On the Functional Equation of the Artin L-functions Robert P ...
On the Functional Equation of the Artin L-functions Robert P ...
On the Functional Equation of the Artin L-functions Robert P ...
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<strong>On</strong> <strong>the</strong> <strong>Functional</strong> <strong>Equation</strong><br />
<strong>of</strong> <strong>the</strong> <strong>Artin</strong> L-<strong>functions</strong><br />
<strong>Robert</strong> P. Langlands
Introduction 2<br />
0. Introduction<br />
1. WeilGroups<br />
2. TheMainTheorem<br />
3. TheLemmas<strong>of</strong>Induction<br />
4. TheLemma<strong>of</strong>Uniqueness<br />
5. AProperty<strong>of</strong> λFunctions<br />
6. AFiltration<strong>of</strong><strong>the</strong>WeilGroup<br />
TABLE OF CONTENTS<br />
7. Consequences<strong>of</strong>Stickelberger’sResult<br />
8. ALemma<strong>of</strong>Lamprecht<br />
9. ALemma<strong>of</strong>Hasse<br />
10. TheFirstMainLemma<br />
11. <strong>Artin</strong>–Schreier<strong>Equation</strong>s<br />
12. TheSecondMainLemma<br />
13. TheThirdMainLemma<br />
14. TheFourthMainLemma<br />
15. Ano<strong>the</strong>rLemma<br />
16. Definition<strong>of</strong><strong>the</strong> λFunctions<br />
17. ASimplification<br />
18. NilpotentGroups<br />
19. Pro<strong>of</strong><strong>of</strong><strong>the</strong>MainTheorem<br />
20. <strong>Artin</strong> LFunctions<br />
21. Pro<strong>of</strong><strong>of</strong><strong>the</strong><strong>Functional</strong><strong>Equation</strong><br />
22. Appendix<br />
23. References
Introduction 3<br />
O. Introduction<br />
InthispaperIwanttoconsidernotjust<strong>the</strong> L<strong>functions</strong>introducedby<strong>Artin</strong>[1]but<strong>the</strong><br />
moregeneral<strong>functions</strong>introducedbyWeil[15].Todefine<strong>the</strong>seoneneeds<strong>the</strong>notion<strong>of</strong>aWeil<br />
groupasdescribedin[3]. Thisnotionwillbeexplainedin<strong>the</strong>firstparagraph. Fornowa<br />
roughideawillsuffice.If Eisaglobalfield,thatisanalgebraicnumberfield<strong>of</strong>finitedegree<br />
over<strong>the</strong>rationalsorafunctionfieldoverafinitefield, CEwillbe<strong>the</strong>idéleclassgroup<strong>of</strong> E.<br />
If Eisalocalfield,thatis<strong>the</strong>completion<strong>of</strong>aglobalfieldatsomeplace[16],archimedeanor<br />
nonarchimedean, CEwillbe<strong>the</strong>multiplicativegroup<strong>of</strong> E.If K/EisafiniteGaloisextension<br />
<strong>the</strong>Weilgroup W K/Eisanextension<strong>of</strong> G(K/E),<strong>the</strong>Galoisgroup<strong>of</strong> K/E,by CK. Itisa<br />
locallycompacttopologicalgroup.<br />
If E ⊆ E ′ ⊆ Kand K/EisfiniteandGalois W K/E ′mayberegardedasasubgroup<strong>of</strong><br />
W K/E.Itisclosedand<strong>of</strong>finiteindex.If E ⊆ K ⊆ L<strong>the</strong>reisacontinuousmap<strong>of</strong> W L/Eonto<br />
W K/E. Thusanyrepresentation<strong>of</strong> W K/Emayberegardedasarepresentation<strong>of</strong> W L/E. In<br />
particular<strong>the</strong>representations ρ1<strong>of</strong> W K1/Eand ρ2<strong>of</strong> W K2/Ewillbecalledequivalentif<strong>the</strong>reis<br />
aGaloisextension L/Econtaining K1/Eand K2/Esuchthat ρ1and ρ2determineequivalent<br />
representations<strong>of</strong> W L/E. Thisallowsustorefertoequivalenceclasses<strong>of</strong>representations<strong>of</strong><br />
<strong>the</strong>Weilgroup<strong>of</strong> Ewithoutmentioninganyparticularextensionfield K.<br />
Inthispaperarepresentation<strong>of</strong> W K/Eisunderstoodtobeacontinuousrepresentation<br />
ρin<strong>the</strong>group<strong>of</strong>invertiblelineartransformations<strong>of</strong>afinitedimensionalcomplexvector<br />
spacewhichissuchthat ρ(w)isdiagonalizable,thatissemisimple,forall win W K/E. Any<br />
onedimensionalrepresentation<strong>of</strong>W K/Ecanbeobtainedbyinflatingaonedimensionalrepresentation<strong>of</strong><br />
W E/E = CE.Thusequivalenceclasses<strong>of</strong>onedimensionalrepresentations<strong>of</strong><strong>the</strong><br />
Weilgroup<strong>of</strong> Ecorrespondtoquasicharacters<strong>of</strong> CE,thatis,tocontinuoushomomorphisms<br />
<strong>of</strong> CEinto C × .<br />
Suppose Eisalocalfield.Thereisastandardway<strong>of</strong>associatingtoeachequivalenceclass<br />
ω<strong>of</strong>onedimensionalrepresentationsameromorphicfunctionL(s, ω).Supposeωcorresponds<br />
to<strong>the</strong>quasicharacter χE. If Eisnonarchimedeanand ̟Eisagenerator<strong>of</strong><strong>the</strong>primeideal<br />
PE<strong>of</strong> OE,<strong>the</strong>ring<strong>of</strong>integersin E,weset<br />
L(s, ω) =<br />
1<br />
1 − χE(̟E) |̟E| s<br />
if χEisunramified.O<strong>the</strong>rwiseweset L(s, ω) = 1.If E = Rand<br />
χE(x) = (sgn x) m |x| r
Introduction 4<br />
with mequalto0or1weset<br />
L(s, ω) = π −1<br />
2 (s+r+m) Γ<br />
s + r + m<br />
If E = Cand z ∈ E<strong>the</strong>n,forus, |z|willbe<strong>the</strong>square<strong>of</strong><strong>the</strong>ordinaryabsolutevalue.If<br />
χE(z) = |z| r z m ¯z n<br />
where mand nareintegerssuchthat m + n ≥ 0, mn = 0,<strong>the</strong>n<br />
2<br />
<br />
.<br />
L(s, ω) = 2 (2π) −(s+r+m+n) Γ(s + r + m + n)<br />
Itisnotdifficulttoverify,andweshalldosolater,thatitispossible,injustoneway,<br />
todefine L(s, ω)forallequivalenceclassessothatithas<strong>the</strong>givenformwhen ωisonedimensional,sothat<br />
L(s, ω1 ⊕ ω2) = L(s, ω1) L(s, ω2)<br />
sothatif E ′ isaseparableextension<strong>of</strong> Eand ωis<strong>the</strong>equivalenceclass<strong>of</strong><strong>the</strong>representation<br />
<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong> Einducedfromarepresentation<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong> E ′ in<strong>the</strong>class Θ<br />
<strong>the</strong>n L(s, ω) = L(s, Θ).<br />
Nowtake Etobeaglobalfieldand ωanequivalenceclass<strong>of</strong>representations<strong>of</strong><strong>the</strong>Weil<br />
group<strong>of</strong> E.Itwillbeseenlaterhow,foreachplace v, ωdeterminesanequivalenceclass ωv<strong>of</strong><br />
representations<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong><strong>the</strong>correspondinglocalfield Ev.Theproduct<br />
<br />
v<br />
L(s, ωv)<br />
whichistakenoverallplaces,including<strong>the</strong>archimedeanones,willconvergeif<strong>the</strong>realpart<strong>of</strong>s<br />
issufficientlylarge.ThefunctionitdefinescanbecontinuedtoafunctionL(s, ω)meromorphic<br />
in<strong>the</strong>wholecomplexplane.Thisis<strong>the</strong><strong>Artin</strong> Lfunctionassociatedtoω.Itisfairlywellknown<br />
thatif ωis<strong>the</strong>classcontragredientto ω<strong>the</strong>reisafunctionalequationconnecting L(s, ω)and<br />
L(1 − s, ω).<br />
Thefactorappearingin<strong>the</strong>functionalequationcanbedescribedinterms<strong>of</strong><strong>the</strong>localdata.<br />
Toseehowthisisdoneweconsiderseparableextensions E<strong>of</strong><strong>the</strong>fixedlocalfield F.If ΨFisa<br />
nontrivialadditivecharacter<strong>of</strong> Flet ψ E/Fbe<strong>the</strong>nontrivialadditivecharacter<strong>of</strong> Edefined<br />
by<br />
ψE/F(x) = ψF(SE/Fx)
Introduction 5<br />
where SE/Fxis<strong>the</strong>trace<strong>of</strong> x. Wewanttoassociatetoeveryquasicharacter χE<strong>of</strong> CEand<br />
everynontrivialadditivecharacter ψE<strong>of</strong> Eanonzerocomplexnumber ∆(χE, ψE).If Eis<br />
nonarchimedean,if Pm Eis<strong>the</strong>conductor<strong>of</strong> χE,andif P −n<br />
E is<strong>the</strong>largestidealonwhich ψEis<br />
trivialchooseany γwith OEγ = P m+n<br />
E andset<br />
<br />
UE<br />
∆(χE, ψE) = χE(γ)<br />
ψE<br />
<br />
α<br />
γ χ −1<br />
E (α)dα<br />
<br />
<br />
χ −1<br />
E (α)dα .<br />
Therightsidedoesnotdependon γ.If E = R,<br />
UE ψE<br />
α<br />
γ<br />
χE(x) = (sgnx) m |x| r<br />
with mequalto0or1,and ψE(x) = e 2πiux <strong>the</strong>n<br />
If E = C, ψC(z) = e 4πi Re(wz) ,and<br />
with m + n ≥ 0, mn = 0<strong>the</strong>n<br />
∆(χE, ψE) = (i sgnu) m |u| r .<br />
χC(z) = |z| r z m ¯z n<br />
∆(χC, ψC) = i m+n χC(w).<br />
Thebulk<strong>of</strong>thispaperistakenupwithapro<strong>of</strong><strong>of</strong><strong>the</strong>following<strong>the</strong>orem.<br />
Theorem A<br />
Suppose Fisagivenlocalfieldand ψFagivennontrivialadditivecharacter<strong>of</strong> F.Itis<br />
possibleinexactlyonewaytoassigntoeachseparableextension E<strong>of</strong> Facomplexnumber<br />
λ(E/F, ψF)andtoeachequivalenceclassω<strong>of</strong>representations<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong>Eacomplex<br />
number ε(ω, Ψ E/F)suchthat<br />
(i)If ωcorrespondsto<strong>the</strong>quasicharacter χE<strong>the</strong>n<br />
(ii)<br />
ε(ω, ψ E/F) = ∆(χE, ψ E/F).<br />
ε(ω1 ⊕ ω2, ψ E/F) = ε(ω1, ψ E/F) ε(ω2, ψ E/F).
Introduction 6<br />
(iii)If ωis<strong>the</strong>equivalenceclass<strong>of</strong><strong>the</strong>representation<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong> Finducedfroma<br />
representation<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong> Ein<strong>the</strong>class θ<strong>the</strong>n<br />
ε(ω, ψF) = λ(E/F, ψF) dim θ ε(θ, ψ E/F).<br />
αs Fwilldenote<strong>the</strong>quasicharacter x −→ |x|s F<strong>of</strong> CFaswellas<strong>the</strong>correspondingequivalenceclass<strong>of</strong>representations.Set<br />
<br />
ε(s, ω, ψF) = ε α s−1<br />
<br />
2 ⊗ ω, ψF .<br />
Theleftsidewillbe<strong>the</strong>product<strong>of</strong>anonzeroconstantandanexponentialfunction.<br />
Nowtake Ftobeaglobalfieldand ωtobeanequivalenceclass<strong>of</strong>representations<strong>of</strong><strong>the</strong>Weil<br />
group<strong>of</strong> F.Let Abe<strong>the</strong>adélegroup<strong>of</strong> Fandlet ψFbeanontrivialcharacter<strong>of</strong> A/F.For<br />
eachplace vlet ψvbe<strong>the</strong>restriction<strong>of</strong> ψFto Fv. ψvisnontrivialforeach vandalmostall<strong>the</strong><br />
<strong>functions</strong> ε(s, ωv, ψv)areidentically1sothatwecanform<strong>the</strong>product<br />
<br />
v ε(s, ωv, ψv).<br />
Itsvaluewillbeindependent<strong>of</strong> ψFandwillbewritten ε(s, ω).<br />
Theorem B<br />
Thefunctionalequation<strong>of</strong><strong>the</strong> Lfunctionassociatedto ωis<br />
L(s, ω) = ε(s, ω) L(1 − s, ω).<br />
This<strong>the</strong>oremisara<strong>the</strong>reasyconsequence<strong>of</strong><strong>the</strong>first<strong>the</strong>oremtoge<strong>the</strong>rwith<strong>the</strong>functional<br />
equations<strong>of</strong><strong>the</strong>Hecke L<strong>functions</strong>.<br />
Forarchimedeanfields<strong>the</strong>first<strong>the</strong>oremsaysverylittle.Fornonarchimedeanfieldsitcan<br />
bereformulatedasacollection<strong>of</strong>identitiesforGaussiansums.Four<strong>of</strong><strong>the</strong>seidentitieswhich<br />
weformulateasourfourmainlemmasarebasic. All<strong>the</strong>o<strong>the</strong>rscanbededucedfrom<strong>the</strong>m<br />
bysimplegroup<strong>the</strong>oreticarguments.Unfortunately<strong>the</strong>onlywayatpresentthatIcanprove<br />
<strong>the</strong>fourbasicidentitiesisbylongandinvolved,althoughra<strong>the</strong>relementary,computations.<br />
HoweverTheoremApromisestobe<strong>of</strong>suchimportancefor<strong>the</strong><strong>the</strong>ory<strong>of</strong>automorphicforms<br />
andgrouprepresentationsthatwecanhopethateventuallyamoreconceptualpro<strong>of</strong><strong>of</strong>itwill<br />
befound.Thefirstand<strong>the</strong>second,whichis<strong>the</strong>mostdifficult,<strong>of</strong><strong>the</strong>fourmainlemmasaredue<br />
toDwork[6].Iamextremelygratefultohimnotonlyforsendingmeacopy<strong>of</strong><strong>the</strong>dissertation<br />
<strong>of</strong>Lakkis[9]inwhichapro<strong>of</strong><strong>of</strong><strong>the</strong>setwolemmasisgivenbutals<strong>of</strong>or<strong>the</strong>interes<strong>the</strong>has<br />
showninthispaper.<br />
F
Chapter1 7<br />
Chapter <strong>On</strong>e.<br />
Weil Groups<br />
TheWeilgroupshavemanyproperties,most<strong>of</strong>whichwillbeusedatsomepointin<strong>the</strong><br />
paper.Itisimpossibletodescribeall<strong>of</strong><strong>the</strong>mwithoutsomeprolixity.Toreduce<strong>the</strong>prolixity<br />
toaminimumIshallintroduce<strong>the</strong>segroupsin<strong>the</strong>language<strong>of</strong>categories.<br />
Consider<strong>the</strong>collection<strong>of</strong>sequences<br />
S : C λ1 µ<br />
−→G −→G<br />
<strong>of</strong>topogicalgroupswhere λisahomeomorphism<strong>of</strong> Cwith<strong>the</strong>kernel<strong>of</strong> µand µinducesa<br />
homeomorphism<strong>of</strong> G/λCwith G.Suppose<br />
S1 : C1<br />
λ1 µ2<br />
−→G1 −→G1<br />
isano<strong>the</strong>rsuchsequence.Twocontinuoushomomorphisms ϕand ψfrom Gto G1whichtake<br />
Cinto C1willbecalledequivalentif<strong>the</strong>reisacin C1suchthat ψ(g) = cϕ(g)c−1forall g<br />
in G. Swillbe<strong>the</strong>categorywhoseobjectsare<strong>the</strong>sequences Sand HomS0 (S, S1)willbe<strong>the</strong><br />
collection<strong>of</strong><strong>the</strong>seequivalenceclasses. Swillbe<strong>the</strong>categorywhoseobjectsare<strong>the</strong>sequences<br />
Sforwhich Cislocallycompactandabelianand Gisfinite;if Sand S1belongto S<br />
HomS(S, S1) = HomS0 (S, S1).<br />
Let P1be<strong>the</strong>functorfrom Sto<strong>the</strong>category<strong>of</strong>locallycompactabeliangroupswhichtakes Sto<br />
Candlet P2be<strong>the</strong>functorfrom Sto<strong>the</strong>category<strong>of</strong>finitegroupswhichtakes Sto G.Wehave<br />
tointroduceonemorecategory S1,0. Theobjects<strong>of</strong> S1willbe<strong>the</strong>sequenceson Sforwhich<br />
Gc ,<strong>the</strong>commutatorsubgroup<strong>of</strong> G,isclosed.Moreover<strong>the</strong>elements<strong>of</strong> HomS1 (S, S1)willbe<br />
<strong>the</strong>equivalenceclassesinHomS(S, S1)all<strong>of</strong>whosemembersdeterminehomeomorphisms<strong>of</strong><br />
Gwithaclosedsubgroupf<strong>of</strong>initeindexin G1.<br />
If Sisin S1let V (S)be<strong>the</strong>topologicalgroup G/Gc . If Φ ∈ HomS1 (S, S1)let ϕbea<br />
homeomorphismin<strong>the</strong>class Φandlet G = ϕ(G). Composing<strong>the</strong>map G1/Gc 1 −→ G/Gc<br />
givenby<strong>the</strong>transferwith<strong>the</strong>map G/G c −→ G/Gcdeterminedby<strong>the</strong>inverse<strong>of</strong> ϕweobtain<br />
amap Φv : V (S1) −→ V (S)whichdependsonlyon Φ. Themap S −→ V (S)becomesa<br />
contravariantfunctorfrom S1to<strong>the</strong>category<strong>of</strong>locallycompactabeliangroups. If Sis<strong>the</strong><br />
sequence<br />
C −→ G −→ G
Chapter1 8<br />
<strong>the</strong>transferfromGtoCdeterminesahomomorphismτfromG/G c to<strong>the</strong>group<strong>of</strong>Ginvariant<br />
elementsin C. τwillsometimesberegardedasamapfrom Gtothissubgroup.<br />
Thecategory Ewillconsist<strong>of</strong>allpairs K/Fwhere Fisaglobalorlocalfieldand Kis<br />
afiniteGaloisextension<strong>of</strong> F.Hom(K/F, L/E)willbeacertaincollection<strong>of</strong>isomorphisms<br />
<strong>of</strong> Kwithasubfield<strong>of</strong> Lunderwhich Fcorrespondstoasubfield<strong>of</strong> E. If<strong>the</strong>fieldsare<strong>of</strong><br />
<strong>the</strong>sametype,thatisallglobaloralllocalwedemandthat Ebefiniteandseparableover<strong>the</strong><br />
image<strong>of</strong> F. If Fisglobaland Eislocalwedemandthat Ebefiniteandseparableover<strong>the</strong><br />
closure<strong>of</strong><strong>the</strong>image<strong>of</strong> F.Iwanttoturn<strong>the</strong>mapwhichassociatestoeach K/F<strong>the</strong>group CK<br />
intoacontravariantfunctorwhichIwilldenoteby C∗ .If ϕ : K/F −→ L/Eand Fand Eare<br />
<strong>of</strong><strong>the</strong>sametypelet K1be<strong>the</strong>image<strong>of</strong> Kin Landlet ϕC∗be<strong>the</strong>composition<strong>of</strong> NL/K1with <strong>the</strong>inverse<strong>of</strong> ϕ.If Fisglobaland Eislocallet K1be<strong>the</strong>closurein L<strong>of</strong><strong>the</strong>image<strong>of</strong> K.As<br />
usual CK1maybeconsideredasubgroup<strong>of</strong><strong>the</strong>group<strong>of</strong>idèles<strong>of</strong> K. ϕC∗is<strong>the</strong>composition <strong>of</strong> NL/K1with<strong>the</strong>projection<strong>of</strong><strong>the</strong>group<strong>of</strong>idèlesonto CK.<br />
If Kisgivenlet E K be<strong>the</strong>subcategory<strong>of</strong> Ewhoseobjectsare<strong>the</strong>extensionswith<strong>the</strong><br />
largerfieldequalto Kandwhosemapsareequalto<strong>the</strong>identityon K.Let C∗be<strong>the</strong>functor<br />
on E K whichtakes K/Fto CF. If Fisgivenlet EFhaveasobjects<strong>the</strong>extensionswith<strong>the</strong><br />
smallerfieldequalto F.Itsmapsaretoequal<strong>the</strong>identityon F.<br />
AWeilgroupisacontravariantfunctor Wfrom Eto Swith<strong>the</strong>followingproperties:<br />
(i) P1 ◦ Wis C ∗ .<br />
(ii) P2 ◦ Wis<strong>the</strong>functor G : L/F −→ G(L/F).<br />
(iii)If ϕ ∈ G(L/F) ⊆ Hom(L/F, L/F)and gisanyelement<strong>of</strong> W L/F,<strong>the</strong>middlegroup<strong>of</strong><br />
<strong>the</strong>sequence W(L/F),whoseimagein G(L/F)is ϕ<strong>the</strong>n<strong>the</strong>map h −→ ghg −1 isin<strong>the</strong><br />
class ϕw.<br />
(iv)Therestriction<strong>of</strong> Wto E K takesvaluesin S1.Moreover,if K/Fbelongsto E K<br />
τ: W K/F/W c K/F<br />
−→ CF<br />
isahomeomorphism.Finally,if ϕ : K/F −→ K/Eis<strong>the</strong>identityon Kand Φ = ϕw<strong>the</strong>n<br />
<strong>the</strong>diagram<br />
W K/F/W c K/F<br />
Φv<br />
−→ WK/E/W c K/E<br />
τ ↓ ↓ τ<br />
CF<br />
−→<br />
ϕC ∗<br />
iscommutativeandif ψ : F/F −→ K/Fis<strong>the</strong>imbedding, ψWis τ.<br />
CE
Chapter1 9<br />
Since<strong>the</strong>functorialproperties<strong>of</strong><strong>the</strong>Weilgrouparenotalldiscussedby<strong>Artin</strong>andTate,<br />
weshouldreview<strong>the</strong>irconstruction<strong>of</strong><strong>the</strong>Weilgrouppointingout,whennecessary,how<br />
<strong>the</strong>functorialpropertiesarise.Thereisassociatedtoeach K/Fafundamentalclass α K/Fin<br />
H 2 (G(K/F), CK).Thegroup W(K/F)isanyextension<strong>of</strong> G(K/F)by CKassociatedtothis<br />
element.Wehavetoshow,atleast,thatif ϕ : K/F −→ L/E<strong>the</strong>diagram<br />
1 −→ CL −→ W L/E −→ G(L/E) −→ 1<br />
↓ ϕC∗ ↓ ϕG<br />
1 −→ CK −→ WK/F −→ G(K/F) −→ 1<br />
canbecompletedtoacommutativediagrambyinserting ϕ : W L/E −→ W K/F.Themap ϕC ∗<br />
commuteswith<strong>the</strong>action<strong>of</strong> G(L/E)on CLand CKsothat ϕexistsifandonlyif ϕC ∗(α L/E)is<br />
<strong>the</strong>restriction ϕ ∗ G (α K/F)<strong>of</strong> ϕ K/Fto G(L/E).Ifthisisso,<strong>the</strong>collection<strong>of</strong>equivalenceclasses<br />
towhich ϕmaybelongisaprincipalhomogeneousspace<strong>of</strong> H 1 (G(L/E), CK).Inparticular,<br />
ifthisgroupistrivial,asitiswhen ϕGisaninjection,<strong>the</strong>class<strong>of</strong> ϕisuniquelydetermined.<br />
Anexamination<strong>of</strong><strong>the</strong>definition<strong>of</strong><strong>the</strong>fundamentalclassandshowsthatitiscanonical.In<br />
o<strong>the</strong>rwords,ifϕisanisomorphism<strong>of</strong>Kand Land<strong>of</strong>FandE,<strong>the</strong>n ϕ ∗ G (α K/F) = ϕ −1 α L/E =<br />
ϕC ∗(α L/E).IfK = Landϕis<strong>the</strong>identityonK,<strong>the</strong>relationϕ ∗ G (α K/F) = α L/E = ϕC ∗(α L/E)<br />
isone<strong>of</strong><strong>the</strong>basicproperties<strong>of</strong><strong>the</strong>fundamentalclass. Thusin<strong>the</strong>setwocases ϕexists<br />
anditsclassisunique. Nowtake Ktobeglobaland Llocal. Supposeatfirstthat Kis<br />
containedin L,thatitsclosureis L,andthat F = K ∩ E. Then,by<strong>the</strong>verydefinition<strong>of</strong><br />
α K/F, ϕ ∗ G (α K/F) = ϕC ∗(α L/E).Moregenerally,ifK1is<strong>the</strong>image<strong>of</strong>KinL,andF1<strong>the</strong>image<br />
<strong>of</strong>FinE,wecanwriteϕasϕ1ϕ2ϕ3whereϕ3 : K/F −→ K1/F1, ϕ2 : K1/F1 −→ K1/K1∩E,<br />
and ϕ1 : K1/K1 ∩E −→ L/E. ϕ3and ϕ2exist.If<strong>the</strong>closure<strong>of</strong> K1is L<strong>the</strong>n ϕ1and<strong>the</strong>refore<br />
ϕ = ϕ3 ϕ2 ϕ1alsoexist.Theclass<strong>of</strong> ϕisuniquelydetermined.<br />
<strong>Artin</strong>andTateshowthat W c K/F isaclosedsubgroup<strong>of</strong> W K/F andthat τisahomeomorphism<strong>of</strong><br />
W K/F/W c K/F and CF. Grantedthis,itiseasytoseethat<strong>the</strong>restriction<strong>of</strong><br />
Wto ξ K takesvaluesin S1. Supposewehave<strong>the</strong>collection<strong>of</strong>fieldsin<strong>the</strong>diagramwith<br />
Land Knormalover Fand Land K ′ normalover F ′ . Let α, β,and νbe<strong>the</strong>imbeddings<br />
α : L/F −→ L/K, β : L/F ′ −→ L/K ′ , ν : L/F −→ L/F ′ .
Chapter1 10<br />
L<br />
K<br />
F<br />
Wehaveshown<strong>the</strong>existence<strong>of</strong> α, β,and ν.Itisclearthat ν β(W L/K ′)iscontainedin α(W L/K).<br />
Thuswehaveanaturalmap<br />
Letusverifythat<strong>the</strong>diagram<br />
K ′<br />
π : ν β(W L/K ′)/ν β(W c L/K ′) −→ α(W L/K)/α(W c L/K ).<br />
WL/K/W c<br />
L/K ′ −→ ν β(WL/K ′)/ν β(W C π<br />
L/K ′) −→ α(WL/K)/α(W c K/K ) −→ WL/K/W c L/K<br />
↓ τ ↓ τ<br />
Ck ′<br />
F ′<br />
N K ′ /K<br />
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→ Ck (A)<br />
iscommutative.Toseethislet W L/K ′be<strong>the</strong>disjointunion<br />
U r i=1 CKhi.<br />
Thenwecanchoose h ′ i , g′ j , 1 ≤ i ≤ r, 1 ≤ j ≤ ssothat W L/Kis<strong>the</strong>disjointunion<br />
r<br />
i=1<br />
s<br />
j=1 CK g ′ jh ′ i<br />
and ν β(h ′ i ) = α(hi). Using<strong>the</strong>secosetrepresentativestocompute<strong>the</strong>transferoneimmediatelyverifies<strong>the</strong>assertion.Weshouldalsoobservethat<strong>the</strong>transitivity<strong>of</strong><strong>the</strong>transferimplies<br />
<strong>the</strong>commutativity<strong>of</strong><strong>the</strong>diagram<br />
W K/F/W c K/F<br />
Φv<br />
−→ WK/F ′/W c K/F ′<br />
τ ↓ ↓<br />
CF<br />
−→<br />
ϕC ∗<br />
CF ′
Chapter1 11<br />
if Φis<strong>the</strong>class<strong>of</strong>animbedding ϕwhere ϕisanimbedding K/F −→ K/F ′ .<br />
Wehavestillnotdefined ϕW forall ϕ. Howeverwehavedefineditwhen ϕisan<br />
isomorphism<strong>of</strong><strong>the</strong>twolargerfieldsorwhen<strong>the</strong>secondlargefieldis<strong>the</strong>closure<strong>of</strong><strong>the</strong>first.<br />
Moreover<strong>the</strong>definitionissuchthat<strong>the</strong>thirdconditionandallparts<strong>of</strong><strong>the</strong>fourthcondition<br />
except<strong>the</strong>lastaresatisfied.Thelastcondition<strong>of</strong>(iv)canbemadeadefinitionwithoutviolating<br />
(i)and(ii). Whatwedonowisshowthat<strong>the</strong>reisoneandonlyoneway<strong>of</strong>extending<strong>the</strong><br />
definition<strong>of</strong> ϕWtoall ϕwithoutviolatingconditions(i)or(ii)and<strong>the</strong>functorialproperty.<br />
Suppose F ⊆ K ⊆ L, K/Fand L/FareGalois,and ψis<strong>the</strong>imbedding L/F −→ L/K.<br />
Itisobservedin<strong>Artin</strong>andTatethat<strong>the</strong>reisoneandonlyclass<strong>of</strong>maps {θ}whichmake<strong>the</strong><br />
followingdiagramcommutative<br />
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<br />
τ ↓ ↓ θ ↓<br />
1 −−−−−−−→ CK −−−−−−−−−−−−−−−−−−−−−−−−→ W K/F −−−−−−−→ G(K/F) −−−−−→ 1.<br />
Thehomomorphismon<strong>the</strong>rightisthatdeducedfrom<br />
W L/F/W L/K = G(L/F)/G(L/K) ≃ G(K/F).<br />
Let ϕ, µ,and νbeimbeddings ϕ : K/F −→ L/F, µ : K/K −→ L/K, ν : K/F −→ K/K.<br />
Then ψ ◦ ϕ = µ ◦ ν,sothat ν ◦ µ = ϕ ◦ ψ. Moreover ν ◦ µis<strong>the</strong>composition<strong>of</strong><strong>the</strong>map<br />
τ : WL/K −→ CKand<strong>the</strong>imbedding<strong>of</strong> CKin WK/F.Thus<strong>the</strong>kernel<strong>of</strong> ϕcontains ψW c L/K<br />
sothat ϕ ◦ ψrestrictedto WL/K/W c L/Kmustbe τand<strong>the</strong>onlypossiblechoicefor ϕis,apart<br />
fromequivalence, θ.Toseethatthischoicedoesnotviolate<strong>the</strong>secondconditionobservethat<br />
<strong>the</strong>restriction<strong>of</strong> τto CLwillbe NL/Kandthat ψis<strong>the</strong>identityon CL.<br />
Denote<strong>the</strong>map θ : W L/F −→ W K/Fby θ L/Kand<strong>the</strong>map τ : W K/F → CKby τ K/F.<br />
Itisclearthat τ K/F ◦ θ L/Kis<strong>the</strong>transferfrom W L/F/W c L/F to ψW L/K/ ψW c L/K followed<br />
by<strong>the</strong>transferfrom ψW L/K/ ψW c L/K to ψCL = CF. By<strong>the</strong>transitivity<strong>of</strong><strong>the</strong>transfer<br />
τ K/F ◦ θ L/K = τ L/F.Itfollowsimmediatelythatif F ⊆ K ⊆ L ⊆ L ′ andallextensionsare<br />
Galois<strong>the</strong>map θ L ′ /Kand θ L/Kθ L ′ /Larein<strong>the</strong>sameclass.<br />
Supposethat ϕisanimbedding K/F −→ K ′ /F ′ andchoose Lsothat K ′ ⊆ Land L/F<br />
isGalois.Let ψ : K ′ /F ′ −→ L/F ′ , µ : K/F −→ L/F, ν : L/F −→ L/F ′ beimbeddings.<br />
Then ψ ◦ ϕ = ν ◦ µsothat µ ◦ ν = ϕ ◦ ψ.If α : L/F −→ L/K, β : L/F ′ −→ L/K ′ are<strong>the</strong>
Chapter1 12<br />
imbeddings<strong>the</strong>n<strong>the</strong>kernel<strong>of</strong> ψis ν βW c L/K ′whichiscontainedin αW c L/K<strong>the</strong>kernel<strong>of</strong> µ.<br />
Thus<strong>the</strong>reisonlyonewaytodefine ϕsothat µ ◦ ν = ϕ ◦ ψ.Thediagram<br />
W L/K ′/W c L/K ′<br />
bβ<br />
−→ W L/F ′/ βW c L/K ′<br />
bψ<br />
−−−−−−−−−−−−−−−−−→ W K ′ /F ′<br />
↓ ν ↓ ϕ<br />
W L/F/ν βW c L/K ′<br />
−→ W L/F/αW c L/K<br />
bµ<br />
−→ W K/F<br />
willbecommutative.Since ψ ◦ β = τL/K ′and µ ◦ α = τK/Fdiagram(A)showsthat ϕhas<br />
<strong>the</strong>requiredeffecton CK.<br />
Todefine ϕWingeneral,weobservethatevery ϕis<strong>the</strong>composition<strong>of</strong>isomorphisms,<br />
imbeddings<strong>of</strong>fields<strong>of</strong><strong>the</strong>sametype,andamap K/F −→ K ′ /F ′ where Kisglobal, K ′ is<br />
local, K ′ is<strong>the</strong>closure<strong>of</strong> K,and F = F ′ ∩ K.Ofcourse<strong>the</strong>identity<br />
(ϕ ◦ ψ)W = ψWϕW<br />
mustbeverified. Iomit<strong>the</strong>verificationwhichiseasyenough. Theuniqueness<strong>of</strong><strong>the</strong>Weil<br />
groupsin<strong>the</strong>sense<strong>of</strong><strong>Artin</strong>andTateimpliesthat<strong>the</strong>functor Wisuniqueuptoisomorphism.<br />
Thesequence<br />
S(n, C) : GL(n, C) id<br />
−→GL(n, C) −→ 1<br />
belongsto S1.If S : C −→ G −→ Gbelongsto S1<strong>the</strong>n<br />
HomS0 (S, S(n, C))<br />
is<strong>the</strong>set<strong>of</strong>equivalenceclasses<strong>of</strong> ndimensionalcomplexrepresentations<strong>of</strong> G.Let Ωn(S)be<br />
<strong>the</strong>set<strong>of</strong>all ΦinHomS0 (S, S(n, C))suchthat,foreach ϕ ∈ Φ, ϕ(g)issemisimpleforall g<br />
in G. Ωn(S)isacontravariantfunctor<strong>of</strong> Sandsois Ω(S) = ∞ n=1 Ωn(S).<strong>On</strong><strong>the</strong>category<br />
S1,itcanbeturnedintoacovariantfunctor. If ψ : S −→ S1,if Φ ∈ Ω(S),andif ϕ ∈ Φ,<br />
let ψassociateto Φ<strong>the</strong>matrixrepresentationscorrespondingto<strong>the</strong>inducedrepresentation<br />
Ind(G1, ψ(G), ϕ ◦ ψ−1 ). Itfollowsfrom<strong>the</strong>transitivity<strong>of</strong><strong>the</strong>inductionprocessthat Ωisa<br />
covariantfunctor<strong>of</strong> S1.<br />
Tobecompleteafur<strong>the</strong>robservationmustbemade.<br />
Lemma 1.1Suppose Hisasubgroup<strong>of</strong>finiteindexin Gand ρisafinitedimensional<br />
complexrepresentation<strong>of</strong> Hsuchthat ρ(L)issemisimpleforall hin H.If<br />
σ = Ind(G, H, ρ)
Chapter1 13<br />
<strong>the</strong>n σ(g)issemisimpleforall g.<br />
Hcontainsasubgroup H1whichisnormaland<strong>of</strong>finiteindexin G,namely,<strong>the</strong>group<br />
<strong>of</strong>elementsactingtriviallyon H\G. Toshowthatanonsingularmatrixissemisimpleone<br />
hasonlytoshowthatsomepower<strong>of</strong>itissemisimple.Since σ n (g) = σ(g n )and g n belongs<br />
to H1forsome nweneedonlyshowthat σ(g)issemisimplefor gin H1.Inthatcase σ(g)is<br />
equivalentto r<br />
i=1<br />
⊕ρ(gigg −1<br />
i )if Gis<strong>the</strong>disjointunion<br />
r<br />
i=1 Hgi<br />
Suppose L/Fand K/Fbelongto EFand ϕ ∈ HomEF (L/F, K/F). Since<strong>the</strong>maps<strong>of</strong><br />
<strong>the</strong>class ϕWalltake WK/Fonto WL/F<strong>the</strong>associatedmap Ω(W(L/F)) −→ Ω(W(K/F))is<br />
injective.Moreoveritisindependent<strong>of</strong>ϕ.IfL1/Fand L2/Fbelongto EF<strong>the</strong>reisanextension<br />
K/Fandmaps ϕ1 ∈ HomEF (L1/F, K/F), ϕ2 ∈ HomEF (L2/F, K/F). ω1in Ω(W(L1/F))<br />
and ω2in Ω(W(L2/F))have<strong>the</strong>sameimagein Ω(W(K/F))foronesuch Kifandonlyif<br />
<strong>the</strong>yhave<strong>the</strong>sameimageforallsuch K. Ifthisissowesaythat ω1and ω2areequivalent.<br />
Thecollection<strong>of</strong>equivalenceclasseswillbedenotedby Ω(F).Itsmembersarereferredtoas<br />
equivalenceclasses<strong>of</strong>representations<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong> F.<br />
Let Fbe<strong>the</strong>categorywhoseobjectsarelocalandglobalfields. If Fand Eare<strong>of</strong><strong>the</strong><br />
sametypeHomF(F, E)consists<strong>of</strong>allisomorphisms<strong>of</strong> Fwithasubfield<strong>of</strong> Eoverwhich<br />
Eisseparable. If Fisglobaland EislocalHomF(F, E)consists<strong>of</strong>allisomorphisms<strong>of</strong> F<br />
withasubfield<strong>of</strong> Eoverwhoseclosure Eisseparable. Ω(F)isclearlyacovariantfunctoron<br />
F.Let Fgℓ,and Flocbe<strong>the</strong>subcategoriesconsisting<strong>of</strong><strong>the</strong>globalandlocalfieldsrespectively.<br />
Suppose Fand Eare<strong>of</strong><strong>the</strong>sametypeand ϕ ∈ HomF(F, E).If ω ∈ Ω(E)choose Ksothat<br />
ωbelongsto Ω(W(K/E)).Wemayassumethat<strong>the</strong>reisan L/Fandanisomorphism ψfrom<br />
Lonto Kwhichagreeswith ϕon F.Then ψW : W K/E −→ W L/Fisaninjection.Let θbe<strong>the</strong><br />
equivalenceclass<strong>of</strong><strong>the</strong>representation<br />
σ = Ind(W L/F, ψW(W K/E), ρ ◦ ψ −1<br />
W )<br />
with ρin ω. Iclaimthat θisindependent<strong>of</strong> Kanddependsonlyon ωand ϕ. Toseethisit<br />
isenoughtoshowthatif L ⊆ L ′ , L ′ /FisGalois, ψ ′ isanisomorphismfrom L ′ to K ′ which<br />
agreeswith ψon L,and ρ ′ isarepresentation<strong>of</strong> W K ′ /Fin ω<strong>the</strong>class<strong>of</strong><br />
σ ′ = Ind(WL ′ /F, ψ ′ W(WK ′ /E), ρ ′ ′ −1<br />
◦ ψ w )<br />
isalso Θ. Suppose µisamapfrom W K ′ /Eto W K/Eassociatedto<strong>the</strong>imbedding K/E −→<br />
K ′ /Eand νisamapfrom W L ′ /F to W L/F associatedto<strong>the</strong>imbedding L/F −→ L ′ /F.
Chapter1 14<br />
Wemaysupposethat ψW ◦ µ = ν ◦ ψ ′ W . Thekernel<strong>of</strong> µis W c K ′ /K if,forsimplicity<strong>of</strong><br />
. Moreover<br />
notation, WK ′ /Kisregardedasasubgroup<strong>of</strong> WK ′ /Eandthat<strong>of</strong> νis W c L ′ /L<br />
ψ ′ W (W c K ′ /K ) = W c L ′ /L .Take ρ′ = ρ ◦ µ.Then σactson<strong>the</strong>space V <strong>of</strong><strong>functions</strong> fon WK/F satisfying f(vw) ≡ ρ(ψ −1<br />
w (h))f(w)for vin ψw(W K/E). Let V ′ be<strong>the</strong>analogousspaceon<br />
which σ ′ acts.Then<br />
V ′ = {f ◦ ν | f ∈ V }.<br />
Theassertionfollows.Thus Ω(F)isacontravariantfunctoron Fgℓand Floc.<br />
Afterthislaboriousandclumsyintroductionwecansettoworkandprove<strong>the</strong>two<br />
<strong>the</strong>orems.ThefirststepistoreformulateTheoremA.
Chapter2 15<br />
Chapter Two.<br />
The Main Theorem<br />
Itwillbeconvenientinthisparagraphandatvariouslatertimestoregard WK/Easa<br />
subgroup<strong>of</strong> W K/Fif F ⊆ E ⊆ K.If F ⊆ E ⊆ L ⊆ Kweshallalsooccasionallytake W L/E<br />
tobe W K/E/W c K/L .<br />
If K/FisfiniteandGalois, P(K/F)willbe<strong>the</strong>set<strong>of</strong>extensions E ′ /Ewith F ⊆ E ⊆<br />
E ′ ⊆ Kand P◦(K/F)willbe<strong>the</strong>set<strong>of</strong>extensionsin P(K/F)with<strong>the</strong>lowerfieldequalto F.<br />
Theorem 2.1<br />
Suppose KisaGaloisextension<strong>of</strong><strong>the</strong>localfield Fand ψFisagivennontrivialadditive<br />
character<strong>of</strong> F. Thereisexactlyonefunction λ(E/F, ψF)definedon P◦(K/F)with<strong>the</strong><br />
followingtwoproperties<br />
(i)<br />
(ii)If E1, . . ., Er, E ′ 1 , . . ., E′ s<br />
λ(F/F, ψF) = 1.<br />
arefieldslyingbetween F and K,if χEi , 1 ≤ i ≤ r,isa<br />
quasicharacter<strong>of</strong> CEi ,if χE ′ j , 1 ≤ j ≤ s,isaquasicharacter<strong>of</strong> CE ′ j ,andif<br />
isequivalentto<br />
<strong>the</strong>n r<br />
isequalto<br />
s<br />
⊕ r i=1 Ind(Wk/F, WK/Ei , χEi )<br />
⊕ s j=1 Ind(W K/F, W K/E ′ j , χ E ′ j )<br />
i=1 ∆(χEi , ψ Ei/F)λ(Ei/F, ψF)<br />
j=1 ∆(χE ′ j , ψ E ′ j /F)λ(E ′ j<br />
/F, ψF).<br />
A<strong>functions</strong>atisfying<strong>the</strong>conditions<strong>of</strong>this<strong>the</strong>oremwillbecalledaλfunction.Itisclear<br />
that<strong>the</strong>functionλ(E/F, ψF)<strong>of</strong>TheoremAwhenrestrictedto P◦(K/F)becomesa λfunction.<br />
Thus<strong>the</strong>uniquenessinthis<strong>the</strong>oremimpliesatleastpart<strong>of</strong><strong>the</strong>uniqueness<strong>of</strong>TheoremA.To<br />
showhowthis<strong>the</strong>oremimpliesall<strong>of</strong>TheoremAwehavetoanticipatesomesimpleresults<br />
whichwillbeprovedinparagraph4.
Chapter2 16<br />
First<strong>of</strong>allaλfunctioncannevertakeon<strong>the</strong>value0. Moreover,if F ⊆ K ⊆ L<strong>the</strong><br />
λfunctionon P◦(K/F)isjust<strong>the</strong>restrictionto P◦(K/F)<strong>of</strong><strong>the</strong> λfunctionon P◦(L/F).Thus<br />
λ(E/F, ψF)isdefinedindependently<strong>of</strong> K.Finallyif E ⊆ E ′ ⊆ E ′′<br />
λ(E ′′ /E, ψE) = λ(E ′′ /E ′ , ψ E ′ /E)λ(E ′ /E, ψE) [E′′ :E ′ ] .<br />
Wealsohavetouseaform<strong>of</strong>Brauer’s<strong>the</strong>orem[4]. If Gisafinitegroup<strong>the</strong>reare<br />
nilpotentsubgroups N1, . . ., Nm,onedimensionalrepresentations χ1, . . ., χm<strong>of</strong> N1, . . ., Nm<br />
respectively,andintegers n1, . . ., nmsuchthat<strong>the</strong>trivialrepresentation<strong>of</strong> Gisequivalentto<br />
⊕ m i=1 niInd(G, Ni, χi).<br />
Themeaning<strong>of</strong>thiswhensome<strong>of</strong><strong>the</strong> niarenegativeisclear<br />
Lemma 2.2<br />
Suppose Fisaglobalorlocalfieldand ρisarepresentation<strong>of</strong> W K/F. Thereareintermediatefields<br />
E1, . . ., Emsuchthat G(K/Ei)isnilpotentfor 1 ≤ i ≤ m,onedimensional<br />
representations χEi <strong>of</strong> W K/Ei ,andintegers n1, . . ., nmsuchthat ρisequivalentto<br />
⊕ m i=1 niInd(WK/F, WK/Ei , χEi ).<br />
Theorem2.1andLemma2.2toge<strong>the</strong>rimply<strong>the</strong>uniqueness<strong>of</strong>TheoremA.Beforeproving<br />
<strong>the</strong>lemmawemustestablishasimpleandwellknownfact.<br />
Lemma 2.3<br />
Suppose Hisasubgroup<strong>of</strong>finiteindexin<strong>the</strong>group G.Suppose τisarepresentation<strong>of</strong><br />
G, σarepresentation<strong>of</strong> H,and ρ<strong>the</strong>restriction<strong>of</strong> τto H.Then<br />
τ ⊗ Ind(G, H, σ) ≃ Ind(G, H, ρ ⊗ σ).<br />
Let τacton V and σon W.ThenInd(G, H, σ)actson X,<strong>the</strong>space<strong>of</strong>all<strong>functions</strong> fon<br />
Gwithvaluesin Wsatisfying<br />
f(hg) = σ(h) f(g)<br />
whileInd(G, H, ρ ⊗ σ)actson Y,<strong>the</strong>space<strong>of</strong>all<strong>functions</strong> fon Gwithvaluesin V ⊗ W<br />
satisfying<br />
f(hg) = (ρ(h) ⊗ σ(h)) f(g).
Chapter2 17<br />
Clearly, V ⊗ Xand Yhave<strong>the</strong>samedimension.Themap<strong>of</strong> V ⊗ Xto Ywhichsends v ⊗ f<br />
to<strong>the</strong>function<br />
f ′ (g) = τ(g)v ⊗ f(g)<br />
is Ginvariant. Ifitwerenotanisomorphism<strong>the</strong>rewouldbeabasis v1, . . ., vn<strong>of</strong> V and<br />
<strong>functions</strong> f1, . . ., fnwhicharenotallzerosuchthat<br />
Thisisclearlyimpossible.<br />
Σ n i=1 τ(g)vi ⊗ fi(g) ≡ 0.<br />
ToproveLemma2.2wetake<strong>the</strong>group G<strong>of</strong>Brauer’s<strong>the</strong>oremtobe G(K/F).Let Fibe<strong>the</strong><br />
fixedfield<strong>of</strong> Niandlet ρibe<strong>the</strong>tensorproduct<strong>of</strong> χi,whichwemayregardasarepresentation<br />
<strong>of</strong> W K/Fi and<strong>the</strong>restriction<strong>of</strong> ρto W K/Fi .Then<br />
ρ ≃ ρ ⊗ 1 ≃ ⊕ m i=1niInd(WK/Fi , ρi).<br />
Thistoge<strong>the</strong>rwith<strong>the</strong>transitivity<strong>of</strong><strong>the</strong>inductionprocessshowsthatinproving<strong>the</strong>lemma<br />
wemaysupposethat G(K/F)isnilpotent.<br />
Weprove<strong>the</strong>lemma,withthisextracondition,byinductionon [K : F]. Weuse<strong>the</strong><br />
symbol ωtodenoteanorbitin<strong>the</strong>set<strong>of</strong>quasicharacters<strong>of</strong> CKunder<strong>the</strong>action<strong>of</strong> G(K/F).<br />
Therestriction<strong>of</strong> ρto CKis<strong>the</strong>directsum<strong>of</strong>onedimensionalrepresentations. If ρactson<br />
V let Vωbe<strong>the</strong>spacespannedby<strong>the</strong>vectorstransformingunder CKaccordingtoaquasicharacterin<br />
ω. V is<strong>the</strong>directsum<strong>of</strong><strong>the</strong>spaces Vωwhichareeachinvariantunder W K/F.<br />
Forourpurposeswemaysupposethat V = Vωforsome ω.Choose χKinthis ωandlet V0be<br />
<strong>the</strong>space<strong>of</strong>vectorstransformingunder CKaccordingto χK. Let Ebe<strong>the</strong>fixedfield<strong>of</strong><strong>the</strong><br />
isotrophygroup<strong>of</strong> χK. V0isinvariantunder W K/E.Let σbe<strong>the</strong>representation<strong>of</strong> W K/Ein<br />
V0.Itiswellknownthat<br />
ρ ≃ Ind(W K/F, W K/E, σ).<br />
Toseethisonehasonlytoverifythat<strong>the</strong>space Xonwhich<strong>the</strong>representationon<strong>the</strong>right<br />
actsand Vhave<strong>the</strong>samedimensionandthat<strong>the</strong>map<br />
f −→ <br />
W K/E\W K/F<br />
ρ(g −1 )f(g)<br />
<strong>of</strong> Xinto Vwhichisclearly W K/Finvarianthasnokernel.Itiseasyenoughtodothis.<br />
If E = F<strong>the</strong>assertion<strong>of</strong><strong>the</strong>lemmafollowsbyinduction.If E = Fchoose Lcontaining<br />
Fsothat K/Liscyclic<strong>of</strong>primedegreeand L/FisGalois.Then ρ(W K/L)isanabeliangroup<br />
and W c K/L iscontainedin<strong>the</strong>kernel<strong>of</strong> ρ.Thus ρmayberegardedasarepresentation<strong>of</strong> W L/F.
Chapter2 18<br />
Theassertionnowfollowsfrom<strong>the</strong>inductionassumptionand<strong>the</strong>concludingremarks<strong>of</strong><strong>the</strong><br />
previousparagraph.<br />
Nowtakealocalfield Eandarepresentation ρ<strong>of</strong> W K/E.Chooseintermediatefields<br />
E1, . . ., Em,onedimensionalrepresentations χEi <strong>of</strong> W K/Ei ,andintegers n1, . . ., nmsothat<br />
If ωis<strong>the</strong>class<strong>of</strong> ρset<br />
ρ ≃ ⊕ m i=1 ni Ind(WK/E, WK/Ei , χEi ).<br />
ε(ω, ψE) = m<br />
i=1 {∆(χEi , Ψ Ei/E)λ(Ei/E, ΨE)} ni .<br />
Theorem2.1showsthat<strong>the</strong>rightsideisindependent<strong>of</strong><strong>the</strong>wayinwhich ρiswrittenasasum<br />
<strong>of</strong>inducedrepresentations.Thefirstandsecondconditions<strong>of</strong>TheoremAareclearlysatisfied.<br />
If ρis<strong>the</strong>representationaboveand σ<strong>the</strong>representation<br />
<strong>the</strong>n<br />
Thusif ω ′ is<strong>the</strong>class<strong>of</strong> σ<br />
while<br />
Ind(W K/F, W K/E, ρ)<br />
σ ≃ ⊕ m i=1 niInd(WK/F, WK/Ei , χEi ).<br />
ε(ω ′ , ψF) = m<br />
ε(ω, ψ E/F) = m<br />
Thethirdpropertyfollowsfrom<strong>the</strong>relations<br />
and<br />
i=1 {∆(χEi , ψ Ei/F)λ(Ei/F, ψF)} ni<br />
i=1 {∆(χEi , ψ Ei/F) λ(Ei/E, ψ E/F)} ni .<br />
dim ω = Σ m i=1 ni [Ei : E]<br />
λ(Ei/F, ψF) = λ(Ei/E, ψ E/F) λ(E/F, ψF) [Ei:E]
Chapter3 19<br />
Chapter Three.<br />
The Lemmas <strong>of</strong> Induction<br />
Inthisparagraphweprovetwosimplebutveryusefullemmas.<br />
Lemma 3.1<br />
Suppose KisaGaloisextension<strong>of</strong><strong>the</strong>localfield F. Suppose<strong>the</strong>subset A<strong>of</strong> P(K/F)<br />
has<strong>the</strong>followingfourproperties.<br />
(i)Forall E,with F ⊆ E ⊆ K, E/E ∈ A.<br />
(ii)If E ′′ /E ′ and E ′ /Ebelongto Asodoes E ′′ /E.<br />
(iii)If L/Ebelongsto P(K/F)and L/Eiscyclic<strong>of</strong>primedegree<strong>the</strong>n L/Ebelongsto A.<br />
(iv)Supposethat L/Ein P(K/F)isaGaloisextension.LetG = G(L/E).SupposeG = H ·C<br />
where H = {1}, H ∩ C = {1},and Cisanontrivialabeliannormalsubgroup<strong>of</strong> Gwhichis<br />
containedineverynontrivialnormalsubgroup<strong>of</strong> G.If E ′ is<strong>the</strong>fixedfield<strong>of</strong> Handifevery<br />
E ′′ /Ein P◦(L/E)forwhich [E ′′ : E] < [E ′ : E ′ ]isin Asois E ′ /E.Then Aisall<strong>of</strong> P(K/F).<br />
Itisconvenienttoproveano<strong>the</strong>rlemmafirst.<br />
Lemma 3.2<br />
Suppose KisaGaloisextension<strong>of</strong><strong>the</strong>localfield Fand F ⊂ = E ⊆ K. Supposethat<strong>the</strong><br />
onlynormalsubfield<strong>of</strong> Kcontaining Eis Kitselfandthat<strong>the</strong>rearen<strong>of</strong>ieldsbetween Fand<br />
E.Let G = G(K/F)andlet Ebe<strong>the</strong>fixedfield<strong>of</strong> H.Let Cbeaminimalnontrivialabelian<br />
normalsubgroup<strong>of</strong> G.Then G = HC, H ∩ C = {1}and Ciscontainedineverynontrivial<br />
normalsubgroup<strong>of</strong> G.Inparticularif H = {1}, G = Cisabelian<strong>of</strong>primeorder.<br />
Hiscontainedinnosubgroupbesidesitselfand Gcontainsnonormalsubgroupbut {1}.<br />
Thusif Hisnormalitis {1}and Ghasnopropersubgroupsandisconsequentlycyclic<strong>of</strong><br />
primeorder.Suppose Hisnotnormal.Since Gissolvableitdoescontainaminimalnontrivial<br />
abeliannormalsubgroup C. Since Cisnotcontainedin H, H ⊂ = HCand G = HC. Since<br />
H ∩ Cisanormalsubgroup<strong>of</strong> Gitis {1}.If Disanontrivialnormalsubgroup<strong>of</strong> Gwhich<br />
doesnotcontain C<strong>the</strong>n D ∩ C = {1}and Discontainedin<strong>the</strong>centralizer Z<strong>of</strong> C. Then<br />
DCisalsoand Zmustmeet Hnontrivially.But Z ∩ Hisanormalsubgroup<strong>of</strong> G.Thisisa<br />
contradictionand<strong>the</strong>lemmaisproved.
Chapter3 20<br />
Thefirstlemmaiscertainlytrueif [K : F] = 1.Suppose [K : F] > 1and<strong>the</strong>lemmais<br />
validforallpairs [K ′ : F ′ ]with [K ′ : F ′ ] < [K : F].If<strong>the</strong>Galoisextension L/Ebelongsto<br />
P(K/F)<strong>the</strong>n A ∩ P(L/E)satisfies<strong>the</strong>condition<strong>of</strong><strong>the</strong>lemmawith Kreplacedby Land F<br />
by E.Thus,byinduction,if [L : E] < [K : F], P(L/E) ⊆ A.Inparticularif E ′ /Eisnotin G<br />
<strong>the</strong>n E = Fand<strong>the</strong>onlynormalsubfield<strong>of</strong> Kcontaining E ′ is Kitself.If Aisnot P(K/F)<br />
<strong>the</strong>namongstallextensionswhicharenotin Gchooseone E/Fforwhich [E : F]isminimal.<br />
Because<strong>of</strong>(ii)<strong>the</strong>rearen<strong>of</strong>ieldsbetween Fand E. Lemma3.2,toge<strong>the</strong>rwith(iii)and(iv),<br />
showthat E/Fisin A.Thisisacontradiction.<br />
Thereisavariant<strong>of</strong>Lemma3.1whichweshallhaveoccasiontouse.<br />
Lemma 3.3<br />
Suppose KisaGaloisextension<strong>of</strong><strong>the</strong>localfield F.Suppose<strong>the</strong>subset A<strong>of</strong> P◦(K/F)<br />
has<strong>the</strong>followingproperties.<br />
(i) F/F ∈ A.<br />
(ii)If L/Fisnormaland L ⊂ = K<strong>the</strong>n P◦(L/F) ⊆ A.<br />
(iii)If F ⊂ E ⊆ E ′ ⊆ Kand E/Fbelongto G<strong>the</strong>n E ′ /Fbelongto A.<br />
(iv)If L/Fin P◦(K/F)iscyclic<strong>of</strong>primedegree<strong>the</strong>n L/F ∈ A.<br />
(v) Supposethat L/Fin P◦(K/F)isGaloisand G = G(L/F). Suppose G = HCwhere<br />
H = {1}, H ∩ C = {1},and Cisanontrivialabeliannormalsubgroup<strong>of</strong> Gwhichis<br />
containedineverynontrivialnormalsubgroup.If Eis<strong>the</strong>fixedfield<strong>of</strong> Handifevery E ′ /F<br />
in P◦(L/F)forwhich [E ′ : F] < [E : F]isin Asois E/F.<br />
Then Ais P◦(K/F).<br />
AgainifAisnot P◦(K/F)<strong>the</strong>reisan E/FnotinAforwhich[E : F]isminimal.Certainly<br />
[E : F] > 1.By(ii)and(iii), Eiscontainedinnopropernormalsubfield<strong>of</strong> Kand<strong>the</strong>reareno<br />
fieldsbetween Eand F.Lemma3.2toge<strong>the</strong>rwith(iv)and(v)leadto<strong>the</strong>contradictionthat<br />
E/Fisin A.
Chapter4 21<br />
Chapter Four.<br />
The Lemma <strong>of</strong> Uniqueness<br />
Suppose K/F isafiniteGaloisextension<strong>of</strong><strong>the</strong>localfield F and ψF isanontrivial<br />
additivecharacter<strong>of</strong> F.Afunction E/F −→ λ(E/F, ψF)on P◦(K/F)willbecalledaweak<br />
λfunctionif<strong>the</strong>followingtwoconditionsaresatisfied.<br />
(i) λ(F/F, ΨF) = 1.<br />
(ii)If E1, . . ., Er, E ′ 1 , . . ., E′ s arefieldslyingbetween Fand K,if µi, 1 ≤ i ≤ r,isaonedimensionalrepresentation<strong>of</strong><br />
G(K/Ei),if νj, 1 ≤ j ≤ s,isaonedimensionalrepresentation<strong>of</strong><br />
G(K/E ′ j ),andif<br />
isequivalentto<br />
r<br />
s<br />
<strong>the</strong>n r<br />
isequalto<br />
s<br />
i=1<br />
j=1<br />
Ind(G(K/F), G(K/Ei), µi)<br />
Ind(G(K/F), G(K/Ej), νj)<br />
i=1 ∆(χEi , ψ Ei/F)λ(Ei/F, ψF)<br />
j=1 ∆(χE ′ j ψ E ′ j /F)λ(E ′ j/F, ψF)<br />
if χEi is<strong>the</strong>character<strong>of</strong> CEi correspondingto µiand χE ′ j is<strong>the</strong>character<strong>of</strong> CE ′ j correspondingto<br />
νj.<br />
Supposingthataweak λfunctionisgivenon P◦(K/F),weshallestablishsome<strong>of</strong>its<br />
properties.<br />
Lemma 4.1<br />
(i)If L/Fin P◦(K/F)isnormal<strong>the</strong>restriction<strong>of</strong> λ(·, ψF)to P◦(L/F)isaweak λfunction.<br />
(ii)If E/Fbelongsto P◦(K/F)and λ(E/F, ψF) = 0<strong>the</strong>functionon P◦(K/E)definedby<br />
isaweak λfunction.<br />
λ(E ′ /E, ψ E/F) = λ(E ′ /F, ψF) λ(E/F, ψF) −[E′ :E]
Chapter4 22<br />
Anyonedimensionalrepresentation µ<strong>of</strong> G(L/E)maybeinflatedtoaonedimensional<br />
representation,againcalled µ,<strong>of</strong> G(K/E)and<br />
isjust<strong>the</strong>inflationto G(K/F)<strong>of</strong><br />
Ind(G(K/F), G(K/E), µ)<br />
Ind(G(L/F), G(L/E), µ).<br />
Thefirstpart<strong>of</strong><strong>the</strong>lemmafollowsimmediatelyfromthisobservation.<br />
Asfor<strong>the</strong>secondpart,<strong>the</strong>relation<br />
λ(E/E, ψ E/F) = 1<br />
isclear.Iffields Ei, 1 ≤ i ≤ r, E ′ j , 1 ≤ j ≤ s,lyingbetween Eand Kandrepresentations<br />
µiand νjaregivenasprescribedandif<br />
isequivalentto<br />
r<br />
s<br />
<strong>the</strong>n r<br />
isequivalentto<br />
i=1 Ind(G(K/E), G(K/Ei), µi) = ρ<br />
j=1 Ind(G(K/E), G(K/E′ j), νj) = σ<br />
s<br />
sothat r<br />
isequalto<br />
s<br />
Since ρand σhave<strong>the</strong>samedimension<br />
sothat r<br />
i=1 Ind(G(K/F), G(K/Ei, µi)<br />
j=1 Ind(G(K/F), G(K/E′ j<br />
), νj)<br />
i=1 ∆(χEi , ψ Ei/F) λ(E1/F, ψF) (A)<br />
j=1 ∆(χE ′ j , ψ E ′ j /F) λ(E ′ j /F, ψF). (B)<br />
Σ r i=1 [Ei : E] = Σ s j=1 [E′ j : E]<br />
i=1 λ(E/F, ΨF) [Ei:E] = s<br />
j=1<br />
λ(E/F, ψF) [E′<br />
j :F] .
Chapter4 23<br />
Dividing(A)by<strong>the</strong>leftside<strong>of</strong>thisequationand(B)by<strong>the</strong>rightandobservingthat<strong>the</strong>results<br />
areequalweobtain<strong>the</strong>relationneededtoprove<strong>the</strong>lemma.<br />
If K/Fisabelian S(K/F)willbe<strong>the</strong>set<strong>of</strong>characters<strong>of</strong> CFwhichare1on N K/FCK.<br />
Lemma 4.2<br />
If K/Fisabelian<br />
λ(K/F, ΨF) = <br />
µF ∈S(K/F) ∆(µF, ψF).<br />
µF determinesaonedimensionalrepresentation<strong>of</strong> G(K/F)whichwealsodenoteby<br />
µF.Thelemmaisanimmediateconsequence<strong>of</strong><strong>the</strong>equivalence<strong>of</strong><br />
and <br />
Lemma 4.3<br />
Ind(G(K/F), G(K/K), 1)<br />
µF ∈S(K/F)<br />
Ind(G(K/F), G(K/F), µF).<br />
Suppose K/Fisnormaland G = G(K/F).Suppose G = HCwhere H ∩ C = {1}and<br />
Cisanontrivialabeliannormalsubgroup.Let Ebe<strong>the</strong>fixedfield<strong>of</strong> Hand Lthat<strong>of</strong> C.Let<br />
Tbeaset<strong>of</strong>representatives<strong>of</strong><strong>the</strong>orbits<strong>of</strong> S(K/L)under<strong>the</strong>action<strong>of</strong> G.If µ ∈ Tlet Bµbe<br />
<strong>the</strong>isotropygroup<strong>of</strong> µandlet Bµ = G(K/Lµ).Then [Lµ : F] < [E : F]and<br />
λ(E/F, ψF) = <br />
µ∈T ∆(µ′ , ψ Lµ/F) λ(Lµ/F, ψF).<br />
Here G(K/Lµ) = G(K/L) · (G(K/Lµ) ∩ G(K/E))and µ ′ is<strong>the</strong>character<strong>of</strong> CLµ associated<br />
to<strong>the</strong>character<strong>of</strong> G(K/Lµ) : g −→ µ(g1)if<br />
g = g1g2, g1 ∈ G(K/L), g2 ∈ G(K/Lµ) ∩ G(K/E),<br />
Wemayaswelldenote<strong>the</strong>givencharacter<strong>of</strong> G(K/Lµ)by µ ′ also.Toprove<strong>the</strong>lemma<br />
weshowthat<br />
Ind(G(K/F), G(K/E), 1) = σ<br />
isequivalentto<br />
<br />
µ∈T Ind(G(K/F), G(K/Lµ), µ ′ ).
Chapter4 24<br />
Since Thasatleasttwoelementsitwillfollowthat<br />
isgreaterthan<br />
[E : F] = dimInd(G(K/F), G(K/E), 1)<br />
[Lµ : F] = dimInd(G(K/F), G(K/Lµ), µ ′ ).<br />
Therepresentation σactson<strong>the</strong>space<strong>of</strong><strong>functions</strong>on H\G.If ν ∈ S(K/L),thatis,isa<br />
character<strong>of</strong> C,let ψν(hc) = ν(c)if h ∈ H, c ∈ C.Theset<br />
{ψν | ν ∈ S(K/L)}<br />
isabasisfor<strong>the</strong><strong>functions</strong>on H\G.If µ ∈ Tlet Sµbeitsorbit;<strong>the</strong>n<br />
Vµ = Σν∈Sµ Cψν<br />
isinvariantandirreducibleunder G.Moreover,if gbelongsto G(K/Lµ)<br />
Since<br />
σ(g)ψµ = µ ′ (g)ψµ.<br />
dim Vµ = [G(K/F), G(K/Lµ)]<br />
<strong>the</strong>Frobeniusreciprocity<strong>the</strong>oremimpliesthat<strong>the</strong>restriction<strong>of</strong> σto Vµisequivalentto<br />
Ind(G(K/F), G(K/Lµ), µ ′ ).<br />
Lemma4.2is<strong>of</strong>courseaspecialcase<strong>of</strong>Lemma4.3.<br />
Lemma 4.4<br />
λ(E/F, ΨF)isdifferentfrom0forall E/Fin P◦(K/F).<br />
Thelemmaisclearif [K : F] = 1.Weproveitbyinductionon [K : F].Let Gbe<strong>the</strong>set<strong>of</strong><br />
E/Fin P◦(K/F)forwhich λ(E/F, ψF) = 0.WemayapplyLemma3.3.Thefirstcondition<br />
<strong>of</strong>thatlemmaisclearlysatisfied.Thesecondfollowsfrom<strong>the</strong>inductionassumptionand<strong>the</strong><br />
firstpart<strong>of</strong>Lemma4.1;<strong>the</strong>thirdfrom<strong>the</strong>inductionassumptionand<strong>the</strong>secondpart<strong>of</strong>Lemma<br />
4.1. ThefourthandfifthfollowfromLemmas4.2and4.3respectively. We<strong>of</strong>courseuse<strong>the</strong><br />
factthat ∆(χE, ΨE),whichisbasicallyaGaussiansumwhen Eisnonarchimedean,isnever<br />
zero.
Chapter4 25<br />
Forevery E ′ /Ein P(K/F)wecandefine λ(E ′ /E, ψ E/F)tobe<br />
Lemma 4.5<br />
λ(E ′ /F, ψF) λ(E/F, ψF) −[E′ :E] .<br />
If E ′′ /E ′ and E ′ /Ebelongto P(K/F)<strong>the</strong>n<br />
Indeed<br />
whichequals<br />
λ(E ′′ /E, ψ E/F) = λ(E ′′ /E ′ , ψ E ′ /F)λ(E ′ /E, ψ E/F) E′′ :E ′ ] .<br />
λ(E ′′ /E, ψ E/F) = λ(E ′′ /F, ψF) λ(E/F, ψF) −[E′′ :E]<br />
λ(E ′′ /F, ψF) λ(E ′ /F, ψF) −[E′′ :E ′ ] λ(E ′ /F, ψF) [E′′ :E ′ ] λ(E/F, ψF) −[E′′ :E] <br />
andthisinturnequals<br />
Lemma 4.6<br />
λ(E ′′ /E ′ , ψ E ′ /F) λ(E ′ /E, ψ E/F) [E′′ :E ′ ] .<br />
If λ1(·, ΨF)and λ2(·, ΨF)aretwoweak λ<strong>functions</strong>on P◦(K/F)<strong>the</strong>n<br />
forall E ′ /Ein P(K/F).<br />
λ1(E ′ /E, ψ E/F) = λ2(E ′ /E, ψ E/F)<br />
WeapplyLemma3.1to<strong>the</strong>collectionG<strong>of</strong>allpairsE ′ /Ein P(K/F)forwhich<strong>the</strong>equality<br />
isvalid.Thefirstcondition<strong>of</strong>thatlemmaisclearlysatisfied.Thesecondisaconsequence<strong>of</strong><br />
<strong>the</strong>previouslemma.Thethirdandfourthareconsequences<strong>of</strong>Lemmas4.2and4.3respectively.<br />
Sinceaλfunctionisalsoaweak λfunction<strong>the</strong>uniqueness<strong>of</strong>Theorem2.1isnowproved.
Chapter5 26<br />
Chapter Five.<br />
A Property <strong>of</strong> λ-Functions<br />
Itfollowsimmediatelyfrom<strong>the</strong>definitionthatif ψ ′ E (x) = ψE(βx)<strong>the</strong>n<br />
∆(χE, ψ ′ E ) = χE(β)∆(χE, ψE).<br />
Associatedtoanyequivalenceclass ω<strong>of</strong>representations<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong><strong>the</strong>field Fisa<br />
onedimensionalrepresentationor,whatis<strong>the</strong>same,aquasicharacter<strong>of</strong> CF. Itisdenoted<br />
detωandisobtainedbytaking<strong>the</strong>determinant<strong>of</strong>anyrepresentationin ω.Suppose ρisin<strong>the</strong><br />
class ωand ρisarepresentation<strong>of</strong> W K/F.T<strong>of</strong>ind<strong>the</strong>value<strong>of</strong><strong>the</strong>quasicharacter detωat β<br />
choose win W K/Fsothat τ K/Fw = β.Thencalculate det(ρ(w))whichequals detω(β).<br />
If F ⊆ E ⊆ K<strong>the</strong>map τ = τ K/F canbeeffectedintwostages. Wefirsttransfer<br />
W K/F/W c K/F into W K/E/W c K/E ;<strong>the</strong>nwetransfer W K/E/W c K/E into CK. If W K/F is<strong>the</strong><br />
disjointunion<br />
r<br />
i=1 W K/Ewi<br />
andif wiw = ui(w)wj(i)<strong>the</strong>n<strong>the</strong>transfer<strong>of</strong> win W K/E/W c K/E is<strong>the</strong>cosettowhich w′ =<br />
r<br />
i=1 ui(w)belongs.<br />
Suppose σisarepresentation<strong>of</strong> W K/Eand<br />
ρ = Ind(W K/F, W K/E, σ).<br />
ρactsonacertainspace V <strong>of</strong><strong>functions</strong>on W K/Fandif Viis<strong>the</strong>collection<strong>of</strong><strong>functions</strong>in V<br />
whichvanishoutside<strong>of</strong> W K/Ewi<strong>the</strong>n<br />
V = r<br />
i=1 Vi.<br />
Wedecompose<strong>the</strong>matrix<strong>of</strong> ρ(w)intocorrespondingblocks ρji(w). ρji(w)is0unless j = j(i)<br />
when ρji(w) = σ(ui, (w)). Thismakesitclearthatif ι E/F is<strong>the</strong>representation<strong>of</strong> W K/F<br />
inducedfrom<strong>the</strong>trivialrepresentation<strong>of</strong> W K/E<br />
or,if θis<strong>the</strong>class<strong>of</strong> σ,<br />
det(ρ(w)) = det(ι E/F(w)) dim σ det(σ(w ′ ))<br />
detω(β) = {det ι E/F(β)} dim θ {det θ(β)}.
Chapter5 27<br />
Lemma 5.1<br />
Suppose F isalocalfieldand E/F −→ λ(E/F, ψF)and ω −→ ε(ω, ψ E/F)satisfy<br />
<strong>the</strong>conditions<strong>of</strong>TheoremAfor<strong>the</strong>character ψF. Let ψ ′ F (x) = ψF(βx)with βin CF. If<br />
E/F −→ λ(E/F, ψ ′ F )and ω −→ ε(ω, ψ′ E/F )satisfy<strong>the</strong>conditions<strong>of</strong>TheoremAfor ψ′ F <strong>the</strong>n<br />
and<br />
λ(E/F, ψ ′ F ) = det ι E/F(β) λ(E/F, ψF)<br />
ε(ω, ψ ′ E/F ) = det ω(β) ε(ω, ψ E/F).<br />
Because<strong>of</strong><strong>the</strong>uniquenessallonehastodoisverifythat<strong>the</strong>expressionson<strong>the</strong>right<br />
satisfy<strong>the</strong>conditions<strong>of</strong><strong>the</strong><strong>the</strong>oremfor<strong>the</strong>character ψ ′ F .Thiscannowbedoneimmediately.
Chapter6 28<br />
Chapter Six.<br />
A Filtration <strong>of</strong> <strong>the</strong> Weil Group<br />
InthisparagraphIwanttoreformulatevariousfactsfoundinSerre’sbook[12]asassertionsaboutafiltration<strong>of</strong><strong>the</strong>Weilgroup.Althoughsome<strong>of</strong><strong>the</strong>lemmast<strong>of</strong>ollowwillbeused<br />
toprove<strong>the</strong>fourmainlemmas,<strong>the</strong>introduction<strong>of</strong><strong>the</strong>filtrationitselfisnotreallynecessary.<br />
Itservesmerelytouniteinaformwhichiseasilyremembered<strong>the</strong>separatelemmas<strong>of</strong>which<br />
wewillactuallybeinneed.<br />
Let KbeafiniteGaloisextension<strong>of</strong><strong>the</strong>nonarchimedeanlocalfield F andlet G =<br />
G(K/F).Let OFbe<strong>the</strong>ring<strong>of</strong>integersin Fandlet pFbe<strong>the</strong>maximalideal<strong>of</strong> OF.If i ≥ −1<br />
isanintegerlet Gibe<strong>the</strong>subgroup<strong>of</strong> Gconsisting<strong>of</strong>thoseelementswhichacttriviallyon<br />
.If u ≥ −1isarealnumberand iis<strong>the</strong>smallestintegergreaterthanorequalto uset<br />
OF/p i+1<br />
F<br />
Gu = Gi.Finallyif u ≥ −1set<br />
ϕ K/F(u) =<br />
u<br />
0<br />
1<br />
[G0 : Gt] dt.<br />
Theintegrandisnotdefinedat1butthatis<strong>of</strong>noconsequence. ϕ K/Fisclearlyapiecewise<br />
linear,continuous,andincreasingmap<strong>of</strong> [−1, ∞)ontoitself.Theinversefunction* ψ K/Fwill<br />
have<strong>the</strong>sameproperties.<br />
WetakefromSerre’sbook<strong>the</strong>followinglemma.<br />
Lemma 6.1<br />
If F ⊆ L ⊆ Kand L/Fisnormal<strong>the</strong>n ϕ K/F = ϕ L/F ◦ ϕ K/Land ψ K/F = ψ K/L ◦ ψ L/F.<br />
Thecircledenotescompositionnotmultiplication.Thislemmaallowsustodefine ϕ E/F<br />
and ψ E/Fforanyfiniteseparableextension E/FbychoosingaGaloisextension L<strong>of</strong> Fwhich<br />
contains Eandsetting<br />
ϕ E/F = ϕ L/F ◦ ψ L/E<br />
ψ E/F = ϕ L/E ◦ ψ L/F<br />
*Inthischapter ψ K/Fdoesnotappearasanadditivecharacter. None<strong>the</strong>less,<strong>the</strong>reisa<br />
regrettableconflict<strong>of</strong>notation.
Chapter6 29<br />
becauseif L ′ isano<strong>the</strong>rsuchextensionwecanchooseaGaloisextension Kcontainingboth L<br />
and L ′ and<br />
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<br />
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.<br />
Ofcourse ψ E/Fis<strong>the</strong>inverse<strong>of</strong> ϕ E/F.<br />
Lemma 6.2<br />
If E ⊆ E ′ ⊆ E ′′ and E ′′ /Eisfiniteandseparable, ϕ E ′′ /E = ϕ E ′ /E ◦ ϕ E ′′ /E ′ and<br />
ψ E ′′ /E = ψ E ′′ /E ′ ◦ ψ E ′ /E.<br />
Each<strong>of</strong><strong>the</strong>serelationscanbeobtainedfrom<strong>the</strong>o<strong>the</strong>rbytakinginverses;weverify<strong>the</strong><br />
second<br />
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.<br />
Itwillbenecessaryforustoknow<strong>the</strong>values<strong>of</strong><strong>the</strong>se<strong>functions</strong>inafewspecialcases.<br />
Lemma 6.3<br />
(i)If K/FisGaloisandunramified ψ K|F(u) ≡ u.<br />
(ii)If K/Fiscyclic<strong>of</strong>primedegree ℓandif G = Gtwhile Gt+1 = {1}where tisanonnegativeinteger<strong>the</strong>n<br />
ψ K/F(u) = u u ≤ t<br />
= t + ℓ(u − t) u ≥ t.<br />
Theseassertionsfollowimmediatelyfrom<strong>the</strong>definitions.<br />
Lemma 6.4<br />
SupposeK/FisGaloisandG = G(K/F)isaproductHCwhere H = {1}, H∩C = {1},<br />
and Cisanontrivialabeliannormalsubgroup<strong>of</strong> Gwhichiscontainedineverynontrivial<br />
normalsubgroup.<br />
(i)If K/Fistamelyramifiedsothat G1 = {1}<strong>the</strong>n G0 = Cisacyclicgroup<strong>of</strong>primeorder<br />
ℓand [G : G0] = [H : 1]divides ℓ −1.If Eis<strong>the</strong>fixedfield<strong>of</strong> H, ψ E/F(u) = ufor u ≤ 0<br />
and ψ E/F(u) = ℓufor u ≥ 0.
Chapter6 30<br />
(ii)If K/Fiswildlyramified<strong>the</strong>reisaninteger t ≥ 1suchthat C = G1 = . . . = Gtwhile<br />
Gt+1 = {1}. [G0 : G1]divides [G1 : 1] − 1andeveryelement<strong>of</strong> Chasorder por1.If E<br />
is<strong>the</strong>fixedfield<strong>of</strong> Hand Lthat<strong>of</strong> C<br />
while<br />
ψ L/F(u) = u u ≤ 0<br />
= [G0 : G1]u u ≥ 0<br />
t<br />
ψE/F(u) = u u ≤<br />
[G0 : G1]<br />
t<br />
=<br />
[G0 : G1] + [G1<br />
<br />
t<br />
t<br />
: 1] u − u ≥<br />
[G0 : G1] [G0 : G1]<br />
Weobservedin<strong>the</strong>thirdparagraphthat Cmustbeitsowncentralizer. G0cannotbe {1}.<br />
Thus C ⊆ G0.Incase(i) G0isabelianandthus G0 = C.Inbothcasesif ℓisaprimedividing<br />
<strong>the</strong>order<strong>of</strong> C<strong>the</strong>set<strong>of</strong>elementsin C<strong>of</strong>order ℓor1isanontrivialnormalsubgroup<strong>of</strong> Gand<br />
thus Citself.Incase(i) Ciscyclicandthus<strong>of</strong>primeorder ℓ.Moreover, Hwhichisisomorphic<br />
to G/G0isabelianand,if h ∈ H, {c ∈ C|hc = ch}isanormalsubgroup<strong>of</strong> Gandhence {1}<br />
or C.If h = 1itmustbe1.Consequentlyeachorbit<strong>of</strong> Hin C − {1}has [H : 1]elementsand<br />
[H : 1]divides ℓ − 1.<br />
Incase(ii) G1isanontrivialnormalsubgroupandhencecontains C. G1and Careboth<br />
pgroups.Thecentralizer<strong>of</strong> G1in Cisnottrivial.Asanormalsubgroup<strong>of</strong> Gitcontains C.<br />
Thereforeitis Cand G1iscontainedin Cwhichisitsowncentralizer.Sinceeach G1, i ≥ 1,<br />
isanormalsubgroup<strong>of</strong> G,itisei<strong>the</strong>r Cor {1}. Thus<strong>the</strong>reisaninteger t ≥ 1suchthat<br />
G1 = Gt = Cwhile Gt+1 = {1}. If i ≥ 0isanintegerlet Ui Kbe<strong>the</strong>group<strong>of</strong>units<strong>of</strong> OK<br />
whicharecongruentto1modulo p i+1<br />
K ;let U(−1)<br />
K = CK,andif U ≥ −1isanyrealnumber<br />
let ibe<strong>the</strong>smallestintegergreaterthanorequalto uandset Uu K = Ui K . If θtis<strong>the</strong>map<strong>of</strong><br />
Gt/Gt+1into pt K /pt+1<br />
K and θ0<strong>the</strong>map<strong>of</strong> G0/G1into U0 K /U1 KintroducedinSerre<strong>the</strong>n,for g<br />
in G0and hin C,<br />
θt(ghg −1 ) = θ0(g) t Θt(h).<br />
If h = 1, ghg −1 = hifandonlyif θ0(g) t = 1and<strong>the</strong>n gbelongsto<strong>the</strong>centralizer<strong>of</strong> C,that<br />
isto G1.Again C − {1}isbrokenupintoorbits,eachwith [G0 : G1]elementsand [G0 : G1]<br />
divides [Gi : 1] − 1.Observethat tmustbeprimeto [G0 : G1].
Chapter6 31<br />
Itfollowsimmediatelyfrom<strong>the</strong>definitionsthat Hu = H ∩ Gu.Incase(i) H0willbe {1}<br />
and ϕ K/E(u)willbeidentically u.Thus ψ E/F = ψ K/Fand,from<strong>the</strong>definition, ψ K/F(u) = u<br />
if u ≤ 0while ψ K/F(u) = [G0 : 1]uif u ≥ 0.Incase(ii), ϕ K/E(u) = uif u ≤ 0and<br />
ϕ K/E(u) =<br />
if u ≥ 0while ψ K/F(u) = uif u ≤ 0and<br />
Thelemmafollows.<br />
Lemma 6.5<br />
u<br />
[H0 : 1] =<br />
u<br />
[G0 : G1]<br />
t<br />
ψK/F(u) = [G0 : G1]u 0 ≤ u ≤<br />
[G0 : G1]<br />
<br />
t<br />
t<br />
= t + [G0 : 1] u −<br />
≤ u.<br />
[G0 : G1] [G0 : G1]<br />
Foreveryseparableextension E ′ /E<strong>the</strong>function ψ E ′ /Eisconvex,andif uisaninteger<br />
sois ψ E ′ /E(u).<br />
Allwehavetodoisprovethat<strong>the</strong>assertionistrueforall E ′ /Ein P(K/F)if Fisan<br />
arbitrarynonarchimedeanlocalfieldand KanarbitraryGaloisextension<strong>of</strong>it. Todothis<br />
wejustcombine<strong>the</strong>previousthreelemmaswithLemma3.1. Wearegoingtouse<strong>the</strong>same<br />
methodtoprove<strong>the</strong>followinglemma.<br />
Lemma 6.6<br />
Foreveryseparableextension E ′ /Eandany u ≥ −1<br />
NE ′ /E (U ψE ′ /E (u)<br />
E ′ ) ⊆ U u E .<br />
Wehavetoverifythat<strong>the</strong>set G<strong>of</strong>all E ′ /Ein P(K/F)forwhich<strong>the</strong>assertionistruesatisfies<br />
<strong>the</strong>conditions<strong>of</strong>Lemma3.1.Thereisnoproblemwith<strong>the</strong>firsttwo.<br />
Lemma 6.7<br />
E ′ /Ebelongsto Gifandonlyifforeveryinteger n ≥ −1<br />
N E ′ /E<br />
<br />
U ψ E ′ /E (n)<br />
E ′<br />
<br />
⊆ U n E
Chapter6 32<br />
and<br />
N E ′ /E<br />
<br />
U ψ E ′ /E (n)+1<br />
E ′<br />
<br />
⊆ U n+1<br />
E .<br />
If E ′ /Ebelongsto Gchoose ε > 0sothat ψ E ′ /E(n + ε) = ψ E ′ /E(n) + 1.Thesmallest<br />
integergreaterthanorequalto n + εisatleast n + 1so<br />
N E ′ /E<br />
<br />
U ψ E ′ /E (n)+1<br />
E ′<br />
<br />
⊆ U n+ε<br />
E<br />
⊆ Un+1<br />
E .<br />
Converselysuppose<strong>the</strong>conditions<strong>of</strong><strong>the</strong>lemmaaresatisfiedand n < u < n + 1. Since<br />
ψ E ′ /E(n)isaninteger<strong>the</strong>smallestintegergreaterthanorequalto ψ E ′ /E(u)isatleast<br />
ψ E ′ /E(n) + 1.Thus<br />
Lemma 6.8<br />
and<br />
N E ′ /E<br />
<br />
U ψ E ′ /E (u)<br />
E ′<br />
<br />
⊆ NE ′ /E<br />
If L/EisGalois<strong>the</strong>n,foreveryinteger n ≥ −1,<br />
N L/E<br />
N L/E<br />
<br />
<br />
U ψ L/E(n)<br />
L<br />
U ψ L/E(n)+1<br />
L<br />
U ψ E ′ /E (n)+1<br />
E ′<br />
<br />
⊆ U n E<br />
<br />
<br />
⊆ U n+1<br />
E .<br />
⊆ U n+1<br />
E = Uu E.<br />
Theassertionisclearif n = −1.Apro<strong>of</strong>for<strong>the</strong>case n ≥ 0and L/Etotallyramifiedisgiven<br />
inSerre’sbook.Sincethatpro<strong>of</strong>worksequallywellforall L/Ewetake<strong>the</strong>lemmaasproved.<br />
Lemma 6.9<br />
Suppose K/FisGaloisand G = G(K/F).Suppose G = HCwhere H = {1}, H ∩ C =<br />
{1},andCisanontrivialabeliannormalsubgroup<strong>of</strong>Gwhichiscontainedineverynontrivial<br />
normalsubgroup<strong>of</strong> G.If Eis<strong>the</strong>fixedfield<strong>of</strong> H<br />
forall u ≥ −1.<br />
N E/F<br />
<br />
U ψ E/F(u)<br />
E<br />
<br />
⊆ U u F
Chapter6 33<br />
Let Lbe<strong>the</strong>fixedfield<strong>of</strong> C.If K/Fistamelyramified K/Eand L/Fareunramifiedso<br />
that ψE/F = ψK/Land Uv E = CE ∩ Uv K , Uv F = CF ∩ Uv Lforevery v ≥ −1.If αbelongsto CE,<br />
<strong>the</strong>ndelete NK/Lα = NE/Fα.Since K/LisGalois<br />
N E/F<br />
<br />
U ψ E/F(u)<br />
E<br />
If K/Fisnottamelyramified<br />
if n ≥ 1and 0 ≤ m < [G0 : G1].Thus<br />
if −1 ≤ v ≤ 0and<br />
if v ≥ 0or,morebriefly,<br />
forall v ≥ −1.In<strong>the</strong>samewaywefind<br />
forall v ≥ −1.Since K/Lisnormal<br />
N E/F<br />
<br />
U ψ E/F(u)<br />
E<br />
<br />
⊆ CF ∩ NK/L (U ψK/L(u) L ) ⊆ CF ∩ U u L = U u F.<br />
p n E = E ∩ p [G0:G1]n−m<br />
k<br />
U v E = GE ∩ U v K<br />
U v E = CE ∩ U [G0:G1]v<br />
K<br />
U v E = CE ∩ U ψ K/E(v)<br />
K<br />
U v F = CF ∩ U ψ L/F(v)<br />
L<br />
<br />
⊆ CF ∩ NK/L Lemma6.6nowfollowsimmediately.<br />
Lemma 6.10<br />
U ψ K/F(u)<br />
K<br />
<br />
⊆ CF ∩ U ψ L/F(u)<br />
L<br />
= U u F.<br />
(a)Suppose K/F isGaloisand G = G(K/F). Suppose t ≥ −1isanintegersuchthat<br />
G = Gt = Gt+1.Then ψ K/F(u) = ufor u ≤ t.Moreover N K/Fdefinesanisomorphism<br />
<strong>of</strong> CK/U t K with CF/U t F andif −1 ≤ u ≤ t<strong>the</strong>inverseimage<strong>of</strong> Uu F /Ut F is Uu K /Ut K .<br />
However<strong>the</strong>map<strong>of</strong> CK/U t+1<br />
K<br />
into CF/U t+1<br />
F definedby<strong>the</strong>normisnotsurjective.<br />
(b)Suppose K/FisGaloisand G = G(K/F).Suppose s ≥ −1isanintegerand G = Gs.If<br />
F ⊆ E ⊆ K, ψ E/F(u) = ufor u ≤ sand N E/Fdefinesanisomorphism<strong>of</strong> CE/U s E and<br />
CF/U s F .If −1 ≤ u ≤ s<strong>the</strong>inverseimage<strong>of</strong> Uu F /Us F is Uu E /Us E .<br />
If t = −1<strong>the</strong>assertions<strong>of</strong>part(a)areclear.If t ≥ 0, K/Fistotallyramified.Therelation<br />
ψ K/F(u) = ufor u ≤ tisanimmediateconsequence<strong>of</strong><strong>the</strong>definition. Since<strong>the</strong>extension
Chapter6 34<br />
istotallyramified N K/Fdefinesanisomorphism<strong>of</strong> U −1<br />
K /U0 K<br />
−1<br />
and UF /U0 F . Itfollowsfrom<br />
PropositionV.9<strong>of</strong>Serre’sbookthatif 0 ≤ n < t<strong>the</strong>associatedmap Un K /Un+1<br />
K −→ Un F /Un+1<br />
F<br />
isanisomorphismbutthat<strong>the</strong>map U t K /Ut+1<br />
K −→ Ut F /Ut+1<br />
F hasanontrivialcokernel.The<br />
firstpart<strong>of</strong><strong>the</strong>lemmaisanimmediateconsequence<strong>of</strong><strong>the</strong>sefacts.<br />
Toprovepart(b)wefirstobservethat<strong>the</strong>reisat ≥ ssuchthat G = Gt = Gt+1. It<br />
<strong>the</strong>nfollowsfrompart(a)that<strong>the</strong>map NK/F determinesanisomorphism<strong>of</strong> CK/U s Kand CF/U s F underwhich Uu K /Us Kand Uu F /Us F correspondif −1 ≤ u ≤ s. Let Ebe<strong>the</strong>fixed<br />
field<strong>of</strong> H. Wehave Hs = H ∩ Gs = H,sothat NK/E determinesonisomorphism<strong>of</strong><br />
CK/U s Kand CE/U s Eunderwhich Uu K /Us Kand Uu E /Us Ecorrespondif −1 ≤ u ≤ s.Moreover<br />
if u ≤ s, ψK/F(u) = ψK/E(u) = usothat ψE/F(u) = u. Part(b)followsfrom<strong>the</strong>se<br />
observationsand<strong>the</strong>relation NK/F = NE/FNK/E. If Eisanynonarchimedeanlocalfieldand u > −1<br />
If αbelongsto CEset<br />
U u E = ∩v
Chapter6 35<br />
If σ = 1bothsidesareinfiniteand<strong>the</strong>assertionisclear.If σ = 1let Ebe<strong>the</strong>fixedfield<br />
<strong>of</strong> 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) =<br />
ϕ E/F(v K/E(σ)).Consequentlyitissufficienttoprove<strong>the</strong>lemmawhen F = E.Theset<br />
S = {τ K/F(w) | σ(w) = σ}<br />
isacoset<strong>of</strong> N K/F(CK)in CFand CFisgeneratedby N K/F(CK)toge<strong>the</strong>rwithanyelement<br />
<strong>of</strong> S.Moreover s = max{vF(β) | β ∈ S}is<strong>the</strong>largestintegersuchthat S ∩ U s F isnotempty.<br />
Since G = Gt = Gt+1<strong>the</strong>precedinglemmashowsthat s = t = ϕ K/F(t).<br />
Lemma 6.12<br />
(a)Forall wand w1in W K/F, v K/F(w) = v K/F(w −1 )and v K/F(w1w w −1<br />
1 ) = v K/F(w).<br />
(b)If F ⊆ E ⊆ Kand wbelongto W K/E<strong>the</strong>n<br />
(c)Forall win WK/F, τK/F(w) ⊂ U vK/F(w) F .<br />
v K/F(w) = ϕ E/F(v K/E(w)).<br />
Thefirsttwoassertionsfollowimmediatelyfrom<strong>the</strong>definitionsand<strong>the</strong>basicproperties<br />
<strong>of</strong><strong>the</strong>Weilgroup. Iproveonly<strong>the</strong>third. Letmefirstobservethatif F ⊆ E ⊆ Kand<br />
w ⊂ W K/E,<strong>the</strong>n<br />
τ K/F(w) = N E/F(τ K/E(w)).<br />
Toseethis,chooseaset<strong>of</strong>representatives w1, . . ., wrfor<strong>the</strong>cosets<strong>of</strong> CKin W K/Eand<strong>the</strong>n<br />
aset<strong>of</strong>representatives v1, . . ., vsfor<strong>the</strong>cosets<strong>of</strong> W K/Ein W K/F.Let wiw = aiw j(i)with ai<br />
in CK;<strong>the</strong>n<br />
τ K/E(w) = r<br />
i=1 ai.<br />
However vjwiw = vjaiv −1<br />
j vjw j(i)sothat<br />
τ K/F(w) = s<br />
j=1<br />
r<br />
i=1<br />
vjaiv −1<br />
j<br />
s<br />
=<br />
j=1 vjτK/E(w)v −1<br />
j = NE/F(τK/E(w)). Inparticular,ifEis<strong>the</strong>fixedfield<strong>of</strong>σ(w), τ K/E(w)iscontained U ψ E/F(v K/F(w))<br />
E<br />
iscontainedin<br />
Lemma 6.13<br />
N E/F (U ψ E/F(v K/F(w))<br />
E<br />
) ⊆ U vK/F(w) F .<br />
and τ K/F(w)
Chapter6 36<br />
If uand vbelongto W K/F<strong>the</strong>n<br />
v K/F(uv) ≥ min{v K/F(u), v K/F(v)}.<br />
Let σ = σ(u)andlet τ = σ(v). Because<strong>of</strong><strong>the</strong>secondassertion<strong>of</strong><strong>the</strong>previouslemma<br />
wemayassumethat σand τgenerate G(K/F).Let Ebe<strong>the</strong>fixedfield<strong>of</strong> στ.If<br />
t = {min(ψ K/F(v K/F(σ)), ψ K/F(v K/F(τ))}<br />
and G = G(K/F)<strong>the</strong>n G = Gt = Gt+1.AccordingtoLemma6.11,if<br />
s = min{v K/F(u), v K/F(v)},<br />
<strong>the</strong>n t ≥ ψ K/F(s)which,byLemma6.10,is<strong>the</strong>reforeequalto s.Since τ K/F(uv) =<br />
τ K/F(u)τ K/F(v), τ K/F(uv)liesin U s F .<strong>On</strong><strong>the</strong>o<strong>the</strong>rhand<br />
τ K/F(uv) = N E/F(τ K/E(uv))<br />
sothat,byLemma6.10again, τ K/E(uv)belongsto U s E and<br />
v K/F(uv) ≥ ϕ E/F(s) = s.<br />
Thus<strong>the</strong>sets W x K/F , x ≥ −1,giveafiltration<strong>of</strong> W K/Fbyacollection<strong>of</strong>normalsubgroups.Thenextsequence<strong>of</strong>lemmasshowthat<strong>the</strong>filtrationisquiteanalogousto<strong>the</strong>upper<br />
filtration<strong>of</strong><strong>the</strong>Galoisgroups.<br />
Lemma 6.14<br />
Foreach x ≥ −1<strong>the</strong>map τ K/F,L/Ftakes G x K/F into Gx L/F .<br />
If wbelongsto W K/Flet ¯w = τ K/F,L/F(w).Wemustshowthat<br />
v L/F( ¯w) ≥ v K/F(w).<br />
Let σ = σ(w)andlet ¯σ = σ( ¯w).If Eis<strong>the</strong>fixedfield<strong>of</strong> σ<strong>the</strong>n E = E ∩ Lis<strong>the</strong>fixedfield<strong>of</strong><br />
¯σ.Since<br />
v L/F( ¯w) = ϕ E/F (v L/E ( ¯w))<br />
and<br />
v K/F(w) = ϕ E/F (v K/E (w))
Chapter6 37<br />
wemaysuppose E = F.Since τ K/F(w) = τ L/F( ¯w),Lemma6.12impliesthat τ L/F( ¯w)liesin<br />
U vK/F(w) F .Thus<br />
v L/F( ¯w) = vF(τ L/F( ¯w)) ≥ v K/F(w).<br />
Ofcourse W F/Fis CFand,if v ≥ −1, W v F/F = Uv F .<br />
Lemma 6.15<br />
Foreach v ≥ −1, τ K/Fmaps W v K/F onto Uv F .<br />
Since v1 ≤ v2implies W v2<br />
K/F<br />
⊆ W v1<br />
K/F<br />
itisenoughtoprove<strong>the</strong>lemmawhen v = nis<br />
aninteger. Thelemmaisclearif [K : F] = 1;soweproceedbyinductionon [K : F]. If<br />
[K : F] > 1,chooseanintermediatenormalextension Lsothat [L : F] = ℓisaprime. Let<br />
G = G(L/F).Lemma6.12impliesthat<br />
W ψ L/F(v)<br />
K/L<br />
= W K/L ∩ W v K/F .<br />
Thereisaninteger t ≥ −1suchthat G = Gtand Gt+1 = {1}. ItisshowninChapterV<strong>of</strong><br />
Serre’sbookthatif n > t <br />
NL/F <br />
= U n F.<br />
Byinduction<br />
τ K/L<br />
<br />
U ψ L/F(n)<br />
L<br />
W ψ L/F(n)<br />
K/L<br />
Since τ K/F(w) = N L/F(τ K/L(w))if wisin W K/L,<br />
<br />
τ K/F(W n K/F ) = Un F<br />
= U ψL/F(n) L .<br />
if n > t.Suppose σgenerates G.Then V L/F(σ) = t. ByHerbrand’s<strong>the</strong>orem<strong>the</strong>reisaσin<br />
G(K/F)with v K/F(σ) = twhoserestrictionto Lis σ.ByLemma6.11<strong>the</strong>reisawin W K/F<br />
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<br />
Serre’sbookagain<br />
<br />
U t F : NL/F U ψ <br />
L/F(t)<br />
F = ℓ<br />
sothat Ut Fisgeneratedby τK/F(w)and NL/F(U ψL/F(t) L )andhenceiscontainedin<strong>the</strong>image<br />
<strong>of</strong> W t K/F . Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong><strong>the</strong>lemmawehaveonlytoobservethatLemma6.10<br />
impliesthat<br />
U n F = U t F N L/F<br />
<br />
U ψ L/F(n)<br />
L
Chapter6 38<br />
if n ≤ t.<br />
Lemma 6.16<br />
Suppose F ⊆ L ⊆ Kand L/Fand K/FareGalois.Then,foreach v ≥ −1, τ K/F,L/F<br />
maps W v K/F onto W v L/F .<br />
If [L : F] = 1thisisjust<strong>the</strong>previouslemmasoweproceedbyinductionon [L : F].We<br />
havetoshowthatif wbelongsto W L/F <strong>the</strong>reisawin W K/F suchthat w = τ K/F,L/F(w)<br />
and v K/F(w) ≥ v L/F(w). Let σ = σ(w)andlet E be<strong>the</strong>fixedfield<strong>of</strong> σ. If E = F<br />
<strong>the</strong>n,by<strong>the</strong>inductionassumption,<strong>the</strong>reisawin W K/Esuchthat τ K/E,L/E (w) = wand<br />
v K/E (w) ≥ v L/E (w).ByLemma6.12, v K/F(w) ≥ v L/F(w).Moreover,wemayassumethat<br />
τ K/E,L/E is<strong>the</strong>restrictionto W K/E <strong>of</strong> τ K/F,L/F.<br />
Suppose E = F.Then v L/F(w) = vF(τ L/F(w)).Choose w1inW K/Fsothat τ K/F(w1) =<br />
τ L/F(w)andv K/F(w1) ≥ vF(τ L/F(w)).Letw1 = τ K/F,L/F(w1)andsetu = w −1<br />
1 w.Certainly<br />
v L/F(u) ≥ v L/F(w).Moreover, τ L/F(u) = 1.Let F ⊆ L1 ⊆ Lwhere L1/Fiscyclic<strong>of</strong>prime<br />
order.If udoesnotbelongto W L/L1 <strong>the</strong>group CFisgeneratedby N L1/F(CL1 )and τ L/F(u),<br />
whichisimpossiblesince τ L/F(u) = 1.Thus ubelongsto W L/L1 and,asobserved,<strong>the</strong>reisa<br />
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<br />
Chapter Seven.<br />
Consequences <strong>of</strong> Stickelberger’s Result<br />
DavenportandHasse[5]haveshownthatStickelberger’sarithmeticcharacterization<strong>of</strong><br />
Gaussiansumsoverafinitefieldcanbeusedtoestablishidentitiesbetween<strong>the</strong>seGaussian<br />
sums. AfterreviewingStickelberger’sresultweshallprove<strong>the</strong>identities<strong>of</strong>Davenportand<br />
Hassetoge<strong>the</strong>rwithsomemorecomplicatedidentities. Howeverfor<strong>the</strong>pro<strong>of</strong><strong>of</strong>Stickelberger’sresultitself,IrefertoDavenportandHasse.<br />
If Z = e 2πi<br />
p and αbelongsto GF(p)<strong>the</strong>meaning<strong>of</strong> Z α isclear.If κisanyfinitefieldand<br />
Sis<strong>the</strong>absolutetrace<strong>of</strong> κlet ψ 0 κ be<strong>the</strong>character<strong>of</strong> κdefinedby ψ0 κ (α) = ZS(α) .If χκisany<br />
character<strong>of</strong> κ ∗ and ψκisanynontrivialadditivecharacter<strong>of</strong> κwewilltake<strong>the</strong>Gaussiansum<br />
τ(χκ, ψκ)tobe<br />
Weabbreviate τ(χκ, ψ0 κ )to τ(χκ).<br />
− <br />
χ−1<br />
α∈κ∗ κ (α)ψκ(α).<br />
Let knbe<strong>the</strong>fieldobtainedbyadjoining<strong>the</strong> n th roots<strong>of</strong>unityto<strong>the</strong>rationalnumbers.<br />
If ̟ = Z − 1<strong>the</strong>nin kp<strong>the</strong>ideal (p)equals (̟ p−1 ).If q = p f and κhas qelements<strong>the</strong>nin<br />
kq−1<strong>the</strong>ideal (p)isaproduct pp ′ . . .where<strong>the</strong>residuefields<strong>of</strong> pp ′ , . . .areisomorphicto κ.<br />
In k p(q−1)<br />
(p) = (p p ′ . . .) p−1<br />
with P = (p, ̟), P ′ = (p ′ , ̟),andsoon.Theresiduefields<strong>of</strong> P, P ′ , . . .arealsoisomorphic<br />
to κ.Chooseone<strong>of</strong><strong>the</strong>seprimeideals,say P.<strong>On</strong>ceanisomorphism<strong>of</strong><strong>the</strong>residuefieldwith<br />
κischosen<strong>the</strong>map<strong>of</strong><strong>the</strong> (q − 1) th roots<strong>of</strong>unityto<strong>the</strong>residuefielddefinesanisomorphism<br />
<strong>of</strong> κ ∗ and<strong>the</strong>group<strong>of</strong> (q − 1) th roots<strong>of</strong>unity.Then χκcanberegardedasacharacter<strong>of</strong><strong>the</strong><br />
lattergroup.Choose α = α(χκ, P)with 0 ≤ α < q − 1sothat χκ(ζ) = ζ α forall (q − 1) th<br />
roots<strong>of</strong>unity.Write<br />
α = α0 + α1p + . . . + αf−1p f−1<br />
Notall<strong>of</strong><strong>the</strong> αicanbeequalto p − 1.Set<br />
σ(α) = α0 + α1 + . . . + αf−1<br />
γ(α = α0!α1! . . .αf−1!<br />
0 ≤ αi < p.<br />
ThefollowinglemmaisStickelberger’sarithmeticalcharacterization<strong>of</strong> τ(χκ).
Chapter7 40<br />
Lemma 7.1<br />
(a) τ(χκ)liesin k p(q−1)andisanalgebraicinteger.<br />
(b)If χκ = 1<strong>the</strong>n τ(χκ) = 1butif χκ = 1<strong>the</strong>absolutevalue<strong>of</strong> τ(χκ)andallitsconjugates<br />
is √ q.<br />
(c)Everyprimedivisor<strong>of</strong> τ(χκ)in k p(q−1)isadivisor<strong>of</strong> p.<br />
(d)If βisanonzeroelement<strong>of</strong><strong>the</strong>primefield<strong>the</strong>n<strong>the</strong>automorphism Z −→ Z β <strong>of</strong> k p(q−1)<br />
over kq−1sends τ(χκ)to χκ(β) τ(χκ).<br />
(e)If Pisaprimedivisor<strong>of</strong> pin k p(q−1)and α = α(χκ, p)<strong>the</strong>multiplicativecongruence<br />
isvalid.<br />
τ(χκ) ≡ ̟σ (α)<br />
γ(α)<br />
(mod ∗ P)<br />
(f)Suppose ℓisaprimedividing q − 1and χκ = χ ′ κχ ′′ κ where<strong>the</strong>order<strong>of</strong> χ′ κ isapower<strong>of</strong><br />
ℓandthat<strong>of</strong> χ ′′ κ isprimeto ℓ.If ℓa is<strong>the</strong>exactpower<strong>of</strong> ℓdividing q − 1and λ = ζ0 − 1<br />
where ζ0isaprimitive ℓ a throot<strong>of</strong>unity<strong>the</strong>n<br />
τ(χκ) ≡ τ(χ ′′ κ) (mod λ).<br />
Beforestating<strong>the</strong>identitiesforGaussiansumswhichareimpliedbythislemma,Ishall<br />
proveafewelementarylemmas.<br />
Lemma 7.2<br />
Suppose 0 ≤ α < p f − 1and<br />
α = α0 + α1p + . . . + αf−1p f−1 0 ≤ αi < p.<br />
Supposealsothat 0 ≤ j0 < j1 < . . . < jr = fandset<br />
If σ = r−1<br />
s=0 βsand γ = r−1<br />
s=0 βs!<strong>the</strong>n<br />
βs = αjs + αjs+1 p + . . . + αj s+1 −1p js+1−js−1 .<br />
̟ σ<br />
γ<br />
= ̟σ(α)<br />
γ(α)<br />
(mod ∗ P).
Chapter7 41<br />
First<strong>of</strong>all,Iremarkonceandforallthatif n ≥ 1, 0 < u ≤ p n − 1,and v = u(modp n )<br />
<strong>the</strong>n v ≡ u(mod ∗ p).Thusif 0 ≤ u ≤ p n − 1and v ≥ 0<br />
Alsoif v ≥ 0<br />
and,byinduction,<br />
(u + vp n )! = (vp n )! u<br />
p n<br />
w=1 (w + vpn ) ≡ u!(vp n )! (mod ∗ p).<br />
w=1 (vpn + w) ≡ (v + 1) p n ! (mod ∗ p)<br />
(vp n )! ≡ v!(p n !) v<br />
(mod ∗ p).<br />
Inparticular p (n+1) ! ≡ p!(p n !) p ≡ (−p) (p n !) p .Applyinductiontoobtain<br />
From<strong>the</strong>relations<br />
and<br />
Weconcludethat<br />
p = p−1<br />
p n ! ≡ (−p) pn −1<br />
p−1 (mod ∗ p).<br />
i=1 (1 − Zi ) = (−̟)<br />
p−1 p−1<br />
i=1<br />
Z i − 1<br />
Z − 1<br />
Z i − 1<br />
Z − 1 = 1 + Z + . . . + Zi−1 ≡ i (mod ∗ p).<br />
p = (p − 1)!(−̟) p−1 ≡ −̟ p−1<br />
(mod ∗ p).<br />
Thelemmaitselfisclearif r = fsoweproceedbyinductiondownwardfrom f. Suppose<br />
r < f, js+2 − js = t > 1,and<strong>the</strong>lemmaisvalidfor<strong>the</strong>sequence j0, j1, . . ., js+1 −<br />
1, js+1, . . .jr.Toproveitfor<strong>the</strong>givensequencewehaveonlytoshowthatif<br />
and y = αjs+1−1<strong>the</strong>n<br />
But<br />
and<br />
x = αjs + αjs+1 p + . . . + αj s+1 −2 p t−2<br />
̟ x+y<br />
x!y!<br />
≡ ̟x+ypt−1<br />
(x + yp t−1 )!<br />
(mod ∗ p).<br />
̟ y(pt−1 −1) ≡ (−p) y pt−1 −1<br />
p−1 (mod ∗ p)<br />
(x + yp t−1 )! = x!y!(p t−1 !) y ≡ x!y!(−p) yt−1 −1<br />
p−1 (mod) ∗ p).
Chapter7 42<br />
Lemma 7.3<br />
Suppose β0, . . ., βr−1and γ0, . . ., γr−1arenonnegativeintegersall<strong>of</strong>whicharelessthan<br />
orequalto q − 1.Supposethat q = p f isaprimepowerand<br />
r−1<br />
i=0 (βi + γi)q i < 2(q r − 1).<br />
Supposealsothat δi, 0 ≤ i ≤ r − 1,aregivensuchthat 0 ≤ δi ≤ q − 1,<br />
r−1<br />
i=0 δiq i < q r − 1<br />
and r−1<br />
i=0 (βi + γi)q i = r−1<br />
i<br />
δiq<br />
i=0<br />
(modq r − 1).<br />
(a) If r−1<br />
i=0 (βi + γi)q i < q r−1 andif νis<strong>the</strong>number<strong>of</strong> k, 1 ≤ k ≤ r,forwhich<br />
k−1<br />
i=0 (βi + γi) ≥ q k <strong>the</strong>n<br />
r−1<br />
i=0 (βi + γi − δi) = ν(q − 1).<br />
(b) If r−1<br />
i=0 (βi + γi)q i ≥ q r − 1andif νis<strong>the</strong>number<strong>of</strong> k, 1 ≤ k ≤ r,forwhich<br />
1 = k−1<br />
i=0 (βi + γi)q i ≥ q k <strong>the</strong>n<br />
r−1<br />
i=0 (βi + γi − δi) = ν(q − 1).<br />
Observeimmediatelythatif 1 ≤ k ≤ r,<strong>the</strong>n 0 ≤ βk−1 + γk−1 ≤ 2(q − 1)and<br />
k−1<br />
i=0 (βi + γi)q i ≤ 2(q − 1) k−1<br />
i=0 qi = 2(q k − 1).<br />
If r = 1<strong>the</strong>n β0 + γ0 = δ0 + ε(q − 1)with εequalto0or1. If ε = 0weareincase(a)and<br />
ν = 0while β0 + γ0 − δ0 = 0.If ε = 1weareincase(b);here ν = 1and β0 + γ0 − δ0 = q − 1.<br />
Suppose<strong>the</strong>nthat r ≥ 2andthatif β ′ 0 , . . ., β′ r−1 , γ′ 0 , . . ., γ′ r−2 , δ′ 0 , . . ., δ′ r−2 ,and ν′ aregiven<br />
asin<strong>the</strong>lemma(with rreplacedby r − 1)<strong>the</strong>n<br />
r−2<br />
i=0 (β′ i + γ ′ i − δ ′ i) = ν ′ (q − 1).
Chapter7 43<br />
Weestablishpart(a)first.Inthiscase<br />
r−1<br />
i=0 (βi + γi)q i = r−1<br />
i=0<br />
and r−2<br />
i=0 (βi + γi)q i = r−2<br />
i=0 δiq i + εq r−1<br />
with ε = δr−1 −βr−1 −γr−1.If εwerenegative<strong>the</strong>leftside<strong>of</strong><strong>the</strong>equationwouldbenegative;<br />
if εweregreaterthan1<strong>the</strong>leftsidewouldbegreaterthan 2(qr−1 − 1).Sincenei<strong>the</strong>r<strong>of</strong><strong>the</strong>se<br />
possibilitiesoccur εis0or1.<br />
Supposefirstthat ε = 0.If r−2 i=0 δiqi < qr−1 −1choose β ′ i = βi, γ ′ i = γi, 0 ≤ i ≤ r −2.<br />
Then δ ′ i = δi, 0 ≤ i ≤ r − 2,and ν ′ = ν.Theassertion<strong>of</strong><strong>the</strong>lemmafollowsinthiscase.If<br />
r−2 1=0 δiqi = qr−1 − 1<strong>the</strong>n δi = q − 1, 0 ≤ i ≤ r − 2.Then β0 + γ0 ≡ q − 1 (modq)and,asa<br />
consequence, β0 + γ0 = q − 1.Weshowbyinductionthat βi + γi = q − 1, 0 ≤ i ≤ r − 2.If<br />
thisiss<strong>of</strong>or i < j<strong>the</strong>n<br />
r−2<br />
i=j (βi + γi)q i = r−2<br />
i=j (q − 1)qi .<br />
Hence βj + γj ≡ q − 1 (modq)and βj + γj = q − 1.Itfollowsimmediatelythat ν = 0and<br />
r−1<br />
i=0 (βi + γi − δi) = 0.<br />
Nowsupposethat ε = 1.If<br />
r−2<br />
δiq i<br />
i=0 (βi + γi)q i = 2(q r−1 − 1)<br />
<strong>the</strong>n βi = γi = q − 1, 0 ≤ i ≤ r − 2, δ0 = q − 2,and δi = q − 1, 1 ≤ i ≤ r − 2. Thus<br />
ν = r − 1and<br />
r−1<br />
Suppose<strong>the</strong>nthat<br />
From<strong>the</strong>relation<br />
i=0 (βi + γi − δi) = 1 + (r − 1) (q − 1) − 1 = (r − 1) (q − 1).<br />
r−2<br />
r−2<br />
i=0 (βi + γi)q i < 2(q r−1 − 1).<br />
i=0 (βi + γi)q i = r−2<br />
i=0 δiq i + 1 + (q r−1 − 1).<br />
Weconcludethat r−2<br />
i=0 δiq i < q r−1 − 1.Thenforsome m,with 0 ≤ m ≤ r − 2, δm < q − 1.<br />
Wechoose<strong>the</strong>minimalvaluefor m.<br />
r−2<br />
i=0 (βi + γi)q i = (δm + 1)q m + r−2<br />
i=m+1 δiq i + (q r−1 − 1).
Chapter7 44<br />
Thusif β ′ i = βi, γ ′ i = γi, 0 ≤ i ≤ r − 2,<strong>the</strong>n δ ′ i = 0, i < m, δ′ m = δm + 1,and δ ′ i = δi, m <<br />
i ≤ r − 2.Arguingbycongruencesasbeforeweseethat βi + γi = q − 1for i < m.Thus<br />
k−1<br />
i=0 (βi + γi)q i = q k − 1<br />
for k ≤ m. However βm + γm = q − 1andthus βm + γm + 1isprimeto q. Moreoverif<br />
r − 1 ≥ k > m<br />
1 + k−1<br />
i=1 (βi + γi)q i ≡ (βm + γm + 1)q m (modq m+1 ).<br />
Thusitisgreaterthanorequalto q k ifandonlyifitisgreaterthanorequalto q k +1.Itfollows<br />
that ν ′ = ν + mandthat<br />
r−2<br />
i=0 (βi + γi − δi) = −m(q − 1) + r−2<br />
i=0 (β′ i + γ′ i − δ′ i ) = ν(q − 1) + 1.<br />
Since βr−1 + γr−1 − δr−1 = −1<strong>the</strong>assertion<strong>of</strong><strong>the</strong>lemmafollows.<br />
and<br />
Nowletustreatpart(b).Inthiscase<br />
r−1<br />
i=0 (βi + γi)q i = r−1<br />
1 + r−2<br />
i=0 (βi + γi)q i = r−2<br />
i=0 δiq i + (q r − 1)<br />
i=0 δiq i + εq r−1<br />
with ε = δr−1 − βr−1 − γr−1 + q.Again εis0or1.If βi = γi = q − 1for 0 ≤ i ≤ r − 2<strong>the</strong>n<br />
ε = 1and δi = q − 1for 0 ≤ i ≤ r − 2.Also ν = rand<br />
r−1<br />
i=0 (βi + γi − δi) = (r − 1) (q − 1) + βr−1 + γr−1 − δr−1 = r(q − 1).<br />
Havingtakencare<strong>of</strong>thiscase,wesupposethat<br />
r−2<br />
i=0 (βi + γi)q i < 2(q r−1 − 1).<br />
Firsttake ε = 0.If δ0 = 0<strong>the</strong>n 1+β0+γ0 ≡ 0 (modq)and β0+γ0 = q−1.Thusone<strong>of</strong><strong>the</strong>mis<br />
lessthan q −1.Bysymmetrywemaysupposeitis β0.Let β ′ 0 = β0 +1, β ′ i = βi, 1 ≤ i ≤ r −2,<br />
and γ ′ i = γi, 0 ≤ i ≤ r − 2.Since δ0 = 0<br />
r−2<br />
i=0 δiq i ≤ q r−1 − q < q r−1 − 1
Chapter7 45<br />
and δ ′ i = δi, 0 ≤ i ≤ r − 1.Also ν = ν ′ + 1sothat<br />
r−1<br />
i=0 (βi + γi − δi) = r−2<br />
i=0 (β′ i + γ′ i − δ′ i ) − 1 + q = ν(q − 1)<br />
asrequired.If δ0 > 0take β ′ i = βiand γ ′ i = γi, 0 ≤ i ≤ r − 2.Then δ ′ 0 = δ0 − 1, δ ′ i = δi, 1 ≤<br />
i ≤ r − 2.Alsoif k ≤ r − 1<br />
k−1<br />
i=0 (βi + γi)q i ≡ δ0 − 1 = −1(modq)<br />
and<strong>the</strong>lefthandsideisgreaterthanorequalto q k ifandonlyifitisgreaterthanorequalto<br />
q k − 1.Itfollowsthat ν = ν ′ + 1.Consequently<br />
r−1<br />
i=0 (βi + γi − δi) = r−2<br />
i=0 (β′ i + γ′ i − δ′ i ) − 1 + q = ν(q − 1).<br />
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<br />
sothat r−1<br />
1=0 (βi + γi − δi) = ν ′ (q − 1) + (βr−1 + γr−1 − δr−1) = ν(q − 1).<br />
Lemma 7.4<br />
Suppose βiand γiaretwoperiodicsequences<strong>of</strong>integerswithperiod r.Thatis βi+r = βi<br />
and γi+r = γiforall iin Z.Suppose 0 ≤ βi ≤ q − 1, 0 ≤ γi ≤ q − 1forall iandthatnone<strong>of</strong><br />
<strong>the</strong>numbers<br />
εk = r−1<br />
i=0 (βi+k + γi+k)q i<br />
isdivisibleby q r − 1.Let<br />
r−1<br />
i=0 (βi + γi)q i ≡ r−1<br />
i=0 δiq i (mod q r − 1)<br />
with 0 ≤ δi ≤ q − 1and r−1<br />
i=0 δiq i < q r − 1.If µis<strong>the</strong>number<strong>of</strong> εk, 1 ≤ k ≤ r,whichare<br />
greaterthanorequalto q r − 1<strong>the</strong>n<br />
r−1<br />
i=0 (βi + γi − δi) = µ(q − 1).<br />
Since ε0 ≤ 2(q r − 1)andisnotdivisibleby q r − 1itislessthan 2(q r − 1).Thusallwe<br />
needdoisshowthat<strong>the</strong> µ<strong>of</strong>thislemmaisequalto<strong>the</strong> ν<strong>of</strong><strong>the</strong>precedinglemma. Observe<br />
first<strong>of</strong>allthat εj ≥ q r − 1ifandonlyif εj ≥ q r .
Chapter7 46<br />
Suppose ε0 < q r .If 1 ≤ k < r<br />
r−1<br />
sothat r−1<br />
Thus,if εk ≥ q r ,<br />
q r ≤ r−1<br />
i=k (βi + γi)q i < q r<br />
i=k (βi + γi)q i−k < q r−k .<br />
i=r−k (βi+k + γi+k) q i + r−k−1<br />
i=0<br />
k−1<br />
r−k<br />
< q<br />
i=0 (βi + γi)q i + q r−k<br />
and k−1<br />
Converselyif 1 ≤ k < rand k−1<br />
<strong>the</strong>n<br />
r−1<br />
Thus µ = νinthiscase.<br />
i=0 (βi + γi)q i ≥ q k .<br />
i=0 (βi + γi)q i ≥ q k ,<br />
i=0 (βi+k + γi+k)q i ≥ r−1<br />
Nowsuppose ε0 ≥ q r .If 1 ≤ k < r<br />
r−1<br />
i=k (βi + γi)q i ≥ q r − k−1<br />
If k−1<br />
<strong>the</strong>n<br />
r−1<br />
i=0 (βi+k + γi+k)q i ≥ r−1<br />
(βi+k + γi+k)q i<br />
i=r−k (βi+k + γi+k)q i<br />
k−1<br />
r−k<br />
= q<br />
i=0 (βi + γi)q i ≥ q r .<br />
i=0 (βi + γi)q i ≥ q r − 2(q k − 1).<br />
i=0 (βi + γi)q i ≥ q k − 1<br />
i=r−k (βi+k + γi+k)q i + r−k−1<br />
i=0<br />
(βi+k + γi+k)q i<br />
= q r−k k−1<br />
i=0 (βi + γi)q i + q −k r−1<br />
i=k (βi + γi)q i<br />
≥ q r−k (q k − 1) + q r−k − 2 + 2q −k .
Chapter7 47<br />
Thus εk ≥ q r − 1andhence εk ≥ q r .Converselyif εk ≥ q r ,<br />
Thus<br />
andagain µ = ν.<br />
Lemma 7.5<br />
k−1<br />
r−k<br />
q<br />
i=0 (βi + γi)q i = r−1<br />
i=r−k (βi+k + γi+k)q i<br />
k−1<br />
≥ q r − r−k−1<br />
i=0<br />
≥ q r − 2(q r−k − 1)<br />
= q r − 2q r−k + 2.<br />
i=0 (βi + γi)q i ≥ q k − 1<br />
(βi+k + γi+k)q i<br />
Supposeq = p f isaprimepower, ℓisapositiveinteger,and(ℓ, q) = 1.Letℓm ≡ 1 (modq)<br />
andif xisanyintegerlet ϕ(x),with 0 ≤ ϕ(x) < q,be<strong>the</strong>remainder<strong>of</strong> xupondivisionby q.<br />
If 0 ≤ β < qandif ψ(x) = ϕ(x)!<br />
ℓβ<br />
β!<br />
ℓ−1<br />
k=0<br />
ψ((β − k)m)<br />
ψ(−km) ≡ 1(mod∗ p).<br />
If ℓ = ℓ1 + uqwith ℓ1 > 0and u ≥ 0<strong>the</strong>n ℓ β ≡ ℓ β<br />
1 (mod∗ p).Moreover<br />
ℓ−1<br />
k=0<br />
and ℓ−1<br />
ψ((β − k)m)<br />
ψ(−km) =<br />
<br />
ℓ1−1<br />
k=0<br />
k=ℓ1<br />
ψ(−km) = { q−1<br />
<br />
ψ((β − k)m) ℓ−1<br />
ψ(−km) k=ℓ1<br />
j=0 j!}u = ℓ−1<br />
k=ℓ1<br />
ψ((β − k)m).<br />
<br />
ψ((β − k)m)<br />
ψ(−km)<br />
Thusitisenoughtoprove<strong>the</strong>lemmawith ℓreplacedby ℓ1.Ino<strong>the</strong>rwordswemaysuppose<br />
that 0 < ℓ < q. Thecase ℓ = 1istrivialandweexcludeitfrom<strong>the</strong>followingdiscussion.<br />
Finallywesupposethat 0 < m < q.
Chapter7 48<br />
Let q − 1 = rℓ + swith 0 < rand 0 ≤ s < ℓ.Arrange<strong>the</strong>integersfrom0to q − 1into<br />
<strong>the</strong>followingarray.<br />
0 1 2 . . . . . . . l − 1<br />
l l + 1 l + 2 . . . . . . . 2l − 1<br />
. . .<br />
. . .<br />
. . β − l + 1 .<br />
. . . β .<br />
. . .<br />
. . .<br />
(r − 1)l (r − 1)l + 1 q − l . rl − 1<br />
rl rl + 1 . . . . . rl + s<br />
Since ℓdoesnotdivide q, rℓ + s = q − 1doesnotliein<strong>the</strong>lastcolumn.Also q − ℓliesin<strong>the</strong><br />
columnfollowingthatinwhich rℓ + slies.<br />
Wereplaceeachnumber jin<strong>the</strong>abovearrayby ϕ(jm). Theresultingarray,whichis<br />
writtenoutbelow,hassomespecialfeatureswhichmustbeexplained. Thefirstcolumnis<br />
explainedby<strong>the</strong>observation xℓm ≡ x(modq). Theo<strong>the</strong>rentries,apartfromthoseat<strong>the</strong><br />
foot<strong>of</strong>eachcolumn,areexplainedby<strong>the</strong>observationthat,when 1 ≤ jand xℓ + jliesin<strong>the</strong><br />
firstarray, ϕ(x + mj) > rwhile 0 < ϕ(mj) + x < q + rsothat ϕ(x + mj) = ϕ(mj) + x.<br />
Theposition<strong>of</strong> q − 1isexplainedby<strong>the</strong>relation m(q − ℓ) ≡ −mℓ ≡ q − 1(mod q). The<br />
o<strong>the</strong>rentriesat<strong>the</strong>feet<strong>of</strong><strong>the</strong>columnsareexplainedby<strong>the</strong>observationthatif 1 ≤ j ≤ ℓ − 1<br />
<strong>the</strong>n ϕ(jm) > r ≥ 1while m(q − k) = m(q − ℓ) + m(ℓ − k) ≡ ϕ((ℓ − k)m) − 1(modq)if<br />
1 ≤ k ≤ ℓ − 1.<br />
0 m . . . . . . . ϕ((l − 1)m)<br />
l m + 1 . . . . . . . ϕ((l − 1)m) + 1<br />
. . .<br />
. . .<br />
. . ϕ((β − l + 1)m) .<br />
. . . ϕ(βm) .<br />
. . .<br />
. . .<br />
r − 1 m + r − 1 q − 1 . ϕ((l − 2 − s)m) − 1<br />
r m + r . . . . ϕ((l − 1)m) − 1<br />
Supposefirst<strong>of</strong>allthat β < ℓ − 1. Then<strong>the</strong>numbers ϕ((β − k)m), 0 ≤ k ≤ ℓ − 1<br />
constitute<strong>the</strong>first β + 1toge<strong>the</strong>rwith<strong>the</strong>last ℓ − β − 1numbersin<strong>the</strong>array.(Theorder<strong>of</strong>
Chapter7 49<br />
<strong>the</strong>numbersin<strong>the</strong>arrayis<strong>the</strong>orderinwhich<strong>the</strong>yappearwhen<strong>the</strong>arrayisreadasthough<br />
itwereaprintedpage.)Thenumbers ϕ(−km), 1 ≤ k ≤ ℓ − 1,are<strong>the</strong>last ℓ − 1numbers<strong>of</strong><br />
<strong>the</strong>array,thatis,<strong>the</strong>numbersafter q − 1.Cancellingin<strong>the</strong>product<strong>of</strong><strong>the</strong>lemma<strong>the</strong>termsin<br />
numeratoranddenominatorcorrespondingto<strong>the</strong>last ℓ − β − 1terms<strong>of</strong><strong>the</strong>array,weobtain<br />
asrequired.<br />
β<br />
j=1<br />
ϕ(jm)!<br />
(ϕ(jm) − 1)!<br />
= β<br />
j=1 ϕ(jm) ≡ mβ β! (mod ∗ p)<br />
Nowtake β ≥ ℓ − 1.Then<strong>the</strong>numbers β, β − 1, . . ., β − (ℓ − 1)occurasindicatedin<strong>the</strong><br />
firstarray.Inparticular<strong>the</strong>reisexactlyoneineachcolumn.Thenumeratorin<strong>the</strong>product<strong>of</strong><br />
<strong>the</strong>lemmais<strong>the</strong>product<strong>of</strong><strong>the</strong>factorials<strong>of</strong><strong>the</strong>correspondingelements<strong>of</strong><strong>the</strong>secondarray.<br />
Thedenominatoris<strong>the</strong>product<strong>of</strong><strong>the</strong>factorials<strong>of</strong><strong>the</strong>elementsappearingafter q − 1. As<br />
indicated tis<strong>the</strong>elementlyingabove q − 1.Thus tislargerthananyelementappearingina<br />
columno<strong>the</strong>rthanthat<strong>of</strong> t.Theproduct<strong>of</strong><strong>the</strong>lemmais t!times<strong>the</strong>product<strong>of</strong><strong>the</strong>factorials<br />
<strong>of</strong><strong>the</strong>o<strong>the</strong>relementson<strong>the</strong>brokenlinedividedby<strong>the</strong>factorials<strong>of</strong><strong>the</strong>elementsat<strong>the</strong>foot<strong>of</strong><br />
<strong>the</strong>columninwhich<strong>the</strong>ylie.Thusitequals t!dividedby<strong>the</strong>product<strong>of</strong>all<strong>the</strong>elementsbelow<br />
<strong>the</strong>brokenlineexceptthosewhichliedirectlybelow t. But t!is<strong>the</strong>product<strong>of</strong>allnumbers<br />
in<strong>the</strong>secondarrayexceptthosewhichliebelow t. Thus<strong>the</strong>quotientis<strong>the</strong>product<strong>of</strong>all<br />
numberswhichlieaboveoron<strong>the</strong>brokenline,thatis,<br />
asrequired.<br />
Lemma 7.6<br />
β<br />
j=1<br />
ϕ(jm) ≡ β<br />
j=1 jm = mβ β! (mod ∗ p)<br />
Supposethat q = p f isaprimepower,that ℓisapositiveintegerdividing q − 1,that<br />
0 ≤ α1 < q − 1,that (ℓ, α1) = 1,andthat<br />
α2 = α1<br />
ℓ<br />
q ℓ − 1<br />
q − 1 .<br />
Then α2isanintegerand 0 ≤ α2 < q − 1.Moreoverif<br />
with 0 ≤ γi ≤ q − 1for 0 ≤ i ≤ ℓ − 1<strong>the</strong>n<br />
α2 = γ0 + γ1q + . . . + γℓ−1 q ℓ−1<br />
ℓ−1<br />
j=1<br />
j=0 γj = α1 + ℓ−1<br />
j ·<br />
q − 1<br />
ℓ
Chapter7 50<br />
and<br />
ℓ α1<br />
ℓ−1<br />
j=1<br />
j=0 γj! ≡ α1! ℓ−1<br />
Certainly 0 ≤ α2 < q ℓ − 1;moreover<br />
<br />
jq − 1!<br />
ℓ<br />
(mod ∗ p).<br />
q ℓ − 1<br />
q − 1 = 1 + q + . . . + qℓ−1 ≡ ℓ (mod ℓ)<br />
sothat α2isaninteger.Let α1 = mℓ + kwith m ≥ 0and 0 ≤ k < ℓandfor 0 ≤ j < ℓlet<br />
(ℓ − 1 − j)k = ij + δjℓ<br />
with δj ≥ 0and 0 ≤ ij < ℓ.Clearly iℓ−1 = δℓ−1 = 0.Also (ℓ − 1)k = ℓ − k + ℓ(k − 1)sothat<br />
i0 = ℓ−kand δ0 = k −1.If j ≥ 1<strong>the</strong>n (δj−1 −δj)ℓ = k+(ij −ij−1).Since −ℓ < ij −ij−1 < ℓ<br />
and 0 < k < ℓ<strong>the</strong>righthandsideisgreaterthan −ℓandlessthan 2ℓsothat δj−1 − δjis0or<br />
1.Ifitis1<strong>the</strong>n k + ij ≥ ℓand ij ≥ ℓ − k.Ifitis0<strong>the</strong>n ij = ij−1 − k < ℓ − k.Recallingthat<br />
i0 = ℓ − kweseethat<br />
and<br />
S = {j|1 ≤ j ≤ ℓ − 1 and δj−1 − δj = 1} = {j|0 ≤ j < ℓ and ij > ℓ − k}.<br />
Weshallprovethat<br />
γj = m + ij<br />
γ0 = m + i0<br />
(q − 1)<br />
ℓ<br />
(q − 1)<br />
ℓ<br />
+ k − δ0<br />
+ δj−1 − δj 1 ≤ j < ℓ.<br />
Since (k, ℓ) = 1<strong>the</strong>numbers ijaredistinctanditwillfollowimmediatelythat<br />
ℓ−1<br />
j=0<br />
j=0 γj = (mℓ + k) + ℓ−1<br />
Moreoverwewillhave<br />
ℓ−1<br />
j=0 γj!<br />
<br />
ℓ−1<br />
= m + j<br />
j=0<br />
Recallthat k − δ0 = 1.Dividing<strong>the</strong>firsttermby<br />
j · q − 1<br />
ℓ = α1 + ℓ−1<br />
j=1<br />
q − 1<br />
j · .<br />
ℓ<br />
<br />
(q − 1) q − 1<br />
! m+i0<br />
ℓ<br />
ℓ +k−δ0<br />
<br />
<br />
<br />
m + 1 + ij<br />
j∈S<br />
ℓ−1<br />
j=0<br />
<br />
j ·<br />
<br />
(q − 1)<br />
!<br />
ℓ<br />
<br />
(q − 1)<br />
.<br />
ℓ
Chapter7 51<br />
weobtain<br />
ℓ−1<br />
j=0<br />
Theproduct<strong>of</strong><strong>the</strong>lasttwotermsis<br />
ℓ−1<br />
j=ℓ−k<br />
m<br />
n=1<br />
<br />
n + j<br />
<br />
m + 1 + j<br />
<br />
(q − 1)<br />
.<br />
ℓ<br />
<br />
(q − 1)<br />
.<br />
ℓ<br />
If 1 ≤ n ≤ mand 0 ≤ j ≤ ℓ − 1<strong>the</strong>n nℓ − j < α1 < qsothat<strong>the</strong>product<strong>of</strong> ℓ mk and<strong>the</strong>first<br />
<strong>of</strong><strong>the</strong>setwoexpressionsismultiplicativelycongruentto<br />
ℓ−1<br />
j=0<br />
m<br />
n=1<br />
(nℓ − j) = (mℓ)!<br />
Moreover,if ℓ − k ≤ j ≤ ℓ − 1,<strong>the</strong>n 0 ≤ (m + 1)ℓ − j ≤ (m + 1)ℓ − (ℓ − k) = α1 < qand<strong>the</strong><br />
second<strong>of</strong><strong>the</strong>seexpressionsuponmultiplicationby ℓ k becomesmultiplicativelycongruentto<br />
ℓ−1<br />
j=ℓ−k<br />
((m + 1)ℓ − j) = k<br />
j=1<br />
Therelationstoge<strong>the</strong>rimply<strong>the</strong>secondidentity<strong>of</strong><strong>the</strong>lemma.<br />
(mℓ + j).<br />
Toverifythat<strong>the</strong> γj, 0 ≤ j < ℓ,have<strong>the</strong>formasserted,westartwith<strong>the</strong>relation<br />
α2 = qℓ − 1<br />
q − 1<br />
Thesecondtermisequalto<br />
Thus<br />
ℓ−1<br />
j=0<br />
α2 =<br />
j−1<br />
i=0 qi<br />
· mℓ + k<br />
ℓ<br />
q − 1<br />
ℓ<br />
= ℓ−1<br />
j=0 qj m + ℓ−1<br />
j=0<br />
q j k<br />
ℓ .<br />
<br />
k + k = k + ℓ−2<br />
j=0 (ℓ − 1 − j)qj q − 1<br />
· · k.<br />
ℓ<br />
<br />
<br />
q − 1<br />
m + (ℓ − 1)k · + k +<br />
ℓ ℓ−1<br />
<br />
<br />
q − 1<br />
m + (ℓ − 1 − j)k · q<br />
j=1<br />
ℓ<br />
j<br />
<br />
<br />
q − 1<br />
= m + i0 · + k − δ0 +<br />
ℓ ℓ−1<br />
<br />
q − 1<br />
m + ij ·<br />
j=1 ℓ + δj−1<br />
<br />
− δj .<br />
Moreover m < q−1<br />
ℓ sothat<br />
0 ≤ m + i0 ·<br />
q − 1<br />
ℓ + k − δ0<br />
q − 1<br />
< ℓ · + 1 = q<br />
ℓ
Chapter7 52<br />
and<br />
0 ≤ m + ij ·<br />
Therequiredrelationsfollowimmediately.<br />
q − 1<br />
ℓ + δj−1<br />
q − 1<br />
− δj < ℓ · + 1 = q.<br />
ℓ<br />
Nowwecanstateandprove<strong>the</strong>promisedidentitiesforGaussiansums.Each<strong>of</strong><strong>the</strong>sewill<br />
amounttoanassertionthatacertainnumberin k p(q−1)is1.Toprovethiswewillshowfirst<br />
that<strong>the</strong>numberisinvariantunderallautomorphisms<strong>of</strong> k p(q−1)over k (q−1)andthusliesin<br />
kq−1.Theonlyprimeidealsoccurringin<strong>the</strong>factorization<strong>of</strong><strong>the</strong>number,whichisnot a priori<br />
analgebraicinteger,intoprimeidealswillbedivisors<strong>of</strong> p.Weshowthateveryconjugate<strong>of</strong><br />
<strong>the</strong>numberhasabsolutevalue1andthatitismultiplicativelycongruentto1moduloevery<br />
divisor<strong>of</strong> p.Itwillfollowthatitisaroot<strong>of</strong>unityin kq−1andhencea(q − 1)throot<strong>of</strong>unityif<br />
qisoddanda2(q − 1)throot<strong>of</strong>unityif qiseven.If qisodd<strong>the</strong>multiplicativecongruences<br />
implythat<strong>the</strong>numberis1.If qiseven<strong>the</strong>yimplythat<strong>the</strong>numberis ±1.Toshowthatitis<br />
actually1somesupplementarydiscussionwillbenecessary.<br />
Stickelberger’sresultisdirectlyapplicableonlyto<strong>the</strong>normalizedGaussiansum τ(χκ).<br />
Weshallhavetouse<strong>the</strong>obviousrelation τ(χκ, Ψκ) = χκ(β)τ(χκ)if ψκ(α) = ψ 0 κ(βα).If κis<br />
anextension<strong>of</strong> λand ψλisgiven,weset<br />
ψ κ/λ(α) = ψλ(S κ/λ(α))<br />
for αin κ.If χλisgiven χ κ/λis<strong>the</strong>characterdefinedby<br />
Lemma 7.7<br />
Set<br />
χ κ/λ(α) = χλ(N κ/λ(α)).<br />
If κisafiniteextension<strong>of</strong><strong>the</strong>finitefieldand χλand ψλaregiven<strong>the</strong>n<br />
τ(χ κ/λ, ψ κ/λ) = {τ(χλ, ψλ)} [κ:λ] .<br />
Since χ κ/λ(β) = χλ(β) [κ:λ] itwillbeenoughtoshowthat<br />
τ(χ κ/λ) = {τ(χλ)} [κ:λ] .<br />
µ =<br />
[χ:λ]<br />
{τ(χλ)}<br />
.<br />
τ(χκ/λ)
Chapter7 53<br />
Let λhave q = p ℓ elementsandlet κhave p k = q f .ItfollowsimmediatelyfromLemma7.1<br />
that<strong>the</strong>absolutevalue<strong>of</strong> µandallitsconjugatesis1,thatitliesin k p(q f −1),thatitisinvariant<br />
underallautomorphisms<strong>of</strong> k p(q f −1)over k q f −1,andthatitsonlyprimefactorsaredivisors<strong>of</strong><br />
p.Themapping β −→ Nκ/λβsends βto β qf −1<br />
q−1.Thusif α = α(χλ, p)and Pdivides p<br />
ApplyingLemmas7.1and7.2,weseethat<br />
and<br />
Consequently<br />
α(χ κ/λ, p) = qf − 1<br />
q − 1 α = α + αq + . . . + αqf−1 .<br />
τ(χλ) ≡ ̟α<br />
α!<br />
τ(χ κ/λ) ≡ ̟fα<br />
(α!) f<br />
(mod ∗ P)<br />
µ ≡ 1 (mod ∗ P).<br />
(mod ∗ P).<br />
Thus µ = 1if qisoddand µ = ±1if qiseven.If χλ = 1<strong>the</strong>n χ κ/λ = 1and,frompart(b)<strong>of</strong><br />
Lemma7.1, µis1.If q = 2<strong>the</strong>n χλ = 1.Suppose<strong>the</strong>n qisevenandgreaterthan2.If χλis<br />
notidentically1chooseaprime rdividing<strong>the</strong>order<strong>of</strong> χλ.Set χλ = χ ′ λ χ′′ λ where<strong>the</strong>order<strong>of</strong><br />
χ ′ λ isapower<strong>of</strong> rand<strong>the</strong>order<strong>of</strong> χ′′ λ isprimeto r.Theanalogousdecomposition<strong>of</strong> χ κ/λis<br />
χ ′ κ/λ χ′′ κ/λ .Ofcourse χλand χ κ/λhave<strong>the</strong>sameorder.Define µ ′ and µ ′′ in<strong>the</strong>obviousway.<br />
Accordingtopart(f)<strong>of</strong>Lemma7.1<br />
µ ≡ µ ′′<br />
(modr).<br />
Since rdoesnotdivide2thisimpliesthat µ = µ ′′ . Thusonecanshowbyinductionon<strong>the</strong><br />
number<strong>of</strong>primesdividing<strong>the</strong>order<strong>of</strong> χλthat µ = 1.<br />
Lemma 7.8<br />
Suppose λisafinitefieldwith qelements, κisafiniteextension<strong>of</strong> λ,and [κ : λ] = f.<br />
Suppose ℓisaprimeand<strong>the</strong>order<strong>of</strong> qmodulo ℓis f.Let Tbeaset<strong>of</strong>representativesfor<strong>the</strong><br />
orbits<strong>of</strong><strong>the</strong>nontrivialcharacters<strong>of</strong> κ ∗ <strong>of</strong>order ℓunder<strong>the</strong>action<strong>of</strong> G(κ/λ)andlet χλbea<br />
character<strong>of</strong> λ ∗ .If ψλisanynontrivialcharacter<strong>of</strong> λ<br />
χλ(ℓ ℓ )τ(χ ℓ λ , ψλ) <br />
µκ∈T τ(µκ, ψ κ/λ) = τ(χλ, ψλ) <br />
µκ∈T τ(χ κ/λ µκ, ψ κ/λ).
Chapter7 54<br />
Since<strong>the</strong>isotropygroup<strong>of</strong>eachpointin Tistrivial<br />
χ ℓ λ(β) <br />
µκ∈T µκ(β) = χλ(β) <br />
andwemaycontentourselveswithshowingthat<br />
µκ∈T χ κ/λ(β) µκ(β)<br />
χλ(ℓ ℓ ) τ(χ ℓ <br />
λ )<br />
µκ∈T τ(µκ) = τ(χλ) <br />
µκ∈T τ(χκ/λ µκ).<br />
Ofcourse χλ(ℓℓ )is<strong>the</strong>value<strong>of</strong> χλat<strong>the</strong>element<strong>of</strong><strong>the</strong>primefieldcorrespondingto ℓℓ .Let<br />
µbe<strong>the</strong>quotient<strong>of</strong><strong>the</strong>rightsideby<strong>the</strong>left.Thecharacters<strong>of</strong> κ∗<strong>of</strong>order ℓare<strong>the</strong>characters<br />
µ k κ , 0 ≤ k < ℓ,definedby<br />
α(µ k κ , P) = k · qf − 1<br />
.<br />
ℓ<br />
Since<strong>the</strong>order<strong>of</strong> qmodulo ℓis f,if T = {µ k κ|k ∈ A}everynontrivialcharacter<strong>of</strong>order ℓ<br />
isrepresentableas µ η(qi k)<br />
κ<br />
with 0 ≤ i < fand k ∈ A. η(q i k)is<strong>the</strong>remainder<strong>of</strong> q i kupon<br />
divisionby ℓ. Thusaswealreadysaw, Thas ℓ−1<br />
f elements. Lemma7.1againshowsthat µ<br />
andallitsconjugateshaveabsolutevalue1andthat µisinvariantunderallautomorphisms<br />
<strong>of</strong> k p(q f −1)over k q f −1.<br />
let<br />
Let α = α(χλ, p)andlet β = α(χℓ λ , p).Then ℓα = β + ν(q − 1)with ν ≥ 0.If 0 ≤ k < ℓ<br />
α(µ k κ , P) = k · qf − 1<br />
ℓ<br />
= f−1<br />
j=0 γk j qj<br />
with 0 ≤ γ k j ≤ q −1.Inparticular, γk 0 is<strong>the</strong>residue<strong>of</strong> k · qf −1<br />
ℓ modulo q.Moreoverif k1 ≡ q i k<br />
(modulo ℓ)<strong>the</strong>n<br />
α(µ k κ, P) = f−1<br />
Itisunderstoodthatif j + i ≥ ℓ<strong>the</strong>n γ k1<br />
j+i<br />
divisionby q,<br />
Certainly<br />
{γ k j<br />
| k ∈ A, 0 ≤ j < f} =<br />
j=0 γk1<br />
j+i qj .<br />
α(χκ/λ µ k κ, P) ≡ qf − 1<br />
q − 1 α + k · qf − 1<br />
ℓ<br />
= γk1<br />
j+i−ℓ .Thusif ϕ(x)is<strong>the</strong>remainder<strong>of</strong> xupon<br />
<br />
ϕ k · qf <br />
− 1<br />
| 0 < k < ℓ .<br />
ℓ<br />
(mod(q f − 1)).
Chapter7 55<br />
Let 0 ≤ k ′ < ℓandlet ν + k ≡ k ′ (modℓ).Since,bydefinition, ℓα = β + ν(q − 1)<br />
qf − 1<br />
q − 1 α + k qf − 1<br />
ℓ ≡ qf − 1<br />
·<br />
ℓ<br />
β<br />
q − 1 + k′ qf − 1<br />
ℓ<br />
(mod(q f − 1)).<br />
Since 0 ≤ β < q −1<strong>the</strong>rightsideisnonnegativeandatmost qf −2.Thusitis α(χκ/λ µ k κ , P).<br />
Let<br />
with 0 ≤ δ k j ≤ q − 1.Thus δk 0 is<strong>the</strong>residue<strong>of</strong><br />
α(χκ/λ µ k f−1<br />
κ , P) =<br />
j=0 δk j qj<br />
q f − 1<br />
ℓ<br />
·<br />
β<br />
q − 1 + k′ · qf − 1<br />
ℓ<br />
modulo q.Since χ κ/λisinvariantunderautomorphisms<strong>of</strong> κ/λ<br />
α(χκ/λ µ k f−1<br />
κ , P) =<br />
j=0 δk1<br />
j+i qj<br />
if k1 ≡ qik (modℓ).Since<strong>the</strong>residue<strong>of</strong> qf −1<br />
αmodulo qis α,<br />
q−1<br />
{α} ∪ {δ k j | 0 ≤ j < ℓ, k ∈ A} =<br />
f q − 1<br />
ϕ ·<br />
ℓ<br />
β<br />
q − 1 + k qf <br />
− 1<br />
| 0 ≤ k < ℓ .<br />
ℓ<br />
Since χλ(ℓℓ ) ≡ ℓαℓ (mod ∗ P)<strong>the</strong>number µismultiplicativelycongruentmodulo Pto<br />
<strong>the</strong>quotient<strong>of</strong><br />
̟α̟ε α! ℓ−1 k∈A j=0 δk j !,<br />
<br />
ε =<br />
k∈A<br />
f−1<br />
j=0 δk j<br />
by<br />
Since<br />
ℓβ ̟β ̟ε′ ℓ−1 β! <br />
k∈A<br />
f−1<br />
j=0<br />
weconcludefromLemma7.4that<br />
fα + f−1<br />
j=0 γk j !, ε′ = <br />
k∈A<br />
f−1<br />
k j<br />
α + γj q ≡<br />
j=0<br />
f−1<br />
j−0 γk j .<br />
j=0 δk j qj ,<br />
k<br />
γj − δ k j = ρ(q − 1),
Chapter7 56<br />
if ρis<strong>the</strong>number<strong>of</strong> i, 0 ≤ i < f,suchthat<br />
Since<br />
<strong>the</strong>number<br />
qf <br />
− 1<br />
α + α µ<br />
q − 1 η(qi <br />
k)<br />
κ , P<br />
≥ q f .<br />
q f − 1<br />
q − 1 · α + α(µ0 κ , P) = qf − 1<br />
q − 1 α,<br />
(ℓ − 1)α + <br />
k∈A<br />
f−1<br />
j=0<br />
is (q − 1)times<strong>the</strong>number<strong>of</strong> k, 0 ≤ k < ℓ,suchthat<br />
qf − 1<br />
q − 1 α + k · qf − 1<br />
ℓ<br />
= qf − 1<br />
q − 1<br />
Thenumber<strong>of</strong>such kis νbecause ν < ℓand<br />
while<br />
Thus<br />
q f − 1<br />
q − 1<br />
If ℓm ≡ 1 (modq)<br />
and<br />
<br />
q f − 1<br />
q − 1<br />
k<br />
γj − δ k j<br />
β<br />
ℓ + (k + ν) qf − 1<br />
ℓ ≥ qf .<br />
β<br />
ℓ + (ℓ − ν + ν) qf − 1<br />
ℓ ≥ 1 + qf − 1 = q f<br />
β<br />
ℓ + (ℓ − 1) qf − 1<br />
ℓ < qf − 1<br />
ℓ + (ℓ − 1) qf − 1<br />
ℓ = qf − 1.<br />
k∈A<br />
f−1<br />
f q − 1<br />
ϕ ·<br />
ℓ<br />
ItfollowsimmediatelyfromLemma7.5that<br />
ℓ β α! <br />
j=0 (γk j − δ k j ) = ν(q − 1) − (ℓ − 1)α = α − β.<br />
k∈A<br />
<br />
ϕ k · qf <br />
− 1<br />
= ϕ(−km)<br />
ℓ<br />
β<br />
q − 1 + k qf <br />
− 1<br />
= ϕ((β − k)m).<br />
ℓ<br />
f−1<br />
j=0 δk j ! ≡ β! <br />
k∈A<br />
f−1<br />
j=0 γk j !<br />
Thus µ = 1if qisoddand µ = ±1if qiseven.If χλ = 1<strong>the</strong>number µisclearly1.This<br />
timetoo,onecanapplypart(f)<strong>of</strong>Lemma7.1andinductionon<strong>the</strong>number<strong>of</strong>primesdividing<br />
<strong>the</strong>order<strong>of</strong> χλtoshowthat µ = 1if qiseven.
Chapter7 57<br />
Lemma 7.9<br />
Let λbeafinitefieldwith qelementsandlet κbeafiniteextension<strong>of</strong> λwith [κ : λ] = ℓ<br />
where ℓisaprimedividing q − 1.Suppose χλisacharacter<strong>of</strong> λ ∗ whoserestrictionto<strong>the</strong> ℓth<br />
roots<strong>of</strong>unityisnottrivialand χκisacharacter<strong>of</strong> κ ∗ suchthat χ ℓ κ = χ κ/λ.If Tis<strong>the</strong>set<strong>of</strong><br />
nontrivialcharacters<strong>of</strong> λ ∗ <strong>of</strong>order ℓ<br />
χλ(ℓ)τ(χλ, ψλ) <br />
µλ∈T τ(µλ, ψλ) = τ(χκ, ψ κ/λ).<br />
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 isacharacter<strong>of</strong>order ℓ.If χσ−1 κ = 1forsome σ = 1<strong>the</strong>nitis1forall σand χκ(α) = 1<br />
if αisa(q − 1)thpower,thatis,if Nκ/λ(α) = 1. Consequently<strong>the</strong>reisacharacter νλ<strong>of</strong> λ∗ suchthat χκ = νκ/λ. Then νℓ λ = χλand χλistrivialon<strong>the</strong> ℓthroots<strong>of</strong>unity,contraryto<br />
assumption.Thus<br />
{χ σ−1<br />
κ | σ = 1} = {µ κ/λ | µλ ∈ T }.<br />
If β ∈ λ ∗ and β = N κ/λ(γ)<strong>the</strong>n<br />
χκ(β) = <br />
σ χκ(γ σ−1 ) = χκ(γ ℓ ) <br />
because µλ(β) = µ κ/λ(γ),anditwillbeenoughtoshowthat<br />
χλ(ℓ) τ(χλ) <br />
σ=1 χσ−1 κ (γ) = χλ(β) <br />
σ=1 µλ(β),<br />
µλ∈T τ(µλ) = τ(χκ).<br />
Let µbe<strong>the</strong>quotient<strong>of</strong><strong>the</strong>leftsideby<strong>the</strong>right.Thus µisanumberin Kp(q−1)and<strong>the</strong><br />
onlyprimesappearingin<strong>the</strong>factorization<strong>of</strong> µaredivisors<strong>of</strong> p.Since χ κ/λisnotidentically<br />
1nei<strong>the</strong>ris χκ.Thus<strong>the</strong>absolutevalue<strong>of</strong> µandallitsconjugatesis1.Let α = α(χλ, p)and<br />
let β = α(χκ, P)where Pdivides p.Then<br />
Since ℓdivides qℓ −1<br />
q−1 wecanwrite<br />
ℓβ ≡ α qℓ − 1<br />
q − 1<br />
β = qℓ − 1<br />
q − 1<br />
(mod(q ℓ − 1)).<br />
α<br />
·<br />
ℓ − j · qℓ − 1<br />
.<br />
ℓ
Chapter7 58<br />
Since<strong>the</strong>restriction<strong>of</strong> χλto<strong>the</strong> ℓthroots<strong>of</strong>unityisnottrivial, α · q−1<br />
ℓ ≡ 0 (mod(q − 1)).<br />
Thus ℓdoesnotdivide α.Forall i ≥ 0<br />
<br />
τ χ qi<br />
<br />
κ = τ(χκ).<br />
Moreover<br />
<br />
α<br />
Since qi −1<br />
q−1<br />
χ qi<br />
κ<br />
<br />
, P ≡ αℓ − 1<br />
q − 1<br />
≡ qℓ − 1<br />
q − 1<br />
α<br />
ℓ − j αℓ − 1<br />
ℓ + (qi − 1) qℓ − 1<br />
q − 1<br />
α<br />
ℓ +<br />
i ℓ q − 1 q − 1<br />
α − j .<br />
q − 1 ℓ<br />
≡ i (modℓ)wechoose isothat iα ≡ j (mod ℓ);<strong>the</strong>n<br />
α(χ qi<br />
κ , P) ≡ qℓ − 1<br />
q − 1<br />
α<br />
ℓ<br />
(mod(q ℓ − 1)).<br />
α<br />
ℓ − j(qi − 1) qℓ − 1<br />
ℓ<br />
Bothsides<strong>of</strong>thiscongruencearenonnegativeandlessthan q ℓ − 1.Thusitisanequalityand<br />
wecanassumethat β = qℓ−1 q−1<br />
definedby<br />
χ −1<br />
κ<br />
· α<br />
ℓ<br />
. Theset Tconsists<strong>of</strong><strong>the</strong>characters µj<br />
λ , 1 ≤ j ≤ ℓ − 1,<br />
α(µ j j<br />
λ , P) =<br />
ℓ<br />
(q − 1).<br />
Under<strong>the</strong>automorphism z −→ z m <strong>of</strong> k p(q ℓ −1)over k q ℓ −1<strong>the</strong>number µismultipliedby<br />
(m) χλ(m) <br />
µλ∈T µλ(m)whichis1because mbelongsto λ.Let<br />
with 0 ≤ γi ≤ q − 1.Then<br />
and<br />
τ(χκ) ≡<br />
χλ(ℓ)τ(χλ) ℓ−1<br />
j=1 τ(µj λ ) ≡<br />
β = γ0 + γ1q + . . . + γℓ−1 q ℓ−1<br />
̟ε<br />
ℓ−1 j=0 γj! (mod∗P), ε = ℓ−1<br />
j=0 γj,<br />
α! ℓ−1<br />
j=1<br />
ℓ α ̟ ε′<br />
j · q−1<br />
ℓ<br />
!<br />
(mod ∗ P), ε ′ = α + ℓ−1<br />
j=0<br />
q − 1<br />
j · .<br />
ℓ<br />
Lemma7.6impliesimmediatelythat µ ≡ 1 (mod ∗ P).Thus µ = 1if qisoddand µ = ±1if q<br />
iseven.If ℓ ′ isaprimedivisor<strong>of</strong> q −1differentfrom ℓ,wewrite χλas χ ′ λ χ′′ λ where<strong>the</strong>order<strong>of</strong>
Chapter7 59<br />
χ ′ λ isapower<strong>of</strong> ℓ′ and<strong>the</strong>order<strong>of</strong> χ ′′ λ isprimeto ℓ.Inasimilarfashionwewrite χκas χ ′ κχ ′′ κ .<br />
Thepair χ ′′ λ and χ′′ κ alsosatisfy<strong>the</strong>conditions<strong>of</strong><strong>the</strong>lemma.Thefinalassertion<strong>of</strong>Lemma7.1<br />
showsthat,if µ ′′ isdefinedin<strong>the</strong>samewayas µ, µ = µ ′′ .Arguingbyinductionweseethatit<br />
isenoughtoverifythat µ = 1when<strong>the</strong>order<strong>of</strong> χλisapower<strong>of</strong> ℓ.Applying<strong>the</strong>lastpart<strong>of</strong><br />
Lemma7.1,againweseethat<strong>the</strong>reisaprime qdividing ℓsuchthat<br />
Since χλ(ℓ)isan ℓ ω throot<strong>of</strong>unityforsome ω,<br />
Thus µ ≡ 1 (modℓ)and µ = 1.<br />
τ(χλ) ≡ τ(χκ) ≡ τ(µ j<br />
λ ) ≡ 1 (modq).<br />
χλ(ℓ) ≡ 1 (modq).
Chapter8 60<br />
Chapter Eight.<br />
A Lemma <strong>of</strong> Lamprecht<br />
Let Fbeanonarchimedeanlocalfieldandlet ψFbeanontrivialcharacter<strong>of</strong> F. n =<br />
n(ψF)is<strong>the</strong>largestintegersuchthat ψF istrivialon P −n<br />
F . If χF isaquasicharacter<strong>of</strong><br />
CF, m = m(χF)is<strong>the</strong>smallestnonnegativeintegersuchthat χFistrivialon Um F .If γin CF<br />
issuchthat γOF = P m+n<br />
F set<br />
<br />
UF<br />
∆1(χF,ψF; γ) =<br />
ψF<br />
<br />
α<br />
γ χ −1<br />
F (α)dα<br />
<br />
<br />
χ −1<br />
F (α)dα .<br />
Then<br />
UF ψF<br />
α<br />
γ<br />
∆(χF,ψF) = χF(γ)∆1(χF, ψF, γ).<br />
AssuggestedbyHasse[8],weshall,in<strong>the</strong>pro<strong>of</strong>s,<strong>of</strong><strong>the</strong>mainlemmas,makeextensive<br />
use<strong>of</strong><strong>the</strong>followinglemmawhichiscentralto<strong>the</strong>paper[10]<strong>of</strong>Lamprecht.<br />
Lemma 8.1<br />
(a)If m = m(χF) = 2dwith dintegralandpositive<strong>the</strong>reisaunit βin OFsuchthat<br />
<br />
βx<br />
ψF = χF(1 + x)<br />
γ<br />
forall xin Pd F .Foranysuch β<br />
∆1(χF,ψF; γ) = ψF<br />
<br />
β<br />
χ<br />
γ<br />
−1<br />
F (β).<br />
(b)If m = m(χF) = 2d + 1with dintegralandpositive<strong>the</strong>reisaunit βin OFsuchthat,for<br />
all xin P d+1<br />
F ,<br />
ψF<br />
Foranysuch β, ∆1(χF, ψF; γ)isequalto<br />
ψF<br />
<br />
β<br />
χ<br />
γ<br />
−1<br />
F (β)<br />
<br />
βx<br />
= χF(1 + x).<br />
γ<br />
<br />
<br />
<br />
OF /PF ψF<br />
<br />
δβx<br />
γ<br />
OF /PF ψF<br />
<br />
δβx<br />
γ<br />
<br />
χ −1<br />
F (1 + δx)dx<br />
<br />
χ −1<br />
F (1 + δx)dx
Chapter8 61<br />
if δOF = P d F .<br />
Let m = 2d + εwith ε = 0incase(a)and ε = 1incase(b).Thefunction ψF<br />
OF, y ∈ P d+ε<br />
F canberegardedasafunctionon<br />
OF/P d F<br />
× Pd+ε<br />
F /Pm F .<br />
<br />
xy<br />
γ , x ∈<br />
Forfixed xitisacharacter<strong>of</strong> P d+ε<br />
F /PmF whichistrivialifandonlyif x ∈ PdF andforfixed y<br />
itisacharacter<strong>of</strong> OF/Pd Fwhichistrivialifandonlyif y ∈ Pm F .Thusitdefinesaduality<strong>of</strong><br />
OF/P d F<br />
and Pd+ε<br />
F /Pm F .Theexistence<strong>of</strong>aβsuchthat<br />
χF(1 + x) = ψF<br />
for xin P d+ε<br />
F followsimmediatelyfrom<strong>the</strong>relation<br />
βx<br />
γ<br />
χF(1 + x) χF(1 + y) = χF(1 + x + y)<br />
whichisvalidfor xin P d+ε<br />
F .Thenumber βmustbeaunitbecause χF(1+x)isdifferentfrom<br />
1forsome xin P m−1<br />
F .<br />
Incase(a)<br />
isequalto<br />
<br />
UF /U d F<br />
ψF<br />
<br />
UF<br />
ψF<br />
<br />
<br />
α<br />
χ<br />
γ<br />
−1<br />
F (α)dα<br />
<br />
α<br />
χ<br />
γ<br />
−1<br />
F (α)<br />
Pd ψF<br />
F<br />
(α − β)x<br />
γ<br />
<br />
dx dα.<br />
Themainintegralis1or0accordingas α − βdoesordoesnotliein Pd F .Thusthisexpression<br />
isequalto<br />
<br />
β<br />
ψF χ<br />
γ<br />
−1<br />
F (β) [UF : U d F] −1 .<br />
Thefirstpart<strong>of</strong><strong>the</strong>lemmafollows.<br />
Incase(b) <br />
isequalto<br />
<br />
UF /U d+1<br />
F<br />
ψF<br />
UF<br />
ψF<br />
<br />
α<br />
χ<br />
γ<br />
−1<br />
F (α)dα<br />
<br />
α<br />
χ<br />
γ<br />
−1<br />
F (α)<br />
P d+1<br />
ψF<br />
F<br />
(α − β)x<br />
γ<br />
<br />
dx dα.
Chapter8 62<br />
Theinnerintegralis0unless α − βliesin P d F whenitis1.Thusthisexpressionisequalto<br />
ψF<br />
<br />
β<br />
χ<br />
γ<br />
−1<br />
F (β) [UF : U d F ]−1<br />
Thesecondpart<strong>of</strong><strong>the</strong>lemmafollows.<br />
<br />
OF /PF<br />
ψF<br />
<br />
δβx<br />
γ<br />
χ −1<br />
F (1 + δx)dx.<br />
Thenumber βisonlydeterminedmodulo Pd F .Whenapplying<strong>the</strong>lemmaweshall,after<br />
choosing β,set<br />
<br />
β<br />
∆2(χF,ψF; γ) = ψF χ<br />
γ<br />
−1<br />
F (β)<br />
and<strong>the</strong>ndefine ∆3(χF, ψF; γ),whichwillbe1when miseven,by<strong>the</strong>equation,<br />
∆1(χF,ψF; γ) = ∆2(χF, ψF; γ) ∆3(χF, ψF; γ).<br />
Whenweneedtomake<strong>the</strong>relationbetween βand χFexplicitwewrite βas β(χF).Tobe<strong>of</strong><br />
anyusetousthislemmamustbesupplementedbysomeo<strong>the</strong>robservations.<br />
If KisafiniteGaloisextension<strong>of</strong> F anyquasicharacter χF <strong>of</strong> CF determinesaonedimensionalrepresentation<strong>of</strong><br />
W K/Fwhoserestrictionto CKisaquasicharacter χ K/F<strong>of</strong> CK.<br />
Thecharacter χ K/Fmaybedefineddirectlyby<br />
χ K/F(α) = χF(N K/Fα).<br />
Moregenerally,if Eisanyfiniteseparableextension<strong>of</strong> Fwedefine χ E/Fby<br />
χ E/F(α) = χF(N E/Fα).<br />
Toapply<strong>the</strong>lemma<strong>of</strong>Lamprechtweshallneedtoknow,insomespecialcases,<strong>the</strong>relation<br />
between β(χF)and β(χ E/F).<br />
Suppose misapositiveintegerand m = 2d + εwhere εis0or1and disapositive<br />
integer.Let m ′ = ψ E/F(m − 1) + 1andlet m ′ = 2d ′ + ε ′ where ε ′ is0or1and d ′ isapositive<br />
integer.*Since ψ E/Fisconvex<br />
ψ E/F<br />
<br />
m − 1<br />
2<br />
≤ 1<br />
2 ψ E/F(m − 1) + 1<br />
2 ψ E/F(0) = 1<br />
2 (m′ − 1) < d ′ + ε ′<br />
*Weareheredealingnotwithanadditivecharacter,butwith<strong>the</strong>function<strong>of</strong>Chapter6!
Chapter8 63<br />
and<br />
d ′ + ε ′ = ψ E/F(u)<br />
with u > m−1<br />
m−1<br />
2 .Since<strong>the</strong>leastintegergreaterthan 2 is d + ε,Lemma6.6impliesthat<br />
Ino<strong>the</strong>rwords,if x ∈ P d′ +ε ′<br />
E<br />
Lemma6.6alsoimpliesthat<br />
<br />
NE/F U d′ +ε ′ <br />
E ≤ U u F<br />
<strong>the</strong>n<br />
if x ∈ Pm′ E .If x ∈ Pd′ +ε ′<br />
E and y ∈ Pm′ E <strong>the</strong>n<br />
iscongruentto<br />
≤ Ud+ε<br />
F .<br />
N E/F(1 + x) − 1 ∈ P d+ε<br />
F .<br />
N E/F(1 + x) − 1 ∈ P m F<br />
N E/F(1 + x + y) − 1 = N E/F(1 + x)N E/F<br />
modulo P m F .Thusif x ∈ Pd′ +ε ′<br />
E<br />
Therightsideis<br />
whichequals<br />
N E/F(1 + x) − 1<br />
and y ∈ P d′ +ε”<br />
E<br />
<br />
1 + y<br />
<br />
− 1<br />
1 + x<br />
sothat xy ∈ P m′<br />
E ,<strong>the</strong>n<br />
N E/F(1 + x + y) − 1 ≡ N E/F(1 + x + y + xy) − 1 (modP m F ).<br />
N E/F(1 + x)N E/F(1 + y) − 1,<br />
{N E/F(1 + x) − 1} + {N E/F(1 + y) − 1} + {(N E/F(1 + x) − 1) (N E/F(1 + y) − 1)}<br />
andthisiscongruentto<br />
modulo P m F .Thus<strong>the</strong>map<br />
{N E/F(1 + x) − 1} + {N E/F(1 + y) − 1}<br />
P E/F : x −→ N E/F(1 + x) − 1
Chapter8 64<br />
isahomomorphismfrom P d′ +ε ′<br />
E<br />
/P m′<br />
E<br />
to Pd+ε<br />
F /Pm F .If E ⊆ E′ wecanreplace Fby E, Eby<br />
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<br />
and<br />
N E ′ /F(1 + x) − 1 = N E/F(1 + (N E ′ /E(1 + x) − 1)) − 1,<br />
<strong>the</strong>relation<br />
isvalid.<br />
P E ′ /F = P E/F ◦ P E ′ /E<br />
If n = n(ψF)and n ′ = n(ψ E/F),choose γFin CFsothat γFOF = P m+n<br />
F<br />
sothat γEOE = P m′ +n ′<br />
E<br />
and γEin CE<br />
. Iapologizeagainfor<strong>the</strong>unfortunateconflict<strong>of</strong>notation. ψ E/Fis<br />
on<strong>the</strong>onehandafunctionon {u ∈ R | u ≥ −1}andon<strong>the</strong>o<strong>the</strong>racharacter<strong>of</strong> E.However,<br />
warnedoneagain,<strong>the</strong>readershouldnotbetooinconveniencedby<strong>the</strong>conflict.Define<br />
by<strong>the</strong>relation<br />
ψF<br />
P ∗ E/F : OF/P d F<br />
−→ OE/P d′<br />
E<br />
<br />
∗<br />
x PE/F(y) PE/F (x)y<br />
= ψE/F .<br />
γF<br />
Itwill<strong>of</strong>tenbenecessarytokeepinmind<strong>the</strong>dependence<strong>of</strong> P ∗ E/F on γFand γE. Thenwe<br />
shallwrite<br />
P ∗ E/F (x) = P ∗ E/F (x; γE, γF).<br />
Itisclearthat<br />
Lemma 8.2<br />
P ∗ E ′ /F (x; γE ′, γF) = P ∗ E ′ /E (P ∗ E/F (x; γE, γF); γE ′, γE).<br />
Let K/Fbeabelianandlet G = G(K/F).Suppose<strong>the</strong>reisaninteger tsuchthat G = Gt<br />
while Gt+1 = {1}.Suppose m ≥ t+1and m > 1and γFischosen.If µFbelongsto S(K/F),<br />
<strong>the</strong>set<strong>of</strong>characters<strong>of</strong> CF/N K/FCK,<strong>the</strong>n m ≥ m(µF)sothatforsome α(µF)in OF<br />
µF(1 + x) = ψF<br />
γE<br />
<br />
α(µF)x<br />
forall xin P d+ε<br />
F .Theelement γKmaybetakenequalto γFandif P ∗ K/F (β) = P ∗ K/F (β; γF, γF)<br />
<strong>the</strong>n<br />
NK/F(P ∗ <br />
K/F (β)) ≡ (β + α(µF)) (modP<br />
µF<br />
d F)<br />
γF
Chapter8 65<br />
forall βin OF.<br />
If t = −1<strong>the</strong>n n(ψF) = n(ψ K/F)and m ′ = msothat γKmaybetakenequalto γF.If<br />
t ≥ 0<strong>the</strong>extensionisramified.Let P δ K/F<br />
K<br />
Bydefinition<br />
be<strong>the</strong>different<strong>of</strong> K/F.Then<br />
n(ψ K/F) = [K : F]n(ψF) + δ K/F.<br />
ψ K/F(m − 1) = t + [K : F](m − t − 1).<br />
ByProposition4<strong>of</strong>paragraphIV.2<strong>of</strong>Serre’sbook δ K/F = ([K : F] − 1)(t + 1).Thus<br />
andwecanagaintake γK = γF.<br />
m ∗ + n(ψ K/F) = [K : F] (m + n(ψF))<br />
Since m(µF) = t + 1wehave m ≥ m(µF)and<br />
µF(1 + x + y) = µF(1 + x) µF(1 + y)<br />
for xand yin P d+ε<br />
F .Thus<strong>the</strong>existence<strong>of</strong> α(µF)isassured.Thelastassertion<strong>of</strong><strong>the</strong>lemma<br />
willbeprovedbyinduction.Wewillneedtoknowthatif x ≡ y(modPd′ K )<strong>the</strong>n<br />
N K/Fx ≡ N K/Fy(modP d F ).<br />
Whenprovingthiswemaysupposethat xOK = P r K with r ≤ d′ andthat y<br />
x<br />
Then<br />
<br />
NK/Fx − NK/Fy = NK/Fx 1 − NK/F <br />
1 +<br />
<br />
y − x<br />
.<br />
x<br />
belongsto OK.<br />
If r ≥ d<strong>the</strong>reisnothingtoprove.Suppose r ≤ d.If d ′ − r = ψK/F(u)and sis<strong>the</strong>smallest<br />
integergreaterthanorequalto u<strong>the</strong>rightsidebelongsto P s+r<br />
F .Since<strong>the</strong>derivative<strong>of</strong> ψK/F isatleastoneeverywhere ψK/F(u + r) ≥ d ′ .But<br />
d ′ ≥ m′ − 1<br />
=<br />
2<br />
1<br />
2 ψK/F(m − 1) + 1<br />
2 ψ <br />
m − 1<br />
K/F(0) ≥ ψK/F .<br />
2<br />
Thus u + r ≥ m−1<br />
2 and s + r ≥ d.<br />
Suppose F ⊆ L ⊆ Kand L/F iscyclic<strong>of</strong>primeorder. Let H = G(K/L)andlet<br />
G = G(L/F).Certainly H = Htwhile Ht+1 = {1}.Since ψ K/F(t) = ψ K/L(t) = t,wehave<br />
ψ L/F(t) = tand,byHerbrand’s<strong>the</strong>orem,<br />
Gt = G t = HG t /H = G/H = G.
Chapter8 66<br />
Moreover t + 1 = ψ L/F(t + δ)with δ > 0sothat<br />
Gt+1 = G t+δ = HG t+δ /H = H/H = {1}.<br />
Finally, ψ L/F(m −1)+1 ≥ t +1sothat L/Fand K/L,with mreplacedby ψ L/F(m −1)+1,<br />
satisfy<strong>the</strong>conditions<strong>of</strong><strong>the</strong>lemma. S(L/F)isasubgroup<strong>of</strong> S(K/F).If µFand νFbelong<br />
to S(K/F)<strong>the</strong>n µ L/F = ν L/Fifandonlyif µFand νFbelongto<strong>the</strong>samecoset<strong>of</strong> S(L/F).<br />
Take Stobeaset<strong>of</strong>representativesfor<strong>the</strong>secosets;<strong>the</strong>n<br />
S(K/L) = {µ L/F | µF | µF ∈ S}.<br />
Wetake α(µFνF) = α(µF) + α(νF)if µF belongsto Sand νF belongsto S(L/F). If µF<br />
belongsto Swetake α(µ L/F)tobe P ∗ L/F (α(µF)). If<strong>the</strong>lemmaisvalidfor K/Land L/F<br />
<strong>the</strong>n<br />
N K/F(P ∗ K/F (β)) = N L/F(N K/L(P ∗ K/L (P ∗ L/F (β)))<br />
whichiscongruentmodulo P d F to<br />
N L/F<br />
or <br />
Thisiscongruentmodulo P d F to<br />
whichequals<br />
<br />
µF ∈S<br />
<br />
µF ∈S (P ∗ L/F (β) + P ∗ L/F (α(µF)))<br />
µF ∈S {N L/F(P ∗ L/F<br />
<br />
<br />
(β + α(µF)))}.<br />
νF ∈S(L/F) {β + α(µF) + α(νF)}<br />
µF ∈S(K/F)<br />
{β + α(µF)}.<br />
Thusitisenoughtoprove<strong>the</strong>lemmawhen K/Fiscyclic<strong>of</strong>primeorder. Inthiscase<br />
morepreciseinformationisneededand<strong>the</strong>assertion<strong>of</strong><strong>the</strong>lemmawillfollowimmediately<br />
fromit.<br />
Lemma 8.3<br />
If K/Fisunramifiedand m ≥ 1wemaytake P ∗ K/F (β) = β.<br />
AccordingtoparagraphV.2<strong>of</strong>Serre’sbook<br />
N K/F(1 + y) − 1 ≡ S K/F(y) (mod P m F )
Chapter8 67<br />
if y ∈ P d′ +ε ′<br />
K .Thus P K/F(y) = S K/F(y)and<br />
Lemma 8.4<br />
ψF<br />
<br />
x PK/Fy xy<br />
= ψK/F γF<br />
Suppose K/Fisabelian,totallyramified,and [K : F] = ℓisanoddprime.If d ≥ t + 1<br />
(β) = β.<br />
wemaytake P ∗ K/L<br />
Therelation<br />
impliesthat m ′ ≡ m(mod 2), ε ′ = ε,and<br />
Since<br />
wehave<br />
Moreover<br />
γF<br />
<br />
.<br />
m ′ = t + 1 + ℓ(m − 1 − t) = ℓm − (t + 1) (ℓ − 1)<br />
d ′ = ℓd +<br />
ℓ − 1<br />
2<br />
ℓ − 1<br />
(ε − t − 1) = d + (m − t − 1).<br />
2<br />
ℓ − 1<br />
(m − t − 1) ≥ m − t − 1 ≥ d + ε<br />
2<br />
d ′ + ε ′ ≥ 2(d + ε) ≥ m.<br />
2(d ′ + ε ′ ) + δ K/F<br />
ℓ<br />
sothatbyLemma5<strong>of</strong>paragraphV.3<strong>of</strong>Serre’sbook<br />
if x ∈ P d′ +ε ′<br />
K .Thelemmafollows.<br />
Let pbe<strong>the</strong>characteristic<strong>of</strong> OF/PF.<br />
Lemma 8.5<br />
≥ m′ + δ K/F<br />
ℓ<br />
= m<br />
N K/F(1 + x) − 1 ≡ S K/F(x) (mod P m F )<br />
Suppose K/Fisabelian,totallyramified,and [K : F] = ℓisanoddprime. Suppose<br />
t+1 ≤ m ≤ 2t+1.Chooseanontrivialcharacter µFin S(K/F).Wemaychoose α = α(µF)<br />
sothat αOF = P v F ,if m = t + 1 + v,sothat α = N K/Fα1forsome α1in OK,andsothat<br />
µF(1 + x) = ψF<br />
<br />
αx<br />
γF
Chapter8 68<br />
for xin Ps t<br />
F . Here sis<strong>the</strong>leastintegergreaterthanorequalto 2 . If ζisa(p − 1)throot<strong>of</strong><br />
unityin F<strong>the</strong>reisauniqueinteger jwith 1 ≤ j ≤ p − 1suchthat ζ − jliesin PF. Set<br />
µ ζ<br />
F = µj<br />
F .Wemaytake α(µζ F )tobe ζα.If βbelongsto OFwecanfindaβ1in OKsuchthat<br />
β ≡ N K/Fβ1(modP d F ).Then<br />
If<br />
P ∗ K/F (β) ≡ β − β1<br />
α<br />
α1<br />
µF(1 + x) = ψF<br />
(mod P d′<br />
K).<br />
<br />
αx<br />
for xin P s F <strong>the</strong>n,necessarily, αOF = P v F . Choose δ1in OKsuchthat δ1OK = P v K andset<br />
δ = N K/Fδ1.Set α = ωδwhere ωisyettobechosen.Wemusthave<br />
µF(1 + x) = ψF<br />
γF<br />
<br />
ωδx<br />
if x ∈ Ps F . Thisequationdetermines<strong>the</strong>unit wmodulo Pr Fif r = t − s. Sinceanyunitis<br />
anormmodulo Pt Fwemaysuppose ω = NK/Fω1. Take α1 = ω2δ1. β1existsforasimilar<br />
reason.<br />
Thenumber ζ − jmustliein pOF.But K/Fiswildlyramified,because 2t + 1 ≥ m ><br />
1, ℓ = pand p = SK/F(1)sothat,byparagraphV.3<strong>of</strong>Serre’sbook, pbelongsto Pu Fif uis<br />
<strong>the</strong>greatestintegerin<br />
(ℓ − 1) t + 1<br />
(t + 1) ≥<br />
ℓ<br />
2 .<br />
However d + ε ≥ ssothat d + ε + u ≥ t + 1and,if xbelongsto P d+ε<br />
F<br />
Thus<br />
and<br />
Since<br />
ψF<br />
γF<br />
<br />
α(ζ − j)x<br />
= µF (1 + (ζ − j)x) = 1<br />
γF<br />
µ j<br />
F (1 + x) = µj<br />
F (1 + x) = ψF<br />
2s + δ K/F<br />
ℓ<br />
<br />
jαx<br />
γF<br />
≥ t + 1 + δ K/F<br />
ℓ<br />
Thelemmas<strong>of</strong>paragraphV.3<strong>of</strong>Serre’sbookimplythat<br />
= ψF<br />
= t + 1.<br />
ζαx<br />
N K/F(1 + x) ≡ 1 + S K/F(x) + N K/F(x) (modP t+1<br />
F )<br />
γF<br />
, (ζ − j)xliesin Pt+1<br />
F .<br />
<br />
.
Chapter8 69<br />
if xbelongsto P s K and<strong>the</strong>n<br />
1 = µF(N K/F(1 + x)) = µF(1 + S K/F(x) + N K/F(x)).<br />
Asweobserved d + ε ≥ s.Moreover d + ε ≤ t + 1sothat<br />
d + ε + δ K/F<br />
ℓ<br />
≥ d + ε<br />
and SK/F(x)and NK/F(x)belongto P d+ε<br />
F if xbelongsto P d+ε<br />
K .Thus,forsuch x,<br />
<br />
αNK/F(x) −αSK/F(x)<br />
.<br />
sothat<br />
Again<br />
if x ∈ P d′ +ε ′<br />
K .Moreover<br />
ψF<br />
γF<br />
= ψF<br />
2(d ′ + ε ′ ) + δ K/F<br />
ℓ<br />
≥ m<br />
N K/F(1 + x) − 1 = S K/F(x) + N K/F(x) (modP m F )<br />
d ′ + ε ′ = d + ε +<br />
ℓ − 1<br />
2<br />
γF<br />
(m − t − 1) ≥ d + ε<br />
sothat NK/F(x)andhence SK/F(x)belongto P d+ε<br />
F .Thus<br />
<br />
β1x<br />
βNK/Fx ≡ αNK/F But β1x/α1belongsto P d′ +ε ′ −v<br />
K<br />
sothat<br />
whichequals<br />
ψF<br />
⎛<br />
ψF<br />
and<br />
d ′ + ε ′ − v = d + ε +<br />
<br />
β PK/F(x)<br />
γF<br />
⎝ β S K/F(x) − αS K/F<br />
γF<br />
= ψF<br />
β1x<br />
α1<br />
α1<br />
ℓ − 1<br />
2<br />
(mod P m F ).<br />
v − v ≥ d + ε<br />
<br />
β SK/F(x) + β NK/F(x) ⎞<br />
⎠ = ψ K/F<br />
γF<br />
<br />
β − αβ1<br />
<br />
x<br />
α1 γF
Chapter8 70<br />
asrequired.<br />
Lemma 8.6<br />
Suppose K/Fisawildlyramifiedquadraticextension, m ≥ t + 1,and m = t + 1 + v.<br />
Let µFbe<strong>the</strong>nontrivialcharacterin S(K/F).If βbelongsto OF<strong>the</strong>reisaβ1in OKandaδ<br />
in Ut Fsuchthat β ≡ δNK/Fβ1(modP d F ).Wecanchoose α = α(µF)sothat<br />
µF(1 + x) = ψF<br />
<br />
αδx<br />
if xisin Ps Fandsothat α = NK/Fα1forsome α1in OK. Here shas<strong>the</strong>samemeaningas<br />
, t + 1 = r + s.With<strong>the</strong>sechoices<br />
before.Thus,if ris<strong>the</strong>integralpart<strong>of</strong> t+1<br />
2<br />
isin U t F<br />
P ∗ β1αδ<br />
K/F (β) ≡ β −<br />
α1<br />
<br />
γF<br />
mod P d′<br />
K<br />
If β = 0<strong>the</strong>existence<strong>of</strong> δand β1isclear.O<strong>the</strong>rwisewecanfindaβ1suchthat NK/Fβ1/β .Wechoose δaccordingly.If m = t + 1 + vand<br />
µF(1 + x) = ψF<br />
<br />
αδx<br />
for xin P s F <strong>the</strong>n OFα = P v F .Choose η1in OKsothat OKη1 = P v K andset η = N K/Fη1.Set<br />
α = ωηwhere ωisyettobechosen.Wemusthave<br />
µF(1 + x) = ψF<br />
γF<br />
<br />
ωηδx<br />
if x ∈ P s F .Thisequationdetermines<strong>the</strong>unit ωmodulo Pr F .Sinceanyunitisanormmodulo<br />
P t F wemaysuppose ω = N K/Fω1.Take α1 = ω1η1.<br />
Since<br />
Since<strong>the</strong>extensionisquadratic<br />
γF<br />
<br />
.<br />
N K/F(1 + x) = 1 + S K/F(x) + N K/F(x).<br />
s + δ K/F<br />
2<br />
= s + t + 1<br />
2<br />
≥ s
Chapter8 71<br />
both S K/F(x)and N K/F(x)arein P s F if xbelongsto Ps K and<br />
ψF<br />
<br />
αδNK/F(x)<br />
γF<br />
= ψF<br />
−αδSK/F(x)<br />
Wehave m ′ = 2m − (t + 1)and d ′ = m − s,sothat d ′ + ε ′ = m − rand d ′ + ε ′ − v = s.Thus<br />
if xbelongsto P d′ +ε ′<br />
K<br />
<br />
β1x<br />
βNK/F(x) ≡ αδNK/F (modP m F )<br />
and β1x/α1liesin P s K .Consequently<br />
asrequired.<br />
Lemma 8.7<br />
ψF<br />
γF<br />
α1<br />
γF<br />
α1<br />
<br />
.<br />
<br />
β PK/Fx<br />
αδ x<br />
= ψK/F β − β1<br />
If K/Fisatamelyramifiedquadraticextensionand m ≥ 2wemaytake P ∗ K/F (β) = β.<br />
Noticethat t + 1 = 1sothat m ≥ t + 1. Inthiscase m ′ = 2m − 1, d ′ = m − 1,and<br />
d ′ + ε ′ = m.If x ∈ P d′ +ε ′<br />
K<br />
iscongruentto<br />
modulo P m F .Thelemmafollows.<br />
N K/F(1 + x) = 1 + S K/F(x) + N K/F(x)<br />
1 + S K/F(x)<br />
Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma8.2wehavetoshowthatif K/Fiscyclic<strong>of</strong>primeorder<br />
NK/F(P ∗ <br />
K/F (β)) ≡<br />
µF ∈S(K/F) (β + α(µF)) (modP d F).<br />
Weconsider<strong>the</strong>casesdiscussedin<strong>the</strong>previouslemmasonebyone. If<strong>the</strong>extensionis<br />
unramifiedwemaytakeall<strong>the</strong>numbers α(µF)tobe0.Thecongruences<strong>the</strong>nreduceto<strong>the</strong><br />
identity β n = β n . Thesameistrueif K/Fiscyclic<strong>of</strong>oddorderand d ≥ t + 1or K/Fis<br />
quadraticand t = 0. If K/Fiscyclic<strong>of</strong>oddorder ℓand t + 1 ≤ m ≤ 2t + 1<strong>the</strong>rightside<br />
becomes<br />
β ℓ − βα ℓ−1 .<br />
γF
Chapter8 72<br />
If β ≡ 0 (modPd F )bothsidesarecongruentto0modulo PdF .Suppose βdoesnotbelongto<br />
Pd Fand βOF = Pu F .Then β1OK = Pu Kand iscongruentto<br />
N K/F<br />
<br />
β − β1<br />
<br />
α<br />
α2<br />
β ℓ − βα ℓ−1 + ℓ−1<br />
i=1 (−1)iβ ℓ E i<br />
<br />
β1α<br />
βα1<br />
modulo Pd F . If x ∈ K<strong>the</strong>n Ei (x)is<strong>the</strong> i<strong>the</strong>lementarysymmetricfunction<strong>of</strong> xandits<br />
(ℓ−1) (v−u)<br />
conjugates.Moreover β1α/βα1belongsto P .If ℓ − 1 ≥ i ≥ 1<br />
Therightsideis<br />
i(ℓ − 1) (v − u) + (ℓ − 1) (t + 1)<br />
ℓ<br />
K<br />
≥<br />
(ℓ − 1)<br />
m − pu ≥ d − ℓu.<br />
ℓ<br />
Theargument<strong>of</strong>paragraphV.3<strong>of</strong>Serre’sbookshowsthat<br />
N K/F<br />
<br />
β − β1<br />
(ℓ − 1) (v + t + 1)<br />
ℓ<br />
<br />
α<br />
≡ β ℓ − βα ℓ−1 (modP d F).<br />
α1<br />
− ℓu.<br />
Forawildlyramifiedquadraticextensionweuse<strong>the</strong>notation<strong>of</strong>Lemma8.6. Theright<br />
side<strong>of</strong><strong>the</strong>congruencemaybetakentobe β2 + βαδ.Theidentityisagainnontrivialonlyif<br />
βOF = Pu Fwith u < d.Then<strong>the</strong>leftsidemaybetakentobe<br />
whichiscongruentto<br />
modulo P d F .Since<br />
wehave<br />
β 2 − β 2 δS K/F<br />
<br />
β1α<br />
+ δ<br />
βα1<br />
2 αNK/Fβ1 β 2 + αβδ − β 2 δS K/F<br />
v − u + t + 1<br />
2<br />
β 2 S K/F<br />
≥ m<br />
2<br />
<br />
β1α<br />
βα1<br />
− u ≥ d − u<br />
<br />
β1α<br />
≡ 0 (modP<br />
βα1<br />
d F).
Chapter8 73<br />
Suppose χFisaquasicharacter<strong>of</strong> CF, m = m(χF),and β = β(χF).If,assometimes<br />
happens, m ′ = m(χ K/F)wecantake β(χ K/F) = P ∗ K/F (β).<br />
Lemma 8.8<br />
Suppose K/FisGaloisand G = G(K/F).Suppose s ≥ 0isanintegerand G s = {1}.If<br />
m = m(χF)and m > s<strong>the</strong>n<br />
m ′ = ψ K/F(m − 1) + 1 = m(ψ K/F).<br />
ItfollowsfromparagraphV.6<strong>of</strong>Serre’sbookthat<br />
NK/F(U ψK/F(v) K ) = U v F<br />
if v ≥ s. Thus χ K/F istrivialon U u K if u > ψ K/F(m − 1)butisnottrivialon U i K if<br />
u = ψ K/F(m − 1).<br />
Wecannowcollecttoge<strong>the</strong>r,withoneortwoadditionalcomments,<strong>the</strong>previousresults<br />
inaformwhichwillbeusefulin<strong>the</strong>pro<strong>of</strong><strong>of</strong><strong>the</strong>firstmainlemma.Weuse<strong>the</strong>samenotation.<br />
Lemma 8.9<br />
SupposeK/Fisacyclicextension<strong>of</strong>primeorderℓ, χFisaquasicharacter<strong>of</strong>CF , m(χF) ≥<br />
t + 1, m(χF) > 1,and m(χ K/F) − 1 = ψ K/F(m(χF) − 1).<br />
(a)If K/Fisunramifiedwemaytake β(χ K/F) = β(χF)and β(µFχF) = β(χF)forall µF<br />
in S(K/F).<br />
(b)If ℓisoddand d ≥ t + 1wemaytake β(χ K/F) = β(χF)and β(µFχF) = β(χF)forall<br />
µFin S(K/F).<br />
(c)If ℓisoddand t + 1 ≤ m ≤ 2t + 1and µFisagivennontrivialcharacterin S(K/F)we<br />
maychoose α = α(µF) = N K/Fα1asinLemma8.5and β = β(χF) = N K/Fβ1forsome<br />
β1in UK.Thenwemaychoose<br />
and<br />
β(χ K/F) = β − β1<br />
α<br />
α1<br />
β(µ ζ<br />
F χF) = N K/F(β1 + ζα1).
Chapter8 74<br />
(d)If ℓis2and K/Fiswildlyramifiedwechoose α = α(µF)asinLemma8.6. Wemay<br />
choose β = β(χF)in<strong>the</strong>form δN K/Fβ1with δin U t F .Thenwemaychoose<br />
and<br />
β(χ K/F) = β − β1<br />
αδ<br />
α1<br />
β(µFχF) = β + αδ.<br />
(e)If ℓis2and K/Fistamelyramifiedwemaytake β(χ K/F) = β(µFχF) = β(χF).<br />
<strong>On</strong>lypart(c)requiresanyfur<strong>the</strong>rverification.Itmustbeshownthat<br />
Theleftsideiscongruentto<br />
Allweneeddoisshowthat<br />
Therightsideis<br />
N K/F(β1 + ζα1) ≡ β + ζα (modP d F ).<br />
N K/F<br />
βN K/F<br />
<br />
1 + ζα1<br />
<br />
.<br />
β1<br />
<br />
1 + ζα1<br />
<br />
≡ 1 +<br />
β1<br />
ζα<br />
β<br />
1 + N K/F<br />
ζα1<br />
β1<br />
<br />
.<br />
(modP d F).<br />
AccordingtoparagraphV.3<strong>of</strong>Serre’sbook<strong>the</strong>congruencewillbesatisfiedif<br />
v + (ℓ − 1) (t + 1)<br />
ℓ<br />
≥ d.<br />
But t + 1 = d + xwith x ≥ 0sothat d + x + v = 2d + εand v = d + ε − x.Thus<br />
v + (ℓ − 1) (t + 1)<br />
ℓ<br />
= d + ε − x + (ℓ − 1) (d + x)<br />
ℓ<br />
= d +<br />
ε + (ℓ − 2)x<br />
ℓ<br />
Theprecedingdiscussionhasnowtoberepeatedwithdifferenthypo<strong>the</strong>sesanddifferent,<br />
butsimilar,conclusions.<br />
Lemma 8.10<br />
≥ d.
Chapter8 75<br />
Suppose K/Fisabelianand G = G(K/F).Suppose<strong>the</strong>reisat ≥ 0suchthat G = Gt<br />
while Gt+1 = {1}.If 2 ≤ m ≤ t+1<strong>the</strong>n m ′ = ψK/F(m−1)+1isjust m.Let t+1 = m+v,let<br />
δbesuchthat δOF = P t+1+n(ψF )<br />
F ,let ε1in OKbesuchthat ε1OK = Pv K ,andlet ε = NK/Fε1. Wemaychoose γF = δ/εand γK = δ/ε1. Let rbe<strong>the</strong>greatestintegerin t+1<br />
andlet 2<br />
s = t + 1 − r.If µFisanontrivialcharacterin S(K/F)<strong>the</strong>n m(µF) = t + 1.Let<br />
for xin P s F .Then<br />
β <br />
µF(1 + x) = ψF<br />
<br />
β(µF)x<br />
µF =1 (βε + β(µF)) ≡ N K/F(P ∗ K/F (β)) (modPd F).<br />
Therelation m ′ = ψ K/F(m − 1) + 1 = misanimmediateconsequence<strong>of</strong><strong>the</strong>definitions.<br />
Since<strong>the</strong>extensionistotallyramified<br />
if n = n(ψF).Thus<br />
and<br />
n(ψ K/F) = [K : F]n + ([K : F] − 1)(t + 1)<br />
m + n = (t + 1 + n) − v<br />
m ′ + n(ψ K/F) = [K : F](t + 1 + n) + (m − t − 1) = [K : F](t + 1 + n) − v.<br />
Consequently γFand γKcanbechosenasasserted.Theresults<strong>of</strong>chapterV<strong>of</strong>Serre’sbook<br />
implythat m(µF) = t + 1if µFisnottrivial.<br />
WesawwhenprovingLemma8.2thatifx ≡ y(modP d′<br />
K )<strong>the</strong>nN K/Fx ≡ N K/Fy(modP d F )<br />
andthatif F ⊆ L ⊆ Kboth L/Fand K/Lsatisfy<strong>the</strong>conditions<strong>of</strong><strong>the</strong>lemma.For L/F, ε1<br />
isreplacedby N K/Lε1and,for K/L, εisreplacedby N K/Lε1.Take Q ∗ L/F tobe P ∗ L/F in<strong>the</strong><br />
specialcasethat m = t + 1and ε1 = 1.Then<br />
ψ L/F<br />
NK/L(ε1)x P ∗ L/F (β)<br />
bydefinition.Therightsideisequalto<br />
δ<br />
ψ L/F<br />
<br />
δ<br />
= ψF<br />
<br />
∗ x QL/F (εβ)<br />
.<br />
δ<br />
<br />
εPL/F(x)β<br />
δ
Chapter8 76<br />
Thus<br />
Q ∗ L/F (εβ) ≡ N K/L(ε1)P ∗ L/F (β) (modPs L ).<br />
If µF belongsto S(K/F)butnotto S(L/F)<strong>the</strong>n m(µ L/F) = m(µF)and β(µ L/F)may<br />
betakentobe Q ∗ L/F (β(µF)). Let S ′ beaset<strong>of</strong>representativesfor<strong>the</strong>cosets<strong>of</strong> S(L/F)in<br />
S(K/F) − S(L/F)andsuppose<strong>the</strong>lemmaistruefor K/Land L/F.Then<br />
iscongruentto<br />
N K/F(P ∗ K/F (β)) = N L/F(N K/L(P ∗ K/L (P ∗ L/F (β))))<br />
NL/F(P ∗ <br />
L/F (β)<br />
µF ∈S ′{NK/L(ε1)P ∗ L/F (β) + Q∗L/F (β(µF))})<br />
modulo P d F .Thisinturniscongruentto<br />
NL/F(P ∗ <br />
L/F (β))<br />
µF ∈S ′ NL/F(Q ∗ L/F (εβ + β(µF))).<br />
Applying<strong>the</strong>inductionhypo<strong>the</strong>sisto<strong>the</strong>firstpartandLemma8.2to<strong>the</strong>second,weseethat<br />
<strong>the</strong>wholeexpressioniscongruentto<br />
β<br />
<br />
modulo P d F asrequired.<br />
νF ∈S(L/F) (εβ + β(νF))<br />
νF =1<br />
<br />
µF ∈S ′ (εβ + β(µF) + β(νF))<br />
νF ∈S(L/F)<br />
<strong>On</strong>ceagainwedevotealemmatocyclicextensions<strong>of</strong>primeorder.<br />
Lemma 8.11<br />
Suppose K/Fiscyclic<strong>of</strong>primeorder ℓand 2 ≤ m ≤ t+1.Chooseanontrivialcharacter<br />
µFin S(K/F).Thereisan α1in UKsuchthatif α = N K/Fα1<br />
µF(1 + x) = ψF<br />
<br />
αx<br />
<br />
δ<br />
for xin P s F .If βbelongsto OF<strong>the</strong>reisaβ1in OKsuchthat β ≡ N K/F(β1) (modP t F ).Then<br />
P ∗ ε<br />
K/F (β) ≡ β<br />
ε1<br />
− β1<br />
α<br />
α1<br />
(mod P d′<br />
K ).
Chapter8 77<br />
Since β(µF)isdeterminedonlymodulo P s F and s ≤ twecantake β(µF) = N K/Fα1for<br />
some α1in UK.Theexistence<strong>of</strong> β1als<strong>of</strong>ollowsasbefore.Since t + 1 ≥ m<br />
and<br />
2(d ′ + ε ′ ) + (ℓ − 1) (t + 1)<br />
ℓ<br />
≥<br />
m + (ℓ − 1) (t + 1)<br />
ℓ<br />
≥ m<br />
N K/F(1 + x) ≡ 1 + S K/F(x) + N K/F(x) (modP m F )<br />
if xbelongsto P d′ +ε ′<br />
K .Thus<br />
<br />
εPK/F(x)β εxβ<br />
ψF<br />
= ψK/F δ<br />
δ<br />
But d ′ + ε ′ + t ≥ t + 1sothat<br />
N K/F(ε1x)β ≡ αN K/F<br />
ε1β1<br />
Since t + 1 = m + v, d ′ + ε ′ + v ≥ sandif y = ε1β1<br />
α1<br />
in P s K .But<br />
sothat<br />
Consequently<br />
Inconclusion<br />
asrequired.<br />
α1<br />
2s + (t + 1)(ℓ − 1)<br />
ℓ<br />
ψF<br />
<br />
· x<br />
NK/F(ε1x)β<br />
δ<br />
(mod P t+1<br />
F )<br />
<br />
.<br />
· x<strong>the</strong>n ywhichliesin P d′ +ε ′ +v<br />
K<br />
≥ t + 1<br />
N K/F(1 + y) ≡ 1 + S K/F(y) + N K/F(y) (mod P t+1<br />
F ).<br />
ψF<br />
ψF<br />
<br />
−αNK/F(y)<br />
<br />
εPK/F(x)β<br />
= ψK/F δ<br />
δ<br />
= ψF<br />
ε1<br />
δ<br />
αSK/F(y)<br />
ε<br />
ε1<br />
δ<br />
· β − α<br />
α1<br />
<br />
.<br />
· β1<br />
<br />
x<br />
alsolies<br />
Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma8.10wehavetoshowthatwhen K/Fiscyclic<strong>of</strong>prime<br />
order, ℓ<br />
β <br />
µF =1 (βε + β(µF)) ≡ N K/F(P ∗ K/F (β)) (mod Pd F )<br />
Since 2 ≤ m ≤ t + 1<strong>the</strong>extensioniswildlyramified, ℓ = p,andonce<strong>the</strong>character µF is<br />
chosenasin<strong>the</strong>previouslemmawecandefine µ ζ<br />
FasinLemma8.5.Theleftsideiscongruent to<br />
β β ℓ−1 ε ℓ−1 + (−1) ℓ α ℓ−1 .
Chapter8 78<br />
If β ∈ Pd Fthisiscongruentto0andsois<strong>the</strong>rightside.Suppose βOF = Pu Fwith u < d.The<br />
rightsideiscongruentto<br />
Since<br />
thisiscongruentto<br />
Since<br />
Weseethat<br />
N K/F<br />
<br />
β ε<br />
ε1<br />
− β1<br />
α1<br />
(ℓ − 1)(u + v) + (ℓ − 1)(t + 1)<br />
ℓ<br />
βα ℓ−1<br />
<br />
βN K/F<br />
<br />
α<br />
≡ βα ℓ−1 <br />
β<br />
NK/F N K/F<br />
β<br />
andthat<strong>the</strong>expression(8.1)iscongruentto<br />
modulo P d F .<br />
Lemma 8.12<br />
β1<br />
≥<br />
α1<br />
α<br />
ℓ − 1<br />
ℓ<br />
β1<br />
α1<br />
α<br />
· (t + 1) ≥<br />
ε<br />
ε1<br />
<br />
− 1 .<br />
t + 1<br />
2<br />
≥ d<br />
<br />
ε<br />
+ (−1) ℓ<br />
<br />
. (8.1)<br />
ε1<br />
−1<br />
t−u<br />
β1 ≡ 1 (mod PF ).<br />
β ℓ+1 N K/Fβ −1<br />
1 ≡ βℓ (mod P t F)<br />
β ℓ ε ℓ−1 + (−1) ℓ βα ℓ−1<br />
SupposeK/Fisabelianand G = G(K/F).Suppose<strong>the</strong>reisaninteger tsuchthatG = Gt<br />
while Gt+1 = {1}. Let χFbeaquasicharacter<strong>of</strong> CFandsuppose 2 ≤ m(χF) ≤ t + 1. If<br />
m(χF) < t + 1<strong>the</strong>n m(χ K/F) = m(χF).If m(χF) = t + 1<strong>the</strong>n m(µFχF) < t + 1forsome<br />
µFin S(K/F)ifandonlyif m(χ K/F) < m(χF).<br />
ItfollowsimmediatelyfromLemma6.7thatif χFisanyquasicharacter<strong>of</strong> CFand Eany<br />
finiteseparableextension<strong>of</strong> F<strong>the</strong>n<br />
m(χ E/F) − 1 ≤ ψ E/F(m(χF) − 1).<br />
In<strong>the</strong>particularcaseunderconsiderationLemma6.10showsthatif m = m(χF) ≤ t<strong>the</strong>n<br />
NK/F : U m−1<br />
K /Ut K −→ Um−1<br />
F /U t F
Chapter8 79<br />
isanisomorphism.Thus χK/F(α)willbedifferentfrom1forsome αin U m−1<br />
K and m(χK/F) willbeatleast m.If m(χF) = t + 1<strong>the</strong>n m(µFχF)islessthan t + 1forsome µFin S(K/F)<br />
ifandonlyif χF istrivialon<strong>the</strong>image<strong>of</strong> Ut K /Ut+1<br />
K in Ut F /Ut+1<br />
F . Thisissoifandonlyif<br />
m(χK/F) ≤ t.<br />
Weshallneed<strong>the</strong>followinglemmain<strong>the</strong>pro<strong>of</strong><strong>of</strong><strong>the</strong>firstmainlemma.<br />
Lemma 8.13<br />
Suppose K/Fiscyclic<strong>of</strong>primeorder, χFisaquasicharacter<strong>of</strong> CFwith m(χF) ≤ t + 1,<br />
and m(χ K/F) = m(χF). Choose α, α1, ε, ε1inLemma8.11. Wemaychoose β = β(χF) =<br />
N K/Fβ1with β1in UKandwemaychoose<br />
β(χ K/F) = β ε<br />
Moreover m(µ ζ<br />
F χF) = t + 1andwemaytake<br />
ε1<br />
− β1<br />
α<br />
.<br />
α1<br />
β(µ ζ<br />
F χF) = N K/F(ζα1 + ε1β1).<br />
Since β(χF)isdeterminedonlymodulo Pd Fand d ≤ t<strong>the</strong>existence<strong>of</strong> β1isclear. Itis<br />
alsoclearthat m(µ ζ<br />
FχF) = t + 1.Theelements β(χF), β(χK/F),and β(µ ζ<br />
FχF)aretosatisfy <strong>the</strong>followingconditions:<br />
(i)If xisin P d F<br />
(ii)If xisin P d′<br />
K<br />
(iii)If xisin P s F<br />
χF(1 + x) = ψF<br />
χ K/F(1 + x) = ψ K/F<br />
µ ζ<br />
F (1 + x)χF(1 + x) = ψF<br />
εβ(χF)x<br />
δ<br />
<br />
.<br />
ε1β(χ K/F)x<br />
Wehavealreadyshownthat β(χ K/F)maybetakentobe<br />
β ε<br />
ε1<br />
α<br />
− β1<br />
α1<br />
δ<br />
<br />
.<br />
<br />
β(µ ζ<br />
FχF)x <br />
.<br />
δ
Chapter8 80<br />
β(µ ζ<br />
F χF)mustbecongruentto ζα + εβmodulo P r F<br />
Since<br />
Therightsideiscongruentto<br />
modulo P r F .∗<br />
N K/F(ζα1 + ε1β1) = ζαN K/F<br />
ν + (ℓ − 1)(t + 1)<br />
ℓ<br />
ζα<br />
<br />
1 + N K/F<br />
∗(1998)Themanuscript<strong>of</strong>Chapter8endshere.<br />
≥<br />
ε1β1<br />
ζα1<br />
<br />
1 + ε1β1<br />
<br />
.<br />
ζα1<br />
ℓ − 1<br />
(t + 1) ≥ r.<br />
ℓ<br />
<br />
= ζα + εβ
Chapter9 81<br />
Chapter Nine.<br />
A Lemma <strong>of</strong> Hasse<br />
Let λ ⊆ κbetw<strong>of</strong>initefieldsandlet G = G(κ/λ).If x ∈ κset<br />
ω κ/λ(x) = x σ1 x σ2<br />
where<strong>the</strong>sumistakenoverallunorderedpairs<strong>of</strong>distinctelements<strong>of</strong> G.Itisclearthat<br />
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).<br />
<strong>On</strong>ereadilyverifiesalsothatif λ ≤ η ≤ κ<strong>the</strong>n<br />
ω κ/λ(x) = ω η/λ(S κ/η(x)) + S η/λ (ω κ/η(x)).<br />
Suppose ψλisanontrivialcharacter<strong>of</strong> λand ϕλisanowherevanishingfunctionon λ<br />
satisfying<strong>the</strong>identity<br />
ϕλ(x + y) = ϕλ(x)ϕλ(y) ψλ(xy).<br />
Define ϕ κ/λon κby<br />
Then ϕ κ/λ(x + y)isequalto<br />
whichis<br />
ϕ κ/λ(x) = ϕλ(S κ/λ(x)) ψλ(−ω κ/λ(x)).<br />
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))<br />
ϕ κ/λ(x)ϕ κ/λ(y)ψ κ/λ(xy).<br />
If<strong>the</strong>fieldshaveoddcharacteristic<strong>the</strong>followinglemmais,basically,aspecialcase<strong>of</strong><br />
Lemma7.7.ThatlemmahasbeenproveninasimpleanddirectmannerbyWeil[14].Weshall<br />
usehismethodtoprove<strong>the</strong>followinglemmawhichincharacteristictwo,whenitcannotbe<br />
reducedto<strong>the</strong>previouslemma,isduetoHasse[8].<br />
Lemma 9.1<br />
Let<br />
σ(ϕλ) = − <br />
x∈λ ϕλ(x)
Chapter9 82<br />
andlet<br />
Then<br />
If<br />
σ(ϕ κ/λ) = − <br />
x∈κ ϕ κ/λ(x).<br />
σ(ϕ κ/λ) = σ(ϕλ) [κ:λ] .<br />
P(X) = X m − aX m−1 + bX m−2 − +...<br />
isanymonicpolynomialwithcoefficientsin λset m(P) = mand<br />
χλ(P) = ϕλ(a) ψλ(−b).<br />
If<strong>the</strong>degree<strong>of</strong><strong>the</strong>polynomialis1, bistakentobe0;if<strong>the</strong>degreeis0both aand baretaken<br />
tobe0.If<br />
P ′ (X) = X m′<br />
− a ′ X m′ −1 ′ m<br />
+ b X ′ −2<br />
− +...<br />
<strong>the</strong>n<br />
and<br />
PP ′ (X) = X m+m′<br />
− (a + a ′ ) X m+m′ −1 ′ ′ m+m<br />
+ (b + b + aa )X ′ −2<br />
− +...<br />
χλ(PP ′ ) = ϕλ(a + a ′ ) ψλ(−b − b ′ − aa ′ ) = χλ(P)χλ(P ′ ).<br />
If tisanindeterminateweintroduce<strong>the</strong>formalseries<br />
Fλ(t) = χλ(P)t m(P) = (1 − χλ(P)t m(P) ) −1 .<br />
Thesumisoverallmonicpolynomialswithcoefficientsin λand<strong>the</strong>productisoverallirreduciblepolynomials<strong>of</strong>positivedegreewithcoefficientsin<br />
λ.If r ≥ 2<br />
sothat<br />
<br />
m(P)=r χλ(P) = 0<br />
Fλ(t) = 1 − σ(ϕλ)t.<br />
Ifwereplace λby κ, ϕλby ϕ κ/λ,and ψλby ψ κ/λ,wecandefine F κ/λ(t)inasimilarway.<br />
If k = [κ : λ]and Tis<strong>the</strong>set<strong>of</strong> kthroots<strong>of</strong>unity<strong>the</strong>problemistoshowthat<br />
<br />
ζ∈T Fλ(ζt) = F κ/λ(t k ).
Chapter9 83<br />
Suppose Pisanirreduciblemonicpolynomialwithcoefficientsin λand P ′ isone<strong>of</strong>its<br />
monicirreduciblefactorsover κ.Let m = m(P)andlet rbe<strong>the</strong>greatestcommondivisor<strong>of</strong><br />
mand k.Thefieldobtainedbyadjoining<strong>the</strong>roots<strong>of</strong> Pto κhasdegree mk<br />
r over λandis<strong>the</strong><br />
sameas<strong>the</strong>fieldobtainedbyadjoining<strong>the</strong>roots<strong>of</strong> P ′ to κ. Thus m(P ′ ) = m<br />
and Psplits<br />
r<br />
into rirreduciblefactorsover κ.Weshallshowthat<br />
Thusif P ′ 1 , . . ., P ′ r<br />
whichequals<br />
χ κ/λ(P ′ ) = {χλ(P)} k<br />
r .<br />
are<strong>the</strong>factors<strong>of</strong> Pand ℓ = k<br />
r<br />
r<br />
Thenecessaryidentityfollows.<br />
i=1 {1 − χ κ/λ(P ′ i)t<br />
<br />
km(P ′<br />
i ) } = {1 − χλ(P) ℓ t ℓm } r<br />
ζ∈T {1 − χλ(P)ζ m t m }.<br />
Let νbe<strong>the</strong>fieldobtainedbyadjoiningaroot x<strong>of</strong> P ′ to κandlet µbe<strong>the</strong>fieldobtained<br />
byadjoining xto λ.If<br />
P(X) = X m − aX m−1 + bX m−2 . . .<br />
<strong>the</strong>n<br />
and<br />
Thus<br />
Since ϕ ν/λ(x)isequalto<br />
whichinturnequals<br />
Weconcludethat<br />
a = S µ/λ(x)<br />
b = ω µ/λ(x).<br />
χλ(P) = ϕλ(S µ/λ(x)) ψλ(−ω µ/λ(x)) = ϕ µ/λ(x).<br />
ϕλ(S κ/λ(S ν/κ(x))) ψλ(−ω κ/λ(S ν/κ(x)) + S κ/λ(ω ν/κ(x)))<br />
Replacing κby µweseethat ϕ ν/λ(x)equals<br />
ϕ κ/λ(S ν/κ(x)) ψ κ/λ(−ω ν/κ(x)).<br />
χ κ/λ(P ′ ) = ϕ ν/λ(x).<br />
ϕ µ/λ(S ν/µ(x)) ψ µ/λ(−ω ν/µ(x)) = ϕ µ/λ(ℓx) ψ µ/λ<br />
<br />
−ℓ<br />
(ℓ − 1)<br />
2<br />
x 2<br />
<br />
.
Chapter9 84<br />
<strong>On</strong>eeasilyshowsbyinductionthatforeveryinteger ℓ<br />
Therelation<br />
follows.<br />
{ϕ µ/λ(x)} ℓ = ϕ µ/λ(ℓx) Ψ µ/λ<br />
<br />
−ℓ<br />
χ κ/λ(P ′ ) = {χλ(P)} ℓ<br />
Taking µ = λin<strong>the</strong>identity(9.1)weseethat<br />
{ϕλ(x)} ℓ = ϕλ(ℓx) ψλ<br />
<br />
−ℓ<br />
(ℓ − 1)<br />
2<br />
(ℓ − 1)<br />
2<br />
x 2<br />
<br />
x 2<br />
<br />
. (9.1)<br />
foreveryinteger ℓ.Moreover {ϕλ(0)} 2 = ϕλ(0)sothat ϕλ(0) = 1.If<strong>the</strong>characteristic p<strong>of</strong><br />
λisoddtake ℓ = ptoseethat {ϕλ(x)} p = 1. If<strong>the</strong>characteristicis2,take ℓ = 4toseethat<br />
{ϕλ(x)} 4 = 1.Suppose ϕ ′ λisano<strong>the</strong>rfunctionon λwhichvanishesnowhereandsatisfies<br />
Then ϕ ′ λ ϕ−1<br />
λ<br />
Ofcourse<br />
Thus<br />
ϕ ′ λ (x + y) = ϕ′ λ (x) ϕ′ λ (y) ψλ(xy).<br />
isacharacterandforsome αin λ<br />
ϕ ′ λ(x) ≡ ϕλ(x) ψλ(αx).<br />
ϕλ(x)ψλ(αx) = ϕλ(x + α)ϕ −1<br />
λ (α).<br />
σ(ϕ ′ λ) = ϕ −1<br />
λ (α) σ(ϕλ).<br />
If aand baretwononzerocomplexnumbersand misapositiveintegerwewrite a ∼m bif,<br />
forsomeinteger r ≥ 0, a<br />
b<br />
Lemma 9.2<br />
m r<br />
= 1.<br />
If α ∈ λ × ,<strong>the</strong>multiplicativegroup<strong>of</strong> λ,let ν(α)be1or1accordingas αisorisnota<br />
squarein λ.Suppose ψ ′ λ (x) = ψλ(αx), ϕλand ϕ ′ λ arenowherevanishing,and<br />
while<br />
ϕλ(x + y) = ϕλ(x)ϕλ(y)ψλ(xy)<br />
ϕ ′ λ(x + y) = ϕ ′ λ(x)ϕ ′ λ(y)ψ ′ λ(xy).
Chapter9 85<br />
Then<br />
Moreover<br />
σ(ϕ ′ λ ) ∼p ν(α)σ(ϕλ).<br />
σ(ϕλ) ∼2p |σ(ϕλ)|.<br />
Supposefirstthat pisodd. By<strong>the</strong>remarkspreceding<strong>the</strong>statement<strong>of</strong><strong>the</strong>lemmaitis<br />
enoughtoprove<strong>the</strong>assertionsforonechoice<strong>of</strong> ϕλand ϕ ′ <br />
λ . Forexamplewecouldtake<br />
2<br />
(x )<br />
ϕλ(x) = Ψλ andif α = β 2<br />
2wecouldtake ϕ ′ 2<br />
(βx)<br />
λ (x) = ψλ .Inthiscaseitisclear<br />
2<br />
that σ(ϕλ) = σ(ϕ ′ λ ).Howeverif αisnotasquare,wetake ϕ′ <br />
2<br />
αx<br />
λ (x) = ψλ .Then<br />
Withthischoice<strong>of</strong> ϕλ,<br />
σ(ϕλ) + σ(ϕ ′ <br />
λ ) = 2<br />
x∈λ ψλ<br />
ϕλ(x) = ψλ<br />
<br />
− x2<br />
<br />
2<br />
<br />
x<br />
<br />
= 0.<br />
2<br />
sothat σ(ϕλ) = ν(−1) σ(ϕλ).Moreoveritiswellknownandeasilyverifiedthat σ(ϕλ) = 0.<br />
Since<br />
{σ(ϕλ)} 4 = {ν(−1)} 2 |σ(ϕλ)| 4 = |σ(ϕλ)| 4<br />
wehave<br />
σ(ϕλ) ∼2p |σ(ϕλ)|.<br />
Theabsolutevalueon<strong>the</strong>rightis<strong>of</strong>course<strong>the</strong>ordinaryabsolutevalue.<br />
Suppose pis2. Againanychoice<strong>of</strong> ϕλand ϕ ′ λwilldo. Inthiscase αisnecessarilya<br />
square.Let α = β2 .Wecantake ϕ ′ λ (x) = ϕλ(βx).Then σ(ϕ ′ λ ) = σ(ϕλ).Itisenoughtoprove<br />
<strong>the</strong>secondassertionforany ψλandany ϕλ.Let φbe<strong>the</strong>primefieldandlet ψφbe<strong>the</strong>unique<br />
nontrivialadditivecharacter<strong>of</strong> φ.Take ψλ = ψλ/φ.Let ϕφ(0) = 1, ϕφ(1) = i.<strong>On</strong>everifies<br />
byinspectionthat<br />
ϕφ(x + y) = ϕφ(x)ϕφ(y)ψφ(xy).<br />
Take ϕλ = ϕ λ/φ.Since<br />
itisenoughtoverifythat<br />
Since σ(ϕφ) = −1 + i,thisisnoproblem.<br />
σ(ϕ λ/φ) = {σ(ϕφ)} [λ:φ] ,<br />
σ(ϕφ) ∼2 |σ(ϕφ)|.<br />
2
Chapter9 86<br />
If aisanonzerocomplexnumberset<br />
A [a] = a<br />
|a| .<br />
Thefollowinglemmaexplainsourinterestin<strong>the</strong>numbers σ(ϕλ).<br />
Lemma 9.3<br />
Suppose Lisanonarchimedeanlocalfieldand χL isaquasicharacter<strong>of</strong> CL with<br />
m = m(χL) = 2d + 1,where disapositiveinteger.Let ψLbeanontrivialadditivecharacter<br />
<strong>of</strong> Landlet n = n(ψL).Let γbesuchthat γOL = P m+n<br />
L andlet βbeaunitsuchthat<br />
<br />
βx<br />
χL(1 + x) = ψL<br />
γ<br />
for xin P d+1<br />
L .Choose δsothat δOL = P d L andlet ψλbe<strong>the</strong>character<strong>of</strong> λ = OL/PLdefined<br />
by<br />
If ϕλisdefinedby<br />
<strong>the</strong>n<br />
and<br />
ψλ(x) = ψL<br />
ϕλ(x) = ψL<br />
<br />
βδx<br />
γ<br />
βδ 2 x<br />
γ<br />
<br />
.<br />
χ −1<br />
L (1 + δx)<br />
ϕλ(x + y) = ϕλ(x)ϕλ(y)ψλ(xy)<br />
∆3(χL, ψL, γ) = A [−σ(ϕλ)].<br />
In<strong>the</strong>statement<strong>of</strong>thislemmawehavenotdistinguished,in<strong>the</strong>notation,betweenan<br />
element<strong>of</strong> OLanditsimagein λ.Thisisconvenientandnottooambiguous.Itwillbedone<br />
again.Theonlyquestionablepart<strong>of</strong><strong>the</strong>lemmais<strong>the</strong>relation<br />
Since<br />
wehave<br />
χ −1<br />
L<br />
ϕλ(x + y) = ϕλ(x)ϕλ(y)ψλ(xy).<br />
(1 + δx) (1 + δy) ≡ (1 + δx + δy) (1 + δ 2 xy) (modP m L )<br />
(1 + δx) χ−1<br />
L<br />
(1 + δy) = χ−1<br />
L (1 + δx + δy) ψL<br />
<br />
− βδ2 <br />
xy<br />
.<br />
γ
Chapter9 87<br />
Therequiredrelationfollowsimmediately.<br />
Thereareafewremarkswhichweshallneedlater. Itisconvenientt<strong>of</strong>ormulate<strong>the</strong>m<br />
explicitlynow.Weretain<strong>the</strong>notation<strong>of</strong><strong>the</strong>previouslemma.<br />
Lemma 9.4<br />
If m(µL) < m(χL)<strong>the</strong>n<br />
∆3(µL χL, ΨL; γ) ∼p ∆3 (χL, ΨL; γ)<br />
andif m(µL) ≤ dwemaytake β(µLχL) = β(χL)and<strong>the</strong>n<br />
∆3(µLχL, ψL; γ) = ∆3(χL, ψL; γ).<br />
Inbothcases m(µLχL) = m(χL).Moreoverif x ∈ P2d L<br />
<br />
β(µLχL)x<br />
ψL<br />
= µL(1 + x) χL(1 + x) = χL(1 + x)<br />
γ<br />
whichinturnequals<br />
Thus<br />
andif<br />
while<br />
ψL<br />
β(χL)x<br />
γ<br />
<br />
.<br />
β(µLχL) ≡ β(χL) (modPL)<br />
ψλ(x) = ψL<br />
ψ ′ λ(x) = ψL<br />
<br />
β(χL)δ2 <br />
x<br />
γ<br />
<br />
β(µLχL)δ2 <br />
x<br />
<strong>the</strong>n ψλ = ψ ′ λ .Thefirstassertion<strong>of</strong><strong>the</strong>lemmanowfollowsfrom<strong>the</strong>previoustwolemmas.<br />
Itisclearthatwecantake β(µLχL) = β(χL)if m(µL) ≤ d.Let<strong>the</strong>commonvalue<strong>of</strong><strong>the</strong>two<br />
numbersbe β.Then<br />
<br />
isequalto<br />
ψL<br />
βδx<br />
γ<br />
ψL<br />
µ −1<br />
L<br />
<br />
βδx<br />
γ<br />
γ<br />
(1 + δx)χ−1 L (1 + δx)<br />
χ −1<br />
L (1 + δx).
Chapter9 88<br />
Weseenowthat<strong>the</strong>secondassertioniscompletelytrivial.<br />
Thereisacorollary<strong>of</strong>thislemmawhichitisconvenienttoobserve.<br />
Lemma 9.5<br />
Suppose m(χL) = 2d + εwhere disapositiveintegerand εis0or1.If m(µL) ≤ dand<br />
µLis<strong>of</strong>order r<strong>the</strong>n<br />
∆(µLχL, ψL) ∼r ∆(χL, ψL).<br />
Choose γin<strong>the</strong>usualwaysothat<br />
and<br />
Itisclearthat<br />
Ifwetake<br />
<strong>the</strong>n,clearly,<br />
∆(χL, ψL) = χL(γ) ∆1(χL, ψL; γ)<br />
∆(µLχL, ψL) = χL(γ)µL(γ) ∆1(µLχL, ψL; γ).<br />
µL(γ) ∼r 1.<br />
β(µLχL) = β(χL)<br />
∆2(µLχL, ψL; γ) ∼r ∆2(χL, ψL, γ).<br />
Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma9.5wehaveonlytoappealtoLemma9.4.<br />
Lemma 9.6<br />
Suppose Kisanunramifiedextension<strong>of</strong> Land χLisaquasicharacter<strong>of</strong> CLwith<br />
m = m(χL) = 2d + 1<br />
where disapositiveinteger.Let ψLbeanontrivialadditivecharacter<strong>of</strong> Fandlet n = n(ψL).<br />
Suppose<br />
<br />
βx<br />
χL(1 + x) = ψL<br />
̟ m+n<br />
<br />
L<br />
for xin P d+1<br />
L .Take<br />
If<br />
ϕλ(x) = ψL<br />
β(χL) = β(χ K/L) = β.<br />
d β̟Lx ̟ m+n<br />
<br />
L<br />
χ −1<br />
L (1 + ̟d Lx)
Chapter9 89<br />
andif<br />
ϕκ(x) = ψ K/L<br />
d β̟Lx ̟ m+n<br />
<br />
L<br />
χ −1<br />
K/L (1 + ̟d L x)<br />
for xin κ = 0K/PK<strong>the</strong>n ϕκ = ϕ κ/λ.Moreoverif [K : L] = ℓ<strong>the</strong>n<br />
∆3(χ K/L, ψ K/L, ̟ m+n<br />
L ) = (−1) ℓ−1 {∆3(χL, ψL, ̟ m+n<br />
L )} ℓ .<br />
<strong>On</strong>ceweprovethat ϕκ = ϕ κ/λthislemmawillfollowfromLemmas9.1and9.3. If x<br />
belongsto Klet E 2 (x)be<strong>the</strong>secondelementarysymmetricfunction<strong>of</strong> xanditsconjugates<br />
over L.If xbelongsto OK<br />
Since<br />
wehave<br />
N K/F(1 + ̟ d L x) ≡ (1 + ̟d L S K/Lx) (1 + ̟ 2d<br />
L E2 (x)) (modP m L ).<br />
E 2 (x) ≡ ω κ/λ(x) (modPF)<br />
ϕκ(x) = ϕλ(S K/Lx)ψλ(−ω κ/λ(x)) = ϕ κ/λ(x).<br />
Nowsuppose Kisaramifiedabelianextension<strong>of</strong> Land [K : L] = ℓisanoddprime.Let<br />
G = G(K/L)andsuppose G = Gtwhile Gt+1 = {1}.Suppose<br />
isgreaterthanorequalto t + 1and<br />
for xin P d+1<br />
L .Let<br />
andif xbelongsto OKset<br />
Supposealsothat<br />
ϕ ′ λ(x) = ψ K/L<br />
m = m(χL) = 2d + 1<br />
χL(1 + x) = ψL<br />
<br />
βx<br />
̟ m+n<br />
<br />
L<br />
d ′ <br />
ℓ − 1<br />
= ℓd − t<br />
2<br />
<br />
β̟d′ Kx ̟ m+n<br />
<br />
L<br />
χ −1<br />
L (1 + S K/L(̟ d′<br />
Kx) + E 2 (̟ d′<br />
Kx)).<br />
̟L = N K/L̟K.
Chapter9 90<br />
Theassumptionslisted,wemaynowstate<strong>the</strong>nextlemma.<br />
Lemma 9.7<br />
If<br />
ε = S K/L<br />
<br />
̟2d′ K<br />
̟2d <br />
L<br />
<strong>the</strong>n εisaunit.Moreover ϕ ′ λ isafunctionon λ = OL/PL + OK/PKwhichsatisfies<br />
if<br />
If<br />
<strong>the</strong>n<br />
Since<br />
<strong>the</strong>number<br />
ϕ ′ λ (x + y) = ϕ′ λ (x)ϕ′ λ (y)ψλ(εxy)<br />
ϕλ(x) = ψL<br />
ψλ(u) = ψL<br />
<br />
βx<br />
̟ d+1+n<br />
L<br />
<br />
βu<br />
̟ 1+n<br />
<br />
.<br />
L<br />
<br />
A[σ(ϕλ)] ℓ = A[σ(ϕ ′ λ )].<br />
d ′ + (ℓ − 1) (t + 1)<br />
ℓ<br />
S K/L (̟ d′<br />
Kx)<br />
χ −1<br />
L (1 + ̟d Lx)<br />
liesin Pd L .Moreover E2 (̟d′ Kx)isasum<strong>of</strong>traces<strong>of</strong>elementsin P2d′ K .Since<br />
2d ′ + (ℓ − 1) (t + 1)<br />
ℓ<br />
itliesin P 2d<br />
L .If xliesin PKitliesin P m L and<br />
liesin P d+1<br />
L because<br />
d ′ + 1 + (ℓ − 1) (t + 1)<br />
ℓ<br />
= 2d +<br />
S K/L(̟ d′<br />
Kx)<br />
= d + 1 +<br />
≥ d<br />
ℓ − 1<br />
ℓ<br />
t(ℓ − 1)<br />
2ℓ<br />
≥ 2d<br />
≥ d + 1.
Chapter9 91<br />
Thusif xbelongsto PK<br />
Since<br />
isequalto<br />
<strong>the</strong>expression<br />
ϕ ′ λ(x) = ψL<br />
β<br />
̟ m+n<br />
L<br />
SK/L (̟ d′<br />
<br />
Kx) χ −1<br />
L (1 + SK/L(̟ d′<br />
Kx)) = 1.<br />
E 2 (̟ d′<br />
K (x + y))<br />
E 2 (̟ d′<br />
Kx) + E 2 (̟ d′<br />
Ky) + S K/L(̟ d′<br />
Kx) S K/L(̟ d′<br />
Ky) − S K/L(̟ 2d′<br />
K xy),<br />
iscongruentmodulo P m L to<strong>the</strong>product<strong>of</strong><br />
and<br />
and<br />
Thus<br />
Since<br />
1 + S K/L(̟ d′<br />
K(x + y)) + E 2 (̟ d′<br />
K(x + y))<br />
1 + S K/L(̟ d′<br />
K x) + E2 (̟ d′<br />
K x)<br />
1 + S K/L(̟ d′<br />
K y) + E2 (̟ d′<br />
K y)<br />
1 − S K/L(̟ 2d′<br />
K xy).<br />
ϕ ′ λ (x + y) = ϕ′ λ (x)ϕ′ λ (y)ψλ(εxy).<br />
2d ′ + (ℓ − 1) (t + 1)<br />
ℓ<br />
≥ 2d<br />
<strong>the</strong>number εisin OL.Weconcludeinparticularthatif ybelongsto PK<strong>the</strong>n<br />
and<br />
ϕ ′ λ(x + y) = ϕ ′ λ(x).<br />
If t = 0let σbeagenerator<strong>of</strong> Gandlet ̟ 1−σ<br />
K = ν.Inthiscase 2d′ = 2dℓand<br />
̟ 2d′<br />
K<br />
̟ 2d<br />
L<br />
<br />
<br />
=<br />
τ∈G ̟1−τ<br />
K<br />
2d<br />
ε ≡ ℓν dℓ(ℓ−1) (modPL)<br />
≡ ν dℓ(ℓ−1) (modPL)
Chapter9 92<br />
isaunit.If t > 0<br />
sothat<br />
̟ 2d′<br />
K<br />
ε ≡ S K/L<br />
≡ ̟2d−t<br />
L ̟t K<br />
t ̟K ̟ t L<br />
(mod PK)<br />
(mod PL).<br />
ItisshowninparagraphV.3<strong>of</strong>Serre’sbookthat<strong>the</strong>rightside<strong>of</strong>thiscongruenceisaunit.<br />
Then<br />
Firsttake poddandlet<br />
σ(ϕλ) = −ψλ<br />
ϕλ(x) = ψλ<br />
x 2<br />
2<br />
2 −α <br />
ψλ<br />
2<br />
<br />
+ αx = ψλ<br />
2 (x + α)<br />
2<br />
2 (x + α)<br />
= −ψλ<br />
2<br />
Makinguse<strong>of</strong><strong>the</strong>calculationsin<strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma9.2weseethat<br />
A[σ(ϕλ)] ℓ = νλ(−1) ℓ−1<br />
2 ψλ<br />
<strong>of</strong> νλis<strong>the</strong>nontrivialquadraticcharacter<strong>of</strong> λ × .<br />
Since<br />
iscongruentto<br />
modulo P m L <strong>the</strong>value<strong>of</strong> ϕ′ λ (x)is<br />
if<br />
Thus<br />
ψλ<br />
1 + S K/L<br />
2 −ℓα<br />
2<br />
ψλ<br />
−α 2<br />
2<br />
<br />
.<br />
2 −α <br />
ψλ<br />
2<br />
<br />
A − <br />
λ ψλ<br />
<br />
̟ d′<br />
Kx <br />
+ E 2 (̟ d′<br />
Kx) {1 + S K/L(̟ d′<br />
Kx)} {1 + E 2 (̟ d′<br />
Kx)}<br />
<br />
(SK/Ly) 2 + 2α SK/Ly − 2E2 <br />
(y)<br />
ϕ ′ λ (x) = ψλ<br />
2<br />
y = ̟d′ K<br />
̟d x.<br />
L<br />
(SK/L(y + α) 2 ) − ℓα 2<br />
2<br />
<br />
.<br />
2 x<br />
2<br />
x 2<br />
2<br />
<br />
.
Chapter9 93<br />
Replacing xby<br />
andsummingwefindthat<br />
σ(ϕ ′ λ) = ψλ<br />
−ℓα 2<br />
2<br />
x − ̟d L<br />
̟d′ α<br />
K<br />
<br />
− <br />
x ψλ<br />
εx 2<br />
Collectingthisinformationtoge<strong>the</strong>rweseethattoprove<strong>the</strong>lemmawhen<strong>the</strong>residual<br />
characteristic pisoddwemustshowthat<br />
νλ(−1) ℓ−1<br />
2 = νλ(ε).<br />
Since ν dℓ(ℓ−1) iscertainlyasquarewehavetoshowthat<br />
νλ(−1) ℓ−1<br />
2 = νλ(ℓ)<br />
when t = 0. If<strong>the</strong>field λis<strong>of</strong>evendegreeover<strong>the</strong>primefieldbothsidesare1. Ifnot,an<br />
oddpower<strong>of</strong> piscongruentto1modulo ℓand<strong>the</strong>relationfollowsfrom<strong>the</strong>law<strong>of</strong>quadratic<br />
reciprocity.If t > 0<strong>the</strong>n<br />
t ̟K ε ≡ SK/L (modPL)<br />
̟ t L<br />
andwecanappealtoparagraphV.3forapro<strong>of</strong>that<br />
ε + u p−1 ≡ 0<br />
hasasolutionin λ.Thus νλ(ε) = νλ(−1)andwehavetoshowthat<br />
νλ(−1) p−3<br />
2 = 1.<br />
2<br />
<br />
.<br />
If p ≡ 1 (mod4)<strong>the</strong>n νλ(−1) = 1andif p ≡ 3 (mod4)<strong>the</strong>exponentiseven.<br />
Beforeconsidering<strong>the</strong>case p = 2,weremarkasimpleconsequence<strong>of</strong><strong>the</strong>preceding<br />
discussion.
Chapter9 94<br />
Lemma 9.8<br />
If pisoddlet<br />
If t = 0and<br />
<strong>the</strong>n<br />
andif t > 0<br />
with<br />
Inbothcases<br />
ϕ ′ λ<br />
ϕ ′ λ(x) = ψλ<br />
ϕλ(x) = ψλ<br />
x 2<br />
2<br />
(ℓ−1)<br />
dℓ<br />
µ = ν 2<br />
<br />
x<br />
= ψλ ℓ<br />
µ<br />
ϕ ′ λ(x) = ψλ<br />
x 2<br />
<br />
+ αx .<br />
2<br />
εx 2<br />
x<br />
<br />
+ αx<br />
<br />
.<br />
<br />
(SK/L (y + α) 2 ) − ℓα2 <br />
y = ̟d′ K<br />
̟d x.<br />
L<br />
If t = 0<strong>the</strong>n y = µx.Thusif xbelongsto OL,aswemayassume,<br />
ϕ ′ λ<br />
<br />
x<br />
= ψλ<br />
µ<br />
If t > 0<strong>the</strong>n ℓ = pisoddand<br />
d ′ ℓd + (ℓ − 1) (t + 1)<br />
ℓ<br />
sothat,if x ∈ OF,<br />
sothat<br />
= 1<br />
ℓ<br />
2 2<br />
ℓ(x + α) − ℓα<br />
2<br />
2<br />
= ψλ<br />
2 x<br />
ℓ<br />
2<br />
<br />
<br />
(ℓ − 1)<br />
(ℓ − 1) (t − 1) − t =<br />
2<br />
1<br />
ℓ<br />
S K/L(y + α) 2 ≡ εx 2<br />
(mod PL).<br />
Nowtake p = 2sothat tisnecessarily0andagainlet<br />
̟ d′<br />
K<br />
̟ d L<br />
(ℓ−1)<br />
dℓ<br />
µ = ν 2<br />
≡ µ (mod PK).<br />
<br />
+ αx .<br />
<br />
<br />
(ℓ − 1)<br />
(ℓ − 1) + t ≥ 1<br />
2
Chapter9 95<br />
If xisin OLand y = x<br />
µ <strong>the</strong>n<br />
isequalto<br />
Since<br />
ψL<br />
<br />
1 + ℓ̟ d Lx +<br />
modulo P m L wehave<br />
whichequals<br />
Moreover<br />
ℓ x<br />
̟ d+1+n<br />
L<br />
ℓ(ℓ − 1)<br />
2<br />
ϕ ′ λ<br />
<br />
ϕ ′ λ<br />
χ −1<br />
L<br />
<br />
x<br />
= ϕ<br />
µ<br />
′ λ (y)<br />
<br />
1 + ℓ ̟ d ℓ(ℓ − 1)<br />
Lx + ̟<br />
2<br />
2d<br />
L x2<br />
<br />
.<br />
̟ 2d<br />
L x 2 ≡ (1 + ℓ̟ d <br />
ℓ(ℓ − 1)<br />
Lx) 1 + ̟<br />
2<br />
2d<br />
L x 2<br />
<br />
<br />
x<br />
= ϕλ(ℓ x) ψλ<br />
µ<br />
{ϕλ(x)} ℓ .<br />
−ℓ(ℓ − 1)<br />
2<br />
x 2<br />
<br />
{ϕλ(x)} 2 = ϕλ(2x) ψλ(−x 2 ) = ψλ(−x 2 ).<br />
Since<strong>the</strong>characteristicis2<strong>the</strong>reisan α = 0suchthat<br />
Then<strong>the</strong>complexconjugate<strong>of</strong> ϕλ(x)is<br />
and<br />
whichequals<br />
isequalto<br />
Consequently<br />
Since<br />
ψλ(x 2 ) = ψλ(αx).<br />
ϕλ(x)ψλ(αx)<br />
σ(ϕλ) = − <br />
− <br />
x ϕλ(x + α)<br />
x ϕλ(x)ϕλ(α)ψλ(αx)<br />
ϕλ(α) σ(ϕλ).<br />
A[σ(ϕλ)] ℓ = ϕλ(α) ℓ−1<br />
2 A[σ(ϕλ)].<br />
{ϕλ(x)} 4 = 1
Chapter9 96<br />
wehave<br />
if ℓ ≡ 1 (mod4)and<br />
if ℓ ≡ 3 (mod4).<br />
Wehavetoshowthat<br />
if ℓ ≡ 1 (mod4)andthat<br />
A[σ(ϕ ′ λ )] = A[σ(ϕλ)]<br />
A[σ(ϕ ′ λ )] = A[σ(ϕλ)] = ϕ −1<br />
λ (α) A[σ(ϕλ)]<br />
ϕλ(α) ℓ−1<br />
2 = 1<br />
ϕλ(α) ℓ+1<br />
2 = 1<br />
if ℓ ≡ 3 (mod4). Theserelationsareclearif ℓiscongruentto1or7modulo8. Ingeneralif<br />
ℓ ≡ 1 (mod 4)<br />
ϕλ(α) ℓ−1<br />
<br />
(ℓ − 1) (ℓ − 3)<br />
2 = ψλ − α<br />
8<br />
2<br />
<br />
andif ℓ ≡ 3 (mod4)<br />
ϕλ(α) ℓ+1<br />
2 = ψλ<br />
<br />
−<br />
(ℓ + 1) (ℓ − 1)<br />
α<br />
8<br />
2<br />
<br />
.<br />
Let φbe<strong>the</strong>primefieldandlet ψφbeitsnontrivialadditivecharacter. Choose α1such<br />
that<br />
Then<br />
and α = α1.Thus<br />
ψλ(x 2 ) = ψ λ/φ<br />
ψ λ/φ(x) = ψλ (α 2 1 x).<br />
2 x<br />
= ψλ/φ α 2 1<br />
x<br />
ψλ(α 2 ) = ψ λ/φ(1).<br />
α1<br />
<br />
= ψλ (α1x)<br />
Therightsideis+1or–1accordingas f = [λ : φ]isevenorodd.But ℓdivides 2 f − 1sothat,<br />
by<strong>the</strong>secondsupplementto<strong>the</strong>law<strong>of</strong>quadraticreciprocity, fisevenif ℓiscongruentto3or<br />
5modulo8.<br />
ThereisacomplementtoLemma9.7.<br />
Lemma 9.9
Chapter9 97<br />
If m(χL) ≥ 2(t + 1)choose β(χK/L) = β(χL) = βin OL.If t + 1 < m(χL) < 2(t + 1)<br />
choose β(χL) = βand<br />
β(χK/L) = β − β1<br />
α<br />
asinLemma8.9.Then m(χ K/L) = 2d ′ + 1and<br />
isequalto<br />
ψK/L<br />
ψ K/L<br />
FromLemma8.8wehave<br />
asrequired.If d ≥ t + 1<strong>the</strong>n<br />
because ℓisodd.Moreover,<br />
Consequently<br />
<br />
β ̟d′ Kx ̟ m+n<br />
<br />
L<br />
α1<br />
<br />
β(χK/L) ̟d′ Kx ̟ m+n<br />
<br />
χ<br />
L<br />
−1<br />
K/L (1 + ̟d′ Kx) m(χ K/L) = 1 + t + ℓ(m − 1 − t) = 2<br />
d ′ ≥<br />
(ℓ + 1)<br />
2<br />
3d ′ + (ℓ − 1) (t + 1)<br />
ℓ<br />
χ −1<br />
L (1 + SK/L (̟ d′<br />
Kx) + E 2 (̟ d′<br />
Kx)).<br />
d +<br />
(ℓ − 1)<br />
2<br />
<br />
(ℓ − 1)<br />
ℓ d − t + 1<br />
2<br />
(d − t) ≥ m<br />
≥ m′ + (ℓ − 1) (t + 1)<br />
ℓ<br />
= m.<br />
N K/L(1 + ̟ d′<br />
Kx) ≡ 1 + S K/L(̟ d′<br />
Kx) + E 2 (̟ d′<br />
Kx) (modP m L )<br />
and<strong>the</strong>lemmaisvalidif m ≥ 2(t + 1).<br />
sothat<br />
If t + 1 < m < 2(t + 1)westillhave<br />
3d ′ + (ℓ − 1) (t + 1)<br />
ℓ<br />
N K/L(1 + ̟ d′<br />
Kx)<br />
≥ m
Chapter9 98<br />
iscongruentto<br />
1 + S K/L(̟ d′<br />
K x) + E2 (̟ d′<br />
K x) + N K/L(̟ d′<br />
K x)<br />
modulo P m L .Since d′ ≥ d + 1thisiscongruentto<br />
modulo P m L .Certainly<br />
Moreover,if m = t + 1 + v<br />
{1 + S K/L(̟ d′<br />
Kx) + E 2 (̟ d′<br />
Kx)} {1 + N K/L(̟ d′<br />
Kx)}<br />
χL(1 + NK/L(̟ d′<br />
Kx)) = ψL<br />
d ′ − v = d +<br />
ℓ − 3<br />
2<br />
<br />
β NK/L(̟d′ Kx) ̟ m+n<br />
<br />
.<br />
L<br />
v ≥ d ≥ s<br />
if sis<strong>the</strong>leastintegergreaterthanorequalto t<br />
2 .Thus,justasin<strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma8.5,<br />
isequalto<br />
ψL<br />
<br />
β NK/L(̟d′ Kx) ̟ m+n<br />
<br />
= ψL<br />
L<br />
ψL<br />
Multiplying<strong>the</strong>inverse<strong>of</strong>thiswith<br />
weobtain<br />
Thelemmafollows.<br />
ψ K/L<br />
⎛<br />
If m = t + 1wemaystillchoose<br />
⎝ −S K/L<br />
⎛<br />
⎝ α N K/L<br />
αβ1<br />
α1 ̟d′<br />
K x<br />
̟ m+n<br />
L<br />
β1<br />
α1 ̟d′<br />
K x<br />
̟ m+n<br />
L<br />
⎞<br />
⎠ .<br />
<br />
β − αβ1<br />
d ̟<br />
α1<br />
′<br />
Kx ̟ m+n<br />
<br />
L<br />
ψ K/L<br />
<br />
β ̟d′ Kx ̟ m+n<br />
<br />
.<br />
L<br />
β(χ K/L) = β − β1<br />
α<br />
α1<br />
⎞<br />
⎠
Chapter9 99<br />
asinLemma8.9.However<strong>the</strong>relationbetween ϕλ(x)and<br />
ϕK(x) = ψ K/L<br />
<br />
β(χK/L) ̟d′ Kx ̟ m+n<br />
<br />
χ<br />
L<br />
−1<br />
K/L (1 + ̟d′ Kx)<br />
willbemorecomplicated.Here x = OK/PKis<strong>the</strong>samefieldas λ = OL/PL.Weintroduce<br />
itonlyfornotationalpurposes.<br />
Since<br />
and<br />
Because m = t + 1<strong>the</strong>number tisatleast2and<br />
<strong>the</strong>expression<br />
iscongruentto<br />
modulo P m L and<br />
isequalto<br />
d = d ′ = t<br />
2 .<br />
3d + (p − 1) (t + 1)<br />
p<br />
d + (p − 1) (t + 1)<br />
p<br />
N K/L(1 + ̟ d K x)<br />
≥ t + 1<br />
≥ d + 1<br />
{1 + S K/L(̟ d Kx) + E 2 (̟ d Kx)} {1 + ̟ d L N K/Lx}<br />
ΨL<br />
AccordingtoNewton’sformulae<br />
Thus<br />
χL(1 + S K/L(̟ d Kx) + E 2 (̟ d Kx))<br />
<br />
β SK/L(̟d Kx) + β E2 (̟d Kx) <br />
.<br />
̟ m+n<br />
F<br />
S K/L(̟ 2d<br />
K x 2 ) − S K/L(̟ d Kx) 2 + 2E 2 (̟ d Kx) = 0.<br />
E 2 (̟ d Kx) ≡ − 1<br />
2 S K/L(̟ 2d<br />
K x 2 ) (mod P m L ).<br />
Observethat pisequalto ℓand<strong>the</strong>refore,in<strong>the</strong>presentcircumstances,odd.
Chapter9 100<br />
with<br />
Certainly<br />
Let µLbeacharacterin S(K/L)asinLemma8.9(c)andlet<br />
ψL<br />
<br />
αx<br />
̟ d+1+n<br />
L<br />
<br />
µ −1<br />
L (1 + ̟d Lx) = ψλ<br />
ρ = α<br />
β .<br />
µL(NK/L(1 + ̟ d Kx)) = 1<br />
if xbelongsto OK.Replacing xby β1<br />
α1 xweseethat<br />
ψL<br />
isequalto<br />
if<br />
<br />
1<br />
̟ m+n<br />
<br />
β1<br />
αSK/L ̟<br />
α1<br />
L<br />
d <br />
Kx − α<br />
2 SK/L z = 1<br />
̟ d L<br />
whichiscongruentto<br />
modulo PL.<br />
equal<br />
Let<br />
If xbelongsto OK<br />
χ −1<br />
L<br />
<br />
β1<br />
SK/L α1<br />
ψλ<br />
<br />
ρ z2<br />
2<br />
̟ d <br />
Kx − 1<br />
2 SK/L ϕλ(x) = ψL<br />
β 2 1<br />
α 2 1<br />
<br />
+ τ z<br />
β 2 1<br />
α 2 1<br />
β<br />
α NK/Lx ≡ β<br />
α xp<br />
<br />
ψλ<br />
1 + ̟ d L N K/Lx = ψλ<br />
β x<br />
̟ d+1+n<br />
L<br />
x 2<br />
2<br />
x 2p<br />
2<br />
<br />
<br />
+ σ x .<br />
<br />
ρ x2<br />
2<br />
<br />
+ τ x<br />
̟ 2d<br />
K x 2<br />
<br />
<br />
+ NK/L β1 ̟ d Kx <br />
̟ 2d<br />
K x 2<br />
<br />
β1<br />
+ NK/L ̟<br />
α1<br />
d <br />
Kx<br />
χ −1<br />
L (1 + ̟d L x)<br />
+ σ xp<br />
<br />
ψL<br />
<br />
−β NK/Lx<br />
.<br />
̟ d+1+n<br />
L
Chapter9 101<br />
Wenowput<strong>the</strong>sefactstoge<strong>the</strong>rt<strong>of</strong>indasuitableexpressionfor ϕκ(x).Wemayaswell<br />
take xin OL.Then ϕκ(x)is<strong>the</strong>product<strong>of</strong><br />
<br />
β ̟<br />
ψK/L d′<br />
Kx ̟ m+n<br />
<br />
L<br />
and<br />
and<br />
and<br />
if<br />
ψL<br />
ψ K/L<br />
<br />
−<br />
<br />
− β1α<br />
α1<br />
β<br />
̟ m+n<br />
L<br />
̟d Kx ̟ m+n<br />
<br />
L<br />
χ −1<br />
L (1 + ̟d L N K/Lx)<br />
<br />
SK/L(̟ d Kx) − 1<br />
2 SK/L(̟ 2d<br />
K x 2 <br />
) .<br />
Thesecond<strong>of</strong><strong>the</strong>sethreeexpressionsisequalto<strong>the</strong>product<strong>of</strong><br />
<br />
2p x −1 x2p<br />
ψλ + σ xp ψλ −ρ<br />
2 2 − ρ−1τ x p<br />
<br />
ψ K/L<br />
<br />
− αβ2 1 ̟2d<br />
K x2<br />
2α2 1 ̟m+n<br />
L<br />
ε = S K/L<br />
Theproduct<strong>of</strong><strong>the</strong>firstandthirdisequalto<br />
2 εx<br />
ψλ<br />
<br />
= ψλ<br />
2d ̟K ̟2d <br />
.<br />
L<br />
2<br />
<br />
.<br />
<br />
− ερβ2 1<br />
2α 2 1<br />
x 2<br />
<br />
AsproveninparagraphV.3<strong>of</strong>Serre’sbook<strong>the</strong>elements<strong>of</strong> ULcongruentto<br />
modulo P t+1<br />
L areallnorms,sothat<br />
Inparticular<br />
ψλ<br />
1 + (εx + x p ) ̟ t L<br />
ψλ(ρ x p ) = ψλ(−ρεx).<br />
<br />
− ερβ2 1<br />
2α2 x<br />
1<br />
2<br />
<br />
= ψλ<br />
<br />
ρ<br />
−1 x2p<br />
∗ (1998)Themanuscript<strong>of</strong>Chapter9endswiththisformula.<br />
2<br />
<br />
. ∗
Chapter11 102<br />
Chapter Eleven.<br />
<strong>Artin</strong>–Schreier <strong>Equation</strong>s<br />
The<strong>the</strong>ory<strong>of</strong><strong>Artin</strong>–SchreierequationsiscentraltoDwork’spro<strong>of</strong><strong>of</strong><strong>the</strong>secondmain<br />
lemma. Wefirstreview<strong>the</strong>basic<strong>the</strong>ory,whichwetakefromMackenzieandWhaples[11],<br />
and<strong>the</strong>nreviewDwork’sra<strong>the</strong>ramazingcalculations.ThesewetakefromLakkis[9].<br />
WestartwithanexercisefromSerre’sbook[12].Suppose Fisanonarchimedeanlocal<br />
fieldand K/FisGalois.Let pbe<strong>the</strong>residualcharacteristic.With<strong>the</strong>convention (0) = P ∞ F<br />
welet<br />
p OF = P e F.<br />
Suppose G = G(K/F)and σ ∈ Giwith i ≥ 1.Let<br />
with ain P i K .Let<br />
̟ σ K = ̟K(1 + a)<br />
ϕ(x) = x σ − x.<br />
ϕisan Flinearoperatoron K.If x = α̟ j<br />
Kbelongsto Pj<br />
K<strong>the</strong>n iscongruentto<br />
ϕ(x) = x σ − x = (α σ − α) ̟ jσ<br />
K<br />
α̟ j<br />
K<br />
modulo P i+j+1<br />
K .Thisinturniscongruentto<br />
modulo P i+j+1<br />
K .<br />
If*<br />
<strong>the</strong>n,asanoperator,<br />
+ α(̟jσ<br />
K<br />
− ̟K)<br />
<br />
̟ j(σ−1)<br />
<br />
K − 1 = α̟ j<br />
K {(1 + a)j − 1}<br />
(α̟ j<br />
K ) (ja) = jax<br />
ψ(x) = x σp<br />
− 1<br />
ψ = (1 + ϕ) p − 1 = p<br />
k=1<br />
<br />
p<br />
<br />
ϕ<br />
k<br />
k .<br />
*Weseemtobedealingwithyetano<strong>the</strong>ruse<strong>of</strong><strong>the</strong>symbol ψ!
Chapter11 103<br />
If xbelongsto P j<br />
K <strong>the</strong>n<br />
and ψ(x)iscongruentto<br />
orto<br />
modulo P i+j+e′ +1<br />
K<br />
ϕ k (x) ≡ j(j + i) . . .(j + (k − 1)i)a k x (modP j+ki+1<br />
K )<br />
if p OK = P e′<br />
K .<br />
Wededuce<strong>the</strong>followingcongruences:<br />
(i)If (p − 1)i > e ′ <strong>the</strong>n<br />
(ii)If (p − 1)i = e ′ <strong>the</strong>n<br />
(iii)If (p − 1)i < e ′ <strong>the</strong>n<br />
pjax + j(j + i) . . .(j + (p − 1)i)a p x<br />
pjax + j(j p−1 − i p−1 )a p x<br />
ψ(x) ≡ pjax (mod P i+j+e′ +1<br />
K ).<br />
ψ(x) ≡ pjax + j(1 − i p−1 )a p x (modP i+j+e′ +1<br />
K ).<br />
ψ(x) ≡ j(1 − i p−1 )a p x (mod P pi+j+1<br />
K ).<br />
Observethatif (j, p) = 1and σbelongsto Gi,with i ≥ 1,<strong>the</strong>n<br />
ϕ(x) ≡ 0 (modP i+j+1<br />
K )<br />
forall xin P j<br />
K ifandonlyif σbelongsto Gi+1.Itfollowsimmediatelythatif σbelongsto G1<br />
and i ≥ 1<strong>the</strong>n<br />
ϕ(x) ≡ 0 (modP i+j<br />
K )<br />
forall xin P j<br />
Konlyif σbelongsto Gi.<br />
If σisreplacedby σP<strong>the</strong>n ϕisreplacedby ψ.If k > e′<br />
p−1 and Gk = {1}<strong>the</strong>n,forsome<br />
i ≥ k, Gi = {1}and Gi+1 = {1}.Taking (j, p) = 1weinferfrom(i)thatif σbelongsto Gi<br />
butnotto Gi+1<strong>the</strong>n σ p isin Gi+e ′butnotin Gi+e ′ +1.Thisisimpossible.Thus Gk = {1}if
Chapter11 104<br />
k > e′<br />
p−1 .If G1 = {1}<strong>the</strong>n pdivides e ′ sothatif (p − 1)i = e ′ <strong>the</strong>number iisalsodivisibleby<br />
p.Thecongruence(ii)reducesto<br />
ψ(x) ≡ j (pa + a p ) (modP i+j+e′ +1<br />
K ).<br />
Thusif σbelongsto Giits pthpower σ p liesin Gi+eandis<strong>the</strong>refore1.Consequently<br />
Letting a = α̟ i K and p = β̟e K wefindthat<br />
pa + a p ≡ 0 (modP pi+1<br />
K ).<br />
α p + βα ≡ 0 (mod PK).<br />
Sincethiscongruencehasonly proots<strong>the</strong>image<strong>of</strong> Θiliesinasubset<strong>of</strong> Ui K /Ui+1<br />
K<br />
elementsand Giisei<strong>the</strong>r {1}orcyclic<strong>of</strong>order p.<br />
with p<br />
If (p − 1) < e ′ and (i, p) = 1<strong>the</strong>congruence(iii)impliesthat σ p belongsto Gpi+1if σ<br />
belongsto Gi.Howeverif (p −1)i < e ′ and pdivides iitshowsthat σ p belongsto Gpibutnot<br />
to Gpi+1if σbelongsto Gibutnotto Gi+1.Thus σ −→ σ p definesaninjection<strong>of</strong> Gi/Gi+1into<br />
Gpi/Gpi+1.If Gi/Gi+1isnottrivialnei<strong>the</strong>ris Gpi/Gpi+1and (p − 1)pi ≤ e ′ .If (p − 1)pi < e ′<br />
wecanrepeat<strong>the</strong>process.Thus,forsomepositiveinteger h, (p − 1)p h i = e ′ and G p h iisnot<br />
trivial.Itis<strong>the</strong>ncyclic<strong>of</strong>order p.AccordingtoPropositionIV.10<strong>of</strong>Serre’sbookthose k ≥ 1<br />
forwhich Gk/Gk+1 = {1}areallcongruentmodulo p.Inparticularif Gk/Gk+1isnottrivial<br />
forsome k ≥ 1divisibleby pitisnottrivialonlywhen kisdivisibleby p. Thepreceding<br />
discussionshowsthatif iis<strong>the</strong>smallestvalue<strong>of</strong> k ≥ 1forwhich Gk/Gk+1isnontrivial<strong>the</strong>n<br />
any σin Gibutnotin Gi+1generates Gi = G1.Ino<strong>the</strong>rwords:<br />
Lemma 11.1<br />
If G1isnotcyclic<strong>the</strong>n (i, p) = 1if i ≥ 1and Gi/Gi+1 = {1}.<br />
Lemma 11.2<br />
Suppose K/Liscyclic<strong>of</strong>primedegreeand G = G(K/L)isequalto GLwith t ≥ 1and<br />
(t, p) = 1.Then<strong>the</strong>reisa∆in Kandan ain Lsuchthat aOL = P −t<br />
L and<br />
∆ p − ∆ = a.<br />
Weobservefirst<strong>of</strong>allthat [K : L]mustbe pandthatif pOK = P e′<br />
K <strong>the</strong>n (p − 1)t < e′ .If<br />
xbelongsto K<strong>the</strong>symbol O(x)willstandforanelementin xOKand<strong>the</strong>symbol o(x)will<br />
standforanelementin xPK.If<br />
x = p−1<br />
i=0 ai ̟ i K
Chapter11 105<br />
with aiin F<strong>the</strong>n<br />
Moreoverif σisagenerator<strong>of</strong> G<br />
andif ̟ σ−1<br />
K = (1 + a̟t K )<br />
for 1 ≤ i < p.Thus<br />
|x| = max<br />
0≤i
Chapter11 106<br />
Toprove<strong>the</strong>lemmaweconstructasequence Λ0, Λ1, Λ2, . . .andasequence Θ0, Θ1, . . .<br />
with<strong>the</strong>followingproperties:<br />
(i) vK(Λn) = −tforall n ≥ 0.<br />
(ii)If σisagivengenerator<strong>of</strong> Gand ζisagiven (p − 1)throot<strong>of</strong>unity<br />
(iii)<br />
(iv)<br />
(v)<br />
and<br />
Λ σ n − Λn = ζ + o(1).<br />
p(Λ σ n ) − p(Λn) = Θ σ n<br />
− Θn<br />
|Θ σ n − Θn| = |̟ t K| |Θn|.<br />
Λn+1 = Λn + Θn.<br />
p(Λ σ n+1 ) − p(Λn+1) = o(p(Λ σ n ) − p(Λn)).<br />
Itwillfollowfrom(iii)and(v)that {Θn}isconvergingto0.Then(iv)impliesthat {Λn}<br />
hasalimit ∆. (i)impliesthat vK(∆) = −tand(v)impliesthat ∆ p − ∆ = abelongsto F.<br />
From(ii)<br />
∆ σ − ∆ = ζ + o(1).<br />
if<br />
Toconstruct Λ0let αbelongto U i K andconsider<br />
ασ ̟<br />
σ t<br />
K<br />
Wecanchoose αsothat<br />
Thenweset<br />
− α<br />
̟ t K<br />
= ασ − α<br />
̟σ t<br />
K<br />
+ α<br />
̟ t K<br />
̟ σ K = ̟K(1 + a̟ t K).<br />
−taα = ζ + o(1).<br />
Λ0 = α<br />
̟t .<br />
K<br />
<br />
̟ t(1−σ)<br />
<br />
K − 1 = −taα + o(1)<br />
Weobserveinpassingthatconditions(i)and(ii)determine Λnmodulo P −t+1<br />
K .
Chapter11 107<br />
Suppose Λ0, . . ., Λnhavebeendefined.Then<br />
whichequals<br />
Choose Θnsothat<br />
and<br />
Then vK(Θn) > −tandif<br />
vK(Λn+1) = −t.Moreover<br />
and<br />
with x = o(Θn).Then<br />
Also<br />
p(Λ σ n ) = p(Λn) + p(ζ + o(1)) + o(1)<br />
p(Λn) + p(ζ) + o(1) = p(Λn) + o(1).<br />
Θ σ n − Θn = p(Λ σ n ) − p(Λn)<br />
|Θ σ n − Θn| = |̟ t K | |Θn|.<br />
Λn+1 = Λn + Θn<br />
Since vK(Θ σ n − Θn)ispositive<strong>the</strong>rightsideis<br />
Thus<br />
whichequals<br />
is<br />
Λ σ n+1 − Λn+1 = Λ σ n − Λn + o(1) = ζ + o(1).<br />
p(Λn+1) = p(Λn) + p(Θn) + x<br />
x σ − x = o(Θ σ n − Θn) = o(p(Λ σ n ) − p(Λn)).<br />
p(Θ σ n) − p(Θn) = p(Θ σ n − Θn) + o(Θ σ n − Θn).<br />
−(Θ σ n − Θn) + o(Θ σ n − Θn).<br />
p(Λ σ n+1 ) − p(Λn+1)<br />
p(Λ σ n ) − p(Λn) − (Θ σ n − Θn) + o(p(Λ σ n ) − p(Λn))<br />
o(p(Λ σ n ) − p(Λn)).
Chapter11 108<br />
Lemma 11.3<br />
Supposer ∆1belongsto K, abelongsto L, vL(a) = −tand<br />
with r ≥ 1.Define ∆ninductivelyby<br />
Then<br />
if n ≥ 2andif r ≥ (e ′ − (p − 1)t)<br />
Moreover<br />
existsand ∆ p − ∆ = a.<br />
∆ p<br />
1 − ∆1 = a + O(̟ r K )<br />
∆n+1 = ∆ p n − a.<br />
∆n+1 − ∆n = o(∆n − ∆n−1)<br />
∆n+1 − ∆n = O(̟ r+(n−1)(e′ −(p−1)t)<br />
K<br />
).<br />
lim<br />
n→∞ ∆n = ∆<br />
Thelastassertionisaconsequence<strong>of</strong><strong>the</strong>first.Itisclearthat<br />
Suppose n ≥ 2,and<br />
Then<br />
isequalto<br />
∆2 − ∆1 = O(̟ r K ).<br />
∆n − ∆n−1 = x = o(1).<br />
∆n+1 − ∆n = ∆ p n − ∆ p<br />
n−1 = (∆n−1 + x) p − ∆n−1<br />
p−1<br />
k=1<br />
<br />
p<br />
<br />
k<br />
whichis o(x)because e ′ − (p − 1)t > 0.If<br />
and r ≥ e ′ − (p − 1)titis<br />
∆ k n−1 xp−k<br />
<br />
+ x p<br />
x = O(̟ r+(n−2)(e′ −(p−1)t)<br />
K<br />
)<br />
O(̟ r+(n−1)(e′ −(p−1)t)<br />
K<br />
).
Chapter11 109<br />
Thelemmahasacouple<strong>of</strong>corollarieswhichshouldberemarked.<br />
Lemma 11.4<br />
If aisin L, vL(a) = −t, ∆p − ∆ = a,and ξisa(p − 1)throot<strong>of</strong>unity<strong>the</strong>reisanumber<br />
∆ξsuchthat<br />
∆ξ = ∆ + ξ + O(̟ e′ −(p−1)t<br />
K )<br />
and<br />
∆ p<br />
ξ − ∆ξ = a.<br />
Relation(11.1)showsthat ∆ + ξsatisfies<strong>the</strong>conditions<strong>of</strong><strong>the</strong>previouslemmawith<br />
r = e ′ − (p − 1)t.<br />
Lemma 11.5<br />
Suppose ∆belongsto K, bbelongsto L, vL(b) = −tand<br />
Thenforany uin U t+1<br />
L <strong>the</strong>equation<br />
hasasolutionin K.<br />
∆ p − ∆ = b.<br />
Λ p − Λ = bu<br />
Take,inLemma11.3, a = buand ∆1 = ∆.Lemma11.5showsthatif Sis<strong>the</strong>set<strong>of</strong>allin<br />
Lwith vL(a) = −tforwhich<strong>the</strong>equation<br />
hasasolutionin K<strong>the</strong>n S = SU t+1<br />
L .<br />
Lemma 11.6<br />
∆ p − ∆ = a<br />
If ℓis<strong>the</strong>integralpart<strong>of</strong> t<br />
<strong>the</strong>number<strong>of</strong>cosets<strong>of</strong> Ut+1<br />
p L in Sis<br />
p − 1<br />
p<br />
[OL : PL] 1+ℓ .<br />
Fixagenerator σ<strong>of</strong> G = G(K/L).If abelongsto S, ∆ p − ∆ = a,and ξisa(p − 1)th<br />
root<strong>of</strong>unity<br />
(ξ∆) p − ξ∆ = ξ a.
Chapter11 110<br />
ByLemma11.4<strong>the</strong>reisa(p − 1)throot<strong>of</strong>unity ζsuchthat<br />
Then<br />
∆ σ = ∆ + ζ + o(1).<br />
(ξ∆) σ = ξ∆ + ξζ + o(1).<br />
Thusif S ′ is<strong>the</strong>set<strong>of</strong> ain Lwith vL(a) = −tforwhich<br />
with<br />
a = ∆ p − ∆<br />
∆ σ = ∆ + 1 + o(1)<br />
<strong>the</strong>number<strong>of</strong>cosetsif U t+1<br />
L in Sis p − 1times<strong>the</strong>number<strong>of</strong>cosets<strong>of</strong> U t+1<br />
L in S ′ .<br />
Choose ∆0,with vK(∆0) = −t,forwhich ∆ p<br />
0 − ∆0 = a0isin Fand<br />
If vK(∆) = −t, ∆ p − ∆isin F,and<br />
<strong>the</strong>n,accordingtoanearlierremark,<br />
with Ω0 = o(∆0).<br />
Since<br />
wehave<br />
Thus<br />
∆ σ 0 = ∆0 + 1 + o(1).<br />
∆ σ = ∆ + 1 + o(1)<br />
∆ = ∆0 + Ω0<br />
Chooseany Ω0 = o(∆0)andset Λ0 = ∆0 + Ω0.Accordingto<strong>the</strong>relation(A)<br />
p−1<br />
σ p<br />
Ω0 − Ωp0<br />
=<br />
i=1<br />
p(Λ0) = p(∆0) + p(Ω0) + o(Ω0).<br />
Ω σ 0 − Ω0 = O(̟ t K Ω0) = o(1)<br />
<br />
p<br />
<br />
i<br />
Ω p−i<br />
0 (Ω σ 0 − Ω0) i = o(Ω σ 0 − Ω0).<br />
p(Λ0) σ − p(Λ0) = Ω σ 0 − Ω0 + o(̟ t KΩ0)
Chapter11 111<br />
and p(Λ0)isin Lonlyif<br />
Ω σ 0 − Ω0 = o(̟ t K Ω0),<br />
thatis,onlyif Ω0 = α0 + o(Ω0)with α0in L.<strong>On</strong><strong>the</strong>o<strong>the</strong>rhandif<br />
Ω σ 0 − Ω0 = o(̟ t K Ω0)<br />
andweconstruct<strong>the</strong>sequence Λ0, Λ1, Λ2, . . .asbeforeandlet<br />
<strong>the</strong>n<br />
∆ = lim<br />
n→∞ Λn<br />
∆ = Λ0 + o(Ω0).<br />
Weconcludethat<strong>the</strong>number<strong>of</strong>cosetsin Ps K /Ps+1<br />
K , s > −t,containingan Ω0suchthat<br />
(∆0 + Ω0) p − (∆0 + Ω0)<br />
isin Lis1if pdoesnotdivide sandis [OL : PL]ifitdoes.<br />
Choose ∆sothat<br />
∆ p − ∆ = a<br />
isin S ′ .If Ωbelongsto Ps K , s > −t,butnotto Ps+1<br />
K and<br />
(∆ + Ω) p − ∆ − Ω = b<br />
isalsoin S ′ <strong>the</strong>n aand bbelongto<strong>the</strong>samecoset<strong>of</strong> U t+1<br />
L ifandonlyif<br />
If s > 0<br />
butif s ≤ 0<br />
and<br />
ifandonlyif s = 0and<br />
b = a + o(1).<br />
p(∆ + Ω) = p(∆) + o(1)<br />
p(∆ + Ω) = p(∆) + Ω p − Ω + o(Ω)<br />
Ω p − Ω + o(Ω) = o(1)<br />
Ω = ξ + o(1)<br />
where ξissome (p − 1)throot<strong>of</strong>unity.Thelemmafollows.
Chapter11 112<br />
let<br />
If x = {x1, . . ., xn}let E i (x)be<strong>the</strong> i<strong>the</strong>lementarysymmetricfunction<strong>of</strong> x1, . . ., xnand<br />
If Zisanindeterminateand<br />
Q(Z) = n<br />
S i (x) = n<br />
<strong>the</strong>n ∞<br />
k=1 xi k.<br />
i=0 (−1)i E i (x)Z i = n<br />
i=1 Si (x)Z i<br />
isclearly −Ztimes<strong>the</strong>logarithmicderivative<strong>of</strong> Q(Z).Thus<br />
∞<br />
i=1 Si (x) Z i<br />
i=1<br />
(1 − xiZ)<br />
n<br />
i=0 (−1)iE i (x)Z i<br />
<br />
= − n<br />
i=0 (−1)i iE i (x)Z i .<br />
ThisidentitywhichwerefertoasNewton’sidentityisequivalentto<strong>the</strong>formulae<strong>of</strong>Newton.<br />
Itimpliesinparticularthat<br />
i−1<br />
j=0 (−1)j S i−j (x) E j (x) = (−1) i+1 iE i (x) (11.2)<br />
if 1 ≤ i ≤ n.WemaydivideNewton’sidentityby Q(Z)and<strong>the</strong>nexpand<strong>the</strong>righthandside<br />
toobtainexpressionsfor<strong>the</strong> S i (x)aspolynominalsin E 1 (x), . . ., E n (x).Thecoefficientsare<br />
necessarilyintegers.Tocalculate<strong>the</strong>mwesupposethat x1, . . ., xnlieinafield<strong>of</strong>characteristic<br />
zero.Let<br />
Q(Z) = 1 + P(Z).<br />
Then<br />
log Q(Z) = − ∞<br />
k=1<br />
(−1) k<br />
Thecoefficient<strong>of</strong> Z i−1 in<strong>the</strong>derivative<strong>of</strong><strong>the</strong>rightsideis<br />
− <br />
k<br />
<br />
α1+...+αn=k<br />
α1+2α2+...+nαn=i<br />
Thisexpressionis<strong>the</strong>reforeequalto −S i (x).<br />
k<br />
i(k − 1)!<br />
α1! . . .αn!<br />
(P(Z)) k .<br />
n<br />
j=1 {Ej (x)} αj .<br />
Suppose K/Lisaramifiedcyclicextension<strong>of</strong>degree pand G = G(K/L).Let G = Gt<br />
and Gt+1 = {1}.Suppose u ≤ t, Λisin K,and<br />
ΛOK = P −u<br />
K .
Chapter11 113<br />
Wetake {x1, . . ., xn}tobe Λanditsconjugatesunder G.Inthiscasewewrite<br />
and<br />
E i (x) = E i K/L (Λ)<br />
S i (x) = S i K/L (Λ).<br />
If 1 ≤ i ≤ p − 1and γiisanyintegerlessthanorequalto<br />
wehave<br />
Wemaytake<br />
−iu + (p − 1) (t + 1)<br />
p<br />
E i K/L (Λ) ≡ 0 (mod Pγi<br />
L ).<br />
γi ≥ − iu<br />
p<br />
(p − 1)t<br />
+ .<br />
p<br />
If iu + tisnotdivisibleby pthisinequalitymaybesupposedstrict.<br />
Suppose α1, . . ., αparenonnegativeintegers,<br />
p<br />
and p<br />
If<br />
γ =<br />
p−1<br />
i=1 αi = k<br />
i=1 iαi = ℓ.<br />
i=1 γiαi<br />
<br />
−uαp.<br />
Then p<br />
i=1 {Ei (Λ)} αi γ<br />
≡ O (̟K ). (11.3)<br />
Wehave<br />
ℓ u<br />
γ ≥ −<br />
p<br />
(p − 1)<br />
+ kt −<br />
p<br />
(p − 1)<br />
αpt.<br />
p<br />
Theinequalityisstrictif αiisnonzer<strong>of</strong>orsome isuchthat iu + tisnotdivisibleby p.<br />
Werecordnowsomeinequalitiesthat γsatisfiesinvariousspecialcases. Theywillbe<br />
neededlater.Weobservefirst<strong>of</strong>allthat,if 1 ≤ i < p, γiisnonnegativeandispositiveunless<br />
pdivides iu + t.
Chapter11 114<br />
(i)If ℓ = pand k = 2<strong>the</strong>n<br />
1 + u + γ ≥ 1 + t.<br />
Inthiscase αp = 0and<strong>the</strong>leftsideisatleast<br />
If pisodd<strong>the</strong>inequalityisstrict.<br />
(ii)If ℓ = pand k = 2<strong>the</strong>n<br />
1 +<br />
2(p − 1)<br />
p<br />
γ ≥ 0.<br />
t ≥ 1 + t.<br />
Moreover<strong>the</strong>inequalityisstrictif pisodd.Thisstatementis<strong>of</strong>courseweakerthanthat<br />
<strong>of</strong>(i).<br />
(iii)If ℓ = p, k ≥ 3,and pisodd,<strong>the</strong>n<br />
αpisagain0.Theleftsideisatleast<br />
−u +<br />
3(p − 1)<br />
p<br />
t = u +<br />
γ ≥ u +<br />
t − u<br />
p<br />
t − u<br />
p<br />
+ 1<br />
p<br />
{(3p − 4)t − (2p − 1)u}.<br />
Thefinaltermisnonnegative. Theinequalityisstrictif u = t. If u = tand pdoesnot<br />
divide uitisagainstrictfor<strong>the</strong>n αi = 0forsome i < p − 1andforsuchan i<strong>the</strong>number<br />
iu + tisnotdivisibleby p.<br />
(iv)If k ≤ p<strong>the</strong>n<br />
(p − 1)u + γ ≥ u +<br />
t − u<br />
p<br />
exceptwhen αp = kor αp = p − 1.Wehavetoshowthat<br />
(p − 2)u + γ +<br />
Theleftsideisatleast<br />
<br />
p − 2 − ℓ<br />
<br />
1 p − 1<br />
+ u +<br />
p p p<br />
u − t<br />
p<br />
≥ 0.<br />
p−1<br />
i=1 αi<br />
<br />
− 1<br />
<br />
t.<br />
p
Chapter11 115<br />
If αp = k<strong>the</strong>coefficient<strong>of</strong> tispositiveandweneedonlyshowthatitisatleastasgreat<br />
as<strong>the</strong>negative<strong>of</strong><strong>the</strong>coefficient<strong>of</strong> uorino<strong>the</strong>rwordsthat<br />
(p − 1)<br />
p−1<br />
i=1 αi<br />
Thisfollowsfrom<strong>the</strong>assumptionthat αp ≤ p − 2.<br />
(v) If k ≤ p − 2and αp = k<strong>the</strong>n<br />
Inthiscase<br />
(p − 1)u + γ ≥ u.<br />
γ ≥ −ku.<br />
<br />
+ (p − 2)p ≥ ℓ.<br />
Therearecircumstancesinwhich<strong>the</strong>estimatesfor γiand<strong>the</strong>reforethosefor γcanbe<br />
substantiallyimproved.Wewilldiscuss<strong>the</strong>mshortly.<br />
Supposenowthat K/FisatotallyramifiedGaloisextensionand G = G(K/F)is<strong>the</strong><br />
directproduct<strong>of</strong>twocyclicgroups<strong>of</strong>order p. ByLemma11.1<strong>the</strong>sequence<strong>of</strong>ramification<br />
groupsis<strong>of</strong><strong>the</strong>form<br />
G = G−1 = G0 = G1 = . . . = Gu = Gu+1 = . . . = Gt = Gt+1 = {1}<br />
with (u, p) = 1and u ≡ t (modp)or<strong>of</strong><strong>the</strong>form<br />
G = G−1 = G0 = G1 = . . . = Gt = Gt+1 = {1}<br />
with (t, p) = 1.In<strong>the</strong>secondcasewetake u = t.In<strong>the</strong>firstcaselet L1be<strong>the</strong>fixedfield<strong>of</strong> Gt<br />
andin<strong>the</strong>secondlet L1beanysubfield<strong>of</strong> K<strong>of</strong>degree pover F.Let L2beanysubfield<strong>of</strong> K<br />
differentfrom L1whichisalso<strong>of</strong>degree pover F.Let G i = G(K/Li)andlet<br />
G i = G i si = Gi si+1 = {1}.<br />
Then s1 = tand s2 = u.AccordingtoPropositionIV.4<strong>of</strong>Serre’sbook<br />
and<br />
and<br />
δ K/F = (p 2 − p) (u + 1) + (p − 1) (t + 1)<br />
δK/L1 = (p − 1) (t + 1)<br />
δK/L2 = (p − 1) (u + 1).
Chapter11 116<br />
Thus<br />
and<br />
If G i = G(Li/F)and<br />
<strong>the</strong>n t1 = uand<br />
Lemma 11.7<br />
δL1/F = 1<br />
p (δK/F − δK/L1 ) = (p − 1) (u + 1)<br />
δL2/F = 1<br />
p (δ (p − 1)<br />
K/F − δK/L2 ) =<br />
p<br />
G i = G i<br />
ti<br />
t2 = u +<br />
Suppose ∆belongsto K, vK(∆) = −u,and<br />
belongsto L2.If Ybelongsto L2<strong>the</strong>n<br />
and<br />
for 1 ≤ i ≤ p − 1.<br />
= Giti+1<br />
= {1}<br />
t − u<br />
p .<br />
∆ p − ∆ = a<br />
((p − 1) (u + 1) + t + 1).<br />
vL1 (SK/L1 (Y ∆i )) ≥ (p − 1)t2 − it1 + vL2 (Y )<br />
vL1 (Ei K/L1 (Y ∆)) ≥ (p − 1)t2 + i(vL2 (Y ) − t1)<br />
Weshowfirstthatif θbelongsto L1and<br />
with Yiin L2<strong>the</strong>n<br />
for 0 ≤ i ≤ p − 1.Since t1 = uand<br />
θ = p−1<br />
i=0<br />
Yi∆ i<br />
vL2 (Yi) ≥ it1 + vL1 (θ)<br />
vK(θ) = min<br />
0≤i≤p−1 {vK(Yi) − iu}<br />
<strong>the</strong>inequalityisclearfor i = 0. Toproveitingeneral,weuseinductionone i. Suppose<br />
0 < j ≤ p − 1and<strong>the</strong>inequalityisvalidfor i < j.
Chapter11 117<br />
Let<br />
pOF = P e F .<br />
Applying<strong>the</strong>exerciseat<strong>the</strong>beginning<strong>of</strong><strong>the</strong>paragraphto<strong>the</strong>extension L2/Fweseethat<br />
pe ≥ (p − 1)t2 = (p − 1)<br />
<br />
u +<br />
<br />
t − u<br />
.<br />
p<br />
If ξisany (p − 1)throot<strong>of</strong>unity<strong>the</strong>n,byLemmas11.3and11.4,<strong>the</strong>reisaσin G 2 suchthat<br />
Wemaywrite<br />
∆ σ = ∆ + ξ + O(̟ p2e−(p−1)u K ).<br />
θ σ − θ = p−1<br />
asalinearcombination p−1<br />
withcoefficientsfrom L2.Since<br />
i=1 Yi(∆ iσ − ∆ i )<br />
i=0<br />
Xi∆ i<br />
vL2 (θσ − θ) ≥ vL1 (θ) + t1<br />
wemayapply<strong>the</strong>inductionassumptiontoseethat<br />
<strong>On</strong><strong>the</strong>o<strong>the</strong>rhand<br />
sothat θ σ − θisequalto<br />
with<br />
Thusif<br />
vL2 (Xj−1) ≥ (j − 1))t1 + vL1 (θσ − θ) ≥ jt1 + vL1 (θ).<br />
∆ iσ − ∆ i = (∆ + ξ) i − ∆ i + O(̟ p2e−(p−1)u−(i−1)u K<br />
)<br />
p−1<br />
i=1 Yi<br />
i−1<br />
k=0<br />
<br />
i<br />
∆<br />
k<br />
k ξ i−k + η<br />
η = O(θ̟ p2e−(p−2)u K ).<br />
η = p−1<br />
i<br />
Zi∆<br />
i=0
Chapter11 118<br />
with<strong>the</strong> Ziin L2wehave<br />
But<br />
whichequals<br />
Since<br />
wehave<br />
vK(Zj−1) ≥ (j − 1)u + vK(Θ) + p 2 e − (p − 2)u.<br />
p 2 e − (p − 2)u + (j − 1)u ≥ p(p − 1)u − (p − 2)u + (j − 1)u<br />
((p − 1) 2 + j)u ≥ pju.<br />
<br />
p−1<br />
Xj−1 =<br />
i=j Yi<br />
vL2<br />
p−1<br />
i=j Yi<br />
<br />
i<br />
j<br />
<br />
i<br />
j<br />
ξ i−j<br />
<br />
+ Zj−1<br />
ξ i−j<br />
<br />
≥ ju + vL1 (θ)<br />
forall ξ.Weobtain<strong>the</strong>requiredestimatefor vL2 (Yj)bysummingover ξ.<br />
Wenowshowthat<br />
vL1 (SK/L1 (Y ∆i )) ≥ (p − 1)t2 − it1 + vL2 (Y )<br />
for Y in L2and 1 ≤ i ≤ p − 1.Allweneeddoisshowthatforany θin<strong>the</strong>inversedifferent<br />
<strong>of</strong> L1/F<br />
orthatif θisin L1and<br />
<strong>the</strong>n<br />
isin OF.<br />
Let<br />
S L1/F(θ̟ −vL 2 (Y )+it1−(p−1)t2<br />
L1<br />
with Yjin L2for 0 ≤ j ≤ p − 1.Then<br />
S K/L1 (Y ∆i )) ∈ OF<br />
vL1 (θ) ≥ −(p − 1) (t1 + 1) + it1 − (p − 1)t2 − vL2 (Y )<br />
θ Y ∆ i = p−1−i<br />
S L1/F(θ S K/L1 (Y ∆i )) = S K/L(θ Y ∆ i ) (11.4)<br />
j=0<br />
θ = p−1<br />
j=0<br />
Yj∆ j<br />
Y Yj∆ j+i + (a + ∆) p−1<br />
j=p−i Y Yj∆ j+i−p .
Chapter11 119<br />
Since<br />
wehave<br />
for 1 ≤ i < p − 1and<br />
Therelations(11.2)implythat<br />
for 1 ≤ i < p − 1andthat<br />
Thus(11.4)isequalto<br />
if i < p − 1andto<strong>the</strong>sum<strong>of</strong>thisand<br />
if i = p − 1.<br />
Weknowthat<br />
foreach j.Thus<br />
∆ p − ∆ = a<br />
E i (∆) = 0<br />
K/L2<br />
E p−1<br />
K/L2 (∆) = (−1)p .<br />
S K/L2 (∆i ) = 0<br />
S K/L2 (∆p−1 ) = p − 1.<br />
(p − 1) S L2/F(Y Yp−1−i) + S L2/F(paY Yp−i)<br />
S L2/F(Y Yp−1)<br />
vL2 (Yj) ≥ jt1 + vL1 (θ)<br />
vL2 (Y Yp−1−i) ≥ (p − 1 − i)t1 − (p − 1) (t1 + 1) + it1 − (p − 1)t2<br />
whichisatleast −(p − 1) (t2 + 1).Sois<br />
If i = p − 1<br />
vL2 (paY Yp−1) ≥ (p − 1)t2 − t1 − (p − 1) (t1 + 1) + it1 + (p − i)t1 − (p − 1)t2.<br />
vL2 (Y Yp−1) ≥ (p − 1)t1 − (p − 1) (t1 + 1) + (p − 1)t1 − (p − 1)t2<br />
isalsoatleast −(p − 1) (t2 + 1).Allweneeddonowisobservethat<br />
S L2/F(P<br />
−(p−1) (t2+1)<br />
L2<br />
) ⊆ OF.
Chapter11 120<br />
Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong><strong>the</strong>lemmawehavetoshowthat<br />
vL1 (Ei K/L1 (Y ∆)) ≥ (p − 1)t2 + i(vL2 (Y ) − t1)<br />
for 1 ≤ i ≤ p − 1. Thishasbeendonefor i = 1;soweproceedbyinduction.Applying<strong>the</strong><br />
relations(11.2)weseethat<br />
(−1) i+1 iE i i−1<br />
(Y ∆) = K/L1 j=0 (−1)j S i−j<br />
(Y ∆) Ej (Y ∆).<br />
K/L1 K/L1<br />
Accordingto<strong>the</strong>inductionassumptionand<strong>the</strong>firstpart<strong>of</strong><strong>the</strong>lemma,with Y replacedby<br />
Y i−j ,atypicaltermin<strong>the</strong>sumon<strong>the</strong>rightis O(̟ v L1 )with<br />
if j > 0and<br />
v = (p − 1)t2 − (i − j)t1 + (i − j)vL2 (Y ) + (p − 1)t2 + j(vL2 (Y ) − t1)<br />
if j = 0.Thelemmafollows.<br />
v = (p − 1)t2 − it1 + ivL2 (Y )<br />
Weapply<strong>the</strong>secondestimatewith Y = 1toimprove,when Λ = ∆, L = L1,andcertain<br />
auxiliaryconditionsaresatisfied,ourestimateson<strong>the</strong>number γappearingin(11.3).<br />
(vi)Suppose pisoddand<br />
If k ≥ ν + 2and αp ≤ k − 2<strong>the</strong>n<br />
If k ≥ ν + 2and αp ≤ k − 1<strong>the</strong>n<br />
andif k ≥ ν + 1and αp ≤ k − 1<br />
In<strong>the</strong>presentcircumstances<br />
ℓ = (p − 1)ν + j + 1.<br />
jt1 + γ ≥ pt2.<br />
jt1 + γ ≥ (p − 1)t2 + t1<br />
jt1 + γ ≥ (p − 1)t2 − t1.<br />
γi ≥ (p − 1)t2 − it1
Chapter11 121<br />
for 1 ≤ i ≤ p − 1.Thus<br />
whichequals<br />
or<br />
If αp ≤ k − 2,thisisatleast<br />
jt1 + γ ≥ jt1 + p−1<br />
i=1 αi((p − 1)t2 − it1) − αpt1<br />
jt1 + (p − 1)kt2 − ℓt1 − (p − 1)αp(t2 − t1)<br />
(p − 1)kt2 − (p − 1)νt1 − t1 − (p − 1)αp(t2 − t1).<br />
2(p − 1)t2 + (p − 1) (k − 2 − ν)t1 − t1<br />
whichinturnisatleast pt2if p ≥ 3and k ≥ ν + 2.If αp ≤ k − 1<br />
Therequiredinequalitiesfollow.<br />
γ ≥ (p − 1)t2 + (p − 1) (k − 1 − ν)t1 − t1.<br />
Weshalluseall<strong>the</strong>seestimatesfor γin<strong>the</strong>nextsequence<strong>of</strong>lemmas.<br />
Lemma 11.8<br />
If ∆isasinLemma11.7and pisodd<strong>the</strong>n<br />
SL1/FNK/L1∆ ≡ SL2/FNK/L2∆ (mod P1+t2<br />
F ).<br />
Theassertion<strong>of</strong><strong>the</strong>lemmamaybereformulatedas<br />
Noticethat<br />
SK/L2NK/L1∆ ≡ SK/L1NK/L2∆ (modP1+pt2 ). L1<br />
pt2 = t + (p − 1)u.<br />
EarlierweappliedNewton’sidentitytoexpress S p<br />
K/L1 (∆)interms<strong>of</strong><strong>the</strong>elementarysymmetric<strong>functions</strong><strong>of</strong><br />
∆anditsconjugates.Since<br />
pe ≥ (p − 1)t2<br />
wecanapply<strong>the</strong>estimates(iii)for γtoseethat<br />
S p<br />
K/L1 (∆)
Chapter11 122<br />
iscongruentto<br />
Since<br />
wehave<br />
p<br />
pNK/L1∆ +<br />
2<br />
p−1<br />
j=1 Ej<br />
K/L1<br />
(∆) Ep−j<br />
K/L1 (∆) + {S K/L1 (∆)}p . (11.5)<br />
∆ p − ∆ = a<br />
SK/L1 (∆) = Sp<br />
K/L1 (∆) − SL2/F(a). AccordingtoLemma11.7<strong>the</strong>leftsidebelongsto P (p−1)t2−t1<br />
.Inparticularitbelongsto PL1 L1<br />
.<br />
Weneedtoknowthatitbelongsto P 1+t2.<br />
Thisisclearif p > 3or t2 > t1. Toproveitin<br />
L1<br />
generalwefirstobservethatalltermsbut<strong>the</strong>lastin(11.5)arecongruentto0modulo P 1+t2.<br />
L1<br />
Themiddletermsaretakencare<strong>of</strong>by<strong>the</strong>estimates(ii)for γ.Totakecare<strong>of</strong><strong>the</strong>firstwehave<br />
toshowthat<br />
pe − u ≥ 1 + t2.<br />
Weknowthat pe ≥ (p − 1)t2andthatif t = u<strong>the</strong>inequalityisstrict.Weneedonlyshowthat<br />
(p − 1)t2 − u ≥ t2<br />
withastrictinequalityif t = u.Thisisclearsince t2 ≥ uand t2 > uif t = u.Thus<br />
Wenowneedonlyshowthat<br />
SK/L1∆ ≡ (SK/L1∆)p − SL2/F(a) (mod P 1+t2).<br />
L1<br />
S L2/F(a) ≡ 0 (modP 1+t2<br />
L1 ).<br />
Theleftsidebelongsto P pb<br />
if bisanyintegerlessthanorequalto<br />
L1<br />
Wemaytake<br />
−u + (p − 1)(t2 + 1)<br />
.<br />
p<br />
b ≥<br />
−u + (p − 1)t2<br />
p<br />
whichisgreaterthanorequalto 1+t2<br />
p exceptwhen p = 3and t = u.Inthiscase,whichis<strong>the</strong><br />
onetoworryabout, t2 = uisprimeto pand<br />
−u + (p − 1)(t2 + 1)<br />
p<br />
= u + 2<br />
3
Chapter11 123<br />
hasintegralpartatleast u+1<br />
3 .<br />
Weapply(11.5)againtoseethat<br />
iscongruentto<br />
SK/L1NK/L2∆ = SK/L1a = Sp (∆) − SK/L1 (∆)<br />
K/L1<br />
−SK/L1∆ + p N p<br />
K/L1∆ +<br />
2<br />
Wehavestilltoconsider<br />
p−2<br />
S K/L2 N K/L1<br />
j=2 Ej<br />
K/L1<br />
(∆) Ep−j<br />
K/L1 (∆).<br />
∆. (11.6)<br />
Therearesomegeneralremarkstobemadefirst.Suppose Λbelongsto Kand<br />
If xand zalsobelongto Kand<br />
and<br />
<strong>the</strong>n<br />
Itisenoughtoshowthat<br />
vK(Λ) = −u.<br />
xΛ j ∈ OK<br />
z ∈ P 1+t+(p2 −1)u<br />
K<br />
N K/L1 (x(Λ + z)j+1 ) ≡ N K/L1 (xΛj+1 ) (modP 1+pt2<br />
L1 ).<br />
N K/L1<br />
<br />
1 + z<br />
j+1 Λ<br />
≡ 1 (mod P 1+t+pu<br />
). L1<br />
ThisfollowsfromLemmaV.5<strong>of</strong>Serre’sbookand<strong>the</strong>relations<br />
and<br />
1 + t + p 2 u ≥ 1 + t + pu<br />
1 + t + p 2 u + (p − 1) (t + 1)<br />
p<br />
= 1 + t + pu.<br />
AccordingtoLemmas11.3and11.4<strong>the</strong>reisforeach σ = 1in G2a (p − 1)throot<strong>of</strong>unity<br />
ξ = ξ(σ)suchthat<br />
∆ σ = (∆ + ξ) p − a + O (̟ 2(p2e−(p−1)u) K ).
Chapter11 124<br />
Wehave<br />
whichequals<br />
p 2 e − (p − 1)u ≥ (p − 1) {(p − 1)u + t} − (p − 1)u<br />
(p − 1)t + (p − 2) (p − 1)u ≥ t + p(p − 2)u.<br />
If t = u<strong>the</strong>first<strong>of</strong><strong>the</strong>seinequalitiesisstrictandif t > u<strong>the</strong>lastis.Thus<br />
and<br />
p 2 e − (p − 1)u ≥ 1 + t + p(p − 2)u<br />
2(p 2 e − (p − 1)u) ≥ {1 + t + (p 2 − 1)u} + {1 + t + ((p − 1) 2 − 2p)u}.<br />
Thesecondtermispositiveunless p = 3.<br />
Theexpression<br />
isequalto<br />
But<br />
and<br />
<br />
p<br />
<br />
=<br />
i<br />
p(p − 1) . . .(p − i + 1)<br />
(∆ + ξ) p − a<br />
∆ + ξ + p−1<br />
i=1<br />
i!<br />
<br />
p<br />
<br />
i<br />
ξ i ∆ p−i .<br />
i+1 p<br />
≡ (−1)<br />
i<br />
2p 2 e − (p − 1)u ≥ 2(p 2 e − (p − 1)u)<br />
(modp 2 )<br />
whichis,aswehavejustseen,atleast 1 + t + (p 2 − 1)u.Thusif p > 3<br />
if<br />
∆ σ ≡ (∆ + ξ) (1 − Z(ξ)) (modP 1+t+(p2−1)u K )<br />
Z(ξ) = p∆p−1<br />
1 + ξ/∆<br />
Expanding<strong>the</strong>denominatorweobtain<br />
If i ≥ p − 1<br />
p−1<br />
i=1<br />
∞<br />
p−1<br />
Z(ξ) = p∆<br />
i=1 ai<br />
p−1<br />
i<br />
ai = (−1)<br />
j=1<br />
1<br />
j<br />
(−1) i<br />
i<br />
i ξ<br />
.<br />
∆<br />
i ξ<br />
.<br />
∆<br />
≡ 0 (mod p).
Chapter11 125<br />
Clearly<br />
If p = 3<br />
Z(ξ) = O(p∆ p−2 ) = O(̟ p2e−(p−2)u K ).<br />
3(p 2 e − (p − 1)u) ≥ {1 + t + (p 2 − 1)u} + {2(1 + t) + (2p 2 − 6p + 1)u}.<br />
Thesecondtermisatleast 3uandinparticular,ispositive.Lemmas11.3and11.4showthat<br />
Therightsideequals<br />
∆ σ ≡ ((∆ + ξ) 3 − a) 3 − a (mod P 1+t+(p2−1)u K ).<br />
(∆ + ξ + 3ξ∆ 2 + 3ξ 2 ∆) 3 − a.<br />
Expanding<strong>the</strong>cubeandignoringalltermsin P 1+t+(p2−1)u K<br />
∆ + ξ + 3∆ 3<br />
<br />
ξ ξ2<br />
+<br />
∆ ∆2 <br />
whichwewriteas<br />
with<br />
and<br />
Since<br />
∞<br />
2<br />
Z(ξ) = 3∆<br />
i=1 ai<br />
(∆ + ξ) (1 − Z(ξ))<br />
+ 9∆ 4 ξ<br />
i ξ ∞<br />
4<br />
+ 9∆<br />
∆<br />
i=1<br />
weobtain<br />
i −ξ<br />
.<br />
∆<br />
2(p 2 e − (p − 2)u) ≥ 2(p 2 e − (p − 1)u) + 2u ≥ 1 + t + p 2 u<br />
p 2 e − (p − 2)u ≥ 1 + t + (p − 1) 2 u ≥ 1 + t + pu<br />
forallodd p,lemmaV.5<strong>of</strong>Serre’sbookshowsthat<br />
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<br />
modulo P 1+t+(p−1)u<br />
L1<br />
if x∆ j liesin OK.<br />
Theexpression(11.6)isequalto<br />
<br />
NK/L1∆ + σ∈G 2<br />
σ=1<br />
N K/L1 ∆σ .
Chapter11 126<br />
Theprecedingremarksshowthat,if p > 3,thisiscongruentto<br />
modulo P 1+t+(p−1)u<br />
.Since<br />
L1<br />
wehave<br />
<br />
NK/L1∆ +<br />
ξ NK/L1 (∆ + ξ) {1 − SK/L1Z(ξ)} 2p 2 e + u + (p − 1)t<br />
p<br />
if i ≥ pandwemayreplace S K/L1 Z(ξ)by<br />
if p > 3.Ofcourse<br />
P 1+pt2<br />
L1<br />
and<br />
and<br />
Since<br />
≥ 2(p − 1)<br />
<br />
u +<br />
<br />
t − u<br />
tu > t + pu<br />
p<br />
N K/L1 (∆ + ξ) S K/L1 (aip ∆ p−1−i ξ i ) ∈ P 1+pt2<br />
L1<br />
p−1<br />
i=1 pξi S K/L1 (ai∆ p−1−i )<br />
p<br />
NK/L1 (∆ + ξ) =<br />
i=0 ξ1−i E i K/L1 (∆).<br />
Putting<strong>the</strong>seobservationstoge<strong>the</strong>rweseethat,if p > 3,(11.6)iscongruentmodulo<br />
to<strong>the</strong>sum<strong>of</strong><br />
NK/L1∆ + (p − 1) {SK/L1∆ + NK/L1∆} −p(p − 1) p−1<br />
i=1 ai S K/L1 (∆p−1−i ) E 1+i<br />
K/L1 (∆)<br />
−p(p − 1) {pap−1 S K/L1 (∆) + ap−2 S K/L1 (∆)}.<br />
pSK/L1 (∆) ∈ P1+pt2<br />
L1<br />
<strong>the</strong>lastexpressionmaybeignoredasmay<strong>the</strong>termin<strong>the</strong>secondcorrespondingto i = p − 2.<br />
Since<br />
ap−1 = p−1<br />
j=1<br />
1<br />
≡ 0 (modp)<br />
j<br />
and<br />
and<br />
S K/L1 (∆0 ) = p<br />
3p 2 e − t2 ≥ 1 + pt2
Chapter11 127<br />
<strong>the</strong>sumin<strong>the</strong>secondexpressionneedonlybetakenfrom1to p−3.Therelation(11.2)implies<br />
that<br />
K/L1 (∆) ≡ (−1)ip(p − 1 − i) E p−1−i<br />
(∆) E1+i<br />
K/L1 (∆)<br />
pS p−1−i<br />
(∆) E1+i<br />
K/L1<br />
K/L1<br />
modulo P 1+pt2<br />
.Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma11.8,for p > 3,weneedonlyshowthat<br />
L1<br />
iai−1 + (p − i) ap−i−1 ≡ (−1) i<br />
for p − 2 ≥ p − i ≥ i ≥ 2.Thisamountstoshowingthat<br />
i i−1<br />
j=1<br />
1<br />
j<br />
+ (p − i) p−i−1<br />
j=1<br />
1<br />
j<br />
(modp)<br />
≡ −1 (modp).<br />
Wemayreplace<strong>the</strong> p − iinfront<strong>of</strong><strong>the</strong>secondsumby −i.Making<strong>the</strong>obviouscancellations<br />
weobtain<br />
−i p−i−1 1 p−i<br />
≡ −1 − i<br />
j=i j j=i<br />
1<br />
j .<br />
If 1<br />
j occursin<strong>the</strong>sumon<strong>the</strong>rightsodoes<br />
1<br />
p−j .<br />
Thepro<strong>of</strong>for p = 3canproceedinexactly<strong>the</strong>samewayprovidedweshowthat<br />
liesin P 1+3t2<br />
L1<br />
for i ≥ 1.Since<br />
andone<strong>of</strong><strong>the</strong>inequalitiesisstrict<br />
Theexpression<br />
9 {N K/L1 (∆ + ξ)} {S K/L1 (ξi ∆ 4−i )} (11.7)<br />
2p 2 e − u ≥ 2(p − 1)t2 − u ≥ 3t2<br />
9NK/L1 (∆ + ξ) ∈ P1+3t2.<br />
L1<br />
ξ i S K/L1 (∆4−i )<br />
isclearlyintegralfor i ≥ 4.By,forexample,Lemma11.7itisalsointegralif iis2or3.Thus<br />
i = 1is<strong>the</strong>onlycasetocauseaproblem.If i = 1wesumover ξtoseethat(11.7)equals<br />
18{E 2 K/L1 (∆) S K/L1 (∆3 ) + S K/L1 (∆3 )}.<br />
Thetermsappearingin<strong>the</strong>expressioninbracketshavebeenshowntoliein OL1 .
Chapter11 128<br />
Thereisonemorelemmatobeprovedbeforewecometo<strong>the</strong>basicfact<strong>of</strong>thisparagraph.<br />
If xisin Kweset<br />
and<br />
g(x) = S L1/F(N K/L1 ∆ S K/L1 (x)) − S L2/F(N K/L2 ∆ S K/L2 (x))<br />
h(x) = S L1/F(N K/L1 (x∆)) − S L2/F(N K/L2 (x∆)).<br />
In<strong>the</strong>followinglemma pissupposedodd.<br />
Lemma 11.9<br />
(a)Suppose xisin L2, 0 ≤ j ≤ p − 1,and x∆jliesin P 1+t2−t1<br />
K .If j = p − 2<strong>the</strong>n<br />
g(x∆ j ) ≡ 0 (modP 1+t2<br />
F )<br />
butif j = p − 2,<strong>the</strong>reisan ωin L2suchthat<br />
and<br />
ωx ≡ −x E p−1<br />
(∆) (mod P1+pt2<br />
K/L1 K )<br />
g(x∆ j ) ≡ −{S L2/Fx − S L2/F(xω)} (mod P 1+t2<br />
F ).<br />
(b)Suppose xisin L2, 0 ≤ j ≤ p − 1,and x∆ j liesin PK.If j = p − 2<br />
butif j = p − 2<br />
modulo P 1+pt2<br />
. L1<br />
Thecongruencesmodulo P 1+t2<br />
F<br />
Westartwithpart(a).If xbelongsto OL2 <strong>the</strong>n<br />
h(x∆ j ) ≡ 0 (modP 1+t2<br />
F )<br />
h(x∆ j ) ≡ (p − 1) (1 − {E p−1<br />
K/L1 (∆)}p )N K/L1 x<br />
are<strong>of</strong>courseequivalenttocongruencesmodulo P 1+pt2<br />
. L1<br />
g(x) = SK/L1 (xSK/L2 (NK/L1∆) − pxa).<br />
Because<strong>of</strong><strong>the</strong>previouslemmathisiscongruentto<br />
S K/L1 (xS L2/Fa − pxa) = S L2/F xS L2/Fa − pS L2/F(xa)
Chapter11 129<br />
modulo P 1+t2<br />
F .Wesawbeforethat<br />
Thesameargumentshowsthat<br />
S L2/Fa ∈ P 1+t2<br />
L1 .<br />
SL2/F(xa) ∈ P 1+t2.<br />
L1<br />
pbelongsto P (p−1)t2<br />
.Since<strong>the</strong>integralpart<strong>of</strong><br />
L1<br />
isatleast<br />
(p − 1) (t2 + 1)<br />
p<br />
(p − 1)t2<br />
,<br />
p<br />
sodoes S L2/Fx.Thistakescare<strong>of</strong><strong>the</strong>case j = 0.<br />
If 1 ≤ j = p − 1<strong>the</strong>n g(x∆ j )isequalto<br />
S L2/F(xS K/L2 (∆j N K/L1 ∆)) − (p − 1)δ S L2/F(xa)<br />
where δ = 0if j = p − 1and δ = 1if j = p − 1.Consider<br />
Itliesin L2andisequalto<br />
Zj = x S K/L2 (∆j N K/L1 ∆).<br />
x∆ j <br />
NK/L1∆ + σ∈G 2 x∆<br />
σ=1<br />
σ j NK/L1∆σ .<br />
Weobservefirst<strong>of</strong>allthatif Λisin K, vK(Λ) = −u, xisin L2, xΛjliesin P 1+t2−t1,<br />
and zliesin P<br />
(p−1) (pt2−t1)<br />
K<br />
<strong>the</strong>n<br />
x(Λ + z) j N K/L1 (Λ + z) ≡ xΛj N K/L1<br />
Λ (mod P1+pt2<br />
K )<br />
provided pisgreaterthan3.Toestablishthiscongruenceweshowthat<br />
<br />
1 + z<br />
j Λ<br />
N K/L1<br />
<br />
1 + z<br />
<br />
Λ<br />
≡ 1 (modP (p−1)t2+(p+1)t1<br />
K ).<br />
K
Chapter11 130<br />
Toshowthis,onehasonlytoobservethat z<br />
Λ<br />
that<br />
whichequals<br />
if p > 3.<br />
andallitsconjugatesliein P(p−1)pt2−(p−2)t1<br />
K<br />
(p − 1)pt2 − (p − 2)t1 ≥ (p − 1)t2 + ((p − 1) 2 − (p − 2))t1<br />
(p − 1)t2 + ((p − 1) (p − 2) + 1)t1 ≥ (p − 1)t2 + (p + 1)t1<br />
Supposefornowthat p > 3.Since<br />
Lemma11.4impliesthat Zjiscongruentto<br />
p 2 e − (p − 1)u ≥ (p − 1) (pt2 − t1).<br />
x∆ j <br />
NK/L1∆ +<br />
ξ x(∆ + ξ)j NK/L1 (∆ + ξ)<br />
modulo P 1+pt2<br />
K<br />
<br />
NK/L1 (∆ + ξ) = ξ 1 + 1<br />
ξ N p−1<br />
K/L1∆ +<br />
i=1 ξ−iE i K/L1 (∆)<br />
<br />
.<br />
AccordingtoLemma11.7thisiscongruentto<br />
ξ + NK/L1∆ + ξ Ep−1<br />
K/L1 (∆)<br />
modulo P pt2<br />
K .Thusif x∆jbelongsto P 1+t2−t1<br />
K<br />
Zj ≡ x∆ j <br />
NK/L1∆ +<br />
ξ x(∆ + ξ)j (ξ + NK/L1∆ + ξ Ep−1<br />
K/L1 (∆))<br />
modulo P 1+pt2<br />
K .Weexpand (∆ + ξ) jandsumover ξtoobtain<br />
if j < p − 2.If j = p − 2weobtain<br />
and<br />
px∆ j NK/L1∆ (11.8)<br />
px∆ j NK/L1∆ + (p − 1)x (1 + Ep−1<br />
K/L1 (∆))
Chapter11 131<br />
andif j = p − 1weobtain<br />
px∆ j NK/L1∆ + (p − 1)x {NK/L1∆ + (p − 1)∆ + (p − 1)∆ Ep−1<br />
K/L1 (∆)}.<br />
Theexpression(11.8)liesin<br />
provided x∆ j liesin PK.<br />
Since<br />
and<br />
wehave<br />
P 1+p2 e−pt1<br />
K<br />
p 2 e − pt1 ≥ p(p − 1)t2 − pt1 ≥ pt2.<br />
L2 ∩ P 1+pt2<br />
K<br />
if 1 ≤ j < p − 2and x∆jliesin P 1+t2−t1<br />
K .<br />
Since<br />
liesin OL1 ,<br />
with<br />
if j = p − 2.Wemaytake ωin L2and<strong>the</strong>n<br />
and<br />
If j = p − 1<strong>the</strong>n<br />
= P 1+t2<br />
L1<br />
S L2/F(P 1+t2<br />
L2 ) ⊆ P 1+t2<br />
F ,<br />
g(x∆ j ) ≡ 0 (modP 1+t2<br />
F )<br />
E p−1<br />
K/L1 (∆)<br />
Zj = (ω − 1)x<br />
ω x = Zj + x ≡ −x E p−1<br />
(∆) (mod P1+pt2<br />
K/L1 K )<br />
g(x∆ j ) ≡ −{S L2/Fx − S L2/F(xω)} (modP 1+t2<br />
F ).<br />
g(x∆ j ) = S L2/F (Zj − (p − 1)xa)<br />
Zj − (p − 1)xa
Chapter11 132<br />
iscongruentto<br />
modulo P 1+pt2<br />
K .Theproduct<br />
liesin P 1+pt2<br />
K<br />
and<br />
Itiseasilyseenthat<br />
isequalto<br />
(p − 1)x {NK/L1∆ + p∆ + (p − 1)∆ Ep−1<br />
K/L1 (∆) − ∆p }<br />
{(p − 1)x} {p∆}<br />
(p − 1) E p−1<br />
(∆) ≡ −Ep−1 (∆) (modP1+pt2<br />
K/L1 K/L1 K ).<br />
∆ p + ∆ E p−1<br />
K/L1 (∆) − N K/L1 ∆<br />
− p−2<br />
i=1 (−1)i ∆ p−i E i K/L1 (∆).<br />
Recallingthat x∆p−1issupposedtoliein PKweappealtoLemma11.7toseethat<strong>the</strong>product<br />
<strong>of</strong>thisexpressionwith xliesin P 1+pt2<br />
K .Thus<br />
with<br />
If<br />
<strong>the</strong>n<br />
If p = 3and ξ = ξ(σ)<strong>the</strong>n<br />
Ifwecanshowthat<br />
itwillfollowthat<br />
g(x∆ j ) ≡ 0 (modP 1+t2<br />
F ).<br />
∆ σ = ∆ + ξ + 3ξ∆ 2 + 3ξ 2 ∆ + z<br />
z = O(̟ 2(p2e−(p−1)u) K ).<br />
∆σ = ∆ + ξ + 3ξ∆ 2<br />
∆ σ = ∆σ + 3ξ 2 ∆ + z.<br />
3ξ 2 ∆ + z = O(̟ (p−1)t2+pt1<br />
K )<br />
x∆ σ j N K/L1 ∆σ ≡ x∆ j σ N K/L1 ∆σ (modP 1+pt2<br />
K ) (11.9)
Chapter11 133<br />
if x∆ j liesin P 1+t2−t1<br />
K<br />
and<br />
because (p − 1) 2 ≥ p.Moreover<br />
whichisatleast<br />
and<br />
3ξ 2 ∆ = O(̟ p2e−t1 K )<br />
p 2 e − t1 ≥ p(p − 1)t2 − t1 ≥ (p − 1)t2 + pt1<br />
2(p 2 e − (p − 1)u) ≥ 2(p(p − 1)t2 − (p − 1)t1)<br />
(p − 1)t2 + ((2p − 1) (p − 1) − 2(p − 1))t1<br />
(2p − 1) (p − 1) − 2(p − 1) = (2p − 3) (p − 1) ≥ p.<br />
Wewanttoreplace N K/L1 ∆σby<br />
NK/L1 (∆ + ξ)<br />
in<strong>the</strong>rightside<strong>of</strong>(11.9).Todothiswehavetoshowthat<br />
Since<br />
and<br />
N K/L1<br />
wehaveonlytoverifythat<br />
andthat<strong>the</strong>integralpart<strong>of</strong><br />
isatleast<br />
<br />
∆σ<br />
∆ + ξ<br />
∆σ<br />
≡ 1 (mod P (p−1)t2+(p+1)t1<br />
K ).<br />
∆ + ξ = 1 + O(̟p2 e−t1<br />
K )<br />
p 2 e − t1 ≥ p(p − 1)t2 − t1,<br />
p{p(p − 1)t2 − t1} ≥ (p − 1)t2 + (p + 1)t1<br />
p(p − 1)t2 − t1 + (p − 1) (t + 1)<br />
p<br />
(p − 1)t2 + (p + 1)t1<br />
.<br />
p<br />
(11.10)<br />
(11.11)
Chapter11 134<br />
Theinequality(11.10)isclear.Since t ≥ t1<strong>the</strong>integralpart<strong>of</strong>(11.11)isatleast<br />
and<br />
p(p − 1)t2 + (p − 2)t1<br />
p<br />
≥ (p − 1)t2 + ((p − 1) 2 + (p − 2))t1<br />
p<br />
(p − 1) 2 + (p − 2) ≥ p + 1.<br />
Justaswhen p > 3wemayreplace NK/L1 (∆ + ξ)in(11.9)by<br />
Thus Zjiscongruentto<br />
modulo P 1+pt2<br />
K<br />
equalto<br />
ξ + NK/L1∆ + ξ Ep−1<br />
K/L1 (∆).<br />
x∆ j <br />
NK/L1∆ +<br />
ξ x(∆ + ξ + 3ξ∆2 ) j (ξ + NK/L1∆ + ξ E2 K/L1 (∆))<br />
andif j = 2itisequalto<br />
if p = 3, jis1or2,and x∆jbelongsto P 1+t2−t1<br />
K .If j = 1thisexpressionis<br />
px∆ j NK/L1∆ + 2x(1 + 3∆2 ) (1 + E 2 (∆)) (11.12)<br />
K/L1<br />
px∆ j NK/L1∆ + 2x{2(1 + 3∆2 )∆(1 + E 2 K/L1 (∆)) + (1 + 3∆2 ) 2 NK/L1∆}. (11.13)<br />
Theterm<br />
px∆ j N K/L1 ∆<br />
canbeignoredasbeforebecauseitliesin P 1+pt2<br />
K .Also<br />
because x∆ j liesin P 1+t2−t1<br />
K<br />
and<br />
3x∆ j+1 = O(̟ 1+p2e+t2−2t1 K )<br />
p 2 e + t2 − 2t1 ≥ p(p − 1)t2 − t1 ≥ pt2.<br />
Wemayalsoreplace<strong>the</strong>factor2in(11.12)by1.Thus(11.12)iscongruentto<br />
−x(1 + E 2 K/L1 (∆))
Chapter11 135<br />
modulo P 1+pt2<br />
K . Atthispointwemayargueaswedidfor p > 3. Tosimplify(11.13),we<br />
observethat<br />
9∆ 4 NK/L1∆ = O(̟2p2 e−7t1<br />
K )<br />
andthat<br />
Moreover<br />
if x∆ 2 belongsto PKand<br />
Thus(11.13)iscongruentto<br />
2p 2 e − 7t1 ≥ 12t2 − 7t1 ≥ 3t2.<br />
3x∆ 2 NK/L1∆ = O(̟1+p2 e−3t1<br />
K )<br />
p 2 e − 3t1 ≥ 6t2 − 3t1 ≥ 3t2.<br />
2x{2∆(1 + E 2 K/L1 (∆)) + N K/L1 ∆}<br />
modulo P 1+3t2<br />
K .Wemayagainargueaswedidfor p > 3.<br />
Weturnto<strong>the</strong>secondpart<strong>of</strong><strong>the</strong>lemma.Weobservefirstthatif xbelongsto L2, ybelongs<br />
to K,and<br />
xy ∈ PK<br />
<strong>the</strong>n<br />
Theleftsideis<br />
h(xy) ≡ h(y) NK/L1 (x) (mod P1+t2<br />
F ).<br />
S L1/F(N K/L1 xN K/L1 y∆) − S L2/F(x p N K/L2 y∆).<br />
Since NK/L2 x = NL2/Fxliesin Fthisequals<br />
{N K/L1 x}h(y) + S L2/F {N K/L2 (y∆) (N L2/Fx − x p )}.<br />
Thesecondtermis<strong>the</strong>tracefrom L2to F<strong>of</strong><br />
{NK/L2 (xy∆)}<br />
<br />
NL2/Fx − xp NK/L2x <br />
if,aswemayaswellassume, x = 0.Allweneeddoisshowthatthisexpressionliesin P 1+t2<br />
L2<br />
for<strong>the</strong>nitstracewillliein P 1+t2<br />
F .Thefirstfactorliesin P 1−t2.Thesecondfactorisequalto<br />
<br />
<br />
σ∈G 2 x σ−1<br />
<br />
− 1.<br />
L2
Chapter11 136<br />
Since p ≥ 3itwillbesufficienttoshowthat<strong>the</strong>image<strong>of</strong><strong>the</strong>homomorphism<br />
x −→ ϕ(x) = <br />
σ∈G 2 x σ−1<br />
<strong>of</strong> CL2into UL2iscontainedin U(p−1)t2.<br />
Let ρbeagenerator<strong>of</strong> G L2<br />
2 andlet P(X)be<strong>the</strong><br />
polynomial<br />
<strong>the</strong>n<br />
Let<br />
p−1<br />
i=1 (Xi − 1) = p−1<br />
If 1 ≤ i ≤ p − 1<strong>the</strong> ithcoefficient<strong>of</strong> Q(X)is<br />
ϕ(x) = x P(ρ) .<br />
Q(X) = (X − 1) p−1 .<br />
p−1−i (p − 1) . . .(p − i)<br />
(−1)<br />
i!<br />
Sinceboth P(X)and Q(X)aredivisibleby X − 1<br />
i=0 Xi − p<br />
≡ 1 (modp).<br />
P(X) = Q(X) + p(X − 1)R(X)<br />
where R(X)isapolynomialwithintegralcoefficients.Forall zin CL2 ,<br />
with w = O(̟ t2<br />
L2 ).Then<br />
and<br />
If a ≥ 1and<br />
<strong>the</strong>n<br />
z ρ−1 = 1 + w<br />
z p(ρ−1) = (1 + w) p ≡ 1 + w p ≡ 1 (modP pt2<br />
L2 )<br />
(1 + w) p−1 =<br />
z p(ρ−1)R(ρ) ∈ U pt2<br />
L2 .<br />
w ∈ P a L2<br />
1 + wp<br />
1 + w = 1 + wp − w<br />
1 + w<br />
≡ 1 (modPa+t2<br />
L2 ).<br />
<strong>On</strong>e<strong>the</strong>nshowseasilybyinductionthat,forall zin CL2andall n ≥ 1,<br />
z (ρ−1)n<br />
∈ U nt2<br />
L2 .
Chapter11 137<br />
If xliesin PKwemaytake y = 1.ApplyingLemma11.8weseethat<br />
h(x) ≡ N L2/F xh(1) ≡ 0 (mod P 1+t2<br />
F ).<br />
If 1 ≤ j ≤ p − 1, xliesin L2,and x∆ j liesin PK,<br />
with<br />
and<br />
Theexpression Pjiscongruentto<br />
N L2/Fx{N K/L1 ∆j+1 + <br />
h(x∆ j ) ≡ Pj − Qj (mod P 1+t2<br />
F )<br />
Pj = N L2/F xS L1/F(N K/L1 ∆j+1 )<br />
Qj = N L2/FxS L2/F(N K/L2 ∆j+1 ).<br />
ξ N K/L1 (∆ + ξ)j+1 {1 − S K/L1 Z(ξ)}j+1 } (11.14)<br />
modulo P 1+pt2.Sinceweareworkingmodulo<br />
P L1<br />
1+pt2<br />
L1<br />
weneedonlyconsider<br />
(1 − S K/L1 Z(ξ))j+1<br />
modulo P pt2+t1<br />
.Supposefirstthat p > 3.Then<br />
L1<br />
and<br />
Moreover<strong>the</strong>integralpart<strong>of</strong><br />
isatleast<br />
Z(ξ) = O(̟ p2e−(p−2)u K )<br />
p 2 e − (p − 2)u ≥ p(p − 1)t2 − (p − 2)u ≥ p(p − 2)t2.<br />
p(p − 2)t2 + (p − 1) (t + 1)<br />
p<br />
(p − 2)t2 −<br />
(p − 1)<br />
t<br />
p<br />
andtwicethisisatleast pt2 + t1.Wereplace(11.15)by<br />
1 − (j + 1) S K/L1 Z(ξ).<br />
(11.15)
Chapter11 138<br />
Since<br />
and<br />
while<br />
p−2<br />
p−1<br />
Z(ξ) ≡ p∆<br />
i=1 ai<br />
i ξ<br />
∆<br />
p 2 = O(̟ 2p2 e<br />
K )<br />
2p 2 e ≥ 2p(p − 1)t2 ≥ p(pt2 + t1),<br />
wemayreplace Z(ξ)by<strong>the</strong>rightside<strong>of</strong>(11.16).ByLemma11.7<br />
if 1 ≤ i ≤ p − 2and<br />
if i ≥ 2.Wereplace Z(ξ)by<br />
Wemaywrite(11.14)as<br />
N L2/FxN K/L1 ∆j+1<br />
Whenweexpand<br />
andsumover ξwewillobtain<br />
whichwewriteas<br />
pSK/L1 (∆p−1−i ) = O(̟ pe+it2)<br />
L1<br />
pe + it2 ≥ (p − 1)t2 + it2 ≥ pt2 + t1<br />
<br />
1 + <br />
ξ N K/L1<br />
N K/L1<br />
N L2/FxN K/L1 ∆j+1<br />
plusasum<strong>of</strong>terms<strong>of</strong><strong>the</strong>form<br />
pa1ξ∆ p−2 .<br />
<br />
1 + ξ<br />
j+1 ∆<br />
<br />
1 + ξ<br />
j+1 ∆<br />
<br />
1 + <br />
<br />
NL2/Fx NK/L1∆j+1 = <br />
αp N L2/FxN K/L1 ∆j+1 E i K/L1<br />
ξ N K/L1<br />
ξ N K/L1<br />
<br />
1<br />
∆<br />
(modp 2 ) (11.16)<br />
{1 − SK/L1 (pa1ξ∆ p−2 <br />
)} .<br />
<br />
1 + ξ<br />
j+1 ∆<br />
(∆ + ξ)j+1<br />
<br />
S K/L1 (∆p−2 )
Chapter11 139<br />
where αisrationalandliesin OFand iisatleast1.Since<br />
for i ≥ 1and<br />
E i K/L1<br />
<strong>the</strong>sesupplementarytermsmaybeignored.<br />
Nowtake p = 3. ∗<br />
<br />
1<br />
= O(̟<br />
∆<br />
t1<br />
L1 )<br />
pS K/L1 (∆p−2 ) = O(̟ pt2<br />
L1 )<br />
∗ (1998)Thisiswhereandhow<strong>the</strong>manuscript<strong>of</strong>Chapter11breaks<strong>of</strong>f.
Chapter12 140<br />
Chapter 12.<br />
The Second Main Lemma<br />
Suppose Kisanormalextension<strong>of</strong><strong>the</strong>localfield F and G = G(K/F)is<strong>the</strong>direct<br />
product<strong>of</strong>twocyclicgroups<strong>of</strong>primeorder ℓ.Let XKbeaquasicharacter<strong>of</strong> CK.If σbelongs<br />
to Gdefine X σ Kby<strong>the</strong>relation X σ σ−1<br />
K (α) = XK(α ).<br />
Supposethat X σ K = XKforall σin Gbutthatfornoquasicharacter XF<strong>of</strong> CFdoes XK =<br />
X K/F.If F ⊆ L ⊆ Kand [K : L] = ℓ<strong>the</strong>n XKcanbeextendedtoaquasicharacter<strong>of</strong> W K/L<br />
because W K/L/CKisisomorphicto G(K/L)whichiscyclic.Ifthisquasicharacteris XL<strong>the</strong>n<br />
XK = X K/L.<br />
Lemma 12.1<br />
Suppose L1and L2aretw<strong>of</strong>ieldslyingbetween Fand Kand [K : L1] = [K : L2] = ℓ.<br />
Suppose XL1isaquasicharacter<strong>of</strong> CL1 , XL2isaquasicharacter<strong>of</strong> CL2 ,and<br />
Then<br />
isequalto<br />
XK = X K/L1 = X K/L2 .<br />
∆(XL1 , ψ L1/F) <br />
∆(XL2 , ψ L2/F) <br />
µF ∈S(L1/F) ∆(µF, ψF)<br />
µF ∈S(L2/F) ∆(µF, ψF).<br />
Because<strong>of</strong><strong>the</strong>assumptionon G(K/F)<strong>the</strong>field Fmustbenonarchimedean.Toprove<br />
<strong>the</strong>lemmaingeneralitisenoughtoproveitforagiven L1andall L2. Therearethree<br />
possibilitiestoconsider.<br />
(i)Thesequence<strong>of</strong>groups<strong>of</strong>ramificationtakes<strong>the</strong>form<br />
G = G−1 = G0 = . . . = Gt = Gt+1 = . . . = {1}.<br />
(ii)Thesequence<strong>of</strong>groups<strong>of</strong>ramificationtakes<strong>the</strong>form<br />
G = G−1 = G0 = G1 = . . . = Gu = Gu+1 = . . . = Gt = Gt+1 = . . . = {1}.
Chapter12 141<br />
(iii)Thesequence<strong>of</strong>groups<strong>of</strong>ramificationtakes<strong>the</strong>form<br />
G = G−1 = G0 = G1 = . . . = Gt = Gt+1 = . . . = {1}.<br />
In<strong>the</strong>firsttwocaseswetake G 1 = G(K/L1)tobe Gt. In<strong>the</strong>thirdcase<strong>the</strong>choice<strong>of</strong> L1is<br />
immaterial.<br />
If<strong>the</strong>relation X σ Li = XLiobtainsforone σdifferentfrom1in Gi = G(Li/F)itobtainsfor<br />
allsuch σand XLiis<strong>of</strong><strong>the</strong>form XLi/Fforsomequasicharacter XF<strong>of</strong> CF.Then XK = XK/F whichiscontrarytoassumption.Thus<strong>the</strong>characters X σ−1<br />
Li with σin Giaredistinct.Theyare clearlytrivialon NK/Li CKso<br />
{X σ−1<br />
Li | σ ∈ G i } = S(K/Li) = {µ Li/F | µF ∈ S(Lj/F)}.<br />
Here jis2or1accordingas iis1or2.<br />
while<br />
Then<br />
Let ti ≥ −1bethatintegerforwhich<br />
G i = G i<br />
ti<br />
G i<br />
ti+1 = {1}.<br />
δi = δ(Li/F) = (ti + 1) (ℓ − 1).<br />
In<strong>the</strong>firstcase L1/Fisunramifiedand L2/Fisramified.Wechoose ̟L2 arbitrarilyandtake<br />
̟K = ̟L2 .Alsoweset<br />
̟L1 = ̟F = N L2/F̟L2 .<br />
In<strong>the</strong>secondandthirdcases K/L1and K/L2areramifiedand K/Fistotallyramified.We<br />
choose ̟kfirstandset<br />
and<br />
̟Li = N K/Li ̟K<br />
̟F = N K/F̟K.<br />
Let mi = m(XLi ).The m(X σ ) = miand<br />
Li<br />
m(X σ−1<br />
Li ) ≤ mi.
Chapter12 142<br />
Thus m(ν) ≤ miif νbelongsto S(K/Li).If G i = G(K/Li)andif<br />
while<br />
G i ui<br />
G i ui+1<br />
= Gi<br />
= {1}<br />
<strong>the</strong>n m(ν) = ui + 1if νisnontrivial.Thus ui + 1 ≤ mi.Since νXLiis<strong>of</strong><strong>the</strong>form X σ Li forall<br />
νin S(K/Li),<br />
m(νXLi ) = m(XLi ).<br />
Lemma8.8and8.12implythat*<br />
This m(XK) = miif K/Liisunramifiedand<br />
m(XK) = ψ K/Li (mi − 1) + 1.<br />
m(XK) = ℓmi − δ(K/Li)<br />
if K/Liisramified.If n = n(ψF)<strong>the</strong>n ni = n(ψ Li/F)is nif Li/Fisunramifiedandis ℓn+δi<br />
if Li/Fisramified.<br />
In<strong>the</strong>firstcase<br />
Therelations<br />
and<br />
implythat t2 = t.Also<br />
sothat<br />
Moreover<br />
δ(K/L2) = δ(L1/F) = 0.<br />
δ(K/F) = δ(K/L1) + ℓδ(L1/F) = δ(K/L1) = (t + 1) (ℓ − 1)<br />
δ(K/F) = δ(K/L2) + δ(L2/F) = δ2 = (t2 + 1) (ℓ − 1)<br />
m(XK) = m2 = ℓm1 − δ(K/L1) = ℓm1 − δ2<br />
m2 + n2 = ℓ(m1 + n1).<br />
XL1 (̟m1+n1<br />
L1<br />
) = XK (̟ m1+n1)<br />
L2<br />
*Weareencounteringonceagain<strong>the</strong>conflictinguses<strong>of</strong><strong>the</strong>symbole ψ.
Chapter12 143<br />
isequalto<br />
XL2 (̟m2+n2<br />
L2<br />
and <br />
isequalto<br />
If<br />
σ∈G2<br />
) = XL2 (̟m1+n1<br />
<br />
F ) XL2<br />
<br />
σ∈G 2 ̟ 1−σ<br />
L2<br />
XL2 (̟1−σ<br />
<br />
) = L2<br />
µF ∈S(L1/F) µ L2/F(̟L2 )<br />
<br />
<strong>the</strong>n <br />
µF ∈S(L1/F) µF(̟F) = (−1) ℓ−1 .<br />
S ′ 1<br />
and <br />
Thuswehavetoshowthat<br />
isequalto<br />
and<br />
S ′ i = S(Li/F) − {1}<br />
µF(̟ t1+1+n<br />
F ) = (−1) n(ℓ−1)<br />
S ′ 2<br />
µF (̟ t2+1+n<br />
F ) = 1.<br />
(−1) m1(ℓ−1)<br />
∆1(XL1 , ψL1/F ̟ m1+n1<br />
F )<br />
∆1(XL2 , ψL2/F, ̟ m1+n1<br />
F ) <br />
S ′ 2<br />
In<strong>the</strong>secondandthirdcases<strong>the</strong>relations<br />
m1+n1<br />
∆1(µF, ψF, ̟ t+1+n<br />
F ).<br />
m(XK) = ℓm1 − δ(K/L1) = ℓm2 − δ(K/L2)<br />
δ(K/F) = δ(K/L1) + ℓδ1 = δ(K/L2) + ℓδ2<br />
implythat m1 + δ1 = m2 + δ2andhencethat m1 + n1 = m2 + n2.Thus<br />
XL1 (̟m1+n2<br />
L1<br />
Since <br />
) = XK (̟ m1+n1<br />
K<br />
S ′ i<br />
µF(̟ ti+1+n<br />
F ) = 1<br />
) = XL2 (̟m2+n2).<br />
L2
Chapter12 144<br />
wehavetoshowthat<br />
isequalto<br />
Suppose X ′ F<br />
isequalto<br />
and<br />
isequalto<br />
∆1(XL1 , ψL1/F, ̟ m1+n1)<br />
L1<br />
<br />
S ′ 1<br />
∆1(XL2 , ψL2/F, ̟ m2+n2)<br />
L2<br />
<br />
S ′ 2<br />
∆1(µF, ψF, ̟ t1+1+n<br />
F )<br />
∆1(µF, ψF, ̟ t2+1+n<br />
F ).<br />
isaquasicharacter<strong>of</strong> CF.AccordingtoLemma10.1<br />
∆(X ′ L1/F , ψ L1/F) <br />
<br />
µF ∈S(L1/F) ∆(µF, ψF)<br />
µF ∈S(L1/F) ∆(µF X ′ F , ψF) (12.1)<br />
∆(X ′ L2/F , ψ L2/F) <br />
<br />
µF ∈S(L2/F) ∆(µF, ψF)<br />
µF ∈S(L2/F) ∆(µF X ′ F, ψF). (12.2)<br />
Suppose m ′ = m(X ′ F ) = 2d′ + ε ′ and d ′ isgreaterthanorequaltoboth 1 + t1and 1 + t2.<br />
Choose γin Fsuchthat<br />
γ OF = P m′ +n<br />
F<br />
and<strong>the</strong>nchoose β = β(XF).ByLemma9.4<strong>the</strong>expression(12.1)isequalto<br />
and(12.2)isequalto<br />
Consequently<br />
{∆(X ′ ℓ<br />
F , ψF)}<br />
{∆(X ′ ℓ<br />
F , ψF)}<br />
∆(X ′ L1/F , ψ L1/F)<br />
<br />
<br />
<br />
µF ∈S(L1/F) µF<br />
µF ∈S(L2/F) µF<br />
µF ∈S(L1/F) µF<br />
<br />
β<br />
γ<br />
<br />
γ<br />
β<br />
<br />
γ<br />
.<br />
β<br />
<br />
∆(µF, ψF)
Chapter12 145<br />
isequalto<br />
∆(X ′ L2/F , ψ L2/F)<br />
<br />
µF ∈S(L2/F) µF<br />
<br />
β<br />
γ<br />
<br />
∆(µF, ψF) .<br />
Supposethatboth m1 = m(XL1 )and m2 = m(XL2 )areatleast2andlet mi = 2di + εi.<br />
Supposethat<br />
for iequalto1and2.Then<br />
If<br />
for xin P di+εi<br />
Li<br />
m(X −1<br />
Li X ′ Li/F ) ≤ di<br />
mi = m(X ′ Li/F ) = ψ Li/F(m ′ − 1) + 1.<br />
<strong>the</strong>n,byLemma9.4again,<br />
X ′ Li/F (1 + x) = ψ Li/F<br />
∆(XLi , ψ Li/F) = X −1<br />
Li X ′ Li/F<br />
<br />
βi<br />
γ<br />
<br />
βix<br />
γ<br />
∆ (X ′ Li/F , ψ Li/F).<br />
ThustoproveLemma12.1in<strong>the</strong>presentcircumstanceswehaveonlytoverifythat<br />
isequalto<br />
X −1<br />
L1 X ′ L1/F<br />
X −1<br />
L2 X ′ L2/F<br />
Supposefirstthat ℓisodd.Then<br />
andweneedonlyverifythat<br />
X −1<br />
L1<br />
γ<br />
β1<br />
<br />
<br />
X ′ L2/F<br />
<br />
β1<br />
<br />
γ µF ∈S(L1/F) µF<br />
<br />
β2<br />
<br />
γ µF ∈S(L2/F) µF<br />
µF ∈S(Li/F) µF<br />
γ<br />
β2<br />
<br />
= X −1<br />
L2<br />
<br />
γ<br />
= 1<br />
β<br />
γ<br />
AccordingtoLemmas8.3and8.4wemaytake β1 = β2 = β.<br />
β2<br />
<br />
<br />
γ<br />
β<br />
<br />
γ<br />
.<br />
β<br />
X ′ L2/F<br />
γ<br />
β2<br />
<br />
.
Chapter12 146<br />
Certainly<br />
X ′ L1/F<br />
<br />
γ<br />
= X<br />
β<br />
′ F<br />
ℓ γ<br />
Since CFis<strong>the</strong>product<strong>of</strong> N L1/FCL1 and N L2/FCL2<br />
Consider<br />
β ℓ<br />
= X ′ L2/F<br />
γ<br />
β = N L1/F δ1 N L2/F δ2.<br />
XLi (N Lj/Fδj) = XK(δj)<br />
where jis1or2accordingas iis2or1.Therightsideequals<br />
Theproductisequalto<br />
whichis1because ℓisodd.<br />
XLj (δℓ j) = XLj (N Lj/F δj) <br />
<br />
<br />
γ<br />
.<br />
β<br />
wemaywrite γ<br />
β asaproduct<br />
σ∈G(Lj/F)<br />
µF ∈S(Li/F) µ Lj/F(δj)<br />
XLj (δ1−σ<br />
j ).<br />
Beforediscussing<strong>the</strong>case ℓ = 2weconsider<strong>the</strong>circumstancesunderwhich,foragiven<br />
XL1 and XL2 ,aquasicharacter X ′ F with<strong>the</strong>propertiesdescribedaboveexists.<br />
Lemma 12.2<br />
(a)If Li/Fisunramified, xbelongsto U ui+1<br />
,and Li<br />
<strong>the</strong>n<br />
N Li/F(x) = 1<br />
XLi (x) = 1.<br />
(b)If Li/Fisramified, K/Liisunramified, xbelongsto U ti+1<br />
,and Li<br />
<strong>the</strong>n<br />
N Li/F(x) = 1<br />
XLi (x) = 1.<br />
(c)If Li/Fand K/Liareramified, xbelongsto U ui+ti+1<br />
,and<br />
Li<br />
N Li/F(x) = 1
Chapter12 147<br />
<strong>the</strong>n<br />
XLi (x) = 1.<br />
Choosesomenontrivial σiin G i = G(Li/F).Then<br />
isanontrivialcharacterin S(K/Li)and<br />
µLi<br />
= X σ−1<br />
i −1<br />
Li<br />
m(µLi ) = ui + 1.<br />
Since Li/Fiscyclic<strong>the</strong>reisayin CLi suchthat<br />
x = y σi−1 .<br />
Weshallshowthat ycanbetakenin U ui+1<br />
.Then<br />
Li<br />
XLi (x) = µLi (y) = 1.<br />
Suppose Li/Fisunramified. Ifwecannotchoose yin U ui+1<br />
<strong>the</strong>reisalargestinteger<br />
Li<br />
a ≥ −1suchthatwecanchoose yin Ua Liwhere ais<strong>of</strong>courselessthan ui + 1.Choosesucha<br />
y.Then aisnot1becausewecanalwaysdivide ybyapower<strong>of</strong> ̟F.If awere0<strong>the</strong>n ycould<br />
notbecongruenttoanelement<strong>of</strong> UF modulo PF. Then yσi−1 1 wouldnotbein UF . Since<br />
ui + 1 > 0in<strong>the</strong>presentsituationthisisimpossible.Let<br />
y = 1 + ε̟ a F .<br />
Then εcannotbecongruenttoanelement<strong>of</strong> OFmodulo PF.Thus<br />
and<br />
isnotin U a+1<br />
.Thisisacontradiction.<br />
Li<br />
ε σi − ε ≡ 0 (modPLi )<br />
y σi−1 ≡ 1 + (ε σi − ε) ̟ a F (modP a+1<br />
Li )<br />
Nowsuppose Li/Fisramifiedand K/Liisunramified.Then ti + 1 ≥ 1and ui + 1 = 0.<br />
Weneedonlyshowthat ycanbetakentobeaunit.Write<br />
y = ε̟ b Li
Chapter12 148<br />
where εisaunit. If biscongruentto0modulo ℓwecandivide ybysomepower<strong>of</strong> ̟Fto<br />
obtainanelement<strong>of</strong> UF = U0 F .Toseethat bmustbecongruentto0modulo ℓwesuppose<strong>the</strong><br />
contrary.Then<br />
y σi−1 σi−1 σi−1<br />
= ε (̟ ) Li<br />
b ≡ (̟ σi−1<br />
) Li<br />
b<br />
(modP ti+1<br />
). Li<br />
If ti = 0<strong>the</strong>residue<strong>of</strong> ̟ σi−1<br />
Li<br />
If ti > 0<strong>the</strong>n<br />
where αisaunit.Thus<br />
modulo PLi isanontrivial ℓthroot<strong>of</strong>unityand<br />
(̟ σi−1<br />
Li ) b ≡ 1 (modPLi ).<br />
̟ σi−1<br />
Li<br />
= 1 + α ̟ti<br />
Li<br />
(̟ σi−1<br />
Li ) b ≡ 1 + αb ̟ ti<br />
Li (mod P ti+1<br />
Li ).<br />
Therightsideisnotcongruentto0modulo P ti+1<br />
. Li<br />
Nowsuppose Li/Fand K/Liarebothramified. Then ℓ = pandboth uiand tiareat<br />
least1.Againsupposethat ycannotbechosenin U ui+1<br />
andlet abe<strong>the</strong>largestintegersuch<br />
Li<br />
that ycanbechosenin Ua .Theargumentjustusedshowsthat a ≥ 0.Since Li/Fisramified<br />
U kp<br />
Li = Uk F Ukp+1<br />
Li .<br />
Therefore aisnotdivisibleby pandinparticularisatleast1.Let<br />
where βisaunit.Then<br />
Let<br />
and<br />
y = 1 + ε̟ a Li<br />
y σi−1 = (1 + ε σi ̟ aσi<br />
Li ) (1 + ε̟a Li )−1 .<br />
ε σi = ε + η ̟ ti+1<br />
Li<br />
̟ σi−1<br />
Li<br />
where αisaunit.Then y σi−1 isequalto<br />
= 1 + α ̟ti<br />
Li<br />
ti+1<br />
1 + (ε + η ̟ ) (1 + α̟ Li ti<br />
Li )a̟ a <br />
a −1<br />
1 + ε̟ Li<br />
Li
Chapter12 149<br />
whichiscongruentto<br />
1 + aαε̟ ti+a<br />
Li<br />
modulo P ti+a+1<br />
.Therefore a ≥ ui + 1.Thisisacontradiction.<br />
Li<br />
Lemma 12.3<br />
and<br />
If L1/Fisunramifiedwecanchoose X ′ F suchthat<br />
m(X −1<br />
L1 X ′ L1/F ) = t + 1<br />
m(X −1<br />
L2 X ′ L2/F ) = t + 1.<br />
If m(XL1 ) > t + 1<strong>the</strong>n m(X ′ F )willequal m(XL1 ).<br />
By<strong>the</strong>previouslemmawecandefineaquasicharacter X ′ F <strong>of</strong><br />
bysetting<br />
N L1/FU u1+1<br />
L1<br />
X ′ F(N L1/Fx) = XL1 (x).<br />
Weextend X ′ Ftoaquasicharacter,whichweagaindenoteby X ′ F ,<strong>of</strong> CF.Then<br />
m(X −1<br />
L1 X ′ L1/F ) ≤ u1 + 1.<br />
However X −1<br />
K X ′ −1<br />
K/F , XL1 X ′ −1<br />
L1/F ,and XL2 X ′ L2/Fsatisfy<strong>the</strong>conditions<strong>of</strong>Lemma12.1.There fore<br />
m(X −1<br />
L1 X ′ L1/F ) ≥ u1 + 1.<br />
Since L1/Fisunramified u1and t2arebo<strong>the</strong>qualto t.Thus<br />
and<br />
m(X −1<br />
L1 X ′ L1/F ) = t + 1<br />
m(X −1<br />
L2 X ′ L2/F ) = ℓ(u1 + 1) − δ2 = ℓ(u1 + 1) − (ℓ − 1) (t2 + 1) = t + 1.<br />
Thelastassertion<strong>of</strong><strong>the</strong>lemmaisclear.<br />
Lemma 12.4
Chapter12 150<br />
If K/Fistotallyramified<strong>the</strong>n<br />
Thereexists X ′ F suchthat<br />
for iequalto1and2.<br />
m(XLi ) ≥ ti + ui + 1.<br />
m(X −1<br />
Li X ′ Li/F ) = ti + ui + 1<br />
In<strong>the</strong>presentcircumstances tiand uiarebothatleast1. Chooseanontrivial σiin G i<br />
andlet<br />
asbefore.Choose yin U ui<br />
Li sothat<br />
Then<br />
Howeverif<br />
where εisaunit<strong>the</strong>n<br />
if<br />
Inparticular<br />
sothat<br />
µLi<br />
= X σ−1<br />
i −1<br />
Li<br />
µLi (y) = 1.<br />
XLi (yσi−1 ) = 1.<br />
y = 1 + ε̟ ui<br />
Li<br />
y σi−1 ≡ 1 + uiα ε̟ ti+ui<br />
Li<br />
̟ σi−1<br />
Li<br />
= 1 + α ̟ti<br />
Li .<br />
y σi−1 ∈ U ti+ui<br />
Li<br />
m(XLi ) ≥ ti + ui + 1.<br />
(modP ti+ui+1<br />
) Li<br />
Justasin<strong>the</strong>previouslemmawecanfindaquasicharacter X ′ F<br />
m(X −1<br />
L1 X ′ L1/F ) = t1 + u1 + 1.<br />
Wehaveseenthat m1 + δ1 = m2 + δ2.Thesameargumentshowsthat<br />
m(X −1<br />
L1 X ′ L1/F ) + δ1 = m(X −1<br />
L2 X ′ L2/F ) + δ2.<br />
<strong>of</strong> CFsuchthat
Chapter12 151<br />
Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong><strong>the</strong>lemmaweshowthat<br />
Since<br />
wehaveonlytoshowthat<br />
t1 + u1 + δ1 = t2 + u2 + δ2.<br />
δi = (ℓ − 1) (ti + 1)<br />
u1 + 1 + ℓ(t1 + 1) = u2 + 1 + ℓ(t2 + 1).<br />
Multiplying<strong>the</strong>leftor<strong>the</strong>rightsideby ℓ − 1weobtain δ(K/F). Theequalityfollows<br />
immediately.<br />
Lemmas12.3and12.4toge<strong>the</strong>rwith<strong>the</strong>remarkswhichprovoked<strong>the</strong>mallowustoprove<br />
Lemma12.1inmany,butbynomeansall,cases. Weshallnothoweverapply<strong>the</strong>selemmas<br />
immediately.Weshallra<strong>the</strong>rbegin<strong>the</strong>systematicexposition<strong>of</strong><strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma12.1taking<br />
up<strong>the</strong>casestowhich<strong>the</strong>selemmasapplyin<strong>the</strong>irturn.<br />
Supposefirstthat L1isunramifiedover F.Asbefore mi = m(XLi ).Then<br />
m2 = m1 + (ℓ − 1) (m1 − t − 1) ≥ m1<br />
because u1 = t.Since<strong>the</strong>number m1isatleast t + 1and t ≥ 0itisatleast1.If m1 = 1<strong>the</strong>n<br />
t = 0and m2 = 1.<strong>On</strong>cewehavetreatedthiscase,asweshallimmediately,wemaysuppose<br />
that m2 ≥ m1 > 1.<br />
andlet<br />
If m2 = 1let<br />
λ = OL2 /PL2 = OF/PF<br />
κ = OK/PK = OL1 /PL1 .<br />
κisanextension<strong>of</strong> λ. Therestriction<strong>of</strong> XL1 to UL1 definesacharacter Xλ<strong>of</strong> λ ∗ and<strong>the</strong><br />
restriction<strong>of</strong> XL2 to UKdefinesacharacter Xκ<strong>of</strong> κ ∗ . Therestriction<strong>of</strong> XKto UKdefinesa<br />
character<strong>of</strong> κ ∗ whichisequalto X κ/λandto X ℓ κ sothat<br />
X ℓ κ = X κ/λ.<br />
As σvariesover G 2 , ̟ σ−1<br />
,takenmodulo PL2 ,variesover<strong>the</strong> ℓthroots<strong>of</strong>unityin λandif<br />
L2<br />
X σ−1 −1<br />
L2<br />
= ν
Chapter12 152<br />
<strong>the</strong>n<br />
Xλ(̟ σ−1<br />
) = ν(̟L2 ).<br />
L2<br />
Therightsideisnot1if σ = 1because νis<strong>the</strong>nnontrivial.Thus<strong>the</strong>restriction<strong>of</strong> Xλto<strong>the</strong> ℓth<br />
roots<strong>of</strong>unityisnottrivial.Toevery µFin S(L2/F)isassociatedacharacter µλ<strong>of</strong> λ ∗ which<br />
is<strong>of</strong>order1if µF = 1and<strong>of</strong>order ℓo<strong>the</strong>rwise.If ψλis<strong>the</strong>additivecharacter<strong>of</strong> λdefinedby<br />
<strong>the</strong>n<br />
if µFisnottrivial.Moreover<br />
and<br />
Finally<br />
ψλ(x) = ψF<br />
<br />
x<br />
̟ 1+n<br />
F<br />
<br />
∆1(µF, ψF, ̟ 1+n<br />
F ) = A [−τ(µλ, ψλ)]<br />
ψ L2/F<br />
<br />
x<br />
̟ 1+n<br />
<br />
= ψλ(ℓ x)<br />
F<br />
∆1(XL2 , ψ L2/F, ̟ 1+n<br />
F ) = A [−Xλ(ℓ)τ(Xλ, ψλ)].<br />
∆1(XL1 , ψ L1/F, ̟ 1+n<br />
F ) = A [−τ(Xκ, ψ κ/λ)].<br />
Thus<strong>the</strong>requiredidentityisaconsequence<strong>of</strong><strong>the</strong>relation<br />
whichweprovedasLemma7.9.<br />
τ(Xκ, ψ κ/λ) = Xλ(ℓ)τ(Xλ, ψλ) <br />
µλ=1 τ(µλ, ψλ)<br />
Retaining<strong>the</strong>assumptionthat L1/Fisunramifiedwenowsupposethat m1 > 1.There<br />
aretwopossibilities.<br />
(a)<br />
(b)<br />
Thesecondpossibilityoccursonlywhen t > 0.<br />
m1 ≥ 2(t + 1)<br />
t + 1 ≤ m1 < 2(t + 1).
Chapter12 153<br />
If mj ≥ 2(t + 1)choose X ′ F sothat<br />
for i = 1and2.Itisclearthat<br />
if mi = 2di + εi.Since m2 ≥ m1wealsohave<br />
Moreover<br />
m(X −1<br />
Li X ′ Li/F ) = t + 1<br />
m(X −1<br />
L1 X ′ L1/F ) ≤ d1<br />
m(X −1<br />
L2 X ′ L2/F ) ≤ d2.<br />
m ′ = m(X ′ F ) = m1<br />
sothat d ′ isgreaterthanorequaltoboth 1 + t2 = 1 + tand 1 + t1 = 0.Lemma12.1for L 1/F<br />
unramifiedand m1 ≥ 2(t + 1),followsimmediatelyif ℓisodd.Suppose ℓ = 2.<br />
If t = 0wecaninvokeLemmas8.3and8.7toseethatif β = β(X ′ F )wemaychoose<br />
β1 = β(χL1/F)and β2 = β(X ′ L2/F )equalto β.If µ(1)<br />
F is<strong>the</strong>nontrivialelement<strong>of</strong> S(L1/F)<br />
and µ (2)<br />
F<br />
isequalto<br />
Certainly<br />
is<strong>the</strong>nontrivialelement<strong>of</strong> S(L2/F)wehaveonlytoshowthat<br />
XL1<br />
XL2<br />
<br />
γ<br />
β<br />
<br />
γ<br />
β<br />
X ′ L1/F<br />
X ′ L1/F<br />
X ′ L2/F<br />
andweneedonlyshowthatif δisin CF<strong>the</strong>n<br />
Wemaywrite<br />
Then<br />
XL1<br />
(δ) µ(1)<br />
F<br />
<br />
β<br />
γ<br />
<br />
β<br />
γ<br />
<br />
β<br />
= X<br />
γ<br />
′ L2/F<br />
µ (1)<br />
F<br />
µ (2)<br />
F<br />
<br />
β<br />
γ<br />
<br />
γ<br />
β<br />
<br />
γ<br />
.<br />
β<br />
(δ) = XL2 (δ) µ(2)<br />
F (δ).<br />
δ = N L1/F δ1 N L2/F δ2.<br />
µ (i)<br />
F (N L1/F δi) = 1
Chapter12 154<br />
and,if jis1or2accordingas iis2or1,<br />
whichequals<br />
XLi (N Lj/F δj) = XK(δj) = XLj (δ2 j )<br />
XLj (N Lj/F δj)µ (i)<br />
Lj/F (δj) = XLj (N Lj/F δj)µ (i)<br />
F (N Lj/F δj).<br />
Therequiredequalityfollowsimmediately.<br />
If tispositivewemaystillchoose β1 = β.If m1 − t − 1 = v<strong>the</strong>n,byLemma8.6,wemay<br />
choose β2in<strong>the</strong>form<br />
β2 = β + η<br />
with ηin Pv .Since v ≥ t + 1<br />
L2<br />
X −1<br />
L2 (β2)X ′ L2/F (β2) = X −1<br />
L2 (β)X ′ L2/F (β).<br />
Thisobservationmade,wecanproceedasbefore.<br />
Somepreparationisnecessarybeforewediscuss<strong>the</strong>secondpossibility.Supposethat tis<br />
positivesothat ℓisequalto<strong>the</strong>residualcharacteristic p.Thefinitefield λ1 = OL1 /PL1isan extension<strong>of</strong>degree p<strong>of</strong> φ = OF/PF.<br />
Themap<br />
x −→ x p − x<br />
isanadditiveendomorphism<strong>of</strong> φwith<strong>the</strong>primefieldaskernel.Chooseayin φwhichisnot<br />
in<strong>the</strong>image<strong>of</strong>thismapandconsider<strong>the</strong>equation<br />
x p − x = y.<br />
If x,insomeextensionfield<strong>of</strong> φ,satisfiesthisequationand φhas qelements<strong>the</strong>n x q = x.<br />
However<br />
(x q − x) p − (x q − x) = (x p − x) q − (x p − x) = y q − y = 0.<br />
So<br />
x q − x = z<br />
where zisanonzeroelement<strong>of</strong><strong>the</strong>primefield.Then<br />
x q2<br />
= (x + z) q = x q + z = x + 2z
Chapter12 155<br />
andingeneral<br />
x qn<br />
= x + nz.<br />
Thus<strong>the</strong>lowestpower n<strong>of</strong> qsuchthat xqn = xis n = pand xdeterminesanextension<strong>of</strong><br />
degree p.Consequently xmaybechosentoliein λ1and<strong>the</strong>n λ1 = φ(x).<br />
wehave<br />
Let E r (x)be<strong>the</strong> r<strong>the</strong>lementarysymmetricfunction<strong>of</strong> xanditsconjugates.Since<br />
if 1 ≤ r < p − i,<br />
and,<strong>of</strong>course,<br />
x p − x + (−1) p N λ1/φx = 0 (12.3)<br />
E r (x) = 0 (12.4)<br />
E p−1 (x) = (−1) p<br />
(12.5)<br />
E p (x) = N λ1/φx. (12.6)<br />
If λisanonzeroelement<strong>of</strong><strong>the</strong>primefieldwecanreplace yby λy.Then xistobereplaced<br />
by λx.Alsowecanreplace xby x + λwithoutchanging y.<br />
Let R(L1)be<strong>the</strong>set<strong>of</strong> (q p − 1)throots<strong>of</strong>unityin L1. Chooseaγin R(L1)whose<br />
imagein λ1is x.Ifwearedealingwithfields<strong>of</strong>powerseries γwillalsosatisfy<strong>the</strong>equations<br />
(12.3),(12.4),(12.5)and(12.6).Letusseehow<strong>the</strong>seequationsaretobemodifiedforfields<strong>of</strong><br />
characteristiczero. Fand L1are<strong>the</strong>nextensions<strong>of</strong><strong>the</strong> padicfield Qp.Let F 0 and L 0 1 be<strong>the</strong><br />
maximalunramifiedextensions<strong>of</strong> Qpcontainedin Fand L1respectively. R(L1)isasubset<br />
<strong>of</strong> L 0 1 and pgenerates<strong>the</strong>ideal P F 0and<strong>the</strong>ideal P L 0 1 .Thus<br />
and<br />
if 1 ≤ r < p − 1while<br />
Let<br />
γ p − γ + (−1) p N L1/F γ ≡ 0 (modp)<br />
E r (γ) ≡ 0 (modp)<br />
E p−1 (γ) ≡ (−1) p<br />
S r (γ) = <br />
(modp).<br />
σ∈G(L1/F) γrσ .<br />
Thefollowingrelationsarespecialcases<strong>of</strong>Newton’sformulae.
Chapter12 156<br />
Weinferthat<br />
if 1 ≤ r < p − 1andthat<br />
S 1 (γ) − E 1 (γ) = 0<br />
S 2 (γ) − E 1 (γ) S 1 (γ) + 2E 2 (γ) = 0<br />
S 3 (γ) − E 1 (γ) S 2 (γ) + E 2 (γ) S 1 (γ) − 3E 3 (γ) = 0<br />
.<br />
S p−1 (γ) − E 1 (γ) S p−2 (γ) + . . . + (−1) p−1 (p − 1)E p−1 (γ) = 0<br />
S p (γ) − E 1 (γ) S p−1 (γ) + . . . + (−1) p p E p (γ) = 0.<br />
S r (γ) ≡ 0 (modp)<br />
S p−1 (γ) ≡ (−1) p (p − 1) E p−1 (γ) (mod p).<br />
Combining<strong>the</strong>first<strong>of</strong><strong>the</strong>secongruenceswithNewton’sformulaeweobtain<br />
if 1 ≤ r ≤ p − 1.If pisodd<br />
Therightsideisequalto<br />
If piseven<br />
Since<br />
wehave<br />
S r (γ) ≡ r(−1) r+1 E r (γ) (modp 2 )<br />
S p (γ) − p E p (γ) ≡ E 1 (γ) S p−1 (γ) − E p−1 (γ) S 1 (γ) (mod p 2 ).<br />
E 1 (γ) (S p−1 (γ) − E p−1 (γ)) ≡ 0 (modp 2 ).<br />
S 2 (γ) + 2E 2 (γ) = {E 1 (γ)} 2 .<br />
E 1 (γ) ≡ 1 (mod2)<br />
S 2 (γ) + 2E 2 (γ) ≡ 1 (mod4).<br />
If σ = 1belongsto G(L1/F)<strong>the</strong>reisa(p − 1)throot<strong>of</strong>unity ζsuchthat<br />
γ σ − γ ≡ ζ (modp).
Chapter12 157<br />
Byasuitablechoice<strong>of</strong> y<strong>the</strong>root ζcanbemadetoequal,foragiven σ,anychosen (p − 1)th<br />
root<strong>of</strong>unity.<br />
Theaboverelationsare<strong>of</strong>coursealsovalidwhen Fisafield<strong>of</strong>powerseries.<br />
Chooseanontrivialcharacter µFin S(L2/F)andchoose αsothat<br />
if xisin P s F<br />
unitywedefine µ ζ<br />
F<br />
µF(1 + x) = ψF<br />
<br />
α x<br />
̟ t+1+n<br />
F<br />
<br />
t+1<br />
.Here sis<strong>the</strong>leastintegergreaterthanorequalto 2 .If ζisa(p − 1)throot<strong>of</strong><br />
tobe µj<br />
F<br />
Asweobservedin<strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma8.5<br />
if xisin P s F .<br />
if jis<strong>the</strong>uniqueintegersuchthat<br />
ζ ≡ j (modp).<br />
µ ζ<br />
F (1 + x) = ψF<br />
Let mi = 2di + εiasusual.If β1in L1issuchthat<br />
for xin P d1+ε1<br />
L1<br />
isequalto<br />
if ζσissuchthat<br />
Thusif<br />
XL1 (1 + x) = ψ L1/F<br />
<br />
αζx<br />
̟ 1+t+n<br />
<br />
F<br />
αβ1x<br />
̟ m1+n1<br />
F<br />
<strong>the</strong>n,if σ = 1belongsto G 1 = G(L1/F)and xbelongsto P d1+ε1,<br />
L1<br />
ψ L1/F<br />
σ α(β1 − β1)x<br />
̟ m1+n2<br />
F<br />
ψ L1/F<br />
X σ−1<br />
L1<br />
<br />
αζσx<br />
̟ 1+t+n<br />
<br />
F<br />
= µζσ<br />
F .<br />
v = m1 − 1 − t<br />
<br />
= X σ−1<br />
L1 (1 + x)
Chapter12 158<br />
wehave<br />
Suppose<br />
Itisclearthat<br />
β σ 1 − β1 ≡ ζσ ̟ v F (mod P d1<br />
L1 ).<br />
ζστ ≡ ζ τ σ + ζσ (modPL1 ).<br />
γ σ − γ ≡ ξσ (modp)<br />
where ξσisalsoa(p − 1)throot<strong>of</strong>unity.Then<br />
Weobservedthatwecouldarrangethat<br />
ξστ ≡ ξ τ σ + ξσ (mod PL1 ).<br />
ξσ = ζσ<br />
foronenontrivial σ. <strong>On</strong>cewedothis<strong>the</strong>equalitywillholdforall σ. Then γpσ − γp ≡<br />
ξσ (mod p)and<br />
(β1 − γ p ̟ v F )σ ≡ β1 − γ p ̟ v F (modP d1<br />
L1 )<br />
forall σbecause,asweobservedin<strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma8.5, pbelongsto Pr if r + s = t + 1<br />
L1<br />
and<br />
2(r + v) ≥ t + 2v = 2m1 − 2 − 2t + t<br />
whichisatleast<br />
(m1 − 1) + (m1 − 1 − t) ≥ m1 − 1<br />
sothat r + v ≥ d1.Since L1/Fisunramified<strong>the</strong>reis<strong>the</strong>reforeaβin Fsuchthat<br />
Wemaysupposethat<br />
β1 − γ p ̟ v F<br />
≡ β (modPd1<br />
L1 ).<br />
β1 = β + γ p ̟ v F .<br />
βisaunitunless v = 0. If v = 0<strong>the</strong>n,byreplacing γbyaroot<strong>of</strong>unitycongruentto γ + 1<br />
moduloPL1 ifnecessary,wecanstillarrangethat βisaunit. βiscongruenttoanorm N L2/Fβ ′<br />
modulo P t F .Since d1 ≤ twe ∗<br />
∗ (1998)At<strong>the</strong>momentthisisallthatcouldbefound<strong>of</strong>Chapter12.
Chapter13 159<br />
Chapter Thirteen.<br />
The Third Main Lemma<br />
Suppose K/FisGaloisand G = G(K/F).Suppose G = HCwhen H = {1}, H ∩ C =<br />
{1},andCisanontrivialabeliannormalsubgroup<strong>of</strong>Gwhichiscontainedineverynontrivial<br />
normalsubgroup<strong>of</strong> G.<br />
Lemma 13.1<br />
Let Ebe<strong>the</strong>fixedfield<strong>of</strong> Handlet XFbeaquasicharacter<strong>of</strong> CF.If m = m(XF)<strong>the</strong>n<br />
Set<br />
m(X E/F) = ψ E/F(m − 1) + 1.<br />
m ′ = m(X E/F) − 1.<br />
Observethat m ′ − 1is<strong>the</strong>greatestlowerbound<strong>of</strong>allrealnumbers v > −1suchthat XE/Fis andthat m − 1is<strong>the</strong>greatestlowerbound<strong>of</strong>allrealnumbers usuchthat XFis<br />
trivialon Uv E<br />
trivialon Uu F .Since<br />
weseeimmediatelythat<br />
Toprove<strong>the</strong>lemmaweneedonlyshowthat<br />
NE/F(U ψE/F(u) E ) ⊆ U u F<br />
m ′ − 1 ≤ ψ E/F(m − 1).<br />
NE/F(U ψE/F(m−1) E ) ⊇ U m−1<br />
F .<br />
Weshowthiswith m − 1replacedbyany u ≥ −1.<br />
ByLemma6.15, τK/Fmaps W u K/Fonto Uu F . Theprojection<strong>of</strong> W u K/Fon Gisanormal<br />
subgroup<strong>of</strong> G.Thusitisei<strong>the</strong>r {1}orasubgroupcontaining C.Ifitis {1}<strong>the</strong>n<br />
and<br />
W u K/F = W u K/F ∩ CK = U ψ K/F(u)<br />
K<br />
U u F = N K/F(U ψ K/F(u)<br />
K<br />
) = NE/F(NK/EU ψK/F(u) K )
Chapter13 160<br />
which,byLemma6.6,iscontainedin<br />
NE/F(U ψE/F(u) E ).<br />
Suppose<strong>the</strong>projectionisnot {1}.If Lis<strong>the</strong>fixedfield<strong>of</strong> C<strong>the</strong>group W u K/F contains<br />
Since Cisgeneratedby<br />
{wvw −1 v −1 | w ∈ W K/F, v ∈ W u K/F ∩ W K/L}. (13.1)<br />
{σρσ −1 ρ −1 | σ ∈ G, ρ ∈ C}<br />
<strong>the</strong>groupgeneratedby<strong>the</strong>set(13.1)containsaset<strong>of</strong>representativesfor<strong>the</strong>cosets<strong>of</strong> CKin<br />
W K/L.Thisgroupclearlyliesin<strong>the</strong>kernel<strong>of</strong> τ K/F.Thuseveryelement<strong>of</strong> W u K/F iscongruent<br />
modulo<strong>the</strong>kernel<strong>of</strong> τ K/Fin W u K/F toanelement<strong>of</strong><br />
and<br />
whichis<br />
andthissetiscontainedin<br />
W u K/F ∩ W K/E = W ψ E/F(u)<br />
K/E<br />
U u F = τK/F(W u K/F ) = τK/F(W ψE/F(u) K/E )<br />
NE/F(τK/E(W ψE/F(u) K/E ))<br />
NE/F(U ψE/F(u) E ).<br />
Suppose F1isnonarchimedean, K1/F1isGaloisand F1 ⊆ E1 ⊆ K1. Let µ1bea<br />
character<strong>of</strong> G(K1/E1).Wemayalsoregard µ1asacharacter<strong>of</strong> CE1 .Let σbeanelement<strong>of</strong><br />
G(K1/F1)anddefine<strong>the</strong>character<strong>of</strong> µ σ 1 <strong>of</strong> G(K1/E σ 1 )by<br />
µ σ 1(ρ) = µ1(σρσ −1 )<br />
for ρ ∈ G(K1/E σ 1 )or,whatamountsto<strong>the</strong>same<br />
for α ∈ CE σ 1 .Since<br />
µ σ 1(α) = µ1(α σ−1<br />
)<br />
ψEσ 1 /F1 (α) = ψE1/F1 (ασ−1)<br />
<strong>the</strong>nextlemmaisacongruence<strong>of</strong><strong>the</strong>definitions.
Chapter13 161<br />
Lemma 13.2<br />
∆(µ σ 1, ψ E σ 1 /F1 ) = ∆(µ1, ψ E1/F1 ).<br />
Wereturnto<strong>the</strong>extension K/Land<strong>the</strong>group G.Let Tbeaset<strong>of</strong>representativesfor<strong>the</strong><br />
orbitsunder G<strong>of</strong><strong>the</strong>nontrivialcharactersin S(K/L).If µ ∈ Tlet Gµbe<strong>the</strong>isotropygroup<br />
<strong>of</strong> µandlet Fµbe<strong>the</strong>fixedfield<strong>of</strong> Gµ.Let Hµ = H ∩Gµ.Since Ciscontainedin Gµwehave<br />
Gµ = Hµ · C.Then µmayalsoberegardedasacharacter<strong>of</strong> C.Let µ ′ be<strong>the</strong>character<strong>of</strong> Gµ<br />
definedby<br />
µ ′ (hc) = µ(c)<br />
if h ∈ Hµand c ∈ C.Eventuallywemustshowthat<br />
isequalto<br />
∆(X E/F, ψ E/F) <br />
∆(XF, ψF) <br />
µ∈T ∆(µ′ , ψ Fµ/F) (13.2)<br />
µ∈T ∆(µ′ X Fµ/F, ψ Fµ/F) (13.3)<br />
if XFisaquasicharacter<strong>of</strong> CF.At<strong>the</strong>momentwecontentourselveswithaspecialcase.The<br />
nextlemmawillbereferredtoas<strong>the</strong>ThirdMainLemma.<br />
Lemma 13.3<br />
If K/Fistamelyramified<strong>the</strong>expressions(13.2)and(13.3)areequal.<br />
AsweobservedinLemma6.4<strong>the</strong>extension L/Fwillbeunramifiedand ℓ = [C : 1]<br />
willbeaprime. Chooseagenerator ̟F<strong>of</strong> PF. Since Fµ/Fisunramifiedwemaychoose<br />
̟Fµ = ̟F.Choose ̟Esothat N E/F ̟E = ̟F.Certainly<br />
while<br />
Since<br />
weconcludethat<br />
δ L/F = δ K/E = 0<br />
δ K/L = ℓ − 1.<br />
δ K/F = δ K/L + ℓ δ L/F = δ K/E + δ E/F<br />
δ E/F = ℓ − 1.
Chapter13 162<br />
Clearly<br />
<br />
µ [Fµ : F] = <br />
µ [H : Hµ] = <br />
µ [G : Gµ]<br />
isjust<strong>the</strong>number<strong>of</strong>nontrivialcharactersin S(K/L),thatis ℓ − 1.Moreover m(µ ′ ) = 1.Let<br />
Eµbe<strong>the</strong>fixedfield<strong>of</strong> Hµ.Then<br />
N Eµ/Fµ (̟E) = N E/F(̟E) = ̟F.<br />
Thus,asanelement<strong>of</strong> CFµ , ̟Fliesin<strong>the</strong>image<strong>of</strong> W K/Eµ under τ K/Fµ andhence µ′ (̟F) =<br />
1.Also<br />
n(ψ E/F) = ℓn(ψF) + δ E/F = ℓn + (ℓ − 1)<br />
while<br />
and<br />
If m = m(XF) = 0<strong>the</strong>n<br />
X E/F(̟ ℓn+ℓ−1<br />
E<br />
sothat<strong>the</strong>lemmaamountsto<strong>the</strong>equality<br />
and<br />
<br />
If m > 0<strong>the</strong>n,byLemma6.4,<br />
Since K/Eisunramified<br />
However<br />
n(ψ Fµ/F) = n.<br />
m(X E/F) = m(X Fµ/F) = 0<br />
) = XF(̟ ℓn+ℓ−1<br />
F ) = XF(̟ n <br />
F )<br />
µ XFµ/F(̟ 1+n<br />
F )<br />
µ ∆1(µ, ψ Fµ/F, ̟ 1+n<br />
F ) = <br />
µ ∆1(µ, ψ Fµ/F, ̟ 1+n<br />
F ).<br />
m(X E/F) = ℓm − (ℓ − 1)<br />
m(X E/F) + n(ψ E/F) = ℓ(m + n).<br />
m(X K/F) = m(X E/F) = ℓm − (ℓ − 1).<br />
X K/F = (µ ′ X Fµ/F) K/Fµ
Chapter13 163<br />
sothat<br />
or<br />
Consequently<br />
ψ K/Fµ (m(µ′ X Fµ/F) − 1) ≥ m(X K/F) − 1 = ℓ(m − 1)<br />
m(µ ′ XFµ/F) − 1 ≥ ϕK/Fµ (ℓ(m − 1)) = m − 1.<br />
m(µ ′ X Fµ/F) ≥ m.<br />
Sinceitisclearlylessthanorequalto mitisequalto m.Because<br />
X E/F(̟ m+n<br />
wehavetoshowthat<br />
isequalto<br />
F ) = XF(̟ ℓ(m+n)<br />
F ) = XF(̟ m+n<br />
F ) <br />
µ XFµ/F(̟ m+n<br />
F )<br />
∆1(X E/F, ψ E/F, ̟ m+n<br />
F ) ∆1(µ ′ , ψ Fµ/F, ̟ m+n<br />
F )<br />
∆1(XF, ψF, ̟ m+n<br />
F ) ∆1(µ ′ X Fµ/F, ψ Fµ/F, ̟ m+n<br />
F ).<br />
Let φbe<strong>the</strong>field OF/PF,let λ = OL/PL,let qbe<strong>the</strong>number<strong>of</strong>elementsin φ,andlet<br />
f = [λ : φ] = [L : F].<br />
Let θbe<strong>the</strong>homomorphism<strong>of</strong> Cinto λ ∗ introducedinChapterIV<strong>of</strong>Serre’sbook.Thus<br />
sothatif h ∈ H<br />
Let h0bethatelement<strong>of</strong> Hsuchthat<br />
θ(c) = ̟ c−1<br />
E<br />
(modPF)<br />
θ(h −1 ch) ≡ (̟ h−1c−h −1<br />
E ) ≡ (̟ c−1<br />
E )h ≡ θ(c) h .<br />
α h0 = α q<br />
if α ∈ λandlet c0beagenerator<strong>of</strong> C.Then θ(c0)hasorder ℓand,since<strong>the</strong>centralizer<strong>of</strong> C<br />
in His {1},<br />
θ(h −r<br />
0 chr 0) = θ(c0) qr<br />
is θ(c0)ifandonlyif fdivides r.<strong>On</strong><strong>the</strong>o<strong>the</strong>rhand,itis θ(c0)ifandonlyif ℓdivides q r − 1.<br />
Thus<strong>the</strong>order<strong>of</strong> qmodulo ℓis f.Wealsoobservethatboth Canditsdualgrouparecyclic<strong>of</strong><br />
primeordersothatanyelement<strong>of</strong> Hwhichfixedanelement<strong>of</strong> Twouldacttriviallyon<strong>the</strong><br />
dualgroupand<strong>the</strong>reforeon Citself.Itfollowsthat Fµ = Lforall µin T.
Chapter13 164<br />
Supposefirstthat m = 1.Let ψφbe<strong>the</strong>character<br />
ψφ(x) = ψF<br />
<br />
x<br />
̟ 1+n<br />
F<br />
<br />
on φ.Since OE/PEisnaturallyisomorphicto OF/PFand<strong>the</strong>map x −→ N E/Fxgives<strong>the</strong><br />
map x −→ x ℓ <strong>of</strong> φintoitselfwhile<strong>the</strong>map x −→ S E/Fxinduces<strong>the</strong>map x −→ ℓx<strong>the</strong><br />
requiredidentityreducesto<strong>the</strong>equality<strong>of</strong><br />
and<br />
Xφ(ℓ ℓ )τ(X ℓ φ, ψφ) <br />
τ(Xφ, ψφ) <br />
ThisequalityhasbeenprovedinLemma7.8.<br />
µ∈T τ(µλ, ψ λ/φ)<br />
µ∈T τ(µλX λ/φ, ψ λ/φ).<br />
Nowlet mbegreaterthan1.Since Fµ = 1forall µwearetryingtoshowthat<br />
isequalto<br />
∆1(X E/F, ψ E/F, ̟ m+n<br />
F ) ∆1(µ, ψ L/F, ̟ m+n<br />
F )<br />
∆1(XF, ψF, ̟ m+n<br />
F ) ∆1(µX L/F, ψ L/F, ̟ m+n<br />
F ).<br />
Since<strong>the</strong>action<strong>of</strong> Hon Cisnottrivial ℓcannotbe2. If µliesin Tand µ −1liesin<strong>the</strong> orbit<strong>of</strong> ν<strong>the</strong>n<br />
∆1(ν, ψL/F, ̟ m+n<br />
F ) = ∆1(µ −1 , ψL/F, ̟ m+n<br />
F )<br />
is µ(−1)times<strong>the</strong>complexconjugate<strong>of</strong><br />
∆1(µ, ψ L/F, ̟ m+n<br />
F ).<br />
Since<strong>the</strong>order<strong>of</strong> µis ℓ, µ(−1) = 1and,if µ = ν,<strong>the</strong>product<strong>of</strong><strong>the</strong>twotermscorresponding<br />
to µand νis1.If<br />
µ −1 = µ qr<br />
with 0 ≤ r < fliesin<strong>the</strong>orbit<strong>of</strong> µ<strong>the</strong>n ℓdivides q r + 1.Thus ℓdivides q 2r − 1and 2r = f.<br />
ByLemma7.1<br />
| τ(µλ, ψ λ/φ) | = √ q f = q r<br />
and<br />
τ(µλ, ψ λ/φ) = −∆1(µ, ψ L/F, ̟ 1+n<br />
F )q r
Chapter13 165<br />
if ψ λ/φhas<strong>the</strong>samemeaningasbeforeand µλis<strong>the</strong>character<strong>of</strong> λ ∗ inducedby µ.Since<br />
δ = ∆1(µ, ψ L/F, ̟ 1+n<br />
F )<br />
isitsowncomplexconjugate,itis ±1.If α ∈ φ<strong>the</strong>n<br />
Since u(α)isan ℓthroot<strong>of</strong>unityitis1.Thus<br />
HoweveritfollowsfromLemma7.1that<br />
µ −1 (α) = µ(α qr<br />
) = µ(α).<br />
τ(µλ, ψ λ/φ) = τ(µλ).<br />
τ(µλ) ≡ 1 (modη)<br />
where ηisanumberin k p(q f −1)whichisnotaunitandwhoseonlyprimedivisorsaredivisors<br />
<strong>of</strong> ℓ.Thus<br />
−δq r = τ(µλ) ≡ 1 (modℓ)<br />
and δ = 1.Wearereducedtoshowingthat<br />
isequalto<br />
∆1(X E/F, ψ E/F, ̟ m+n<br />
F )<br />
∆1(XF, ψF, ̟ m+n<br />
F ) ∆1(µX L/F, ψ L/F, ̟ m+n<br />
F )<br />
Let β = β(XF).Byrepeatedapplications<strong>of</strong>Lemma8.9weseethatwemaytake<br />
If β(X E/F)ischosenwecouldalsotake<br />
Thusif<br />
wehave<br />
β(X L/F) = β(X K/F) = β(µX L/F) = β.<br />
β(X K/F) = β(X E/F).<br />
m ′ = m(X E/F) = m(X K/F) = 2d ′ + ε ′<br />
β ≡ β(X E/F) (mod P d′<br />
K).
Chapter13 166<br />
Sincebothsides<strong>of</strong><strong>the</strong>congruenceliein E<br />
andwemaytake<br />
Then<br />
while<br />
isequalto<br />
β ≡ β(X E/F) (modP d′<br />
E )<br />
β(X E/F) = β.<br />
∆2(X E/F, ψ E/F, ̟ m+n<br />
F ) = ψF<br />
<br />
ℓβ<br />
̟ m+n<br />
<br />
F<br />
X −1<br />
F (βℓ )<br />
∆2(XF, ψF, ̟ m+n<br />
F ) <br />
µ∈T ∆2(µXF, ψ L/F, ̟ m+n<br />
F )<br />
ψF<br />
<br />
ℓβ<br />
̟ m+n<br />
<br />
F<br />
X −1<br />
F (βℓ ).<br />
Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong><strong>the</strong>lemmawehavetoshowthat<br />
isequalto<br />
∆3(XF, ψF, ̟ m+n<br />
F ) <br />
µ∈T ∆3(µXF, ψ L/F, ̟ m+n<br />
F ) (13.4)<br />
∆3(X E/F, ψ E/F, ̟ m+n<br />
F )<br />
whenone,andhenceboth,<strong>of</strong> mand m ′ isodd.<br />
AsremarkedinLemma9.4<br />
∆3(µX L/F, ψ L/F, ̟ m+n<br />
F ) = ∆3(X L/F, ψ L/F, ̟ m+n<br />
F ).<br />
AccordingtoLemma9.6<strong>the</strong>rightsideisequalto<br />
ε∆3(XF, ψF, ̟ m+n<br />
F ) [L:F]<br />
where εis1if f = [L : F]isoddand1ifitiseven.Thus(13.4)isequalto<br />
Asbefore<br />
ε ℓ−1<br />
f {∆3(XF, ψF, ̟ m+n<br />
F )}.<br />
φ = OF/PF = OE/PE.
Chapter13 167<br />
Let ϕφbe<strong>the</strong>functionon φdefinedby<br />
ϕφ(x) = ψF<br />
<br />
βx<br />
̟ d+1+n<br />
F<br />
if m = 2d + 1.Then m ′ = 2ℓd + 1sothat d ′ = ℓd.Let ϕ ′ φ<br />
ϕ ′ φ(x) = ψ E/F<br />
<br />
βx<br />
̟ d+1+n<br />
F<br />
<br />
<br />
X −1<br />
F (1 + ̟d F x)<br />
be<strong>the</strong>functionon φdefinedby<br />
X −1<br />
E/F (1 + ̟d Fx).<br />
Because<strong>of</strong>Lemma9.3,tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong><strong>the</strong>lemmawehaveonlytoshowthat<br />
wehave<br />
Since d ′ ≥ mand<br />
ε ℓ−1<br />
f A[σ(ϕφ) ℓ ] = A[σ(ϕ ′ φ )].<br />
3d ′ + ℓ − 1<br />
ℓ<br />
≥ m<br />
N K/L(1 + ̟ d F x) ≡ 1 + ̟d F S K/Lx + ̟ 2d<br />
F E2 K/L (x) (modPm L )<br />
if E2 K/L (x)is<strong>the</strong>secondelementarysymmetricfunction<strong>of</strong> xanditsconjugatesover L.Thus<br />
Thisinturniscongruentto<br />
Thus<br />
if<br />
or<br />
N E/F(1 + ̟ d Fx) ≡ 1 + ̟ d FS E/Fx + ̟ 2d<br />
F E 2 E/F (x) (modPm F ).<br />
(1 + ̟ d F S E/Fx) (1 + ̟ 2d<br />
F E2 K/F (x)).<br />
ϕ ′ φ(x) = ϕφ(ℓx)ψφ(−E 2 λ/φ (x))<br />
ϕ ′ φ (x) = ϕφ(ℓx)ψφ<br />
ψφ(x) = ϕF<br />
<br />
βx<br />
̟ 1+n<br />
<br />
F<br />
−ℓ(ℓ − 1)<br />
2<br />
x 2<br />
<br />
= {ϕφ(x)} ℓ .
Chapter13 168<br />
sothat<br />
Supposefirstthat pisoddandlet<br />
ϕ ′ φ(x) = ψφ<br />
ϕφ(x) = ψφ<br />
2 ℓx − 2ℓαx<br />
2<br />
2 x − 2αx<br />
= ψφ<br />
2<br />
2 ℓ(x − α)<br />
2<br />
ψφ<br />
−ℓα 2<br />
Referringto<strong>the</strong>observationsinparagraph9weseethatwemustshowthat<br />
ε ℓ−1<br />
f νφ(−1) ℓ−1<br />
<br />
2<br />
2<br />
−ℓα −ℓα<br />
2 ψφ = νφ(ℓ)ψφ<br />
2<br />
2<br />
or<br />
ε ℓ−1<br />
f = νφ(−1) ℓ−1<br />
2 νφ(ℓ)<br />
if νφis<strong>the</strong>quadraticcharacter<strong>of</strong> φ ∗ . Let qbe<strong>the</strong>number<strong>of</strong>elementsin φ. If qisaneven<br />
power<strong>of</strong> p<strong>the</strong>rightsideis1andif qisanoddpower<strong>of</strong> p<strong>the</strong>rightsideis,by<strong>the</strong>law<strong>of</strong><br />
quadraticreciprocity, ω(p)if ωis<strong>the</strong>quadraticcharacter<strong>of</strong><strong>the</strong>fieldwith ℓelements.Thusin<br />
allcases<strong>the</strong>rightsideis ω(q).If fisodd<strong>the</strong>n q f isaquadraticresidue<strong>of</strong> ℓifandonlyif qis.<br />
Since<br />
q f − 1 ≡ 0 (modℓ).<br />
qisaquadraticresidueandbothsides<strong>of</strong><strong>the</strong>equationare1.If fisodd<strong>the</strong>leftsideis (−1) ℓ−1<br />
f .<br />
Since fis<strong>the</strong>order<strong>of</strong> qmodulo ℓthisis ω(q).<br />
Nowsupposethat p = 2.If<br />
ψφ(−x 2 ) = ψφ(αx)<br />
<strong>the</strong>n,by<strong>the</strong>remarksin<strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma9.7,wehavetoshowthat<br />
if ℓ ≡ 1(mod4)andthat<br />
ε ℓ−1<br />
f ϕφ(α) ℓ−1<br />
2 = 1<br />
ε ℓ−1<br />
f ϕφ(α) ℓ+1<br />
2 = 1<br />
if ℓ ≡ 3 (mod4).Wealsosawinparagraph9that<br />
{ϕφ(α)} 2 = ψφ(α 2 )<br />
was+1or1accordingas qisorisnotanevenpower<strong>of</strong> p.By<strong>the</strong>secondsupplementto<strong>the</strong><br />
law<strong>of</strong>quadraticreciprocity<br />
ϕφ(α) ℓ−1<br />
2 = ω(q)<br />
2<br />
<br />
.
Chapter13 169<br />
if ℓ ≡ 1 (mod4)and<br />
if ℓ ≡ 3 (mod4).Wehavejustseenthat<br />
Thelemmaisproved.<br />
ϕφ(α) ℓ+1<br />
2 = ω(q)<br />
ε ℓ−1<br />
f = ω(q).
Chapter14 170<br />
Chapter Fourteen.<br />
The Fourth Main Lemma.<br />
In<strong>the</strong>previousparagraphwesaidthatwewouldeventuallyhavetoshowthat<br />
isequalto<br />
∆(χ E/F,ψ E/F) <br />
∆(χF, ψF) <br />
µ∈T ∆(µ′ , ψ Fµ/F) (14.1)<br />
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F). (14.2)<br />
Howeverweverifiedthat<strong>the</strong>twoexpressionsareequalonlywhen K/Fistamelyramified.<br />
Inthisparagraphweshallshowthat<strong>the</strong>yareequalifTheorem2.1isvalidforallpairs K ′ /F ′<br />
in P(K/F)forwhich [K ′ : F ′ ] < [K : F].<br />
Lemma 14.1<br />
Suppose K/FiswildlyramifiedandTheorem2.1isvalidforallpairs K ′ /F ′ in P(K/F)<br />
forwhich [K ′ : F ′ ] < [K : F].If χFisanyquasicharacter<strong>of</strong> CF<strong>the</strong>expressions(14.1)and<br />
(14.2)areequal.<br />
If aand baretwononzerocomplexnumbersand misapositiveintegerweagainwrite<br />
a ∼m bif,forsomenonnegativeinteger r, a<br />
b isan mr throot<strong>of</strong>unity. Define<strong>the</strong>nonzero<br />
complexnumber ρbydemandingthat<br />
beequalto<br />
∆(χ E/F,ψ E/F) <br />
ρ∆(χF, ψF) <br />
µ∈T ∆(µ′ , ψ Fµ/F)<br />
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F).<br />
Wehavetoshowthat ρ = 1. Lemma14.1willbeaneasyconsequence<strong>of</strong><strong>the</strong>followingfour<br />
lemmas.<br />
Lemma 14.2<br />
If m(χF)is0or1<strong>the</strong>n ρ = 1andinallcases p ∼2p 1.<br />
Lemma 14.3<br />
If [G : G1]isapower<strong>of</strong>2<strong>the</strong>n ρ ∼p 1.
Chapter14 171<br />
Lemma 14.4<br />
If<strong>the</strong>inductionassumptionisvalid,if F ⊆ F ′ ⊆ L,if F ′ /Fisnormal,andif [F ′ : F] = ℓ<br />
isaprime<strong>the</strong>n p ∼ℓ 1.<br />
Lemma 14.5<br />
Suppose H = H1H2where H2isacyclicnormalsubgroup<strong>of</strong> H, [H2 : 1]isapower<strong>of</strong>a<br />
prime ℓ,and [H1 : 1]isprimeto ℓ.If<strong>the</strong>inductionassumptionisvalid ρ ∼ℓ 1.<br />
Grant<strong>the</strong>sefourlemmasforamomentandobservethatif mand narerelativelyprime<br />
<strong>the</strong>n ρ ∼m 1and ρ ∼n 1implythat ρ = 1. If ℓisaprimewhichdivides [G : G0]<strong>the</strong>reis<br />
afield F ′ containing Fandcontainedin Lsothat F ′ /Fisnormaland [F ′ : F] = ℓ. Thus<br />
Lemma14.1followsfromLemma14.4unless [G : G0]isaprimepower.Lemma14.1follows<br />
fromLemmas14.2and14.4unless [G : G0]isapower<strong>of</strong>2or p.Suppose [G : G0]isapower<br />
<strong>of</strong>2or p.Then ρ ∼ [G:G0] 1exceptperhapswhen [G : G0] = 1.If ℓisaprimewhichdoesnot<br />
divide [G : G0]butdoesdivide [G0 : G1]let H2be<strong>the</strong> ℓSylowsubgroup<strong>of</strong> G0/G1. H2isa<br />
normalsubgroup<strong>of</strong> G/G1whichwemayidentifywith Hand H/H2hasorderprimeto H2.<br />
Thus,byawellknown<strong>the</strong>orem<strong>of</strong>Schur[7], H = H1H2where H1 ∩ H2 = {1}and H1has<br />
orderprimeto H2.ItfollowsfromLemma14.5that ρ = 1unless [G : G0] = 1or [G : G1]is<br />
apower<strong>of</strong>2or p.If [G : G0] = 1and ℓisaprimedividing [G0 : G1]<strong>the</strong>reisafield F ′ with<br />
F ⊆ F ′ ⊆ Lsuchthat F ′ /Fisnormaland [F ′ : F] = ℓ.Thusif [G : G0] = 1itfollowsfrom<br />
Lemma14.4that ρ = 1unless [G : G1]isapower<strong>of</strong>2. Howeverif [G : G1]isapower<strong>of</strong>2<br />
<strong>the</strong>recertainlyisan F ′ in Lwith [F ′ : F] = 2.ItfollowsfromLemmas14.3and14.4that ρ = 1<br />
inthiscaseunless p = 2.If [G : G1]isapower<strong>of</strong> p<strong>the</strong>n G0 = G1and G/G1isabelian.By<br />
assumption<strong>the</strong>abelian pgroup G/G1actson<strong>the</strong> pgroup C = G1faithfullyandirreducibly.<br />
Thisisimpossible.<br />
WeproveLemma14.2first.Let t ≥ 1besuchthat C = Gtwhile Gt+1 = {1}.Let θtbe<br />
<strong>the</strong>homomorphism<strong>of</strong> Gtinto Pt K /Pt+1<br />
K and θ0<strong>the</strong>homomorphism<strong>of</strong> G0/G1into U0 K /U1 K<br />
introducedinSerre’sbook.If σ ∈ G0and γ ∈ Gt<strong>the</strong>n<br />
θt(σγσ −1 γ −1 ) = (θ t 0 (σ) − 1)θt(γ).<br />
If σisnotin G1<strong>the</strong>n θ t 0 (σ)isnot1and γ → σγσ−1 γ −1 isaonetoonemap<strong>of</strong> Contoitself.<br />
Thus,if σ ∈ G0,<br />
µ(σγσ −1 ) = µ(γ)<br />
implies µ = 1or σ ∈ G1. Consequentlyif µ = 1, Gµ ∩ G0 = G1and L/Fµisunramified.<br />
Since µ = µ ′ L/Fµ ,<br />
m(µ ′ ) = m(µ) = t + 1.
Chapter14 172<br />
Observealsothat tmustberelativelyprimeto [G0 : G1].Inparticularif tiseven [G0 : G1]is<br />
odd.<br />
and<br />
Therelations<br />
δ K/L = ([Gt : 1] − 1) (t + 1)<br />
δ L/F = [G0 : G1] − 1<br />
δ K/E = [G0 : G1] − 1<br />
δ K/F = δ K/L + [G1 : 1] δ L/F = δ K/E + [G0 : G1]δ E/F<br />
obtainedfromProposition4<strong>of</strong>ChapterIV<strong>of</strong>Serre’sbook,implythat<br />
If n = n(ψF)<strong>the</strong>n<br />
and<br />
<br />
t<br />
δE/F = ([G1; 1] − 1)<br />
[G0 : G1]<br />
<br />
+ 1 .<br />
n(ψ Fµ/F) = [G0 : G1]n + [G0 : G1] − 1<br />
n ′ <br />
t<br />
= n(ψE/F) = [G1 : 1]n + ([G1 : 1] − 1)<br />
[G0 : G1]<br />
<br />
+ 1 .<br />
Chooseagenerator ̟K<strong>of</strong> PKandagenerator ̟E<strong>of</strong> PE.Thenset ̟L = N K/L̟Kand<br />
̟F = N E/F̟E.Thereisaunit δin Ksuchthat<br />
̟ 1+n<br />
F<br />
̟ 1+n′<br />
E<br />
= δ ̟t K<br />
̟t .<br />
L<br />
Taking<strong>the</strong>normfrom Kto L<strong>of</strong>bothsidesweseethatif q = [G1 : 1]and k = [G0 : G1]<strong>the</strong>n<br />
̟ t(q−1)<br />
L<br />
̟ t(q−1)<br />
k<br />
F<br />
= N K/Lδ.<br />
t<br />
Let m = m(χF).If m − 1isequalto [G0:G1] <strong>the</strong>n [G0 : G1]divides tand [G0 : G1]is1.<br />
Supposethat<br />
t<br />
m < + 1.<br />
[G0 : G1]
Chapter14 173<br />
Then<br />
m(χ Fµ/F) < ψ Fµ/F<br />
However ψ Fµ/F = ϕ L/Fµ ◦ ψ L/F = ψ L/Fsothat<br />
<br />
t<br />
+ 1.<br />
[G0 : G1]<br />
ψ Fµ/F(u) = [G0 : G1]u<br />
if u ≥ 0.Thus m(χFµ/F) < t + 1and m(µ ′ χFµ/F) = t + 1.Moreover,byLemmas13.1and<br />
6.4, m ′ = m(χE/F) = m.Chooseagenerator ̟Fµ <strong>of</strong> PFµ .Then<br />
NFµ/F(̟ t Fµ ) = γµ ̟ t[Fµ:F]<br />
k<br />
F<br />
where γµisaunit.Theorder<strong>of</strong> ̟ 1+n<br />
F ̟t Fµ in Fµis 1 + t + n(ψ Fµ/F).Observingthat<br />
weseethat<br />
isequalto<br />
<br />
µ [Fµ : F] = q − 1<br />
∆1(χE/F,ψ E/F ′̟ m′ +n ′<br />
E ) <br />
µ∈T ∆1(µ ′ , ψFµ/F, ̟ 1+n<br />
F ̟t Fµ )<br />
<br />
<br />
p<br />
µ χF(γµ)<br />
<br />
{∆1(χF,ψF, ̟ m+n<br />
<br />
<br />
F }<br />
µ ∆1(µ ′ χFµ/F, ψFµ/F, ̟ 1+n<br />
F ̟t Fµ )<br />
<br />
.<br />
Itisnowclearthat ρ = 1if m = 0.<br />
If<br />
sothatinparticular m ≥ 2,<strong>the</strong>n<br />
m ≥<br />
t<br />
+ 1<br />
[G0 : G1]<br />
m ′ <br />
t<br />
= m(χE/F) = [G1 : 1]m − ([G1 : 1] − 1)<br />
[G0 : G1]<br />
isalsogreaterthanorequalto2and<br />
Since m ′ ≥ 2and K/Eistamelyramified<br />
m ′ + n ′ = [G1 : 1] (m + n).<br />
<br />
+ 1<br />
m(χ K/F) = ψ K/E(m(χ E/F) − 1) + 1 = ψ K/F(m − 1) + 1.
Chapter14 174<br />
Since<br />
and<br />
wehave<br />
However<br />
sothat<br />
isatmost<br />
Thus<br />
Consequently<br />
m(χ Fµ/F) ≤ ψ Fµ/F(m − 1) + 1<br />
ψ Fµ/F(m − 1) + 1 ≥ ψ Fµ/F<br />
<br />
t<br />
+ 1 = t + 1<br />
k<br />
m(µ ′ χFµ/F) ≤ ψFµ (m − 1) + 1.<br />
χ K/F = (µ ′ χ Fµ/F) K/Fµ<br />
ψ K/Fµ (ψ Fµ/F(m − 1)) + 1 = ψ K/F(m − 1) + 1 = m(χ K/F)<br />
ψ K/F(m(µ ′ χ Fµ/F) − 1) + 1.<br />
m(µ ′ χ Fµ/F) = ψ Fµ/F(m − 1) + 1.<br />
m(µ ′ χ Fµ/F) + n(ψ Fµ/F) = [G0 : G1] (m + n).<br />
Since<strong>the</strong>range<strong>of</strong>each µ ′ liesin<strong>the</strong>group<strong>of</strong> qthroots<strong>of</strong>unity<br />
isequalto<br />
with σ ∼p ρ.<br />
∆1(χ E/F, ψ E/F, ̟ m+n<br />
F ) <br />
µ ∆1(µ ′ , ψ Fµ/F, ̟ n+1<br />
F ̟t Fµ )<br />
σ∆1(χF, ψF, ̟ m+n<br />
F ) <br />
µ ∆1(µ ′ χFµ/F, ψFµ/F, ̟ m+n<br />
F )<br />
Thenextstepin<strong>the</strong>pro<strong>of</strong><strong>of</strong><strong>the</strong>lemmaistoestablishasimpleidentity.Asusuallet rbe<br />
<strong>the</strong>integralpart<strong>of</strong> t+1<br />
2 andlet r + s = t + 1.Choose β(µ′ )sothat<br />
ψ Fµ/F<br />
<br />
β(µ ′ )x<br />
̟ 1+n<br />
F ̟t Fµ<br />
<br />
= µ ′ (1 + x)<br />
for xin P s Fµ .Thereisaunit αµin Lsuchthat αµ̟ t Fµ = ̟t L .Then<br />
<br />
αµβ(µ<br />
ψL/F ′ <br />
)x<br />
= µ(1 + x)<br />
̟ 1+n<br />
F ̟t L
Chapter14 175<br />
for xin P s L .Wetake β(µ) = αµβ(µ ′ ).If σ ∈ Gapossiblechoicefor β(µ σ )is<br />
β(µ) σ ̟t L<br />
σ t ̟L Let φ = OF/PF = OE/PEandlet ψφbe<strong>the</strong>additivecharacter<strong>of</strong> φdefinedby<br />
Thereisaunique αin φsuchthat<br />
Finallylet<br />
Iwanttoshowthat<br />
in φ.<br />
<br />
Let λ = OL/PL = OK/PK.If<br />
ψφ(x) = ψF<br />
.<br />
<br />
x<br />
̟ 1+n<br />
<br />
.<br />
F<br />
ψφ(αx) = ψφ(x q ).<br />
ω1 = S E/F<br />
µ γµ = ωq 1<br />
α q<br />
<br />
<br />
ω = S K/L<br />
<strong>the</strong>n ω1 = δωin λ.Weneed<strong>the</strong>followinglemma.<br />
Lemma 14.6<br />
̟ 1+n<br />
F<br />
̟ 1+n′<br />
E<br />
<br />
.<br />
µ N Fµ/Fβ(µ ′ ) (14.3)<br />
t ̟K Suppose K ′ /F ′ isanabelianextensionand G ′ = G(K ′ /F ′ ). Suppose<strong>the</strong>reisat ≥ 1<br />
suchthat G ′ = G ′ tand G′ t+1 = {1}.Let ̟K ′beagenerator<strong>of</strong> P′ K ,let ̟F ′ = NK ′ /F ′(̟K ′),<br />
andlet<br />
t ̟K ′<br />
ω = SK ′ /F ′<br />
̟t F ′<br />
<br />
.<br />
Alsolet q = [K ′ : F ′ ].Therearenumbers a, . . ., fin OF ′suchthatforall xin OF ′<br />
̟ t L<br />
N K ′ /F ′(1 + x̟ t K ′)
Chapter14 176<br />
iscongruentto<br />
modulo P t+1<br />
F ′ .<br />
1 + (x q + ax q<br />
p + . . . + fx p + ωx)̟ t F ′<br />
Suppose F ′ ⊆ L ′ ⊆ K ′ and<strong>the</strong>lemmaistruefor K ′ /L ′ and L ′ /F ′ .Thelemmafor K ′ /F ′<br />
followsfrom<strong>the</strong>relations<br />
[K ′ : F ′ ] = [K ′ : L ′ ] [L ′ : F ′ ]<br />
and<br />
and<br />
N K ′ /F ′(1 + x̟ t K ′) = N L ′ /F ′(N K ′ /L ′(1 + x̟t K ′))<br />
S K ′ /F ′<br />
t ̟K ′<br />
̟ t F ′<br />
≡ S L ′ /F ′<br />
t ̟L ′<br />
̟ t F ′<br />
S K ′ /L ′<br />
̟ t K ′<br />
Thelemmaforextensions<strong>of</strong>primeorderisprovedinSerre’sbook.<br />
Suppose<strong>the</strong>n<br />
for xin OL.Since<br />
whichinturnequals<br />
weconcludethat<br />
̟ t L ′<br />
<br />
.<br />
N K/L(1 + x̟ t K ) ≡ 1 + (xq + . . . + ̟x)̟ t L (modPt+1<br />
L )<br />
ψ λ/φ(αx) = ψφ(αS λ/φ(x)) = ψφ((S λ/φ(x)) q )<br />
ψφ(S λ/φx q ) = ψ λ/φ(x q )<br />
ψ λ/φ(y(x q + . . . + ωx)) = ψ λ/φ((αy 1<br />
q + . . . + ωy)x).<br />
Also <br />
αy 1 q q + . . . + ωy = α q y + . . . + ω q y q<br />
isapolynomial Q(y)in y.<br />
Foreach νin S(K/L),wechoose β1(ν)sothat<br />
<br />
β1(ν)x<br />
ψL/F = ν(1 + x)<br />
for xin P s L .Since kp = kin λ<br />
̟ 1+n<br />
F ̟t L<br />
ψ λ/φ(kβ1(ν)(x q + . . . + ωx) = ψ λ/φ(β1(ν) [(kx) q + . . . + ωkx])
Chapter14 177<br />
if xisin OF.Theleftsideisalsoequalto<br />
<br />
β1(ν) PK/L(x̟ ψL/F t K )<br />
<br />
= 1.<br />
̟ 1+n<br />
F ̟t L<br />
Thus Q(kβ1(ν)) = 0.Since β1(ν1) = β2(ν2) (modPL)implies ν1 = ν2wehavefoundall<strong>the</strong><br />
roots<strong>of</strong> Q(y) = 0.Thus<br />
αq <br />
≡<br />
ωq ν=1 kβ1(ν) ≡ <br />
ν=1 β1(ν) (mod PL).<br />
Let Mµbeaset<strong>of</strong>representativesfor<strong>the</strong>cosets<strong>of</strong> Gµin G.Then<br />
iscongruentto<br />
<br />
µ∈T NFµ/Fβ(µ ′ ) = <br />
<br />
<br />
ν=1 β1(ν)<br />
<br />
<br />
µ∈T<br />
µ∈T<br />
<br />
<br />
σ∈Mµ<br />
α<br />
σ∈Mµ<br />
σ µ<br />
modulo PL.Toverify<strong>the</strong>identity(14.3)wehavetoshowthat<br />
Since<br />
<br />
<strong>the</strong>congruencereducesto<br />
µ<br />
<br />
σ∈Mµ<br />
γµ =<br />
α σ µ<br />
̟t L<br />
σ t<br />
L<br />
̟<br />
<br />
̟ t(q−1)<br />
L<br />
(q−1)<br />
t k ̟F σ∈Mµ<br />
<br />
≡ δ q<br />
whichisvalidbecause<strong>the</strong>leftsideis N K/Lδand<br />
N K/Lδ ≡ δ q<br />
σ t<br />
L<br />
ασ µ<br />
̟<br />
µ γµ<br />
<br />
̟<br />
<br />
≡ δ q<br />
β(µ ′ ) σ<br />
̟t −1 L<br />
σ t<br />
L<br />
̟<br />
[Fµ:F]<br />
−t k<br />
F<br />
(mod PK)<br />
(modPK).<br />
(modPK).<br />
If m = 1<strong>the</strong>n m(χ Fµ/F) ≤ 1andwecantake β(µ ′ χ Fµ/F) = β(µ ′ ).Then<br />
∆2(µ ′ , ψ Fµ/F, ̟ 1+n<br />
F ̟t Fµ )
Chapter14 178<br />
isequalto<br />
Lemma9.4impliesthat<br />
If xbelongsto OE<strong>the</strong>n<br />
χF(N Fµ/F(β(µ ′ ))) ∆2(µ ′ χ Fµ/F, ψ Fµ/F, ̟ 1+n<br />
F ̟t Fµ ).<br />
∆3(µ ′ , ψ Fµ/F, ̟ 1+n<br />
F ̟t Fµ ) = ∆3(µ ′ χ Fµ/F, ψ Fµ/F, ̟ 1+n<br />
F ̟t Fµ ).<br />
ψ E/F<br />
<br />
x<br />
̟ 1+n′<br />
E<br />
If χφis<strong>the</strong>character<strong>of</strong> φ ∗ determinedby χF<strong>the</strong>n<br />
and<br />
<br />
= ψφ(ω1x).<br />
∆1(χF, ψF, ̟ 1+n<br />
F ) = A[−τ(χφ, ψφ)]<br />
∆1(χE/F, ψE/F, ̟ 1+n′<br />
E ) = χφ(ω q<br />
1 ) A[−τ(χq φ , ψφ)].<br />
Therightside<strong>of</strong>thisexpressionisequalto<br />
χφ(ω q<br />
1 α−q ) A[−τ(χφ, ψφ)].<br />
Theidentity(14.3)nowshowsthat ρ = 1when m = 1.<br />
Supposethat<br />
Let βbeagivenchoice<strong>of</strong> β(χF).Then<br />
if m ′ = 2d ′ + ε ′ .<strong>On</strong><strong>the</strong>o<strong>the</strong>rhand<br />
1 < m <<br />
t<br />
+ 1.<br />
[G0 : G1]<br />
β(χE/F) ≡ P ∗ E/F (β, ̟m′ +n ′<br />
E , ̟ m+n<br />
F ) (mod P d′<br />
E )<br />
ψ L/F(m − 1) + 1 + n(ψ L/F) = [G0 : G1](m − 1) + 1 + [G : G0]n + [G0 : G1] − 1<br />
whichequals [G0 : G1](m + n)andLemmas8.3,8.4and8.7implythat<br />
if<br />
P ∗ L/F (β, ̟m+n<br />
F , ̟ m+n<br />
F ) ≡ β (modP d1<br />
L )<br />
ψ L/F(m − 1) + 1 = 2d1 + ε1.
Chapter14 179<br />
If<br />
<strong>the</strong>n<br />
Thus<br />
If<br />
and<br />
<strong>the</strong>n<br />
Let<br />
iscongruentto<br />
modulo P d′<br />
1<br />
K .<br />
andthat<br />
Itisclearthat<br />
Ifwechoose<br />
<strong>the</strong>n<br />
P ∗ K/E (β(χ E/F,̟ m′ +n ′<br />
E<br />
ψ K/F(m − 1) + 1 = 2d ′ 1 + ε′ 1<br />
, ̟ m′ +n ′<br />
E<br />
β(χE/F) ≡ P ∗ K/F (β, ̟m′ +n ′<br />
E , ̟ m+n<br />
) ≡ β(χ E/F) (modP d′<br />
1<br />
K ).<br />
F ) (mod P d′<br />
1<br />
K ).<br />
v = t + 1 − (ψ L/F(m − 1) + 1) = t − [G0 : G1] (m − 1).<br />
γ = ̟t−v<br />
L<br />
̟ m−1<br />
F<br />
γ ′ = ̟m′ +n ′<br />
E ̟v K<br />
̟ 1+n<br />
F ̟t L<br />
P ∗ K/F (β, ̟m′ +n ′<br />
E , ̟ m+n<br />
F ) ≡ P ∗ K/L (β, ̟m′ +n ′<br />
E<br />
γ ′ P ∗ K/L (γβ, ̟1+n<br />
F ̟t L̟ −v<br />
K<br />
, ̟ m+n<br />
F ) (modP d′<br />
1<br />
, ̟1+n<br />
F ̟t−v<br />
L )<br />
∆2(χE/F, ψE/F, ̟ m′ +n ′<br />
E ) ∼p χ −1<br />
F (NE/F(β(χ E/F)))<br />
∆2(µ ′ , ψ Fµ/F, ̟ n+1<br />
F ̟t Fµ ) ∼p 1.<br />
β(µ ′ χ Fµ/F) = β(µ ′ ) + β ̟t Fµ<br />
̟ m−1<br />
F<br />
∆2(µ ′ χ Fµ/F, ψ Fµ/F, ̟ n+1<br />
F ̟t Fµ ) ∼p χ −1<br />
F<br />
<br />
N Fµ/F<br />
<br />
K )<br />
β(µ ′ ) + β ̟t Fµ<br />
̟ m−1<br />
F<br />
<br />
.
Chapter14 180<br />
Moreover<br />
Let<br />
equal<br />
Since<br />
∆2(χF, ψF, ̟ m+n<br />
F ) ∼p χ −1<br />
F (β).<br />
∆2(χE/F,ψ E/F, ̟ m′ +n ′<br />
E ) <br />
µ ∆2(µ ′ , ψFµ/F, ̟ n+1<br />
F ̟t Fµ )<br />
τ∆2(χF,ψF, ̟ m+n<br />
<br />
<br />
F )<br />
µ ∆2(µ ′ χFµ/F, ψFµ/F, ̟ n+1<br />
F ̟t Fµ )<br />
<br />
<br />
µ χF(γµ)<br />
<br />
.<br />
χF(u) ∼p 1<br />
if u ∈ U 1 F ,allweneeddotoshowthat τ ∼p 1isprovethat<br />
iscongruentto<br />
modulo PF.<br />
β <br />
µ N Fµ/F<br />
<br />
β(µ ′ ) + β ̟t Fµ<br />
̟ m−1<br />
F<br />
N E/F(β(χ E/F)) <br />
µ γµ<br />
Asbeforewechoose β(µ) = αµ(β(µ ′ ).If ν = µ σ apossiblechoicefor β(ν)is<br />
Wecanalsochoose<br />
α σ µβ(µ ′ ) σ ̟t L<br />
̟σ t<br />
L<br />
β(µχ L/F) = αµβ(µ ′ ) + β ̟t L<br />
̟ m−1<br />
F<br />
Thenapossiblechoicefor B(µ σ χ L/F)is<br />
α σ µ<br />
<br />
β(µ ′ ) + β ̟t Fµ<br />
̟ m−1<br />
F<br />
σ ̟ t L<br />
WeapplyLemma8.10with Freplaceby L,<br />
.<br />
= αµ<br />
<br />
<br />
β(µ ′ ) + β ̟t Fµ<br />
̟ m−1<br />
F<br />
<br />
̟σ t = α<br />
L<br />
σ µ(β(µ ′ ) σ ̟t L<br />
̟σ t + β<br />
L<br />
̟t L<br />
̟ m−1<br />
F<br />
δ = ̟ 1+n<br />
F ̟t L<br />
.<br />
.
Chapter14 181<br />
and ε1 = ̟ v K .Itimpliesthat<br />
iscongruentto<br />
γβ<br />
<br />
µ∈T<br />
<br />
σ∈Mµ<br />
N K/L<br />
<br />
α σ <br />
µ<br />
modulo PL.Thelastexpressionisequalto<br />
γβ<br />
<br />
µ∈T N Fµ/F<br />
andwehavetoshowthat<br />
γNK/L(γ ′ <br />
<br />
)<br />
<br />
β(µ ′ ) + β ̟t Fµ<br />
̟ t F<br />
µ γµ<br />
iscongruentto1modulo PL.First<strong>of</strong>all<br />
Since<br />
<strong>the</strong>requiredrelationfollows.<br />
equalto<br />
Define ηbysetting<br />
β(χE/F)<br />
<br />
γ ′<br />
<br />
β(µ ′ ) + β ̟t Fµ<br />
̟ m−1<br />
F<br />
<br />
µ∈T<br />
<br />
µ∈T<br />
NK/L(γ ′ ) = ̟ m′ +n ′ −q(1+n)<br />
F ̟ −qt+v<br />
L = γ<br />
γµ =<br />
<br />
σ∈Mµ<br />
<br />
η∆3 χE/F,ψ E/F, ̟ m′ +n ′ <br />
E<br />
ασ −1 µ<br />
σ t ̟L µ∈T ∆3<br />
̟<br />
σ<br />
<br />
α<br />
σ∈Mµ<br />
σ µ<br />
̟t L<br />
σ t<br />
L<br />
̟<br />
<br />
α<br />
σ∈Mµ<br />
σ µ<br />
̟t L<br />
σ t<br />
L<br />
̟<br />
(q−1)t<br />
k<br />
−1 ̟ F<br />
[Fµ:F]<br />
−t k<br />
F<br />
̟ (q−1)t<br />
L<br />
<br />
.<br />
̟t L<br />
σ t<br />
L<br />
̟<br />
<br />
µ ′ , ψFµ/F, ̟ 1+n<br />
F ̟t <br />
Fµ<br />
∆3(χF, ψF, ̟ m+n<br />
F ) <br />
µ∈T ∆3<br />
<br />
µ ′ χFµ/F, ψFµ/F, ̟ 1+n<br />
F ̟t <br />
. Fµ<br />
Wenowknowthat η ∼p ρ.Weshallshowthat η ∼p 1.Thiswillprovenotonly<strong>the</strong>assertion<br />
<strong>of</strong>Lemma14.2butalsothat<strong>of</strong>Lemma14.3,provided<strong>of</strong>coursethat<br />
m − 1 <<br />
t<br />
[G0 : G1] .
Chapter14 182<br />
Lemmas9.2and9.3implydirectlythat η ∼p 1if pis2.<br />
Suppose pisodd.Lemma9.4impliesthat<br />
∆3(µ ′ , ψ Fµ/F, ̟ 1+n<br />
F ̟t Fµ ) ∼p ∆3(µ ′ χ Fµ/F, ψ Fµ/F, ̟ 1+n<br />
F ̟t Fµ ).<br />
Since m ′ = mallweneeddoisshowthat<br />
∆3(χE/F,ψ E/F, ̟ m′ +n ′<br />
E ) ∼p ∆3(χF, ψF, ̟ m+n<br />
F )<br />
when misodd.Let φ = OF/PF = OE/PE.Let β ′ = β(χ E/F)andlet β = β(χF).If ψ ′ φ is<br />
<strong>the</strong>character<strong>of</strong> φdefinedby<br />
and ψ ′′<br />
φis<strong>the</strong>character<strong>of</strong> φdefinedby<br />
ψ ′ φ (x) = ψF<br />
ψ ′′<br />
φ (x) = ψ E/F<br />
<br />
β x<br />
̟ n+1<br />
<br />
F<br />
<br />
β ′ x<br />
̟ n′ +1<br />
E<br />
andif ψ ′′<br />
φ (x) = ψ′ φ (δx)<strong>the</strong>n,byLemmas9.2and9.3,allwehavetodoisshowthat δisasquare<br />
in φ.If<br />
ω1 = S E/F<br />
<br />
̟ n+1<br />
F<br />
̟ n′ +1<br />
E<br />
<strong>the</strong>n δ = ω1 β′<br />
β in φ.Toshowthat δisasquareweshowthat δq isasquare.<br />
Wesawthat<br />
in φ.But<br />
N E/F(β ′ )<br />
β<br />
<br />
<br />
δ q ≡ NE/Fα ≡ β 1−q NE/F(β ′ )<br />
ω<br />
β<br />
q<br />
1 .<br />
<br />
<br />
=<br />
µ γµ<br />
−1 <br />
β ̟t Fµ<br />
̟ m−1<br />
F<br />
µ N Fµ/F<br />
<br />
≡ 0 (modPL)<br />
β(µ ′ ) + β ̟t Fµ<br />
̟ m−1<br />
F
Chapter14 183<br />
because t > [G0 : G1](m − 1).Wealsosawthat<br />
<br />
µ γµ<br />
−1 <br />
<br />
µ NFµ/Fβ(µ ′ <br />
)<br />
= αq<br />
̟ q<br />
1<br />
in φ.Since β 1−q isclearlyasquare,weneedonlycheckthat αisasquare.Thecharacter<br />
isidentically1,sothat<strong>the</strong>kernel<strong>of</strong><strong>the</strong>map<br />
isnontrivial.Thus α = x q−1 forsome xin φ.<br />
Nowsupposethat<br />
x −→ ψφ (x q − αx)<br />
x −→ x q − αx<br />
m − 1 ≥<br />
t<br />
[G0 : G1] .<br />
Wehavetoshowthat<strong>the</strong>complexnumber σdefinedat<strong>the</strong>beginning<strong>of</strong><strong>the</strong>pro<strong>of</strong>satisfies<br />
σ ∼2p 1.ToproveLemma14.3wewillhavetoshowthat σ ∼p 1if [G : G1]isapower<strong>of</strong>2.<br />
Given β = β(χF)wemaychoose β(χ L/F) = β(χ Fµ/F) = β.Moreover<br />
β(χE/F) ≡ P ∗ E/F (β, ̟m+n<br />
F , ̟ m+n<br />
F ) (modP d′<br />
E)<br />
if m ′ = m(χ E/F) = 2d ′ + ε ′ .ByLemmas8.3,8.4and8.7<br />
modulo P d′<br />
1<br />
K if<br />
Thus<br />
If<br />
and<br />
P ∗ K/F (β, ̟m+n<br />
F , ̟ m+n<br />
F ) ≡ P ∗ K/E (β(χE/F), ̟ m+n<br />
F , ̟ m+n<br />
F ) ≡ β(χE/F )<br />
ψ K/F(m − 1) + 1 = 2d ′ 1 + ε ′ 1.<br />
β(χE/F ) ≡ P ∗ K/L (β, ̟m+n<br />
F , ̟ m+n<br />
F ) (modP d′<br />
1<br />
ψ Fµ/F(m − 1) + 1 = 2dµ + εµ<br />
ψ Fµ/F<br />
′ α(µ )x<br />
̟ m+n<br />
<br />
= µ<br />
F<br />
′ (1 + x)<br />
K ).
Chapter14 184<br />
for xin P dµ+εµ<br />
Fµ<br />
If<br />
<strong>the</strong>n<br />
wemaytake<br />
for xin P d1+ε1<br />
L .If ν = µ σ<strong>the</strong>n β(µ ′ χ Fµ/F) = β + α(µ ′ ).<br />
ψ L/F(m − 1) + 1 = 2d1 + ε1<br />
µ(1 + x) = ψ L/F<br />
ν(1 + x) = ψ L/F<br />
for xin P d1+ε1<br />
L .Lemma8.2impliesthat<br />
iscongruentto<br />
′ α(µ )x<br />
̟ m+n<br />
<br />
F<br />
′ σ α(µ ) x<br />
̟ m+n<br />
<br />
F<br />
N E/F(β(χ E/F)) = N K/L(β(χ E/F))<br />
β <br />
µ∈T<br />
<br />
modulo PL.Thelastexpressionisequalto<br />
Moreover<br />
β <br />
σ∈Mµ<br />
(β + α(µ ′ ) σ )<br />
µ∈T N Fµ/F(β + α(µ ′ )).<br />
∆2(χF, ψF, ̟ m+n<br />
F ) ∼p χ −1<br />
F (β)<br />
∆2(µ ′ , ψ Fµ/F,̟ 1+n<br />
F ̟t Fµ ) ∼p 1<br />
∆2(χ E/F,ψ E/F, ̟ m+n<br />
F ) ∼p χ −1<br />
F (N E/F(β(χ E/F)))<br />
∆2(µ ′ χ Fµ/F, ψ Fµ/F, ̟ m+n<br />
F ) ∼p χ −1<br />
F (N Fµ/F(β + α(µ ′ ))).<br />
Define τbydemandingthat<br />
beequalto<br />
∆3(χ E/F,ψ E/F, ̟ m+n<br />
F ) <br />
µ ∆3(µ ′ , ψ Fµ/F, ̟ 1+n<br />
F ̟t Fµ )<br />
τ ∆3(χF, ψF, ̟ m+n<br />
F ) <br />
µ ∆3(µ ′ , χ Fµ/F, ψ Fµ/F, ̟ m+n<br />
F ).
Chapter14 185<br />
Since χF(u) ∼p 1if u ∈ U 1 F <strong>the</strong>precedingdiscussionshowsthat σ ∼p τ.Lemmas9.2and9.3<br />
showthat τ ∼2p 1.Lemma14.2isnowcompletelyproved.ToproveLemma14.3wehaveto<br />
showthat τ ∼p 1if [G : G1]isapower<strong>of</strong>2.Wemaysupposethat pisodd.<br />
Thereareanumber<strong>of</strong>possibilities.<br />
(i.a) tisevenand misodd. [G0 : G1]mustbeoddandhence1,forwearenowassumingthat<br />
[G : G1]isapower<strong>of</strong>2.Since<br />
and<br />
m(χ Fµ/F) = [G0 : G1] (m − 1) + 1<br />
m(χ E/F) = [G1 : 1] (m − 1) − ([G1 : 1] − 1)t<br />
[G0 : G1]<br />
both m(χ Fµ/F)and m(χ E/F)areodd.<br />
(i.b) tisevenand miseven.Again [G0 : G1]is1.Thistimeboth m(χ Fµ/F)and m(χ E/F)are<br />
even.<br />
(ii.a) tisoddand misodd.Then m(χ Fµ/F)isodd.If<br />
[G1 : 1] − 1<br />
[G0 : G1]<br />
iseven m(χ E/F)isodd.O<strong>the</strong>rwiseitiseven.<br />
(ii.b) tisoddand miseven.If [G0 : G1]isodd,thatis1,<strong>the</strong>n m(χ Fµ/F)isoddand m(χ E/F)<br />
iseven.If [G0 : G1]iseven<strong>the</strong>n m(χ Fµ/F)isoddand m(χ E/F)isevenoroddaccording<br />
as<br />
[G1 : 1] − 1<br />
[G0 : G1]<br />
isevenorodd.<br />
If tisodd<strong>the</strong>nclearly<br />
<br />
µ ∆3(µ ′ , ψ Fµ/F, ̟ 1+n<br />
F ̟t Fµ ) ∼p 1.<br />
Wearegoingtoshowthatthisisalsotrueif tiseven. Then L/F,andhence Fµ/F,is<br />
unramified.Let φµ = OFµ /PFµ .If<br />
ψφµ (x) = ψ Fµ/F<br />
′ β(µ )x<br />
̟ 1+n<br />
<br />
F<br />
+ 1
Chapter14 186<br />
andif ϕφµ isanowherevanishingfunctionon φµsatisfying<br />
<strong>the</strong>n<br />
ϕφµ (x + y) = ϕφµ (x) ϕφµ (y)ψφµ (xy)<br />
∆3(µ ′ , ψ Fµ/F, ̟ 1+n<br />
F ̟t Fµ ) ∼p A[−σ(ϕφµ )].<br />
If αbelongsto φ ∗ µ let νφµ (α)equal+1or1accordingas αisorisnotasquarein φµ. If<br />
φ = OF/PF<strong>the</strong>n<br />
νφµ (α) = νφ (N φµ/φ(α)).<br />
If<br />
<strong>the</strong>n,accordingtoparagraph9,<br />
ψφ(x) = ψF<br />
<br />
x<br />
̟ 1+n<br />
F<br />
<br />
∆3(µ ′ , ψ Fµ/F,̟ 1+n<br />
F ̟t Fµ ) ∼p νφ(N Fµ/F(β(µ ′ ))) {A[−σ(ϕφ)]} [Fµ:F]<br />
if ϕφisanynowherevanishingfunctionon φsatisfying<br />
Thusif ais<strong>the</strong>number<strong>of</strong> µin T<br />
isequalto<br />
ϕφ(x + y) = ϕφ(x) ϕφ(x) ψφ(xy).<br />
<br />
η(−1) a νφ<br />
where η ∼p 1and q = [G1 : 1].<br />
Wesawinparagraph9that<br />
µ ∆3(µ ′ , ψ Fµ/F, ̟ 1+n<br />
F ̟t Fµ )<br />
<br />
µ NFµ/F(β(µ ′ <br />
)) A[σ(ϕφ) q−1 ]<br />
A[σ(ϕφ) 2 ] ∼p νφ(−1) A[|σ(ϕφ)| 2 ] = νφ(−1).<br />
Since tiseven G0 = G1and G/G1 = G/G0isabelian.If σ ∈ G<br />
{µ ∈ S(K/L) | µ = µ σ }
Chapter14 187<br />
isasubgroup<strong>of</strong> S(K/L)invariantunder G.Itisnecessarilyei<strong>the</strong>r S(K/L)or {1}.If σisnot<br />
in G1itisnot S(K/L). Thus Gµ,<strong>the</strong>isotropygroup<strong>of</strong> µ,is G1forall µin Tand Fµ = L.<br />
Moreover <br />
µ NFµ/F(β(µ ′ )) = <br />
β(µ<br />
µ<br />
′ ) σ .<br />
σ∈G/G1<br />
Wemayregard C = G1asavectorspaceover<strong>the</strong>fieldwith pelements.If σ ∈ G/G1and<br />
<strong>the</strong>order<strong>of</strong> σdivides p − 1<strong>the</strong>nall<strong>the</strong>eigenvalues<strong>of</strong><strong>the</strong>lineartransformation c −→ σcσ −1<br />
liein<strong>the</strong>primefield. Since<strong>the</strong>lineartransformationalsohasorderdividing p − 1itis<br />
diagonalizable.Since G/G1isabelianandactsirreduciblyon C<strong>the</strong>lineartransformationisa<br />
multiple<strong>of</strong><strong>the</strong>identity.Inparticularif σ0is<strong>the</strong>uniqueelement<strong>of</strong>order2<strong>the</strong>n σ0cσ −1<br />
0<br />
forall c.Asaconsequence µ σ0 = µ −1 and<br />
β(µ ′ ) σ0 ≡ −β(µ ′ ) (modPL)<br />
= c−1<br />
ifwechoose,aswemaysince Fµ/Fisunramified, ̟Fµ = ̟F.If Dis<strong>the</strong>group {1, σ0}and<br />
Misaset<strong>of</strong>representativesfor<strong>the</strong>cosets<strong>of</strong> Din G/G1<strong>the</strong>n<br />
if<br />
Clearly<br />
<br />
µ∈T<br />
<br />
γ = <br />
If χis<strong>the</strong>nontrivialcharacter<strong>of</strong> Dand<br />
is<strong>the</strong>transfer<strong>the</strong>n<br />
σ∈G/G1<br />
µ∈T<br />
<br />
γγ σ0 = (1−) q−1<br />
2 γ 2<br />
β(µ ′ ) σ = γγ σ0<br />
σ∈M β(µ′ ) σ .<br />
v : G/G0 −→ D<br />
γ σ = χ(v(σ)) a γ<br />
(mod PL).<br />
forall σin G/G0. νφ(γ 2 ) = 1ifandonlyif χ(v(σ)) a is1forall σ.If σisagenerator<strong>of</strong> G/G0<br />
<strong>the</strong>n<br />
v(σ) = σ [G:G 0 ]<br />
2 = σ0<br />
sothat νφ(γ 2 ) = (−1) a .Puttingall<strong>the</strong>sefactstoge<strong>the</strong>rweseethat<br />
<br />
µ ∆3(µ ′ , ψ Fµ/F, ̟ 1+n<br />
F ̟t Fµ ) ∼p 1.
Chapter14 188<br />
Observethatifwehadtaken ̟Fµ tobe δµ̟F<strong>the</strong>n N Fµ/Fβ(µ ′ )wouldhavetobemultiplied<br />
by<br />
{N Fµ/Fδµ} t<br />
whichisasquaremodulo PFbecause tiseven.Thus<strong>the</strong>resultisvalidforallchoices<strong>of</strong> ̟Fµ .<br />
Eventuallywewillhavetodiscuss<strong>the</strong>variouspossibilitiesseparately.Therearehowever<br />
anumber<strong>of</strong>commentsweshouldmakefirst.If misoddand m(χ Fµ/F)isodd<strong>the</strong>n<br />
if<br />
∆3(µ ′ χ Fµ/F, ψ Fµ/F, ̟ m+n<br />
F ) ∼p νφµ (β + α(µ′ )) A[−σ(ϕφµ )]<br />
ψφµ (x) = ψ Fµ/F<br />
<br />
x<br />
̟ 1+n<br />
<br />
.<br />
F<br />
Observethat,because misodd,wemaytake<strong>the</strong>number δinLemma9.3tobe ̟ m−1<br />
2<br />
F . Of<br />
course ϕφµ isanyfunctionon φµwhichvanishesnowhereandsatisfies<br />
ApplyingLemma9.1weseethat<br />
if k = [G0 : G1],if<br />
ϕφµ (x + y) = ϕφµ (x) ϕφµ (y)ψφµ (xy).<br />
A[−σ(ϕφµ )] ∼p −νφ<br />
<br />
ψφ(x) = ψF<br />
k [Fµ:F]<br />
k<br />
<br />
<br />
x<br />
̟ 1+n<br />
F<br />
<br />
A σ(ϕφ) [Fµ:F] <br />
k<br />
andif ϕφbears<strong>the</strong>usualrelationto ψφ.Weuse,<strong>of</strong>course,<strong>the</strong>relation<br />
Observealsothat<br />
If misodd<br />
kS φµ/φ(x) = S Fµ/F(x).<br />
νφµ (β + α(µ′ )) = νφ(N φµ/φ(β + α(µ ′ ))).<br />
∆3(χF, ψF, ̟ m+n<br />
F ) ∼p −νφ(β) A[σ(ϕφ)].<br />
Ifboth mand m ′ = m(χ E/F)areoddandif β ′ = β(χ E/F)<strong>the</strong>n<br />
∆3(χ E/F, ψ E/F, ̟ m+n<br />
F ) ∼p −νφ(β ′ ) A[σ(ϕ ′ φ )]
if ϕ ′ φ bears<strong>the</strong>usualrelationto<strong>the</strong>character<br />
ψ ′ φ (x) = ψ E/F<br />
⎛<br />
⎝<br />
̟ 1+n<br />
F<br />
⎞<br />
x<br />
⎠ . (q−1)t<br />
k ̟<br />
Thereisaunit εin OKsuchthat ̟E = ε̟ q<br />
K .If σ ∈ C<strong>the</strong>n<br />
̟ σ−1<br />
E = εσ−1̟ (σ−1)q<br />
K ≡ 1 (mod PK)<br />
because t ≥ 1.Thus<strong>the</strong>multiplicativecongruence<br />
issatisfiedand<br />
If<br />
asbefore,<strong>the</strong>n<br />
Since<br />
wehave<br />
̟ q<br />
E ≡ N E/F̟E = ̟F (mod ∗ PE)<br />
1<br />
̟ (q−1)t<br />
k<br />
E<br />
≡ ̟1+n<br />
F<br />
̟ 1+n′<br />
E<br />
ω1 = S<br />
<br />
ψ ′ φ(x) = ψF<br />
̟ 1+n<br />
F<br />
̟ 1+n′<br />
E<br />
E<br />
(mod ∗ PE).<br />
<br />
<br />
ω1x<br />
̟ 1+n<br />
<br />
.<br />
F<br />
νφ(β ′ ) = νφ(β ′ ) q = νφ(N E/Fβ ′ )<br />
∆3(χ E/F, ψ E/F, ̟ m+n<br />
F ) ∼p −νφ(N E/Fβ ′ ) νφ(ω1) A[σ(ϕφ)].<br />
Define ηbydemandingthat<br />
beequalto<br />
∆3(χ E/F,ψ E/F, ̟ m+n<br />
F )<br />
η∆3(χF, ψF, ̟ m+n<br />
F ) <br />
µ ∆3 (µ ′ χ Fµ/F, ψ Fµ/F, ̟ m+n<br />
F ).<br />
Wehavetoshowthat η ∼p 1.Ifboth tand mareeventhisisclear.If tisevenand misodd<br />
wearetoshowthat<br />
νφ(N E/Fβ ′ )νφ(ω1) ∼p (−1) a νφ(−1) q−1<br />
2k νφ(k) q−1<br />
k νφ(β) <br />
µ νφ(N φµ/φ(β + α(µ ′ )))
Chapter14 190<br />
if ais<strong>the</strong>number<strong>of</strong>elementsin T.Since tiseven kis1.Asbefore<br />
β <br />
µ N φµ/φ(β + α(µ ′ )) = β <br />
µ<br />
<br />
σ∈G/G1<br />
iscongruentto N E/Fβ ′ modulo PK.Allweneeddoisshowthat<br />
νφ(ω1) = (−1) a νφ(−1) q−1<br />
2 .<br />
(β + α(µ ′ ) σ )<br />
Since tiseveneach γµisasquarein φ.Applying<strong>the</strong>identity(14.3)weseethat<br />
νφ(ω1) = νφ(ω q<br />
<br />
−1<br />
1 ) = νφ(α)νφ µ NFµ/Fβ(µ ′ <br />
) .<br />
Wehaveseenthat αisasquarein φsothat νφ(α) = 1.Wealsosawthat<br />
νφ<br />
<br />
µ NFµ/Fβ(µ ′ <br />
)<br />
when tiseven.Therequiredrelationfollows.<br />
= (−1) a νφ(−1) q−1<br />
2<br />
Wesupposehenceforththat tisodd.Thediscussionwillbefairlycomplicated.Suppose<br />
firstthat misalsoodd.Then<br />
[G0 : G1](m − 1) = t<br />
and<br />
sothat<br />
and<br />
<br />
m − 1 ><br />
t<br />
[G0 : G1]<br />
β + α(µ ′ ) ≡ β (mod PL)<br />
µ ∆3(µ ′ χ Fµ/F, ψ Fµ/F, ̟ m+n<br />
F ) ∼p (−1) a νφ<br />
Thusif q−1<br />
k isoddwehavetoshowthat<br />
andif q−1<br />
k isevenwehavetoshowthat<br />
<br />
k q−1<br />
k<br />
<br />
νφ<br />
<br />
β q−1<br />
k<br />
<br />
A[σ(ϕφ)] q−1<br />
k .<br />
(−1) a+1 νφ(k) νφ(−1) q−1 1<br />
2k + 2 ∼p 1 (14.4)<br />
νφ(ω1) ∼p (−1) a νφ(−1) q−1<br />
2k . (14.5)
Chapter14 191<br />
Nowsuppose miseven. If [G0 : G1]is1<strong>the</strong>reisnothingtoprove. If k = [G0 : G1]is<br />
even<strong>the</strong>n<br />
t<br />
m − 1 ><br />
[G0 : G1]<br />
and<br />
If<br />
β + α(µ ′ ) ≡ β (modPL).<br />
ψ ′ φµ (x) = ψ Fµ/F<br />
⎛<br />
⎝β ̟k(m−1)<br />
Fµ<br />
̟ m+n<br />
F<br />
⎞<br />
x<br />
⎠<br />
and ϕ ′ isafunctionon φµwhichvanishesnowhereandsatisfies<br />
φµ<br />
<strong>the</strong>n<br />
If<br />
<strong>the</strong>n εµisaunitand<br />
if,asbefore,<br />
ByLemma9.1, A[−σ(ϕφµ )]isequalto<br />
−νφ<br />
<br />
ϕ ′ φµ (x + y) = ϕ′ φµ (x) ϕ′ φµ (y) ψ′ φµ (xy)<br />
∆3(µ ′ χ Fµ/F, ψ Fµ/F, ̟ m+n<br />
F ) ∼p A[−σ(ϕ ′ φµ )].<br />
k [Fµ:F]<br />
k<br />
<br />
νφ<br />
If q−1<br />
k isevenwehavetoshowthat<br />
(−1) a νφ<br />
εµ̟ m−1<br />
F<br />
= ̟ k(m−1)<br />
Fµ<br />
ψφµ (x) = ψ φµ/φ(kβεµx)<br />
ψφ(x) = ψF<br />
<br />
<br />
β [Fµ:F]<br />
k<br />
<br />
µ N φµ/φεµ<br />
If q−1<br />
k isodd<strong>the</strong>n m ′ = m(χ E/F)isodd.If<br />
ψ ′′<br />
φ (x) = ψ E/F<br />
<br />
x<br />
̟ 1+n<br />
<br />
.<br />
F<br />
νφ(N φµ/φεµ) A[σ(ϕφ)] [Fµ:F]<br />
k .<br />
<br />
<br />
νφ(−1) q−1<br />
2k ∼p 1 (14.6)<br />
β ′ ̟m′ −1<br />
E<br />
̟ m+n<br />
F<br />
<br />
x
Chapter14 192<br />
and ϕ ′′ φ<br />
bears<strong>the</strong>usualrelationto ψ′′<br />
φ <strong>the</strong>n<br />
Now νφ(β ′ ) = νφ(β ′ ) q and<br />
whichinturniscongruentto<br />
modulo PK.Let<br />
and,asbefore,<br />
<strong>the</strong>n<br />
Wesawthat<br />
sothat<br />
Thuswehavetoshowthat<br />
∆3(χ E/F,ψ E/F, ̟ m+n<br />
F ) ∼p A[−σ(ϕ ′′ φ )].<br />
β <br />
µ∈T<br />
(β ′ ) q ≡ N E/Fβ ′<br />
<br />
σ∈Mµ<br />
ε1̟ m+n<br />
F<br />
ω1 = S E/F<br />
(β + α(µ ′ ) σ ) ≡ β q<br />
= ̟ q(m+n)<br />
E<br />
<br />
̟ 1+n<br />
F<br />
̟ 1+n′<br />
E<br />
A[−σ(ϕ ′′ φ )] ∼p νφ(ω1) νφ(ε1) νφ(β) A[−σ(ϕφ)].<br />
<br />
̟ q<br />
E ≡ ̟F (mod ∗ PE)<br />
ε1 ≡ 1 (modPE).<br />
νφ(ω1) ∼p (−1) a+1 νφ(k) νφ(−1) q−1<br />
2k −1<br />
2 νφ<br />
<br />
µ N φµ/φεµ<br />
<br />
. (14.7)<br />
Thefouridentities(14.4),(14.5),(14.6),and(14.7)lookra<strong>the</strong>rinnocuous. Howeverto<br />
prove<strong>the</strong>misnotanentirelytrivialmatter.Wefirstconsider<strong>the</strong>casethat G/G1isabelian.If<br />
σ0 ∈ G/G1is<strong>of</strong>order2,<strong>the</strong>argumentusedbeforeshowsthat σ0cσ −1<br />
0 = c−1 forall cin C.<br />
Since<strong>the</strong>representation<strong>of</strong> G/G1on Cisfaithful G/G1hasonlyoneelement<strong>of</strong>order2andis<br />
<strong>the</strong>reforecyclic.Inthiscase Fµ = Lforall µand a = q−1<br />
[G:G1] .Wemaychoose ̟Fµ = ̟L.If<br />
<strong>the</strong>n γµ = γ t and<br />
N L/F̟L = γ̟ [G:G0]<br />
F<br />
<br />
µ γµ = γ at .
Chapter14 193<br />
If [G0 : G1] = 1wemaychoose ̟L = ̟Fsothat γ = 1. Theargumentusedbefore<br />
showsthat <br />
<br />
β(µ ′ ) σ<br />
<br />
= (−1) a νφ(−1) q−1<br />
2 .<br />
νφ<br />
µ∈T<br />
Theidentity(14.3)showsthat<br />
σ∈G/G1<br />
νφ(ω1) = (−1) a νφ(−1) q−1<br />
2k .<br />
Theidentity(14.5)whichis<strong>the</strong>onlyone<strong>of</strong>concernherebecomes<br />
whichisclearbecause k = [G0 : G1] = 1.<br />
νφ(−1) q−1<br />
2 = νφ(−1) q−1<br />
2k<br />
Nowtake [G : G0] = 1.Wemaychoose ̟E = N K/E̟Ksothat ̟F = N L/F̟Land γ<br />
isagain1.Itisperhapsworthpointingout<strong>the</strong>sespecialchoicesarenotinconsistentwithany<br />
choicesyetmadeinthisparagraph.Thisisnecessarybecause<strong>the</strong>argumentsappearingin<strong>the</strong><br />
<strong>functions</strong> ∆2mustbe<strong>the</strong>sameasthoseappearingin<strong>the</strong><strong>functions</strong> ∆3.Wepreviouslydefined<br />
andshowedthat<br />
Observethat<br />
because<br />
̟ k L<br />
̟F<br />
= <br />
σ∈G/G1<br />
δ = ̟1+n<br />
F<br />
̟ 1+n′<br />
E<br />
· ̟t L<br />
̟ t K<br />
NK/Lδ = ̟t(q−1)<br />
L<br />
̟ t(q−1)<br />
k<br />
F<br />
.<br />
̟ 1−σ<br />
L ≡ θ0(σ) −1 ≡ −1 (modPL)<br />
{θ0(σ) | σ ∈ G/G1}<br />
isjust<strong>the</strong>set<strong>of</strong> kthroots<strong>of</strong>unityin φand kisapower<strong>of</strong>2.Itisnot1because<br />
Since<br />
wehave<br />
[G0 : G1] = [G : G1] > 1.<br />
δ q ≡ N K/Lδ (modPK)<br />
νφ(δ) = νφ(−1) q−1<br />
k .
Chapter14 194<br />
Ifasbefore<br />
ψ L/F<br />
β1(ν)x<br />
̟ 1+n<br />
F ̟t L<br />
<br />
= ν(1 + x)<br />
for xin Ps Land νin S(K/L)<strong>the</strong>n,aswesawwhenproving<strong>the</strong>identity(14.3),<br />
Thus<br />
ω q<br />
1 ≡ δqα q<br />
<br />
<br />
ν=1 β1(ν)<br />
<br />
νφ(ω1) = νφ(−1) q−1<br />
k<br />
<br />
(mod PL).<br />
ν=1 νφ(β1(ν)).<br />
Wecanchoose q−1<br />
p−1elements νiin S(K/L)sothateverynontrivialelement<strong>of</strong> S(K/L)is<strong>of</strong><br />
<strong>the</strong>form ν j<br />
i , 0 < j < p.Then<br />
because<br />
When miseven<br />
<br />
ν=1 νφ(β1(ν)) = νφ<br />
νφ(εµ) = νφ<br />
q−1<br />
p−1<br />
i=1<br />
j=1 jβ(νi)<br />
<br />
p−1<br />
νφ(β(νi)) p−1 = 1.<br />
k ̟L = νφ(−1).<br />
̟F<br />
= νφ(−1) q−1<br />
p−1<br />
Since a = q−1<br />
k <strong>the</strong>identities(14.4),(14.5),(14.6)and(14.7)become<br />
νφ(k) νφ(−1) q−1 1<br />
2k + 2 = 1 (14.4 ′ )<br />
νφ(−1) q−1<br />
p−1 = νφ(−1) q−1<br />
2k (14.5 ′ )<br />
νφ(−1) q−1<br />
2k = 1 (14.6 ′ )<br />
νφ(−1) q−1<br />
k νφ(−1) q−1<br />
p−1 = νφ(k) νφ(−1) q−1 1<br />
2k + 2 νφ(−1) q−1<br />
k . (14.7 ′ )<br />
If p ≡ 1 (mod4)<strong>the</strong>identities(14.5 ′ )and(14.6 ′ )areclearlyvalid. Moreoverfor(14.5 ′ )and<br />
(14.6 ′ )<strong>the</strong>number q−1<br />
k iseven. Since kisapositivepower<strong>of</strong>2, qisanevenpower<strong>of</strong> pif<br />
p ≡ 3 (mod4).If q = p2f<strong>the</strong>n q − 1<br />
p − 1 = 1 + p + . . . + p2f−1 ≡ 0 (mod4)
Chapter14 195<br />
and<strong>the</strong>leftside<strong>of</strong>(14.5 ′ )is1. If q−1<br />
2k iseven(14.5′ )and(14.6 ′ )arenowclear. Ifitisodd,4<br />
divides kbecause8divides q − 1.But {θ0(σ) | σ ∈ G0/G1}is<strong>the</strong>set<strong>of</strong> kthroots<strong>of</strong>unityin<br />
OL/PL = φso1isasquarein φ, νφ(−1) = 1,and<strong>the</strong>relationsarevalidinthiscasetoo.The<br />
relations(14.4 ′ )and(14.7 ′ )areobviousif<strong>the</strong>degree<strong>of</strong> φover<strong>the</strong>primefield φ0iseven.Since<br />
φ × contains<strong>the</strong> kthroots<strong>of</strong>unityand kisapower<strong>of</strong>2<strong>the</strong>degreecanbeoddonlyif kdivides<br />
p − 1.Since<br />
q − 1<br />
k<br />
q − 1 p − 1<br />
= ·<br />
p − 1 k<br />
and q−1<br />
k isnowodd p−1<br />
k mustalsobeoddandbyquadraticreciprocity<br />
because<br />
νφ(k) = νφ0 (k) = νφ0 (−1) νφ0<br />
<br />
p ≡ 1 mod<br />
<br />
p − 1<br />
k<br />
<br />
p − 1<br />
.<br />
k<br />
p−1<br />
= νφ0 (−1) νφ0 (−1) 2k −1<br />
2<br />
If p ≡ 1 (mod4)<strong>the</strong>tworelationsarenowclear.If p ≡ 3 (mod 4)and q = p f<br />
q − 1<br />
p − 1<br />
= 1 + p + . . . + pf−1<br />
mustbeodd.Itis<strong>the</strong>reforecongruentto1modulo4.(14.4 ′ )becomes<br />
and(14.7 ′ )becomes<br />
p−1<br />
νφ0 (−1) νφ0 (−1) k = 1<br />
νφ0 (−1) = νφ0 (−1).<br />
Thereisnoquestionthatboth<strong>the</strong>serelationsarevalid.<br />
Wehavestilltotreat<strong>the</strong>casethat G/G1isabelianwhilenei<strong>the</strong>r [G : G0]nor [G0 : G1]is<br />
1.Then<br />
NFµ/Fβ(µ ′ <br />
<br />
) ≡ β(µ ′ k ) (modPFµ )<br />
σ∈G/G0<br />
isasquarein φand<strong>the</strong>identity(14.3)impliesthat<br />
νφ(ω1) = νφ(γ a )<br />
CF/N L/FCLiscyclic<strong>of</strong>order [G : G1]. Ithasageneratorwhichcontainsanelement<strong>of</strong><strong>the</strong><br />
form γ1̟F.Moreover<strong>the</strong>coset<strong>of</strong><br />
(γ1̟F) [G:G0] NL/F̟ −1<br />
L<br />
= γ[G:G0]<br />
1 γ −1
Chapter14 196<br />
isagenerator<strong>of</strong>UF/UF ∩N L/FCL.Theorder<strong>of</strong>thisgroupisapower<strong>of</strong>2andpisoddsoevery<br />
element<strong>of</strong> UF ∩ N L/FCLisasquare.Consequently γcannotbeasquareand νφ(γ) = −1.If<br />
misevenand F ′ is<strong>the</strong>fixedfield<strong>of</strong> G0<strong>the</strong>n<br />
εµ =<br />
whichiscongruentto<br />
̟ k L<br />
̟F<br />
m−1<br />
=<br />
modulo PL.Since [F ′ : F]iseven<br />
and<br />
Because<br />
NL/F ′̟L<br />
−<br />
̟F<br />
N φµ/φ εµ = N F ′ /F εµ =<br />
m−1 <br />
m−1 NL/F ′̟L<br />
̟F<br />
σ∈G0/G1<br />
m−1 NL/F̟L<br />
̟ [G:G0]<br />
F<br />
νφ(N φµ/φ εµ) = νφ(γ) = −1.<br />
a =<br />
q − 1<br />
[G : G1]<br />
̟ 1−σ<br />
m−1 L<br />
= γ m−1<br />
isintegral, q−1<br />
k iseven,andweneedonlyworryabout<strong>the</strong>identities(14.5)and(14.6). They<br />
bothreduceto<br />
νφ(−1) q−1<br />
2k = 1.<br />
Toprovethisweshowthat q−1<br />
2k isevenif νφ(−1) = −1.Since<br />
k = [UF : UF ∩ N L/FCL]<br />
andthisindexmustdivide<strong>the</strong>order<strong>of</strong> φ ∗ <strong>the</strong>number νφ(−1)is1onlyif k = 2.Ofcourse<br />
pwillbecongruentto3modulo4. Since4divides q − 1, qisanevenpower<strong>of</strong> pand<br />
q ≡ 1 (mod8).Thus<br />
iseven.<br />
q − 1<br />
2k<br />
= q − 1<br />
4<br />
Nowsupposethat G/G1isnotabelian.Let σ −→ x(σ)beagivenisomorphism<strong>of</strong> G0/G1<br />
with Z/kZandlet x −→ σ(x)beitsinverse.Let τ −→ λ(τ)bethathomomorphism<strong>of</strong> G/G0<br />
into<strong>the</strong>units<strong>of</strong> Z/kZwhichsatisfies<br />
x(τστ −1 ) = λ(τ) x(σ).
Chapter14 197<br />
Thereispreciselyoneelement<strong>of</strong>order2in G0/G1,namely σ <br />
k<br />
2 ,anditliesin<strong>the</strong>center<strong>of</strong><br />
G/G1.Since G/G0iscyclic, G/G1isnonabelianonlyif k > 2.Chooseafixed σ0in Gwhich<br />
generates G/G0andset<br />
µ0 = λ(σ0)<br />
and<br />
y0 = x(σ [G:G0]<br />
0 ).<br />
Weshallsometimesregard Casavectorspaceover<strong>the</strong>fieldwith pelements.If σbelongsto<br />
G/G1let π(σ)be<strong>the</strong>lineartransformation<br />
c −→ σcσ −1 .<br />
Thedualspacewillbeidentifiedwith S(K/L)and π ∗ willbe<strong>the</strong>representationcontragredient<br />
to π.<br />
Therelation<br />
NFµ/Fβ(µ ′ <br />
<br />
) ≡<br />
toge<strong>the</strong>rwith<strong>the</strong>identity(14.3)impliesthat<br />
Moreoverif misevenand F ′ µ<br />
εµ =<br />
̟ k Fµ<br />
whichiscongruentto<br />
modulo PFµ .Since<br />
whichequals<br />
̟F<br />
m−1<br />
νφ(ω1) = νφ<br />
σ∈G/GµG0<br />
<br />
µ γµ<br />
is<strong>the</strong>fixedfield<strong>of</strong> GµG0<br />
=<br />
NFµ/F ′ µ ̟Fµ<br />
̟F<br />
{N φµ/φ εµ} t = N F ′ µ /F<br />
β(µ ′ ) σ<br />
k <br />
.<br />
m−1 <br />
<br />
− NFµ/F ′ µ ̟Fµ<br />
m−1 ̟F<br />
σ∈G0/G1<br />
<br />
− NFµ/F ′ µ ̟Fµ<br />
(m−1)t<br />
̟F<br />
(−1) t[φµ:φ] γ m−1<br />
µ<br />
̟ 1−σ<br />
m−1 Fµ
Chapter14 198<br />
and tisodd<br />
νφ(N φµ/φ εµ) = νφ(−1) [φµ:φ] νφ(γµ).<br />
Theserelationswillbeusedfrequentlyandwithoutcomment.<br />
Iwanttodiscuss<strong>the</strong>case [G : G0] = 2and µ0 ≡ −1 (mod4)first.Since<br />
wemusthave<br />
or,if k > 4,<br />
Then<br />
or<br />
Since<br />
(−µ0) 2 = µ 2 0 = λ(σ 2 0) ≡ 1 (modk)<br />
−µ0 ≡ 1 (modk)<br />
−µ0 ≡ k<br />
+ 1 (mod k).<br />
2<br />
µ0 ≡ −1 (modk)<br />
µ0 ≡ k<br />
− 1 (mod k).<br />
2<br />
µ0 − 1 ≡ 2 (mod 4)<br />
<strong>the</strong>centralizer<strong>of</strong> σ0in G0/G1consists<strong>of</strong><strong>the</strong>identityand σ <br />
k 2<br />
2 .Thus x(σ0)is0or k<br />
2 .<br />
Suppose µ0 ≡ −1 (modk)and x(σ2 k<br />
0 ) = 0 = σ−1<br />
and (σ0σ) 2 = σ2 0 . Thus σ <br />
k<br />
is<strong>the</strong>onlyelement<strong>of</strong>order2in G/G1. If σbelongsto G/G1<br />
2<br />
<strong>the</strong>n σhasanontrivialfixedpointin S(K/L)ifandonlyif π(σ)has1asaneigenvalue. If<br />
σ = 1<strong>the</strong>reisaninteger nsuchthat σnhasorder2.Then π(σn )alsohas1asaneigenvalue.<br />
Thusifanynontrivialelement<strong>of</strong> G/G1hasanontrivialfixedpoint<strong>the</strong>reisanelement τ<strong>of</strong><br />
order2suchthat π(τ)has1asaneigenvalue.Theusualargumentshowsthat<br />
<br />
π σ<br />
2 . If σbelongsto G0/G1<strong>the</strong>n σ0σσ −1<br />
<br />
k<br />
= −I<br />
2<br />
sothat,in<strong>the</strong>caseunderconsideration,only<strong>the</strong>identityhasfixedpoints.Then<br />
a =<br />
q − 1<br />
[G : G1] .
Chapter14 199<br />
Inparticular q−1<br />
k iseven.Wechoose ̟Fµ = ̟Landlet<br />
γ̟ [G:G0]<br />
F<br />
= N L/F̟L.<br />
<strong>On</strong>lyidentities(14.5)and(14.6)aretobeconsidered.(14.5)reducesto<br />
and(14.6)reducesto<br />
νφ(γ) at = (−1) a νφ(−1) q−1<br />
2k<br />
(−1) a νφ(−1) a[G:G0] νφ(γ) at νφ(−1) q−1<br />
2k = 1.<br />
Since [G : G0]iseven<strong>the</strong>yareequivalent.Suppose φhas relements.If x ∈ λ = OL/PL<strong>the</strong>n<br />
xσ0 r = x f<br />
forsome f.If σbelongsto G0/G1<strong>the</strong>n<br />
Thus<br />
and<br />
f<br />
µ0r<br />
θ0(σ)<br />
= θ0(σ0σσ −1<br />
0 )rf<br />
µ rf<br />
0<br />
≡<br />
≡ 1 (modk)<br />
r ≡ −1 (mod4)<br />
sothat νφ(−1) = −1.Since,in<strong>the</strong>presentcase,<br />
<strong>the</strong>identitiesbecome<br />
Themap<br />
a =<br />
q − 1<br />
2k<br />
νφ(γ) at = 1.<br />
τ L/F : W L/F −→ CF<br />
σ0σ<br />
̟L ̟ σ0<br />
f −1<br />
r σo ≡ θ0(σ).<br />
L<br />
determinesamap<strong>of</strong> G/G1onto CF/N L/FCL. Theimage<strong>of</strong> σ0containsanelement<strong>of</strong><strong>the</strong><br />
form γ1̟Fwhere γ1isaunit.Theimage<strong>of</strong> σ 2 0 is1because<strong>the</strong>commutatorsubgroupcontains<br />
{σ((µ0 − 1)x)} = {σ(x) | x ≡ 0 (mod2)}<br />
andinparticularcontains σ 2 0 .Since [G : G0] = 2<strong>the</strong>number γγ −2<br />
1 liesin UF ∩ NL/FCL.The<br />
index<strong>of</strong><strong>the</strong>commutatorsubgroup<strong>of</strong> G/G1in G/G1is4so<br />
[UF : UF ∩ N L/FCL] = 2.
Chapter14 200<br />
Consequently γγ −2<br />
1 and γarebothsquaresand νφ(γ) = 1.<br />
Nowsuppose µ0 ≡ −1 (modk)and x(σ 2 0 ) = 0. Everyelement<strong>of</strong><strong>the</strong>form σ0σ, σ ∈<br />
G0/G1,hasorder2.If π(σ0σ) = −I<strong>the</strong>n σ0σliesin<strong>the</strong>center<strong>of</strong> G/G1whichisimpossible.<br />
Thus π(σ0σ)has1asaneigenvalue.If τ ∈ G0/G1<strong>the</strong>n<br />
τ −1 σ0στ = σ0στ 2<br />
so<strong>the</strong>rearetwoconjugacyclassesin<strong>the</strong>set σ0G0/G1.<strong>On</strong>ehas σ0asrepresentativeand<strong>the</strong><br />
o<strong>the</strong>rhas σ1 = σ0σ(1).<br />
Let V beanontrivialsubspace<strong>of</strong> S(K/L)invariantandirreducibleunder<strong>the</strong>action<strong>of</strong><br />
G0/G1. Supposefirstthat V isalsoinvariantunder π ∗ (σ0)sothat V = S(K/L). Choose<br />
v0 = 0sothat π ∗ (σ0)v0 = v0.Let λ ′ be<strong>the</strong>fieldobtainedbyadjoining<strong>the</strong> kthroots<strong>of</strong>unity<br />
to<strong>the</strong>primefield.Certainly λ ′ ⊆ λand,since<br />
θ0(σ) σ0 = θ0(σ −1<br />
0 σσ0),<br />
λ ′ isnotcontainedin φ.Let φ ′ = φ∩λ ′ .Wemayregard {1, σ0}as G(λ ′ /φ ′ ).Themap ϕwhich<br />
sends σin G0/G1to (θ −1<br />
0 (σ), 1)and σσ0to (θ −1<br />
0 (σ), σ0)isanisomorphism<strong>of</strong> G/G1with<strong>the</strong><br />
semidirectproduct<strong>of</strong><strong>the</strong> kthroots<strong>of</strong>unityin λ ′ and G(λ ′ /φ ′ ).Thereisauniquemap,again<br />
denotedby ϕ,<strong>of</strong> Vonto λ ′ suchthat ϕ(v0) = 1while<br />
ϕ(π ∗ (τ)v) = ϕ(τ)ϕ(v)<br />
for τin G/G1. Ofcourse<strong>the</strong> kthroots<strong>of</strong>unityacton λ ′ byleftmultiplication. TheGalois<br />
groupactsby σ0α = ασ−1 0 . Putting<strong>the</strong>actionstoge<strong>the</strong>rwegetanaction<strong>of</strong><strong>the</strong>semidirect<br />
product.Tostudy<strong>the</strong>action<strong>of</strong> G/G1on V westudy<strong>the</strong>equivalentaction<strong>of</strong><strong>the</strong>semidirect<br />
productin λ ′ .<br />
Itisbesttoconsideramoregeneralsituation.Suppose φ ′ isafinitefieldwith pfelements, λ ′ isanextension<strong>of</strong> φ ′ with pℓelementsand Γis<strong>the</strong>semidirectproduct<strong>of</strong><strong>the</strong>group<strong>of</strong> kth<br />
roots<strong>of</strong>unity,where kdivides pℓ − 1,and G(λ ′ /φ ′ ). Γactson λ ′ asbefore. Let ℓ = nf. If<br />
0 ≤ j1 < n, j = (j1, n),and ρis<strong>the</strong>automorphism x −→ xpf<strong>of</strong> λ ′ /φ ′ <strong>the</strong>n<strong>the</strong>number<strong>of</strong><br />
elements<strong>of</strong> λ ′ fixedbyamember<strong>of</strong> Γ<strong>of</strong><strong>the</strong>form (α, ρji )where αisakthroot<strong>of</strong>unityis<strong>the</strong><br />
sameas<strong>the</strong>number<strong>of</strong>elementsfixedbysomeo<strong>the</strong>rmember<strong>of</strong><strong>the</strong>form (β, ρ−j ).Indeedif<br />
b j1<br />
j<br />
<br />
≡ −1 mod n<br />
<br />
j
Chapter14 201<br />
and bisprimeto<strong>the</strong>order<strong>of</strong> (α, ρ j1 )wecantake<br />
(β, ρ −j ) = (α, ρ j1 ) b .<br />
Let θbeagenerator<strong>of</strong><strong>the</strong>multiplicativegroup<strong>of</strong> λ ′ .Theequation<br />
β θ mρj<br />
= θ m<br />
canbesolvedfor βifandonlyif θ m(pjf −1 )hasorderdividing k,thatis,ifandonlyif p ℓ − 1<br />
divides km(p jf − 1)or,if<br />
u = pℓ − 1<br />
k<br />
ifandonlyif udivides m(pjf − 1).Let u(j)be<strong>the</strong>greatestcommondivisor<strong>of</strong> uand pjf − 1.<br />
udivides m(pjf −1)ifandonlyif u<br />
u(j) divides m.Thenumber<strong>of</strong>such mwith 0 ≤ m < pℓ −1<br />
is<br />
u(j)<br />
u (pℓ − 1) = u(j)k.<br />
<strong>On</strong>ce mand jarechosen αisdetermined.Thenumber<strong>of</strong>nonzero xin λ ′ whicharefixedby<br />
some (β, ρ −j )where jdivides nbutbyno (β, ρ −i )where iproperlydivides jis<br />
<br />
i|j µ<br />
<br />
j<br />
i<br />
u(i)k<br />
if µ(·)is<strong>the</strong>Möbiusfunction.Thenumber<strong>of</strong>orbitsformedbysuch xis<br />
1<br />
jk<br />
<br />
i|j µ<br />
<br />
j<br />
i<br />
u(i)k<br />
sothat<strong>the</strong>totalnumber<strong>of</strong>orbits<strong>of</strong> Γin<strong>the</strong>multiplicativegroup<strong>of</strong> λ ′ is<br />
whichequals<br />
Theproductisoverprimes.<br />
Lemma 14.7<br />
a = <br />
<br />
i|n<br />
i|n<br />
<br />
π<br />
<br />
<br />
n<br />
<br />
n<br />
i<br />
j<br />
i<br />
µ(j)<br />
ij u(i)<br />
<br />
1 − 1<br />
<br />
u(i)<br />
.<br />
π i
Chapter14 202<br />
because<br />
If pℓ −1<br />
k isodd<strong>the</strong>n<br />
Theidentity<strong>of</strong><strong>the</strong>lemmaisequivalentto<br />
(−1) a+1 νφ ′(k) νφ ′(−1) pℓ−1 1<br />
2k + 2 = 1.<br />
(−1) a+1 u−1<br />
νφ ′(u) νφ ′(−1) 2 = 1<br />
νφ ′(k) = νφ ′(−1) νφ ′(u).<br />
By<strong>the</strong>law<strong>of</strong>quadraticreciprocity<strong>the</strong>leftside<strong>of</strong><strong>the</strong>identityisequalto<br />
(−1) a+1 (p f |u)<br />
if (p f |u)isJacobi’ssymbol.If u = 1<strong>the</strong>reisonlyoneorbitso<br />
(−1) a+1 = 1.<br />
Ofcourse (p f |1) = 1so<strong>the</strong>identityisclearinthiscase.<br />
Weproveitingeneralbyinductionon<strong>the</strong>number<strong>of</strong>primefactors<strong>of</strong> u. Let π0bea<br />
primefactor<strong>of</strong> uandlet u = π x 0 vwith vprimeto π0.Let v(j)be<strong>the</strong>analogue<strong>of</strong> u(j).Then<br />
u(j) = π x(j)<br />
0 v(j).Leb bbe<strong>the</strong>analogue<strong>of</strong> a.Then<br />
ν = a − b = <br />
i|n<br />
<br />
π<br />
<br />
n<br />
i<br />
<br />
1 − 1<br />
<br />
v(i)<br />
(π<br />
π i<br />
x(i)<br />
0 − 1).<br />
Observethat π0andall v(i)areodd.Toprove<strong>the</strong>lemmabyinductionwemustshowthat<br />
Let<br />
(−1) ν (p f | π x 0 ) = 1. (14.8)<br />
n = 2 y n1<br />
with n1odd.Therearetwopossibilitiestobeconsidered.<br />
(i)<br />
π0 ≡ 1 (mod 2 y+1 ).
Chapter14 203<br />
(ii)<br />
Since<strong>the</strong>order<strong>of</strong> p f modulo π0divides n<strong>the</strong>quotient<strong>of</strong> π0 − 1bythisorderisevenand<br />
p f isaquadraticresidue<strong>of</strong> π0.Alsoif idivides n<br />
π x(i)<br />
0<br />
− 1<br />
i<br />
isdivisible,in<strong>the</strong>2adicfield,by4if2divides n<br />
i<br />
evenand(14.8)isvalid.<br />
π0 = 1 + 2 c w<br />
with c ≤ yand wodd.Let i = n1divide n1andconsider<br />
y<br />
j=0<br />
<br />
π<br />
<br />
n<br />
2 j i<br />
<br />
1 − 1<br />
j v(2 i)<br />
π 2ji andisalwaysdivisibleby2.Thus νis<br />
(π x(2j i)<br />
0 − 1). (14.9)<br />
If x(2 y i) = 0<strong>the</strong>sumiszero. If x(2 y i) = 0let zbe<strong>the</strong>smallestintegerforwhich<br />
x(2 z i) = 0.If j < z<strong>the</strong>n x(2 j i) = 0.If j ≥ z<br />
p 2j if − 1 =<br />
<br />
p 2z <br />
if<br />
− 1<br />
2 j−z −1<br />
d=0<br />
Theresidue<strong>of</strong><strong>the</strong>summodulo π0is 2 j−z .Thus<br />
if j ≥ zand(14.9)isequalto<br />
Wewrite<br />
as<br />
1<br />
i<br />
<br />
π<br />
<br />
n1 i<br />
x(2 j i) = x(2 z i)<br />
<br />
1 − 1<br />
y v(2 i) y−1<br />
+<br />
π 2y j=z<br />
v(2yi) y−1<br />
+<br />
2y j=z<br />
v(2zi) y<br />
+<br />
2z j=z+1<br />
v(2 j i)<br />
2 j+1<br />
p 2z <br />
cif<br />
.<br />
v(2ji) − v(2j−1i) 2j .<br />
v(2ji) 2j+1 π x(2y <br />
i)<br />
0 − 1 .<br />
If kisreplacedby pℓ −1<br />
v <strong>the</strong>number<strong>of</strong>elements<strong>of</strong> λ ∗ fixedbysome (α, ρ −2j i )butbyno<br />
(α, ρ −2j−1 i )is<br />
{v(2 j i) − v(2 j−1 i)} pℓ − 1<br />
.<br />
v
Chapter14 204<br />
Thecollection<strong>of</strong>suchelementsisinvariantunder<strong>the</strong>groupobtainedbyreplacing kby pℓ −1<br />
v<br />
and φ ′ by<strong>the</strong>fieldwith p if elements.Theisotropygroup<strong>of</strong>eachsuchpointhasagenerator<br />
<strong>of</strong><strong>the</strong>form (α, ρ −2j i )and,<strong>the</strong>refore,hasorder n<br />
sothat 2 j divides<br />
2jiandindex 2j (p ℓ −1)<br />
v<br />
{v(2 j i) − v(2 j−1 i)} pℓ − 1<br />
v<br />
v(2 j i) − v(2 j−1 i).<br />
.Thus 2j (p ℓ −1)<br />
v<br />
divides<br />
Since n1<br />
i isdivisiblebyatleastoneprime,<strong>the</strong>expression(14.9)iscongruent,in<strong>the</strong>2adicfield,<br />
to<br />
1<br />
i<br />
<br />
π<br />
<br />
n1 i<br />
<br />
1 − 1<br />
z v(2 i)<br />
π 2z <br />
π x(2y <br />
i)<br />
− 1<br />
modulo4.Since z ≤ cand<strong>the</strong>productisnotemptythisiscongruent,in<strong>the</strong>2adicfieldagain,<br />
to0modulo2.Thus νisevenoroddaccordingas<br />
y<br />
j=0<br />
<br />
<br />
π2y−j<br />
<br />
1 − 1<br />
j v(2 n1)<br />
π 2j <br />
π<br />
n1<br />
x(2j <br />
n1)<br />
0 − 1<br />
isorisnotdivisibleby2in<strong>the</strong>2adicfield.Consequently<br />
ν ≡ y<br />
j=0<br />
<br />
<br />
π<br />
2y−j <br />
1 − 1<br />
j v(2 n1)<br />
π 2j (π x(2j n1)<br />
0<br />
− 1) (mod2).<br />
Ofcourse x(2 y n1) = x = 0. Let zagainbe<strong>the</strong>smallestintegerfor x(2 z n1) = 0. Then<br />
z ≤ cand<br />
x(2 j n1) = x(2 z n1)<br />
if j ≥ z.Thesumaboveisequalto<br />
z v(2 n1)<br />
+ y<br />
2 z<br />
j=z+1<br />
Asbeforethisiscongruentmodulo2to<br />
v(2 z n1)<br />
2 z<br />
v(2jn1) − v(2j−1n1) 2j <br />
(π x 0 − 1).<br />
(π x 0<br />
− 1).<br />
If z < cthisisevenand<strong>the</strong>order<strong>of</strong> pmodulo π0divides π0−1<br />
2 sothat (p|π0) = 1. If z = c<br />
<strong>the</strong>n<br />
πx 0<br />
2z − 1<br />
1<br />
=<br />
2c x<br />
i=1<br />
<br />
x<br />
<br />
i<br />
(2 c w) i ≡ x (mod2)
Chapter14 205<br />
sothat ν ≡ x (mod 2).However<strong>the</strong>order<strong>of</strong> p f modulo π0isdivisibleby 2 z sothatitdoes<br />
notdivide π0−1<br />
2 and<br />
Therelation(14.8)isnoweasilyverified.<br />
<br />
f x x<br />
p | π0 = (−1) .<br />
Wereturnto<strong>the</strong>originalproblem. Since λ ′ isaquadraticextension<strong>of</strong> φ ′ and λ ′ isnot<br />
containedin φ<strong>the</strong>degree<strong>of</strong> φover φ ′ isodd.Since Vand λ ′ have<strong>the</strong>samenumber<strong>of</strong>elements<br />
q = p ℓ .If q−1<br />
k isodd<strong>the</strong>relation(14.4)followsimmediatelyfrom<strong>the</strong>equality<br />
and<strong>the</strong>precedinglemma.<br />
(−1) a+1 νφ(k) νφ(−1) q−1 1<br />
2k + 2 = (−1) a+1 νφ<br />
q−1<br />
′(k) νφ ′(−1) 2k<br />
Thenumber<strong>of</strong> µin Twithisotropygroup<strong>of</strong>order2is u(1)and<strong>the</strong>number<strong>of</strong> µwith<br />
trivialisotropygroupis u(2)−u(1)<br />
.Forpoints<strong>of</strong><strong>the</strong>secondtype [φµ : φ] = 2andforpoints<strong>of</strong><br />
2<br />
<strong>the</strong>firsttype [φµ : φ] = 1.Since,asweverifiedearlier,<br />
and<br />
<strong>the</strong>identity(14.7)reducesto<br />
νφ(ω1) = νφ<br />
<br />
µ γµ<br />
νφ(N φµ/φ εµ) = νφ(−1) [φµ:φ] νφ(γµ)<br />
u(1) ≡ 1 (mod 2)<br />
whichistruebecause u(1)divides u = pℓ −1<br />
k which,when(14.7)isunderconsideration,isodd<br />
byassumption.<br />
Theidentity(14.5)maybeformulatedas<br />
and(14.6)as<br />
νφ<br />
<br />
νφ<br />
µ γµ<br />
<br />
µ γµ<br />
For<strong>the</strong>setwoidentities q−1<br />
k iseven.Again<br />
<br />
<br />
<br />
= (−1) a νφ(−1) q−1<br />
2k<br />
νφ(−1) P<br />
µ [φµ:φ] = (−1) a νφ(−1) q−1<br />
2k .<br />
[φµ : φ] ≡ u(1) (mod2).<br />
+ 1<br />
2
Chapter14 206<br />
But<br />
and<br />
u(2) = u = pℓ − 1<br />
k<br />
2a = u(1) + u(2)<br />
so u(1)iseven.Itwillbeenoughtoverify(14.5).<br />
= q − 1<br />
k<br />
Wemaychoose Tsothatif µisin T<strong>the</strong>nitsisotropygroupistrivialorcontainsone<strong>of</strong> σ0<br />
or σ1.If σ0liesin<strong>the</strong>isotropygroup<strong>of</strong> µand νin<strong>the</strong>orbit<strong>of</strong> µcorrespondsto θ m in λ ′ <strong>the</strong>n<br />
α 2 θ mρ = θ m<br />
forsome kthroot<strong>of</strong>unity α.Thisispossibleifandonlyif p ℓ − 1divides km<br />
2 (pf − 1)or 2u<br />
divides m(p f − 1).Thisis<strong>the</strong>sameasrequiringthat 2u<br />
u(1) divide m(pf −1)<br />
u(1) . Thenumber uis<br />
even.Wehavealreadyobservedthatif ris<strong>the</strong>number<strong>of</strong>elementsin φsothat x σ0 = x r for x<br />
in φ<strong>the</strong>n<br />
µ0r ≡ 1 (modk)<br />
andinparticular<br />
µ0r ≡ 1 (mod4).<br />
Since [φ : φ ′ ]isoddand µ0 ≡ −1 (mod4)<strong>the</strong>highestpower<strong>of</strong>2dividing p f − 1is2.Thus<br />
2u<br />
u(1) and pf −1<br />
u(1)<br />
arerelativelyprimesothat 2u<br />
u(1) divides m(pf −1)<br />
u(1) ifandonlyif 2u<br />
u(1)<br />
divides m.<br />
Thereare u(1)k<br />
2 such mwith 0 ≤ m < pℓ − 1. Thecorrespondingcharacters νfallinto u(1)<br />
2<br />
orbits. Thus<strong>the</strong>reare u(1)<br />
2 elementsin Twhoseisotropygroupcontains σ0and u(1)<br />
2 whose<br />
isotropygroupcontains σ1.Let L0be<strong>the</strong>fixedfield<strong>of</strong> σ0and L1<strong>the</strong>fixedfield<strong>of</strong> σ1.Let ̟L0<br />
and ̟L1generate PL0and PL1respectivelyandlet Wehavetoshowthat<br />
νφ<br />
<br />
γ u(1)<br />
2<br />
N L0/F̟L0 = γ0 ̟F<br />
NL1/F̟L1 = γ1̟F<br />
NL/F̟L = γ̟ 2 F .<br />
0 γ u(1)<br />
2<br />
1 γ u(2)−u(1)<br />
<br />
2<br />
= (−1) a νφ(−1) q−1<br />
2k .<br />
Firstweprovealemma,specialcases<strong>of</strong>whichwehavealreadyseen.
Chapter14 207<br />
Lemma 14.8<br />
SupposeL/Fisnormalbutnonabelianand[L : F]isapower<strong>of</strong>2.SupposeH = G(L/F)<br />
and<strong>the</strong>firstramificationgroup H1is {1}but [H : H0] > 1and [H0 : H1] > 1.Let ̟Lgenerate<br />
<strong>the</strong>primeideal<strong>of</strong> OL,let ̟Fgenerate<strong>the</strong>primeideal<strong>of</strong> OF,andlet<br />
Then γisasquarein UF.<br />
NL/F ̟L = γ̟ [H:H0]<br />
F .<br />
Thehypo<strong>the</strong>sesimplythat<strong>the</strong>residuefieldhasoddcharacteristic.Let Abe<strong>the</strong>fixedfield<br />
<strong>of</strong> H0and L ′ be<strong>the</strong>fixedfield<strong>of</strong><strong>the</strong>commutatorsubgroup<strong>of</strong> H.Then A ⊆ L ′ andif<br />
<strong>the</strong>n<br />
̟L ′ = N L/L ′̟L<br />
N L ′ /F̟L<br />
′ = γ̟[A:F]<br />
F<br />
Ofcourse [A : F] = [H : H0]. Since Hisnilpotentbutnotabelian L ′ cannotbeacyclic<br />
extension.If γisnotasquarein UF<strong>the</strong>n γ −1 generates UF/UF ∩ N L ′ /FCL.Since<br />
̟ [A:F]<br />
F<br />
≡ γ −1<br />
(modN L ′ /FCL ′).<br />
̟Fwould<strong>the</strong>ngenerate CF/N L ′ /FCL ′whichisimpossible.<br />
Returningto<strong>the</strong>problemathand,weobservethat<strong>the</strong>quotient<strong>of</strong> G/G1by<strong>the</strong>squares<br />
in G0/G1isagroup<strong>of</strong>order4inwhicheverysquareis1.Thefixedfield F ′ <strong>of</strong>thisgroupis<br />
<strong>the</strong>composite<strong>of</strong>allquadraticextensions<strong>of</strong> F. F0 = F ′ ∩ L0and F1 = F ′ ∩ L1are<strong>the</strong>two<br />
differentramifiedquadraticextensions<strong>of</strong> F.Define<br />
and<br />
Then<br />
and<br />
̟F0 = N L0/F0 ̟L0<br />
̟F1 = NL1/F1̟L1 .<br />
N F0/F̟F0<br />
= γ0̟F<br />
NF1/F̟F1 = γ1̟F.<br />
.
Chapter14 208<br />
Weneedtoshowthat<br />
νφ(γ0γ1) = νφ<br />
γ0<br />
γ1<br />
<br />
= −1.<br />
Ifnot, γ0<br />
γ1 isasquareandthusin NF1/FCF1 .Then γ0̟Fbelongsto<br />
N F0/FCF0 ∩ N F1/FCF1 = N F ′ /FCF ′.<br />
Thisisimpossiblebecause F ′ containsanunramifiedextension.<br />
Weobservedbeforethatsince<br />
µ0 = λ(σ0) ≡ −1 (mod 4)<br />
<strong>the</strong>number νφ(−1)is −1.Theidentity(14.5)reducesto<br />
Since<br />
and<br />
thisrelationisclearlyvalid.<br />
(−1) u(1)<br />
2 = (−1) a (−1) q−1<br />
2k .<br />
a = u(1)<br />
2<br />
u(2) =<br />
+ u(2)<br />
2<br />
q − 1<br />
k<br />
Wecontinuetosupposethat µ0 ≡ −1 (mod k)andthat σ 2 0 = 1butnowwesupposethat<br />
Visnotinvariantunder π ∗ (σ0).Since π ∗ (σ0)V ∩Vand π ∗ (σ0)V +Varebothinvariantunder<br />
G/G1<strong>the</strong>firstis0and<strong>the</strong>secondis S(K/L)sothat S(K/L)is<strong>the</strong>directsum V ⊕ π ∗ (σ0)V.<br />
Let V have p ℓ elementssothat q = p 2ℓ .If λ ′ isagain<strong>the</strong>fieldgeneratedover<strong>the</strong>primefield<br />
by<strong>the</strong> kthroots<strong>of</strong>unity λ ′ has p ℓ elements.If φ ′ = λ ′ ∩ φhas p f elements<strong>the</strong>n p ℓ = p 2f so<br />
that p ℓ ≡ 1 (mod8).Also kdivides p ℓ − 1sothat<br />
iseven.<br />
q − 1<br />
k =<br />
ℓ p − 1<br />
(p<br />
k<br />
ℓ + 1)<br />
If σ ∈ G0/G1<strong>the</strong>nonzer<strong>of</strong>ixedpoints<strong>of</strong> σ0σare<strong>the</strong>elements<strong>of</strong><strong>the</strong>form<br />
v ⊕ π ∗ (σ0σ)v
Chapter14 209<br />
with v = 0. Thereare (p ℓ − 1)k<strong>of</strong><strong>the</strong>maltoge<strong>the</strong>rand<strong>the</strong>yfallinto p ℓ − 1orbits. The<br />
remaining<br />
(p 2ℓ − 1) − (p ℓ − 1)k<br />
elementsfallinto<br />
orbits.Thus<br />
1<br />
2k {(p2ℓ − 1) − (p ℓ − 1)k}<br />
a = pℓ − 1<br />
2 + p2ℓ − 1<br />
.<br />
2k<br />
Since,for<strong>the</strong>samereasonsasbefore, νφ(−1) = −1<strong>the</strong>identity(14.5)becomes<br />
while(14.6)becomes<br />
νφ<br />
<br />
νφ<br />
µ γµ<br />
<br />
<br />
µ γµ<br />
<br />
= (−1) pℓ −1<br />
2 (14.10)<br />
νφ(−1) Σ[φµ:φ] = (−1) p ℓ −1<br />
2 .<br />
Since [φµ : φ] ≡ p ℓ − 1 ≡ 0 (mod2)<br />
only(14.5)needbeproved.(14.4)and(14.7)arenottobeconsideredbecause q−1<br />
k iseven.<br />
Weproceedasbefore. Thepointsin Tcanbechosensothat<strong>the</strong>irisotropygroupsare<br />
ei<strong>the</strong>rtrivialorcontain σ0or σ1. pℓ −1<br />
2 willhaveisotropygroupscontaining σ0and pℓ −1<br />
2 will<br />
haveisotropygroupscontaining σ1. Theargumentusedaboveshowsthat<strong>the</strong>leftside<strong>of</strong><br />
(14.10)isequalto<br />
asdesired.<br />
Nowsuppose k ≥ 8and<br />
(−1) pℓ−1 2<br />
µ0 = λ(σ0) ≡ k<br />
− 1 (modk).<br />
2<br />
Weare<strong>of</strong>coursestillsupposingthat [G : G0] = 2.If σbelongsto G0/G1<strong>the</strong>n<br />
and<br />
σσ0σ −1 = σ0σ k<br />
2 −2<br />
(σ0σ) 2 = σ 2 0σ k<br />
2 .
Chapter14 210<br />
Thus<br />
x((σ0σ) 2 ) = x(σ 2 k<br />
0 ) +<br />
2 x(σ).<br />
Since x(σ 2 0)is0or k<br />
2 ,wecanmake<strong>the</strong>sumon<strong>the</strong>right0.Replacing σ0by σ0σifnecessary,<br />
wesupposethat σ 2 0 = 1.Then (σ0σ) 2 = 1ifandonlyif<br />
whichissoifandonlyif σ0σisconjugateto σ.<br />
k<br />
2<br />
x(σ) ≡ 0 (modk)<br />
Take Vin S(K/L)asbefore.If Visinvariantunder π ∗ (σ0)and λ ′ with p ℓ elementsand<br />
φ ′ with p f elementshave<strong>the</strong>samemeaningasbefore<strong>the</strong>n<br />
forsomeinteger wsothat<br />
and<br />
p f = k<br />
− 1 + wk<br />
2<br />
q − 1 = p 2f − 1 = k · k<br />
<br />
k<br />
− k + 2wk − 1 + (wk)<br />
4 2 2<br />
q − 1<br />
k<br />
k<br />
≡ − 1 (mod 2)<br />
4<br />
isodd.Thus<strong>the</strong>identities(14.5)and(14.6)arenottobeconsidered.Theidentities(14.4)and<br />
(14.7)followfromLemma14.7exactlyasabove.<br />
and<br />
Suppose<strong>the</strong>n S(K/L)is<strong>the</strong>directsum V ⊕ π ∗ (σ0)V.If Vhas p ℓ elements<strong>the</strong>n q = p 2ℓ<br />
q − 1<br />
k = pℓ − 1<br />
k<br />
(p ℓ + 1)<br />
isevenbecause kdivides p ℓ − 1.Thenonzeroelements<strong>of</strong> S(K/L)whicharefixedpoints<strong>of</strong><br />
some σ0σwith σasquarein G0/G1are<strong>the</strong>elements<br />
v ⊕ π ∗ (σ0σ)v<br />
with v = 0.Thereare (p ℓ − 1) k<br />
2 suchelementsand<strong>the</strong>yfallinto pℓ −1<br />
2 orbits.Theremaining<br />
(p 2ℓ − 1) − (p ℓ − 1) k<br />
2
Chapter14 211<br />
nonzeroelementshavetrivialisotropygroupandfallinto<br />
orbits.Thus<br />
1<br />
2k<br />
<br />
2ℓ ℓ k<br />
(p − 1) − (p − 1)<br />
2<br />
a = p2ℓ − 1<br />
2k + pℓ − 1<br />
.<br />
4<br />
Since,asbefore, νφ(−1) = −1<strong>the</strong>identity(14.5)becomes<br />
while(14.6)becomes<br />
Again<br />
νφ<br />
<br />
νφ<br />
<br />
µ γµ<br />
<br />
µ γµ<br />
<br />
[φµ : φ] ≡ pℓ − 1<br />
2<br />
= (−1) pℓ −1<br />
4 (14.11)<br />
(−1) Σ[φµ:φ] = (−1) p ℓ −1<br />
4 .<br />
≡ 0 (mod2)<br />
sothatitisenoughtoprove(14.11).Theidentities(14.4)and(14.7)neednotbeconsidered.<br />
that<br />
If λ ′ and φ ′ aredefinedasbeforeand φ ′ has p f elements<strong>the</strong>n λ ′ has p ℓ = p 2f elementsso<br />
p ℓ ≡ 1 (mod 8)<br />
and pℓ −1<br />
4 iseven.Wemaysupposethateach µin Tei<strong>the</strong>rhastrivialisotropygrouporisfixed<br />
by σ0. Lemma14.8showsthatthose µwithtrivialisotropygroupcontributenothingto<strong>the</strong><br />
leftside<strong>of</strong>(14.11).If L0is<strong>the</strong>fixedfield<strong>of</strong> σ0and<br />
<strong>the</strong>leftside<strong>of</strong> Kis<br />
N L0/F̟L0<br />
νφ(γ0) pℓ−1 2<br />
whichis1.Thetruth<strong>of</strong><strong>the</strong>identityisnowclear.<br />
= γ0̟F<br />
Wereturnto<strong>the</strong>generalcasesothat [G : G0]maybegreaterthan2and µ0maybe<br />
congruentto1modulo4.Ofcourse [G : G0]isstilleven.Let<br />
<br />
λ0 = λ σ 1<br />
2 [G:G0]<br />
0
Chapter14 212<br />
sothat<br />
If [G : G0] > 2<strong>the</strong>n<br />
λ0 = µ 1<br />
2 [G:G0]<br />
0<br />
λ0 ≡ 1 (mod4).<br />
If [G : G0] = 2<strong>the</strong>n λ0 = µ0. Since<strong>the</strong>casethat [G : G0] = 2and µ0 ≡ −1 (mod 4)is<br />
completelysettledwemaysupposethat λ0 ≡ 1 (mod4).Set<br />
τ0 = σ 1<br />
2 [G:G0]<br />
0<br />
Anyelement<strong>of</strong> G/G1whichdoesnotliein G0/G1andwhosesquareis1is<strong>of</strong><strong>the</strong>form<br />
σ(x)τ0.If<br />
= σ(y0)<br />
<strong>the</strong>n<br />
Since G/G1isnotcyclic y0iseven.Since<br />
σ [G:G0]<br />
0<br />
(σ(x)τ0) 2 = σ((λ0 + 1)x)τ 2 0 = σ(y0 + (λ0 + 1)x).<br />
<strong>the</strong>reareexactlytwosolutions<strong>of</strong><strong>the</strong>equation<br />
Let x0beone<strong>of</strong><strong>the</strong>m.Then x0 + k<br />
2<br />
Set<br />
λ0 + 1 ≡ 2 (mod 4)<br />
y0 + (λ0 + 1)x ≡ 0 (modk).<br />
.<br />
.<br />
is<strong>the</strong>o<strong>the</strong>r.Wemaysupposethat kdoesnotdivide x0.<br />
ρ0 = σ(x0)τ0.<br />
Weobservedbeforethatif σ = 1belongsto G/G1and π∗ (σ)hasanonzer<strong>of</strong>ixedpoint<strong>the</strong>n<br />
somepower<strong>of</strong> σis<strong>of</strong>order2andhasanonzer<strong>of</strong>ixedpoint. Since σ <br />
k<br />
hasnononzero<br />
2<br />
fixedpointthispowermustbe ρ0or σ <br />
k<br />
2 ρ0.Since σ <br />
k<br />
2 liesin<strong>the</strong>center<strong>of</strong> G/G1, σmustlie<br />
in<strong>the</strong>centralizer<strong>of</strong> ρ0.<br />
Thegroup<br />
<br />
k k<br />
1, σ , ρ0, σ<br />
2 2<br />
is<strong>of</strong>order4andeveryelementinitis<strong>of</strong>order2soitcannotbecontainedin<strong>the</strong>center<strong>of</strong><br />
G/G1. Howeveritisanormalsubgroupanditscentralizer H ∗ hasindex2in G/G1. G/G1<br />
ρ0
Chapter14 213<br />
maybeidentifiedwith H. Everyelement σ<strong>of</strong> Hsuchthat π ∗ (σ)hasanonzer<strong>of</strong>ixedpoint<br />
liesin H ∗ . S(K/L)is<strong>the</strong>directsum<strong>of</strong> Vand Wwhere<br />
V = {v | π ∗ (ρ0)v = v}<br />
W = {w | π ∗ (ρ0)w = −w}.<br />
If σin Hdoesnotbelongto H ∗ <strong>the</strong>n π ∗ (σ)V = Wand π ∗ (σ)W = V.Thenumber<strong>of</strong>nonzero<br />
orbits<strong>of</strong> Hin V ∪ Wis<strong>the</strong>sameas<strong>the</strong>number a ′ <strong>of</strong>nonzeroorbits<strong>of</strong> H ∗ in V.If V has p ℓ<br />
elementssothat q = p 2ℓ <strong>the</strong>number<strong>of</strong>nonzeroorbits<strong>of</strong> Hin V ⊕ W − (V ∪ W)is<br />
a ′′ = (pℓ − 1) 2<br />
[G : G1] = pℓ − 1<br />
·<br />
k<br />
pℓ − 1<br />
[G : G0] .<br />
Theaction<strong>of</strong> H ∗ on V mustbeirreduciblealthoughitisnotfaithful.However<strong>the</strong>action<strong>of</strong><br />
H ∗ ∩ H0 = H ∗ 0 isfaithful.<br />
Let F ′ be<strong>the</strong>fixedfield<strong>of</strong> H ∗ in Lor,whatis<strong>the</strong>same,<strong>of</strong> H ∗ Cin K. Let C 1 ⊆ Cbe<br />
<strong>the</strong>orthogonalcomplement<strong>of</strong> V andlet H 1 be<strong>the</strong>subgroup<strong>of</strong> Hwhichactstriviallyon V.<br />
H 1 C 1 isanormalsubgroup<strong>of</strong> H ∗ Canditsfixedfield K ′ isnormalover F ′ .If H ′ = H ∗ /H 1<br />
and C ′ = C/C 1 <strong>the</strong>n G ′ = G(K ′ /F ′ ) = H ′ C ′ . Moreover H ′ ∩ C ′ = {1}and H ′ = {1}<br />
because σ <br />
k<br />
1 ′ ′ ′<br />
2 doesnotliein H . Since<strong>the</strong>action<strong>of</strong> H on C isfaithfulandirreducible C<br />
iscontainedineverynontrivialnormalsubgroup<strong>of</strong> G ′ . Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong><strong>the</strong>four<br />
identities(14.4),(14.5),(14.6),and(14.7)weuseinductionon [K : F].<br />
Let k ′ be<strong>the</strong>order<strong>of</strong> H ′ 0 andlet φ′ = OF ′/PF ′.If K/Fisreplacedby K′ /F ′ <strong>the</strong>identity<br />
(14.4)becomes<br />
(−1) a′ +1 νφ ′(k′ ) νφ ′(−1) pℓ −1<br />
2k ′ + 1<br />
2 = 1. (14.4 ′′ )<br />
Tistobereplacedby T ′ ,aset<strong>of</strong>representativesfor<strong>the</strong>nonzeroorbits<strong>of</strong> H ′ or H ∗ in V,<br />
whichmaybeidentifiedwith<strong>the</strong>charactergroup<strong>of</strong> C ′ .Wemaysupposethat T ′ isasubset<br />
<strong>of</strong> T.Because H ′ 0 = {1}<strong>the</strong>identity(14.5)for<strong>the</strong>field K′ /F ′ maybewrittenas<br />
Ofcourse<br />
νφ ′<br />
<br />
µ∈T ′ γ′ µ<br />
<br />
= (−1) a′<br />
NFµ/F ′(̟t Fµ ) = γ′ µ ̟<br />
t[Fµ:F ′ ]<br />
k ′<br />
νφ ′(−1) pℓ −1<br />
2k ′ . (14.5 ′′ )<br />
Recallthat tisodd.ByPropositionIV.3<strong>of</strong>Serre’sbook, thas<strong>the</strong>samesignificancefor K ′ /F ′<br />
asithadfor K/F.Theidentity(14.6)maybewrittenas<br />
(−1) a′<br />
νφ ′<br />
<br />
µ∈T ′ γ′ µ<br />
<br />
F ′<br />
.<br />
νφ ′(−1)Σ µ∈T ′[φµ:φ ′ ] νφ ′(−1) pℓ −1<br />
2k ′ = 1. (14.6 ′′ )
Chapter14 214<br />
and(14.7)as<br />
(−1) a′ +1 νφ ′(k′ ) νφ ′(−1) pℓ −1<br />
2k ′ − 1<br />
2 νφ ′(−1)Σ µ∈T ′ [φµ:φ ′ ] = 1 (14.7 ′′ )<br />
Assuming(14.4 ′′ ),(14.5 ′′ ),(14.6 ′′ ),and(14.7 ′′ )wearegoingtoprove(14.4),(14.5),(14.6),and<br />
(14.7).<br />
Since H ′ 0 isisomorphicto H∗ 0 ei<strong>the</strong>r k′ = kor k ′ = k<br />
2<br />
k ′ = k<br />
2 forifnot<br />
q − 1<br />
k =<br />
ℓ p − 1<br />
(p<br />
k<br />
ℓ + 1)<br />
.Supposefirstthat q−1<br />
k isodd.Then<br />
wouldbeeven.Thus H0 = G0/G1isnotcontainedin H ∗ and F ′ /Fisramifiedsothat φ ′ = φ.<br />
Since<br />
<strong>the</strong>number pℓ −1<br />
k ′<br />
if<br />
q − 1<br />
k =<br />
ℓ ℓ p − 1 (p + 1)<br />
isodd.Toprove(14.4)wehavetoshowthat<br />
(−1) a′′<br />
k ′<br />
2<br />
νφ(2) νφ(−1) δ = 1<br />
δ = p2ℓ − 1<br />
2k − pℓ − 1<br />
k = pℓ − 1<br />
k<br />
p ℓ + 1<br />
2<br />
<br />
− 1 .<br />
Since G0/G1isnotcontainedin H ∗ , τ0doesnotcommutewith G0/G1and<strong>the</strong>map λ<strong>of</strong> G/G0<br />
into<strong>the</strong>units<strong>of</strong> Z/k Zisfaithful.Thus<br />
But<br />
λ0 ≡ 1 (modk).<br />
λ0 ≡ 1 (mod4)<br />
sothat k ≥ 8.Ingeneralif k ≥ 4,<strong>the</strong>group<strong>of</strong>units<strong>of</strong> Z/k Zis<strong>the</strong>product<strong>of</strong> {1, −1}and<br />
If<br />
with xoddand 4 ≤ 2 b ≤ k<strong>the</strong>n<br />
{α | α ≡ 1 (mod4)}.<br />
α = 1 + 2 b x<br />
α 2 = 1 + 2 b+1 y
Chapter14 215<br />
with yodd.<strong>On</strong>eshowseasilybyinductionthat<strong>the</strong>order<strong>of</strong> αis 2 −b ksothat<br />
{α | α ≡ 1 (mod 4)}<br />
iscyclic<strong>of</strong>order k<br />
4 . Thisimpliesin<strong>the</strong>particularcaseunderconsiderationthat [G : G0]<br />
divides k<br />
4 .Write<br />
a ′′ isoddifandonlyif<br />
(i)<br />
(ii)<br />
a ′′ ℓ p − 1<br />
=<br />
k ′<br />
ℓ p − 1<br />
.<br />
2[G : G0]<br />
2[G : G0] = k ′ .<br />
Weconsidervariouscasesseparately.Asbefore µ0 = λ(σ0).If φhas p f elements<strong>the</strong>n<br />
Then<br />
µ0p f ≡ 1 (mod 8)<br />
µ0 ≡ 1 (mod 8).<br />
νφ(2) = νφ(−1) = 1<br />
and<strong>the</strong>order<strong>of</strong> µ0in<strong>the</strong>units<strong>of</strong> Z/k Zwhichisequalto [G : G0]divides k<br />
8 .Thus a′′ is<br />
even.Theidentity(14.4)follows.<br />
Then<br />
µ0 ≡ 3 (mod 8).<br />
νφ(2) = νφ(−1) = −1.<br />
Since µ0 ≡ 3 (mod8)<strong>the</strong>numbers µ0and λ0aredifferent.Thus λ0isasquareandhence<br />
congruentto1modulo8.Then k > 8and<br />
Then<br />
δ ≡ pℓ + 1<br />
4<br />
p ℓ ≡ 1 (mod 8).<br />
− 1<br />
2<br />
≡ 0 (mod2).<br />
Since µ0 = λ0<strong>the</strong>index [G : G0]isnot2.Thus<strong>the</strong>order<strong>of</strong> µ0isatleast4andis<strong>the</strong>refore<br />
<strong>the</strong>order<strong>of</strong> −µ0.Since −µ0 ≡ 5 (mod8)itsorderis k<br />
4 and<br />
[G : G0] = k<br />
4 .
Chapter14 216<br />
(iii)<br />
(iv)<br />
Consequently a ′′ isodd.Again(14.4)issatisfied.<br />
µ0 ≡ 5 (mod 8).<br />
Then νφ(2) = −1while νφ(−1) = 1.Theorder<strong>of</strong> µ0whichequals [G : G0]isagain k<br />
4 so<br />
that a ′′ isoddand(14.4)issatisfied.<br />
µ0 ≡ 7 (mod 8).<br />
Then νφ(2) = 1while νφ(−1) = −1.Again k > 8and<br />
δ ≡ 0 (mod2).<br />
Theorder<strong>of</strong> µ0isagainatleast4and<strong>the</strong>reforeequalto<strong>the</strong>order<strong>of</strong> −µ0andthatdivides<br />
k<br />
8<br />
.Thus [G : G0]divides k<br />
8 and a′′ iseven.(14.4)followsoncemore.<br />
Since φ ′ = φallweneedtoprove(14.7)once(14.4)and(14.7 ′′ )aregrantedisshowthat<br />
<br />
µ∈T −T ′ [φµ : φ] ≡ 0 (mod2).<br />
Thisisclearbecause,for<strong>the</strong>se µ, Fµ = Land φµ = OL/PLis<strong>of</strong>evendegreeover φ.<br />
Finallywehavetoassumethat q−1<br />
k isevenandprove(14.5)and(14.6).Firstalemma.<br />
Lemma 14.9<br />
If q−1<br />
k iseven,<br />
<br />
λ σ 1<br />
2 [G:G0]<br />
0<br />
and G/G0actsfaithfullyon G0/G1,<strong>the</strong>n<br />
<br />
≡ 1 (mod4),<br />
(−1) a νφ(−1) q−1<br />
2k = 1.<br />
Since<strong>the</strong>actionisfaithful G0/G1isnotcontainedin H ∗ and k ′ = k<br />
2 . Asbefore λ0 ≡<br />
1 (mod4)and λ0 ≡ 1 (modk)toge<strong>the</strong>rimplythat k ≥ 8and k ′ ≥ 4.Since k ′ divides p ℓ − 1,<br />
and pℓ +1<br />
2 isodd.Since<br />
q − 1<br />
k =<br />
p ℓ ≡ 1 (mod 4)<br />
<br />
ℓ ℓ p − 1 p + 1<br />
k ′<br />
2
Chapter14 217<br />
<strong>the</strong>number pℓ −1<br />
k ′<br />
sothat<br />
iseven.<br />
If σbelongsto H ∗ and σactstriviallyon H ∗ 0 <strong>the</strong>n<br />
<br />
λ(σ) ≡ 1 mod k<br />
<br />
2<br />
λ(σ 2 ) ≡ 1(modk)<br />
and σ 2 belongsto H0.Thus σbelongsto ρ0H0 ∪ H0.Since ρ0belongsto H 1 <strong>the</strong>image<strong>of</strong> σin<br />
H ′ liesin H ′ 0 .Thus G′ /G ′ 0 actsfaithfullyon G′ 0 /G′ 1 .If σbelongsto H1 <strong>the</strong>n σactstrivially<br />
on H ∗ 0 because<strong>the</strong>representation<strong>of</strong> H∗ 0 on Visfaithful.Thus H1 iscontainedin ρ0 H0 ∪ H0<br />
andis<strong>the</strong>reforejust {ρ0, 1}.Thus<br />
Supposethat<br />
[G ′ : G ′ 0 ] = [H′ : H ′ 0 ] = [H∗ : H ∗ 0 H1 ] = 1<br />
2<br />
Since φ ′ = φand,because k ′ ≥ 4divides p ℓ − 1,<br />
q − 1<br />
2k =<br />
[G : G0].<br />
(−1) a νφ ′(−1) pℓ −1<br />
2k ′ = 1. (14.12)<br />
ℓ p − 1<br />
2k ′<br />
ℓ ℓ p + 1 p − 1<br />
≡<br />
2 2k ′<br />
<br />
allweneeddotoestablish<strong>the</strong>lemmaistoshowthat<br />
Asbefore [G : G0]divides k<br />
4 .If<br />
<strong>the</strong>n<br />
a ′′ = 1<br />
k<br />
a ′′ ≡ 0 (mod2).<br />
k<br />
4<br />
= n[G : G0]<br />
(p ℓ − 1) 2<br />
[G : G0]<br />
iscertainlyevenbecause 2k ′ divides p ℓ − 1.<br />
If [G : G0] ≥ 4let<br />
λ ′ 0<br />
<br />
= λ σ 1<br />
ℓ p − 1<br />
= n<br />
k ′<br />
4 [G:G0]<br />
0<br />
<br />
.<br />
2<br />
(mod2),
Chapter14 218<br />
If<br />
λ ′ 0<br />
≡ 1 (mod4)<br />
wemaysupposethat(14.12)istruebyinduction.If [G : G0] = 4and<br />
orif [G : G0] = 2wemustestablishitdirectly.<br />
λ ′ 0 ≡ 3 (mod4)<br />
Supposefirstthat [G : G0] = 2.If φhas p f elements<strong>the</strong>n<br />
sothat νφ(−1) = 1.Itisclearthatinthiscase<br />
a ′ isthusevenand(14.12)isvalid.<br />
λ0 ≡ µ0 ≡ p f ≡ 1 (mod4)<br />
a ′ = pℓ − 1<br />
k ′<br />
.<br />
Nowsuppose [G : G0] = 4sothat [G ′ : G ′ 0 ] = 2. If σ′ 0 generates G′ modulo G ′ 0 <strong>the</strong>n<br />
λ ′ 0is<strong>the</strong>image<strong>of</strong> σ′ 0in<strong>the</strong>group<strong>of</strong>units<strong>of</strong> Z/k′ Z.Wehavealreadystudied<strong>the</strong>casethat<br />
≡ 3 (mod4)intensively.Let<br />
λ ′ 0<br />
x : σ ′ −→ x(σ ′ )<br />
be<strong>the</strong>map<strong>of</strong> G ′ 0 /G′ 1 onto Z/k′ Z. If λ ′ 0 ≡ −1 (modk′ )and x((σ ′ 0 )2 ) = k′<br />
2 weshowed,<br />
incidentally,that(14.12)isvalid.If λ ′ 0 ≡ −1 (mod k′ ), x((σ ′ 0 )2 ) = 0,and<strong>the</strong>action<strong>of</strong> H ′ 0on S(K ′ /L ′ )isreduciblewesawthat pℓisasquare p2ℓ′ andthat<strong>the</strong>leftside<strong>of</strong> (L)is<br />
But<strong>the</strong>fieldwith p ℓ′<br />
and(14.12)isagainvalid.If k ′ ≥ 8,<br />
(−1) pℓ′ −1<br />
2 .<br />
elementsmustcontains<strong>the</strong> k ′ throots<strong>of</strong>unityand k ′ ≡ 0 (mod 4).Thus<br />
p ℓ′<br />
− 1 ≡ 0 (mod4)<br />
λ ′ 0 ≡ k′<br />
2 − 1 (mod k′ )<br />
and<strong>the</strong>action<strong>of</strong> H ′ 0 on S(K′ /L ′ )isreducible,<strong>the</strong>leftside<strong>of</strong>(14.12)is<br />
(−1) pℓ′ −1<br />
4 .
Chapter14 219<br />
Thistime<br />
p ℓ′<br />
− 1 ≡ 0 (mod8).<br />
Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong><strong>the</strong>lemmaweshowthatin<strong>the</strong>caseunderconsideration<strong>the</strong><br />
on Visreducible.Ifnot<strong>the</strong><br />
action<strong>of</strong> H ′ 0and S(K′ /L ′ )or,whatis<strong>the</strong>same,<strong>the</strong>action<strong>of</strong> H∗ 0<br />
fieldgeneratedover<strong>the</strong>primefieldby<strong>the</strong> k ′ throots<strong>of</strong>unityhas pℓelements.Thus p ℓ ≡ 1 (mod4).<br />
Howeveraswehaveobservedrepeatedly,<strong>the</strong>number<strong>of</strong>elementsin φiscongruentto3<br />
modulo4.Thus ℓiseven.Let ℓ = 2ℓ ′ .Ei<strong>the</strong>r pℓ′ − 1or pℓ′ + 1iscongruentto2modulo4.If<br />
pℓ′ + 1 ≡ 2 (mod4)<strong>the</strong>n k ′ divides pℓ′ − 1because<br />
p ℓ − 1<br />
k ′<br />
=<br />
<br />
pℓ′ − 1<br />
k ′<br />
<br />
iseven.Since k ′ cannotdivide pℓ′ − 1wehave<br />
p ℓ′<br />
(p ℓ′<br />
≡ 3 (mod4)<br />
and ℓ ′ isodd.Indeeditis1butthatdoesnotmatter.Since kdivides p ℓ − 1,<strong>the</strong> kthroots<strong>of</strong><br />
unityarecontainedin<strong>the</strong>fieldwith p ℓ elements.Adjoining<strong>the</strong>mto φ = OF/PFweobtaina<br />
quadraticextensionbecause4doesnotdivide ℓ.Thereforeif σbelongsto G0/G1<br />
sothat<br />
+ 1)<br />
θ0(σ) = θ0(σ) σ−2<br />
0 = θ0(σ) λ(σ2<br />
0 )<br />
λ(σ 2 0) ≡ 1 (modk).<br />
Thiscontradicts<strong>the</strong>assumptionthat G/G0actsfaithfullyon G0/G1.<br />
Returningto<strong>the</strong>pro<strong>of</strong><strong>of</strong>(14.5),wesupposefirstthat H0isnotcontainedin H∗sothat<strong>the</strong> action<strong>of</strong> G/G0on G0/G1isfaithful.Because<strong>of</strong>Lemma14.9<strong>the</strong>identity(14.5)isequivalent<br />
to<br />
<br />
= 1.<br />
νφ<br />
µ∈T N Fµ/F γµ<br />
If µbelongsto Tbutnotto T ′ <strong>the</strong>n Fµ = Land,byLemma14.8,<br />
νφ(N Fµ/F γµ) = 1.
Chapter14 220<br />
If µbelongsto T ′ <strong>the</strong>n Gµiscontainedin H ∗ Csothat Fµcontains F ′ .Moreoverwedonot<br />
change Fµifwereplace K/Fby K ′ /F ′ .Let ̟F ′generate PF ′andtake ̟F = N F ′ /F̟F ′.If<br />
E ′ is<strong>the</strong>fixedfield<strong>of</strong> H ∗ wemaysupposethat<br />
andthat<br />
Then<br />
asrequired.Let<br />
Then<br />
̟F ′ = NE ′ /F ′̟E ′<br />
̟E = NE ′ /E̟E ′.<br />
̟F = N E/F̟E<br />
N Fµ/F, ̟ t Fµ = γ′ µ ̟ t[Fµ:F ′ ]<br />
k ′ .<br />
γµ = N F ′ /F γ ′ µ .<br />
Since F ′ /Fisramified γµisasquarein UFand(14.5)isproved. Toprove(14.6)wehaveto<br />
showthat<br />
νφ(−1) Σµ∈T [φµ:φ] = νφ ′(−1)Σ µ∈T ′[φµ:φ ′ ] = 1.<br />
But pℓ −1<br />
k ′<br />
isevenandthisfollowsfrom<strong>the</strong>simultaneousvalidity<strong>of</strong>(14.5 ′′ )and(14.6 ′′ ).<br />
Wehaveyettotreat<strong>the</strong>casethat q−1<br />
k isevenand H0iscontainedin H ∗ .Then F ′ /Fis<br />
unramifiedand k ′ = k.Supposefirst<strong>of</strong>allthat pℓ −1<br />
k isalsoeven.Then<br />
q − 1<br />
2k =<br />
<br />
ℓ ℓ p − 1 p + 1<br />
k<br />
iseven. H0iscontainedin H∗and Hisgeneratedby σ0and H0. Consequently σ0isnot<br />
containedin H∗and σ0ρ0σ −1<br />
<br />
k<br />
0 = σ ρ0.<br />
2<br />
Since ρ0 = σ(x0)τ0<br />
if µ0 = λ(σ0).If<br />
(µ0 − 1)x0 ≡ k<br />
2<br />
<br />
y0 = x<br />
σ [G:G0]<br />
0<br />
2<br />
(modk)
Chapter14 221<br />
and mis<strong>the</strong>greatestcommondivisor<strong>of</strong> y0and k<strong>the</strong>nby<strong>the</strong>definition<strong>of</strong> x0<strong>the</strong>greatest<br />
commondivisor<strong>of</strong> x0and kis m<br />
2<br />
k.Inparticular m < k.Theorder<strong>of</strong> σ0in His<br />
.Therefore k<br />
m is<strong>the</strong>greatestcommondivisor<strong>of</strong> µ0 − 1and<br />
k<br />
[G : G0].<br />
m<br />
Therefore [G : G0]divides k<br />
2m [G : G0]and H ∗ containsacyclicsubgroup<strong>of</strong>order<br />
k<br />
[G : G0].<br />
2m<br />
If σis<strong>the</strong>element<strong>of</strong>order2inthissubgroup,<strong>the</strong>n σbelongsto H0and π ∗ (σ)doesnothave<br />
1asaneigenvalue. Thusnononzeroelement<strong>of</strong> V isfixedbyanyelement<strong>of</strong>thiscyclic<br />
subgroupand<br />
Inparticular [G : G0]divides p ℓ − 1and<br />
p ℓ <br />
− 1 ≡ 0 mod k<br />
2m<br />
<br />
[G : G0] .<br />
a ′′ <br />
ℓ ℓ p − 1 p − 1<br />
=<br />
k [G : G0]<br />
iseven.Asbefore νφ(γµ) = 1if µbelongsto Tand Fµ = L.If Fµ = L<strong>the</strong>n µbelongsto T ′<br />
and Gµliesin H ∗ Csothat Fµcontains F ′ .In<strong>the</strong>presentsituation F ′ /Fisunramifiedand<br />
wemaytake ̟F ′ = ̟F.If<br />
<strong>the</strong>n<br />
Theidentity(14.5)reducesto<br />
or<br />
NFµ/F ′̟t Fµ = γ′ t[Fµ:F<br />
µ̟<br />
′ ]<br />
k<br />
F<br />
NFµ/F̟ t Fµ = (NF ′ /F γ ′ t[Fµ:F]<br />
k<br />
µ ) ̟ .<br />
νφ<br />
<br />
νφ ′<br />
µ∈T ′ N F ′ /F γ ′ µ<br />
<br />
µ∈T ′ γ′ µ<br />
<br />
F<br />
<br />
= (−1) a′<br />
= (−1) a′<br />
.<br />
Since φ ′ isaquadraticextension<strong>of</strong> φ<strong>the</strong>number νφ ′(−1)is1andthisrelationisequivalentto<br />
(14.5 ′′ ).Toprove(14.6)wehavetoshowthat<br />
νφ(−1) Σµ∈T [φµ:φ] = 1.
Chapter14 222<br />
Thisisclearbecause2divideseach<strong>of</strong><strong>the</strong>degrees [φµ : φ].<br />
Finallywehavetosupposethat pℓ −1<br />
k isodd. Since [φ ′ : φ] = 2<strong>the</strong>relation(14.4 ′′ )<br />
amountsto<br />
(−1) a′ +1 = 1.<br />
Again<br />
νφ<br />
<br />
µ∈T γµ<br />
<br />
= νφ ′<br />
<br />
µ∈T ′ γ′ µ<br />
<br />
. (14.13)<br />
If µbelongsto T ′ and σ = 1belongsto Gµ<strong>the</strong>nsomepower<strong>of</strong> σwillequal ρ0.Since<br />
isoddand<br />
p ℓ − 1<br />
k<br />
= <br />
µ∈T ′<br />
[Fµ : F ′ ]<br />
k<br />
[Fµ : F ′ ]<br />
k<br />
isapower<strong>of</strong>2<strong>the</strong>reisatleastone µin T ′ forwhich [Fµ : F ′ ] = k.Then Gµmustcontainan<br />
element<strong>of</strong><strong>the</strong>form σ(z0)σ 2 0 .Then<br />
ρ0 = σ(x0)τ0 = (σ(z0)σ 2 0) 1<br />
4 [G:G0] = σ<br />
µ 2 0<br />
<br />
µ 1<br />
2 [G:G0]<br />
0<br />
Thus <br />
µ 1<br />
2 [G:G0]<br />
<br />
0 − 1<br />
z0 ≡ x0 (modk).<br />
− 1<br />
Let<br />
Since<br />
1<br />
4 [G : G0] = 2 b .<br />
µ 2 0<br />
≡ 1 (mod4)<br />
µ 2 0<br />
− 1<br />
− 1<br />
z0τ0<br />
and,asbefore,<strong>the</strong>greatestcommondivisor<strong>of</strong> x0and kis m<br />
2 if<strong>the</strong>greatestcommondivisor<strong>of</strong><br />
y0and kis m,weinferthat<br />
µ 1<br />
2 [G:G0]<br />
0<br />
µ 2 0<br />
ismultiplicativelycongruentto<br />
− 1<br />
=<br />
− 1<br />
b<br />
j=1<br />
− 1<br />
µ 2j<br />
b<br />
=<br />
− 1<br />
µ2j+1 0<br />
·<br />
0<br />
[G : G0]<br />
4<br />
j=1 µ2j<br />
0 + 1<br />
.
Chapter14 223<br />
modulo2andthat<strong>the</strong>greatestcommondivisor<strong>of</strong> z0and kis<br />
2m<br />
[G : G0] .<br />
Inparticular [G:G0]<br />
2 divides m. z0isoddifandonlyif<br />
m = 1<br />
2<br />
[G : G0].<br />
If µ0 ≡ 1 (mod 4)<strong>the</strong>order<strong>of</strong> µ0in<strong>the</strong>group<strong>of</strong>units<strong>of</strong> Z/k Zis mbecauseasweobserved<br />
whentreating<strong>the</strong>casethat pℓ−1 k iseven,<strong>the</strong>greatestcommondivisor<strong>of</strong> µ0 − 1and kis k<br />
m .<br />
However<br />
andinthiscase mdivides 1 [G : G0].Thus<br />
2<br />
if µ0 ≡ 1 (mod 4).<br />
µ 1<br />
2 [G:G0]<br />
0 ≡ λ(τ0) ≡ 1 (mod k)<br />
m = 1<br />
[G : G0]<br />
2<br />
Weshalldefineasequence<strong>of</strong>fields F (i) , L (i) , K (i) , 1 ≤ i ≤ n. nisanintegertobe<br />
specified. Wewillhave F (i) ⊆ L (i) ⊆ K (i) and K (i) /F (i) and L (i) /F (i) willbeGalois. Let<br />
G (i) = G(K (i) /F (i) )and C (i) = G (L (i) /F (i) ). Therewillbeasubgroup H (i) <strong>of</strong> G (i) such<br />
that H (i) = {1}, H (i) ∩ C (i) = {1},and G (i) = H (i) C (i) . C (i) willbeanontrivialabelian<br />
normalsubgroup<strong>of</strong> G (i) whichiscontainedineveryo<strong>the</strong>rnontrivialnormalsubgroup. H (n)<br />
willbeabelianbut H (i) willbenonabelianif i < n.Moreover k (i) = [H (i)<br />
0 : 1]willbeatleast<br />
4forall iand k (i) willequal 2k (i+1) if i < n.If xisanisomorphism<strong>of</strong> H (i)<br />
0 with Z/k(i) Zand<br />
σbelongsto H (i) let<br />
x(στσ −1 ) = λ (i) (σ)x(τ).<br />
Then λ (i) (σ)willbecongruentto1modulo8if i < n.If q (i) is<strong>the</strong>number<strong>of</strong>elementsin C (i)<br />
<strong>the</strong>n<br />
q (i) − 1<br />
k (i)<br />
willbeodd.<br />
F ′ and K ′ havealreadybeendefined. L ′ isjust<strong>the</strong>fixedfield<strong>of</strong> C ′ .<br />
q ′ − 1<br />
k ′<br />
= pℓ − 1<br />
k
Chapter14 224<br />
isodd.If σ ′ in H ′ is<strong>the</strong>image<strong>of</strong> σin H ∗ C<strong>the</strong>n<br />
Since σisasquaremodulo H0<br />
λ ′ (σ ′ ) ≡ λ(σ) (modk).<br />
λ ′ (σ ′ ) ≡ 1 (modk).<br />
IfF (i) , L (i) ,andK (i) havebeendefinedandH (i) isnotabelianwecandefineF (i+1) , L (i+1) ,<br />
K (i+1) by<strong>the</strong>processweusedtopassfrom F, L, Kto F ′ , L ′ , K ′ .Wehaveseenthatif<br />
isodd<strong>the</strong>n<br />
isalsooddandthat<br />
q (i) − 1<br />
k (i)<br />
q (i+1) − 1<br />
k (i+1)<br />
k (i) = 2k (i+1) .<br />
Wehavealsoseenthat k (i) ≥ 8if H (i) isnotabelian.If H (i) isabelianwetake n = i.<br />
Whenwepassfrom<strong>the</strong> ithstageto<strong>the</strong> (i+1)thwebreakup T (i) ,<strong>the</strong>analogue<strong>of</strong> T,into<br />
T (i+1) andacomplementaryset U (i) . Wemaythink<strong>of</strong> T (i) aslyingin T. If σ (i)<br />
0 generates<br />
H (i) modulo H (i)<br />
0 <strong>the</strong>n<br />
λ ′ (σ (i)<br />
0 ) ≡ 1 (mod8).<br />
Wesawthatthisimpliesthat U (i) hasanevennumber<strong>of</strong>elements.If µbelongsto U (i) <strong>the</strong>n<br />
Fµisequalto L (i) .Thuswemaysupposethat<br />
νφ ′<br />
<br />
µ∈U (i) γ′ µ<br />
<br />
= 1.<br />
Moreover L (i) /F (i) isnonabelianand<strong>the</strong>refore L (i) /F (i) isnottotallyramified. Thus µis<br />
notin U (i) if [Fµ : F ′ ] = k.<br />
Since L (n) /F (n) isabelian<strong>the</strong>isotropygroupin H (n) <strong>of</strong>any µin T (n) istrivialsothat<br />
Fµ = L (n) forsuch µ.Since<br />
<br />
µ∈T ′<br />
[Fµ : F ′ ]<br />
k<br />
≡ <br />
µ∈T (n)<br />
[L (n) : F ′ ]<br />
k<br />
(mod2).
Chapter14 225<br />
Thereareanoddnumber<strong>of</strong>elementsin T (n) and<br />
[L (n) : F ′ ] = k.<br />
Choose z0sothat σ(z0)σ 2 0 liesin G(L/L(n) ).It<strong>the</strong>nfixeseach µin T (n) .<br />
Since L (n) /F ′ mustbetotallyramified<strong>the</strong>reisaδin UFsuchthat<br />
N L (n) /F ̟ L (n) = δ̟ 2 F .<br />
Therightside<strong>of</strong> (14.13)isequalto νφ(δ). L (n) iscontainedin L.Choose w0in W L/Fsothat<br />
τ L/F(w0) = ̟F. Wemaysupposethat σ0hasbeenchosentobe σ(w0).Let L0be<strong>the</strong>fixed<br />
field<strong>of</strong> H0.Choose u0in W L/L0 sothat σ(u0) = σ(z0)andsothat τ L/L0 (u0)isaunit.Clearly<br />
z0isevenifandonlyif τ L/L0 (u0)or<br />
isasquare.Since σ(z0)σ 2 0 liesin G(L/L(n) ),<br />
liesin W L/L (n).Wemaytake<br />
Then<br />
<br />
NL0/F τL/L0 (u0) = τL/F(u0) u0w 2 0<br />
̟ L (n) = τ L/L (n) (u0w 2 0).<br />
N L (n) /F (̟ L (n)) = τ L/F (u0w 2 0 ) = τ L/F̟ 2 F<br />
and δ = τ L/F(u0)isasquareifandonlyif z0iseven.<br />
Since (−1) a′ +1 = 1<strong>the</strong>relation(14.5)amountsto<br />
(1) a′′ −1 νφ(−1) q−1<br />
2k = (−1) z0 .<br />
(14.6)isequivalentto(14.5)becauseeach [φµ : φ] = 2[φµ : φ ′ ]iseven.Recallthat<br />
andthat<br />
a ′′ <br />
ℓ ℓ p − 1 p − 1<br />
=<br />
≡<br />
k [G : G0]<br />
pℓ − 1<br />
[G : G0]<br />
q − 1<br />
2k =<br />
<br />
ℓ ℓ p − 1 p + 1<br />
k<br />
2<br />
≡ pℓ + 1<br />
2<br />
(mod2)<br />
(mod2).
Chapter14 226<br />
If<br />
µ0 ≡ 1 (mod4)<br />
<strong>the</strong>n νφ(−1) = 1and,asweobservedearlier, z0isodd. Wehavetoshowthat a ′′ iseven.<br />
Weshowedbeforethat H∗hastocontainacyclicsubgroup<strong>of</strong>order k [G : G0]andthat<br />
2m<br />
k<br />
2m [G : G0]hastodivide pℓ − 1.But k<br />
mis<strong>the</strong>greatestcommondivisor<strong>of</strong> µ0 − 1and k.Since<br />
4divides µ0 − 1and kitdivides k<br />
mand 2[G : G0]divides pℓ − 1.Thus a ′′ iseven.<br />
If<br />
µ0 ≡ 3 (mod4)<br />
<strong>the</strong>n νφ(−1) = −1.Moreover k > 2sothat p ℓ ≡ 1 (mod 4)and<br />
Wehavetoshowthat a ′′ isoddif<br />
q − 1<br />
2k ≡ pℓ + 1<br />
2<br />
m = 1<br />
[G : G0]<br />
2<br />
andeveno<strong>the</strong>rwise.But µ0 ≡ 3 (mod4)sothat k<br />
m<br />
andonlyif [G : G0] = k.If [G : G0] = k<strong>the</strong>n<br />
a ′′ ≡ pℓ − 1<br />
k<br />
isodd.O<strong>the</strong>rwise 2[G : G0]divides kand a ′′ iseven.<br />
≡ 1 (mod2).<br />
= 2and m = k<br />
2 .Thus [G : G0] = 2mif<br />
(mod2)<br />
Lemma14.3isnowcompletelyproved,soweturntoLemma14.4. In<strong>the</strong>pro<strong>of</strong><strong>of</strong>both<br />
Lemma14.4and14.5,wewillcombine<strong>the</strong>inductionassumptionwithLemma15.1whichis<br />
statedandprovedinparagraph15,<strong>the</strong>followingparagraph.Suppose F ⊆ F ′ ⊆ Land F ′ /F<br />
iscyclic<strong>of</strong>primedegree ℓ.Let G(K/F ′ )be H ′ Cwhere H ′ ⊆ Handlet E ′ be<strong>the</strong>fixedfield<strong>of</strong><br />
H ′ .Then E ′ /Eiscyclic<strong>of</strong>primeorder ℓ.If S(F ′ /F)is<strong>the</strong>set<strong>of</strong>characters<strong>of</strong> CF/N F ′ /FCF ′<br />
<strong>the</strong>n<br />
S(E ′ /E) = {ν E/F | νF ∈ S(F ′ /F)}.<br />
FromLemma15.1weseethatforanyquasicharacter χF,<br />
Therefore<br />
Ind(W K/E, W K/E ′, χ E ′ /E) ≃ ⊕ νF ∈S(F ′ /F) ν E/F χ E/F.<br />
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<br />
whichisequivalentto<br />
⊕νF {(⊕µ∈T Ind(W K/F, W K/Fµ , µ′ ν Fµ/F χ Fµ/F)) ⊕ νFχF }.<br />
If T ′ isaset<strong>of</strong>representativesfor<strong>the</strong>nontrivialorbits<strong>of</strong> H ′ in S(K/L)<strong>the</strong>n<br />
isequivalentto<br />
Moreover<br />
Ind(W K/F ′, W K/E ′, χ E ′ /F) = σ<br />
(⊕µ∈T ′Ind(W K/F ′, W K/F ′ µ , µ′ χ F ′ µ /F)) ⊕ χ F ′ /F.<br />
Ind(W K/F, W K/F ′, σ) ≃ Ind(W K/F, W K/E ′, χ E ′ /E).<br />
Applying<strong>the</strong>inductionassumptionto L/Fweseethat<br />
<br />
isequalto<br />
νF<br />
<br />
<br />
∆(νF, χF, ψF)<br />
νF<br />
{∆(χ F ′ /F, ψ F ′ /F)λ(F ′ /F, ψF)}<br />
<br />
µ∈T ∆(µ′ <br />
νFµ/F χFµ/F, ψFµ/F)λ(F/F, ψF)<br />
<br />
Theapplicationislegitimatebecause<strong>the</strong>fields F ′ , Fµ,and F ′ µ<br />
Lemma4.5<br />
Also<br />
µ∈T ′ ∆(µ′ χ F ′ µ /F, ψ F ′ µ /F)λ(F ′ µ /F, ψF). (14.14)<br />
λ(F ′ µ /F, ψF) = λ(F ′ µ /F ′ , ψF ′ /F) λ(F ′ ′<br />
[F<br />
/F, ψF) µ :F ′ ]<br />
.<br />
λ(F ′ <br />
<br />
/F, ψF)<br />
µ∈T ′ λ(F ′ /F, ψF)<br />
′<br />
[F µ :F ′ <br />
]<br />
allliebetween Fand L. By<br />
= λ(F ′ /F, ψF) [E′ :F ′ ] .<br />
Since<strong>the</strong>fields F ′ and F ′ µ liebetween F ′ and Kwecanapply<strong>the</strong>inductionassumptionto<br />
K/F ′ toseethat(14.14)isequalto<strong>the</strong>product<strong>of</strong><br />
and<br />
λ(F ′ /F, ψF) [E′ :F ′ ]<br />
∆(χ E ′ /F, ψ E ′ /F) λ(E ′ /F ′ , ψ F ′ /F).<br />
Applying<strong>the</strong>inductionassumptionto K/Eweseethat<br />
∆(χ E ′ /F, ψ E ′ /F)
Chapter14 228<br />
isequalto<br />
<br />
Weconcludethat<strong>the</strong>quotient<br />
<br />
νF<br />
νF ∈S(F ′ /F) ∆(ν E/F χ E/F, ψ E/F)<br />
<br />
λ(E ′ /E, ψ E/F) −1 .<br />
<br />
∆(νF χF,ψF) <br />
µ∈T ∆(µ′ <br />
νFµ/F χFµ/F, ψFµ/F ∆(νE/F χE/F,ψ E/F)<br />
isindependent<strong>of</strong> χF.Taking χFtobetrivialweseethatitequals<br />
Thus<br />
<br />
νF<br />
∆(νF, ψF) <br />
µ∈T ∆(µ′ ν Fµ/F, ψ Fµ/F)<br />
∆(ν E/F,ψ E/F)<br />
Itiseasilyseenthat<strong>the</strong>complexconjugate<strong>of</strong> ∆(νF, ψF)is<br />
If ℓisodd<strong>the</strong>rightsideis1.Since<br />
and νF = ν −1<br />
F<br />
if ℓisodd,<strong>the</strong>product<br />
For<strong>the</strong>samereasons <br />
However,if ℓis2<br />
νF(−1) ∆(ν −1<br />
F , ψF).<br />
∆(νF,ψF) ∆(ν −1<br />
F , ψF) = νF(−1).<br />
<br />
∆(1, ψF) = 1<br />
νF ∈S(L/F) ∆(νF, ψF) = 1.<br />
νF ∈S(L/F) ∆(ν E/F, ψ E/F) = 1.<br />
∆(νF, ψF) = ∆(ν −1<br />
F , ψF)<br />
hassquare ±1andis<strong>the</strong>reforeafourthroot<strong>of</strong>unity.Thus<br />
<br />
ν∈S(L/F) ∆(νF, ψF) ∼ <br />
ν∈S(L/F) ∆(ν E/F,ψ E/F) ∼2 1.<br />
(14.15)<br />
<br />
. (14.16)<br />
<strong>On</strong><strong>the</strong>o<strong>the</strong>rhand, m(µ ′ ) = t + 1 ≥ 2while m(ν Fµ/F) ≤ 1.ThusLemma9.5showsthat<br />
∆(µ ′ ν Fµ/F,ψ Fµ/F) ∼ℓ ∆(µ ′ , ψ Fµ/F).
Chapter14 229<br />
Thus<strong>the</strong>expression(14.16)and<strong>the</strong>refore<strong>the</strong>expression(14.15)isequalto<br />
where η ∼ℓ 1.<br />
<br />
<br />
η<br />
µ∈T ∆(µ′ ℓ , ψFµ/F) If m(χF)is0or1,Lemma14.4isaconsequence<strong>of</strong>Lemma14.2. Wesuppose<strong>the</strong>refore<br />
that m(χF) ≥ 2.InthiscaseLemma9.5impliesthat<br />
andthat <br />
νF<br />
<br />
νF<br />
∆(νF χF,ψF) ∼ℓ ∆(χF,ψF) ℓ<br />
∆(ν E/F χ E/F,ψ E/F) ∼ℓ ∆(χ E/F, ψ E/F) ℓ .<br />
Wealsosawin<strong>the</strong>beginning<strong>of</strong><strong>the</strong>paragraphthat,inallcases, m(µ ′ χ Fµ/F) ≥ 2.Thus<br />
∆(µ ′ ν Fµ/Fχ Fµ/F,ψ Fµ/F) ∼ℓ ∆(µ ′ χ Fµ/F, ψ Fµ/F).<br />
Putting<strong>the</strong>sefactstoge<strong>the</strong>rweseethatif<br />
isequalto<br />
<strong>the</strong>n σ ∼ℓ 1.Since σ = ρ ℓ weconcludethat<br />
<br />
σ ∆ (χF, ψF) <br />
µ∈T ∆(µ′ ℓ χFµ/F, ψFµ/F) <br />
∆(χE/F,ψE/F) <br />
µ∈T ∆(µ′ ℓ , ψFµ/F)<br />
ρ ∼ℓ 1.<br />
FinallywehavetoproveLemma14.5.Let F ′ be<strong>the</strong>fixedfield<strong>of</strong> H1Candlet L ′ be<strong>the</strong><br />
fixedfield<strong>of</strong> H2C.Let E ′ be<strong>the</strong>fixedfield<strong>of</strong> H1andlet K ′ be<strong>the</strong>fixedfield<strong>of</strong> H2.Let Pbe<br />
aset<strong>of</strong>representativesfor<strong>the</strong>orbitsunder G(L/F)<strong>of</strong><strong>the</strong>charactersin S(L/L ′ ).If νisone<br />
<strong>of</strong><strong>the</strong>serepresentatives,let HνH2Cwith Hνand H1beitsisotropygroupandlet Fνbe<strong>the</strong><br />
fixedfield<strong>of</strong> HνH2C.Applying<strong>the</strong>inductionassumptionandLemma15.1to<strong>the</strong>extension<br />
L/Fweseethat<br />
∆(χ F ′ /F, ψ F ′ /F) ρ (F ′ /F, ψF)
Chapter14 230<br />
isequalto<br />
Let<br />
<br />
ν∈P ∆(ν′ χ Fν/F, ψ Fν/F) λ(Fν/F, ψF). (14.17)<br />
R = {ν ∈ P | Fν = F }<br />
andlet Sbe<strong>the</strong>complement<strong>of</strong> Rin P. Rconsists<strong>of</strong><strong>the</strong>elements<strong>of</strong> S(L/L ′ )fixedbyeach<br />
element<strong>of</strong> G(L/F).Itisasubgroup<strong>of</strong> S(L/L ′ )anditsorder rmust<strong>the</strong>reforebeapower<strong>of</strong><br />
ℓ.Theexpression(14.17)maybewrittenas<br />
<br />
<br />
ν∈R ∆(ν′ <br />
<br />
χF, ψF)<br />
ν∈S ∆(ν′ <br />
χFν/F, ψFν/F) λ(Fν/F, ψF) .<br />
If Fisreplacedby Eand F ′ by E ′ <strong>the</strong>n Pisreplacedby<br />
{νK ′ /L ′ | ν = νL ′ ∈ P }.<br />
Also Fνisreplacedby Eν,<strong>the</strong>fixedfield<strong>of</strong> HνH2,and ν ′ isreplacedby ν ′ Eν/Fν .Applying<strong>the</strong><br />
inductionassumptionto K/Eweseethat<br />
isequalto<strong>the</strong>product<strong>of</strong><br />
and <br />
∆(χ E ′ /F,ψ E ′ /F) λ(E ′ /E, ψ E/F)<br />
<br />
ν∈R ∆(ν′ E/F χ E/F,ψ E/F<br />
ν∈S ∆(ν′ Eν/Fν χ Eν/Fν , ψ Eν/F) λ(Eν/E, ψ E/F<br />
Thisequalitywillbereferredtoasrelation(14.18).<br />
Toderivethisequalitywehaveusednotonly<strong>the</strong>inductionassumptionbutalsoLemma<br />
15.1whichimpliesthat<br />
Ind(W K/E, W K/E ′, χ E ′ /F)<br />
isequivalentto<br />
Thus<br />
{⊕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)}.<br />
Ind(W K/F, W K/E ′, χ E ′ /F)<br />
<br />
<br />
.
Chapter14 231<br />
willbeequivalentto<strong>the</strong>directsum<strong>of</strong><br />
and<br />
⊕R Ind(W K/F, W K/E, ν ′ E/F χ E/F)<br />
⊕S Ind(W K/F, W K/Eν , ν′ Eν/Fν χ Eν/F).<br />
If νisin RwecanapplyLemma15.1toseethat<br />
isequivalentto<br />
Wecanobtain<br />
Ind (W K/F, W K/E, ν ′ E/F χ E/F)<br />
{⊕µ∈T Ind(W K/F, W K/Fµ , µ′ ν ′ Fµ/F χ Fµ/F)} ⊕ ν ′ χF.<br />
Ind (W K/F, W K/Fν , ν′ Eν/Fν χ Eν/F)<br />
byfirstinducingfrom W K/Eν to W K/Fν and<strong>the</strong>nfrom W K/Fν to W K/F.<br />
If Tνisaset<strong>of</strong>representativesfor<strong>the</strong>orbits<strong>of</strong> S(K/L)under<strong>the</strong>action<strong>of</strong> G(K/Fν)and<br />
Fν,µis<strong>the</strong>fixedfield<strong>of</strong><strong>the</strong>isotropygroup<strong>of</strong> µin Tν<strong>the</strong>n,byLemma15.1again,<br />
isequivalentto<br />
Ind (W K/Fν , W K/Eν , ν′ Eν/Fν χ Eν/F)<br />
⊕Tν Ind (W K/Fν , W K/Fν,µ , µ′ ν ′ F ν,µ/Fν χ Fν,µ/F).<br />
Since [K : Fν] < [K : F]if νbelongsto S,wecanapply<strong>the</strong>inductionassumptiontoseethat<br />
isequalto<br />
<br />
∆(ν ′ Eν/Fν χ Eν/F, ψ Eν/F) λ(Eν/Fν, ψ Fν/F)<br />
∆(µ<br />
µ∈Tν<br />
′ ν ′ Fν,µ/FνχFν,µ/F, ψFν/F) λ(Fν,µ/Fν, ψFν/F). Thisequalitywillbereferredtoasrelation(14.19).<br />
Itals<strong>of</strong>ollowsthat<br />
isequivalentto<br />
Ind(W K/F, W K/Eν , ν′ Eν/Fν χ Eν/F)<br />
⊕µ∈Tν Ind(W K/F, W K/Fν,µ , µ′ ν ′ Fν,µ/Fν χ Fν,µ/F).
Chapter14 232<br />
Thefields Fνand Fν,µallliebetween Fand L.Thuswehaveexpressed<br />
asadirectsum<strong>of</strong>terms<strong>of</strong><strong>the</strong>form<br />
Ind(W K/F, W K/E ′, χ E ′ /F) (14.20)<br />
Ind(W K/F, W K/M, χM) (14.21)<br />
where Mliesbetween Fand L.Moreoversucharepresentationisinfactarepresentation<strong>of</strong><br />
W K/Fobtainedbyinflatingarepresentation<strong>of</strong> W L/F,namely,byinflating<br />
Ind(W L/F, W L/M, χM).<br />
Thusanyo<strong>the</strong>rexpression<strong>of</strong>(14.20)asasum<strong>of</strong>representations<strong>of</strong><strong>the</strong>form(14.21)willlead,<br />
byanapplication<strong>of</strong><strong>the</strong>inductionassumptionto L/F,toanidentitybetween<strong>the</strong>numbers<br />
∆(χM, ψ M/F).<br />
Toobtainano<strong>the</strong>rsuchexpression,weobservethat<strong>the</strong>representation(14.20)canbe<br />
obtainedbyfirstinducingfrom W K/E ′to W K/F ′and<strong>the</strong>nfrom W K/F ′to W K/F. If T ′ is<br />
aset<strong>of</strong>representativesfor<strong>the</strong>orbits<strong>of</strong>nontrivialcharactersin S(K/L)under<strong>the</strong>action<strong>of</strong><br />
G(K/F ′ )and F ′ µ is<strong>the</strong>fixedfield<strong>of</strong><strong>the</strong>isotropygroupin G(K/F ′ )<strong>of</strong> µin T ′ <strong>the</strong>n<br />
isequivalentto<br />
Ind(W K/F ′, W K/E ′, χ E ′ /F)<br />
{⊕µ∈T ′ Ind(W K/F ′, W K/F ′ µ , µ′ χ F ′ µ /F } ⊕ χ F ′ /F.<br />
Thus(14.20)isequivalentto<strong>the</strong>directsum<strong>of</strong><br />
and<br />
Ind(W K/F, W K/F ′, χ F ′ /F)<br />
⊕µ∈T ′ Ind(W K/F, W K/F ′ µ , µ ′ χ F ′ µ /F).<br />
Weshalldescribe<strong>the</strong>resultantidentityinamoment.Wefirstapply<strong>the</strong>inductionassumption<br />
to<strong>the</strong>extension K/F ′ toseethat<br />
isequalto<br />
∆(χ F ′ /F, ψ F ′ /F) <br />
∆(χ E ′ /F, ψ E ′ /F)λ(E ′ /F ′ , ψ F ′ /F)<br />
µ∈T ′ ∆(µ′ χ F ′ µ /F, ψ F ′ µ /F) λ(F ′ µ /F ′ , ψ F ′ /F).
Chapter14 233<br />
Thisequalitywillberelation(14.22).<br />
Thetwoexpressionsfor<strong>the</strong>representation(14.20)leadto<strong>the</strong>conclusionthat<strong>the</strong>product<br />
<strong>of</strong> <br />
ν∈R ∆(ν′ χF,ψF) (14.23)<br />
and <br />
and <br />
ν∈S<br />
ν∈R<br />
<br />
isequalto<strong>the</strong>product<strong>of</strong><br />
<br />
µ∈Tν<br />
and <br />
µ∈T ∆(µ′ ν ′ Fµ/F χ Fµ/F, ψ Fµ/F) λ(Fµ/F, ψF) (14.24)<br />
∆(µ ′ ν ′ Fν,µ/Fν χ Fν,µ/F, ψ Fν,µ/F) λ(Fν,µ/F, ψF) (14.25)<br />
∆(χ F ′ /F, ψ F ′ /F) λ(F ′ /F, ψF)<br />
µ∈T ′ ∆(µ′ χ F ′ µ /F,ψ F ′ µ /F) λ(F ′ µ/F, ψF).<br />
Applyingrelation(14.22)andLemma4.5weseethat<strong>the</strong>second<strong>of</strong><strong>the</strong>setwoproductsisequal<br />
to<br />
∆(χ E ′ /F, ψ E ′ /F) λ(E ′ /F ′ , ψ F ′ /F) λ(F ′ /F, ψF) [E′ :F ′ ] .<br />
Accordingto<strong>the</strong>relation(14.18)thisexpressionis<strong>the</strong>product<strong>of</strong><br />
<br />
<br />
ν∈R ∆(ν′ E/F χ <br />
<br />
E/F,ψ E/F)<br />
ν∈S ∆(ν′ E/F χ <br />
E/F,ψ E/F)<br />
and <br />
and<br />
ν∈S λ(Eν/E, ψ E/F)<br />
λ(E ′ /E, ψ E/F) −1 λ(E ′ /F ′ , ψ F ′ /F) λ(F ′ /F, ψF) [E′ :F ′ ] . (14.26)<br />
Equatingthisfinalproductto<strong>the</strong>product<strong>of</strong>(14.23),(14.24),and(14.25)and<strong>the</strong>nmaking<br />
certaincancellationsbymeans<strong>of</strong>(14.19),weseethat<strong>the</strong>product<strong>of</strong>(14.23)and(14.24)and<br />
<br />
isequalto<strong>the</strong>product<strong>of</strong><br />
ν∈S<br />
<br />
µ∈Tν<br />
<br />
λ −1 (Fν,µ/Fν, ψ Fν/F) λ(Fν,µ/F, ψF)<br />
ν∈R ∆(ν′ E/F χ E/F,ψ E/F)
Chapter14 234<br />
and <br />
and<strong>the</strong>expression(14.26).<br />
Inparticular,<strong>the</strong>expression<br />
<br />
ν∈R<br />
ν∈S λ−1 (Eν/Fν, ψ Fν/F) λ(Eν/E, ψ E/F)<br />
∆(ν ′ χF,ψF) <br />
isindependent<strong>of</strong> χF.Taking χFtobetrivialweseethat<br />
isequalto<br />
Theset<br />
<br />
ν∈R<br />
µ∈T ∆(µ′ ν ′ Fµ/F χ Fµ/F, ψ Fµ/F)<br />
∆(ν ′ E/F χ E/F,ψ E/F)<br />
′ ∆(ν χF,ψF) <br />
µ∈T ∆(µ′ ν ′ Fµ/F χFµ/F, ψFµ/F) ∆(ν ′ E/F χE/F, ψE/F) <br />
µ∈T ∆(µ′ ν ′ Fµ/F , ψ <br />
Fµ/F)<br />
<br />
ν∈R<br />
∆(ν ′ , ψF)<br />
∆(ν ′ E/F , ψ E/F) .<br />
R ′ = {ν ′ | ν ∈ R}<br />
isagroup<strong>of</strong>characters<strong>of</strong> CFor<strong>of</strong> H.Regardedascharacters<strong>of</strong> H<strong>the</strong>elements<strong>of</strong> R ′ arejust<br />
thosecharacterswhicharetrivialon H1.Asagroup R ′ iscyclicanditsorderisapower<strong>of</strong> ℓ.<br />
Theargumentusedin<strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma14.4showsthat<br />
and <br />
<br />
ν∈R ∆(ν′ , ψF) ∼ℓ 1<br />
ν∈R ∆(ν′ E/F , ψ E/F) ∼ℓ 1.<br />
If m(χF)is0or1,Lemma14.5isaconsequence<strong>of</strong>Lemma14.2.Wemayaswellsuppose<br />
<strong>the</strong>reforethat m(χF) > 1. If νbelongsto R<strong>the</strong>n ν ′ is1on N L/FCL. Therefore m(ν ′ ),as<br />
wellas m(ν ′ Fµ/F )and m(ν′ E/F )isatmost1.Wesawin<strong>the</strong>beginning<strong>of</strong>thisparagraphthat<br />
m(χ E/F)wouldalsobeatleast2. Wealsosawthat m(µ ′ χ Fµ/F)wouldbeei<strong>the</strong>r t + 1or
Chapter14 235<br />
ψ Fµ/F(m − 1) + 1. Inanycaseitisatleast2. Also m(µ ′ ) = t + 1isatleast2. Lemma9.5<br />
<strong>the</strong>reforeimplies<strong>the</strong>followingrelations:<br />
Weconcludefinallythat<br />
∆(ν ′ χF,ψF) ∼ℓ ∆(χF, ψF)<br />
∆(ν ′ E/F χ E/F,ψ E/F ∼ℓ ∆(χ E/F, ψ E/F)<br />
∆(µ ′ ν ′ Fµ/F χ Fµ/F, ψ Fµ/F ∼ℓ ∆(µ ′ χ Fµ/F, ψ Fµ/F)<br />
∆(µ ′ ν ′ Fµ/F , ψ Fµ/F) ∼ℓ ∆(µ ′ , ψ Fµ/F).<br />
<br />
∆(χF, ψF) <br />
µ∈T ∆(µ′ χFµ/F, ψFµ/F) ∆(χE/F,ψ E/F) <br />
µ∈T ∆(µ′ , ψFµ/F) if ris<strong>the</strong>number<strong>of</strong>elementsin R.Thelemmafollows.<br />
r<br />
∼ℓ 1
Chapter15 236<br />
Chapter Fifteen.<br />
Ano<strong>the</strong>r Lemma<br />
Suppose K/Fisnormaland G = G(K/F).Suppose Hisasubgroup<strong>of</strong> Gand Cisan<br />
abeliannormalsubgroup<strong>of</strong> G.Let Ebe<strong>the</strong>fixedfield<strong>of</strong> Hand Lthat<strong>of</strong> C.If µisacharacter<br />
<strong>of</strong> Cand hbelongsto H,define µ h by<br />
µ h (c) = µ(hch −1 ).<br />
Theset<strong>of</strong>characters<strong>of</strong> Cmaybeidentifiedwith S(K/L).If αbelongsto CL<br />
µ h (α) = µ(h(α)).<br />
Theset<strong>of</strong>elementsin S(K/L)whicharetrivialon H ∩ Cisinvariantunder H.Let Tbeaset<br />
<strong>of</strong>representativesfor<strong>the</strong>orbits<strong>of</strong> Hinthisset.If µ ∈ Tlet Hµbe<strong>the</strong>isotropygroup<strong>of</strong> µ,let<br />
Gµ = HµCandlet Fµbe<strong>the</strong>fixedfield<strong>of</strong> Gµ.Defineacharacter µ ′ <strong>of</strong> Gµby<br />
µ ′ (hc) = µ(c)<br />
if h ∈ Hµand c ∈ C. µ ′ mayberegardedasacharacter<strong>of</strong> CFµ .<br />
Lemma 15.1<br />
If χFisaquasicharacter<strong>of</strong> CF<strong>the</strong>n<br />
isequivalentto<br />
ρ = Ind(W K/F, W K/E, χ E/F)<br />
⊕µ∈T Ind(W K/F, W K/Fµ , µ′ χ Fµ/F).<br />
Let G ′ = HCandlet F ′ be<strong>the</strong>fixedfield<strong>of</strong> G ′ . F ′ iscontainedin Eandin<strong>the</strong>fields Fµ.<br />
Because<strong>of</strong><strong>the</strong>transitivity<strong>of</strong><strong>the</strong>inductionprocessitisenoughtoshowthat<br />
isequivalentto<br />
Ind(W K/F ′, W K/E, χ E/F)<br />
⊕µ∈TInd(W K/F ′, W K/Fµ , µ′ χ Fµ/F).
Chapter15 237<br />
If<br />
<strong>the</strong>n<br />
and<br />
χ ′ F ′ = χ F ′ /F<br />
χ E/F = χ ′ E/F ′<br />
χ Fµ/F = χ ′ Fµ/F ′.<br />
Consequentlywemaysuppose,withnoloss<strong>of</strong>generality,that F ′ is F.<br />
by<br />
If K ′ is<strong>the</strong>fixedfield<strong>of</strong> H ∩Cand ν ∈ S(K ′ /L)let ϕνbe<strong>the</strong>functionon W K/Fdefined<br />
ϕν(hc) = χF(τ K/F(hc)) ν(τ K/L(c))<br />
for hin W K/E, cin W K/L. ρactson<strong>the</strong>space<strong>of</strong>all<strong>functions</strong> ϕon W K/Fsatisfying<br />
forall hin W K/Eandall gin W K/F.Theset<br />
isabasisforthisspace.Clearly<br />
if cbelongsto W K/Land<br />
ϕ(hg) = χF(τ K/F(h)) ϕ(g)<br />
{ϕν | ν ∈ S(K ′ /L)}<br />
ρ(c)ϕν = χF(τ K/F(c)) ν(τ K/L(c))ϕν<br />
ρ(h)ϕν = χF(τ K/F(h))ϕν ′,<br />
with ν ′ = νh−1,if hbelongsto WK/E.Thusif Risanorbit<strong>of</strong> Hin S(K ′ /L)<br />
isaninvariantsubspace.<br />
⊕ν∈RCϕν<br />
= V<br />
Let µbe<strong>the</strong>elementcommonto Tand Randconsider<br />
If W K/Fis<strong>the</strong>disjointunion<br />
σ = Ind(W K/F, W K/Fµ , µ′ χ Fµ/F).<br />
r<br />
i=1 W K/Fµ hi
Chapter15 238<br />
andif ϕi(w) = 0unless<br />
while<br />
for win W K/Fµ <strong>the</strong>n<br />
w ∈ W K/Fµ hi<br />
ϕi(whi) = µ ′ χ Fµ/F(τ K/Fµ (w))<br />
{ϕi | 1 ≤ i ≤ r}<br />
isabasisfor<strong>the</strong>space Uonwhich σacts.If νi = µ hi andif λis<strong>the</strong>mapfrom Uto V which<br />
sends ϕito χ −1<br />
F (τ K/F(hi))ϕνi <strong>the</strong>n,asoneverifieseasily,<br />
forall win W K/F.Thelemmafollows.<br />
Thelemmahasacorollary.<br />
Lemma 15.2<br />
IfTheorem2.1isvalidfor K/F<strong>the</strong>n<br />
isequalto<br />
IfTheorem2.1isvalid<br />
isequalto<br />
Taking χF = 1,weseethat<br />
<br />
λσ(w) = ρ(w)λ<br />
∆(χ E/F,ψ E/F) <br />
<br />
µ∈T ∆(µ′ , ψ Fµ/F)<br />
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F).<br />
∆(χ E/F, ψ E/F) λ(E/F, ψF)<br />
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F) λ(Fµ/F, ψF).<br />
λ(E/F, ψF) = <br />
µ∈T ∆(µ′ , ψ Fµ/F) λ(Fµ/F, ψF).<br />
Substitutingthisinto<strong>the</strong>firstequalityandcancelling<strong>the</strong>nonzer<strong>of</strong>actor<br />
weobtain<strong>the</strong>lemma.<br />
<br />
µ∈T λ(Fµ/F, ψF)
Chapter15 239<br />
Todefine<strong>the</strong> λfunctionweshallneed<strong>the</strong>followinglemma.<br />
Lemma 15.3<br />
SupposeTheorem2.1isvalidforallGaloisextensions K1/F1with F ⊆ F1 ⊆ K1 ⊆ K<br />
and [K1 : F1] < [K : F].Then<br />
isequalto<br />
∆(χ E/F,ψ E/F) <br />
<br />
µ∈T ∆(µ′ , ψ Fµ/F)<br />
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F).<br />
Theconclusion<strong>of</strong>thislemmais<strong>the</strong>sameasthat<strong>of</strong><strong>the</strong>previousone.Thereishowevera<br />
criticaldifferencein<strong>the</strong>assumptions.<br />
Let F ′ be<strong>the</strong>fixedfield<strong>of</strong> HC.If<br />
<strong>the</strong>nforallseparableextensions E ′ <strong>of</strong> F ′<br />
ψ ′ F ′ = ψ F ′ /F<br />
ψ ′ E ′ /F ′ = ψ E ′ /F.<br />
If [K : F ′ ] < [K : F]<strong>the</strong>relation<strong>of</strong><strong>the</strong>lemmaisaconsequence<strong>of</strong><strong>the</strong>inductionassumption<br />
and<strong>the</strong>previouslemma.Wethussupposethat F = F ′ and G = HC.<br />
Supposeinadditionthat<strong>the</strong>reisasubgroup C1<strong>of</strong> C,whichisnei<strong>the</strong>r Cnor {1},whose<br />
normalizercontains H. C1is<strong>the</strong>nanormalsubgroup<strong>of</strong> G.Let F1be<strong>the</strong>fixedfield<strong>of</strong> HC1<br />
and L1<strong>the</strong>fixedfield<strong>of</strong> C1.Lemma15.1appliesto<strong>the</strong>extension K/F1.Thus<strong>the</strong>rearefields<br />
A1, . . ., Arlyingbetween F1and L1andquasicharacters χA1 , . . ., χArsuchthat isequivalentto<br />
Theinductionassumption<strong>the</strong>nimpliesthat<br />
Ind(W K/F1 , W K/E, χ E/F)<br />
⊕ r i=1 Ind(WK/F1 , WK/Ai , χAi ).<br />
∆(χ E/F, ψ E/F) λ(E/F1, ψ F1/F) (15.1)
Chapter15 240<br />
isequalto<br />
Thus<br />
r<br />
i=1 ∆(χAi , ψ Ai/F) λ(Ai/F1, ψ F1/F). (15.2)<br />
Inducing<strong>the</strong>first<strong>of</strong><strong>the</strong>setworepresentationsfrom W K/F1 to W K/F,weobtain<br />
isequivalentto<br />
Ind(W K/F, W K/E, ψ E/F).<br />
⊕µ∈T Ind(W K/F, W K/Fµ , µ′ χ Fµ/F) (15.3)<br />
Werecallthat<strong>the</strong>reexistsurjectivehomomorphisms<br />
⊕ r i=1 Ind(WK/F, WK/Ai , χAi ). (15.4)<br />
τ K/F,L1/F : W K/F −→ W L1/F<br />
τ K/Ai,L1/Ai : W K/Ai −→ W L1/Ai<br />
τ K/Fµ,L1/Fµ : W K/Fµ −→ W L1/Fµ<br />
whosekernelsareallequalto<strong>the</strong>commutatorsubgroup W c K/L1 <strong>of</strong> WK/L1 . Moreover<strong>the</strong><br />
diagrams<br />
and<br />
W K/Ai −→ W L1/Ai<br />
↓ ↓<br />
W K/F −→ W L1/F<br />
W K/Fµ −→ W L1/Fµ<br />
↓ ↓<br />
W K/F −→ W L1/F<br />
maybesupposedcommutative. Since W c K/L1 liesin<strong>the</strong>kernel<strong>of</strong> χAi and µ′ χ Fµ/F <strong>the</strong><br />
equivalence<strong>of</strong>(15.3)and(15.4)amountsto<strong>the</strong>equivalence<strong>of</strong><br />
and<br />
⊕µ∈TInd(W L1/F, W L1/Fµ , µ′ χ Fµ/F)<br />
⊕ r i=1Ind(WL1/F, WL1/Ai , χAi ).
Chapter15 241<br />
Theinductionassumptionappliedto<strong>the</strong>extension L1/Fimpliesthat<br />
r<br />
i=1 ∆(χAi , ψ Ai/F) λ(Ai/F, ψF)<br />
isequalto<br />
<br />
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F) λ(Fµ/F, ψF).<br />
Italsoimpliesthat<br />
λ(Ai/F, ψF) = λ(Ai/F1, ψ F1/F) λ(F1/F, ψF) [Ai:F1] .<br />
Since <br />
i [Ai : F1] = [E : F1]<br />
weinferfrom<strong>the</strong>equality<strong>of</strong>(15.1)and(15.2)that<br />
isequalto<br />
∆(χ E/F,ψ E/F) λ(E/F1, ψ F1/F) λ(F1/F, ψF) [E:F1]<br />
<br />
µ∈T ∆(µ′ χ Fµ/F, ψ Fµ/F) λ(Fµ/F, ψF).<br />
Taking χF = 1t<strong>of</strong>ind<strong>the</strong>value<strong>of</strong><br />
λ(E/F1, ψ F1/F) λ(F1/F, ψF) [E:F1]<br />
and<strong>the</strong>nsubstituting<strong>the</strong>resultinto<strong>the</strong>equationandcancelling<strong>the</strong>commonfactorsweobtain<br />
<strong>the</strong>assertion<strong>of</strong><strong>the</strong>lemma.<br />
Nowsupposethat Hcontainsanormalsubgroup H1 = {1}whichliesin<strong>the</strong>centralizer<br />
<strong>of</strong> C. H1isanormalsubgroup<strong>of</strong> Gif,asweareassuming, G = HC. K1,<strong>the</strong>fixedfield<strong>of</strong><br />
H1,contains Eandall<strong>the</strong>fields Fµ.Lemma15.1toge<strong>the</strong>rwith<strong>the</strong>argumentjustappliedto<br />
L1showsthat<br />
Ind(W K1/F, W K1/E, χE)<br />
isequivalentto<br />
⊕µ∈TInd(W K1/F,W K1/Fµ , µ′ χ Fµ/F).<br />
Inthiscase<strong>the</strong>assertion<strong>of</strong><strong>the</strong>lemmafollowsfrom<strong>the</strong>inductionassumptionappliedtoK1/F.<br />
Wehavefinallytosupposethat G = HC, Ccontainsnopropersubgroupinvariant<br />
under H,and Hcontainsnonormalsubgrouplyingin<strong>the</strong>centralizer<strong>of</strong> C. Inparticular<br />
H ∩C = {1}.If Zis<strong>the</strong>centralizer<strong>of</strong> C<strong>the</strong>n Z = (Z ∩H)Cand Z ∩Hisanormalsubgroup<br />
<strong>of</strong> H.Consequently Z = C.If Disanormalsubgroup<strong>of</strong> Gand Ddoesnotcontain C<strong>the</strong>n<br />
D ∩ C = {1}.<br />
Thisimpliesthat Discontainedin Z.Thus Discontainedin Cand D = {1}.If H = {1}<strong>the</strong><br />
assertion<strong>of</strong><strong>the</strong>lemmaisthat<strong>of</strong><strong>the</strong>thirdandfourthmainlemmas.If H = {1}<strong>the</strong>n G = C<br />
and Ciscyclic<strong>of</strong>primeordersothat<strong>the</strong>assertionisthat<strong>of</strong><strong>the</strong>firstmainlemma.
Chapter16 242<br />
Chapter Sixteen.<br />
Definition <strong>of</strong> <strong>the</strong> λ Functions<br />
Inthisand<strong>the</strong>nextthreeparagraphs,wetakeafixedGaloisextension K/F,assume<br />
thatTheorem2.1isvalidforallGaloisextensions K ′ /F ′ with F ⊆ F ′ ⊆ K ′ ⊆ Kand<br />
[K ′ : F ′ ] < [K : F],andprovethatitisvalidfor K/Fitself. Thefirststepistodefineand<br />
establishsomesimpleproperties<strong>of</strong><strong>the</strong>functionwhichwillserveas<strong>the</strong> λfunction.<br />
Lemma 16.1<br />
Suppose<br />
E/F ′ −→ λ(E/F ′ , ψF ′)<br />
isaweak λfunctionon P0(K ′ /F ′ ).If σ ∈ G(K ′ /F ′ )let<br />
Then<br />
E σ = {σ −1 (α) | α ∈ E}.<br />
λ(E σ /F ′ , ψF ′) = λ(E/F ′ , ψF ′).<br />
If µisacharacter<strong>of</strong> G(K/E)let µ σ be<strong>the</strong>character<strong>of</strong> G(K/E σ )definedby<br />
AccordingtoLemma13.2<br />
Therepresentation<br />
µ σ (ρ) = µ(σρσ −1 ).<br />
∆(µ σ , ψ E σ /F ′) = ∆(µ, ψ E/F ′).<br />
Ind(G(K ′ /F ′ ), G(K ′ /E), µ)<br />
actson<strong>the</strong>space U<strong>of</strong><strong>functions</strong> ϕon G(K ′ /F ′ )satisfying<br />
ϕ(ρτ) = µ(ρ) ϕ(τ)<br />
forall τin G(K ′ /F ′ )andall ρin G(K ′ /E).Themap ϕ −→ ψwith<br />
ψ(τ) = ϕ(στ)
Chapter16 243<br />
isaG(K ′ /F ′ )isomorphism<strong>of</strong> Uwith<strong>the</strong>spaceonwhich<br />
Ind(G(K ′ /F ′ ), G(K ′ /E σ ), µ σ )<br />
acts.Thus<strong>the</strong>tworepresentationsareequivalent.<br />
If<br />
isequivalentto<br />
<strong>the</strong>n<br />
isequivalentto<br />
⊕ r i=1 Ind(G(K′ /F ′ ), G(K ′ /Ei), µi)<br />
⊕ s j=1Ind(G(K ′ /F ′ ), G(K ′ /Fj), νj)<br />
⊕ r i=1 Ind(G(K′ /F ′ ), G(K ′ /E σ i ), µσ i )<br />
⊕ s j=1 Ind(G(K′ /F ′ ), G(K ′ /F σ j ), νσ j )<br />
and,with<strong>the</strong>conventions<strong>of</strong><strong>the</strong>fourthparagraph,<br />
isequalto<br />
Since<br />
and<br />
r<br />
s<br />
Weconcludethat r<br />
isequalto<br />
Ino<strong>the</strong>rwords<br />
i=1 (χE σ i , ψ E σ i /F ′) λ(Eσ i /F ′ , ψF ′)<br />
j=1 ∆(χF σ j , ψ F σ j /F ′) λ(F σ j /F ′ , ψF ′).<br />
∆(χF σ j , ψ F σ j /F ′) = ∆(χFj , ψ Fj/F ′)<br />
∆(χE σ i , ψ E σ i /F ′) = ∆(χEi , ψ Ei/F ′).<br />
s<br />
i=1 ∆(χEi , ψ Ei/F ′) λ(Eσ i /F ′ , ψF ′)<br />
j=1 ∆(χFj , ψ Fj/F ′) λ(F σ j /F ′ , ψF ′).<br />
E/F ′ −→ λ(E σ /F ′ , ψF ′)<br />
isaweakλfunctionon P0(K ′ /F ′ ).Lemma16.1followsfrom<strong>the</strong>uniqueness<strong>of</strong>such<strong>functions</strong>.<br />
Wereturnto<strong>the</strong>problem<strong>of</strong>definingaλfunctionon P0(K/F). Chooseanontrivial<br />
abeliannormalsubgroup C<strong>of</strong> G = G(K/F)andlet Lbe<strong>the</strong>fixedfield<strong>of</strong> C. If Eisany
Chapter16 244<br />
fieldlyingbetween Fand Klet Hbe<strong>the</strong>correspondingsubgroup<strong>of</strong> G. Choose<strong>the</strong>set T<br />
<strong>of</strong>charactersand<strong>the</strong>fields Fµasin<strong>the</strong>previousparagraph. Since Fµ ⊆ L<strong>the</strong>numbers<br />
λ(Fµ/F, ψF)aredefined.<br />
Lemma 16.2<br />
Suppose F ⊆ E ⊆ K1 ⊂ = Kwith K1/Fnormalsothat λ(E/F, ψF)isdefined.Then<br />
λ(E/F, ψF) = <br />
µ∈T ∆(µ′ , ψ Fµ/F) λ(Fµ/F, ψF).<br />
Let K1be<strong>the</strong>fixedfield<strong>of</strong> H1.If H1 ∩ C = {1}wemayenlarge K1andreplace H1by<br />
H1 ∩C.Thuswemaysupposethatei<strong>the</strong>r H1iscontainedin Cor H1 ∩C = {1}.Inei<strong>the</strong>rcase<br />
H1iscontainedin<strong>the</strong>centralizer<strong>of</strong> C. Wesawin<strong>the</strong>previousparagraphthatunder<strong>the</strong>se<br />
circumstance<br />
Consequently<br />
Ingeneral,wedefine<br />
Ind(W K1/F, W K1/E, 1) ≃ ⊕µ∈TInd(W K1/F, W K1/Fµ , µ′ ).<br />
λ(E/F, ψF) = <br />
λ(E/F, ψF) = <br />
µ∈T ∆(µ′ , ψ Fµ/F) λ(Fµ/F, ψF).<br />
µ∈T ∆(µ′ , ψ Fµ/F) λ(Fµ/F, ψF)<br />
if E/Fisin P0(K/F) Tis,<strong>of</strong>course,notalwaysuniquelydetermined.Wemayreplaceany<br />
µin Tby µ σ with σin H.Then Hµand Gµarereplacedby σ −1 Hµσand σ −1 Gµσwhile Fµis<br />
replacedby F σ µ and µ′ isreplacedby (µ ′ ) σ .Since<br />
∆(µ ′ , ψ Fµ/F) λ(Fµ/F, ψF) = ∆((µ ′ ) σ , ψ F σ µ /F) λ(F σ µ /F, ψF)<br />
<strong>the</strong>number λ(E/F, ψF)doesnotdependon T. A priori,itmaydependon Cbutthatis<br />
unimportantsince Cisfixedand,<strong>the</strong>uniquenesshavingbeenproved,weareinterestedonly<br />
in<strong>the</strong>existence<strong>of</strong>aλfunction.<br />
Weshallneedonlyoneproperty<strong>of</strong><strong>the</strong>functionjustdefined.<br />
Lemma 16.3
Chapter16 245<br />
If F ⊆ E ⊆ E ′ ⊆ K<strong>the</strong>n<br />
If E = F<strong>the</strong>n<br />
andif E = F<br />
λ(E ′ /F, ψF) = λ(E ′ /E, ψ E/F) λ(E/F, ψF) [E′ :E] .<br />
λ(E ′ /E, ψ E/F) = λ(E ′ /F, ψF)<br />
λ(E ′ /E, ψ E/F)<br />
is<strong>the</strong>value<strong>of</strong><strong>the</strong> λfunction<strong>of</strong> P(K/E),whichisdefinedbyassumption,at E ′ /E.Since<br />
λ(F/F, ψF) = 1<br />
<strong>the</strong>assertionisclearif E = F.Itisalsoclearif E = E ′ .<br />
Let Ebe<strong>the</strong>fixedfield<strong>of</strong> Hasbeforeandlet F ′ be<strong>the</strong>fixedfield<strong>of</strong> HC. Wesuppose<br />
that H = G.Lemma4.5and<strong>the</strong>inductionassumptionimplythat<br />
Therelation<br />
impliesthat<br />
λ(Fµ/F, ψF) = λ(Fµ/F ′ , ψ F ′ /F) λ(F ′ /F, ψF) [Fµ:F ′ ] .<br />
[E : F ′ ] = [Fµ : F ′ ]<br />
λ(E/F, ψF) = λ(E/F ′ , ψ F ′ /F) λ(F ′ /F, ψF) [E:F ′ ] .<br />
Thereisasimilarformulafor λ(E ′ /F, ψF).If F ′ = F<strong>the</strong>inductionassumptionimpliesthat<br />
Since<br />
λ(E ′ /E, ψ E/F) λ(E/F ′ , ψ F ′ /F) [E′ :E] = λ(E ′ /F ′ , ψF ′ /F).<br />
[E ′ : F ′ ] = [E ′ : E] [E : F ′ ]<br />
<strong>the</strong>assertion<strong>of</strong><strong>the</strong>lemmaisprovedsimplybymultiplyingbothsides<strong>of</strong>thisequationby<br />
λ(F ′ /F, ψF) [E′ :F ′ ] .<br />
Nowsupposethat G = HCand H ∩ C = {1}. Let E ′ be<strong>the</strong>fixedfield<strong>of</strong> H ′ andlet<br />
F ′ be<strong>the</strong>fixedfield<strong>of</strong> H ′ C = G ′ .Eachcharacter<strong>of</strong> H ′ maybeidentifiedwithacharacter<strong>of</strong>
Chapter16 246<br />
CE ′/N K/E ′CKandeachcharacter<strong>of</strong> G ′ maybeidentifiedwithacharacter<strong>of</strong> CF ′/N K/F ′CK.<br />
Anycharacter χE ′<strong>of</strong> H′ maybeextendedtoacharacter χF ′<strong>of</strong> G′ bysetting<br />
if ρ ∈ H ′ and σ ∈ C.Then<br />
χF ′(ρσ) = χ E ′ (ρ)<br />
χE ′ = χ E ′ /F ′.<br />
ItfollowsfromLemma15.1that<strong>the</strong>rearefields<strong>of</strong> Fi(E ′ ), 1 ≤ i ≤ m(E ′ ),lyingbetween F ′<br />
and Landcharacters µ Fi(E ′ )suchthat<br />
isequivalentto<br />
Ind(W K/F ′, W K/E ′, χE ′)<br />
⊕ m(E′ )<br />
i=1 Ind(W K/F ′, W K/Fi(E ′ ), µ Fi(E ′ )χ Fi(E ′ )/F ′).<br />
If E = E ′ sothat F = F ′ <strong>the</strong>inductionassumptionimpliesthat<br />
isequalto m(E ′ )<br />
∆(χE ′, ψ E ′ /F) λ(E ′ /F ′ , ψ F ′ /F)<br />
i=1 ∆(µ Fi(E ′ )χ Fi(E ′ )/F ′, ψ Fi(E ′ )/F) λ(Fi(E)/F ′ , ψ F ′ /F).<br />
Wehaveseenthat<strong>the</strong>lemmaisvalidforanypair E ′ , Eforwhich HC = G.Inparticular,<br />
itisvalidfor<strong>the</strong>pair E ′ , F ′ and<strong>the</strong>pairs Fi(E ′ ), F ′ .Multiplying<strong>the</strong>equalityjustobtained<br />
by<br />
λ(F ′ /F, ψF) [E′ :F ′ ]<br />
weseethat<br />
isequalto<br />
m(E ′ )<br />
∆(χE ′, ψ E ′ /F) λ(E ′ /F, ψF) (16.1)<br />
i=1 ∆(µ Fi(E ′ )χ Fi(E ′ )/F ′, ψ Fi(E ′ )/F) λ(Fi(E ′ )/F, ψF). (16.2)<br />
If F ′ = F<strong>the</strong>equality<strong>of</strong>(16.1)and(16.2),forasuitablechoice<strong>of</strong><strong>the</strong>fields Fi(E ′ ),resultsfrom<br />
Lemma15.1,Lemma15.3,and<strong>the</strong>definition<strong>of</strong><br />
λ(E ′ /F, ψF).<br />
Inanycase<strong>the</strong>equalityisvalidforallfieldslyingbetween Eand K.
Chapter16 247<br />
Suppose E1, . . ., Er, E ′ 1 , . . ., E′ s aresuchfields, χEi isacharacter<strong>of</strong> CEi /N K/Ei CK, χE ′ j<br />
isacharacter<strong>of</strong> C E ′ j /N K/E ′ j CK,and<br />
isequivalentto<br />
Then r<br />
⊕ r i=1Ind(WK/E, WK/Ei , χEi )<br />
⊕ s j=1Ind(W K/E, W K/E ′ j , χE ′ j ).<br />
i=1 [Ei : E] = s<br />
and,by<strong>the</strong>transitivity<strong>of</strong><strong>the</strong>inductionprocess,<br />
isequivalentto<br />
j=1 [E′ j<br />
⊕ r i=1 ⊕ m(Ei)<br />
k=1 Ind(W K/F, W K/Fk(Ei), µ Fk(Ei) χ Fk(Ei)/Fi )<br />
⊕ s j=1 ⊕m(E′<br />
j )<br />
ℓ=1 Ind(WK/F, WK/Fℓ(E ′ j ), µ Fℓ(E ′ j ) χFℓ(E ′ j )/F ′ j ).<br />
: E] (16.3)<br />
If Eiis<strong>the</strong>fixedfield<strong>of</strong> Hiand E ′ j<strong>the</strong>fixedfield<strong>of</strong> H′ j<strong>the</strong>n Fiand F ′ jare<strong>the</strong>fixedfields<strong>of</strong> HiCand H ′ jC.Thisequivalenceand<strong>the</strong>inductionassumptionfor L/Fimplythat<br />
isequalto<br />
r<br />
s<br />
i=1<br />
j=1<br />
m(Ei)<br />
k=1 ∆(µ Fk(Ei) χ Fk(Ei)/Fi , ψ Fk(Ei)/F) λ(Fk(Ei)/F, ψF)<br />
m(E ′<br />
j )<br />
ℓ=1 ∆(µ Fℓ(E ′ j ) χ Fℓ(E ′ j )/F ′ j , ψ Fℓ(E ′ j )/F) λ(Fℓ(E ′ j<br />
)/F, ψF).<br />
Thisequality,<strong>the</strong>equality<strong>of</strong>(16.1)and(16.2),and<strong>the</strong>relation(16.3)implythat<br />
isequalto<br />
Consequently<br />
r<br />
s<br />
i=1 ∆(χEi , ψ Ei/F) λ(Ei/F, ψF) λ(E/F, ψF) −[Ei:E]<br />
j=1 ∆(χ E ′ j , ψ E ′ j /F) λ(E ′ j /F, ψF) λ(E/F, ψF) −[E′<br />
E ′ −→ λ(E ′ /F, ψF) λ(E/F, ψF) −[E′ :E]<br />
isaweak λfunctionon P0(K/E).Thelemma<strong>of</strong>uniquenessimpliesthat<br />
λ(E ′ /F, ψF) λ(E/F, ψF) −[E′ :E] = λ(E ′ /E, ψE/F).<br />
j :E] .
Chapter16 248<br />
Thisis,<strong>of</strong>course,<strong>the</strong>assertion<strong>of</strong><strong>the</strong>lemma.<br />
Atthispoint,wehaveproved<strong>the</strong>lemmawhenvarioussupplementaryconditionsare<br />
satisfied.Beforeprovingit,ingeneral,wemakeanobservation.Suppose<br />
F ⊆ E ⊆ E ′ ⊆ E ′′ ⊆ K<br />
and<strong>the</strong>assertion<strong>of</strong><strong>the</strong>lemmaisvalidfor E ′′ /E ′ and E ′ /E.Then<br />
and<br />
Moreover,byinduction,<br />
λ(E ′′ /F, ψF) = λ(E ′′ /E ′ , ψ E ′ /F) λ(E ′ /F, ψF) [E′′ :E ′ ]<br />
λ(E ′ /F, ψF) = λ(E ′ /E, ψ E/F) λ(E/F, ψF) [E′ :E] .<br />
λ(E ′′ /E, ψ E/F) = λ(E ′′ /E ′ , ψ E ′ /F) λ(E ′ /E, ψ E/F) [E′′ :E ′ ] .<br />
Theassertionfor E ′′ /Eisobtainedbysubstituting<strong>the</strong>secondrelationin<strong>the</strong>firstandsimplifyingaccordingto<strong>the</strong>third.<br />
If<strong>the</strong>lemmaisfalseingeneral,choseamongstall<strong>the</strong>extensionsin P(K/F)forwhichit<br />
isfalseone E ′ /Eforwhich [E ′ : E]isaminimum.Let Ebe<strong>the</strong>fixedfield<strong>of</strong> Hand E ′ that<br />
<strong>of</strong> H ′ .Accordingto<strong>the</strong>previousdiscussion G = HC, H ∩ C = {1},and<strong>the</strong>rearen<strong>of</strong>ields<br />
lyingbetween Eand E ′ . If H ′ ∩ C = H ∩ C,whichisanormalsubgroup<strong>of</strong> G,<strong>the</strong>fields<br />
F, E,and E ′ arecontainedin<strong>the</strong>fixedfield<strong>of</strong> H ∩ Cand<strong>the</strong>assertionisaconsequence<strong>of</strong><br />
<strong>the</strong>inductionassumption.Thus H ′ isapropersubgroup<strong>of</strong> H ′ (H ∩ C).Because<strong>the</strong>reareno<br />
intermediatefields H = H ′ (H ∩ C).<br />
Aswehaveseen<strong>the</strong>rearefields E1, . . ., Erlyingbetween Eand<strong>the</strong>fixedfield K1<strong>of</strong><br />
, . . ., µErsuchthat H ∩ Candcharacters µE1<br />
isequivalentto<br />
Then<br />
λ(E ′ /E, ψ E/F) = r<br />
Ind(W K/E, W K/E ′, 1)<br />
⊕ r i=1Ind(WK/E, WK/Ei , µEi ).<br />
By<strong>the</strong>inductionassumption,appliedto K1/F,<br />
i=1 ∆(µEi , ψ Ei/E) λ(Ei/E, ψ E/F).<br />
λ(Ei/E, ψ E/F) λ(E/F, ψF) [Ei:E] = λ(Ei/F, ψF).
Chapter16 249<br />
Thus<br />
isequalto<br />
λ(E ′ /E, ψ E/F) λ(E/F, ψF) [E′ :E]<br />
r<br />
Moreover,by<strong>the</strong>transitivity<strong>of</strong><strong>the</strong>inductionprocess,<br />
isequivalentto<br />
i=1 ∆(µEi , ψ Ei/E) λ(Ei/F, ψF). (16.4)<br />
Ind(W K/F, W K/E ′, 1) (16.5)<br />
⊕ r i=1Ind(WK/F, WK/Ei , µEi ). (16.6)<br />
<strong>On</strong><strong>the</strong>o<strong>the</strong>rhand,<strong>the</strong>rearefields F1, . . ., Fscontainedin Landcharacters νF1 , . . ., νFssuch that(16.5)isequivalentto<br />
⊕ s j=1Ind(WK/F, WK/Fj , νFj ) (16.7)<br />
andsuchthat,bydefinition,<br />
λ(E ′ /F, ψF) = s<br />
j=1 ∆(νFj , ψ Fj/F) λ(Fj/F, ψF). (16.8)<br />
Since<strong>the</strong>representations(16.6)and(16.7)areequivalent<strong>the</strong>inductionassumption,appliedto<br />
K1/F,showsthat(16.4)isequalto<strong>the</strong>rightside<strong>of</strong>(16.8).Thisisacontradiction.
Chapter17 250<br />
Chapter Seventeen.<br />
A Simplification.<br />
Weshalluse<strong>the</strong>symbol Ωtodenoteanorbitin<strong>the</strong>set<strong>of</strong>quasicharacters<strong>of</strong> CKunder<br />
<strong>the</strong>action<strong>of</strong> G(K/F)or,whatis<strong>the</strong>same,under<strong>the</strong>action<strong>of</strong> W K/F on CKbymeans<strong>of</strong><br />
innerautomorphisms. If χKisaquasicharacter<strong>of</strong> CKitsorbitwillbedenoted Ω(χK). If<br />
ρisarepresentation<strong>of</strong> W K/F<strong>the</strong>restriction<strong>of</strong> ρto CKis<strong>the</strong>directsum<strong>of</strong>onedimensional<br />
representations.Let S(ρ)be<strong>the</strong>collection<strong>of</strong>quasicharacterstowhich<strong>the</strong>seonedimensional<br />
representationscorrespond.<br />
Suppose<br />
Let W K/Fbe<strong>the</strong>disjointunion<br />
Define<strong>the</strong>function ϕiby<br />
ρ = Ind(W K/F, W K/E, χE).<br />
m<br />
i=1 W K/Ewi.<br />
ϕi(wwj) = 0 w ∈ W K/E, j = i<br />
ϕi(wwi) = χE(τ K/Ew) w ∈ W K/E.<br />
{ϕ1, . . ., ϕm}isabasisfor<strong>the</strong>space<strong>of</strong><strong>functions</strong><strong>of</strong>which ρacts.If a ∈ CK<strong>the</strong>n<br />
wwja = w(wjaw −1<br />
j )wj<br />
and wjaw −1<br />
j belongsto CKwhich,<strong>of</strong>course,liesin W K/E.Thus<br />
if σiis<strong>the</strong>image<strong>of</strong> wiin G(K/F).Thus<br />
Suppose E1, . . ., Er, E ′ 1 , . . ., E′ s<br />
χE ′ j isaquasicharacter<strong>of</strong> E′ j .Let<br />
ρ(a)ϕi = χE(τ K/E(wiaw −1<br />
i ))ϕi = χ σi<br />
K/E (a)ϕi<br />
S(ρ) = Ω(χ K/E).<br />
liebetween Fand K, χEi isaquasicharacter<strong>of</strong> Ei,and<br />
ρi = Ind(WK/F, WK/Ei , χEi )
Chapter17 251<br />
andlet<br />
Suppose ρiactson Viand ρ ′ j<br />
ρ ′ j = Ind(W K/F, W K/E ′ j , χ E ′ j ).<br />
′ actson V j .Thedirectsum<strong>of</strong><strong>the</strong>representations ρiactson<br />
V = ⊕ r i=1 Vi<br />
and<strong>the</strong>directsum<strong>of</strong><strong>the</strong>representations ρ ′ j actson<br />
Let<br />
V ′ = ⊕ s ′<br />
j=1V j .<br />
VΩ = ⊕ {i | χK/Ei ∈Ω}Vi<br />
V ′ Ω = ⊕ {i | χK/E ′ ∈Ω}V<br />
j<br />
′<br />
j .<br />
Anyisomorphism<strong>of</strong> Vwith V ′ whichcommuteswith<strong>the</strong>action<strong>of</strong> W K/Ftakes VΩto V ′ Ω .<br />
and<br />
If χ K/Ei ∈ Ω(χK)<strong>the</strong>reisaσin G(K/F)suchthat χK = χ σ K/Ei .Then<br />
ρi ≃ Ind(W K/F, W K/E σ i , χ σ Ei )<br />
∆(χEi , ψ Ei/F) λ(Ei/F, ψF) = ∆(χ σ Ei , ψ E σ i /F) λ(E σ i /F, ψF).<br />
WeconcludethatTheorem2.1isaconsequence<strong>of</strong><strong>the</strong>followinglemma.<br />
Lemma 17.1<br />
Suppose χKisaquasicharacter<strong>of</strong> CK.Suppose E1, . . ., Er, E ′ 1 , . . ., E′ s<br />
and K, χEi isaquasicharacter<strong>of</strong> CEi , χE ′ j isaquasicharacter<strong>of</strong> CE ′ j ,and<br />
isequivalentto<br />
If χ K/Ei = χ K/E ′ j = χKforall iand j<strong>the</strong>n<br />
ρ = ⊕ r i=1Ind(WK/F, WK/Ei , χEi )<br />
ρ ′ = ⊕ s j=1 Ind(W K/F, W K/E ′ j , χ E ′ j ).<br />
r<br />
i=1 ∆(χEi , ψ Ei/F) λ(Ei/F, ψF)<br />
liebetween F
Chapter17 252<br />
isequalto<br />
s<br />
j=1 ∆(χ E ′ j , ψ E ′ j /F), λ(E ′ j<br />
/F, ψF).<br />
Let F(χK)be<strong>the</strong>fixedfield<strong>of</strong><strong>the</strong>isotropygroup<strong>of</strong> χK.Let ρacton Vandlet ρ ′ acton<br />
V ′ .Let<br />
V (χK) = {v ∈ V | ρ(a)v = χK(a)v forall a in CK}.<br />
Define V ′ (χK)inasimilarfashion. Itisclearthatanyisomorphism<strong>of</strong> V with V ′ which<br />
commuteswith<strong>the</strong>action<strong>of</strong> W K/Ftakes V (χK)to V ′ (χK).Thegroup W K/F(χK)leavesboth<br />
V (χK)and V ′ (χK)invariantanditsrepresentationson<strong>the</strong>setwospacesareequivalent.<br />
Let<br />
Ind(WK/F, WK/Ei , χEi )<br />
actson Vianddefine Vi(χK)in<strong>the</strong>obviousmanner.Then<br />
Defining V ′<br />
j<br />
V (χK) = ⊕ r i=1 Vi(χK).<br />
and V ′<br />
j (χK)inasimilarmanner,wehave<br />
V ′ (χK) = ⊕ s ′<br />
j=1V j (χK).<br />
Itisclearthat<strong>the</strong>representation<strong>of</strong> WK/F(χK)on Vi(χK)isequivalentto<br />
Thus<br />
isequivalentto<br />
Ind(WK/F(χK), WK/Ei , χEi ).<br />
⊕ r i=1Ind(WK/F(χK), WK/Ei , χEi )<br />
⊕ s j=1 Ind(W K/F(χK), W K/E ′ j , χE ′ j ).<br />
If F(χK) = F<strong>the</strong>assertion<strong>of</strong><strong>the</strong>lemmafollowsfrom<strong>the</strong>inductionassumptionandLemma<br />
16.3.
Chapter18 253<br />
Chapter Eighteen.<br />
Nilpotent Groups.<br />
InthisparagraphweproveLemma17.1assumingthatF = F(χK)andthatG = G(K/F)<br />
isnilpotent.<br />
Lemma 18.1<br />
Suppose Disanormalsubgroup<strong>of</strong> G<strong>of</strong>primeorder ℓwhichiscontainedin<strong>the</strong>center<br />
<strong>of</strong> G.Let Mbe<strong>the</strong>fixedfield<strong>of</strong> D.Suppose F ⊆ E ⊆ Kand χEisaquasicharacter<strong>of</strong> CE.<br />
Supposealsothat F(χ K/E) = F.<br />
(a)Therearefields F1, . . ., Frcontainedin Mandquasicharacters χF1 , . . ., χFrsuchthat χK/Fi = χK/Eandsuchthat isequivalentto<br />
Ind(W K/F, W K/E, χE)<br />
⊕ r i=1Ind(WK/F, WK/Fi , χFi ).<br />
(b)IfTheorem2.1isvalidforallGaloisextensionsK ′ /F ′ in P(K/F)with[K ′ : F ′ ] < [K : F]<br />
<strong>the</strong>n<br />
∆(χE, ψ E/F) λ(E/F, ψF)<br />
isequalto<br />
r<br />
i=1 ∆(χFi , ψ Fi/F) λ(Fi/F, ψF).<br />
Weprove<strong>the</strong>lemmabyinductionon [K : F].Let Hbe<strong>the</strong>subgroup<strong>of</strong> Gcorresponding<br />
to E;let G ′ = HDandlet F ′ be<strong>the</strong>fixedfield<strong>of</strong> G ′ . If F ′ = F<strong>the</strong>inductionassumption<br />
impliesthat<strong>the</strong>rearefields F1, . . ., Frcontainedin Mandquasicharacters χF1 , . . ., χFrsuch that χK/Fi = χK/Eforeach iandsuchthat<br />
isequivalentto<br />
Ind(W K/F, W K/E, χE)<br />
⊕ r i=1Ind(WK/F ′, WK/Fi , χFi ).
Thefirstpart<strong>of</strong><strong>the</strong>lemmafollowsfrom<strong>the</strong>transitivity<strong>of</strong><strong>the</strong>inductionprocess.Thesecond<br />
partfollowsfromLemma16.3and<strong>the</strong>assumedvalidity<strong>of</strong>Theorem2.1for<strong>the</strong>extension<br />
K/F ′ .<br />
Wesupposenowthat G = HD. Supposethat Hcontainsanormalsubgroup H1<strong>of</strong> G<br />
whichisdifferentfrom {1}andsupposethat,if K1is<strong>the</strong>fixedfield<strong>of</strong> H1, F(χK1/E) = F.<br />
If M1is<strong>the</strong>fixedfield<strong>of</strong> H1D<strong>the</strong>n,accordingto<strong>the</strong>inductionassumption,<strong>the</strong>rearefields<br />
F1, . . ., Frcontainedin M1andquasicharacters χF1 , . . ., χFrsuchthat andsuchthat<br />
isequivalentto<br />
Itfollowsimmediatelythat<br />
andthat<br />
isequivalentto<br />
χ K1/Fi = χ K1/E<br />
Ind(W K1/F, W K1/E, χE)<br />
⊕ r i=1Ind(WK1/F, WK1/Fi , χFi ).<br />
χ K/Fi = χ K/E<br />
Ind(W K/F, W K/E, χE)<br />
⊕ r i=1Ind(WK/F, WK/Fi , χFi ).<br />
Theequality<strong>of</strong>(b)isaconsequence<strong>of</strong><strong>the</strong>assumedvalidity<strong>of</strong>Theorem2.1for K1/F.<br />
Weassumenowthat G = HDandthatif H1isanormalsubgroup<strong>of</strong> Hdifferentfrom<br />
{1}withfixedfield K1<strong>the</strong>field F(χ K1/E)isnot F.If w1belongsto W K/Eand w2belongsto<br />
W K/M<strong>the</strong>n<br />
Let χK = χ K/E.Since F(χK) = F<br />
χK(w1w2w −1<br />
w1w2w −1<br />
1 w−1 2 ∈ CK.<br />
1 w−1<br />
−1<br />
2 ) = χK(w2w1 w−1 2 w1) = χK(w −1<br />
1 w−1 2 w1w2) = χK(w −1 −1<br />
2 w1w2w1 ).<br />
Denote<strong>the</strong>commonvalue<strong>of</strong><strong>the</strong>expressionsby ω(w1, w2).Then ω(v1w1, w2)isequalto<br />
Therightsideis<br />
χK(v1w1w2w −1<br />
1 v−1<br />
1 w−1 2<br />
−1 −1<br />
) = χK(w2 w1w2w1 v−1 1 w−1 2 v1w2).<br />
ω(v1, w2) ω(w1, w2).
Chapter18 255<br />
In<strong>the</strong>sameway ω(w1, v2w2)is<br />
whichequals<br />
χK(w1v2w2w −1<br />
1 w−1 2 v−1 2<br />
Ifei<strong>the</strong>r w1or w2belongto CK,wehave<br />
Thus,foreach w2,<br />
) = χK(w −1<br />
ω(w1, v2) ω(w1, w2).<br />
ω(w1, w2) = 1.<br />
w1 −→ ω(w1, w2)<br />
1 v−1<br />
−1<br />
2 w1v2w2w1 w−1 2 w1)<br />
isahomomorphism<strong>of</strong> H = W K/E/CKinto C ∗ and,foreach w1,<br />
w2 −→ ω(w1, w2)<br />
isahomomorphism<strong>of</strong> D = W K/M/CKinto C ∗ .If wbelongsto W K/F<strong>the</strong>n<br />
ω(ww1w −1 , ww2w −1 ) = ω(w1, w2).<br />
Thus<strong>the</strong>reisanormalextension K1containing Esuchthat<br />
W K/K1 = {w1 | ω(w1, w2) = 1 forall w1 ∈ W K/M}.<br />
But F(χ K1/E)willbe Fsothat K1mustbe K.<br />
Itfollowsimmediatelythat Hisisomorphictoasubgroup<strong>of</strong><strong>the</strong>dualgroup<strong>of</strong> D.Thus<br />
H = {1}or Hiscyclic<strong>of</strong>order ℓ. Inei<strong>the</strong>rcase Hmustliein<strong>the</strong>centralizer<strong>of</strong> Dsothat<br />
E/Fisnormaland G(E/F)isisomorphicto D.If H = {1}<strong>the</strong>n χEmaybeextendedfrom<br />
CE = CKtoaquasicharacter<strong>of</strong> W E/F.Ino<strong>the</strong>rwords,<strong>the</strong>reisaquasicharacter χF<strong>of</strong> CF<br />
suchthat χE = χ E/F.Then<br />
Ind(W K/F, W K/E, χE)<br />
isequivalentto<br />
⊕ µF ∈S(E/F) Ind(W K/F, W K/F, µFχF).<br />
Suppose H = {1}.Since W K/M/CKiscyclic<strong>the</strong>reisaquasicharacter χM<strong>of</strong> CMsuch<br />
that χK = χ K/M. If w1belongsto W K/Elet χw1 be<strong>the</strong>character<strong>of</strong> W K/Mor,whatis<strong>the</strong><br />
same,<strong>of</strong> CMdefinedby<br />
χw1 (w2) = ω(w1, w2)
Chapter18 256<br />
andif w2belongsto W K/Mlet<br />
Clearly<br />
and<br />
χw2 (w1) = ω(w1, w2).<br />
{χw1 | w1 ∈ W K/E} = S(K/M)<br />
{χw2 | w2 ∈ W K/M} = S(K/E).<br />
If σ1is<strong>the</strong>image<strong>of</strong> w1in Hand σ2<strong>the</strong>image<strong>of</strong> w2in D<strong>the</strong>n<br />
and<br />
χ σ2<br />
E (w1) = χE(w2w1w −1<br />
2 w−1<br />
1 w1) = χ −1<br />
w2 (w1) χE(w1)<br />
χ σ1<br />
M (w2) = χM(w1w2w −1<br />
1 w−1<br />
2 w2) = χw1 (w2) χM(w2).<br />
Let W K/Fbe<strong>the</strong>disjointunion<br />
ℓ<br />
i=1 W K/Evi<br />
with viin W K/M.Define<strong>the</strong>function ϕion W K/Fby<br />
if w ∈ W K/Eand j = iandby<br />
if w ∈ W K/E.Then<br />
isabasisfor<strong>the</strong>space Uonwhich<br />
ϕi(wvj) = 0<br />
ϕi(wvi) = χE(w)<br />
{ϕi | 1 ≤ i ≤ ℓ}<br />
Ind(W K/F, W K/E, χE)<br />
acts.Let ψi, 1 ≤ i ≤ ℓbe<strong>the</strong>function W K/Fdefinedby<br />
ψi(w2w1) = χM(w2)χ σ(vi)<br />
E (w1)<br />
if w1belongsto W K/Eand w2belongsto W K/M.Here σ(vi)is<strong>the</strong>image<strong>of</strong> viin G(K/F).It<br />
isnecessary,buteasy,toverifythat ψiiswelldefined.Thecollection<br />
{ψi | 1 ≤ i ≤ ℓ}
Chapter18 257<br />
isabasisfor<strong>the</strong>space Vonwhich<br />
Ind(W K/F, W K/M, χM)<br />
acts.Itiseasilyverifiedthat<strong>the</strong>isomorphism<strong>of</strong> Uwith V whichsends χM(vi)ϕito ψiisan<br />
isomorphism.Thus<br />
Ind(W K/F, W K/E, χE) ≃ Ind(W K/F, W K/M, χM).<br />
Thistakescare<strong>of</strong><strong>the</strong>firstpart<strong>of</strong><strong>the</strong>lemma.<br />
Whe<strong>the</strong>r H = {1}ornot,<br />
isequivalentto<br />
Ind(WK/F, WK/E, 1)<br />
⊕ µF ∈S(E/F) Ind(W K/F, W K/F, µF).<br />
If H = 1wemayapplyTheorem2.1to E/Ftoseethat<br />
λ(E/F, ψF) = <br />
µF ∈S(E/F) ∆(µF, ψF).<br />
If H = {1}thisequalityisjust<strong>the</strong>definition<strong>of</strong><strong>the</strong>leftside.Inthiscase<strong>the</strong>secondpart<strong>of</strong><strong>the</strong><br />
lemmaassertsthat<br />
∆(χE, ψ E/F) <br />
µF ∈S(E/F) ∆(µF, ψF) (18.1)<br />
isequalto<br />
<br />
µF ∈S(E/F) ∆(µFχF, ψF)<br />
where χE = χ E/F.Thisisaconsequence<strong>of</strong><strong>the</strong>firstmainlemma.If H = {1},Theorem2.1<br />
appliedto M/F,showsthat<br />
λ(M/F, ψF) = <br />
µF ∈S(M/F) ∆(µF, ψF)<br />
and<strong>the</strong>secondpart<strong>of</strong><strong>the</strong>lemmaassertsthat(18.1)isequalto<br />
∆(χM, ψ M/F) <br />
Thisisaconsequence<strong>of</strong><strong>the</strong>secondmainlemma.<br />
µF ∈S(M/F) ∆(µF, ψF).<br />
Anontrivialnilpotentgroupalwayscontainsasubgroup Dsatisfying<strong>the</strong>conditions<strong>of</strong><br />
<strong>the</strong>previouslemma.Lemma17.1isclearif K = F.If K = Fand G(K/F)isnilpotentitisa<br />
consequence<strong>of</strong><strong>the</strong>followinglemma.
Chapter18 258<br />
Lemma 18.2<br />
Suppose K/F isnormalandTheorem2.1isvalidforallnormalextensions K ′ /F ′ in<br />
P(K/F)with [K ′ : F ′ ] < [K : F]. Suppose F ⊆ M ⊂ = Kand M/F isnormal. Suppose<br />
E1, . . ., Er,<br />
E ′ 1, . . ., E ′ sliebetween Fand M, χEiisaquasicharacter<strong>of</strong> CEi , χE ′ j isaquasicharacter<strong>of</strong><br />
CE ′ j ,and<br />
⊕ r i=1Ind(WK/F, WK/Ei , χEi )<br />
isequivalentto<br />
Then r<br />
isequalto<br />
Therepresentation<br />
s<br />
⊕ s j=1 Ind(W K/F, W K/E ′ j , χ E ′ j ).<br />
i=1 ∆(χEi , ψ Ei/F) λ(Ei/F, ψF)<br />
j=1 ∆(χE ′ j , ψ E ′ j /F) λ(E ′ j<br />
canbeobtainedbyinflating<strong>the</strong>representation<br />
Ind(WK/F, WK/Ei , χEi )<br />
Ind(WM/F, WM/Ei , χEi )<br />
/F, ψF).<br />
from W M/Fto W K/F.Asimilarremarkappliesto<strong>the</strong>representationsinducedfrom<strong>the</strong> χE ′ j .<br />
Thus<br />
isequivalentto<br />
⊕ r i=1Ind(WM/F, WM/Ei , χEi )<br />
⊕ s j=1Ind(W M/F, W M/E ′ j , χE ′ j ).<br />
ApplyingTheorem2.1to<strong>the</strong>extension M/Fweobtain<strong>the</strong>lemma.
Chapter19 259<br />
Chapter Nineteen.<br />
Pro<strong>of</strong> <strong>of</strong> <strong>the</strong> Main Theorem.<br />
WeshallfirstproveLemma17.1when<strong>the</strong>reisaquasicharacter χF <strong>of</strong> CF suchthat<br />
χK = χ K/F.Implicitin<strong>the</strong>statement<strong>of</strong><strong>the</strong>followinglemmaasinthat<strong>of</strong>Lemma17.1,is<strong>the</strong><br />
assumptionthatTheorem2.1isvalidforallpairs K ′ /F ′ in P(K/F)forwhich [K ′ : F ′ ] <<br />
[K : F].Recallthatwehavefixedanontrivialabeliannormalsubgroup C<strong>of</strong> G = G(K/F)<br />
andthat Lisitsfixedfield.<br />
Lemma 19.1<br />
Suppose F ⊆ E ⊆ K, χFisaquasicharacter<strong>of</strong> CF, χEisaquasicharacterif CE,and<br />
, 1 ≤ i ≤ r,<br />
χ K/E = χ K/F.Therearefields F1, . . ., Frcontainedin Landquasicharacters χFi<br />
suchthat χK/Fi<br />
isequivalentto<br />
and<br />
isequalto<br />
= χK/F,<br />
r<br />
Ind(W K/F, W K/E, χE)<br />
⊕ r i=1Ind(WK/F, WK/Fi , χFi )<br />
∆(χE, ψ E/F) λ(E/F, ψF)<br />
i=1 ∆(χFi , ψ Fi/F) λ(Fi/F, ψF).<br />
Weprove<strong>the</strong>lemmabyinductionon [K : F].Let Ebe<strong>the</strong>fixedfield<strong>of</strong> Handlet F ′ be<br />
<strong>the</strong>fixedfield<strong>of</strong> HC.If F ′ = F<strong>the</strong>n,byinduction,<strong>the</strong>rearefields F1, . . ., Frlyingbetween<br />
F ′ and Landquasicharacters χF1 , . . ., χFr suchthat χ K/Fi = χ K/Fand<br />
isequivalentto<br />
Ind(WK/F ′, WK/E, χE)<br />
⊕ r i=1Ind(WK/F ′, WK/Fi , χFi ).<br />
Inthiscase<strong>the</strong>lemmafollowsfrom<strong>the</strong>transitivity<strong>of</strong><strong>the</strong>inductionprocess,<strong>the</strong>assumed<br />
validity<strong>of</strong>Theorem2.1for K/F ′ andLemma16.3.
Chapter19 260<br />
Wesupposehenceforththat G = HC. Thereisacharacter θEin S(K/E)suchthat<br />
χE = θEχ E/F. θEmayberegardedasacharacter<strong>of</strong> H. If H ∩ C = {1}wemaydefinea<br />
character θF<strong>of</strong> Gbysetting<br />
θF(hc) = θE(h)<br />
if hisin Hand cisin C. θFmayberegardedasacharacter<strong>of</strong> CFand θE = θ E/F.Replacing<br />
χFby θF χFwesupposethat χE = χ E/F.Thenin<strong>the</strong>notation<strong>of</strong>Lemma15.1,wemaytake<br />
andif Fi = Fµ,<br />
{F1, . . ., Fr} = {Fµ | µ ∈ T }<br />
χFi = µ′ χ Fµ/F.<br />
Theassertions<strong>of</strong><strong>the</strong>lemmaareconsequences<strong>of</strong>Lemmas15.1and15.3.<br />
Wesupposenownotonlythat G = HCbutalsothat H ∩ C = {1}. Let Sbe<strong>the</strong>set<br />
<strong>of</strong>charactersin S(K/L)whoserestrictionto H ∩ Cagreeswith<strong>the</strong>restriction<strong>of</strong> θE. Sis<br />
invariantunder<strong>the</strong>action<strong>of</strong> Hon S(K/L).If νbelongsto Slet ϕνbe<strong>the</strong>functionon W K/F<br />
definedby<br />
ϕν(wv) = χE(w) χ L/F(v) ν(v)<br />
if wisin W K/Eand visin W K/L. νisacharacter<strong>of</strong> Candmay<strong>the</strong>reforeberegardedasa<br />
character<strong>of</strong> W K/Lor<strong>of</strong> CL.Itiseasytoverifythat ϕνiswelldefined.If<br />
<strong>the</strong>n<br />
ρ = Ind(W K/F, W K/E, χE)<br />
{ϕν | ν ∈ S}<br />
isabasisfor<strong>the</strong>space<strong>of</strong><strong>functions</strong>onwhich ρacts.If wbelongsto W K/E<br />
ρ(w)ϕν = χE(w)ϕν ′<br />
with ν ′ = νσ−1is σis<strong>the</strong>image<strong>of</strong> win G(K/F).If vbelongsto WK/L ρ(v)ϕν = χ L/F(v) ν(v)ϕν.<br />
Thusif Risanorbitin Sunder<strong>the</strong>action<strong>of</strong> H,<strong>the</strong>space<br />
VR = <br />
ν∈R Cϕν<br />
isinvariantunder W K/Fand ρis<strong>the</strong>directsum<strong>of</strong>itsrestrictionsto<strong>the</strong>spaces VR.
Chapter19 261<br />
If µbelongsto Rlet Hµbe<strong>the</strong>isotropygroup<strong>of</strong> µ,let Gµ = HµC,andlet Fµbe<strong>the</strong>fixed<br />
field<strong>of</strong> Gµ.Extend µtoacharacter µ ′ <strong>of</strong> Gµbysetting<br />
µ ′ (hc) = θE(h) µ(c)<br />
if hisin Hµand cisin C. µ ′ ,whichiseasilyseentobewelldefined,mayberegardedasa<br />
character<strong>of</strong> W K/Fµ <strong>of</strong> CFµ .Let W K/Fbe<strong>the</strong>disjointunion<br />
s<br />
i=1 W K/Fµ wi<br />
with wiin W K/Eandlet σibe<strong>the</strong>image<strong>of</strong> wiin G(K/F).Let ϕibe<strong>the</strong>function<strong>of</strong> W K/F<br />
definedby<br />
Thecollection<br />
ϕi(wwj) = 0 w ∈ WK/Fµ , j = i<br />
ϕi(wwi) = µ ′ (w)χFµ/F(w) w ∈ WK/Fµ .<br />
{ϕi | 1 ≤ i ≤ s}<br />
isabasisfor<strong>the</strong>space Vµonwhich<strong>the</strong>representation<br />
acts.Let<br />
If wbelongsto W K/L<br />
σµ = Ind(W K/F, W K/Fµ , µ′ χ Fµ/F)<br />
ψi = χE(wi)ϕi.<br />
σµ(w)ψi = µ σi (w) χL/F(w)ψi.<br />
If wbelongsto W K/Eand wjw = vwiwith vin W K/Fµ <strong>the</strong>n<br />
σµ(w)ψi = χE(w)ψj.<br />
Thus<strong>the</strong>isomorphism<strong>of</strong> Vµwith VRwhichtakes ψito ϕµ σ i commuteswith<strong>the</strong>action<strong>of</strong><br />
W K/F.If Tisaset<strong>of</strong>representativesfor<strong>the</strong>orbitsin S<br />
ρ ≃ ⊕µ∈Tσµ.<br />
If K1is<strong>the</strong>fixedfield<strong>of</strong> H ∩ C<strong>the</strong>n K1/Fisnormaland ρis<strong>the</strong>inflationto W K/F<strong>of</strong><br />
Ind(W K1/F, W K1/E, χE) (19.1)
Chapter19 262<br />
and σµis<strong>the</strong>inflation<strong>of</strong><br />
Thus<strong>the</strong>representation(19.1)isequivalentto<br />
ApplyingTheorem2.1to K1/Fweseethat<br />
isequalto<br />
Ind(W K1/F, W K1/Fµ , µ′ χ Fµ/F).<br />
⊕µ∈TInd(W K1/F,W K1/Fµ , µ′ χ Fµ/F).<br />
<br />
∆(χE, ψ E/F) λ(E/F, ψF)<br />
µ∈T (µ′ χ Fµ/F, ψ Fµ/F) λ(Fµ/F, ψF).<br />
If<strong>the</strong>reisaquasicharacter χFsuchthat χK = χK/F,Lemma17.1followsfromLemma 18.2and<strong>the</strong>lemmajustproved. Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong>Theorem2.1wehavetoprove<br />
Lemma17.1when F = F(χK), Gisnotnilpotent,and<strong>the</strong>reisnoquasicharacter χF<strong>of</strong> CF<br />
suchthat χK = χK/F.Inthiscasenone<strong>of</strong><strong>the</strong>fields E1, . . ., Er, E ′ 1, . . ., E ′ sisequalto Fand<br />
Theorem2.1maybeappliedto K/Eiand K/E ′ j .<br />
Lemma 19.2<br />
Suppose Aand Bliebetween Fand K.Suppose χAand χBarequasicharacters<strong>of</strong> CA<br />
and CBrespectively.TherearefieldsA1, . . ., Amlyingbetween AandK,fieldsB1, . . ., BmlyingbetweenBandK,elementsσ1,<br />
. . ., σminG,andquasicharactersχA1 , . . ., χAm , χB1 , . . ., χBm<br />
suchthat Bi = A σi<br />
i , χBi = χσi<br />
Ai ,andsuchthat<strong>the</strong>tensorproduct<br />
isequivalentto<br />
andto<br />
Let<br />
Ind(W K/F, W K/A, χA) ⊗ Ind(W K/F, W K/B, χB)<br />
⊕ m i=1Ind(WK/F, WK/Ai , χAi )<br />
⊕ m i=1Ind(WK/F, WK/Bi , χBi ).<br />
ρ = Ind(W K/F, W K/A, χA)<br />
σ = Ind(W K/F, W K/B, χB).
Chapter19 263<br />
Let αbe<strong>the</strong>restriction<strong>of</strong> σto W K/Aand β<strong>the</strong>restriction<strong>of</strong> ρto W K/B.ByLemma2.3<br />
and<br />
Let W K/Fbe<strong>the</strong>disjointunion<br />
ρ ⊗ σ ≃ Ind(W K/F, W K/A, χA ⊗ α)<br />
ρ ⊗ σ ≃ Ind(W K/F, W K/B, χB ⊗ β).<br />
m<br />
i=1 W K/AwiW K/B.<br />
If Uiis<strong>the</strong>space<strong>of</strong><strong>functions</strong>in U,<strong>the</strong>spaceonwhich ρacts,whicharezerooutside<strong>of</strong><strong>the</strong><br />
doublecoset W K/AwiW K/B<strong>the</strong>n Uiisinvariantunder β.Define<strong>the</strong>field Bibydemanding<br />
that<br />
W K/Bi = W K/B ∩ w −1<br />
i W K/Awi.<br />
If σiis<strong>the</strong>image<strong>of</strong> wiin G(K/F)let χ ′ Bi<br />
<strong>of</strong><strong>functions</strong>onwhich<br />
Ind(WK/B, WK/Bi , χ′ Bi )<br />
be<strong>the</strong>restriction<strong>of</strong> χσi<br />
A to W K/Bi .If U ′ i is<strong>the</strong>space<br />
acts,<strong>the</strong>map<strong>of</strong> Uito U ′ i whichsends ϕto<strong>the</strong>function ϕ′ definedby<br />
ϕ ′ (w) = ϕ(wiw)<br />
if wisin W K/Bisanisomorphismwhichcommuteswith<strong>the</strong>action<strong>of</strong> W K/B.Thus<br />
and,if χBi = χ Bi/B χ ′ Bi ,<br />
β ≃ ⊕ m i=1 Ind(W K/B, W K/Bi , χ′ Bi )<br />
χB ⊗ β ≃ ⊕ m i=1Ind(WK/B, WK/Bi , χBi ).<br />
Similarconsiderationsapplyif<strong>the</strong>roles<strong>of</strong> Aand Bareinterchanged.Thedoublecoset<br />
decompositionbecomes<br />
m<br />
i=1 WK/Bw −1<br />
i WK/A and<br />
W K/Ai = W K/A ∩ wiW K/Bw −1<br />
i = wiW K/Bi w−1<br />
i .<br />
Thus Bi = A σi<br />
i .Itisalsoclearthat χBi = χσi<br />
Ai .
Chapter19 264<br />
Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma17.1weuseBrauer’s<strong>the</strong>oremin<strong>the</strong>followingform.<br />
Therearefields F1, . . ., Fnlyingbetween Fand Ksuchthat G(K/Fk)isnilpotentforeach k,<br />
characters χFk <strong>of</strong> CFk /N K/Fk CK,andintegers m1, . . ., mnsuchthat<br />
1 ≃ ⊕ n k=1mk Ind(WK/F, WK/Fk , χFk ).<br />
Sinceweareassumingthat Gisnotnilpotentnone<strong>of</strong><strong>the</strong> Fkareequalto Fandwemayapply<br />
Theorem2.1toeach<strong>of</strong><strong>the</strong>extensions K/Fk.<br />
Weshallapply<strong>the</strong>previouslemmawith A = Ei, B = Fkandwith A = E ′ j , B = Fk. m<br />
and Bℓwillbedenoted<br />
willbedenotedby m(ik)or m ′ (jℓ). Aℓwillbedenotedby Eikℓor E ′ jkℓ<br />
by Fikℓor F ′ jkℓ .Observethat<br />
isequalto<br />
andthat<br />
isequalto<br />
∆(χEikℓ , ψ Eikℓ/F λ(Eikℓ/F, ψF) (19.2)<br />
∆(χFikℓ , ψ Fikℓ/F) λ(Fikℓ/F, ψF) (19.3)<br />
∆(χ E ′ jkℓ , ψ E ′ jkℓ /F) λ(E ′ jkℓ /F, ψF) (19.4)<br />
∆(χF ′ jkℓ , ψ F ′ jkℓ /F) λ(F ′ jkℓ /F, ψF). (19.5)<br />
χEi mayberegardedasaonedimensionalrepresentation<strong>of</strong> W K/Ei andassuchis<br />
equivalentto<br />
Therefore<br />
and<br />
isequalto<br />
n<br />
k=1<br />
⊕ n k=1 ⊕m(ik)<br />
ℓ=1 mk Ind(WK/Ei , WK/Eikℓ , χEikℓ ).<br />
m(ik)<br />
1 = n<br />
k=1<br />
m(ik)<br />
ℓ=1 mk [Eikℓ : Ei]<br />
∆(χEi , ψ Ei/F)<br />
ℓ=1 {∆(χEikℓ , ψ Eikℓ/F) λ(Eikℓ/Ei, ψ Ei/F)} mk .<br />
Multiplyingboth<strong>of</strong><strong>the</strong>seexpressionsby λ(Ei/F, ψF),weseethat<br />
∆(χEi , ψ Ei/F) λ(Ei/F, ψF) (19.6)
Chapter19 265<br />
isequalto<br />
n<br />
m(ik)<br />
k=1 ℓ=1 {∆(χEikℓ , ψEikℓ/F) λ(Eikℓ/F, ψF)} mk . (19.7)<br />
Thesameargumentestablishesthat<br />
isequalto<br />
n<br />
k=1<br />
m ′ (jk)<br />
ℓ=1<br />
∆(χ E ′ j , ψ E ′ j /F) λ(E ′ j /F, ψF) (19.8)<br />
{∆(χ E ′ jkℓ , ψ E ′ jkℓ /F) λ(E ′ jkℓ /F, ψF)} mk . (19.9)<br />
Wearetryingtoshowthat<strong>the</strong>productover i<strong>of</strong><strong>the</strong>expressions(19.6)isequalto<strong>the</strong><br />
productover j<strong>of</strong><strong>the</strong>expressions(19.8). Itwillbeenoughtoshowthat<strong>the</strong>product<strong>of</strong><strong>the</strong><br />
expressions(19.7)isequalto<strong>the</strong>product<strong>of</strong><strong>the</strong>expressions(19.9).<br />
and<br />
Therepresentations<br />
areequivalent.Therefore<br />
r<br />
i=1<br />
⊕ r i=1 ⊕ m(ik)<br />
ℓ=1 Ind(WK/Fk , WK/Fikℓ , χFikℓ )<br />
⊕ s j=1 ⊕m′ (jk)<br />
ℓ=1 Ind(W K/Fk , W K/F ′ jkℓ , χF ′ jkℓ )<br />
m(ik)<br />
ℓ=1 [Fikℓ : Fk] = s<br />
m<br />
j=1<br />
′ (jk)<br />
[F<br />
ℓ=1<br />
′ jkℓ<br />
Denote<strong>the</strong>commonvalue<strong>of</strong><strong>the</strong>seexpressionsby N(k).Moreover<br />
isequalto<br />
r<br />
s<br />
i=1<br />
j=1<br />
m(ik)<br />
: Fk].<br />
ℓ=1 ∆(χFikℓ , ψ Fikℓ/F) λ(Fikℓ/Fk, ψ Fk/F)<br />
m ′ (jk)<br />
Multiplyingboth<strong>of</strong><strong>the</strong>seexpressionsby<br />
weseethat<br />
isequalto<br />
r<br />
s<br />
i=1<br />
j=1<br />
ℓ=1 ∆(χF ′ jkℓ , ψ F ′ jkℓ /F)λ(F ′ jkℓ/F, ψ Fk/F).<br />
m(ik)<br />
λ(Fk/F, ψF) N(k)<br />
ℓ=1 ∆(χFikℓ , ψ Fikℓ/F) λ(Fikℓ/F, ψF) (19.10)<br />
m ′ (jk)<br />
ℓ=1 ∆(χF ′ jkℓ , ψ F ′ jkℓ /F) λ(F ′ jkℓ/F, ψF). (19.11)<br />
Because<strong>of</strong><strong>the</strong>equality<strong>of</strong>(19.2)and(19.3)<strong>the</strong>productover i<strong>of</strong><strong>the</strong>expressions(19.7)isequal<br />
to<strong>the</strong>productover k<strong>of</strong><strong>of</strong><strong>the</strong> mkthpowers<strong>of</strong><strong>the</strong>expressions(19.10). Theproductover j<br />
<strong>of</strong><strong>the</strong>expressions(19.9)isequalto<strong>the</strong>productover k<strong>of</strong><strong>the</strong> mkthpowers<strong>of</strong><strong>the</strong>expressions<br />
(19.11).Lemma17.1,andwithitTheorem2.1,isnowcompletelyproved.
Chapter20 266<br />
Chapter Twenty.<br />
<strong>Artin</strong> L-<strong>functions</strong>.<br />
Suppose ω isanequivalenceclass<strong>of</strong>representations<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong><strong>the</strong>nonarchimedeanlocalfield<br />
F. Let KbeaGaloisextension<strong>of</strong> F andlet σbearepresentation<br />
<strong>of</strong> W K/Fin<strong>the</strong>class ω. Suppose σactson V. Let V 0 be<strong>the</strong>subspace<strong>of</strong> V fixedbyevery<br />
element<strong>of</strong> W 0 K/F .Since W 0 K/F isanormalsubgroup<strong>of</strong> W K/F<strong>the</strong>space V 0 isinvariantunder<br />
WK/Fandon V 0wegetarepresentation σ0 .Since W 0 −1<br />
K/F = τK/F (u0F )<strong>the</strong>class<strong>of</strong> σ0depends onlyon w. σ0breaksupinto<strong>the</strong>directsum<strong>of</strong>1dimensionalrepresentationscorresponding tounramifiedgeneralizedcharacters µ1, . . ., µr<strong>of</strong> CF.Weset<br />
L(s, w) = r<br />
i=1<br />
1<br />
1 − µi(πF) |πF | s.<br />
Thiswetakeas<strong>the</strong>localfunction. Itisclearthatwhen wisonedimensional<strong>the</strong>present<br />
definitionagreeswiththat<strong>of</strong><strong>the</strong>introductionandthat<strong>of</strong> ω = ω1 ⊕ ω2.Then<br />
L(s, ω) = L(s, ω1) ⊕ L(s, ω2).<br />
Suppose F ⊆ E ⊆ K, ρisarepresentation<strong>of</strong> W K/E,and<br />
Wehavetoshowthatif θis<strong>the</strong>class<strong>of</strong> ρ<strong>the</strong>n<br />
σ = Ind(W K/F, W K/E, ρ).<br />
L(s, ω) = L(s, θ).<br />
Let ρacton W.Then Vis<strong>the</strong>space<strong>of</strong><strong>functions</strong> fon W K/Fwithvaluesin Wwhichsatisfy<br />
f(uv) = ρ(u)f(v)<br />
for uin W K/Eand vin W K/F.If fliesin V0and uliesin W 0 K/E <strong>the</strong>n<br />
ρ(u)f(v) = f(uv) = f(vv −1 uv) = f(v)<br />
because v −1 liesin W 0 K/F .Thus ftakesvaluesin W 0 .Ino<strong>the</strong>rwords,wemayaswellassume<br />
that W = W 0 .Indeedwemayaswellg<strong>of</strong>ur<strong>the</strong>randassumethat W = W 0 hasdimension<br />
one.
Chapter20 267<br />
Let N E/F πE = επ f<br />
F where εisaunitandchoose w0in W K/Fsothat τ K/Fw0 = πF.<br />
Then w f = u0v0with u0in W 0 K/F and v0in W K/Esuchthat τ K/Ev0 = πE. Clearly, V 0<br />
consists<strong>of</strong><strong>the</strong><strong>functions</strong> fwithvaluesin Wwhichsatisfy f(uw) = f(w)for uin W 0 K/F and<br />
f(uw) = µ(τ K/Eu)f(w)if ρisassociatedto<strong>the</strong>generalizedcharacter µ<strong>of</strong> CE.Takeasbasis<br />
<strong>of</strong> V 0 <strong>the</strong><strong>functions</strong> ϕ0, . . ., ϕf−1definedby<br />
ϕi(uvw j<br />
0 ) = µ(τ K/Ev)δ j<br />
i x<br />
where xisanonzerovectorin W, ubelongsto W 0 K/F , vbelongsto W K/E, 0 ≤ j < f,and δ j<br />
i<br />
isKronecker’sdelta.Thematrix<strong>of</strong> σ(w0)withrespecttothisbasisis<br />
and<br />
since |πF | f = |πE|.<br />
L(s, ω) =<br />
⎛<br />
0 ·<br />
⎜<br />
1 0<br />
⎜<br />
A = ⎜ 1<br />
⎜<br />
⎝<br />
· ·<br />
. ..<br />
. .. . ..<br />
⎞<br />
µ(τK/Ev0) · ⎟<br />
· ⎟<br />
· ⎠<br />
0 1 0<br />
1<br />
det(I − A |πF | s ) =<br />
1<br />
= L(s, θ)<br />
fs 1 − µ(πE) |πF |<br />
Forarchimedeanfieldsweproceedinadifferentmanner.Ifwewrite ω,aswemay,asa<br />
sum<strong>of</strong>irreduciblerepresentations<strong>the</strong>componentsareuniqueuptoorder.If ω = r i=1 ⊕ωi,<br />
wewillhavetohave<br />
L(s, ω) = r<br />
L(s, ωi).<br />
Thusitisamatter<strong>of</strong>defining L(s, ω)forirreducible ω.If ωisonedimensionalthiswasdone<br />
in<strong>the</strong>introduction.If ωisnotonedimensional<strong>the</strong>n Fmustbe R.Let σbearepresentation<br />
<strong>of</strong> W C/Rin<strong>the</strong>class ω. W C/Risanextension<strong>of</strong><strong>the</strong>group<strong>of</strong>order2by C ∗ . Let W C/R =<br />
C × ∪ w0C × . If σactson V <strong>the</strong>reisanonzerovector xin V andageneralizedcharacter µ<br />
<strong>of</strong> C × suchthat σ(a)x = µ(a)xforall ain C × . Then<strong>the</strong>spacespannedby {x, σ(w0)x}is<br />
invariantand<strong>the</strong>reforeall<strong>of</strong> V.Since Visnotonedimensional σ(w0)xisnotamultiple<strong>of</strong> x.<br />
Noticethat σ(a)σ(w0)x = σ(w0)σ(w −1<br />
0 aw0)x = µ(a)σ(w0)x.If<br />
i=1<br />
µ(z) = |z| r zm z n<br />
|z| m+n<br />
2
Chapter20 268<br />
with m + n ≥ 0, mn = 0weset<br />
<br />
m+n<br />
−(s+r+<br />
L(s, ω) = 2(2π) 2 ) Γ s + r +<br />
<br />
m + n<br />
.<br />
2<br />
Theinitialchoice<strong>of</strong> µis<strong>of</strong>coursenotuniquelydetermined.Howeverif µ0isonechoice<strong>the</strong><br />
onlyo<strong>the</strong>rchoiceis<strong>the</strong>charactera → µ0(a).Thus<strong>the</strong>resultinglocalLfunctionisindependent<br />
<strong>of</strong><strong>the</strong>choice.<br />
Theonlypointtobecheckedisthat<strong>the</strong>localLfunctionbehavesproperlyunderinduction.<br />
Wehavetoverifythatif ρisarepresentation<strong>of</strong> C ∗ = W C/Cin<strong>the</strong>class θand<br />
σ = Ind(W C/R, W C/C, ρ)<br />
isin<strong>the</strong>class w<strong>the</strong>n L(s, w) = L(s, θ). Wemayaswellassumethat ρisirreducibleand<br />
<strong>the</strong>reforeonedimensional.Letitcorrespondto<strong>the</strong>generalizedcharacter ν.If σisirreducible<br />
wecouldchoose<strong>the</strong>generalizedcharacter µabovetobe νand<strong>the</strong>equality<strong>of</strong><strong>the</strong>two L<strong>functions</strong>becomesamatter<strong>of</strong>definition.<br />
If σisirreducibleitbreaksupinto<strong>the</strong>sum<strong>of</strong>two<br />
onedimensionalrepresentatives.Itfollowseasilythat ν(a) = ν(a)forall a.Thus νis<strong>of</strong><strong>the</strong><br />
form ν(a) = |a| r and<br />
L(s, θ) = 2(2π) −(s+r) Γ(s + r).<br />
If µR = µis<strong>the</strong>generalizedcharacter x → |x| r <strong>of</strong> R × <strong>the</strong>n ν = µ C/Rand,aswesawinchapter<br />
10,<strong>the</strong>representation σisequivalentto<strong>the</strong>directsum<strong>of</strong><strong>the</strong>onedimensionalrepresentations<br />
correspondingto µandto µ ′ where µ ′ (x) = sgnxµ(x).Thus<br />
L(s, ω) =<br />
<br />
π −1<br />
2 (s+r) Γ<br />
s + r<br />
2<br />
<br />
π −1<br />
2 (s+r+1) Γ<br />
s + r + 1<br />
Therequiredresultisthusaconsequence<strong>of</strong><strong>the</strong>familiarduplicationformula<br />
2 2z−1 Γ(z) Γ(z + 1/2) = π 1/2 Γ(2z).<br />
If Fisaglobalfieldand ωisanequivalenceclass<strong>of</strong>representations<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong><br />
F,wedefineasin<strong>the</strong>introduction,<strong>the</strong>global Lfunctiontobe<br />
L(s, ω) = <br />
p<br />
L(s, ωp).<br />
Irepreatthat<strong>the</strong>productistakenoverallprimes,includingthoseatinfinity.Itisnotdifficult<br />
toseethat<strong>the</strong>productconvergesinahalfplaneRe s > c. <strong>On</strong>eneedonlyverifyitfor ω<br />
2<br />
<br />
.
Chapter20 269<br />
irreducible. ChooseaGaloisextension K<strong>of</strong> Fsothat<strong>the</strong>reisarepresentation σ<strong>of</strong> WK/F in<strong>the</strong>class ω. Therestriction<strong>of</strong> σto CKisequivalentto<strong>the</strong>directsum<strong>of</strong>1dimensional<br />
representationscorrespondingtogeneralizedcharacters µ (1) , . . ., µ (r) <strong>of</strong> CK.Foreach iand<br />
j<strong>the</strong>reisaσin G(K/F)suchthat µ j (a) ≡ µ i (σ(a)).Then |µ −1<br />
i µj(a)| = |µ i (a−1σ(a))| = 1<br />
because a −1 σ(a)belongsto<strong>the</strong>compactgroup<strong>of</strong> iidèleclasses<strong>of</strong>norm1.Let |µ ′ (a)| = |a| r .<br />
Let νF be<strong>the</strong>generalizedcharacter a → |a| r <strong>of</strong> CF. Replacing σby ν −1<br />
F<br />
⊗ σwereplace<br />
L(s, wp)by L(s − r, wp)and µ (i) by |µ (i) | −1 µ (i) . Thuswemayaswellsupposethatall µ (i)<br />
areordinarycharacters.Since CKis<strong>of</strong>finiteindexin W K/F<strong>the</strong>eigenvalues<strong>of</strong> σ(w)willall<br />
haveabsolute1forany win W K/Fandatanynonarchimedeanprime<strong>the</strong>local Lfunction<br />
willbe<strong>of</strong><strong>the</strong>form<br />
s<br />
i=1<br />
1<br />
1 − αi |πFp |s<br />
with s ≤ dimwand |αi| = 1, 1 ≤ i ≤ s. Therequiredresultfollowsfrom<strong>the</strong>wellknown<br />
factthat<br />
1 1<br />
1 − |πFp |s<br />
convergesfromRe s > 1.Thisproductistakenonlyover<strong>the</strong>nonarchimedeanprimes.
Chapter21 270<br />
Chapter Twenty-one.<br />
Pro<strong>of</strong> <strong>of</strong> <strong>the</strong> <strong>Functional</strong> <strong>Equation</strong>.<br />
Chooseanontrivialcharacter ψF<strong>of</strong> AF/F.Beforewecanwritedown<strong>the</strong>factorappearingin<strong>the</strong>functionalequation<strong>of</strong><strong>the</strong>global<br />
Lfunctionwehavetoverifythat ε(s, ωγ, ψFγ ) = 1<br />
forallbutafinitenumber<strong>of</strong> γ.<br />
Let ωberealizedasarepresentation σ<strong>of</strong> W K/F andlet<strong>the</strong>restriction<strong>of</strong> σto CKbe<br />
equivalentto<strong>the</strong>directsum<strong>of</strong>1dimensionalrepresentationscorrespondingto<strong>the</strong>generalized<br />
characters µ (1) , . . ., µ (r) .Allbutfinitelymanyprimes pwillsatisfy<strong>the</strong>followingconditions.<br />
(i) pisnonarchimedean.<br />
(ii) n(ψFp ) = 1.<br />
(iii) pdoesnotramifyin K.<br />
(iv) m(µ (i)<br />
P ) = 0forall Pdividing pandall i.<br />
Chooseonesuch pandlet Pdivide p.Correspondingto<strong>the</strong>map K/F −→ KP/Fpisa<br />
map ϕp : WKP/Fp −→ WK/F. ωpis<strong>the</strong>class<strong>of</strong> σp = σ ◦ ϕp.Thekernel<strong>of</strong> σpcontains UKP .<br />
Since KP/Fpisunramified<strong>the</strong>quotient<strong>of</strong> WKP/Fpby UKPisabelianand σpis<strong>the</strong>direct<br />
sum<strong>of</strong>onedimensionalrepresentations. Let<strong>the</strong>mcorrespondto<strong>the</strong>generalizedcharacters<br />
ν (1)<br />
p , . . ., ν (r)<br />
p <strong>of</strong> CFp .Since τKP/Fptakes UKPonto UFpeach<strong>of</strong><strong>the</strong>secharactersisunramified. Thus<br />
r<br />
ε(s, ωp, ψFp ) =<br />
i=1 ∆<br />
<br />
α s−1 2<br />
Fp ν(i)<br />
p , ψFp<br />
<br />
= 1.<br />
Ifψ ′ Fisano<strong>the</strong>rnontrivialcharacter<strong>of</strong> AF/F<strong>the</strong>reisaβinF ∗suchthatψ ′ F (x) ≡ ψF(βx).<br />
AccordingtoLemma5.1<br />
Since <br />
<strong>the</strong>function<br />
ε(s, ω, ψFp<br />
isindeedindependent<strong>of</strong> ψF.<br />
p<br />
2 ) = αs−1<br />
Fp (β)det ωp(β) ε(s, ω, ψFp ).<br />
αFp (β)s−1 2 det ωp(β) = |β| s−1 2 det ω(β) = 1<br />
ε(s, ω) = <br />
p ε(s, ωp, ψFp )
Chapter21 271<br />
WecaninferfromTate’s<strong>the</strong>sisnotonlythat L(s, ω)ismeromorphicin<strong>the</strong>wholecomplex<br />
planeif ωisonedimensionalbutalsothatitsatisfies<strong>the</strong>functionalequation<br />
L(s, ω) = ε(s, ω) L(1 − s, ω)<br />
if ωiscontragredientto ω.Asiswellknown,Lemma2.2<strong>the</strong>nimpliesthat L(s, ω)ismeromorphicin<strong>the</strong>wholecomplexplaneforany<br />
ω.Inanycase,TheoremBistrueforonedimensional<br />
ωand,grantingthis,wehavetoestablishitingeneral.<br />
Firstweneedalemma.<br />
Lemma 21.1<br />
Suppose Fisaglobalfield, KisaGaloisextension<strong>of</strong> F, Eisafieldlyingbetween Fand<br />
K, χisageneralizedcharacter<strong>of</strong> CEand<br />
σ = Ind(W K/F, W K/E, χ).<br />
If ωis<strong>the</strong>class<strong>of</strong> σand,foreachprime q<strong>of</strong> E, χqis<strong>the</strong>restriction<strong>of</strong> χto CEq<strong>the</strong>nforeach prime p<strong>of</strong> F<br />
<br />
ε(s, ωp, ψFp ) = {ε(s, χq, ψEq/Fp ) ρ (Eq/Fp, ψFp )}.<br />
q|p<br />
Let Pbeaprime<strong>of</strong> Kdividing p.Thefirststepist<strong>of</strong>indaset<strong>of</strong>representativesfor<strong>the</strong><br />
doublecosets W K/E w W KP/Fp . Since CK ⊆ W K/Eisanormalsubgroup<strong>of</strong> W K/Fwecan<br />
factorout CKandmerelyfindaset<strong>of</strong>representativesfor<strong>the</strong>doublecosets<br />
G(K/E)σ G(KP/Fp).<br />
Let P1, . . ., Prbe<strong>the</strong>primes<strong>of</strong> Kdividing pandlet P1divide qiin E. G(K/F)is<strong>the</strong>disjoint<br />
union r<br />
i=1 σiG(KP/Fp)<br />
where σi(P) = Pi.If σiand σjbelongto<strong>the</strong>samedoublecoset qi = qj.Conversely,if qi = qj<br />
<strong>the</strong>reisaρin G(K/E)suchthat ρ(Pi) = Pj.Then ρσi(P) = σj(P)and<br />
ρσi ∈ σj G(KP/Fp).<br />
Thuswemaywrite G(K/F)as<strong>the</strong>disjointunion<br />
<br />
τ∈S<br />
G(K/E)τ G(KP/Fp)
Chapter21 272<br />
sothatif Pdivides qin E<strong>the</strong>collection {τ(q) | τ ∈ S}is<strong>the</strong>collection<strong>of</strong>distinctprimesin E<br />
dividing p.<br />
Foreach τin Schoosearepresentative w(τ)in W K/F.Foreach τin S<strong>the</strong>restriction<strong>of</strong> σ<br />
to W KP/Fp leavesinvariant<strong>the</strong>space<strong>of</strong><strong>functions</strong> fon<strong>the</strong>doublecoset W K/Ew(τ)W KP|Fp<br />
whichsatisfy f(vw) = χ(τ K/E(v))f(w)forall vin W K/E.Therepresentation<strong>of</strong> W KP/Fp on<br />
thisspaceisequivalentto<br />
Ind(W KP/Fp , W KP/E τ q τ , χ′ q τ<br />
if E τ = τ −1 (E)and<br />
Thus<br />
whichis<strong>of</strong>courseequalto<br />
Weset<br />
χ ′ qτ(a) = χ(τ(a)).<br />
<br />
ε(s, ωq, ψFp ) =<br />
τ∈S ε(s, χ′ qτ, ψEτ qτ /Fp ) ρ(Eτ qτ/Fp, ψFp )<br />
<br />
τ∈S ε(s, χq, ψ Eq/Fp ) ρ(Eq/Fp, ψFp ).<br />
ρ(E/F) = <br />
q<br />
<br />
q|p ρ(Eq/Fp, ψFp ).<br />
Theprecedingdiscussiontoge<strong>the</strong>rwithLemma5.1showsthatitdoesnotdependon ψF.<br />
Howeverthatdoesnotreallymattersinceweareabouttoshowthatforanychoice<strong>of</strong> ψFitis<br />
1.Observefirst<strong>of</strong>allthat<strong>the</strong>previouslemmaimpliesimmediatelythatif ωis<strong>the</strong>class<strong>of</strong><br />
<strong>the</strong>n<br />
σ = Ind(W K/F, W K/E, χ)<br />
ε(s, ω) = ε(s, χ) ρ(E/F).<br />
Givenanarbitraryclass ωrealizableasarepresentation<strong>of</strong> WK/F wecanfindfields<br />
E1, . . ., ErlyingbetweenFandK,generalizedcharactersχE1 , . . ., χEr ,andintegersm1, . . ., mr<br />
suchthat r<br />
i=1 ⊕mi Ind(WK/F, WK/Ei , χEi )<br />
isin<strong>the</strong>class ω.Then<br />
ε(s, ω) = r<br />
i=1 {ε(s, χEi )mi ρ(Ei/F) mi }.
Chapter21 273<br />
and<br />
Since<br />
wehave<br />
<strong>On</strong><strong>the</strong>o<strong>the</strong>rhand<br />
because ωcontains r<br />
Consequently<br />
L(s, ω) = r<br />
L(s, ω) = r<br />
i=1<br />
i=1<br />
L(s, χEi )mi<br />
L(s, χ−1<br />
Ei )mi .<br />
L(s, χEi ) = ε(s, χEi ) L(1 − s, χ−1<br />
Ei )<br />
L(s, ω) = r<br />
i=1 ε(s, χEi )mi L(1 − s, ω)<br />
i=1 ⊕mi Ind(W K/F, W K/Ei<br />
r<br />
i=1<br />
ε(s, χEi )mi<br />
, χ−1<br />
Ei ).<br />
dependsonlyon ωandnoton<strong>the</strong>particularwayitiswrittenasasum<strong>of</strong>inducedrepresentations.Thus<br />
r<br />
mi ρ(Ei/F)<br />
i=1<br />
alsodependsonlyon ω.Wecallit H(ω).ItisclearthattoproveTheoremBwehavetoshow<br />
that H(ω) = 1forall ωor,whatis<strong>the</strong>same,that ρ(E/F) = 1forall Eand F.<br />
q ′ .Then<br />
Suppose F ⊆ E ⊆ E ′ .Denote<strong>the</strong>primes<strong>of</strong> Fby p,those<strong>of</strong> Eby q,andthose<strong>of</strong> E ′ by<br />
ρ(E ′ /F) = <br />
p<br />
<br />
ApplyLemma4.5toseethat<strong>the</strong>rightsideequals<br />
<br />
p<br />
<br />
q|p<br />
<br />
q ′ |q<br />
q ′ |p ρ(E′ q ′/Fp, ψFp ).<br />
<br />
′<br />
ρ(E q ′/Eq, ψFq/Fp ) ρ(Eq/Fp, ψFp )[E′ q :Eq] .<br />
Since <br />
[E ′ q ′ : Eq] = [E ′ : E]<br />
thismaybewrittenas<br />
<br />
q<br />
<br />
q ′ |q<br />
q ′ |q ρ(E′ q ′/Eq, ψ Fq/Fp )<br />
<br />
p<br />
<br />
q|p ρ(Eq/Fp, ψFp )<br />
′<br />
[E :E]
Chapter21 274<br />
whichis<strong>of</strong>course<br />
ρ(E ′ /E) ρ(E/F) [E′ :E] . (20.1)<br />
Suppose E/Fisanabelianextensionand ωis<strong>the</strong>class<strong>of</strong><strong>the</strong>representation<strong>of</strong> W E/F<br />
inducedfrom<strong>the</strong>trivialrepresentation<strong>of</strong> CE = W E/E. Then H(ω) = ρ(E/F). <strong>On</strong><strong>the</strong><br />
o<strong>the</strong>rhand, ωis<strong>the</strong>directsum<strong>of</strong> [E : F]onedimensionalrepresentations; so H(ω) =<br />
ρ(F/F) [E:F] = 1. Itfollowsimmediatelynotonlythat ρ(E/F) = 1if E/Fisabelianbut<br />
alsothat ρ(E/F) = 1if Ecanbeobtainedfrom Fbyasuccession<strong>of</strong>abelianextensions. In<br />
particularif F ⊆ E ⊆ Land L/Fisnilpotent ρ(E/F) = 1.<br />
Observethat(20.1)toge<strong>the</strong>rwithLemma2.2and<strong>the</strong>transitivity<strong>of</strong>inductionimplythat<br />
if ωis<strong>the</strong>class<strong>of</strong><br />
σ = Ind(W K/F, W K/E, ρ)<br />
and θis<strong>the</strong>class<strong>of</strong> ρ<strong>the</strong>n<br />
H(ω) = H(θ) ρ(E/F) dim θ .<br />
Tocomplete<strong>the</strong>pro<strong>of</strong>wewillshowthat H(ω1 ⊗ ω2) = H(ω2) dim ω1 forall ω1and ω2.Taking<br />
ω2 = 1wefind H(ω1) = 1. Itisenoughtoprove<strong>the</strong>equalitywhen ω1and ω2areboth<br />
realizableasrepresentations<strong>of</strong> W K/F and<strong>the</strong>reisafield Elyingbetween Eand Kwith<br />
G(K/E)nilpotentandageneralizedcharacter χEsuchthat ω2is<strong>the</strong>class<strong>of</strong><br />
Ind(W K/F, W K/E, χE).<br />
Then H(ω2) = ρ(E/F). If ρisarepresentationin<strong>the</strong>restriction<strong>of</strong> ω1to W K/E<strong>the</strong>n,by<br />
Lemma2.3, ω1 ⊗ ω2is<strong>the</strong>class<strong>of</strong><br />
Ind(W K/F, W K/E, ρ ⊗ χE).<br />
Let θbe<strong>the</strong>class<strong>of</strong> ρ ⊗ χE. H(θ)is<strong>of</strong><strong>the</strong>form<br />
r<br />
i=1<br />
where E ⊆ Ei ⊆ Kandis<strong>the</strong>refore1.Thus<br />
asrequired.<br />
ρ(Ei/E) mi<br />
H(ω1 ⊗ ω2) = H(θ)ρ(E/F) dim θ = ρ(E/F) dim ω1 .
Chapter22 275<br />
Chapter Twenty-two.<br />
Appendix.<br />
Thereisclearlynotmuchtobesaidabout<strong>the</strong><strong>functions</strong>ε(s, ω, ψF)whenFisarchimedean.<br />
Howeverfornonarchimedean F<strong>the</strong>irpropertiesaremoreobscure.Inthisappendixweshall<br />
describeandprovesomepropertieswhichwerenotneededin<strong>the</strong>pro<strong>of</strong>s<strong>of</strong><strong>the</strong>main<strong>the</strong>orems<br />
ands<strong>of</strong>oundnoplacein<strong>the</strong>mainbody<strong>of</strong><strong>the</strong>paperbutwhichwillbeusedelsewhere.<br />
Thefirststepistodefine<strong>the</strong><strong>Artin</strong>conductor<strong>of</strong> ω. Wefollowawelltroddenpath. If<br />
KisafiniteGaloisextension<strong>of</strong><strong>the</strong>localfield F<strong>the</strong>n W 0 K/Fcontains UKasasubgroup<strong>of</strong><br />
finiteindexandis<strong>the</strong>reforecompact. Itis,infact,amaximalcompactsubgroup<strong>of</strong> WK/F. ChoosethatHaarmeasure dwon WK/Fwhichassigns<strong>the</strong>measure1to W 0 K/F .If fisalocally<br />
constantfunctionon WK/Fand uisanonnegativerealnumberset<br />
<br />
ˆf(u) =<br />
W u<br />
dw<br />
K/F<br />
−1 <br />
W u<br />
{f(1) − f(w)}dw.<br />
K/F<br />
Since W u K/Fisanopensubgroup<strong>of</strong> WK/Fitismeaningfultorestrict dwtoit. ˆ f(u)isbounded,<br />
continuousfrom<strong>the</strong>left,and0for usufficientlylarge. Since W u K/F = W 0 K/Ffor 1 < u ≤ 0<br />
wehave ˆ f(u) = ˆ f(0)forsuch u.Theintegral<br />
iswelldefined.<br />
∞<br />
−1<br />
ˆf(u)du<br />
Therearesomesimplelemmastobeverified.<br />
Lemma 22.1<br />
Suppose F ⊆ K ⊆ Land L/FisalsoaGaloisextension.Define gon W L/Fby g(w) =<br />
f (τL/F ,K/F(w)).Then ˆg(u) = ˆ f(u)forall u.<br />
ThisisimmediatebecausebyLemma6.16, τ L/F,K/Fmaps W u L/F onto W u K/F<br />
Whenwewanttomake<strong>the</strong>roles<strong>of</strong> Kand Fexplicitwewrite ˆ f(u) = ˆ f K/F(u).<br />
Lemma 22.2<br />
forevery u.
Chapter22 276<br />
Suppose F ⊆ E ⊆ Kand gisafunctionon W K/Esatisfying g(wzw −1 ) = g(z)forall z<br />
and win W K/E.Regard W K/Easasubgroup<strong>of</strong> W K/Fandset<br />
If NE/F πEisaunittimes π f E/F<br />
F<br />
∞<br />
−1<br />
f(w) = <br />
and P δ E/F<br />
E<br />
ˆf K/F(u)du = f E/F<br />
z∈W K/E\W K/F<br />
∞<br />
g(z −1 wz).<br />
is<strong>the</strong>different<strong>of</strong> E/F<br />
−1<br />
ˆg K/E(u)du + f E/Fδ E/Fg(1).<br />
Let dw K/Fbe<strong>the</strong>normalizedHaarmeasureon W K/Fandlet dw K/Ebe<strong>the</strong>normalized<br />
Haarmeasureon W K/E.<strong>On</strong> W K/E<br />
dw K/E = [W 0 K/F : W 0 K/E ]dw K/F.<br />
Supposeatfirstthat g(1) = 0.Denotealsoby g<strong>the</strong>functionon W K/Fwhichequals<strong>the</strong>given<br />
gon W K/Ebutis0outside<strong>of</strong> W K/E.Then<br />
Since W u K/F ∩ W K/E = W v K/E if v = ψ E/F(u)<br />
Recallthat<br />
Moreover<br />
ˆf K/F(u) = [W K/F : W K/E] ˆg K/F(u).<br />
ˆg K/F(u) = −[W 0 K/F : W u K/F ]<br />
= − [W 0 K/F : W u K/F ]<br />
[W 0 K/F : W 0 K/E ]<br />
=<br />
1<br />
[W 0 K/F : W 0 K/E ]<br />
<br />
W u<br />
K/F<br />
<br />
W v<br />
K/E<br />
f E/F = [W K/F : W K/E]<br />
[W 0 K/F : W 0 K/E ].<br />
g(w)dw K/F<br />
g(w)dw K/E<br />
[W 0 K/F : W u K/F ]<br />
[W 0 K/E : W v K/E ] ˆg K/E(v).<br />
[W 0 K/F : W u K/F ]<br />
[W 0 K/E : W v K/E ] = [W 0 K/F : W u K/F U0 K ]<br />
[W 0 K/E : W v K/E U0 K ] · [U0 K : U0 K ∩ W u K/F ]<br />
[U 0 K : U0 K ∩ W v K/E ]
Chapter22 277<br />
and U 0 K ∩ W u K/F = U0 K ∩ W v K/E .ByLemma6.11<strong>the</strong>firstterminthisproductisequalto<br />
if G = G(K/F)and H = G(K/E).But<br />
and<br />
while<br />
Thus<br />
∞<br />
−1<br />
[G 0 : G u ]<br />
[H 0 : H v ]<br />
[G 0 : G u ] = ψ ′ K/F (u)<br />
[H 0 : H v ] = ψ ′ K/E (v)<br />
ψ ′ K/F (u) = ψ′ K/E (v)ψ′ E/F (u).<br />
ˆf K/F(u)du = f E/F<br />
= f E/F<br />
∞<br />
−1<br />
∞<br />
−1<br />
ˆg K/E(ψ E/F(u)) ψ ′ E/F (u)du<br />
ˆg K/E(v)dv.<br />
Tocomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong><strong>the</strong>lemma,wehavetoshowthatif g(w) = 1sothat ˆg K/E(u) ≡ 0<br />
<strong>the</strong>n ∞<br />
Inthiscase<br />
−1<br />
ˆf K/F(u) = [G : H] −<br />
ˆf K/F(u)du = f E/F δ E/F.<br />
[G : H]<br />
[G 0 : H 0 ]<br />
if v = ψ E/F(u).Aftersomesimplerearrangingthisbecomes<br />
Thefactor<br />
[G : H]<br />
[G u : 1] {[Gu : 1] − [H v : 1]} =<br />
and,fromparagraphIV.2<strong>of</strong> [ ],<br />
∞<br />
1<br />
[G 0 : G u ]<br />
[H 0 : H v ]<br />
[G : H]<br />
[G u : 1] {([Gu : 1] − 1) − ([H v : 1] − 1)}.<br />
[G : H]<br />
[G u : 1] = [G : G0 ]<br />
[H : 1] ψ′ K/F (u)<br />
([G u : 1] − 1)ψ ′ K/F (u)du =<br />
∞<br />
−1<br />
([Gx : 1] − 1)dx = δ K/F
Chapter22 278<br />
while ∞<br />
−1<br />
Thus ∞<br />
because<br />
([H v : 1] − 1) ψ ′ K/F (u)du =<br />
−1<br />
∞<br />
−1<br />
([H v : 1] − 1) ψ ′ K/E (v)dv = δ K/E.<br />
ˆf K/F(u)du = [G : G0 ]<br />
[H : 1] (δ K/F − δ K/E) = f E/F δ E/F<br />
δ K/F = δ K/E + [H0 : 1] δ E/F.<br />
Suppose ωisanequivalenceclass<strong>of</strong>representations<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong> Fand σisa<br />
representation<strong>of</strong> WK/Fin<strong>the</strong>class<strong>of</strong> ω.Let fσbe<strong>the</strong>character<strong>of</strong> σ.ItfollowsfromLemma<br />
22.1that<strong>the</strong>value<strong>of</strong> ∞<br />
ˆfσ(u)du<br />
−1<br />
dependsonlyon ωandnoton σ.Wecallit<strong>the</strong>order<strong>of</strong> ωanddenoteitby m(ω).Since ˆ fσ(u)is<br />
clearlynonnegativeforall uandvanishesidenticallyifandonlyif W 0 K/F iscontainedin<strong>the</strong><br />
kernel<strong>of</strong> σ,<strong>the</strong>order m(ω)isalwaysnonnegativeandequalszeroifandonlyif<strong>the</strong>kernel<strong>of</strong><br />
eachrealization σ<strong>of</strong> ωcontains W 0 K/F .<br />
Lemma 22.3<br />
(a)If ω = ω1 ⊕ ω2<strong>the</strong>n m(ω) = m(ω1) + m(ω2).<br />
(b)If<br />
<strong>the</strong>n<br />
(c) m(ω)isanonnegativeinteger.<br />
ω = Ind(W K/F, W K/E, ν)<br />
m(ω) = f E/Fm(ν) = f E/F δ E/F dimν.<br />
Thefirstpropertyisimmediate.Thesecondisaconsequence<strong>of</strong>Lemma22.2. Toverify<br />
<strong>the</strong>thirdwemerelyhavetoshowthat m(ω)isintegral. If ω = µ ⊕ νand<strong>the</strong>assertionis<br />
trueforanytwo<strong>of</strong> µ, νand ωitistruefor<strong>the</strong>third. Thisobservation,toge<strong>the</strong>rwithpart<br />
(b)andLemma2.2,showsthatitisenoughtoverify(c)when ωis<strong>the</strong>onedimensionalclass<br />
correspondingtoageneralizedcharacter χF<strong>of</strong> CF.Todothisweshowthat m(ω) = m(χF).<br />
If f(a) = χF(a)for ain CF = WF/F<strong>the</strong>n ˆ f(u) = ˆ f(m)for m − 1 < u ≤ mand<br />
<br />
{1 − χF(a)}da.<br />
ˆf(m) = [U 0 F : Um F ]<br />
U m F
Chapter22 279<br />
Therightsideis1if m < m(χF)and0if m ≥ m(χF).Thus<br />
m(ω) =<br />
m(χF )−1<br />
−1<br />
du = m(χF).<br />
Thefunction ω −→ m(ω)ischaracterizedby(a)and(b)toge<strong>the</strong>rwith<strong>the</strong>factthat<br />
m(ω) = m(χF)if ωis<strong>the</strong>class<strong>of</strong> χF.<br />
Lemma 22.4<br />
If ωisanequivalenceclass<strong>of</strong>representations<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong><strong>the</strong>nonarchimedean<br />
localfield Fand ψFisanontrivialadditivecharacter<strong>of</strong> Fset m ′ (ω) = m(ω) +n(ψF) dimω.<br />
Thereisanonzerocomplexconstant a(ω)suchthat,asafunction<strong>of</strong> s,<br />
ε(s, ω, ψF) = a(ω) |πF | m′ (ω)s .<br />
If ω = µ ⊕ νand<strong>the</strong>lemmaistrueforanytwo<strong>of</strong> µ, ν,and ω,itistruefor<strong>the</strong>third.<br />
ApplyingLemma2.2weseethatitisenoughtoverifyitwhen ωcontainsarepresentation<br />
Then<br />
Clearly<br />
But<br />
∆(α s−1<br />
2<br />
αE<br />
Ind(W K/F, W K/E, χE).<br />
ε(s, ω, ψF) = ∆(α s−1<br />
2<br />
E χE, ψ E/F) ρ(E/F, ψF).<br />
E χE, ψE/F) = α s−1<br />
<br />
2<br />
E π m(χE)+δE/F E<br />
<br />
π m(χE)+δ E/F<br />
E<br />
π n(ψF)<br />
<br />
F = αF NE/F and<strong>the</strong>argumenton<strong>the</strong>rightis<strong>the</strong>product<strong>of</strong>aunitand<br />
Thelemmafollows.<br />
π f E/F(m(χE)+δ E/F)+n(ψF ) dim ω<br />
F<br />
π n(ψF)<br />
<br />
F ∆(χE, ψE/F). π m(χE)+δ E/F<br />
E<br />
= π m′ (ω)<br />
F .<br />
π n(ψF<br />
<br />
)<br />
F<br />
Thenextlemmaisra<strong>the</strong>rtechnicalandtoproveitwewillhavetouse<strong>the</strong>notationsand<br />
results<strong>of</strong>paragraphs8and9.<br />
Lemma 22.5
Chapter22 280<br />
Let ωbeanequivalenceclass<strong>of</strong>representations<strong>of</strong><strong>the</strong>Weilgroup<strong>of</strong><strong>the</strong>nonarchimedean<br />
localfield F and m1apositiveinteger. Thereisapositiveinteger m2suchthatif χF and<br />
µ1, . . ., µrwith r = dimω,aregeneralizedcharacters<strong>of</strong> CF and m(χF) ≥ m2, m(µi) ≤<br />
m1, 1 ≤ i ≤ r,while r<br />
i=1 µi = detω<br />
<strong>the</strong>nforanynontrivialadditivecharacter ψF<br />
ε(s, χF ⊗ σ, ψF) = r<br />
i=1 ε(s, µiχF, ψF).<br />
Choose,asastart, m2 ≥ 2m1+1.If µFisageneralizedcharacter<strong>of</strong> CFand m(µF) ≤ m1<br />
while m(χF) ≥ m2<strong>the</strong>n m(µFχF) = m(χF) = m. Let n = n(ψF)andchoose γsothat<br />
OFγ = P m+n<br />
F . If β = β(χF)wemaychoose β(µFχF) = β. AppealingtoLemmas8.1and<br />
9.4weseethat<br />
Inparticular<br />
r<br />
If ω = µ ⊕ ν<strong>the</strong>n<br />
ε(s, µFχF,ψF) = ∆(α s−1 2<br />
F µFχF, ψF)<br />
= (α s−1 2<br />
F µF)<br />
<br />
γ<br />
∆(χF, ψF).<br />
β<br />
i=1 ε(s, µiχF,ψF) = α r(s−1<br />
F<br />
2 )<br />
<br />
γ<br />
detω<br />
β<br />
<br />
γ<br />
β<br />
{∆(χF, ψF)} r .<br />
ε(s, χF ⊗ ω, ψF) = ε(s, χF ⊗ µ, ψF) ε(s, χF ⊗ ν, ψF)<br />
andallthreetermsaredifferentfromzero.Thusif<strong>the</strong>lemmaistruefortwo<strong>of</strong> µ, νand ωitis<br />
truefor<strong>the</strong>third.UsingLemma2.2onceagain,weseethatitisenoughtoprove<strong>the</strong>lemma<br />
when<strong>the</strong>reisanintermediatefield Eandageneralizedcharacter µE<strong>of</strong> CEsuchthat ωis<strong>the</strong><br />
class<strong>of</strong><br />
Ind(W K/F, W K/E, µE).<br />
Then χF ⊗ ωis<strong>the</strong>class<strong>of</strong><br />
and<br />
Ind(W K/F, W K/E, µE χ E/F)<br />
ε(s, χF ⊗ ω, ψF) = ∆(α s−1<br />
2<br />
E µEχ E/F, ψ E/F) ρ(E/F, ψF).<br />
Therearetwosimplelemmaswhichweneedbeforewecanproceedfur<strong>the</strong>randwe<br />
digresstoprove<strong>the</strong>m.
Chapter22 281<br />
Lemma 22.6<br />
Let Ebeaseparableextension<strong>of</strong> F.If missufficientlylarge<br />
if e E/Fis<strong>the</strong>index<strong>of</strong>ramification<strong>of</strong> Fin E.<br />
ψ E/F(m − 1) + 1 = me E/F − δ E/F<br />
Suppose F ⊆ E ⊆ Kwhere K/FisGaloisand<strong>the</strong>assertionistruefor K/Fand K/E.<br />
Subtracting1frombothsides<strong>of</strong><strong>the</strong>equation,applying ψ K/E,and<strong>the</strong>nadding1,weobtain<br />
<strong>the</strong>equivalentequation<br />
Byassumption,<strong>the</strong>leftsideequals<br />
and<strong>the</strong>rightsideequals<br />
ψ K/F(m − 1) + 1 = ψ K/E(me E/F − δ E/F − 1) + 1.<br />
me K/F − δ K/F<br />
(me E/F − δ E/F)e K/E − δ K/E.<br />
Since e K/F = e K/Ee E/Fand δ K/F = δ K/E + e K/Eδ E/F<strong>the</strong>setwoexpressionsareequaland<br />
wehaveonlytoprove<strong>the</strong>lemmaforGaloisextensions.<br />
Suppose F ⊆ K ⊆ Land L/Fand K/FareGalois.Supposealsothat<strong>the</strong>lemmaistrue<br />
for L/Kand K/F.Then<br />
ψ L/F(m − 1) + 1 = ψ L/F(ψ K/F(m − 1)) + 1<br />
= ψ L/F(me K/F − δ K/F − 1) + 1<br />
= (me K/F − δ K/F) e L/K − δ L/K<br />
= me L/F − δ L/F<br />
asbefore. Thus,ifweuseinduction,weneedonlyverify<strong>the</strong>lemmadirectlyforaGalois<br />
extension K/F<strong>of</strong>primedegree.<br />
WeapplyLemma6.3.If K/Fisunramified, e K/F = 1and δ K/F = 0while ψ K/F(m −<br />
1) = m − 1; so<strong>the</strong>relationfollows. If K/F isramified<strong>the</strong>reisaninteger tsuchthat<br />
δ K/F = ([K : F] − 1)(t + 1)while ψ K/F(m − 1) + 1 = [K : F]m − ([K : F] − 1)(t + 1)for<br />
m − 1 ≥ t.Since e K/F = [K : F]<strong>the</strong>relationfollowsagain.
Chapter22 282<br />
If n = n(ψF)<strong>the</strong>n<br />
n ′ = n(ψ E/F) = n e E/F + δ E/F.<br />
Thusif missufficientlylargeand m ′ = ψ E/F(m − 1) + 1<br />
andif OFγ = P m+n<br />
F<br />
asinparagraph8.<br />
Lemma 22.7<br />
m ′ + n ′ = (m + n) e E/F<br />
<strong>the</strong>n OEγ = P m′ +n ′<br />
E .Wedefine<br />
P ∗ E/F (x) = P ∗ E/F (x; γ, γ)<br />
If m1isagivenpositiveinteger<strong>the</strong>nfor msufficientlylarge<br />
P ∗ E/F (x) ≡ x (mod Pm1<br />
E ).<br />
Asinparagraph8,let dbe<strong>the</strong>integralpart<strong>of</strong> m<br />
2 , d′ <strong>the</strong>integralpart<strong>of</strong> m′<br />
2 ,andlet<br />
m = 2d + ε, m ′ = 2d ′ + ε ′ . P ∗ E/F (x)dependsonlyon<strong>the</strong>residue<strong>of</strong> xmodulo Pd F andis<br />
onlydeterminedmodulo P d′<br />
E .Recallthatif<br />
for yin P d′ +ε ′<br />
E<br />
<strong>the</strong>n<br />
Toshowthat P ∗ E/F<br />
ψ E/F<br />
P E/F(y) = N E/F(1 + y) − 1<br />
<br />
∗ PE/F (x)y<br />
γ<br />
= ψF<br />
x PE/F(y)<br />
(x) ≡ x (modPm1 E )when missufficientlylarge,wehavetoshowthat<br />
ψF<br />
for yin P m′ −m1<br />
E .Todothisweshowthat<br />
<br />
x PE/F(y)<br />
= ψE/F γ<br />
γ<br />
<br />
xy<br />
γ<br />
P E/F(y) ≡ S E/F(y) (modP m F )<br />
when missufficientlylargeand yisin P m′ −m1<br />
E .<br />
<br />
.
Chapter22 283<br />
Toputitano<strong>the</strong>rway,wehavetoshowthatif K/FisanyGaloisextension<strong>the</strong>assertion<br />
istrueforallintermediatefields E.Forthisweuseinductionon [K : F]toge<strong>the</strong>rwithLemma<br />
3.3.Therearethreefactstoverify:<br />
(i)If E/FisaGaloisextension<strong>of</strong>primedegree<strong>the</strong>n<br />
P E/F(y) ≡ S E/F(y) (modP m F )<br />
when missufficientlylargeand yisin P m′ −m1<br />
E .<br />
(ii)Suppose F ⊆ E ⊆ Kand K/FisGalois.Let G = G(K/F)andlet Ebe<strong>the</strong>fixedfield<br />
<strong>of</strong> H.Suppose H = {1}and G = HCwhere H ∩ C = {1}and Cisanontrivialabelian<br />
normalsubgroup<strong>of</strong> Gwhichiscontainedineveryo<strong>the</strong>rnontrivialnormalsubgroup.If<br />
<strong>the</strong>inductionassumptionisvalid<br />
P E/F(y) ≡ S E/F(y) (modP m F )<br />
when missufficientlylargeand yisin P m′ −m1<br />
E .<br />
(iii)Suppose F ⊆ E ⊆ E ′ ⊆ Kand m ′′ = ψ E ′ /F(m − 1) + 1.If,foranychoice<strong>of</strong> m1,<br />
P E/F(y) ≡ S E/F(y) (modP m F )<br />
when missufficientlylargeand yisin P m′ −m1<br />
E<br />
and,foranychoice<strong>of</strong> m ′ 1 ,<br />
P E ′ /E(y) ≡ S E ′ /E(y) (modP m′<br />
E )<br />
when m,or m ′ ,issufficientlylargeand yisin P m′′ −m ′<br />
1<br />
E ′ <strong>the</strong>n,foranychoice<strong>of</strong> m ′ 1 ,<br />
P E ′ /F(y) ≡ S E ′ /F(y) (modP m F )<br />
if missufficientlylargeand yisin P m′′ −m ′<br />
1.<br />
Wefirstverify(i)for E/Funramified.ByparagraphV.2<strong>of</strong>[12]<br />
E ′<br />
P E/F(y) = N E/F(1 + y) − 1 ≡ S E/F(y) (modP m F )<br />
if ybelongsto P d′ +ε ′<br />
E .Inthiscase m = m ′ andwetake m > 2m1sothat m ′ −m1 > d ′ +ε ′ .If<br />
E/Fisramifiedand<strong>of</strong>degree ℓweagainchoose msufficientlylargethat m ′ > 2m1.If m > t<br />
2(m ′ − m1) + (ℓ − 1)(t + 1)<br />
ℓ<br />
≥ m′ + (ℓ − 1)(t + 1)<br />
ℓ<br />
= m
Chapter22 284<br />
sothatbyChapterV<strong>of</strong>[12],<br />
if ybelongsto P m′ −m1<br />
E<br />
P E/F(y) ≡ S E/F(y) + N E/F(y) (modP m F )<br />
. t<strong>of</strong>coursehasitsusualmeaning.Since N E/F(y)belongsto P m′ −m1<br />
F<br />
allwehavetodoisarrangethat m ′ − m1 ≥ m.Since<br />
m ′ − m1 = ℓm − (ℓ − 1) (t + 1) − m1<br />
and ℓ ≥ 2,thiscancertainlybedonebychoosing msufficientlylarge.<br />
Toverify<strong>the</strong>secondfactlet Lbe<strong>the</strong>fixedfield<strong>of</strong> C. Wecanassumethat<strong>the</strong>required<br />
assertionistruefor<strong>the</strong>extension K/L.Let ℓ = ψ L/F(m − 1) + 1and ℓ ′ = ψ K/F(m − 1) + 1.<br />
If missufficientlylargeand H0is<strong>the</strong>inertialgroup<strong>of</strong> H<br />
and<br />
Thus<br />
andif ℓ1 = [H0 : 1]m1<strong>the</strong>n<br />
ℓ = [H0 : 1]m − ([H0 : 1] − 1)<br />
ℓ ′ = [H0 : 1]m ′ − ([H0 : 1] − 1).<br />
P ℓ′ −ℓ1<br />
K<br />
If mand<strong>the</strong>refore ℓissufficientlylarge<br />
P ℓ L ∩ F = Pm F<br />
∩ E = P m′ −m1<br />
E .<br />
P K/L(y) = N K/L(1 + y) − 1 ≡ S K/L(y) (modP ℓ L )<br />
if ybelongsto P ℓ′ −ℓ1<br />
K .Thusif ybelongsto P m′ −m1<br />
E<br />
P E/F(y) = P K/L(y) ≡ S K/L(y) = S E/F(y) (mod P m F ).<br />
Toverify<strong>the</strong>thirdfactwechoose,once m ′ 1<br />
If missufficientlylarge<br />
S E ′ /E<br />
<br />
P −δE ′ /E−m′ 1<br />
E ′<br />
isgiven, m1sothat<br />
<br />
= P −m1<br />
E .<br />
m ′′ = m ′ e E ′ /E − δ E ′ /E
Chapter22 285<br />
andif ybelongsto P m′′ −m ′<br />
1<br />
E ′<br />
Takinginevenlargerifnecessary,wehave<br />
SE ′ /E(y) ∈ P m′ −m1<br />
E .<br />
P E ′ /F(y) ≡ P E/F(P E ′ /E(y))<br />
≡ P E/F(S E ′ /E(y))<br />
≡ S E/F(S E ′ /E(y))<br />
≡ S E ′ /F(y) (modP m F ).<br />
Returningto<strong>the</strong>pro<strong>of</strong><strong>of</strong>Lemma22.5,wechoose<br />
β ′ = β(χ E/F) = P ∗ E/F (β).<br />
If m(χF)and<strong>the</strong>refore m(χ E/F)issufficientlylarge,<br />
∆(α s−1 2<br />
E µE χE/F,ψ E/F) = α s−1 2<br />
E<br />
Both βand β ′ areunitsand<strong>the</strong>refore<br />
α s−1<br />
2<br />
E<br />
<br />
γ<br />
β ′<br />
<br />
= α<br />
If m(χF)issufficientlylarge<br />
1 r− 2<br />
E<br />
<br />
γ<br />
β<br />
β ′ ≡ β<br />
= α s−1<br />
2<br />
F<br />
<br />
γ<br />
β ′<br />
<br />
and µE(β ′ ) = µE(β).Inparagraph5wesawthat<br />
det ω<br />
<br />
γ<br />
= µE<br />
β<br />
<br />
µE<br />
<br />
N E/F<br />
modP m(µE)<br />
E<br />
<br />
γ<br />
β ′<br />
<br />
<br />
γ<br />
β<br />
<br />
<br />
γ<br />
det ιE/F β<br />
<br />
γ<br />
,<br />
β<br />
∆(χ E/F, ψ E/F).<br />
= α r(s−1<br />
2 )<br />
F<br />
<br />
γ<br />
.<br />
β<br />
if ι E/Fis<strong>the</strong>representation<strong>of</strong> W K/Finducedfrom<strong>the</strong>trivialrepresentation<strong>of</strong> W K/E. We<br />
arereducedtoshowingthat<br />
detι E/F<br />
<br />
γ<br />
{∆(χF, ψF)}<br />
β<br />
r = ∆(χE/F, ψE/F) ρ(E/F, ψF) (22.1)<br />
if m(χF)issufficientlylarge.Ofcourse r = [E : F].
Chapter22 286<br />
WhatwedoisshowthatforeachGaloisextension K/F<strong>the</strong>relation(22.1)istrueforall<br />
fields Elyingbetween Kand F.Forthisweuseinductionon [K : F].Let G = G(K/F)and<br />
let Cbeanontrivialabeliannormalsubgroup<strong>of</strong> G.Let Lbe<strong>the</strong>fixedfield<strong>of</strong> C.Wesawin<br />
Chapter13that<strong>the</strong>rearefields F1, . . ., Fslyingbetween Fand Landgeneralizedcharacters<br />
µ1, . . ., µs<strong>of</strong> CF1 , . . ., CFsrespectivelysuchthat Then<br />
ιE/F ≃ ⊕ s i=1Ind(WK/F, WK/Fi , µi).<br />
χF ⊗ ι E/F ≃ ⊕ s i=1Ind(W K/F, W K/Ei , µi χ Fi/F)<br />
andbyTheorem2.1,<strong>the</strong>MainTheorem,<strong>the</strong>rightside<strong>of</strong>(22.1)isequalto<br />
s<br />
i=1 ∆(µi χ Fi/F, ψ Fi/F) ρ(Fi/F, ψF).<br />
Wejustsawthatif m(χF)issufficientlylargethisisequalto<br />
s<br />
i=1 µi<br />
Since s<br />
<br />
γ s<br />
β i=1 ∆(χFi/F, ψFi/F) ρ(Fi/F, ψF)}.<br />
i=1 [Fi : F] = [E : F]<br />
weseeuponapplying<strong>the</strong>inductionassumptionto L/Fthatthisequals<br />
s<br />
i=1 µi<br />
<br />
γ<br />
det ιFi/F β<br />
<br />
γ<br />
{∆(χF, ψF)}<br />
β<br />
r .<br />
Wecomplete<strong>the</strong>pro<strong>of</strong><strong>of</strong>(22.1)byappealingtoChapter5toseethat<br />
det ι E/F = s<br />
i=1 µidet ι Fi/F.
References 287<br />
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