma in which delta to the power of prime takes multiset values comma

as follows period .... Put .... Q to the power of prime = Q times open brace 0 comma 1 closing brace .... and ....Capital Sigma to the power of prime = Capital Sigma sub p period .... For .... open parenthesis q comma i closingparenthesis in Q to the power of prime .... and s in

open brace 0 comma period period period comma p minus 1 closing brace comma we include open parenthesisq to the power of prime comma 0 closing parenthesis open parenthesis resp period open parenthesis q to the powerof prime comma 1 closing parenthesis closing parenthesis in delta to the power of prime open parenthesis openparenthesis q comma i closing parenthesis comma s closing parenthesis once for each

pair open parenthesis t comma u closing parenthesis in open brace 0 comma period period period comma p minus1 closing brace times open brace 0 comma period period period comma p minus 1 closing brace with t plus u plus iless p open parenthesis resp period t plus u plus i greater equal p closing parenthesis

and t plus u plus i equiv s open parenthesis mod p closing parenthesis such that delta open parenthesis q commaopen parenthesis t comma u closing parenthesis closing parenthesis = q to the power of prime semicolon for openparenthesis q comma i closing parenthesis in Q to the power of prime and s

equal to the radix point comma we include open parenthesis q to the power of prime comma i closing parenthesisin delta open parenthesis open parenthesis q comma i closing parenthesis comma s closing parenthesis if delta openparenthesis q comma open parenthesis s comma s closing parenthesis closing parenthesis = q to the power of prime

open parenthesis and we never include open parenthesis q to the power of prime comma 1 minus i closingparenthesis closing parenthesis period Put q sub 0 to the power of prime = open parenthesis q 0 comma 0 closingparenthesis and put F to the power of prime = F times open brace 0 closing brace period

Suppose w i s a string which comma upon removal of leading and trailing zeroes commabecomes the reversed valid base p expansion of some z in S sub p period .. Then thenumber of accepting paths of w .. in M to the power of prime .. is .. equal to the number of pairsopen parenthesis w sub 1 comma w sub 2 closing parenthesis in S which sum to w .. with .. its leading .. and

trailing zeroes .. under

Finite automata and algebraic extensions of function fields 4 1 1

Proof . The claim for p− quasi - automatic series follows from the claimfor p− automatic series by Lemma 2 . 3 . 6 , so we may as well assume thatx, y are p− automatic . Write x =

∑xit

i and y =∑yit

i

; then by assumption, the functions i 7→ xi and i 7→ yi on Sp are p− automatic . Hence thefunctioni 7→ (xi, yi) i s also p− automatic , as then i s the function i 7→ xi + yi. Thisyields the claim . �

We next tackle the stickier subject of multiplication of automatic series.Lemma 7 . 2 . 2 . Let x, y ∈ Fq((tQ)) be p− automatic ( resp .p−quasi - automa -tic ) . Then xy is p− automatic ( resp .p− quasi - automatic ) .

Beware that this proof works primarily with reversed base p expansions; the notions of leading and trailing zeroes will be in terms of the reversedexpansions , so leading zeroes are in the least significant places .Proof . Again , it suffices to treat the automatic case . Write each of xand y as an Fq− linear combination of generalized power series of the form∑i∈S t

i

for some S ⊂ Sp; by Lemma 7 . 2 . 1 , if the product of two series of thisform i s always p− automatic , then so i s xy.

That is , we may assume without loss of generality that x =∑i∈A1

ti andy =

∑i∈A2

ti. Let S be the subset of Σ∗p × Σ∗p consisting of pairs (w1, w2)

with the following properties .( a ) w1 and w2 have the same length .

( b ) w1 and w2 each end with 0 .( c ) w1 and w2 each have a single radix point , and both are in the same

position .( d ) After removing leading and trailing zeroes , w1 and w2 become

the reversed valid base p expansions of some i, j ∈ Sp.( e ) The pair (i, j) belongs to A1 ×A2.

By ( a ) , we may view S as a language over Σp ×Σp; it i s straightforwardto verify that this language i s in fact regular . Let M = (Q,Σ, δ, q0, F ) be aDFA accepting S, with Σ = Σp × Σp.

Define an NFA M ′ = (Q′,Σ′, δ′, q′0,F′), in which δ′ takes multiset values ,

as follows . Put Q′ = Q× {0, 1} and Σ′ = Σp. For (q, i) ∈ Q′ ands ∈{0, ..., p−1}, we include (q′, 0) ( resp . (q′, 1)) in δ′((q, i), s) once for each pair(t, u) ∈ {0, ..., p− 1} × {0, ..., p− 1} with t+ u+ i < p( resp .t+ u+ i ≥ p)and t+ u+ i ≡ s ( mod p) such that δ(q, (t, u)) = q′; for (q, i) ∈ Q′ and sequal to the radix point , we include (q′, i) in δ((q, i), s) if δ(q, (s, s)) = q′

( and we never include (q′, 1− i)). Put q′0 = (q0, 0) and put F ′ = F × {0}.Suppose w i s a string which , upon removal of leading and trailing zeroes

, becomes the reversed valid base p expansion of some z ∈ Sp. Thenthenumber of accepting paths of w in M ′ is equal to the number of pairs(w1, w2) ∈ S which sum to w with its leading and trailing zeroesunder

\noindent 4 1 2 \quad Kiran S . Kedlaya

\noindent ord inary base $ p $ add i t i on with c a r r i e s . By Lemma 2 . 2 . 2 , tak ing t h i s numbermodulo $ p $ y i e l d s a f i n i t e − s t a t e func t i on .

Suppose f u r t h e r that $ w $ beg ins with $ m $ l ead ing z e r o e s , f o r $ m $ g r e a t e r thanthe number o f s t a t e s o f $ M . $ We cla im that f o r any pa i r $ ( w { 1 } ,

w { 2 } ) \ in S $ whichsums to $ w , $ both $ w { 1 }$ and $ w { 2 }$ must begin with a l ead ing zero . Namely , i f t h i swere not the case , then in p r o c e s s i n g $ ( w { 1 } , w { 2 } ) \ in S $

under $ M , $ some s t a t emust be repeated with in the f i r s t $ m $ d i g i t s . Let $ ( b { 1 } , b { 2 }

) $ be the s t r i n g s thatl ead to the f i r s t a r r i v a l at such repeated s t a t e , l e t $ ( m { 1 } , m { 2 }

) $ be the s t r i n g sbetween \quad the f i r s t \quad and \quad second \quad a r r i v a l s , \quad and l e t \quad

$ ( e { 1 } , e { 2 } ) $ \quad be the remainings t r i n g s . Then

\ [ b { i } e { i } , b { i } m { i } e { i } , b { i } m { i } m { i }e { i } , . . . , \ ]

\noindent r e p r e s e n t the r eve r s ed base $ p $ expans ions ( with p o s s i b l e t r a i l i n g z e r o e s , but

\noindent no l ead ing z e r o e s ) o f some elements $ z { 0 i } , z { 1 i } ,. . . $ \quad o f $ A { i } . $ Moreover , the num −

bers $ z { 0 i } , z { 1 i } , . . . $ \quad are a l l d i s t i n c t : \quad i f$ b { i }$ i s nonempty , \quad then $ b { i }$ beg ins with a

nonzero d i g i t , whi l e i f $ b { i }$ i s empty , then $ m { i }$ beg ins with a nonzero d i g i t .

\hspace ∗{\ f i l l }We next v e r i f y that $ z { j 0 } + z { j 1 } = z $ f o r each$ j . $ We are g iven t h i s a s s e r t i o n

\noindent f o r $ j = 1 ; $ \quad in doing that \quad add i t i on , \quad the s t r e t c h during which$ w { 1 }$ \quad and $ w { 2 }$

are added beg ins with an incoming carry and ends with an outgoing carry .Moreover , a l l d i g i t s produced be f o r e and during the s t r e t c h are z e r o e s . Thuswe may remove t h i s s t r e t c h , or repeat i t at w i l l , without changing the base

$ p $ number r epre s ented by the sum \quad ( though the number o f l e ad ing z e r o e s

\noindent w i l l change ) .

\hspace ∗{\ f i l l } Since $ z { j 0 } + z { j 1 } = z $ f o r a l l $ j , $one o f the sequences $ z { 0 i } , z { 1 i } , . . . $ \quad must be

\noindent s t r i c t l y dec r ea s ing . This y i e l d s a c o n t r a d i c t i o n , implying that both $ w { 1 }$and

$ w { 2 }$ had to begin with l ead ing z e r o e s . We conclude from t h i s that the number

\noindent o f accept ing paths o f $ w $ in $ M ˆ{ \prime }$ i s pre se rved by adding a l ead ing zero to$ w $

provided that $ w $ a l ready has $ m $ l ead ing z e r o e s .

I t f o l l o w s that the func t i on that , g iven the r eve r s ed v a l i d base $ p $ expansiono f a number $ k \ in S { p } , $ computes the mod $ p $ reduct ion o f the number o f ways

\noindent to wr i t e $ k = i + j $ with $ i \ in A { 1 }$ \quad and $ j\ in A { 2 } , $ i s a f i n i t e − s t a t e func t i on : wemay compute i t by appending $ m $ l ead ing z e r o e s and one t r a i l i n g zero to

\noindent the r eve r s ed expansion and then running the r e s u l t through $ M ˆ{ \prime }. $ Thus $ xy $

i s $ p − $ automatic , as d e s i r e d $ . \ square $

\hspace ∗{\ f i l l }Div i s i on \quad seems \quad even more compl icated to handle d i r e c t l y \quad ( though we

\noindent suspec t i t i s p o s s i b l e to do so ) ; we t r e a t i t here with an i n d i r e c t approach

\noindent ( v ia the ‘ ‘ automatic i m p l i e s a l g e b r a i c ’ ’ d i r e c t i o n o f Theorem 4 . 1 . 3 ) .

\noindent Lemma 7 . 2 . 3 . \ h f i l l I f $ x \ in F { q } ( t ) , $ \ h f i l l then$ x ( $ viewed as an element o f $ F { q } ( ( t ˆ{ Q } ) ) ) $ i s

\begin { a l i g n ∗}p − automatic .\end{ a l i g n ∗}

\noindent Proof . \quad There i s no l o s s o f g e n e r a l i t y in assuming $ x \ in F { q }\ l l b r a c k e t t \ r rb ra cke t . $ Writing $ x = $

$ \sum ˆ{ \ infty } { i = 0 } c { i } t ˆ{ i } , $ we see that the sequence$ \{ c { i } \} $ i s l i n e a r r e c u r r e n t over $ F { q } , $ hence

even tua l l y p e r i o d i c . By [ 2 , Theorem $ 5 . 4 . 2 ] , x $ i s $ p− $ automatic $ . \ square $

4 1 2 .. Kiran S period Kedlayaordinary base p addition with carries period By Lemma 2 period 2 period 2 comma taking this numbermodulo p yields a finite hyphen state function periodSuppose further that w begins with m leading zeroes comma for m greater thanthe number of states of M period We claim that for any pair open parenthesis w sub 1 comma w sub 2 closing

parenthesis in S whichsums to w comma both w sub 1 and w sub 2 must begin with a leading zero period Namely comma if thiswere not the case comma then in processing open parenthesis w sub 1 comma w sub 2 closing parenthesis in S

under M comma some statemust be repeated within the first m digits period Let open parenthesis b sub 1 comma b sub 2 closing parenthesis

be the strings thatlead to the first arrival at such repeated state comma let open parenthesis m sub 1 comma m sub 2 closing

parenthesis be the stringsbetween .. the first .. and .. second .. arrivals comma .. and let .. open parenthesis e sub 1 comma e sub 2

closing parenthesis .. be the remainingstrings period Thenb sub i e sub i comma b sub i m sub i e sub i comma b sub i m sub i m sub i e sub i comma period period period

commarepresent the reversed base p expansions open parenthesis with possible trailing zeroes comma butno leading zeroes closing parenthesis of some elements z sub 0 i comma z sub 1 i comma period period period ..

of A sub i period Moreover comma the num hyphenbers z sub 0 i comma z sub 1 i comma period period period .. are all distinct : .. if b sub i i s nonempty comma

.. then b sub i begins with anonzero digit comma while if b sub i i s empty comma then m sub i begins with a nonzero digit periodWe next verify that z sub j 0 plus z sub j 1 = z for each j period We are given this assertionfor j = 1 semicolon .. in doing that .. addition comma .. the stretch during which w sub 1 .. and w sub 2are added begins with an incoming carry and ends with an outgoing carry periodMoreover comma all digits produced before and during the stretch are zeroes period Thuswe may remove this stretch comma or repeat it at will comma without changing the basep number represented by the sum .. open parenthesis though the number of leading zeroeswill change closing parenthesis periodSince z sub j 0 plus z sub j 1 = z for all j comma one of the sequences z sub 0 i comma z sub 1 i comma period

period period .. must bestrictly decreasing period This yields a contradiction comma implying that both w sub 1 andw sub 2 had to begin with leading zeroes period We conclude from this that the numberof accepting paths of w in M to the power of prime i s preserved by adding a leading zero to wprovided that w already has m leading zeroes periodIt follows that the function that comma given the reversed valid base p expansionof a number k in S sub p comma computes the mod p reduction of the number of waysto write k = i plus j with i in A sub 1 .. and j in A sub 2 comma i s a finite hyphen state function : wemay compute it by appending m leading zeroes and one trailing zero tothe reversed expansion and then running the result through M to the power of prime period Thus xyi s p hyphen automatic comma as desired period squareDivision .. seems .. even more complicated to handle directly .. open parenthesis though wesuspect it i s possible to do so closing parenthesis semicolon we treat it here with an indirect approachopen parenthesis via the quotedblleft automatic implies algebraic quotedblright direction of Theorem 4 period 1

period 3 closing parenthesis periodLemma 7 period 2 period 3 period .... If x in F sub q open parenthesis t closing parenthesis comma .... then

x open parenthesis viewed as an element of F sub q open parenthesis open parenthesis t to the power of Q closingparenthesis closing parenthesis closing parenthesis is

p hyphen automatic periodProof period .. There i s no loss of generality in assuming x in F sub q llbracket t rrbracket period Writing x =sum sub i = 0 to the power of infinity c sub i t to the power of i comma we see that the sequence open brace c

sub i closing brace i s linear recurrent over F sub q comma henceeventually periodic period By open square bracket 2 comma Theorem 5 period 4 period 2 closing square bracket

comma x is p hyphen automatic period square

4 1 2 Kiran S . Kedlaya

ordinary base p addition with carries . By Lemma 2 . 2 . 2 , taking thisnumber modulo p yields a finite - state function .

Suppose further that w begins with m leading zeroes , for m greater thanthe number of states of M. We claim that for any pair (w1, w2) ∈ S whichsums to w, both w1 and w2 must begin with a leading zero . Namely , if thiswere not the case , then in processing (w1, w2) ∈ S under M, some statemust be repeated within the first m digits . Let (b1, b2) be the strings thatlead to the first arrival at such repeated state , let (m1,m2) be the stringsbetween the first and second arrivals , and let (e1, e2) be theremaining strings . Then

biei, bimiei, bimimiei, ...,

represent the reversed base p expansions ( with possible trailing zeroes , butno leading zeroes ) of some elements z0i, z1i, ... of Ai. Moreover , the numbers z0i, z1i, ... are all distinct : if bi i s nonempty , then bi beginswith a nonzero digit , while if bi i s empty , then mi begins with a nonzerodigit .

We next verify that zj0 + zj1 = z for each j. We are given this assertionfor j = 1; in doing that addition , the stretch during which w1

and w2 are added begins with an incoming carry and ends with an outgoingcarry . Moreover , all digits produced before and during the stretch arezeroes . Thus we may remove this stretch , or repeat it at will , withoutchanging the base p number represented by the sum ( though the numberof leading zeroeswill change ) .

Since zj0 + zj1 = z for all j, one of the sequences z0i, z1i, ... must bestrictly decreasing . This yields a contradiction , implying that both w1

and w2 had to begin with leading zeroes . We conclude from this that thenumberof accepting paths of w in M ′ i s preserved by adding a leading zero to wprovided that w already has m leading zeroes .

It follows that the function that , given the reversed valid base p expansionof a number k ∈ Sp, computes the mod p reduction of the number of waysto write k = i + j with i ∈ A1 and j ∈ A2, i s a finite - state function : wemay compute it by appending m leading zeroes and one trailing zero tothe reversed expansion and then running the result through M ′. Thus xy is p− automatic , as desired . �

Division seems even more complicated to handle directly (though wesuspect it i s possible to do so ) ; we treat it here with an indirect approach( via the “ automatic implies algebraic ” direction of Theorem 4 . 1 . 3 ) .Lemma 7 . 2 . 3 . If x ∈ Fq(t), then x ( viewed as an element ofFq((tQ))) is

p− automatic.

Proof . There i s no loss of generality in assuming x ∈ Fqt. Writing x =∑∞i=0 cit

i, we see that the sequence {ci} i s linear recurrent over Fq, henceeventually periodic . By [ 2 , Theorem 5.4.2], x is p− automatic . �


Proposition 7 . 2 . 4 . The set of p− quasi - automatics eries is a subfield ofFq((tQ)) contained in the integral c lo sure of Fq(t).Proof . Let S be the set of p− quasi - automatic series . By Lemmas 7 .2 . 1 and 7 . 2 . 2 ,S i s a subring of Fq((tQ)). By Lemma 7.2.3, S contains Fq(t); by Propo - sition5 . 2 . 7 , each element of S i s algebraic over Fq(t). Hence given x ∈ S,there exists a polynomial P (z) = c0 + c1z + · · · + cnz

n over Fq(t) withP (x) = 0, and we may assume without loss of generality that c0, cn 6= 0.We can then write

x−1 = −c−10 (c1 + c2x+ · · ·+ cnxn−1),

and the right side i s contained in S. We conclude that S i s closed undertaking reciprocals , and hence is a subfield of Fq((tQ))contained in theintegral closure of Fq(t), as desired . �7 . 3 . Newton ’ s algorithm . To complete the “ concrete ”proof of the “ al - gebraic implies automatic ” direction of Theorem4 . 1 . 3 , we must explain why the field of p− quasi - automaticseries is closed under extraction of roots of polynomials . The argument wegive below implicitly performs a positive characteristic variant of Newton ’s algorithm ; most of the work has already been carried out in Chapter 6 .( By contrast , a direct adaptation of New - ton ’ s algorithm to generalizedpower series gives a transfinite process ; see [ 1 2 , Proposition 1 ] . )Before proceeding further , we explicitly check that the class of p− quasi -automatic series i s closed under the formation of Artin - Schreier extensions.Lemma 7 . 3 . 1 . For any p− quasi - automatic x =

∑xit

i ∈ Fq((tQ))supportedwithin (−∞, 0), there exists a p− quasi - automatic s eries y ∈ Fq((tQ))such

thatyp − y = x.

Proof . We can take

y = x1/p + x1/p2

+ · · ·

once we show that this generalized power series i s p− quasi - automatic .There i s no loss of generality in assuming that xta is p− automatic forsome non - negative integer a; in fact , by decimating , we may reduce tothe case where

a = 1.

Writey =∑

yiti

. Since

yi = xip + xip2 + · · ·,

for any fixed j, the series

\centerline{F i n i t e automata and a l g e b r a i c ex t en s i on s o f func t i on f i e l d s \quad 4 1 3 }

\noindent Propos i t i on \quad 7 . 2 . 4 . \quad The \quad s e t \quad o f $ p − $ quas i − automatic \quad s e r i e s \quad i s \quad a \quad s u b f i e l d o f

\noindent $ F { q } ( ( t ˆ{ Q } ) ) $ \quad conta ined in the i n t e g r a l c l o sure o f$ F { q } ( t ) . $

\noindent Proof . \ h f i l l Let $ S $ be the s e t o f $ p − $ quas i − automatic s e r i e s . By Lemmas 7 . 2 . 1 and 7 . 2 . 2 ,

\noindent $ S $ i s a subr ing o f $ F { q } ( ( t ˆ{ Q } ) ) . $ By Lemma$ 7 . 2 . 3 , S $ conta in s $ F { q } ( t ) ; $ by Propo −

s i t i o n 5 . 2 . 7 , each element o f $ S $ i s a l g e b r a i c over $ F { q } ( t ). $ Hence g iven $ x \ in S , $

the re e x i s t s a polynomial $ P ( z ) = c { 0 } + c { 1 } z + \cdot\cdot \cdot + c { n } z ˆ{ n }$ \quad over $ F { q } ( t ) $ \quad with

$ P ( x ) = 0 , $ and we may assume without l o s s o f g e n e r a l i t y that $ c { 0 }, c { n } \not= 0 . $

\noindent We can then wr i t e

\ [ x ˆ{ − 1 } = − c ˆ{ − 1 } { 0 } ( c { 1 } + c { 2 } x + \cdot\cdot \cdot + c { n } x ˆ{ n − 1 } ) , \ ]

\noindent and the r i g h t s i d e i s conta ined in $ S . $ We conclude that $ S $ i s c l o s e d under

\noindent tak ing \quad r e c i p r o c a l s , \quad and \quad hence \quad i s \quad a \quad s u b f i e l d \quad o f$ F { q } ( ( t ˆ{ Q } ) ) $ \quad conta ined \quad in the

\noindent i n t e g r a l c l o s u r e o f $ F { q } ( t ) , $ as d e s i r e d $ . \ square $

\noindent 7 . 3 . \quad Newton ’ s a lgor i thm . \quad To complete the ‘ ‘ conc r e t e ’ ’ p roo f o f the ‘ ‘ a l −gebra i c i m p l i e s \quad automatic ’ ’ \quad d i r e c t i o n o f Theorem 4 . 1 . 3 , \quad we must \quad exp la inwhy the f i e l d o f $ p − $ quas i − automatic s e r i e s i s c l o s e d under e x t r a c t i o n o f r oo t so f po lynomia ls . The argument we g ive below i m p l i c i t l y performs a p o s i t i v ec h a r a c t e r i s t i c va r i ant o f Newton ’ s a lgor i thm ; most o f the work has a l r eadybeen c a r r i e d out in Chapter 6 . \quad ( By cont ra s t , a d i r e c t adaptat ion o f New −ton ’ s a lgor i thm to g e n e r a l i z e d power s e r i e s g i v e s a t r a n s f i n i t e p roce s s ; s e e[ 1 2 , Propos i t i on 1 ] . )

\noindent Before proceed ing f u r t h e r , we e x p l i c i t l y check that the c l a s s o f $ p − $quas i −

automatic s e r i e s i s c l o s e d under the format ion o f Artin − S c h r e i e r ex t en s i on s .

\noindent Lemma 7 . 3 . 1 . For any $ p − $ quas i − automatic $ x = \sum x { i }t ˆ{ i } \ in F { q } ( ( t ˆ{ Q } ) ) $ supported

\noindent with in $ ( − \ infty , 0 ) , $ \quad the re e x i s t s a $ p − $quas i − automatic s e r i e s $ y \ in F { q } ( ( t ˆ{ Q } ) ) $ \quad such

\begin { a l i g n ∗}that y ˆ{ p } − y = x .\end{ a l i g n ∗}

\noindent Proof . \quad We can take

\ [ y = x ˆ{ 1 / p } + x ˆ{ 1 / p ˆ{ 2 }} + \cdot \cdot \cdot \ ]

\noindent once we show that t h i s g e n e r a l i z e d power s e r i e s i s $ p − $ quas i − automatic . Therei s no l o s s o f g e n e r a l i t y in assuming that $ xt ˆ{ a }$ i s $ p − $ automatic f o r some non −negat ive i n t e g e r $ a ; $ in f a c t , by decimating , we may reduce to the case where

\begin { a l i g n ∗}a = 1 . \\ Write y = \sum y i ˆ{ t ˆ{ i }} { . } Since \\ y i =

x { ip } + x { ip ˆ{ 2 }} + \cdot \cdot \cdot ,\end{ a l i g n ∗}

\noindent f o r any f i x e d $ j , $ the s e r i e s

\ [\ begin { a l i gned } \sum y i ˆ{ t ˆ{ i }}\\i < − p ˆ{ − j }\end{ a l i gned }\ ]

\noindent i s $ p − $ automatic . Also , f o r $ i \ in [ − 1 , 0 ), $ the sequence

\ [ x { i } , x { i / p } , x { i / p } 2 , . . . \ ]

Finite automata and algebraic extensions of function fields .. 4 1 3Proposition .. 7 period 2 period 4 period .. The .. set .. of p hyphen quasi hyphen automatic .. s eries .. is .. a

.. subfield ofF sub q open parenthesis open parenthesis t to the power of Q closing parenthesis closing parenthesis .. contained

in the integral c lo sure of F sub q open parenthesis t closing parenthesis periodProof period .... Let S be the set of p hyphen quasi hyphen automatic series period By Lemmas 7 period 2 period

1 and 7 period 2 period 2 commaS i s a subring of F sub q open parenthesis open parenthesis t to the power of Q closing parenthesis closing

parenthesis period By Lemma 7 period 2 period 3 comma S contains F sub q open parenthesis t closing parenthesissemicolon by Propo hyphen

sition 5 period 2 period 7 comma each element of S i s algebraic over F sub q open parenthesis t closing parenthesisperiod Hence given x in S comma

there exists a polynomial P open parenthesis z closing parenthesis = c sub 0 plus c sub 1 z plus times times timesplus c sub n z to the power of n .. over F sub q open parenthesis t closing parenthesis .. with

P open parenthesis x closing parenthesis = 0 comma and we may assume without loss of generality that c sub 0comma c sub n negationslash-equal 0 period

We can then writex to the power of minus 1 = minus c sub 0 to the power of minus 1 open parenthesis c sub 1 plus c sub 2 x plus

times times times plus c sub n x to the power of n minus 1 closing parenthesis commaand the right side i s contained in S period We conclude that S i s closed undertaking .. reciprocals comma .. and .. hence .. is .. a .. subfield .. of F sub q open parenthesis open parenthesis t

to the power of Q closing parenthesis closing parenthesis .. contained .. in theintegral closure of F sub q open parenthesis t closing parenthesis comma as desired period square7 period 3 period .. Newton quoteright s algorithm period .. To complete the quotedblleft concrete quotedblright

proof of the quotedblleft al hyphengebraic implies .. automatic quotedblright .. direction of Theorem 4 period 1 period 3 comma .. we must ..

explainwhy the field of p hyphen quasi hyphen automatic series is closed under extraction of rootsof polynomials period The argument we give below implicitly performs a positivecharacteristic variant of Newton quoteright s algorithm semicolon most of the work has alreadybeen carried out in Chapter 6 period .. open parenthesis By contrast comma a direct adaptation of New hyphenton quoteright s algorithm to generalized power series gives a transfinite process semicolon seeopen square bracket 1 2 comma Proposition 1 closing square bracket period closing parenthesisBefore proceeding further comma we explicitly check that the class of p hyphen quasi hyphenautomatic series i s closed under the formation of Artin hyphen Schreier extensions periodLemma 7 period 3 period 1 period For any p hyphen quasi hyphen automatic x = sum x sub i t to the power of

i in F sub q open parenthesis open parenthesis t to the power of Q closing parenthesis closing parenthesis supportedwithin open parenthesis minus infinity comma 0 closing parenthesis comma .. there exists a p hyphen quasi

hyphen automatic s eries y in F sub q open parenthesis open parenthesis t to the power of Q closing parenthesisclosing parenthesis .. such

that y to the power of p minus y = x periodProof period .. We can takey = x to the power of 1 slash p plus x to the power of 1 slash p to the power of 2 plus times times timesonce we show that this generalized power series i s p hyphen quasi hyphen automatic period Therei s no loss of generality in assuming that xt to the power of a is p hyphen automatic for some non hyphennegative integer a semicolon in fact comma by decimating comma we may reduce to the case wherea = 1 period Write y = sum y i to the power of t to the power of i sub period Since y i = x sub ip plus x sub ip

to the power of 2 plus times times times commafor any fixed j comma the seriesLine 1 sum y i to the power of t to the power of i Line 2 i less minus p to the power of minus ji s p hyphen automatic period Also comma for i in open square bracket minus 1 comma 0 closing parenthesis

comma the sequencex sub i comma x sub i slash p comma x sub i slash p 2 comma period period period

∑yit

i

i < −p−j

i s p− automatic . Also , for i ∈ [−1, 0), the sequence

xi, xi/p, xi/p2, ...

\noindent 4 14 \quad Kiran S . Kedlayai s generated by input t ing the base $ p $ expans ions o f $ 1 + i , 1 +

i / p , 1 + i / p ˆ{ 2 } , . . . $

\noindent i n t o a f i x e d f i n i t e automaton ; hence the re e x i s t i n t e g e r s $ m $ and $ n $such that

\begin { a l i g n ∗}f o r any i \ in [ − 1 , 0 ) , y { ip } − m = y { ip }− n { . }\end{ a l i g n ∗}

From t h i s we can cons t ruc t a f i n i t e automaton that , upon r e c e i v i n g $ 1 + i $at \quad input , \quad r e tu rn s \quad $ y i . $ \quad Namely , \quad the \quad automat i c i ty o f

$ \sum { i < − p ˆ{ − n }} y i ˆ{ t ˆ{ i }}$ \quad g i v e s \quad anautomaton that r e tu rn s $ y i $ i f $ i < − p ˆ{ − n } . $ I f $ i \geq− p ˆ{ − n } , $ then $ 1 + i \geq 1 − p ˆ{ − n } , $

so the base $ p $ expansion beg ins with $ n $ d i g i t s equal to $ p − 1 , $and conve r s e l y .We can thus loop back to wherever we were a f t e r $ m $ d i g i t s equal to $ p − 1

; $s i n c e $ y { ip } − m = y { ip } − n , $ we end up computing the r i g h t c o e f f i c i e n t . Hence

$ y $ i s

\noindent $ p − $ quas i − automatic , as d e s i r e d $ . \ square $

\noindent Lemma 7 . 3 . 2 . \ h f i l l Let $ R { q } \subset F { q } ( ( t ˆ{ Q }) ) $ \ h f i l l be the c l o s u r e , \ h f i l l under the va luat i on $ v , $

\noindent o f the f i e l d o f $ p − $ quas i − automatic s e r i e s . \quad Then f o r any$ a , b \ in R { q }$ with $ a \ne 0 , $

the equat ion

\ [ z ˆ{ p } − az = b \ ]

\noindent has $ p $ d i s t i n c t s o l u t i o n s in $ R { q } \prime $ f o r s ome power$ q ˆ{ \prime }$ o f $ q . $

\noindent Proof . \ h f i l l We f i r s t check the case $ b = 0 , $ i . e . , we show that$ z ˆ{ p − 1 } = a $ has $ p − 1 $

\noindent d i s t i n c t s o l u t i o n s in $ R { q } \prime $ f o r some $ q ˆ{ \prime } . $Write $ a = a { 0 } t ˆ{ i } ( 1 + u ) , $ where $ a { 0 } \ inF { q } , $

\noindent $ i \ in Q , $ and $ v ( u ) > 0 . $ Choose $ q ˆ{ \prime }$so that $ a { 0 }$ has a f u l l s e t o f $ ( p − 1 ) − $ s t r oo t s

in $ F { q } \prime ; $ we can then take

\ [ z = 1{ 0 { a }}ˆ{ / ( p − 1 ) } t ˆ{ i / ( p − 1 ) } \sum ˆ{ j= 0 } { \ infty } \ l e f t (\ begin { array }{ c} 1 / ( p − 1 ) \\ j \end{ array }\ right )u ˆ{ j }\ ]

\noindent f o r any $ ( p − 1 ) − $ s t root $1{ 0 { a }}ˆ{ / ( p −1 ) }$ o f $ a { 0 } . $

\hspace ∗{\ f i l l }We now proceed to the case o f g ene ra l $ b . $ \quad By the above argument , we

\noindent can reduce to the case $ a = 1 . $ We can then s p l i t $ b = b { − }+ b { + } , $ where $ b { − }$ i s

supported on $ ( − \ infty , 0 ) $ and $ b { + }$ i s supported on $ [0 , \ infty ) , $ and t r e a t the ca s e s

$ b = b { − }$ \quad and $ b = b { + }$ s e p a r a t e l y . The former case i s p r e c i s e l y Lemma 7 . 3 . 1 .As f o r the l a t t e r case , l e t $ b { 0 }$ be the constant c o e f f i c i e n t o f $ b { + }

, $ and choose$ q ˆ{ \prime }$ so that the equat ion $ z ˆ{ p } − z = b { 0 }$ has d i s t i n c t r oo t s

$ c { 1 } , . . . , c { p } \ in F { q } \prime { . }$ Wemay then take

\ [\ begin { a l i gned } \ infty \\z = c { i } − \sum ( b { + } − b { 0 } ) ˆ{ p ˆ{ j }}\\j = 0 \end{ a l i gned }\ ]

\noindent to obta in the d e s i r e d s o l u t i o n s $ . \ square $

\noindent Propos i t i on 7 . 3 . 3 . \quad Let $ R { q } \subset F { q } ( (t ˆ{ Q } ) ) $ \quad be the c l o sure , under the va lua t i on

$ v , $ o f the f i e l d o f $ p − $ quas i − automatic s e r i e s , and l e t $ R $be the union o f the r i n g s

$ R { q } \prime $ \quad over a l l powers $ q ˆ{ \prime }$ \quad o f $ q . $\quad Then $ R ( $ which i s a l s o a f i e l d ) i s a l g e b r a i c a l l y

c l o s ed .

\noindent Proof . \quad By Lemma 3 . 3 . 4 , \quad i t s u f f i c e s to show that f o r every monic tw i s t edpolynomial $ P ( F ) $ over $ R { q } , $ the polynomial $ P ( F

) ( z ) $ f a c t o r s complete ly over

4 14 .. Kiran S period Kedlayai s generated by inputting the base p expansions of 1 plus i comma 1 plus i slash p comma 1 plus i slash p to the

power of 2 comma period period periodinto a fixed finite automaton semicolon hence there exist integers m and n such thatfor any i in open square bracket minus 1 comma 0 closing parenthesis comma y sub ip minus m = y sub ip minus

n sub periodFrom this we can construct a finite automaton that comma upon receiving 1 plus iat .. input comma .. returns .. y i period .. Namely comma .. the .. automaticity of sum sub i less minus p to

the power of minus n y i to the power of t to the power of i .. gives .. anautomaton that returns y i if i less minus p to the power of minus n period If i greater equal minus p to the

power of minus n comma then 1 plus i greater equal 1 minus p to the power of minus n commaso the base p expansion begins with n digits equal to p minus 1 comma and conversely periodWe can thus loop back to wherever we were after m digits equal to p minus 1 semicolonsince y sub ip minus m = y sub ip minus n comma we end up computing the right coefficient period Hence y i sp hyphen quasi hyphen automatic comma as desired period squareLemma 7 period 3 period 2 period .... Let R sub q subset F sub q open parenthesis open parenthesis t to the

power of Q closing parenthesis closing parenthesis .... be the c losure comma .... under the valuation v commaof the field of p hyphen quasi hyphen automatic s eries period .. Then for any a comma b in R sub q with a

equal-negationslash 0 commathe equationz to the power of p minus az = bhas p distinct s olutions in R sub q prime for s ome power q to the power of prime of q periodProof period .... We first check the case b = 0 comma i period e period comma we show that z to the power of

p minus 1 = a has p minus 1distinct solutions in R sub q prime for some q to the power of prime period Write a = a sub 0 t to the power of

i open parenthesis 1 plus u closing parenthesis comma where a sub 0 in F sub q commai in Q comma and v open parenthesis u closing parenthesis greater 0 period Choose q to the power of prime so

that a sub 0 has a full set of open parenthesis p minus 1 closing parenthesis hyphen st rootsin F sub q prime semicolon we can then takez = 1 0 a to the power of slash open parenthesis p minus 1 closing parenthesis t to the power of i slash open

parenthesis p minus 1 closing parenthesis sum from j = 0 to infinity Row 1 1 slash open parenthesis p minus 1 closingparenthesis Row 2 j . u to the power of j

for any open parenthesis p minus 1 closing parenthesis hyphen st root 1 0 a to the power of slash open parenthesisp minus 1 closing parenthesis of a sub 0 period

We now proceed to the case of general b period .. By the above argument comma wecan reduce to the case a = 1 period We can then split b = b sub minus plus b sub plus comma where b sub

minus i ssupported on open parenthesis minus infinity comma 0 closing parenthesis and b sub plus i s supported on open

square bracket 0 comma infinity closing parenthesis comma and treat the casesb = b sub minus .. and b = b sub plus separately period The former case i s precisely Lemma 7 period 3 period

1 periodAs for the latter case comma let b sub 0 be the constant coefficient of b sub plus comma and chooseq to the power of prime so that the equation z to the power of p minus z = b sub 0 has distinct roots c sub 1

comma period period period comma c sub p in F sub q prime sub period Wemay then takeLine 1 infinity Line 2 z = c sub i minus sum open parenthesis b sub plus minus b sub 0 closing parenthesis to

the power of p to the power of j Line 3 j = 0to obtain the desired solutions period squareProposition 7 period 3 period 3 period .. Let R sub q subset F sub q open parenthesis open parenthesis t to the

power of Q closing parenthesis closing parenthesis .. be the c lo sure comma under the valuationv comma of the field of p hyphen quasi hyphen automatic s eries comma and let R be the union of the ringsR sub q prime .. over all powers q to the power of prime .. of q period .. Then R open parenthesis which is also

a field closing parenthesis is algebraicallyc los ed periodProof period .. By Lemma 3 period 3 period 4 comma .. it suffices to show that for every monic twistedpolynomial P open parenthesis F closing parenthesis over R sub q comma the polynomial P open parenthesis F

closing parenthesis open parenthesis z closing parenthesis factors completely over

4 14 Kiran S . Kedlaya i s generated by inputting the base p expansions of1 + i, 1 + i/p, 1 + i/p2, ...into a fixed finite automaton ; hence there exist integers m and n such that

foranyi ∈ [−1, 0), yip −m = yip − n.

From this we can construct a finite automaton that , upon receiving1 + i at input , returns yi. Namely , the automaticity of∑i<−p−n yi

ti gives an automaton that returns yi if i < −p−n. If i ≥ −p−n,then 1 + i ≥ 1 − p−n, so the base p expansion begins with n digits equal top− 1, and conversely . We can thus loop back to wherever we were after mdigits equal to p − 1; since yip −m = yip − n, we end up computing the rightcoefficient . Hence y i sp− quasi - automatic , as desired . �Lemma 7 . 3 . 2 . Let Rq ⊂ Fq((tQ)) be the c losure , under thevaluation v,of the field of p− quasi - automatic s eries . Then for any a, b ∈ Rqwith a 6= 0, the equation

zp − az = b

has p distinct s olutions in Rq′ for s ome power q′ of q.Proof . We first check the case b = 0, i . e . , we show that zp−1 = a hasp− 1distinct solutions in Rq′ for some q′. Write a = a0t

i(1 + u), where a0 ∈ Fq,i ∈ Q, and v(u) > 0. Choose q′ so that a0 has a full set of (p− 1)− st roots inFq′; we can then take

z = 10a/(p−1)ti/(p−1)

j=0∑∞

(1/(p− 1)

j

)uj

for any (p− 1)− st root 10a/(p−1) of a0.

We now proceed to the case of general b. By the above argument , wecan reduce to the case a = 1. We can then split b = b− + b+, where b− i ssupported on (−∞, 0) and b+ i s supported on [0,∞), and treat the casesb = b− and b = b+ separately . The former case i s precisely Lemma 7 . 3. 1 . As for the latter case , let b0 be the constant coefficient of b+, andchoose q′ so that the equation zp− z = b0 has distinct roots c1, ..., cp ∈ Fq′. Wemay then take

∞

z = ci −∑

(b+ − b0)pj

j = 0

to obtain the desired solutions . �Proposition 7 . 3 . 3 . Let Rq ⊂ Fq((tQ)) be the c lo sure , underthe valuation v, of the field of p− quasi - automatic s eries , and let Rbe the union of the rings Rq′ over all powers q′ of q. Then R (which is also a field ) is algebraically c los ed .Proof . By Lemma 3 . 3 . 4 , it suffices to show that for every monictwisted polynomial P (F ) over Rq, the polynomial P (F )(z) factors completelyover


\noindent $ R { q } \prime $ f o r some $ q ˆ{ \prime } . $ S ince $ R { q }$i s p e r f e c t , we may f a c t o r o f f $ F $ on the r i g h t i f i t

\noindent appears , to reduce to the case where $ P $ has nonzero constant c o e f f i c i e n t , ore q u i v a l e n t l y ( by Lemma 6 . 1 . 4 ) where $ P ( F ) ( z ) $ has no repeated roo t s .

By Coro l l a ry 6 . 4 . 6 , we can wr i t e $ P = Q 1 \cdot \cdot \cdot Q { n }$f o r some monic l i n e a r

tw i s t ed polynomia l s $ Q 1 , . . . , Q { n }$ over $ R { q } \prime $\quad f o r some $ q ˆ{ \prime } . $ Write $ Q i = F − c { i }$

\noindent ( where $ c { i } \ne 0 ) ; $ the proce s s o f f i n d i n g roo t s o f $ P( F ) ( z ) $ can then be de s c r ibed

as the proce s s o f f i n d i n g s o l u t i o n s o f the system o f equat ions

\ [\ begin { a l i gned }z{ p } { 1 } − c { 1 } z { 1 } = 0 \\z{ p } { 2 } − c { 2 } z { 2 } = z { 1 }\end{ a l i gned }\ ]

\noindent ...

\ [ z ˆ{ p } { n } − c { n } z { n } = z { n − 1 }\ ]

\noindent ( the roo t s o f $ P ( F ) ( z ) $ being p r e c i s e l y the p o s s i b l e va lue s o f$ z { n } ) . $ By repeated

a p p l i c a t i o n s o f Lemma 7 . 3 . 2 , the system has $ p ˆ{ n }$ d i s t i n c t s o l u t i o n s , and so$ P ( F ) ( z ) $ s p l i t s complete ly as d e s i r e d $ . \ square $

\hspace ∗{\ f i l l }At t h i s po int , we can now deduce that \quad ‘ ‘ a l g e b r a i c i m p l i e s automatic ’ ’ in

\noindent much the same manner as in Propos i t i on 5 . 2 . 7 . This means in p a r t i c u l a rthat again we w i l l need to invoke C h r i s t o l ’ s theorem .

\noindent Propos i t i on \ h f i l l 7 . 3 . 4 . \ h f i l l Let $ x = \sum x { i } t ˆ{ i }\ in F { q } ( ( t ˆ{ Q } ) ) $ \ h f i l l be \ h f i l l a g e n e r a l i z e d power

\noindent s e r i e s which i s a l g e b r a i c over $ F { q } ( t ) . $ \quad Then $ x $i s $ p − $ quas i − automatic .

\noindent Proof . \quad There e x i s t s a polynomial $ P ( z ) $ over $ F { q }( t ˆ{ 1 / p ˆ{ m }} ) , $ f o r some nonnegat ive

i n t e g e r $ m , $ \quad such that $ P $ has $ x $ as a root with m u l t i p l i c i t y 1 ; by r e p l a c i n g

\noindent $ x $ with $ x ˆ{ p ˆ{ m }} , $ we may reduce without l o s s o f g e n e r a l i t y to the case where$ m = 1 . $ Choose $ c \ in Q $ such that $ c > v ( x − x ˆ{ \prime }

) $ f o r any root $ x ˆ{ \prime } \not= x $ o f $ P . $

\hspace ∗{\ f i l l }By Propos i t i on 7 . 3 . 3 , the re e x i s t s a power $ q ˆ{ \prime }$ o f$ q $ and a $ p − $ quas i − automatic

\noindent s e r i e s $ y $ over $ F { q } \prime ( ( t ˆ{ Q } ) ) $ \quad such that$ v ( x − y ) \geq c . $ \quad The polynomial $ P ( z + y) $

then has exac t l y one root o f s l ope at l e a s t $ c , $ namely $ x − y . $

\hspace ∗{\ f i l l }By Propos i t i on $ 5 . 1 . 2 , y $ i s a l g e b r a i c over $ F { q }( t ) . $ Let $ K $ be the f i n i t e ex −

\noindent t en s i on o f $ F { q } \prime ( ( t ) ) $ obta ined by ad jo in ing$ y . $ Then $ K $ i s complete under $ v ; $

\noindent by Coro l l a ry 6 . 3 . 3 , we may s p l i t o f f the unique f a c t o r o f $ P ( z+ y ) $ o f s l ope

at l e a s t $ c . $ In other words $ , x − y \ in K , $ and so $ x\ in K . $

At t h i s po int the argument p a r a l l e l s that o f Propos i t i on 5 . 2 . 7 . Let $ m $ bethe degree o f the minimal polynomial o f $ y $ over $ F { q } ( t ) . $ For

$ j = 0 , . . . , m − 1 , $wr i t e

\ [\ begin { a l i gned } m − 1 \\y ˆ{ pj } = \sum a { i j } y ˆ{ i }\\i = 0 \end{ a l i gned }\ ]

\noindent with $ a { i j } \ in F { q } ( ( t ) ) ; $ then the $ a { i j }$are a l g e b r a i c over $ F { q } ( t ) . $ Choose $ n $ minimal

such that $ x , x ˆ{ p } , . . . , x ˆ{ p ˆ{ n }}$ \quad are l i n e a r l y dependent over$ F { q } ( ( t ) ) . $ \quad Write $ x ˆ{ p ˆ{ j }} = $

Finite automata and algebraic extensions of function fields .. 4 1 5R sub q prime for some q to the power of prime period Since R sub q i s perfect comma we may factor off F on

the right if itappears comma to reduce to the case where P has nonzero constant coefficient comma orequivalently open parenthesis by Lemma 6 period 1 period 4 closing parenthesis where P open parenthesis F

closing parenthesis open parenthesis z closing parenthesis has no repeated roots periodBy Corollary 6 period 4 period 6 comma we can write P = Q 1 times times times Q sub n for some monic lineartwisted polynomials Q 1 comma period period period comma Q sub n over R sub q prime .. for some q to the

power of prime period Write Q i = F minus c sub iopen parenthesis where c sub i equal-negationslash 0 closing parenthesis semicolon the process of finding roots

of P open parenthesis F closing parenthesis open parenthesis z closing parenthesis can then be describedas the process of finding solutions of the system of equationsLine 1 z p sub 1 minus c sub 1 z sub 1 = 0 Line 2 z p sub 2 minus c sub 2 z sub 2 = z sub 1periodperiodperiodz sub n to the power of p minus c sub n z sub n = z sub n minus 1open parenthesis the roots of P open parenthesis F closing parenthesis open parenthesis z closing parenthesis

being precisely the possible values of z sub n closing parenthesis period By repeatedapplications of Lemma 7 period 3 period 2 comma the system has p to the power of n distinct solutions comma

and soP open parenthesis F closing parenthesis open parenthesis z closing parenthesis splits completely as desired period

squareAt this point comma we can now deduce that .. quotedblleft algebraic implies automatic quotedblright inmuch the same manner as in Proposition 5 period 2 period 7 period This means in particularthat again we will need to invoke Christol quoteright s theorem periodProposition .... 7 period 3 period 4 period .... Let x = sum x sub i t to the power of i in F sub q open parenthesis

open parenthesis t to the power of Q closing parenthesis closing parenthesis .... be .... a generalized powers eries which is algebraic over F sub q open parenthesis t closing parenthesis period .. Then x is p hyphen quasi

hyphen automatic periodProof period .. There exists a polynomial P open parenthesis z closing parenthesis over F sub q open parenthesis

t to the power of 1 slash p to the power of m closing parenthesis comma for some nonnegativeinteger m comma .. such that P has x as a root with multiplicity 1 semicolon by replacingx with x to the power of p to the power of m comma we may reduce without loss of generality to the case wherem = 1 period Choose c in Q such that c greater v open parenthesis x minus x to the power of prime closing

parenthesis for any root x to the power of prime negationslash-equal x of P periodBy Proposition 7 period 3 period 3 comma there exists a power q to the power of prime of q and a p hyphen

quasi hyphen automaticseries y over F sub q prime open parenthesis open parenthesis t to the power of Q closing parenthesis closing

parenthesis .. such that v open parenthesis x minus y closing parenthesis greater equal c period .. The polynomialP open parenthesis z plus y closing parenthesis

then has exactly one root of slope at least c comma namely x minus y periodBy Proposition 5 period 1 period 2 comma y i s algebraic over F sub q open parenthesis t closing parenthesis

period Let K be the finite ex hyphentension of F sub q prime open parenthesis open parenthesis t closing parenthesis closing parenthesis obtained by

adjoining y period Then K i s complete under v semicolonby Corollary 6 period 3 period 3 comma we may split off the unique factor of P open parenthesis z plus y closing

parenthesis of slopeat least c period In other words comma x minus y in K comma and so x in K periodAt this point the argument parallels that of Proposition 5 period 2 period 7 period Let m bethe degree of the minimal polynomial of y over F sub q open parenthesis t closing parenthesis period For j = 0

comma period period period comma m minus 1 commawriteLine 1 m minus 1 Line 2 y to the power of pj = sum a sub ij y to the power of i Line 3 i = 0with a sub ij in F sub q open parenthesis open parenthesis t closing parenthesis closing parenthesis semicolon

then the a sub ij are algebraic over F sub q open parenthesis t closing parenthesis period Choose n minimalsuch that x comma x to the power of p comma period period period comma x to the power of p to the power of

n .. are linearly dependent over F sub q open parenthesis open parenthesis t closing parenthesis closing parenthesisperiod .. Write x to the power of p to the power of j =


Rq′ for some q′. Since Rq i s perfect , we may factor off F on the right if itappears , to reduce to the case where P has nonzero constant coefficient , orequivalently ( by Lemma 6 . 1 . 4 ) where P (F )(z) has no repeated roots .

By Corollary 6 . 4 . 6 , we can write P = Q1 · · ·Qn for some monic lineartwisted polynomials Q1, ..., Qn over Rq′ for some q′. Write Qi = F − ci( where ci 6= 0); the process of finding roots of P (F )(z) can then be describedas the process of finding solutions of the system of equations

zp1 − c1z1 = 0

zp2 − c2z2 = z1. . .

zpn − cnzn = zn−1

( the roots of P (F )(z) being precisely the possible values of zn). By repeatedapplications of Lemma 7 . 3 . 2 , the system has pn distinct solutions , andso P (F )(z) splits completely as desired . �

At this point , we can now deduce that “ algebraic implies automatic” inmuch the same manner as in Proposition 5 . 2 . 7 . This means in particularthat again we will need to invoke Christol ’ s theorem .Proposition 7 . 3 . 4 . Let x =

∑xit

i ∈ Fq((tQ)) be ageneralized powers eries which is algebraic over Fq(t). Then x is p− quasi - automatic .Proof . There exists a polynomial P (z) over Fq(t1/p

m

), for some non-negative integer m, such that P has x as a root with multiplicity 1 ; byreplacingx with xp

m

, we may reduce without loss of generality to the case where m = 1.Choose c ∈ Q such that c > v(x− x′) for any root x′ 6= x of P.

By Proposition 7 . 3 . 3 , there exists a power q′ of q and a p− quasi -automaticseries y over Fq′((tQ)) such that v(x− y) ≥ c. The polynomial P (z + y)then has exactly one root of slope at least c, namely x− y.

By Proposition 5.1.2, y i s algebraic over Fq(t). Let K be the finite ex -tension of Fq′((t)) obtained by adjoining y. Then K i s complete under v;by Corollary 6 . 3 . 3 , we may split off the unique factor of P (z+ y) of slopeat least c. In other words , x− y ∈ K, and so x ∈ K.

At this point the argument parallels that of Proposition 5 . 2 . 7 . Let mbe the degree of the minimal polynomial of y over Fq(t). For j = 0, ...,m− 1,write

m− 1

ypj =∑

aijyi

i = 0

with aij ∈ Fq((t)); then the aij are algebraic over Fq(t). Choose n minimalsuch that x, xp, ..., xp

n

are linearly dependent over Fq((t)). Write xpj

=

\noindent 4 1 6 \quad Kiran S . Kedlaya$ \sum ˆ{ m − 1 } { i = 0 } b { i j } y ˆ{ i }$ with $ b { i j } \ in

F { q } ( ( t ) ) ; $ we then have the equat ions

\ [\ begin { a l i gned } m − 1 \\b { i ( j + 1 ) } = \sum b−p { l j ˆ{ a { i l }}} ( j = 0

, . . . , n − 1 ) . \\l = 0 \end{ a l i gned }\ ]

\noindent Let $ c { 0 } x + \cdot \cdot \cdot + c { n } x ˆ{ p ˆ{ n }}= 0 $ be a l i n e a r r e l a t i o n over $ F { q } ( ( t ) ) , $ which we may

normal ize by s e t t i n g $ c { 0 } = 1 ; $ \quad then the \quad $ c { i }$ \quad are a l g e b r a i c over$ F { q } ( t ) . $ \quad Also ,

\noindent wr i t i ng $ x = − c { 1 } x ˆ{ p } − \cdot \cdot \cdot −c { n } ( x ˆ{ p ˆ{ n − 1 }} ) ˆ{ p } , $ we have

\ [\ begin { a l i gned } m − 1 n − 1 \\\sum b { i 0 } y ˆ{ i } = \sum − c { j + 1 } ( x ˆ{ p ˆ{ j }}

) ˆ{ p }\\i = 0 j = 0 \\

m − 1 n − 1 \\= \sum \sum − c { j + 1 }ˆ{ b−p } { l j } y ˆ{ pl }\\l = 0 j = 0 \\

m − 1 n − 1 m − 1 \\= \sum \sum \sum − c { j + 1 }ˆ{ b−p } { l j } a { i l } y ˆ{ i }

, \\i = 0 j = 0 l = 0 \end{ a l i gned }\ ]

\noindent and hence

\ [\ begin { a l i gned } n − 1 m − 1 \\b { i 0 } = \sum \sum − c { j + 1 } b ˆ{ p } { l j } a { i l } . \\j = 0 l = 0 \end{ a l i gned }\ ]

\noindent We thus have a system o f equat ions in the $ b { i j }$ as in Lemma 3 . 3 . 5 ( with$ B $

equal to the i d e n t i t y matrix ) ; t h i s a l l ows us to conclude that each $ b { i j }$\quad i s

\begin { a l i g n ∗}a l g e b r a i c over F { q } ( t ) .\end{ a l i g n ∗}

\hspace ∗{\ f i l l }By C h r i s t o l ’ s theorem ( Theorem 4 . 1 . 1 ) , each $ b { i j } \ inF { q } ( ( t ) ) $ i s $ p − $ automatic .

\noindent Hence $ x = \sum ˆ{ m − 1 } { i = 0 } b { i 0 } y ˆ{ i }$i s $ p − $ quas i − automatic , as d e s i r e d $ . \ square $

\centerline{ $ 8 . F−u $ r the r que s t i on s }

We end with some f u r t h e r que s t i on s \quad about \quad automata and g e n e r a l i z e dpower s e r i e s , o f both an a l go r i thmi c and a t h e o r e t i c a l nature .

\noindent 8 . 1 . \quad Algor i thmics \quad o f automatic \quad s e r i e s . \quad I t seems that one should beab le to make e x p l i c i t c a l c u l a t i o n s in the a l g e b r a i c c l o s u r e o f $ F { q } ( t

) $ \quad us ing

\noindent automatic power s e r i e s ; we have not made any sys temat i c attempt to doso , but i t i s worth r e co rd ing some obse rva t i on s here f o r the b e n e f i t o f anyonec o n s i d e r i n g doing so in the fu tu r e .

\hspace ∗{\ f i l l }The i s s u e o f computing in the a l g e b r a i c c l o s u r e o f $ F { q } ( t) $ us ing g e n e r a l i z e d

\noindent power s e r i e s re sembles that o f computing in the a l g e b r a i c c l o s u r e o f $ Q $us ing

complex approximations o f a l g e b r a i c numbers , \quad and some l e s s o n s may bep r o f i t a b l y drawn from that case . In p a r t i c u l a r , i t may be worth computing

4 1 6 .. Kiran S period Kedlayasum sub i = 0 to the power of m minus 1 b sub ij y to the power of i with b sub ij in F sub q open parenthesis

open parenthesis t closing parenthesis closing parenthesis semicolon we then have the equationsLine 1 m minus 1 Line 2 b sub i open parenthesis j plus 1 closing parenthesis = sum b-p sub lj to the power of a

sub il open parenthesis j = 0 comma period period period comma n minus 1 closing parenthesis period Line 3 l = 0Let c sub 0 x plus times times times plus c sub n x to the power of p to the power of n = 0 be a linear relation

over F sub q open parenthesis open parenthesis t closing parenthesis closing parenthesis comma which we maynormalize by setting c sub 0 = 1 semicolon .. then the .. c sub i .. are algebraic over F sub q open parenthesis t

closing parenthesis period .. Also commawriting x = minus c sub 1 x to the power of p minus times times times minus c sub n open parenthesis x to the

power of p to the power of n minus 1 closing parenthesis to the power of p comma we haveLine 1 m minus 1 n minus 1 Line 2 sum b sub i 0 y to the power of i = sum minus c sub j plus 1 open parenthesis

x to the power of p to the power of j closing parenthesis to the power of p Line 3 i = 0 j = 0 Line 4 m minus 1 nminus 1 Line 5 = sum sum minus c sub j plus 1 to the power of b-p sub lj y to the power of pl Line 6 l = 0 j = 0Line 7 m minus 1 n minus 1 m minus 1 Line 8 = sum sum sum minus c sub j plus 1 to the power of b-p sub lj a subil y to the power of i comma Line 9 i = 0 j = 0 l = 0

and henceLine 1 n minus 1 m minus 1 Line 2 b sub i 0 = sum sum minus c sub j plus 1 b sub lj to the power of p a sub il

period Line 3 j = 0 l = 0We thus have a system of equations in the b sub ij as in Lemma 3 period 3 period 5 open parenthesis with Bequal to the identity matrix closing parenthesis semicolon this allows us to conclude that each b sub ij .. i salgebraic over F sub q open parenthesis t closing parenthesis periodBy Christol quoteright s theorem open parenthesis Theorem 4 period 1 period 1 closing parenthesis comma each

b sub ij in F sub q open parenthesis open parenthesis t closing parenthesis closing parenthesis i s p hyphen automaticperiod

Hence x = sum sub i = 0 to the power of m minus 1 b sub i 0 y to the power of i i s p hyphen quasi hyphenautomatic comma as desired period square

8 period F-u rther questionsWe end with some further questions .. about .. automata and generalizedpower series comma of both an algorithmic and a theoretical nature period8 period 1 period .. Algorithmics .. of automatic .. series period .. It seems that one should beable to make explicit calculations in the algebraic closure of F sub q open parenthesis t closing parenthesis ..

usingautomatic power series semicolon we have not made any systematic attempt to doso comma but it i s worth recording some observations here for the benefit of anyoneconsidering doing so in the future periodThe issue of computing in the algebraic closure of F sub q open parenthesis t closing parenthesis using generalizedpower series resembles that of computing in the algebraic closure of Q usingcomplex approximations of algebraic numbers comma .. and some lessons may beprofitably drawn from that case period In particular comma it may be worth computing

4 1 6 Kiran S . Kedlaya∑m−1i=0 bijy

i with bij ∈ Fq((t)); we then have the equations

m− 1

bi(j+1) =∑

b− pljail (j = 0, ..., n− 1).

l = 0

Let c0x + · · · + cnxpn = 0 be a linear relation over Fq((t)), which we may

normalize by setting c0 = 1; then the ci are algebraic over Fq(t).Also ,writing x = −c1xp − · · · − cn(xp

n−1

)p, we have

m− 1 n− 1∑bi0y

i =∑−cj+1(xp

j

)p

i = 0 j = 0

m− 1n− 1

=∑∑

−cb−pj+1ljypl

l = 0j = 0

m− 1n− 1m− 1

=∑∑∑

−cb−pj+1ljailyi,

i = 0j = 0l = 0

and hence

n− 1m− 1

bi0 =∑∑

−cj+1bpljail.

j = 0l = 0

We thus have a system of equations in the bij as in Lemma 3 . 3 . 5 ( withB equal to the identity matrix ) ; this allows us to conclude that each bij is

algebraicoverFq(t).

By Christol ’ s theorem ( Theorem 4 . 1 . 1 ) , each bij ∈ Fq((t)) i s p−automatic .Hence x =

∑m−1i=0 bi0y

i i s p− quasi - automatic , as desired . �8. F− u rther questions

We end with some further questions about automata and generalizedpower series , of both an algorithmic and a theoretical nature .8 . 1 . Algorithmics of automatic series . It seems thatone should be able to make explicit calculations in the algebraic closure ofFq(t) usingautomatic power series ; we have not made any systematic attempt to do so, but it i s worth recording some observations here for the benefit of anyoneconsidering doing so in the future .

The issue of computing in the algebraic closure of Fq(t) usinggeneralized

power series resembles that of computing in the algebraic closure of Q usingcomplex approximations of algebraic numbers , and some lessons may beprofitably drawn from that case . In particular , it may be worth computing


\noindent ‘ ‘ approximately ’ ’ and not exac t l y with automatic power s e r i e s , in a form o f‘ ‘ i n t e r v a l $\ l e f t . a r i thmet i c \begin { a l i gned } & ’ ’ \\& . \end{ a l i gned }\ right . $

In con t ra s t with the $ Q − $ analogue , however , the re are two ways to t runcatea computation with automata : one can ignore l a r g e powers o f $ t ( $ analogous

\noindent to working with a complex approximation o f an a l g e b r a i c number ) , \quad butone can a l s o prune the automata by i gno r i ng s t a t e s that cannot be reachedin some p a r t i c u l a r number o f s t ep s from the i n i t i a l s t a t e . The l a t t e r maybe c r u c i a l f o r making m u l t i p l i c a t i o n o f automatic \quad s e r i e s e f f i c i e n t , \quad as themethods we have de s c r ibed seem to e n t a i l exponent i a l growth in the numbero f s t a t e s over the course o f a sequence o f a r i thmet i c ope ra t i on s . Furthera n a l y s i s w i l l be needed , however , to determine how much pruning one canget away with , and how e a s i l y one can recove r the miss ing p r e c i s i o n in casei t i s needed again .

Our techn iques \quad seem to depend ra the r badly on the s i z e o f the f i n i t ef i e l d $ F { q }$ under c o n s i d e r a t i o n , but i t i s p o s s i b l e t h i s dependence can be ame −

\noindent l i o r a t e d . For in s t ance , \ h f i l l some o f t h i s dependence ( l i k e the complexity o f a

\noindent product ) \ h f i l l i s r e a l l y not on $ q $ but on the c h a r a c t e r i s t i c $ p, $ \ h f i l l and so i s not \ h f i l l so

\noindent much o f an i s sue when working over a f i e l d o f smal l c h a r a c t e r i s t i c ( as long

\noindent as one decomposes everyth ing over a b a s i s o f $ F { q }$ over $ F { p }) . $ Also , i t may be

\noindent p o s s i b l e to work even in l a r g e c h a r a c t e r i s t i c by wr i t i ng everyth ing in termso f a smal l a d d i t i v e b a s i s o f the f i n i t e f i e l d \quad ( e . g . , \quad f o r $ F { p }

, $ \quad use the powers

\noindent o f 2 l e s s than $ p ) , $ \quad at l e a s t i f one i s w i l l i n g to t runcate as in the prev iousparagraph .

\hspace ∗{\ f i l l }One a d d i t i o n a l concern that a r i s e s when the c h a r a c t e r i s t i c i s not smal l i s

\noindent that our method f o r e x t r a c t i n g roo t s o f po lynomia ls r e q u i r e s working witha d d i t i v e po lynomia l s . Given an ord inary polynomial $ P ( z ) $ o f degree $ n

, $ one

\noindent e a s i l y obta ins an a d d i t i v e polynomial o f degree at most $ p ˆ{ n }$ which has$ P ( z ) $

\noindent as a f a c t o r ( by reduc ing $ z , z ˆ{ p } , z ˆ{ p ˆ{ 2 }} , .. . $ \quad modulo $ P ) ; $ however , the complexity

o f the c o e f f i c i e n t s o f the new polynomial grows e x p o n e n t i a l l y in $ n . $ I t wouldbe o f some i n t e r e s t to develop a form o f Newton ’ s a lgor i thm to dea l d i r e c t l ywith ord inary polynomia l s .

\noindent 8 . 2 . \quad Mult i va r i a t e \quad s e r i e s \quad and \quad automata . \quad We conclude by mentioninga m u l t i v a r i a t e ve r s i o n o f C h r i s t o l ’ s theorem and c o n j e c t u r i n g a g e n e r a l i z e dpower s e r i e s analogue . Fol lowing [ 2 , Chapter 14 ] , we r e s t r i c t our not ion o f‘ ‘ m u l t i v a r i a t e ’ ’ \quad to \quad ‘ ‘ b i v a r i a t e ’ ’ \quad f o r n o t a t i o n a l s i m p l i c i t y , and l eave i t to thereader ’ s imag inat ion to come up with f u l l m u l t i v a r i a t e analogues .

For $ b $ an i n t e g e r g r e a t e r than \quad 1 , \quad a v a l i d pa i r o f base $ b $ \quad expans ions i s apa i r o f s t r i n g s \quad $ ( s { 1 } . . . s { n } , t { 1 } . . .

t { n } ) $ \quad o f equal l ength over the a lphabet \quad $ \Sigma = $

\noindent $ \{ 0 , . . . , b − 1 , . \} $ such that $ s { 1 }$and $ t { 1 }$ are not both $ 0 , s { n }$ and $ t { n }$ are not both

0 , each o f the s t r i n g s $ s { 1 } . . . s { n }$ and $ t { 1 } . .. t { n }$ conta in s exac t l y one rad ix po int ,and those rad ix po in t s occur at the same index $ k . $ We d e f i n e the value o f

Finite automata and algebraic extensions of function fields .. 4 1 7quotedblleft approximately quotedblright and not exactly with automatic power series comma in a form ofquotedblleft interval Case 1 quotedblright Case 2 periodIn contrast with the Q hyphen analogue comma however comma there are two ways to truncatea computation with automata : one can ignore large powers of t open parenthesis analogousto working with a complex approximation of an algebraic number closing parenthesis comma .. butone can also prune the automata by ignoring states that cannot be reachedin some particular number of steps from the initial state period The latter maybe crucial for making multiplication of automatic .. series efficient comma .. as themethods we have described seem to entail exponential growth in the numberof states over the course of a sequence of arithmetic operations period Furtheranalysis will be needed comma however comma to determine how much pruning one canget away with comma and how easily one can recover the missing precision in caseit is needed again periodOur techniques .. seem to depend rather badly on the size of the finitefield F sub q under consideration comma but it is possible this dependence can be ame hyphenliorated period For instance comma .... some of this dependence open parenthesis like the complexity of aproduct closing parenthesis .... i s really not on q but on the characteristic p comma .... and so i s not .... somuch of an i ssue when working over a field of small characteristic open parenthesis as longas one decomposes everything over a basis of F sub q over F sub p closing parenthesis period Also comma it may

bepossible to work even in large characteristic by writing everything in termsof a small additive basis of the finite field .. open parenthesis e period g period comma .. for F sub p comma ..

use the powersof 2 less than p closing parenthesis comma .. at least if one i s willing to truncate as in the previousparagraph periodOne additional concern that arises when the characteristic is not small isthat our method for extracting roots of polynomials requires working withadditive polynomials period Given an ordinary polynomial P open parenthesis z closing parenthesis of degree n

comma oneeasily obtains an additive polynomial of degree at most p to the power of n which has P open parenthesis z

closing parenthesisas a factor open parenthesis by reducing z comma z to the power of p comma z to the power of p to the power

of 2 comma period period period .. modulo P closing parenthesis semicolon however comma the complexityof the coefficients of the new polynomial grows exponentially in n period It wouldbe of some interest to develop a form of Newton quoteright s algorithm to deal directlywith ordinary polynomials period8 period 2 period .. Multivariate .. series .. and .. automata period .. We conclude by mentioninga multivariate version of Christol quoteright s theorem and conjecturing a generalizedpower series analogue period Following open square bracket 2 comma Chapter 14 closing square bracket comma

we restrict our notion ofquotedblleft multivariate quotedblright .. to .. quotedblleft bivariate quotedblright .. for notational simplicity

comma and leave it to thereader quoteright s imagination to come up with full multivariate analogues periodFor b an integer greater than .. 1 comma .. a valid pair of base b .. expansions i s apair of strings .. open parenthesis s sub 1 period period period s sub n comma t sub 1 period period period t sub

n closing parenthesis .. of equal length over the alphabet .. Capital Sigma =open brace 0 comma period period period comma b minus 1 comma period closing brace such that s sub 1 and

t sub 1 are not both 0 comma s sub n and t sub n are not both0 comma each of the strings s sub 1 period period period s sub n and t sub 1 period period period t sub n contains

exactly one radix point commaand those radix points occur at the same index k period We define the value of


“ approximately ” and not exactly with automatic power series , in a form

of “ interval arithmetic′′

.

In contrast with the Q− analogue , however , there are two ways totruncate a computation with automata : one can ignore large powers of t(analogousto working with a complex approximation of an algebraic number ) , butone can also prune the automata by ignoring states that cannot be reachedin some particular number of steps from the initial state . The latter maybe crucial for making multiplication of automatic series efficient , as themethods we have described seem to entail exponential growth in the numberof states over the course of a sequence of arithmetic operations . Furtheranalysis will be needed , however , to determine how much pruning one canget away with , and how easily one can recover the missing precision in caseit is needed again .

Our techniques seem to depend rather badly on the size of the finitefield Fq under consideration , but it is possible this dependence can be ame-liorated . For instance , some of this dependence ( like the complexity of aproduct ) i s really not on q but on the characteristic p, and so i s not somuch of an i ssue when working over a field of small characteristic ( as longas one decomposes everything over a basis of Fq over Fp). Also , it may bepossible to work even in large characteristic by writing everything in termsof a small additive basis of the finite field ( e . g . , for Fp, use thepowersof 2 less than p), at least if one i s willing to truncate as in the previousparagraph .

One additional concern that arises when the characteristic is not smallisthat our method for extracting roots of polynomials requires working withadditive polynomials . Given an ordinary polynomial P (z) of degree n, oneeasily obtains an additive polynomial of degree at most pn which has P (z)

as a factor ( by reducing z, zp, zp2

, ... modulo P ); however , the complexityof the coefficients of the new polynomial grows exponentially in n. It wouldbe of some interest to develop a form of Newton ’ s algorithm to deal directlywith ordinary polynomials .8 . 2 . Multivariate series and automata . Weconclude by mentioning a multivariate version of Christol ’ s theorem andconjecturing a generalized power series analogue . Following [ 2 , Chapter14 ] , we restrict our notion of “ multivariate ” to “ bivariate ” fornotational simplicity , and leave it to the reader ’ s imagination to come upwith full multivariate analogues .

For b an integer greater than 1 , a valid pair of base b expansionsi s a pair of strings (s1...sn, t1...tn) of equal length over the alphabetΣ ={0, ..., b− 1, .} such that s1 and t1 are not both 0, sn and tn are not both 0 ,each of the strings s1...sn and t1...tn contains exactly one radix point , andthose radix points occur at the same index k. We define the value of

\noindent 4 1 8 \quad Kiran S . Kedlayasuch a pa i r to be the pa i r

\ [\ l e f t (\ begin { array }{ cc cc cc } k − 1 & n & k − 1 & n \\ \sum s { i }b ˆ{ k − 1 − i } + & \sum & s { i } b ˆ{ k − i } , & \sum t { i }b ˆ{ k − 1 − i } + & \sum & t { i } b ˆ{ k − i }\\ i = 1 &i = k + 1 & i = 1 & i = k + 1 \end{ array }\ right ) ; \ ]

\noindent then the value func t i on g i v e s a b $ i j $ e c t i o n between the s e t o f v a l i d p a i r s o fbase $ b $ expans ions and $ S { b } \times S { b } . $ Let $ s $ denote the i n v e r s e func t i on . \quad ( Notethat i t may happen that one s t r i n g or the other has some l ead ing and / ort r a i l i n g z e r o e s , s i n c e we are f o r c i n g them to have the same length . )

We may i d e n t i f y p a i r s o f s t r i n g s over $ \Sigma { b }$ o f equal l ength with s t r i n g s over$ \Sigma { b } \times \Sigma { b }$ in the obvious f a s h i o n , and under t h i s i d e n t i f i c a t i o n , the s e t o f v a l i d

p a i r s o f base $ b $ expans ions are seen to form a r e g u l a r language . With thatin mind , we d e c l a r e a func t i on $ f : S { b } \times S { b } \rightarrow\Delta $ to be $ b − $ automatic i f the r e

e x i s t s a DFAO with input a lphabet $ \Sigma = \Sigma { b } \times \Sigma { b }$and output alphabet $ \Delta $

such that f o r any pa i r $ ( v , w ) \ in S { b } \times S { b } ,f ( v , w ) = f M ( s ( v , w ) ) . $ We d e c l a r ea double sequence $ \{ a { i , j } \} ˆ{ \ infty } { i , j = 0 }$ over

$ \Delta $ to be $ b − $ automatic i f f o r some $ \ star element−s l a s h \Delta, $

\noindent the func t i on $ f : S { b } \times S { b } \rightarrow \Delta\cup \{ \ star \} $ given by

\ [ f ( v , w ) = \ l e f t \{\ begin { a l i gned } & a { v , w } v , w\ in Z \\

& 0 otherwi se \end{ a l i gned }\ right . \ ]

\noindent i s $ b − $ automatic .

In t h i s language , the b i v a r i a t e ve r s i on o f C h r i s t o l ’ s theorem i s the f o l −lowing r e s u l t , due to Salon \quad [ 1 6 ] , \quad [ 1 7 ] \quad ( s ee a l s o \quad [ 2 , Theorem 14 . 4 . 1 ] ) . Here$ F { q } ( t , u ) $ denotes the f r a c t i o n f i e l d o f the polynomial r i ng

$ F { q } [ t , u ] ; $ t h i s f i e l d

\noindent i s conta ined in the f r a c t i o n f i e l d o f the power s e r i e s r i ng $ F { q } \ l l b r a c k e tt , u \ r rb ra cke t . $

\noindent Theorem 8 . 2 . 1 . Let $ q $ be a power o f the prime number $ p , $and l e t $ \{ a { i , j } \} ˆ{ \ infty } { i , j = 0 }$

\noindent be \ h f i l l a \ h f i l l double \ h f i l l sequence \ h f i l l over $ F { q } . $ \ h f i l l Then \ h f i l l the \ h f i l l double \ h f i l l s e r i e s \ h f i l l$ \sum ˆ{ \ infty } { i , j = 0 } a { i , j } t ˆ{ i } u ˆ{ j } \ in $

\noindent $ F { q } \ l l b r a c k e t t , u \ r rb ra cke t $ i s a l g e b r a i c over $ F { q }( t , u ) $ i f and only i f the double sequence $ \{ a { i , j } \} ˆ{ \ infty } { i, j = 0 }$

\noindent i s $ p − $ automatic .

To even formulate a g e n e r a l i z e d power s e r i e s analogue o f Theorem 8 . 2 . 1 ,we must dec ide what we mean by \quad ‘ ‘ g e n e r a l i z e d power s e r i e s in two v a r i −ab l e s ’ ’ . The c on s t r u c t i on below i s natura l enough , but we are not aware o f

\noindent a p r i o r appearance in the l i t e r a t u r e .

Let \quad $ G $ be \quad a p a r t i a l l y \quad ordered \quad abe l i an \quad group \quad ( wr i t t en \quad a d d i t i v e l y ) \quad withi d e n t i t y element 0 ; that i s $ , G $ i s an abe l i an group equipped with a binaryr e l a t i o n $ > $ such that f o r a l l $ a , b , c \ in G , $

\ [\ begin { a l i gned } a greate r−n e g a t i o n s l a s h a \\a > b , b > c \Rightarrow a > c \\a > b \Leftrightarrow a + c > b + c . \end{ a l i gned }\ ]

\noindent Let $ P $ be the s e t o f $ a \ in G $ with $ a > 0 ; P $ i s again c a l l e d the p o s i t i v e coneo f $ G . $ We wr i t e $ a \geq b $ to mean $ a > b $ or $ a = b

, $ and $ a \ leq b $ to mean $ b \geq a . $

4 1 8 .. Kiran S period Kedlayasuch a pair to be the pairRow 1 k minus 1 n k minus 1 n Row 2 sum s sub i b to the power of k minus 1 minus i plus sum s sub i b to

the power of k minus i comma sum t sub i b to the power of k minus 1 minus i plus sum t sub i b to the power of kminus i Row 3 i = 1 i = k plus 1 i = 1 i = k plus 1 . semicolon

then the value function gives a b ij ection between the set of valid pairs ofbase b expansions and S sub b times S sub b period Let s denote the inverse function period .. open parenthesis

Notethat it may happen that one string or the other has some leading and slash ortrailing zeroes comma since we are forcing them to have the same length period closing parenthesisWe may identify pairs of strings over Capital Sigma sub b of equal length with strings overCapital Sigma sub b times Capital Sigma sub b in the obvious fashion comma and under this identification

comma the set of validpairs of base b expansions are seen to form a regular language period With thatin mind comma we declare a function f : S sub b times S sub b right arrow Capital Delta to be b hyphen

automatic if thereexists a DFAO with input alphabet Capital Sigma = Capital Sigma sub b times Capital Sigma sub b and output

alphabet Capital Deltasuch that for any pair open parenthesis v comma w closing parenthesis in S sub b times S sub b comma f open

parenthesis v comma w closing parenthesis = f M open parenthesis s open parenthesis v comma w closing parenthesisclosing parenthesis period We declare

a double sequence open brace a sub i comma j closing brace sub i comma j = 0 to the power of infinity overCapital Delta to be b hyphen automatic if for some big star element-slash Capital Delta comma

the function f : S sub b times S sub b right arrow Capital Delta cup open brace big star closing brace given byf open parenthesis v comma w closing parenthesis = Case 1 a sub v comma w v comma w in Z Case 2 0 otherwisei s b hyphen automatic periodIn this language comma the bivariate version of Christol quoteright s theorem is the fol hyphenlowing result comma due to Salon .. open square bracket 1 6 closing square bracket comma .. open square bracket

1 7 closing square bracket .. open parenthesis see also .. open square bracket 2 comma Theorem 14 period 4 period1 closing square bracket closing parenthesis period Here

F sub q open parenthesis t comma u closing parenthesis denotes the fraction field of the polynomial ring F subq open square bracket t comma u closing square bracket semicolon this field

i s contained in the fraction field of the power series ring F sub q llbracket t comma u rrbracket periodTheorem 8 period 2 period 1 period Let q be a power of the prime number p comma and let open brace a sub i

comma j closing brace sub i comma j = 0 to the power of infinitybe .... a .... double .... sequence .... over F sub q period .... Then .... the .... double .... series .... sum sub i

comma j = 0 to the power of infinity a sub i comma j t to the power of i u to the power of j inF sub q llbracket t comma u rrbracket is algebraic over F sub q open parenthesis t comma u closing parenthesis if

and only if the double sequence open brace a sub i comma j closing brace sub i comma j = 0 to the power of infinityis p hyphen automatic periodTo even formulate a generalized power series analogue of Theorem 8 period 2 period 1 commawe must decide what we mean by .. quotedblleft generalized power series in two vari hyphenables quotedblright period The construction below i s natural enough comma but we are not aware ofa prior appearance in the literature periodLet .. G be .. a partially .. ordered .. abelian .. group .. open parenthesis written .. additively closing parenthesis

.. withidentity element 0 semicolon that i s comma G i s an abelian group equipped with a binaryrelation greater such that for all a comma b comma c in G commaLine 1 a greater-negationslash a Line 2 a greater b comma b greater c double stroke right arrow a greater c Line

3 a greater b Leftrightarrow a plus c greater b plus c periodLet P be the set of a in G with a greater 0 semicolon P i s again called the positive coneof G period We write a greater equal b to mean a greater b or a = b comma and a less or equal b to mean b

greater equal a period

4 1 8 Kiran S . Kedlaya such a pair to be the pair k − 1 n k − 1 n∑sib

k−1−i+∑

sibk−i,

∑tib

k−1−i+∑

tibk−i

i = 1 i = k + 1 i = 1 i = k + 1

;

then the value function gives a b ij ection between the set of valid pairs ofbase b expansions and Sb×Sb. Let s denote the inverse function . ( Notethat it may happen that one string or the other has some leading and / ortrailing zeroes , since we are forcing them to have the same length . )

We may identify pairs of strings over Σb of equal length with strings overΣb×Σb in the obvious fashion , and under this identification , the set of validpairs of base b expansions are seen to form a regular language . With thatin mind , we declare a function f : Sb × Sb → ∆ to be b− automatic if thereexists a DFAO with input alphabet Σ = Σb×Σb and output alphabet ∆ suchthat for any pair (v, w) ∈ Sb × Sb, f(v, w) = fM(s(v, w)). We declare a doublesequence {ai,j}∞i,j=0 over ∆ to be b− automatic if for some ?element− slash∆,

the function f : Sb × Sb → ∆ ∪ {?} given by

f(v, w) =

{av,w v, w ∈ Z0 otherwise

i s b− automatic .In this language , the bivariate version of Christol ’ s theorem is the fol -

lowing result , due to Salon [ 1 6 ] , [ 1 7 ] ( see also [ 2 , Theorem14 . 4 . 1 ] ) . Here Fq(t, u) denotes the fraction field of the polynomial ringFq[t, u]; this fieldi s contained in the fraction field of the power series ring Fqt, u.Theorem 8 . 2 . 1 . Let q be a power of the prime number p, and let{ai,j}∞i,j=0

be a double sequence over Fq. Then the double series∑∞i,j=0 ai,jt

iuj ∈Fqt, u is algebraic over Fq(t, u) if and only if the double sequence {ai,j}∞i,j=0

is p− automatic .To even formulate a generalized power series analogue of Theorem 8 . 2

. 1 , we must decide what we mean by “ generalized power series in twovari - ables ” . The construction below i s natural enough , but we are notaware ofa prior appearance in the literature .

Let G be a partially ordered abelian group ( writtenadditively ) with identity element 0 ; that i s , G i s an abelian groupequipped with a binary relation > such that for all a, b, c ∈ G,

agreater − negationslashaa > b, b > c⇒ a > c

a > b⇔ a+ c > b+ c.

Let P be the set of a ∈ G with a > 0;P i s again called the positive cone ofG. We write a ≥ b to mean a > b or a = b, and a ≤ b to mean b ≥ a.


\noindent Then one has the f o l l o w i n g analogue o f Lemma 3 . 1 . 2 , whose proo f we l eaveto the reader .

\noindent Lemma 8 . 2 . 2 . \quad Let $ S $ be a subset o f $ G . $ \quad Then the f o l l o w i n g two c o n d i t i o n sare equ iva l en t .

\hspace ∗{\ f i l l }( a ) \quad Every nonempty subset o f $ S $ has at l e a s t one , but only f i n i t e l y many ,

\hspace ∗{\ f i l l }minimal e lements . ( An e lement $ x \ in S $ i s minimal i f $ y\ in S $ and $ y \ leq x $

\centerline{ imply $ y = x . $ Note that \quad ‘ ‘ minimal ’ ’ does not mean \quad ‘ ‘ s m a l l e s t ’ ’ . ) }

\hspace ∗{\ f i l l }( b ) \quad Any sequence $ s { 1 } , s { 2 } , . . . $ \quad over$ S $ conta in s an i n f i n i t e weakly i n c r e a s −

\ [ ing subsequence s { i { 1 }} \ leq s { i { 2 }} \ leq \cdot \cdot\cdot . \ ]

A subset $ S $ o f $ G $ i s we l l − p a r t i a l l y − ordered i f i t s a t i s f i e s e i t h e r o f the equiv −a l e n t c o n d i t i o n s o f Lemma 8 . 2 . 2 . \quad ( This cond i t i on has cropped up repeat −ed ly in combinator i a l s i t u a t i o n s ; s e e [ 14 ] f o r a survey , i f a somewhat datedone . ) \quad Then f o r any r ing $ R , $ the s e t o f f u n c t i o n s \quad $ f : G\rightarrow R $ which have

we l l − p a r t i a l l y − ordered support forms a r ing under termwise add i t i on and

\noindent convo lut ion .

For $ G = Q \times Q , $ l e t $ R ( ( t ˆ{ Q } , u ˆ{ Q } ) ) $denote the r ing j u s t cons t ruc ted ; we r e f e r

to i t s e lements as \quad ‘ ‘ g e n e r a l i z e d double Laurent \quad s e r i e s ’ ’ . \quad Then by analogy

\noindent with Theorem 4 . 1 . 3 , we formulate the f o l l o w i n g c on j e c tu r e .

\noindent Conjecture 8 . 2 . 3 . \quad Let $ q $ be a power o f the prime $ p , $and l e t $ f : Q \times Q \rightarrow F { q }$

be \quad a func t i on \quad whose \quad support $ S $ \quad i s \quad we l l − p a r t i a l l y − ordered . \quad Then \quad the \quad co r r e −

\noindent sponding g e n e r a l i z e d double Laurent s e r i e s $ \sum { i , j } f (i , j ) t ˆ{ i } u ˆ{ j } \ in F { q } ( ( t ˆ{ Q } , u ˆ{ Q } )) $ i s

\noindent a l g e b r a i c over $ F { q } ( t , u ) $ i f and only i f the f o l l o w i n g c o n d i t i o n s hold .

( a ) \quad For some p o s i t i v e i n t e g e r s $ a $ and $ b , $ the s e t $ aS + b= \{ ( a i + b , a j + b ) : $

$ ( i , j ) \ in S \} $ i s conta ined in $ S { p } \times S { p }. $

\hspace ∗{\ f i l l }( b ) \quad For some $ a , b $ f o r which ( a ) ho lds , \quad the func t i on$ f { a , } b : S { p } \times S { p } \rightarrow F { q }$

\centerline{ given by $ f { a , } b ( x , y ) = f ( ( x −b ) / a , ( y − b ) / a ) $ \quad i s $ p − $ automatic . }

\noindent Moreover , \ h f i l l i f the se \ h f i l l c o n d i t i o n s hold , \ h f i l l then $ f { a, } b $ \ h f i l l i s $ p − $ automatic f o r any non −

\noindent negat ive i n t e g e r s $ a , b $ f o r which ( a ) ho lds .

\hspace ∗{\ f i l l }An a f f i r m a t i v e answer to Conjecture 8 . 2 . 3 would imply that i f the gener −

\noindent a l i z e d double Laurent s e r i e s $ \sum { i , j } c { i , j } t ˆ{ i }u ˆ{ j }$ i s a l g e b r a i c over $ F { q } ( t , u ) , $ then the

\noindent d iagona l s e r i e s $ \sum { i } c { i , i } t ˆ{ i }$ i s a l g e b r a i c over$ F { q } ( t ) ; $ f o r ord inary Laurent s e r i e s ,

\noindent t h i s r e s u l t i s due to Del igne [ 6 ] .

De l igne ’ s r e s u l t a c t u a l l y a l l ows an a r b i t r a r y f i e l d o f p o s i t i v e c h a r a c t e r i s t i cin p lace o f $ F { q } , $ \quad but \quad h i s proo f i s \quad in the context \quad o f s o p h i s t i c a t e d \quad a lgebro −

\noindent geometr ic \quad machinery \quad ( van i sh ing c y c l e s \quad in \quad \ ’{ e} t a l e \quad cohomology ) ; \quad a \quad proo f o fDel igne ’ s g ene ra l r e s u l t in the s p i r i t o f automata − t h e o r e t i c methods wasg iven by S h a r i f and Woodcock [ 1 9 ] . I t may be p o s s i b l e to s t a t e and provean analogous a s s e r t i o n in the g e n e r a l i z e d Laurent s e r i e s s e t t i n g by g iv ing as u i t a b l e m u l t i v a r i a t e ex tens i on o f the r e s u l t s o f [ 1 1 ] .

Finite automata and algebraic extensions of function fields .. 4 1 9Then one has the following analogue of Lemma 3 period 1 period 2 comma whose proof we leaveto the reader periodLemma 8 period 2 period 2 period .. Let S be a subset of G period .. Then the following two conditionsare equivalent periodopen parenthesis a closing parenthesis .. Every nonempty subset of S has at least one comma but only finitely

many commaminimal e lements period open parenthesis An e lement x in S is minimal if y in S and y less or equal ximply y = x period Note that .. quotedblleft minimal quotedblright does not mean .. quotedblleft smallest

quotedblright period closing parenthesisopen parenthesis b closing parenthesis .. Any sequence s sub 1 comma s sub 2 comma period period period ..

over S contains an infinite weakly increas hyphening subsequence s sub i sub 1 less or equal s sub i sub 2 less or equal times times times periodA subset S of G i s well hyphen partially hyphen ordered if it satisfies either of the equiv hyphenalent conditions of Lemma 8 period 2 period 2 period .. open parenthesis This condition has cropped up repeat

hyphenedly in combinatorial situations semicolon see open square bracket 14 closing square bracket for a survey comma

if a somewhat datedone period closing parenthesis .. Then for any ring R comma the set of functions .. f : G right arrow R which

havewell hyphen partially hyphen ordered support forms a ring under termwise addition andconvolution periodFor G = Q times Q comma let R open parenthesis open parenthesis t to the power of Q comma u to the power

of Q closing parenthesis closing parenthesis denote the ring just constructed semicolon we referto it s elements as .. quotedblleft generalized double Laurent .. series quotedblright period .. Then by analogywith Theorem 4 period 1 period 3 comma we formulate the following conjecture periodConjecture 8 period 2 period 3 period .. Let q be a power of the prime p comma and let f : Q times Q right

arrow F sub qbe .. a function .. whose .. support S .. is .. well hyphen partially hyphen ordered period .. Then .. the .. corre

hyphensponding generalized double Laurent series sum sub i comma j f open parenthesis i comma j closing parenthesis

t to the power of i u to the power of j in F sub q open parenthesis open parenthesis t to the power of Q comma uto the power of Q closing parenthesis closing parenthesis is

algebraic over F sub q open parenthesis t comma u closing parenthesis if and only if the following conditions holdperiod

open parenthesis a closing parenthesis .. For some positive integers a and b comma the set aS plus b = openbrace open parenthesis ai plus b comma aj plus b closing parenthesis :

open parenthesis i comma j closing parenthesis in S closing brace is contained in S sub p times S sub p periodopen parenthesis b closing parenthesis .. For some a comma b for which open parenthesis a closing parenthesis

holds comma .. the function f sub a comma b : S sub p times S sub p right arrow F sub qgiven by f sub a comma b open parenthesis x comma y closing parenthesis = f open parenthesis open parenthesis x

minus b closing parenthesis slash a comma open parenthesis y minus b closing parenthesis slash a closing parenthesis.. is p hyphen automatic period

Moreover comma .... if these .... conditions hold comma .... then f sub a comma b .... is p hyphen automatic forany non hyphen

negative integers a comma b for which open parenthesis a closing parenthesis holds periodAn affirmative answer to Conjecture 8 period 2 period 3 would imply that if the gener hyphenalized double Laurent series sum sub i comma j c sub i comma j t to the power of i u to the power of j i s algebraic

over F sub q open parenthesis t comma u closing parenthesis comma then thediagonal series sum sub i c sub i comma i t to the power of i i s algebraic over F sub q open parenthesis t closing

parenthesis semicolon for ordinary Laurent series commathis result i s due to Deligne open square bracket 6 closing square bracket periodDeligne quoteright s result actually allows an arbitrary field of positive characteristicin place of F sub q comma .. but .. his proof is .. in the context .. of sophisticated .. algebro hyphengeometric .. machinery .. open parenthesis vanishing cycles .. in .. eacutetale .. cohomology closing parenthesis

semicolon .. a .. proof ofDeligne quoteright s general result in the spirit of automata hyphen theoretic methods wasgiven by Sharif and Woodcock open square bracket 1 9 closing square bracket period It may be possible to state

and provean analogous assertion in the generalized Laurent series setting by giving asuitable multivariate extension of the results of open square bracket 1 1 closing square bracket period


Then one has the following analogue of Lemma 3 . 1 . 2 , whose proof weleave to the reader .Lemma 8 . 2 . 2 . Let S be a subset of G. Then the following twoconditions are equivalent .

( a ) Every nonempty subset of S has at least one , but only finitelymany ,

minimal e lements . ( An e lement x ∈ S is minimal if y ∈ S andy ≤ x

imply y = x. Note that “ minimal ” does not mean “ smallest ” . )( b ) Any sequence s1, s2, ... over S contains an infinite weakly

increas -

ingsubsequencesi1 ≤ si2 ≤ · · ·.

A subset S of G i s well - partially - ordered if it satisfies either of theequiv - alent conditions of Lemma 8 . 2 . 2 . ( This condition hascropped up repeat - edly in combinatorial situations ; see [ 14 ] for a survey, if a somewhat dated one . ) Then for any ring R, the set of functionsf : G → R which have well - partially - ordered support forms a ringunder termwise addition andconvolution .

For G = Q × Q, let R((tQ, uQ)) denote the ring just constructed ; we referto it s elements as “ generalized double Laurent series ” . Then byanalogywith Theorem 4 . 1 . 3 , we formulate the following conjecture .Conjecture 8 . 2 . 3 . Let q be a power of the prime p, andlet f : Q×Q→ Fq be a function whose support S is well -partially - ordered . Then the corre -sponding generalized double Laurent series

∑i,j f(i, j)tiuj ∈ Fq((tQ, uQ)) is

algebraic over Fq(t, u) if and only if the following conditions hold .( a ) For some positive integers a and b, the set aS+b = {(ai+b, aj+b) :

(i, j) ∈ S} is contained in Sp × Sp.( b ) For some a, b for which ( a ) holds , the function

fa,b : Sp × Sp → Fqgiven by fa,b(x, y) = f((x− b)/a, (y − b)/a) is p− automatic .

Moreover , if these conditions hold , then fa,b is p− automatic forany non -negative integers a, b for which ( a ) holds .

An affirmative answer to Conjecture 8 . 2 . 3 would imply that if thegener -alized double Laurent series

∑i,j ci,jt

iuj i s algebraic over Fq(t, u), then thediagonal series

∑i ci,it

i i s algebraic over Fq(t); for ordinary Laurent series ,this result i s due to Deligne [ 6 ] .

Deligne ’ s result actually allows an arbitrary field of positive characteristicin place of Fq, but his proof is in the context of sophisticatedalgebro -geometric machinery ( vanishing cycles in etale cohomology ) ;a proof of Deligne ’ s general result in the spirit of automata - theoreticmethods was given by Sharif and Woodcock [ 1 9 ] . It may be possibleto state and prove an analogous assertion in the generalized Laurent seriessetting by giving a suitable multivariate extension of the results of [ 1 1 ] .


\centerline{Refe rences }

[ 1 ] \quad S . Abhyankar , Two notes on formal power s e r i e s . Proc . Amer . Math . Soc . 7 ( 1 956 ) , 903 −−905 .

[ 2 ] \quad J . − P . Al louche , J . S h a l l i t , Automatic Sequences : Theory , App l i ca t i on s , G e n e r a l i z a t i o n s .Cambridge Univ . Press , 2003 .

[ 3 ] \quad C . Cheval ley , In t roduc t i on to the Theory o f Algebra i c Functions o f One Var iab le . Amer .Math . Soc . , 1 95 1 .

[ 4 ] \quad G . C h r i s t o l , Ensembles presque p e r i o d i q u e s $ k − $ r e c o n n a i s s a b l e s . Theoret . Comput . Sc i . 9( 1 979 ) , 141 −− 145 .

[ 5 ] \quad G . C h r i s t o l , T . Kamae , M . Mend $ \grave{e} $ s France , G . Rauzy , S u i t e s a l g \ ’{ e} br ique s , automateset s u b s t i t u t i o n s . Bul l . Soc . Math . France 1 8 ( 1 980 ) , 401 −− 41 9 .

\centerline { [ 6 ] \quad P . Del igne , Int \ ’{ e} g ra t i on sur un c y c l e \ ’{ e} vanescent . Invent . Math . 76 ( 1 984 ) , 1 29 −− 143 . }

\centerline { [ 7 ] \quad H . Furstenberg , A lgebra i c f u n c t i o n s over f i n i t e f i e l d s . J . Alg . 7 ( 1 967 ) , 271 −− 277 . }

[ 8 ] \quad H . Hahn $ , \ddot{U} $ ber d i e n i chtarch imed i s che Gr\”{o} \ ss ensysteme ( 1 907 ) . Gesammelte AbhandlungenI , Spr inger − Verlag , 1 995 .

[ 9 ] \quad D . R . Hayes , A b r i e f i n t r o d u c t i o n to D r i n f e l ’ d modules . The Arithmet ic o f Function F i e l d s( ed i t ed by D . Goss , D . R . Hayes , and M . I . Rosen ) , 1 −− 32 , de Gruyter , 1 992 .[ 10 ] \quad I . Kaplansky , Maximal f i e l d s with v a lu a t i on s . Duke Math . J . 9 ( 1 942 ) , 303 −− 321 .[ 1 1 ] \quad K . S . Kedlaya , The a l g e b r a i c c l o s u r e o f the power s e r i e s f i e l d in p o s i t i v e c h a r a c t e r i s t i c .Proc . Amer . Math . Soc . 1 29 ( 200 1 ) , 346 1 −− 3470 .[ 1 2 ] \quad K . S . Kedlaya , Power s e r i e s and $ p − $ ad ic a l g e b r a i c c l o s u r e s . J . Number Theory 89 ( 2001 ) ,324 −− 339 .

\noindent [ 13 ] \quad K . S . Kedlaya , Algebra i c g e n e r a l i z e d power s e r i e s and automata . arXiv p r ep r i n t math .AC / 0 1 10089 , 2001 .[ 14 ] \quad J . B . Kruskal , The theory o f we l l − quas i − orde r ing : a f r e q u e n t l y d i s covered concept . J . Comb .Theory Ser . A 1 3 ( 1 972 ) , 297 −− 305 .[ 1 5 ] \quad D . S . Pas sman , The Algebra i c St ructure o f Group Rings . Wiley , 1 977 .[ 1 6 ] \quad O . Salon , S u i t e s automatiques $ \grave{a} $ mult i − i n d i c e s et a l g \ ’{ e} b r i c i t \ ’{ e} . C . R . Acad . Sc i . Par i s S\ ’{ e} r . IMath . 305 ( 1 987 ) , 501 −− 504 .[ 1 7 ] \quad O . Salon , S u i t e s automatiques $ \grave{a} $ mult i − i n d i c e s ( with an appendix by J . S h a l l i t ) . Sem .Th\ ’{ e} o r i e Nombres Bordeaux 4 ( 1 986 −− 1 987 ) , 1 −− 27 .[ 18 ] \quad J . − P . Ser re , Local F i e l d s ( t r a n s l a t e d by M . J . Greenberg ) . Spr inger − Verlag , 1 979 .[ 1 9 ] \quad H . S h a r i f , C . F . Woodcock , Algebra i c f u n c t i o n s over a f i e l d o f p o s i t i v e c h a r a c t e r i s t i c andHadamard products . J . London Math . Soc . 37 ( 1 988 ) , 395 −− 403 .

\noindent Kiran S . Kedlaya

\noindent Department o f MathematicsMassachusetts I n s t i t u t e o f TechnologyCambridge , MA 2 139 , USAE − mail : kedlaya $ @ $ math . mit . eduURL : http : / / math . mit . edu / \ t e x t a s c i i t i l d e kedlaya

420 .. Kiran S period KedlayaReferencesopen square bracket 1 closing square bracket .. S period Abhyankar comma Two notes on formal power series

period Proc period Amer period Math period Soc period 7 open parenthesis 1 956 closing parenthesis comma 903endash

905 periodopen square bracket 2 closing square bracket .. J period hyphen P period Allouche comma J period Shallit

comma Automatic Sequences : Theory comma Applications comma Generalizations periodCambridge Univ period Press comma 2003 periodopen square bracket 3 closing square bracket .. C period Chevalley comma Introduction to the Theory of Algebraic

Functions of One Variable period Amer periodMath period Soc period comma 1 95 1 periodopen square bracket 4 closing square bracket .. G period Christol comma Ensembles presque periodiques k

hyphen reconnaissables period Theoret period Comput period Sci period 9open parenthesis 1 979 closing parenthesis comma 141 endash 145 periodopen square bracket 5 closing square bracket .. G period Christol comma T period Kamae comma M period

Mend egrave s France comma G period Rauzy comma Suites algeacutebriques comma automateset substitutions period Bull period Soc period Math period France 1 8 open parenthesis 1 980 closing parenthesis

comma 401 endash 41 9 periodopen square bracket 6 closing square bracket .. P period Deligne comma Inteacutegration sur un cycle eacute-

vanescent period Invent period Math period 76 open parenthesis 1 984 closing parenthesis comma 1 29 endash 143period

open square bracket 7 closing square bracket .. H period Furstenberg comma Algebraic functions over finite fieldsperiod J period Alg period 7 open parenthesis 1 967 closing parenthesis comma 271 endash 277 period

open square bracket 8 closing square bracket .. H period Hahn comma Udieresis ber die nichtarchimedischeGrodieresis germandbls ensysteme open parenthesis 1 907 closing parenthesis period Gesammelte Abhandlungen

I comma Springer hyphen Verlag comma 1 995 periodopen square bracket 9 closing square bracket .. D period R period Hayes comma A brief introduction to Drinfel

quoteright d modules period The Arithmetic of Function Fieldsopen parenthesis edited by D period Goss comma D period R period Hayes comma and M period I period Rosen

closing parenthesis comma 1 endash 32 comma de Gruyter comma 1 992 periodopen square bracket 10 closing square bracket .. I period Kaplansky comma Maximal fields with valuations

period Duke Math period J period 9 open parenthesis 1 942 closing parenthesis comma 303 endash 321 periodopen square bracket 1 1 closing square bracket .. K period S period Kedlaya comma The algebraic closure of the

power series field in positive characteristic periodProc period Amer period Math period Soc period 1 29 open parenthesis 200 1 closing parenthesis comma 346 1

endash 3470 periodopen square bracket 1 2 closing square bracket .. K period S period Kedlaya comma Power series and p hyphen

adic algebraic closures period J period Number Theory 89 open parenthesis 2001 closing parenthesis comma324 endash 339 periodopen square bracket 13 closing square bracket .. K period S period Kedlaya comma Algebraic generalized power

series and automata period arXiv preprint math periodAC slash 0 1 10089 comma 2001 periodopen square bracket 14 closing square bracket .. J period B period Kruskal comma The theory of well hyphen

quasi hyphen ordering : a frequently dis covered concept period J period Comb periodTheory Ser period A 1 3 open parenthesis 1 972 closing parenthesis comma 297 endash 305 periodopen square bracket 1 5 closing square bracket .. D period S period Pas sman comma The Algebraic Structure

of Group Rings period Wiley comma 1 977 periodopen square bracket 1 6 closing square bracket .. O period Salon comma Suites automatiques agrave multi hyphen

indices et alg eacutebriciteacute period C period R period Acad period Sci period Paris Seacuter period IMath period 305 open parenthesis 1 987 closing parenthesis comma 501 endash 504 periodopen square bracket 1 7 closing square bracket .. O period Salon comma Suites automatiques agrave multi hyphen

indices open parenthesis with an appendix by J period Shallit closing parenthesis period Sem periodTheacuteorie Nombres Bordeaux 4 open parenthesis 1 986 endash 1 987 closing parenthesis comma 1 endash 27

periodopen square bracket 18 closing square bracket .. J period hyphen P period Serre comma Local Fields open

parenthesis translated by M period J period Greenberg closing parenthesis period Springer hyphen Verlag comma 1979 period

open square bracket 1 9 closing square bracket .. H period Sharif comma C period F period Woodcock commaAlgebraic functions over a field of positive characteristic and

Hadamard products period J period London Math period Soc period 37 open parenthesis 1 988 closing parenthesiscomma 395 endash 403 period

Kiran S period KedlayaDepartment of MathematicsMassachusetts Institute of TechnologyCambridge comma MA 2 139 comma USAE hyphen mail : kedlaya at math period mit period eduURL : http : slash slash math period mit period edu slash asciitilde kedlaya


References[ 1 ] S . Abhyankar , Two notes on formal power series . Proc . Amer . Math . Soc . 7 ( 1

956 ) , 903 – 905 .

[ 2 ] J . - P . Allouche , J . Shallit , Automatic Sequences : Theory , Applications ,

Generalizations . Cambridge Univ . Press , 2003 .

[ 3 ] C . Chevalley , Introduction to the Theory of Algebraic Functions of One Variable . Amer

. Math . Soc . , 1 95 1 .

[ 4 ] G . Christol , Ensembles presque periodiques k− reconnaissables . Theoret . Comput .

Sci . 9 ( 1 979 ) , 141 – 145 .

[ 5 ] G . Christol , T . Kamae , M . Mend e s France , G . Rauzy , Suites algebriques ,

automates et substitutions . Bull . Soc . Math . France 1 8 ( 1 980 ) , 401 – 41 9 .

[ 6 ] P . Deligne , Integration sur un cycle evanescent . Invent . Math . 76 ( 1 984 ) , 1 29 – 143 .

[ 7 ] H . Furstenberg , Algebraic functions over finite fields . J . Alg . 7 ( 1 967 ) , 271 – 277 .

[ 8 ] H . Hahn , U ber die nichtarchimedische Gro ßensysteme ( 1 907 ) . Gesammelte

Abhandlungen I , Springer - Verlag , 1 995 .

[ 9 ] D . R . Hayes , A brief introduction to Drinfel ’ d modules . The Arithmetic of Function

Fields ( edited by D . Goss , D . R . Hayes , and M . I . Rosen ) , 1 – 32 , de Gruyter , 1 992 . [ 10

] I . Kaplansky , Maximal fields with valuations . Duke Math . J . 9 ( 1 942 ) , 303 – 321 . [ 1 1

] K . S . Kedlaya , The algebraic closure of the power series field in positive characteristic . Proc

. Amer . Math . Soc . 1 29 ( 200 1 ) , 346 1 – 3470 . [ 1 2 ] K . S . Kedlaya , Power series and

p− adic algebraic closures . J . Number Theory 89 ( 2001 ) , 324 – 339 .

[ 13 ] K . S . Kedlaya , Algebraic generalized power series and automata . arXiv preprint math .

AC / 0 1 10089 , 2001 . [ 14 ] J . B . Kruskal , The theory of well - quasi - ordering : a frequently

dis covered concept . J . Comb . Theory Ser . A 1 3 ( 1 972 ) , 297 – 305 . [ 1 5 ] D . S .

Pas sman , The Algebraic Structure of Group Rings . Wiley , 1 977 . [ 1 6 ] O . Salon , Suites

automatiques a multi - indices et alg ebricite . C . R . Acad . Sci . Paris Ser . I Math . 305 ( 1 987

) , 501 – 504 . [ 1 7 ] O . Salon , Suites automatiques a multi - indices ( with an appendix by J

. Shallit ) . Sem . Theorie Nombres Bordeaux 4 ( 1 986 – 1 987 ) , 1 – 27 . [ 18 ] J . - P . Serre ,

Local Fields ( translated by M . J . Greenberg ) . Springer - Verlag , 1 979 . [ 1 9 ] H . Sharif , C

. F . Woodcock , Algebraic functions over a field of positive characteristic and Hadamard products

. J . London Math . Soc . 37 ( 1 988 ) , 395 – 403 .

Kiran S . Kedlaya

Department of Mathematics Massachusetts Institute of Technology Cambridge , MA 2 139 , USA E -

mail : kedlaya @ math . mit . edu URL : http : / / math . mit . edu / ~kedlaya

Finite automata and algebraic extensions of function fields

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