On the Functional Equation of the Artin L-functions Robert P ...

On the Functional Equation
of the Artin L-functions
Robert P. Langlands

Introduction 2
0. Introduction
1. WeilGroups
2. TheMainTheorem
3. TheLemmasofInduction
4. TheLemmaofUniqueness
5. APropertyof λ­Functions
6. AFiltrationoftheWeilGroup
TABLE OF CONTENTS
7. ConsequencesofStickelberger’sResult
8. ALemmaofLamprecht
9. ALemmaofHasse
10. TheFirstMainLemma
11. Artin–SchreierEquations
12. TheSecondMainLemma
13. TheThirdMainLemma
14. TheFourthMainLemma
15. AnotherLemma
16. Definitionofthe λ­Functions
17. ASimplification
18. NilpotentGroups
19. ProofoftheMainTheorem
20. Artin L­Functions
21. ProofoftheFunctionalEquation
22. Appendix
23. References

Introduction 3
O. Introduction
InthispaperIwanttoconsidernotjustthe L­functionsintroducedbyArtin[1]butthe
moregeneralfunctionsintroducedbyWeil[15].TodefinetheseoneneedsthenotionofaWeil
groupasdescribedin[3]. Thisnotionwillbeexplainedinthefirstparagraph. Fornowa
roughideawillsuffice.If Eisaglobalfield,thatisanalgebraicnumberfieldoffinitedegree
overtherationalsorafunctionfieldoverafinitefield, CEwillbetheidéleclassgroupof E.
If Eisalocalfield,thatisthecompletionofaglobalfieldatsomeplace[16],archimedeanor
nonarchimedean, CEwillbethemultiplicativegroupof E.If K/EisafiniteGaloisextension
theWeilgroup W K/Eisanextensionof G(K/E),theGaloisgroupof K/E,by CK. Itisa
locallycompacttopologicalgroup.
If E ⊆ E ′ ⊆ Kand K/EisfiniteandGalois W K/E ′mayberegardedasasubgroupof
W K/E.Itisclosedandoffiniteindex.If E ⊆ K ⊆ Lthereisacontinuousmapof W L/Eonto
W K/E. Thusanyrepresentationof W K/Emayberegardedasarepresentationof W L/E. In
particulartherepresentations ρ1of W K1/Eand ρ2of W K2/Ewillbecalledequivalentifthereis
aGaloisextension L/Econtaining K1/Eand K2/Esuchthat ρ1and ρ2determineequivalent
representationsof W L/E. Thisallowsustorefertoequivalenceclassesofrepresentationsof
theWeilgroupof Ewithoutmentioninganyparticularextensionfield K.
Inthispaperarepresentationof W K/Eisunderstoodtobeacontinuousrepresentation
ρinthegroupofinvertiblelineartransformationsofafinite­dimensionalcomplexvector
spacewhichissuchthat ρ(w)isdiagonalizable,thatissemisimple,forall win W K/E. Any
one­dimensionalrepresentationofW K/Ecanbeobtainedbyinflatingaone­dimensionalrepresentationof
W E/E = CE.Thusequivalenceclassesofone­dimensionalrepresentationsofthe
Weilgroupof Ecorrespondtoquasi­charactersof CE,thatis,tocontinuoushomomorphisms
of CEinto C × .
Suppose Eisalocalfield.Thereisastandardwayofassociatingtoeachequivalenceclass
ωofone­dimensionalrepresentationsameromorphicfunctionL(s, ω).Supposeωcorresponds
tothequasi­character χE. If Eisnonarchimedeanand ̟Eisageneratoroftheprimeideal
PEof OE,theringofintegersin E,weset
L(s, ω) =
1
1 − χE(̟E) |̟E| s
if χEisunramified.Otherwiseweset L(s, ω) = 1.If E = Rand
χE(x) = (sgn x) m |x| r

Introduction 4
with mequalto0or1weset
L(s, ω) = π −1
2 (s+r+m) Γ
s + r + m
If E = Cand z ∈ Ethen,forus, |z|willbethesquareoftheordinaryabsolutevalue.If
χE(z) = |z| r z m ¯z n
where mand nareintegerssuchthat m + n ≥ 0, mn = 0,then
2
.
L(s, ω) = 2 (2π) −(s+r+m+n) Γ(s + r + m + n)
Itisnotdifficulttoverify,andweshalldosolater,thatitispossible,injustoneway,
todefine L(s, ω)forallequivalenceclassessothatithasthegivenformwhen ωisonedimensional,sothat
L(s, ω1 ⊕ ω2) = L(s, ω1) L(s, ω2)
sothatif E ′ isaseparableextensionof Eand ωistheequivalenceclassoftherepresentation
oftheWeilgroupof EinducedfromarepresentationoftheWeilgroupof E ′ intheclass Θ
then L(s, ω) = L(s, Θ).
Nowtake Etobeaglobalfieldand ωanequivalenceclassofrepresentationsoftheWeil
groupof E.Itwillbeseenlaterhow,foreachplace v, ωdeterminesanequivalenceclass ωvof
representationsoftheWeilgroupofthecorrespondinglocalfield Ev.Theproduct
v
L(s, ωv)
whichistakenoverallplaces,includingthearchimedeanones,willconvergeiftherealpartofs
issufficientlylarge.ThefunctionitdefinescanbecontinuedtoafunctionL(s, ω)meromorphic
inthewholecomplexplane.ThisistheArtin L­functionassociatedtoω.Itisfairlywell­known
thatif ωistheclasscontragredientto ωthereisafunctionalequationconnecting L(s, ω)and
L(1 − s, ω).
Thefactorappearinginthefunctionalequationcanbedescribedintermsofthelocaldata.
Toseehowthisisdoneweconsiderseparableextensions Eofthefixedlocalfield F.If ΨFisa
non­trivialadditivecharacterof Flet ψ E/Fbethenon­trivialadditivecharacterof Edefined
by
ψE/F(x) = ψF(SE/Fx)

Introduction 5
where SE/Fxisthetraceof x. Wewanttoassociatetoeveryquasi­character χEof CEand
everynon­trivialadditivecharacter ψEof Eanon­zerocomplexnumber ∆(χE, ψE).If Eis
nonarchimedean,if Pm Eistheconductorof χE,andif P −n
E isthelargestidealonwhich ψEis
trivialchooseany γwith OEγ = P m+n
E andset
UE
∆(χE, ψE) = χE(γ)
ψE
α
γ χ −1
E (α)dα
χ −1
E (α)dα .
Therightsidedoesnotdependon γ.If E = R,
UE ψE
α
γ
χE(x) = (sgnx) m |x| r
with mequalto0or1,and ψE(x) = e 2πiux then
If E = C, ψC(z) = e 4πi Re(wz) ,and
with m + n ≥ 0, mn = 0then
∆(χE, ψE) = (i sgnu) m |u| r .
χC(z) = |z| r z m ¯z n
∆(χC, ψC) = i m+n χC(w).
Thebulkofthispaperistakenupwithaproofofthefollowingtheorem.
Theorem A
Suppose Fisagivenlocalfieldand ψFagivennon­trivialadditivecharacterof F.Itis
possibleinexactlyonewaytoassigntoeachseparableextension Eof Facomplexnumber
λ(E/F, ψF)andtoeachequivalenceclassωofrepresentationsoftheWeilgroupofEacomplex
number ε(ω, Ψ E/F)suchthat
(i)If ωcorrespondstothequasi­character χEthen
(ii)
ε(ω, ψ E/F) = ∆(χE, ψ E/F).
ε(ω1 ⊕ ω2, ψ E/F) = ε(ω1, ψ E/F) ε(ω2, ψ E/F).

Introduction 6
(iii)If ωistheequivalenceclassoftherepresentationoftheWeilgroupof Finducedfroma
representationoftheWeilgroupof Eintheclass θthen
ε(ω, ψF) = λ(E/F, ψF) dim θ ε(θ, ψ E/F).
αs Fwilldenotethequasi­character x −→ |x|s Fof CFaswellasthecorrespondingequivalenceclassofrepresentations.Set
ε(s, ω, ψF) = ε α s−1
2 ⊗ ω, ψF .
Theleftsidewillbetheproductofanon­zeroconstantandanexponentialfunction.
Nowtake Ftobeaglobalfieldand ωtobeanequivalenceclassofrepresentationsoftheWeil
groupof F.Let Abetheadélegroupof Fandlet ψFbeanon­trivialcharacterof A/F.For
eachplace vlet ψvbetherestrictionof ψFto Fv. ψvisnon­trivialforeach vandalmostallthe
functions ε(s, ωv, ψv)areidentically1sothatwecanformtheproduct
v ε(s, ωv, ψv).
Itsvaluewillbeindependentof ψFandwillbewritten ε(s, ω).
Theorem B
Thefunctionalequationofthe L­functionassociatedto ωis
L(s, ω) = ε(s, ω) L(1 − s, ω).
Thistheoremisarathereasyconsequenceofthefirsttheoremtogetherwiththefunctional
equationsoftheHecke L­functions.
Forarchimedeanfieldsthefirsttheoremsaysverylittle.Fornonarchimedeanfieldsitcan
bereformulatedasacollectionofidentitiesforGaussiansums.Fouroftheseidentitieswhich
weformulateasourfourmainlemmasarebasic. Alltheotherscanbededucedfromthem
bysimplegroup­theoreticarguments.UnfortunatelytheonlywayatpresentthatIcanprove
thefourbasicidentitiesisbylongandinvolved,althoughratherelementary,computations.
HoweverTheoremApromisestobeofsuchimportanceforthetheoryofautomorphicforms
andgrouprepresentationsthatwecanhopethateventuallyamoreconceptualproofofitwill
befound.Thefirstandthesecond,whichisthemostdifficult,ofthefourmainlemmasaredue
toDwork[6].Iamextremelygratefultohimnotonlyforsendingmeacopyofthedissertation
ofLakkis[9]inwhichaproofofthesetwolemmasisgivenbutalsofortheinteresthehas
showninthispaper.
F

Chapter1 7
Chapter One.
Weil Groups
TheWeilgroupshavemanyproperties,mostofwhichwillbeusedatsomepointinthe
paper.Itisimpossibletodescribeallofthemwithoutsomeprolixity.Toreducetheprolixity
toaminimumIshallintroducethesegroupsinthelanguageofcategories.
Considerthecollectionofsequences
S : C λ1 µ
−→G −→G
oftopogicalgroupswhere λisahomeomorphismof Cwiththekernelof µand µinducesa
homeomorphismof G/λCwith G.Suppose
S1 : C1
λ1 µ2
−→G1 −→G1
isanothersuchsequence.Twocontinuoushomomorphisms ϕand ψfrom Gto G1whichtake
Cinto C1willbecalledequivalentifthereisacin C1suchthat ψ(g) = cϕ(g)c−1forall g
in G. Swillbethecategorywhoseobjectsarethesequences Sand HomS0 (S, S1)willbethe
collectionoftheseequivalenceclasses. Swillbethecategorywhoseobjectsarethesequences
Sforwhich Cislocallycompactandabelianand Gisfinite;if Sand S1belongto S
HomS(S, S1) = HomS0 (S, S1).
Let P1bethefunctorfrom Stothecategoryoflocallycompactabeliangroupswhichtakes Sto
Candlet P2bethefunctorfrom Stothecategoryoffinitegroupswhichtakes Sto G.Wehave
tointroduceonemorecategory S1,0. Theobjectsof S1willbethesequenceson Sforwhich
Gc ,thecommutatorsubgroupof G,isclosed.Moreovertheelementsof HomS1 (S, S1)willbe
theequivalenceclassesinHomS(S, S1)allofwhosemembersdeterminehomeomorphismsof
Gwithaclosedsubgroupfofiniteindexin G1.
If Sisin S1let V (S)bethetopologicalgroup G/Gc . If Φ ∈ HomS1 (S, S1)let ϕbea
homeomorphismintheclass Φandlet G = ϕ(G). Composingthemap G1/Gc 1 −→ G/Gc
givenbythetransferwiththemap G/G c −→ G/Gcdeterminedbytheinverseof ϕweobtain
amap Φv : V (S1) −→ V (S)whichdependsonlyon Φ. Themap S −→ V (S)becomesa
contravariantfunctorfrom S1tothecategoryoflocallycompactabeliangroups. If Sisthe
sequence
C −→ G −→ G

Chapter1 8
thetransferfromGtoCdeterminesahomomorphismτfromG/G c tothegroupofG­invariant
elementsin C. τwillsometimesberegardedasamapfrom Gtothissubgroup.
Thecategory Ewillconsistofallpairs K/Fwhere Fisaglobalorlocalfieldand Kis
afiniteGaloisextensionof F.Hom(K/F, L/E)willbeacertaincollectionofisomorphisms
of Kwithasubfieldof Lunderwhich Fcorrespondstoasubfieldof E. Ifthefieldsareof
thesametype,thatisallglobaloralllocalwedemandthat Ebefiniteandseparableoverthe
imageof F. If Fisglobaland Eislocalwedemandthat Ebefiniteandseparableoverthe
closureoftheimageof F.Iwanttoturnthemapwhichassociatestoeach K/Fthegroup CK
intoacontravariantfunctorwhichIwilldenoteby C∗ .If ϕ : K/F −→ L/Eand Fand Eare
ofthesametypelet K1betheimageof Kin Landlet ϕC∗bethecompositionof NL/K1with theinverseof ϕ.If Fisglobaland Eislocallet K1betheclosurein Loftheimageof K.As
usual CK1maybeconsideredasubgroupofthegroupofidèlesof K. ϕC∗isthecomposition of NL/K1withtheprojectionofthegroupofidèlesonto CK.
If Kisgivenlet E K bethesubcategoryof Ewhoseobjectsaretheextensionswiththe
largerfieldequalto Kandwhosemapsareequaltotheidentityon K.Let C∗bethefunctor
on E K whichtakes K/Fto CF. If Fisgivenlet EFhaveasobjectstheextensionswiththe
smallerfieldequalto F.Itsmapsaretoequaltheidentityon F.
AWeilgroupisacontravariantfunctor Wfrom Eto Swiththefollowingproperties:
(i) P1 ◦ Wis C ∗ .
(ii) P2 ◦ Wisthefunctor G : L/F −→ G(L/F).
(iii)If ϕ ∈ G(L/F) ⊆ Hom(L/F, L/F)and gisanyelementof W L/F,themiddlegroupof
thesequence W(L/F),whoseimagein G(L/F)is ϕthenthemap h −→ ghg −1 isinthe
class ϕw.
(iv)Therestrictionof Wto E K takesvaluesin S1.Moreover,if K/Fbelongsto E K
τ: W K/F/W c K/F
−→ CF
isahomeomorphism.Finally,if ϕ : K/F −→ K/Eistheidentityon Kand Φ = ϕwthen
thediagram
W K/F/W c K/F
Φv
−→ WK/E/W c K/E
τ ↓ ↓ τ
CF
−→
ϕC ∗
iscommutativeandif ψ : F/F −→ K/Fistheimbedding, ψWis τ.
CE

Chapter1 9
SincethefunctorialpropertiesoftheWeilgrouparenotalldiscussedbyArtinandTate,
weshouldreviewtheirconstructionoftheWeilgrouppointingout,whennecessary,how
thefunctorialpropertiesarise.Thereisassociatedtoeach K/Fafundamentalclass α K/Fin
H 2 (G(K/F), CK).Thegroup W(K/F)isanyextensionof G(K/F)by CKassociatedtothis
element.Wehavetoshow,atleast,thatif ϕ : K/F −→ L/Ethediagram
1 −→ CL −→ W L/E −→ G(L/E) −→ 1
↓ ϕC∗ ↓ ϕG
1 −→ CK −→ WK/F −→ G(K/F) −→ 1
canbecompletedtoacommutativediagrambyinserting ϕ : W L/E −→ W K/F.Themap ϕC ∗
commuteswiththeactionof G(L/E)on CLand CKsothat ϕexistsifandonlyif ϕC ∗(α L/E)is
therestriction ϕ ∗ G (α K/F)of ϕ K/Fto G(L/E).Ifthisisso,thecollectionofequivalenceclasses
towhich ϕmaybelongisaprincipalhomogeneousspaceof H 1 (G(L/E), CK).Inparticular,
ifthisgroupistrivial,asitiswhen ϕGisaninjection,theclassof ϕisuniquelydetermined.
Anexaminationofthedefinitionofthefundamentalclassandshowsthatitiscanonical.In
otherwords,ifϕisanisomorphismofKand LandofFandE,then ϕ ∗ G (α K/F) = ϕ −1 α L/E =
ϕC ∗(α L/E).IfK = LandϕistheidentityonK,therelationϕ ∗ G (α K/F) = α L/E = ϕC ∗(α L/E)
isoneofthebasicpropertiesofthefundamentalclass. Thusinthesetwocases ϕexists
anditsclassisunique. Nowtake Ktobeglobaland Llocal. Supposeatfirstthat Kis
containedin L,thatitsclosureis L,andthat F = K ∩ E. Then,bytheverydefinitionof
α K/F, ϕ ∗ G (α K/F) = ϕC ∗(α L/E).Moregenerally,ifK1istheimageofKinL,andF1theimage
ofFinE,wecanwriteϕasϕ1ϕ2ϕ3whereϕ3 : K/F −→ K1/F1, ϕ2 : K1/F1 −→ K1/K1∩E,
and ϕ1 : K1/K1 ∩E −→ L/E. ϕ3and ϕ2exist.Iftheclosureof K1is Lthen ϕ1andtherefore
ϕ = ϕ3 ϕ2 ϕ1alsoexist.Theclassof ϕisuniquelydetermined.
ArtinandTateshowthat W c K/F isaclosedsubgroupof W K/F andthat τisahomeomorphismof
W K/F/W c K/F and CF. Grantedthis,itiseasytoseethattherestrictionof
Wto ξ K takesvaluesin S1. Supposewehavethecollectionoffieldsinthediagramwith
Land Knormalover Fand Land K ′ normalover F ′ . Let α, β,and νbetheimbeddings
α : L/F −→ L/K, β : L/F ′ −→ L/K ′ , ν : L/F −→ L/F ′ .

Chapter1 10
L
K
F
Wehaveshowntheexistenceof α, β,and ν.Itisclearthat ν β(W L/K ′)iscontainedin α(W L/K).
Thuswehaveanaturalmap
Letusverifythatthediagram
K ′
π : ν β(W L/K ′)/ν β(W c L/K ′) −→ α(W L/K)/α(W c L/K ).
WL/K/W c
L/K ′ −→ ν β(WL/K ′)/ν β(W C π
L/K ′) −→ α(WL/K)/α(W c K/K ) −→ WL/K/W c L/K
↓ τ ↓ τ
Ck ′
F ′
N K ′ /K
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→ Ck (A)
iscommutative.Toseethislet W L/K ′bethedisjointunion
U r i=1 CKhi.
Thenwecanchoose h ′ i , g′ j , 1 ≤ i ≤ r, 1 ≤ j ≤ ssothat W L/Kisthedisjointunion
r
i=1
s
j=1 CK g ′ jh ′ i
and ν β(h ′ i ) = α(hi). Usingthesecosetrepresentativestocomputethetransferoneimmediatelyverifiestheassertion.Weshouldalsoobservethatthetransitivityofthetransferimplies
thecommutativityofthediagram
W K/F/W c K/F
Φv
−→ WK/F ′/W c K/F ′
τ ↓ ↓
CF
−→
ϕC ∗
CF ′

Chapter1 11
if Φistheclassofanimbedding ϕwhere ϕisanimbedding K/F −→ K/F ′ .
Wehavestillnotdefined ϕW forall ϕ. Howeverwehavedefineditwhen ϕisan
isomorphismofthetwolargerfieldsorwhenthesecondlargefieldistheclosureofthefirst.
Moreoverthedefinitionissuchthatthethirdconditionandallpartsofthefourthcondition
exceptthelastaresatisfied.Thelastconditionof(iv)canbemadeadefinitionwithoutviolating
(i)and(ii). Whatwedonowisshowthatthereisoneandonlyonewayofextendingthe
definitionof ϕWtoall ϕwithoutviolatingconditions(i)or(ii)andthefunctorialproperty.
Suppose F ⊆ K ⊆ L, K/Fand L/FareGalois,and ψistheimbedding L/F −→ L/K.
ItisobservedinArtinandTatethatthereisoneandonlyclassofmaps {θ}whichmakethe
followingdiagramcommutative
1 −→ W L/K/W c L/K −→ ψW L/K/ ψW c L/K −→ W L/F/ ψW c L/K −→ W L/F/ ψW L/K −→ 1
τ ↓ ↓ θ ↓
1 −−−−−−−→ CK −−−−−−−−−−−−−−−−−−−−−−−−→ W K/F −−−−−−−→ G(K/F) −−−−−→ 1.
Thehomomorphismontherightisthatdeducedfrom
W L/F/W L/K = G(L/F)/G(L/K) ≃ G(K/F).
Let ϕ, µ,and νbeimbeddings ϕ : K/F −→ L/F, µ : K/K −→ L/K, ν : K/F −→ K/K.
Then ψ ◦ ϕ = µ ◦ ν,sothat ν ◦ µ = ϕ ◦ ψ. Moreover ν ◦ µisthecompositionofthemap
τ : WL/K −→ CKandtheimbeddingof CKin WK/F.Thusthekernelof ϕcontains ψW c L/K
sothat ϕ ◦ ψrestrictedto WL/K/W c L/Kmustbe τandtheonlypossiblechoicefor ϕis,apart
fromequivalence, θ.Toseethatthischoicedoesnotviolatethesecondconditionobservethat
therestrictionof τto CLwillbe NL/Kandthat ψistheidentityon CL.
Denotethemap θ : W L/F −→ W K/Fby θ L/Kandthemap τ : W K/F → CKby τ K/F.
Itisclearthat τ K/F ◦ θ L/Kisthetransferfrom W L/F/W c L/F to ψW L/K/ ψW c L/K followed
bythetransferfrom ψW L/K/ ψW c L/K to ψCL = CF. Bythetransitivityofthetransfer
τ K/F ◦ θ L/K = τ L/F.Itfollowsimmediatelythatif F ⊆ K ⊆ L ⊆ L ′ andallextensionsare
Galoisthemap θ L ′ /Kand θ L/Kθ L ′ /Lareinthesameclass.
Supposethat ϕisanimbedding K/F −→ K ′ /F ′ andchoose Lsothat K ′ ⊆ Land L/F
isGalois.Let ψ : K ′ /F ′ −→ L/F ′ , µ : K/F −→ L/F, ν : L/F −→ L/F ′ beimbeddings.
Then ψ ◦ ϕ = ν ◦ µsothat µ ◦ ν = ϕ ◦ ψ.If α : L/F −→ L/K, β : L/F ′ −→ L/K ′ arethe

Chapter1 12
imbeddingsthenthekernelof ψis ν βW c L/K ′whichiscontainedin αW c L/Kthekernelof µ.
Thusthereisonlyonewaytodefine ϕsothat µ ◦ ν = ϕ ◦ ψ.Thediagram
W L/K ′/W c L/K ′
bβ
−→ W L/F ′/ βW c L/K ′
bψ
−−−−−−−−−−−−−−−−−→ W K ′ /F ′
↓ ν ↓ ϕ
W L/F/ν βW c L/K ′
−→ W L/F/αW c L/K
bµ
−→ W K/F
willbecommutative.Since ψ ◦ β = τL/K ′and µ ◦ α = τK/Fdiagram(A)showsthat ϕhas
therequiredeffecton CK.
Todefine ϕWingeneral,weobservethatevery ϕisthecompositionofisomorphisms,
imbeddingsoffieldsofthesametype,andamap K/F −→ K ′ /F ′ where Kisglobal, K ′ is
local, K ′ istheclosureof K,and F = F ′ ∩ K.Ofcoursetheidentity
(ϕ ◦ ψ)W = ψWϕW
mustbeverified. Iomittheverificationwhichiseasyenough. TheuniquenessoftheWeil
groupsinthesenseofArtinandTateimpliesthatthefunctor Wisuniqueuptoisomorphism.
Thesequence
S(n, C) : GL(n, C) id
−→GL(n, C) −→ 1
belongsto S1.If S : C −→ G −→ Gbelongsto S1then
HomS0 (S, S(n, C))
isthesetofequivalenceclassesof n­dimensionalcomplexrepresentationsof G.Let Ωn(S)be
thesetofall ΦinHomS0 (S, S(n, C))suchthat,foreach ϕ ∈ Φ, ϕ(g)issemi­simpleforall g
in G. Ωn(S)isacontravariantfunctorof Sandsois Ω(S) = ∞ n=1 Ωn(S).Onthecategory
S1,itcanbeturnedintoacovariantfunctor. If ψ : S −→ S1,if Φ ∈ Ω(S),andif ϕ ∈ Φ,
let ψassociateto Φthematrixrepresentationscorrespondingtotheinducedrepresentation
Ind(G1, ψ(G), ϕ ◦ ψ−1 ). Itfollowsfromthetransitivityoftheinductionprocessthat Ωisa
covariantfunctorof S1.
Tobecompleteafurtherobservationmustbemade.
Lemma 1.1Suppose Hisasubgroupoffiniteindexin Gand ρisafinite­dimensional
complexrepresentationof Hsuchthat ρ(L)issemi­simpleforall hin H.If
σ = Ind(G, H, ρ)

Chapter1 13
then σ(g)issemi­simpleforall g.
Hcontainsasubgroup H1whichisnormalandoffiniteindexin G,namely,thegroup
ofelementsactingtriviallyon H\G. Toshowthatanon­singularmatrixissemi­simpleone
hasonlytoshowthatsomepowerofitissemi­simple.Since σ n (g) = σ(g n )and g n belongs
to H1forsome nweneedonlyshowthat σ(g)issemi­simplefor gin H1.Inthatcase σ(g)is
equivalentto r
i=1
⊕ρ(gigg −1
i )if Gisthedisjointunion
r
i=1 Hgi
Suppose L/Fand K/Fbelongto EFand ϕ ∈ HomEF (L/F, K/F). Sincethemapsof
theclass ϕWalltake WK/Fonto WL/Ftheassociatedmap Ω(W(L/F)) −→ Ω(W(K/F))is
injective.Moreoveritisindependentofϕ.IfL1/Fand L2/Fbelongto EFthereisanextension
K/Fandmaps ϕ1 ∈ HomEF (L1/F, K/F), ϕ2 ∈ HomEF (L2/F, K/F). ω1in Ω(W(L1/F))
and ω2in Ω(W(L2/F))havethesameimagein Ω(W(K/F))foronesuch Kifandonlyif
theyhavethesameimageforallsuch K. Ifthisissowesaythat ω1and ω2areequivalent.
Thecollectionofequivalenceclasseswillbedenotedby Ω(F).Itsmembersarereferredtoas
equivalenceclassesofrepresentationsoftheWeilgroupof F.
Let Fbethecategorywhoseobjectsarelocalandglobalfields. If Fand Eareofthe
sametypeHomF(F, E)consistsofallisomorphismsof Fwithasubfieldof Eoverwhich
Eisseparable. If Fisglobaland EislocalHomF(F, E)consistsofallisomorphismsof F
withasubfieldof Eoverwhoseclosure Eisseparable. Ω(F)isclearlyacovariantfunctoron
F.Let Fgℓ,and Flocbethesubcategoriesconsistingoftheglobalandlocalfieldsrespectively.
Suppose Fand Eareofthesametypeand ϕ ∈ HomF(F, E).If ω ∈ Ω(E)choose Ksothat
ωbelongsto Ω(W(K/E)).Wemayassumethatthereisan L/Fandanisomorphism ψfrom
Lonto Kwhichagreeswith ϕon F.Then ψW : W K/E −→ W L/Fisaninjection.Let θbethe
equivalenceclassoftherepresentation
σ = Ind(W L/F, ψW(W K/E), ρ ◦ ψ −1
W )
with ρin ω. Iclaimthat θisindependentof Kanddependsonlyon ωand ϕ. Toseethisit
isenoughtoshowthatif L ⊆ L ′ , L ′ /FisGalois, ψ ′ isanisomorphismfrom L ′ to K ′ which
agreeswith ψon L,and ρ ′ isarepresentationof W K ′ /Fin ωtheclassof
σ ′ = Ind(WL ′ /F, ψ ′ W(WK ′ /E), ρ ′ ′ −1
◦ ψ w )
isalso Θ. Suppose µisamapfrom W K ′ /Eto W K/Eassociatedtotheimbedding K/E −→
K ′ /Eand νisamapfrom W L ′ /F to W L/F associatedtotheimbedding L/F −→ L ′ /F.

Chapter1 14
Wemaysupposethat ψW ◦ µ = ν ◦ ψ ′ W . Thekernelof µis W c K ′ /K if,forsimplicityof
. Moreover
notation, WK ′ /Kisregardedasasubgroupof WK ′ /Eandthatof νis W c L ′ /L
ψ ′ W (W c K ′ /K ) = W c L ′ /L .Take ρ′ = ρ ◦ µ.Then σactsonthespace V offunctions fon WK/F satisfying f(vw) ≡ ρ(ψ −1
w (h))f(w)for vin ψw(W K/E). Let V ′ betheanalogousspaceon
which σ ′ acts.Then
V ′ = {f ◦ ν | f ∈ V }.
Theassertionfollows.Thus Ω(F)isacontravariantfunctoron Fgℓand Floc.
Afterthislaboriousandclumsyintroductionwecansettoworkandprovethetwo
theorems.ThefirststepistoreformulateTheoremA.

Chapter2 15
Chapter Two.
The Main Theorem
Itwillbeconvenientinthisparagraphandatvariouslatertimestoregard WK/Easa
subgroupof W K/Fif F ⊆ E ⊆ K.If F ⊆ E ⊆ L ⊆ Kweshallalsooccasionallytake W L/E
tobe W K/E/W c K/L .
If K/FisfiniteandGalois, P(K/F)willbethesetofextensions E ′ /Ewith F ⊆ E ⊆
E ′ ⊆ Kand P◦(K/F)willbethesetofextensionsin P(K/F)withthelowerfieldequalto F.
Theorem 2.1
Suppose KisaGaloisextensionofthelocalfield Fand ψFisagivennon­trivialadditive
characterof F. Thereisexactlyonefunction λ(E/F, ψF)definedon P◦(K/F)withthe
followingtwoproperties
(i)
(ii)If E1, . . ., Er, E ′ 1 , . . ., E′ s
λ(F/F, ψF) = 1.
arefieldslyingbetween F and K,if χEi , 1 ≤ i ≤ r,isa
quasi­characterof CEi ,if χE ′ j , 1 ≤ j ≤ s,isaquasi­characterof CE ′ j ,andif
isequivalentto
then r
isequalto
s
⊕ r i=1 Ind(Wk/F, WK/Ei , χEi )
⊕ s j=1 Ind(W K/F, W K/E ′ j , χ E ′ j )
i=1 ∆(χEi , ψ Ei/F)λ(Ei/F, ψF)
j=1 ∆(χE ′ j , ψ E ′ j /F)λ(E ′ j
/F, ψF).
Afunctionsatisfyingtheconditionsofthistheoremwillbecalledaλ­function.Itisclear
thatthefunctionλ(E/F, ψF)ofTheoremAwhenrestrictedto P◦(K/F)becomesa λ­function.
ThustheuniquenessinthistheoremimpliesatleastpartoftheuniquenessofTheoremA.To
showhowthistheoremimpliesallofTheoremAwehavetoanticipatesomesimpleresults
whichwillbeprovedinparagraph4.

Chapter2 16
Firstofallaλ­functioncannevertakeonthevalue0. Moreover,if F ⊆ K ⊆ Lthe
λ­functionon P◦(K/F)isjusttherestrictionto P◦(K/F)ofthe λ­functionon P◦(L/F).Thus
λ(E/F, ψF)isdefinedindependentlyof K.Finallyif E ⊆ E ′ ⊆ E ′′
λ(E ′′ /E, ψE) = λ(E ′′ /E ′ , ψ E ′ /E)λ(E ′ /E, ψE) [E′′ :E ′ ] .
WealsohavetouseaformofBrauer’stheorem[4]. If Gisafinitegroupthereare
nilpotentsubgroups N1, . . ., Nm,one­dimensionalrepresentations χ1, . . ., χmof N1, . . ., Nm
respectively,andintegers n1, . . ., nmsuchthatthetrivialrepresentationof Gisequivalentto
⊕ m i=1 niInd(G, Ni, χi).
Themeaningofthiswhensomeofthe niarenegativeisclear
Lemma 2.2
Suppose Fisaglobalorlocalfieldand ρisarepresentationof W K/F. Thereareintermediatefields
E1, . . ., Emsuchthat G(K/Ei)isnilpotentfor 1 ≤ i ≤ m,one­dimensional
representations χEi of W K/Ei ,andintegers n1, . . ., nmsuchthat ρisequivalentto
⊕ m i=1 niInd(WK/F, WK/Ei , χEi ).
Theorem2.1andLemma2.2togetherimplytheuniquenessofTheoremA.Beforeproving
thelemmawemustestablishasimpleandwell­knownfact.
Lemma 2.3
Suppose Hisasubgroupoffiniteindexinthegroup G.Suppose τisarepresentationof
G, σarepresentationof H,and ρtherestrictionof τto H.Then
τ ⊗ Ind(G, H, σ) ≃ Ind(G, H, ρ ⊗ σ).
Let τacton V and σon W.ThenInd(G, H, σ)actson X,thespaceofallfunctions fon
Gwithvaluesin Wsatisfying
f(hg) = σ(h) f(g)
whileInd(G, H, ρ ⊗ σ)actson Y,thespaceofallfunctions fon Gwithvaluesin V ⊗ W
satisfying
f(hg) = (ρ(h) ⊗ σ(h)) f(g).

Chapter2 17
Clearly, V ⊗ Xand Yhavethesamedimension.Themapof V ⊗ Xto Ywhichsends v ⊗ f
tothefunction
f ′ (g) = τ(g)v ⊗ f(g)
is G­invariant. Ifitwerenotanisomorphismtherewouldbeabasis v1, . . ., vnof V and
functions f1, . . ., fnwhicharenotallzerosuchthat
Thisisclearlyimpossible.
Σ n i=1 τ(g)vi ⊗ fi(g) ≡ 0.
ToproveLemma2.2wetakethegroup GofBrauer’stheoremtobe G(K/F).Let Fibethe
fixedfieldof Niandlet ρibethetensorproductof χi,whichwemayregardasarepresentation
of W K/Fi andtherestrictionof ρto W K/Fi .Then
ρ ≃ ρ ⊗ 1 ≃ ⊕ m i=1niInd(WK/Fi , ρi).
Thistogetherwiththetransitivityoftheinductionprocessshowsthatinprovingthelemma
wemaysupposethat G(K/F)isnilpotent.
Weprovethelemma,withthisextracondition,byinductionon [K : F]. Weusethe
symbol ωtodenoteanorbitinthesetofquasi­charactersof CKundertheactionof G(K/F).
Therestrictionof ρto CKisthedirectsumofone­dimensionalrepresentations. If ρactson
V let Vωbethespacespannedbythevectorstransformingunder CKaccordingtoaquasicharacterin
ω. V isthedirectsumofthespaces Vωwhichareeachinvariantunder W K/F.
Forourpurposeswemaysupposethat V = Vωforsome ω.Choose χKinthis ωandlet V0be
thespaceofvectorstransformingunder CKaccordingto χK. Let Ebethefixedfieldofthe
isotrophygroupof χK. V0isinvariantunder W K/E.Let σbetherepresentationof W K/Ein
V0.Itiswell­knownthat
ρ ≃ Ind(W K/F, W K/E, σ).
Toseethisonehasonlytoverifythatthespace Xonwhichtherepresentationontheright
actsand Vhavethesamedimensionandthatthemap
f −→
W K/E\W K/F
ρ(g −1 )f(g)
of Xinto Vwhichisclearly W K/F­invarianthasnokernel.Itiseasyenoughtodothis.
If E = Ftheassertionofthelemmafollowsbyinduction.If E = Fchoose Lcontaining
Fsothat K/Liscyclicofprimedegreeand L/FisGalois.Then ρ(W K/L)isanabeliangroup
and W c K/L iscontainedinthekernelof ρ.Thus ρmayberegardedasarepresentationof W L/F.

Chapter2 18
Theassertionnowfollowsfromtheinductionassumptionandtheconcludingremarksofthe
previousparagraph.
Nowtakealocalfield Eandarepresentation ρof W K/E.Chooseintermediatefields
E1, . . ., Em,one­dimensionalrepresentations χEi of W K/Ei ,andintegers n1, . . ., nmsothat
If ωistheclassof ρset
ρ ≃ ⊕ m i=1 ni Ind(WK/E, WK/Ei , χEi ).
ε(ω, ψE) = m
i=1 {∆(χEi , Ψ Ei/E)λ(Ei/E, ΨE)} ni .
Theorem2.1showsthattherightsideisindependentofthewayinwhich ρiswrittenasasum
ofinducedrepresentations.ThefirstandsecondconditionsofTheoremAareclearlysatisfied.
If ρistherepresentationaboveand σtherepresentation
then
Thusif ω ′ istheclassof σ
while
Ind(W K/F, W K/E, ρ)
σ ≃ ⊕ m i=1 niInd(WK/F, WK/Ei , χEi ).
ε(ω ′ , ψF) = m
ε(ω, ψ E/F) = m
Thethirdpropertyfollowsfromtherelations
and
i=1 {∆(χEi , ψ Ei/F)λ(Ei/F, ψF)} ni
i=1 {∆(χEi , ψ Ei/F) λ(Ei/E, ψ E/F)} ni .
dim ω = Σ m i=1 ni [Ei : E]
λ(Ei/F, ψF) = λ(Ei/E, ψ E/F) λ(E/F, ψF) [Ei:E]

Chapter3 19
Chapter Three.
The Lemmas of Induction
Inthisparagraphweprovetwosimplebutveryusefullemmas.
Lemma 3.1
Suppose KisaGaloisextensionofthelocalfield F. Supposethesubset Aof P(K/F)
hasthefollowingfourproperties.
(i)Forall E,with F ⊆ E ⊆ K, E/E ∈ A.
(ii)If E ′′ /E ′ and E ′ /Ebelongto Asodoes E ′′ /E.
(iii)If L/Ebelongsto P(K/F)and L/Eiscyclicofprimedegreethen L/Ebelongsto A.
(iv)Supposethat L/Ein P(K/F)isaGaloisextension.LetG = G(L/E).SupposeG = H ·C
where H = {1}, H ∩ C = {1},and Cisanon­trivialabeliannormalsubgroupof Gwhichis
containedineverynon­trivialnormalsubgroupof G.If E ′ isthefixedfieldof Handifevery
E ′′ /Ein P◦(L/E)forwhich [E ′′ : E] < [E ′ : E ′ ]isin Asois E ′ /E.Then Aisallof P(K/F).
Itisconvenienttoproveanotherlemmafirst.
Lemma 3.2
Suppose KisaGaloisextensionofthelocalfield Fand F ⊂ = E ⊆ K. Supposethatthe
onlynormalsubfieldof Kcontaining Eis Kitselfandthattherearenofieldsbetween Fand
E.Let G = G(K/F)andlet Ebethefixedfieldof H.Let Cbeaminimalnon­trivialabelian
normalsubgroupof G.Then G = HC, H ∩ C = {1}and Ciscontainedineverynon­trivial
normalsubgroupof G.Inparticularif H = {1}, G = Cisabelianofprimeorder.
Hiscontainedinnosubgroupbesidesitselfand Gcontainsnonormalsubgroupbut {1}.
Thusif Hisnormalitis {1}and Ghasnopropersubgroupsandisconsequentlycyclicof
primeorder.Suppose Hisnotnormal.Since Gissolvableitdoescontainaminimalnon­trivial
abeliannormalsubgroup C. Since Cisnotcontainedin H, H ⊂ = HCand G = HC. Since
H ∩ Cisanormalsubgroupof Gitis {1}.If Disanon­trivialnormalsubgroupof Gwhich
doesnotcontain Cthen D ∩ C = {1}and Discontainedinthecentralizer Zof C. Then
DCisalsoand Zmustmeet Hnon­trivially.But Z ∩ Hisanormalsubgroupof G.Thisisa
contradictionandthelemmaisproved.

Chapter3 20
Thefirstlemmaiscertainlytrueif [K : F] = 1.Suppose [K : F] > 1andthelemmais
validforallpairs [K ′ : F ′ ]with [K ′ : F ′ ] < [K : F].IftheGaloisextension L/Ebelongsto
P(K/F)then A ∩ P(L/E)satisfiestheconditionofthelemmawith Kreplacedby Land F
by E.Thus,byinduction,if [L : E] < [K : F], P(L/E) ⊆ A.Inparticularif E ′ /Eisnotin G
then E = Fandtheonlynormalsubfieldof Kcontaining E ′ is Kitself.If Aisnot P(K/F)
thenamongstallextensionswhicharenotin Gchooseone E/Fforwhich [E : F]isminimal.
Becauseof(ii)therearenofieldsbetween Fand E. Lemma3.2,togetherwith(iii)and(iv),
showthat E/Fisin A.Thisisacontradiction.
ThereisavariantofLemma3.1whichweshallhaveoccasiontouse.
Lemma 3.3
Suppose KisaGaloisextensionofthelocalfield F.Supposethesubset Aof P◦(K/F)
hasthefollowingproperties.
(i) F/F ∈ A.
(ii)If L/Fisnormaland L ⊂ = Kthen P◦(L/F) ⊆ A.
(iii)If F ⊂ E ⊆ E ′ ⊆ Kand E/Fbelongto Gthen E ′ /Fbelongto A.
(iv)If L/Fin P◦(K/F)iscyclicofprimedegreethen L/F ∈ A.
(v) Supposethat L/Fin P◦(K/F)isGaloisand G = G(L/F). Suppose G = HCwhere
H = {1}, H ∩ C = {1},and Cisanon­trivialabeliannormalsubgroupof Gwhichis
containedineverynon­trivialnormalsubgroup.If Eisthefixedfieldof Handifevery E ′ /F
in P◦(L/F)forwhich [E ′ : F] < [E : F]isin Asois E/F.
Then Ais P◦(K/F).
AgainifAisnot P◦(K/F)thereisan E/FnotinAforwhich[E : F]isminimal.Certainly
[E : F] > 1.By(ii)and(iii), Eiscontainedinnopropernormalsubfieldof Kandthereareno
fieldsbetween Eand F.Lemma3.2togetherwith(iv)and(v)leadtothecontradictionthat
E/Fisin A.

Chapter4 21
Chapter Four.
The Lemma of Uniqueness
Suppose K/F isafiniteGaloisextensionofthelocalfield F and ψF isanon­trivial
additivecharacterof F.Afunction E/F −→ λ(E/F, ψF)on P◦(K/F)willbecalledaweak
λ­functionifthefollowingtwoconditionsaresatisfied.
(i) λ(F/F, ΨF) = 1.
(ii)If E1, . . ., Er, E ′ 1 , . . ., E′ s arefieldslyingbetween Fand K,if µi, 1 ≤ i ≤ r,isaonedimensionalrepresentationof
G(K/Ei),if νj, 1 ≤ j ≤ s,isaone­dimensionalrepresentationof
G(K/E ′ j ),andif
isequivalentto
r
s
then r
isequalto
s
i=1
j=1
Ind(G(K/F), G(K/Ei), µi)
Ind(G(K/F), G(K/Ej), νj)
i=1 ∆(χEi , ψ Ei/F)λ(Ei/F, ψF)
j=1 ∆(χE ′ j ψ E ′ j /F)λ(E ′ j/F, ψF)
if χEi isthecharacterof CEi correspondingto µiand χE ′ j isthecharacterof CE ′ j correspondingto
νj.
Supposingthataweak λ­functionisgivenon P◦(K/F),weshallestablishsomeofits
properties.
Lemma 4.1
(i)If L/Fin P◦(K/F)isnormaltherestrictionof λ(·, ψF)to P◦(L/F)isaweak λ­function.
(ii)If E/Fbelongsto P◦(K/F)and λ(E/F, ψF) = 0thefunctionon P◦(K/E)definedby
isaweak λ­function.
λ(E ′ /E, ψ E/F) = λ(E ′ /F, ψF) λ(E/F, ψF) −[E′ :E]

Chapter4 22
Anyone­dimensionalrepresentation µof G(L/E)maybeinflatedtoaone­dimensional
representation,againcalled µ,of G(K/E)and
isjusttheinflationto G(K/F)of
Ind(G(K/F), G(K/E), µ)
Ind(G(L/F), G(L/E), µ).
Thefirstpartofthelemmafollowsimmediatelyfromthisobservation.
Asforthesecondpart,therelation
λ(E/E, ψ E/F) = 1
isclear.Iffields Ei, 1 ≤ i ≤ r, E ′ j , 1 ≤ j ≤ s,lyingbetween Eand Kandrepresentations
µiand νjaregivenasprescribedandif
isequivalentto
r
s
then r
isequivalentto
i=1 Ind(G(K/E), G(K/Ei), µi) = ρ
j=1 Ind(G(K/E), G(K/E′ j), νj) = σ
s
sothat r
isequalto
s
Since ρand σhavethesamedimension
sothat r
i=1 Ind(G(K/F), G(K/Ei, µi)
j=1 Ind(G(K/F), G(K/E′ j
), νj)
i=1 ∆(χEi , ψ Ei/F) λ(E1/F, ψF) (A)
j=1 ∆(χE ′ j , ψ E ′ j /F) λ(E ′ j /F, ψF). (B)
Σ r i=1 [Ei : E] = Σ s j=1 [E′ j : E]
i=1 λ(E/F, ΨF) [Ei:E] = s
j=1
λ(E/F, ψF) [E′
j :F] .

Chapter4 23
Dividing(A)bytheleftsideofthisequationand(B)bytherightandobservingthattheresults
areequalweobtaintherelationneededtoprovethelemma.
If K/Fisabelian S(K/F)willbethesetofcharactersof CFwhichare1on N K/FCK.
Lemma 4.2
If K/Fisabelian
λ(K/F, ΨF) =
µF ∈S(K/F) ∆(µF, ψF).
µF determinesaone­dimensionalrepresentationof G(K/F)whichwealsodenoteby
µF.Thelemmaisanimmediateconsequenceoftheequivalenceof
and
Lemma 4.3
Ind(G(K/F), G(K/K), 1)
µF ∈S(K/F)
Ind(G(K/F), G(K/F), µF).
Suppose K/Fisnormaland G = G(K/F).Suppose G = HCwhere H ∩ C = {1}and
Cisanon­trivialabeliannormalsubgroup.Let Ebethefixedfieldof Hand Lthatof C.Let
Tbeasetofrepresentativesoftheorbitsof S(K/L)undertheactionof G.If µ ∈ Tlet Bµbe
theisotropygroupof µandlet Bµ = G(K/Lµ).Then [Lµ : F] < [E : F]and
λ(E/F, ψF) =
µ∈T ∆(µ′ , ψ Lµ/F) λ(Lµ/F, ψF).
Here G(K/Lµ) = G(K/L) · (G(K/Lµ) ∩ G(K/E))and µ ′ isthecharacterof CLµ associated
tothecharacterof G(K/Lµ) : g −→ µ(g1)if
g = g1g2, g1 ∈ G(K/L), g2 ∈ G(K/Lµ) ∩ G(K/E),
Wemayaswelldenotethegivencharacterof G(K/Lµ)by µ ′ also.Toprovethelemma
weshowthat
Ind(G(K/F), G(K/E), 1) = σ
isequivalentto
µ∈T Ind(G(K/F), G(K/Lµ), µ ′ ).

Chapter4 24
Since Thasatleasttwoelementsitwillfollowthat
isgreaterthan
[E : F] = dimInd(G(K/F), G(K/E), 1)
[Lµ : F] = dimInd(G(K/F), G(K/Lµ), µ ′ ).
Therepresentation σactsonthespaceoffunctionson H\G.If ν ∈ S(K/L),thatis,isa
characterof C,let ψν(hc) = ν(c)if h ∈ H, c ∈ C.Theset
{ψν | ν ∈ S(K/L)}
isabasisforthefunctionson H\G.If µ ∈ Tlet Sµbeitsorbit;then
Vµ = Σν∈Sµ Cψν
isinvariantandirreducibleunder G.Moreover,if gbelongsto G(K/Lµ)
Since
σ(g)ψµ = µ ′ (g)ψµ.
dim Vµ = [G(K/F), G(K/Lµ)]
theFrobeniusreciprocitytheoremimpliesthattherestrictionof σto Vµisequivalentto
Ind(G(K/F), G(K/Lµ), µ ′ ).
Lemma4.2isofcourseaspecialcaseofLemma4.3.
Lemma 4.4
λ(E/F, ΨF)isdifferentfrom0forall E/Fin P◦(K/F).
Thelemmaisclearif [K : F] = 1.Weproveitbyinductionon [K : F].Let Gbethesetof
E/Fin P◦(K/F)forwhich λ(E/F, ψF) = 0.WemayapplyLemma3.3.Thefirstcondition
ofthatlemmaisclearlysatisfied.Thesecondfollowsfromtheinductionassumptionandthe
firstpartofLemma4.1;thethirdfromtheinductionassumptionandthesecondpartofLemma
4.1. ThefourthandfifthfollowfromLemmas4.2and4.3respectively. Weofcourseusethe
factthat ∆(χE, ΨE),whichisbasicallyaGaussiansumwhen Eisnon­archimedean,isnever
zero.

Chapter4 25
Forevery E ′ /Ein P(K/F)wecandefine λ(E ′ /E, ψ E/F)tobe
Lemma 4.5
λ(E ′ /F, ψF) λ(E/F, ψF) −[E′ :E] .
If E ′′ /E ′ and E ′ /Ebelongto P(K/F)then
Indeed
whichequals
λ(E ′′ /E, ψ E/F) = λ(E ′′ /E ′ , ψ E ′ /F)λ(E ′ /E, ψ E/F) E′′ :E ′ ] .
λ(E ′′ /E, ψ E/F) = λ(E ′′ /F, ψF) λ(E/F, ψF) −[E′′ :E]
λ(E ′′ /F, ψF) λ(E ′ /F, ψF) −[E′′ :E ′ ] λ(E ′ /F, ψF) [E′′ :E ′ ] λ(E/F, ψF) −[E′′ :E]
andthisinturnequals
Lemma 4.6
λ(E ′′ /E ′ , ψ E ′ /F) λ(E ′ /E, ψ E/F) [E′′ :E ′ ] .
If λ1(·, ΨF)and λ2(·, ΨF)aretwoweak λ­functionson P◦(K/F)then
forall E ′ /Ein P(K/F).
λ1(E ′ /E, ψ E/F) = λ2(E ′ /E, ψ E/F)
WeapplyLemma3.1tothecollectionGofallpairsE ′ /Ein P(K/F)forwhichtheequality
isvalid.Thefirstconditionofthatlemmaisclearlysatisfied.Thesecondisaconsequenceof
thepreviouslemma.ThethirdandfourthareconsequencesofLemmas4.2and4.3respectively.
Sinceaλ­functionisalsoaweak λ­functiontheuniquenessofTheorem2.1isnowproved.

Chapter5 26
Chapter Five.
A Property of λ-Functions
Itfollowsimmediatelyfromthedefinitionthatif ψ ′ E (x) = ψE(βx)then
∆(χE, ψ ′ E ) = χE(β)∆(χE, ψE).
Associatedtoanyequivalenceclass ωofrepresentationsoftheWeilgroupofthefield Fisa
one­dimensionalrepresentationor,whatisthesame,aquasi­characterof CF. Itisdenoted
detωandisobtainedbytakingthedeterminantofanyrepresentationin ω.Suppose ρisinthe
class ωand ρisarepresentationof W K/F.Tofindthevalueofthequasi­character detωat β
choose win W K/Fsothat τ K/Fw = β.Thencalculate det(ρ(w))whichequals detω(β).
If F ⊆ E ⊆ Kthemap τ = τ K/F canbeeffectedintwostages. Wefirsttransfer
W K/F/W c K/F into W K/E/W c K/E ;thenwetransfer W K/E/W c K/E into CK. If W K/F isthe
disjointunion
r
i=1 W K/Ewi
andif wiw = ui(w)wj(i)thenthetransferof win W K/E/W c K/E isthecosettowhich w′ =
r
i=1 ui(w)belongs.
Suppose σisarepresentationof W K/Eand
ρ = Ind(W K/F, W K/E, σ).
ρactsonacertainspace V offunctionson W K/Fandif Viisthecollectionoffunctionsin V
whichvanishoutsideof W K/Ewithen
V = r
i=1 Vi.
Wedecomposethematrixof ρ(w)intocorrespondingblocks ρji(w). ρji(w)is0unless j = j(i)
when ρji(w) = σ(ui, (w)). Thismakesitclearthatif ι E/F istherepresentationof W K/F
inducedfromthetrivialrepresentationof W K/E
or,if θistheclassof σ,
det(ρ(w)) = det(ι E/F(w)) dim σ det(σ(w ′ ))
detω(β) = {det ι E/F(β)} dim θ {det θ(β)}.

Chapter5 27
Lemma 5.1
Suppose F isalocalfieldand E/F −→ λ(E/F, ψF)and ω −→ ε(ω, ψ E/F)satisfy
theconditionsofTheoremAforthecharacter ψF. Let ψ ′ F (x) = ψF(βx)with βin CF. If
E/F −→ λ(E/F, ψ ′ F )and ω −→ ε(ω, ψ′ E/F )satisfytheconditionsofTheoremAfor ψ′ F then
and
λ(E/F, ψ ′ F ) = det ι E/F(β) λ(E/F, ψF)
ε(ω, ψ ′ E/F ) = det ω(β) ε(ω, ψ E/F).
Becauseoftheuniquenessallonehastodoisverifythattheexpressionsontheright
satisfytheconditionsofthetheoremforthecharacter ψ ′ F .Thiscannowbedoneimmediately.

Chapter6 28
Chapter Six.
A Filtration of the Weil Group
InthisparagraphIwanttoreformulatevariousfactsfoundinSerre’sbook[12]asassertionsaboutafiltrationoftheWeilgroup.Althoughsomeofthelemmastofollowwillbeused
toprovethefourmainlemmas,theintroductionofthefiltrationitselfisnotreallynecessary.
Itservesmerelytouniteinaformwhichiseasilyrememberedtheseparatelemmasofwhich
wewillactuallybeinneed.
Let KbeafiniteGaloisextensionofthenon­archimedeanlocalfield F andlet G =
G(K/F).Let OFbetheringofintegersin Fandlet pFbethemaximalidealof OF.If i ≥ −1
isanintegerlet Gibethesubgroupof Gconsistingofthoseelementswhichacttriviallyon
.If u ≥ −1isarealnumberand iisthesmallestintegergreaterthanorequalto uset
OF/p i+1
F
Gu = Gi.Finallyif u ≥ −1set
ϕ K/F(u) =
u
0
1
[G0 : Gt] dt.
Theintegrandisnotdefinedat­1butthatisofnoconsequence. ϕ K/Fisclearlyapiecewise
linear,continuous,andincreasingmapof [−1, ∞)ontoitself.Theinversefunction* ψ K/Fwill
havethesameproperties.
WetakefromSerre’sbookthefollowinglemma.
Lemma 6.1
If F ⊆ L ⊆ Kand L/Fisnormalthen ϕ K/F = ϕ L/F ◦ ϕ K/Land ψ K/F = ψ K/L ◦ ψ L/F.
Thecircledenotescompositionnotmultiplication.Thislemmaallowsustodefine ϕ E/F
and ψ E/Fforanyfiniteseparableextension E/FbychoosingaGaloisextension Lof Fwhich
contains Eandsetting
ϕ E/F = ϕ L/F ◦ ψ L/E
ψ E/F = ϕ L/E ◦ ψ L/F
*Inthischapter ψ K/Fdoesnotappearasanadditivecharacter. Nonetheless,thereisa
regrettableconflictofnotation.

Chapter6 29
becauseif L ′ isanothersuchextensionwecanchooseaGaloisextension Kcontainingboth L
and L ′ and
ϕ L/F ◦ ψ L/E = ϕ L/F ◦ ϕ K/L ◦ ψ K/L ◦ ψ L/E = ϕ K/F ◦ ψ K/E = ϕ L ′ /F ◦ ψ L ′ /E
ϕ L/E ◦ ψ L/F = ϕ L/E ◦ ϕ K/L ◦ ψ K/L ◦ ψ L/F = ϕ K/E ◦ ψ K/F = ϕ L ′ /E ◦ ψ L ′ /F.
Ofcourse ψ E/Fistheinverseof ϕ E/F.
Lemma 6.2
If E ⊆ E ′ ⊆ E ′′ and E ′′ /Eisfiniteandseparable, ϕ E ′′ /E = ϕ E ′ /E ◦ ϕ E ′′ /E ′ and
ψ E ′′ /E = ψ E ′′ /E ′ ◦ ψ E ′ /E.
Eachoftheserelationscanbeobtainedfromtheotherbytakinginverses;weverifythe
second
ψ E ′′ /E ′ ◦ ψ E ′ /E = ϕ L/E ′′ ◦ ψ L/E ′ ◦ ϕ L/E ′ ◦ ψ L/E = ϕ L/E ′′ ◦ ψ L/E = ψ E ′′ /E.
Itwillbenecessaryforustoknowthevaluesofthesefunctionsinafewspecialcases.
Lemma 6.3
(i)If K/FisGaloisandunramified ψ K|F(u) ≡ u.
(ii)If K/Fiscyclicofprimedegree ℓandif G = Gtwhile Gt+1 = {1}where tisanonnegativeintegerthen
ψ K/F(u) = u u ≤ t
= t + ℓ(u − t) u ≥ t.
Theseassertionsfollowimmediatelyfromthedefinitions.
Lemma 6.4
SupposeK/FisGaloisandG = G(K/F)isaproductHCwhere H = {1}, H∩C = {1},
and Cisanon­trivialabeliannormalsubgroupof Gwhichiscontainedineverynon­trivial
normalsubgroup.
(i)If K/Fistamelyramifiedsothat G1 = {1}then G0 = Cisacyclicgroupofprimeorder
ℓand [G : G0] = [H : 1]divides ℓ −1.If Eisthefixedfieldof H, ψ E/F(u) = ufor u ≤ 0
and ψ E/F(u) = ℓufor u ≥ 0.

Chapter6 30
(ii)If K/Fiswildlyramifiedthereisaninteger t ≥ 1suchthat C = G1 = . . . = Gtwhile
Gt+1 = {1}. [G0 : G1]divides [G1 : 1] − 1andeveryelementof Chasorder por1.If E
isthefixedfieldof Hand Lthatof C
while
ψ L/F(u) = u u ≤ 0
= [G0 : G1]u u ≥ 0
t
ψE/F(u) = u u ≤
[G0 : G1]
t
=
[G0 : G1] + [G1
t
t
: 1] u − u ≥
[G0 : G1] [G0 : G1]
Weobservedinthethirdparagraphthat Cmustbeitsowncentralizer. G0cannotbe {1}.
Thus C ⊆ G0.Incase(i) G0isabelianandthus G0 = C.Inbothcasesif ℓisaprimedividing
theorderof Cthesetofelementsin Coforder ℓor1isanon­trivialnormalsubgroupof Gand
thus Citself.Incase(i) Ciscyclicandthusofprimeorder ℓ.Moreover, Hwhichisisomorphic
to G/G0isabelianand,if h ∈ H, {c ∈ C|hc = ch}isanormalsubgroupof Gandhence {1}
or C.If h = 1itmustbe1.Consequentlyeachorbitof Hin C − {1}has [H : 1]elementsand
[H : 1]divides ℓ − 1.
Incase(ii) G1isanon­trivialnormalsubgroupandhencecontains C. G1and Careboth
p­groups.Thecentralizerof G1in Cisnottrivial.Asanormalsubgroupof Gitcontains C.
Thereforeitis Cand G1iscontainedin Cwhichisitsowncentralizer.Sinceeach G1, i ≥ 1,
isanormalsubgroupof G,itiseither Cor {1}. Thusthereisaninteger t ≥ 1suchthat
G1 = Gt = Cwhile Gt+1 = {1}. If i ≥ 0isanintegerlet Ui Kbethegroupofunitsof OK
whicharecongruentto1modulo p i+1
K ;let U(−1)
K = CK,andif U ≥ −1isanyrealnumber
let ibethesmallestintegergreaterthanorequalto uandset Uu K = Ui K . If θtisthemapof
Gt/Gt+1into pt K /pt+1
K and θ0themapof G0/G1into U0 K /U1 KintroducedinSerrethen,for g
in G0and hin C,
θt(ghg −1 ) = θ0(g) t Θt(h).
If h = 1, ghg −1 = hifandonlyif θ0(g) t = 1andthen gbelongstothecentralizerof C,that
isto G1.Again C − {1}isbrokenupintoorbits,eachwith [G0 : G1]elementsand [G0 : G1]
divides [Gi : 1] − 1.Observethat tmustbeprimeto [G0 : G1].

Chapter6 31
Itfollowsimmediatelyfromthedefinitionsthat Hu = H ∩ Gu.Incase(i) H0willbe {1}
and ϕ K/E(u)willbeidentically u.Thus ψ E/F = ψ K/Fand,fromthedefinition, ψ K/F(u) = u
if u ≤ 0while ψ K/F(u) = [G0 : 1]uif u ≥ 0.Incase(ii), ϕ K/E(u) = uif u ≤ 0and
ϕ K/E(u) =
if u ≥ 0while ψ K/F(u) = uif u ≤ 0and
Thelemmafollows.
Lemma 6.5
u
[H0 : 1] =
u
[G0 : G1]
t
ψK/F(u) = [G0 : G1]u 0 ≤ u ≤
[G0 : G1]
t
t
= t + [G0 : 1] u −
≤ u.
[G0 : G1] [G0 : G1]
Foreveryseparableextension E ′ /Ethefunction ψ E ′ /Eisconvex,andif uisaninteger
sois ψ E ′ /E(u).
Allwehavetodoisprovethattheassertionistrueforall E ′ /Ein P(K/F)if Fisan
arbitrarynon­archimedeanlocalfieldand KanarbitraryGaloisextensionofit. Todothis
wejustcombinethepreviousthreelemmaswithLemma3.1. Wearegoingtousethesame
methodtoprovethefollowinglemma.
Lemma 6.6
Foreveryseparableextension E ′ /Eandany u ≥ −1
NE ′ /E (U ψE ′ /E (u)
E ′ ) ⊆ U u E .
Wehavetoverifythattheset Gofall E ′ /Ein P(K/F)forwhichtheassertionistruesatisfies
theconditionsofLemma3.1.Thereisnoproblemwiththefirsttwo.
Lemma 6.7
E ′ /Ebelongsto Gifandonlyifforeveryinteger n ≥ −1
N E ′ /E
U ψ E ′ /E (n)
E ′
⊆ U n E

Chapter6 32
and
N E ′ /E
U ψ E ′ /E (n)+1
E ′
⊆ U n+1
E .
If E ′ /Ebelongsto Gchoose ε > 0sothat ψ E ′ /E(n + ε) = ψ E ′ /E(n) + 1.Thesmallest
integergreaterthanorequalto n + εisatleast n + 1so
N E ′ /E
U ψ E ′ /E (n)+1
E ′
⊆ U n+ε
E
⊆ Un+1
E .
Converselysupposetheconditionsofthelemmaaresatisfiedand n < u < n + 1. Since
ψ E ′ /E(n)isanintegerthesmallestintegergreaterthanorequalto ψ E ′ /E(u)isatleast
ψ E ′ /E(n) + 1.Thus
Lemma 6.8
and
N E ′ /E
U ψ E ′ /E (u)
E ′
⊆ NE ′ /E
If L/EisGaloisthen,foreveryinteger n ≥ −1,
N L/E
N L/E
U ψ L/E(n)
L
U ψ L/E(n)+1
L
U ψ E ′ /E (n)+1
E ′
⊆ U n E
⊆ U n+1
E .
⊆ U n+1
E = Uu E.
Theassertionisclearif n = −1.Aproofforthecase n ≥ 0and L/Etotallyramifiedisgiven
inSerre’sbook.Sincethatproofworksequallywellforall L/Ewetakethelemmaasproved.
Lemma 6.9
Suppose K/FisGaloisand G = G(K/F).Suppose G = HCwhere H = {1}, H ∩ C =
{1},andCisanon­trivialabeliannormalsubgroupofGwhichiscontainedineverynon­trivial
normalsubgroupof G.If Eisthefixedfieldof H
forall u ≥ −1.
N E/F
U ψ E/F(u)
E
⊆ U u F

Chapter6 33
Let Lbethefixedfieldof C.If K/Fistamelyramified K/Eand L/Fareunramifiedso
that ψE/F = ψK/Land Uv E = CE ∩ Uv K , Uv F = CF ∩ Uv Lforevery v ≥ −1.If αbelongsto CE,
thendelete NK/Lα = NE/Fα.Since K/LisGalois
N E/F
U ψ E/F(u)
E
If K/Fisnottamelyramified
if n ≥ 1and 0 ≤ m < [G0 : G1].Thus
if −1 ≤ v ≤ 0and
if v ≥ 0or,morebriefly,
forall v ≥ −1.Inthesamewaywefind
forall v ≥ −1.Since K/Lisnormal
N E/F
U ψ E/F(u)
E
⊆ CF ∩ NK/L (U ψK/L(u) L ) ⊆ CF ∩ U u L = U u F.
p n E = E ∩ p [G0:G1]n−m
k
U v E = GE ∩ U v K
U v E = CE ∩ U [G0:G1]v
K
U v E = CE ∩ U ψ K/E(v)
K
U v F = CF ∩ U ψ L/F(v)
L
⊆ CF ∩ NK/L Lemma6.6nowfollowsimmediately.
Lemma 6.10
U ψ K/F(u)
K
⊆ CF ∩ U ψ L/F(u)
L
= U u F.
(a)Suppose K/F isGaloisand G = G(K/F). Suppose t ≥ −1isanintegersuchthat
G = Gt = Gt+1.Then ψ K/F(u) = ufor u ≤ t.Moreover N K/Fdefinesanisomorphism
of CK/U t K with CF/U t F andif −1 ≤ u ≤ ttheinverseimageof Uu F /Ut F is Uu K /Ut K .
Howeverthemapof CK/U t+1
K
into CF/U t+1
F definedbythenormisnotsurjective.
(b)Suppose K/FisGaloisand G = G(K/F).Suppose s ≥ −1isanintegerand G = Gs.If
F ⊆ E ⊆ K, ψ E/F(u) = ufor u ≤ sand N E/Fdefinesanisomorphismof CE/U s E and
CF/U s F .If −1 ≤ u ≤ stheinverseimageof Uu F /Us F is Uu E /Us E .
If t = −1theassertionsofpart(a)areclear.If t ≥ 0, K/Fistotallyramified.Therelation
ψ K/F(u) = ufor u ≤ tisanimmediateconsequenceofthedefinition. Sincetheextension

Chapter6 34
istotallyramified N K/Fdefinesanisomorphismof U −1
K /U0 K
−1
and UF /U0 F . Itfollowsfrom
PropositionV.9ofSerre’sbookthatif 0 ≤ n < ttheassociatedmap Un K /Un+1
K −→ Un F /Un+1
F
isanisomorphismbutthatthemap U t K /Ut+1
K −→ Ut F /Ut+1
F hasanon­trivialcokernel.The
firstpartofthelemmaisanimmediateconsequenceofthesefacts.
Toprovepart(b)wefirstobservethatthereisat ≥ ssuchthat G = Gt = Gt+1. It
thenfollowsfrompart(a)thatthemap NK/F determinesanisomorphismof CK/U s Kand CF/U s F underwhich Uu K /Us Kand Uu F /Us F correspondif −1 ≤ u ≤ s. Let Ebethefixed
fieldof H. Wehave Hs = H ∩ Gs = H,sothat NK/E determinesonisomorphismof
CK/U s Kand CE/U s Eunderwhich Uu K /Us Kand Uu E /Us Ecorrespondif −1 ≤ u ≤ s.Moreover
if u ≤ s, ψK/F(u) = ψK/E(u) = usothat ψE/F(u) = u. Part(b)followsfromthese
observationsandtherelation NK/F = NE/FNK/E. If Eisanynon­archimedeanlocalfieldand u > −1
If αbelongsto CEset
U u E = ∩v

Chapter6 35
If σ = 1bothsidesareinfiniteandtheassertionisclear.If σ = 1let Ebethefixedfield
of σ. If σ(w) = σ, wbelongsto W K/Eand v K/F(w) = ϕ E/F(v K/E(w)). Also v K/F(σ) =
ϕ E/F(v K/E(σ)).Consequentlyitissufficienttoprovethelemmawhen F = E.Theset
S = {τ K/F(w) | σ(w) = σ}
isacosetof N K/F(CK)in CFand CFisgeneratedby N K/F(CK)togetherwithanyelement
of S.Moreover s = max{vF(β) | β ∈ S}isthelargestintegersuchthat S ∩ U s F isnotempty.
Since G = Gt = Gt+1theprecedinglemmashowsthat s = t = ϕ K/F(t).
Lemma 6.12
(a)Forall wand w1in W K/F, v K/F(w) = v K/F(w −1 )and v K/F(w1w w −1
1 ) = v K/F(w).
(b)If F ⊆ E ⊆ Kand wbelongto W K/Ethen
(c)Forall win WK/F, τK/F(w) ⊂ U vK/F(w) F .
v K/F(w) = ϕ E/F(v K/E(w)).
Thefirsttwoassertionsfollowimmediatelyfromthedefinitionsandthebasicproperties
oftheWeilgroup. Iproveonlythethird. Letmefirstobservethatif F ⊆ E ⊆ Kand
w ⊂ W K/E,then
τ K/F(w) = N E/F(τ K/E(w)).
Toseethis,chooseasetofrepresentatives w1, . . ., wrforthecosetsof CKin W K/Eandthen
asetofrepresentatives v1, . . ., vsforthecosetsof W K/Ein W K/F.Let wiw = aiw j(i)with ai
in CK;then
τ K/E(w) = r
i=1 ai.
However vjwiw = vjaiv −1
j vjw j(i)sothat
τ K/F(w) = s
j=1
r
i=1
vjaiv −1
j
s
=
j=1 vjτK/E(w)v −1
j = NE/F(τK/E(w)). Inparticular,ifEisthefixedfieldofσ(w), τ K/E(w)iscontained U ψ E/F(v K/F(w))
E
iscontainedin
Lemma 6.13
N E/F (U ψ E/F(v K/F(w))
E
) ⊆ U vK/F(w) F .
and τ K/F(w)

Chapter6 36
If uand vbelongto W K/Fthen
v K/F(uv) ≥ min{v K/F(u), v K/F(v)}.
Let σ = σ(u)andlet τ = σ(v). Becauseofthesecondassertionofthepreviouslemma
wemayassumethat σand τgenerate G(K/F).Let Ebethefixedfieldof στ.If
t = {min(ψ K/F(v K/F(σ)), ψ K/F(v K/F(τ))}
and G = G(K/F)then G = Gt = Gt+1.AccordingtoLemma6.11,if
s = min{v K/F(u), v K/F(v)},
then t ≥ ψ K/F(s)which,byLemma6.10,isthereforeequalto s.Since τ K/F(uv) =
τ K/F(u)τ K/F(v), τ K/F(uv)liesin U s F .Ontheotherhand
τ K/F(uv) = N E/F(τ K/E(uv))
sothat,byLemma6.10again, τ K/E(uv)belongsto U s E and
v K/F(uv) ≥ ϕ E/F(s) = s.
Thusthesets W x K/F , x ≥ −1,giveafiltrationof W K/Fbyacollectionofnormalsubgroups.Thenextsequenceoflemmasshowthatthefiltrationisquiteanalogoustotheupper
filtrationoftheGaloisgroups.
Lemma 6.14
Foreach x ≥ −1themap τ K/F,L/Ftakes G x K/F into Gx L/F .
If wbelongsto W K/Flet ¯w = τ K/F,L/F(w).Wemustshowthat
v L/F( ¯w) ≥ v K/F(w).
Let σ = σ(w)andlet ¯σ = σ( ¯w).If Eisthefixedfieldof σthen E = E ∩ Listhefixedfieldof
¯σ.Since
v L/F( ¯w) = ϕ E/F (v L/E ( ¯w))
and
v K/F(w) = ϕ E/F (v K/E (w))

Chapter6 37
wemaysuppose E = F.Since τ K/F(w) = τ L/F( ¯w),Lemma6.12impliesthat τ L/F( ¯w)liesin
U vK/F(w) F .Thus
v L/F( ¯w) = vF(τ L/F( ¯w)) ≥ v K/F(w).
Ofcourse W F/Fis CFand,if v ≥ −1, W v F/F = Uv F .
Lemma 6.15
Foreach v ≥ −1, τ K/Fmaps W v K/F onto Uv F .
Since v1 ≤ v2implies W v2
K/F
⊆ W v1
K/F
itisenoughtoprovethelemmawhen v = nis
aninteger. Thelemmaisclearif [K : F] = 1;soweproceedbyinductionon [K : F]. If
[K : F] > 1,chooseanintermediatenormalextension Lsothat [L : F] = ℓisaprime. Let
G = G(L/F).Lemma6.12impliesthat
W ψ L/F(v)
K/L
= W K/L ∩ W v K/F .
Thereisaninteger t ≥ −1suchthat G = Gtand Gt+1 = {1}. ItisshowninChapterVof
Serre’sbookthatif n > t
NL/F
= U n F.
Byinduction
τ K/L
U ψ L/F(n)
L
W ψ L/F(n)
K/L
Since τ K/F(w) = N L/F(τ K/L(w))if wisin W K/L,
τ K/F(W n K/F ) = Un F
= U ψL/F(n) L .
if n > t.Suppose σgenerates G.Then V L/F(σ) = t. ByHerbrand’stheoremthereisaσin
G(K/F)with v K/F(σ) = twhoserestrictionto Lis σ.ByLemma6.11thereisawin W K/F
suchthat σ = σ(w)and v K/F(w) = t.Then τ K/F(w)liesin U t F butnotin N L/F(CL).From
Serre’sbookagain
U t F : NL/F U ψ
L/F(t)
F = ℓ
sothat Ut Fisgeneratedby τK/F(w)and NL/F(U ψL/F(t) L )andhenceiscontainedintheimage
of W t K/F . TocompletetheproofofthelemmawehaveonlytoobservethatLemma6.10
impliesthat
U n F = U t F N L/F
U ψ L/F(n)
L

Chapter6 38
if n ≤ t.
Lemma 6.16
Suppose F ⊆ L ⊆ Kand L/Fand K/FareGalois.Then,foreach v ≥ −1, τ K/F,L/F
maps W v K/F onto W v L/F .
If [L : F] = 1thisisjustthepreviouslemmasoweproceedbyinductionon [L : F].We
havetoshowthatif wbelongsto W L/F thereisawin W K/F suchthat w = τ K/F,L/F(w)
and v K/F(w) ≥ v L/F(w). Let σ = σ(w)andlet E bethefixedfieldof σ. If E = F
then,bytheinductionassumption,thereisawin W K/Esuchthat τ K/E,L/E (w) = wand
v K/E (w) ≥ v L/E (w).ByLemma6.12, v K/F(w) ≥ v L/F(w).Moreover,wemayassumethat
τ K/E,L/E istherestrictionto W K/E of τ K/F,L/F.
Suppose E = F.Then v L/F(w) = vF(τ L/F(w)).Choose w1inW K/Fsothat τ K/F(w1) =
τ L/F(w)andv K/F(w1) ≥ vF(τ L/F(w)).Letw1 = τ K/F,L/F(w1)andsetu = w −1
1 w.Certainly
v L/F(u) ≥ v L/F(w).Moreover, τ L/F(u) = 1.Let F ⊆ L1 ⊆ Lwhere L1/Fiscyclicofprime
order.If udoesnotbelongto W L/L1 thegroup CFisgeneratedby N L1/F(CL1 )and τ L/F(u),
whichisimpossiblesince τ L/F(u) = 1.Thus ubelongsto W L/L1 and,asobserved,thereisa
uin W K/L1 suchthat τ K/F,L/F(u) = τ K/L1,L/L1 (u) = u.Then τ K/F,L/F(uw1) = w.

Chapter7 39
Chapter Seven.
Consequences of Stickelberger’s Result
DavenportandHasse[5]haveshownthatStickelberger’sarithmeticcharacterizationof
GaussiansumsoverafinitefieldcanbeusedtoestablishidentitiesbetweentheseGaussian
sums. AfterreviewingStickelberger’sresultweshallprovetheidentitiesofDavenportand
Hassetogetherwithsomemorecomplicatedidentities. HoweverfortheproofofStickelberger’sresultitself,IrefertoDavenportandHasse.
If Z = e 2πi
p and αbelongsto GF(p)themeaningof Z α isclear.If κisanyfinitefieldand
Sistheabsolutetraceof κlet ψ 0 κ bethecharacterof κdefinedby ψ0 κ (α) = ZS(α) .If χκisany
characterof κ ∗ and ψκisanynon­trivialadditivecharacterof κwewilltaketheGaussiansum
τ(χκ, ψκ)tobe
Weabbreviate τ(χκ, ψ0 κ )to τ(χκ).
−
χ−1
α∈κ∗ κ (α)ψκ(α).
Let knbethefieldobtainedbyadjoiningthe n th rootsofunitytotherationalnumbers.
If ̟ = Z − 1thenin kptheideal (p)equals (̟ p−1 ).If q = p f and κhas qelementsthenin
kq−1theideal (p)isaproduct pp ′ . . .wheretheresiduefieldsof pp ′ , . . .areisomorphicto κ.
In k p(q−1)
(p) = (p p ′ . . .) p−1
with P = (p, ̟), P ′ = (p ′ , ̟),andsoon.Theresiduefieldsof P, P ′ , . . .arealsoisomorphic
to κ.Chooseoneoftheseprimeideals,say P.Onceanisomorphismoftheresiduefieldwith
κischosenthemapofthe (q − 1) th rootsofunitytotheresiduefielddefinesanisomorphism
of κ ∗ andthegroupof (q − 1) th rootsofunity.Then χκcanberegardedasacharacterofthe
lattergroup.Choose α = α(χκ, P)with 0 ≤ α < q − 1sothat χκ(ζ) = ζ α forall (q − 1) th
rootsofunity.Write
α = α0 + α1p + . . . + αf−1p f−1
Notallofthe αicanbeequalto p − 1.Set
σ(α) = α0 + α1 + . . . + αf−1
γ(α = α0!α1! . . .αf−1!
0 ≤ αi < p.
ThefollowinglemmaisStickelberger’sarithmeticalcharacterizationof τ(χκ).

Chapter7 40
Lemma 7.1
(a) τ(χκ)liesin k p(q−1)andisanalgebraicinteger.
(b)If χκ = 1then τ(χκ) = 1butif χκ = 1theabsolutevalueof τ(χκ)andallitsconjugates
is √ q.
(c)Everyprimedivisorof τ(χκ)in k p(q−1)isadivisorof p.
(d)If βisanon­zeroelementoftheprimefieldthentheautomorphism Z −→ Z β of k p(q−1)
over kq−1sends τ(χκ)to χκ(β) τ(χκ).
(e)If Pisaprimedivisorof pin k p(q−1)and α = α(χκ, p)themultiplicativecongruence
isvalid.
τ(χκ) ≡ ̟σ (α)
γ(α)
(mod ∗ P)
(f)Suppose ℓisaprimedividing q − 1and χκ = χ ′ κχ ′′ κ wheretheorderof χ′ κ isapowerof
ℓandthatof χ ′′ κ isprimeto ℓ.If ℓa istheexactpowerof ℓdividing q − 1and λ = ζ0 − 1
where ζ0isaprimitive ℓ a ­throotofunitythen
τ(χκ) ≡ τ(χ ′′ κ) (mod λ).
BeforestatingtheidentitiesforGaussiansumswhichareimpliedbythislemma,Ishall
proveafewelementarylemmas.
Lemma 7.2
Suppose 0 ≤ α < p f − 1and
α = α0 + α1p + . . . + αf−1p f−1 0 ≤ αi < p.
Supposealsothat 0 ≤ j0 < j1 < . . . < jr = fandset
If σ = r−1
s=0 βsand γ = r−1
s=0 βs!then
βs = αjs + αjs+1 p + . . . + αj s+1 −1p js+1−js−1 .
̟ σ
γ
= ̟σ(α)
γ(α)
(mod ∗ P).

Chapter7 41
Firstofall,Iremarkonceandforallthatif n ≥ 1, 0 < u ≤ p n − 1,and v = u(modp n )
then v ≡ u(mod ∗ p).Thusif 0 ≤ u ≤ p n − 1and v ≥ 0
Alsoif v ≥ 0
and,byinduction,
(u + vp n )! = (vp n )! u
p n
w=1 (w + vpn ) ≡ u!(vp n )! (mod ∗ p).
w=1 (vpn + w) ≡ (v + 1) p n ! (mod ∗ p)
(vp n )! ≡ v!(p n !) v
(mod ∗ p).
Inparticular p (n+1) ! ≡ p!(p n !) p ≡ (−p) (p n !) p .Applyinductiontoobtain
Fromtherelations
and
Weconcludethat
p = p−1
p n ! ≡ (−p) pn −1
p−1 (mod ∗ p).
i=1 (1 − Zi ) = (−̟)
p−1 p−1
i=1
Z i − 1
Z − 1
Z i − 1
Z − 1 = 1 + Z + . . . + Zi−1 ≡ i (mod ∗ p).
p = (p − 1)!(−̟) p−1 ≡ −̟ p−1
(mod ∗ p).
Thelemmaitselfisclearif r = fsoweproceedbyinductiondownwardfrom f. Suppose
r < f, js+2 − js = t > 1,andthelemmaisvalidforthesequence j0, j1, . . ., js+1 −
1, js+1, . . .jr.Toproveitforthegivensequencewehaveonlytoshowthatif
and y = αjs+1−1then
But
and
x = αjs + αjs+1 p + . . . + αj s+1 −2 p t−2
̟ x+y
x!y!
≡ ̟x+ypt−1
(x + yp t−1 )!
(mod ∗ p).
̟ y(pt−1 −1) ≡ (−p) y pt−1 −1
p−1 (mod ∗ p)
(x + yp t−1 )! = x!y!(p t−1 !) y ≡ x!y!(−p) yt−1 −1
p−1 (mod) ∗ p).

Chapter7 42
Lemma 7.3
Suppose β0, . . ., βr−1and γ0, . . ., γr−1arenon­negativeintegersallofwhicharelessthan
orequalto q − 1.Supposethat q = p f isaprimepowerand
r−1
i=0 (βi + γi)q i < 2(q r − 1).
Supposealsothat δi, 0 ≤ i ≤ r − 1,aregivensuchthat 0 ≤ δi ≤ q − 1,
r−1
i=0 δiq i < q r − 1
and r−1
i=0 (βi + γi)q i = r−1
i
δiq
i=0
(modq r − 1).
(a) If r−1
i=0 (βi + γi)q i < q r−1 andif νisthenumberof k, 1 ≤ k ≤ r,forwhich
k−1
i=0 (βi + γi) ≥ q k then
r−1
i=0 (βi + γi − δi) = ν(q − 1).
(b) If r−1
i=0 (βi + γi)q i ≥ q r − 1andif νisthenumberof k, 1 ≤ k ≤ r,forwhich
1 = k−1
i=0 (βi + γi)q i ≥ q k then
r−1
i=0 (βi + γi − δi) = ν(q − 1).
Observeimmediatelythatif 1 ≤ k ≤ r,then 0 ≤ βk−1 + γk−1 ≤ 2(q − 1)and
k−1
i=0 (βi + γi)q i ≤ 2(q − 1) k−1
i=0 qi = 2(q k − 1).
If r = 1then β0 + γ0 = δ0 + ε(q − 1)with εequalto0or1. If ε = 0weareincase(a)and
ν = 0while β0 + γ0 − δ0 = 0.If ε = 1weareincase(b);here ν = 1and β0 + γ0 − δ0 = q − 1.
Supposethenthat r ≥ 2andthatif β ′ 0 , . . ., β′ r−1 , γ′ 0 , . . ., γ′ r−2 , δ′ 0 , . . ., δ′ r−2 ,and ν′ aregiven
asinthelemma(with rreplacedby r − 1)then
r−2
i=0 (β′ i + γ ′ i − δ ′ i) = ν ′ (q − 1).

Chapter7 43
Weestablishpart(a)first.Inthiscase
r−1
i=0 (βi + γi)q i = r−1
i=0
and r−2
i=0 (βi + γi)q i = r−2
i=0 δiq i + εq r−1
with ε = δr−1 −βr−1 −γr−1.If εwerenegativetheleftsideoftheequationwouldbenegative;
if εweregreaterthan1theleftsidewouldbegreaterthan 2(qr−1 − 1).Sinceneitherofthese
possibilitiesoccur εis0or1.
Supposefirstthat ε = 0.If r−2 i=0 δiqi < qr−1 −1choose β ′ i = βi, γ ′ i = γi, 0 ≤ i ≤ r −2.
Then δ ′ i = δi, 0 ≤ i ≤ r − 2,and ν ′ = ν.Theassertionofthelemmafollowsinthiscase.If
r−2 1=0 δiqi = qr−1 − 1then δi = q − 1, 0 ≤ i ≤ r − 2.Then β0 + γ0 ≡ q − 1 (modq)and,asa
consequence, β0 + γ0 = q − 1.Weshowbyinductionthat βi + γi = q − 1, 0 ≤ i ≤ r − 2.If
thisissofor i < jthen
r−2
i=j (βi + γi)q i = r−2
i=j (q − 1)qi .
Hence βj + γj ≡ q − 1 (modq)and βj + γj = q − 1.Itfollowsimmediatelythat ν = 0and
r−1
i=0 (βi + γi − δi) = 0.
Nowsupposethat ε = 1.If
r−2
δiq i
i=0 (βi + γi)q i = 2(q r−1 − 1)
then βi = γi = q − 1, 0 ≤ i ≤ r − 2, δ0 = q − 2,and δi = q − 1, 1 ≤ i ≤ r − 2. Thus
ν = r − 1and
r−1
Supposethenthat
Fromtherelation
i=0 (βi + γi − δi) = 1 + (r − 1) (q − 1) − 1 = (r − 1) (q − 1).
r−2
r−2
i=0 (βi + γi)q i < 2(q r−1 − 1).
i=0 (βi + γi)q i = r−2
i=0 δiq i + 1 + (q r−1 − 1).
Weconcludethat r−2
i=0 δiq i < q r−1 − 1.Thenforsome m,with 0 ≤ m ≤ r − 2, δm < q − 1.
Wechoosetheminimalvaluefor m.
r−2
i=0 (βi + γi)q i = (δm + 1)q m + r−2
i=m+1 δiq i + (q r−1 − 1).

Chapter7 44
Thusif β ′ i = βi, γ ′ i = γi, 0 ≤ i ≤ r − 2,then δ ′ i = 0, i < m, δ′ m = δm + 1,and δ ′ i = δi, m <
i ≤ r − 2.Arguingbycongruencesasbeforeweseethat βi + γi = q − 1for i < m.Thus
k−1
i=0 (βi + γi)q i = q k − 1
for k ≤ m. However βm + γm = q − 1andthus βm + γm + 1isprimeto q. Moreoverif
r − 1 ≥ k > m
1 + k−1
i=1 (βi + γi)q i ≡ (βm + γm + 1)q m (modq m+1 ).
Thusitisgreaterthanorequalto q k ifandonlyifitisgreaterthanorequalto q k +1.Itfollows
that ν ′ = ν + mandthat
r−2
i=0 (βi + γi − δi) = −m(q − 1) + r−2
i=0 (β′ i + γ′ i − δ′ i ) = ν(q − 1) + 1.
Since βr−1 + γr−1 − δr−1 = −1theassertionofthelemmafollows.
and
Nowletustreatpart(b).Inthiscase
r−1
i=0 (βi + γi)q i = r−1
1 + r−2
i=0 (βi + γi)q i = r−2
i=0 δiq i + (q r − 1)
i=0 δiq i + εq r−1
with ε = δr−1 − βr−1 − γr−1 + q.Again εis0or1.If βi = γi = q − 1for 0 ≤ i ≤ r − 2then
ε = 1and δi = q − 1for 0 ≤ i ≤ r − 2.Also ν = rand
r−1
i=0 (βi + γi − δi) = (r − 1) (q − 1) + βr−1 + γr−1 − δr−1 = r(q − 1).
Havingtakencareofthiscase,wesupposethat
r−2
i=0 (βi + γi)q i < 2(q r−1 − 1).
Firsttake ε = 0.If δ0 = 0then 1+β0+γ0 ≡ 0 (modq)and β0+γ0 = q−1.Thusoneofthemis
lessthan q −1.Bysymmetrywemaysupposeitis β0.Let β ′ 0 = β0 +1, β ′ i = βi, 1 ≤ i ≤ r −2,
and γ ′ i = γi, 0 ≤ i ≤ r − 2.Since δ0 = 0
r−2
i=0 δiq i ≤ q r−1 − q < q r−1 − 1

Chapter7 45
and δ ′ i = δi, 0 ≤ i ≤ r − 1.Also ν = ν ′ + 1sothat
r−1
i=0 (βi + γi − δi) = r−2
i=0 (β′ i + γ′ i − δ′ i ) − 1 + q = ν(q − 1)
asrequired.If δ0 > 0take β ′ i = βiand γ ′ i = γi, 0 ≤ i ≤ r − 2.Then δ ′ 0 = δ0 − 1, δ ′ i = δi, 1 ≤
i ≤ r − 2.Alsoif k ≤ r − 1
k−1
i=0 (βi + γi)q i ≡ δ0 − 1 = −1(modq)
andtheleft­handsideisgreaterthanorequalto q k ifandonlyifitisgreaterthanorequalto
q k − 1.Itfollowsthat ν = ν ′ + 1.Consequently
r−1
i=0 (βi + γi − δi) = r−2
i=0 (β′ i + γ′ i − δ′ i ) − 1 + q = ν(q − 1).
If ε = 1take γ ′ i = γiand β ′ i = βi, 0 ≤ i ≤ r − 2.Then δ ′ i = δi, 0 ≤ i ≤ r − 2,and ν = ν ′ + 1
sothat r−1
1=0 (βi + γi − δi) = ν ′ (q − 1) + (βr−1 + γr−1 − δr−1) = ν(q − 1).
Lemma 7.4
Suppose βiand γiaretwoperiodicsequencesofintegerswithperiod r.Thatis βi+r = βi
and γi+r = γiforall iin Z.Suppose 0 ≤ βi ≤ q − 1, 0 ≤ γi ≤ q − 1forall iandthatnoneof
thenumbers
εk = r−1
i=0 (βi+k + γi+k)q i
isdivisibleby q r − 1.Let
r−1
i=0 (βi + γi)q i ≡ r−1
i=0 δiq i (mod q r − 1)
with 0 ≤ δi ≤ q − 1and r−1
i=0 δiq i < q r − 1.If µisthenumberof εk, 1 ≤ k ≤ r,whichare
greaterthanorequalto q r − 1then
r−1
i=0 (βi + γi − δi) = µ(q − 1).
Since ε0 ≤ 2(q r − 1)andisnotdivisibleby q r − 1itislessthan 2(q r − 1).Thusallwe
needdoisshowthatthe µofthislemmaisequaltothe νoftheprecedinglemma. Observe
firstofallthat εj ≥ q r − 1ifandonlyif εj ≥ q r .

Chapter7 46
Suppose ε0 < q r .If 1 ≤ k < r
r−1
sothat r−1
Thus,if εk ≥ q r ,
q r ≤ r−1
i=k (βi + γi)q i < q r
i=k (βi + γi)q i−k < q r−k .
i=r−k (βi+k + γi+k) q i + r−k−1
i=0
k−1
r−k
< q
i=0 (βi + γi)q i + q r−k
and k−1
Converselyif 1 ≤ k < rand k−1
then
r−1
Thus µ = νinthiscase.
i=0 (βi + γi)q i ≥ q k .
i=0 (βi + γi)q i ≥ q k ,
i=0 (βi+k + γi+k)q i ≥ r−1
Nowsuppose ε0 ≥ q r .If 1 ≤ k < r
r−1
i=k (βi + γi)q i ≥ q r − k−1
If k−1
then
r−1
i=0 (βi+k + γi+k)q i ≥ r−1
(βi+k + γi+k)q i
i=r−k (βi+k + γi+k)q i
k−1
r−k
= q
i=0 (βi + γi)q i ≥ q r .
i=0 (βi + γi)q i ≥ q r − 2(q k − 1).
i=0 (βi + γi)q i ≥ q k − 1
i=r−k (βi+k + γi+k)q i + r−k−1
i=0
(βi+k + γi+k)q i
= q r−k k−1
i=0 (βi + γi)q i + q −k r−1
i=k (βi + γi)q i
≥ q r−k (q k − 1) + q r−k − 2 + 2q −k .

Chapter7 47
Thus εk ≥ q r − 1andhence εk ≥ q r .Converselyif εk ≥ q r ,
Thus
andagain µ = ν.
Lemma 7.5
k−1
r−k
q
i=0 (βi + γi)q i = r−1
i=r−k (βi+k + γi+k)q i
k−1
≥ q r − r−k−1
i=0
≥ q r − 2(q r−k − 1)
= q r − 2q r−k + 2.
i=0 (βi + γi)q i ≥ q k − 1
(βi+k + γi+k)q i
Supposeq = p f isaprimepower, ℓisapositiveinteger,and(ℓ, q) = 1.Letℓm ≡ 1 (modq)
andif xisanyintegerlet ϕ(x),with 0 ≤ ϕ(x) < q,betheremainderof xupondivisionby q.
If 0 ≤ β < qandif ψ(x) = ϕ(x)!
ℓβ
β!
ℓ−1
k=0
ψ((β − k)m)
ψ(−km) ≡ 1(mod∗ p).
If ℓ = ℓ1 + uqwith ℓ1 > 0and u ≥ 0then ℓ β ≡ ℓ β
1 (mod∗ p).Moreover
ℓ−1
k=0
and ℓ−1
ψ((β − k)m)
ψ(−km) =
ℓ1−1
k=0
k=ℓ1
ψ(−km) = { q−1
ψ((β − k)m) ℓ−1
ψ(−km) k=ℓ1
j=0 j!}u = ℓ−1
k=ℓ1
ψ((β − k)m).
ψ((β − k)m)
ψ(−km)
Thusitisenoughtoprovethelemmawith ℓreplacedby ℓ1.Inotherwordswemaysuppose
that 0 < ℓ < q. Thecase ℓ = 1istrivialandweexcludeitfromthefollowingdiscussion.
Finallywesupposethat 0 < m < q.

Chapter7 48
Let q − 1 = rℓ + swith 0 < rand 0 ≤ s < ℓ.Arrangetheintegersfrom0to q − 1into
thefollowingarray.
0 1 2 . . . . . . . l − 1
l l + 1 l + 2 . . . . . . . 2l − 1
. . .
. . .
. . β − l + 1 .
. . . β .
. . .
. . .
(r − 1)l (r − 1)l + 1 q − l . rl − 1
rl rl + 1 . . . . . rl + s
Since ℓdoesnotdivide q, rℓ + s = q − 1doesnotlieinthelastcolumn.Also q − ℓliesinthe
columnfollowingthatinwhich rℓ + slies.
Wereplaceeachnumber jintheabovearrayby ϕ(jm). Theresultingarray,whichis
writtenoutbelow,hassomespecialfeatureswhichmustbeexplained. Thefirstcolumnis
explainedbytheobservation xℓm ≡ x(modq). Theotherentries,apartfromthoseatthe
footofeachcolumn,areexplainedbytheobservationthat,when 1 ≤ jand xℓ + jliesinthe
firstarray, ϕ(x + mj) > rwhile 0 < ϕ(mj) + x < q + rsothat ϕ(x + mj) = ϕ(mj) + x.
Thepositionof q − 1isexplainedbytherelation m(q − ℓ) ≡ −mℓ ≡ q − 1(mod q). The
otherentriesatthefeetofthecolumnsareexplainedbytheobservationthatif 1 ≤ j ≤ ℓ − 1
then ϕ(jm) > r ≥ 1while m(q − k) = m(q − ℓ) + m(ℓ − k) ≡ ϕ((ℓ − k)m) − 1(modq)if
1 ≤ k ≤ ℓ − 1.
0 m . . . . . . . ϕ((l − 1)m)
l m + 1 . . . . . . . ϕ((l − 1)m) + 1
. . .
. . .
. . ϕ((β − l + 1)m) .
. . . ϕ(βm) .
. . .
. . .
r − 1 m + r − 1 q − 1 . ϕ((l − 2 − s)m) − 1
r m + r . . . . ϕ((l − 1)m) − 1
Supposefirstofallthat β < ℓ − 1. Thenthenumbers ϕ((β − k)m), 0 ≤ k ≤ ℓ − 1
constitutethefirst β + 1togetherwiththelast ℓ − β − 1numbersinthearray.(Theorderof

Chapter7 49
thenumbersinthearrayistheorderinwhichtheyappearwhenthearrayisreadasthough
itwereaprintedpage.)Thenumbers ϕ(−km), 1 ≤ k ≤ ℓ − 1,arethelast ℓ − 1numbersof
thearray,thatis,thenumbersafter q − 1.Cancellingintheproductofthelemmathetermsin
numeratoranddenominatorcorrespondingtothelast ℓ − β − 1termsofthearray,weobtain
asrequired.
β
j=1
ϕ(jm)!
(ϕ(jm) − 1)!
= β
j=1 ϕ(jm) ≡ mβ β! (mod ∗ p)
Nowtake β ≥ ℓ − 1.Thenthenumbers β, β − 1, . . ., β − (ℓ − 1)occurasindicatedinthe
firstarray.Inparticularthereisexactlyoneineachcolumn.Thenumeratorintheproductof
thelemmaistheproductofthefactorialsofthecorrespondingelementsofthesecondarray.
Thedenominatoristheproductofthefactorialsoftheelementsappearingafter q − 1. As
indicated tistheelementlyingabove q − 1.Thus tislargerthananyelementappearingina
columnotherthanthatof t.Theproductofthelemmais t!timestheproductofthefactorials
oftheotherelementsonthebrokenlinedividedbythefactorialsoftheelementsatthefootof
thecolumninwhichtheylie.Thusitequals t!dividedbytheproductofalltheelementsbelow
thebrokenlineexceptthosewhichliedirectlybelow t. But t!istheproductofallnumbers
inthesecondarrayexceptthosewhichliebelow t. Thusthequotientistheproductofall
numberswhichlieaboveoronthebrokenline,thatis,
asrequired.
Lemma 7.6
β
j=1
ϕ(jm) ≡ β
j=1 jm = mβ β! (mod ∗ p)
Supposethat q = p f isaprimepower,that ℓisapositiveintegerdividing q − 1,that
0 ≤ α1 < q − 1,that (ℓ, α1) = 1,andthat
α2 = α1
ℓ
q ℓ − 1
q − 1 .
Then α2isanintegerand 0 ≤ α2 < q − 1.Moreoverif
with 0 ≤ γi ≤ q − 1for 0 ≤ i ≤ ℓ − 1then
α2 = γ0 + γ1q + . . . + γℓ−1 q ℓ−1
ℓ−1
j=1
j=0 γj = α1 + ℓ−1
j ·
q − 1
ℓ

Chapter7 50
and
ℓ α1
ℓ−1
j=1
j=0 γj! ≡ α1! ℓ−1
Certainly 0 ≤ α2 < q ℓ − 1;moreover
jq − 1!
ℓ
(mod ∗ p).
q ℓ − 1
q − 1 = 1 + q + . . . + qℓ−1 ≡ ℓ (mod ℓ)
sothat α2isaninteger.Let α1 = mℓ + kwith m ≥ 0and 0 ≤ k < ℓandfor 0 ≤ j < ℓlet
(ℓ − 1 − j)k = ij + δjℓ
with δj ≥ 0and 0 ≤ ij < ℓ.Clearly iℓ−1 = δℓ−1 = 0.Also (ℓ − 1)k = ℓ − k + ℓ(k − 1)sothat
i0 = ℓ−kand δ0 = k −1.If j ≥ 1then (δj−1 −δj)ℓ = k+(ij −ij−1).Since −ℓ < ij −ij−1 < ℓ
and 0 < k < ℓtheright­handsideisgreaterthan −ℓandlessthan 2ℓsothat δj−1 − δjis0or
1.Ifitis1then k + ij ≥ ℓand ij ≥ ℓ − k.Ifitis0then ij = ij−1 − k < ℓ − k.Recallingthat
i0 = ℓ − kweseethat
and
S = {j|1 ≤ j ≤ ℓ − 1 and δj−1 − δj = 1} = {j|0 ≤ j < ℓ and ij > ℓ − k}.
Weshallprovethat
γj = m + ij
γ0 = m + i0
(q − 1)
ℓ
(q − 1)
ℓ
+ k − δ0
+ δj−1 − δj 1 ≤ j < ℓ.
Since (k, ℓ) = 1thenumbers ijaredistinctanditwillfollowimmediatelythat
ℓ−1
j=0
j=0 γj = (mℓ + k) + ℓ−1
Moreoverwewillhave
ℓ−1
j=0 γj!
ℓ−1
= m + j
j=0
Recallthat k − δ0 = 1.Dividingthefirsttermby
j · q − 1
ℓ = α1 + ℓ−1
j=1
q − 1
j · .
ℓ
(q − 1) q − 1
! m+i0
ℓ
ℓ +k−δ0
m + 1 + ij
j∈S
ℓ−1
j=0
j ·
(q − 1)
!
ℓ
(q − 1)
.
ℓ

Chapter7 51
weobtain
ℓ−1
j=0
Theproductofthelasttwotermsis
ℓ−1
j=ℓ−k
m
n=1
n + j
m + 1 + j
(q − 1)
.
ℓ
(q − 1)
.
ℓ
If 1 ≤ n ≤ mand 0 ≤ j ≤ ℓ − 1then nℓ − j < α1 < qsothattheproductof ℓ mk andthefirst
ofthesetwoexpressionsismultiplicativelycongruentto
ℓ−1
j=0
m
n=1
(nℓ − j) = (mℓ)!
Moreover,if ℓ − k ≤ j ≤ ℓ − 1,then 0 ≤ (m + 1)ℓ − j ≤ (m + 1)ℓ − (ℓ − k) = α1 < qandthe
secondoftheseexpressionsuponmultiplicationby ℓ k becomesmultiplicativelycongruentto
ℓ−1
j=ℓ−k
((m + 1)ℓ − j) = k
j=1
Therelationstogetherimplythesecondidentityofthelemma.
(mℓ + j).
Toverifythatthe γj, 0 ≤ j < ℓ,havetheformasserted,westartwiththerelation
α2 = qℓ − 1
q − 1
Thesecondtermisequalto
Thus
ℓ−1
j=0
α2 =
j−1
i=0 qi
· mℓ + k
ℓ
q − 1
ℓ
= ℓ−1
j=0 qj m + ℓ−1
j=0
q j k
ℓ .
k + k = k + ℓ−2
j=0 (ℓ − 1 − j)qj q − 1
· · k.
ℓ
q − 1
m + (ℓ − 1)k · + k +
ℓ ℓ−1
q − 1
m + (ℓ − 1 − j)k · q
j=1
ℓ
j
q − 1
= m + i0 · + k − δ0 +
ℓ ℓ−1
q − 1
m + ij ·
j=1 ℓ + δj−1
− δj .
Moreover m < q−1
ℓ sothat
0 ≤ m + i0 ·
q − 1
ℓ + k − δ0
q − 1
< ℓ · + 1 = q
ℓ

Chapter7 52
and
0 ≤ m + ij ·
Therequiredrelationsfollowimmediately.
q − 1
ℓ + δj−1
q − 1
− δj < ℓ · + 1 = q.
ℓ
NowwecanstateandprovethepromisedidentitiesforGaussiansums.Eachofthesewill
amounttoanassertionthatacertainnumberin k p(q−1)is1.Toprovethiswewillshowfirst
thatthenumberisinvariantunderallautomorphismsof k p(q−1)over k (q−1)andthusliesin
kq−1.Theonlyprimeidealsoccurringinthefactorizationofthenumber,whichisnot a priori
analgebraicinteger,intoprimeidealswillbedivisorsof p.Weshowthateveryconjugateof
thenumberhasabsolutevalue1andthatitismultiplicativelycongruentto1moduloevery
divisorof p.Itwillfollowthatitisarootofunityin kq−1andhencea(q − 1)throotofunityif
qisoddanda2(q − 1)throotofunityif qiseven.If qisoddthemultiplicativecongruences
implythatthenumberis1.If qiseventheyimplythatthenumberis ±1.Toshowthatitis
actually1somesupplementarydiscussionwillbenecessary.
Stickelberger’sresultisdirectlyapplicableonlytothenormalizedGaussiansum τ(χκ).
Weshallhavetousetheobviousrelation τ(χκ, Ψκ) = χκ(β)τ(χκ)if ψκ(α) = ψ 0 κ(βα).If κis
anextensionof λand ψλisgiven,weset
ψ κ/λ(α) = ψλ(S κ/λ(α))
for αin κ.If χλisgiven χ κ/λisthecharacterdefinedby
Lemma 7.7
Set
χ κ/λ(α) = χλ(N κ/λ(α)).
If κisafiniteextensionofthefinitefieldand χλand ψλaregiventhen
τ(χ κ/λ, ψ κ/λ) = {τ(χλ, ψλ)} [κ:λ] .
Since χ κ/λ(β) = χλ(β) [κ:λ] itwillbeenoughtoshowthat
τ(χ κ/λ) = {τ(χλ)} [κ:λ] .
µ =
[χ:λ]
{τ(χλ)}
.
τ(χκ/λ)

Chapter7 53
Let λhave q = p ℓ elementsandlet κhave p k = q f .ItfollowsimmediatelyfromLemma7.1
thattheabsolutevalueof µandallitsconjugatesis1,thatitliesin k p(q f −1),thatitisinvariant
underallautomorphismsof k p(q f −1)over k q f −1,andthatitsonlyprimefactorsaredivisorsof
p.Themapping β −→ Nκ/λβsends βto β qf −1
q−1.Thusif α = α(χλ, p)and Pdivides p
ApplyingLemmas7.1and7.2,weseethat
and
Consequently
α(χ κ/λ, p) = qf − 1
q − 1 α = α + αq + . . . + αqf−1 .
τ(χλ) ≡ ̟α
α!
τ(χ κ/λ) ≡ ̟fα
(α!) f
(mod ∗ P)
µ ≡ 1 (mod ∗ P).
(mod ∗ P).
Thus µ = 1if qisoddand µ = ±1if qiseven.If χλ = 1then χ κ/λ = 1and,frompart(b)of
Lemma7.1, µis1.If q = 2then χλ = 1.Supposethen qisevenandgreaterthan2.If χλis
notidentically1chooseaprime rdividingtheorderof χλ.Set χλ = χ ′ λ χ′′ λ wheretheorderof
χ ′ λ isapowerof randtheorderof χ′′ λ isprimeto r.Theanalogousdecompositionof χ κ/λis
χ ′ κ/λ χ′′ κ/λ .Ofcourse χλand χ κ/λhavethesameorder.Define µ ′ and µ ′′ intheobviousway.
Accordingtopart(f)ofLemma7.1
µ ≡ µ ′′
(modr).
Since rdoesnotdivide2thisimpliesthat µ = µ ′′ . Thusonecanshowbyinductiononthe
numberofprimesdividingtheorderof χλthat µ = 1.
Lemma 7.8
Suppose λisafinitefieldwith qelements, κisafiniteextensionof λ,and [κ : λ] = f.
Suppose ℓisaprimeandtheorderof qmodulo ℓis f.Let Tbeasetofrepresentativesforthe
orbitsofthenon­trivialcharactersof κ ∗ oforder ℓundertheactionof G(κ/λ)andlet χλbea
characterof λ ∗ .If ψλisanynon­trivialcharacterof λ
χλ(ℓ ℓ )τ(χ ℓ λ , ψλ)
µκ∈T τ(µκ, ψ κ/λ) = τ(χλ, ψλ)
µκ∈T τ(χ κ/λ µκ, ψ κ/λ).

Chapter7 54
Sincetheisotropygroupofeachpointin Tistrivial
χ ℓ λ(β)
µκ∈T µκ(β) = χλ(β)
andwemaycontentourselveswithshowingthat
µκ∈T χ κ/λ(β) µκ(β)
χλ(ℓ ℓ ) τ(χ ℓ
λ )
µκ∈T τ(µκ) = τ(χλ)
µκ∈T τ(χκ/λ µκ).
Ofcourse χλ(ℓℓ )isthevalueof χλattheelementoftheprimefieldcorrespondingto ℓℓ .Let
µbethequotientoftherightsidebytheleft.Thecharactersof κ∗oforder ℓarethecharacters
µ k κ , 0 ≤ k < ℓ,definedby
α(µ k κ , P) = k · qf − 1
.
ℓ
Sincetheorderof qmodulo ℓis f,if T = {µ k κ|k ∈ A}everynon­trivialcharacteroforder ℓ
isrepresentableas µ η(qi k)
κ
with 0 ≤ i < fand k ∈ A. η(q i k)istheremainderof q i kupon
divisionby ℓ. Thusaswealreadysaw, Thas ℓ−1
f elements. Lemma7.1againshowsthat µ
andallitsconjugateshaveabsolutevalue1andthat µisinvariantunderallautomorphisms
of k p(q f −1)over k q f −1.
let
Let α = α(χλ, p)andlet β = α(χℓ λ , p).Then ℓα = β + ν(q − 1)with ν ≥ 0.If 0 ≤ k < ℓ
α(µ k κ , P) = k · qf − 1
ℓ
= f−1
j=0 γk j qj
with 0 ≤ γ k j ≤ q −1.Inparticular, γk 0 istheresidueof k · qf −1
ℓ modulo q.Moreoverif k1 ≡ q i k
(modulo ℓ)then
α(µ k κ, P) = f−1
Itisunderstoodthatif j + i ≥ ℓthen γ k1
j+i
divisionby q,
Certainly
{γ k j
| k ∈ A, 0 ≤ j < f} =
j=0 γk1
j+i qj .
α(χκ/λ µ k κ, P) ≡ qf − 1
q − 1 α + k · qf − 1
ℓ
= γk1
j+i−ℓ .Thusif ϕ(x)istheremainderof xupon
ϕ k · qf
− 1
| 0 < k < ℓ .
ℓ
(mod(q f − 1)).

Chapter7 55
Let 0 ≤ k ′ < ℓandlet ν + k ≡ k ′ (modℓ).Since,bydefinition, ℓα = β + ν(q − 1)
qf − 1
q − 1 α + k qf − 1
ℓ ≡ qf − 1
·
ℓ
β
q − 1 + k′ qf − 1
ℓ
(mod(q f − 1)).
Since 0 ≤ β < q −1therightsideisnon­negativeandatmost qf −2.Thusitis α(χκ/λ µ k κ , P).
Let
with 0 ≤ δ k j ≤ q − 1.Thus δk 0 istheresidueof
α(χκ/λ µ k f−1
κ , P) =
j=0 δk j qj
q f − 1
ℓ
·
β
q − 1 + k′ · qf − 1
ℓ
modulo q.Since χ κ/λisinvariantunderautomorphismsof κ/λ
α(χκ/λ µ k f−1
κ , P) =
j=0 δk1
j+i qj
if k1 ≡ qik (modℓ).Sincetheresidueof qf −1
αmodulo qis α,
q−1
{α} ∪ {δ k j | 0 ≤ j < ℓ, k ∈ A} =
f q − 1
ϕ ·
ℓ
β
q − 1 + k qf
− 1
| 0 ≤ k < ℓ .
ℓ
Since χλ(ℓℓ ) ≡ ℓαℓ (mod ∗ P)thenumber µismultiplicativelycongruentmodulo Pto
thequotientof
̟α̟ε α! ℓ−1 k∈A j=0 δk j !,
ε =
k∈A
f−1
j=0 δk j
by
Since
ℓβ ̟β ̟ε′ ℓ−1 β!
k∈A
f−1
j=0
weconcludefromLemma7.4that
fα + f−1
j=0 γk j !, ε′ =
k∈A
f−1
k j
α + γj q ≡
j=0
f−1
j−0 γk j .
j=0 δk j qj ,
k
γj − δ k j = ρ(q − 1),

Chapter7 56
if ρisthenumberof i, 0 ≤ i < f,suchthat
Since
thenumber
qf
− 1
α + α µ
q − 1 η(qi
k)
κ , P
≥ q f .
q f − 1
q − 1 · α + α(µ0 κ , P) = qf − 1
q − 1 α,
(ℓ − 1)α +
k∈A
f−1
j=0
is (q − 1)timesthenumberof k, 0 ≤ k < ℓ,suchthat
qf − 1
q − 1 α + k · qf − 1
ℓ
= qf − 1
q − 1
Thenumberofsuch kis νbecause ν < ℓand
while
Thus
q f − 1
q − 1
If ℓm ≡ 1 (modq)
and
q f − 1
q − 1
k
γj − δ k j
β
ℓ + (k + ν) qf − 1
ℓ ≥ qf .
β
ℓ + (ℓ − ν + ν) qf − 1
ℓ ≥ 1 + qf − 1 = q f
β
ℓ + (ℓ − 1) qf − 1
ℓ < qf − 1
ℓ + (ℓ − 1) qf − 1
ℓ = qf − 1.
k∈A
f−1
f q − 1
ϕ ·
ℓ
ItfollowsimmediatelyfromLemma7.5that
ℓ β α!
j=0 (γk j − δ k j ) = ν(q − 1) − (ℓ − 1)α = α − β.
k∈A
ϕ k · qf
− 1
= ϕ(−km)
ℓ
β
q − 1 + k qf
− 1
= ϕ((β − k)m).
ℓ
f−1
j=0 δk j ! ≡ β!
k∈A
f−1
j=0 γk j !
Thus µ = 1if qisoddand µ = ±1if qiseven.If χλ = 1thenumber µisclearly1.This
timetoo,onecanapplypart(f)ofLemma7.1andinductiononthenumberofprimesdividing
theorderof χλtoshowthat µ = 1if qiseven.

Chapter7 57
Lemma 7.9
Let λbeafinitefieldwith qelementsandlet κbeafiniteextensionof λwith [κ : λ] = ℓ
where ℓisaprimedividing q − 1.Suppose χλisacharacterof λ ∗ whoserestrictiontothe ℓth
rootsofunityisnottrivialand χκisacharacterof κ ∗ suchthat χ ℓ κ = χ κ/λ.If Tisthesetof
non­trivialcharactersof λ ∗ oforder ℓ
χλ(ℓ)τ(χλ, ψλ)
µλ∈T τ(µλ, ψλ) = τ(χκ, ψ κ/λ).
If σ ∈ G(κ/λ)define χσ κby χσκ(α) = χκ(ασ−1). Since χσ κ/λ = χκ/λ, χσℓ κ = χκ/λand χσ−1 κ isacharacteroforder ℓ.If χσ−1 κ = 1forsome σ = 1thenitis1forall σand χκ(α) = 1
if αisa(q − 1)thpower,thatis,if Nκ/λ(α) = 1. Consequentlythereisacharacter νλof λ∗ suchthat χκ = νκ/λ. Then νℓ λ = χλand χλistrivialonthe ℓthrootsofunity,contraryto
assumption.Thus
{χ σ−1
κ | σ = 1} = {µ κ/λ | µλ ∈ T }.
If β ∈ λ ∗ and β = N κ/λ(γ)then
χκ(β) =
σ χκ(γ σ−1 ) = χκ(γ ℓ )
because µλ(β) = µ κ/λ(γ),anditwillbeenoughtoshowthat
χλ(ℓ) τ(χλ)
σ=1 χσ−1 κ (γ) = χλ(β)
σ=1 µλ(β),
µλ∈T τ(µλ) = τ(χκ).
Let µbethequotientoftheleftsidebytheright.Thus µisanumberin Kp(q−1)andthe
onlyprimesappearinginthefactorizationof µaredivisorsof p.Since χ κ/λisnotidentically
1neitheris χκ.Thustheabsolutevalueof µandallitsconjugatesis1.Let α = α(χλ, p)and
let β = α(χκ, P)where Pdivides p.Then
Since ℓdivides qℓ −1
q−1 wecanwrite
ℓβ ≡ α qℓ − 1
q − 1
β = qℓ − 1
q − 1
(mod(q ℓ − 1)).
α
·
ℓ − j · qℓ − 1
.
ℓ

Chapter7 58
Sincetherestrictionof χλtothe ℓthrootsofunityisnottrivial, α · q−1
ℓ ≡ 0 (mod(q − 1)).
Thus ℓdoesnotdivide α.Forall i ≥ 0
τ χ qi
κ = τ(χκ).
Moreover
α
Since qi −1
q−1
χ qi
κ
, P ≡ αℓ − 1
q − 1
≡ qℓ − 1
q − 1
α
ℓ − j αℓ − 1
ℓ + (qi − 1) qℓ − 1
q − 1
α
ℓ +
i ℓ q − 1 q − 1
α − j .
q − 1 ℓ
≡ i (modℓ)wechoose isothat iα ≡ j (mod ℓ);then
α(χ qi
κ , P) ≡ qℓ − 1
q − 1
α
ℓ
(mod(q ℓ − 1)).
α
ℓ − j(qi − 1) qℓ − 1
ℓ
Bothsidesofthiscongruencearenon­negativeandlessthan q ℓ − 1.Thusitisanequalityand
wecanassumethat β = qℓ−1 q−1
definedby
χ −1
κ
· α
ℓ
. Theset Tconsistsofthecharacters µj
λ , 1 ≤ j ≤ ℓ − 1,
α(µ j j
λ , P) =
ℓ
(q − 1).
Undertheautomorphism z −→ z m of k p(q ℓ −1)over k q ℓ −1thenumber µismultipliedby
(m) χλ(m)
µλ∈T µλ(m)whichis1because mbelongsto λ.Let
with 0 ≤ γi ≤ q − 1.Then
and
τ(χκ) ≡
χλ(ℓ)τ(χλ) ℓ−1
j=1 τ(µj λ ) ≡
β = γ0 + γ1q + . . . + γℓ−1 q ℓ−1
̟ε
ℓ−1 j=0 γj! (mod∗P), ε = ℓ−1
j=0 γj,
α! ℓ−1
j=1
ℓ α ̟ ε′
j · q−1
ℓ
!
(mod ∗ P), ε ′ = α + ℓ−1
j=0
q − 1
j · .
ℓ
Lemma7.6impliesimmediatelythat µ ≡ 1 (mod ∗ P).Thus µ = 1if qisoddand µ = ±1if q
iseven.If ℓ ′ isaprimedivisorof q −1differentfrom ℓ,wewrite χλas χ ′ λ χ′′ λ wheretheorderof

Chapter7 59
χ ′ λ isapowerof ℓ′ andtheorderof χ ′′ λ isprimeto ℓ.Inasimilarfashionwewrite χκas χ ′ κχ ′′ κ .
Thepair χ ′′ λ and χ′′ κ alsosatisfytheconditionsofthelemma.ThefinalassertionofLemma7.1
showsthat,if µ ′′ isdefinedinthesamewayas µ, µ = µ ′′ .Arguingbyinductionweseethatit
isenoughtoverifythat µ = 1whentheorderof χλisapowerof ℓ.Applyingthelastpartof
Lemma7.1,againweseethatthereisaprime qdividing ℓsuchthat
Since χλ(ℓ)isan ℓ ω throotofunityforsome ω,
Thus µ ≡ 1 (modℓ)and µ = 1.
τ(χλ) ≡ τ(χκ) ≡ τ(µ j
λ ) ≡ 1 (modq).
χλ(ℓ) ≡ 1 (modq).

Chapter8 60
Chapter Eight.
A Lemma of Lamprecht
Let Fbeanon­archimedeanlocalfieldandlet ψFbeanon­trivialcharacterof F. n =
n(ψF)isthelargestintegersuchthat ψF istrivialon P −n
F . If χF isaquasi­characterof
CF, m = m(χF)isthesmallestnon­negativeintegersuchthat χFistrivialon Um F .If γin CF
issuchthat γOF = P m+n
F set
UF
∆1(χF,ψF; γ) =
ψF
α
γ χ −1
F (α)dα
χ −1
F (α)dα .
Then
UF ψF
α
γ
∆(χF,ψF) = χF(γ)∆1(χF, ψF, γ).
AssuggestedbyHasse[8],weshall,intheproofs,ofthemainlemmas,makeextensive
useofthefollowinglemmawhichiscentraltothepaper[10]ofLamprecht.
Lemma 8.1
(a)If m = m(χF) = 2dwith dintegralandpositivethereisaunit βin OFsuchthat
βx
ψF = χF(1 + x)
γ
forall xin Pd F .Foranysuch β
∆1(χF,ψF; γ) = ψF
β
χ
γ
−1
F (β).
(b)If m = m(χF) = 2d + 1with dintegralandpositivethereisaunit βin OFsuchthat,for
all xin P d+1
F ,
ψF
Foranysuch β, ∆1(χF, ψF; γ)isequalto
ψF
β
χ
γ
−1
F (β)
βx
= χF(1 + x).
γ
OF /PF ψF
δβx
γ
OF /PF ψF
δβx
γ
χ −1
F (1 + δx)dx
χ −1
F (1 + δx)dx

Chapter8 61
if δOF = P d F .
Let m = 2d + εwith ε = 0incase(a)and ε = 1incase(b).Thefunction ψF
OF, y ∈ P d+ε
F canberegardedasafunctionon
OF/P d F
× Pd+ε
F /Pm F .
xy
γ , x ∈
Forfixed xitisacharacterof P d+ε
F /PmF whichistrivialifandonlyif x ∈ PdF andforfixed y
itisacharacterof OF/Pd Fwhichistrivialifandonlyif y ∈ Pm F .Thusitdefinesadualityof
OF/P d F
and Pd+ε
F /Pm F .Theexistenceofaβsuchthat
χF(1 + x) = ψF
for xin P d+ε
F followsimmediatelyfromtherelation
βx
γ
χF(1 + x) χF(1 + y) = χF(1 + x + y)
whichisvalidfor xin P d+ε
F .Thenumber βmustbeaunitbecause χF(1+x)isdifferentfrom
1forsome xin P m−1
F .
Incase(a)
isequalto
UF /U d F
ψF
UF
ψF
α
χ
γ
−1
F (α)dα
α
χ
γ
−1
F (α)
Pd ψF
F
(α − β)x
γ
dx dα.
Themainintegralis1or0accordingas α − βdoesordoesnotliein Pd F .Thusthisexpression
isequalto
β
ψF χ
γ
−1
F (β) [UF : U d F] −1 .
Thefirstpartofthelemmafollows.
Incase(b)
isequalto
UF /U d+1
F
ψF
UF
ψF
α
χ
γ
−1
F (α)dα
α
χ
γ
−1
F (α)
P d+1
ψF
F
(α − β)x
γ
dx dα.

Chapter8 62
Theinnerintegralis0unless α − βliesin P d F whenitis1.Thusthisexpressionisequalto
ψF
β
χ
γ
−1
F (β) [UF : U d F ]−1
Thesecondpartofthelemmafollows.
OF /PF
ψF
δβx
γ
χ −1
F (1 + δx)dx.
Thenumber βisonlydeterminedmodulo Pd F .Whenapplyingthelemmaweshall,after
choosing β,set
β
∆2(χF,ψF; γ) = ψF χ
γ
−1
F (β)
andthendefine ∆3(χF, ψF; γ),whichwillbe1when miseven,bytheequation,
∆1(χF,ψF; γ) = ∆2(χF, ψF; γ) ∆3(χF, ψF; γ).
Whenweneedtomaketherelationbetween βand χFexplicitwewrite βas β(χF).Tobeof
anyusetousthislemmamustbesupplementedbysomeotherobservations.
If KisafiniteGaloisextensionof F anyquasi­character χF of CF determinesaonedimensionalrepresentationof
W K/Fwhoserestrictionto CKisaquasi­character χ K/Fof CK.
Thecharacter χ K/Fmaybedefineddirectlyby
χ K/F(α) = χF(N K/Fα).
Moregenerally,if Eisanyfiniteseparableextensionof Fwedefine χ E/Fby
χ E/F(α) = χF(N E/Fα).
ToapplythelemmaofLamprechtweshallneedtoknow,insomespecialcases,therelation
between β(χF)and β(χ E/F).
Suppose misapositiveintegerand m = 2d + εwhere εis0or1and disapositive
integer.Let m ′ = ψ E/F(m − 1) + 1andlet m ′ = 2d ′ + ε ′ where ε ′ is0or1and d ′ isapositive
integer.*Since ψ E/Fisconvex
ψ E/F
m − 1
2
≤ 1
2 ψ E/F(m − 1) + 1
2 ψ E/F(0) = 1
2 (m′ − 1) < d ′ + ε ′
*Weareheredealingnotwithanadditivecharacter,butwiththefunctionofChapter6!

Chapter8 63
and
d ′ + ε ′ = ψ E/F(u)
with u > m−1
m−1
2 .Sincetheleastintegergreaterthan 2 is d + ε,Lemma6.6impliesthat
Inotherwords,if x ∈ P d′ +ε ′
E
Lemma6.6alsoimpliesthat
NE/F U d′ +ε ′
E ≤ U u F
then
if x ∈ Pm′ E .If x ∈ Pd′ +ε ′
E and y ∈ Pm′ E then
iscongruentto
≤ Ud+ε
F .
N E/F(1 + x) − 1 ∈ P d+ε
F .
N E/F(1 + x) − 1 ∈ P m F
N E/F(1 + x + y) − 1 = N E/F(1 + x)N E/F
modulo P m F .Thusif x ∈ Pd′ +ε ′
E
Therightsideis
whichequals
N E/F(1 + x) − 1
and y ∈ P d′ +ε”
E
1 + y
− 1
1 + x
sothat xy ∈ P m′
E ,then
N E/F(1 + x + y) − 1 ≡ N E/F(1 + x + y + xy) − 1 (modP m F ).
N E/F(1 + x)N E/F(1 + y) − 1,
{N E/F(1 + x) − 1} + {N E/F(1 + y) − 1} + {(N E/F(1 + x) − 1) (N E/F(1 + y) − 1)}
andthisiscongruentto
modulo P m F .Thusthemap
{N E/F(1 + x) − 1} + {N E/F(1 + y) − 1}
P E/F : x −→ N E/F(1 + x) − 1

Chapter8 64
isahomomorphismfrom P d′ +ε ′
E
/P m′
E
to Pd+ε
F /Pm F .If E ⊆ E′ wecanreplace Fby E, Eby
E ′ , mby m ′ ,and m ′ by ψ E ′ /E(m ′ − 1) + 1,anddefine P E ′ /E.Since ψ E ′ /F = ψ E ′ /E ◦ ψ E/F
and
N E ′ /F(1 + x) − 1 = N E/F(1 + (N E ′ /E(1 + x) − 1)) − 1,
therelation
isvalid.
P E ′ /F = P E/F ◦ P E ′ /E
If n = n(ψF)and n ′ = n(ψ E/F),choose γFin CFsothat γFOF = P m+n
F
sothat γEOE = P m′ +n ′
E
and γEin CE
. Iapologizeagainfortheunfortunateconflictofnotation. ψ E/Fis
ontheonehandafunctionon {u ∈ R | u ≥ −1}andontheotheracharacterof E.However,
warnedoneagain,thereadershouldnotbetooinconveniencedbytheconflict.Define
bytherelation
ψF
P ∗ E/F : OF/P d F
−→ OE/P d′
E
∗
x PE/F(y) PE/F (x)y
= ψE/F .
γF
Itwilloftenbenecessarytokeepinmindthedependenceof P ∗ E/F on γFand γE. Thenwe
shallwrite
P ∗ E/F (x) = P ∗ E/F (x; γE, γF).
Itisclearthat
Lemma 8.2
P ∗ E ′ /F (x; γE ′, γF) = P ∗ E ′ /E (P ∗ E/F (x; γE, γF); γE ′, γE).
Let K/Fbeabelianandlet G = G(K/F).Supposethereisaninteger tsuchthat G = Gt
while Gt+1 = {1}.Suppose m ≥ t+1and m > 1and γFischosen.If µFbelongsto S(K/F),
thesetofcharactersof CF/N K/FCK,then m ≥ m(µF)sothatforsome α(µF)in OF
µF(1 + x) = ψF
γE
α(µF)x
forall xin P d+ε
F .Theelement γKmaybetakenequalto γFandif P ∗ K/F (β) = P ∗ K/F (β; γF, γF)
then
NK/F(P ∗
K/F (β)) ≡ (β + α(µF)) (modP
µF
d F)
γF

Chapter8 65
forall βin OF.
If t = −1then n(ψF) = n(ψ K/F)and m ′ = msothat γKmaybetakenequalto γF.If
t ≥ 0theextensionisramified.Let P δ K/F
K
Bydefinition
bethedifferentof K/F.Then
n(ψ K/F) = [K : F]n(ψF) + δ K/F.
ψ K/F(m − 1) = t + [K : F](m − t − 1).
ByProposition4ofparagraphIV.2ofSerre’sbook δ K/F = ([K : F] − 1)(t + 1).Thus
andwecanagaintake γK = γF.
m ∗ + n(ψ K/F) = [K : F] (m + n(ψF))
Since m(µF) = t + 1wehave m ≥ m(µF)and
µF(1 + x + y) = µF(1 + x) µF(1 + y)
for xand yin P d+ε
F .Thustheexistenceof α(µF)isassured.Thelastassertionofthelemma
willbeprovedbyinduction.Wewillneedtoknowthatif x ≡ y(modPd′ K )then
N K/Fx ≡ N K/Fy(modP d F ).
Whenprovingthiswemaysupposethat xOK = P r K with r ≤ d′ andthat y
x
Then
NK/Fx − NK/Fy = NK/Fx 1 − NK/F
1 +
y − x
.
x
belongsto OK.
If r ≥ dthereisnothingtoprove.Suppose r ≤ d.If d ′ − r = ψK/F(u)and sisthesmallest
integergreaterthanorequalto utherightsidebelongsto P s+r
F .Sincethederivativeof ψK/F isatleastoneeverywhere ψK/F(u + r) ≥ d ′ .But
d ′ ≥ m′ − 1
=
2
1
2 ψK/F(m − 1) + 1
2 ψ
m − 1
K/F(0) ≥ ψK/F .
2
Thus u + r ≥ m−1
2 and s + r ≥ d.
Suppose F ⊆ L ⊆ Kand L/F iscyclicofprimeorder. Let H = G(K/L)andlet
G = G(L/F).Certainly H = Htwhile Ht+1 = {1}.Since ψ K/F(t) = ψ K/L(t) = t,wehave
ψ L/F(t) = tand,byHerbrand’stheorem,
Gt = G t = HG t /H = G/H = G.

Chapter8 66
Moreover t + 1 = ψ L/F(t + δ)with δ > 0sothat
Gt+1 = G t+δ = HG t+δ /H = H/H = {1}.
Finally, ψ L/F(m −1)+1 ≥ t +1sothat L/Fand K/L,with mreplacedby ψ L/F(m −1)+1,
satisfytheconditionsofthelemma. S(L/F)isasubgroupof S(K/F).If µFand νFbelong
to S(K/F)then µ L/F = ν L/Fifandonlyif µFand νFbelongtothesamecosetof S(L/F).
Take Stobeasetofrepresentativesforthesecosets;then
S(K/L) = {µ L/F | µF | µF ∈ S}.
Wetake α(µFνF) = α(µF) + α(νF)if µF belongsto Sand νF belongsto S(L/F). If µF
belongsto Swetake α(µ L/F)tobe P ∗ L/F (α(µF)). Ifthelemmaisvalidfor K/Land L/F
then
N K/F(P ∗ K/F (β)) = N L/F(N K/L(P ∗ K/L (P ∗ L/F (β)))
whichiscongruentmodulo P d F to
N L/F
or
Thisiscongruentmodulo P d F to
whichequals
µF ∈S
µF ∈S (P ∗ L/F (β) + P ∗ L/F (α(µF)))
µF ∈S {N L/F(P ∗ L/F
(β + α(µF)))}.
νF ∈S(L/F) {β + α(µF) + α(νF)}
µF ∈S(K/F)
{β + α(µF)}.
Thusitisenoughtoprovethelemmawhen K/Fiscyclicofprimeorder. Inthiscase
morepreciseinformationisneededandtheassertionofthelemmawillfollowimmediately
fromit.
Lemma 8.3
If K/Fisunramifiedand m ≥ 1wemaytake P ∗ K/F (β) = β.
AccordingtoparagraphV.2ofSerre’sbook
N K/F(1 + y) − 1 ≡ S K/F(y) (mod P m F )

Chapter8 67
if y ∈ P d′ +ε ′
K .Thus P K/F(y) = S K/F(y)and
Lemma 8.4
ψF
x PK/Fy xy
= ψK/F γF
Suppose K/Fisabelian,totallyramified,and [K : F] = ℓisanoddprime.If d ≥ t + 1
(β) = β.
wemaytake P ∗ K/L
Therelation
impliesthat m ′ ≡ m(mod 2), ε ′ = ε,and
Since
wehave
Moreover
γF
.
m ′ = t + 1 + ℓ(m − 1 − t) = ℓm − (t + 1) (ℓ − 1)
d ′ = ℓd +
ℓ − 1
2
ℓ − 1
(ε − t − 1) = d + (m − t − 1).
2
ℓ − 1
(m − t − 1) ≥ m − t − 1 ≥ d + ε
2
d ′ + ε ′ ≥ 2(d + ε) ≥ m.
2(d ′ + ε ′ ) + δ K/F
ℓ
sothatbyLemma5ofparagraphV.3ofSerre’sbook
if x ∈ P d′ +ε ′
K .Thelemmafollows.
Let pbethecharacteristicof OF/PF.
Lemma 8.5
≥ m′ + δ K/F
ℓ
= m
N K/F(1 + x) − 1 ≡ S K/F(x) (mod P m F )
Suppose K/Fisabelian,totallyramified,and [K : F] = ℓisanoddprime. Suppose
t+1 ≤ m ≤ 2t+1.Chooseanon­trivialcharacter µFin S(K/F).Wemaychoose α = α(µF)
sothat αOF = P v F ,if m = t + 1 + v,sothat α = N K/Fα1forsome α1in OK,andsothat
µF(1 + x) = ψF
αx
γF

Chapter8 68
for xin Ps t
F . Here sistheleastintegergreaterthanorequalto 2 . If ζisa(p − 1)throotof
unityin Fthereisauniqueinteger jwith 1 ≤ j ≤ p − 1suchthat ζ − jliesin PF. Set
µ ζ
F = µj
F .Wemaytake α(µζ F )tobe ζα.If βbelongsto OFwecanfindaβ1in OKsuchthat
β ≡ N K/Fβ1(modP d F ).Then
If
P ∗ K/F (β) ≡ β − β1
α
α1
µF(1 + x) = ψF
(mod P d′
K).
αx
for xin P s F then,necessarily, αOF = P v F . Choose δ1in OKsuchthat δ1OK = P v K andset
δ = N K/Fδ1.Set α = ωδwhere ωisyettobechosen.Wemusthave
µF(1 + x) = ψF
γF
ωδx
if x ∈ Ps F . Thisequationdeterminestheunit wmodulo Pr Fif r = t − s. Sinceanyunitis
anormmodulo Pt Fwemaysuppose ω = NK/Fω1. Take α1 = ω2δ1. β1existsforasimilar
reason.
Thenumber ζ − jmustliein pOF.But K/Fiswildlyramified,because 2t + 1 ≥ m >
1, ℓ = pand p = SK/F(1)sothat,byparagraphV.3ofSerre’sbook, pbelongsto Pu Fif uis
thegreatestintegerin
(ℓ − 1) t + 1
(t + 1) ≥
ℓ
2 .
However d + ε ≥ ssothat d + ε + u ≥ t + 1and,if xbelongsto P d+ε
F
Thus
and
Since
ψF
γF
α(ζ − j)x
= µF (1 + (ζ − j)x) = 1
γF
µ j
F (1 + x) = µj
F (1 + x) = ψF
2s + δ K/F
ℓ
jαx
γF
≥ t + 1 + δ K/F
ℓ
ThelemmasofparagraphV.3ofSerre’sbookimplythat
= ψF
= t + 1.
ζαx
N K/F(1 + x) ≡ 1 + S K/F(x) + N K/F(x) (modP t+1
F )
γF
, (ζ − j)xliesin Pt+1
F .
.

Chapter8 69
if xbelongsto P s K andthen
1 = µF(N K/F(1 + x)) = µF(1 + S K/F(x) + N K/F(x)).
Asweobserved d + ε ≥ s.Moreover d + ε ≤ t + 1sothat
d + ε + δ K/F
ℓ
≥ d + ε
and SK/F(x)and NK/F(x)belongto P d+ε
F if xbelongsto P d+ε
K .Thus,forsuch x,
αNK/F(x) −αSK/F(x)
.
sothat
Again
if x ∈ P d′ +ε ′
K .Moreover
ψF
γF
= ψF
2(d ′ + ε ′ ) + δ K/F
ℓ
≥ m
N K/F(1 + x) − 1 = S K/F(x) + N K/F(x) (modP m F )
d ′ + ε ′ = d + ε +
ℓ − 1
2
γF
(m − t − 1) ≥ d + ε
sothat NK/F(x)andhence SK/F(x)belongto P d+ε
F .Thus
β1x
βNK/Fx ≡ αNK/F But β1x/α1belongsto P d′ +ε ′ −v
K
sothat
whichequals
ψF
⎛
ψF
and
d ′ + ε ′ − v = d + ε +
β PK/F(x)
γF
⎝ β S K/F(x) − αS K/F
γF
= ψF
β1x
α1
α1
ℓ − 1
2
(mod P m F ).
v − v ≥ d + ε
β SK/F(x) + β NK/F(x) ⎞
⎠ = ψ K/F
γF
β − αβ1
x
α1 γF

Chapter8 70
asrequired.
Lemma 8.6
Suppose K/Fisawildlyramifiedquadraticextension, m ≥ t + 1,and m = t + 1 + v.
Let µFbethenon­trivialcharacterin S(K/F).If βbelongsto OFthereisaβ1in OKandaδ
in Ut Fsuchthat β ≡ δNK/Fβ1(modP d F ).Wecanchoose α = α(µF)sothat
µF(1 + x) = ψF
αδx
if xisin Ps Fandsothat α = NK/Fα1forsome α1in OK. Here shasthesamemeaningas
, t + 1 = r + s.Withthesechoices
before.Thus,if ristheintegralpartof t+1
2
isin U t F
P ∗ β1αδ
K/F (β) ≡ β −
α1
γF
mod P d′
K
If β = 0theexistenceof δand β1isclear.Otherwisewecanfindaβ1suchthat NK/Fβ1/β .Wechoose δaccordingly.If m = t + 1 + vand
µF(1 + x) = ψF
αδx
for xin P s F then OFα = P v F .Choose η1in OKsothat OKη1 = P v K andset η = N K/Fη1.Set
α = ωηwhere ωisyettobechosen.Wemusthave
µF(1 + x) = ψF
γF
ωηδx
if x ∈ P s F .Thisequationdeterminestheunit ωmodulo Pr F .Sinceanyunitisanormmodulo
P t F wemaysuppose ω = N K/Fω1.Take α1 = ω1η1.
Since
Sincetheextensionisquadratic
γF
.
N K/F(1 + x) = 1 + S K/F(x) + N K/F(x).
s + δ K/F
2
= s + t + 1
2
≥ s

Chapter8 71
both S K/F(x)and N K/F(x)arein P s F if xbelongsto Ps K and
ψF
αδNK/F(x)
γF
= ψF
−αδSK/F(x)
Wehave m ′ = 2m − (t + 1)and d ′ = m − s,sothat d ′ + ε ′ = m − rand d ′ + ε ′ − v = s.Thus
if xbelongsto P d′ +ε ′
K
β1x
βNK/F(x) ≡ αδNK/F (modP m F )
and β1x/α1liesin P s K .Consequently
asrequired.
Lemma 8.7
ψF
γF
α1
γF
α1
.
β PK/Fx
αδ x
= ψK/F β − β1
If K/Fisatamelyramifiedquadraticextensionand m ≥ 2wemaytake P ∗ K/F (β) = β.
Noticethat t + 1 = 1sothat m ≥ t + 1. Inthiscase m ′ = 2m − 1, d ′ = m − 1,and
d ′ + ε ′ = m.If x ∈ P d′ +ε ′
K
iscongruentto
modulo P m F .Thelemmafollows.
N K/F(1 + x) = 1 + S K/F(x) + N K/F(x)
1 + S K/F(x)
TocompletetheproofofLemma8.2wehavetoshowthatif K/Fiscyclicofprimeorder
NK/F(P ∗
K/F (β)) ≡
µF ∈S(K/F) (β + α(µF)) (modP d F).
Weconsiderthecasesdiscussedinthepreviouslemmasonebyone. Iftheextensionis
unramifiedwemaytakeallthenumbers α(µF)tobe0.Thecongruencesthenreducetothe
identity β n = β n . Thesameistrueif K/Fiscyclicofoddorderand d ≥ t + 1or K/Fis
quadraticand t = 0. If K/Fiscyclicofoddorder ℓand t + 1 ≤ m ≤ 2t + 1therightside
becomes
β ℓ − βα ℓ−1 .
γF

Chapter8 72
If β ≡ 0 (modPd F )bothsidesarecongruentto0modulo PdF .Suppose βdoesnotbelongto
Pd Fand βOF = Pu F .Then β1OK = Pu Kand iscongruentto
N K/F
β − β1
α
α2
β ℓ − βα ℓ−1 + ℓ−1
i=1 (−1)iβ ℓ E i
β1α
βα1
modulo Pd F . If x ∈ Kthen Ei (x)isthe ithelementarysymmetricfunctionof xandits
(ℓ−1) (v−u)
conjugates.Moreover β1α/βα1belongsto P .If ℓ − 1 ≥ i ≥ 1
Therightsideis
i(ℓ − 1) (v − u) + (ℓ − 1) (t + 1)
ℓ
K
≥
(ℓ − 1)
m − pu ≥ d − ℓu.
ℓ
TheargumentofparagraphV.3ofSerre’sbookshowsthat
N K/F
β − β1
(ℓ − 1) (v + t + 1)
ℓ
α
≡ β ℓ − βα ℓ−1 (modP d F).
α1
− ℓu.
ForawildlyramifiedquadraticextensionweusethenotationofLemma8.6. Theright
sideofthecongruencemaybetakentobe β2 + βαδ.Theidentityisagainnon­trivialonlyif
βOF = Pu Fwith u < d.Thentheleftsidemaybetakentobe
whichiscongruentto
modulo P d F .Since
wehave
β 2 − β 2 δS K/F
β1α
+ δ
βα1
2 αNK/Fβ1 β 2 + αβδ − β 2 δS K/F
v − u + t + 1
2
β 2 S K/F
≥ m
2
β1α
βα1
− u ≥ d − u
β1α
≡ 0 (modP
βα1
d F).

Chapter8 73
Suppose χFisaquasi­characterof CF, m = m(χF),and β = β(χF).If,assometimes
happens, m ′ = m(χ K/F)wecantake β(χ K/F) = P ∗ K/F (β).
Lemma 8.8
Suppose K/FisGaloisand G = G(K/F).Suppose s ≥ 0isanintegerand G s = {1}.If
m = m(χF)and m > sthen
m ′ = ψ K/F(m − 1) + 1 = m(ψ K/F).
ItfollowsfromparagraphV.6ofSerre’sbookthat
NK/F(U ψK/F(v) K ) = U v F
if v ≥ s. Thus χ K/F istrivialon U u K if u > ψ K/F(m − 1)butisnottrivialon U i K if
u = ψ K/F(m − 1).
Wecannowcollecttogether,withoneortwoadditionalcomments,thepreviousresults
inaformwhichwillbeusefulintheproofofthefirstmainlemma.Weusethesamenotation.
Lemma 8.9
SupposeK/Fisacyclicextensionofprimeorderℓ, χFisaquasi­characterofCF , m(χF) ≥
t + 1, m(χF) > 1,and m(χ K/F) − 1 = ψ K/F(m(χF) − 1).
(a)If K/Fisunramifiedwemaytake β(χ K/F) = β(χF)and β(µFχF) = β(χF)forall µF
in S(K/F).
(b)If ℓisoddand d ≥ t + 1wemaytake β(χ K/F) = β(χF)and β(µFχF) = β(χF)forall
µFin S(K/F).
(c)If ℓisoddand t + 1 ≤ m ≤ 2t + 1and µFisagivennon­trivialcharacterin S(K/F)we
maychoose α = α(µF) = N K/Fα1asinLemma8.5and β = β(χF) = N K/Fβ1forsome
β1in UK.Thenwemaychoose
and
β(χ K/F) = β − β1
α
α1
β(µ ζ
F χF) = N K/F(β1 + ζα1).

Chapter8 74
(d)If ℓis2and K/Fiswildlyramifiedwechoose α = α(µF)asinLemma8.6. Wemay
choose β = β(χF)intheform δN K/Fβ1with δin U t F .Thenwemaychoose
and
β(χ K/F) = β − β1
αδ
α1
β(µFχF) = β + αδ.
(e)If ℓis2and K/Fistamelyramifiedwemaytake β(χ K/F) = β(µFχF) = β(χF).
Onlypart(c)requiresanyfurtherverification.Itmustbeshownthat
Theleftsideiscongruentto
Allweneeddoisshowthat
Therightsideis
N K/F(β1 + ζα1) ≡ β + ζα (modP d F ).
N K/F
βN K/F
1 + ζα1
.
β1
1 + ζα1
≡ 1 +
β1
ζα
β
1 + N K/F
ζα1
β1
.
(modP d F).
AccordingtoparagraphV.3ofSerre’sbookthecongruencewillbesatisfiedif
v + (ℓ − 1) (t + 1)
ℓ
≥ d.
But t + 1 = d + xwith x ≥ 0sothat d + x + v = 2d + εand v = d + ε − x.Thus
v + (ℓ − 1) (t + 1)
ℓ
= d + ε − x + (ℓ − 1) (d + x)
ℓ
= d +
ε + (ℓ − 2)x
ℓ
Theprecedingdiscussionhasnowtoberepeatedwithdifferenthypothesesanddifferent,
butsimilar,conclusions.
Lemma 8.10
≥ d.

Chapter8 75
Suppose K/Fisabelianand G = G(K/F).Supposethereisat ≥ 0suchthat G = Gt
while Gt+1 = {1}.If 2 ≤ m ≤ t+1then m ′ = ψK/F(m−1)+1isjust m.Let t+1 = m+v,let
δbesuchthat δOF = P t+1+n(ψF )
F ,let ε1in OKbesuchthat ε1OK = Pv K ,andlet ε = NK/Fε1. Wemaychoose γF = δ/εand γK = δ/ε1. Let rbethegreatestintegerin t+1
andlet 2
s = t + 1 − r.If µFisanon­trivialcharacterin S(K/F)then m(µF) = t + 1.Let
for xin P s F .Then
β
µF(1 + x) = ψF
β(µF)x
µF =1 (βε + β(µF)) ≡ N K/F(P ∗ K/F (β)) (modPd F).
Therelation m ′ = ψ K/F(m − 1) + 1 = misanimmediateconsequenceofthedefinitions.
Sincetheextensionistotallyramified
if n = n(ψF).Thus
and
n(ψ K/F) = [K : F]n + ([K : F] − 1)(t + 1)
m + n = (t + 1 + n) − v
m ′ + n(ψ K/F) = [K : F](t + 1 + n) + (m − t − 1) = [K : F](t + 1 + n) − v.
Consequently γFand γKcanbechosenasasserted.TheresultsofchapterVofSerre’sbook
implythat m(µF) = t + 1if µFisnottrivial.
WesawwhenprovingLemma8.2thatifx ≡ y(modP d′
K )thenN K/Fx ≡ N K/Fy(modP d F )
andthatif F ⊆ L ⊆ Kboth L/Fand K/Lsatisfytheconditionsofthelemma.For L/F, ε1
isreplacedby N K/Lε1and,for K/L, εisreplacedby N K/Lε1.Take Q ∗ L/F tobe P ∗ L/F inthe
specialcasethat m = t + 1and ε1 = 1.Then
ψ L/F
NK/L(ε1)x P ∗ L/F (β)
bydefinition.Therightsideisequalto
δ
ψ L/F
δ
= ψF
∗ x QL/F (εβ)
.
δ
εPL/F(x)β
δ

Chapter8 76
Thus
Q ∗ L/F (εβ) ≡ N K/L(ε1)P ∗ L/F (β) (modPs L ).
If µF belongsto S(K/F)butnotto S(L/F)then m(µ L/F) = m(µF)and β(µ L/F)may
betakentobe Q ∗ L/F (β(µF)). Let S ′ beasetofrepresentativesforthecosetsof S(L/F)in
S(K/F) − S(L/F)andsupposethelemmaistruefor K/Land L/F.Then
iscongruentto
N K/F(P ∗ K/F (β)) = N L/F(N K/L(P ∗ K/L (P ∗ L/F (β))))
NL/F(P ∗
L/F (β)
µF ∈S ′{NK/L(ε1)P ∗ L/F (β) + Q∗L/F (β(µF))})
modulo P d F .Thisinturniscongruentto
NL/F(P ∗
L/F (β))
µF ∈S ′ NL/F(Q ∗ L/F (εβ + β(µF))).
ApplyingtheinductionhypothesistothefirstpartandLemma8.2tothesecond,weseethat
thewholeexpressioniscongruentto
β
modulo P d F asrequired.
νF ∈S(L/F) (εβ + β(νF))
νF =1
µF ∈S ′ (εβ + β(µF) + β(νF))
νF ∈S(L/F)
Onceagainwedevotealemmatocyclicextensionsofprimeorder.
Lemma 8.11
Suppose K/Fiscyclicofprimeorder ℓand 2 ≤ m ≤ t+1.Chooseanon­trivialcharacter
µFin S(K/F).Thereisan α1in UKsuchthatif α = N K/Fα1
µF(1 + x) = ψF
αx
δ
for xin P s F .If βbelongsto OFthereisaβ1in OKsuchthat β ≡ N K/F(β1) (modP t F ).Then
P ∗ ε
K/F (β) ≡ β
ε1
− β1
α
α1
(mod P d′
K ).

Chapter8 77
Since β(µF)isdeterminedonlymodulo P s F and s ≤ twecantake β(µF) = N K/Fα1for
some α1in UK.Theexistenceof β1alsofollowsasbefore.Since t + 1 ≥ m
and
2(d ′ + ε ′ ) + (ℓ − 1) (t + 1)
ℓ
≥
m + (ℓ − 1) (t + 1)
ℓ
≥ m
N K/F(1 + x) ≡ 1 + S K/F(x) + N K/F(x) (modP m F )
if xbelongsto P d′ +ε ′
K .Thus
εPK/F(x)β εxβ
ψF
= ψK/F δ
δ
But d ′ + ε ′ + t ≥ t + 1sothat
N K/F(ε1x)β ≡ αN K/F
ε1β1
Since t + 1 = m + v, d ′ + ε ′ + v ≥ sandif y = ε1β1
α1
in P s K .But
sothat
Consequently
Inconclusion
asrequired.
α1
2s + (t + 1)(ℓ − 1)
ℓ
ψF
· x
NK/F(ε1x)β
δ
(mod P t+1
F )
.
· xthen ywhichliesin P d′ +ε ′ +v
K
≥ t + 1
N K/F(1 + y) ≡ 1 + S K/F(y) + N K/F(y) (mod P t+1
F ).
ψF
ψF
−αNK/F(y)
εPK/F(x)β
= ψK/F δ
δ
= ψF
ε1
δ
αSK/F(y)
ε
ε1
δ
· β − α
α1
.
· β1
x
alsolies
TocompletetheproofofLemma8.10wehavetoshowthatwhen K/Fiscyclicofprime
order, ℓ
β
µF =1 (βε + β(µF)) ≡ N K/F(P ∗ K/F (β)) (mod Pd F )
Since 2 ≤ m ≤ t + 1theextensioniswildlyramified, ℓ = p,andoncethecharacter µF is
chosenasinthepreviouslemmawecandefine µ ζ
FasinLemma8.5.Theleftsideiscongruent to
β β ℓ−1 ε ℓ−1 + (−1) ℓ α ℓ−1 .

Chapter8 78
If β ∈ Pd Fthisiscongruentto0andsoistherightside.Suppose βOF = Pu Fwith u < d.The
rightsideiscongruentto
Since
thisiscongruentto
Since
Weseethat
N K/F
β ε
ε1
− β1
α1
(ℓ − 1)(u + v) + (ℓ − 1)(t + 1)
ℓ
βα ℓ−1
βN K/F
α
≡ βα ℓ−1
β
NK/F N K/F
β
andthattheexpression(8.1)iscongruentto
modulo P d F .
Lemma 8.12
β1
≥
α1
α
ℓ − 1
ℓ
β1
α1
α
· (t + 1) ≥
ε
ε1
− 1 .
t + 1
2
≥ d
ε
+ (−1) ℓ
. (8.1)
ε1
−1
t−u
β1 ≡ 1 (mod PF ).
β ℓ+1 N K/Fβ −1
1 ≡ βℓ (mod P t F)
β ℓ ε ℓ−1 + (−1) ℓ βα ℓ−1
SupposeK/Fisabelianand G = G(K/F).Supposethereisaninteger tsuchthatG = Gt
while Gt+1 = {1}. Let χFbeaquasi­characterof CFandsuppose 2 ≤ m(χF) ≤ t + 1. If
m(χF) < t + 1then m(χ K/F) = m(χF).If m(χF) = t + 1then m(µFχF) < t + 1forsome
µFin S(K/F)ifandonlyif m(χ K/F) < m(χF).
ItfollowsimmediatelyfromLemma6.7thatif χFisanyquasi­characterof CFand Eany
finiteseparableextensionof Fthen
m(χ E/F) − 1 ≤ ψ E/F(m(χF) − 1).
IntheparticularcaseunderconsiderationLemma6.10showsthatif m = m(χF) ≤ tthen
NK/F : U m−1
K /Ut K −→ Um−1
F /U t F

Chapter8 79
isanisomorphism.Thus χK/F(α)willbedifferentfrom1forsome αin U m−1
K and m(χK/F) willbeatleast m.If m(χF) = t + 1then m(µFχF)islessthan t + 1forsome µFin S(K/F)
ifandonlyif χF istrivialontheimageof Ut K /Ut+1
K in Ut F /Ut+1
F . Thisissoifandonlyif
m(χK/F) ≤ t.
Weshallneedthefollowinglemmaintheproofofthefirstmainlemma.
Lemma 8.13
Suppose K/Fiscyclicofprimeorder, χFisaquasi­characterof CFwith m(χF) ≤ t + 1,
and m(χ K/F) = m(χF). Choose α, α1, ε, ε1inLemma8.11. Wemaychoose β = β(χF) =
N K/Fβ1with β1in UKandwemaychoose
β(χ K/F) = β ε
Moreover m(µ ζ
F χF) = t + 1andwemaytake
ε1
− β1
α
.
α1
β(µ ζ
F χF) = N K/F(ζα1 + ε1β1).
Since β(χF)isdeterminedonlymodulo Pd Fand d ≤ ttheexistenceof β1isclear. Itis
alsoclearthat m(µ ζ
FχF) = t + 1.Theelements β(χF), β(χK/F),and β(µ ζ
FχF)aretosatisfy thefollowingconditions:
(i)If xisin P d F
(ii)If xisin P d′
K
(iii)If xisin P s F
χF(1 + x) = ψF
χ K/F(1 + x) = ψ K/F
µ ζ
F (1 + x)χF(1 + x) = ψF
εβ(χF)x
δ
.
ε1β(χ K/F)x
Wehavealreadyshownthat β(χ K/F)maybetakentobe
β ε
ε1
α
− β1
α1
δ
.
β(µ ζ
FχF)x
.
δ

Chapter8 80
β(µ ζ
F χF)mustbecongruentto ζα + εβmodulo P r F
Since
Therightsideiscongruentto
modulo P r F .∗
N K/F(ζα1 + ε1β1) = ζαN K/F
ν + (ℓ − 1)(t + 1)
ℓ
ζα
1 + N K/F
∗(1998)ThemanuscriptofChapter8endshere.
≥
ε1β1
ζα1
1 + ε1β1
.
ζα1
ℓ − 1
(t + 1) ≥ r.
ℓ
= ζα + εβ

Chapter9 81
Chapter Nine.
A Lemma of Hasse
Let λ ⊆ κbetwofinitefieldsandlet G = G(κ/λ).If x ∈ κset
ω κ/λ(x) = x σ1 x σ2
wherethesumistakenoverallunorderedpairsofdistinctelementsof G.Itisclearthat
ω κ/λ(x + y) = ω κ/λ(x) + ω κ/λ(y) + S κ/λ(x)S κ/λ(y) − S κ/λ(xy).
Onereadilyverifiesalsothatif λ ≤ η ≤ κthen
ω κ/λ(x) = ω η/λ(S κ/η(x)) + S η/λ (ω κ/η(x)).
Suppose ψλisanon­trivialcharacterof λand ϕλisanowherevanishingfunctionon λ
satisfyingtheidentity
ϕλ(x + y) = ϕλ(x)ϕλ(y) ψλ(xy).
Define ϕ κ/λon κby
Then ϕ κ/λ(x + y)isequalto
whichis
ϕ κ/λ(x) = ϕλ(S κ/λ(x)) ψλ(−ω κ/λ(x)).
ϕλ(S κ/λ(x + y)) ψλ (−ω κ/λ(x) − ω κ/λ(y) − S κ/λ(x)S κ/λ(y) + S κ/λ(xy))
ϕ κ/λ(x)ϕ κ/λ(y)ψ κ/λ(xy).
Ifthefieldshaveoddcharacteristicthefollowinglemmais,basically,aspecialcaseof
Lemma7.7.ThatlemmahasbeenproveninasimpleanddirectmannerbyWeil[14].Weshall
usehismethodtoprovethefollowinglemmawhichincharacteristictwo,whenitcannotbe
reducedtothepreviouslemma,isduetoHasse[8].
Lemma 9.1
Let
σ(ϕλ) = −
x∈λ ϕλ(x)

Chapter9 82
andlet
Then
If
σ(ϕ κ/λ) = −
x∈κ ϕ κ/λ(x).
σ(ϕ κ/λ) = σ(ϕλ) [κ:λ] .
P(X) = X m − aX m−1 + bX m−2 − +...
isanymonicpolynomialwithcoefficientsin λset m(P) = mand
χλ(P) = ϕλ(a) ψλ(−b).
Ifthedegreeofthepolynomialis1, bistakentobe0;ifthedegreeis0both aand baretaken
tobe0.If
P ′ (X) = X m′
− a ′ X m′ −1 ′ m
+ b X ′ −2
− +...
then
and
PP ′ (X) = X m+m′
− (a + a ′ ) X m+m′ −1 ′ ′ m+m
+ (b + b + aa )X ′ −2
− +...
χλ(PP ′ ) = ϕλ(a + a ′ ) ψλ(−b − b ′ − aa ′ ) = χλ(P)χλ(P ′ ).
If tisanindeterminateweintroducetheformalseries
Fλ(t) = χλ(P)t m(P) = (1 − χλ(P)t m(P) ) −1 .
Thesumisoverallmonicpolynomialswithcoefficientsin λandtheproductisoverallirreduciblepolynomialsofpositivedegreewithcoefficientsin
λ.If r ≥ 2
sothat
m(P)=r χλ(P) = 0
Fλ(t) = 1 − σ(ϕλ)t.
Ifwereplace λby κ, ϕλby ϕ κ/λ,and ψλby ψ κ/λ,wecandefine F κ/λ(t)inasimilarway.
If k = [κ : λ]and Tisthesetof kthrootsofunitytheproblemistoshowthat
ζ∈T Fλ(ζt) = F κ/λ(t k ).

Chapter9 83
Suppose Pisanirreduciblemonicpolynomialwithcoefficientsin λand P ′ isoneofits
monicirreduciblefactorsover κ.Let m = m(P)andlet rbethegreatestcommondivisorof
mand k.Thefieldobtainedbyadjoiningtherootsof Pto κhasdegree mk
r over λandisthe
sameasthefieldobtainedbyadjoiningtherootsof P ′ to κ. Thus m(P ′ ) = m
and Psplits
r
into rirreduciblefactorsover κ.Weshallshowthat
Thusif P ′ 1 , . . ., P ′ r
whichequals
χ κ/λ(P ′ ) = {χλ(P)} k
r .
arethefactorsof Pand ℓ = k
r
r
Thenecessaryidentityfollows.
i=1 {1 − χ κ/λ(P ′ i)t
km(P ′
i ) } = {1 − χλ(P) ℓ t ℓm } r
ζ∈T {1 − χλ(P)ζ m t m }.
Let νbethefieldobtainedbyadjoiningaroot xof P ′ to κandlet µbethefieldobtained
byadjoining xto λ.If
P(X) = X m − aX m−1 + bX m−2 . . .
then
and
Thus
Since ϕ ν/λ(x)isequalto
whichinturnequals
Weconcludethat
a = S µ/λ(x)
b = ω µ/λ(x).
χλ(P) = ϕλ(S µ/λ(x)) ψλ(−ω µ/λ(x)) = ϕ µ/λ(x).
ϕλ(S κ/λ(S ν/κ(x))) ψλ(−ω κ/λ(S ν/κ(x)) + S κ/λ(ω ν/κ(x)))
Replacing κby µweseethat ϕ ν/λ(x)equals
ϕ κ/λ(S ν/κ(x)) ψ κ/λ(−ω ν/κ(x)).
χ κ/λ(P ′ ) = ϕ ν/λ(x).
ϕ µ/λ(S ν/µ(x)) ψ µ/λ(−ω ν/µ(x)) = ϕ µ/λ(ℓx) ψ µ/λ
−ℓ
(ℓ − 1)
2
x 2
.

Chapter9 84
Oneeasilyshowsbyinductionthatforeveryinteger ℓ
Therelation
follows.
{ϕ µ/λ(x)} ℓ = ϕ µ/λ(ℓx) Ψ µ/λ
−ℓ
χ κ/λ(P ′ ) = {χλ(P)} ℓ
Taking µ = λintheidentity(9.1)weseethat
{ϕλ(x)} ℓ = ϕλ(ℓx) ψλ
−ℓ
(ℓ − 1)
2
(ℓ − 1)
2
x 2
x 2
. (9.1)
foreveryinteger ℓ.Moreover {ϕλ(0)} 2 = ϕλ(0)sothat ϕλ(0) = 1.Ifthecharacteristic pof
λisoddtake ℓ = ptoseethat {ϕλ(x)} p = 1. Ifthecharacteristicis2,take ℓ = 4toseethat
{ϕλ(x)} 4 = 1.Suppose ϕ ′ λisanotherfunctionon λwhichvanishesnowhereandsatisfies
Then ϕ ′ λ ϕ−1
λ
Ofcourse
Thus
ϕ ′ λ (x + y) = ϕ′ λ (x) ϕ′ λ (y) ψλ(xy).
isacharacterandforsome αin λ
ϕ ′ λ(x) ≡ ϕλ(x) ψλ(αx).
ϕλ(x)ψλ(αx) = ϕλ(x + α)ϕ −1
λ (α).
σ(ϕ ′ λ) = ϕ −1
λ (α) σ(ϕλ).
If aand baretwonon­zerocomplexnumbersand misapositiveintegerwewrite a ∼m bif,
forsomeinteger r ≥ 0, a
b
Lemma 9.2
m r
= 1.
If α ∈ λ × ,themultiplicativegroupof λ,let ν(α)be1or­1accordingas αisorisnota
squarein λ.Suppose ψ ′ λ (x) = ψλ(αx), ϕλand ϕ ′ λ arenowherevanishing,and
while
ϕλ(x + y) = ϕλ(x)ϕλ(y)ψλ(xy)
ϕ ′ λ(x + y) = ϕ ′ λ(x)ϕ ′ λ(y)ψ ′ λ(xy).

Chapter9 85
Then
Moreover
σ(ϕ ′ λ ) ∼p ν(α)σ(ϕλ).
σ(ϕλ) ∼2p |σ(ϕλ)|.
Supposefirstthat pisodd. Bytheremarksprecedingthestatementofthelemmaitis
enoughtoprovetheassertionsforonechoiceof ϕλand ϕ ′
λ . Forexamplewecouldtake
2
(x )
ϕλ(x) = Ψλ andif α = β 2
2wecouldtake ϕ ′ 2
(βx)
λ (x) = ψλ .Inthiscaseitisclear
2
that σ(ϕλ) = σ(ϕ ′ λ ).Howeverif αisnotasquare,wetake ϕ′
2
αx
λ (x) = ψλ .Then
Withthischoiceof ϕλ,
σ(ϕλ) + σ(ϕ ′
λ ) = 2
x∈λ ψλ
ϕλ(x) = ψλ
− x2
2
x
= 0.
2
sothat σ(ϕλ) = ν(−1) σ(ϕλ).Moreoveritiswellknownandeasilyverifiedthat σ(ϕλ) = 0.
Since
{σ(ϕλ)} 4 = {ν(−1)} 2 |σ(ϕλ)| 4 = |σ(ϕλ)| 4
wehave
σ(ϕλ) ∼2p |σ(ϕλ)|.
Theabsolutevalueontherightisofcoursetheordinaryabsolutevalue.
Suppose pis2. Againanychoiceof ϕλand ϕ ′ λwilldo. Inthiscase αisnecessarilya
square.Let α = β2 .Wecantake ϕ ′ λ (x) = ϕλ(βx).Then σ(ϕ ′ λ ) = σ(ϕλ).Itisenoughtoprove
thesecondassertionforany ψλandany ϕλ.Let φbetheprimefieldandlet ψφbetheunique
non­trivialadditivecharacterof φ.Take ψλ = ψλ/φ.Let ϕφ(0) = 1, ϕφ(1) = i.Oneverifies
byinspectionthat
ϕφ(x + y) = ϕφ(x)ϕφ(y)ψφ(xy).
Take ϕλ = ϕ λ/φ.Since
itisenoughtoverifythat
Since σ(ϕφ) = −1 + i,thisisnoproblem.
σ(ϕ λ/φ) = {σ(ϕφ)} [λ:φ] ,
σ(ϕφ) ∼2 |σ(ϕφ)|.
2

Chapter9 86
If aisanon­zerocomplexnumberset
A [a] = a
|a| .
Thefollowinglemmaexplainsourinterestinthenumbers σ(ϕλ).
Lemma 9.3
Suppose Lisanon­archimedeanlocalfieldand χL isaquasi­characterof CL with
m = m(χL) = 2d + 1,where disapositiveinteger.Let ψLbeanon­trivialadditivecharacter
of Landlet n = n(ψL).Let γbesuchthat γOL = P m+n
L andlet βbeaunitsuchthat
βx
χL(1 + x) = ψL
γ
for xin P d+1
L .Choose δsothat δOL = P d L andlet ψλbethecharacterof λ = OL/PLdefined
by
If ϕλisdefinedby
then
and
ψλ(x) = ψL
ϕλ(x) = ψL
βδx
γ
βδ 2 x
γ
.
χ −1
L (1 + δx)
ϕλ(x + y) = ϕλ(x)ϕλ(y)ψλ(xy)
∆3(χL, ψL, γ) = A [−σ(ϕλ)].
Inthestatementofthislemmawehavenotdistinguished,inthenotation,betweenan
elementof OLanditsimagein λ.Thisisconvenientandnottooambiguous.Itwillbedone
again.Theonlyquestionablepartofthelemmaistherelation
Since
wehave
χ −1
L
ϕλ(x + y) = ϕλ(x)ϕλ(y)ψλ(xy).
(1 + δx) (1 + δy) ≡ (1 + δx + δy) (1 + δ 2 xy) (modP m L )
(1 + δx) χ−1
L
(1 + δy) = χ−1
L (1 + δx + δy) ψL
− βδ2
xy
.
γ

Chapter9 87
Therequiredrelationfollowsimmediately.
Thereareafewremarkswhichweshallneedlater. Itisconvenienttoformulatethem
explicitlynow.Weretainthenotationofthepreviouslemma.
Lemma 9.4
If m(µL) < m(χL)then
∆3(µL χL, ΨL; γ) ∼p ∆3 (χL, ΨL; γ)
andif m(µL) ≤ dwemaytake β(µLχL) = β(χL)andthen
∆3(µLχL, ψL; γ) = ∆3(χL, ψL; γ).
Inbothcases m(µLχL) = m(χL).Moreoverif x ∈ P2d L
β(µLχL)x
ψL
= µL(1 + x) χL(1 + x) = χL(1 + x)
γ
whichinturnequals
Thus
andif
while
ψL
β(χL)x
γ
.
β(µLχL) ≡ β(χL) (modPL)
ψλ(x) = ψL
ψ ′ λ(x) = ψL
β(χL)δ2
x
γ
β(µLχL)δ2
x
then ψλ = ψ ′ λ .Thefirstassertionofthelemmanowfollowsfromtheprevioustwolemmas.
Itisclearthatwecantake β(µLχL) = β(χL)if m(µL) ≤ d.Letthecommonvalueofthetwo
numbersbe β.Then
isequalto
ψL
βδx
γ
ψL
µ −1
L
βδx
γ
γ
(1 + δx)χ−1 L (1 + δx)
χ −1
L (1 + δx).

Chapter9 88
Weseenowthatthesecondassertioniscompletelytrivial.
Thereisacorollaryofthislemmawhichitisconvenienttoobserve.
Lemma 9.5
Suppose m(χL) = 2d + εwhere disapositiveintegerand εis0or1.If m(µL) ≤ dand
µLisoforder rthen
∆(µLχL, ψL) ∼r ∆(χL, ψL).
Choose γintheusualwaysothat
and
Itisclearthat
Ifwetake
then,clearly,
∆(χL, ψL) = χL(γ) ∆1(χL, ψL; γ)
∆(µLχL, ψL) = χL(γ)µL(γ) ∆1(µLχL, ψL; γ).
µL(γ) ∼r 1.
β(µLχL) = β(χL)
∆2(µLχL, ψL; γ) ∼r ∆2(χL, ψL, γ).
TocompletetheproofofLemma9.5wehaveonlytoappealtoLemma9.4.
Lemma 9.6
Suppose Kisanunramifiedextensionof Land χLisaquasi­characterof CLwith
m = m(χL) = 2d + 1
where disapositiveinteger.Let ψLbeanon­trivialadditivecharacterof Fandlet n = n(ψL).
Suppose
βx
χL(1 + x) = ψL
̟ m+n
L
for xin P d+1
L .Take
If
ϕλ(x) = ψL
β(χL) = β(χ K/L) = β.
d β̟Lx ̟ m+n
L
χ −1
L (1 + ̟d Lx)

Chapter9 89
andif
ϕκ(x) = ψ K/L
d β̟Lx ̟ m+n
L
χ −1
K/L (1 + ̟d L x)
for xin κ = 0K/PKthen ϕκ = ϕ κ/λ.Moreoverif [K : L] = ℓthen
∆3(χ K/L, ψ K/L, ̟ m+n
L ) = (−1) ℓ−1 {∆3(χL, ψL, ̟ m+n
L )} ℓ .
Onceweprovethat ϕκ = ϕ κ/λthislemmawillfollowfromLemmas9.1and9.3. If x
belongsto Klet E 2 (x)bethesecondelementarysymmetricfunctionof xanditsconjugates
over L.If xbelongsto OK
Since
wehave
N K/F(1 + ̟ d L x) ≡ (1 + ̟d L S K/Lx) (1 + ̟ 2d
L E2 (x)) (modP m L ).
E 2 (x) ≡ ω κ/λ(x) (modPF)
ϕκ(x) = ϕλ(S K/Lx)ψλ(−ω κ/λ(x)) = ϕ κ/λ(x).
Nowsuppose Kisaramifiedabelianextensionof Land [K : L] = ℓisanoddprime.Let
G = G(K/L)andsuppose G = Gtwhile Gt+1 = {1}.Suppose
isgreaterthanorequalto t + 1and
for xin P d+1
L .Let
andif xbelongsto OKset
Supposealsothat
ϕ ′ λ(x) = ψ K/L
m = m(χL) = 2d + 1
χL(1 + x) = ψL
βx
̟ m+n
L
d ′
ℓ − 1
= ℓd − t
2
β̟d′ Kx ̟ m+n
L
χ −1
L (1 + S K/L(̟ d′
Kx) + E 2 (̟ d′
Kx)).
̟L = N K/L̟K.

Chapter9 90
Theassumptionslisted,wemaynowstatethenextlemma.
Lemma 9.7
If
ε = S K/L
̟2d′ K
̟2d
L
then εisaunit.Moreover ϕ ′ λ isafunctionon λ = OL/PL + OK/PKwhichsatisfies
if
If
then
Since
thenumber
ϕ ′ λ (x + y) = ϕ′ λ (x)ϕ′ λ (y)ψλ(εxy)
ϕλ(x) = ψL
ψλ(u) = ψL
βx
̟ d+1+n
L
βu
̟ 1+n
.
L
A[σ(ϕλ)] ℓ = A[σ(ϕ ′ λ )].
d ′ + (ℓ − 1) (t + 1)
ℓ
S K/L (̟ d′
Kx)
χ −1
L (1 + ̟d Lx)
liesin Pd L .Moreover E2 (̟d′ Kx)isasumoftracesofelementsin P2d′ K .Since
2d ′ + (ℓ − 1) (t + 1)
ℓ
itliesin P 2d
L .If xliesin PKitliesin P m L and
liesin P d+1
L because
d ′ + 1 + (ℓ − 1) (t + 1)
ℓ
= 2d +
S K/L(̟ d′
Kx)
= d + 1 +
≥ d
ℓ − 1
ℓ
t(ℓ − 1)
2ℓ
≥ 2d
≥ d + 1.

Chapter9 91
Thusif xbelongsto PK
Since
isequalto
theexpression
ϕ ′ λ(x) = ψL
β
̟ m+n
L
SK/L (̟ d′
Kx) χ −1
L (1 + SK/L(̟ d′
Kx)) = 1.
E 2 (̟ d′
K (x + y))
E 2 (̟ d′
Kx) + E 2 (̟ d′
Ky) + S K/L(̟ d′
Kx) S K/L(̟ d′
Ky) − S K/L(̟ 2d′
K xy),
iscongruentmodulo P m L totheproductof
and
and
Thus
Since
1 + S K/L(̟ d′
K(x + y)) + E 2 (̟ d′
K(x + y))
1 + S K/L(̟ d′
K x) + E2 (̟ d′
K x)
1 + S K/L(̟ d′
K y) + E2 (̟ d′
K y)
1 − S K/L(̟ 2d′
K xy).
ϕ ′ λ (x + y) = ϕ′ λ (x)ϕ′ λ (y)ψλ(εxy).
2d ′ + (ℓ − 1) (t + 1)
ℓ
≥ 2d
thenumber εisin OL.Weconcludeinparticularthatif ybelongsto PKthen
and
ϕ ′ λ(x + y) = ϕ ′ λ(x).
If t = 0let σbeageneratorof Gandlet ̟ 1−σ
K = ν.Inthiscase 2d′ = 2dℓand
̟ 2d′
K
̟ 2d
L
=
τ∈G ̟1−τ
K
2d
ε ≡ ℓν dℓ(ℓ−1) (modPL)
≡ ν dℓ(ℓ−1) (modPL)

Chapter9 92
isaunit.If t > 0
sothat
̟ 2d′
K
ε ≡ S K/L
≡ ̟2d−t
L ̟t K
t ̟K ̟ t L
(mod PK)
(mod PL).
ItisshowninparagraphV.3ofSerre’sbookthattherightsideofthiscongruenceisaunit.
Then
Firsttake poddandlet
σ(ϕλ) = −ψλ
ϕλ(x) = ψλ
x 2
2
2 −α
ψλ
2
+ αx = ψλ
2 (x + α)
2
2 (x + α)
= −ψλ
2
MakinguseofthecalculationsintheproofofLemma9.2weseethat
A[σ(ϕλ)] ℓ = νλ(−1) ℓ−1
2 ψλ
of νλisthenon­trivialquadraticcharacterof λ × .
Since
iscongruentto
modulo P m L thevalueof ϕ′ λ (x)is
if
Thus
ψλ
1 + S K/L
2 −ℓα
2
ψλ
−α 2
2
.
2 −α
ψλ
2
A −
λ ψλ
̟ d′
Kx
+ E 2 (̟ d′
Kx) {1 + S K/L(̟ d′
Kx)} {1 + E 2 (̟ d′
Kx)}
(SK/Ly) 2 + 2α SK/Ly − 2E2
(y)
ϕ ′ λ (x) = ψλ
2
y = ̟d′ K
̟d x.
L
(SK/L(y + α) 2 ) − ℓα 2
2
.
2 x
2
x 2
2
.

Chapter9 93
Replacing xby
andsummingwefindthat
σ(ϕ ′ λ) = ψλ
−ℓα 2
2
x − ̟d L
̟d′ α
K
−
x ψλ
εx 2
Collectingthisinformationtogetherweseethattoprovethelemmawhentheresidual
characteristic pisoddwemustshowthat
νλ(−1) ℓ−1
2 = νλ(ε).
Since ν dℓ(ℓ−1) iscertainlyasquarewehavetoshowthat
νλ(−1) ℓ−1
2 = νλ(ℓ)
when t = 0. Ifthefield λisofevendegreeovertheprimefieldbothsidesare1. Ifnot,an
oddpowerof piscongruentto1modulo ℓandtherelationfollowsfromthelawofquadratic
reciprocity.If t > 0then
t ̟K ε ≡ SK/L (modPL)
̟ t L
andwecanappealtoparagraphV.3foraproofthat
ε + u p−1 ≡ 0
hasasolutionin λ.Thus νλ(ε) = νλ(−1)andwehavetoshowthat
νλ(−1) p−3
2 = 1.
2
.
If p ≡ 1 (mod4)then νλ(−1) = 1andif p ≡ 3 (mod4)theexponentiseven.
Beforeconsideringthecase p = 2,weremarkasimpleconsequenceofthepreceding
discussion.

Chapter9 94
Lemma 9.8
If pisoddlet
If t = 0and
then
andif t > 0
with
Inbothcases
ϕ ′ λ
ϕ ′ λ(x) = ψλ
ϕλ(x) = ψλ
x 2
2
(ℓ−1)
dℓ
µ = ν 2
x
= ψλ ℓ
µ
ϕ ′ λ(x) = ψλ
x 2
+ αx .
2
εx 2
x
+ αx
.
(SK/L (y + α) 2 ) − ℓα2
y = ̟d′ K
̟d x.
L
If t = 0then y = µx.Thusif xbelongsto OL,aswemayassume,
ϕ ′ λ
x
= ψλ
µ
If t > 0then ℓ = pisoddand
d ′ ℓd + (ℓ − 1) (t + 1)
ℓ
sothat,if x ∈ OF,
sothat
= 1
ℓ
2 2
ℓ(x + α) − ℓα
2
2
= ψλ
2 x
ℓ
2
(ℓ − 1)
(ℓ − 1) (t − 1) − t =
2
1
ℓ
S K/L(y + α) 2 ≡ εx 2
(mod PL).
Nowtake p = 2sothat tisnecessarily0andagainlet
̟ d′
K
̟ d L
(ℓ−1)
dℓ
µ = ν 2
≡ µ (mod PK).
+ αx .
(ℓ − 1)
(ℓ − 1) + t ≥ 1
2

Chapter9 95
If xisin OLand y = x
µ then
isequalto
Since
ψL
1 + ℓ̟ d Lx +
modulo P m L wehave
whichequals
Moreover
ℓ x
̟ d+1+n
L
ℓ(ℓ − 1)
2
ϕ ′ λ
ϕ ′ λ
χ −1
L
x
= ϕ
µ
′ λ (y)
1 + ℓ ̟ d ℓ(ℓ − 1)
Lx + ̟
2
2d
L x2
.
̟ 2d
L x 2 ≡ (1 + ℓ̟ d
ℓ(ℓ − 1)
Lx) 1 + ̟
2
2d
L x 2
x
= ϕλ(ℓ x) ψλ
µ
{ϕλ(x)} ℓ .
−ℓ(ℓ − 1)
2
x 2
{ϕλ(x)} 2 = ϕλ(2x) ψλ(−x 2 ) = ψλ(−x 2 ).
Sincethecharacteristicis2thereisan α = 0suchthat
Thenthecomplexconjugateof ϕλ(x)is
and
whichequals
isequalto
Consequently
Since
ψλ(x 2 ) = ψλ(αx).
ϕλ(x)ψλ(αx)
σ(ϕλ) = −
−
x ϕλ(x + α)
x ϕλ(x)ϕλ(α)ψλ(αx)
ϕλ(α) σ(ϕλ).
A[σ(ϕλ)] ℓ = ϕλ(α) ℓ−1
2 A[σ(ϕλ)].
{ϕλ(x)} 4 = 1

Chapter9 96
wehave
if ℓ ≡ 1 (mod4)and
if ℓ ≡ 3 (mod4).
Wehavetoshowthat
if ℓ ≡ 1 (mod4)andthat
A[σ(ϕ ′ λ )] = A[σ(ϕλ)]
A[σ(ϕ ′ λ )] = A[σ(ϕλ)] = ϕ −1
λ (α) A[σ(ϕλ)]
ϕλ(α) ℓ−1
2 = 1
ϕλ(α) ℓ+1
2 = 1
if ℓ ≡ 3 (mod4). Theserelationsareclearif ℓiscongruentto1or7modulo8. Ingeneralif
ℓ ≡ 1 (mod 4)
ϕλ(α) ℓ−1
(ℓ − 1) (ℓ − 3)
2 = ψλ − α
8
2
andif ℓ ≡ 3 (mod4)
ϕλ(α) ℓ+1
2 = ψλ
−
(ℓ + 1) (ℓ − 1)
α
8
2
.
Let φbetheprimefieldandlet ψφbeitsnon­trivialadditivecharacter. Choose α1such
that
Then
and α = α1.Thus
ψλ(x 2 ) = ψ λ/φ
ψ λ/φ(x) = ψλ (α 2 1 x).
2 x
= ψλ/φ α 2 1
x
ψλ(α 2 ) = ψ λ/φ(1).
α1
= ψλ (α1x)
Therightsideis+1or–1accordingas f = [λ : φ]isevenorodd.But ℓdivides 2 f − 1sothat,
bythesecondsupplementtothelawofquadraticreciprocity, fisevenif ℓiscongruentto3or
5modulo8.
ThereisacomplementtoLemma9.7.
Lemma 9.9

Chapter9 97
If m(χL) ≥ 2(t + 1)choose β(χK/L) = β(χL) = βin OL.If t + 1 < m(χL) < 2(t + 1)
choose β(χL) = βand
β(χK/L) = β − β1
α
asinLemma8.9.Then m(χ K/L) = 2d ′ + 1and
isequalto
ψK/L
ψ K/L
FromLemma8.8wehave
asrequired.If d ≥ t + 1then
because ℓisodd.Moreover,
Consequently
β ̟d′ Kx ̟ m+n
L
α1
β(χK/L) ̟d′ Kx ̟ m+n
χ
L
−1
K/L (1 + ̟d′ Kx) m(χ K/L) = 1 + t + ℓ(m − 1 − t) = 2
d ′ ≥
(ℓ + 1)
2
3d ′ + (ℓ − 1) (t + 1)
ℓ
χ −1
L (1 + SK/L (̟ d′
Kx) + E 2 (̟ d′
Kx)).
d +
(ℓ − 1)
2
(ℓ − 1)
ℓ d − t + 1
2
(d − t) ≥ m
≥ m′ + (ℓ − 1) (t + 1)
ℓ
= m.
N K/L(1 + ̟ d′
Kx) ≡ 1 + S K/L(̟ d′
Kx) + E 2 (̟ d′
Kx) (modP m L )
andthelemmaisvalidif m ≥ 2(t + 1).
sothat
If t + 1 < m < 2(t + 1)westillhave
3d ′ + (ℓ − 1) (t + 1)
ℓ
N K/L(1 + ̟ d′
Kx)
≥ m

Chapter9 98
iscongruentto
1 + S K/L(̟ d′
K x) + E2 (̟ d′
K x) + N K/L(̟ d′
K x)
modulo P m L .Since d′ ≥ d + 1thisiscongruentto
modulo P m L .Certainly
Moreover,if m = t + 1 + v
{1 + S K/L(̟ d′
Kx) + E 2 (̟ d′
Kx)} {1 + N K/L(̟ d′
Kx)}
χL(1 + NK/L(̟ d′
Kx)) = ψL
d ′ − v = d +
ℓ − 3
2
β NK/L(̟d′ Kx) ̟ m+n
.
L
v ≥ d ≥ s
if sistheleastintegergreaterthanorequalto t
2 .Thus,justasintheproofofLemma8.5,
isequalto
ψL
β NK/L(̟d′ Kx) ̟ m+n
= ψL
L
ψL
Multiplyingtheinverseofthiswith
weobtain
Thelemmafollows.
ψ K/L
⎛
If m = t + 1wemaystillchoose
⎝ −S K/L
⎛
⎝ α N K/L
αβ1
α1 ̟d′
K x
̟ m+n
L
β1
α1 ̟d′
K x
̟ m+n
L
⎞
⎠ .
β − αβ1
d ̟
α1
′
Kx ̟ m+n
L
ψ K/L
β ̟d′ Kx ̟ m+n
.
L
β(χ K/L) = β − β1
α
α1
⎞
⎠

Chapter9 99
asinLemma8.9.Howevertherelationbetween ϕλ(x)and
ϕK(x) = ψ K/L
β(χK/L) ̟d′ Kx ̟ m+n
χ
L
−1
K/L (1 + ̟d′ Kx)
willbemorecomplicated.Here x = OK/PKisthesamefieldas λ = OL/PL.Weintroduce
itonlyfornotationalpurposes.
Since
and
Because m = t + 1thenumber tisatleast2and
theexpression
iscongruentto
modulo P m L and
isequalto
d = d ′ = t
2 .
3d + (p − 1) (t + 1)
p
d + (p − 1) (t + 1)
p
N K/L(1 + ̟ d K x)
≥ t + 1
≥ d + 1
{1 + S K/L(̟ d Kx) + E 2 (̟ d Kx)} {1 + ̟ d L N K/Lx}
ΨL
AccordingtoNewton’sformulae
Thus
χL(1 + S K/L(̟ d Kx) + E 2 (̟ d Kx))
β SK/L(̟d Kx) + β E2 (̟d Kx)
.
̟ m+n
F
S K/L(̟ 2d
K x 2 ) − S K/L(̟ d Kx) 2 + 2E 2 (̟ d Kx) = 0.
E 2 (̟ d Kx) ≡ − 1
2 S K/L(̟ 2d
K x 2 ) (mod P m L ).
Observethat pisequalto ℓandtherefore,inthepresentcircumstances,odd.

Chapter9 100
with
Certainly
Let µLbeacharacterin S(K/L)asinLemma8.9(c)andlet
ψL
αx
̟ d+1+n
L
µ −1
L (1 + ̟d Lx) = ψλ
ρ = α
β .
µL(NK/L(1 + ̟ d Kx)) = 1
if xbelongsto OK.Replacing xby β1
α1 xweseethat
ψL
isequalto
if
1
̟ m+n
β1
αSK/L ̟
α1
L
d
Kx − α
2 SK/L z = 1
̟ d L
whichiscongruentto
modulo PL.
equal
Let
If xbelongsto OK
χ −1
L
β1
SK/L α1
ψλ
ρ z2
2
̟ d
Kx − 1
2 SK/L ϕλ(x) = ψL
β 2 1
α 2 1
+ τ z
β 2 1
α 2 1
β
α NK/Lx ≡ β
α xp
ψλ
1 + ̟ d L N K/Lx = ψλ
β x
̟ d+1+n
L
x 2
2
x 2p
2
+ σ x .
ρ x2
2
+ τ x
̟ 2d
K x 2
+ NK/L β1 ̟ d Kx
̟ 2d
K x 2
β1
+ NK/L ̟
α1
d
Kx
χ −1
L (1 + ̟d L x)
+ σ xp
ψL
−β NK/Lx
.
̟ d+1+n
L

Chapter9 101
Wenowputthesefactstogethertofindasuitableexpressionfor ϕκ(x).Wemayaswell
take xin OL.Then ϕκ(x)istheproductof
β ̟
ψK/L d′
Kx ̟ m+n
L
and
and
and
if
ψL
ψ K/L
−
− β1α
α1
β
̟ m+n
L
̟d Kx ̟ m+n
L
χ −1
L (1 + ̟d L N K/Lx)
SK/L(̟ d Kx) − 1
2 SK/L(̟ 2d
K x 2
) .
Thesecondofthesethreeexpressionsisequaltotheproductof
2p x −1 x2p
ψλ + σ xp ψλ −ρ
2 2 − ρ−1τ x p
ψ K/L
− αβ2 1 ̟2d
K x2
2α2 1 ̟m+n
L
ε = S K/L
Theproductofthefirstandthirdisequalto
2 εx
ψλ
= ψλ
2d ̟K ̟2d
.
L
2
.
− ερβ2 1
2α 2 1
x 2
AsproveninparagraphV.3ofSerre’sbooktheelementsof ULcongruentto
modulo P t+1
L areallnorms,sothat
Inparticular
ψλ
1 + (εx + x p ) ̟ t L
ψλ(ρ x p ) = ψλ(−ρεx).
− ερβ2 1
2α2 x
1
2
= ψλ
ρ
−1 x2p
∗ (1998)ThemanuscriptofChapter9endswiththisformula.
2
. ∗

Chapter11 102
Chapter Eleven.
Artin–Schreier Equations
ThetheoryofArtin–SchreierequationsiscentraltoDwork’sproofofthesecondmain
lemma. Wefirstreviewthebasictheory,whichwetakefromMackenzieandWhaples[11],
andthenreviewDwork’sratheramazingcalculations.ThesewetakefromLakkis[9].
WestartwithanexercisefromSerre’sbook[12].Suppose Fisanon­archimedeanlocal
fieldand K/FisGalois.Let pbetheresidualcharacteristic.Withtheconvention (0) = P ∞ F
welet
p OF = P e F.
Suppose G = G(K/F)and σ ∈ Giwith i ≥ 1.Let
with ain P i K .Let
̟ σ K = ̟K(1 + a)
ϕ(x) = x σ − x.
ϕisan F­linearoperatoron K.If x = α̟ j
Kbelongsto Pj
Kthen iscongruentto
ϕ(x) = x σ − x = (α σ − α) ̟ jσ
K
α̟ j
K
modulo P i+j+1
K .Thisinturniscongruentto
modulo P i+j+1
K .
If*
then,asanoperator,
+ α(̟jσ
K
− ̟K)
̟ j(σ−1)
K − 1 = α̟ j
K {(1 + a)j − 1}
(α̟ j
K ) (ja) = jax
ψ(x) = x σp
− 1
ψ = (1 + ϕ) p − 1 = p
k=1
p
ϕ
k
k .
*Weseemtobedealingwithyetanotheruseofthesymbol ψ!

Chapter11 103
If xbelongsto P j
K then
and ψ(x)iscongruentto
orto
modulo P i+j+e′ +1
K
ϕ k (x) ≡ j(j + i) . . .(j + (k − 1)i)a k x (modP j+ki+1
K )
if p OK = P e′
K .
Wededucethefollowingcongruences:
(i)If (p − 1)i > e ′ then
(ii)If (p − 1)i = e ′ then
(iii)If (p − 1)i < e ′ then
pjax + j(j + i) . . .(j + (p − 1)i)a p x
pjax + j(j p−1 − i p−1 )a p x
ψ(x) ≡ pjax (mod P i+j+e′ +1
K ).
ψ(x) ≡ pjax + j(1 − i p−1 )a p x (modP i+j+e′ +1
K ).
ψ(x) ≡ j(1 − i p−1 )a p x (mod P pi+j+1
K ).
Observethatif (j, p) = 1and σbelongsto Gi,with i ≥ 1,then
ϕ(x) ≡ 0 (modP i+j+1
K )
forall xin P j
K ifandonlyif σbelongsto Gi+1.Itfollowsimmediatelythatif σbelongsto G1
and i ≥ 1then
ϕ(x) ≡ 0 (modP i+j
K )
forall xin P j
Konlyif σbelongsto Gi.
If σisreplacedby σPthen ϕisreplacedby ψ.If k > e′
p−1 and Gk = {1}then,forsome
i ≥ k, Gi = {1}and Gi+1 = {1}.Taking (j, p) = 1weinferfrom(i)thatif σbelongsto Gi
butnotto Gi+1then σ p isin Gi+e ′butnotin Gi+e ′ +1.Thisisimpossible.Thus Gk = {1}if

Chapter11 104
k > e′
p−1 .If G1 = {1}then pdivides e ′ sothatif (p − 1)i = e ′ thenumber iisalsodivisibleby
p.Thecongruence(ii)reducesto
ψ(x) ≡ j (pa + a p ) (modP i+j+e′ +1
K ).
Thusif σbelongsto Giits pthpower σ p liesin Gi+eandistherefore1.Consequently
Letting a = α̟ i K and p = β̟e K wefindthat
pa + a p ≡ 0 (modP pi+1
K ).
α p + βα ≡ 0 (mod PK).
Sincethiscongruencehasonly prootstheimageof Θiliesinasubsetof Ui K /Ui+1
K
elementsand Giiseither {1}orcyclicoforder p.
with p
If (p − 1) < e ′ and (i, p) = 1thecongruence(iii)impliesthat σ p belongsto Gpi+1if σ
belongsto Gi.Howeverif (p −1)i < e ′ and pdivides iitshowsthat σ p belongsto Gpibutnot
to Gpi+1if σbelongsto Gibutnotto Gi+1.Thus σ −→ σ p definesaninjectionof Gi/Gi+1into
Gpi/Gpi+1.If Gi/Gi+1isnottrivialneitheris Gpi/Gpi+1and (p − 1)pi ≤ e ′ .If (p − 1)pi < e ′
wecanrepeattheprocess.Thus,forsomepositiveinteger h, (p − 1)p h i = e ′ and G p h iisnot
trivial.Itisthencyclicoforder p.AccordingtoPropositionIV.10ofSerre’sbookthose k ≥ 1
forwhich Gk/Gk+1 = {1}areallcongruentmodulo p.Inparticularif Gk/Gk+1isnottrivial
forsome k ≥ 1divisibleby pitisnottrivialonlywhen kisdivisibleby p. Thepreceding
discussionshowsthatif iisthesmallestvalueof k ≥ 1forwhich Gk/Gk+1isnon­trivialthen
any σin Gibutnotin Gi+1generates Gi = G1.Inotherwords:
Lemma 11.1
If G1isnotcyclicthen (i, p) = 1if i ≥ 1and Gi/Gi+1 = {1}.
Lemma 11.2
Suppose K/Liscyclicofprimedegreeand G = G(K/L)isequalto GLwith t ≥ 1and
(t, p) = 1.Thenthereisa∆in Kandan ain Lsuchthat aOL = P −t
L and
∆ p − ∆ = a.
Weobservefirstofallthat [K : L]mustbe pandthatif pOK = P e′
K then (p − 1)t < e′ .If
xbelongsto Kthesymbol O(x)willstandforanelementin xOKandthesymbol o(x)will
standforanelementin xPK.If
x = p−1
i=0 ai ̟ i K

Chapter11 105
with aiin Fthen
Moreoverif σisageneratorof G
andif ̟ σ−1
K = (1 + a̟t K )
for 1 ≤ i < p.Thus
|x| = max
0≤i

Chapter11 106
Toprovethelemmaweconstructasequence Λ0, Λ1, Λ2, . . .andasequence Θ0, Θ1, . . .
withthefollowingproperties:
(i) vK(Λn) = −tforall n ≥ 0.
(ii)If σisagivengeneratorof Gand ζisagiven (p − 1)throotofunity
(iii)
(iv)
(v)
and
Λ σ n − Λn = ζ + o(1).
p(Λ σ n ) − p(Λn) = Θ σ n
− Θn
|Θ σ n − Θn| = |̟ t K| |Θn|.
Λn+1 = Λn + Θn.
p(Λ σ n+1 ) − p(Λn+1) = o(p(Λ σ n ) − p(Λn)).
Itwillfollowfrom(iii)and(v)that {Θn}isconvergingto0.Then(iv)impliesthat {Λn}
hasalimit ∆. (i)impliesthat vK(∆) = −tand(v)impliesthat ∆ p − ∆ = abelongsto F.
From(ii)
∆ σ − ∆ = ζ + o(1).
if
Toconstruct Λ0let αbelongto U i K andconsider
ασ ̟
σ t
K
Wecanchoose αsothat
Thenweset
− α
̟ t K
= ασ − α
̟σ t
K
+ α
̟ t K
̟ σ K = ̟K(1 + a̟ t K).
−taα = ζ + o(1).
Λ0 = α
̟t .
K
̟ t(1−σ)
K − 1 = −taα + o(1)
Weobserveinpassingthatconditions(i)and(ii)determine Λnmodulo P −t+1
K .

Chapter11 107
Suppose Λ0, . . ., Λnhavebeendefined.Then
whichequals
Choose Θnsothat
and
Then vK(Θn) > −tandif
vK(Λn+1) = −t.Moreover
and
with x = o(Θn).Then
Also
p(Λ σ n ) = p(Λn) + p(ζ + o(1)) + o(1)
p(Λn) + p(ζ) + o(1) = p(Λn) + o(1).
Θ σ n − Θn = p(Λ σ n ) − p(Λn)
|Θ σ n − Θn| = |̟ t K | |Θn|.
Λn+1 = Λn + Θn
Since vK(Θ σ n − Θn)ispositivetherightsideis
Thus
whichequals
is
Λ σ n+1 − Λn+1 = Λ σ n − Λn + o(1) = ζ + o(1).
p(Λn+1) = p(Λn) + p(Θn) + x
x σ − x = o(Θ σ n − Θn) = o(p(Λ σ n ) − p(Λn)).
p(Θ σ n) − p(Θn) = p(Θ σ n − Θn) + o(Θ σ n − Θn).
−(Θ σ n − Θn) + o(Θ σ n − Θn).
p(Λ σ n+1 ) − p(Λn+1)
p(Λ σ n ) − p(Λn) − (Θ σ n − Θn) + o(p(Λ σ n ) − p(Λn))
o(p(Λ σ n ) − p(Λn)).

Chapter11 108
Lemma 11.3
Supposer ∆1belongsto K, abelongsto L, vL(a) = −tand
with r ≥ 1.Define ∆ninductivelyby
Then
if n ≥ 2andif r ≥ (e ′ − (p − 1)t)
Moreover
existsand ∆ p − ∆ = a.
∆ p
1 − ∆1 = a + O(̟ r K )
∆n+1 = ∆ p n − a.
∆n+1 − ∆n = o(∆n − ∆n−1)
∆n+1 − ∆n = O(̟ r+(n−1)(e′ −(p−1)t)
K
).
lim
n→∞ ∆n = ∆
Thelastassertionisaconsequenceofthefirst.Itisclearthat
Suppose n ≥ 2,and
Then
isequalto
∆2 − ∆1 = O(̟ r K ).
∆n − ∆n−1 = x = o(1).
∆n+1 − ∆n = ∆ p n − ∆ p
n−1 = (∆n−1 + x) p − ∆n−1
p−1
k=1
p
k
whichis o(x)because e ′ − (p − 1)t > 0.If
and r ≥ e ′ − (p − 1)titis
∆ k n−1 xp−k
+ x p
x = O(̟ r+(n−2)(e′ −(p−1)t)
K
)
O(̟ r+(n−1)(e′ −(p−1)t)
K
).

Chapter11 109
Thelemmahasacoupleofcorollarieswhichshouldberemarked.
Lemma 11.4
If aisin L, vL(a) = −t, ∆p − ∆ = a,and ξisa(p − 1)throotofunitythereisanumber
∆ξsuchthat
∆ξ = ∆ + ξ + O(̟ e′ −(p−1)t
K )
and
∆ p
ξ − ∆ξ = a.
Relation(11.1)showsthat ∆ + ξsatisfiestheconditionsofthepreviouslemmawith
r = e ′ − (p − 1)t.
Lemma 11.5
Suppose ∆belongsto K, bbelongsto L, vL(b) = −tand
Thenforany uin U t+1
L theequation
hasasolutionin K.
∆ p − ∆ = b.
Λ p − Λ = bu
Take,inLemma11.3, a = buand ∆1 = ∆.Lemma11.5showsthatif Sisthesetofallin
Lwith vL(a) = −tforwhichtheequation
hasasolutionin Kthen S = SU t+1
L .
Lemma 11.6
∆ p − ∆ = a
If ℓistheintegralpartof t
thenumberofcosetsof Ut+1
p L in Sis
p − 1
p
[OL : PL] 1+ℓ .
Fixagenerator σof G = G(K/L).If abelongsto S, ∆ p − ∆ = a,and ξisa(p − 1)th
rootofunity
(ξ∆) p − ξ∆ = ξ a.

Chapter11 110
ByLemma11.4thereisa(p − 1)throotofunity ζsuchthat
Then
∆ σ = ∆ + ζ + o(1).
(ξ∆) σ = ξ∆ + ξζ + o(1).
Thusif S ′ isthesetof ain Lwith vL(a) = −tforwhich
with
a = ∆ p − ∆
∆ σ = ∆ + 1 + o(1)
thenumberofcosetsif U t+1
L in Sis p − 1timesthenumberofcosetsof U t+1
L in S ′ .
Choose ∆0,with vK(∆0) = −t,forwhich ∆ p
0 − ∆0 = a0isin Fand
If vK(∆) = −t, ∆ p − ∆isin F,and
then,accordingtoanearlierremark,
with Ω0 = o(∆0).
Since
wehave
Thus
∆ σ 0 = ∆0 + 1 + o(1).
∆ σ = ∆ + 1 + o(1)
∆ = ∆0 + Ω0
Chooseany Ω0 = o(∆0)andset Λ0 = ∆0 + Ω0.Accordingtotherelation(A)
p−1
σ p
Ω0 − Ωp0
=
i=1
p(Λ0) = p(∆0) + p(Ω0) + o(Ω0).
Ω σ 0 − Ω0 = O(̟ t K Ω0) = o(1)
p
i
Ω p−i
0 (Ω σ 0 − Ω0) i = o(Ω σ 0 − Ω0).
p(Λ0) σ − p(Λ0) = Ω σ 0 − Ω0 + o(̟ t KΩ0)

Chapter11 111
and p(Λ0)isin Lonlyif
Ω σ 0 − Ω0 = o(̟ t K Ω0),
thatis,onlyif Ω0 = α0 + o(Ω0)with α0in L.Ontheotherhandif
Ω σ 0 − Ω0 = o(̟ t K Ω0)
andweconstructthesequence Λ0, Λ1, Λ2, . . .asbeforeandlet
then
∆ = lim
n→∞ Λn
∆ = Λ0 + o(Ω0).
Weconcludethatthenumberofcosetsin Ps K /Ps+1
K , s > −t,containingan Ω0suchthat
(∆0 + Ω0) p − (∆0 + Ω0)
isin Lis1if pdoesnotdivide sandis [OL : PL]ifitdoes.
Choose ∆sothat
∆ p − ∆ = a
isin S ′ .If Ωbelongsto Ps K , s > −t,butnotto Ps+1
K and
(∆ + Ω) p − ∆ − Ω = b
isalsoin S ′ then aand bbelongtothesamecosetof U t+1
L ifandonlyif
If s > 0
butif s ≤ 0
and
ifandonlyif s = 0and
b = a + o(1).
p(∆ + Ω) = p(∆) + o(1)
p(∆ + Ω) = p(∆) + Ω p − Ω + o(Ω)
Ω p − Ω + o(Ω) = o(1)
Ω = ξ + o(1)
where ξissome (p − 1)throotofunity.Thelemmafollows.

Chapter11 112
let
If x = {x1, . . ., xn}let E i (x)bethe ithelementarysymmetricfunctionof x1, . . ., xnand
If Zisanindeterminateand
Q(Z) = n
S i (x) = n
then ∞
k=1 xi k.
i=0 (−1)i E i (x)Z i = n
i=1 Si (x)Z i
isclearly −Ztimesthelogarithmicderivativeof Q(Z).Thus
∞
i=1 Si (x) Z i
i=1
(1 − xiZ)
n
i=0 (−1)iE i (x)Z i
= − n
i=0 (−1)i iE i (x)Z i .
ThisidentitywhichwerefertoasNewton’sidentityisequivalenttotheformulaeofNewton.
Itimpliesinparticularthat
i−1
j=0 (−1)j S i−j (x) E j (x) = (−1) i+1 iE i (x) (11.2)
if 1 ≤ i ≤ n.WemaydivideNewton’sidentityby Q(Z)andthenexpandtheright­handside
toobtainexpressionsforthe S i (x)aspolynominalsin E 1 (x), . . ., E n (x).Thecoefficientsare
necessarilyintegers.Tocalculatethemwesupposethat x1, . . ., xnlieinafieldofcharacteristic
zero.Let
Q(Z) = 1 + P(Z).
Then
log Q(Z) = − ∞
k=1
(−1) k
Thecoefficientof Z i−1 inthederivativeoftherightsideis
−
k
α1+...+αn=k
α1+2α2+...+nαn=i
Thisexpressionisthereforeequalto −S i (x).
k
i(k − 1)!
α1! . . .αn!
(P(Z)) k .
n
j=1 {Ej (x)} αj .
Suppose K/Lisaramifiedcyclicextensionofdegree pand G = G(K/L).Let G = Gt
and Gt+1 = {1}.Suppose u ≤ t, Λisin K,and
ΛOK = P −u
K .

Chapter11 113
Wetake {x1, . . ., xn}tobe Λanditsconjugatesunder G.Inthiscasewewrite
and
E i (x) = E i K/L (Λ)
S i (x) = S i K/L (Λ).
If 1 ≤ i ≤ p − 1and γiisanyintegerlessthanorequalto
wehave
Wemaytake
−iu + (p − 1) (t + 1)
p
E i K/L (Λ) ≡ 0 (mod Pγi
L ).
γi ≥ − iu
p
(p − 1)t
+ .
p
If iu + tisnotdivisibleby pthisinequalitymaybesupposedstrict.
Suppose α1, . . ., αparenon­negativeintegers,
p
and p
If
γ =
p−1
i=1 αi = k
i=1 iαi = ℓ.
i=1 γiαi
−uαp.
Then p
i=1 {Ei (Λ)} αi γ
≡ O (̟K ). (11.3)
Wehave
ℓ u
γ ≥ −
p
(p − 1)
+ kt −
p
(p − 1)
αpt.
p
Theinequalityisstrictif αiisnon­zeroforsome isuchthat iu + tisnotdivisibleby p.
Werecordnowsomeinequalitiesthat γsatisfiesinvariousspecialcases. Theywillbe
neededlater.Weobservefirstofallthat,if 1 ≤ i < p, γiisnon­negativeandispositiveunless
pdivides iu + t.

Chapter11 114
(i)If ℓ = pand k = 2then
1 + u + γ ≥ 1 + t.
Inthiscase αp = 0andtheleftsideisatleast
If pisoddtheinequalityisstrict.
(ii)If ℓ = pand k = 2then
1 +
2(p − 1)
p
γ ≥ 0.
t ≥ 1 + t.
Moreovertheinequalityisstrictif pisodd.Thisstatementisofcourseweakerthanthat
of(i).
(iii)If ℓ = p, k ≥ 3,and pisodd,then
αpisagain0.Theleftsideisatleast
−u +
3(p − 1)
p
t = u +
γ ≥ u +
t − u
p
t − u
p
+ 1
p
{(3p − 4)t − (2p − 1)u}.
Thefinaltermisnon­negative. Theinequalityisstrictif u = t. If u = tand pdoesnot
divide uitisagainstrictforthen αi = 0forsome i < p − 1andforsuchan ithenumber
iu + tisnotdivisibleby p.
(iv)If k ≤ pthen
(p − 1)u + γ ≥ u +
t − u
p
exceptwhen αp = kor αp = p − 1.Wehavetoshowthat
(p − 2)u + γ +
Theleftsideisatleast
p − 2 − ℓ
1 p − 1
+ u +
p p p
u − t
p
≥ 0.
p−1
i=1 αi
− 1
t.
p

Chapter11 115
If αp = kthecoefficientof tispositiveandweneedonlyshowthatitisatleastasgreat
asthenegativeofthecoefficientof uorinotherwordsthat
(p − 1)
p−1
i=1 αi
Thisfollowsfromtheassumptionthat αp ≤ p − 2.
(v) If k ≤ p − 2and αp = kthen
Inthiscase
(p − 1)u + γ ≥ u.
γ ≥ −ku.
+ (p − 2)p ≥ ℓ.
Therearecircumstancesinwhichtheestimatesfor γiandthereforethosefor γcanbe
substantiallyimproved.Wewilldiscussthemshortly.
Supposenowthat K/FisatotallyramifiedGaloisextensionand G = G(K/F)isthe
directproductoftwocyclicgroupsoforder p. ByLemma11.1thesequenceoframification
groupsisoftheform
G = G−1 = G0 = G1 = . . . = Gu = Gu+1 = . . . = Gt = Gt+1 = {1}
with (u, p) = 1and u ≡ t (modp)oroftheform
G = G−1 = G0 = G1 = . . . = Gt = Gt+1 = {1}
with (t, p) = 1.Inthesecondcasewetake u = t.Inthefirstcaselet L1bethefixedfieldof Gt
andinthesecondlet L1beanysubfieldof Kofdegree pover F.Let L2beanysubfieldof K
differentfrom L1whichisalsoofdegree pover F.Let G i = G(K/Li)andlet
G i = G i si = Gi si+1 = {1}.
Then s1 = tand s2 = u.AccordingtoPropositionIV.4ofSerre’sbook
and
and
δ K/F = (p 2 − p) (u + 1) + (p − 1) (t + 1)
δK/L1 = (p − 1) (t + 1)
δK/L2 = (p − 1) (u + 1).

Chapter11 116
Thus
and
If G i = G(Li/F)and
then t1 = uand
Lemma 11.7
δL1/F = 1
p (δK/F − δK/L1 ) = (p − 1) (u + 1)
δL2/F = 1
p (δ (p − 1)
K/F − δK/L2 ) =
p
G i = G i
ti
t2 = u +
Suppose ∆belongsto K, vK(∆) = −u,and
belongsto L2.If Ybelongsto L2then
and
for 1 ≤ i ≤ p − 1.
= Giti+1
= {1}
t − u
p .
∆ p − ∆ = a
((p − 1) (u + 1) + t + 1).
vL1 (SK/L1 (Y ∆i )) ≥ (p − 1)t2 − it1 + vL2 (Y )
vL1 (Ei K/L1 (Y ∆)) ≥ (p − 1)t2 + i(vL2 (Y ) − t1)
Weshowfirstthatif θbelongsto L1and
with Yiin L2then
for 0 ≤ i ≤ p − 1.Since t1 = uand
θ = p−1
i=0
Yi∆ i
vL2 (Yi) ≥ it1 + vL1 (θ)
vK(θ) = min
0≤i≤p−1 {vK(Yi) − iu}
theinequalityisclearfor i = 0. Toproveitingeneral,weuseinductionone i. Suppose
0 < j ≤ p − 1andtheinequalityisvalidfor i < j.

Chapter11 117
Let
pOF = P e F .
Applyingtheexerciseatthebeginningoftheparagraphtotheextension L2/Fweseethat
pe ≥ (p − 1)t2 = (p − 1)
u +
t − u
.
p
If ξisany (p − 1)throotofunitythen,byLemmas11.3and11.4,thereisaσin G 2 suchthat
Wemaywrite
∆ σ = ∆ + ξ + O(̟ p2e−(p−1)u K ).
θ σ − θ = p−1
asalinearcombination p−1
withcoefficientsfrom L2.Since
i=1 Yi(∆ iσ − ∆ i )
i=0
Xi∆ i
vL2 (θσ − θ) ≥ vL1 (θ) + t1
wemayapplytheinductionassumptiontoseethat
Ontheotherhand
sothat θ σ − θisequalto
with
Thusif
vL2 (Xj−1) ≥ (j − 1))t1 + vL1 (θσ − θ) ≥ jt1 + vL1 (θ).
∆ iσ − ∆ i = (∆ + ξ) i − ∆ i + O(̟ p2e−(p−1)u−(i−1)u K
)
p−1
i=1 Yi
i−1
k=0
i
∆
k
k ξ i−k + η
η = O(θ̟ p2e−(p−2)u K ).
η = p−1
i
Zi∆
i=0

Chapter11 118
withthe Ziin L2wehave
But
whichequals
Since
wehave
vK(Zj−1) ≥ (j − 1)u + vK(Θ) + p 2 e − (p − 2)u.
p 2 e − (p − 2)u + (j − 1)u ≥ p(p − 1)u − (p − 2)u + (j − 1)u
((p − 1) 2 + j)u ≥ pju.
p−1
Xj−1 =
i=j Yi
vL2
p−1
i=j Yi
i
j
i
j
ξ i−j
+ Zj−1
ξ i−j
≥ ju + vL1 (θ)
forall ξ.Weobtaintherequiredestimatefor vL2 (Yj)bysummingover ξ.
Wenowshowthat
vL1 (SK/L1 (Y ∆i )) ≥ (p − 1)t2 − it1 + vL2 (Y )
for Y in L2and 1 ≤ i ≤ p − 1.Allweneeddoisshowthatforany θintheinversedifferent
of L1/F
orthatif θisin L1and
then
isin OF.
Let
S L1/F(θ̟ −vL 2 (Y )+it1−(p−1)t2
L1
with Yjin L2for 0 ≤ j ≤ p − 1.Then
S K/L1 (Y ∆i )) ∈ OF
vL1 (θ) ≥ −(p − 1) (t1 + 1) + it1 − (p − 1)t2 − vL2 (Y )
θ Y ∆ i = p−1−i
S L1/F(θ S K/L1 (Y ∆i )) = S K/L(θ Y ∆ i ) (11.4)
j=0
θ = p−1
j=0
Yj∆ j
Y Yj∆ j+i + (a + ∆) p−1
j=p−i Y Yj∆ j+i−p .

Chapter11 119
Since
wehave
for 1 ≤ i < p − 1and
Therelations(11.2)implythat
for 1 ≤ i < p − 1andthat
Thus(11.4)isequalto
if i < p − 1andtothesumofthisand
if i = p − 1.
Weknowthat
foreach j.Thus
∆ p − ∆ = a
E i (∆) = 0
K/L2
E p−1
K/L2 (∆) = (−1)p .
S K/L2 (∆i ) = 0
S K/L2 (∆p−1 ) = p − 1.
(p − 1) S L2/F(Y Yp−1−i) + S L2/F(paY Yp−i)
S L2/F(Y Yp−1)
vL2 (Yj) ≥ jt1 + vL1 (θ)
vL2 (Y Yp−1−i) ≥ (p − 1 − i)t1 − (p − 1) (t1 + 1) + it1 − (p − 1)t2
whichisatleast −(p − 1) (t2 + 1).Sois
If i = p − 1
vL2 (paY Yp−1) ≥ (p − 1)t2 − t1 − (p − 1) (t1 + 1) + it1 + (p − i)t1 − (p − 1)t2.
vL2 (Y Yp−1) ≥ (p − 1)t1 − (p − 1) (t1 + 1) + (p − 1)t1 − (p − 1)t2
isalsoatleast −(p − 1) (t2 + 1).Allweneeddonowisobservethat
S L2/F(P
−(p−1) (t2+1)
L2
) ⊆ OF.

Chapter11 120
Tocompletetheproofofthelemmawehavetoshowthat
vL1 (Ei K/L1 (Y ∆)) ≥ (p − 1)t2 + i(vL2 (Y ) − t1)
for 1 ≤ i ≤ p − 1. Thishasbeendonefor i = 1;soweproceedbyinduction.Applyingthe
relations(11.2)weseethat
(−1) i+1 iE i i−1
(Y ∆) = K/L1 j=0 (−1)j S i−j
(Y ∆) Ej (Y ∆).
K/L1 K/L1
Accordingtotheinductionassumptionandthefirstpartofthelemma,with Y replacedby
Y i−j ,atypicalterminthesumontherightis O(̟ v L1 )with
if j > 0and
v = (p − 1)t2 − (i − j)t1 + (i − j)vL2 (Y ) + (p − 1)t2 + j(vL2 (Y ) − t1)
if j = 0.Thelemmafollows.
v = (p − 1)t2 − it1 + ivL2 (Y )
Weapplythesecondestimatewith Y = 1toimprove,when Λ = ∆, L = L1,andcertain
auxiliaryconditionsaresatisfied,ourestimatesonthenumber γappearingin(11.3).
(vi)Suppose pisoddand
If k ≥ ν + 2and αp ≤ k − 2then
If k ≥ ν + 2and αp ≤ k − 1then
andif k ≥ ν + 1and αp ≤ k − 1
Inthepresentcircumstances
ℓ = (p − 1)ν + j + 1.
jt1 + γ ≥ pt2.
jt1 + γ ≥ (p − 1)t2 + t1
jt1 + γ ≥ (p − 1)t2 − t1.
γi ≥ (p − 1)t2 − it1

Chapter11 121
for 1 ≤ i ≤ p − 1.Thus
whichequals
or
If αp ≤ k − 2,thisisatleast
jt1 + γ ≥ jt1 + p−1
i=1 αi((p − 1)t2 − it1) − αpt1
jt1 + (p − 1)kt2 − ℓt1 − (p − 1)αp(t2 − t1)
(p − 1)kt2 − (p − 1)νt1 − t1 − (p − 1)αp(t2 − t1).
2(p − 1)t2 + (p − 1) (k − 2 − ν)t1 − t1
whichinturnisatleast pt2if p ≥ 3and k ≥ ν + 2.If αp ≤ k − 1
Therequiredinequalitiesfollow.
γ ≥ (p − 1)t2 + (p − 1) (k − 1 − ν)t1 − t1.
Weshallusealltheseestimatesfor γinthenextsequenceoflemmas.
Lemma 11.8
If ∆isasinLemma11.7and pisoddthen
SL1/FNK/L1∆ ≡ SL2/FNK/L2∆ (mod P1+t2
F ).
Theassertionofthelemmamaybereformulatedas
Noticethat
SK/L2NK/L1∆ ≡ SK/L1NK/L2∆ (modP1+pt2 ). L1
pt2 = t + (p − 1)u.
EarlierweappliedNewton’sidentitytoexpress S p
K/L1 (∆)intermsoftheelementarysymmetricfunctionsof
∆anditsconjugates.Since
pe ≥ (p − 1)t2
wecanapplytheestimates(iii)for γtoseethat
S p
K/L1 (∆)

Chapter11 122
iscongruentto
Since
wehave
p
pNK/L1∆ +
2
p−1
j=1 Ej
K/L1
(∆) Ep−j
K/L1 (∆) + {S K/L1 (∆)}p . (11.5)
∆ p − ∆ = a
SK/L1 (∆) = Sp
K/L1 (∆) − SL2/F(a). AccordingtoLemma11.7theleftsidebelongsto P (p−1)t2−t1
.Inparticularitbelongsto PL1 L1
.
Weneedtoknowthatitbelongsto P 1+t2.
Thisisclearif p > 3or t2 > t1. Toproveitin
L1
generalwefirstobservethatalltermsbutthelastin(11.5)arecongruentto0modulo P 1+t2.
L1
Themiddletermsaretakencareofbytheestimates(ii)for γ.Totakecareofthefirstwehave
toshowthat
pe − u ≥ 1 + t2.
Weknowthat pe ≥ (p − 1)t2andthatif t = utheinequalityisstrict.Weneedonlyshowthat
(p − 1)t2 − u ≥ t2
withastrictinequalityif t = u.Thisisclearsince t2 ≥ uand t2 > uif t = u.Thus
Wenowneedonlyshowthat
SK/L1∆ ≡ (SK/L1∆)p − SL2/F(a) (mod P 1+t2).
L1
S L2/F(a) ≡ 0 (modP 1+t2
L1 ).
Theleftsidebelongsto P pb
if bisanyintegerlessthanorequalto
L1
Wemaytake
−u + (p − 1)(t2 + 1)
.
p
b ≥
−u + (p − 1)t2
p
whichisgreaterthanorequalto 1+t2
p exceptwhen p = 3and t = u.Inthiscase,whichisthe
onetoworryabout, t2 = uisprimeto pand
−u + (p − 1)(t2 + 1)
p
= u + 2
3

Chapter11 123
hasintegralpartatleast u+1
3 .
Weapply(11.5)againtoseethat
iscongruentto
SK/L1NK/L2∆ = SK/L1a = Sp (∆) − SK/L1 (∆)
K/L1
−SK/L1∆ + p N p
K/L1∆ +
2
Wehavestilltoconsider
p−2
S K/L2 N K/L1
j=2 Ej
K/L1
(∆) Ep−j
K/L1 (∆).
∆. (11.6)
Therearesomegeneralremarkstobemadefirst.Suppose Λbelongsto Kand
If xand zalsobelongto Kand
and
then
Itisenoughtoshowthat
vK(Λ) = −u.
xΛ j ∈ OK
z ∈ P 1+t+(p2 −1)u
K
N K/L1 (x(Λ + z)j+1 ) ≡ N K/L1 (xΛj+1 ) (modP 1+pt2
L1 ).
N K/L1
1 + z
j+1 Λ
≡ 1 (mod P 1+t+pu
). L1
ThisfollowsfromLemmaV.5ofSerre’sbookandtherelations
and
1 + t + p 2 u ≥ 1 + t + pu
1 + t + p 2 u + (p − 1) (t + 1)
p
= 1 + t + pu.
AccordingtoLemmas11.3and11.4thereisforeach σ = 1in G2a (p − 1)throotofunity
ξ = ξ(σ)suchthat
∆ σ = (∆ + ξ) p − a + O (̟ 2(p2e−(p−1)u) K ).

Chapter11 124
Wehave
whichequals
p 2 e − (p − 1)u ≥ (p − 1) {(p − 1)u + t} − (p − 1)u
(p − 1)t + (p − 2) (p − 1)u ≥ t + p(p − 2)u.
If t = uthefirstoftheseinequalitiesisstrictandif t > uthelastis.Thus
and
p 2 e − (p − 1)u ≥ 1 + t + p(p − 2)u
2(p 2 e − (p − 1)u) ≥ {1 + t + (p 2 − 1)u} + {1 + t + ((p − 1) 2 − 2p)u}.
Thesecondtermispositiveunless p = 3.
Theexpression
isequalto
But
and
p
=
i
p(p − 1) . . .(p − i + 1)
(∆ + ξ) p − a
∆ + ξ + p−1
i=1
i!
p
i
ξ i ∆ p−i .
i+1 p
≡ (−1)
i
2p 2 e − (p − 1)u ≥ 2(p 2 e − (p − 1)u)
(modp 2 )
whichis,aswehavejustseen,atleast 1 + t + (p 2 − 1)u.Thusif p > 3
if
∆ σ ≡ (∆ + ξ) (1 − Z(ξ)) (modP 1+t+(p2−1)u K )
Z(ξ) = p∆p−1
1 + ξ/∆
Expandingthedenominatorweobtain
If i ≥ p − 1
p−1
i=1
∞
p−1
Z(ξ) = p∆
i=1 ai
p−1
i
ai = (−1)
j=1
1
j
(−1) i
i
i ξ
.
∆
i ξ
.
∆
≡ 0 (mod p).

Chapter11 125
Clearly
If p = 3
Z(ξ) = O(p∆ p−2 ) = O(̟ p2e−(p−2)u K ).
3(p 2 e − (p − 1)u) ≥ {1 + t + (p 2 − 1)u} + {2(1 + t) + (2p 2 − 6p + 1)u}.
Thesecondtermisatleast 3uandinparticular,ispositive.Lemmas11.3and11.4showthat
Therightsideequals
∆ σ ≡ ((∆ + ξ) 3 − a) 3 − a (mod P 1+t+(p2−1)u K ).
(∆ + ξ + 3ξ∆ 2 + 3ξ 2 ∆) 3 − a.
Expandingthecubeandignoringalltermsin P 1+t+(p2−1)u K
∆ + ξ + 3∆ 3
ξ ξ2
+
∆ ∆2
whichwewriteas
with
and
Since
∞
2
Z(ξ) = 3∆
i=1 ai
(∆ + ξ) (1 − Z(ξ))
+ 9∆ 4 ξ
i ξ ∞
4
+ 9∆
∆
i=1
weobtain
i −ξ
.
∆
2(p 2 e − (p − 2)u) ≥ 2(p 2 e − (p − 1)u) + 2u ≥ 1 + t + p 2 u
p 2 e − (p − 2)u ≥ 1 + t + (p − 1) 2 u ≥ 1 + t + pu
forallodd p,lemmaV.5ofSerre’sbookshowsthat
N K/L1 (x(∆ + ξ)j+1 (1 − Z(ξ)) j+1 ) ≡ {N K/L1 (x(∆ + ξ)j+1 )} {1 − S K/L1 Z(ξ)}j+1
modulo P 1+t+(p−1)u
L1
if x∆ j liesin OK.
Theexpression(11.6)isequalto
NK/L1∆ + σ∈G 2
σ=1
N K/L1 ∆σ .

Chapter11 126
Theprecedingremarksshowthat,if p > 3,thisiscongruentto
modulo P 1+t+(p−1)u
.Since
L1
wehave
NK/L1∆ +
ξ NK/L1 (∆ + ξ) {1 − SK/L1Z(ξ)} 2p 2 e + u + (p − 1)t
p
if i ≥ pandwemayreplace S K/L1 Z(ξ)by
if p > 3.Ofcourse
P 1+pt2
L1
and
and
Since
≥ 2(p − 1)
u +
t − u
tu > t + pu
p
N K/L1 (∆ + ξ) S K/L1 (aip ∆ p−1−i ξ i ) ∈ P 1+pt2
L1
p−1
i=1 pξi S K/L1 (ai∆ p−1−i )
p
NK/L1 (∆ + ξ) =
i=0 ξ1−i E i K/L1 (∆).
Puttingtheseobservationstogetherweseethat,if p > 3,(11.6)iscongruentmodulo
tothesumof
NK/L1∆ + (p − 1) {SK/L1∆ + NK/L1∆} −p(p − 1) p−1
i=1 ai S K/L1 (∆p−1−i ) E 1+i
K/L1 (∆)
−p(p − 1) {pap−1 S K/L1 (∆) + ap−2 S K/L1 (∆)}.
pSK/L1 (∆) ∈ P1+pt2
L1
thelastexpressionmaybeignoredasmaytheterminthesecondcorrespondingto i = p − 2.
Since
ap−1 = p−1
j=1
1
≡ 0 (modp)
j
and
and
S K/L1 (∆0 ) = p
3p 2 e − t2 ≥ 1 + pt2

Chapter11 127
thesuminthesecondexpressionneedonlybetakenfrom1to p−3.Therelation(11.2)implies
that
K/L1 (∆) ≡ (−1)ip(p − 1 − i) E p−1−i
(∆) E1+i
K/L1 (∆)
pS p−1−i
(∆) E1+i
K/L1
K/L1
modulo P 1+pt2
.TocompletetheproofofLemma11.8,for p > 3,weneedonlyshowthat
L1
iai−1 + (p − i) ap−i−1 ≡ (−1) i
for p − 2 ≥ p − i ≥ i ≥ 2.Thisamountstoshowingthat
i i−1
j=1
1
j
+ (p − i) p−i−1
j=1
1
j
(modp)
≡ −1 (modp).
Wemayreplacethe p − iinfrontofthesecondsumby −i.Makingtheobviouscancellations
weobtain
−i p−i−1 1 p−i
≡ −1 − i
j=i j j=i
1
j .
If 1
j occursinthesumontherightsodoes
1
p−j .
Theprooffor p = 3canproceedinexactlythesamewayprovidedweshowthat
liesin P 1+3t2
L1
for i ≥ 1.Since
andoneoftheinequalitiesisstrict
Theexpression
9 {N K/L1 (∆ + ξ)} {S K/L1 (ξi ∆ 4−i )} (11.7)
2p 2 e − u ≥ 2(p − 1)t2 − u ≥ 3t2
9NK/L1 (∆ + ξ) ∈ P1+3t2.
L1
ξ i S K/L1 (∆4−i )
isclearlyintegralfor i ≥ 4.By,forexample,Lemma11.7itisalsointegralif iis2or3.Thus
i = 1istheonlycasetocauseaproblem.If i = 1wesumover ξtoseethat(11.7)equals
18{E 2 K/L1 (∆) S K/L1 (∆3 ) + S K/L1 (∆3 )}.
Thetermsappearingintheexpressioninbracketshavebeenshowntoliein OL1 .

Chapter11 128
Thereisonemorelemmatobeprovedbeforewecometothebasicfactofthisparagraph.
If xisin Kweset
and
g(x) = S L1/F(N K/L1 ∆ S K/L1 (x)) − S L2/F(N K/L2 ∆ S K/L2 (x))
h(x) = S L1/F(N K/L1 (x∆)) − S L2/F(N K/L2 (x∆)).
Inthefollowinglemma pissupposedodd.
Lemma 11.9
(a)Suppose xisin L2, 0 ≤ j ≤ p − 1,and x∆jliesin P 1+t2−t1
K .If j = p − 2then
g(x∆ j ) ≡ 0 (modP 1+t2
F )
butif j = p − 2,thereisan ωin L2suchthat
and
ωx ≡ −x E p−1
(∆) (mod P1+pt2
K/L1 K )
g(x∆ j ) ≡ −{S L2/Fx − S L2/F(xω)} (mod P 1+t2
F ).
(b)Suppose xisin L2, 0 ≤ j ≤ p − 1,and x∆ j liesin PK.If j = p − 2
butif j = p − 2
modulo P 1+pt2
. L1
Thecongruencesmodulo P 1+t2
F
Westartwithpart(a).If xbelongsto OL2 then
h(x∆ j ) ≡ 0 (modP 1+t2
F )
h(x∆ j ) ≡ (p − 1) (1 − {E p−1
K/L1 (∆)}p )N K/L1 x
areofcourseequivalenttocongruencesmodulo P 1+pt2
. L1
g(x) = SK/L1 (xSK/L2 (NK/L1∆) − pxa).
Becauseofthepreviouslemmathisiscongruentto
S K/L1 (xS L2/Fa − pxa) = S L2/F xS L2/Fa − pS L2/F(xa)

Chapter11 129
modulo P 1+t2
F .Wesawbeforethat
Thesameargumentshowsthat
S L2/Fa ∈ P 1+t2
L1 .
SL2/F(xa) ∈ P 1+t2.
L1
pbelongsto P (p−1)t2
.Sincetheintegralpartof
L1
isatleast
(p − 1) (t2 + 1)
p
(p − 1)t2
,
p
sodoes S L2/Fx.Thistakescareofthecase j = 0.
If 1 ≤ j = p − 1then g(x∆ j )isequalto
S L2/F(xS K/L2 (∆j N K/L1 ∆)) − (p − 1)δ S L2/F(xa)
where δ = 0if j = p − 1and δ = 1if j = p − 1.Consider
Itliesin L2andisequalto
Zj = x S K/L2 (∆j N K/L1 ∆).
x∆ j
NK/L1∆ + σ∈G 2 x∆
σ=1
σ j NK/L1∆σ .
Weobservefirstofallthatif Λisin K, vK(Λ) = −u, xisin L2, xΛjliesin P 1+t2−t1,
and zliesin P
(p−1) (pt2−t1)
K
then
x(Λ + z) j N K/L1 (Λ + z) ≡ xΛj N K/L1
Λ (mod P1+pt2
K )
provided pisgreaterthan3.Toestablishthiscongruenceweshowthat
1 + z
j Λ
N K/L1
1 + z
Λ
≡ 1 (modP (p−1)t2+(p+1)t1
K ).
K

Chapter11 130
Toshowthis,onehasonlytoobservethat z
Λ
that
whichequals
if p > 3.
andallitsconjugatesliein P(p−1)pt2−(p−2)t1
K
(p − 1)pt2 − (p − 2)t1 ≥ (p − 1)t2 + ((p − 1) 2 − (p − 2))t1
(p − 1)t2 + ((p − 1) (p − 2) + 1)t1 ≥ (p − 1)t2 + (p + 1)t1
Supposefornowthat p > 3.Since
Lemma11.4impliesthat Zjiscongruentto
p 2 e − (p − 1)u ≥ (p − 1) (pt2 − t1).
x∆ j
NK/L1∆ +
ξ x(∆ + ξ)j NK/L1 (∆ + ξ)
modulo P 1+pt2
K
NK/L1 (∆ + ξ) = ξ 1 + 1
ξ N p−1
K/L1∆ +
i=1 ξ−iE i K/L1 (∆)
.
AccordingtoLemma11.7thisiscongruentto
ξ + NK/L1∆ + ξ Ep−1
K/L1 (∆)
modulo P pt2
K .Thusif x∆jbelongsto P 1+t2−t1
K
Zj ≡ x∆ j
NK/L1∆ +
ξ x(∆ + ξ)j (ξ + NK/L1∆ + ξ Ep−1
K/L1 (∆))
modulo P 1+pt2
K .Weexpand (∆ + ξ) jandsumover ξtoobtain
if j < p − 2.If j = p − 2weobtain
and
px∆ j NK/L1∆ (11.8)
px∆ j NK/L1∆ + (p − 1)x (1 + Ep−1
K/L1 (∆))

Chapter11 131
andif j = p − 1weobtain
px∆ j NK/L1∆ + (p − 1)x {NK/L1∆ + (p − 1)∆ + (p − 1)∆ Ep−1
K/L1 (∆)}.
Theexpression(11.8)liesin
provided x∆ j liesin PK.
Since
and
wehave
P 1+p2 e−pt1
K
p 2 e − pt1 ≥ p(p − 1)t2 − pt1 ≥ pt2.
L2 ∩ P 1+pt2
K
if 1 ≤ j < p − 2and x∆jliesin P 1+t2−t1
K .
Since
liesin OL1 ,
with
if j = p − 2.Wemaytake ωin L2andthen
and
If j = p − 1then
= P 1+t2
L1
S L2/F(P 1+t2
L2 ) ⊆ P 1+t2
F ,
g(x∆ j ) ≡ 0 (modP 1+t2
F )
E p−1
K/L1 (∆)
Zj = (ω − 1)x
ω x = Zj + x ≡ −x E p−1
(∆) (mod P1+pt2
K/L1 K )
g(x∆ j ) ≡ −{S L2/Fx − S L2/F(xω)} (modP 1+t2
F ).
g(x∆ j ) = S L2/F (Zj − (p − 1)xa)
Zj − (p − 1)xa

Chapter11 132
iscongruentto
modulo P 1+pt2
K .Theproduct
liesin P 1+pt2
K
and
Itiseasilyseenthat
isequalto
(p − 1)x {NK/L1∆ + p∆ + (p − 1)∆ Ep−1
K/L1 (∆) − ∆p }
{(p − 1)x} {p∆}
(p − 1) E p−1
(∆) ≡ −Ep−1 (∆) (modP1+pt2
K/L1 K/L1 K ).
∆ p + ∆ E p−1
K/L1 (∆) − N K/L1 ∆
− p−2
i=1 (−1)i ∆ p−i E i K/L1 (∆).
Recallingthat x∆p−1issupposedtoliein PKweappealtoLemma11.7toseethattheproduct
ofthisexpressionwith xliesin P 1+pt2
K .Thus
with
If
then
If p = 3and ξ = ξ(σ)then
Ifwecanshowthat
itwillfollowthat
g(x∆ j ) ≡ 0 (modP 1+t2
F ).
∆ σ = ∆ + ξ + 3ξ∆ 2 + 3ξ 2 ∆ + z
z = O(̟ 2(p2e−(p−1)u) K ).
∆σ = ∆ + ξ + 3ξ∆ 2
∆ σ = ∆σ + 3ξ 2 ∆ + z.
3ξ 2 ∆ + z = O(̟ (p−1)t2+pt1
K )
x∆ σ j N K/L1 ∆σ ≡ x∆ j σ N K/L1 ∆σ (modP 1+pt2
K ) (11.9)

Chapter11 133
if x∆ j liesin P 1+t2−t1
K
and
because (p − 1) 2 ≥ p.Moreover
whichisatleast
and
3ξ 2 ∆ = O(̟ p2e−t1 K )
p 2 e − t1 ≥ p(p − 1)t2 − t1 ≥ (p − 1)t2 + pt1
2(p 2 e − (p − 1)u) ≥ 2(p(p − 1)t2 − (p − 1)t1)
(p − 1)t2 + ((2p − 1) (p − 1) − 2(p − 1))t1
(2p − 1) (p − 1) − 2(p − 1) = (2p − 3) (p − 1) ≥ p.
Wewanttoreplace N K/L1 ∆σby
NK/L1 (∆ + ξ)
intherightsideof(11.9).Todothiswehavetoshowthat
Since
and
N K/L1
wehaveonlytoverifythat
andthattheintegralpartof
isatleast
∆σ
∆ + ξ
∆σ
≡ 1 (mod P (p−1)t2+(p+1)t1
K ).
∆ + ξ = 1 + O(̟p2 e−t1
K )
p 2 e − t1 ≥ p(p − 1)t2 − t1,
p{p(p − 1)t2 − t1} ≥ (p − 1)t2 + (p + 1)t1
p(p − 1)t2 − t1 + (p − 1) (t + 1)
p
(p − 1)t2 + (p + 1)t1
.
p
(11.10)
(11.11)

Chapter11 134
Theinequality(11.10)isclear.Since t ≥ t1theintegralpartof(11.11)isatleast
and
p(p − 1)t2 + (p − 2)t1
p
≥ (p − 1)t2 + ((p − 1) 2 + (p − 2))t1
p
(p − 1) 2 + (p − 2) ≥ p + 1.
Justaswhen p > 3wemayreplace NK/L1 (∆ + ξ)in(11.9)by
Thus Zjiscongruentto
modulo P 1+pt2
K
equalto
ξ + NK/L1∆ + ξ Ep−1
K/L1 (∆).
x∆ j
NK/L1∆ +
ξ x(∆ + ξ + 3ξ∆2 ) j (ξ + NK/L1∆ + ξ E2 K/L1 (∆))
andif j = 2itisequalto
if p = 3, jis1or2,and x∆jbelongsto P 1+t2−t1
K .If j = 1thisexpressionis
px∆ j NK/L1∆ + 2x(1 + 3∆2 ) (1 + E 2 (∆)) (11.12)
K/L1
px∆ j NK/L1∆ + 2x{2(1 + 3∆2 )∆(1 + E 2 K/L1 (∆)) + (1 + 3∆2 ) 2 NK/L1∆}. (11.13)
Theterm
px∆ j N K/L1 ∆
canbeignoredasbeforebecauseitliesin P 1+pt2
K .Also
because x∆ j liesin P 1+t2−t1
K
and
3x∆ j+1 = O(̟ 1+p2e+t2−2t1 K )
p 2 e + t2 − 2t1 ≥ p(p − 1)t2 − t1 ≥ pt2.
Wemayalsoreplacethefactor2in(11.12)by1.Thus(11.12)iscongruentto
−x(1 + E 2 K/L1 (∆))

Chapter11 135
modulo P 1+pt2
K . Atthispointwemayargueaswedidfor p > 3. Tosimplify(11.13),we
observethat
9∆ 4 NK/L1∆ = O(̟2p2 e−7t1
K )
andthat
Moreover
if x∆ 2 belongsto PKand
Thus(11.13)iscongruentto
2p 2 e − 7t1 ≥ 12t2 − 7t1 ≥ 3t2.
3x∆ 2 NK/L1∆ = O(̟1+p2 e−3t1
K )
p 2 e − 3t1 ≥ 6t2 − 3t1 ≥ 3t2.
2x{2∆(1 + E 2 K/L1 (∆)) + N K/L1 ∆}
modulo P 1+3t2
K .Wemayagainargueaswedidfor p > 3.
Weturntothesecondpartofthelemma.Weobservefirstthatif xbelongsto L2, ybelongs
to K,and
xy ∈ PK
then
Theleftsideis
h(xy) ≡ h(y) NK/L1 (x) (mod P1+t2
F ).
S L1/F(N K/L1 xN K/L1 y∆) − S L2/F(x p N K/L2 y∆).
Since NK/L2 x = NL2/Fxliesin Fthisequals
{N K/L1 x}h(y) + S L2/F {N K/L2 (y∆) (N L2/Fx − x p )}.
Thesecondtermisthetracefrom L2to Fof
{NK/L2 (xy∆)}
NL2/Fx − xp NK/L2x
if,aswemayaswellassume, x = 0.Allweneeddoisshowthatthisexpressionliesin P 1+t2
L2
forthenitstracewillliein P 1+t2
F .Thefirstfactorliesin P 1−t2.Thesecondfactorisequalto
σ∈G 2 x σ−1
− 1.
L2

Chapter11 136
Since p ≥ 3itwillbesufficienttoshowthattheimageofthehomomorphism
x −→ ϕ(x) =
σ∈G 2 x σ−1
of CL2into UL2iscontainedin U(p−1)t2.
Let ρbeageneratorof G L2
2 andlet P(X)bethe
polynomial
then
Let
p−1
i=1 (Xi − 1) = p−1
If 1 ≤ i ≤ p − 1the ithcoefficientof Q(X)is
ϕ(x) = x P(ρ) .
Q(X) = (X − 1) p−1 .
p−1−i (p − 1) . . .(p − i)
(−1)
i!
Sinceboth P(X)and Q(X)aredivisibleby X − 1
i=0 Xi − p
≡ 1 (modp).
P(X) = Q(X) + p(X − 1)R(X)
where R(X)isapolynomialwithintegralcoefficients.Forall zin CL2 ,
with w = O(̟ t2
L2 ).Then
and
If a ≥ 1and
then
z ρ−1 = 1 + w
z p(ρ−1) = (1 + w) p ≡ 1 + w p ≡ 1 (modP pt2
L2 )
(1 + w) p−1 =
z p(ρ−1)R(ρ) ∈ U pt2
L2 .
w ∈ P a L2
1 + wp
1 + w = 1 + wp − w
1 + w
≡ 1 (modPa+t2
L2 ).
Onethenshowseasilybyinductionthat,forall zin CL2andall n ≥ 1,
z (ρ−1)n
∈ U nt2
L2 .

Chapter11 137
If xliesin PKwemaytake y = 1.ApplyingLemma11.8weseethat
h(x) ≡ N L2/F xh(1) ≡ 0 (mod P 1+t2
F ).
If 1 ≤ j ≤ p − 1, xliesin L2,and x∆ j liesin PK,
with
and
Theexpression Pjiscongruentto
N L2/Fx{N K/L1 ∆j+1 +
h(x∆ j ) ≡ Pj − Qj (mod P 1+t2
F )
Pj = N L2/F xS L1/F(N K/L1 ∆j+1 )
Qj = N L2/FxS L2/F(N K/L2 ∆j+1 ).
ξ N K/L1 (∆ + ξ)j+1 {1 − S K/L1 Z(ξ)}j+1 } (11.14)
modulo P 1+pt2.Sinceweareworkingmodulo
P L1
1+pt2
L1
weneedonlyconsider
(1 − S K/L1 Z(ξ))j+1
modulo P pt2+t1
.Supposefirstthat p > 3.Then
L1
and
Moreovertheintegralpartof
isatleast
Z(ξ) = O(̟ p2e−(p−2)u K )
p 2 e − (p − 2)u ≥ p(p − 1)t2 − (p − 2)u ≥ p(p − 2)t2.
p(p − 2)t2 + (p − 1) (t + 1)
p
(p − 2)t2 −
(p − 1)
t
p
andtwicethisisatleast pt2 + t1.Wereplace(11.15)by
1 − (j + 1) S K/L1 Z(ξ).
(11.15)

Chapter11 138
Since
and
while
p−2
p−1
Z(ξ) ≡ p∆
i=1 ai
i ξ
∆
p 2 = O(̟ 2p2 e
K )
2p 2 e ≥ 2p(p − 1)t2 ≥ p(pt2 + t1),
wemayreplace Z(ξ)bytherightsideof(11.16).ByLemma11.7
if 1 ≤ i ≤ p − 2and
if i ≥ 2.Wereplace Z(ξ)by
Wemaywrite(11.14)as
N L2/FxN K/L1 ∆j+1
Whenweexpand
andsumover ξwewillobtain
whichwewriteas
pSK/L1 (∆p−1−i ) = O(̟ pe+it2)
L1
pe + it2 ≥ (p − 1)t2 + it2 ≥ pt2 + t1
1 +
ξ N K/L1
N K/L1
N L2/FxN K/L1 ∆j+1
plusasumoftermsoftheform
pa1ξ∆ p−2 .
1 + ξ
j+1 ∆
1 + ξ
j+1 ∆
1 +
NL2/Fx NK/L1∆j+1 =
αp N L2/FxN K/L1 ∆j+1 E i K/L1
ξ N K/L1
ξ N K/L1
1
∆
(modp 2 ) (11.16)
{1 − SK/L1 (pa1ξ∆ p−2
)} .
1 + ξ
j+1 ∆
(∆ + ξ)j+1
S K/L1 (∆p−2 )

Chapter11 139
where αisrationalandliesin OFand iisatleast1.Since
for i ≥ 1and
E i K/L1
thesesupplementarytermsmaybeignored.
Nowtake p = 3. ∗
1
= O(̟
∆
t1
L1 )
pS K/L1 (∆p−2 ) = O(̟ pt2
L1 )
∗ (1998)ThisiswhereandhowthemanuscriptofChapter11breaksoff.

Chapter12 140
Chapter 12.
The Second Main Lemma
Suppose Kisanormalextensionofthelocalfield F and G = G(K/F)isthedirect
productoftwocyclicgroupsofprimeorder ℓ.Let XKbeaquasi­characterof CK.If σbelongs
to Gdefine X σ Kbytherelation X σ σ−1
K (α) = XK(α ).
Supposethat X σ K = XKforall σin Gbutthatfornoquasi­character XFof CFdoes XK =
X K/F.If F ⊆ L ⊆ Kand [K : L] = ℓthen XKcanbeextendedtoaquasi­characterof W K/L
because W K/L/CKisisomorphicto G(K/L)whichiscyclic.Ifthisquasi­characteris XLthen
XK = X K/L.
Lemma 12.1
Suppose L1and L2aretwofieldslyingbetween Fand Kand [K : L1] = [K : L2] = ℓ.
Suppose XL1isaquasi­characterof CL1 , XL2isaquasi­characterof CL2 ,and
Then
isequalto
XK = X K/L1 = X K/L2 .
∆(XL1 , ψ L1/F)
∆(XL2 , ψ L2/F)
µF ∈S(L1/F) ∆(µF, ψF)
µF ∈S(L2/F) ∆(µF, ψF).
Becauseoftheassumptionon G(K/F)thefield Fmustbenon­archimedean.Toprove
thelemmaingeneralitisenoughtoproveitforagiven L1andall L2. Therearethree
possibilitiestoconsider.
(i)Thesequenceofgroupsoframificationtakestheform
G = G−1 = G0 = . . . = Gt = Gt+1 = . . . = {1}.
(ii)Thesequenceofgroupsoframificationtakestheform
G = G−1 = G0 = G1 = . . . = Gu = Gu+1 = . . . = Gt = Gt+1 = . . . = {1}.

Chapter12 141
(iii)Thesequenceofgroupsoframificationtakestheform
G = G−1 = G0 = G1 = . . . = Gt = Gt+1 = . . . = {1}.
Inthefirsttwocaseswetake G 1 = G(K/L1)tobe Gt. Inthethirdcasethechoiceof L1is
immaterial.
Iftherelation X σ Li = XLiobtainsforone σdifferentfrom1in Gi = G(Li/F)itobtainsfor
allsuch σand XLiisoftheform XLi/Fforsomequasi­character XFof CF.Then XK = XK/F whichiscontrarytoassumption.Thusthecharacters X σ−1
Li with σin Giaredistinct.Theyare clearlytrivialon NK/Li CKso
{X σ−1
Li | σ ∈ G i } = S(K/Li) = {µ Li/F | µF ∈ S(Lj/F)}.
Here jis2or1accordingas iis1or2.
while
Then
Let ti ≥ −1bethatintegerforwhich
G i = G i
ti
G i
ti+1 = {1}.
δi = δ(Li/F) = (ti + 1) (ℓ − 1).
Inthefirstcase L1/Fisunramifiedand L2/Fisramified.Wechoose ̟L2 arbitrarilyandtake
̟K = ̟L2 .Alsoweset
̟L1 = ̟F = N L2/F̟L2 .
Inthesecondandthirdcases K/L1and K/L2areramifiedand K/Fistotallyramified.We
choose ̟kfirstandset
and
̟Li = N K/Li ̟K
̟F = N K/F̟K.
Let mi = m(XLi ).The m(X σ ) = miand
Li
m(X σ−1
Li ) ≤ mi.

Chapter12 142
Thus m(ν) ≤ miif νbelongsto S(K/Li).If G i = G(K/Li)andif
while
G i ui
G i ui+1
= Gi
= {1}
then m(ν) = ui + 1if νisnon­trivial.Thus ui + 1 ≤ mi.Since νXLiisoftheform X σ Li forall
νin S(K/Li),
m(νXLi ) = m(XLi ).
Lemma8.8and8.12implythat*
This m(XK) = miif K/Liisunramifiedand
m(XK) = ψ K/Li (mi − 1) + 1.
m(XK) = ℓmi − δ(K/Li)
if K/Liisramified.If n = n(ψF)then ni = n(ψ Li/F)is nif Li/Fisunramifiedandis ℓn+δi
if Li/Fisramified.
Inthefirstcase
Therelations
and
implythat t2 = t.Also
sothat
Moreover
δ(K/L2) = δ(L1/F) = 0.
δ(K/F) = δ(K/L1) + ℓδ(L1/F) = δ(K/L1) = (t + 1) (ℓ − 1)
δ(K/F) = δ(K/L2) + δ(L2/F) = δ2 = (t2 + 1) (ℓ − 1)
m(XK) = m2 = ℓm1 − δ(K/L1) = ℓm1 − δ2
m2 + n2 = ℓ(m1 + n1).
XL1 (̟m1+n1
L1
) = XK (̟ m1+n1)
L2
*Weareencounteringonceagaintheconflictingusesofthesymbole ψ.

Chapter12 143
isequalto
XL2 (̟m2+n2
L2
and
isequalto
If
σ∈G2
) = XL2 (̟m1+n1
F ) XL2
σ∈G 2 ̟ 1−σ
L2
XL2 (̟1−σ
) = L2
µF ∈S(L1/F) µ L2/F(̟L2 )
then
µF ∈S(L1/F) µF(̟F) = (−1) ℓ−1 .
S ′ 1
and
Thuswehavetoshowthat
isequalto
and
S ′ i = S(Li/F) − {1}
µF(̟ t1+1+n
F ) = (−1) n(ℓ−1)
S ′ 2
µF (̟ t2+1+n
F ) = 1.
(−1) m1(ℓ−1)
∆1(XL1 , ψL1/F ̟ m1+n1
F )
∆1(XL2 , ψL2/F, ̟ m1+n1
F )
S ′ 2
Inthesecondandthirdcasestherelations
m1+n1
∆1(µF, ψF, ̟ t+1+n
F ).
m(XK) = ℓm1 − δ(K/L1) = ℓm2 − δ(K/L2)
δ(K/F) = δ(K/L1) + ℓδ1 = δ(K/L2) + ℓδ2
implythat m1 + δ1 = m2 + δ2andhencethat m1 + n1 = m2 + n2.Thus
XL1 (̟m1+n2
L1
Since
) = XK (̟ m1+n1
K
S ′ i
µF(̟ ti+1+n
F ) = 1
) = XL2 (̟m2+n2).
L2

Chapter12 144
wehavetoshowthat
isequalto
Suppose X ′ F
isequalto
and
isequalto
∆1(XL1 , ψL1/F, ̟ m1+n1)
L1
S ′ 1
∆1(XL2 , ψL2/F, ̟ m2+n2)
L2
S ′ 2
∆1(µF, ψF, ̟ t1+1+n
F )
∆1(µF, ψF, ̟ t2+1+n
F ).
isaquasi­characterof CF.AccordingtoLemma10.1
∆(X ′ L1/F , ψ L1/F)
µF ∈S(L1/F) ∆(µF, ψF)
µF ∈S(L1/F) ∆(µF X ′ F , ψF) (12.1)
∆(X ′ L2/F , ψ L2/F)
µF ∈S(L2/F) ∆(µF, ψF)
µF ∈S(L2/F) ∆(µF X ′ F, ψF). (12.2)
Suppose m ′ = m(X ′ F ) = 2d′ + ε ′ and d ′ isgreaterthanorequaltoboth 1 + t1and 1 + t2.
Choose γin Fsuchthat
γ OF = P m′ +n
F
andthenchoose β = β(XF).ByLemma9.4theexpression(12.1)isequalto
and(12.2)isequalto
Consequently
{∆(X ′ ℓ
F , ψF)}
{∆(X ′ ℓ
F , ψF)}
∆(X ′ L1/F , ψ L1/F)
µF ∈S(L1/F) µF
µF ∈S(L2/F) µF
µF ∈S(L1/F) µF
β
γ
γ
β
γ
.
β
∆(µF, ψF)

Chapter12 145
isequalto
∆(X ′ L2/F , ψ L2/F)
µF ∈S(L2/F) µF
β
γ
∆(µF, ψF) .
Supposethatboth m1 = m(XL1 )and m2 = m(XL2 )areatleast2andlet mi = 2di + εi.
Supposethat
for iequalto1and2.Then
If
for xin P di+εi
Li
m(X −1
Li X ′ Li/F ) ≤ di
mi = m(X ′ Li/F ) = ψ Li/F(m ′ − 1) + 1.
then,byLemma9.4again,
X ′ Li/F (1 + x) = ψ Li/F
∆(XLi , ψ Li/F) = X −1
Li X ′ Li/F
βi
γ
βix
γ
∆ (X ′ Li/F , ψ Li/F).
ThustoproveLemma12.1inthepresentcircumstanceswehaveonlytoverifythat
isequalto
X −1
L1 X ′ L1/F
X −1
L2 X ′ L2/F
Supposefirstthat ℓisodd.Then
andweneedonlyverifythat
X −1
L1
γ
β1
X ′ L2/F
β1
γ µF ∈S(L1/F) µF
β2
γ µF ∈S(L2/F) µF
µF ∈S(Li/F) µF
γ
β2
= X −1
L2
γ
= 1
β
γ
AccordingtoLemmas8.3and8.4wemaytake β1 = β2 = β.
β2
γ
β
γ
.
β
X ′ L2/F
γ
β2
.

Chapter12 146
Certainly
X ′ L1/F
γ
= X
β
′ F
ℓ γ
Since CFistheproductof N L1/FCL1 and N L2/FCL2
Consider
β ℓ
= X ′ L2/F
γ
β = N L1/F δ1 N L2/F δ2.
XLi (N Lj/Fδj) = XK(δj)
where jis1or2accordingas iis2or1.Therightsideequals
Theproductisequalto
whichis1because ℓisodd.
XLj (δℓ j) = XLj (N Lj/F δj)
γ
.
β
wemaywrite γ
β asaproduct
σ∈G(Lj/F)
µF ∈S(Li/F) µ Lj/F(δj)
XLj (δ1−σ
j ).
Beforediscussingthecase ℓ = 2weconsiderthecircumstancesunderwhich,foragiven
XL1 and XL2 ,aquasi­character X ′ F withthepropertiesdescribedaboveexists.
Lemma 12.2
(a)If Li/Fisunramified, xbelongsto U ui+1
,and Li
then
N Li/F(x) = 1
XLi (x) = 1.
(b)If Li/Fisramified, K/Liisunramified, xbelongsto U ti+1
,and Li
then
N Li/F(x) = 1
XLi (x) = 1.
(c)If Li/Fand K/Liareramified, xbelongsto U ui+ti+1
,and
Li
N Li/F(x) = 1

Chapter12 147
then
XLi (x) = 1.
Choosesomenon­trivial σiin G i = G(Li/F).Then
isanon­trivialcharacterin S(K/Li)and
µLi
= X σ−1
i −1
Li
m(µLi ) = ui + 1.
Since Li/Fiscyclicthereisayin CLi suchthat
x = y σi−1 .
Weshallshowthat ycanbetakenin U ui+1
.Then
Li
XLi (x) = µLi (y) = 1.
Suppose Li/Fisunramified. Ifwecannotchoose yin U ui+1
thereisalargestinteger
Li
a ≥ −1suchthatwecanchoose yin Ua Liwhere aisofcourselessthan ui + 1.Choosesucha
y.Then aisnot­1becausewecanalwaysdivide ybyapowerof ̟F.If awere0then ycould
notbecongruenttoanelementof UF modulo PF. Then yσi−1 1 wouldnotbein UF . Since
ui + 1 > 0inthepresentsituationthisisimpossible.Let
y = 1 + ε̟ a F .
Then εcannotbecongruenttoanelementof OFmodulo PF.Thus
and
isnotin U a+1
.Thisisacontradiction.
Li
ε σi − ε ≡ 0 (modPLi )
y σi−1 ≡ 1 + (ε σi − ε) ̟ a F (modP a+1
Li )
Nowsuppose Li/Fisramifiedand K/Liisunramified.Then ti + 1 ≥ 1and ui + 1 = 0.
Weneedonlyshowthat ycanbetakentobeaunit.Write
y = ε̟ b Li

Chapter12 148
where εisaunit. If biscongruentto0modulo ℓwecandivide ybysomepowerof ̟Fto
obtainanelementof UF = U0 F .Toseethat bmustbecongruentto0modulo ℓwesupposethe
contrary.Then
y σi−1 σi−1 σi−1
= ε (̟ ) Li
b ≡ (̟ σi−1
) Li
b
(modP ti+1
). Li
If ti = 0theresidueof ̟ σi−1
Li
If ti > 0then
where αisaunit.Thus
modulo PLi isanon­trivial ℓthrootofunityand
(̟ σi−1
Li ) b ≡ 1 (modPLi ).
̟ σi−1
Li
= 1 + α ̟ti
Li
(̟ σi−1
Li ) b ≡ 1 + αb ̟ ti
Li (mod P ti+1
Li ).
Therightsideisnotcongruentto0modulo P ti+1
. Li
Nowsuppose Li/Fand K/Liarebothramified. Then ℓ = pandboth uiand tiareat
least1.Againsupposethat ycannotbechosenin U ui+1
andlet abethelargestintegersuch
Li
that ycanbechosenin Ua .Theargumentjustusedshowsthat a ≥ 0.Since Li/Fisramified
U kp
Li = Uk F Ukp+1
Li .
Therefore aisnotdivisibleby pandinparticularisatleast1.Let
where βisaunit.Then
Let
and
y = 1 + ε̟ a Li
y σi−1 = (1 + ε σi ̟ aσi
Li ) (1 + ε̟a Li )−1 .
ε σi = ε + η ̟ ti+1
Li
̟ σi−1
Li
where αisaunit.Then y σi−1 isequalto
= 1 + α ̟ti
Li
ti+1
1 + (ε + η ̟ ) (1 + α̟ Li ti
Li )a̟ a
a −1
1 + ε̟ Li
Li

Chapter12 149
whichiscongruentto
1 + aαε̟ ti+a
Li
modulo P ti+a+1
.Therefore a ≥ ui + 1.Thisisacontradiction.
Li
Lemma 12.3
and
If L1/Fisunramifiedwecanchoose X ′ F suchthat
m(X −1
L1 X ′ L1/F ) = t + 1
m(X −1
L2 X ′ L2/F ) = t + 1.
If m(XL1 ) > t + 1then m(X ′ F )willequal m(XL1 ).
Bythepreviouslemmawecandefineaquasi­character X ′ F of
bysetting
N L1/FU u1+1
L1
X ′ F(N L1/Fx) = XL1 (x).
Weextend X ′ Ftoaquasi­character,whichweagaindenoteby X ′ F ,of CF.Then
m(X −1
L1 X ′ L1/F ) ≤ u1 + 1.
However X −1
K X ′ −1
K/F , XL1 X ′ −1
L1/F ,and XL2 X ′ L2/FsatisfytheconditionsofLemma12.1.There fore
m(X −1
L1 X ′ L1/F ) ≥ u1 + 1.
Since L1/Fisunramified u1and t2arebothequalto t.Thus
and
m(X −1
L1 X ′ L1/F ) = t + 1
m(X −1
L2 X ′ L2/F ) = ℓ(u1 + 1) − δ2 = ℓ(u1 + 1) − (ℓ − 1) (t2 + 1) = t + 1.
Thelastassertionofthelemmaisclear.
Lemma 12.4

Chapter12 150
If K/Fistotallyramifiedthen
Thereexists X ′ F suchthat
for iequalto1and2.
m(XLi ) ≥ ti + ui + 1.
m(X −1
Li X ′ Li/F ) = ti + ui + 1
Inthepresentcircumstances tiand uiarebothatleast1. Chooseanon­trivial σiin G i
andlet
asbefore.Choose yin U ui
Li sothat
Then
Howeverif
where εisaunitthen
if
Inparticular
sothat
µLi
= X σ−1
i −1
Li
µLi (y) = 1.
XLi (yσi−1 ) = 1.
y = 1 + ε̟ ui
Li
y σi−1 ≡ 1 + uiα ε̟ ti+ui
Li
̟ σi−1
Li
= 1 + α ̟ti
Li .
y σi−1 ∈ U ti+ui
Li
m(XLi ) ≥ ti + ui + 1.
(modP ti+ui+1
) Li
Justasinthepreviouslemmawecanfindaquasi­character X ′ F
m(X −1
L1 X ′ L1/F ) = t1 + u1 + 1.
Wehaveseenthat m1 + δ1 = m2 + δ2.Thesameargumentshowsthat
m(X −1
L1 X ′ L1/F ) + δ1 = m(X −1
L2 X ′ L2/F ) + δ2.
of CFsuchthat

Chapter12 151
Tocompletetheproofofthelemmaweshowthat
Since
wehaveonlytoshowthat
t1 + u1 + δ1 = t2 + u2 + δ2.
δi = (ℓ − 1) (ti + 1)
u1 + 1 + ℓ(t1 + 1) = u2 + 1 + ℓ(t2 + 1).
Multiplyingtheleftortherightsideby ℓ − 1weobtain δ(K/F). Theequalityfollows
immediately.
Lemmas12.3and12.4togetherwiththeremarkswhichprovokedthemallowustoprove
Lemma12.1inmany,butbynomeansall,cases. Weshallnothoweverapplytheselemmas
immediately.WeshallratherbeginthesystematicexpositionoftheproofofLemma12.1taking
upthecasestowhichtheselemmasapplyintheirturn.
Supposefirstthat L1isunramifiedover F.Asbefore mi = m(XLi ).Then
m2 = m1 + (ℓ − 1) (m1 − t − 1) ≥ m1
because u1 = t.Sincethenumber m1isatleast t + 1and t ≥ 0itisatleast1.If m1 = 1then
t = 0and m2 = 1.Oncewehavetreatedthiscase,asweshallimmediately,wemaysuppose
that m2 ≥ m1 > 1.
andlet
If m2 = 1let
λ = OL2 /PL2 = OF/PF
κ = OK/PK = OL1 /PL1 .
κisanextensionof λ. Therestrictionof XL1 to UL1 definesacharacter Xλof λ ∗ andthe
restrictionof XL2 to UKdefinesacharacter Xκof κ ∗ . Therestrictionof XKto UKdefinesa
characterof κ ∗ whichisequalto X κ/λandto X ℓ κ sothat
X ℓ κ = X κ/λ.
As σvariesover G 2 , ̟ σ−1
,takenmodulo PL2 ,variesoverthe ℓthrootsofunityin λandif
L2
X σ−1 −1
L2
= ν

Chapter12 152
then
Xλ(̟ σ−1
) = ν(̟L2 ).
L2
Therightsideisnot1if σ = 1because νisthennon­trivial.Thustherestrictionof Xλtothe ℓth
rootsofunityisnottrivial.Toevery µFin S(L2/F)isassociatedacharacter µλof λ ∗ which
isoforder1if µF = 1andoforder ℓotherwise.If ψλistheadditivecharacterof λdefinedby
then
if µFisnottrivial.Moreover
and
Finally
ψλ(x) = ψF
x
̟ 1+n
F
∆1(µF, ψF, ̟ 1+n
F ) = A [−τ(µλ, ψλ)]
ψ L2/F
x
̟ 1+n
= ψλ(ℓ x)
F
∆1(XL2 , ψ L2/F, ̟ 1+n
F ) = A [−Xλ(ℓ)τ(Xλ, ψλ)].
∆1(XL1 , ψ L1/F, ̟ 1+n
F ) = A [−τ(Xκ, ψ κ/λ)].
Thustherequiredidentityisaconsequenceoftherelation
whichweprovedasLemma7.9.
τ(Xκ, ψ κ/λ) = Xλ(ℓ)τ(Xλ, ψλ)
µλ=1 τ(µλ, ψλ)
Retainingtheassumptionthat L1/Fisunramifiedwenowsupposethat m1 > 1.There
aretwopossibilities.
(a)
(b)
Thesecondpossibilityoccursonlywhen t > 0.
m1 ≥ 2(t + 1)
t + 1 ≤ m1 < 2(t + 1).

Chapter12 153
If mj ≥ 2(t + 1)choose X ′ F sothat
for i = 1and2.Itisclearthat
if mi = 2di + εi.Since m2 ≥ m1wealsohave
Moreover
m(X −1
Li X ′ Li/F ) = t + 1
m(X −1
L1 X ′ L1/F ) ≤ d1
m(X −1
L2 X ′ L2/F ) ≤ d2.
m ′ = m(X ′ F ) = m1
sothat d ′ isgreaterthanorequaltoboth 1 + t2 = 1 + tand 1 + t1 = 0.Lemma12.1for L 1/F
unramifiedand m1 ≥ 2(t + 1),followsimmediatelyif ℓisodd.Suppose ℓ = 2.
If t = 0wecaninvokeLemmas8.3and8.7toseethatif β = β(X ′ F )wemaychoose
β1 = β(χL1/F)and β2 = β(X ′ L2/F )equalto β.If µ(1)
F isthenon­trivialelementof S(L1/F)
and µ (2)
F
isequalto
Certainly
isthenon­trivialelementof S(L2/F)wehaveonlytoshowthat
XL1
XL2
γ
β
γ
β
X ′ L1/F
X ′ L1/F
X ′ L2/F
andweneedonlyshowthatif δisin CFthen
Wemaywrite
Then
XL1
(δ) µ(1)
F
β
γ
β
γ
β
= X
γ
′ L2/F
µ (1)
F
µ (2)
F
β
γ
γ
β
γ
.
β
(δ) = XL2 (δ) µ(2)
F (δ).
δ = N L1/F δ1 N L2/F δ2.
µ (i)
F (N L1/F δi) = 1

Chapter12 154
and,if jis1or2accordingas iis2or1,
whichequals
XLi (N Lj/F δj) = XK(δj) = XLj (δ2 j )
XLj (N Lj/F δj)µ (i)
Lj/F (δj) = XLj (N Lj/F δj)µ (i)
F (N Lj/F δj).
Therequiredequalityfollowsimmediately.
If tispositivewemaystillchoose β1 = β.If m1 − t − 1 = vthen,byLemma8.6,wemay
choose β2intheform
β2 = β + η
with ηin Pv .Since v ≥ t + 1
L2
X −1
L2 (β2)X ′ L2/F (β2) = X −1
L2 (β)X ′ L2/F (β).
Thisobservationmade,wecanproceedasbefore.
Somepreparationisnecessarybeforewediscussthesecondpossibility.Supposethat tis
positivesothat ℓisequaltotheresidualcharacteristic p.Thefinitefield λ1 = OL1 /PL1isan extensionofdegree pof φ = OF/PF.
Themap
x −→ x p − x
isanadditiveendomorphismof φwiththeprimefieldaskernel.Chooseayin φwhichisnot
intheimageofthismapandconsidertheequation
x p − x = y.
If x,insomeextensionfieldof φ,satisfiesthisequationand φhas qelementsthen x q = x.
However
(x q − x) p − (x q − x) = (x p − x) q − (x p − x) = y q − y = 0.
So
x q − x = z
where zisanon­zeroelementoftheprimefield.Then
x q2
= (x + z) q = x q + z = x + 2z

Chapter12 155
andingeneral
x qn
= x + nz.
Thusthelowestpower nof qsuchthat xqn = xis n = pand xdeterminesanextensionof
degree p.Consequently xmaybechosentoliein λ1andthen λ1 = φ(x).
wehave
Let E r (x)bethe rthelementarysymmetricfunctionof xanditsconjugates.Since
if 1 ≤ r < p − i,
and,ofcourse,
x p − x + (−1) p N λ1/φx = 0 (12.3)
E r (x) = 0 (12.4)
E p−1 (x) = (−1) p
(12.5)
E p (x) = N λ1/φx. (12.6)
If λisanon­zeroelementoftheprimefieldwecanreplace yby λy.Then xistobereplaced
by λx.Alsowecanreplace xby x + λwithoutchanging y.
Let R(L1)bethesetof (q p − 1)throotsofunityin L1. Chooseaγin R(L1)whose
imagein λ1is x.Ifwearedealingwithfieldsofpowerseries γwillalsosatisfytheequations
(12.3),(12.4),(12.5)and(12.6).Letusseehowtheseequationsaretobemodifiedforfieldsof
characteristiczero. Fand L1arethenextensionsofthe p­adicfield Qp.Let F 0 and L 0 1 bethe
maximalunramifiedextensionsof Qpcontainedin Fand L1respectively. R(L1)isasubset
of L 0 1 and pgeneratestheideal P F 0andtheideal P L 0 1 .Thus
and
if 1 ≤ r < p − 1while
Let
γ p − γ + (−1) p N L1/F γ ≡ 0 (modp)
E r (γ) ≡ 0 (modp)
E p−1 (γ) ≡ (−1) p
S r (γ) =
(modp).
σ∈G(L1/F) γrσ .
ThefollowingrelationsarespecialcasesofNewton’sformulae.

Chapter12 156
Weinferthat
if 1 ≤ r < p − 1andthat
S 1 (γ) − E 1 (γ) = 0
S 2 (γ) − E 1 (γ) S 1 (γ) + 2E 2 (γ) = 0
S 3 (γ) − E 1 (γ) S 2 (γ) + E 2 (γ) S 1 (γ) − 3E 3 (γ) = 0
.
S p−1 (γ) − E 1 (γ) S p−2 (γ) + . . . + (−1) p−1 (p − 1)E p−1 (γ) = 0
S p (γ) − E 1 (γ) S p−1 (γ) + . . . + (−1) p p E p (γ) = 0.
S r (γ) ≡ 0 (modp)
S p−1 (γ) ≡ (−1) p (p − 1) E p−1 (γ) (mod p).
CombiningthefirstofthesecongruenceswithNewton’sformulaeweobtain
if 1 ≤ r ≤ p − 1.If pisodd
Therightsideisequalto
If piseven
Since
wehave
S r (γ) ≡ r(−1) r+1 E r (γ) (modp 2 )
S p (γ) − p E p (γ) ≡ E 1 (γ) S p−1 (γ) − E p−1 (γ) S 1 (γ) (mod p 2 ).
E 1 (γ) (S p−1 (γ) − E p−1 (γ)) ≡ 0 (modp 2 ).
S 2 (γ) + 2E 2 (γ) = {E 1 (γ)} 2 .
E 1 (γ) ≡ 1 (mod2)
S 2 (γ) + 2E 2 (γ) ≡ 1 (mod4).
If σ = 1belongsto G(L1/F)thereisa(p − 1)throotofunity ζsuchthat
γ σ − γ ≡ ζ (modp).

Chapter12 157
Byasuitablechoiceof ytheroot ζcanbemadetoequal,foragiven σ,anychosen (p − 1)th
rootofunity.
Theaboverelationsareofcoursealsovalidwhen Fisafieldofpowerseries.
Chooseanon­trivialcharacter µFin S(L2/F)andchoose αsothat
if xisin P s F
unitywedefine µ ζ
F
µF(1 + x) = ψF
α x
̟ t+1+n
F
t+1
.Here sistheleastintegergreaterthanorequalto 2 .If ζisa(p − 1)throotof
tobe µj
F
AsweobservedintheproofofLemma8.5
if xisin P s F .
if jistheuniqueintegersuchthat
ζ ≡ j (modp).
µ ζ
F (1 + x) = ψF
Let mi = 2di + εiasusual.If β1in L1issuchthat
for xin P d1+ε1
L1
isequalto
if ζσissuchthat
Thusif
XL1 (1 + x) = ψ L1/F
αζx
̟ 1+t+n
F
αβ1x
̟ m1+n1
F
then,if σ = 1belongsto G 1 = G(L1/F)and xbelongsto P d1+ε1,
L1
ψ L1/F
σ α(β1 − β1)x
̟ m1+n2
F
ψ L1/F
X σ−1
L1
αζσx
̟ 1+t+n
F
= µζσ
F .
v = m1 − 1 − t
= X σ−1
L1 (1 + x)

Chapter12 158
wehave
Suppose
Itisclearthat
β σ 1 − β1 ≡ ζσ ̟ v F (mod P d1
L1 ).
ζστ ≡ ζ τ σ + ζσ (modPL1 ).
γ σ − γ ≡ ξσ (modp)
where ξσisalsoa(p − 1)throotofunity.Then
Weobservedthatwecouldarrangethat
ξστ ≡ ξ τ σ + ξσ (mod PL1 ).
ξσ = ζσ
foronenon­trivial σ. Oncewedothistheequalitywillholdforall σ. Then γpσ − γp ≡
ξσ (mod p)and
(β1 − γ p ̟ v F )σ ≡ β1 − γ p ̟ v F (modP d1
L1 )
forall σbecause,asweobservedintheproofofLemma8.5, pbelongsto Pr if r + s = t + 1
L1
and
2(r + v) ≥ t + 2v = 2m1 − 2 − 2t + t
whichisatleast
(m1 − 1) + (m1 − 1 − t) ≥ m1 − 1
sothat r + v ≥ d1.Since L1/Fisunramifiedthereisthereforeaβin Fsuchthat
Wemaysupposethat
β1 − γ p ̟ v F
≡ β (modPd1
L1 ).
β1 = β + γ p ̟ v F .
βisaunitunless v = 0. If v = 0then,byreplacing γbyarootofunitycongruentto γ + 1
moduloPL1 ifnecessary,wecanstillarrangethat βisaunit. βiscongruenttoanorm N L2/Fβ ′
modulo P t F .Since d1 ≤ twe ∗
∗ (1998)AtthemomentthisisallthatcouldbefoundofChapter12.

Chapter13 159
Chapter Thirteen.
The Third Main Lemma
Suppose K/FisGaloisand G = G(K/F).Suppose G = HCwhen H = {1}, H ∩ C =
{1},andCisanon­trivialabeliannormalsubgroupofGwhichiscontainedineverynon­trivial
normalsubgroupof G.
Lemma 13.1
Let Ebethefixedfieldof Handlet XFbeaquasi­characterof CF.If m = m(XF)then
Set
m(X E/F) = ψ E/F(m − 1) + 1.
m ′ = m(X E/F) − 1.
Observethat m ′ − 1isthegreatestlowerboundofallrealnumbers v > −1suchthat XE/Fis andthat m − 1isthegreatestlowerboundofallrealnumbers usuchthat XFis
trivialon Uv E
trivialon Uu F .Since
weseeimmediatelythat
Toprovethelemmaweneedonlyshowthat
NE/F(U ψE/F(u) E ) ⊆ U u F
m ′ − 1 ≤ ψ E/F(m − 1).
NE/F(U ψE/F(m−1) E ) ⊇ U m−1
F .
Weshowthiswith m − 1replacedbyany u ≥ −1.
ByLemma6.15, τK/Fmaps W u K/Fonto Uu F . Theprojectionof W u K/Fon Gisanormal
subgroupof G.Thusitiseither {1}orasubgroupcontaining C.Ifitis {1}then
and
W u K/F = W u K/F ∩ CK = U ψ K/F(u)
K
U u F = N K/F(U ψ K/F(u)
K
) = NE/F(NK/EU ψK/F(u) K )

Chapter13 160
which,byLemma6.6,iscontainedin
NE/F(U ψE/F(u) E ).
Supposetheprojectionisnot {1}.If Listhefixedfieldof Cthegroup W u K/F contains
Since Cisgeneratedby
{wvw −1 v −1 | w ∈ W K/F, v ∈ W u K/F ∩ W K/L}. (13.1)
{σρσ −1 ρ −1 | σ ∈ G, ρ ∈ C}
thegroupgeneratedbytheset(13.1)containsasetofrepresentativesforthecosetsof CKin
W K/L.Thisgroupclearlyliesinthekernelof τ K/F.Thuseveryelementof W u K/F iscongruent
modulothekernelof τ K/Fin W u K/F toanelementof
and
whichis
andthissetiscontainedin
W u K/F ∩ W K/E = W ψ E/F(u)
K/E
U u F = τK/F(W u K/F ) = τK/F(W ψE/F(u) K/E )
NE/F(τK/E(W ψE/F(u) K/E ))
NE/F(U ψE/F(u) E ).
Suppose F1isnon­archimedean, K1/F1isGaloisand F1 ⊆ E1 ⊆ K1. Let µ1bea
characterof G(K1/E1).Wemayalsoregard µ1asacharacterof CE1 .Let σbeanelementof
G(K1/F1)anddefinethecharacterof µ σ 1 of G(K1/E σ 1 )by
µ σ 1(ρ) = µ1(σρσ −1 )
for ρ ∈ G(K1/E σ 1 )or,whatamountstothesame
for α ∈ CE σ 1 .Since
µ σ 1(α) = µ1(α σ−1
)
ψEσ 1 /F1 (α) = ψE1/F1 (ασ−1)
thenextlemmaisacongruenceofthedefinitions.

Chapter13 161
Lemma 13.2
∆(µ σ 1, ψ E σ 1 /F1 ) = ∆(µ1, ψ E1/F1 ).
Wereturntotheextension K/Landthegroup G.Let Tbeasetofrepresentativesforthe
orbitsunder Gofthenon­trivialcharactersin S(K/L).If µ ∈ Tlet Gµbetheisotropygroup
of µandlet Fµbethefixedfieldof Gµ.Let Hµ = H ∩Gµ.Since Ciscontainedin Gµwehave
Gµ = Hµ · C.Then µmayalsoberegardedasacharacterof C.Let µ ′ bethecharacterof Gµ
definedby
µ ′ (hc) = µ(c)
if h ∈ Hµand c ∈ C.Eventuallywemustshowthat
isequalto
∆(X E/F, ψ E/F)
∆(XF, ψF)
µ∈T ∆(µ′ , ψ Fµ/F) (13.2)
µ∈T ∆(µ′ X Fµ/F, ψ Fµ/F) (13.3)
if XFisaquasi­characterof CF.Atthemomentwecontentourselveswithaspecialcase.The
nextlemmawillbereferredtoastheThirdMainLemma.
Lemma 13.3
If K/Fistamelyramifiedtheexpressions(13.2)and(13.3)areequal.
AsweobservedinLemma6.4theextension L/Fwillbeunramifiedand ℓ = [C : 1]
willbeaprime. Chooseagenerator ̟Fof PF. Since Fµ/Fisunramifiedwemaychoose
̟Fµ = ̟F.Choose ̟Esothat N E/F ̟E = ̟F.Certainly
while
Since
weconcludethat
δ L/F = δ K/E = 0
δ K/L = ℓ − 1.
δ K/F = δ K/L + ℓ δ L/F = δ K/E + δ E/F
δ E/F = ℓ − 1.

Chapter13 162
Clearly
µ [Fµ : F] =
µ [H : Hµ] =
µ [G : Gµ]
isjustthenumberofnon­trivialcharactersin S(K/L),thatis ℓ − 1.Moreover m(µ ′ ) = 1.Let
Eµbethefixedfieldof Hµ.Then
N Eµ/Fµ (̟E) = N E/F(̟E) = ̟F.
Thus,asanelementof CFµ , ̟Fliesintheimageof W K/Eµ under τ K/Fµ andhence µ′ (̟F) =
1.Also
n(ψ E/F) = ℓn(ψF) + δ E/F = ℓn + (ℓ − 1)
while
and
If m = m(XF) = 0then
X E/F(̟ ℓn+ℓ−1
E
sothatthelemmaamountstotheequality
and
If m > 0then,byLemma6.4,
Since K/Eisunramified
However
n(ψ Fµ/F) = n.
m(X E/F) = m(X Fµ/F) = 0
) = XF(̟ ℓn+ℓ−1
F ) = XF(̟ n
F )
µ XFµ/F(̟ 1+n
F )
µ ∆1(µ, ψ Fµ/F, ̟ 1+n
F ) =
µ ∆1(µ, ψ Fµ/F, ̟ 1+n
F ).
m(X E/F) = ℓm − (ℓ − 1)
m(X E/F) + n(ψ E/F) = ℓ(m + n).
m(X K/F) = m(X E/F) = ℓm − (ℓ − 1).
X K/F = (µ ′ X Fµ/F) K/Fµ

Chapter13 163
sothat
or
Consequently
ψ K/Fµ (m(µ′ X Fµ/F) − 1) ≥ m(X K/F) − 1 = ℓ(m − 1)
m(µ ′ XFµ/F) − 1 ≥ ϕK/Fµ (ℓ(m − 1)) = m − 1.
m(µ ′ X Fµ/F) ≥ m.
Sinceitisclearlylessthanorequalto mitisequalto m.Because
X E/F(̟ m+n
wehavetoshowthat
isequalto
F ) = XF(̟ ℓ(m+n)
F ) = XF(̟ m+n
F )
µ XFµ/F(̟ m+n
F )
∆1(X E/F, ψ E/F, ̟ m+n
F ) ∆1(µ ′ , ψ Fµ/F, ̟ m+n
F )
∆1(XF, ψF, ̟ m+n
F ) ∆1(µ ′ X Fµ/F, ψ Fµ/F, ̟ m+n
F ).
Let φbethefield OF/PF,let λ = OL/PL,let qbethenumberofelementsin φ,andlet
f = [λ : φ] = [L : F].
Let θbethehomomorphismof Cinto λ ∗ introducedinChapterIVofSerre’sbook.Thus
sothatif h ∈ H
Let h0bethatelementof Hsuchthat
θ(c) = ̟ c−1
E
(modPF)
θ(h −1 ch) ≡ (̟ h−1c−h −1
E ) ≡ (̟ c−1
E )h ≡ θ(c) h .
α h0 = α q
if α ∈ λandlet c0beageneratorof C.Then θ(c0)hasorder ℓand,sincethecentralizerof C
in His {1},
θ(h −r
0 chr 0) = θ(c0) qr
is θ(c0)ifandonlyif fdivides r.Ontheotherhand,itis θ(c0)ifandonlyif ℓdivides q r − 1.
Thustheorderof qmodulo ℓis f.Wealsoobservethatboth Canditsdualgrouparecyclicof
primeordersothatanyelementof Hwhichfixedanelementof Twouldacttriviallyonthe
dualgroupandthereforeon Citself.Itfollowsthat Fµ = Lforall µin T.

Chapter13 164
Supposefirstthat m = 1.Let ψφbethecharacter
ψφ(x) = ψF
x
̟ 1+n
F
on φ.Since OE/PEisnaturallyisomorphicto OF/PFandthemap x −→ N E/Fxgivesthe
map x −→ x ℓ of φintoitselfwhilethemap x −→ S E/Fxinducesthemap x −→ ℓxthe
requiredidentityreducestotheequalityof
and
Xφ(ℓ ℓ )τ(X ℓ φ, ψφ)
τ(Xφ, ψφ)
ThisequalityhasbeenprovedinLemma7.8.
µ∈T τ(µλ, ψ λ/φ)
µ∈T τ(µλX λ/φ, ψ λ/φ).
Nowlet mbegreaterthan1.Since Fµ = 1forall µwearetryingtoshowthat
isequalto
∆1(X E/F, ψ E/F, ̟ m+n
F ) ∆1(µ, ψ L/F, ̟ m+n
F )
∆1(XF, ψF, ̟ m+n
F ) ∆1(µX L/F, ψ L/F, ̟ m+n
F ).
Sincetheactionof Hon Cisnottrivial ℓcannotbe2. If µliesin Tand µ −1liesinthe orbitof νthen
∆1(ν, ψL/F, ̟ m+n
F ) = ∆1(µ −1 , ψL/F, ̟ m+n
F )
is µ(−1)timesthecomplexconjugateof
∆1(µ, ψ L/F, ̟ m+n
F ).
Sincetheorderof µis ℓ, µ(−1) = 1and,if µ = ν,theproductofthetwotermscorresponding
to µand νis1.If
µ −1 = µ qr
with 0 ≤ r < fliesintheorbitof µthen ℓdivides q r + 1.Thus ℓdivides q 2r − 1and 2r = f.
ByLemma7.1
| τ(µλ, ψ λ/φ) | = √ q f = q r
and
τ(µλ, ψ λ/φ) = −∆1(µ, ψ L/F, ̟ 1+n
F )q r

Chapter13 165
if ψ λ/φhasthesamemeaningasbeforeand µλisthecharacterof λ ∗ inducedby µ.Since
δ = ∆1(µ, ψ L/F, ̟ 1+n
F )
isitsowncomplexconjugate,itis ±1.If α ∈ φthen
Since u(α)isan ℓthrootofunityitis1.Thus
HoweveritfollowsfromLemma7.1that
µ −1 (α) = µ(α qr
) = µ(α).
τ(µλ, ψ λ/φ) = τ(µλ).
τ(µλ) ≡ 1 (modη)
where ηisanumberin k p(q f −1)whichisnotaunitandwhoseonlyprimedivisorsaredivisors
of ℓ.Thus
−δq r = τ(µλ) ≡ 1 (modℓ)
and δ = 1.Wearereducedtoshowingthat
isequalto
∆1(X E/F, ψ E/F, ̟ m+n
F )
∆1(XF, ψF, ̟ m+n
F ) ∆1(µX L/F, ψ L/F, ̟ m+n
F )
Let β = β(XF).ByrepeatedapplicationsofLemma8.9weseethatwemaytake
If β(X E/F)ischosenwecouldalsotake
Thusif
wehave
β(X L/F) = β(X K/F) = β(µX L/F) = β.
β(X K/F) = β(X E/F).
m ′ = m(X E/F) = m(X K/F) = 2d ′ + ε ′
β ≡ β(X E/F) (mod P d′
K).

Chapter13 166
Sincebothsidesofthecongruenceliein E
andwemaytake
Then
while
isequalto
β ≡ β(X E/F) (modP d′
E )
β(X E/F) = β.
∆2(X E/F, ψ E/F, ̟ m+n
F ) = ψF
ℓβ
̟ m+n
F
X −1
F (βℓ )
∆2(XF, ψF, ̟ m+n
F )
µ∈T ∆2(µXF, ψ L/F, ̟ m+n
F )
ψF
ℓβ
̟ m+n
F
X −1
F (βℓ ).
Tocompletetheproofofthelemmawehavetoshowthat
isequalto
∆3(XF, ψF, ̟ m+n
F )
µ∈T ∆3(µXF, ψ L/F, ̟ m+n
F ) (13.4)
∆3(X E/F, ψ E/F, ̟ m+n
F )
whenone,andhenceboth,of mand m ′ isodd.
AsremarkedinLemma9.4
∆3(µX L/F, ψ L/F, ̟ m+n
F ) = ∆3(X L/F, ψ L/F, ̟ m+n
F ).
AccordingtoLemma9.6therightsideisequalto
ε∆3(XF, ψF, ̟ m+n
F ) [L:F]
where εis1if f = [L : F]isoddand­1ifitiseven.Thus(13.4)isequalto
Asbefore
ε ℓ−1
f {∆3(XF, ψF, ̟ m+n
F )}.
φ = OF/PF = OE/PE.

Chapter13 167
Let ϕφbethefunctionon φdefinedby
ϕφ(x) = ψF
βx
̟ d+1+n
F
if m = 2d + 1.Then m ′ = 2ℓd + 1sothat d ′ = ℓd.Let ϕ ′ φ
ϕ ′ φ(x) = ψ E/F
βx
̟ d+1+n
F
X −1
F (1 + ̟d F x)
bethefunctionon φdefinedby
X −1
E/F (1 + ̟d Fx).
BecauseofLemma9.3,tocompletetheproofofthelemmawehaveonlytoshowthat
wehave
Since d ′ ≥ mand
ε ℓ−1
f A[σ(ϕφ) ℓ ] = A[σ(ϕ ′ φ )].
3d ′ + ℓ − 1
ℓ
≥ m
N K/L(1 + ̟ d F x) ≡ 1 + ̟d F S K/Lx + ̟ 2d
F E2 K/L (x) (modPm L )
if E2 K/L (x)isthesecondelementarysymmetricfunctionof xanditsconjugatesover L.Thus
Thisinturniscongruentto
Thus
if
or
N E/F(1 + ̟ d Fx) ≡ 1 + ̟ d FS E/Fx + ̟ 2d
F E 2 E/F (x) (modPm F ).
(1 + ̟ d F S E/Fx) (1 + ̟ 2d
F E2 K/F (x)).
ϕ ′ φ(x) = ϕφ(ℓx)ψφ(−E 2 λ/φ (x))
ϕ ′ φ (x) = ϕφ(ℓx)ψφ
ψφ(x) = ϕF
βx
̟ 1+n
F
−ℓ(ℓ − 1)
2
x 2
= {ϕφ(x)} ℓ .

Chapter13 168
sothat
Supposefirstthat pisoddandlet
ϕ ′ φ(x) = ψφ
ϕφ(x) = ψφ
2 ℓx − 2ℓαx
2
2 x − 2αx
= ψφ
2
2 ℓ(x − α)
2
ψφ
−ℓα 2
Referringtotheobservationsinparagraph9weseethatwemustshowthat
ε ℓ−1
f νφ(−1) ℓ−1
2
2
−ℓα −ℓα
2 ψφ = νφ(ℓ)ψφ
2
2
or
ε ℓ−1
f = νφ(−1) ℓ−1
2 νφ(ℓ)
if νφisthequadraticcharacterof φ ∗ . Let qbethenumberofelementsin φ. If qisaneven
powerof ptherightsideis1andif qisanoddpowerof ptherightsideis,bythelawof
quadraticreciprocity, ω(p)if ωisthequadraticcharacterofthefieldwith ℓelements.Thusin
allcasestherightsideis ω(q).If fisoddthen q f isaquadraticresidueof ℓifandonlyif qis.
Since
q f − 1 ≡ 0 (modℓ).
qisaquadraticresidueandbothsidesoftheequationare1.If fisoddtheleftsideis (−1) ℓ−1
f .
Since fistheorderof qmodulo ℓthisis ω(q).
Nowsupposethat p = 2.If
ψφ(−x 2 ) = ψφ(αx)
then,bytheremarksintheproofofLemma9.7,wehavetoshowthat
if ℓ ≡ 1(mod4)andthat
ε ℓ−1
f ϕφ(α) ℓ−1
2 = 1
ε ℓ−1
f ϕφ(α) ℓ+1
2 = 1
if ℓ ≡ 3 (mod4).Wealsosawinparagraph9that
{ϕφ(α)} 2 = ψφ(α 2 )
was+1or­1accordingas qisorisnotanevenpowerof p.Bythesecondsupplementtothe
lawofquadraticreciprocity
ϕφ(α) ℓ−1
2 = ω(q)
2
.

Chapter13 169
if ℓ ≡ 1 (mod4)and
if ℓ ≡ 3 (mod4).Wehavejustseenthat
Thelemmaisproved.
ϕφ(α) ℓ+1
2 = ω(q)
ε ℓ−1
f = ω(q).

Chapter14 170
Chapter Fourteen.
The Fourth Main Lemma.
Inthepreviousparagraphwesaidthatwewouldeventuallyhavetoshowthat
isequalto
∆(χ E/F,ψ E/F)
∆(χF, ψF)
µ∈T ∆(µ′ , ψ Fµ/F) (14.1)
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F). (14.2)
Howeverweverifiedthatthetwoexpressionsareequalonlywhen K/Fistamelyramified.
InthisparagraphweshallshowthattheyareequalifTheorem2.1isvalidforallpairs K ′ /F ′
in P(K/F)forwhich [K ′ : F ′ ] < [K : F].
Lemma 14.1
Suppose K/FiswildlyramifiedandTheorem2.1isvalidforallpairs K ′ /F ′ in P(K/F)
forwhich [K ′ : F ′ ] < [K : F].If χFisanyquasi­characterof CFtheexpressions(14.1)and
(14.2)areequal.
If aand baretwonon­zerocomplexnumbersand misapositiveintegerweagainwrite
a ∼m bif,forsomenon­negativeinteger r, a
b isan mr throotofunity. Definethenon­zero
complexnumber ρbydemandingthat
beequalto
∆(χ E/F,ψ E/F)
ρ∆(χF, ψF)
µ∈T ∆(µ′ , ψ Fµ/F)
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F).
Wehavetoshowthat ρ = 1. Lemma14.1willbeaneasyconsequenceofthefollowingfour
lemmas.
Lemma 14.2
If m(χF)is0or1then ρ = 1andinallcases p ∼2p 1.
Lemma 14.3
If [G : G1]isapowerof2then ρ ∼p 1.

Chapter14 171
Lemma 14.4
Iftheinductionassumptionisvalid,if F ⊆ F ′ ⊆ L,if F ′ /Fisnormal,andif [F ′ : F] = ℓ
isaprimethen p ∼ℓ 1.
Lemma 14.5
Suppose H = H1H2where H2isacyclicnormalsubgroupof H, [H2 : 1]isapowerofa
prime ℓ,and [H1 : 1]isprimeto ℓ.Iftheinductionassumptionisvalid ρ ∼ℓ 1.
Grantthesefourlemmasforamomentandobservethatif mand narerelativelyprime
then ρ ∼m 1and ρ ∼n 1implythat ρ = 1. If ℓisaprimewhichdivides [G : G0]thereis
afield F ′ containing Fandcontainedin Lsothat F ′ /Fisnormaland [F ′ : F] = ℓ. Thus
Lemma14.1followsfromLemma14.4unless [G : G0]isaprimepower.Lemma14.1follows
fromLemmas14.2and14.4unless [G : G0]isapowerof2or p.Suppose [G : G0]isapower
of2or p.Then ρ ∼ [G:G0] 1exceptperhapswhen [G : G0] = 1.If ℓisaprimewhichdoesnot
divide [G : G0]butdoesdivide [G0 : G1]let H2bethe ℓ­Sylowsubgroupof G0/G1. H2isa
normalsubgroupof G/G1whichwemayidentifywith Hand H/H2hasorderprimeto H2.
Thus,byawell­knowntheoremofSchur[7], H = H1H2where H1 ∩ H2 = {1}and H1has
orderprimeto H2.ItfollowsfromLemma14.5that ρ = 1unless [G : G0] = 1or [G : G1]is
apowerof2or p.If [G : G0] = 1and ℓisaprimedividing [G0 : G1]thereisafield F ′ with
F ⊆ F ′ ⊆ Lsuchthat F ′ /Fisnormaland [F ′ : F] = ℓ.Thusif [G : G0] = 1itfollowsfrom
Lemma14.4that ρ = 1unless [G : G1]isapowerof2. Howeverif [G : G1]isapowerof2
therecertainlyisan F ′ in Lwith [F ′ : F] = 2.ItfollowsfromLemmas14.3and14.4that ρ = 1
inthiscaseunless p = 2.If [G : G1]isapowerof pthen G0 = G1and G/G1isabelian.By
assumptiontheabelian p­group G/G1actsonthe p­group C = G1faithfullyandirreducibly.
Thisisimpossible.
WeproveLemma14.2first.Let t ≥ 1besuchthat C = Gtwhile Gt+1 = {1}.Let θtbe
thehomomorphismof Gtinto Pt K /Pt+1
K and θ0thehomomorphismof G0/G1into U0 K /U1 K
introducedinSerre’sbook.If σ ∈ G0and γ ∈ Gtthen
θt(σγσ −1 γ −1 ) = (θ t 0 (σ) − 1)θt(γ).
If σisnotin G1then θ t 0 (σ)isnot1and γ → σγσ−1 γ −1 isaone­to­onemapof Contoitself.
Thus,if σ ∈ G0,
µ(σγσ −1 ) = µ(γ)
implies µ = 1or σ ∈ G1. Consequentlyif µ = 1, Gµ ∩ G0 = G1and L/Fµisunramified.
Since µ = µ ′ L/Fµ ,
m(µ ′ ) = m(µ) = t + 1.

Chapter14 172
Observealsothat tmustberelativelyprimeto [G0 : G1].Inparticularif tiseven [G0 : G1]is
odd.
and
Therelations
δ K/L = ([Gt : 1] − 1) (t + 1)
δ L/F = [G0 : G1] − 1
δ K/E = [G0 : G1] − 1
δ K/F = δ K/L + [G1 : 1] δ L/F = δ K/E + [G0 : G1]δ E/F
obtainedfromProposition4ofChapterIVofSerre’sbook,implythat
If n = n(ψF)then
and
t
δE/F = ([G1; 1] − 1)
[G0 : G1]
+ 1 .
n(ψ Fµ/F) = [G0 : G1]n + [G0 : G1] − 1
n ′
t
= n(ψE/F) = [G1 : 1]n + ([G1 : 1] − 1)
[G0 : G1]
+ 1 .
Chooseagenerator ̟Kof PKandagenerator ̟Eof PE.Thenset ̟L = N K/L̟Kand
̟F = N E/F̟E.Thereisaunit δin Ksuchthat
̟ 1+n
F
̟ 1+n′
E
= δ ̟t K
̟t .
L
Takingthenormfrom Kto Lofbothsidesweseethatif q = [G1 : 1]and k = [G0 : G1]then
̟ t(q−1)
L
̟ t(q−1)
k
F
= N K/Lδ.
t
Let m = m(χF).If m − 1isequalto [G0:G1] then [G0 : G1]divides tand [G0 : G1]is1.
Supposethat
t
m < + 1.
[G0 : G1]

Chapter14 173
Then
m(χ Fµ/F) < ψ Fµ/F
However ψ Fµ/F = ϕ L/Fµ ◦ ψ L/F = ψ L/Fsothat
t
+ 1.
[G0 : G1]
ψ Fµ/F(u) = [G0 : G1]u
if u ≥ 0.Thus m(χFµ/F) < t + 1and m(µ ′ χFµ/F) = t + 1.Moreover,byLemmas13.1and
6.4, m ′ = m(χE/F) = m.Chooseagenerator ̟Fµ of PFµ .Then
NFµ/F(̟ t Fµ ) = γµ ̟ t[Fµ:F]
k
F
where γµisaunit.Theorderof ̟ 1+n
F ̟t Fµ in Fµis 1 + t + n(ψ Fµ/F).Observingthat
weseethat
isequalto
µ [Fµ : F] = q − 1
∆1(χE/F,ψ E/F ′̟ m′ +n ′
E )
µ∈T ∆1(µ ′ , ψFµ/F, ̟ 1+n
F ̟t Fµ )
p
µ χF(γµ)
{∆1(χF,ψF, ̟ m+n
F }
µ ∆1(µ ′ χFµ/F, ψFµ/F, ̟ 1+n
F ̟t Fµ )
.
Itisnowclearthat ρ = 1if m = 0.
If
sothatinparticular m ≥ 2,then
m ≥
t
+ 1
[G0 : G1]
m ′
t
= m(χE/F) = [G1 : 1]m − ([G1 : 1] − 1)
[G0 : G1]
isalsogreaterthanorequalto2and
Since m ′ ≥ 2and K/Eistamelyramified
m ′ + n ′ = [G1 : 1] (m + n).
+ 1
m(χ K/F) = ψ K/E(m(χ E/F) − 1) + 1 = ψ K/F(m − 1) + 1.

Chapter14 174
Since
and
wehave
However
sothat
isatmost
Thus
Consequently
m(χ Fµ/F) ≤ ψ Fµ/F(m − 1) + 1
ψ Fµ/F(m − 1) + 1 ≥ ψ Fµ/F
t
+ 1 = t + 1
k
m(µ ′ χFµ/F) ≤ ψFµ (m − 1) + 1.
χ K/F = (µ ′ χ Fµ/F) K/Fµ
ψ K/Fµ (ψ Fµ/F(m − 1)) + 1 = ψ K/F(m − 1) + 1 = m(χ K/F)
ψ K/F(m(µ ′ χ Fµ/F) − 1) + 1.
m(µ ′ χ Fµ/F) = ψ Fµ/F(m − 1) + 1.
m(µ ′ χ Fµ/F) + n(ψ Fµ/F) = [G0 : G1] (m + n).
Sincetherangeofeach µ ′ liesinthegroupof qthrootsofunity
isequalto
with σ ∼p ρ.
∆1(χ E/F, ψ E/F, ̟ m+n
F )
µ ∆1(µ ′ , ψ Fµ/F, ̟ n+1
F ̟t Fµ )
σ∆1(χF, ψF, ̟ m+n
F )
µ ∆1(µ ′ χFµ/F, ψFµ/F, ̟ m+n
F )
Thenextstepintheproofofthelemmaistoestablishasimpleidentity.Asusuallet rbe
theintegralpartof t+1
2 andlet r + s = t + 1.Choose β(µ′ )sothat
ψ Fµ/F
β(µ ′ )x
̟ 1+n
F ̟t Fµ
= µ ′ (1 + x)
for xin P s Fµ .Thereisaunit αµin Lsuchthat αµ̟ t Fµ = ̟t L .Then
αµβ(µ
ψL/F ′
)x
= µ(1 + x)
̟ 1+n
F ̟t L

Chapter14 175
for xin P s L .Wetake β(µ) = αµβ(µ ′ ).If σ ∈ Gapossiblechoicefor β(µ σ )is
β(µ) σ ̟t L
σ t ̟L Let φ = OF/PF = OE/PEandlet ψφbetheadditivecharacterof φdefinedby
Thereisaunique αin φsuchthat
Finallylet
Iwanttoshowthat
in φ.
Let λ = OL/PL = OK/PK.If
ψφ(x) = ψF
.
x
̟ 1+n
.
F
ψφ(αx) = ψφ(x q ).
ω1 = S E/F
µ γµ = ωq 1
α q
ω = S K/L
then ω1 = δωin λ.Weneedthefollowinglemma.
Lemma 14.6
̟ 1+n
F
̟ 1+n′
E
.
µ N Fµ/Fβ(µ ′ ) (14.3)
t ̟K Suppose K ′ /F ′ isanabelianextensionand G ′ = G(K ′ /F ′ ). Supposethereisat ≥ 1
suchthat G ′ = G ′ tand G′ t+1 = {1}.Let ̟K ′beageneratorof P′ K ,let ̟F ′ = NK ′ /F ′(̟K ′),
andlet
t ̟K ′
ω = SK ′ /F ′
̟t F ′
.
Alsolet q = [K ′ : F ′ ].Therearenumbers a, . . ., fin OF ′suchthatforall xin OF ′
̟ t L
N K ′ /F ′(1 + x̟ t K ′)

Chapter14 176
iscongruentto
modulo P t+1
F ′ .
1 + (x q + ax q
p + . . . + fx p + ωx)̟ t F ′
Suppose F ′ ⊆ L ′ ⊆ K ′ andthelemmaistruefor K ′ /L ′ and L ′ /F ′ .Thelemmafor K ′ /F ′
followsfromtherelations
[K ′ : F ′ ] = [K ′ : L ′ ] [L ′ : F ′ ]
and
and
N K ′ /F ′(1 + x̟ t K ′) = N L ′ /F ′(N K ′ /L ′(1 + x̟t K ′))
S K ′ /F ′
t ̟K ′
̟ t F ′
≡ S L ′ /F ′
t ̟L ′
̟ t F ′
S K ′ /L ′
̟ t K ′
ThelemmaforextensionsofprimeorderisprovedinSerre’sbook.
Supposethen
for xin OL.Since
whichinturnequals
weconcludethat
̟ t L ′
.
N K/L(1 + x̟ t K ) ≡ 1 + (xq + . . . + ̟x)̟ t L (modPt+1
L )
ψ λ/φ(αx) = ψφ(αS λ/φ(x)) = ψφ((S λ/φ(x)) q )
ψφ(S λ/φx q ) = ψ λ/φ(x q )
ψ λ/φ(y(x q + . . . + ωx)) = ψ λ/φ((αy 1
q + . . . + ωy)x).
Also
αy 1 q q + . . . + ωy = α q y + . . . + ω q y q
isapolynomial Q(y)in y.
Foreach νin S(K/L),wechoose β1(ν)sothat
β1(ν)x
ψL/F = ν(1 + x)
for xin P s L .Since kp = kin λ
̟ 1+n
F ̟t L
ψ λ/φ(kβ1(ν)(x q + . . . + ωx) = ψ λ/φ(β1(ν) [(kx) q + . . . + ωkx])

Chapter14 177
if xisin OF.Theleftsideisalsoequalto
β1(ν) PK/L(x̟ ψL/F t K )
= 1.
̟ 1+n
F ̟t L
Thus Q(kβ1(ν)) = 0.Since β1(ν1) = β2(ν2) (modPL)implies ν1 = ν2wehavefoundallthe
rootsof Q(y) = 0.Thus
αq
≡
ωq ν=1 kβ1(ν) ≡
ν=1 β1(ν) (mod PL).
Let Mµbeasetofrepresentativesforthecosetsof Gµin G.Then
iscongruentto
µ∈T NFµ/Fβ(µ ′ ) =
ν=1 β1(ν)
µ∈T
µ∈T
σ∈Mµ
α
σ∈Mµ
σ µ
modulo PL.Toverifytheidentity(14.3)wehavetoshowthat
Since
thecongruencereducesto
µ
σ∈Mµ
γµ =
α σ µ
̟t L
σ t
L
̟
̟ t(q−1)
L
(q−1)
t k ̟F σ∈Mµ
≡ δ q
whichisvalidbecausetheleftsideis N K/Lδand
N K/Lδ ≡ δ q
σ t
L
ασ µ
̟
µ γµ
̟
≡ δ q
β(µ ′ ) σ
̟t −1 L
σ t
L
̟
[Fµ:F]
−t k
F
(mod PK)
(modPK).
(modPK).
If m = 1then m(χ Fµ/F) ≤ 1andwecantake β(µ ′ χ Fµ/F) = β(µ ′ ).Then
∆2(µ ′ , ψ Fµ/F, ̟ 1+n
F ̟t Fµ )

Chapter14 178
isequalto
Lemma9.4impliesthat
If xbelongsto OEthen
χF(N Fµ/F(β(µ ′ ))) ∆2(µ ′ χ Fµ/F, ψ Fµ/F, ̟ 1+n
F ̟t Fµ ).
∆3(µ ′ , ψ Fµ/F, ̟ 1+n
F ̟t Fµ ) = ∆3(µ ′ χ Fµ/F, ψ Fµ/F, ̟ 1+n
F ̟t Fµ ).
ψ E/F
x
̟ 1+n′
E
If χφisthecharacterof φ ∗ determinedby χFthen
and
= ψφ(ω1x).
∆1(χF, ψF, ̟ 1+n
F ) = A[−τ(χφ, ψφ)]
∆1(χE/F, ψE/F, ̟ 1+n′
E ) = χφ(ω q
1 ) A[−τ(χq φ , ψφ)].
Therightsideofthisexpressionisequalto
χφ(ω q
1 α−q ) A[−τ(χφ, ψφ)].
Theidentity(14.3)nowshowsthat ρ = 1when m = 1.
Supposethat
Let βbeagivenchoiceof β(χF).Then
if m ′ = 2d ′ + ε ′ .Ontheotherhand
1 < m <
t
+ 1.
[G0 : G1]
β(χE/F) ≡ P ∗ E/F (β, ̟m′ +n ′
E , ̟ m+n
F ) (mod P d′
E )
ψ L/F(m − 1) + 1 + n(ψ L/F) = [G0 : G1](m − 1) + 1 + [G : G0]n + [G0 : G1] − 1
whichequals [G0 : G1](m + n)andLemmas8.3,8.4and8.7implythat
if
P ∗ L/F (β, ̟m+n
F , ̟ m+n
F ) ≡ β (modP d1
L )
ψ L/F(m − 1) + 1 = 2d1 + ε1.

Chapter14 179
If
then
Thus
If
and
then
Let
iscongruentto
modulo P d′
1
K .
andthat
Itisclearthat
Ifwechoose
then
P ∗ K/E (β(χ E/F,̟ m′ +n ′
E
ψ K/F(m − 1) + 1 = 2d ′ 1 + ε′ 1
, ̟ m′ +n ′
E
β(χE/F) ≡ P ∗ K/F (β, ̟m′ +n ′
E , ̟ m+n
) ≡ β(χ E/F) (modP d′
1
K ).
F ) (mod P d′
1
K ).
v = t + 1 − (ψ L/F(m − 1) + 1) = t − [G0 : G1] (m − 1).
γ = ̟t−v
L
̟ m−1
F
γ ′ = ̟m′ +n ′
E ̟v K
̟ 1+n
F ̟t L
P ∗ K/F (β, ̟m′ +n ′
E , ̟ m+n
F ) ≡ P ∗ K/L (β, ̟m′ +n ′
E
γ ′ P ∗ K/L (γβ, ̟1+n
F ̟t L̟ −v
K
, ̟ m+n
F ) (modP d′
1
, ̟1+n
F ̟t−v
L )
∆2(χE/F, ψE/F, ̟ m′ +n ′
E ) ∼p χ −1
F (NE/F(β(χ E/F)))
∆2(µ ′ , ψ Fµ/F, ̟ n+1
F ̟t Fµ ) ∼p 1.
β(µ ′ χ Fµ/F) = β(µ ′ ) + β ̟t Fµ
̟ m−1
F
∆2(µ ′ χ Fµ/F, ψ Fµ/F, ̟ n+1
F ̟t Fµ ) ∼p χ −1
F
N Fµ/F
K )
β(µ ′ ) + β ̟t Fµ
̟ m−1
F
.

Chapter14 180
Moreover
Let
equal
Since
∆2(χF, ψF, ̟ m+n
F ) ∼p χ −1
F (β).
∆2(χE/F,ψ E/F, ̟ m′ +n ′
E )
µ ∆2(µ ′ , ψFµ/F, ̟ n+1
F ̟t Fµ )
τ∆2(χF,ψF, ̟ m+n
F )
µ ∆2(µ ′ χFµ/F, ψFµ/F, ̟ n+1
F ̟t Fµ )
µ χF(γµ)
.
χF(u) ∼p 1
if u ∈ U 1 F ,allweneeddotoshowthat τ ∼p 1isprovethat
iscongruentto
modulo PF.
β
µ N Fµ/F
β(µ ′ ) + β ̟t Fµ
̟ m−1
F
N E/F(β(χ E/F))
µ γµ
Asbeforewechoose β(µ) = αµ(β(µ ′ ).If ν = µ σ apossiblechoicefor β(ν)is
Wecanalsochoose
α σ µβ(µ ′ ) σ ̟t L
̟σ t
L
β(µχ L/F) = αµβ(µ ′ ) + β ̟t L
̟ m−1
F
Thenapossiblechoicefor B(µ σ χ L/F)is
α σ µ
β(µ ′ ) + β ̟t Fµ
̟ m−1
F
σ ̟ t L
WeapplyLemma8.10with Freplaceby L,
.
= αµ
β(µ ′ ) + β ̟t Fµ
̟ m−1
F
̟σ t = α
L
σ µ(β(µ ′ ) σ ̟t L
̟σ t + β
L
̟t L
̟ m−1
F
δ = ̟ 1+n
F ̟t L
.
.

Chapter14 181
and ε1 = ̟ v K .Itimpliesthat
iscongruentto
γβ
µ∈T
σ∈Mµ
N K/L
α σ
µ
modulo PL.Thelastexpressionisequalto
γβ
µ∈T N Fµ/F
andwehavetoshowthat
γNK/L(γ ′
)
β(µ ′ ) + β ̟t Fµ
̟ t F
µ γµ
iscongruentto1modulo PL.Firstofall
Since
therequiredrelationfollows.
equalto
Define ηbysetting
β(χE/F)
γ ′
β(µ ′ ) + β ̟t Fµ
̟ m−1
F
µ∈T
µ∈T
NK/L(γ ′ ) = ̟ m′ +n ′ −q(1+n)
F ̟ −qt+v
L = γ
γµ =
σ∈Mµ
η∆3 χE/F,ψ E/F, ̟ m′ +n ′
E
ασ −1 µ
σ t ̟L µ∈T ∆3
̟
σ
α
σ∈Mµ
σ µ
̟t L
σ t
L
̟
α
σ∈Mµ
σ µ
̟t L
σ t
L
̟
(q−1)t
k
−1 ̟ F
[Fµ:F]
−t k
F
̟ (q−1)t
L
.
̟t L
σ t
L
̟
µ ′ , ψFµ/F, ̟ 1+n
F ̟t
Fµ
∆3(χF, ψF, ̟ m+n
F )
µ∈T ∆3
µ ′ χFµ/F, ψFµ/F, ̟ 1+n
F ̟t
. Fµ
Wenowknowthat η ∼p ρ.Weshallshowthat η ∼p 1.Thiswillprovenotonlytheassertion
ofLemma14.2butalsothatofLemma14.3,providedofcoursethat
m − 1 <
t
[G0 : G1] .

Chapter14 182
Lemmas9.2and9.3implydirectlythat η ∼p 1if pis2.
Suppose pisodd.Lemma9.4impliesthat
∆3(µ ′ , ψ Fµ/F, ̟ 1+n
F ̟t Fµ ) ∼p ∆3(µ ′ χ Fµ/F, ψ Fµ/F, ̟ 1+n
F ̟t Fµ ).
Since m ′ = mallweneeddoisshowthat
∆3(χE/F,ψ E/F, ̟ m′ +n ′
E ) ∼p ∆3(χF, ψF, ̟ m+n
F )
when misodd.Let φ = OF/PF = OE/PE.Let β ′ = β(χ E/F)andlet β = β(χF).If ψ ′ φ is
thecharacterof φdefinedby
and ψ ′′
φisthecharacterof φdefinedby
ψ ′ φ (x) = ψF
ψ ′′
φ (x) = ψ E/F
β x
̟ n+1
F
β ′ x
̟ n′ +1
E
andif ψ ′′
φ (x) = ψ′ φ (δx)then,byLemmas9.2and9.3,allwehavetodoisshowthat δisasquare
in φ.If
ω1 = S E/F
̟ n+1
F
̟ n′ +1
E
then δ = ω1 β′
β in φ.Toshowthat δisasquareweshowthat δq isasquare.
Wesawthat
in φ.But
N E/F(β ′ )
β
δ q ≡ NE/Fα ≡ β 1−q NE/F(β ′ )
ω
β
q
1 .
=
µ γµ
−1
β ̟t Fµ
̟ m−1
F
µ N Fµ/F
≡ 0 (modPL)
β(µ ′ ) + β ̟t Fµ
̟ m−1
F

Chapter14 183
because t > [G0 : G1](m − 1).Wealsosawthat
µ γµ
−1
µ NFµ/Fβ(µ ′
)
= αq
̟ q
1
in φ.Since β 1−q isclearlyasquare,weneedonlycheckthat αisasquare.Thecharacter
isidentically1,sothatthekernelofthemap
isnon­trivial.Thus α = x q−1 forsome xin φ.
Nowsupposethat
x −→ ψφ (x q − αx)
x −→ x q − αx
m − 1 ≥
t
[G0 : G1] .
Wehavetoshowthatthecomplexnumber σdefinedatthebeginningoftheproofsatisfies
σ ∼2p 1.ToproveLemma14.3wewillhavetoshowthat σ ∼p 1if [G : G1]isapowerof2.
Given β = β(χF)wemaychoose β(χ L/F) = β(χ Fµ/F) = β.Moreover
β(χE/F) ≡ P ∗ E/F (β, ̟m+n
F , ̟ m+n
F ) (modP d′
E)
if m ′ = m(χ E/F) = 2d ′ + ε ′ .ByLemmas8.3,8.4and8.7
modulo P d′
1
K if
Thus
If
and
P ∗ K/F (β, ̟m+n
F , ̟ m+n
F ) ≡ P ∗ K/E (β(χE/F), ̟ m+n
F , ̟ m+n
F ) ≡ β(χE/F )
ψ K/F(m − 1) + 1 = 2d ′ 1 + ε ′ 1.
β(χE/F ) ≡ P ∗ K/L (β, ̟m+n
F , ̟ m+n
F ) (modP d′
1
ψ Fµ/F(m − 1) + 1 = 2dµ + εµ
ψ Fµ/F
′ α(µ )x
̟ m+n
= µ
F
′ (1 + x)
K ).

Chapter14 184
for xin P dµ+εµ
Fµ
If
then
wemaytake
for xin P d1+ε1
L .If ν = µ σthen β(µ ′ χ Fµ/F) = β + α(µ ′ ).
ψ L/F(m − 1) + 1 = 2d1 + ε1
µ(1 + x) = ψ L/F
ν(1 + x) = ψ L/F
for xin P d1+ε1
L .Lemma8.2impliesthat
iscongruentto
′ α(µ )x
̟ m+n
F
′ σ α(µ ) x
̟ m+n
F
N E/F(β(χ E/F)) = N K/L(β(χ E/F))
β
µ∈T
modulo PL.Thelastexpressionisequalto
Moreover
β
σ∈Mµ
(β + α(µ ′ ) σ )
µ∈T N Fµ/F(β + α(µ ′ )).
∆2(χF, ψF, ̟ m+n
F ) ∼p χ −1
F (β)
∆2(µ ′ , ψ Fµ/F,̟ 1+n
F ̟t Fµ ) ∼p 1
∆2(χ E/F,ψ E/F, ̟ m+n
F ) ∼p χ −1
F (N E/F(β(χ E/F)))
∆2(µ ′ χ Fµ/F, ψ Fµ/F, ̟ m+n
F ) ∼p χ −1
F (N Fµ/F(β + α(µ ′ ))).
Define τbydemandingthat
beequalto
∆3(χ E/F,ψ E/F, ̟ m+n
F )
µ ∆3(µ ′ , ψ Fµ/F, ̟ 1+n
F ̟t Fµ )
τ ∆3(χF, ψF, ̟ m+n
F )
µ ∆3(µ ′ , χ Fµ/F, ψ Fµ/F, ̟ m+n
F ).

Chapter14 185
Since χF(u) ∼p 1if u ∈ U 1 F theprecedingdiscussionshowsthat σ ∼p τ.Lemmas9.2and9.3
showthat τ ∼2p 1.Lemma14.2isnowcompletelyproved.ToproveLemma14.3wehaveto
showthat τ ∼p 1if [G : G1]isapowerof2.Wemaysupposethat pisodd.
Thereareanumberofpossibilities.
(i.a) tisevenand misodd. [G0 : G1]mustbeoddandhence1,forwearenowassumingthat
[G : G1]isapowerof2.Since
and
m(χ Fµ/F) = [G0 : G1] (m − 1) + 1
m(χ E/F) = [G1 : 1] (m − 1) − ([G1 : 1] − 1)t
[G0 : G1]
both m(χ Fµ/F)and m(χ E/F)areodd.
(i.b) tisevenand miseven.Again [G0 : G1]is1.Thistimeboth m(χ Fµ/F)and m(χ E/F)are
even.
(ii.a) tisoddand misodd.Then m(χ Fµ/F)isodd.If
[G1 : 1] − 1
[G0 : G1]
iseven m(χ E/F)isodd.Otherwiseitiseven.
(ii.b) tisoddand miseven.If [G0 : G1]isodd,thatis1,then m(χ Fµ/F)isoddand m(χ E/F)
iseven.If [G0 : G1]iseventhen m(χ Fµ/F)isoddand m(χ E/F)isevenoroddaccording
as
[G1 : 1] − 1
[G0 : G1]
isevenorodd.
If tisoddthenclearly
µ ∆3(µ ′ , ψ Fµ/F, ̟ 1+n
F ̟t Fµ ) ∼p 1.
Wearegoingtoshowthatthisisalsotrueif tiseven. Then L/F,andhence Fµ/F,is
unramified.Let φµ = OFµ /PFµ .If
ψφµ (x) = ψ Fµ/F
′ β(µ )x
̟ 1+n
F
+ 1

Chapter14 186
andif ϕφµ isanowherevanishingfunctionon φµsatisfying
then
ϕφµ (x + y) = ϕφµ (x) ϕφµ (y)ψφµ (xy)
∆3(µ ′ , ψ Fµ/F, ̟ 1+n
F ̟t Fµ ) ∼p A[−σ(ϕφµ )].
If αbelongsto φ ∗ µ let νφµ (α)equal+1or­1accordingas αisorisnotasquarein φµ. If
φ = OF/PFthen
νφµ (α) = νφ (N φµ/φ(α)).
If
then,accordingtoparagraph9,
ψφ(x) = ψF
x
̟ 1+n
F
∆3(µ ′ , ψ Fµ/F,̟ 1+n
F ̟t Fµ ) ∼p νφ(N Fµ/F(β(µ ′ ))) {A[−σ(ϕφ)]} [Fµ:F]
if ϕφisanynowherevanishingfunctionon φsatisfying
Thusif aisthenumberof µin T
isequalto
ϕφ(x + y) = ϕφ(x) ϕφ(x) ψφ(xy).
η(−1) a νφ
where η ∼p 1and q = [G1 : 1].
Wesawinparagraph9that
µ ∆3(µ ′ , ψ Fµ/F, ̟ 1+n
F ̟t Fµ )
µ NFµ/F(β(µ ′
)) A[σ(ϕφ) q−1 ]
A[σ(ϕφ) 2 ] ∼p νφ(−1) A[|σ(ϕφ)| 2 ] = νφ(−1).
Since tiseven G0 = G1and G/G1 = G/G0isabelian.If σ ∈ G
{µ ∈ S(K/L) | µ = µ σ }

Chapter14 187
isasubgroupof S(K/L)invariantunder G.Itisnecessarilyeither S(K/L)or {1}.If σisnot
in G1itisnot S(K/L). Thus Gµ,theisotropygroupof µ,is G1forall µin Tand Fµ = L.
Moreover
µ NFµ/F(β(µ ′ )) =
β(µ
µ
′ ) σ .
σ∈G/G1
Wemayregard C = G1asavectorspaceoverthefieldwith pelements.If σ ∈ G/G1and
theorderof σdivides p − 1thenalltheeigenvaluesofthelineartransformation c −→ σcσ −1
lieintheprimefield. Sincethelineartransformationalsohasorderdividing p − 1itis
diagonalizable.Since G/G1isabelianandactsirreduciblyon Cthelineartransformationisa
multipleoftheidentity.Inparticularif σ0istheuniqueelementoforder2then σ0cσ −1
0
forall c.Asaconsequence µ σ0 = µ −1 and
β(µ ′ ) σ0 ≡ −β(µ ′ ) (modPL)
= c−1
ifwechoose,aswemaysince Fµ/Fisunramified, ̟Fµ = ̟F.If Disthegroup {1, σ0}and
Misasetofrepresentativesforthecosetsof Din G/G1then
if
Clearly
µ∈T
γ =
If χisthenon­trivialcharacterof Dand
isthetransferthen
σ∈G/G1
µ∈T
γγ σ0 = (1−) q−1
2 γ 2
β(µ ′ ) σ = γγ σ0
σ∈M β(µ′ ) σ .
v : G/G0 −→ D
γ σ = χ(v(σ)) a γ
(mod PL).
forall σin G/G0. νφ(γ 2 ) = 1ifandonlyif χ(v(σ)) a is1forall σ.If σisageneratorof G/G0
then
v(σ) = σ [G:G 0 ]
2 = σ0
sothat νφ(γ 2 ) = (−1) a .Puttingallthesefactstogetherweseethat
µ ∆3(µ ′ , ψ Fµ/F, ̟ 1+n
F ̟t Fµ ) ∼p 1.

Chapter14 188
Observethatifwehadtaken ̟Fµ tobe δµ̟Fthen N Fµ/Fβ(µ ′ )wouldhavetobemultiplied
by
{N Fµ/Fδµ} t
whichisasquaremodulo PFbecause tiseven.Thustheresultisvalidforallchoicesof ̟Fµ .
Eventuallywewillhavetodiscussthevariouspossibilitiesseparately.Therearehowever
anumberofcommentsweshouldmakefirst.If misoddand m(χ Fµ/F)isoddthen
if
∆3(µ ′ χ Fµ/F, ψ Fµ/F, ̟ m+n
F ) ∼p νφµ (β + α(µ′ )) A[−σ(ϕφµ )]
ψφµ (x) = ψ Fµ/F
x
̟ 1+n
.
F
Observethat,because misodd,wemaytakethenumber δinLemma9.3tobe ̟ m−1
2
F . Of
course ϕφµ isanyfunctionon φµwhichvanishesnowhereandsatisfies
ApplyingLemma9.1weseethat
if k = [G0 : G1],if
ϕφµ (x + y) = ϕφµ (x) ϕφµ (y)ψφµ (xy).
A[−σ(ϕφµ )] ∼p −νφ
ψφ(x) = ψF
k [Fµ:F]
k
x
̟ 1+n
F
A σ(ϕφ) [Fµ:F]
k
andif ϕφbearstheusualrelationto ψφ.Weuse,ofcourse,therelation
Observealsothat
If misodd
kS φµ/φ(x) = S Fµ/F(x).
νφµ (β + α(µ′ )) = νφ(N φµ/φ(β + α(µ ′ ))).
∆3(χF, ψF, ̟ m+n
F ) ∼p −νφ(β) A[σ(ϕφ)].
Ifboth mand m ′ = m(χ E/F)areoddandif β ′ = β(χ E/F)then
∆3(χ E/F, ψ E/F, ̟ m+n
F ) ∼p −νφ(β ′ ) A[σ(ϕ ′ φ )]

if ϕ ′ φ bearstheusualrelationtothecharacter
ψ ′ φ (x) = ψ E/F
⎛
⎝
̟ 1+n
F
⎞
x
⎠ . (q−1)t
k ̟
Thereisaunit εin OKsuchthat ̟E = ε̟ q
K .If σ ∈ Cthen
̟ σ−1
E = εσ−1̟ (σ−1)q
K ≡ 1 (mod PK)
because t ≥ 1.Thusthemultiplicativecongruence
issatisfiedand
If
asbefore,then
Since
wehave
̟ q
E ≡ N E/F̟E = ̟F (mod ∗ PE)
1
̟ (q−1)t
k
E
≡ ̟1+n
F
̟ 1+n′
E
ω1 = S
ψ ′ φ(x) = ψF
̟ 1+n
F
̟ 1+n′
E
E
(mod ∗ PE).
ω1x
̟ 1+n
.
F
νφ(β ′ ) = νφ(β ′ ) q = νφ(N E/Fβ ′ )
∆3(χ E/F, ψ E/F, ̟ m+n
F ) ∼p −νφ(N E/Fβ ′ ) νφ(ω1) A[σ(ϕφ)].
Define ηbydemandingthat
beequalto
∆3(χ E/F,ψ E/F, ̟ m+n
F )
η∆3(χF, ψF, ̟ m+n
F )
µ ∆3 (µ ′ χ Fµ/F, ψ Fµ/F, ̟ m+n
F ).
Wehavetoshowthat η ∼p 1.Ifboth tand mareeventhisisclear.If tisevenand misodd
wearetoshowthat
νφ(N E/Fβ ′ )νφ(ω1) ∼p (−1) a νφ(−1) q−1
2k νφ(k) q−1
k νφ(β)
µ νφ(N φµ/φ(β + α(µ ′ )))

Chapter14 190
if aisthenumberofelementsin T.Since tiseven kis1.Asbefore
β
µ N φµ/φ(β + α(µ ′ )) = β
µ
σ∈G/G1
iscongruentto N E/Fβ ′ modulo PK.Allweneeddoisshowthat
νφ(ω1) = (−1) a νφ(−1) q−1
2 .
(β + α(µ ′ ) σ )
Since tiseveneach γµisasquarein φ.Applyingtheidentity(14.3)weseethat
νφ(ω1) = νφ(ω q
−1
1 ) = νφ(α)νφ µ NFµ/Fβ(µ ′
) .
Wehaveseenthat αisasquarein φsothat νφ(α) = 1.Wealsosawthat
νφ
µ NFµ/Fβ(µ ′
)
when tiseven.Therequiredrelationfollows.
= (−1) a νφ(−1) q−1
2
Wesupposehenceforththat tisodd.Thediscussionwillbefairlycomplicated.Suppose
firstthat misalsoodd.Then
[G0 : G1](m − 1) = t
and
sothat
and
m − 1 >
t
[G0 : G1]
β + α(µ ′ ) ≡ β (mod PL)
µ ∆3(µ ′ χ Fµ/F, ψ Fµ/F, ̟ m+n
F ) ∼p (−1) a νφ
Thusif q−1
k isoddwehavetoshowthat
andif q−1
k isevenwehavetoshowthat
k q−1
k
νφ
β q−1
k
A[σ(ϕφ)] q−1
k .
(−1) a+1 νφ(k) νφ(−1) q−1 1
2k + 2 ∼p 1 (14.4)
νφ(ω1) ∼p (−1) a νφ(−1) q−1
2k . (14.5)

Chapter14 191
Nowsuppose miseven. If [G0 : G1]is1thereisnothingtoprove. If k = [G0 : G1]is
eventhen
t
m − 1 >
[G0 : G1]
and
If
β + α(µ ′ ) ≡ β (modPL).
ψ ′ φµ (x) = ψ Fµ/F
⎛
⎝β ̟k(m−1)
Fµ
̟ m+n
F
⎞
x
⎠
and ϕ ′ isafunctionon φµwhichvanishesnowhereandsatisfies
φµ
then
If
then εµisaunitand
if,asbefore,
ByLemma9.1, A[−σ(ϕφµ )]isequalto
−νφ
ϕ ′ φµ (x + y) = ϕ′ φµ (x) ϕ′ φµ (y) ψ′ φµ (xy)
∆3(µ ′ χ Fµ/F, ψ Fµ/F, ̟ m+n
F ) ∼p A[−σ(ϕ ′ φµ )].
k [Fµ:F]
k
νφ
If q−1
k isevenwehavetoshowthat
(−1) a νφ
εµ̟ m−1
F
= ̟ k(m−1)
Fµ
ψφµ (x) = ψ φµ/φ(kβεµx)
ψφ(x) = ψF
β [Fµ:F]
k
µ N φµ/φεµ
If q−1
k isoddthen m ′ = m(χ E/F)isodd.If
ψ ′′
φ (x) = ψ E/F
x
̟ 1+n
.
F
νφ(N φµ/φεµ) A[σ(ϕφ)] [Fµ:F]
k .
νφ(−1) q−1
2k ∼p 1 (14.6)
β ′ ̟m′ −1
E
̟ m+n
F
x

Chapter14 192
and ϕ ′′ φ
bearstheusualrelationto ψ′′
φ then
Now νφ(β ′ ) = νφ(β ′ ) q and
whichinturniscongruentto
modulo PK.Let
and,asbefore,
then
Wesawthat
sothat
Thuswehavetoshowthat
∆3(χ E/F,ψ E/F, ̟ m+n
F ) ∼p A[−σ(ϕ ′′ φ )].
β
µ∈T
(β ′ ) q ≡ N E/Fβ ′
σ∈Mµ
ε1̟ m+n
F
ω1 = S E/F
(β + α(µ ′ ) σ ) ≡ β q
= ̟ q(m+n)
E
̟ 1+n
F
̟ 1+n′
E
A[−σ(ϕ ′′ φ )] ∼p νφ(ω1) νφ(ε1) νφ(β) A[−σ(ϕφ)].
̟ q
E ≡ ̟F (mod ∗ PE)
ε1 ≡ 1 (modPE).
νφ(ω1) ∼p (−1) a+1 νφ(k) νφ(−1) q−1
2k −1
2 νφ
µ N φµ/φεµ
. (14.7)
Thefouridentities(14.4),(14.5),(14.6),and(14.7)lookratherinnocuous. Howeverto
provethemisnotanentirelytrivialmatter.Wefirstconsiderthecasethat G/G1isabelian.If
σ0 ∈ G/G1isoforder2,theargumentusedbeforeshowsthat σ0cσ −1
0 = c−1 forall cin C.
Sincetherepresentationof G/G1on Cisfaithful G/G1hasonlyoneelementoforder2andis
thereforecyclic.Inthiscase Fµ = Lforall µand a = q−1
[G:G1] .Wemaychoose ̟Fµ = ̟L.If
then γµ = γ t and
N L/F̟L = γ̟ [G:G0]
F
µ γµ = γ at .

Chapter14 193
If [G0 : G1] = 1wemaychoose ̟L = ̟Fsothat γ = 1. Theargumentusedbefore
showsthat
β(µ ′ ) σ
= (−1) a νφ(−1) q−1
2 .
νφ
µ∈T
Theidentity(14.3)showsthat
σ∈G/G1
νφ(ω1) = (−1) a νφ(−1) q−1
2k .
Theidentity(14.5)whichistheonlyoneofconcernherebecomes
whichisclearbecause k = [G0 : G1] = 1.
νφ(−1) q−1
2 = νφ(−1) q−1
2k
Nowtake [G : G0] = 1.Wemaychoose ̟E = N K/E̟Ksothat ̟F = N L/F̟Land γ
isagain1.Itisperhapsworthpointingoutthesespecialchoicesarenotinconsistentwithany
choicesyetmadeinthisparagraph.Thisisnecessarybecausetheargumentsappearinginthe
functions ∆2mustbethesameasthoseappearinginthefunctions ∆3.Wepreviouslydefined
andshowedthat
Observethat
because
̟ k L
̟F
=
σ∈G/G1
δ = ̟1+n
F
̟ 1+n′
E
· ̟t L
̟ t K
NK/Lδ = ̟t(q−1)
L
̟ t(q−1)
k
F
.
̟ 1−σ
L ≡ θ0(σ) −1 ≡ −1 (modPL)
{θ0(σ) | σ ∈ G/G1}
isjustthesetof kthrootsofunityin φand kisapowerof2.Itisnot1because
Since
wehave
[G0 : G1] = [G : G1] > 1.
δ q ≡ N K/Lδ (modPK)
νφ(δ) = νφ(−1) q−1
k .

Chapter14 194
Ifasbefore
ψ L/F
β1(ν)x
̟ 1+n
F ̟t L
= ν(1 + x)
for xin Ps Land νin S(K/L)then,aswesawwhenprovingtheidentity(14.3),
Thus
ω q
1 ≡ δqα q
ν=1 β1(ν)
νφ(ω1) = νφ(−1) q−1
k
(mod PL).
ν=1 νφ(β1(ν)).
Wecanchoose q−1
p−1elements νiin S(K/L)sothateverynon­trivialelementof S(K/L)isof
theform ν j
i , 0 < j < p.Then
because
When miseven
ν=1 νφ(β1(ν)) = νφ
νφ(εµ) = νφ
q−1
p−1
i=1
j=1 jβ(νi)
p−1
νφ(β(νi)) p−1 = 1.
k ̟L = νφ(−1).
̟F
= νφ(−1) q−1
p−1
Since a = q−1
k theidentities(14.4),(14.5),(14.6)and(14.7)become
νφ(k) νφ(−1) q−1 1
2k + 2 = 1 (14.4 ′ )
νφ(−1) q−1
p−1 = νφ(−1) q−1
2k (14.5 ′ )
νφ(−1) q−1
2k = 1 (14.6 ′ )
νφ(−1) q−1
k νφ(−1) q−1
p−1 = νφ(k) νφ(−1) q−1 1
2k + 2 νφ(−1) q−1
k . (14.7 ′ )
If p ≡ 1 (mod4)theidentities(14.5 ′ )and(14.6 ′ )areclearlyvalid. Moreoverfor(14.5 ′ )and
(14.6 ′ )thenumber q−1
k iseven. Since kisapositivepowerof2, qisanevenpowerof pif
p ≡ 3 (mod4).If q = p2fthen q − 1
p − 1 = 1 + p + . . . + p2f−1 ≡ 0 (mod4)

Chapter14 195
andtheleftsideof(14.5 ′ )is1. If q−1
2k iseven(14.5′ )and(14.6 ′ )arenowclear. Ifitisodd,4
divides kbecause8divides q − 1.But {θ0(σ) | σ ∈ G0/G1}isthesetof kthrootsofunityin
OL/PL = φso­1isasquarein φ, νφ(−1) = 1,andtherelationsarevalidinthiscasetoo.The
relations(14.4 ′ )and(14.7 ′ )areobviousifthedegreeof φovertheprimefield φ0iseven.Since
φ × containsthe kthrootsofunityand kisapowerof2thedegreecanbeoddonlyif kdivides
p − 1.Since
q − 1
k
q − 1 p − 1
= ·
p − 1 k
and q−1
k isnowodd p−1
k mustalsobeoddandbyquadraticreciprocity
because
νφ(k) = νφ0 (k) = νφ0 (−1) νφ0
p ≡ 1 mod
p − 1
k
p − 1
.
k
p−1
= νφ0 (−1) νφ0 (−1) 2k −1
2
If p ≡ 1 (mod4)thetworelationsarenowclear.If p ≡ 3 (mod 4)and q = p f
q − 1
p − 1
= 1 + p + . . . + pf−1
mustbeodd.Itisthereforecongruentto1modulo4.(14.4 ′ )becomes
and(14.7 ′ )becomes
p−1
νφ0 (−1) νφ0 (−1) k = 1
νφ0 (−1) = νφ0 (−1).
Thereisnoquestionthatboththeserelationsarevalid.
Wehavestilltotreatthecasethat G/G1isabelianwhileneither [G : G0]nor [G0 : G1]is
1.Then
NFµ/Fβ(µ ′
) ≡ β(µ ′ k ) (modPFµ )
σ∈G/G0
isasquarein φandtheidentity(14.3)impliesthat
νφ(ω1) = νφ(γ a )
CF/N L/FCLiscyclicoforder [G : G1]. Ithasageneratorwhichcontainsanelementofthe
form γ1̟F.Moreoverthecosetof
(γ1̟F) [G:G0] NL/F̟ −1
L
= γ[G:G0]
1 γ −1

Chapter14 196
isageneratorofUF/UF ∩N L/FCL.Theorderofthisgroupisapowerof2andpisoddsoevery
elementof UF ∩ N L/FCLisasquare.Consequently γcannotbeasquareand νφ(γ) = −1.If
misevenand F ′ isthefixedfieldof G0then
εµ =
whichiscongruentto
̟ k L
̟F
m−1
=
modulo PL.Since [F ′ : F]iseven
and
Because
NL/F ′̟L
−
̟F
N φµ/φ εµ = N F ′ /F εµ =
m−1
m−1 NL/F ′̟L
̟F
σ∈G0/G1
m−1 NL/F̟L
̟ [G:G0]
F
νφ(N φµ/φ εµ) = νφ(γ) = −1.
a =
q − 1
[G : G1]
̟ 1−σ
m−1 L
= γ m−1
isintegral, q−1
k iseven,andweneedonlyworryabouttheidentities(14.5)and(14.6). They
bothreduceto
νφ(−1) q−1
2k = 1.
Toprovethisweshowthat q−1
2k isevenif νφ(−1) = −1.Since
k = [UF : UF ∩ N L/FCL]
andthisindexmustdividetheorderof φ ∗ thenumber νφ(−1)is­1onlyif k = 2.Ofcourse
pwillbecongruentto3modulo4. Since4divides q − 1, qisanevenpowerof pand
q ≡ 1 (mod8).Thus
iseven.
q − 1
2k
= q − 1
4
Nowsupposethat G/G1isnotabelian.Let σ −→ x(σ)beagivenisomorphismof G0/G1
with Z/kZandlet x −→ σ(x)beitsinverse.Let τ −→ λ(τ)bethathomomorphismof G/G0
intotheunitsof Z/kZwhichsatisfies
x(τστ −1 ) = λ(τ) x(σ).

Chapter14 197
Thereispreciselyoneelementoforder2in G0/G1,namely σ
k
2 ,anditliesinthecenterof
G/G1.Since G/G0iscyclic, G/G1isnon­abelianonlyif k > 2.Chooseafixed σ0in Gwhich
generates G/G0andset
µ0 = λ(σ0)
and
y0 = x(σ [G:G0]
0 ).
Weshallsometimesregard Casavectorspaceoverthefieldwith pelements.If σbelongsto
G/G1let π(σ)bethelineartransformation
c −→ σcσ −1 .
Thedualspacewillbeidentifiedwith S(K/L)and π ∗ willbetherepresentationcontragredient
to π.
Therelation
NFµ/Fβ(µ ′
) ≡
togetherwiththeidentity(14.3)impliesthat
Moreoverif misevenand F ′ µ
εµ =
̟ k Fµ
whichiscongruentto
modulo PFµ .Since
whichequals
̟F
m−1
νφ(ω1) = νφ
σ∈G/GµG0
µ γµ
isthefixedfieldof GµG0
=
NFµ/F ′ µ ̟Fµ
̟F
{N φµ/φ εµ} t = N F ′ µ /F
β(µ ′ ) σ
k
.
m−1
− NFµ/F ′ µ ̟Fµ
m−1 ̟F
σ∈G0/G1
− NFµ/F ′ µ ̟Fµ
(m−1)t
̟F
(−1) t[φµ:φ] γ m−1
µ
̟ 1−σ
m−1 Fµ

Chapter14 198
and tisodd
νφ(N φµ/φ εµ) = νφ(−1) [φµ:φ] νφ(γµ).
Theserelationswillbeusedfrequentlyandwithoutcomment.
Iwanttodiscussthecase [G : G0] = 2and µ0 ≡ −1 (mod4)first.Since
wemusthave
or,if k > 4,
Then
or
Since
(−µ0) 2 = µ 2 0 = λ(σ 2 0) ≡ 1 (modk)
−µ0 ≡ 1 (modk)
−µ0 ≡ k
+ 1 (mod k).
2
µ0 ≡ −1 (modk)
µ0 ≡ k
− 1 (mod k).
2
µ0 − 1 ≡ 2 (mod 4)
thecentralizerof σ0in G0/G1consistsoftheidentityand σ
k 2
2 .Thus x(σ0)is0or k
2 .
Suppose µ0 ≡ −1 (modk)and x(σ2 k
0 ) = 0 = σ−1
and (σ0σ) 2 = σ2 0 . Thus σ
k
istheonlyelementoforder2in G/G1. If σbelongsto G/G1
2
then σhasanon­trivialfixedpointin S(K/L)ifandonlyif π(σ)has1asaneigenvalue. If
σ = 1thereisaninteger nsuchthat σnhasorder2.Then π(σn )alsohas1asaneigenvalue.
Thusifanynon­trivialelementof G/G1hasanon­trivialfixedpointthereisanelement τof
order2suchthat π(τ)has1asaneigenvalue.Theusualargumentshowsthat
π σ
2 . If σbelongsto G0/G1then σ0σσ −1
k
= −I
2
sothat,inthecaseunderconsideration,onlytheidentityhasfixedpoints.Then
a =
q − 1
[G : G1] .

Chapter14 199
Inparticular q−1
k iseven.Wechoose ̟Fµ = ̟Landlet
γ̟ [G:G0]
F
= N L/F̟L.
Onlyidentities(14.5)and(14.6)aretobeconsidered.(14.5)reducesto
and(14.6)reducesto
νφ(γ) at = (−1) a νφ(−1) q−1
2k
(−1) a νφ(−1) a[G:G0] νφ(γ) at νφ(−1) q−1
2k = 1.
Since [G : G0]iseventheyareequivalent.Suppose φhas relements.If x ∈ λ = OL/PLthen
xσ0 r = x f
forsome f.If σbelongsto G0/G1then
Thus
and
f
µ0r
θ0(σ)
= θ0(σ0σσ −1
0 )rf
µ rf
0
≡
≡ 1 (modk)
r ≡ −1 (mod4)
sothat νφ(−1) = −1.Since,inthepresentcase,
theidentitiesbecome
Themap
a =
q − 1
2k
νφ(γ) at = 1.
τ L/F : W L/F −→ CF
σ0σ
̟L ̟ σ0
f −1
r σo ≡ θ0(σ).
L
determinesamapof G/G1onto CF/N L/FCL. Theimageof σ0containsanelementofthe
form γ1̟Fwhere γ1isaunit.Theimageof σ 2 0 is1becausethecommutatorsubgroupcontains
{σ((µ0 − 1)x)} = {σ(x) | x ≡ 0 (mod2)}
andinparticularcontains σ 2 0 .Since [G : G0] = 2thenumber γγ −2
1 liesin UF ∩ NL/FCL.The
indexofthecommutatorsubgroupof G/G1in G/G1is4so
[UF : UF ∩ N L/FCL] = 2.

Chapter14 200
Consequently γγ −2
1 and γarebothsquaresand νφ(γ) = 1.
Nowsuppose µ0 ≡ −1 (modk)and x(σ 2 0 ) = 0. Everyelementoftheform σ0σ, σ ∈
G0/G1,hasorder2.If π(σ0σ) = −Ithen σ0σliesinthecenterof G/G1whichisimpossible.
Thus π(σ0σ)has1asaneigenvalue.If τ ∈ G0/G1then
τ −1 σ0στ = σ0στ 2
sotherearetwoconjugacyclassesintheset σ0G0/G1.Onehas σ0asrepresentativeandthe
otherhas σ1 = σ0σ(1).
Let V beanon­trivialsubspaceof S(K/L)invariantandirreducibleundertheactionof
G0/G1. Supposefirstthat V isalsoinvariantunder π ∗ (σ0)sothat V = S(K/L). Choose
v0 = 0sothat π ∗ (σ0)v0 = v0.Let λ ′ bethefieldobtainedbyadjoiningthe kthrootsofunity
totheprimefield.Certainly λ ′ ⊆ λand,since
θ0(σ) σ0 = θ0(σ −1
0 σσ0),
λ ′ isnotcontainedin φ.Let φ ′ = φ∩λ ′ .Wemayregard {1, σ0}as G(λ ′ /φ ′ ).Themap ϕwhich
sends σin G0/G1to (θ −1
0 (σ), 1)and σσ0to (θ −1
0 (σ), σ0)isanisomorphismof G/G1withthe
semi­directproductofthe kthrootsofunityin λ ′ and G(λ ′ /φ ′ ).Thereisauniquemap,again
denotedby ϕ,of Vonto λ ′ suchthat ϕ(v0) = 1while
ϕ(π ∗ (τ)v) = ϕ(τ)ϕ(v)
for τin G/G1. Ofcoursethe kthrootsofunityacton λ ′ byleftmultiplication. TheGalois
groupactsby σ0α = ασ−1 0 . Puttingtheactionstogetherwegetanactionofthesemi­direct
product.Tostudytheactionof G/G1on V westudytheequivalentactionofthesemi­direct
productin λ ′ .
Itisbesttoconsideramoregeneralsituation.Suppose φ ′ isafinitefieldwith pfelements, λ ′ isanextensionof φ ′ with pℓelementsand Γisthesemi­directproductofthegroupof kth
rootsofunity,where kdivides pℓ − 1,and G(λ ′ /φ ′ ). Γactson λ ′ asbefore. Let ℓ = nf. If
0 ≤ j1 < n, j = (j1, n),and ρistheautomorphism x −→ xpfof λ ′ /φ ′ thenthenumberof
elementsof λ ′ fixedbyamemberof Γoftheform (α, ρji )where αisakthrootofunityisthe
sameasthenumberofelementsfixedbysomeothermemberoftheform (β, ρ−j ).Indeedif
b j1
j
≡ −1 mod n
j

Chapter14 201
and bisprimetotheorderof (α, ρ j1 )wecantake
(β, ρ −j ) = (α, ρ j1 ) b .
Let θbeageneratorofthemultiplicativegroupof λ ′ .Theequation
β θ mρj
= θ m
canbesolvedfor βifandonlyif θ m(pjf −1 )hasorderdividing k,thatis,ifandonlyif p ℓ − 1
divides km(p jf − 1)or,if
u = pℓ − 1
k
ifandonlyif udivides m(pjf − 1).Let u(j)bethegreatestcommondivisorof uand pjf − 1.
udivides m(pjf −1)ifandonlyif u
u(j) divides m.Thenumberofsuch mwith 0 ≤ m < pℓ −1
is
u(j)
u (pℓ − 1) = u(j)k.
Once mand jarechosen αisdetermined.Thenumberofnon­zero xin λ ′ whicharefixedby
some (β, ρ −j )where jdivides nbutbyno (β, ρ −i )where iproperlydivides jis
i|j µ
j
i
u(i)k
if µ(·)istheMöbiusfunction.Thenumberoforbitsformedbysuch xis
1
jk
i|j µ
j
i
u(i)k
sothatthetotalnumberoforbitsof Γinthemultiplicativegroupof λ ′ is
whichequals
Theproductisoverprimes.
Lemma 14.7
a =
i|n
i|n
π
n
n
i
j
i
µ(j)
ij u(i)
1 − 1
u(i)
.
π i

Chapter14 202
because
If pℓ −1
k isoddthen
Theidentityofthelemmaisequivalentto
(−1) a+1 νφ ′(k) νφ ′(−1) pℓ−1 1
2k + 2 = 1.
(−1) a+1 u−1
νφ ′(u) νφ ′(−1) 2 = 1
νφ ′(k) = νφ ′(−1) νφ ′(u).
Bythelawofquadraticreciprocitytheleftsideoftheidentityisequalto
(−1) a+1 (p f |u)
if (p f |u)isJacobi’ssymbol.If u = 1thereisonlyoneorbitso
(−1) a+1 = 1.
Ofcourse (p f |1) = 1sotheidentityisclearinthiscase.
Weproveitingeneralbyinductiononthenumberofprimefactorsof u. Let π0bea
primefactorof uandlet u = π x 0 vwith vprimeto π0.Let v(j)betheanalogueof u(j).Then
u(j) = π x(j)
0 v(j).Leb bbetheanalogueof a.Then
ν = a − b =
i|n
π
n
i
1 − 1
v(i)
(π
π i
x(i)
0 − 1).
Observethat π0andall v(i)areodd.Toprovethelemmabyinductionwemustshowthat
Let
(−1) ν (p f | π x 0 ) = 1. (14.8)
n = 2 y n1
with n1odd.Therearetwopossibilitiestobeconsidered.
(i)
π0 ≡ 1 (mod 2 y+1 ).

Chapter14 203
(ii)
Sincetheorderof p f modulo π0divides nthequotientof π0 − 1bythisorderisevenand
p f isaquadraticresidueof π0.Alsoif idivides n
π x(i)
0
− 1
i
isdivisible,inthe2­adicfield,by4if2divides n
i
evenand(14.8)isvalid.
π0 = 1 + 2 c w
with c ≤ yand wodd.Let i = n1divide n1andconsider
y
j=0
π
n
2 j i
1 − 1
j v(2 i)
π 2ji andisalwaysdivisibleby2.Thus νis
(π x(2j i)
0 − 1). (14.9)
If x(2 y i) = 0thesumiszero. If x(2 y i) = 0let zbethesmallestintegerforwhich
x(2 z i) = 0.If j < zthen x(2 j i) = 0.If j ≥ z
p 2j if − 1 =
p 2z
if
− 1
2 j−z −1
d=0
Theresidueofthesummodulo π0is 2 j−z .Thus
if j ≥ zand(14.9)isequalto
Wewrite
as
1
i
π
n1 i
x(2 j i) = x(2 z i)
1 − 1
y v(2 i) y−1
+
π 2y j=z
v(2yi) y−1
+
2y j=z
v(2zi) y
+
2z j=z+1
v(2 j i)
2 j+1
p 2z
cif
.
v(2ji) − v(2j−1i) 2j .
v(2ji) 2j+1 π x(2y
i)
0 − 1 .
If kisreplacedby pℓ −1
v thenumberofelementsof λ ∗ fixedbysome (α, ρ −2j i )butbyno
(α, ρ −2j−1 i )is
{v(2 j i) − v(2 j−1 i)} pℓ − 1
.
v

Chapter14 204
Thecollectionofsuchelementsisinvariantunderthegroupobtainedbyreplacing kby pℓ −1
v
and φ ′ bythefieldwith p if elements.Theisotropygroupofeachsuchpointhasagenerator
oftheform (α, ρ −2j i )and,therefore,hasorder n
sothat 2 j divides
2jiandindex 2j (p ℓ −1)
v
{v(2 j i) − v(2 j−1 i)} pℓ − 1
v
v(2 j i) − v(2 j−1 i).
.Thus 2j (p ℓ −1)
v
divides
Since n1
i isdivisiblebyatleastoneprime,theexpression(14.9)iscongruent,inthe2­adicfield,
to
1
i
π
n1 i
1 − 1
z v(2 i)
π 2z
π x(2y
i)
− 1
modulo4.Since z ≤ candtheproductisnotemptythisiscongruent,inthe2­adicfieldagain,
to0modulo2.Thus νisevenoroddaccordingas
y
j=0
π2y−j
1 − 1
j v(2 n1)
π 2j
π
n1
x(2j
n1)
0 − 1
isorisnotdivisibleby2inthe2­adicfield.Consequently
ν ≡ y
j=0
π
2y−j
1 − 1
j v(2 n1)
π 2j (π x(2j n1)
0
− 1) (mod2).
Ofcourse x(2 y n1) = x = 0. Let zagainbethesmallestintegerfor x(2 z n1) = 0. Then
z ≤ cand
x(2 j n1) = x(2 z n1)
if j ≥ z.Thesumaboveisequalto
z v(2 n1)
+ y
2 z
j=z+1
Asbeforethisiscongruentmodulo2to
v(2 z n1)
2 z
v(2jn1) − v(2j−1n1) 2j
(π x 0 − 1).
(π x 0
− 1).
If z < cthisisevenandtheorderof pmodulo π0divides π0−1
2 sothat (p|π0) = 1. If z = c
then
πx 0
2z − 1
1
=
2c x
i=1
x
i
(2 c w) i ≡ x (mod2)

Chapter14 205
sothat ν ≡ x (mod 2).Howevertheorderof p f modulo π0isdivisibleby 2 z sothatitdoes
notdivide π0−1
2 and
Therelation(14.8)isnoweasilyverified.
f x x
p | π0 = (−1) .
Wereturntotheoriginalproblem. Since λ ′ isaquadraticextensionof φ ′ and λ ′ isnot
containedin φthedegreeof φover φ ′ isodd.Since Vand λ ′ havethesamenumberofelements
q = p ℓ .If q−1
k isoddtherelation(14.4)followsimmediatelyfromtheequality
andtheprecedinglemma.
(−1) a+1 νφ(k) νφ(−1) q−1 1
2k + 2 = (−1) a+1 νφ
q−1
′(k) νφ ′(−1) 2k
Thenumberof µin Twithisotropygroupoforder2is u(1)andthenumberof µwith
trivialisotropygroupis u(2)−u(1)
.Forpointsofthesecondtype [φµ : φ] = 2andforpointsof
2
thefirsttype [φµ : φ] = 1.Since,asweverifiedearlier,
and
theidentity(14.7)reducesto
νφ(ω1) = νφ
µ γµ
νφ(N φµ/φ εµ) = νφ(−1) [φµ:φ] νφ(γµ)
u(1) ≡ 1 (mod 2)
whichistruebecause u(1)divides u = pℓ −1
k which,when(14.7)isunderconsideration,isodd
byassumption.
Theidentity(14.5)maybeformulatedas
and(14.6)as
νφ
νφ
µ γµ
µ γµ
Forthesetwoidentities q−1
k iseven.Again
= (−1) a νφ(−1) q−1
2k
νφ(−1) P
µ [φµ:φ] = (−1) a νφ(−1) q−1
2k .
[φµ : φ] ≡ u(1) (mod2).
+ 1
2

Chapter14 206
But
and
u(2) = u = pℓ − 1
k
2a = u(1) + u(2)
so u(1)iseven.Itwillbeenoughtoverify(14.5).
= q − 1
k
Wemaychoose Tsothatif µisin Tthenitsisotropygroupistrivialorcontainsoneof σ0
or σ1.If σ0liesintheisotropygroupof µand νintheorbitof µcorrespondsto θ m in λ ′ then
α 2 θ mρ = θ m
forsome kthrootofunity α.Thisispossibleifandonlyif p ℓ − 1divides km
2 (pf − 1)or 2u
divides m(p f − 1).Thisisthesameasrequiringthat 2u
u(1) divide m(pf −1)
u(1) . Thenumber uis
even.Wehavealreadyobservedthatif risthenumberofelementsin φsothat x σ0 = x r for x
in φthen
µ0r ≡ 1 (modk)
andinparticular
µ0r ≡ 1 (mod4).
Since [φ : φ ′ ]isoddand µ0 ≡ −1 (mod4)thehighestpowerof2dividing p f − 1is2.Thus
2u
u(1) and pf −1
u(1)
arerelativelyprimesothat 2u
u(1) divides m(pf −1)
u(1) ifandonlyif 2u
u(1)
divides m.
Thereare u(1)k
2 such mwith 0 ≤ m < pℓ − 1. Thecorrespondingcharacters νfallinto u(1)
2
orbits. Thusthereare u(1)
2 elementsin Twhoseisotropygroupcontains σ0and u(1)
2 whose
isotropygroupcontains σ1.Let L0bethefixedfieldof σ0and L1thefixedfieldof σ1.Let ̟L0
and ̟L1generate PL0and PL1respectivelyandlet Wehavetoshowthat
νφ
γ u(1)
2
N L0/F̟L0 = γ0 ̟F
NL1/F̟L1 = γ1̟F
NL/F̟L = γ̟ 2 F .
0 γ u(1)
2
1 γ u(2)−u(1)
2
= (−1) a νφ(−1) q−1
2k .
Firstweprovealemma,specialcasesofwhichwehavealreadyseen.

Chapter14 207
Lemma 14.8
SupposeL/Fisnormalbutnon­abelianand[L : F]isapowerof2.SupposeH = G(L/F)
andthefirstramificationgroup H1is {1}but [H : H0] > 1and [H0 : H1] > 1.Let ̟Lgenerate
theprimeidealof OL,let ̟Fgeneratetheprimeidealof OF,andlet
Then γisasquarein UF.
NL/F ̟L = γ̟ [H:H0]
F .
Thehypothesesimplythattheresiduefieldhasoddcharacteristic.Let Abethefixedfield
of H0and L ′ bethefixedfieldofthecommutatorsubgroupof H.Then A ⊆ L ′ andif
then
̟L ′ = N L/L ′̟L
N L ′ /F̟L
′ = γ̟[A:F]
F
Ofcourse [A : F] = [H : H0]. Since Hisnilpotentbutnotabelian L ′ cannotbeacyclic
extension.If γisnotasquarein UFthen γ −1 generates UF/UF ∩ N L ′ /FCL.Since
̟ [A:F]
F
≡ γ −1
(modN L ′ /FCL ′).
̟Fwouldthengenerate CF/N L ′ /FCL ′whichisimpossible.
Returningtotheproblemathand,weobservethatthequotientof G/G1bythesquares
in G0/G1isagroupoforder4inwhicheverysquareis1.Thefixedfield F ′ ofthisgroupis
thecompositeofallquadraticextensionsof F. F0 = F ′ ∩ L0and F1 = F ′ ∩ L1arethetwo
differentramifiedquadraticextensionsof F.Define
and
Then
and
̟F0 = N L0/F0 ̟L0
̟F1 = NL1/F1̟L1 .
N F0/F̟F0
= γ0̟F
NF1/F̟F1 = γ1̟F.
.

Chapter14 208
Weneedtoshowthat
νφ(γ0γ1) = νφ
γ0
γ1
= −1.
Ifnot, γ0
γ1 isasquareandthusin NF1/FCF1 .Then γ0̟Fbelongsto
N F0/FCF0 ∩ N F1/FCF1 = N F ′ /FCF ′.
Thisisimpossiblebecause F ′ containsanunramifiedextension.
Weobservedbeforethatsince
µ0 = λ(σ0) ≡ −1 (mod 4)
thenumber νφ(−1)is −1.Theidentity(14.5)reducesto
Since
and
thisrelationisclearlyvalid.
(−1) u(1)
2 = (−1) a (−1) q−1
2k .
a = u(1)
2
u(2) =
+ u(2)
2
q − 1
k
Wecontinuetosupposethat µ0 ≡ −1 (mod k)andthat σ 2 0 = 1butnowwesupposethat
Visnotinvariantunder π ∗ (σ0).Since π ∗ (σ0)V ∩Vand π ∗ (σ0)V +Varebothinvariantunder
G/G1thefirstis0andthesecondis S(K/L)sothat S(K/L)isthedirectsum V ⊕ π ∗ (σ0)V.
Let V have p ℓ elementssothat q = p 2ℓ .If λ ′ isagainthefieldgeneratedovertheprimefield
bythe kthrootsofunity λ ′ has p ℓ elements.If φ ′ = λ ′ ∩ φhas p f elementsthen p ℓ = p 2f so
that p ℓ ≡ 1 (mod8).Also kdivides p ℓ − 1sothat
iseven.
q − 1
k =
ℓ p − 1
(p
k
ℓ + 1)
If σ ∈ G0/G1thenon­zerofixedpointsof σ0σaretheelementsoftheform
v ⊕ π ∗ (σ0σ)v

Chapter14 209
with v = 0. Thereare (p ℓ − 1)kofthemaltogetherandtheyfallinto p ℓ − 1orbits. The
remaining
(p 2ℓ − 1) − (p ℓ − 1)k
elementsfallinto
orbits.Thus
1
2k {(p2ℓ − 1) − (p ℓ − 1)k}
a = pℓ − 1
2 + p2ℓ − 1
.
2k
Since,forthesamereasonsasbefore, νφ(−1) = −1theidentity(14.5)becomes
while(14.6)becomes
νφ
νφ
µ γµ
µ γµ
= (−1) pℓ −1
2 (14.10)
νφ(−1) Σ[φµ:φ] = (−1) p ℓ −1
2 .
Since [φµ : φ] ≡ p ℓ − 1 ≡ 0 (mod2)
only(14.5)needbeproved.(14.4)and(14.7)arenottobeconsideredbecause q−1
k iseven.
Weproceedasbefore. Thepointsin Tcanbechosensothattheirisotropygroupsare
eithertrivialorcontain σ0or σ1. pℓ −1
2 willhaveisotropygroupscontaining σ0and pℓ −1
2 will
haveisotropygroupscontaining σ1. Theargumentusedaboveshowsthattheleftsideof
(14.10)isequalto
asdesired.
Nowsuppose k ≥ 8and
(−1) pℓ−1 2
µ0 = λ(σ0) ≡ k
− 1 (modk).
2
Weareofcoursestillsupposingthat [G : G0] = 2.If σbelongsto G0/G1then
and
σσ0σ −1 = σ0σ k
2 −2
(σ0σ) 2 = σ 2 0σ k
2 .

Chapter14 210
Thus
x((σ0σ) 2 ) = x(σ 2 k
0 ) +
2 x(σ).
Since x(σ 2 0)is0or k
2 ,wecanmakethesumontheright0.Replacing σ0by σ0σifnecessary,
wesupposethat σ 2 0 = 1.Then (σ0σ) 2 = 1ifandonlyif
whichissoifandonlyif σ0σisconjugateto σ.
k
2
x(σ) ≡ 0 (modk)
Take Vin S(K/L)asbefore.If Visinvariantunder π ∗ (σ0)and λ ′ with p ℓ elementsand
φ ′ with p f elementshavethesamemeaningasbeforethen
forsomeinteger wsothat
and
p f = k
− 1 + wk
2
q − 1 = p 2f − 1 = k · k
k
− k + 2wk − 1 + (wk)
4 2 2
q − 1
k
k
≡ − 1 (mod 2)
4
isodd.Thustheidentities(14.5)and(14.6)arenottobeconsidered.Theidentities(14.4)and
(14.7)followfromLemma14.7exactlyasabove.
and
Supposethen S(K/L)isthedirectsum V ⊕ π ∗ (σ0)V.If Vhas p ℓ elementsthen q = p 2ℓ
q − 1
k = pℓ − 1
k
(p ℓ + 1)
isevenbecause kdivides p ℓ − 1.Thenon­zeroelementsof S(K/L)whicharefixedpointsof
some σ0σwith σasquarein G0/G1aretheelements
v ⊕ π ∗ (σ0σ)v
with v = 0.Thereare (p ℓ − 1) k
2 suchelementsandtheyfallinto pℓ −1
2 orbits.Theremaining
(p 2ℓ − 1) − (p ℓ − 1) k
2

Chapter14 211
non­zeroelementshavetrivialisotropygroupandfallinto
orbits.Thus
1
2k
2ℓ ℓ k
(p − 1) − (p − 1)
2
a = p2ℓ − 1
2k + pℓ − 1
.
4
Since,asbefore, νφ(−1) = −1theidentity(14.5)becomes
while(14.6)becomes
Again
νφ
νφ
µ γµ
µ γµ
[φµ : φ] ≡ pℓ − 1
2
= (−1) pℓ −1
4 (14.11)
(−1) Σ[φµ:φ] = (−1) p ℓ −1
4 .
≡ 0 (mod2)
sothatitisenoughtoprove(14.11).Theidentities(14.4)and(14.7)neednotbeconsidered.
that
If λ ′ and φ ′ aredefinedasbeforeand φ ′ has p f elementsthen λ ′ has p ℓ = p 2f elementsso
p ℓ ≡ 1 (mod 8)
and pℓ −1
4 iseven.Wemaysupposethateach µin Teitherhastrivialisotropygrouporisfixed
by σ0. Lemma14.8showsthatthose µwithtrivialisotropygroupcontributenothingtothe
leftsideof(14.11).If L0isthefixedfieldof σ0and
theleftsideof Kis
N L0/F̟L0
νφ(γ0) pℓ−1 2
whichis1.Thetruthoftheidentityisnowclear.
= γ0̟F
Wereturntothegeneralcasesothat [G : G0]maybegreaterthan2and µ0maybe
congruentto1modulo4.Ofcourse [G : G0]isstilleven.Let
λ0 = λ σ 1
2 [G:G0]
0

Chapter14 212
sothat
If [G : G0] > 2then
λ0 = µ 1
2 [G:G0]
0
λ0 ≡ 1 (mod4).
If [G : G0] = 2then λ0 = µ0. Sincethecasethat [G : G0] = 2and µ0 ≡ −1 (mod 4)is
completelysettledwemaysupposethat λ0 ≡ 1 (mod4).Set
τ0 = σ 1
2 [G:G0]
0
Anyelementof G/G1whichdoesnotliein G0/G1andwhosesquareis1isoftheform
σ(x)τ0.If
= σ(y0)
then
Since G/G1isnotcyclic y0iseven.Since
σ [G:G0]
0
(σ(x)τ0) 2 = σ((λ0 + 1)x)τ 2 0 = σ(y0 + (λ0 + 1)x).
thereareexactlytwosolutionsoftheequation
Let x0beoneofthem.Then x0 + k
2
Set
λ0 + 1 ≡ 2 (mod 4)
y0 + (λ0 + 1)x ≡ 0 (modk).
.
.
istheother.Wemaysupposethat kdoesnotdivide x0.
ρ0 = σ(x0)τ0.
Weobservedbeforethatif σ = 1belongsto G/G1and π∗ (σ)hasanon­zerofixedpointthen
somepowerof σisoforder2andhasanon­zerofixedpoint. Since σ
k
hasnonon­zero
2
fixedpointthispowermustbe ρ0or σ
k
2 ρ0.Since σ
k
2 liesinthecenterof G/G1, σmustlie
inthecentralizerof ρ0.
Thegroup
k k
1, σ , ρ0, σ
2 2
isoforder4andeveryelementinitisoforder2soitcannotbecontainedinthecenterof
G/G1. Howeveritisanormalsubgroupanditscentralizer H ∗ hasindex2in G/G1. G/G1
ρ0

Chapter14 213
maybeidentifiedwith H. Everyelement σof Hsuchthat π ∗ (σ)hasanon­zerofixedpoint
liesin H ∗ . S(K/L)isthedirectsumof Vand Wwhere
V = {v | π ∗ (ρ0)v = v}
W = {w | π ∗ (ρ0)w = −w}.
If σin Hdoesnotbelongto H ∗ then π ∗ (σ)V = Wand π ∗ (σ)W = V.Thenumberofnon­zero
orbitsof Hin V ∪ Wisthesameasthenumber a ′ ofnon­zeroorbitsof H ∗ in V.If V has p ℓ
elementssothat q = p 2ℓ thenumberofnon­zeroorbitsof Hin V ⊕ W − (V ∪ W)is
a ′′ = (pℓ − 1) 2
[G : G1] = pℓ − 1
·
k
pℓ − 1
[G : G0] .
Theactionof H ∗ on V mustbeirreduciblealthoughitisnotfaithful.Howevertheactionof
H ∗ ∩ H0 = H ∗ 0 isfaithful.
Let F ′ bethefixedfieldof H ∗ in Lor,whatisthesame,of H ∗ Cin K. Let C 1 ⊆ Cbe
theorthogonalcomplementof V andlet H 1 bethesubgroupof Hwhichactstriviallyon V.
H 1 C 1 isanormalsubgroupof H ∗ Canditsfixedfield K ′ isnormalover F ′ .If H ′ = H ∗ /H 1
and C ′ = C/C 1 then G ′ = G(K ′ /F ′ ) = H ′ C ′ . Moreover H ′ ∩ C ′ = {1}and H ′ = {1}
because σ
k
1 ′ ′ ′
2 doesnotliein H . Sincetheactionof H on C isfaithfulandirreducible C
iscontainedineverynon­trivialnormalsubgroupof G ′ . Tocompletetheproofofthefour
identities(14.4),(14.5),(14.6),and(14.7)weuseinductionon [K : F].
Let k ′ betheorderof H ′ 0 andlet φ′ = OF ′/PF ′.If K/Fisreplacedby K′ /F ′ theidentity
(14.4)becomes
(−1) a′ +1 νφ ′(k′ ) νφ ′(−1) pℓ −1
2k ′ + 1
2 = 1. (14.4 ′′ )
Tistobereplacedby T ′ ,asetofrepresentativesforthenon­zeroorbitsof H ′ or H ∗ in V,
whichmaybeidentifiedwiththecharactergroupof C ′ .Wemaysupposethat T ′ isasubset
of T.Because H ′ 0 = {1}theidentity(14.5)forthefield K′ /F ′ maybewrittenas
Ofcourse
νφ ′
µ∈T ′ γ′ µ
= (−1) a′
NFµ/F ′(̟t Fµ ) = γ′ µ ̟
t[Fµ:F ′ ]
k ′
νφ ′(−1) pℓ −1
2k ′ . (14.5 ′′ )
Recallthat tisodd.ByPropositionIV.3ofSerre’sbook, thasthesamesignificancefor K ′ /F ′
asithadfor K/F.Theidentity(14.6)maybewrittenas
(−1) a′
νφ ′
µ∈T ′ γ′ µ
F ′
.
νφ ′(−1)Σ µ∈T ′[φµ:φ ′ ] νφ ′(−1) pℓ −1
2k ′ = 1. (14.6 ′′ )

Chapter14 214
and(14.7)as
(−1) a′ +1 νφ ′(k′ ) νφ ′(−1) pℓ −1
2k ′ − 1
2 νφ ′(−1)Σ µ∈T ′ [φµ:φ ′ ] = 1 (14.7 ′′ )
Assuming(14.4 ′′ ),(14.5 ′′ ),(14.6 ′′ ),and(14.7 ′′ )wearegoingtoprove(14.4),(14.5),(14.6),and
(14.7).
Since H ′ 0 isisomorphicto H∗ 0 either k′ = kor k ′ = k
2
k ′ = k
2 forifnot
q − 1
k =
ℓ p − 1
(p
k
ℓ + 1)
.Supposefirstthat q−1
k isodd.Then
wouldbeeven.Thus H0 = G0/G1isnotcontainedin H ∗ and F ′ /Fisramifiedsothat φ ′ = φ.
Since
thenumber pℓ −1
k ′
if
q − 1
k =
ℓ ℓ p − 1 (p + 1)
isodd.Toprove(14.4)wehavetoshowthat
(−1) a′′
k ′
2
νφ(2) νφ(−1) δ = 1
δ = p2ℓ − 1
2k − pℓ − 1
k = pℓ − 1
k
p ℓ + 1
2
− 1 .
Since G0/G1isnotcontainedin H ∗ , τ0doesnotcommutewith G0/G1andthemap λof G/G0
intotheunitsof Z/k Zisfaithful.Thus
But
λ0 ≡ 1 (modk).
λ0 ≡ 1 (mod4)
sothat k ≥ 8.Ingeneralif k ≥ 4,thegroupofunitsof Z/k Zistheproductof {1, −1}and
If
with xoddand 4 ≤ 2 b ≤ kthen
{α | α ≡ 1 (mod4)}.
α = 1 + 2 b x
α 2 = 1 + 2 b+1 y

Chapter14 215
with yodd.Oneshowseasilybyinductionthattheorderof αis 2 −b ksothat
{α | α ≡ 1 (mod 4)}
iscyclicoforder k
4 . Thisimpliesintheparticularcaseunderconsiderationthat [G : G0]
divides k
4 .Write
a ′′ isoddifandonlyif
(i)
(ii)
a ′′ ℓ p − 1
=
k ′
ℓ p − 1
.
2[G : G0]
2[G : G0] = k ′ .
Weconsidervariouscasesseparately.Asbefore µ0 = λ(σ0).If φhas p f elementsthen
Then
µ0p f ≡ 1 (mod 8)
µ0 ≡ 1 (mod 8).
νφ(2) = νφ(−1) = 1
andtheorderof µ0intheunitsof Z/k Zwhichisequalto [G : G0]divides k
8 .Thus a′′ is
even.Theidentity(14.4)follows.
Then
µ0 ≡ 3 (mod 8).
νφ(2) = νφ(−1) = −1.
Since µ0 ≡ 3 (mod8)thenumbers µ0and λ0aredifferent.Thus λ0isasquareandhence
congruentto1modulo8.Then k > 8and
Then
δ ≡ pℓ + 1
4
p ℓ ≡ 1 (mod 8).
− 1
2
≡ 0 (mod2).
Since µ0 = λ0theindex [G : G0]isnot2.Thustheorderof µ0isatleast4andistherefore
theorderof −µ0.Since −µ0 ≡ 5 (mod8)itsorderis k
4 and
[G : G0] = k
4 .

Chapter14 216
(iii)
(iv)
Consequently a ′′ isodd.Again(14.4)issatisfied.
µ0 ≡ 5 (mod 8).
Then νφ(2) = −1while νφ(−1) = 1.Theorderof µ0whichequals [G : G0]isagain k
4 so
that a ′′ isoddand(14.4)issatisfied.
µ0 ≡ 7 (mod 8).
Then νφ(2) = 1while νφ(−1) = −1.Again k > 8and
δ ≡ 0 (mod2).
Theorderof µ0isagainatleast4andthereforeequaltotheorderof −µ0andthatdivides
k
8
.Thus [G : G0]divides k
8 and a′′ iseven.(14.4)followsoncemore.
Since φ ′ = φallweneedtoprove(14.7)once(14.4)and(14.7 ′′ )aregrantedisshowthat
µ∈T −T ′ [φµ : φ] ≡ 0 (mod2).
Thisisclearbecause,forthese µ, Fµ = Land φµ = OL/PLisofevendegreeover φ.
Finallywehavetoassumethat q−1
k isevenandprove(14.5)and(14.6).Firstalemma.
Lemma 14.9
If q−1
k iseven,
λ σ 1
2 [G:G0]
0
and G/G0actsfaithfullyon G0/G1,then
≡ 1 (mod4),
(−1) a νφ(−1) q−1
2k = 1.
Sincetheactionisfaithful G0/G1isnotcontainedin H ∗ and k ′ = k
2 . Asbefore λ0 ≡
1 (mod4)and λ0 ≡ 1 (modk)togetherimplythat k ≥ 8and k ′ ≥ 4.Since k ′ divides p ℓ − 1,
and pℓ +1
2 isodd.Since
q − 1
k =
p ℓ ≡ 1 (mod 4)
ℓ ℓ p − 1 p + 1
k ′
2

Chapter14 217
thenumber pℓ −1
k ′
sothat
iseven.
If σbelongsto H ∗ and σactstriviallyon H ∗ 0 then
λ(σ) ≡ 1 mod k
2
λ(σ 2 ) ≡ 1(modk)
and σ 2 belongsto H0.Thus σbelongsto ρ0H0 ∪ H0.Since ρ0belongsto H 1 theimageof σin
H ′ liesin H ′ 0 .Thus G′ /G ′ 0 actsfaithfullyon G′ 0 /G′ 1 .If σbelongsto H1 then σactstrivially
on H ∗ 0 becausetherepresentationof H∗ 0 on Visfaithful.Thus H1 iscontainedin ρ0 H0 ∪ H0
andisthereforejust {ρ0, 1}.Thus
Supposethat
[G ′ : G ′ 0 ] = [H′ : H ′ 0 ] = [H∗ : H ∗ 0 H1 ] = 1
2
Since φ ′ = φand,because k ′ ≥ 4divides p ℓ − 1,
q − 1
2k =
[G : G0].
(−1) a νφ ′(−1) pℓ −1
2k ′ = 1. (14.12)
ℓ p − 1
2k ′
ℓ ℓ p + 1 p − 1
≡
2 2k ′
allweneeddotoestablishthelemmaistoshowthat
Asbefore [G : G0]divides k
4 .If
then
a ′′ = 1
k
a ′′ ≡ 0 (mod2).
k
4
= n[G : G0]
(p ℓ − 1) 2
[G : G0]
iscertainlyevenbecause 2k ′ divides p ℓ − 1.
If [G : G0] ≥ 4let
λ ′ 0
= λ σ 1
ℓ p − 1
= n
k ′
4 [G:G0]
0
.
2
(mod2),

Chapter14 218
If
λ ′ 0
≡ 1 (mod4)
wemaysupposethat(14.12)istruebyinduction.If [G : G0] = 4and
orif [G : G0] = 2wemustestablishitdirectly.
λ ′ 0 ≡ 3 (mod4)
Supposefirstthat [G : G0] = 2.If φhas p f elementsthen
sothat νφ(−1) = 1.Itisclearthatinthiscase
a ′ isthusevenand(14.12)isvalid.
λ0 ≡ µ0 ≡ p f ≡ 1 (mod4)
a ′ = pℓ − 1
k ′
.
Nowsuppose [G : G0] = 4sothat [G ′ : G ′ 0 ] = 2. If σ′ 0 generates G′ modulo G ′ 0 then
λ ′ 0istheimageof σ′ 0inthegroupofunitsof Z/k′ Z.Wehavealreadystudiedthecasethat
≡ 3 (mod4)intensively.Let
λ ′ 0
x : σ ′ −→ x(σ ′ )
bethemapof G ′ 0 /G′ 1 onto Z/k′ Z. If λ ′ 0 ≡ −1 (modk′ )and x((σ ′ 0 )2 ) = k′
2 weshowed,
incidentally,that(14.12)isvalid.If λ ′ 0 ≡ −1 (mod k′ ), x((σ ′ 0 )2 ) = 0,andtheactionof H ′ 0on S(K ′ /L ′ )isreduciblewesawthat pℓisasquare p2ℓ′ andthattheleftsideof (L)is
Butthefieldwith p ℓ′
and(14.12)isagainvalid.If k ′ ≥ 8,
(−1) pℓ′ −1
2 .
elementsmustcontainsthe k ′ throotsofunityand k ′ ≡ 0 (mod 4).Thus
p ℓ′
− 1 ≡ 0 (mod4)
λ ′ 0 ≡ k′
2 − 1 (mod k′ )
andtheactionof H ′ 0 on S(K′ /L ′ )isreducible,theleftsideof(14.12)is
(−1) pℓ′ −1
4 .

Chapter14 219
Thistime
p ℓ′
− 1 ≡ 0 (mod8).
Tocompletetheproofofthelemmaweshowthatinthecaseunderconsiderationthe
on Visreducible.Ifnotthe
actionof H ′ 0and S(K′ /L ′ )or,whatisthesame,theactionof H∗ 0
fieldgeneratedovertheprimefieldbythe k ′ throotsofunityhas pℓelements.Thus p ℓ ≡ 1 (mod4).
Howeveraswehaveobservedrepeatedly,thenumberofelementsin φiscongruentto3
modulo4.Thus ℓiseven.Let ℓ = 2ℓ ′ .Either pℓ′ − 1or pℓ′ + 1iscongruentto2modulo4.If
pℓ′ + 1 ≡ 2 (mod4)then k ′ divides pℓ′ − 1because
p ℓ − 1
k ′
=
pℓ′ − 1
k ′
iseven.Since k ′ cannotdivide pℓ′ − 1wehave
p ℓ′
(p ℓ′
≡ 3 (mod4)
and ℓ ′ isodd.Indeeditis1butthatdoesnotmatter.Since kdivides p ℓ − 1,the kthrootsof
unityarecontainedinthefieldwith p ℓ elements.Adjoiningthemto φ = OF/PFweobtaina
quadraticextensionbecause4doesnotdivide ℓ.Thereforeif σbelongsto G0/G1
sothat
+ 1)
θ0(σ) = θ0(σ) σ−2
0 = θ0(σ) λ(σ2
0 )
λ(σ 2 0) ≡ 1 (modk).
Thiscontradictstheassumptionthat G/G0actsfaithfullyon G0/G1.
Returningtotheproofof(14.5),wesupposefirstthat H0isnotcontainedin H∗sothatthe actionof G/G0on G0/G1isfaithful.BecauseofLemma14.9theidentity(14.5)isequivalent
to
= 1.
νφ
µ∈T N Fµ/F γµ
If µbelongsto Tbutnotto T ′ then Fµ = Land,byLemma14.8,
νφ(N Fµ/F γµ) = 1.

Chapter14 220
If µbelongsto T ′ then Gµiscontainedin H ∗ Csothat Fµcontains F ′ .Moreoverwedonot
change Fµifwereplace K/Fby K ′ /F ′ .Let ̟F ′generate PF ′andtake ̟F = N F ′ /F̟F ′.If
E ′ isthefixed­fieldof H ∗ wemaysupposethat
andthat
Then
asrequired.Let
Then
̟F ′ = NE ′ /F ′̟E ′
̟E = NE ′ /E̟E ′.
̟F = N E/F̟E
N Fµ/F, ̟ t Fµ = γ′ µ ̟ t[Fµ:F ′ ]
k ′ .
γµ = N F ′ /F γ ′ µ .
Since F ′ /Fisramified γµisasquarein UFand(14.5)isproved. Toprove(14.6)wehaveto
showthat
νφ(−1) Σµ∈T [φµ:φ] = νφ ′(−1)Σ µ∈T ′[φµ:φ ′ ] = 1.
But pℓ −1
k ′
isevenandthisfollowsfromthesimultaneousvalidityof(14.5 ′′ )and(14.6 ′′ ).
Wehaveyettotreatthecasethat q−1
k isevenand H0iscontainedin H ∗ .Then F ′ /Fis
unramifiedand k ′ = k.Supposefirstofallthat pℓ −1
k isalsoeven.Then
q − 1
2k =
ℓ ℓ p − 1 p + 1
k
iseven. H0iscontainedin H∗and Hisgeneratedby σ0and H0. Consequently σ0isnot
containedin H∗and σ0ρ0σ −1
k
0 = σ ρ0.
2
Since ρ0 = σ(x0)τ0
if µ0 = λ(σ0).If
(µ0 − 1)x0 ≡ k
2
y0 = x
σ [G:G0]
0
2
(modk)

Chapter14 221
and misthegreatestcommondivisorof y0and kthenbythedefinitionof x0thegreatest
commondivisorof x0and kis m
2
k.Inparticular m < k.Theorderof σ0in His
.Therefore k
m isthegreatestcommondivisorof µ0 − 1and
k
[G : G0].
m
Therefore [G : G0]divides k
2m [G : G0]and H ∗ containsacyclicsubgroupoforder
k
[G : G0].
2m
If σistheelementoforder2inthissubgroup,then σbelongsto H0and π ∗ (σ)doesnothave
1asaneigenvalue. Thusnonon­zeroelementof V isfixedbyanyelementofthiscyclic
subgroupand
Inparticular [G : G0]divides p ℓ − 1and
p ℓ
− 1 ≡ 0 mod k
2m
[G : G0] .
a ′′
ℓ ℓ p − 1 p − 1
=
k [G : G0]
iseven.Asbefore νφ(γµ) = 1if µbelongsto Tand Fµ = L.If Fµ = Lthen µbelongsto T ′
and Gµliesin H ∗ Csothat Fµcontains F ′ .Inthepresentsituation F ′ /Fisunramifiedand
wemaytake ̟F ′ = ̟F.If
then
Theidentity(14.5)reducesto
or
NFµ/F ′̟t Fµ = γ′ t[Fµ:F
µ̟
′ ]
k
F
NFµ/F̟ t Fµ = (NF ′ /F γ ′ t[Fµ:F]
k
µ ) ̟ .
νφ
νφ ′
µ∈T ′ N F ′ /F γ ′ µ
µ∈T ′ γ′ µ
F
= (−1) a′
= (−1) a′
.
Since φ ′ isaquadraticextensionof φthenumber νφ ′(−1)is1andthisrelationisequivalentto
(14.5 ′′ ).Toprove(14.6)wehavetoshowthat
νφ(−1) Σµ∈T [φµ:φ] = 1.

Chapter14 222
Thisisclearbecause2divideseachofthedegrees [φµ : φ].
Finallywehavetosupposethat pℓ −1
k isodd. Since [φ ′ : φ] = 2therelation(14.4 ′′ )
amountsto
(−1) a′ +1 = 1.
Again
νφ
µ∈T γµ
= νφ ′
µ∈T ′ γ′ µ
. (14.13)
If µbelongsto T ′ and σ = 1belongsto Gµthensomepowerof σwillequal ρ0.Since
isoddand
p ℓ − 1
k
=
µ∈T ′
[Fµ : F ′ ]
k
[Fµ : F ′ ]
k
isapowerof2thereisatleastone µin T ′ forwhich [Fµ : F ′ ] = k.Then Gµmustcontainan
elementoftheform σ(z0)σ 2 0 .Then
ρ0 = σ(x0)τ0 = (σ(z0)σ 2 0) 1
4 [G:G0] = σ
µ 2 0
µ 1
2 [G:G0]
0
Thus
µ 1
2 [G:G0]
0 − 1
z0 ≡ x0 (modk).
− 1
Let
Since
1
4 [G : G0] = 2 b .
µ 2 0
≡ 1 (mod4)
µ 2 0
− 1
− 1
z0τ0
and,asbefore,thegreatestcommondivisorof x0and kis m
2 ifthegreatestcommondivisorof
y0and kis m,weinferthat
µ 1
2 [G:G0]
0
µ 2 0
ismultiplicativelycongruentto
− 1
=
− 1
b
j=1
− 1
µ 2j
b
=
− 1
µ2j+1 0
·
0
[G : G0]
4
j=1 µ2j
0 + 1
.

Chapter14 223
modulo2andthatthegreatestcommondivisorof z0and kis
2m
[G : G0] .
Inparticular [G:G0]
2 divides m. z0isoddifandonlyif
m = 1
2
[G : G0].
If µ0 ≡ 1 (mod 4)theorderof µ0inthegroupofunitsof Z/k Zis mbecauseasweobserved
whentreatingthecasethat pℓ−1 k iseven,thegreatestcommondivisorof µ0 − 1and kis k
m .
However
andinthiscase mdivides 1 [G : G0].Thus
2
if µ0 ≡ 1 (mod 4).
µ 1
2 [G:G0]
0 ≡ λ(τ0) ≡ 1 (mod k)
m = 1
[G : G0]
2
Weshalldefineasequenceoffields F (i) , L (i) , K (i) , 1 ≤ i ≤ n. nisanintegertobe
specified. Wewillhave F (i) ⊆ L (i) ⊆ K (i) and K (i) /F (i) and L (i) /F (i) willbeGalois. Let
G (i) = G(K (i) /F (i) )and C (i) = G (L (i) /F (i) ). Therewillbeasubgroup H (i) of G (i) such
that H (i) = {1}, H (i) ∩ C (i) = {1},and G (i) = H (i) C (i) . C (i) willbeanon­trivialabelian
normalsubgroupof G (i) whichiscontainedineveryothernon­trivialnormalsubgroup. H (n)
willbeabelianbut H (i) willbenon­abelianif i < n.Moreover k (i) = [H (i)
0 : 1]willbeatleast
4forall iand k (i) willequal 2k (i+1) if i < n.If xisanisomorphismof H (i)
0 with Z/k(i) Zand
σbelongsto H (i) let
x(στσ −1 ) = λ (i) (σ)x(τ).
Then λ (i) (σ)willbecongruentto1modulo8if i < n.If q (i) isthenumberofelementsin C (i)
then
q (i) − 1
k (i)
willbeodd.
F ′ and K ′ havealreadybeendefined. L ′ isjustthefixedfieldof C ′ .
q ′ − 1
k ′
= pℓ − 1
k

Chapter14 224
isodd.If σ ′ in H ′ istheimageof σin H ∗ Cthen
Since σisasquaremodulo H0
λ ′ (σ ′ ) ≡ λ(σ) (modk).
λ ′ (σ ′ ) ≡ 1 (modk).
IfF (i) , L (i) ,andK (i) havebeendefinedandH (i) isnotabelianwecandefineF (i+1) , L (i+1) ,
K (i+1) bytheprocessweusedtopassfrom F, L, Kto F ′ , L ′ , K ′ .Wehaveseenthatif
isoddthen
isalsooddandthat
q (i) − 1
k (i)
q (i+1) − 1
k (i+1)
k (i) = 2k (i+1) .
Wehavealsoseenthat k (i) ≥ 8if H (i) isnotabelian.If H (i) isabelianwetake n = i.
Whenwepassfromthe ithstagetothe (i+1)thwebreakup T (i) ,theanalogueof T,into
T (i+1) andacomplementaryset U (i) . Wemaythinkof T (i) aslyingin T. If σ (i)
0 generates
H (i) modulo H (i)
0 then
λ ′ (σ (i)
0 ) ≡ 1 (mod8).
Wesawthatthisimpliesthat U (i) hasanevennumberofelements.If µbelongsto U (i) then
Fµisequalto L (i) .Thuswemaysupposethat
νφ ′
µ∈U (i) γ′ µ
= 1.
Moreover L (i) /F (i) isnon­abelianandtherefore L (i) /F (i) isnottotallyramified. Thus µis
notin U (i) if [Fµ : F ′ ] = k.
Since L (n) /F (n) isabeliantheisotropygroupin H (n) ofany µin T (n) istrivialsothat
Fµ = L (n) forsuch µ.Since
µ∈T ′
[Fµ : F ′ ]
k
≡
µ∈T (n)
[L (n) : F ′ ]
k
(mod2).

Chapter14 225
Thereareanoddnumberofelementsin T (n) and
[L (n) : F ′ ] = k.
Choose z0sothat σ(z0)σ 2 0 liesin G(L/L(n) ).Itthenfixeseach µin T (n) .
Since L (n) /F ′ mustbetotallyramifiedthereisaδin UFsuchthat
N L (n) /F ̟ L (n) = δ̟ 2 F .
Therightsideof (14.13)isequalto νφ(δ). L (n) iscontainedin L.Choose w0in W L/Fsothat
τ L/F(w0) = ̟F. Wemaysupposethat σ0hasbeenchosentobe σ(w0).Let L0bethefixed
fieldof H0.Choose u0in W L/L0 sothat σ(u0) = σ(z0)andsothat τ L/L0 (u0)isaunit.Clearly
z0isevenifandonlyif τ L/L0 (u0)or
isasquare.Since σ(z0)σ 2 0 liesin G(L/L(n) ),
liesin W L/L (n).Wemaytake
Then
NL0/F τL/L0 (u0) = τL/F(u0) u0w 2 0
̟ L (n) = τ L/L (n) (u0w 2 0).
N L (n) /F (̟ L (n)) = τ L/F (u0w 2 0 ) = τ L/F̟ 2 F
and δ = τ L/F(u0)isasquareifandonlyif z0iseven.
Since (−1) a′ +1 = 1therelation(14.5)amountsto
(1) a′′ −1 νφ(−1) q−1
2k = (−1) z0 .
(14.6)isequivalentto(14.5)becauseeach [φµ : φ] = 2[φµ : φ ′ ]iseven.Recallthat
andthat
a ′′
ℓ ℓ p − 1 p − 1
=
≡
k [G : G0]
pℓ − 1
[G : G0]
q − 1
2k =
ℓ ℓ p − 1 p + 1
k
2
≡ pℓ + 1
2
(mod2)
(mod2).

Chapter14 226
If
µ0 ≡ 1 (mod4)
then νφ(−1) = 1and,asweobservedearlier, z0isodd. Wehavetoshowthat a ′′ iseven.
Weshowedbeforethat H∗hastocontainacyclicsubgroupoforder k [G : G0]andthat
2m
k
2m [G : G0]hastodivide pℓ − 1.But k
misthegreatestcommondivisorof µ0 − 1and k.Since
4divides µ0 − 1and kitdivides k
mand 2[G : G0]divides pℓ − 1.Thus a ′′ iseven.
If
µ0 ≡ 3 (mod4)
then νφ(−1) = −1.Moreover k > 2sothat p ℓ ≡ 1 (mod 4)and
Wehavetoshowthat a ′′ isoddif
q − 1
2k ≡ pℓ + 1
2
m = 1
[G : G0]
2
andevenotherwise.But µ0 ≡ 3 (mod4)sothat k
m
andonlyif [G : G0] = k.If [G : G0] = kthen
a ′′ ≡ pℓ − 1
k
isodd.Otherwise 2[G : G0]divides kand a ′′ iseven.
≡ 1 (mod2).
= 2and m = k
2 .Thus [G : G0] = 2mif
(mod2)
Lemma14.3isnowcompletelyproved,soweturntoLemma14.4. Intheproofofboth
Lemma14.4and14.5,wewillcombinetheinductionassumptionwithLemma15.1whichis
statedandprovedinparagraph15,thefollowingparagraph.Suppose F ⊆ F ′ ⊆ Land F ′ /F
iscyclicofprimedegree ℓ.Let G(K/F ′ )be H ′ Cwhere H ′ ⊆ Handlet E ′ bethefixedfieldof
H ′ .Then E ′ /Eiscyclicofprimeorder ℓ.If S(F ′ /F)isthesetofcharactersof CF/N F ′ /FCF ′
then
S(E ′ /E) = {ν E/F | νF ∈ S(F ′ /F)}.
FromLemma15.1weseethatforanyquasi­character χF,
Therefore
Ind(W K/E, W K/E ′, χ E ′ /E) ≃ ⊕ νF ∈S(F ′ /F) ν E/F χ E/F.
Ind(W K/F, W K/E ′, χ E ′ /E) ≃ ⊕νF Ind(W K/F, W K/E, ν E/F χ E/F)

Chapter14 227
whichisequivalentto
⊕νF {(⊕µ∈T Ind(W K/F, W K/Fµ , µ′ ν Fµ/F χ Fµ/F)) ⊕ νFχF }.
If T ′ isasetofrepresentativesforthenon­trivialorbitsof H ′ in S(K/L)then
isequivalentto
Moreover
Ind(W K/F ′, W K/E ′, χ E ′ /F) = σ
(⊕µ∈T ′Ind(W K/F ′, W K/F ′ µ , µ′ χ F ′ µ /F)) ⊕ χ F ′ /F.
Ind(W K/F, W K/F ′, σ) ≃ Ind(W K/F, W K/E ′, χ E ′ /E).
Applyingtheinductionassumptionto L/Fweseethat
isequalto
νF
∆(νF, χF, ψF)
νF
{∆(χ F ′ /F, ψ F ′ /F)λ(F ′ /F, ψF)}
µ∈T ∆(µ′
νFµ/F χFµ/F, ψFµ/F)λ(F/F, ψF)
Theapplicationislegitimatebecausethefields F ′ , Fµ,and F ′ µ
Lemma4.5
Also
µ∈T ′ ∆(µ′ χ F ′ µ /F, ψ F ′ µ /F)λ(F ′ µ /F, ψF). (14.14)
λ(F ′ µ /F, ψF) = λ(F ′ µ /F ′ , ψF ′ /F) λ(F ′ ′
[F
/F, ψF) µ :F ′ ]
.
λ(F ′
/F, ψF)
µ∈T ′ λ(F ′ /F, ψF)
′
[F µ :F ′
]
allliebetween Fand L. By
= λ(F ′ /F, ψF) [E′ :F ′ ] .
Sincethefields F ′ and F ′ µ liebetween F ′ and Kwecanapplytheinductionassumptionto
K/F ′ toseethat(14.14)isequaltotheproductof
and
λ(F ′ /F, ψF) [E′ :F ′ ]
∆(χ E ′ /F, ψ E ′ /F) λ(E ′ /F ′ , ψ F ′ /F).
Applyingtheinductionassumptionto K/Eweseethat
∆(χ E ′ /F, ψ E ′ /F)

Chapter14 228
isequalto
Weconcludethatthequotient
νF
νF ∈S(F ′ /F) ∆(ν E/F χ E/F, ψ E/F)
λ(E ′ /E, ψ E/F) −1 .
∆(νF χF,ψF)
µ∈T ∆(µ′
νFµ/F χFµ/F, ψFµ/F ∆(νE/F χE/F,ψ E/F)
isindependentof χF.Taking χFtobetrivialweseethatitequals
Thus
νF
∆(νF, ψF)
µ∈T ∆(µ′ ν Fµ/F, ψ Fµ/F)
∆(ν E/F,ψ E/F)
Itiseasilyseenthatthecomplexconjugateof ∆(νF, ψF)is
If ℓisoddtherightsideis1.Since
and νF = ν −1
F
if ℓisodd,theproduct
Forthesamereasons
However,if ℓis2
νF(−1) ∆(ν −1
F , ψF).
∆(νF,ψF) ∆(ν −1
F , ψF) = νF(−1).
∆(1, ψF) = 1
νF ∈S(L/F) ∆(νF, ψF) = 1.
νF ∈S(L/F) ∆(ν E/F, ψ E/F) = 1.
∆(νF, ψF) = ∆(ν −1
F , ψF)
hassquare ±1andisthereforeafourthrootofunity.Thus
ν∈S(L/F) ∆(νF, ψF) ∼
ν∈S(L/F) ∆(ν E/F,ψ E/F) ∼2 1.
(14.15)
. (14.16)
Ontheotherhand, m(µ ′ ) = t + 1 ≥ 2while m(ν Fµ/F) ≤ 1.ThusLemma9.5showsthat
∆(µ ′ ν Fµ/F,ψ Fµ/F) ∼ℓ ∆(µ ′ , ψ Fµ/F).

Chapter14 229
Thustheexpression(14.16)andthereforetheexpression(14.15)isequalto
where η ∼ℓ 1.
η
µ∈T ∆(µ′ ℓ , ψFµ/F) If m(χF)is0or1,Lemma14.4isaconsequenceofLemma14.2. Wesupposetherefore
that m(χF) ≥ 2.InthiscaseLemma9.5impliesthat
andthat
νF
νF
∆(νF χF,ψF) ∼ℓ ∆(χF,ψF) ℓ
∆(ν E/F χ E/F,ψ E/F) ∼ℓ ∆(χ E/F, ψ E/F) ℓ .
Wealsosawinthebeginningoftheparagraphthat,inallcases, m(µ ′ χ Fµ/F) ≥ 2.Thus
∆(µ ′ ν Fµ/Fχ Fµ/F,ψ Fµ/F) ∼ℓ ∆(µ ′ χ Fµ/F, ψ Fµ/F).
Puttingthesefactstogetherweseethatif
isequalto
then σ ∼ℓ 1.Since σ = ρ ℓ weconcludethat
σ ∆ (χF, ψF)
µ∈T ∆(µ′ ℓ χFµ/F, ψFµ/F)
∆(χE/F,ψE/F)
µ∈T ∆(µ′ ℓ , ψFµ/F)
ρ ∼ℓ 1.
FinallywehavetoproveLemma14.5.Let F ′ bethefixedfieldof H1Candlet L ′ bethe
fixedfieldof H2C.Let E ′ bethefixedfieldof H1andlet K ′ bethefixedfieldof H2.Let Pbe
asetofrepresentativesfortheorbitsunder G(L/F)ofthecharactersin S(L/L ′ ).If νisone
oftheserepresentatives,let HνH2Cwith Hνand H1beitsisotropygroupandlet Fνbethe
fixedfieldof HνH2C.ApplyingtheinductionassumptionandLemma15.1totheextension
L/Fweseethat
∆(χ F ′ /F, ψ F ′ /F) ρ (F ′ /F, ψF)

Chapter14 230
isequalto
Let
ν∈P ∆(ν′ χ Fν/F, ψ Fν/F) λ(Fν/F, ψF). (14.17)
R = {ν ∈ P | Fν = F }
andlet Sbethecomplementof Rin P. Rconsistsoftheelementsof S(L/L ′ )fixedbyeach
elementof G(L/F).Itisasubgroupof S(L/L ′ )anditsorder rmustthereforebeapowerof
ℓ.Theexpression(14.17)maybewrittenas
ν∈R ∆(ν′
χF, ψF)
ν∈S ∆(ν′
χFν/F, ψFν/F) λ(Fν/F, ψF) .
If Fisreplacedby Eand F ′ by E ′ then Pisreplacedby
{νK ′ /L ′ | ν = νL ′ ∈ P }.
Also Fνisreplacedby Eν,thefixedfieldof HνH2,and ν ′ isreplacedby ν ′ Eν/Fν .Applyingthe
inductionassumptionto K/Eweseethat
isequaltotheproductof
and
∆(χ E ′ /F,ψ E ′ /F) λ(E ′ /E, ψ E/F)
ν∈R ∆(ν′ E/F χ E/F,ψ E/F
ν∈S ∆(ν′ Eν/Fν χ Eν/Fν , ψ Eν/F) λ(Eν/E, ψ E/F
Thisequalitywillbereferredtoasrelation(14.18).
ToderivethisequalitywehaveusednotonlytheinductionassumptionbutalsoLemma
15.1whichimpliesthat
Ind(W K/E, W K/E ′, χ E ′ /F)
isequivalentto
Thus
{⊕R Ind(W K/E, W K/E, ν ′ E/F χ E/F)} ⊕ {⊕S Ind(W K/E, W K/Eν , ν′ Eν/Fν χ Eν/F)}.
Ind(W K/F, W K/E ′, χ E ′ /F)
.

Chapter14 231
willbeequivalenttothedirectsumof
and
⊕R Ind(W K/F, W K/E, ν ′ E/F χ E/F)
⊕S Ind(W K/F, W K/Eν , ν′ Eν/Fν χ Eν/F).
If νisin RwecanapplyLemma15.1toseethat
isequivalentto
Wecanobtain
Ind (W K/F, W K/E, ν ′ E/F χ E/F)
{⊕µ∈T Ind(W K/F, W K/Fµ , µ′ ν ′ Fµ/F χ Fµ/F)} ⊕ ν ′ χF.
Ind (W K/F, W K/Fν , ν′ Eν/Fν χ Eν/F)
byfirstinducingfrom W K/Eν to W K/Fν andthenfrom W K/Fν to W K/F.
If Tνisasetofrepresentativesfortheorbitsof S(K/L)undertheactionof G(K/Fν)and
Fν,µisthefixedfieldoftheisotropygroupof µin Tνthen,byLemma15.1again,
isequivalentto
Ind (W K/Fν , W K/Eν , ν′ Eν/Fν χ Eν/F)
⊕Tν Ind (W K/Fν , W K/Fν,µ , µ′ ν ′ F ν,µ/Fν χ Fν,µ/F).
Since [K : Fν] < [K : F]if νbelongsto S,wecanapplytheinductionassumptiontoseethat
isequalto
∆(ν ′ Eν/Fν χ Eν/F, ψ Eν/F) λ(Eν/Fν, ψ Fν/F)
∆(µ
µ∈Tν
′ ν ′ Fν,µ/FνχFν,µ/F, ψFν/F) λ(Fν,µ/Fν, ψFν/F). Thisequalitywillbereferredtoasrelation(14.19).
Italsofollowsthat
isequivalentto
Ind(W K/F, W K/Eν , ν′ Eν/Fν χ Eν/F)
⊕µ∈Tν Ind(W K/F, W K/Fν,µ , µ′ ν ′ Fν,µ/Fν χ Fν,µ/F).

Chapter14 232
Thefields Fνand Fν,µallliebetween Fand L.Thuswehaveexpressed
asadirectsumoftermsoftheform
Ind(W K/F, W K/E ′, χ E ′ /F) (14.20)
Ind(W K/F, W K/M, χM) (14.21)
where Mliesbetween Fand L.Moreoversucharepresentationisinfactarepresentationof
W K/Fobtainedbyinflatingarepresentationof W L/F,namely,byinflating
Ind(W L/F, W L/M, χM).
Thusanyotherexpressionof(14.20)asasumofrepresentationsoftheform(14.21)willlead,
byanapplicationoftheinductionassumptionto L/F,toanidentitybetweenthenumbers
∆(χM, ψ M/F).
Toobtainanothersuchexpression,weobservethattherepresentation(14.20)canbe
obtainedbyfirstinducingfrom W K/E ′to W K/F ′andthenfrom W K/F ′to W K/F. If T ′ is
asetofrepresentativesfortheorbitsofnon­trivialcharactersin S(K/L)undertheactionof
G(K/F ′ )and F ′ µ isthefixedfieldoftheisotropygroupin G(K/F ′ )of µin T ′ then
isequivalentto
Ind(W K/F ′, W K/E ′, χ E ′ /F)
{⊕µ∈T ′ Ind(W K/F ′, W K/F ′ µ , µ′ χ F ′ µ /F } ⊕ χ F ′ /F.
Thus(14.20)isequivalenttothedirectsumof
and
Ind(W K/F, W K/F ′, χ F ′ /F)
⊕µ∈T ′ Ind(W K/F, W K/F ′ µ , µ ′ χ F ′ µ /F).
Weshalldescribetheresultantidentityinamoment.Wefirstapplytheinductionassumption
totheextension K/F ′ toseethat
isequalto
∆(χ F ′ /F, ψ F ′ /F)
∆(χ E ′ /F, ψ E ′ /F)λ(E ′ /F ′ , ψ F ′ /F)
µ∈T ′ ∆(µ′ χ F ′ µ /F, ψ F ′ µ /F) λ(F ′ µ /F ′ , ψ F ′ /F).

Chapter14 233
Thisequalitywillberelation(14.22).
Thetwoexpressionsfortherepresentation(14.20)leadtotheconclusionthattheproduct
of
ν∈R ∆(ν′ χF,ψF) (14.23)
and
and
ν∈S
ν∈R
isequaltotheproductof
µ∈Tν
and
µ∈T ∆(µ′ ν ′ Fµ/F χ Fµ/F, ψ Fµ/F) λ(Fµ/F, ψF) (14.24)
∆(µ ′ ν ′ Fν,µ/Fν χ Fν,µ/F, ψ Fν,µ/F) λ(Fν,µ/F, ψF) (14.25)
∆(χ F ′ /F, ψ F ′ /F) λ(F ′ /F, ψF)
µ∈T ′ ∆(µ′ χ F ′ µ /F,ψ F ′ µ /F) λ(F ′ µ/F, ψF).
Applyingrelation(14.22)andLemma4.5weseethatthesecondofthesetwoproductsisequal
to
∆(χ E ′ /F, ψ E ′ /F) λ(E ′ /F ′ , ψ F ′ /F) λ(F ′ /F, ψF) [E′ :F ′ ] .
Accordingtotherelation(14.18)thisexpressionistheproductof
ν∈R ∆(ν′ E/F χ
E/F,ψ E/F)
ν∈S ∆(ν′ E/F χ
E/F,ψ E/F)
and
and
ν∈S λ(Eν/E, ψ E/F)
λ(E ′ /E, ψ E/F) −1 λ(E ′ /F ′ , ψ F ′ /F) λ(F ′ /F, ψF) [E′ :F ′ ] . (14.26)
Equatingthisfinalproducttotheproductof(14.23),(14.24),and(14.25)andthenmaking
certaincancellationsbymeansof(14.19),weseethattheproductof(14.23)and(14.24)and
isequaltotheproductof
ν∈S
µ∈Tν
λ −1 (Fν,µ/Fν, ψ Fν/F) λ(Fν,µ/F, ψF)
ν∈R ∆(ν′ E/F χ E/F,ψ E/F)

Chapter14 234
and
andtheexpression(14.26).
Inparticular,theexpression
ν∈R
ν∈S λ−1 (Eν/Fν, ψ Fν/F) λ(Eν/E, ψ E/F)
∆(ν ′ χF,ψF)
isindependentof χF.Taking χFtobetrivialweseethat
isequalto
Theset
ν∈R
µ∈T ∆(µ′ ν ′ Fµ/F χ Fµ/F, ψ Fµ/F)
∆(ν ′ E/F χ E/F,ψ E/F)
′ ∆(ν χF,ψF)
µ∈T ∆(µ′ ν ′ Fµ/F χFµ/F, ψFµ/F) ∆(ν ′ E/F χE/F, ψE/F)
µ∈T ∆(µ′ ν ′ Fµ/F , ψ
Fµ/F)
ν∈R
∆(ν ′ , ψF)
∆(ν ′ E/F , ψ E/F) .
R ′ = {ν ′ | ν ∈ R}
isagroupofcharactersof CForof H.Regardedascharactersof Htheelementsof R ′ arejust
thosecharacterswhicharetrivialon H1.Asagroup R ′ iscyclicanditsorderisapowerof ℓ.
TheargumentusedintheproofofLemma14.4showsthat
and
ν∈R ∆(ν′ , ψF) ∼ℓ 1
ν∈R ∆(ν′ E/F , ψ E/F) ∼ℓ 1.
If m(χF)is0or1,Lemma14.5isaconsequenceofLemma14.2.Wemayaswellsuppose
thereforethat m(χF) > 1. If νbelongsto Rthen ν ′ is1on N L/FCL. Therefore m(ν ′ ),as
wellas m(ν ′ Fµ/F )and m(ν′ E/F )isatmost1.Wesawinthebeginningofthisparagraphthat
m(χ E/F)wouldalsobeatleast2. Wealsosawthat m(µ ′ χ Fµ/F)wouldbeeither t + 1or

Chapter14 235
ψ Fµ/F(m − 1) + 1. Inanycaseitisatleast2. Also m(µ ′ ) = t + 1isatleast2. Lemma9.5
thereforeimpliesthefollowingrelations:
Weconcludefinallythat
∆(ν ′ χF,ψF) ∼ℓ ∆(χF, ψF)
∆(ν ′ E/F χ E/F,ψ E/F ∼ℓ ∆(χ E/F, ψ E/F)
∆(µ ′ ν ′ Fµ/F χ Fµ/F, ψ Fµ/F ∼ℓ ∆(µ ′ χ Fµ/F, ψ Fµ/F)
∆(µ ′ ν ′ Fµ/F , ψ Fµ/F) ∼ℓ ∆(µ ′ , ψ Fµ/F).
∆(χF, ψF)
µ∈T ∆(µ′ χFµ/F, ψFµ/F) ∆(χE/F,ψ E/F)
µ∈T ∆(µ′ , ψFµ/F) if risthenumberofelementsin R.Thelemmafollows.
r
∼ℓ 1

Chapter15 236
Chapter Fifteen.
Another Lemma
Suppose K/Fisnormaland G = G(K/F).Suppose Hisasubgroupof Gand Cisan
abeliannormalsubgroupof G.Let Ebethefixedfieldof Hand Lthatof C.If µisacharacter
of Cand hbelongsto H,define µ h by
µ h (c) = µ(hch −1 ).
Thesetofcharactersof Cmaybeidentifiedwith S(K/L).If αbelongsto CL
µ h (α) = µ(h(α)).
Thesetofelementsin S(K/L)whicharetrivialon H ∩ Cisinvariantunder H.Let Tbeaset
ofrepresentativesfortheorbitsof Hinthisset.If µ ∈ Tlet Hµbetheisotropygroupof µ,let
Gµ = HµCandlet Fµbethefixedfieldof Gµ.Defineacharacter µ ′ of Gµby
µ ′ (hc) = µ(c)
if h ∈ Hµand c ∈ C. µ ′ mayberegardedasacharacterof CFµ .
Lemma 15.1
If χFisaquasi­characterof CFthen
isequivalentto
ρ = Ind(W K/F, W K/E, χ E/F)
⊕µ∈T Ind(W K/F, W K/Fµ , µ′ χ Fµ/F).
Let G ′ = HCandlet F ′ bethefixedfieldof G ′ . F ′ iscontainedin Eandinthefields Fµ.
Becauseofthetransitivityoftheinductionprocessitisenoughtoshowthat
isequivalentto
Ind(W K/F ′, W K/E, χ E/F)
⊕µ∈TInd(W K/F ′, W K/Fµ , µ′ χ Fµ/F).

Chapter15 237
If
then
and
χ ′ F ′ = χ F ′ /F
χ E/F = χ ′ E/F ′
χ Fµ/F = χ ′ Fµ/F ′.
Consequentlywemaysuppose,withnolossofgenerality,that F ′ is F.
by
If K ′ isthefixedfieldof H ∩Cand ν ∈ S(K ′ /L)let ϕνbethefunctionon W K/Fdefined
ϕν(hc) = χF(τ K/F(hc)) ν(τ K/L(c))
for hin W K/E, cin W K/L. ρactsonthespaceofallfunctions ϕon W K/Fsatisfying
forall hin W K/Eandall gin W K/F.Theset
isabasisforthisspace.Clearly
if cbelongsto W K/Land
ϕ(hg) = χF(τ K/F(h)) ϕ(g)
{ϕν | ν ∈ S(K ′ /L)}
ρ(c)ϕν = χF(τ K/F(c)) ν(τ K/L(c))ϕν
ρ(h)ϕν = χF(τ K/F(h))ϕν ′,
with ν ′ = νh−1,if hbelongsto WK/E.Thusif Risanorbitof Hin S(K ′ /L)
isaninvariantsubspace.
⊕ν∈RCϕν
= V
Let µbetheelementcommonto Tand Randconsider
If W K/Fisthedisjointunion
σ = Ind(W K/F, W K/Fµ , µ′ χ Fµ/F).
r
i=1 W K/Fµ hi

Chapter15 238
andif ϕi(w) = 0unless
while
for win W K/Fµ then
w ∈ W K/Fµ hi
ϕi(whi) = µ ′ χ Fµ/F(τ K/Fµ (w))
{ϕi | 1 ≤ i ≤ r}
isabasisforthespace Uonwhich σacts.If νi = µ hi andif λisthemapfrom Uto V which
sends ϕito χ −1
F (τ K/F(hi))ϕνi then,asoneverifieseasily,
forall win W K/F.Thelemmafollows.
Thelemmahasacorollary.
Lemma 15.2
IfTheorem2.1isvalidfor K/Fthen
isequalto
IfTheorem2.1isvalid
isequalto
Taking χF = 1,weseethat
λσ(w) = ρ(w)λ
∆(χ E/F,ψ E/F)
µ∈T ∆(µ′ , ψ Fµ/F)
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F).
∆(χ E/F, ψ E/F) λ(E/F, ψF)
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F) λ(Fµ/F, ψF).
λ(E/F, ψF) =
µ∈T ∆(µ′ , ψ Fµ/F) λ(Fµ/F, ψF).
Substitutingthisintothefirstequalityandcancellingthenon­zerofactor
weobtainthelemma.
µ∈T λ(Fµ/F, ψF)

Chapter15 239
Todefinethe λ­functionweshallneedthefollowinglemma.
Lemma 15.3
SupposeTheorem2.1isvalidforallGaloisextensions K1/F1with F ⊆ F1 ⊆ K1 ⊆ K
and [K1 : F1] < [K : F].Then
isequalto
∆(χ E/F,ψ E/F)
µ∈T ∆(µ′ , ψ Fµ/F)
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F).
Theconclusionofthislemmaisthesameasthatofthepreviousone.Thereishowevera
criticaldifferenceintheassumptions.
Let F ′ bethefixedfieldof HC.If
thenforallseparableextensions E ′ of F ′
ψ ′ F ′ = ψ F ′ /F
ψ ′ E ′ /F ′ = ψ E ′ /F.
If [K : F ′ ] < [K : F]therelationofthelemmaisaconsequenceoftheinductionassumption
andthepreviouslemma.Wethussupposethat F = F ′ and G = HC.
Supposeinadditionthatthereisasubgroup C1of C,whichisneither Cnor {1},whose
normalizercontains H. C1isthenanormalsubgroupof G.Let F1bethefixedfieldof HC1
and L1thefixedfieldof C1.Lemma15.1appliestotheextension K/F1.Thustherearefields
A1, . . ., Arlyingbetween F1and L1andquasi­characters χA1 , . . ., χArsuchthat isequivalentto
Theinductionassumptionthenimpliesthat
Ind(W K/F1 , W K/E, χ E/F)
⊕ r i=1 Ind(WK/F1 , WK/Ai , χAi ).
∆(χ E/F, ψ E/F) λ(E/F1, ψ F1/F) (15.1)

Chapter15 240
isequalto
Thus
r
i=1 ∆(χAi , ψ Ai/F) λ(Ai/F1, ψ F1/F). (15.2)
Inducingthefirstofthesetworepresentationsfrom W K/F1 to W K/F,weobtain
isequivalentto
Ind(W K/F, W K/E, ψ E/F).
⊕µ∈T Ind(W K/F, W K/Fµ , µ′ χ Fµ/F) (15.3)
Werecallthatthereexistsurjectivehomomorphisms
⊕ r i=1 Ind(WK/F, WK/Ai , χAi ). (15.4)
τ K/F,L1/F : W K/F −→ W L1/F
τ K/Ai,L1/Ai : W K/Ai −→ W L1/Ai
τ K/Fµ,L1/Fµ : W K/Fµ −→ W L1/Fµ
whosekernelsareallequaltothecommutatorsubgroup W c K/L1 of WK/L1 . Moreoverthe
diagrams
and
W K/Ai −→ W L1/Ai
↓ ↓
W K/F −→ W L1/F
W K/Fµ −→ W L1/Fµ
↓ ↓
W K/F −→ W L1/F
maybesupposedcommutative. Since W c K/L1 liesinthekernelof χAi and µ′ χ Fµ/F the
equivalenceof(15.3)and(15.4)amountstotheequivalenceof
and
⊕µ∈TInd(W L1/F, W L1/Fµ , µ′ χ Fµ/F)
⊕ r i=1Ind(WL1/F, WL1/Ai , χAi ).

Chapter15 241
Theinductionassumptionappliedtotheextension L1/Fimpliesthat
r
i=1 ∆(χAi , ψ Ai/F) λ(Ai/F, ψF)
isequalto
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F) λ(Fµ/F, ψF).
Italsoimpliesthat
λ(Ai/F, ψF) = λ(Ai/F1, ψ F1/F) λ(F1/F, ψF) [Ai:F1] .
Since
i [Ai : F1] = [E : F1]
weinferfromtheequalityof(15.1)and(15.2)that
isequalto
∆(χ E/F,ψ E/F) λ(E/F1, ψ F1/F) λ(F1/F, ψF) [E:F1]
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F) λ(Fµ/F, ψF).
Taking χF = 1tofindthevalueof
λ(E/F1, ψ F1/F) λ(F1/F, ψF) [E:F1]
andthensubstitutingtheresultintotheequationandcancellingthecommonfactorsweobtain
theassertionofthelemma.
Nowsupposethat Hcontainsanormalsubgroup H1 = {1}whichliesinthecentralizer
of C. H1isanormalsubgroupof Gif,asweareassuming, G = HC. K1,thefixedfieldof
H1,contains Eandallthefields Fµ.Lemma15.1togetherwiththeargumentjustappliedto
L1showsthat
Ind(W K1/F, W K1/E, χE)
isequivalentto
⊕µ∈TInd(W K1/F,W K1/Fµ , µ′ χ Fµ/F).
InthiscasetheassertionofthelemmafollowsfromtheinductionassumptionappliedtoK1/F.
Wehavefinallytosupposethat G = HC, Ccontainsnopropersubgroupinvariant
under H,and Hcontainsnonormalsubgrouplyinginthecentralizerof C. Inparticular
H ∩C = {1}.If Zisthecentralizerof Cthen Z = (Z ∩H)Cand Z ∩Hisanormalsubgroup
of H.Consequently Z = C.If Disanormalsubgroupof Gand Ddoesnotcontain Cthen
D ∩ C = {1}.
Thisimpliesthat Discontainedin Z.Thus Discontainedin Cand D = {1}.If H = {1}the
assertionofthelemmaisthatofthethirdandfourthmainlemmas.If H = {1}then G = C
and Ciscyclicofprimeordersothattheassertionisthatofthefirstmainlemma.

Chapter16 242
Chapter Sixteen.
Definition of the λ Functions
Inthisandthenextthreeparagraphs,wetakeafixedGaloisextension K/F,assume
thatTheorem2.1isvalidforallGaloisextensions K ′ /F ′ with F ⊆ F ′ ⊆ K ′ ⊆ Kand
[K ′ : F ′ ] < [K : F],andprovethatitisvalidfor K/Fitself. Thefirststepistodefineand
establishsomesimplepropertiesofthefunctionwhichwillserveasthe λ­function.
Lemma 16.1
Suppose
E/F ′ −→ λ(E/F ′ , ψF ′)
isaweak λ­functionon P0(K ′ /F ′ ).If σ ∈ G(K ′ /F ′ )let
Then
E σ = {σ −1 (α) | α ∈ E}.
λ(E σ /F ′ , ψF ′) = λ(E/F ′ , ψF ′).
If µisacharacterof G(K/E)let µ σ bethecharacterof G(K/E σ )definedby
AccordingtoLemma13.2
Therepresentation
µ σ (ρ) = µ(σρσ −1 ).
∆(µ σ , ψ E σ /F ′) = ∆(µ, ψ E/F ′).
Ind(G(K ′ /F ′ ), G(K ′ /E), µ)
actsonthespace Uoffunctions ϕon G(K ′ /F ′ )satisfying
ϕ(ρτ) = µ(ρ) ϕ(τ)
forall τin G(K ′ /F ′ )andall ρin G(K ′ /E).Themap ϕ −→ ψwith
ψ(τ) = ϕ(στ)

Chapter16 243
isaG(K ′ /F ′ )isomorphismof Uwiththespaceonwhich
Ind(G(K ′ /F ′ ), G(K ′ /E σ ), µ σ )
acts.Thusthetworepresentationsareequivalent.
If
isequivalentto
then
isequivalentto
⊕ r i=1 Ind(G(K′ /F ′ ), G(K ′ /Ei), µi)
⊕ s j=1Ind(G(K ′ /F ′ ), G(K ′ /Fj), νj)
⊕ r i=1 Ind(G(K′ /F ′ ), G(K ′ /E σ i ), µσ i )
⊕ s j=1 Ind(G(K′ /F ′ ), G(K ′ /F σ j ), νσ j )
and,withtheconventionsofthefourthparagraph,
isequalto
Since
and
r
s
Weconcludethat r
isequalto
Inotherwords
i=1 (χE σ i , ψ E σ i /F ′) λ(Eσ i /F ′ , ψF ′)
j=1 ∆(χF σ j , ψ F σ j /F ′) λ(F σ j /F ′ , ψF ′).
∆(χF σ j , ψ F σ j /F ′) = ∆(χFj , ψ Fj/F ′)
∆(χE σ i , ψ E σ i /F ′) = ∆(χEi , ψ Ei/F ′).
s
i=1 ∆(χEi , ψ Ei/F ′) λ(Eσ i /F ′ , ψF ′)
j=1 ∆(χFj , ψ Fj/F ′) λ(F σ j /F ′ , ψF ′).
E/F ′ −→ λ(E σ /F ′ , ψF ′)
isaweakλ­functionon P0(K ′ /F ′ ).Lemma16.1followsfromtheuniquenessofsuchfunctions.
Wereturntotheproblemofdefiningaλ­functionon P0(K/F). Chooseanon­trivial
abeliannormalsubgroup Cof G = G(K/F)andlet Lbethefixedfieldof C. If Eisany

Chapter16 244
fieldlyingbetween Fand Klet Hbethecorrespondingsubgroupof G. Choosetheset T
ofcharactersandthefields Fµasinthepreviousparagraph. Since Fµ ⊆ Lthenumbers
λ(Fµ/F, ψF)aredefined.
Lemma 16.2
Suppose F ⊆ E ⊆ K1 ⊂ = Kwith K1/Fnormalsothat λ(E/F, ψF)isdefined.Then
λ(E/F, ψF) =
µ∈T ∆(µ′ , ψ Fµ/F) λ(Fµ/F, ψF).
Let K1bethefixedfieldof H1.If H1 ∩ C = {1}wemayenlarge K1andreplace H1by
H1 ∩C.Thuswemaysupposethateither H1iscontainedin Cor H1 ∩C = {1}.Ineithercase
H1iscontainedinthecentralizerof C. Wesawinthepreviousparagraphthatunderthese
circumstance
Consequently
Ingeneral,wedefine
Ind(W K1/F, W K1/E, 1) ≃ ⊕µ∈TInd(W K1/F, W K1/Fµ , µ′ ).
λ(E/F, ψF) =
λ(E/F, ψF) =
µ∈T ∆(µ′ , ψ Fµ/F) λ(Fµ/F, ψF).
µ∈T ∆(µ′ , ψ Fµ/F) λ(Fµ/F, ψF)
if E/Fisin P0(K/F) Tis,ofcourse,notalwaysuniquelydetermined.Wemayreplaceany
µin Tby µ σ with σin H.Then Hµand Gµarereplacedby σ −1 Hµσand σ −1 Gµσwhile Fµis
replacedby F σ µ and µ′ isreplacedby (µ ′ ) σ .Since
∆(µ ′ , ψ Fµ/F) λ(Fµ/F, ψF) = ∆((µ ′ ) σ , ψ F σ µ /F) λ(F σ µ /F, ψF)
thenumber λ(E/F, ψF)doesnotdependon T. A priori,itmaydependon Cbutthatis
unimportantsince Cisfixedand,theuniquenesshavingbeenproved,weareinterestedonly
intheexistenceofaλ­function.
Weshallneedonlyonepropertyofthefunctionjustdefined.
Lemma 16.3

Chapter16 245
If F ⊆ E ⊆ E ′ ⊆ Kthen
If E = Fthen
andif E = F
λ(E ′ /F, ψF) = λ(E ′ /E, ψ E/F) λ(E/F, ψF) [E′ :E] .
λ(E ′ /E, ψ E/F) = λ(E ′ /F, ψF)
λ(E ′ /E, ψ E/F)
isthevalueofthe λ­functionof P(K/E),whichisdefinedbyassumption,at E ′ /E.Since
λ(F/F, ψF) = 1
theassertionisclearif E = F.Itisalsoclearif E = E ′ .
Let Ebethefixedfieldof Hasbeforeandlet F ′ bethefixedfieldof HC. Wesuppose
that H = G.Lemma4.5andtheinductionassumptionimplythat
Therelation
impliesthat
λ(Fµ/F, ψF) = λ(Fµ/F ′ , ψ F ′ /F) λ(F ′ /F, ψF) [Fµ:F ′ ] .
[E : F ′ ] = [Fµ : F ′ ]
λ(E/F, ψF) = λ(E/F ′ , ψ F ′ /F) λ(F ′ /F, ψF) [E:F ′ ] .
Thereisasimilarformulafor λ(E ′ /F, ψF).If F ′ = Ftheinductionassumptionimpliesthat
Since
λ(E ′ /E, ψ E/F) λ(E/F ′ , ψ F ′ /F) [E′ :E] = λ(E ′ /F ′ , ψF ′ /F).
[E ′ : F ′ ] = [E ′ : E] [E : F ′ ]
theassertionofthelemmaisprovedsimplybymultiplyingbothsidesofthisequationby
λ(F ′ /F, ψF) [E′ :F ′ ] .
Nowsupposethat G = HCand H ∩ C = {1}. Let E ′ bethefixedfieldof H ′ andlet
F ′ bethefixedfieldof H ′ C = G ′ .Eachcharacterof H ′ maybeidentifiedwithacharacterof

Chapter16 246
CE ′/N K/E ′CKandeachcharacterof G ′ maybeidentifiedwithacharacterof CF ′/N K/F ′CK.
Anycharacter χE ′of H′ maybeextendedtoacharacter χF ′of G′ bysetting
if ρ ∈ H ′ and σ ∈ C.Then
χF ′(ρσ) = χ E ′ (ρ)
χE ′ = χ E ′ /F ′.
ItfollowsfromLemma15.1thattherearefieldsof Fi(E ′ ), 1 ≤ i ≤ m(E ′ ),lyingbetween F ′
and Landcharacters µ Fi(E ′ )suchthat
isequivalentto
Ind(W K/F ′, W K/E ′, χE ′)
⊕ m(E′ )
i=1 Ind(W K/F ′, W K/Fi(E ′ ), µ Fi(E ′ )χ Fi(E ′ )/F ′).
If E = E ′ sothat F = F ′ theinductionassumptionimpliesthat
isequalto m(E ′ )
∆(χE ′, ψ E ′ /F) λ(E ′ /F ′ , ψ F ′ /F)
i=1 ∆(µ Fi(E ′ )χ Fi(E ′ )/F ′, ψ Fi(E ′ )/F) λ(Fi(E)/F ′ , ψ F ′ /F).
Wehaveseenthatthelemmaisvalidforanypair E ′ , Eforwhich HC = G.Inparticular,
itisvalidforthepair E ′ , F ′ andthepairs Fi(E ′ ), F ′ .Multiplyingtheequalityjustobtained
by
λ(F ′ /F, ψF) [E′ :F ′ ]
weseethat
isequalto
m(E ′ )
∆(χE ′, ψ E ′ /F) λ(E ′ /F, ψF) (16.1)
i=1 ∆(µ Fi(E ′ )χ Fi(E ′ )/F ′, ψ Fi(E ′ )/F) λ(Fi(E ′ )/F, ψF). (16.2)
If F ′ = Ftheequalityof(16.1)and(16.2),forasuitablechoiceofthefields Fi(E ′ ),resultsfrom
Lemma15.1,Lemma15.3,andthedefinitionof
λ(E ′ /F, ψF).
Inanycasetheequalityisvalidforallfieldslyingbetween Eand K.

Chapter16 247
Suppose E1, . . ., Er, E ′ 1 , . . ., E′ s aresuchfields, χEi isacharacterof CEi /N K/Ei CK, χE ′ j
isacharacterof C E ′ j /N K/E ′ j CK,and
isequivalentto
Then r
⊕ r i=1Ind(WK/E, WK/Ei , χEi )
⊕ s j=1Ind(W K/E, W K/E ′ j , χE ′ j ).
i=1 [Ei : E] = s
and,bythetransitivityoftheinductionprocess,
isequivalentto
j=1 [E′ j
⊕ r i=1 ⊕ m(Ei)
k=1 Ind(W K/F, W K/Fk(Ei), µ Fk(Ei) χ Fk(Ei)/Fi )
⊕ s j=1 ⊕m(E′
j )
ℓ=1 Ind(WK/F, WK/Fℓ(E ′ j ), µ Fℓ(E ′ j ) χFℓ(E ′ j )/F ′ j ).
: E] (16.3)
If Eiisthefixedfieldof Hiand E ′ jthefixedfieldof H′ jthen Fiand F ′ jarethefixedfieldsof HiCand H ′ jC.Thisequivalenceandtheinductionassumptionfor L/Fimplythat
isequalto
r
s
i=1
j=1
m(Ei)
k=1 ∆(µ Fk(Ei) χ Fk(Ei)/Fi , ψ Fk(Ei)/F) λ(Fk(Ei)/F, ψF)
m(E ′
j )
ℓ=1 ∆(µ Fℓ(E ′ j ) χ Fℓ(E ′ j )/F ′ j , ψ Fℓ(E ′ j )/F) λ(Fℓ(E ′ j
)/F, ψF).
Thisequality,theequalityof(16.1)and(16.2),andtherelation(16.3)implythat
isequalto
Consequently
r
s
i=1 ∆(χEi , ψ Ei/F) λ(Ei/F, ψF) λ(E/F, ψF) −[Ei:E]
j=1 ∆(χ E ′ j , ψ E ′ j /F) λ(E ′ j /F, ψF) λ(E/F, ψF) −[E′
E ′ −→ λ(E ′ /F, ψF) λ(E/F, ψF) −[E′ :E]
isaweak λ­functionon P0(K/E).Thelemmaofuniquenessimpliesthat
λ(E ′ /F, ψF) λ(E/F, ψF) −[E′ :E] = λ(E ′ /E, ψE/F).
j :E] .

Chapter16 248
Thisis,ofcourse,theassertionofthelemma.
Atthispoint,wehaveprovedthelemmawhenvarioussupplementaryconditionsare
satisfied.Beforeprovingit,ingeneral,wemakeanobservation.Suppose
F ⊆ E ⊆ E ′ ⊆ E ′′ ⊆ K
andtheassertionofthelemmaisvalidfor E ′′ /E ′ and E ′ /E.Then
and
Moreover,byinduction,
λ(E ′′ /F, ψF) = λ(E ′′ /E ′ , ψ E ′ /F) λ(E ′ /F, ψF) [E′′ :E ′ ]
λ(E ′ /F, ψF) = λ(E ′ /E, ψ E/F) λ(E/F, ψF) [E′ :E] .
λ(E ′′ /E, ψ E/F) = λ(E ′′ /E ′ , ψ E ′ /F) λ(E ′ /E, ψ E/F) [E′′ :E ′ ] .
Theassertionfor E ′′ /Eisobtainedbysubstitutingthesecondrelationinthefirstandsimplifyingaccordingtothethird.
Ifthelemmaisfalseingeneral,choseamongstalltheextensionsin P(K/F)forwhichit
isfalseone E ′ /Eforwhich [E ′ : E]isaminimum.Let Ebethefixedfieldof Hand E ′ that
of H ′ .Accordingtothepreviousdiscussion G = HC, H ∩ C = {1},andtherearenofields
lyingbetween Eand E ′ . If H ′ ∩ C = H ∩ C,whichisanormalsubgroupof G,thefields
F, E,and E ′ arecontainedinthefixedfieldof H ∩ Candtheassertionisaconsequenceof
theinductionassumption.Thus H ′ isapropersubgroupof H ′ (H ∩ C).Becausethereareno
intermediatefields H = H ′ (H ∩ C).
Aswehaveseentherearefields E1, . . ., Erlyingbetween Eandthefixedfield K1of
, . . ., µErsuchthat H ∩ Candcharacters µE1
isequivalentto
Then
λ(E ′ /E, ψ E/F) = r
Ind(W K/E, W K/E ′, 1)
⊕ r i=1Ind(WK/E, WK/Ei , µEi ).
Bytheinductionassumption,appliedto K1/F,
i=1 ∆(µEi , ψ Ei/E) λ(Ei/E, ψ E/F).
λ(Ei/E, ψ E/F) λ(E/F, ψF) [Ei:E] = λ(Ei/F, ψF).

Chapter16 249
Thus
isequalto
λ(E ′ /E, ψ E/F) λ(E/F, ψF) [E′ :E]
r
Moreover,bythetransitivityoftheinductionprocess,
isequivalentto
i=1 ∆(µEi , ψ Ei/E) λ(Ei/F, ψF). (16.4)
Ind(W K/F, W K/E ′, 1) (16.5)
⊕ r i=1Ind(WK/F, WK/Ei , µEi ). (16.6)
Ontheotherhand,therearefields F1, . . ., Fscontainedin Landcharacters νF1 , . . ., νFssuch that(16.5)isequivalentto
⊕ s j=1Ind(WK/F, WK/Fj , νFj ) (16.7)
andsuchthat,bydefinition,
λ(E ′ /F, ψF) = s
j=1 ∆(νFj , ψ Fj/F) λ(Fj/F, ψF). (16.8)
Sincetherepresentations(16.6)and(16.7)areequivalenttheinductionassumption,appliedto
K1/F,showsthat(16.4)isequaltotherightsideof(16.8).Thisisacontradiction.

Chapter17 250
Chapter Seventeen.
A Simplification.
Weshallusethesymbol Ωtodenoteanorbitinthesetofquasi­charactersof CKunder
theactionof G(K/F)or,whatisthesame,undertheactionof W K/F on CKbymeansof
innerautomorphisms. If χKisaquasi­characterof CKitsorbitwillbedenoted Ω(χK). If
ρisarepresentationof W K/Ftherestrictionof ρto CKisthedirectsumofone­dimensional
representations.Let S(ρ)bethecollectionofquasi­characterstowhichtheseone­dimensional
representationscorrespond.
Suppose
Let W K/Fbethedisjointunion
Definethefunction ϕiby
ρ = Ind(W K/F, W K/E, χE).
m
i=1 W K/Ewi.
ϕi(wwj) = 0 w ∈ W K/E, j = i
ϕi(wwi) = χE(τ K/Ew) w ∈ W K/E.
{ϕ1, . . ., ϕm}isabasisforthespaceoffunctionsofwhich ρacts.If a ∈ CKthen
wwja = w(wjaw −1
j )wj
and wjaw −1
j belongsto CKwhich,ofcourse,liesin W K/E.Thus
if σiistheimageof wiin G(K/F).Thus
Suppose E1, . . ., Er, E ′ 1 , . . ., E′ s
χE ′ j isaquasi­characterof E′ j .Let
ρ(a)ϕi = χE(τ K/E(wiaw −1
i ))ϕi = χ σi
K/E (a)ϕi
S(ρ) = Ω(χ K/E).
liebetween Fand K, χEi isaquasi­characterof Ei,and
ρi = Ind(WK/F, WK/Ei , χEi )

Chapter17 251
andlet
Suppose ρiactson Viand ρ ′ j
ρ ′ j = Ind(W K/F, W K/E ′ j , χ E ′ j ).
′ actson V j .Thedirectsumoftherepresentations ρiactson
V = ⊕ r i=1 Vi
andthedirectsumoftherepresentations ρ ′ j actson
Let
V ′ = ⊕ s ′
j=1V j .
VΩ = ⊕ {i | χK/Ei ∈Ω}Vi
V ′ Ω = ⊕ {i | χK/E ′ ∈Ω}V
j
′
j .
Anyisomorphismof Vwith V ′ whichcommuteswiththeactionof W K/Ftakes VΩto V ′ Ω .
and
If χ K/Ei ∈ Ω(χK)thereisaσin G(K/F)suchthat χK = χ σ K/Ei .Then
ρi ≃ Ind(W K/F, W K/E σ i , χ σ Ei )
∆(χEi , ψ Ei/F) λ(Ei/F, ψF) = ∆(χ σ Ei , ψ E σ i /F) λ(E σ i /F, ψF).
WeconcludethatTheorem2.1isaconsequenceofthefollowinglemma.
Lemma 17.1
Suppose χKisaquasi­characterof CK.Suppose E1, . . ., Er, E ′ 1 , . . ., E′ s
and K, χEi isaquasi­characterof CEi , χE ′ j isaquasi­characterof CE ′ j ,and
isequivalentto
If χ K/Ei = χ K/E ′ j = χKforall iand jthen
ρ = ⊕ r i=1Ind(WK/F, WK/Ei , χEi )
ρ ′ = ⊕ s j=1 Ind(W K/F, W K/E ′ j , χ E ′ j ).
r
i=1 ∆(χEi , ψ Ei/F) λ(Ei/F, ψF)
liebetween F

Chapter17 252
isequalto
s
j=1 ∆(χ E ′ j , ψ E ′ j /F), λ(E ′ j
/F, ψF).
Let F(χK)bethefixedfieldoftheisotropygroupof χK.Let ρacton Vandlet ρ ′ acton
V ′ .Let
V (χK) = {v ∈ V | ρ(a)v = χK(a)v forall a in CK}.
Define V ′ (χK)inasimilarfashion. Itisclearthatanyisomorphismof V with V ′ which
commuteswiththeactionof W K/Ftakes V (χK)to V ′ (χK).Thegroup W K/F(χK)leavesboth
V (χK)and V ′ (χK)invariantanditsrepresentationsonthesetwospacesareequivalent.
Let
Ind(WK/F, WK/Ei , χEi )
actson Vianddefine Vi(χK)intheobviousmanner.Then
Defining V ′
j
V (χK) = ⊕ r i=1 Vi(χK).
and V ′
j (χK)inasimilarmanner,wehave
V ′ (χK) = ⊕ s ′
j=1V j (χK).
Itisclearthattherepresentationof WK/F(χK)on Vi(χK)isequivalentto
Thus
isequivalentto
Ind(WK/F(χK), WK/Ei , χEi ).
⊕ r i=1Ind(WK/F(χK), WK/Ei , χEi )
⊕ s j=1 Ind(W K/F(χK), W K/E ′ j , χE ′ j ).
If F(χK) = FtheassertionofthelemmafollowsfromtheinductionassumptionandLemma
16.3.

Chapter18 253
Chapter Eighteen.
Nilpotent Groups.
InthisparagraphweproveLemma17.1assumingthatF = F(χK)andthatG = G(K/F)
isnilpotent.
Lemma 18.1
Suppose Disanormalsubgroupof Gofprimeorder ℓwhichiscontainedinthecenter
of G.Let Mbethefixedfieldof D.Suppose F ⊆ E ⊆ Kand χEisaquasi­characterof CE.
Supposealsothat F(χ K/E) = F.
(a)Therearefields F1, . . ., Frcontainedin Mandquasi­characters χF1 , . . ., χFrsuchthat χK/Fi = χK/Eandsuchthat isequivalentto
Ind(W K/F, W K/E, χE)
⊕ r i=1Ind(WK/F, WK/Fi , χFi ).
(b)IfTheorem2.1isvalidforallGaloisextensionsK ′ /F ′ in P(K/F)with[K ′ : F ′ ] < [K : F]
then
∆(χE, ψ E/F) λ(E/F, ψF)
isequalto
r
i=1 ∆(χFi , ψ Fi/F) λ(Fi/F, ψF).
Weprovethelemmabyinductionon [K : F].Let Hbethesubgroupof Gcorresponding
to E;let G ′ = HDandlet F ′ bethefixedfieldof G ′ . If F ′ = Ftheinductionassumption
impliesthattherearefields F1, . . ., Frcontainedin Mandquasi­characters χF1 , . . ., χFrsuch that χK/Fi = χK/Eforeach iandsuchthat
isequivalentto
Ind(W K/F, W K/E, χE)
⊕ r i=1Ind(WK/F ′, WK/Fi , χFi ).

Thefirstpartofthelemmafollowsfromthetransitivityoftheinductionprocess.Thesecond
partfollowsfromLemma16.3andtheassumedvalidityofTheorem2.1fortheextension
K/F ′ .
Wesupposenowthat G = HD. Supposethat Hcontainsanormalsubgroup H1of G
whichisdifferentfrom {1}andsupposethat,if K1isthefixedfieldof H1, F(χK1/E) = F.
If M1isthefixedfieldof H1Dthen,accordingtotheinductionassumption,therearefields
F1, . . ., Frcontainedin M1andquasi­characters χF1 , . . ., χFrsuchthat andsuchthat
isequivalentto
Itfollowsimmediatelythat
andthat
isequivalentto
χ K1/Fi = χ K1/E
Ind(W K1/F, W K1/E, χE)
⊕ r i=1Ind(WK1/F, WK1/Fi , χFi ).
χ K/Fi = χ K/E
Ind(W K/F, W K/E, χE)
⊕ r i=1Ind(WK/F, WK/Fi , χFi ).
Theequalityof(b)isaconsequenceoftheassumedvalidityofTheorem2.1for K1/F.
Weassumenowthat G = HDandthatif H1isanormalsubgroupof Hdifferentfrom
{1}withfixedfield K1thefield F(χ K1/E)isnot F.If w1belongsto W K/Eand w2belongsto
W K/Mthen
Let χK = χ K/E.Since F(χK) = F
χK(w1w2w −1
w1w2w −1
1 w−1 2 ∈ CK.
1 w−1
−1
2 ) = χK(w2w1 w−1 2 w1) = χK(w −1
1 w−1 2 w1w2) = χK(w −1 −1
2 w1w2w1 ).
Denotethecommonvalueoftheexpressionsby ω(w1, w2).Then ω(v1w1, w2)isequalto
Therightsideis
χK(v1w1w2w −1
1 v−1
1 w−1 2
−1 −1
) = χK(w2 w1w2w1 v−1 1 w−1 2 v1w2).
ω(v1, w2) ω(w1, w2).

Chapter18 255
Inthesameway ω(w1, v2w2)is
whichequals
χK(w1v2w2w −1
1 w−1 2 v−1 2
Ifeither w1or w2belongto CK,wehave
Thus,foreach w2,
) = χK(w −1
ω(w1, v2) ω(w1, w2).
ω(w1, w2) = 1.
w1 −→ ω(w1, w2)
1 v−1
−1
2 w1v2w2w1 w−1 2 w1)
isahomomorphismof H = W K/E/CKinto C ∗ and,foreach w1,
w2 −→ ω(w1, w2)
isahomomorphismof D = W K/M/CKinto C ∗ .If wbelongsto W K/Fthen
ω(ww1w −1 , ww2w −1 ) = ω(w1, w2).
Thusthereisanormalextension K1containing Esuchthat
W K/K1 = {w1 | ω(w1, w2) = 1 forall w1 ∈ W K/M}.
But F(χ K1/E)willbe Fsothat K1mustbe K.
Itfollowsimmediatelythat Hisisomorphictoasubgroupofthedualgroupof D.Thus
H = {1}or Hiscyclicoforder ℓ. Ineithercase Hmustlieinthecentralizerof Dsothat
E/Fisnormaland G(E/F)isisomorphicto D.If H = {1}then χEmaybeextendedfrom
CE = CKtoaquasi­characterof W E/F.Inotherwords,thereisaquasi­character χFof CF
suchthat χE = χ E/F.Then
Ind(W K/F, W K/E, χE)
isequivalentto
⊕ µF ∈S(E/F) Ind(W K/F, W K/F, µFχF).
Suppose H = {1}.Since W K/M/CKiscyclicthereisaquasi­character χMof CMsuch
that χK = χ K/M. If w1belongsto W K/Elet χw1 bethecharacterof W K/Mor,whatisthe
same,of CMdefinedby
χw1 (w2) = ω(w1, w2)

Chapter18 256
andif w2belongsto W K/Mlet
Clearly
and
χw2 (w1) = ω(w1, w2).
{χw1 | w1 ∈ W K/E} = S(K/M)
{χw2 | w2 ∈ W K/M} = S(K/E).
If σ1istheimageof w1in Hand σ2theimageof w2in Dthen
and
χ σ2
E (w1) = χE(w2w1w −1
2 w−1
1 w1) = χ −1
w2 (w1) χE(w1)
χ σ1
M (w2) = χM(w1w2w −1
1 w−1
2 w2) = χw1 (w2) χM(w2).
Let W K/Fbethedisjointunion
ℓ
i=1 W K/Evi
with viin W K/M.Definethefunction ϕion W K/Fby
if w ∈ W K/Eand j = iandby
if w ∈ W K/E.Then
isabasisforthespace Uonwhich
ϕi(wvj) = 0
ϕi(wvi) = χE(w)
{ϕi | 1 ≤ i ≤ ℓ}
Ind(W K/F, W K/E, χE)
acts.Let ψi, 1 ≤ i ≤ ℓbethefunction W K/Fdefinedby
ψi(w2w1) = χM(w2)χ σ(vi)
E (w1)
if w1belongsto W K/Eand w2belongsto W K/M.Here σ(vi)istheimageof viin G(K/F).It
isnecessary,buteasy,toverifythat ψiiswell­defined.Thecollection
{ψi | 1 ≤ i ≤ ℓ}

Chapter18 257
isabasisforthespace Vonwhich
Ind(W K/F, W K/M, χM)
acts.Itiseasilyverifiedthattheisomorphismof Uwith V whichsends χM(vi)ϕito ψiisan
isomorphism.Thus
Ind(W K/F, W K/E, χE) ≃ Ind(W K/F, W K/M, χM).
Thistakescareofthefirstpartofthelemma.
Whether H = {1}ornot,
isequivalentto
Ind(WK/F, WK/E, 1)
⊕ µF ∈S(E/F) Ind(W K/F, W K/F, µF).
If H = 1wemayapplyTheorem2.1to E/Ftoseethat
λ(E/F, ψF) =
µF ∈S(E/F) ∆(µF, ψF).
If H = {1}thisequalityisjustthedefinitionoftheleftside.Inthiscasethesecondpartofthe
lemmaassertsthat
∆(χE, ψ E/F)
µF ∈S(E/F) ∆(µF, ψF) (18.1)
isequalto
µF ∈S(E/F) ∆(µFχF, ψF)
where χE = χ E/F.Thisisaconsequenceofthefirstmainlemma.If H = {1},Theorem2.1
appliedto M/F,showsthat
λ(M/F, ψF) =
µF ∈S(M/F) ∆(µF, ψF)
andthesecondpartofthelemmaassertsthat(18.1)isequalto
∆(χM, ψ M/F)
Thisisaconsequenceofthesecondmainlemma.
µF ∈S(M/F) ∆(µF, ψF).
Anon­trivialnilpotentgroupalwayscontainsasubgroup Dsatisfyingtheconditionsof
thepreviouslemma.Lemma17.1isclearif K = F.If K = Fand G(K/F)isnilpotentitisa
consequenceofthefollowinglemma.

Chapter18 258
Lemma 18.2
Suppose K/F isnormalandTheorem2.1isvalidforallnormalextensions K ′ /F ′ in
P(K/F)with [K ′ : F ′ ] < [K : F]. Suppose F ⊆ M ⊂ = Kand M/F isnormal. Suppose
E1, . . ., Er,
E ′ 1, . . ., E ′ sliebetween Fand M, χEiisaquasi­characterof CEi , χE ′ j isaquasi­characterof
CE ′ j ,and
⊕ r i=1Ind(WK/F, WK/Ei , χEi )
isequivalentto
Then r
isequalto
Therepresentation
s
⊕ s j=1 Ind(W K/F, W K/E ′ j , χ E ′ j ).
i=1 ∆(χEi , ψ Ei/F) λ(Ei/F, ψF)
j=1 ∆(χE ′ j , ψ E ′ j /F) λ(E ′ j
canbeobtainedbyinflatingtherepresentation
Ind(WK/F, WK/Ei , χEi )
Ind(WM/F, WM/Ei , χEi )
/F, ψF).
from W M/Fto W K/F.Asimilarremarkappliestotherepresentationsinducedfromthe χE ′ j .
Thus
isequivalentto
⊕ r i=1Ind(WM/F, WM/Ei , χEi )
⊕ s j=1Ind(W M/F, W M/E ′ j , χE ′ j ).
ApplyingTheorem2.1totheextension M/Fweobtainthelemma.

Chapter19 259
Chapter Nineteen.
Proof of the Main Theorem.
WeshallfirstproveLemma17.1whenthereisaquasi­character χF of CF suchthat
χK = χ K/F.ImplicitinthestatementofthefollowinglemmaasinthatofLemma17.1,isthe
assumptionthatTheorem2.1isvalidforallpairs K ′ /F ′ in P(K/F)forwhich [K ′ : F ′ ] <
[K : F].Recallthatwehavefixedanon­trivialabeliannormalsubgroup Cof G = G(K/F)
andthat Lisitsfixedfield.
Lemma 19.1
Suppose F ⊆ E ⊆ K, χFisaquasi­characterof CF, χEisaquasi­characterif CE,and
, 1 ≤ i ≤ r,
χ K/E = χ K/F.Therearefields F1, . . ., Frcontainedin Landquasi­characters χFi
suchthat χK/Fi
isequivalentto
and
isequalto
= χK/F,
r
Ind(W K/F, W K/E, χE)
⊕ r i=1Ind(WK/F, WK/Fi , χFi )
∆(χE, ψ E/F) λ(E/F, ψF)
i=1 ∆(χFi , ψ Fi/F) λ(Fi/F, ψF).
Weprovethelemmabyinductionon [K : F].Let Ebethefixedfieldof Handlet F ′ be
thefixedfieldof HC.If F ′ = Fthen,byinduction,therearefields F1, . . ., Frlyingbetween
F ′ and Landquasi­characters χF1 , . . ., χFr suchthat χ K/Fi = χ K/Fand
isequivalentto
Ind(WK/F ′, WK/E, χE)
⊕ r i=1Ind(WK/F ′, WK/Fi , χFi ).
Inthiscasethelemmafollowsfromthetransitivityoftheinductionprocess,theassumed
validityofTheorem2.1for K/F ′ andLemma16.3.

Chapter19 260
Wesupposehenceforththat G = HC. Thereisacharacter θEin S(K/E)suchthat
χE = θEχ E/F. θEmayberegardedasacharacterof H. If H ∩ C = {1}wemaydefinea
character θFof Gbysetting
θF(hc) = θE(h)
if hisin Hand cisin C. θFmayberegardedasacharacterof CFand θE = θ E/F.Replacing
χFby θF χFwesupposethat χE = χ E/F.TheninthenotationofLemma15.1,wemaytake
andif Fi = Fµ,
{F1, . . ., Fr} = {Fµ | µ ∈ T }
χFi = µ′ χ Fµ/F.
TheassertionsofthelemmaareconsequencesofLemmas15.1and15.3.
Wesupposenownotonlythat G = HCbutalsothat H ∩ C = {1}. Let Sbetheset
ofcharactersin S(K/L)whoserestrictionto H ∩ Cagreeswiththerestrictionof θE. Sis
invariantundertheactionof Hon S(K/L).If νbelongsto Slet ϕνbethefunctionon W K/F
definedby
ϕν(wv) = χE(w) χ L/F(v) ν(v)
if wisin W K/Eand visin W K/L. νisacharacterof Candmaythereforeberegardedasa
characterof W K/Lorof CL.Itiseasytoverifythat ϕνiswell­defined.If
then
ρ = Ind(W K/F, W K/E, χE)
{ϕν | ν ∈ S}
isabasisforthespaceoffunctionsonwhich ρacts.If wbelongsto W K/E
ρ(w)ϕν = χE(w)ϕν ′
with ν ′ = νσ−1is σistheimageof win G(K/F).If vbelongsto WK/L ρ(v)ϕν = χ L/F(v) ν(v)ϕν.
Thusif Risanorbitin Sundertheactionof H,thespace
VR =
ν∈R Cϕν
isinvariantunder W K/Fand ρisthedirectsumofitsrestrictionstothespaces VR.

Chapter19 261
If µbelongsto Rlet Hµbetheisotropygroupof µ,let Gµ = HµC,andlet Fµbethefixed
fieldof Gµ.Extend µtoacharacter µ ′ of Gµbysetting
µ ′ (hc) = θE(h) µ(c)
if hisin Hµand cisin C. µ ′ ,whichiseasilyseentobewell­defined,mayberegardedasa
characterof W K/Fµ of CFµ .Let W K/Fbethedisjointunion
s
i=1 W K/Fµ wi
with wiin W K/Eandlet σibetheimageof wiin G(K/F).Let ϕibethefunctionof W K/F
definedby
Thecollection
ϕi(wwj) = 0 w ∈ WK/Fµ , j = i
ϕi(wwi) = µ ′ (w)χFµ/F(w) w ∈ WK/Fµ .
{ϕi | 1 ≤ i ≤ s}
isabasisforthespace Vµonwhichtherepresentation
acts.Let
If wbelongsto W K/L
σµ = Ind(W K/F, W K/Fµ , µ′ χ Fµ/F)
ψi = χE(wi)ϕi.
σµ(w)ψi = µ σi (w) χL/F(w)ψi.
If wbelongsto W K/Eand wjw = vwiwith vin W K/Fµ then
σµ(w)ψi = χE(w)ψj.
Thustheisomorphismof Vµwith VRwhichtakes ψito ϕµ σ i commuteswiththeactionof
W K/F.If Tisasetofrepresentativesfortheorbitsin S
ρ ≃ ⊕µ∈Tσµ.
If K1isthefixedfieldof H ∩ Cthen K1/Fisnormaland ρistheinflationto W K/Fof
Ind(W K1/F, W K1/E, χE) (19.1)

Chapter19 262
and σµistheinflationof
Thustherepresentation(19.1)isequivalentto
ApplyingTheorem2.1to K1/Fweseethat
isequalto
Ind(W K1/F, W K1/Fµ , µ′ χ Fµ/F).
⊕µ∈TInd(W K1/F,W K1/Fµ , µ′ χ Fµ/F).
∆(χE, ψ E/F) λ(E/F, ψF)
µ∈T (µ′ χ Fµ/F, ψ Fµ/F) λ(Fµ/F, ψF).
Ifthereisaquasi­character χFsuchthat χK = χK/F,Lemma17.1followsfromLemma 18.2andthelemmajustproved. TocompletetheproofofTheorem2.1wehavetoprove
Lemma17.1when F = F(χK), Gisnotnilpotent,andthereisnoquasi­character χFof CF
suchthat χK = χK/F.Inthiscasenoneofthefields E1, . . ., Er, E ′ 1, . . ., E ′ sisequalto Fand
Theorem2.1maybeappliedto K/Eiand K/E ′ j .
Lemma 19.2
Suppose Aand Bliebetween Fand K.Suppose χAand χBarequasi­charactersof CA
and CBrespectively.TherearefieldsA1, . . ., Amlyingbetween AandK,fieldsB1, . . ., BmlyingbetweenBandK,elementsσ1,
. . ., σminG,andquasi­charactersχA1 , . . ., χAm , χB1 , . . ., χBm
suchthat Bi = A σi
i , χBi = χσi
Ai ,andsuchthatthetensorproduct
isequivalentto
andto
Let
Ind(W K/F, W K/A, χA) ⊗ Ind(W K/F, W K/B, χB)
⊕ m i=1Ind(WK/F, WK/Ai , χAi )
⊕ m i=1Ind(WK/F, WK/Bi , χBi ).
ρ = Ind(W K/F, W K/A, χA)
σ = Ind(W K/F, W K/B, χB).

Chapter19 263
Let αbetherestrictionof σto W K/Aand βtherestrictionof ρto W K/B.ByLemma2.3
and
Let W K/Fbethedisjointunion
ρ ⊗ σ ≃ Ind(W K/F, W K/A, χA ⊗ α)
ρ ⊗ σ ≃ Ind(W K/F, W K/B, χB ⊗ β).
m
i=1 W K/AwiW K/B.
If Uiisthespaceoffunctionsin U,thespaceonwhich ρacts,whicharezerooutsideofthe
doublecoset W K/AwiW K/Bthen Uiisinvariantunder β.Definethefield Bibydemanding
that
W K/Bi = W K/B ∩ w −1
i W K/Awi.
If σiistheimageof wiin G(K/F)let χ ′ Bi
offunctionsonwhich
Ind(WK/B, WK/Bi , χ′ Bi )
betherestrictionof χσi
A to W K/Bi .If U ′ i isthespace
acts,themapof Uito U ′ i whichsends ϕtothefunction ϕ′ definedby
ϕ ′ (w) = ϕ(wiw)
if wisin W K/Bisanisomorphismwhichcommuteswiththeactionof W K/B.Thus
and,if χBi = χ Bi/B χ ′ Bi ,
β ≃ ⊕ m i=1 Ind(W K/B, W K/Bi , χ′ Bi )
χB ⊗ β ≃ ⊕ m i=1Ind(WK/B, WK/Bi , χBi ).
Similarconsiderationsapplyiftherolesof Aand Bareinterchanged.Thedoublecoset
decompositionbecomes
m
i=1 WK/Bw −1
i WK/A and
W K/Ai = W K/A ∩ wiW K/Bw −1
i = wiW K/Bi w−1
i .
Thus Bi = A σi
i .Itisalsoclearthat χBi = χσi
Ai .

Chapter19 264
TocompletetheproofofLemma17.1weuseBrauer’stheoreminthefollowingform.
Therearefields F1, . . ., Fnlyingbetween Fand Ksuchthat G(K/Fk)isnilpotentforeach k,
characters χFk of CFk /N K/Fk CK,andintegers m1, . . ., mnsuchthat
1 ≃ ⊕ n k=1mk Ind(WK/F, WK/Fk , χFk ).
Sinceweareassumingthat Gisnotnilpotentnoneofthe Fkareequalto Fandwemayapply
Theorem2.1toeachoftheextensions K/Fk.
Weshallapplythepreviouslemmawith A = Ei, B = Fkandwith A = E ′ j , B = Fk. m
and Bℓwillbedenoted
willbedenotedby m(ik)or m ′ (jℓ). Aℓwillbedenotedby Eikℓor E ′ jkℓ
by Fikℓor F ′ jkℓ .Observethat
isequalto
andthat
isequalto
∆(χEikℓ , ψ Eikℓ/F λ(Eikℓ/F, ψF) (19.2)
∆(χFikℓ , ψ Fikℓ/F) λ(Fikℓ/F, ψF) (19.3)
∆(χ E ′ jkℓ , ψ E ′ jkℓ /F) λ(E ′ jkℓ /F, ψF) (19.4)
∆(χF ′ jkℓ , ψ F ′ jkℓ /F) λ(F ′ jkℓ /F, ψF). (19.5)
χEi mayberegardedasaone­dimensionalrepresentationof W K/Ei andassuchis
equivalentto
Therefore
and
isequalto
n
k=1
⊕ n k=1 ⊕m(ik)
ℓ=1 mk Ind(WK/Ei , WK/Eikℓ , χEikℓ ).
m(ik)
1 = n
k=1
m(ik)
ℓ=1 mk [Eikℓ : Ei]
∆(χEi , ψ Ei/F)
ℓ=1 {∆(χEikℓ , ψ Eikℓ/F) λ(Eikℓ/Ei, ψ Ei/F)} mk .
Multiplyingbothoftheseexpressionsby λ(Ei/F, ψF),weseethat
∆(χEi , ψ Ei/F) λ(Ei/F, ψF) (19.6)

Chapter19 265
isequalto
n
m(ik)
k=1 ℓ=1 {∆(χEikℓ , ψEikℓ/F) λ(Eikℓ/F, ψF)} mk . (19.7)
Thesameargumentestablishesthat
isequalto
n
k=1
m ′ (jk)
ℓ=1
∆(χ E ′ j , ψ E ′ j /F) λ(E ′ j /F, ψF) (19.8)
{∆(χ E ′ jkℓ , ψ E ′ jkℓ /F) λ(E ′ jkℓ /F, ψF)} mk . (19.9)
Wearetryingtoshowthattheproductover ioftheexpressions(19.6)isequaltothe
productover joftheexpressions(19.8). Itwillbeenoughtoshowthattheproductofthe
expressions(19.7)isequaltotheproductoftheexpressions(19.9).
and
Therepresentations
areequivalent.Therefore
r
i=1
⊕ r i=1 ⊕ m(ik)
ℓ=1 Ind(WK/Fk , WK/Fikℓ , χFikℓ )
⊕ s j=1 ⊕m′ (jk)
ℓ=1 Ind(W K/Fk , W K/F ′ jkℓ , χF ′ jkℓ )
m(ik)
ℓ=1 [Fikℓ : Fk] = s
m
j=1
′ (jk)
[F
ℓ=1
′ jkℓ
Denotethecommonvalueoftheseexpressionsby N(k).Moreover
isequalto
r
s
i=1
j=1
m(ik)
: Fk].
ℓ=1 ∆(χFikℓ , ψ Fikℓ/F) λ(Fikℓ/Fk, ψ Fk/F)
m ′ (jk)
Multiplyingbothoftheseexpressionsby
weseethat
isequalto
r
s
i=1
j=1
ℓ=1 ∆(χF ′ jkℓ , ψ F ′ jkℓ /F)λ(F ′ jkℓ/F, ψ Fk/F).
m(ik)
λ(Fk/F, ψF) N(k)
ℓ=1 ∆(χFikℓ , ψ Fikℓ/F) λ(Fikℓ/F, ψF) (19.10)
m ′ (jk)
ℓ=1 ∆(χF ′ jkℓ , ψ F ′ jkℓ /F) λ(F ′ jkℓ/F, ψF). (19.11)
Becauseoftheequalityof(19.2)and(19.3)theproductover ioftheexpressions(19.7)isequal
totheproductover kofofthe mkthpowersoftheexpressions(19.10). Theproductover j
oftheexpressions(19.9)isequaltotheproductover kofthe mkthpowersoftheexpressions
(19.11).Lemma17.1,andwithitTheorem2.1,isnowcompletelyproved.

Chapter20 266
Chapter Twenty.
Artin L-functions.
Suppose ω isanequivalenceclassofrepresentationsoftheWeilgroupofthenonarchimedeanlocalfield
F. Let KbeaGaloisextensionof F andlet σbearepresentation
of W K/Fintheclass ω. Suppose σactson V. Let V 0 bethesubspaceof V fixedbyevery
elementof W 0 K/F .Since W 0 K/F isanormalsubgroupof W K/Fthespace V 0 isinvariantunder
WK/Fandon V 0wegetarepresentation σ0 .Since W 0 −1
K/F = τK/F (u0F )theclassof σ0depends onlyon w. σ0breaksupintothedirectsumof1­dimensionalrepresentationscorresponding tounramifiedgeneralizedcharacters µ1, . . ., µrof CF.Weset
L(s, w) = r
i=1
1
1 − µi(πF) |πF | s.
Thiswetakeasthelocalfunction. Itisclearthatwhen wisone­dimensionalthepresent
definitionagreeswiththatoftheintroductionandthatof ω = ω1 ⊕ ω2.Then
L(s, ω) = L(s, ω1) ⊕ L(s, ω2).
Suppose F ⊆ E ⊆ K, ρisarepresentationof W K/E,and
Wehavetoshowthatif θistheclassof ρthen
σ = Ind(W K/F, W K/E, ρ).
L(s, ω) = L(s, θ).
Let ρacton W.Then Visthespaceoffunctions fon W K/Fwithvaluesin Wwhichsatisfy
f(uv) = ρ(u)f(v)
for uin W K/Eand vin W K/F.If fliesin V0and uliesin W 0 K/E then
ρ(u)f(v) = f(uv) = f(vv −1 uv) = f(v)
because v −1 liesin W 0 K/F .Thus ftakesvaluesin W 0 .Inotherwords,wemayaswellassume
that W = W 0 .Indeedwemayaswellgofurtherandassumethat W = W 0 hasdimension
one.

Chapter20 267
Let N E/F πE = επ f
F where εisaunitandchoose w0in W K/Fsothat τ K/Fw0 = πF.
Then w f = u0v0with u0in W 0 K/F and v0in W K/Esuchthat τ K/Ev0 = πE. Clearly, V 0
consistsofthefunctions fwithvaluesin Wwhichsatisfy f(uw) = f(w)for uin W 0 K/F and
f(uw) = µ(τ K/Eu)f(w)if ρisassociatedtothegeneralizedcharacter µof CE.Takeasbasis
of V 0 thefunctions ϕ0, . . ., ϕf−1definedby
ϕi(uvw j
0 ) = µ(τ K/Ev)δ j
i x
where xisanon­zerovectorin W, ubelongsto W 0 K/F , vbelongsto W K/E, 0 ≤ j < f,and δ j
i
isKronecker’sdelta.Thematrixof σ(w0)withrespecttothisbasisis
and
since |πF | f = |πE|.
L(s, ω) =
⎛
0 ·
⎜
1 0
⎜
A = ⎜ 1
⎜
⎝
· ·
. ..
. .. . ..
⎞
µ(τK/Ev0) · ⎟
· ⎟
· ⎠
0 1 0
1
det(I − A |πF | s ) =
1
= L(s, θ)
fs 1 − µ(πE) |πF |
Forarchimedeanfieldsweproceedinadifferentmanner.Ifwewrite ω,aswemay,asa
sumofirreduciblerepresentationsthecomponentsareuniqueuptoorder.If ω = r i=1 ⊕ωi,
wewillhavetohave
L(s, ω) = r
L(s, ωi).
Thusitisamatterofdefining L(s, ω)forirreducible ω.If ωisone­dimensionalthiswasdone
intheintroduction.If ωisnotone­dimensionalthen Fmustbe R.Let σbearepresentation
of W C/Rintheclass ω. W C/Risanextensionofthegroupoforder2by C ∗ . Let W C/R =
C × ∪ w0C × . If σactson V thereisanon­zerovector xin V andageneralizedcharacter µ
of C × suchthat σ(a)x = µ(a)xforall ain C × . Thenthespacespannedby {x, σ(w0)x}is
invariantandthereforeallof V.Since Visnotone­dimensional σ(w0)xisnotamultipleof x.
Noticethat σ(a)σ(w0)x = σ(w0)σ(w −1
0 aw0)x = µ(a)σ(w0)x.If
i=1
µ(z) = |z| r zm z n
|z| m+n
2

Chapter20 268
with m + n ≥ 0, mn = 0weset
m+n
−(s+r+
L(s, ω) = 2(2π) 2 ) Γ s + r +
m + n
.
2
Theinitialchoiceof µisofcoursenotuniquelydetermined.Howeverif µ0isonechoicethe
onlyotherchoiceisthecharactera → µ0(a).ThustheresultinglocalL­functionisindependent
ofthechoice.
TheonlypointtobecheckedisthatthelocalL­functionbehavesproperlyunderinduction.
Wehavetoverifythatif ρisarepresentationof C ∗ = W C/Cintheclass θand
σ = Ind(W C/R, W C/C, ρ)
isintheclass wthen L(s, w) = L(s, θ). Wemayaswellassumethat ρisirreducibleand
thereforeone­dimensional.Letitcorrespondtothegeneralizedcharacter ν.If σisirreducible
wecouldchoosethegeneralizedcharacter µabovetobe νandtheequalityofthetwo Lfunctionsbecomesamatterofdefinition.
If σisirreducibleitbreaksupintothesumoftwo
one­dimensionalrepresentatives.Itfollowseasilythat ν(a) = ν(a)forall a.Thus νisofthe
form ν(a) = |a| r and
L(s, θ) = 2(2π) −(s+r) Γ(s + r).
If µR = µisthegeneralizedcharacter x → |x| r of R × then ν = µ C/Rand,aswesawinchapter
10,therepresentation σisequivalenttothedirectsumoftheone­dimensionalrepresentations
correspondingto µandto µ ′ where µ ′ (x) = sgnxµ(x).Thus
L(s, ω) =
π −1
2 (s+r) Γ
s + r
2
π −1
2 (s+r+1) Γ
s + r + 1
Therequiredresultisthusaconsequenceofthefamiliarduplicationformula
2 2z−1 Γ(z) Γ(z + 1/2) = π 1/2 Γ(2z).
If Fisaglobalfieldand ωisanequivalenceclassofrepresentationsoftheWeilgroupof
F,wedefineasintheintroduction,theglobal L­functiontobe
L(s, ω) =
p
L(s, ωp).
Irepreatthattheproductistakenoverallprimes,includingthoseatinfinity.Itisnotdifficult
toseethattheproductconvergesinahalf­planeRe s > c. Oneneedonlyverifyitfor ω
2
.

Chapter20 269
irreducible. ChooseaGaloisextension Kof Fsothatthereisarepresentation σof WK/F intheclass ω. Therestrictionof σto CKisequivalenttothedirectsumof1­dimensional
representationscorrespondingtogeneralizedcharacters µ (1) , . . ., µ (r) of CK.Foreach iand
jthereisaσin G(K/F)suchthat µ j (a) ≡ µ i (σ(a)).Then |µ −1
i µj(a)| = |µ i (a−1σ(a))| = 1
because a −1 σ(a)belongstothecompactgroupof iidèleclassesofnorm1.Let |µ ′ (a)| = |a| r .
Let νF bethegeneralizedcharacter a → |a| r of CF. Replacing σby ν −1
F
⊗ σwereplace
L(s, wp)by L(s − r, wp)and µ (i) by |µ (i) | −1 µ (i) . Thuswemayaswellsupposethatall µ (i)
areordinarycharacters.Since CKisoffiniteindexin W K/Ftheeigenvaluesof σ(w)willall
haveabsolute1forany win W K/Fandatanynon­archimedeanprimethelocal L­function
willbeoftheform
s
i=1
1
1 − αi |πFp |s
with s ≤ dimwand |αi| = 1, 1 ≤ i ≤ s. Therequiredresultfollowsfromthewell­known
factthat
1 1
1 − |πFp |s
convergesfromRe s > 1.Thisproductistakenonlyoverthenon­archimedeanprimes.

Chapter21 270
Chapter Twenty-one.
Proof of the Functional Equation.
Chooseanon­trivialcharacter ψFof AF/F.Beforewecanwritedownthefactorappearinginthefunctionalequationoftheglobal
L­functionwehavetoverifythat ε(s, ωγ, ψFγ ) = 1
forallbutafinitenumberof γ.
Let ωberealizedasarepresentation σof W K/F andlettherestrictionof σto CKbe
equivalenttothedirectsumof1­dimensionalrepresentationscorrespondingtothegeneralized
characters µ (1) , . . ., µ (r) .Allbutfinitelymanyprimes pwillsatisfythefollowingconditions.
(i) pisnon­archimedean.
(ii) n(ψFp ) = 1.
(iii) pdoesnotramifyin K.
(iv) m(µ (i)
P ) = 0forall Pdividing pandall i.
Chooseonesuch pandlet Pdivide p.Correspondingtothemap K/F −→ KP/Fpisa
map ϕp : WKP/Fp −→ WK/F. ωpistheclassof σp = σ ◦ ϕp.Thekernelof σpcontains UKP .
Since KP/Fpisunramifiedthequotientof WKP/Fpby UKPisabelianand σpisthedirect
sumofone­dimensionalrepresentations. Letthemcorrespondtothegeneralizedcharacters
ν (1)
p , . . ., ν (r)
p of CFp .Since τKP/Fptakes UKPonto UFpeachofthesecharactersisunramified. Thus
r
ε(s, ωp, ψFp ) =
i=1 ∆
α s−1 2
Fp ν(i)
p , ψFp
= 1.
Ifψ ′ Fisanothernon­trivialcharacterof AF/FthereisaβinF ∗suchthatψ ′ F (x) ≡ ψF(βx).
AccordingtoLemma5.1
Since
thefunction
ε(s, ω, ψFp
isindeedindependentof ψF.
p
2 ) = αs−1
Fp (β)det ωp(β) ε(s, ω, ψFp ).
αFp (β)s−1 2 det ωp(β) = |β| s−1 2 det ω(β) = 1
ε(s, ω) =
p ε(s, ωp, ψFp )

Chapter21 271
WecaninferfromTate’sthesisnotonlythat L(s, ω)ismeromorphicinthewholecomplex
planeif ωisone­dimensionalbutalsothatitsatisfiesthefunctionalequation
L(s, ω) = ε(s, ω) L(1 − s, ω)
if ωiscontragredientto ω.Asiswell­known,Lemma2.2thenimpliesthat L(s, ω)ismeromorphicinthewholecomplexplaneforany
ω.Inanycase,TheoremBistrueforone­dimensional
ωand,grantingthis,wehavetoestablishitingeneral.
Firstweneedalemma.
Lemma 21.1
Suppose Fisaglobalfield, KisaGaloisextensionof F, Eisafieldlyingbetween Fand
K, χisageneralizedcharacterof CEand
σ = Ind(W K/F, W K/E, χ).
If ωistheclassof σand,foreachprime qof E, χqistherestrictionof χto CEqthenforeach prime pof F
ε(s, ωp, ψFp ) = {ε(s, χq, ψEq/Fp ) ρ (Eq/Fp, ψFp )}.
q|p
Let Pbeaprimeof Kdividing p.Thefirststepistofindasetofrepresentativesforthe
doublecosets W K/E w W KP/Fp . Since CK ⊆ W K/Eisanormalsubgroupof W K/Fwecan
factorout CKandmerelyfindasetofrepresentativesforthedoublecosets
G(K/E)σ G(KP/Fp).
Let P1, . . ., Prbetheprimesof Kdividing pandlet P1divide qiin E. G(K/F)isthedisjoint
union r
i=1 σiG(KP/Fp)
where σi(P) = Pi.If σiand σjbelongtothesamedoublecoset qi = qj.Conversely,if qi = qj
thereisaρin G(K/E)suchthat ρ(Pi) = Pj.Then ρσi(P) = σj(P)and
ρσi ∈ σj G(KP/Fp).
Thuswemaywrite G(K/F)asthedisjointunion
τ∈S
G(K/E)τ G(KP/Fp)

Chapter21 272
sothatif Pdivides qin Ethecollection {τ(q) | τ ∈ S}isthecollectionofdistinctprimesin E
dividing p.
Foreach τin Schoosearepresentative w(τ)in W K/F.Foreach τin Stherestrictionof σ
to W KP/Fp leavesinvariantthespaceoffunctions fonthedoublecoset W K/Ew(τ)W KP|Fp
whichsatisfy f(vw) = χ(τ K/E(v))f(w)forall vin W K/E.Therepresentationof W KP/Fp on
thisspaceisequivalentto
Ind(W KP/Fp , W KP/E τ q τ , χ′ q τ
if E τ = τ −1 (E)and
Thus
whichisofcourseequalto
Weset
χ ′ qτ(a) = χ(τ(a)).
ε(s, ωq, ψFp ) =
τ∈S ε(s, χ′ qτ, ψEτ qτ /Fp ) ρ(Eτ qτ/Fp, ψFp )
τ∈S ε(s, χq, ψ Eq/Fp ) ρ(Eq/Fp, ψFp ).
ρ(E/F) =
q
q|p ρ(Eq/Fp, ψFp ).
TheprecedingdiscussiontogetherwithLemma5.1showsthatitdoesnotdependon ψF.
Howeverthatdoesnotreallymattersinceweareabouttoshowthatforanychoiceof ψFitis
1.Observefirstofallthatthepreviouslemmaimpliesimmediatelythatif ωistheclassof
then
σ = Ind(W K/F, W K/E, χ)
ε(s, ω) = ε(s, χ) ρ(E/F).
Givenanarbitraryclass ωrealizableasarepresentationof WK/F wecanfindfields
E1, . . ., ErlyingbetweenFandK,generalizedcharactersχE1 , . . ., χEr ,andintegersm1, . . ., mr
suchthat r
i=1 ⊕mi Ind(WK/F, WK/Ei , χEi )
isintheclass ω.Then
ε(s, ω) = r
i=1 {ε(s, χEi )mi ρ(Ei/F) mi }.

Chapter21 273
and
Since
wehave
Ontheotherhand
because ωcontains r
Consequently
L(s, ω) = r
L(s, ω) = r
i=1
i=1
L(s, χEi )mi
L(s, χ−1
Ei )mi .
L(s, χEi ) = ε(s, χEi ) L(1 − s, χ−1
Ei )
L(s, ω) = r
i=1 ε(s, χEi )mi L(1 − s, ω)
i=1 ⊕mi Ind(W K/F, W K/Ei
r
i=1
ε(s, χEi )mi
, χ−1
Ei ).
dependsonlyon ωandnotontheparticularwayitiswrittenasasumofinducedrepresentations.Thus
r
mi ρ(Ei/F)
i=1
alsodependsonlyon ω.Wecallit H(ω).ItisclearthattoproveTheoremBwehavetoshow
that H(ω) = 1forall ωor,whatisthesame,that ρ(E/F) = 1forall Eand F.
q ′ .Then
Suppose F ⊆ E ⊆ E ′ .Denotetheprimesof Fby p,thoseof Eby q,andthoseof E ′ by
ρ(E ′ /F) =
p
ApplyLemma4.5toseethattherightsideequals
p
q|p
q ′ |q
q ′ |p ρ(E′ q ′/Fp, ψFp ).
′
ρ(E q ′/Eq, ψFq/Fp ) ρ(Eq/Fp, ψFp )[E′ q :Eq] .
Since
[E ′ q ′ : Eq] = [E ′ : E]
thismaybewrittenas
q
q ′ |q
q ′ |q ρ(E′ q ′/Eq, ψ Fq/Fp )
p
q|p ρ(Eq/Fp, ψFp )
′
[E :E]

Chapter21 274
whichisofcourse
ρ(E ′ /E) ρ(E/F) [E′ :E] . (20.1)
Suppose E/Fisanabelianextensionand ωistheclassoftherepresentationof W E/F
inducedfromthetrivialrepresentationof CE = W E/E. Then H(ω) = ρ(E/F). Onthe
otherhand, ωisthedirectsumof [E : F]one­dimensionalrepresentations; so H(ω) =
ρ(F/F) [E:F] = 1. Itfollowsimmediatelynotonlythat ρ(E/F) = 1if E/Fisabelianbut
alsothat ρ(E/F) = 1if Ecanbeobtainedfrom Fbyasuccessionofabelianextensions. In
particularif F ⊆ E ⊆ Land L/Fisnilpotent ρ(E/F) = 1.
Observethat(20.1)togetherwithLemma2.2andthetransitivityofinductionimplythat
if ωistheclassof
σ = Ind(W K/F, W K/E, ρ)
and θistheclassof ρthen
H(ω) = H(θ) ρ(E/F) dim θ .
Tocompletetheproofwewillshowthat H(ω1 ⊗ ω2) = H(ω2) dim ω1 forall ω1and ω2.Taking
ω2 = 1wefind H(ω1) = 1. Itisenoughtoprovetheequalitywhen ω1and ω2areboth
realizableasrepresentationsof W K/F andthereisafield Elyingbetween Eand Kwith
G(K/E)nilpotentandageneralizedcharacter χEsuchthat ω2istheclassof
Ind(W K/F, W K/E, χE).
Then H(ω2) = ρ(E/F). If ρisarepresentationintherestrictionof ω1to W K/Ethen,by
Lemma2.3, ω1 ⊗ ω2istheclassof
Ind(W K/F, W K/E, ρ ⊗ χE).
Let θbetheclassof ρ ⊗ χE. H(θ)isoftheform
r
i=1
where E ⊆ Ei ⊆ Kandistherefore1.Thus
asrequired.
ρ(Ei/E) mi
H(ω1 ⊗ ω2) = H(θ)ρ(E/F) dim θ = ρ(E/F) dim ω1 .

Chapter22 275
Chapter Twenty-two.
Appendix.
Thereisclearlynotmuchtobesaidaboutthefunctionsε(s, ω, ψF)whenFisarchimedean.
Howeverfornon­archimedean Ftheirpropertiesaremoreobscure.Inthisappendixweshall
describeandprovesomepropertieswhichwerenotneededintheproofsofthemaintheorems
andsofoundnoplaceinthemainbodyofthepaperbutwhichwillbeusedelsewhere.
ThefirststepistodefinetheArtinconductorof ω. Wefollowawell­troddenpath. If
KisafiniteGaloisextensionofthelocalfield Fthen W 0 K/Fcontains UKasasubgroupof
finiteindexandisthereforecompact. Itis,infact,amaximalcompactsubgroupof WK/F. ChoosethatHaarmeasure dwon WK/Fwhichassignsthemeasure1to W 0 K/F .If fisalocally
constantfunctionon WK/Fand uisanon­negativerealnumberset
ˆf(u) =
W u
dw
K/F
−1
W u
{f(1) − f(w)}dw.
K/F
Since W u K/Fisanopensubgroupof WK/Fitismeaningfultorestrict dwtoit. ˆ f(u)isbounded,
continuousfromtheleft,and0for usufficientlylarge. Since W u K/F = W 0 K/Ffor 1 < u ≤ 0
wehave ˆ f(u) = ˆ f(0)forsuch u.Theintegral
iswell­defined.
∞
−1
ˆf(u)du
Therearesomesimplelemmastobeverified.
Lemma 22.1
Suppose F ⊆ K ⊆ Land L/FisalsoaGaloisextension.Define gon W L/Fby g(w) =
f (τL/F ,K/F(w)).Then ˆg(u) = ˆ f(u)forall u.
ThisisimmediatebecausebyLemma6.16, τ L/F,K/Fmaps W u L/F onto W u K/F
Whenwewanttomaketherolesof Kand Fexplicitwewrite ˆ f(u) = ˆ f K/F(u).
Lemma 22.2
forevery u.

Chapter22 276
Suppose F ⊆ E ⊆ Kand gisafunctionon W K/Esatisfying g(wzw −1 ) = g(z)forall z
and win W K/E.Regard W K/Easasubgroupof W K/Fandset
If NE/F πEisaunittimes π f E/F
F
∞
−1
f(w) =
and P δ E/F
E
ˆf K/F(u)du = f E/F
z∈W K/E\W K/F
∞
g(z −1 wz).
isthedifferentof E/F
−1
ˆg K/E(u)du + f E/Fδ E/Fg(1).
Let dw K/FbethenormalizedHaarmeasureon W K/Fandlet dw K/Ebethenormalized
Haarmeasureon W K/E.On W K/E
dw K/E = [W 0 K/F : W 0 K/E ]dw K/F.
Supposeatfirstthat g(1) = 0.Denotealsoby gthefunctionon W K/Fwhichequalsthegiven
gon W K/Ebutis0outsideof W K/E.Then
Since W u K/F ∩ W K/E = W v K/E if v = ψ E/F(u)
Recallthat
Moreover
ˆf K/F(u) = [W K/F : W K/E] ˆg K/F(u).
ˆg K/F(u) = −[W 0 K/F : W u K/F ]
= − [W 0 K/F : W u K/F ]
[W 0 K/F : W 0 K/E ]
=
1
[W 0 K/F : W 0 K/E ]
W u
K/F
W v
K/E
f E/F = [W K/F : W K/E]
[W 0 K/F : W 0 K/E ].
g(w)dw K/F
g(w)dw K/E
[W 0 K/F : W u K/F ]
[W 0 K/E : W v K/E ] ˆg K/E(v).
[W 0 K/F : W u K/F ]
[W 0 K/E : W v K/E ] = [W 0 K/F : W u K/F U0 K ]
[W 0 K/E : W v K/E U0 K ] · [U0 K : U0 K ∩ W u K/F ]
[U 0 K : U0 K ∩ W v K/E ]

Chapter22 277
and U 0 K ∩ W u K/F = U0 K ∩ W v K/E .ByLemma6.11thefirstterminthisproductisequalto
if G = G(K/F)and H = G(K/E).But
and
while
Thus
∞
−1
[G 0 : G u ]
[H 0 : H v ]
[G 0 : G u ] = ψ ′ K/F (u)
[H 0 : H v ] = ψ ′ K/E (v)
ψ ′ K/F (u) = ψ′ K/E (v)ψ′ E/F (u).
ˆf K/F(u)du = f E/F
= f E/F
∞
−1
∞
−1
ˆg K/E(ψ E/F(u)) ψ ′ E/F (u)du
ˆg K/E(v)dv.
Tocompletetheproofofthelemma,wehavetoshowthatif g(w) = 1sothat ˆg K/E(u) ≡ 0
then ∞
Inthiscase
−1
ˆf K/F(u) = [G : H] −
ˆf K/F(u)du = f E/F δ E/F.
[G : H]
[G 0 : H 0 ]
if v = ψ E/F(u).Aftersomesimplerearrangingthisbecomes
Thefactor
[G : H]
[G u : 1] {[Gu : 1] − [H v : 1]} =
and,fromparagraphIV.2of [ ],
∞
1
[G 0 : G u ]
[H 0 : H v ]
[G : H]
[G u : 1] {([Gu : 1] − 1) − ([H v : 1] − 1)}.
[G : H]
[G u : 1] = [G : G0 ]
[H : 1] ψ′ K/F (u)
([G u : 1] − 1)ψ ′ K/F (u)du =
∞
−1
([Gx : 1] − 1)dx = δ K/F

Chapter22 278
while ∞
−1
Thus ∞
because
([H v : 1] − 1) ψ ′ K/F (u)du =
−1
∞
−1
([H v : 1] − 1) ψ ′ K/E (v)dv = δ K/E.
ˆf K/F(u)du = [G : G0 ]
[H : 1] (δ K/F − δ K/E) = f E/F δ E/F
δ K/F = δ K/E + [H0 : 1] δ E/F.
Suppose ωisanequivalenceclassofrepresentationsoftheWeilgroupof Fand σisa
representationof WK/Fintheclassof ω.Let fσbethecharacterof σ.ItfollowsfromLemma
22.1thatthevalueof ∞
ˆfσ(u)du
−1
dependsonlyon ωandnoton σ.Wecallittheorderof ωanddenoteitby m(ω).Since ˆ fσ(u)is
clearlynon­negativeforall uandvanishesidenticallyifandonlyif W 0 K/F iscontainedinthe
kernelof σ,theorder m(ω)isalwaysnon­negativeandequalszeroifandonlyifthekernelof
eachrealization σof ωcontains W 0 K/F .
Lemma 22.3
(a)If ω = ω1 ⊕ ω2then m(ω) = m(ω1) + m(ω2).
(b)If
then
(c) m(ω)isanon­negativeinteger.
ω = Ind(W K/F, W K/E, ν)
m(ω) = f E/Fm(ν) = f E/F δ E/F dimν.
Thefirstpropertyisimmediate.ThesecondisaconsequenceofLemma22.2. Toverify
thethirdwemerelyhavetoshowthat m(ω)isintegral. If ω = µ ⊕ νandtheassertionis
trueforanytwoof µ, νand ωitistrueforthethird. Thisobservation,togetherwithpart
(b)andLemma2.2,showsthatitisenoughtoverify(c)when ωistheone­dimensionalclass
correspondingtoageneralizedcharacter χFof CF.Todothisweshowthat m(ω) = m(χF).
If f(a) = χF(a)for ain CF = WF/Fthen ˆ f(u) = ˆ f(m)for m − 1 < u ≤ mand
{1 − χF(a)}da.
ˆf(m) = [U 0 F : Um F ]
U m F

Chapter22 279
Therightsideis1if m < m(χF)and0if m ≥ m(χF).Thus
m(ω) =
m(χF )−1
−1
du = m(χF).
Thefunction ω −→ m(ω)ischaracterizedby(a)and(b)togetherwiththefactthat
m(ω) = m(χF)if ωistheclassof χF.
Lemma 22.4
If ωisanequivalenceclassofrepresentationsoftheWeilgroupofthenon­archimedean
localfield Fand ψFisanon­trivialadditivecharacterof Fset m ′ (ω) = m(ω) +n(ψF) dimω.
Thereisanon­zerocomplexconstant a(ω)suchthat,asafunctionof s,
ε(s, ω, ψF) = a(ω) |πF | m′ (ω)s .
If ω = µ ⊕ νandthelemmaistrueforanytwoof µ, ν,and ω,itistrueforthethird.
ApplyingLemma2.2weseethatitisenoughtoverifyitwhen ωcontainsarepresentation
Then
Clearly
But
∆(α s−1
2
αE
Ind(W K/F, W K/E, χE).
ε(s, ω, ψF) = ∆(α s−1
2
E χE, ψ E/F) ρ(E/F, ψF).
E χE, ψE/F) = α s−1
2
E π m(χE)+δE/F E
π m(χE)+δ E/F
E
π n(ψF)
F = αF NE/F andtheargumentontherightistheproductofaunitand
Thelemmafollows.
π f E/F(m(χE)+δ E/F)+n(ψF ) dim ω
F
π n(ψF)
F ∆(χE, ψE/F). π m(χE)+δ E/F
E
= π m′ (ω)
F .
π n(ψF
)
F
Thenextlemmaisrathertechnicalandtoproveitwewillhavetousethenotationsand
resultsofparagraphs8and9.
Lemma 22.5

Chapter22 280
Let ωbeanequivalenceclassofrepresentationsoftheWeilgroupofthenon­archimedean
localfield F and m1apositiveinteger. Thereisapositiveinteger m2suchthatif χF and
µ1, . . ., µrwith r = dimω,aregeneralizedcharactersof CF and m(χF) ≥ m2, m(µi) ≤
m1, 1 ≤ i ≤ r,while r
i=1 µi = detω
thenforanynon­trivialadditivecharacter ψF
ε(s, χF ⊗ σ, ψF) = r
i=1 ε(s, µiχF, ψF).
Choose,asastart, m2 ≥ 2m1+1.If µFisageneralizedcharacterof CFand m(µF) ≤ m1
while m(χF) ≥ m2then m(µFχF) = m(χF) = m. Let n = n(ψF)andchoose γsothat
OFγ = P m+n
F . If β = β(χF)wemaychoose β(µFχF) = β. AppealingtoLemmas8.1and
9.4weseethat
Inparticular
r
If ω = µ ⊕ νthen
ε(s, µFχF,ψF) = ∆(α s−1 2
F µFχF, ψF)
= (α s−1 2
F µF)
γ
∆(χF, ψF).
β
i=1 ε(s, µiχF,ψF) = α r(s−1
F
2 )
γ
detω
β
γ
β
{∆(χF, ψF)} r .
ε(s, χF ⊗ ω, ψF) = ε(s, χF ⊗ µ, ψF) ε(s, χF ⊗ ν, ψF)
andallthreetermsaredifferentfromzero.Thusifthelemmaistruefortwoof µ, νand ωitis
trueforthethird.UsingLemma2.2onceagain,weseethatitisenoughtoprovethelemma
whenthereisanintermediatefield Eandageneralizedcharacter µEof CEsuchthat ωisthe
classof
Ind(W K/F, W K/E, µE).
Then χF ⊗ ωistheclassof
and
Ind(W K/F, W K/E, µE χ E/F)
ε(s, χF ⊗ ω, ψF) = ∆(α s−1
2
E µEχ E/F, ψ E/F) ρ(E/F, ψF).
Therearetwosimplelemmaswhichweneedbeforewecanproceedfurtherandwe
digresstoprovethem.

Chapter22 281
Lemma 22.6
Let Ebeaseparableextensionof F.If missufficientlylarge
if e E/Fistheindexoframificationof Fin E.
ψ E/F(m − 1) + 1 = me E/F − δ E/F
Suppose F ⊆ E ⊆ Kwhere K/FisGaloisandtheassertionistruefor K/Fand K/E.
Subtracting1frombothsidesoftheequation,applying ψ K/E,andthenadding1,weobtain
theequivalentequation
Byassumption,theleftsideequals
andtherightsideequals
ψ K/F(m − 1) + 1 = ψ K/E(me E/F − δ E/F − 1) + 1.
me K/F − δ K/F
(me E/F − δ E/F)e K/E − δ K/E.
Since e K/F = e K/Ee E/Fand δ K/F = δ K/E + e K/Eδ E/Fthesetwoexpressionsareequaland
wehaveonlytoprovethelemmaforGaloisextensions.
Suppose F ⊆ K ⊆ Land L/Fand K/FareGalois.Supposealsothatthelemmaistrue
for L/Kand K/F.Then
ψ L/F(m − 1) + 1 = ψ L/F(ψ K/F(m − 1)) + 1
= ψ L/F(me K/F − δ K/F − 1) + 1
= (me K/F − δ K/F) e L/K − δ L/K
= me L/F − δ L/F
asbefore. Thus,ifweuseinduction,weneedonlyverifythelemmadirectlyforaGalois
extension K/Fofprimedegree.
WeapplyLemma6.3.If K/Fisunramified, e K/F = 1and δ K/F = 0while ψ K/F(m −
1) = m − 1; sotherelationfollows. If K/F isramifiedthereisaninteger tsuchthat
δ K/F = ([K : F] − 1)(t + 1)while ψ K/F(m − 1) + 1 = [K : F]m − ([K : F] − 1)(t + 1)for
m − 1 ≥ t.Since e K/F = [K : F]therelationfollowsagain.

Chapter22 282
If n = n(ψF)then
n ′ = n(ψ E/F) = n e E/F + δ E/F.
Thusif missufficientlylargeand m ′ = ψ E/F(m − 1) + 1
andif OFγ = P m+n
F
asinparagraph8.
Lemma 22.7
m ′ + n ′ = (m + n) e E/F
then OEγ = P m′ +n ′
E .Wedefine
P ∗ E/F (x) = P ∗ E/F (x; γ, γ)
If m1isagivenpositiveintegerthenfor msufficientlylarge
P ∗ E/F (x) ≡ x (mod Pm1
E ).
Asinparagraph8,let dbetheintegralpartof m
2 , d′ theintegralpartof m′
2 ,andlet
m = 2d + ε, m ′ = 2d ′ + ε ′ . P ∗ E/F (x)dependsonlyontheresidueof xmodulo Pd F andis
onlydeterminedmodulo P d′
E .Recallthatif
for yin P d′ +ε ′
E
then
Toshowthat P ∗ E/F
ψ E/F
P E/F(y) = N E/F(1 + y) − 1
∗ PE/F (x)y
γ
= ψF
x PE/F(y)
(x) ≡ x (modPm1 E )when missufficientlylarge,wehavetoshowthat
ψF
for yin P m′ −m1
E .Todothisweshowthat
x PE/F(y)
= ψE/F γ
γ
xy
γ
P E/F(y) ≡ S E/F(y) (modP m F )
when missufficientlylargeand yisin P m′ −m1
E .
.

Chapter22 283
Toputitanotherway,wehavetoshowthatif K/FisanyGaloisextensiontheassertion
istrueforallintermediatefields E.Forthisweuseinductionon [K : F]togetherwithLemma
3.3.Therearethreefactstoverify:
(i)If E/FisaGaloisextensionofprimedegreethen
P E/F(y) ≡ S E/F(y) (modP m F )
when missufficientlylargeand yisin P m′ −m1
E .
(ii)Suppose F ⊆ E ⊆ Kand K/FisGalois.Let G = G(K/F)andlet Ebethefixedfield
of H.Suppose H = {1}and G = HCwhere H ∩ C = {1}and Cisanon­trivialabelian
normalsubgroupof Gwhichiscontainedineveryothernon­trivialnormalsubgroup.If
theinductionassumptionisvalid
P E/F(y) ≡ S E/F(y) (modP m F )
when missufficientlylargeand yisin P m′ −m1
E .
(iii)Suppose F ⊆ E ⊆ E ′ ⊆ Kand m ′′ = ψ E ′ /F(m − 1) + 1.If,foranychoiceof m1,
P E/F(y) ≡ S E/F(y) (modP m F )
when missufficientlylargeand yisin P m′ −m1
E
and,foranychoiceof m ′ 1 ,
P E ′ /E(y) ≡ S E ′ /E(y) (modP m′
E )
when m,or m ′ ,issufficientlylargeand yisin P m′′ −m ′
1
E ′ then,foranychoiceof m ′ 1 ,
P E ′ /F(y) ≡ S E ′ /F(y) (modP m F )
if missufficientlylargeand yisin P m′′ −m ′
1.
Wefirstverify(i)for E/Funramified.ByparagraphV.2of[12]
E ′
P E/F(y) = N E/F(1 + y) − 1 ≡ S E/F(y) (modP m F )
if ybelongsto P d′ +ε ′
E .Inthiscase m = m ′ andwetake m > 2m1sothat m ′ −m1 > d ′ +ε ′ .If
E/Fisramifiedandofdegree ℓweagainchoose msufficientlylargethat m ′ > 2m1.If m > t
2(m ′ − m1) + (ℓ − 1)(t + 1)
ℓ
≥ m′ + (ℓ − 1)(t + 1)
ℓ
= m

Chapter22 284
sothatbyChapterVof[12],
if ybelongsto P m′ −m1
E
P E/F(y) ≡ S E/F(y) + N E/F(y) (modP m F )
. tofcoursehasitsusualmeaning.Since N E/F(y)belongsto P m′ −m1
F
allwehavetodoisarrangethat m ′ − m1 ≥ m.Since
m ′ − m1 = ℓm − (ℓ − 1) (t + 1) − m1
and ℓ ≥ 2,thiscancertainlybedonebychoosing msufficientlylarge.
Toverifythesecondfactlet Lbethefixedfieldof C. Wecanassumethattherequired
assertionistruefortheextension K/L.Let ℓ = ψ L/F(m − 1) + 1and ℓ ′ = ψ K/F(m − 1) + 1.
If missufficientlylargeand H0istheinertialgroupof H
and
Thus
andif ℓ1 = [H0 : 1]m1then
ℓ = [H0 : 1]m − ([H0 : 1] − 1)
ℓ ′ = [H0 : 1]m ′ − ([H0 : 1] − 1).
P ℓ′ −ℓ1
K
If mandtherefore ℓissufficientlylarge
P ℓ L ∩ F = Pm F
∩ E = P m′ −m1
E .
P K/L(y) = N K/L(1 + y) − 1 ≡ S K/L(y) (modP ℓ L )
if ybelongsto P ℓ′ −ℓ1
K .Thusif ybelongsto P m′ −m1
E
P E/F(y) = P K/L(y) ≡ S K/L(y) = S E/F(y) (mod P m F ).
Toverifythethirdfactwechoose,once m ′ 1
If missufficientlylarge
S E ′ /E
P −δE ′ /E−m′ 1
E ′
isgiven, m1sothat
= P −m1
E .
m ′′ = m ′ e E ′ /E − δ E ′ /E

Chapter22 285
andif ybelongsto P m′′ −m ′
1
E ′
Takinginevenlargerifnecessary,wehave
SE ′ /E(y) ∈ P m′ −m1
E .
P E ′ /F(y) ≡ P E/F(P E ′ /E(y))
≡ P E/F(S E ′ /E(y))
≡ S E/F(S E ′ /E(y))
≡ S E ′ /F(y) (modP m F ).
ReturningtotheproofofLemma22.5,wechoose
β ′ = β(χ E/F) = P ∗ E/F (β).
If m(χF)andtherefore m(χ E/F)issufficientlylarge,
∆(α s−1 2
E µE χE/F,ψ E/F) = α s−1 2
E
Both βand β ′ areunitsandtherefore
α s−1
2
E
γ
β ′
= α
If m(χF)issufficientlylarge
1 r− 2
E
γ
β
β ′ ≡ β
= α s−1
2
F
γ
β ′
and µE(β ′ ) = µE(β).Inparagraph5wesawthat
det ω
γ
= µE
β
µE
N E/F
modP m(µE)
E
γ
β ′
γ
β
γ
det ιE/F β
γ
,
β
∆(χ E/F, ψ E/F).
= α r(s−1
2 )
F
γ
.
β
if ι E/Fistherepresentationof W K/Finducedfromthetrivialrepresentationof W K/E. We
arereducedtoshowingthat
detι E/F
γ
{∆(χF, ψF)}
β
r = ∆(χE/F, ψE/F) ρ(E/F, ψF) (22.1)
if m(χF)issufficientlylarge.Ofcourse r = [E : F].

Chapter22 286
WhatwedoisshowthatforeachGaloisextension K/Ftherelation(22.1)istrueforall
fields Elyingbetween Kand F.Forthisweuseinductionon [K : F].Let G = G(K/F)and
let Cbeanon­trivialabeliannormalsubgroupof G.Let Lbethefixedfieldof C.Wesawin
Chapter13thattherearefields F1, . . ., Fslyingbetween Fand Landgeneralizedcharacters
µ1, . . ., µsof CF1 , . . ., CFsrespectivelysuchthat Then
ιE/F ≃ ⊕ s i=1Ind(WK/F, WK/Fi , µi).
χF ⊗ ι E/F ≃ ⊕ s i=1Ind(W K/F, W K/Ei , µi χ Fi/F)
andbyTheorem2.1,theMainTheorem,therightsideof(22.1)isequalto
s
i=1 ∆(µi χ Fi/F, ψ Fi/F) ρ(Fi/F, ψF).
Wejustsawthatif m(χF)issufficientlylargethisisequalto
s
i=1 µi
Since s
γ s
β i=1 ∆(χFi/F, ψFi/F) ρ(Fi/F, ψF)}.
i=1 [Fi : F] = [E : F]
weseeuponapplyingtheinductionassumptionto L/Fthatthisequals
s
i=1 µi
γ
det ιFi/F β
γ
{∆(χF, ψF)}
β
r .
Wecompletetheproofof(22.1)byappealingtoChapter5toseethat
det ι E/F = s
i=1 µidet ι Fi/F.

References 287
References
1. Artin,E.,ZurTheorieder L­ReihenmitallgemeinenGruppencharakteren,Collected
Papers,Addison–Wesley
2. Artin,E.,DiegruppentheoretischeStrukturderDiskriminantenalgebraischerZahlkörper,CollectedPapers,Addison–Wesley
3. Artin,E.andTate,J.,ClassFieldTheory,W.ABenjamin,NY
4. Brauer,R.andTate,J,Onthecharactersoffinitegroups,Ann.ofMath. 62(1955)
5. Davenport,H.andHasse,H.,DieNullstellenderKongruenzzetafunktioningewissen
zyklischenFällen,Jour.fürMath. 172(1935)
6. Dwork,B.,OntheArtinrootnumber,Amer.Jour.Math. 78(1956)
7. Hall,M.,TheTheoryofGroups,Macmillan,N.Y.(1959)
8. Hasse,H.,ArtinscheFührer,Artinsche L­FunktionenundGauss’scheSummen
überendlich­algebraischenZahlkörpern,ActaSalmanticensia(1954)
9. Lakkis,K.,DiegaloisschenGauss’schenSummenvonHasse,Dissertation,
Hamburg(1964)
10. Lamprecht,E.,AllgemeineTheoriederGauss’schenSummeninendlichenkommutativen
Ringen,Math.Nach. 9(1953)
11. Mackenzie,E.andWhaples,G.,Artin–Schreierequationsincharacteristiczero,Amer.
Jour.Math 78(1956)
12. Serre,J.­P.,Corpslocaux,Hermann.(1962)
13. Tate,J.,FourierAnalysisinNumberFieldsandHecke’sZeta­functions,Thesis,
Princeton(1950)
14. Weil,A.,Numbersofsolutionsofequationsinfinitefields,Bull.Amer.Math.
Soc. 55(1949)
15. Weil,A.,Surlathéorieducorpsdeclasses,Jour.Math.Soc.Japan, 3(1951)
16. Weil,A.,BasicNumberTheory,Springer­Verlag,(1967)