Table Of ContentA polynomial with Galois group SL (F )
2 16
7
0
JohanBosman∗
0
2
February 2, 2008
n
a
J
6
1 Abstract
] Inthispaperweshowthatthepolynomialx17−5x16+12x15−28x14+72x13−132x12+116x11−
T 74x9+90x8−28x7−12x6 +24x5 −12x4 −4x3 −3x−1 ∈ Q[x]hasGaloisgroupSL2(F16),
N fillinginagapinthetablesofJu¨rgenKlu¨nersandGuntherMalle(see[12]). Thecomputationofthis
. polynomialusesmodularformsandtheirGaloisrepresentations.
h
t
a
m 1 Introduction
[
ItisacomputationalchallengetoconstructpolynomialswithaprescribedGaloisgroup,see[12]formeth-
1 odsandexamples. Here, bythe Galois groupofa polynomialf ∈ Q[x] we meanthe Galoisgroupofa
v
splitting field of f overQ togetherwith its naturalaction onthe rootsof f in this splitting field. Ju¨rgen
2
Klu¨nersinformedmeaboutaninterestinggroupforwhichapolynomialhadnotbeenfoundyet, namely
4
4 SL2(F16)withitsnaturalactiononP1(F16). Thisactionisfaithfulbecauseofchar(F16) = 2. Itmustbe
1 notedthattheexistenceofsuchapolynomialwasalreadyknowntoMestre(unpublished).Inthispaperwe
0 willgiveanexplicitexample.
7
0 Proposition1. Thepolynomial
/
h
t P(x):= x17−5x16+12x15−28x14+72x13−132x12+116x11
a
m −74x9+90x8−28x7−12x6+24x5−12x4−4x3−3x−1∈Q[x]
:
v hasGaloisgroupisomorphictoSL (F )withitsnaturalactiononP1(F ).
2 16 16
i
X
WhatisstillunknowniswhetherthereexistsaregularextensionofQ(T)withGaloisgroupisomorphicto
r
a SL2(F16);regularheremeansthatitcontainsnoalgebraicelementsoverQapartfromQitself. Insection
2wewillsaysomewordsaboutthecalculationofthepolynomialandtheconnectionwithmodularforms.
We’ll indicate how one can verify that it has the claimed Galois group in section 3 using computational
Galoistheory. We willshowinsection4thatthispolynomialgivesa Galoisrepresentationassociatedto
anexplicitlygivenmodularform.
2 Computation of the polynomial
Inthissectionwewillbrieflyindicatehowonecanfindapolynomialasinproposition1. Wewillmakeuse
ofmodularforms. Foranoverviewaswellasmanyfurtherreferencesonthissubjectthereaderisrefered
to[6].
Let N be a positive integer and consider the space S (Γ (N)) of holomorphic cusp forms of weight 2
2 0
for Γ (N). A newform f ∈ S (Γ (N)) has a q-expansion f = a qn where the coefficients a
0 2 0 n≥1 n n
are in a number field. The smallest number field containing all thPe coefficients is denoted by Kf. To a
∗PartiallysupportedbytheDutchscientificorganisationNWO. E-mail:[email protected]
1
given prime number ℓ and a place λ of K above ℓ one can attach a semi-simple Galois representation
f
ρ : Gal(Q/Q) → GL (F ) unramified outside Nℓ satisfying the following property: for each prime
f 2 λ
p∤NℓandanyFrobeniuselementFrob ∈Gal(Q/Q)attachedtopwehave
p
Tr(ρ (Frob ))≡a modλ and Det(ρ (Frob ))≡pmodλ. (1)
f p p f p
Therepresentationρ isuniqueuptoisomorphism.ThefixedfieldofKer(ρ )inQisGaloisoverQwith
f f
GaloisgroupisomorphictoIm(ρ ). Forℓ=2andanyλaboveℓequation(1)togetherwithChebotarev’s
f
density theoremimply that Im(ρ ) is containedin SL (F ). So to show that there is an extensionof Q
f 2 λ
with Galoisgroupisomorphicto SL (F )it sufficesto findan N anda newformf ∈ S (Γ (N)) such
2 16 2 0
thatthereisaprimeλofdegree4above2inK andIm(ρ )isthefullgroupSL (F ). Usingmodular
f f 2 λ
symbols we can calculate the coefficients of f, hence traces of matrices that occur in the image of ρ .
f
For a survey paper on how this works, see [18]. A subgroupΓ of SL (F ) contains elements of every
2 16
traceifandonlyifΓequalsSL (F );thiscanbeshowninseveralways,eitherbyadirectcalculationor
2 16
byinvokinga moregeneralclassificationresultlike[20, Thm. III.6.25]. Withthisinmind,afterasmall
computersearch in which we check the occurringvaluesof Tr(ρ (Frob )) up to some moderatebound
f p
ofp,onefindsthatasuitablemodularformf existsinS2(Γ0(137)). ItturnsoutthatwehaveKf ∼=Q(α)
withtheminimalpolynomialofαequaltox4+3x3−4x−1andthatf istheformwhoseq-expansion
startswith
f =q+αq2+(α3+α2−3α−2)q3+(α2−2)q4+··· .
Now the next question comes in: knowing this modular form, how does one produce a polynomial? In
general, one can use the Jacobian J (N) to construct ρ . In this particular case we can do that in the
0 f
followingway. WeobservethatK isofdegree4andthattheprime2isinertinit. Furthermorewecan
f
verifythatthesubspaceofS (Γ (137))fixedbytheAtkin-Lehneroperatorw isexactlythesubspace
2 0 137
generatedbyallthecomplexconjugatesoff. Theseobservationsimplythatρ isisomorphictotheaction
f
ofGal(Q/Q)onJac(X (137)/hw i)[2], wherewe givethislatter space anF -vectorspace structure
0 137 16
viatheactionoftheHeckeoperators.NotethatIm(ρ )=SL (F )impliessurjectivityofthenaturalmap
f 2 16
T → OK,f/(2) ∼= F16,whereTistheHeckealgebraattachedtoS2(Γ0(N)). Themethodsdescribedin
[8]allowusnowtogivecomplexapproximationsofthe2-torsionpointsofJac(X (137)/hw i)toahigh
0 137
precision. Thispartofthecalculationtookbyfarthemosteffort;theauthorwillwritemoredetailsabout
howthisworksinafuturepaper(orthesis). Weusethistogivearealapproximationofapolynomialwith
GaloisgroupisomorphictoSL (F ). Thepaper[8]does,atleastimplicitly,giveatheoreticalupperbound
2 16
fortheheightofthecoefficientsofthepolynomialhenceanupperboundforthecalculationprecisiontoget
anexactresult. Thoughthisupperboundissmallinthesensethatitleadstoapolynomialtimealgorithm,
itisstillfartoohightobeofuseinpractice.Howeveritturnsoutthatwecanuseamuchsmallerprecision
toobtainourpolynomial,theonlydrawbackbeingthatthisdoesnotgiveusaproofofitscorrectness,so
wehavetoverifythisafterwards.
ThepolynomialP′obtainedinthiswayhascoefficientsofabout200digitssowewanttofindapolynomial
ofsmallerheightdefiningthesamenumberfieldK. Todothis,wefirstcomputetheringofintegersO
K
ofK. In[2]analgorithmtodothisisdescribed,providedthatoneknowsthesquarefreefactorisationof
Disc(f)andevenifwedon’tknowthesquafrefreefactorisationofthediscriminant,thealgorithmproduces
a ’good’orderin K. AssumingthatourpolynomialP′ is correctwe knowthatK is unramifiedoutside
2·137so we caneasily calculatethesquarefreefactorisationofDisc(f)andhenceapplythealgorithm.
HavingdonethisweobtainanorderinK withadiscriminantsmallenoughtobeabletofactorandhence
weknowthatthisisindeedthemaximalorderO . Explicitly,thediscriminantisequalto
K
Disc(O )=230·1378. (2)
K
We embed O as a lattice into C[K:Q] in the natural way and use lattice basis reduction, see [13], to
K
computeashortvectorα∈O −Z.Theminimalpolynomialofαhassmallcoefficients.Inourparticular
K
case [K : Q] is equal to 17, which is a prime number, hence this new polynomial must define the full
fieldK. ThismethodgivesusalsoawayofexpressingαasanelementofQ(x)/(P′(x)).
2
3 Verification of the Galois group
NowthatwehavecomputedapolynomialP(x),wewanttoverifythatitsGaloisgroupGal(P)isreally
isomorphictoSL (F )andthatwecanidentifythesetΩ(P)ofrootsofP withP1(F )insucha way
2 16 16
thattheactionofGal(P)onΩ(P)isidentifiedwiththeactionofSL (F )onP1(F ).
2 16 16
ForcompletenessletusremarkthatitiseasytoverifythatP(x)isirreduciblesinceitisirreduciblemodulo
5.TheirreducibilityofP impliesthatGal(P)isatransitivepermutationgroupofdegree17.Thetransitive
permutationgroupsof degree17havebeenclassified, see forexample[17]. From[20, Thm. III.6.25]it
follows that up to conjugacy there is only one subgroup of index 17 in SL (F ), namely the group of
2 16
uppertriangularmatrices. Thisimpliesthatupto conjugacythereisexactlyonetransitiveG < S that
17
is isomorphicto SL (F ). Hence if Gal(P) ∼= SL (F ) is an isomorphismof groupsthen there is an
2 16 2 16
identificationof Ω(P)with P1(F ) suchthatthe groupactionsbecomecompatible. Itfollowsfromthe
16
classification in [17] that if the order of a transitive G < S is divisible by 5, then G must contain a
17
transitivesubgroupisomorphictoSL (F ). Toshowthat5 | #Gal(P)weusethefactthatforaprime
2 16
p ∤ Disc(P)wethedecompositiontypeofP modulopisequaltothecycletypeofanyFrob ∈ Gal(P)
p
attachedtop. Onecanverifythatmodulo7wegetthefollowingdecompositionintoirreducibles:
P =(x−3)(x−5)(x15+3x14+4x12+6x11+3x10+x9+5x8+x6+x5+2x4+4x3+4x2+3x+6),
showingthatindeed5|#Gal(P)henceGal(P)containsSL (F )asasubgroup.
2 16
ToshowthatGal(P)cannotbebiggerthanSL (F )itseemsinevitabletouseheavycomputercalcula-
2 16
tions. Itwouldbeinterestingtoseeamethodwhichdoesnotusethis.
NotethattheactionofSL (F )onP1(F )issharply3-transitive. So firstweshowthatGal(P) isnot
2 16 16
4-transitivetoprovethatitdoesnotcontainA . Todothiswestartwithcalculatingthepolynomial
17
Q(x):= (X −α −α −α −α ), (3)
1 2 3 4
{α1,α2,αY3,α4}⊂Ω(P)
wheretheproductrunsoverallsubsetsof{1,...,17}consistingofexactly4elements. Thisimpliesthat
deg(Q) = 2380. OnecancalculateQ(x)bysymbolicmethods,see[5]. SupposethatGal(P)actingon
Ω(P)is4-transitive. ThentheactiononΩ(Q)istransitivehenceQ(x)isirreducible. Soifwecanshow
thatQ(x)isreducible,wehaveshownthatGal(P)isnot4-transitive.
We have two ways to find a nontrivialfactor of Q(x): the first way is use a factorisation algorithm and
thesecondwayistoproduceacandidatefactorourselves. Analgorithmthatworksverywellforourtype
of polynomial is Van Hoeij’s algorithm, see [10]. One finds that Q(x) is the product of 3 polynomials
of degrees 340, 1020 and 1020 respectively. A more direct way to produce a candidate factorisation is
as follows. The methodfrom[8] givesa bijectionbetweenthe set ofapproximatedcomplexrootsof P′
and the set P1(F ) such thatthe action of Gal(P′) on Ω(P′) correspondsto the action of SL (F ) on
16 2 16
P1(F ), assumingthe outcomeis correct. From theprevioussectionwe knowhowto expressthe roots
16
ofP asrationalexpressionsintherootsofP hencethisgivesusabijectionbetweenΩ(P)andP1(F ),
16
conjecturallycompatiblewiththegroupactionsofGal(P)andSL (F )respectively.Acalculationshows
2 16
thattheactionofSL (F )onthesetofunorderedfour-tuplesofelementsofP1(F )has3orbits,ofsize
2 16 16
340,1020and1020respectively.Usingapproximationstoahighprecisionoftheroots,weusetheseorbits
toproducesub-productsof(3), roundoffthecoefficientstothenearestintegerandverifyafterwardsthat
theobtainedpolynomialsareindeedfactorsofQ(x).
Let us remark that SL (F )⋊Aut(F ) with its natural action on P1(F ) is a transitive permutation
2 16 16 16
groupofdegree17,hencealsoitssubgroupG:=SL (F )⋊hFrob2i. Furthermore,itiswell-knownthat
2 16 2
SL (F )⋊Aut(F )isisomorpictoAut(SL (F ))(whereSL (F )actsbyconjugationandAut(F )
2 16 16 2 16 2 16 16
acts on matrix entries) and actually inside S this group is the normaliser of both SL (F ) and itself.
17 2 16
3
Accordingto the classifiation of transitive permutationgroupsof degree17in [17] these two groupsare
theonlyonesthatliestrictlybetweenSL (F )andA . OncewehavefixedSL (F )insideS ,these
2 16 17 2 16 17
twogroupsareactuallyuniquesubgroupsofS ,notjustuptoconjugacy.
17
Notethattheindex[A :Aut(SL (F ))]ishuge,namely10897286400.IfwecanverifythatGisnota
17 2 16
subgroupofGal(P),thenwearedone. ThefactthattheindexissmallandtheuniquenessofGmakean
algorithmofGeisslerandKlu¨ners,see[9],verysuitabletodecidewhetherG < Gal(P). Itturnsoutthat
thisisnotthecase,henceGal(P)∼=SL (F ).
2 16
4 Does P indeed define ρ ?
f
SonowthatwehaveshownthatGal(P)∼=SL (F )wecanwonderwhetherwecanprovethatP comes
2 16
fromthemodularformf weusedtoconstructitwith. OnceanisomorphismofGal(P)withSL (F )is
2 16
given,thepolynomialP definesarepresentationρ :Gal(Q/Q)→SL (F ). Abovewementionedthat
P 2 16
thatOut(SL (F ))isisomorphictoAut(F )actingonmatrixentries.Hence,uptoanautomorphismof
2 16 16
F ,themapsendingσ ∈Gal(Q/Q)tothecharacteristicpolynomialofρ inF [x]isdeterminedbyP
16 P 16
andinfacttheisomorphismclassofρ iswell-defineduptoanautomorphismofF . Moreconcretely,
P 16
wehavetoshowthatthesplittingfieldofP,whichwewilldenotebyL,isthefixedfieldofKer(ρ ). In
f
thissectionwewillbeusingbasicpropertiesoflocalfieldsascanbefoundin[15].
A continuousrepresentationρ : Gal(Q/Q) → GL (F ) hasa levelN(ρ) anda weight k(ρ). Insteadof
2 ℓ
repeatingthefulldefinitionshere,whicharelengthy(atleastfortheweight)andcanbefoundin[16](see
also[7]foradiscussiononthedefinitionoftheweight),wewilljustsaythattheyaredefinedintermsof
the local representationsρ : Gal(Q /Q ) → GL (F ) obtained from ρ. The level is defined in terms
p p p 2 ℓ
of the representationsρ with p 6= ℓ andthe weightis definedin termsof ρ . Serre states the following
p ℓ
conjecturein[16].
Conjecture1. Letℓbeaprimeandletρ:Gal(Q/Q)→GL (F )beacontinuousoddirreducibleGalois
2 ℓ
representation(arepresentationiscalledoddiftheimageofacomplexconjugationhasdeterminant−1).
Thenthereexistsamodularformf oflevelN(ρ)andweightk(ρ)whichisanormalisedeigenformanda
primeλ|ℓofK suchthatρandρ becomeisomorphicafterasuitableembeddingofF intoF .
f f,λ λ ℓ
Recently,KhareandWintenbergerprovedin[11]thefollowingpartofconjecture1.
Theorem1. Conjecture1holdsineachofthefollowingcases:
• N(ρ)isoddandℓ>2.
• ℓ=2andk(ρ)=2.
Withtheorem1inminditissufficienttoprovethatarepresentationρ = ρ attachedtoP haslevel137
P
andweight2,whicharethelevelandweightofthemodularformf weusedtoconstructitwithandthat
ofalleigenformsinS (Γ (137)),theformf isonewhichgivesrisetoρ . Therefore,intheremainderof
2 1 P
thissectionwewillverifythefollowingproposition.
Proposition2. Letf bethecuspformfromsection2. UptoanautomorphismofF ,therepresentations
16
ρ andρ areisomorphic.Inparticular,therepresentationρ hasSerre-level137andSerre-weight2.
P f,(2) P
4.1 Verification ofthelevel
Thelevelisthe easiest ofthe two to verify. Herewe have to dolocalcomputationsin p-adicfieldswith
p 6= 2. AccordingtothedefinitionofN(ρ)in[16]itsufficestoverifythatρisunramifiedoutside2and
137, tamely ramified at 137 and the local inertia subgroupI at 137 leaves exactly one pointof P1(F )
16
fixed.Thatρ isunramifiedoutside2and137followsimmediatelyfrom(2).
P
4
From(2)andthefactthat1378kDisc(P)itfollowsthatthemonogeneousorderdefinedbyP ismaximal
at137.Modulo137,thepolynomialP factorsas
P =(x+14)(x2+6x+101)2(x2+88x+97)2(x2+106x+112)2(x2+133x+110)2.
Letvbeanyprimeabove137inL.Fromtheabovefactorisationitfollowsthattheprime137decomposes
in K as a product of 5 primes; one of them has its inertial and ramification degree equal to 1 and the
other four ones have their inertial and ramification degrees equal to 2. Thus deg(v) is a power of 2, as
L is obtained by succesively adjoining roots of P and in each step the relative inertial and ramification
degreesoftheprimebelowv arebothatmost2. Inparticular,Gal(L /Q )isasubgroupofSL (F )
v 137 2 16
whose order is a power of 2. Now, { 1 ∗ } is a Sylow 2-subgroup of SL (F ), so Gal(L /Q ) is,
0 1 2 16 v 137
up to conjugacy, a subgroup of { 1 ∗(cid:0)}. H(cid:1)ence I is also conjugate to a subgroup of { 1 ∗ } and it is
0 1 0 1
actually nontrivial because 137 ra(cid:0)mifie(cid:1)s in L (so I of order 2 since the tame inertia gro(cid:0)up o(cid:1)f any finite
Galoisextensionoflocalfieldsiscyclic).
Itisimmediatethatρistamelyramifiedat137asnopowerof2isdivisibleby137. Also,itisclearthat
I has exactly one fixed point in P1(F ) since [ ∗ ] is the only fixed point of any nontrivial element of
16 0
{ 1 ∗ }. ThisestablishestheverificationofN(ρ(cid:0))=(cid:1) 137.
0 1
(cid:0) (cid:1)
4.2 Verification oftheweight
Because the weight is defined in terms of the induced local representation Gal(Q /Q ), we will try to
2 2
computesomerelevantpropertiesofthesplittingfieldL ofP overQ ,wherevisanyplaceofLabove2.
v 2
Inp-adicfieldsonecanonlydocalculationswithacertainprecision,butthisdoesnotgiveanyproblems
since practically all properties one needs to know can be verified rigorously using a bounded precision
calculationandtheerrorboundsinthecalculationscanbekepttrackofexactly.
ThepolynomialP doesnotdefineanorderwhichismaximalattheprime2.Insteadweusethepolynomial
R=x17−11x16+64x15−322x14+916x13+276x12−5380x11+2748x10+6904x9−23320x8
+131500x7−140744x6−16288x5−39752x4−48840x3+102352x2+234466x−1518,
whichistheminimalpolynomialof
36863+22144α+123236α2+154875α3−416913α4+436074α5+229905α6−1698406α7
(cid:0)+1857625α8−467748α9−2289954α10+2838473α11−1565993α12+605054α13−263133α14
+112104α15−22586α16 /8844,
(cid:1)
whereαisarootofP. We canfactorRoverQ andseethatithasonerootinQ whichhappenstobe
2 2
odd,andanEisensteinfactorofdegree16,whichwewillcallE. Thistypeofdecompositioncanberead
offfromtheNewtonpolygonofRanditalsoshowsthattheorderdefinedbyRisindeedmaximalat2.
Fromtheoddnessoftherootand(2)itfollowsthat.
v (Disc(E))=30. (4)
2
For the action of Gal(Q /Q ) on P1(F ) the factorisation means that there is one fixed point and one
2 2 16
orbitofdegree16. Ifwe adjoina rootβ ofE to Q andfactorE overQ (β) thenwe see thatit hasan
2 2
irreduciblefactorofdegree15;in[4]onecanfindmethodsforfactorisationandirreducibilitytestingthat
canbeusedtoverifythis. Thismeansthat[L :Q ]isatleast240.
v 2
AsubgroupofSL (F )thatfixesapointhastobeconjugatetoasubgroupofthegroup
2 16
∗ ∗
H := ⊂SL (F ),
2 16
0 ∗
n(cid:16) (cid:17)o
5
whichisthestabilisersubgroupof[ ∗ ]. From#H =240itfollowsthatGal(L /Q )isisomorphictoH
0 v 2
andfromnowonwewillidentifyth(cid:0)es(cid:1)etwogroupswitheachother. WecanfilterH bynormalsubgroups:
H ⊃I ⊃I ⊃{e},
2
where I is the inertia subgroup and I is the wild ramification subgroup, which is the unique Sylow 2-
2
subgroupofI. WewishtodeterminethegroupsI andI . Letk(v)betheresidueclassfieldofL . The
2 v
groupH/I is isomorphicto Gal(k(v)/F ) andI/I isisomorphicto a subgroupof k(v)∗. In particular
2 2
[I :I ]|(2[H:I]−1)follows. ThegroupH hastheniceproperty
2
1 ∗
[H,H]= ∼=F ,
16
(cid:26)(cid:18)0 1(cid:19)(cid:27)
which is its unique Sylow 2-subgroup. As H/I is abelian, we see that [H,H] ⊂ I. We conclude that
I = [H,H], since above we remarked that I is the unique Sylow 2-subgroup of I. The restriction
2 2
[I :I ]|(2[H:I]−1)leavesonlyonepossibilityforI,namelyI =I .
2 2
Let L′ be the subextension of L /Q fixed by I. Then L′ is the maximal unramified subextension as
v v 2 v
wellasthemaximaltamelyramifiedsubextension. ItisinfactisomorphictoQ215,theuniqueunramified
extension of Q of degree 15 and the Eisenstein polynomial E from above, being irreducible over any
2
unramifiedextensionofQ2,isadefiningpolynomialfortheextensionLv/Q215.Accordingto[14,Thm.3]
wecanrelatethediscriminantofL tok(ρ)asfollows:
v
240· 15 =450 ifk(ρ)=2
v (Disc(L ))= 8
2 v (cid:26) 240· 19 =570 ifk(ρ)6=2
8
Itfollowsfrom(4)thatv (Disc(L /Q ))=30·15=450,soindeedk(ρ)=2.
2 v 2
4.3 Verification oftheform f
NowweknowN(ρ )=137andk(ρ )=2theorem1showsthatthereisaneigenformg ∈S (Γ (137))
P P 2 1
givingrisetoρ . Using[3, Cor. 2.7]weseethatifsuchag exists, thenthereactuallyexistssuchag of
P
trivial Nebentypus, i.e. g ∈ S (Γ (137)) (as SL (F ) is non-solvableρ cannotbe an induced Hecke
2 0 2 16 P
characterfromQ(i)).
AmodularsymbolscalculationshowsthatthereexisttwoGaloisorbitsofnewformsinS (Γ (137)): the
2 0
formf weusedforourcalculationsandanotherform,gsay. Theprime2decomposesinK asaproduct
g
λ3µ,whereλhasinertialdegree1andµhasinertialdegree4.Soitcouldbethatgmodµgivesrisetoρ .
P
Wewillshownowthatfmod(2)andgmodµactuallygivethesamerepresentation. Thecompletionsof
O andO attheprimes(2)andµrespectivelyarebothisomorphictoZ , theunramifiedextension
Kf Kg 16
ofZ ofdegree4. AfterachoiceofembeddingsofO andO intoZ weobtaintwomodularforms
2 Kf Kg 16
f′ andg′ with coefficientsin Z andwe wish to show thata suitable choiceof embeddingsexists such
16
thattheyarecongruentmodulo2. Accordingto[19,Thm.1],itsufficestocheckthereisasuitablechoice
of embeddingsthat gives a (f′) ≡ a (g′)mod2 for all n ≤ [SL (Z) : Γ (137)]/6 = 23 (in [19] this
n n 2 0
theoremisformulatedformodularformswithcoefficientsintheringofintegersofanumberfield,butthe
proofalsoworksforp-adicrings).Usingamodularsymbolscalculation,thiscanbeeasilyverified.
5 Acknowledgements
I would like to thank Ju¨rgen Klu¨ners for proposing this computational challenge and explaining some
computationalGalois theory to me. Furthermore I want to thank Bas Edixhovenfor teaching me about
modularformsandthecalculationoftheircoefficients.AllthecalculationsweredonewithMAGMA(see
[1]), many of them on the MEDICIS cluster (http://medicis.polytechnique.fr). For being
abletomakeuseoftheclusterIwanttothankMarcGiustiandPierreLafon.
6
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