Picard curves with small conductor

0.22 MB·

by  Michel Börner

Picard curves with small conductor
MichelBo¨rner,IreneI.Bouw,andStefanWewers
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Abstract We study the conductorof Picard curves over Q, which is a product of

] local factors. Our results are based on previous results on stable reduction of su-
G
perellipticcurvesthatallowtocomputetheconductorexponent f attheprimes p
p
A
ofbadreduction.Acarefulanalysisofthepossibilitiesofthestablereductionat p
. yieldsrestrictionsontheconductorexponent f .We provethatPicardcurvesover
h p
t Q always have bad reductionat p=3, with f3 4. As an applicationwe discuss
a ≥
thequestionoffindingPicardcurveswithsmallconductor.
m
[
Keywords: 2010MathematicsSubjectClassification.Primary14H25.Secondary:

1 11G30,14H45.
v
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9 1 Introduction
1
0
. LetY bea smoothprojectivecurveofgenusgovera numberfield K. Tosimplify
1
0 the exposition, let us assume that K =Q. WithY we can associate an L-function
7 L(Y,s)andaconductorN N.Conjecturally,theL-functionsatisfiesafunctional
Y
1 equationoftheform ∈
v: L (Y,s)= L (Y,2 s),
± −
i
X where
r L (Y,s):=√NYs (2p )−gs G (s)g L(Y,s).
a · · ·
Bydefinition,bothL(Y,s)andN areaproductoflocalfactors.Inthispaperweare
Y
reallyonlyconcernedwiththeconductor,whichcanbewrittenas
N =(cid:213) pfp.
Y
p
Institutfu¨rReineMathematik,Universita¨tUlm
e-mail:[email protected],e-mail:[email protected]
1
2 Bo¨rner,Bouw,Wewers
Theexponent f iscalledtheconductorexponentofY at p.Itisknownthat f only
p p
depends on the ramification of the local Galois representation associated with Y.
Inparticular,ifY hasgoodreductionat pthen f =0.IfY hasbadreductionat p
p
thenthecomputationof f canbequitedifficult.Untilrecently,aneffectivemethod
p
forcomputing f wasonlyknownforellipticcurves([24], IV.10)andforgenus2
p
§
curvesif p=2([11]).
6
Itwasshownin[2]that f caneffectivelybecomputedfromthestablereduction
p
ofY atp.Moreover,forcertainfamiliesofcurves(thesuperellipticcurves)wegave
a rathersimple recipe forcomputingthe stable reduction.The latter resultneeded
theassumptionthatpdoesnotdividethedegreen.In[18]thisrestrictionisremoved
forsuperellipticcurvesofprimedegree.
InthepresentpaperwesystematicallystudythecaseofPicardcurves.Theseare
superellipticcurvesofgenus3anddegree3,givenbyanequationoftheform
Y : y3= f(x)=x4+a x3+a x2+a x+a ,
3 2 1 0
with f Q[x]separable.Picardcurvesforminsomesensethenextfamilyofcurves
∈
tostudyafterhyperellipticcurves.Theyareinterestingformanyreasonsandhave
beenintensivelystudied,seee.g.[14],[8],[9],and[16].
Our main results classify all possible configurationsfor the stable reduction of
aPicardcurveataprime p,andusethistodeterminerestrictionsontheconductor
exponents.Forinstance,weprovethefollowing.
Theorem1.1.LetY beaPicardcurveoverQ.
(a)ThenY hasbadreductionat p=3,and f 4.
3
≥
(b)For p=2wehave f =1.
2
6
(c)For p 5wehave f 0,2,4,6 .
p
≥ ∈{ }
Theorem3.6isasomewhatstrongerversionofthefirststatement.Theorem4.4
contains the last two statements. We also give explicit examples, showing that at
least part of our results are sharp. Our result can be seen as a complement, for
Picardcurves,toaresultofBrumer–Kramer([3],Theorem6.2),whoproveanupper
boundfor f forabelianvarietiesoffixeddimension.Sincetheconductorofacurve
p
coincides with that of its Jacobian, the result applies to our situation, as well. A
morecarefulcase-by-caseanalysis,combinedwithideasfrom[3],couldprobably
beusedtoobtainamorepreciselistofpossiblevaluesfortheconductorexponent
at p=2,3,aswell.
In the last section we discuss the problem of constructing Picard curves with
smallconductor.AsaconsequenceoftheShafarevichconjecture(akaFaltings’The-
orem),thereareatmostafinitenumberofnonisomorphiccurvesofgivengenusand
of boundedconductor.But except in very special cases, no effective proof of this
theoremisknown.
InhisrecentPhDthesis,thefirstnamedauthorhasmadeanextensivesearchfor
Picardcurveswithgoodreductionoutsideasmallsetofsmallprimes,andcomputed
theirconductor.ThePicardcurvewiththesmallestconductorthatwasfoundisthe
curve
Picardcurveswithsmallconductor 3
Y : y3=x4 1,
−
whichhasconductor
N =2636=46656.
Y
Weproposeasasubjectforfurtherresearchtoeitherprovethattheaboveexample
is thePicard curveoverQ with thesmallest possibleconductor,orto find (oneor
all) counterexamples.We believethat the methodspresentedin thispaper maybe
veryhelpfultoachievethisgoal.
2 Semistable reduction
We first introduce the general setup concerning the stable reduction and the con-
ductorexponentsofPicardcurves.Asexplainedintheintroduction,theconductor
exponentisalocalinvariant,encodinginformationabouttheramificationofthelo-
calGaloisrepresentationassociatedwiththecurve.Therefore,wemayreplacethe
number field K by its strict henselization. In other words, we may work from the
startoverahenselianfieldofmixedcharacteristicwithalgebraicallyclosedresidue
field.
2.1 Setup andnotation
ThroughoutSection2-4theletterKwilldenoteafieldofcharacteristiczerothatis
henselianwithrespecttoadiscretevaluation.WedenotethevaluationringbyO ,
K
themaximalidealofO bypandtheresiduefieldbyk=O /p.Weassumethatk
K K
isalgebraicallyclosedofcharacteristicp>0.Themostimportantexampleforusis
whenK=Qnr isthemaximallyunramifiedextensionofthe p-adicnumbers.Then
p
p=(p)andk=F¯ .
p
LetY/KbeaPicardcurve,givenbytheequation
Y : y3= f(x), (1)
where f K[x]isaseparablepolynomialofdegree4.WesetX :=P1 andinterpret
(1)asafi∈nitecoverf :Y X, (x,y) x,ofdegree3. K
→ 7→
By the Semistable Reduction Theorem (see [5]), there exists a finite extension
L/KsuchthatthecurveY :=Y Lhassemistablereduction.Sinceg(Y)=3 2,
L K
thereevenexistsa(unique)distin⊗guishedsemistablemodelY SpecO ofY ≥,the
L L
stablemodel([5],Corollary2.7).ThespecialfiberY¯ :=Y ofY→iscalledthestable
s
reductionofY.Itisastablecurveoverk([5], 1),anditonlydependsonY,upto
§
uniqueisomorphism.
ItisnorestrictiontoassumethattheextensionL/KisGaloisandcontainsathird
rootofunityz L. Thenthe coverf :Y X (thebasechangeoff to L) isa
3 L L L
∈ →
4 Bo¨rner,Bouw,Wewers
Galois cover. Its Galois group G is cyclic of order 3, generated by the element s
whichisdeterminedby
s (y)=z y.
3
Let G :=Gal(L/K) denote the Galois group of the extension L/K. The group G
acts faithfully and in a natural way on the schemeY =Y L. We denote by G˜
L K
thesubgroupofAut(Y )generatedbyGandtheimageofG⊗.By definition,G˜ isa
L
semidirectproduct,
G˜ =G⋊G .
TheactionofG onGviaconjugationisdeterminedbythefollowingformula:fort
inG wehave
s ift (z )=z ,
tst 1= 3 3 (2)
− (s 2 ift (z 3)=z 32.
Becauseoftheuniquenesspropertiesofthestablemodel,theactionofG˜ onY
L
extendstoanactiononY.Byrestriction,weseethatG˜ hasanatural,k-linearaction
onY¯.ThisactionwillplayadecisiveroleinouranalysisofthestablereductionY¯.
FortherestofthissubsectionwefocusontheactionofthesubgroupG G˜. The
roleofthesubgroupG G˜ willbecomeimportantlater. ⊂
⊂
Remark2.1. (a)ThequotientschemeX :=Y/GisasemistablemodelofX =P1,
L L
seee.g.[17],Cor.1.3.3.i.SincethemapY X isfiniteandY isnormal,Y
is the normalization of X in the function→field of Y . This means that Y is
L
uniquelydeterminedbyasuitablesemistablemodelX ofX .
L
(b)LetX¯ :=X kdenotethespecialfiberofX andf¯ :Y¯ X¯ theinducedmap.
Wenotethat⊗f¯ isafiniteG-invariantmap.Itisnottruein→generalthatY¯/G=X¯.
However,the naturalmapY¯/G X¯ is radicialandin particulara homeomor-
→
phism(seee.g.[17],Prop.2.4.11).
(c)EveryirreduciblecomponentW Y¯ issmooth.Toseethisnotethatthequotient
ofW byitsstabilizerinGishom⊂eomorphictoanirreduciblecomponentZ X¯,
whichisasmoothcurveofgenus0.IfW hasasingularpoint,thens actso⊂nW
andpermutesthetwobranchesofW passingthroughthispoint.Butsinces has
order3,thisisimpossible.
LetD Y¯ denotethecomponentgraphofY¯:theverticesaretheirreduciblecomponents
ofY¯ andtheedgescorrespondto thesingularpoints.ThestabilityconditionforY¯
means that an irreducible componentof genus0 correspondsto a vertex of D Y¯ of
degree≥3.ThenumberofloopsofD Y¯ isgivenbythewellknownformula
g (Y¯):=dimQH1(D Y¯,Q)=r−s+1, (3)
whereristhenumberofedgesandsthenumberverticesofD Y¯.
ThecurveX¯ isalsosemistable,butingeneralnotstable.SinceX¯ hasarithmetic
genus0,thecomponentgraphD X¯ isatree,andeveryvertexcorrespondstoasmooth
curveofgenus0.ItfollowsfromRemark2.1thatD X¯ =D Y¯/G.
Lemma2.2.IfW Y¯ isanirreduciblecomponent,thens (W)=W.
⊂
Picardcurveswithsmallconductor 5
Proof. To deriveacontradiction,weassumethatW ,W ,W Y¯ arethreedistinct
1 2 3
componentsthatforma singleG-orbit.ThenWi ∼ Z:=f¯(W⊂i). SinceZ isa com-
ponentofX¯,weconcludethatg(W)=0,fori=→1,2,3.Thestabilityconditionon
i
Y¯ implies that each W contains at least three singular points of Y¯. Hence Z also
i
containsatleastthreesingularpointsofX¯.
LetY¯ Y¯ denote the unique morphism which contracts all componentsofY¯
0
→
except the W and which is an isomorphism on the intersection of W with the
i i i
smoothlocusofY¯.Similarly,letX¯ X¯ bethemapcontractingallc∪omponentsof
0
X¯ exceptZ.Thesemapsfitintoaco→mmutativediagram
Y¯ //Y¯
0
(cid:15)(cid:15) (cid:15)(cid:15)
X¯ //X¯ ,
0
wheretheverticalarrowsarequotientmapsbythegroupG(atleastfortheunder-
lyingtopologicalspaces).Also,X¯0∼=Z.
Letx¯ Z beoneofthesingularpointofX¯ lyingonZ,andletT X¯ theclosed
subset w∈hich is contracted to x¯ Z =X¯ . Then T is a nonempty⊂and connected
0
unionofirreduciblecomponents∈ofX¯ andhenceasemistablecurveofgenus0.In
particular,thecomponentgraphofT isatree.LetZ T beatailcomponent.Asa
′
componentofX¯,Z intersectstherestofX¯ inatmost⊂twopoints.LetW Y¯ bean
′ ′
irreduciblecomponentlyingaboveZ .ThestabilityofY¯ impliesthats (⊂W )=W
′ ′ ′
andthattheactionofs onW isnontrivial.(OtherwiseW wouldbehomeomorphic
′ ′
to Z, and henceW would be a component of genus 0 intersecting the rest ofY¯
′ ′
in atmosttwo points.) Itfollowsthatthe inverseimageS Y¯ ofT isconnected.
⊂
NotethatS meetsthe componentW inthe uniquepointonW abovex¯. SinceS is
i i
connected,itfollowsthatthemapY¯ Y¯ contractsStoasinglepoint.
0
WeconcludethatthecurveY¯ has→atleastthreedistinctsingularpointswhereall
0
three componentsW meet. Equation(3) impliesthatg (Y¯ ) is atleast 1. It follows
i 0
thatthearithmeticgenusofY¯ is 4,andhenceg(Y¯) 4aswell.Thisisacontra-
0
≥ ≥
diction,andthelemmafollows.
⊓⊔
2.2 Theconductor exponent
Letcp be the conductorof the Gal(K¯/K)-representationHe1t(YK¯,Qℓ), see [22]. By
definition,thisisanidealofO oftheform
K
c =pfp,
p
with f 0.Theinteger f iscalledtheconductorexponentofY/K.1
p p
≥
1Whenworkinginalocalcontext, f isoftensimplycalledtheconductorofY.
p
6 Bo¨rner,Bouw,Wewers
We recall from [2] an explicit formula for f , in terms of the action of G =
p
Gal(L/K)onY¯.ForthisweletG u G ,foru 0,denotetheuthhigherramification
group(inthe uppernumbering).W⊂e setY¯u:≥=Y¯/G u.NotethatY¯u isasemistable
curveforallu.NotealsothatG =G 0 becausetheresiduefieldk isassumedtobe
algebraicallyclosed.
Proposition2.3.TheconductorexponentofthecurveY/Kisgivenby
f =e +d , (4)
p
where
e :=6 dimH1(Y¯0,Q ) (5)
− et ℓ
and
¥
d := 6 2g(Y¯u) du. (6)
0 −
Z
(cid:0) (cid:1)
Proof. See[2],Theorem2.9and[1],Corollary2.14.
⊓⊔
Thee´talecohomologygroupH1(Y¯u,Q )decomposesas
et ℓ
He1t(Y¯u,Qℓ)=⊕WHe1t(W,Qℓ)⊕H1(D Y¯u,Qℓ),
where the first sum runs overthe set of irreduciblecomponentsW of the normal-
izationofY¯u andD Y¯u isthegraphofcomponentsofY¯u.(See[2],Lemma2.7.(1).)
Therefore,thesecondtermin(5)canbewrittenas
dimHe1t(Y¯0,Qℓ)=(cid:229) dimHe1t(W,Qℓ) + dimH1(D Y¯0). (7)
W
ThearithmeticgenusofY¯u,whichoccursin(6),isgivenbytheformula
g(Y¯u)=(cid:229) g(W) + dimH1(D Y¯u). (8)
W
For future reference we note that dimH1(W,Q )=2g(W). The integer g (Y¯0):=
et ℓ
dimH1(D Y¯0) can be interpreted as the number of loops of the graph D Y¯0. It is
boundedbyg(Y¯0),andhencebyg(Y)=3.
Lemma2.4.Thefollowingstatementsareequivalent.
(a)d =0.
(b)G u actstriviallyonY¯,forallu>0.
(c)ThecurveY hassemistablereductionoveratamelyramifiedextensionofK.
Proof. Assumethatd =0.By(6)thismeansthat3=g(Y¯)=g(Y¯u)forallu>0.
Using(8)oneeasilyshowsthatthismeansthatG u actstriviallyonthecomponent
IgtrafoplhloDwY¯sothfaY¯t.GMuoarcetosvterirv,ifaolrlyevoenryY¯.coWmephoanveenptrWov⊂edY¯thweeimhapvliecagt(iWon)(=a)g(W(b)/.GTuh)e.
⇒
implication(b) (c)followsfrom[12],Theorem4.44.Theimplication(c) (a)fol-
lowsimmediate⇒lyfromthedefinitionofd . ⇒
⊓⊔
Picardcurveswithsmallconductor 7
3 The wildcase: p=3
Inthissectionweassumethat p=3.Wefirstanalyzethespecialfiberofthestable
modelofY ,andshowthatthereareessentiallyfivereductiontypes.From 3.2we
L
§
consider the case where K is absolutely unramified, and derive a lower boundfor
theconductorexponent f .
3
3.1 Thestablemodel
Wekeepallthenotationintroducedin 2.Inaddition,weassumethatp=3.Lemma
§
2.2impliesthatwecandistinguishbetweentwotypesofirreduciblecomponentsof
Y¯.
Definition3.1.An irreducible componentW Y¯ is called e´tale if the restriction
s Aut (W)isnontrivial.Ifs istheiden⊂tity,thenW iscalledaninseparable
W k W
| ∈ |
component.
LetW Y¯ be an irreduciblecomponent,and let Z :=f¯(W) X¯ be its image.
Then Z is⊂an irreducible component of X¯ and hence a smooth⊂curve of genus 0.
Lemma2.2showsthats (W)=W.ItfollowsthatW/G Z isahomeomorphism.
→
IfW isaninseparablecomponent,thenW Z isapurelyinseparablehomeomor-
→
phism(sinceW Zhasdegree3,thiscanonlyhappenwhenp=3).Itfollowsthat
→
everyinseparablecomponenthasgenuszero.
IfW isane´talecomponent,thenZ=W/G,andW ZisaG-Galoiscover.For
∼ →
future reference we recall that the Riemann–Hurwitz formula for wildly ramified
Galoiscoversofcurvesyields
(cid:229)
2g(W) 2= 2 3+ 2(h +1), (9)
z
− − ·
z
wherethesumrunsoverthebranchpointsofW Zandh isthe(unique)jumpin
z
→
thefiltrationofthehigherramificationgroupsinthelowernumbering.Wehavethat
h 1isprimeto p([21], IV.2,Cor.2toProp.9).
z
≥ §
Theorem3.2.Weareinexactlyoneofthefollowingfivecases.
(a)ThecurveY¯ issmoothandirreducible.
(b)ThereareexactlytwocomponentsW ,W whicharebothe´tale,meetinasingle
1 2
point,andhavegenusg(W )=2,g(W )=1.
1 2
(c)Therearethreee´talecomponentsW ,W ,W ofgenusone,andoneinseparable
1 2 3
componentW ofgenuszero.Fori=1,2,3,W intersectsW inauniquepoint,
0 i 0
andtheseintersectionpointsarepreciselythesingularpointsofY¯.
(d)TherearetwocomponentsW ,W whicharee´taleofgenusg(W )=1,g(W )=
1 2 1 2
0.Thereareexactlythreesingularpoints,whichformanorbitundertheaction
ofG,andwhereW andW meet.
1 2
8 Bo¨rner,Bouw,Wewers
(e)Therearethree componentsW ,W ,W ,whichare e´taleandofgenusg(W )=
1 2 3 1
g(W )=0andg(W )=1,andfoursingularpoints.Threeofthesingularpoints
2 3
are pointsofintersection ofW andW , andform anorbitunderthe actionof
1 2
G.ThefourthsingularpointisthepointofintersectionofW andW .
2 3
Proof. Letr (resp.s )bethenumberofsingularpoints(resp.irreduciblecompo-
1 1
nents) ofY¯ which are fixed by s , and let r (resp. s ) be the number of orbits of
2 2
singular point (resp. irreducible components) of Y¯ of length 3. Lemma 2.2 states
thats =0.Therefore,(3)becomes
2
g (Y¯)=r s+1=r +3r s+1. (10)
1 2
− −
BecauseD X¯ =D Y¯/Gisatree,wehave
g (X¯):=dimH1(D X¯)=r1+r2−s+1=0. (11)
Combining(10)and(11)weobtain
g (Y¯)=2r . (12)
2
Since0 g (Y¯) 3,weconcludethatg (Y¯) 0,2 andr 0,1 .
2
≤ ≤ ∈{ } ∈{ }
Case1:r =0andg (Y¯)=0.
2
InthiscaseD Y¯ isatree,andthesumofthegeneraofallirreduciblecomponentsis
3.Inparticular,thereareatmost3componentsofgenus>0.Moreover,thestability
conditionimpliesthateverycomponentofgenuszerocontainsatleastthreesingular
pointsofY¯.Itisaneasycombinatorialexercisetoseethatthisleavesuswithexactly
fourpossibilitiesforthetreeD Y¯.Goingthroughthesefourcaseswewillseethatone
ofthemisexcluded,whiletheremainingthreecorrespondtoCase(a),(b),and(c)
ofTheorem3.2.
ThefirstcaseiswhenY¯ hasauniqueirreduciblecomponent.ThenY¯ issmooth.
ThisisCase(a)ofthelemma.Secondly,theremaybetwoirreduciblecomponents,
ofgenus1and2,andauniquesingularpoint.ThiscorrespondstoCase(b).
Thirdly, there may be three irreducible components, each of genus 1, and two
singularpoints.Weclaimthatthiscasecannotoccur.Indeed,oneofthethreecom-
ponentswouldcontaintwosingularpoints,andeachofthesetwopointsmustbea
fixedpointofs .ItfollowsthattheG-coverW Z=W/Gisramifiedinatleast
→
twopoints.TheRiemann–Hurwitzformula(9)impliesthatg(W) 2.Thisyieldsa
≥
contradiction,andweconcludethatthiscasedoesnotoccur.
Finally,inthelastcase,therearefoursingularpointsandfourirreduciblecompo-
nents.Threeofthemhavegenus1andonehasgenuszero.Thecomponentofgenus
zeronecessarilycontainsallthreesingularpoints.Asimilarargumentasinthepre-
viouscaseshowsthatthegenus-0componentcannotbee´tale.Thiscorrespondsto
Case(c).
Case2:g (Y¯)=2andr =1.
2
Inthiscasethesumofthegeneraofallcomponentsisequalto1.Therefore,there
Picardcurveswithsmallconductor 9
mustbeauniquecomponentofgenus1,andallothercomponentshavegenus0.Let
W andW betwocomponentswhichmeetinasingularpointy¯suchthats (y¯)=y¯.
1 2
Since s (W)=W for i=1,2 (Lemma 2.2),W andW are e´tale components6and
i i 1 2
intersecteachotherinexactlythreepoints(theG-orbitofy¯).
Iftherearenofurthercomponents,weareinCase(d).Assumethatthereexists
athirdcomponentW .LetT Y¯ bethemaximalconnectedunionofcomponents
3
⊂
whichcontainsW butneitherW norW .ThenT containsauniquecomponentW
3 1 2 0
which meets eitherW orW in a singular point. The component graph of T is a
1 2
tree,andweconsiderW asitsroot.Bythestabilitycondition,everytailcomponent
0
of T must have positive genus, so T has a unique tail. IfW is not this tail, it has
0
genus0andintersectstherestofY¯ inexactly2points.Thiscontradictsthestability
condition. We conclude that Y¯ has exactly three components, of genus g(W )=
1
g(W )=0andg(W )=1.ThisisCase(e)ofthelemma.Nowtheproofiscomplete.
2 3
⊓⊔
3.2 Alower bound for f
3
We continue with the assumptions from the previous subsection. In addition, we
assume that K is absolutelyunramified.By this we mean thatp=(3). Under this
assumption,we provea lowerboundfortheconductorexponent f := f . Infact,
3 p
we will give a lower bound for e , where f =e +d is the decomposition from
3
Proposition2.3.IfL/Kisatmosttamelyramified,thend =0(Lemma2.4).Inthis
case,ourboundsaresharp.
SinceK isabsolutelyunramified,thethirdrootofunityz Lisnotcontained
3
in K. Therefore,there exists an elementt G =Gal(L/K) su∈ch thatt (z )=z 2.
Letmbetheorderoft .Afterreplacingt b∈yasuitableoddpowerofitself3wema3y
assumethatmisapowerof2.Wekeepthisnotationfixedfortherestofthispaper.
RecallthatthesemidirectproductG˜ =G⋊G actsonY¯ inanaturalway.
Thefollowingobservationiscrucialforouranalysisoftheconductorexponent.
Lemma3.3.LetW Y¯ be an e´tale componentsuch that t (W)=W. Then inside
⊂
theautomorphismgroupofW wehave
t s t 1=s 2=s . (13)
−
◦ ◦ 6
Inparticular,t isnontrivial.
W
|
Proof. ThestatementfollowsimmediatelyfromEquation(2)andDefinition3.1.
⊓⊔
Despiteitssimplicity,Lemma3.3hasthefollowingstrikingconsequence.Note
that we consider potentially good but not good reduction as bad reduction in this
paper.
Proposition3.4.Assume thatp=(3). Thenevery PicardcurveY overK hasbad
reduction.
10 Bo¨rner,Bouw,Wewers
Proof. Lemma3.3impliesthatY acquiressemistable reductiononlyafterpassing
to a ramified extensionL z . ThereforeY/K doesnothavegoodreduction.The
3
fact that f = 0 follows f∋rom Proposition 2.3, together with the fact that t acts
3
nontriviallyo6 neachirreduciblecomponentofY¯ (Lemma3.3).
⊓⊔
Inordertoprovemorepreciselowerboundsfor f ,weneedtoanalyzetheaction
3
ofs andt onY¯ inmoredetail.
Lemma3.5.Let W Y¯ be an e´tale component. Then one of the following cases
⊂
occurs:
g(W) r h g(W/G 0)
0 1 1 0
1 1 2 0
2 2 (1,1) 1
3 1 4 0
Here r isthe numberoframificationpointsofthe G-coverW Z:=W/G andh
→
liststhesetoflowerjumps.Thefourthcolumngivesanupperboundforthegenus
ofW/G 0.
Proof. Recallthatwehaveassumedthattheordermoft isapowerof2.
TheRiemann–Hurwitzformula(9)immediatelyyieldsthecasesforg(W),r,and
h stated in the lemma, together with one additional possibility: the curve W has
genus3 andf :W Z =P1 isbranchedattwo points,with lowerjump1and 2,
→ ∼
respectively.Weclaimthatthiscasedoesnotoccur.
AssumethatW isane´talecomponentofY¯ suchthatf :W Z isbranchedat2
points.Lemma3.3impliesthatt actsnontriviallyonW.Since→t normalizess and
thetwo ramificationpointshavedifferentlowerjumps,itfollowsthatt fixesboth
ramificationpointsw off .WeconcludethatH:= s ,t actsonW asanonabelian
i
h i
groupoforder6fixingthev.
i
Wewriteh forlowerjumpofw.Lemma2.6of[15]impliesthatgcd(h,m)isthe
i i i
orderoftheprime-to-3partofthecentralizerofH.Sincegcd(h ,m)=gcd(h ,m)
1 2
6
weobtainacontradiction,andconcludethatthiscasedoesnotoccur.
We compute an upper bound for the genus ofW/ t in each of the remaining
cases.Thisisalsoanupperboundforg(W/G 0). h i
In the case thatg(W)=0 there is nothingto prove.In the case thatg(W)=1,
theautomorphismt fixestheuniqueramificationpointoff ,henceg(W/G 0)=0.
Assumethatg(W)=2.TheRiemann–Hurwitzformulaimmediatelyimpliesthat
thatg(W/ t ) 1.
h i ≤
Finally,weconsiderthecasethatg(W)=3,i.e.Y haspotentiallygoodreduction.
Asbefore,wehavethatt fixestheuniquefixedpointofs .PutH= s ,t .Lemma
3.3togetherwiththeassumptionthattheordermoft isapowerofh2imipliesthat
the order of the prime-to-p centralizer of H is gcd(h =4,m)= m/2. It follows
that m=8. Since t has at least one fixed point onW, namely the point at ¥ , the
Riemann–Hurwitz formula implies that g(W/ t )=0. This finishes the proof of
h i
thelemma.
⊓⊔