Problems on Finite Extensions of Local Function Fields

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Problems on Finite Extensions of Local Function Fields
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Huynh, Duc V
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Doctorate ( Ph.D.)
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University of Florida
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Mathematics
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KEATING,KEVIN P
Committee Co-Chair:
CREW,RICHARD MALCOLM
Committee Members:
GARVAN,FRANCIS G
ALLADI,KRISHNASWAMI
SITHARAM,MEERA

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aiba -- amano -- associated -- basis -- break -- byott -- classes -- conjugacy -- conjugate -- cyclic -- eisenstein -- enumerate -- fields -- finite -- free -- full -- function -- galois -- group -- isomorphism -- jump -- k-isomorphism -- keating -- klopsch -- lettl -- local -- modules -- nottingham -- orders -- polynomial -- ramification -- ramified -- totally
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Abstract:
Let $K$ be a local field with finite residue field $k$ and characteristic $p$, and let $K_s$ be a separable closure of $K$. This work deals with the study of totally ramified extensions $L/K$ in $K_s$ of degree $p$ with a fixed ramification break $\lambda \in \mathbb{Q}$, and it is a combination of two problems. The first problem is an extension of a work done by Akira Aiba, while the second problem deals with the enumeration of $k$-isomorphism classes of totally ramified extensions in $K_s$ of degree $p$. Let $L/K$ be a Galois extension of degree $p$ with Galois group $G$ and ramification break $\lambda$, and let $\nu_K$ be the valuation of $K_s$ normalized on $K$. The first problem is regarding the determination of a basis for the Galois module $M_L^r = \{ x \in L : \nu_L(x) \geq \frac{r}{p} \}$ over the generalized associated order $A_{r,r}(L/K) = \{ \varphi \in KG : \varphi \cdot M_L^r \subset M_L^r \}$, and the criterion for which $M_L^r$ is a free $A_{r,r}(L/K)$-module. Say two extensions $L_1/K$ and$L_2/K$ of degree $p$ with ramification break $\lambda$ are $k$-isomorphic if there exists an isomorphism $\psi: L_1 \rightarrow L_2$ such that $\psi(K) = K$ and $\psi|_{k} = $ id$_k$. Let $\mathcal{E}_{\lambda}$ to be the set of all totally ramified extensions $N/K$ of degree $p$ with ramification break $\lambda$. The second problem is regarding the enumeration of $k$-isomorphism classes of $\mathcal{E}_{\lambda}$.
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by Duc V Huynh.
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Thesis (Ph.D.)--University of Florida, 2014.
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Adviser: KEATING,KEVIN P.
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Co-adviser: CREW,RICHARD MALCOLM.

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PROBLEMSONFINITEEXTENSIONSOFLOCALFUNCTIONFIELDSByDUCVANHUYNHADISSERTATIONPRESENTEDTOTHEGRADUATESCHOOLOFTHEUNIVERSITYOFFLORIDAINPARTIALFULFILLMENTOFTHEREQUIREMENTSFORTHEDEGREEOFDOCTOROFPHILOSOPHYUNIVERSITYOFFLORIDA2014

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c2014DucVanHuynh 2

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Idedicatethisworktoeveryonewhonevergiveuponchasingtheirdream. 3

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ACKNOWLEDGMENTS Iwouldliketobeginbythankingmyadvisor,Dr.KevinKeating.Youhavegivenmesomuchofyourpatience,timeandeffortinhelpingmealongmyjourney.Youhavetaughtmesomuchmathematics,andhavehelpedmebecomeabettermathematician.Ireallyappreciateeverythingyouhavedoneforme,anditisanhonortobeyourstudent.SomeofthetoughestyearsingraduateschoolwerewhiletakingDr.RichardCrew'salgebraclasses.Thoughyourlectureswerealwayswonderful,yourhomeworkassignmentsandexamsweredifcult.However,throughthesestruggles,Ihavegainedavastknowledgeandagoodunderstandingofmathematics.Furthermore,Iwouldliketothankyouforbeingamemberofmycommittee.IamproudtohavebeenastudentofDr.KrishnaswamiAlladiinanalyticnumbertheory.Ihavealwaysstruggledinanalyticnumbertheory,butyouhavemadeyourlecturesenjoyableandincrediblyawless.Thankyouforinvitingallofustoyourbeautifulhome,wherewewereabletoenjoydeliciousfoodandconversewithgreatmathematicians.Thankyouforbeingamemberofmycommitteeandforprovidinghelpfulremarks.Iwouldliketosaythankyoutomyothercommitteemembers:Dr.FrankGarvan,Dr.MeeraSitharamandDr.MyThai.Thankyouforallthecommentsyouhavemadeandalltheadviceyouhavegiven.Iappreciateforyourtimeandconsideration.Youeachhavehelpedmebecomeabettermathematician.WiththetravelgrantfromCenterforAppliedMathematics,IwasabletopresentmyworkattheSoutheastRegionalMeetingonNumbersinWesternCarolinaUniversity.Throughthetalk,Igainedsomeinsightintomywork,andmetmanygreatmathematicianswhomImaydoresearchwithinthenearfuture.IappreciatethecommitteeofCAMforawardingmethissupport. 4

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ThepastsevenyearsattheUniversityofFloridaDepartmentofMathematicshasbeendelightful.Thedepartmenthasbeensupportiveandhasprovidedmeastimulatingresearchenvironment.Besidelearningtobeabettermathematician,thedepartmenthasalsoshowmehowtobeabetterteacher.Beingapartofthedepartmenthasgivenmeanopportunitytolearnfromsomeofthegreatestprofessors,makefriendswithwonderfulpeople,andmatureasbothamathematicianandaperson.Iamgratefulforeverythingthatthedepartmenthasgivenme.IwouldliketothanktheJournalofNumberTheoryforacceptingmywork.Therefereeformypapergavemanyinsightfulremarks,whichallowmetoseemyworkbetter.Thereferee'ssuggestionsonthestructureofmypaperhaveimprovedmywritingtremendously. 5

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TABLEOFCONTENTS page ACKNOWLEDGMENTS .................................. 4 LISTOFTABLES ...................................... 7 ABSTRACT ......................................... 8 CHAPTER 1INTRODUCTIONANDRESULTS .......................... 9 1.1Introduction ................................... 9 1.2GaloisModulesOverGeneralizedAssociatedOrders ........... 9 1.3EnumerationofIsomorphismClasses .................... 16 2DEFINITIONSANDPRELIMINARIES ....................... 18 2.1DiscreteValuationFields ........................... 18 2.2CompleteDiscreteValuationFields ...................... 20 2.3UniqueRepresentation ............................ 22 2.4ExtensionsofDiscreteValuationFields ................... 23 2.5TotallyRamiedExtensions .......................... 25 2.6RamicationBreak ............................... 26 3GALOISMODULESOVERGENERALIZEDASSOCIATEDORDERS ..... 28 3.1AFewLemmas ................................. 28 3.2TheFirstStep .................................. 30 3.3PropertyA ................................... 35 3.4ProofofTheorem 1.2.2 ............................ 40 4ENUMERATIONOFISOMORPHISMCLASSES ................. 42 4.1AFewLemmas ................................. 42 4.2TheWorkofAmano .............................. 42 4.3TheFullGroup ................................. 52 4.4ActiononAmano'sPolynomials ........................ 54 4.5AnotherLookatGaloisExtensions ...................... 58 APPENDIX:C++PROGRAMUSEDTOGENERATETABLE 1-1 ........... 60 REFERENCES ....................................... 62 BIOGRAPHICALSKETCH ................................ 63 6

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LISTOFTABLES Table page 1-1Galoismodulesovergeneralizedassociatedorders,p=29. ........... 15 7

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AbstractofDissertationPresentedtotheGraduateSchooloftheUniversityofFloridainPartialFulllmentoftheRequirementsfortheDegreeofDoctorofPhilosophyPROBLEMSONFINITEEXTENSIONSOFLOCALFUNCTIONFIELDSByDucVanHuynhMay2014Chair:KevinKeatingMajor:Mathematics LetKbealocaleldwithniteresidueeldkandcharacteristicp,andletKsbeaseparableclosureofK.ThisworkdealswiththestudyoftotallyramiedextensionsL=KinKsofdegreepwithaxedramicationbreak2Q,anditisacombinationoftwoproblems.TherstproblemisanextensionofaworkdonebyAkiraAiba,whilethesecondproblemdealswiththeenumerationofk-isomorphismclassesoftotallyramiedextensionsinKsofdegreep. LetL=KbeaGaloisextensionofdegreepwithGaloisgroupGandramicationbreak,andletKbethevaluationofKsnormalizedonK.TherstproblemisregardingthedeterminationofabasisfortheGaloismoduleMrL=fx2L:L(x)r pgoverthegeneralizedassociatedorderAr;r(L=K)=f'2K[G]:'MrLMrLg,andthecriterionforwhichMrLisafreeAr;r(L=K)-module. SaytwoextensionsL1=KandL2=Kofdegreepwithramicationbreakarek-isomorphicifthereexistsanisomorphism :L1!L2suchthat (K)=Kand jk=idk.LetEtobethesetofalltotallyramiedextensionsN=Kofdegreepwithramicationbreak.Thesecondproblemisregardingtheenumerationofk-isomorphismclassesofE. 8

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CHAPTER1INTRODUCTIONANDRESULTS 1.1Introduction LetKbealocaleldwithniteresidueeldkandcharacteristicp.LetKbethediscretevaluationofKsuchthatK(K)=1foraxedprimeelementKinK,andletKsbeaseparableclosureofK.ThisworkdealswiththestudyoftotallyramiedextensionsL=KinKsofdegreepwithaxedramicationbreak2Q,anditisacombinationoftwoproblems.TherstproblemisanextensionofaworkdonebyAkiraAibaregardingGaloismodulesoverassociatedorders,whilethesecondproblemdealswiththeenumerationofk-isomorphismclassesoftotallyramiedextensionsinKsofdegreep. 1.2GaloisModulesOverGeneralizedAssociatedOrders LetL=KbetotallyramiedandGaloisoforderpwithGaloisgroupGanduniqueramicationbreak2Z(seeSection 2.6 ).Usingthesamenotation,letKbethediscretevaluationonLdenedbyK(K)=1,letOLdenotethevaluationringofL,thatis,OL=fx2L:K(x)0g.LetMrL=fx2L:K(x)r pg.NotethatM0L=OL.In[ 1 ],A.AibaconstructedanexplicitbasisforA(L=K)=fx2K[G]:xOLOLgoverOK.Furthermore,heshowedthatOLisafreeA(L=K)-moduleundercertainconditions.FormoreontheresultsandcurrentprogressofGaloismodulesoverassociatedorders,see[ 15 ]. Inthesamefashionas[ 1 ],wewillconstructanexplicitbasisforAr;b(L=K)=fx2K[G]:xMrLMbLg: NotethatA0;0=A(L=K).Whenb=r,Ar;risaringwith1,andsoMrLisamoduleoverAr;r.WewillstatethecriterionbelowinTheorem 1.2.2 forwhichMrLisafreemoduleoverAr;r.Fromnowon,setAr=Ar;r. 9

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Writer=pf+r0,where0r0p)]TJ /F6 11.955 Tf 11.22 0 Td[(1.Thenforx2Ar,wehavexMrLMrLifandonlyifx(Mr0L)Mr0Lsincex(MrL)=Tf(xMr0L).Hence,wemayassumethat0r
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Remark1.2.1. Foreachm0,takingb=r,wehavem;r;r=1ifthereexistsuwithmup)]TJ /F6 11.955 Tf 12.09 0 Td[(1suchthatNu;m;r2.IfM0LisA0-free,thenfromTheorem 1.2.2 (ii),eithers=p p)]TJ /F3 11.955 Tf 11.96 0 Td[(n1orsp 2n1.Ifsp 2n1,thenp2n1s,butthisinequalityhappensonlywhens=n1=1,which,inthiscase,wehavesjp)]TJ /F6 11.955 Tf 11.98 0 Td[(1.Nowifs=p p)]TJ /F3 11.955 Tf 11.96 0 Td[(n1,then sp p)]TJ /F3 11.955 Tf 11.96 0 Td[(n1
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whichimplies p)]TJ /F3 11.955 Tf 13.16 8.08 Td[(p sn11,wehaven1>p 2.ByTheorem 1.2.2 (ii),MrLisAr-freewhenr=s)]TJ /F11 11.955 Tf 11.96 16.86 Td[(p p)]TJ /F3 11.955 Tf 11.95 0 Td[(n1=0sincen1=p)]TJ /F3 11.955 Tf 13.15 8.09 Td[(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1 s. Remark1.2.4. Fromthecorollaryabove,wehaveM0LisA0-freeifandonlyifsj(p)]TJ /F6 11.955 Tf 12.07 0 Td[(1).ThisresulthasalreadybeenprovenbyAibaandLettlin[ 1 ]andin[ 12 ].ThesameresultwasalsoshowntoholdwhenLhascharacteristic0byBertrandiasandFertonin[ 7 ]. Remark1.2.5. Writethefractions pasthecontinuedfraction[0;a1;a2;:::;am],thatis,s p=1 a1+1 a21 ...+1 am In[ 8 ],FertonstudiedthecasewhenL=Kiscyclicofdegreep,whereKisalocaleldofcharacteristic0.Whens=0,whichdoesnothappeninthecharacteristicpcase,sheshowedthatMrLisAr-freeforall0rp)]TJ /F6 11.955 Tf 12.48 0 Td[(1.Whens=1and
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showedthatArisfreeifandonlyifrp+3 2.Whens2and0,wededucethatxm=sandym=p. 13

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Firstsupposethatmiseven.From( 1 )wehaveym)]TJ /F7 7.97 Tf 6.58 0 Td[(1sp)]TJ /F6 11.955 Tf 12.61 0 Td[(1(modp).Since0
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Table1-1. Galoismodulesovergeneralizedassociatedorders,p=29. s 12345678910111213141516171819202122232425262728 r 0000000000000000000000 0 000000000000000000000000000 1 00000000000000000000000000 2 00000000000000000000000000 3 0000000000000000000000000 4 00000000000000000000000000 5 00000000000000000000000000 6 0000000000000000000000000 7 000000000000000000000000 8 0000000000000000000000000 9 00000000000000000000000 10 00000000000000000000000000 11 0000000000000000000000000 12 0000000000000000000000000 13 00000000000000000000000000 14 00000000000000000000000000 15 0000000000000000000000000 16 00000000000000000000000000 17 000000000000000000000000 18 000000000000000000000000 19 000000000000000000000000 20 0000000000000000000000000 21 00000000000000000000000000 22 0000000000000000000000000 23 000000000000000000000000 24 00000000000000000000000000 25 0000000000000000000000000 26 00000000000000000000000000 27 00000000000000000000000000 28 15

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1.3EnumerationofIsomorphismClasses IdentifyKwithk((K)).LetEtobethesetofalltotallyramiedeldextensionsofKinKswithdegreepandramicationbreak.IfL1;L22E,thenwesayL1=KandL2=Karek-isomorphicifthereexistsanisomorphism:L1!L2suchthat(K)=Kandjk=idk.LetSgtobethesetofk-isomorphismclassesofGaloisextensionsofEandSngtobethesetofk-isomorphismclassesofnon-GaloisextensionsofE.WewillcomputethecardinalityofSgandSng. InSection 4.2 ,wendthat(p)]TJ /F6 11.955 Tf 12.32 0 Td[(1)2Z.Hence,Eisemptyunless(p)]TJ /F6 11.955 Tf 12.32 0 Td[(1)2Z,andsoSgandSngareemptywhen(p)]TJ /F6 11.955 Tf 11.13 0 Td[(1)isnotaninteger.Furthermore,whenisnotaninteger,Sgisempty. Theorem1.3.1. Let21 p)]TJ /F6 11.955 Tf 11.95 0 Td[(1N,andletdbethedenominatorofwhenitisinreducedform.DeneStobethesetofk-isomorphismclassesE.DeneSg(resp.Sng)tobethesetofk-isomorphismclassesofdegreeptotallyramiedGalois(resp.non-Galois)extensionswithramicationbreak.ThenwehavejSj=(p)]TJ /F6 11.955 Tf 11.95 0 Td[(1)gcdq)]TJ /F6 11.955 Tf 11.96 0 Td[(1 p)]TJ /F6 11.955 Tf 11.95 0 Td[(1;d: Specically, (i) jSgj=gcdq)]TJ /F6 11.955 Tf 11.96 0 Td[(1 p)]TJ /F6 11.955 Tf 11.96 0 Td[(1;and (ii) jSngj=(p)]TJ /F6 11.955 Tf 11.95 0 Td[(2)gcdq)]TJ /F6 11.955 Tf 11.95 0 Td[(1 p)]TJ /F6 11.955 Tf 11.95 0 Td[(1;ifisaninteger,while (iii) jSngj=(p)]TJ /F6 11.955 Tf 11.95 0 Td[(1)gcdq)]TJ /F6 11.955 Tf 11.95 0 Td[(1 p)]TJ /F6 11.955 Tf 11.95 0 Td[(1;difisnotaninteger. LetU1;Kbethegroupof1-unitsofKandwrite=(m)]TJ /F6 11.955 Tf 11.95 0 Td[(1)p+n p)]TJ /F6 11.955 Tf 11.96 0 Td[(1with1n
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extensionsofdegreep:oneviatheworksofKlopsch(seeRemark 4.4.3 )andoneviaArtin-SchreierTheory(seeSection 4.5 ). 17

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CHAPTER2DEFINITIONSANDPRELIMINARIES 2.1DiscreteValuationFields Denition2.1.1. LetKbeaeld.Amap:K!Q[f1gwiththeproperties (i) ()=1ifandonlyif=0 (ii) ()=()+() (iii) (+)min(();) issaidtobeadiscretevaluationonK.If(K)=0,theniscalledthetrivialvaluationonK.TheeldKiscalledadiscretevaluationeldifisnon-trivial. Lemma2.1.2. If()6=(),then(+)=min(();()). Proof. Notethat()]TJ /F3 11.955 Tf 9.29 0 Td[()=().Assumewithoutlossofgeneralitythat()<().Thenwehave()=(+)]TJ /F3 11.955 Tf 11.95 0 Td[()min((+);())=(+)min(();()): Hence,(+)=()=min(();()). LetO=f2K:()0gandM=f2K:()>0g.TheringMcoincideswiththesetofnon-invertibleelementsofO,whichimpliesthatOisalocalringwithuniquemaximalidealM.TheringOiscalledtheringofintegersofKwithrespectto.Theeld K=O=MiscalledtheresidueeldofK.Notethatifchar(K)6=char( K),thenchar(K)=0andchar( K)=p>0.ThemultiplicativegroupU=O)]TJ /F3 11.955 Tf 12.07 0 Td[(Miscalledthegroupofunits;itisthesetconsistingoftheinvertibleelementsofO. Henceforth,wewillassumethatKisadiscretevaluationeld.AnelementK2Oissaidtobeaprimeelement(oruniformizingelement)if(K)generatesthecyclicgroup(K).Wewillassumewithoutlossofgeneralitythat(K)=Z;insuchacase,wesaythatisnormalizedonK. Lemma2.1.3. (SeeLemma3.3of[ 6 ,ch.1])LetKbeadiscretevaluationeld,andKbeaprimeelement.ThentheringofintegersOisaprincipalidealring,andevery 18

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properidealofOcanbewrittenasnKOforsomen>0.Inparticular,M=KO.TheintersectionofallproperidealsofOisthezeroideal. Lemma2.1.4. (SeeLemma3.4of[ 6 ,ch.1])IfKisadiscretevaluationeld,thenanyelement2KcanbeuniquelywrittenasnKforsomen2Zand2U. Denition2.1.5. Amapjj:K!RissaidtobeanormonKifthefollowingthreepropertiesaresatised: (i) jj>0if6=0,j0j=0, (ii) jj=jjjj, (iii) j+jjj+jj(triangleinequality) Letcbearealnumberwith0
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Proposition2.1.6. (see[ 13 ,ch.2])Theultrametricspace(K;d)possessesthefollowingproperties: (a) Anypointofaballisapossiblecenteroftheball. (b) Iftwoballshaveacommonpoint,thenoneiscontainedintheother. (c) Asequence(n)n0isaCauchysequencepreciselywhenlimn!1d(n;n+1)=0. Example2.1.7. 1. Foranon-zerointegerm,letp(m)bethemaximalintegersuchthatpp(m)dividesm.Extendptorationalnumbersbyputtingp(m=n)=p(m))]TJ /F3 11.955 Tf 11.95 0 Td[(p(n)andp(0)=1: Thenormdenedbyjjp=p)]TJ /F4 7.97 Tf 6.59 0 Td[(p() isap-adicnorm.AnaturalquestioniswhetherifthereisanyothernormonQbesidetheEuclideannormandp(seeTheorem 2.1.8 ). 2. ConsiderthefunctioneldK(X).Forf(X)=Pki=maiXi2K[X]witham6=0,deneX(f(X))=m.SinceK(X)istheeldoffractionsofK[X],wecanextendXtoK(X)inthesamefashionasptoQ.ThenitfollowsthatthefunctionjjXdenedbyjf(X)jX=cX(f(X)) isanormonK(X)foranyrealcwith0
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discretevaluationeldwithitsdiscretevaluationbdenedbyb((n))=limn!1(n)foraCauchysequence(an)n0. AdiscretevaluationeldKiscalledacompletediscretevaluationeldifeveryCauchysequence(n)n0convergestosomeelement2K.AeldbKwithadiscretevaluationbiscalledacompletionofKifitiscomplete,therestrictionofbtoKisequalto,andKisadensesubeldofbKwithrespecttob. Proposition2.2.2. (seeProposition4.1of[ 6 ,ch.1])EverydiscretevaluationeldKhasacompletionwhichisuniqueuptoanisomorphismoverK. WeshalldenotethecompletionofKwithrespecttobybK.WealreadyknowthatKisdenseinbK.TheringofintegersOisdenseinOb,andthemaximalidealMisdenseinMb.Furthermore,theresidueeldofKcoincideswiththeresidueeldofbK(seeLemma4.3of[ 6 ,ch.1]). EventhoughbKisanextensionofinnitedegreeoverK,(K)=b(bK),thatis,thevaluationgroupofbKdoesnotenlarge.Indeed,let2bK.Let(n)n0beasequenceofelementsinKthatconvergesto.Chooseanintegerisuchthatb()]TJ /F3 11.955 Tf 11.95 0 Td[(i)>b().Then,(i)=b(i)=b(i)]TJ /F3 11.955 Tf 11.95 0 Td[(+)=b().Hence,b()2(K). Example2.2.3. 1. ThecompletionofQwithrespecttop(seeExample 2.1.7 )isthep-adicnumbersQp. 2. ThecompletionofK(X)withrespecttoX(seeExample 2.1.7 )istheformalLaurentseriesK((X)). WesaythataeldKisalocaleldifKiscompletewithrespecttoadiscretevaluationandifitsresidueeldisnite.AlocaleldKisisomorphictoeitheraniteextensionofQporaniteextensionofK((X))(see[ 14 ]). IfKisisomorphictoaniteextensionofQp,thenwesaythatKisalocalnumbereld.IfKisisomorphicaniteextensionofK((X)),thenwesaythatKisalocalfunctioneld. 21

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Remark2.2.4. AlocaleldKisanultrametricspace.Itiswellknownthatiftheseries1Xi=0nconvergesinK,thenlimn!1n=0.Conversely,whenKisalocaleld,iflimn!1n=0,thentheseries1Xi=0nconvergesinK.ThisisduetoapropertymentionedinProposition 2.1.6 :asequence(n)n0ofelementsinKisCauchyifandonlyiflimn!1d(n;n+1)=0.ThisisnotnecessarilythecaseinArchimedeanmetricspace.Forexample,limn!11 n=0inRbuttheseries1Xi=11 ndiverges. 2.3UniqueRepresentation AsetRissaidtobeasetofrepresentativesforKifROv,02RandRismappedbijectivelyto KunderthecanonicalmapOv!Ov=Mv= K. Lemma2.3.1. (Hensel'sLemma)SupposeKisalocaleldandf(X)isamonicpolynomialinOv[X].If f(X)2 K[X]hasasimpleroot2 K,thenf(X)hasasimpleroot2Ovsuchthat =. Suppose Khascardinalityq,ap-power.Thenthepolynomial f(X)=Xq)]TJ /F3 11.955 Tf 12.06 0 Td[(Xsplitsin K.ByHensel'sLemma,f(X)=Xq)]TJ /F3 11.955 Tf 12.13 0 Td[(XsplitsinK.Hence,wemaytakeRtobetherootsoff(X).ThesesetofrepresentativesarecalledTeichmuller'srepresentatives,andareinfactisomorphicto K. Proposition2.3.2. (See[ 6 ,ch.1,x5])IfKisalocalfunctioneld,theneveryelement2Kcanbewrittenuniquelyas=1Xi=jaiiK; wherej2Zandai2R. Example2.3.3. TheresidueeldofQpisisomorphictoZ=pZ.Infact,R=f0;1;:::;p)]TJ /F6 11.955 Tf -453.81 -23.91 Td[(1g.Forexample,takep=7.Theelement2+3(7)+73+5(710)+4(7100)+:::isanelementofQ7. Denition2.3.4. ThegroupsUi;=1+iKOfori1arecalledthegroupsofprincipalunits. 22

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Lemma2.3.5. (SeeCorollary5.5of[ 6 ,ch.1])IfKisalocaleldandlisanintegernotdivisiblebychar( K),thenwehaveUli;=Ui;: Proof. Let2Ui;nUi+1;.Letf(X)=Xl)]TJ /F3 11.955 Tf 12.53 0 Td[(.Then f(X)hasasimplerootin K.ByHensel'sLemma,f(X)hasasimplerootin2K.Itfollowsthat(l)]TJ /F6 11.955 Tf 11.95 0 Td[(1)=()]TJ /F6 11.955 Tf 11.95 0 Td[(1)=i.Writel)]TJ /F6 11.955 Tf 11.4 0 Td[(1=()]TJ /F6 11.955 Tf 11.39 0 Td[(1).Sincel(modF),weseethat()]TJ /F6 11.955 Tf 11.39 0 Td[(1)=i,whichimpliesthat2Ui;nUi+1;. 2.4ExtensionsofDiscreteValuationFields LetL=Kbeaniteextensionofdegreen.Let!beanormalizeddiscretevaluationofL.Wesaythat!isanextensionofthevaluationif!0=!jKandinduceequivalenttopologies,where!jKistherestrictionof!toK.Laterwewillseethatthereexistspreciselyonesuchdiscretevaluation!extending. Theintegere(L=K;!)=[!(L):!0(K)]iscalledtheramicationindexofL=K.Theintegerf(L=K;!)=[ L: K]iscalledtheresiduedegreeofL=K. Lemma2.4.1. (SeeLemma2.1of[ 6 ,ch.2,x2])LetLbeanextensionofalocaleldKandlet!beavaluationofL.LetKML.Thene(L=K;!)=e(L=M;!)e(M=K;!0);f(L=K;!)=f(L=M;!)f(M=K;!0): Proposition2.4.2. (SeeProposition2.4of[ 6 ,ch.2,x2])IfLisanextensionofdegreenoveralocaleldK,thenn=e(L=K;!)f(L=K;!): ForaniteextensionL=K,denotesbyianembeddingofKintoKs,aseparableclosureofK.DenethemapNL=K:L!KbyNL=K()=Yi(); 23

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wheretheproductisoveralldistinctembeddingsofLintoKs.ThemapNL=KiscalledtheeldnormofL=K(see[ 11 ,ch.6,x5]). Theorem2.4.3. (SeeTheorem2.5of[ 6 ,ch.2])LetL=KbeaniteextensionoveralocaleldK.Thenthereexistspreciselyoneextension!onLofthevaluationand!=1 f(L=K;!)NL=K.Moreover,theeldLiscompletewithrespectto!. Remark2.4.4. LetKbealocaleldandlet2K.LetL=Kbeaniteextensionofdegreen.Thenwehave!()=n f(L=K;!)()=e(L=K;!)():Hence,weseethat!0=e(L=K;!). ThefactthatKiscompleteisanecessityforTheorem 2:4:3 .ItisnaturaltoponderofthecasewhenKisnotcomplete.SupposeL=K()isaniteextensionofKwithavaluation!.IfisanembeddingofLintoKs,then!0=!isavaluationonLthatmaynotbeequivalentto!.Hence,weobservethatnumberofvaluationsonLisdependentonthenumberofembeddingsofLintoKs. Theorem2.4.5. (SeeTheorem2.6of[ 6 ,ch.2])LetKbeaeldwithdiscretevaluation.LetbKbethecompletionofK,andbthediscretevaluationofbK.SupposeL=K(1)isaniteextensionofKandf(X)theminimalpolynomialofoverK.Letf(X)=Qki=1gi(X)eibethedecompositionofthepolynomialf(X)intoirreduciblemonicfactorsinbK[X].LetLi=bK(i),whereiisarootofgi(X).Letb!ibetheuniquediscretevaluationonLiextendingb. ThenLisembeddednaturallyasadensesubeldinthecompletediscretevaluationeldLi,andtherestriction!iofb!ionLisadiscretevaluationonLwhichextends.Thevaluations!iaredistinctandeverydiscretevaluationwhichisanextensionoftoLcoincideswithsome!ifor1ik. Example2.4.6. (See[ 4 ,ch.5])LetK=Q(),whereisarootoftheirreduciblepolynomialf(X)=X3)]TJ /F6 11.955 Tf 12.24 0 Td[(2X+2.Thepolynomialf(X)splitscompletelyinQ33,thatis,f(X)=(X)]TJ /F3 11.955 Tf 12.2 0 Td[()(X)]TJ /F3 11.955 Tf 12.21 0 Td[()(X)]TJ /F3 11.955 Tf 12.2 0 Td[()forsome;;2Q33,where =3; =9; =11.ByTheorem 2.4.5 ,therearethreedistinctdiscretevaluationonKextending33. 24

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2.5TotallyRamiedExtensions Henceforth,assumethatKisalocaleldwithaxedprimeelementKandanormalizeddiscretevaluationK.WeshallwriteOK=OK;MK=MK,andUK=UF.ForanalgebraicextensionL=K,writee(L=K)=e(L=K;L)andf(L=K)=f(L=K;L). AniteextensionL=Kiscalledunramiedif L= Kisseparableande(L=K)=1.AniteextensionL=Kiscalledtotallyramiediff(L=K)=1.Aniteextensioniscalledtamelyramiedif L= Kisseparableandp-e(L=K)whenp=char( K). Inthiswork,wefocusontotallyramiedextensions,mainlytotallyramiedextensionsofdegreep.Forinformationregardingunramiedextensions,see[ 6 ,ch.2,x3]. ToenumerateEinTheorem 1.3.1 ,wewillinvestigatetheactionsofAonacertainsetofpolynomialsthatinducetotallyramiedextensionsofdegreepwithramicationbreak,whereAisthefullgroup(seeSection 4.3 ).Hence,weneedtoknowtheminimalpolynomialsfortotallyramiedextensions. Proposition2.5.1. (SeeProposition3.5of[ 6 ,ch.2])IfL=Kistotallytamelyramiedextensionofdegreen,thenL=K(L),whereLisaprimeelementofLsatisfyingtheequationXn)]TJ /F3 11.955 Tf 11.96 0 Td[(Kforsome2UF. Apolynomialf(X)=Xn+an)]TJ /F7 7.97 Tf 6.59 0 Td[(1Xn)]TJ /F7 7.97 Tf 6.59 0 Td[(1++a02OK[X]iscalledanEisensteinpolynomialifK(ai)1for0in)]TJ /F6 11.955 Tf 11.95 0 Td[(1,andK(a0)=1. Proposition2.5.2. (SeeProposition3.6of[ 6 ,ch.2,x3] (i) TheEisensteinpolynomialf(X)isirreducibleoverK.Ifisarootoff(X),thenK()=Kisatotallyramiedextensionofdegreen,andisaprimeelementinK(). (ii) LetL=Kbeaseparablytotallyramiedextensionofdegreen,andletLbeaprimeelementinL.ThenLisarootofanEisensteinpolynomialoverKofdegreen. WhenL=Kistotallyramiedofdegreepwithramicationbreak2Q,Amanogivesanexplicitpolynomial(seeSection 4.2 ).Write=(m)]TJ /F6 11.955 Tf 11.95 0 Td[(1)p+n p)]TJ /F6 11.955 Tf 11.96 0 Td[(1,where1n
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HeshowsthatL=K=K[X]=(Xp)]TJ /F3 11.955 Tf 12.12 0 Td[(!mKXn)]TJ /F3 11.955 Tf 12.12 0 Td[(uK),forsomerepresentative!of Kandu2UK. 2.6RamicationBreak LetL=KbeaniteGaloisextensionwithGaloisgroupG.Foranintegeri)]TJ /F6 11.955 Tf 21.92 0 Td[(1putGi=f2G:()(modi+1L)forall2OLg: WehaveG)]TJ /F7 7.97 Tf 6.58 0 Td[(1=GandeachGiisanormalsubgroupofG(see[ 6 ,ch.2,x4]).Moreover,Gi+1Gi. Denition2.6.1. LetL=KbeaniteGaloisextensionwithaseparableresidueeldextensions.LetG=Gal(L=K).IntegersisuchthatGi6=Gi+1arecalledramicationbreaksofL=K. WhenGhasorderp,ithasauniqueramicationbreak,namely,=L(L) L)]TJ /F6 11.955 Tf 11.96 0 Td[(1; foranygeneratorofG.WecanstilldeneramicationbreakwhenL=KistotallyramiedextensionofpbutnotGalois(seeLemma1of[ 3 ]).Indeed,letLbethevaluationofKsnormalizedonL.ThenL=Khastheuniqueramicationbreak=L'(L) L)]TJ /F6 11.955 Tf 11.95 0 Td[(1; foranynon-identityembedding'ofLintoKs. Theorem2.6.2. (Artin-SchreierTheory,seeTheorem6.4of[ 11 ,ch.6])LetKbeaeldofcharacteristicp. (i) LetLbeacyclicextensionofKofdegreep.Thenthereexists2LsuchthatL=K(),whereisarootofthepolynomialf(X)=Xp)]TJ /F3 11.955 Tf 11.96 0 Td[(X)]TJ /F3 11.955 Tf 11.95 0 Td[(aforsomea2K. (ii) Conversely,givena2K,thepolynomialf(X)=Xp)]TJ /F3 11.955 Tf 12.08 0 Td[(X)]TJ /F3 11.955 Tf 12.08 0 Td[(aeitherhasallrootsinK,oritisirreducibleoverK.Inthelattercase,ifisaroot,thenK()iscyclicofdegreepoverK. 26

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WesaythatL=KisanArtin-SchreierextensionifL=KisGaloisofdegreep.ByTheorem 2.6.2 ,L=K(),whereisarootoff(X)=Xp)]TJ /F3 11.955 Tf 11.96 0 Td[(X)]TJ /F3 11.955 Tf 11.96 0 Td[(aforsomea2K.Letibeanintegerwith1i
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CHAPTER3GALOISMODULESOVERGENERALIZEDASSOCIATEDORDERS 3.1AFewLemmas Lemma3.1.1. (Nakayama,Lemma4.2of[ 11 ])LetAbealocalringandmitsmaximalideal.LetEbeanitelygeneratedA-module,andletFbeasubmoduleofE.IfE=F+mE,thenE=F. LetFbeaeldofcharacteristicp.SetFalgtobethealgebraicclosureofF,andset=(Falg),themultiplicativegroupofFalg.ForanyniteabeliangroupBoforderrelativelyprimetop,letbBbethesetofgrouphomomorphismsfromBtoB,thegroupofjBj-throotsofunityof.Infact,bBisagroup.Thisistheanalogtothegroupofcharacterswithcomplexvalues. Lemma3.1.2. (Lemma2of[ 2 ])LetH=HpH1beaniteabeliangroup,whereHpistheSylowp-subgroupofH.LetfbeafunctionfromHintoaeldFofcharacteristicp.Thendet(f()]TJ /F7 7.97 Tf 6.59 0 Td[(1));2H=Y2cH10@X2Hp X2H1()f()!1AjHpj; wherecH1isasdenedabove. Remark3.1.3. ThelemmaaboveisjustanapplicationoftheGroupDeterminantFormula(Lemma5.26of[ 16 ])andadaptingtothecaseofpositivecharacteristic. Corollary3.1.4. SupposeNisanabelianextensionofKsuchthatH=Gal(N=K)isap-group.Then,for2N,det()]TJ /F9 5.978 Tf 5.76 0 Td[(1);2H=)]TJ /F3 11.955 Tf 5.48 -9.68 Td[(TrN=K()jHj: Proof. SetH=Hp;H1=1andf()=(),andapplythelemmaabove. Fromnowon,putdet(b)=det(b)]TJ /F9 5.978 Tf 5.76 0 Td[(1);2H,forb2MrN.Callanelementa2MrNminimalifdet(a)dividesdet(b)forallb2MrN. 28

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Lemma3.1.5. LetNbeaniteGaloisextensionofK;H=Gal(N=K)andAr=Ar(N=K).SupposethatMrNisafreeAr-module,whichisnecessarilyofrankone.Then,Ara=MrNifandonlyifaisminimal. Proof. SupposeAra=MrN.NotethatsinceGaislinearlyindependentifandonlyifdet(a)6=0,andthattherankofAraoverOKis[N:K],weseethatthesetHaislinearlyindependentoverOK,sodet(a)6=0.Letb2MrN.Ifdet(b)=0,thenthereisnothingtoprove.So,wemayassumethatHbislinearlyindependentoverOK.Thenwehave[MrN:(OKH)b]=[MrN:Arb][Arb:(OKH)b]=[MrN:Arb][Ar:OKH]and[MrN:(OKH)b]=[Ara:(OKH)a][(OKH)a:(OKH)b]=[Ar:OKH][(OKH)a:(OKH)b]: HencewehavethefollowingequalityofidealsinOK(2.4of[ 9 ,ch.3]): det(a)2[MrN:Arb]2=det(a)2[(OKH)a:(OKH)b]2=det(b)2:(3) So,det(a)dividesdet(b)forallb2MrN. Conversely,supposeaisminimal,thatis,det(a)dividesdet(b)forallb2MrN,andArc=MrNforsomec2MrN.Then,inparticular,det(a)dividesdet(c).SinceAraisasubmoduleofMrN,wehavedet(c)dividesdet(a)by( 3 ).Therefore,Arahasindex1inArcandsoAra=Arc=MrN. 29

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Lemma3.1.6. (Lemma3of[ 2 ])Let1;:::;tbearbitrarynumbersinL.Setn=tXi=1ni,f(x)=tYi=1(1)]TJ /F3 11.955 Tf 11.95 0 Td[(ix)andg(x)=1Xi=1nxn.Theng(x)=)]TJ /F3 11.955 Tf 9.3 0 Td[(xf0(x) f(x); wheref0(x)istheformalderivativeoff(x). Lemma3.1.7. Let=p)]TJ /F7 7.97 Tf 6.58 0 Td[(1t(p)]TJ /F7 7.97 Tf 6.59 0 Td[(1)+r(p)]TJ /F7 7.97 Tf 6.58 0 Td[(1)K2MrL.IfMrLisAr-free,thenMrL=Ar. Proof. Letf(x)=XpP(1=X)=1)]TJ /F3 11.955 Tf 12.23 0 Td[(Xp)]TJ /F7 7.97 Tf 6.58 0 Td[(1)]TJ /F3 11.955 Tf 12.23 0 Td[(Xp=p)]TJ /F7 7.97 Tf 6.59 0 Td[(1Yi=0(1)]TJ /F3 11.955 Tf 12.23 0 Td[(iX),wherei=+i.ByApplyingLemma 3.1.6 ,wehaveTrL=K(i)=8><>:)]TJ /F6 11.955 Tf 9.29 0 Td[(1ifi=p)]TJ /F6 11.955 Tf 11.95 0 Td[(10if0i
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Lemma3.2.1. MrL=p)]TJ /F7 7.97 Tf 6.59 0 Td[(1Xu=0OKutu+r(u)K wherer(u)=)]TJ /F11 11.955 Tf 11.29 13.27 Td[(j)]TJ /F11 11.955 Tf 11.29 13.27 Td[(su+r pk,and=pt+swhere1sp)]TJ /F6 11.955 Tf 11.96 0 Td[(1. Proof. Notethatwehavevutu+r(u)K=)]TJ /F3 11.955 Tf 10.49 8.09 Td[(pt+s pu+tu+r(u)=)]TJ /F3 11.955 Tf 10.49 8.09 Td[(su p)]TJ /F11 11.955 Tf 11.96 16.86 Td[()]TJ /F11 11.955 Tf 11.29 16.86 Td[(su+r p: However,)]TJ /F3 11.955 Tf 10.5 8.09 Td[(su p)]TJ /F11 11.955 Tf 11.96 16.85 Td[()]TJ /F11 11.955 Tf 11.3 16.85 Td[(su+r p=)]TJ /F3 11.955 Tf 10.5 8.09 Td[(su+r p)]TJ /F11 11.955 Tf 11.96 16.85 Td[()]TJ /F11 11.955 Tf 11.29 16.85 Td[(su+r p+r p=)]TJ /F3 11.955 Tf 10.49 8.09 Td[(us+r p+r p Weseethatthevaluationsofthepelementsutu+r(u)K(0up)]TJ /F6 11.955 Tf 12.14 0 Td[(1)aredistinctandliebetweenr pandp)]TJ /F7 7.97 Tf 6.59 0 Td[(1+r p.Hence,wehavethedesiredresult. Lemma3.2.2. For0u;mp)]TJ /F6 11.955 Tf 11.95 0 Td[(1,'muvK=8>><>>:0um,'muvK=vK mt+0r;b(m)Ku)]TJ /F7 7.97 Tf 6.59 0 Td[(1Xu1=m)]TJ /F7 7.97 Tf 6.59 0 Td[(1uu1u1)]TJ /F7 7.97 Tf 6.58 0 Td[(1Xu2=m)]TJ /F7 7.97 Tf 6.58 0 Td[(2u1u2um)]TJ /F9 5.978 Tf 5.75 0 Td[(1)]TJ /F7 7.97 Tf 6.59 0 Td[(1Xum=0um)]TJ /F7 7.97 Tf 6.59 0 Td[(1umum Proof. Thecasesum. 31

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Theorem3.2.3. LetL=K()beawildlyramiedcyclicextensionofKofdegreep,whereisarootofXp)]TJ /F3 11.955 Tf 11.96 0 Td[(X)]TJ /F3 11.955 Tf 11.96 0 Td[(=0; withv()=)]TJ /F3 11.955 Tf 9.3 0 Td[(and=pt+swith0Nn;m;r:(3) Remark3.2.4. NotethattherstassertionofthetheoremaboveisexactlytheassertioninTheorem1.3(i),anditisageneralizationofthemaintheoremofAibain[ 1 ].Rewriting( 3 )fromTheorem 3.2.3 ,wehaves)]TJ /F3 11.955 Tf 11.96 0 Td[(r p(k+1)s p>(j+1)s)]TJ /F3 11.955 Tf 11.96 0 Td[(r p; wheren=p)]TJ /F6 11.955 Tf 12.07 0 Td[(1)]TJ /F3 11.955 Tf 12.08 0 Td[(j;m=p)]TJ /F6 11.955 Tf 12.07 0 Td[(1)]TJ /F3 11.955 Tf 12.08 0 Td[(k,and0jkp)]TJ /F6 11.955 Tf 12.07 0 Td[(1.Hence,thesecondpropertyfromTheorem 3.2.3 canbestatedas: MrLisafreeAr(L=K)-moduleifandonlyiftheredonotexistintegers1mnp)]TJ /F6 11.955 Tf 11.96 0 Td[(1suchthat s)]TJ /F3 11.955 Tf 11.96 0 Td[(r pns p>ms)]TJ /F3 11.955 Tf 11.96 0 Td[(r p:(3) Ifweletr=0,wehaveexactlythesamepropertyasin[ 1 ]. In[ 1 ],AibashowedthatM0L=OLisafreeA0;0-moduleifandonlyifshasthepropertythattherearenointegersm;nsuchthat1mns p>ms p: 32

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Weregretthatwehaveaccidentallychosenthevariablesm;noppositethatofAibain[ 1 ].Aibaincorrectlystatedtheinequalityas1
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Foreachm,thereexistsusuchthatb pv('m(utu+r(u)K))<1+b p.Indeed,ifv('m(utu+r(u)K))=Nu;m;r+m;r;b1+b p; forall0up)]TJ /F6 11.955 Tf 11.98 0 Td[(1,thenm;r;bisnotthesmallestpossible,contradictingthedenitionofm;r;b. Now,ifv('m(utu+r(u)K))v('m0(utu+r(u)K))mod1,thenms p)]TJ /F3 11.955 Tf 11.05 0 Td[(0r;b(m)m0s p)]TJ /F3 11.955 Tf -458.7 -28.38 Td[(0(m0)mod1.Butthatimpliesms pm0s pmod1.So,m=m0,since0m;m0p)]TJ /F6 11.955 Tf 12.24 0 Td[(1.Hence,for0mp)]TJ /F6 11.955 Tf 12.03 0 Td[(1,alltheelements'm(utu+r(u)K)havedistinctvaluations.Letuibeanintegerwith1uip)]TJ /F6 11.955 Tf 12.11 0 Td[(1suchthatb pv('i(uitui+r(ui)K))<1+b p.Itfollowsthatv('(uitui+r(ui)K))=minfv(am'm(uitui+r(ui)K))gp)]TJ /F7 7.97 Tf 6.59 0 Td[(1m=0v(ai'i(uitui+r(ui)K))<1+b p; acontradiction.Hence,am2KOK.Therefore,wehaveAr;b(L=K)=p)]TJ /F7 7.97 Tf 6.58 0 Td[(1Xk=0'mOK: NowweprovethesecondpartofTheorem 3.2.3 .SupposethatMrLisafreeAr-module.Assumeonthecontrarythattheredoexistintegers0mnp)]TJ /F6 11.955 Tf 12.05 0 Td[(1suchthat( 3 )issatised.Recallthat=p)]TJ /F7 7.97 Tf 6.58 0 Td[(1t(p)]TJ /F7 7.97 Tf 6.59 0 Td[(1)+r(p)]TJ /F7 7.97 Tf 6.58 0 Td[(1)K2MrL.FromRemark 1.2.1 ,wehaveNu;m;r+m;r;r=Nu;m;r+1forallusuchthat0mup)]TJ /F6 11.955 Tf 11.7 0 Td[(1.Then,fromtherstinequalityof( 3 ),wehave(a)v('m)=Np)]TJ /F7 7.97 Tf 6.59 0 Td[(1;m;r+m;r;rr p+1: Fromthesecondinequalityof( 3 ),wehave(b)v('m(ntn+r(n)K))=Nn;m;r+m;r;r
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Itfollowsfrom(a)that(1=K)'m2MrL.ByCorollary 3.1.8 ,(1=K)'m2Ar.However,from(b),(1=K)'m(ntn+r(n)K)62MrL,acontradiction. Nowassumethattheredonotexistintegers0mnp)]TJ /F6 11.955 Tf 12.26 0 Td[(1suchthat( 3 )issatised.Itfollowsthatforany0mp)]TJ /F6 11.955 Tf 11.98 0 Td[(1suchthatNp)]TJ /F7 7.97 Tf 6.58 0 Td[(1;m;rr p,theredoesnotexistannwith0mnp)]TJ /F6 11.955 Tf 12.24 0 Td[(1suchthatNn;m;r
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sayapair(r;s)haspropertyAiftheredonotexistintegers1mnp)]TJ /F6 11.955 Tf 11.96 0 Td[(1suchthats)]TJ /F3 11.955 Tf 11.96 0 Td[(r pns p>ms)]TJ /F3 11.955 Tf 11.95 0 Td[(r p: Hence,foraxeds,MrLisAr-freeifandonlyif(r;s)haspropertyA. Sinceit'sclearthatanypairoftheform(s;s)haspropertyA,wewillassumethatr6=s.Let1nip)]TJ /F6 11.955 Tf 12.49 0 Td[(1betheleastpositiveintegersuchthatnisi(modp).SetI=(n1;n2;:::;nk)andM=max(I),wherek=s)]TJ /F3 11.955 Tf 12.69 0 Td[(rifs>randk=p+s)]TJ /F3 11.955 Tf 12.69 0 Td[(rifs1.Toprovethesecondstatement,notethat ni+njni+j(modp)foralli;jk:(3) (i)Ifn1nk.Notethatnk+p)]TJ /F3 11.955 Tf 10.07 0 Td[(n1
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Fromtherstequalityof(ii),weobtainn1=nk+(k)]TJ /F6 11.955 Tf 12.3 0 Td[(1)(p)]TJ /F3 11.955 Tf 12.3 0 Td[(n1).Substitutingthisbackintotherstequalityof(ii),weobtainni=nk+(k)]TJ /F3 11.955 Tf 11.96 0 Td[(i)(p)]TJ /F3 11.955 Tf 11.95 0 Td[(n1)=nk+(k)]TJ /F6 11.955 Tf 11.96 0 Td[(1)(p)]TJ /F3 11.955 Tf 11.96 0 Td[(n1))]TJ /F6 11.955 Tf 11.95 0 Td[((i)]TJ /F6 11.955 Tf 11.95 0 Td[(1)(p)]TJ /F3 11.955 Tf 11.96 0 Td[(n1)=in1)]TJ /F6 11.955 Tf 11.95 0 Td[((i)]TJ /F6 11.955 Tf 11.96 0 Td[(1)p Lemma3.3.2. Thepair(r;s)haspropertyAifandonlyif i pms)]TJ /F3 11.955 Tf 11.95 0 Td[(r pforalli;mwith1ik;1mni:(3) Proof. Writes)]TJ /F3 11.955 Tf 11.95 0 Td[(r p=k p.ThenpropertyAcanberestatedas:Foreachik,theredonotexistintegers1mnip)]TJ /F6 11.955 Tf 11.96 0 Td[(1suchthat k pi p>ms)]TJ /F3 11.955 Tf 11.96 0 Td[(r p:(3) Thelemmanowfollows. Lemma3.3.3. Thepair(r;s)haspropertyAifandonlyifM
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Combining( 3 )fromLemma 3.3.2 with( 3 ),propertyAcanberestatedas: i ps(nk)]TJ /F6 11.955 Tf 11.95 0 Td[(1+m) pforalli;mwith1ik;1mni:(3) SupposethatM1since( 3 )isclearlysatisedfori=1. IfIisincreasing,thennzniforallzs,then(r;s)haspropertyAifandonlyifs=1andrp+3 2. Proof. (i) Thiscaseisclear. (ii) Certainly,foranyrp+3 2,thepair(r;1)haspropertyA.Conversely,suppose(r;s)haspropertyAwithr>s.ByLemma 3.3.2 ,wehave nks pms)]TJ /F3 11.955 Tf 11.96 0 Td[(r pforallmnk:(3) 38

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Puttingm=nkinto( 3 ),weseethatk pk)]TJ /F3 11.955 Tf 11.95 0 Td[(r p; sor>kandhencer>p+s 2. Notethatk=p)]TJ /F3 11.955 Tf 12.82 0 Td[(r+ss+1.Weseethatns+1>1,andns+1=2ifandonlyifs=1.Supposes>1sothats+13.Thens(ns+1)]TJ /F6 11.955 Tf 12.64 0 Td[(1)1(modp),whichimpliesn1=ns+1)]TJ /F6 11.955 Tf 12.22 0 Td[(1.SinceIisstrictlymonotonicandn11.Hence,wemusthaves=1.Therefore,rp+3 2. Theorem3.3.5. Supposerk)]TJ /F6 11.955 Tf 11.95 0 Td[(1 kp.Solvingforr,wehave r>s)]TJ /F3 11.955 Tf 26.32 8.09 Td[(p p)]TJ /F3 11.955 Tf 11.96 0 Td[(n1:(3) 39

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Combiningequations( 3 )and( 3 ),wehave r=s)]TJ /F3 11.955 Tf 26.32 8.09 Td[(p p)]TJ /F3 11.955 Tf 11.95 0 Td[(n1=s)]TJ /F11 11.955 Tf 11.95 16.86 Td[(p p)]TJ /F3 11.955 Tf 11.96 0 Td[(n1:(3) Conversely,supposersatises( 3 ).WewillrstshowthatIisdecreasing.Sinceequation( 3 )issatised,equation( 3 )isalsosatised,whichimpliesn1>k)]TJ /F6 11.955 Tf 11.96 0 Td[(1 kp.Notethenthatkn1)]TJ /F6 11.955 Tf 11.99 0 Td[((k)]TJ /F6 11.955 Tf 11.99 0 Td[(1)p>0.Hence,wehaveni=in1)]TJ /F6 11.955 Tf 11.99 0 Td[((i)]TJ /F6 11.955 Tf 12 0 Td[(1)pforik,whichimpliesIisdecreasing.Furthermore,wealsohavethatequation( 3 )issatised,whichimpliesMs,weprovedinTheorem 3.3.4 thatMrLisAr-freeifandonlyifs=1andrp+3 2.Whenr
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or s)]TJ /F6 11.955 Tf 13.15 8.08 Td[(1 2p n1rp 2,equation( 3 )yieldsr=sandanrsuchthatrs)]TJ /F6 11.955 Tf 12.76 0 Td[(2,andequation( 3 )yieldsr=s.Ontheotherhand,whenn1


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CHAPTER4ENUMERATIONOFISOMORPHISMCLASSES 4.1AFewLemmas Lemma4.1.1. (Hensel'sLemma)Supposef(X)isamonicpolynomialinOK[X].If f(X)hasasimpleroot2 K,thenf(X)hasasimpleroot2OFsuchthat =. Lemma4.1.2. (Krasner'sLemma)LetL=Kbeatotallyramiedextensionofdegreep.LetLbeaprimeelementofLwithconjugatesL=(1)L;(2)L;:::;(p)LinKs.IfA2KssatisestheconditionK(A)]TJ /F3 11.955 Tf 11.96 0 Td[(L)>K(L)]TJ /F3 11.955 Tf 11.95 0 Td[((i)L)forall1j (4)=jj (4)=q)]TJ /F6 11.955 Tf 11.95 0 Td[(1 gcd(e;): (4) 4.2TheWorkofAmano Inthissection,wewilldiscusssomeworkofAmanoin[ 3 ].LetJbealocaleldwithcharacteristic0.Inthepaper,AmanoenumeratestheJ-isomorphismclassesoftotallyramiedextensionsofdegreepoverJwithramcationbreak.Toenumerate,AmanoshowsthatifH=Jistotallyramiedofdegreep,thenH=J[X]=(f(X)),wheref(X)isanEisensteinpolynomialoftheformXp)]TJ /F3 11.955 Tf 12.15 0 Td[(!mJXn)]TJ /F3 11.955 Tf 12.14 0 Td[(JuorXp)]TJ /F3 11.955 Tf 12.14 0 Td[(Ju,forsomeu2U1;Jand!inasetofrepresentativesoftheresidueeldofJ.Sinceoureldof 42

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interesthascharacteristicp,f(X)isoftheformertype.WewilladaptAmano'sproofstocharacteristicp,thoughtheyareessentiallythesameasincharacteristic0. LetKbealocaleldofcharacteristicpwithniteresidueeldkoforderq.SupposeL=Kistotallyramiedofdegreepwithramicationbreak.AprimeelementLofLsatisesanEisensteinpolynomialf(X)overK,where f(X)=Xp)]TJ /F4 7.97 Tf 13.04 15.43 Td[(p)]TJ /F7 7.97 Tf 6.58 0 Td[(1Xi=1aiXi)]TJ /F3 11.955 Tf 11.96 0 Td[(Ka0(4) withK(ao)=0andK(ai)1for1ip)]TJ /F6 11.955 Tf 13.38 0 Td[(1.NotethatpLKa0NL=K(L)(modp+1L),whereNL=KisthenormofLoverK.Writea0=!0u,forsome!02kandu2U1;K.ByreplacingLwith!)]TJ /F7 7.97 Tf 6.59 0 Td[(1=p0Lwemayassumea02U1;K. LetLtobethevaluationofKsnormalizedonL.ForA;B2L,thecongruenceABmeansL(A)]TJ /F3 11.955 Tf 12.19 0 Td[(B)>L(A)=L(B).LetLbeaprimeelementofL.ThenL=Khasauniqueramicationbreak =L0L L)]TJ /F6 11.955 Tf 11.96 0 Td[(1;(4) where0L6=LisanyconjugateofLinKs(seeLemma1of[ 3 ]). Letm=minfK(a1);:::;K(ap)]TJ /F7 7.97 Tf 6.59 0 Td[(1)g.DenotebyntheleastpositiveintegersuchthatK(an)=m.Let!2kbesuchthatan!mK.WesaythattheEisensteinpolynomialf(X)isofthetype(n;m;!).ThenbytheproofofLemma1of[ 3 ],theramicationbreakofL=Kcanbewrittenas =(m)]TJ /F6 11.955 Tf 11.95 0 Td[(1)p+n p)]TJ /F6 11.955 Tf 11.96 0 Td[(1:(4) Since1np)]TJ /F6 11.955 Tf 11.95 0 Td[(1,thepairn;maretheuniqueintegerssuchthat( 4 )issatised. Lemma4.2.1. (SeeLemma2of[ 3 ])LetA2L.Then,foranyconjugateA0(6=A)ofA,L(A)]TJ /F3 11.955 Tf 11.96 0 Td[(A0)8>><>>:=+L(A)ifL(A)60(modp);>+L(A)ifL(A)0(modp): 43

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Proof. WriteA=p)]TJ /F7 7.97 Tf 6.59 0 Td[(1Xj=0aijLandA0=p)]TJ /F7 7.97 Tf 6.58 0 Td[(1Xj=0aj(i)jL,whereai2Kand(i)L(6=L)isaconjugateofL.Thenwehave L(A)=min0jp)]TJ /F7 7.97 Tf 6.59 0 Td[(1L(ajjL);(4) and L(A)]TJ /F3 11.955 Tf 11.95 0 Td[(A0)=min1jp)]TJ /F7 7.97 Tf 6.58 0 Td[(1L(aj(jL)]TJ /F3 11.955 Tf 11.95 0 Td[((i)jL)):(4) NotethatpjL(A)exactlywhenL(A)=L(a0).Since(i)L=L(1+Yi)withL(Yi)=,itfollowsthat (i)jL=jL(1+Yi)jjL(1+jYi);(4) whichimplies L(aj(jL)]TJ /F3 11.955 Tf 11.95 0 Td[((i)jL))=L(ajjLYi)=+L(ajjL)(4) for1jp)]TJ /F6 11.955 Tf 11.95 0 Td[(1.Hence, L(A)]TJ /F3 11.955 Tf 11.95 0 Td[(A0)=+min1jp)]TJ /F7 7.97 Tf 6.59 0 Td[(1L(ajjL)+L(A);(4) withequalityexactlywhenp-L(A). LetA2LsuchthatL(A)=randp-r.Let g(X)=Xp)]TJ /F4 7.97 Tf 13.05 15.43 Td[(p)]TJ /F7 7.97 Tf 6.59 0 Td[(1Xi=1biXi)]TJ /F3 11.955 Tf 11.95 0 Td[(b0(bi2Kfor0ip)]TJ /F6 11.955 Tf 11.96 0 Td[(1)(4) bethepolynomialsuchthatg(A)=0,andletA=A0;A1;:::;Ap)]TJ /F7 7.97 Tf 6.58 0 Td[(1bealltherootsoftheequationg(X)=0.Since g0(A)=p)]TJ /F7 7.97 Tf 6.58 0 Td[(1Yi=1(A)]TJ /F3 11.955 Tf 11.95 0 Td[(Ai)=)]TJ /F4 7.97 Tf 12.38 15.44 Td[(p)]TJ /F7 7.97 Tf 6.59 0 Td[(1Xi=1ibiAi)]TJ /F7 7.97 Tf 6.59 0 Td[(1;(4) wehavebyLemma 4.2.1 that L(g0(A))=(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)(+r)=min1ip)]TJ /F7 7.97 Tf 6.58 0 Td[(1L(ibiAi)]TJ /F7 7.97 Tf 6.59 0 Td[(1)=L(bl)+(l)]TJ /F6 11.955 Tf 11.96 0 Td[(1)r(4) 44

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forsomelwith1lp)]TJ /F6 11.955 Tf 11.95 0 Td[(1.ThenwehaveL(bl)=(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)(+r))]TJ /F6 11.955 Tf 11.96 0 Td[((l)]TJ /F6 11.955 Tf 11.95 0 Td[(1)r (4)=(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+r(p)]TJ /F3 11.955 Tf 11.95 0 Td[(l) (4)(p)]TJ /F6 11.955 Tf 11.95 0 Td[(1)+r: (4) Sincebl2K,lisdeterminedby rl(p)]TJ /F6 11.955 Tf 11.95 0 Td[(1)(modp);1lp)]TJ /F6 11.955 Tf 11.95 0 Td[(1:(4) Lemma4.2.2. (SeeLemma3of[ 3 ])LetL=Kbeatotallyramiedextensionoftype(n;m;!).LetA2LsuchthatL(A)=r>0,wherep-r.ThenL(NL=K(1+A))]TJ /F6 11.955 Tf 11.95 0 Td[(1)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(A))(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+r: Theequalitysignholdsexactlywhenr)]TJ /F3 11.955 Tf 21.92 0 Td[(n(modp). Proof. SinceNL=K(1+A)=()]TJ /F6 11.955 Tf 9.3 0 Td[(1)pp)]TJ /F7 7.97 Tf 6.58 0 Td[(1Yi=0()]TJ /F6 11.955 Tf 9.3 0 Td[(1)]TJ /F3 11.955 Tf 11.95 0 Td[(Ai) (4)=()]TJ /F6 11.955 Tf 9.3 0 Td[(1)pg()]TJ /F6 11.955 Tf 9.3 0 Td[(1)=1+NL=K(A)+p)]TJ /F7 7.97 Tf 6.58 0 Td[(1Xi=1()]TJ /F6 11.955 Tf 9.3 0 Td[(1)ibi; (4) andfori6=lwehaveL(bi)=L(biAi))]TJ /F3 11.955 Tf 11.96 0 Td[(L(Ai) (4)>L(blAl))]TJ /F3 11.955 Tf 11.96 0 Td[(L(Ai) (4)=(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+r(p)]TJ /F3 11.955 Tf 11.96 0 Td[(i) (4)(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+r; (4) itfollowsthat L(NL=K(1+A))]TJ /F6 11.955 Tf 11.95 0 Td[(1)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(A))min1ip)]TJ /F7 7.97 Tf 6.59 0 Td[(1L(bi)(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+r:(4) 45

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By( 4 ),wehaveequalityexactlywhenl=p)]TJ /F6 11.955 Tf 12.1 0 Td[(1,andinthiscase,wehaveby( 4 )thatthisisequivalenttorl)]TJ /F3 11.955 Tf 21.92 0 Td[((modp) (4)r(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)(p)]TJ /F6 11.955 Tf 11.95 0 Td[(1)(modp) (4))]TJ /F3 11.955 Tf 9.3 0 Td[(rn(modp): (4) Lemma4.2.3. (SeeLemma4andLemma5of[ 3 ])LetL=Kbeatotallyramiedextensionoftype(n;m;!).LetA2LwithL(A)=r>0.Thenthefollowingholds: (i) L(Ap)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(A))(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+pr; withequalitysignholdsexactlywhenp-r,and (ii) L((1+A)p)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(1+A))(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+r; withequalityholdsexactlywhenr)]TJ /F3 11.955 Tf 21.92 0 Td[(n(modp). Proof. Firstweassumethatp-r.Lettheminimalpolynomialg(X)ofAbegivenby( 4 ),fromwhichwehave L(Ap)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(A))=min1ip)]TJ /F7 7.97 Tf 6.59 0 Td[(1L(biAi):(4) By( 4 ),wehave L(Ap)]TJ /F3 11.955 Tf 11.95 0 Td[(NL=K(A))=(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)(+r)+r=(p)]TJ /F6 11.955 Tf 11.95 0 Td[(1)+pr:(4) Thisproves(i)inthecasep-r.Tosee(ii)inthecasep-rnotethat (1+A)p)]TJ /F3 11.955 Tf 11.95 0 Td[(NL=K(1+A)=(Ap)]TJ /F3 11.955 Tf 11.95 0 Td[(NL=K(A)))]TJ /F6 11.955 Tf 11.96 0 Td[((NL=K(1+A))]TJ /F6 11.955 Tf 11.95 0 Td[(1)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(A)):(4) TheresultfollowsfromLemma 4.2.2 and(i)inthecasep-r. 46

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Nowassumethatpjr.WriteA=a+A0,wherea2K;L(a)=L(A)pr+(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1); (4) where( 4 )isdueto(ii)inthecasep-r.Thisproves(i)inthecasepjr. Similarly,tosee(ii)inthecasepjr,writeA=a+A0,wherea2K;L(a)=L(A)(p)]TJ /F6 11.955 Tf 11.95 0 Td[(1)+r; (4) where( 4 )isdueto(ii)inthecasep-r. Theorem4.2.4. (SeeTheorem1of[ 3 ])LetL=Kbeatotallyramiedextensionwithramicationbreak.Iff(X)andg(X)areEisensteinpolynomialssuchthatL=K=K[X]=(f(X))=K[X]=(g(X)),thenf(X)andg(X)areEisensteinpolynomialsofthesametype. Proof. Supposef(X)hastype(n;m;!)andg(X)hastype(n;m;!1).Let;12Lsuchthatf()=0andg(1)=0.Since pKp1;(4) 47

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itfollowsthat1=uforsomeu2U1;L.Bydenitionofthetype,wehave p)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K()!mKn(4) and p1)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(1)!1mKn1!1mKn:(4) Recallthat=p(m)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+n p)]TJ /F6 11.955 Tf 11.95 0 Td[(1,sowehave L(!mKn)=L(!1mKn1)=(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+p:(4) Since1=u,wehave (p1)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(1)))]TJ /F6 11.955 Tf 11.95 0 Td[((p)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K())=(up)]TJ /F6 11.955 Tf 11.95 0 Td[(1)(p)]TJ /F3 11.955 Tf 11.95 0 Td[(NL=K())+NL=K()(up)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(u));(4) andbyapplyingLemma 4.2.3 (i)andLemma 4.2.3 (ii),weobtain L((up)]TJ /F6 11.955 Tf 11.96 0 Td[(1)(p)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K()))=L(up)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+(p)]TJ /F6 11.955 Tf 11.95 0 Td[(1)+p>(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+p;(4) L(NL=K()(up)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(u)))p+(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+L(u)]TJ /F6 11.955 Tf 11.96 0 Td[(1)>(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+p:(4) Hence, L((!1)]TJ /F3 11.955 Tf 11.96 0 Td[(!)mKn)>(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+p=L(!mKn);(4) whichimplies!=!1since!;!12k. Henceforth,wesaythatanextensionL=Khastype(n;m;!)ifL=K=K[X]=(g(X))forsomeEisensteinpolynomialoftype(n;m;!). Lemma4.2.5. (SeeLemma7of[ 3 ])LetL=Kbeatotallyramiedextensionoftype(n;m;!).ForeachprimeelementofL,dene()=p)]TJ /F3 11.955 Tf 12.27 0 Td[(!mKn)]TJ /F3 11.955 Tf 12.28 0 Td[(NL=K(),whereNL=KisthenormofL=K.If(1)6=0foraprimeelement1ofL,thenthereexistsaprimeelement2ofLsuchthatL((2))>L((1)). 48

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Proof. SupposeL=Kisoftype.Let f(X)=Xp)]TJ /F4 7.97 Tf 13.05 15.43 Td[(p)]TJ /F7 7.97 Tf 6.59 0 Td[(1Xi=1aiXi)]TJ /F3 11.955 Tf 11.95 0 Td[(Ka(4) betheEisensteinpolynomialfor1.Since1isarootoff(X)wehave p1)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(1)=!mKn1(b11)]TJ /F4 7.97 Tf 6.58 0 Td[(n1+:::bn++bp)]TJ /F7 7.97 Tf 6.58 0 Td[(1p)]TJ /F7 7.97 Tf 6.58 0 Td[(1)]TJ /F4 7.97 Tf 6.59 0 Td[(n1);(4) wherebi2MKfor1i0.Substitutingbn=1+into( 4 ),weobtain p1)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(1)=!mKn1(1+b11)]TJ /F4 7.97 Tf 6.58 0 Td[(n1++++bp)]TJ /F7 7.97 Tf 6.58 0 Td[(1p)]TJ /F7 7.97 Tf 6.59 0 Td[(1)]TJ /F4 7.97 Tf 6.59 0 Td[(n1):(4) SettingY=b11)]TJ /F4 7.97 Tf 6.59 0 Td[(n1++++bp)]TJ /F7 7.97 Tf 6.59 0 Td[(1p)]TJ /F7 7.97 Tf 6.59 0 Td[(1)]TJ /F4 7.97 Tf 6.59 0 Td[(n1,wehave L(Y)=minfpK();pK(bi)+i)]TJ /F3 11.955 Tf 11.95 0 Td[(n:1ip)]TJ /F6 11.955 Tf 11.96 0 Td[(1;i6=ng;(4) whichimpliesL(Y)>0andL(Y)6n(modp). Letr=L(Y).Equation( 4 )yields L((1))=L(!mKn1Y)=pm+n+r=(p)]TJ /F6 11.955 Tf 11.95 0 Td[(1)+p+r:(4) Since0
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Sowehave(2)=p2)]TJ /F3 11.955 Tf 11.96 0 Td[(!mKn2)]TJ /F3 11.955 Tf 11.95 0 Td[(NL=K(2) (4)=(p2)]TJ /F3 11.955 Tf 11.96 0 Td[(p1))]TJ /F6 11.955 Tf 11.95 0 Td[((NL=K(2))]TJ /F3 11.955 Tf 11.95 0 Td[(NL=K(1)) (4)=(Z1)p)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(1)(NL=K(1+Z))]TJ /F6 11.955 Tf 11.95 0 Td[(1) (4)=(Z1)p)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(Z1))]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(1)(NL=K(1+Z))]TJ /F6 11.955 Tf 11.95 0 Td[(1)]TJ /F3 11.955 Tf 11.96 0 Td[(NL=K(Z)): (4) Sincer6=0andr6)]TJ /F3 11.955 Tf 21.92 0 Td[(n(modp),Lemma 4.2.2 yieldsthestrictinequality L(NL=K(1)(NL=K(1+Z))]TJ /F6 11.955 Tf 11.96 0 Td[(1)]TJ /F3 11.955 Tf 11.95 0 Td[(NL=K(Z)))>(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+p+r:(4) ByLemma 4.2.3 (i),wehave L)]TJ /F6 11.955 Tf 5.48 -9.68 Td[((Z1)p)]TJ /F3 11.955 Tf 11.95 0 Td[(NL=K(Z1)(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)+p(r+1):(4) Hence,weobtain L((2))>(p)]TJ /F6 11.955 Tf 11.95 0 Td[(1)+p+r=L((1)):(4) Theorem4.2.6. (SeeTheorem4of[ 3 ])SupposeL=Kisanextensionoftype(n;m;!).ThenthereexistsaprimeelementLofLthatisarootofapolynomialoftheform A!;u(X)=Xp)]TJ /F3 11.955 Tf 11.96 0 Td[(!mKXn)]TJ /F3 11.955 Tf 11.96 0 Td[(Ku;(4) whereu2U1;K. Proof. ByLemma 4.2.5 ,thereexistsaprimeelementofLsuchthatL(())>p(+1).LetNL=K()=Ka,forsomea2U1;K.For1ip,let(i)betherootsofA!;a(X)=Xp)]TJ /F3 11.955 Tf 11.96 0 Td[(!mKXn)]TJ /F3 11.955 Tf 11.96 0 Td[(Ka. Since ()=A!;a()=pYi=1()]TJ /F3 11.955 Tf 11.96 0 Td[((i));(4) 50

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wehave pXi=1L()]TJ /F3 11.955 Tf 11.95 0 Td[((i))=L(())>p(+1):(4) Hence,L()]TJ /F3 11.955 Tf 12.51 0 Td[((j))>+1forsomej.SinceLK((j))byKrasner'sLemmaandK((j))=Kisanextensionofdegreep,wehaveL=K((j)).ThenL=(j). Corollary4.2.7. LetL=Kbeanextensionoftype(n;m;!).ThenL=KisGaloisifandonlyisanintegerandn!2(k)p)]TJ /F7 7.97 Tf 6.58 0 Td[(1. Proof. IfL=KisGalois,thenthelowerramicationbreakisaninteger.Wewillshowthatifisaninteger,thenL=KisGaloisifandonlyifn!2(k)p)]TJ /F7 7.97 Tf 6.58 0 Td[(1.WriteL=K(1),where1isarootofanAmanopolynomialA!;u(X).Let2beaconjugateof1differentfrom1.Wewillndthecriterionfor22L. Since=L(2)]TJ /F7 7.97 Tf 6.58 0 Td[(11)]TJ /F6 11.955 Tf 12.01 0 Td[(1),wecanwrite2=1(1+1Y)forsomeunitY2Ks.Then22LifandonlyifY2L. SinceA!;u(2)=0,wehave p1(1+1Y)p)]TJ /F3 11.955 Tf 11.96 0 Td[(!mKn1(1+1Y)n)]TJ /F3 11.955 Tf 11.95 0 Td[(uK=0:(4) Rewriting( 4 ),weobtain p(+1)1Yp)]TJ /F3 11.955 Tf 11.95 0 Td[(!mKnXi=1nii+n1Yi+p1)]TJ /F3 11.955 Tf 11.95 0 Td[(!mKn1)]TJ /F3 11.955 Tf 11.95 0 Td[(uK=0:(4) ButA!;u(1)=p1)]TJ /F3 11.955 Tf 11.95 0 Td[(!mKn1)]TJ /F3 11.955 Tf 11.96 0 Td[(uK=0,so p(+1)1Yp)]TJ /F3 11.955 Tf 11.95 0 Td[(!mKnXi=1nii+n1Yi=0:(4) Dividing( 4 )byp(+1)1andusingn=p()]TJ /F3 11.955 Tf 11.95 0 Td[(m+1))]TJ /F3 11.955 Tf 11.96 0 Td[(,weobtain Yp)]TJ /F4 7.97 Tf 18.02 14.95 Td[(nXi=1ni!mK(i)]TJ /F7 7.97 Tf 6.59 0 Td[(1))]TJ /F4 7.97 Tf 6.59 0 Td[(pm1Yi=0:(4) 51

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SinceL()]TJ /F4 7.97 Tf 5.48 -4.38 Td[(ni!mK(i)]TJ /F7 7.97 Tf 6.58 0 Td[(1))]TJ /F4 7.97 Tf 6.59 0 Td[(pm1)=(i)]TJ /F6 11.955 Tf 11.95 0 Td[(1),itfollowsthat Yp)]TJ /F3 11.955 Tf 11.96 0 Td[(n!Y0(mod1):(4) ByHensel'sLemma,( 4 )hasanonzerosolutionifandonlyif( 4 )hasanonzerosolution.But( 3 )hasanonzerosolutionifandonlyifn!2(L)p)]TJ /F7 7.97 Tf 6.59 0 Td[(1\k=(k)p)]TJ /F7 7.97 Tf 6.59 0 Td[(1. 4.3TheFullGroup LetA=Autk(K),thegroupofautomorphismsofKxingk.ThegroupAisisomorphictotk+t2k[[t]],thefullgroupofformalpowerseriesoverkundersubstitution(see[ 10 ]).WewillcallAthefullgroupofKaswell. SinceK=k((K)),anyparticularelement'2AisdeterminedbyitsactiononK.Furthermore,'actsonKby '(K)=av'K;(4) wherea2kandv'2U1;K.ThegroupAactstransitivelyontheprimeelementsofK. SetP=fXp)]TJ /F3 11.955 Tf 12.1 0 Td[(!mKXn)]TJ /F3 11.955 Tf 12.1 0 Td[(Ku:!2k;u2U1;Kg.Forf(X);g(X)2P,wewritef(X)g(X)if K[X]=(f(X))=K[X]=(g(X)):(4) LetP=betheequivalenceclassesofPundertherelation.Forf(X)2P,denotetheequivalenceclassoff(X)inP=by[f(X)]. For'2AandA!1;u1(X)=Xp)]TJ /F3 11.955 Tf 11.99 0 Td[(!1mKXn)]TJ /F3 11.955 Tf 11.99 0 Td[(Ku12P,wewillshowinSection 4.4 thatthereexists!22k;u22U1;KsuchthatK[X]=('(A!1;u1(X)))=K[X]=(Xp)]TJ /F3 11.955 Tf 11.96 0 Td[('(!mK)Xn)]TJ /F3 11.955 Tf 11.96 0 Td[('(Ku)) (4)=K[X]=(A!2;u2(X)): (4) DeneanactionofAonP=by '[A!1;u1(X)]=[A!2;u2(X)]:(4) 52

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ForA!;u(X)2P,letA[A!;u(X)]betheorbitof[A!;u(X)]undertheactionofA.Itfollowsthatenumeratingthek-isomorphismclassesofEisequivalenttoenumeratingthesetoforbits A(P=)=fA[A!;u(X)]:A!;u(X)2Pg:(4) TheactionofAonP=denedin( 4 )iswell-dened.Indeed,letA!01;u01(X)2PsuchthatA!01;u01(X)A!1;u1(X),andlet :K[X]=(A!1;u1(X))!K[X]=(A!01;u01(X))(4) beaK-isomorphism.LetA!02;u02(X)2Psuchthat K[X]=('(A!01;u01(X)))=K[X]=(A!02;u02(X)):(4) Denetheringisomorphism'X:K[X]!K[X]by 'X nXi=0aiXi!=nXi=0'(ai)Xi:(4) ThenwehaveA!2;u2(X)A!02;u02(X),inducedbytheK-isomorphism='X')]TJ /F7 7.97 Tf 6.59 0 Td[(1X,whichimplies[A!2;u2(X)]=[A!02;u02(X)].Hence,theactiondenedin( 4 )iswell-dened. Letnbeapositiveintegerrelativelyprimetop.BeforeenumeratingthesetA(P=),wewillenumeratethesetA(Tn=),where Tn=fXn)]TJ /F3 11.955 Tf 11.95 0 Td[(Ku:u2kU1;Kg:(4) ByProposition3:4,pg.53of[ 6 ],ifL=KistotallytamelyramiedextensionofdegreenoverK,thenL=K=K[X]=(g(X)),whereg(X)=Xn)]TJ /F3 11.955 Tf 12.52 0 Td[(0K,forsomeprimeelement0KofK.However,AactstransitivelyontheprimeelementsofK.Hence,itfollowsthatthenumberofk-isomorphismclassesoftotallytamelyramiedextensionofdegreenisequalto1. 53

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4.4ActiononAmano'sPolynomials Inthissection,wewillseehowanarbitraryelement'2AactsonA!;u(X).Thenwewillshowthatthereexist!02kandu02U1;Ksuchthat K[X]=('(A!;u(X)))=K[X]=(A!0;u0(X)):(4) LetN=f2A:(K)2KU1;Kg.ThesetNisasubgroupofindexjkjinAandisisomorphictotheNottinghamGroup(orthegroupofwildautomorphisms)(see[ 10 ]).LetZpbethegroupofunitsofthep-adicintegers.Thenwehavethefollowinglemma: Lemma4.4.1. Letu2U1;K,andh2Zp.ThenwehaveU1;K=((u)(K) Kh:2N) Proof. Letv=u1=h2U1;KThenw=vKisaprimeelementofK.WehaveU1;K=(w) w:2N (4)=v(w) w:2N (4)=(vK) K:2N (4)=(u)1 h(K) K:2N: (4) Sinceh2Zp,weseethat Uh1;K=U1;K:(4) Hence,byraisingtothepowerh,weobtain U1;K=((u)(K) Kh:2N):(4) 54

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Lemma4.4.2. Letc2kandletNc=NcbetherightcosetofNinArepresentedbycwherec(K)=cK.Thenforu2U1;Kandh2Zp,wehaveU1;K=f'(u)vh':'2Ncg: Proof. For2Ndenedby(K)=vKforsomev2U1;K,notethat (u)(K) Kh=(u)vh:(4) Letu0=c(u)2U1;K.Thenf'(u)vh':'2Ncg=fc(u)vh:2Ng (4)=f(u0)vh:2Ng (4)=U1;K; (4) where( 4 )isdueto( 4 )andLemma 4.4.1 Letting'actonA!;u(X),weget '(A!;u(X))=Xp)]TJ /F3 11.955 Tf 11.95 0 Td[(!cmvm'mKXn)]TJ /F3 11.955 Tf 11.95 0 Td[(c'(u)v'K:(4) Leta2ksuchthatap=c,andletX=avm p)]TJ /F14 5.978 Tf 5.76 0 Td[(n'Y.Thenwehave a)]TJ /F4 7.97 Tf 6.59 0 Td[(pv)]TJ /F14 5.978 Tf 5.76 0 Td[(pm p)]TJ /F14 5.978 Tf 5.76 0 Td[(n''(A!;u(X))=Yp)]TJ /F3 11.955 Tf 11.96 0 Td[(!a(p)]TJ /F7 7.97 Tf 6.58 0 Td[(1)mKYn)]TJ /F3 11.955 Tf 11.96 0 Td[('(u)vh'K;(4) whereh=p)]TJ /F3 11.955 Tf 11.95 0 Td[(n)]TJ /F3 11.955 Tf 11.95 0 Td[(pm p)]TJ /F3 11.955 Tf 11.96 0 Td[(n.Now,sinceavm p)]TJ /F14 5.978 Tf 5.75 0 Td[(n'2K,itfollowsthat K[X]=('(A!;u(X)))=K[x]=(A!a(p)]TJ /F9 5.978 Tf 5.76 0 Td[(1);'(u)vh'(X));(4) whichimplies A[A!;u(X)]=A[A!a(p)]TJ /F9 5.978 Tf 5.76 0 Td[(1);'(u)vh'(X)]:(4) By( 4 )andLemma 4.4.1 A[A!;u(X)]=A[A!a(p)]TJ /F9 5.978 Tf 5.75 0 Td[(1);v(X)];(4) 55

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forallv2U1;K.Theelementcisarbitrary,soaisalsoarbitrary.Hence, A[A!;u(X)]=A[A!a(p)]TJ /F9 5.978 Tf 5.75 0 Td[(1);v(X)];(4) forallv2U1;Kandforalla2k. WewillnowcomputethecardinalityofSg.ByCorollary 4.2.7 ,theextensionK[X]=(A!;u(X))isGaloisifandonlyifisanintegerandn!2(k)p)]TJ /F7 7.97 Tf 6.59 0 Td[(1,sowecanwrite!=n)]TJ /F7 7.97 Tf 6.59 0 Td[(1tp)]TJ /F7 7.97 Tf 6.59 0 Td[(1forsomet2k.Since A[A!;u(X)]=A[An)]TJ /F9 5.978 Tf 5.76 0 Td[(1tp)]TJ /F9 5.978 Tf 5.76 0 Td[(1a(p)]TJ /F9 5.978 Tf 5.76 0 Td[(1);v(X)](4) foralla2kandv2U1;K,itfollowsthatenumeratingtheorbitsofA(P=)isequivalenttocalculatingthecardinalityof(k)p)]TJ /F7 7.97 Tf 6.58 0 Td[(1=(k)(p)]TJ /F7 7.97 Tf 6.59 0 Td[(1),thatis, jSgj=j(k)p)]TJ /F7 7.97 Tf 6.59 0 Td[(1=(k)(p)]TJ /F7 7.97 Tf 6.59 0 Td[(1)j:(4) But (k)p)]TJ /F7 7.97 Tf 6.59 0 Td[(1=(k)(p)]TJ /F7 7.97 Tf 6.58 0 Td[(1)=k=(Fp(k)):(4) Hence,jSgj=k=(Fp(k)) (4)=gcdq)]TJ /F6 11.955 Tf 11.95 0 Td[(1 p)]TJ /F6 11.955 Tf 11.95 0 Td[(1;; (4) wherethesecondequalityfollowsfromLemma 4.1.3 Remark4.4.3. ThecardinalityofSghadalreadybeencalculatedbyKlopschin[ 10 ].LetL=k((L))bealocalfunctioneld.SayL=K1andL=K2arek-isomorphicifthereexists'2F=Autk(L)suchthat'(K1)=K2.Thenitfollowsthatenumeratingthek-isomorphismclassesofdegreepGaloissub-extensionsofLwithramicationbreakisequivalenttoenumeratingSg. Henceforth,allmentionedsub-extensionsofLareGaloisofdegreepwithramicationbreak.SupposethatL=K1andL=K2arek-isomorphicinducedby 56

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'2Fwith'(K2)=K1.LetGi=Gal(L=Ki)fori=1;2.Letg12G1beanelementoforderp.Theng2=')]TJ /F7 7.97 Tf 6.58 0 Td[(1g1'2G2andithasorderp,whichimplies')]TJ /F7 7.97 Tf 6.59 0 Td[(1G1'=G2.ItfollowsthatL=K1andL=K2arek-isomorphicifandonlyG1isconjugatetoG2inF.Hence,ourdesiredenumerationisthenumberofconjugacyclassesofsubgroupsofFoforderpandwithramicationbreak. Sinceg1hasramicationbreak,wecanwrite g1(K)K+a+1+1K(mod+2K):(4) Thenfor1i
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But (k)p)]TJ /F7 7.97 Tf 6.59 0 Td[(1=(k)(p)]TJ /F7 7.97 Tf 6.58 0 Td[(1)=k=)]TJ /F5 11.955 Tf 5.48 -9.68 Td[(Fp(k):(4) Hence,ifisaninteger,wehavejSngj=(p)]TJ /F6 11.955 Tf 11.96 0 Td[(2)jkj Fp(k) (4)=(p)]TJ /F6 11.955 Tf 11.96 0 Td[(2)gcdq)]TJ /F6 11.955 Tf 11.95 0 Td[(1 p)]TJ /F6 11.955 Tf 11.95 0 Td[(1;: (4) Whenisnotaninteger,L=KisnotGalois,hence!=n)]TJ /F7 7.97 Tf 6.59 0 Td[(1ritp)]TJ /F7 7.97 Tf 6.58 0 Td[(1forsomet2kand1ip)]TJ /F6 11.955 Tf 12.07 0 Td[(1.Letd=p)]TJ /F6 11.955 Tf 11.96 0 Td[(1 gcd(p)]TJ /F6 11.955 Tf 11.95 0 Td[(1;(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)),thatis,disthedenominatorofwhenisinreducedform.Itfollowssimilarlythatwhenisnotaninteger,wehavejSngj=(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)jkj Fp(k)d (4)=(p)]TJ /F6 11.955 Tf 11.96 0 Td[(1)gcdq)]TJ /F6 11.955 Tf 11.95 0 Td[(1 p)]TJ /F6 11.955 Tf 11.96 0 Td[(1;d: (4) 4.5AnotherLookatGaloisExtensions Intheprevioussection,weshowedthatjSgj=jk=(k)jusingtheworksofAmanoin[ 3 ].Inthissection,wewillusebasicArtin-SchreierTheorytocomputejSgj. LetL=KbeGaloisofdegreepwithramicationbreak.ByArtin-SchreierTheory,L=K(),whereisarootof g(X)=Xp)]TJ /F3 11.955 Tf 11.95 0 Td[(X)]TJ /F3 11.955 Tf 11.95 0 Td[(=0;(4) and2K.ByProposition2:4,pg.75of[ 6 ],theelementcanbechosensothat()=)]TJ /F3 11.955 Tf 9.29 0 Td[(andthatisrelativelyprimep.Write=c)]TJ /F4 7.97 Tf 6.59 0 Td[(K,where2U1;Kandc2k.Sinceisrelativelyprimetop,wehave U1;K=U1;K:(4) Hence,wemaywrite=c(K))]TJ /F4 7.97 Tf 6.59 0 Td[(forsome2U1;K. 58

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Notethatforanyintegerisuchthat1ip)]TJ /F6 11.955 Tf 12.97 0 Td[(1,wehavegi(i)=0andK()=K(i).Hence,forallintegersiwith1ip)]TJ /F6 11.955 Tf 13.07 0 Td[(1,wehavethatallthepolynomialsgi(X)yieldexactlythesameextensionsoverK.Hence,wehave ggiforalli2Fp:(4) Foranarbitraryautomorphism'2A,wehave '(g(X))=Xp)]TJ /F3 11.955 Tf 11.95 0 Td[(X)]TJ /F3 11.955 Tf 11.96 0 Td[(c('(K)))]TJ /F4 7.97 Tf 6.58 0 Td[((4) Leta2kand2U1;KbearbitraryelementsofK.BydenitionofAinSection 4.3 ,wemaychoose'2Asuchthat '(K)=aK:(4) Applying'tog(X),weobtain '(g(X))=Xp)]TJ /F3 11.955 Tf 11.96 0 Td[(X)]TJ /F3 11.955 Tf 11.95 0 Td[(ca)]TJ /F4 7.97 Tf 6.58 0 Td[((K))]TJ /F4 7.97 Tf 6.59 0 Td[(:(4) Sinceisrelativelyprimetopandisarbitrary,isarbitrary.Itfollowsthat g(X)gca)]TJ /F14 5.978 Tf 5.75 0 Td[(u)]TJ /F14 5.978 Tf 5.76 0 Td[(K(X)(4) foralla2kandu2U1;K.Combining( 4 )and( 4 ),weobtainjSgj=jk=(k)j=gcdq)]TJ /F6 11.955 Tf 11.96 0 Td[(1 p)]TJ /F6 11.955 Tf 11.95 0 Td[(1;: 59

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APPENDIX:C++PROGRAMUSEDTOGENERATETABLE 1-1 #includeintmain(void){intr,s,p,m,n,j,k,a,b,c,d;printf("Enterprimep.\n");scanf("%i",&p);a=1;b=0;c=0;while(a=10){printf("%i",a%10);}if(a==p-1){printf("\n");}a++;}for(r=0;r=r){j=s-r;}else{j=p+s-r;}m=1;while(m=((n*s)%p))&&(((n*s)%p)>k)){ 60

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printf("O");n=p;m=p;}else{n++;}}m++;}if((n==p)&&(m==p)){printf("~");b++;}if(s==p-1){printf("%i\n",r);}}}printf("Thenumberofsolutionsis%i.\n",b);} 61

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REFERENCES [1] A.Aiba,Artin-SchreierextensionsandGaloismodulestructure,J.NumberTheory102(2003)118-124 [2] A.Aiba,CarlitzModulesandGaloisModuleStructure,J.NumberTheory62(1997)213-219 [3] S.Amano,EisensteinEquationsofdegreepinap-adiceld,J.Fac.Sci.Univ.TokyoSect.IAMath.18(1971),1-21 [4] G.Bachman,Introductiontop-AdicNumbersandValuationTheory,(AcademicPress,1964) [5] N.P.Byott,Someself-duallocalringsofintegersnotfreeovertheirassociatedorders,Math.Proc.CambridgePhilos.Soc.110(1991)5-10 [6] I.B.FesenkoandS.V.Vostokov,LocalFieldsandTheirExtensions,TranslationofMathematicalMonographsV.121,(AMS,2002) [7] F.BertrandiasandM.Ferton,C.R.Acad.Scri.Parisser.A-B274(1972),A1330-A1333 [8] M.Ferton,Surlesideauxd'uneextensioncycliquededegrepremierd'uncorpslocal,C.R.Acad.Sc.Paris,t.276(June4,1973)SerieA1483-1486 [9] A.FrohlichandM.J.Taylor,AlgebraicNumberTheory,Cambridgestudiesinadvancedmathematics27(CambridgeUniversityPress,1991) [10] B.Klopsch,AutomorphismsoftheNottinghamGroup,JournalofAlgebra223(2000)37-56 [11] S.Lang,Algebra,GraduateTextsinMathematicsno.211(Springer-Verlag,2002) [12] G.Lettl,NoteonatheoremofA.Aiba,J.NumberTheory115(2005)87-88 [13] A.Robert,ACourseinp-adicAnalysis,GraduateTextsinMathmeaticsno.198(Springer-Verlag,2000) [14] J.P.Serre,LocalFields,GraduateTextsinMathematicsno.67(Springer-Verlag,1979) [15] L.Thomas,OnGaloisModuleStructureofExtensionsofLocalFields,PublicationsMathematiquesdeBesancon,2010 [16] L.Washington,CyclotomicFields,GraduateTextsinMathematicsno.83(Springer-Verlag,1982) 62

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BIOGRAPHICALSKETCH DucVanHuynhwasbornin1984inSaigon,Vietnam.HecamewithhisfamilytoAmericafromVietnaminAprilof1993.HegrewupinSavannah,Georgia,wheretheweatherwasalwaysbeautifulandthefoodwasalwaysgood.Duringhighschool,heguredthathelovesmathematics,althoughhewasnotverygoodatit.Aftergraduatingfromhighschool,heattendedArmstrongAtlanticStateUniversitytostudymathematics.There,heattendedmanywonderfulclassesandmetmanyamazingprofessorswholovetoanswerhismathematicsquestionsduringtheinquisitiveyears.ItwashisloveformathematicsandtheencouragementfromhisundergraduateprofessorsthathelpedhimtodecidetopursueadoctoratefromtheUniversityofFlorida.InMayof2014,heearnedhisdoctorateinmathematicsundertheguidanceofDr.KevinKeating. 63