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A 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 Itmustbe 1 notedthattheexistenceofsuchapolynomialwasalreadyknowntoMestre(unpublished).Inthispaperwe 0 willgiveanexplicitexample. 7 0 Proposition1.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 References [1] W.Bosma,J.J.Cannon,C.E.Playoust,ThemagmaalgebrasystemI:theuserlanguage,J.Symbolic Comput.24(1997),no.3/4,235–265. [2] J.A. Buchmann and H.W. 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