196.TheTheoryofHardy’sZ-Function.ByA. Ivi ´ c 197.InducedRepresentationsofLocallyCompactGroups.ByE.KANIUTHandK.F. Taylor 198.TopicsinCriticalPointTheory.ByK.PERERAandM. Schechter 199.CombinatoricsofMinusculeRepresentations.ByR.M. Green 200.SingularitiesoftheMinimalModelProgram.ByJ. Koll ´ ar 201.CoherenceinThree-DimensionalCategoryTheory.ByN. Gurski 202.CanonicalRamseyTheoryonPolishSpaces.ByV. Kanovei,M. Sabok,andJ. Zapletal 203.APrimerontheDirichletSpace.ByO.EL-Fallah,K. Kellay,J. Mashreghi,and T. Ransford
InChapter4weprovethatcertaindiophantineequationsintwovariables haveatmostfinitelymanysolutions,usingtheauxiliarypolynomialpioneered byRunge.Themethodenablesallsolutionstobefoundinprinciple.Atypical exampleisthatthereareatmostfinitelymanyintegers x, y with x(x 3 2y 3 ) = y.
Or,comingfromCassels’swell-knownresultontheCatalanEquationrecently solvedcompletelybyMih ˘ ailescu,thereareatmostfinitelymanyintegers x, y with x5 y7 = 1provided y isnotdivisibleby5(wedonotproveMih ˘ ailescu’s Theoremhere).Ofcourseequationslike
x 3 2y 3 = m
forfixed m aremorenatural,andthesewillbeconsideredinChapter12.For theproofshere,weneedtoknowthatthelargecomplexsolutionsaregiven byPuiseux(orbetterLaurent)series.Itseemsthatthisisnotsoeasytofind intheliterature,especiallyregardingthecrucialconvergenceproperties,sowe providequiteafewdetails.
TheninChapter5weprovesomeresultssimilartotheclassicalHilbert IrreducibilityTheorem,usuallyabbreviatedtoHIT,byusingthemachineryof theprecedingchapter.TheyarenotsogeneralasHIT,butwhentheydowork, theydelivermoreinformation.TheresultswerefirstfoundbySprindzhukalso usingauxiliarypolynomials,butinamoreelaborateway.Nowadaysthissort ofthingcanbedonewithheightsmachinery,butthatisnotsoelementary.A typicalexample,relatedtothatofthepreviouschapter,isthatthereareatmost finitelymanyintegers y suchthatthepolynomial
X (X 3 2y 3 ) y
in Q[X ] isreducibleovertherationals,andinprinciplethesecanallbefound. AliteralapplicationofHITwouldshowonlythatthereareinfinitelymany rational y suchthatthepolynomialisnotreducible.Sosometimeswegeta StrongHilbertIrreducibilityTheorem;butwerefrainfromabbreviatingthis. Hereweneedresultants;thesecanbefoundalmostanywhere,butbecausewe usethemfrequentlyinthisbookweprovideaself-containedaccount.
InChapter6wejumptoadifferenttopic.Weprovethatthenumber N of pointsmoduloaprime p onanaffineellipticcurvesatisfies
ThereisafamousresultofP ´ olyaonentirefunctionsmappingthenatural numberstotherationalintegers;thismayhaveinfluencedGelfondinhispioneeringworkonthetranscendenceof α β (seeChapter19).P ´ olya’soriginal proofusedinterpolationformulaeandgavethebestpossibleconstant.Much laterWaldschmidtgaveaversionbyauxiliarypolynomials,whichsadlygives aworseconstant.Theproofisneverthelessilluminating;itneedsbinomial coefficientstoavoidfactorials,oneofthekeyideasinThue’sfamousproof (seeChapter12).Moreprecisely,weshowinChapter9thatanentirefunction f with
allin Z mustbeapolynomialif |f (z)| growsoforderatmost C |z| foracertain C > 1.P ´ olyacouldtakeany C < 2;andthestandardexample2z showsthat nothingbetterispossible.Orreformulated:ifanon-polynomialentirefunction f hasthisgrowth,thenatleastoneof f (0), f (1), f (2), ... mustbenon-integral. Gelfond’sstepfromnon-integralitytotranscendenceneededmanymoreideas, allofwhichwillbedevelopedinthisbook.
TherathernaturalgeneralizationtotheGaussianintegers G = Z + Zi with f mapping G intoitselfalsoplayedasimilarhistoricalrole;forexampleit probablydirectlyinspiredGelfond’sproofofthetranscendenceof eπ .Butthe bestpossibleconstantdidnotappearuntilarelativelyrecentpaperofGramain; paradoxicallyenough,hisproofinvolvesanauxiliarypolynomial(orbetteran auxiliaryfunction).Moreprecisely, f itselfmustbeapolynomialif |f (z)| now growsoforderatmost C|z|2 foracertain C > 1.Gelfondconsideredthis problemtoo,andobtainedthenotoriousvalue
C = exp π
(modestlynotmentionedinhisbook).Inthelate1970s,Iobtainedaconstant, extremelydifficulttocompute,whichlaterturnedouttobeabout1.181;and Iconjecturedthatthebestpossibleconstantwasexp( π 2e ) about1.782.This Gramainproved,andsodoweinChapter10.
Hereismaybeoneofthesimplestexamples. Thereisanoldchestnutwhichoftenturnsupinproblem-solvingsessions: givenapolynomial F inavariable X ,canonealwaysmultiplyitbyanon-zero polynomialtogetaproductinvolvingonlypowers X p for p prime?
Forexamplewith F = X 100 + 1wehave X 3 F = X 103 + X 3 .Butwhatabout F = X 100 + X 3 ?Heremultiplyingbysome P = aX d willnotdo.However (X 111 X 14 )F = X 11 (X 100 X 3 )F = X 11 (X 200 X 6 ) = X 211 X 17 .
Atfirstsightitappearstobeadifficultproblemaboutprimes,possiblyin arithmeticprogressions.Sowhatabout F = X 1000 + X 100 + X 3 ?
Letusconsidermultiplying F bysome
forunknown L andundeterminedcoefficients pi (notnecessarilyprimes,but theymightbe).Then PF hasdegreeatmost L + 1000,anditscoefficientsare linearformsinthe pi .Wewouldliketoeliminatetheterms X n for n notprime with0 ≤ n ≤ L + 1000.Thereare
L + 1001 π(L + 1000) homogeneouslinearequationsinthe L + 1unknowns pi .Bylinearalgebrathissystemissolvablenon-trivially,provided L + 1 > L + 1001 π(L + 1000); thatis, π(L + 1000)> 1000.Everyschoolgirlknowsthatthereareinfinitely manyprimenumbers,so π(x) tendstoinfinitywith x andthereexistssuchan L;forexample L = 6927(withMaple).Sotheanswerisyesforthis F ;the troubleofcourseisthatwehavetosolve6927equationsin6928unknownsto get P explicitly.
Thereadermaynowseefirstthatthisworksforany F ,andsecondthatthe primesareirrelevant,inthesensethatwemaydemandonlypowers X m in PF with m inanyprescribedinfiniteset;forexampletheelementsofthesequence 4,27,3125,823543, ... ofall m = pp .
Howcanweeliminate t ?Commonsense,orageneralconsiderationoftranscendencedegree,showsthattheremustbeanalgebraicrelationbetween x, y notinvolving t .Andindeedamoment’sthoughtgives
0.(1.3)
Butwhatabout
= t 3 + t , y = t 4 + t ?
Wecouldsolvethefirstequationbyradicalsfor t ,andthensubstituteintothe secondequation,andfinallysomehowcleartheradicals.Weget
Thereadermayseefirstthatthisworksforanytwopolynomials F , G in t insteadof(1.5),andsecondthatitgeneralizestomorevariables;forexample betweenanythreepolynomialsintwovariablesthereisanon-trivialalgebraicrelation(aswouldfollowmoresimplybyconsiderationoftranscendence degree).
Thisexampleisperhapsmoretypicalofthosetofollowinthesepages.After thesubstitution(1.5)wemayregardthefunction P(x, y) ashavingalargeorder ofvanishingat t = 0;solarge,indeed,thatitmustvanishidentically.
whosesizeisthenumber n ofunknowns.Suchdeterminantsalreadyhave n! terms,sotheirvaluesarelikelytobesomewhatlarger.Thusitwouldbe surprisingiftheintegersinourrelationweresubstantiallylessthanthefactorial 10491121!.Anditcanbeseenthatsomeoftheentriesofthedeterminantsare almostaslargeas36476 (asin Exercise1.17).Thismeansthatwecouldexpect somecoefficientsin P tohavethirtythousandmillion(Americanthirtybillion, SwissdreissigMilliarden)decimaldigits(seehowever Exercise8.13).Soitis doubtfulif P couldeverbeexpressedexplicitly.
Exercises
1.1 Showthatthereis P = 0in C[X ] suchthat P(X )(X 1000 + X 100 + X 3 ) has theform N n=0 an X n2 .
1.2 Let F bein C[X ] withdegreeatmost D.Showthatthereis P = 0in C[X ] withdegreeatmost D2 D suchthat PF hastheform N n=0 an X n2
1.3 Let F bein C[X ].Find P = 0in C[X ] suchthat PF hastheform N n=0 an X 2n .
1.4 Let F bein C[t ] withdegreeatmost D ≥ 1andlet G bein C[t ] with degreeatmost E ≥ 1.Showthatthereis P = 0in C[X , Y ] withdegreeatmost D + E 1ineachvariablesuchthat P(F , G) = 0.
1.5 Let F bein C[t ] withdegreeatmost D ≥ 1andlet G bein C[t ] with degreeatmost E ≥ 1.Showthatthereis P = 0in C[X , Y ] withdegreeatmost E in X anddegreeatmost DE E + 1in Y suchthat P(F , G) = 0.
1.6 Let t , u beindependentvariables,andlet F , G, H bein C[t , u].Showthat thereis P = 0in C[X , Y , Z ] suchthat P(F , G, H ) = 0.
1.7 Showthatthereisanabsoluteconstant c (thatis,notdependingonany parameters)withthefollowingproperty.Let F bein C[X ] withdegreeatmost D ≥ 2.Thenthereis P = 0in C[X ] withdegreeatmost cD log D suchthat PF hastheform p ap X p ,where p runsoverthesetofprimes.
1.8 Let F bein C[X ].Find P = 0in C[X ] suchthat PF hastheform N n=0 an X 3n .
1.9 Let F bein Fp [X ].Showthatthereis P = 0in Fp [X ] suchthat G = PF satisfies G(X1 + X2 ) = G(X1 ) + G(X2 ) in Fp [X1 , X2 ]
1.10 Let F bein C[t ] withdegreeatmost D ≥ 1andlet G bein C[t ] with degreeatmost E ≥ 1.Showthatthereis P = 0in C[X , Y ] withdegreeatmost E in X anddegreeatmost D in Y suchthat P(F , G) = 0[Hint:resultants].
1.11 Let P = 0in C[X , Y ] besuchthat P(t1948 + t 666 + 1, t 1291 + t 163 + t ) = 0.Showthat P hasdegreeatleast1291in X anddegreeatleast1948in Y (compare Exercise5.7).
1.12 Canoneessentiallyimprovethe D 2 D in Exercise1.2?Idon’tknow.
1.13 Let F , G bein C(t ) (rationalfunctions).Showthatthereis P = 0in C[X , Y ] suchthat P(F , G) = 0.
1.14 Let F = 256 (t 2 t + 1)3 t 2 (t 1)2 , G = 256 (t 2 + t + 1)3 t 2 (t + 1)2 .
Showthat P(F , G) = 0for P = X 3 Y 2X 2 Y 2 + XY 3 1728(X 3 + Y 3 ) + 1216(X 2 Y + XY 2 ) +3538944(X 2 + Y 2 ) 2752512XY 2415919104(X + Y ) + 549755813888.
1.15 Let F = tu(t 10 + 11t 5 u 5 u 10 ), G =−t 20 u 20 + 228(t 15 u 5 t 5 u 15 ) 494t 10 u 10 , H = t 30 + u 30 + 522(t 25 u 5 t 5 u 25 ) 10005(t 20 u 10 + t 10 u 20 )
Showthat G3 + H 2 = 1728F 5 .(Thisisrelatedtotheicosahedron–seeKlein, 1956.)
1.16 Let F = 1728 u3 u3 v2 , G =−1728 u2 v u3 v2 , H =−288 u(tuv 3u3 4v2 ) u3 v2 , K =−24 3t 2 u2 v 18tu4 24tuv2 + 95u3 v + 16v3 u3 v2 .
Itisprovedinanyelementarytextonnumbertheorythat e isirrational; theproofisbasedontherapidconvergenceoftheseriestogetherwiththe reasonablebehaviourofthedenominators.Wegiveaproofnevertheless.
Considerthetruncation
for n = 0,1,2, .Thefirsttermontheextremeright-handsidedominates andindeed
Thus0 < fn < 2/(n + 1)! and
Nowif s isadenominatorfortherational e,thenmultiplyingby s and making n tendtoinfinitygivesacontradictiontotheso-calledFundamental TheoremofTranscendencethateverynon-zerointegerhasabsolutevalueat least1.
Theproofisslightlyeasierforthealternatingseries
becausewenolongerneedthedominance.
Theproofsextendtogivethelinearindependenceover Q of1, e, e 1 ,which amountstothefactthat e cannotbequadraticover Q.Inparticular e2 is irrational.Butthereisaminorsnag.Weassumethat r + se + te 1 = 0for integers r , s, t notallzero,andthenfor
weget
Hencethe n!fn areintegerstendingtozeroas n tendstoinfinity.Butweno longerknowthattheseintegersarenon-zeroasin(2.1).
Infactitisnottoohardtoshowthat
isimpossibleforany n.Namely,
andso
Thus(2.2)wouldimply s = t = 0so r = 0too,acontradiction.
(1 t )n dt (see Exercise2.14).It canalsobedonewithauxiliarypolynomialsofthetypementionedinChapter 1,mostefficientlybyintroducingderivativesasinthedifferentialequation (ez ) = ez .Howeversomeextraarithmeticandanalyticmachineryisneeded. Thiswillbeintroducedstepbystepinthefollowingchapters.Bythesemeans wewillshowinChapter13that
Theproofworksiftheestimatetendstozeroas n tendstoinfinity(assuming wecanruleoutthesnag fn = 0asin(2.2)above).Unfortunatelythisisthe caseonlyfor a = 1.
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