The Riemann Hypothesis for the Goss zeta function

TheRiemannHypothesisfortheGosszetafunctionforFq[T]JereyT.SheatsDepartmentofMathematics,UniversityofArizona,Tucson,Arizona85721December30,1997AbstractLetqbeapowerofaprimep.WeproveanassertionofCarlitzwhichtakesqasaparameter.Diaz-Vargas’proofoftheRiemannHypothesisfortheGosszetafunctionforFp[T]dependsonhisvericationofCarlitz’sassertionforthespeciccaseq=p[D-V].OurproofofthegeneralcaseallowsustoextendDiaz-Vargas’prooftoFq[T].In[Gos1]Gosspresentszetafunctionsforfunctioneldsofnitecharacter-isticwhichpossessmanyinterestingpropertiesincludinganalogsofpropertiesoftheRiemannzetafunction.InthispaperweproveananalogoftheRiemannHypothesisfortheGosszetafunctionforFq[T]whereqisapowerofaprimep.ThestatementoftheanalogappearsinTheorem1.1below.Forthecaseq=ptheanalogwasprovenbyD.Wan[Wan].Ourprooffollowsanotherproofforthecaseq=precentlyputforwardbyDiaz-Vargas[D-V].AkeypartofDiaz-Vargas’proofinvolvesaresultofCarlitzconcerningthevanishingofcertainpowersums[Car3].TheproofthatCarlitzsketchedforhisresultdependsonacombinator-ialassertionwhichtakesqasaparameter.Carlitzgivesnojusticationforthisassertion.ArestatementofCarlitz’sassertionappearsasourTheorem1.2.Diaz-Vargasprovedtheassertionforq=p.ThisenabledhimtoconstructhiselegantproofoftheRiemannHypothesisforFp[T].PoonenprovedCarlitz’sassertionforthecaseq=4[Po].OurmainresultisaproofofCarlitz’sassertionforallq.Thisproofwasinspiredby[Po]andusessomeofthesameideas.Carlitzuseshisassertionintheproofsoftworesults.Astheyarenowfullyjustied,werestatetheminTheorem1.4.InSection1wedenetheGosszetafunctionforFq[T]andthenstateTheorem1.1:theanalogoftheRiemannHypothesis.NextwestateCarlitz’sresultandTheorem1.2:thecombinatorialassertiononwhichtheresultdepends.AbriefoutlineoftheproofofTheorem1.2isgivenattheendofSection1.InSection2weproveTheorem1.1.Sections3through7aredevotedtotheproofofTheorem1.2.Theauthor(acombinatorialist)wishestothankDineshThakurandDavidGossfortheirhelpwiththenumbertheory.TheauthorthanksDineshThakurforbringingCarlitz’sassertiontohisattention.Specialthanksgototherefereewhomadevaluablestylisticsuggestionsandalsofoundmanymistakesinearlierversionsofthispaper.1.MainResultsInthispaperNdenotesthesetofnonnegativeintegersandwesetZ+:=Nnf0g.Letpbeprimeandsetq:=ps.LetvdenotetheT1-adicvaluationonK:=Fq(T).ThentheeldofLaurentseriesK1:=Fq((T1))isthecompletionofKwithrespecttov.DenotebyA+thesetofmonicpolynomialsinA:=Fq[T].LetZpdenotethep-adicintegersandletbethecompletionofanalgebraicclosureofK1.TheanalogywithcharacteristiczeroisgivenbyA$Z,A+$Z+,K$Q,K1$R,and$C.TheGosszetafunctionforFq[T]isdenedas(z):=Xn2A+nzwherezistakenfromZp.Exponentiationisdenedasfollows:foramonicpolynomialnsethni:=nTdegn;thenforz=(x;y)2ZpGossdenesnz:=xdegnhniy.Thetermhniyiswelldenedsincehni1(modT1).Onedrawsananalogybe-tweenthisdenitionandcomplexexponentiationwhenCisregardedcanonicallyasRR:forpositiveintegersnwehaven(x;y)=(ex)logn(eilogn)y.Gossshowed2thatbygroupingtogethertermsofthesamedegreebecomeswelldenedoverallZp:(z):=(x;y):=Xm0xm0@Xn2A+,degn=mhniy1A:Fromthedenitionofexponentiationwehaven(Tk;k)=nkforanyintegerk:Dene(k):=(Tk;k)fork2Z.Thuswhenk0wehave(k)=Pn2A+nk.Thevaluesofonthesetof\integersf(Tk;k):k2ZgareanalogoustospecialvaluesoftheRiemannzetafunction:CarlitzshowedthefollowinganalogtoatheoremofEuler.Fork0wehave((q1)k)=B(q1)k~(q1)kg(q1)kwhere~2isananalogto2i,theB(q1)k2KareanalogstotheevenBernoullinumbers,andtheg(q1)k2Aareanalogstofactorials[Car1][Car2].Gossshowed(ingeneral,notjustfortheFq[T]case)thatfork0wehave(k)2Aand(k)=0whenk0(modq1)[Gos1].Inanalogy,theRiemannzetafunctionisrationalonthenegativeintegersandzeroonthenegativeevens.Consult[Gos2]formoreinterestingpropertiesofGoss’zetafunctionsincludingtheirconnectionwithcyclotomicextensionsandwithDrinfeldmodules.InSection2weprovethefollowinganalogtotheRiemannHypothesis.Itstatesthatforxedythezerosof(x;y)aresimpleandalllieonthesame\realline.Foracompleteexplanationoftheanalogsee[Gos2].Theorem1.1.Fixy2Zp.Asafunctionofx,thezerosof(x;y)aresimpleandlieinK1.InfacttheylieinthesubeldFp((T1)).In[Car3]Carlitzinvestigated,amongotherthings,thevanishingofthepowersumsS0k(N):=Xn2A+,degn=knN3forpositiveintegersN.HestatedthatS0k(N)6=0ifandonlyifthereexistsa(k+1)-tuple(r0;r1;:::;rk)2Nk+1whosetermssumtoNandsatisfythefollowingtwoconditions:(i)thereisnocarryoverofp-adicdigitsinthesumN=Xrj;(ii)rj0and(ps1)jrjfor0jk1.(Notethat(ii)putsnoconditiononrk.)LetUk+1(N)bethecollectionofallsuch(k+1)-tuples.ThenCarlitzclaimedthatS0k(N)6=0ifandonlyifUk+1(N)6=;.Hisproofwentasfollows.Foramonicn2A+setn=a0+a1T1++ak1Tk1+Tk.ThenS0k(N)=XNr0;:::;rkXar00ark1k1Tr1+2r2++krk(1.1)wheretheoutersumisoverall(k+1)-tuplesr=(r0;:::;rk)suchthatPrj=Nandtheinnersumisoverall(a0;:::;ak1)2(Fq)k.NotethatthesumPa2Fqahequals1whenhisapositivemultipleofq1,andequals0otherwise.Thus(1.1)becomesS0k(N)=XNr0;:::;rk(1)kTr1+2r2++krk(1.2)wherethesumisoverall(k+1)-tuples(r0;r1;:::;rk)suchthatPrj=Nandalsosatisfycondition(ii)above.AwellknownresultofLucasstatesthatt