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ABOOKOF ABSTRACTALGEBRA SecondEdition CharlesC.Pinter ProfessorofMathematics BucknellUniversity DoverPublications,Inc.,Mineola,NewYork Copyright Copyright©1982,1990byCharlesC.Pinter Allrightsreserved. BibliographicalNote ThisDoveredition,firstpublishedin2010,isanunabridgedrepublicationofthe1990secondeditionoftheworkoriginallypublishedin1982 bytheMcGraw-HillPublishingCompany,Inc.,NewYork. LibraryofCongressCataloging-in-PublicationData Pinter,CharlesC,1932– Abookofabstractalgebra/CharlesC.Pinter.—Dovered. p.cm. Originallypublished:2nded.NewYork:McGraw-Hill,1990. Includesbibliographicalreferencesandindex. ISBN-13:978-0-486-47417-5 ISBN-10:0-486-47417-8 1.Algebra,Abstract.I.Title. QA162.P562010 512′.02—dc22 2009026228 ManufacturedintheUnitedStatesbyCourierCorporation 47417803 www.doverpublications.com Tomywife,Donna, andmysons, Nicholas,Marco, Andrés,andAdrian CONTENTS* Preface Chapter 1 WhyAbstractAlgebra? HistoryofAlgebra.NewAlgebras.AlgebraicStructures.AxiomsandAxiomaticAlgebra. AbstractioninAlgebra. Chapter 2 Operations OperationsonaSet.PropertiesofOperations. Chapter 3 TheDefinitionofGroups Groups.ExamplesofInfiniteandFiniteGroups.ExamplesofAbelianandNonabelian Groups.GroupTables. TheoryofCoding:Maximum-LikelihoodDecoding. Chapter 4 ElementaryPropertiesofGroups UniquenessofIdentityandInverses.PropertiesofInverses. DirectProductofGroups. Chapter 5 Subgroups DefinitionofSubgroup.GeneratorsandDefiningRelations. CayleyDiagrams.CenterofaGroup.GroupCodes;HammingCode. Chapter 6 Functions Injective,Surjective,BijectiveFunction.CompositeandInverseofFunctions. Finite-StateMachines.AutomataandTheirSemigroups. Chapter 7 GroupsofPermutations SymmetricGroups.DihedralGroups. AnApplicationofGroupstoAnthropology. Chapter 8 PermutationsofaFiniteSet DecompositionofPermutationsintoCycles.Transpositions.EvenandOddPermutations. AlternatingGroups. Chapter 9 Isomorphism TheConceptofIsomorphisminMathematics.IsomorphicandNonisomorphicGroups. Cayley’sTheorem. GroupAutomorphisms. Chapter 10 OrderofGroupElements Powers/MultiplesofGroupElements.LawsofExponents.PropertiesoftheOrderofGroup Elements. Chapter 11 CyclicGroups FiniteandInfiniteCyclicGroups.IsomorphismofCyclicGroups.SubgroupsofCyclic Groups. Chapter 12 PartitionsandEquivalenceRelations Chapter 13 CountingCosets Lagrange’sTheoremandElementaryConsequences. SurveyofGroupsofOrder≤10. NumberofConjugateElements.GroupActingonaSet. Chapter 14 Homomorphisms ElementaryPropertiesofHomomorphisms.NormalSubgroups.KernelandRange. InnerDirectProducts.ConjugateSubgroups. Chapter 15 QuotientGroups QuotientGroupConstruction.ExamplesandApplications. TheClassEquation.InductionontheOrderofaGroup. Chapter 16 TheFundamentalHomomorphismTheorem FundamentalHomomorphismTheoremandSomeConsequences. TheIsomorphismTheorems.TheCorrespondenceTheorem.Cauchy’sTheorem.Sylow Subgroups.Sylow’sTheorem.DecompositionTheoremforFiniteAbelianGroups. Chapter 17 Rings:DefinitionsandElementaryProperties CommutativeRings.Unity.InvertiblesandZero-Divisors.IntegralDomain.Field. Chapter 18 IdealsandHomomorphisms Chapter 19 QuotientRings ConstructionofQuotientRings.Examples.FundamentalHomomorphismTheoremand SomeConsequences.PropertiesofPrimeandMaximalIdeals. Chapter 20 IntegralDomains CharacteristicofanIntegralDomain.PropertiesoftheCharacteristic.FiniteFields. ConstructionoftheFieldofQuotients. Chapter 21 TheIntegers OrderedIntegralDomains.Well-ordering.Characterizationof UptoIsomorphism. MathematicalInduction.DivisionAlgorithm. Chapter 22 FactoringintoPrimes Idealsof .PropertiesoftheGCD.RelativelyPrimeIntegers.Primes.Euclid’sLemma. UniqueFactorization. Chapter 23 ElementsofNumberTheory(Optional) PropertiesofCongruence.TheoremsofFermâtandEuler.SolutionsofLinearCongruences. ChineseRemainderTheorem. Wilson’sTheoremandConsequences.QuadraticResidues.TheLegendreSymbol. PrimitiveRoots. Chapter 24 RingsofPolynomials MotivationandDefinitions.DomainofPolynomialsoveraField.DivisionAlgorithm. PolynomialsinSeveralVariables.FieldsofPolynomialQuotients. Chapter 25 FactoringPolynomials IdealsofF[x].PropertiesoftheGCD.IrreduciblePolynomials.Uniquefactorization. EuclideanAlgorithm. Chapter 26 SubstitutioninPolynomials RootsandFactors.PolynomialFunctions.Polynomialsover .Eisenstein’sIrreducibility Criterion.PolynomialsovertheReals.PolynomialInterpolation. Chapter 27 ExtensionsofFields AlgebraicandTranscendentalElements.TheMinimumPolynomial.BasicTheoremon FieldExtensions. Chapter 28 VectorSpaces ElementaryPropertiesofVectorSpaces.LinearIndependence.Basis.Dimension.Linear Transformations. Chapter 29 DegreesofFieldExtensions SimpleandIteratedExtensions.DegreeofanIteratedExtension. FieldsofAlgebraicElements.AlgebraicNumbers.AlgebraicClosure. Chapter 30 RulerandCompass ConstructiblePointsandNumbers.ImpossibleConstructions. ConstructibleAnglesandPolygons. Chapter 31 GaloisTheory:Preamble MultipleRoots.RootField.ExtensionofaField.Isomorphism. RootsofUnity.SeparablePolynomials.NormalExtensions. Chapter 32 GaloisTheory:TheHeartoftheMatter FieldAutomorphisms.TheGaloisGroup.TheGaloisCorrespondence.Fundamental TheoremofGaloisTheory. ComputingGaloisGroups. Chapter 33 SolvingEquationsbyRadicals RadicalExtensions.AbelianExtensions.SolvableGroups.InsolvabilityoftheQuintic. AppendixA ReviewofSetTheory AppendixB ReviewoftheIntegers AppendixC ReviewofMathematicalInduction AnswerstoSelectedExercises Index *Italicheadingsindicatetopicsdiscussedintheexercisesections. PREFACE Once,whenIwasastudentstrugglingtounderstandmodernalgebra,Iwastoldtoviewthissubjectasan intellectualchessgame,withconventionalmovesandprescribedrulesofplay.Iwasillservedbythisbit ofextemporaneousadvice,andvowednevertoperpetuatethefalsehoodthatmathematicsispurely—or primarily—aformalism.Mypledgehasstronglyinfluencedtheshapeandstyleofthisbook. While giving due emphasis to the deductive aspect of modern algebra, I have endeavored here to present modern algebra as a lively branch of mathematics, having considerable imaginative appeal and restingonsomefirm,clear,andfamiliarintuitions.Ihavedevotedagreatdealofattentiontobringingout themeaningfulnessofalgebraicconcepts,bytracingtheseconceptstotheiroriginsinclassicalalgebra and at the same time exploring their connections with other parts of mathematics, especially geometry, numbertheory,andaspectsofcomputationandequationsolving. InanintroductorychapterentitledWhyAbstractAlgebra?,aswellasinnumeroushistoricalasides, concepts of abstract algebra are traced to the historic context in which they arose. I have attempted to showthattheyarosewithoutartifice,asanaturalresponsetoparticularneeds,inthecourseofanatural processofevolution.Furthermore,Ihaveendeavoredtobringtolight,explicitly,theintuitivecontentof thealgebraicconceptsusedinthisbook.Conceptsaremoremeaningfultostudentswhenthestudentsare able to represent those concepts in their minds by clear and familiar mental images. Accordingly, the processofconcreteconcept-formationisdevelopedwithcarethroughoutthisbook. Ihavedeliberatelyavoidedarigidconventionalformat,withitssuccessionofdefinition, theorem, proof,corollary,example.Inmyexperience,thatkindofformatencouragessomestudentstobelievethat mathematical concepts have a merely conventional character, and may encourage rote memorization. Instead, each chapter has the form of a discussion with the student, with the accent on explaining and motivating. Inanefforttoavoidfragmentationofthesubjectmatterintolooselyrelateddefinitionsandresults, eachchapterisbuiltaroundacentralthemeandremainsanchoredtothisfocalpoint.Inthelaterchapters especially,thisfocalpointisaspecificapplicationoruse.Detailsofeverytopicarethenwovenintothe generaldiscussion,soastokeepanaturalflowofideasrunningthrougheachchapter. The arrangement of topics is designed to avoid tedious proofs and long-winded explanations. Routine arguments are worked into the discussion whenever this seems natural and appropriate, and proofstotheoremsareseldommorethanafewlineslong.(Thereare,ofcourse,afewexceptionstothis.) Elementary background material is filled in as it is needed. For example, a brief chapter on functions precedes the discussion of permutation groups, and a chapter on equivalence relations and partitions pavesthewayforLagrange’stheorem. This book addresses itself especially to the average student, to enable him or her to learn and understand as much algebra as possible. In scope and subject-matter coverage, it is no different from manyotherstandardtexts.Itbeginswiththepromiseofdemonstratingtheunsolvabilityofthequinticand endswiththatpromisefulfilled.Standardtopicsarediscussedintheirusualorder,andmanyadvanced and peripheral subjects are introduced in the exercises, accompanied by ample instruction and commentary. Ihaveincludedacopioussupplyofexercises—probablymoreexercisesthaninotherbooksatthis level. They are designed to offer a wide range of experiences to students at different levels of ability. Thereissomenoveltyinthewaytheexercisesareorganized:attheendofeachchapter,theexercisesare groupedintoexercisesets,eachsetcontainingaboutsixtoeightexercisesandheadedbyadescriptive title.Eachsettouchesuponanideaorskillcoveredinthechapter. Thefirstfewexercisesetsineachchaptercontainproblemswhichareessentiallycomputationalor manipulative.Then,therearetwoorthreesetsofsimpleproof-typequestions,whichrequiremainlythe ability to put together definitions and results with understanding of their meaning. After that, I have endeavoredtomaketheexercisesmoreinterestingbyarrangingthemsothatineachsetanewresultis proved,ornewlightisshedonthesubjectofthechapter. As a rule, all the exercises have the same weight: very simple exercises are grouped together as partsofasingleproblem,andconversely,problemswhichrequireacomplexargumentarebrokeninto severalsubproblemswhichthestudentmaytackleinturn.Ihaveselectedmainlyproblemswhichhave intrinsic relevance, and are not merely drill, on the premise that this is much more satisfying to the student. CHANGESINTHESECONDEDITION During the seven years that have elapsed since publication of the first edition of A Book of Abstract Algebra,Ihavereceivedlettersfrommanyreaderswithcommentsandsuggestions.Moreover,anumber ofreviewershavegoneoverthetextwiththeaimoffindingwaystoincreaseitseffectivenessandappeal asateachingtool.Inpreparingthesecondedition,Ihavetakenaccountofthemanysuggestionsthatwere made,andofmyownexperiencewiththebookinmyclasses. Inadditiontonumeroussmallchangesthatshouldmakethebookeasiertoread,thefollowingmajor changesshouldbenoted: EXERCISES Many of the exercises have been refined or reworded—and a few of the exercise sets reorganized—in order to enhance their clarity or, in some cases, to make them more mathematically interesting.Inaddition,severalnewexericsesetshavebeenincludedwhichtouchuponapplicationsof algebraandarediscussednext: APPLICATIONSThequestionofincluding“applications”ofabstractalgebrainanundergraduatecourse (especiallyaone-semestercourse)isatouchyone.Eitheronerunstheriskofmakingavisiblyweakcase for the applicability of the notions of abstract algebra, or on the other hand—by including substantive applications—onemayenduphavingtoomitalotofimportantalgebra.IhaveadoptedwhatIbelieveis areasonablecompromisebyaddinganelementarydiscussionofafewapplicationareas(chieflyaspects ofcodingandautomatatheory)onlyintheexercisesections,inconnectionwithspecificexercise.These exercisesmaybeeitherstressed,de-emphasized,oromittedaltogether. PRELIMINARIESItmaywellbearguedthat,inordertoguaranteethesmootheflowandcontinuityofa course in abstract algebra, the course should begin with a review of such preliminaries as set theory, inductionandthepropertiesofintegers.Inordertoprovidematerialforteacherswhoprefertostartthe courseinthisfashion,IhaveaddedanAppendixwiththreebriefchaptersonSets,IntegersandInduction, respectively,eachwithitsownsetofexercises. SOLUTIONSTOSELECTEDEXERCISESAfewexercisesineachchapteraremarkedwiththesymbol #.Thisindicatesthatapartialsolution,orsometimesmerelyadecisivehint,aregivenattheendofthe bookinthesectiontitledSolutionstoSelectedExercises. ACKNOWLEDGMENTS Iwouldliketoexpressmythanksforthemanyusefulcommentsandsuggestionsprovidedbycolleagues whoreviewedthistextduringthecourseofthisrevision,especiallytoJ.RichardByrne,PortlandState University: D. R. LaTorre, Clemson University; Kazem Mahdavi, State University College at Potsdam; ThomasN.Roe,SouthDakotaStateUniversity;andArmondE.Spencer,StateUniversityofNewYorkPotsdam.Inparticular,IwouldliketothankRobertWeinstein,mathematicseditoratMcGraw-Hillduring the preparation of the second edition of this book. I am indebted to him for his guidance, insight, and steadyencouragement. CharlesC.Pinter CHAPTER ONE WHYABSTRACTALGEBRA? Whenweopenatextbookofabstractalgebraforthefirsttimeandperusethetableofcontents,weare struckbytheunfamiliarityofalmosteverytopicweseelisted.Algebraisasubjectweknowwell,but hereitlookssurprisinglydifferent.Whatarethesedifferences,andhowfundamentalarethey? First,thereisamajordifferenceinemphasis.Inelementaryalgebrawelearnedthebasicsymbolism andmethodologyofalgebra;wecametoseehowproblemsoftherealworldcanbereducedtosetsof equations and how these equations can be solved to yield numerical answers. This technique for translating complicated problems into symbols is the basis for all further work in mathematics and the exactsciences,andisoneofthetriumphsofthehumanmind.However,algebraisnotonlyatechnique,it isalsoabranchoflearning,adiscipline,likecalculusorphysicsorchemistry.Itisacoherentandunified bodyofknowledgewhichmaybestudiedsystematically,startingfromfirstprinciplesandbuildingup.So the first difference between the elementary and the more advanced course in algebra is that, whereas earlierweconcentratedontechnique,wewillnowdevelopthatbranchofmathematicscalledalgebrain asystematicway.Ideasandgeneralprincipleswilltakeprecedenceoverproblemsolving.(Bytheway, this does not mean that modern algebra has no applications—quite the opposite is true, as we will see soon.) Algebraatthemoreadvancedlevelisoftendescribedasmodernorabstractalgebra.Infact,both ofthesedescriptionsarepartlymisleading.Someofthegreatdiscoveriesintheupperreachesofpresentdayalgebra(forexample,theso-calledGaloistheory)wereknownmanyyearsbeforetheAmericanCivil War;andthebroadaimsofalgebratodaywereclearlystatedbyLeibnizintheseventeenthcentury.Thus, “modern”algebraisnotsoverymodern,afterall!Towhatextentisitabstract?Well,abstractionisall relative; one person’s abstraction is another person’s bread and butter. The abstract tendency in mathematicsisalittlelikethesituationofchangingmoralcodes,orchangingtastesinmusic:Whatshocks onegenerationbecomesthenorminthenext.Thishasbeentruethroughoutthehistoryofmathematics. Forexample,1000yearsagonegativenumberswereconsideredtobeanoutrageousidea.Afterall, itwassaid,numbersareforcounting:wemayhaveoneorange,ortwooranges,ornoorangesatall;but how can we have minus an orange? The logisticians, or professional calculators, of those days used negativenumbersasanaidintheircomputations;theyconsideredthesenumberstobeausefulfiction,for ifyoubelieveinthemtheneverylinearequationax+b=0hasasolution(namelyx=−b/a,provideda≠ 0).EventhegreatDiophantusoncedescribedthesolutionof4x+6=2asanabsurdnumber.Theideaof a system of numeration which included negative numbers was far too abstract for many of the learned headsofthetenthcentury! The history of the complex numbers (numbers which involve ) is very much the same. For hundredsofyears,mathematiciansrefusedtoacceptthembecausetheycouldn’tfindconcreteexamplesor applications.(Theyarenowabasictoolofphysics.) Settheorywasconsideredtobehighlyabstractafewyearsago,andsowereothercommonplacesof today. Many of the abstractions of modern algebra are already being used by scientists, engineers, and computerspecialistsintheireverydaywork.Theywillsoonbecommonfare,respectably“concrete,”and bythentherewillbenew“abstractions.” Laterinthischapterwewilltakeacloserlookattheparticularbrandofabstractionusedinalgebra. Wewillconsiderhowitcameaboutandwhyitisuseful. Algebrahasevolvedconsiderably,especiallyduringthepast100years.Itsgrowthhasbeenclosely linked with the development of other branches of mathematics, and it has been deeply influenced by philosophicalideasonthenatureofmathematicsandtheroleoflogic.Tohelpusunderstandthenature andspiritofmodernalgebra,weshouldtakeabrieflookatitsorigins. ORIGINS Theorderinwhichsubjectsfolloweachotherinourmathematicaleducationtendstorepeatthehistorical stages in the evolution of mathematics. In this scheme, elementary algebra corresponds to the great classicalageofalgebra,whichspansabout300yearsfromthesixteenththroughtheeighteenthcenturies. It was during these years that the art of solving equations became highly developed and modern symbolismwasinvented. The word “algebra”—al jebr in Arabic—was first used by Mohammed of Kharizm, who taught mathematicsinBaghdadduringtheninthcentury.Thewordmayberoughlytranslatedas“reunion,”and describeshismethodforcollectingthetermsofanequationinordertosolveit.Itisanamusingfactthat the word “algebra” was first used in Europe in quite another context. In Spain barbers were called algebristas,orbonesetters(theyreunitedbrokenbones),becausemedievalbarbersdidbonesettingand bloodlettingasasidelinetotheirusualbusiness. Theoriginofthewordclearlyreflectstheactualcontextofalgebraatthattime,foritwasmainly concerned with ways of solving equations. In fact, Omar Khayyam, who is best remembered for his brilliantversesonwine,song,love,andfriendshipwhicharecollectedintheRubaiyat—but who was alsoagreatmathematician—explicitlydefinedalgebraasthescienceofsolvingequations. Thus, as we enter upon the threshold of the classical age of algebra, its central theme is clearly identifiedasthatofsolvingequations.Methodsofsolvingthelinearequationax+b=0andthequadratic ax2+bx+c=0werewellknownevenbeforetheGreeks.Butnobodyhadyetfoundageneralsolution forcubicequations x3+ax2+bx=c orquartic(fourth-degree)equations x4+ax3+bx2+cx=d Thisgreataccomplishmentwasthetriumphofsixteenthcenturyalgebra. The setting is Italy and the time is the Renaissance—an age of high adventure and brilliant achievement,whenthewideworldwasreawakeningafterthelongausterityoftheMiddleAges.America hadjustbeendiscovered,classicalknowledgehadbeenbroughttolight,andprosperityhadreturnedto thegreatcitiesofEurope.Itwasaheadyagewhennothingseemedimpossibleandeventheoldbarriers of birth and rank could be overcome. Courageous individuals set out for great adventures in the far cornersoftheearth,whileothers,nowconfidentonceagainofthepowerofthehumanmind,wereboldly exploringthelimitsofknowledgeinthesciencesandthearts.Theidealwastobeboldandmany-faceted, to“knowsomethingofeverything,andeverythingofatleastonething.”Thegreattraderswerepatronsof thearts,thefinestmindsinsciencewereadeptsatpoliticalintrigueandhighfinance.Thestudyofalgebra wasreborninthislivelymilieu. Thosemenwhobroughtalgebratoahighlevelofperfectionatthebeginningofitsclassicalage—all typical products of the Italian Renaissanee —were as colorful and extraordinary a lot as have ever appeared in a chapter of history. Arrogant and unscrupulous, brilliant, flamboyant, swaggering, and remarkable,theylivedtheirlivesastheydidtheirwork:withstyleandpanache,inbrilliantdashesand inspiredleapsoftheimagination. The spirit of scholarship was not exactly as it is today. These men, instead of publishing their discoveries, kept them as well-guarded secrets to be used against each other in problem-solving competitions. Such contests were a popular attraction: heavy bets were made on the rival parties, and theirreputations(aswellasasubstantialpurse)dependedontheoutcome. One of the most remarkable of these men was Girolamo Cardan. Cardan was born in 1501 as the illegitimatesonofafamousjuristofthecityofPavia.Amanofpassionatecontrasts,hewasdestinedto becomefamousasaphysician,astrologer,andmathematician—andnotoriousasacompulsivegambler, scoundrel,andheretic.Afterhegraduatedinmedicine,hiseffortstobuildupamedicalpracticewereso unsuccessfulthatheandhiswifewereforcedtoseekrefugeinthepoorhouse.Withthehelpoffriendshe became a lecturer in mathematics, and, after he cured the child of a senator from Milan, his medical careeralsopickedup.Hewasfinallyadmittedtothecollegeofphysiciansandsoonbecameitsrector.A brilliantdoctor,hegavethefirstclinicaldescriptionoftyphusfever,andashisfamespreadhebecame thepersonalphysicianofmanyofthehighandmightyofhisday. Cardan’searlyinterestinmathematicswasnotwithoutapracticalside.Asaninveterategamblerhe was fascinated by what he recognized to be the laws of chance. He wrote a gamblers’ manual entitled Book on Games of Chance, which presents the first systematic computations of probabilities. He also needed mathematics as a tool in casting horoscopes, for his fame as an astrologer was great and his predictionswerehighlyregardedandsoughtafter.Hismostimportantachievementwasthepublicationof a book called Ars Magna (The Great Art), in which he presented systematically all the algebraic knowledgeofhistime.However,asalreadystated,muchofthisknowledgewasthepersonalsecretofits practitioners, and had to be wheedled out of them by cunning and deceit. The most important accomplishment of the day, the general solution of the cubic equation which had been discovered by Tartaglia,wasobtainedinthatfashion. Tartaglia’slifewasasturbulentasanyinthosedays.BornwiththenameofNiccoloFontanaabout 1500,hewaspresentattheoccupationofBresciabytheFrenchin1512.Heandhisfatherfledwithmany othersintoacathedralforsanctuary,butintheheatofbattlethesoldiersmassacredthehaplesscitizens eveninthatholyplace.Thefatherwaskilled,andtheboy,withasplitskullandadeepsabercutacross hisjawsandpalate,wasleftfordead.Atnighthismotherstoleintothecathedralandmanagedtocarry himoff;miraculouslyhesurvived.Thehorrorofwhathehadwitnessedcausedhimtostammerforthe restofhislife,earninghimthenicknameTartaglia,“thestammerer,”whichheeventuallyadopted. Tartaglia received no formal schooling, for that was a privilege of rank and wealth. However, he taught himself mathematics and became one of the most gifted mathematicians of his day. He translated Euclid and Archimedes and may be said to have originated the science of ballistics, for he wrote a treatiseongunnerywhichwasapioneeringeffortonthelawsoffallingbodies. In1535Tartagliafoundawayofsolvinganycubicequationoftheformx3+ax2=b(thatis,without an x term). When be announced his accomplishment (without giving any details, of course), he was challenged to an algebra contest by a certain Antonio Fior, a pupil of the celebrated professor of mathematicsScipiodelFerro.Scipiohadalreadyfoundamethodforsolvinganycubicequationofthe formx3+ax=b(thatis,withoutanx2term),andhadconfidedhissecrettohispupilFior.Itwasagreed thateachcontestantwastodrawup30problemsandhandthelisttohisopponent.Whoeversolvedthe greaternumberofproblemswouldreceiveasumofmoneydepositedwithalawyer.Afewdaysbefore thecontest,Tartagliafoundawayofextendinghismethodsoastosolveanycubicequation.Inlessthan2 hours he solved all his opponent’s problems, while his opponent failed to solve even one of those proposedbyTartaglia. For some time Tartaglia kept his method for solving cubic equations to himself, but in the end he succumbedtoCardan’saccomplishedpowersofpersuasion.InfluencedbyCardan’spromisetohelphim becomeartilleryadvisertotheSpanisharmy,herevealedthedetailsofhismethodtoCardanunderthe promiseofstrictsecrecy.Afewyearslater,toTartaglia’sunbelievingamazementandindignation,Cardan published Tartaglia’s method in his book Ars Magna. Even though he gave Tartaglia full credit as the originatorofthemethod,therecanbenodoubtthathebrokehissolemnpromise.Abitterdisputearose betweenthemathematicians,fromwhichTartagliawasperhapsluckytoescapealive.Helosthisposition aspubliclectureratBrescia,andlivedouthisremainingyearsinobscurity. Thenextgreatstepintheprogressofalgebrawasmadebyanothermemberofthesamecircle.Itwas Ludovico Ferrari who discovered the general method for solving quartic equations—equations of the form x4+ax3+bx2+cx=d Ferrari was Cardan’s personal servant. As a boy in Cardan’s service he learned Latin, Greek, and mathematics.HewonfameafterdefeatingTartagliainacontestin1548,andreceivedanappointmentas supervisoroftaxassessmentsinMantua.Thispositionbroughthimwealthandinfluence,buthewasnot abletodominatehisownviolentdisposition.HequarreledwiththeregentofMantua,losthisposition, anddiedattheageof43.Traditionhasitthathewaspoisonedbyhissister. As for Cardan, after a long career of brilliant and unscrupulous achievement, his luck finally abandoned him. Cardan’s son poisoned his unfaithful wife and was executed in 1560. Ten years later, Cardan was arrested for heresy because he published a horoscope of Christ’s life. He spent several monthsinjailandwasreleasedafterrenouncinghisheresyprivately,butlosthisuniversitypositionand the right to publish books. He was left with a small pension which had been granted to him, for some unaccountablereason,bythePope. Asthiscolorfultimedrawstoaclose,algebraemergesasamajorbranchofmathematics.Itbecame clear that methods can be found to solve many different types of equations. In particular, formulas had been discovered which yielded the roots of all cubic and quartic equations. Now the challenge was clearlyouttotakethenextstep,namely,tofindaformulafortherootsofequationsofdegree5orhigher (inotherwords,equationswithanx5term,oranx6term,orhigher).Duringthenext200years,therewas hardlyamathematicianofdistinctionwhodidnottrytosolvethisproblem,butnonesucceeded.Progress was made in new parts of algebra, and algebra was linked to geometry with the invention of analytic geometry.Buttheproblemofsolvingequationsofdegreehigherthan4remainedunsettled.Itwas,inthe expressionofLagrange,“achallengetothehumanmind.” Itwasthereforeagreatsurprisetoallmathematicianswhenin1824theworkofayoungNorwegian prodigynamedNielsAbelcametolight.Inhiswork,Abelshowedthattheredoesnotexistanyformula (intheconventionalsensewehaveinmind)fortherootsofanalgebraicequationwhosedegreeis5or greater. This sensational discovery brings to a close what is called the classical age of algebra. Throughoutthisagealgebrawasconceivedessentiallyasthescienceofsolvingequations,andnowthe outerlimitsofthisquesthadapparentlybeenreached.Intheyearsahead,algebrawastostrikeoutinnew directions. THEMODERNAGE About the time Niels Abel made his remarkable discovery, several mathematicians, working independentlyindifferentpartsofEurope,beganraisingquestionsaboutalgebrawhichhadneverbeen considered before. Their researches in different branches of mathematics had led them to investigate “algebras”ofaveryunconventionalkind—andinconnectionwiththesealgebrastheyhadtofindanswers toquestionswhichhadnothingtodowithsolvingequations.Theirworkhadimportantapplications,and wassoontocompelmathematicianstogreatlyenlargetheirconceptionofwhatalgebraisabout. The new varieties of algebra arose as a perfectly natural development in connection with the applicationofmathematicstopracticalproblems.Thisiscertainlytruefortheexampleweareaboutto lookatfirst. TheAlgebraofMatrices Amatrixisarectangulararrayofnumberssuchas Such arrays come up naturally in many situations, for example, in the solution of simultaneous linear equations.Theabovematrix,forinstance,isthematrixofcoefficientsofthepairofequations Sincethesolutionofthispairofequationsdependsonlyonthecoefficients,wemaysolveitbyworking onthematrixofcoefficientsaloneandignoringeverythingelse. Wemayconsidertheentriesofamatrixtobearrangedinrowsandcolumns;theabovematrixhas tworowswhichare (211−3)and(90.54) andthreecolumnswhichare Itisa2×3matrix. Tosimplifyourdiscussion,wewillconsideronly2×2matricesintheremainderofthissection. Matricesareaddedbyaddingcorrespondingentries: Thematrix iscalledthezeromatrixandbehaves,underaddition,likethenumberzero. Themultiplicationofmatricesisalittlemoredifficult.First,letusrecallthatthedotproductoftwo vectors(a,b)and(a′,b′)is (a,b)·(a′,b′)=aa′+bb′ thatis,wemultiplycorrespondingcomponentsandadd.Now,supposewewanttomultiplytwomatrices AandB;weobtaintheproductABasfollows: TheentryinthefirstrowandfirstcolumnofAB,thatis,inthisposition isequaltothedotproductofthefirstrowofAbythefirstcolumnofB.Theentryinthefirstrowand secondcolumnofAB,inotherwords,thisposition isequaltothedotproductofthefirstrowofAbythesecondcolumnofB.Andsoon.Forexample, Sofinally, The rules of algebra for matrices are very different from the rules of “conventional” algebra. For instance,thecommutativelawofmultplica-tion,AB=BA,isnottrue.Hereisasimpleexample: IfAisarealnumberandA2=0,thennecessarilyA=0;butthisisnottrueofmatrices.Forexample, thatis,A2=0althoughA≠0. Inthealgebraofnumbers,ifAB=ACwhereA≠0,wemaycancelAandconcludethatB = C. In matrixalgebrawecannot.Forexample, thatis,AB=AC,A≠0,yetB≠C. Theidentitymatrix correspondsinmatrixmultiplicationtothenumber1;forwehaveAI=IA=Aforevery2×2matrixA. IfAisanumberandA2=1,weconcludethatA=±1Matricesdonotobeythisrule.Forexample, thatis,A2=I,andyetAisneitherInor−I. Nomorewillbesaidaboutthealgebraofmatricesatthispoint,exceptthatwemustbeaware,once again,thatitisanewgamewhoserulesarequitedifferentfromthoseweapplyinconventionalalgebra. BooleanAlgebra An even more bizarre kind of algebra was developed in the mid-nineteenth century by an Englishman named George Boole. This algebra—subsequently named boolean algebra after its inventor—has a myriadofapplicationstoday.Itisformallythesameasthealgebraofsets. If S is a set, we may consider union and intersection to be operations on the subsets of 5. Let us agreeprovisionallytowrite A+B for A∪B and A·B for A∩B (Thisconventionisnotunusual.)Then, andsoon. Theseidentitiesareanalogoustotheonesweuseinelementaryalgebra.Butthefollowingidentities arealsotrue,andtheyhavenocounterpartinconventionalalgebra: andsoon. This unusual algebra has become a familiar tool for people who work with electrical networks, computersystems,codes,andsoon.Itisasdifferentfromthealgebraofnumbersasitisfromthealgebra ofmatrices. Other exotic algebras arose in a variety of contexts, often in connection with scientific problems. Therewere“complex”and“hypercomplex”algebras,algebrasofvectorsandtensors,andmanyothers. Todayitisestimatedthatover200differentkindsofalgebraicsystemshavebeenstudied,eachofwhich aroseinconnectionwithsomeapplicationorspecificneed. AlgebraicStructures As legions of new algebras began to occupy the attention of mathematicians, the awareness grew that algebracannolongerbeconceivedmerelyasthescienceofsolvingequations.Ithadtobeviewedmuch morebroadlyasabranchofmathematicscapableofrevealinggeneralprincipleswhichapplyequallyto allknownandallpossiblealgebras. Whatisitthatallalgebrashaveincommon?Whattraitdotheysharewhichletsusrefertoallof themas“algebras”?Inthemostgeneralsense,everyalgebraconsistsofaset(asetofnumbers,asetof matrices,asetofswitchingcomponents,oranyotherkindofset)andcertainoperationsonthatset.An operationissimplyawayofcombininganytwomembersofasettoproduceauniquethirdmemberofthe sameset. Thus,weareledtothemodernnotionofalgebraicstructure.Analgebraicstructureisunderstoodto be an arbitrary set, with one or more operations defined on it. And algebra, then, is defined to be the studyofalgebraicstructures. Itisimportantthatwebeawakenedtothefullgeneralityofthenotionofalgebraicstructure.Wemust makeanefforttodiscardallourpreconceivednotionsofwhatanalgebrais,andlookatthisnewnotion ofalgebraicstructureinitsnakedsimplicity.Anyset,witharule(orrules)forcombiningitselements,is already an algebraic structure. There does not need to be any connection with known mathematics. For example,considerthesetofallcolors(purecolorsaswellascolorcombinations),andtheoperationof mixinganytwocolorstoproduceanewcolor.Thismaybeconceivedasanalgebraicstructure.Itobeys certainrules,suchasthecommutativelaw(mixingredandblueisthesameasmixingblueandred).Ina similar vein, consider the set of all musical sounds with the operation of combining any two sounds to produceanew(harmoniousordisharmonious)combination. Asanotherexample,imaginethattheguestsatafamilyreunionhavemadeuparuleforpickingthe closestcommonrelativeofanytwopersonspresentatthereunion(andsupposethat,foranytwopeople at the reunion, their closest common relative is also present at the reunion). This too, is an algebraic structure:wehaveaset(namelythesetofpersonsatthereunion)andanoperationonthatset(namelythe “closestcommonrelative”operation). Asthegeneralnotionofalgebraicstructurebecamemorefamiliar(itwasnotfullyaccepteduntilthe early part of the twentieth century), it was bound to have a profound influence on what mathematicians perceived algebra to be. In the end it became clear that the purpose of algebra is to study algebraic structures,andnothinglessthanthat.Ideallyitshouldaimtobeageneralscienceofalgebraicstructures whoseresultsshouldhaveapplicationstoparticularcases,therebymakingcontactwiththeolderpartsof algebra.Beforewetakeacloserlookatthisprogram,wemustbrieflyexamineanotheraspectofmodern mathematics,namely,theincreasinguseoftheaxiomaticmethod. AXIOMS Theaxiomaticmethodisbeyonddoubtthemostremarkableinventionofantiquity,andinasensethemost puzzling. It appeared suddenly in Greek geometry in a highly developed form—already sophisticated, elegant,andthoroughlymoderninstyle.Nothingseemstohaveforeshadoweditanditwasunknownto ancientmathematiciansbeforetheGreeks.Itappearsforthefirsttimeinthelightofhistoryinthegreat textbook of early geometry, Euclid’s Elements. Its origins—the first tentative experiments in formal deductivereasoningwhichmusthaveprecededit—remainsteepedinmystery. Euclid’s Elements embodies the axiomatic method in its purest form. This amazing book contains 465 geometric propositions, some fairly simple, some of astounding complexity. What is really remarkable,though,isthatthe465propositions,formingthelargestbodyofscientificknowledgeinthe ancientworld,arederivedlogicallyfromonly10premiseswhichwouldpassastrivialobservationsof commonsense.Typicalofthepremisesarethefollowing: Thingsequaltothesamethingareequaltoeachother. Thewholeisgreaterthanthepart. Astraightlinecanbedrawnthroughanytwopoints. Allrightanglesareequal. So great was the impression made by Euclid’s Elements on following generations that it became the modelofcorrectmathematicalformandremainssotothisday. Itwouldbewrongtobelievetherewasnonotionofdemonstrativemathematicsbeforethetimeof Euclid. There is evidence that the earliest geometers of the ancient Middle East used reasoning to discovergeometricprinciples.Theyfoundproofsandmusthavehituponmanyofthesameproofswefind in Euclid. The difference is that Egyptian and Babylonian mathematicians considered logical demonstrationtobeanauxiliaryprocess,likethepreliminarysketchmadebyartists—aprivatemental processwhichguidedthemtoaresultbutdidnotdeservetoberecorded.Suchanattitudeshowslittle understandingofthetruenatureofgeometryanddoesnotcontaintheseedsoftheaxiomaticmethod. It is also known today that many—maybe most—of the geometric theorems in Euclid’s Elements came from more ancient times, and were probably borrowed by Euclid from Egyptian and Babylonian sources.However,thisdoesnotdetractfromthegreatnessofhiswork.Importantasarethecontentsof the Elements, what has proved far more important for posterity is the formal manner in which Euclid presented these contents. The heart of the matter was the way he organized geometric facts—arranged themintoalogicalsequencewhereeachtheorembuildsonprecedingtheoremsandthenformsthelogical basisforothertheorems. (Wemustcarefullynotethattheaxiomaticmethodisnotawayofdiscoveringfactsbutoforganizing them.Newfactsinmathematicsarefound,asoftenasnot,byinspiredguessesorexperiencedintuition. Tobeaccepted,however,theyshouldbesupportedbyproofinanaxiomaticsystem.) Euclid’sElementshasstoodthroughouttheagesasthemodeloforganized,rationalthoughtcarried toitsultimateperfection.Mathematiciansandphilosophersineverygenerationhavetriedtoimitateits lucid perfection and flawless simplicity. Descartes and Leibniz dreamed of organizing all human knowledge into an axiomatic system, and Spinoza created a deductive system of ethics patterned after Euclid’sgeometry.Whilemanyofthesedreamshaveprovedtobeimpractical,themethodpopularizedby Euclidhasbecometheprototypeofmodernmathematicalform.Sincethemiddleofthenineteenthcentury, theaxiomaticmethodhasbeenacceptedastheonlycorrectwayoforganizingmathematicalknowledge. To perceive why the axiomatic method is truly central to mathematics, we must keep one thing in mind: mathematics by its nature is essentially abstract. For example, in geometry straight lines are not stretchedthreads,butaconceptobtainedbydisregardingallthepropertiesofstretchedthreadsexceptthat ofextendinginonedirection.Similarly,theconceptofageometricfigureistheresultofidealizingfrom allthepropertiesofactualobjectsandretainingonlytheirspatialrelationships.Now,sincetheobjectsof mathematicsareabstractions,itstandstoreasonthatwemustacquireknowledgeaboutthembylogicand notbyobservationorexperiment(forhowcanoneexperimentwithanabstractthought?). Thisremarkappliesveryaptlytomodernalgebra.Thenotionofalgebraicstructureisobtainedby idealizing from all particular, concrete systems of algebra. We choose to ignore the properties of the actualobjectsinasystemofalgebra(theymaybenumbers,ormatrices,orwhatever—wedisregardwhat theyare),andweturnourattentionsimplytothewaytheycombineunderthegivenoperations.Infact, justaswedisregardwhattheobjectsinasystemare,wealsodisregardwhattheoperationsdotothem. We retain only the equations and inequalities which hold in the system, for only these are relevant to algebra.Everythingelsemaybediscarded.Finally,equationsandinequalitiesmaybededucedfromone anotherlogically,justasspatialrelationshipsarededucedfromeachotheringeometry. THEAXIOMATICSOFALGEBRA Letusrememberthatinthemid-nineteenthcentury,wheneccentricnewalgebrasseemedtoshowupat everyturninmathematicalresearch,itwasfinallyunderstoodthatsacrosanctlawssuchastheidentities ab = ba and a(bc) = (ab)c are not inviolable—for there are algebras in which they do not hold. By varyingordeletingsomeoftheseidentities,orbyreplacingthembynewones,anenormousvarietyof newsystemscanbecreated. Mostimportantly,mathematiciansslowlydiscoveredthatallthealgebraiclawswhichholdinany systemcanbederivedfromafewsimple,basicones.Thisisagenuinelyremarkablefact,foritparallels thediscoverymadebyEuclidthatafewverysimplegeometricpostulatesaresufficienttoproveallthe theorems of geometry. As it turns out, then, we have the same phenomenon in algebra: a few simple algebraicequationsofferthemselvesnaturallyasaxioms,andfromthemallotherfactsmaybeproved. These basic algebraic laws are familiar to most high school students today. We list them here for reference.WeassumethatAisanysetandthereisanoperationonAwhichwedesignatewiththesymbol * a*b=b*a (1) IfEquation(1)istrueforanytwoelementsaandbinA,wesaythattheoperation*iscommutative.What itmeans,ofcourse,isthatthevalueofa*b(orb*a)isindependentoftheorderinwhichaandb are taken. a*(b*c)=(a*b)*c (2) If Equation (2) is true for any three elements a, b, and c in A, we say the operation * is associative. Remember that an operation is a rule for combining any two elements, so if we want to combine three elements, we can do so in different ways. If we want to combine a, b, and c without changing their order,wemayeithercombineawiththeresultofcombiningbandc,whichproducesa*(b*c);orwe mayfirstcombineawithb,andthencombinetheresultwithc,producing(a*b)*c.Theassociativelaw assertsthatthesetwopossiblewaysofcombiningthreeelements(withoutchangingtheirorder)yieldthe sameresult. ThereexistsanelementeinAsuchthat e*a=a and a*e=a foreveryainA (3) IfsuchanelementeexistsinA,wecallitanidentityelementfortheoperation*.Anidentityelementis sometimescalleda“neutral”element,foritmaybecombinedwithanyelementawithoutalteringa.For example,0isanidentityelementforaddition,and1isanidentityelementformultiplication. ForeveryelementainA,thereisanelementa−l(“ainverse”)inAsuchthat a*a−l=e and a−1*a=e (4) Ifstatement(4)istrueinasystemofalgebra,wesaythateveryelementhasaninversewithrespecttothe operation*.Themeaningoftheinverseshouldbeclear:thecombinationofanyelementwithitsinverse producestheneutralelement(onemightroughlysaythattheinverseofa“neutralizes”a).Forexample,if A is a set of numbers and the operation is addition, then the inverse of any number a is (−a); if the operationismultiplication,theinverseofanya≠0is1/a. Let us assume now that the same set A has a second operation, symbolized by ⊥, as well as the operation*: a*(b⊥c)=(a*b)⊥(a*c) (5) IfEquation(5)holdsforanythreeelementsa,b,andcinA,wesaythat*isdistributiveover⊥.Ifthere are two operations in a system, they must interact in some way; otherwise there would be no need to considerthemtogether.Thedistributivelawisthemostcommonway(butnottheonlypossibleone)for twooperationstoberelatedtooneanother. There are other “basic” laws besides the five we have just seen, but these are the most common ones.Themostimportantalgebraicsystemshaveaxiomschosenfromamongthem.Forexample,whena mathematiciannowadaysspeaksofaring,themathematicianisreferringtoasetAwithtwooperations, usuallysymbolizedby+and·,havingthefollowingaxioms: Additioniscommutativeandassociative,ithasaneutralelementcommonlysymbolizedby0,and everyelementahasaninverse–awithrespecttoaddition.Multiplicationisassociative,hasa neutralelement1,andisdistributiveoveraddition. Matrixalgebraisaparticularexampleofaring,andallthelawsofmatrixalgebramaybeprovedfrom the preceding axioms. However, there are many other examples of rings: rings of numbers, rings of functions,ringsofcode“words,”ringsofswitchingcomponents,andagreatmanymore.Everyalgebraic lawwhichcanbeprovedinaring(fromtheprecedingaxioms)istrueineveryexampleofaring.Inother words, instead of proving the same formula repeatedly—once for numbers, once for matrices, once for switchingcomponents,andsoon—itissufficientnowadaystoproveonlythattheformulaholdsinrings, andthenofnecessityitwillbetrueinallthehundredsofdifferentconcreteexamplesofrings. Byvaryingthepossiblechoicesofaxioms,wecankeepcreatingnewaxiomaticsystemsofalgebra endlessly.Wemaywellask:isitlegitimatetostudyany axiomatic system, with any choice of axioms, regardlessofusefulness,relevance,orapplicability?Thereare“radicals”inmathematicswhoclaimthe freedom for mathematicians to study any system they wish, without the need to justify it. However, the practiceinestablishedmathematicsismoreconservative:particularaxiomaticsystemsareinvestigated onaccountoftheirrelevancetonewandtraditionalproblemsandotherpartsofmathematics,orbecause theycorrespondtoparticularapplications. Inpractice,howisaparticularchoiceofalgebraicaxiomsmade?Verysimply:whenmathematicians look at different parts of algebra and notice that a common pattern of proofs keeps recurring, and essentiallythesameassumptionsneedtobemadeeachtime,theyfinditnaturaltosingleoutthischoiceof assumptionsastheaxiomsforanewsystem.Alltheimportantnewsystemsofalgebrawerecreatedin thisfashion. ABSTRACTIONREVISITED Another important aspect of axiomatic mathematics is this: when we capture mathematical facts in an axiomaticsystem,wenevertrytoreproducethefactsinfull,butonlythatsideofthemwhichisimportant orrelevantinaparticularcontext.Thisprocessofselectingwhatisrelevantanddisregardingeverything elseistheveryessenceofabstraction. Thiskindofabstractionissonaturaltousashumanbeingsthatwepracticeitallthetimewithout beingawareofdoingso.LiketheBourgeoisGentlemaninMolière’splaywhowasamazedtolearnthat hespokeinprose,someofusmaybesurprisedtodiscoverhowmuchwethinkinabstractions.Nature presentsuswithamyriadofinterwovenfactsandsensations,andwearechallengedateveryinstantto singleoutthosewhichareimmediatelyrelevantanddiscardtherest.Inordertomakeoursurroundings comprehensible,wemustcontinuallypickoutcertaindataandseparatethemfromeverythingelse. Fornaturalscientists,thisprocessistheverycoreandessenceofwhattheydo.Natureisnotmade up of forces, velocities, and moments of inertia. Nature is a whole—nature simply is! The physicist isolatescertainaspectsofnaturefromtherestandfindsthelawswhichgoverntheseabstractions. Itisthesamewithmathematics.Forexample,thesystemoftheintegers(wholenumbers),asknown byourintuition,isacomplexrealitywithmanyfacets.Themathematicianseparatesthesefacetsfromone anotherandstudiesthemindividually.Fromonepointofviewthesetoftheintegers,withadditionand multiplication,formsaring(thatis,itsatisfiestheaxiomsstatedpreviously).Fromanotherpointofview itisanorderedset,andsatisfiesspecialaxiomsofordering.Onadifferentlevel,thepositiveintegers form the basis of “recursion theory,” which singles out the particular way positive integers may be constructed,beginningwith1andadding1eachtime. It therefore happens that the traditional subdivision of mathematics into subject matters has been radicallyaltered.Nolongeraretheintegersonesubject,complexnumbersanother,matricesanother,and so on; instead, particular aspects of these systems are isolated, put in axiomatic form, and studied abstractly without reference to any specific objects. The other side of the coin is that each aspect is sharedbymanyofthetraditionalsystems:forexample,algebraicallytheintegersformaring,andsodo thecomplexnumbers,matrices,andmanyotherkindsofobjects. Thereisnothingintrinsicallynewaboutthisprocessofdivorcingpropertiesfromtheactualobjects havingtheproperties;aswehaveseen,itispreciselywhatgeometryhasdoneformorethan2000years. Somehow,ittooklongerforthisprocesstotakeholdinalgebra. Themovementtowardaxiomaticsandabstractioninmodernalgebrabeganaboutthe1830sandwas completed 100 years later. The movement was tentative at first, not quite conscious of its aims, but it gainedmomentumasitconvergedwithsimilartrendsinotherpartsofmathematics.Thethinkingofmany great mathematicians played a decisive role, but none left a deeper or longer lasting impression than a veryyoungFrenchmanbythenameofÉvaristeGalois. ThestoryofÉvaristeGaloisisprobablythemostfantasticandtragicinthehistoryofmathematics.A sensitiveandprodigiouslygiftedyoungman,hewaskilledinaduelattheageof20,endingalifewhich initsbriefspanhadofferedhimnothingbuttragedyandfrustration.Whenhewasonlyayouthhisfather commitedsuicide,andGaloiswaslefttofendforhimselfinthelabyrinthineworldofFrenchuniversity life and student politics. He was twice refused admittance to the Ecole Polytechnique, the most prestigiousscientificestablishmentofitsday,probablybecausehisanswerstotheentranceexamination were too original and unorthodox. When he presented an early version of his important discoveries in algebratothegreatacademicianCauchy,thisgentlemandidnotreadtheyoungstudent’spaper,butlostit. Later,GaloisgavehisresultstoFourierinthehopeofwinningthemathematicsprizeoftheAcademyof Sciences. But Fourier died, and that paper, too, was lost. Another paper submitted to Poisson was eventuallyreturnedbecausePoissondidnothavetheinteresttoreaditthrough. Galois finally gained admittance to the École Normale, another focal point of research in mathematics,buthewassoonexpelledforwritinganessaywhichattackedtheking.Hewasjailedtwice for political agitation in the student world of Paris. In the midst of such a turbulent life, it is hard to believethatGaloisfoundtimetocreatehiscolossallyoriginaltheoriesonalgebra. WhatGaloisdidwastotieintheproblemoffindingtherootsofequationswithnewdiscoverieson groupsofpermutations.Heexplainedexactlywhichequationsofdegree5orhigherhavesolutionsofthe traditional kind—and which others do not. Along the way, he introduced some amazingly original and powerfulconcepts,whichformtheframeworkofmuchalgebraicthinkingtothisday.AlthoughGaloisdid notworkexplicitlyinaxiomaticalgebra(whichwasunknowninhisday),theabstractnotionofalgebraic structureisclearlyprefiguredinhiswork. In1832,whenGaloiswasonly20yearsold,hewaschallengedtoaduel.Whatargumentledtothe challengeisnotclear:somesaytheissuewaspolitical,whileothersmaintaintheduelwasfoughtovera fickle lady’s wavering love. The truth may never be known, but the turbulent, brilliant, and idealistic Galoisdiedofhiswounds.Fortunatelyformathematics,thenightbeforetheduelhewrotedownhismain mathematical results and entrusted them to a friend. This time, they weren’t lost—but they were only published15yearsafterhisdeath.Themathematicalworldwasnotreadyforthembeforethen! Algebratodayisorganizedaxiomatically,andassuchitisabstract.Mathematiciansstudyalgebraic structuresfromageneralpointofview,comparedifferentstructures,andfindrelationshipsbetweenthem. Thisabstractionandgeneralizationmightappeartobehopelesslyimpractical—butitisnot!Thegeneral approach in algebra has produced powerful new methods for “algebraizing” different parts of mathematics and science, formulating problems which could never have been formulated before, and findingentirelynewkindsofsolutions. Suchexcursionsintopuremathematicalfancyhaveanoddwayofrunningaheadofphysicalscience, providing a theoretical framework to account for facts even before those facts are fully known. This pattern is so characteristic that many mathematicians see themselves as pioneers in a world of possibilitiesratherthanfacts.Mathematiciansstudystructureindependentlyofcontent,andtheirscience isavoyageofexplorationthroughallthekindsofstructureandorderwhichthehumanmindiscapableof discerning. CHAPTER TWO OPERATIONS Addition, subtraction, multiplication, division—these and many others are familiar examples of operationsonappropriatesetsofnumbers. Intuitively,anoperationonasetAisawayofcombininganytwoelementsofAtoproduceanother elementinthesamesetA. Everyoperationisdenotedbyasymbol,suchas+,×,or÷Inthisbookwewilllookatoperations fromaloftyperspective;wewilldiscoverfactspertainingtooperationsgenerallyratherthantospecific operationsonspecificsets.Accordingly,wewillsometimesmakeupoperationsymbolssuchas*and torefertoarbitraryoperationsonarbitrarysets. LetusnowdefineformallywhatwemeanbyanoperationonsetA.LetAbeanyset: Anoperation*onAisarulewhichassignstoeachorderedpair(a,b)ofelementsofAexactly oneelementa*binA. Therearethreeaspectsofthisdefinitionwhichneedtobestressed: 1. a * b is defined for every ordered pair (a, b) of elements of A. There are many rules which look deceptivelylikeoperationsbutarenot,becausethisconditionfails.Oftena*bisdefinedforallthe obviouschoicesofaandb,butremainsundefinedinafewexceptionalcases.Forexample,division doesnotqualifyasanoperationontheset oftherealnumbers,forthereareorderedpairssuchas(3, 0)whosequotient3/0isundefined.Inordertobeanoperationon ,divisionwouldhavetoassociatea realnumberalbwitheveryorderedpair(a,b)ofelementsof .Noexceptionsallowed! 2. a*bmustbeuniquelydefined.Inotherwords,thevalueofa*bmustbegivenunambiguously.For example,onemightattempttodefineanoperation□ontheset oftherealnumbersbylettinga□bbe thenumberwhosesquareisab.Obviouslythisisambiguousbecause2□8,letussay,maybeeither4 or-4.Thus,□doesnotqualifyasanoperationon ! 3. IfaandbareinA,a*bmustbeinA.ThisconditionisoftenexpressedbysayingthatA is closed undertheoperation*.Ifweproposetodefineanoperation*onasetA,wemusttakecarethat*,when appliedtoelementsofA,doesnottakeusoutofA.Forexample,divisioncannotberegardedasan operationonthesetoftheintegers,fortherearepairsofintegerssuchas(3,4)whosequotient3/4is notaninteger. Ontheotherhand,divisiondoesqualifyasanoperationonthesetofallthepositivereal numbers,forthequotientofanytwopositiverealnumbersisauniquelydeterminedpositivereal number. AnoperationisanyrulewhichassignstoeachorderedpairofelementsofAauniqueelementinA. Therefore it is obvious that there are, in general, many possible operations on a given set A. If, for example, A is a set consisting of just two distinct elements, say a and b, each operation on A may be describedbyatablesuchasthisone: IntheleftcolumnarelistedthefourpossibleorderedpairsofelementsofA,andtotherightofeachpair (x,y)isthevalueofx*y.Hereareafewofthepossibleoperations: EachofthesetablesdescribesadifferentoperationonA.Eachtablehasfourrows,andeachrowmaybe filledwitheitheranaorab;hencethereare16possiblewaysoffillingthetable,correspondingto16 possibleoperationsonthesetA. WehavealreadyseenthatanyoperationonasetAcomeswithcertain“options.”Anoperation*may becommutative,thatis,itmaysatisfy a*b=b*a (1) foranytwoelementsaandbinA.Itmaybeassociative,thatis,itmaysatisfytheequation (a*b)*c=a*(b*c) (2) foranythreeelementsa,b,andcinA. Tounderstandtheimportanceoftheassociativelaw,wemustrememberthatanoperationisawayof combiningtwoelements;soifwewanttocombinethreeelements,wecandosoindifferentways.Ifwe wanttocombinea,b,andcwithoutchangingtheirorder, we may either combine a with the result of combiningbandc,whichproducesa*(b*c);orwemayfirstcombineawithb,andthencombinethe result with c, producing (a * b) * c. The associative law asserts that these two possible ways of combiningthreeelements(withoutchangingtheirorder)producethesameresult. Forexample,theadditionofrealnumbersisassociativebecausea+(b+c)=(a+b)+c.However, divisionofrealnumbersisnotassociative:forinstance,3/(4/5)is15/4,whereas(3/4)/5is3/20. IfthereisanelementeinAwiththepropertythat e*a=a and a*e=a foreveryelementainA (3) then e is called an identity or “neutral” element with respect to the operation *. Roughly speaking, Equation(3)tellsusthatwheneiscombinedwithanyelementa,itdoesnotchangea.Forexample,inthe set oftherealnumbers,0isaneutralelementforaddition,and1isaneutralelementformultiplication. IfaisanyelementofA,andxisanelementofAsuchthat a*x=e and x*a=e (4) thenxiscalledaninverseofa.Roughlyspeaking,Equation(4)tellsusthatwhenanelementiscombined withitsinverseitproducestheneutralelement.Forexample,intheset oftherealnumbers, −a is the inverseofawithrespecttoaddition;ifa≠0,then1/aistheinverseofawithrespecttomultiplication. The inverse of a is often denoted by the symbol a−l. (The symbol a−l is usually pronounced “a inverse.”) EXERCISES Throughoutthisbook,theexercisesaregroupedintoexercisesets,eachsetbeingidentifiedbyaletterA, B, C, etc, and headed by a descriptive title. Each exercise set contains six to ten exercises, numbered consecutively. Generally, the exercises in each set are independent of each other and may be done separately.However,whentheexercisesinasetarerelated,withsomeexercisesbuildingonpreceding onessothattheymustbedoneinsequence,thisisindicatedwithasymboltinthemargintotheleftofthe heading. Thesymbol#nexttoanexercisenumberindicatesthatapartialsolutiontothatexerciseisgivenin theAnswerssectionattheendofthebook. A.ExamplesofOperations Whichofthefollowingrulesareoperationsontheindicatedset?( designatesthesetoftheintegers, therationalnumbers,and therealnumbers.)Foreachrulewhichisnotanoperation,explainwhyitis not. Example ,ontheset . SolutionThisisnotanoperationon .Thereareintegersaandbsuchthat{a+b)/abisnotaninteger.(For example, isnotaninteger.)Thus, isnotclosedunder*. 1 ,ontheset . 2 a*b=alnb,ontheset{x∈ :x>0}. 3 a*bisarootoftheequationx2−a2b2=0,ontheset . 4 Subtraction,ontheset . 5 Subtraction,ontheset{n∈ :≥0}. 6 a*b=|a−b,ontheset{n∈ :≥0}. B.PropertiesofOperations Eachofthefollowingisanoperation*onU.Indicatewhetherornot (i) itiscommutative, (ii) itisassociative, (iii) hasanidentityelementwithrespectto*, (iv) everyx∈ hasaninversewithrespectto*. InstructionsFor(i),computex*yandy*x,andverifywhetherornottheyareequal.For(ii),computex *(y*z)and(x*y)*z,andverifywhetherornottheyareequal.For(iii),firstsolvetheequationx*e= x for e; if the equation cannot be solved, there is no identity element. If it can be solved, it is still necessarytocheckthate*x=x*e=xforanyx .Ifitchecks,theneisanidentityelement.For(iv),first note that if there is no identity element, there can be no inverses. If there is an identity element e, first solvetheequationx*x′=eforx′iftheequationcannotbesolved,xdoesnothaveaninverse.Ifitcanbe solved,checktomakesurethatx*x′=x′*x=x′*x=e.Ifthischecks,x′istheinverseofx. Examplex*y=x+y+1 (i) x*y=x+y+1;y*x=y+x+1=x+y+1. (Thus,*iscommutative.) (ii) x*(y*z)=x*(y+z+l)=x+(y+z+l)+l=x+y+z+2. (x*y)*z=(x+y+1)*z=(x+y+1)+z+1=x+y+z+2. (*isassociative.) (iii) Solvex*e=xfore:x*e=x+e+1=x;therefore,e=−1. Check:x*(−1)=x+(−1)+1=x;(−1)*x=(−1)+x+1=x. Therefore,−1istheidentityelement. (*hasanidentityelement.) (iv) Solvex*x′= −1forx′: x*x′=x+x′+1= −1;thereforex′= −x −2.Check:x*(−x −2)=x+ (−x−2)+1=−1;(−x−2)*x=(−x−2)+x+l=−l.Therefore,−x−2istheinverseofx. (Everyelementhasaninverse.) 1 x*y=x+2y+4 (i) x*y=x+2y+4;y*x= (ii) x*(y*z)=x*( )= (x*y)*z=( )*z= (iii) Solvex*e=xfore.Check. (iv) Solvex*x′=eforx′.Check. 2 x*y=x+2y−xy 3 x*y=|x+y| 4 x*y=|x−y| 5 x*y=xy+1 6 x*y=max{x,y}=thelargerofthetwonumbersxandy 7 (onthesetofpositiverealnumbers) C.OperationsonaTwo-ElementSet LetAbethetwo-elementsetA={a,b}. 1 Writethetablesofall16operationsonA.(Usetheformatexplainedonpage20.) Labeltheseoperations0lto016.Then: 2 Identifywhichoftheoperations0lto016arecommutative. 3 Identifywhichoperations,among0lto016,areassociative. 4 Forwhichoftheoperations0lto016isthereanidentityelement? 5 Forwhichoftheoperations0lto016doeseveryelementhaveaninverse? D.Automata:TheAlgebraofInput/OutputSequences Digital computers and related machines process information which is received in the form of input sequences.AninputsequenceisafinitesequenceofsymbolsfromsomealphabetA.Forinstance,ifA= {0,1}(thatis,ifthealphabetconsistsofonlythetwosymbols0and1),thenexamplesofinputsequences are011010and10101111.IfA={a,b,c},thenexamplesofinputsequencesarebabbcacandcccabaa. Outputsequencesaredefinedinthesamewayasinputsequences.Thesetofallsequencesofsymbolsin thealphabetAisdenotedbyA*. ThereisanoperationonA*calledconcatenation:IfaandbareinA*,saya=a1a2...anandb= blb2…bm,then ab=a1a2…anbb2...bm Inotherwords,thesequenceabconsistsofthetwosequencesa and b end to end. For example, in the alphabetA={0,1},ifa=1001andb=010,thenab=1001010. 1 2 3 Thesymbolλdenotestheemptysequence. Provethattheoperationdefinedaboveisassociative. Explainwhytheoperationisnotcommutative. Provethatthereisanidentityelementforthisoperation. CHAPTER THREE THEDEFINITIONOFGROUPS Oneofthesimplestandmostbasicofallalgebraicstructuresisthegroup.Agroupisdefinedtobeaset withanoperation(letuscallit*)whichisassociative,hasaneutralelement,andforwhicheachelement hasaninverse.Moreformally, ByagroupwemeanasetGwithanoperation*whichsatisfiestheaxioms: (G1) *isassociative. (G2) ThereisanelementeinGsuchthata*e=aande*a=aforeveryelementainG. (G3) ForeveryelementainG,thereisanelementa−linGsuchthata*a−1=eanda−1*a=e. The group we have just defined may be represented by the symbol 〈G, *〉. This notation makes it explicit that the group consists of the set G and the operation *. (Remember that, in general, there are otherpossibleoperationsonG,soitmaynotalwaysbeclearwhichisthegroup’soperationunlesswe indicateit.)Ifthereisnodangerofconfusion,weshalldenotethegroupsimplywiththeletterG. Thegroupswhichcometomindmostreadilyarefoundinourfamiliarnumbersystems.Herearea fewexamples. isthesymbolcustomarilyusedtodenotetheset {…,−3,−2,−1,0,1,2,3,…} oftheintegers.Theset ,withtheoperationofaddition,isobviouslyagroup.Itiscalledtheadditive groupoftheintegersandisrepresentedbythesymbol〈 ,+〉.Mostly,wedenoteitsimplybythesymbol . designatesthesetoftherationalnumbers(thatis,quotientsm/nofintegers,wheren≠0).Thisset, withtheoperationofaddition,iscalledtheadditivegroupoftherationalnumbers,〈 ,+〉.Mostoften wedenoteitsimplyby . Thesymbol representsthesetoftherealnumbers. ,withtheoperationofaddition,iscalledthe additivegroupoftherealnumbers,andisrepresentedby〈 ,+〉,orsimply . The set of all the nonzero rational numbers is represented by *. This set, with the operation of multiplication,isthegroup〈 *,·〉,orsimply *.Similarly,thesetofallthenonzero real numbers is representedby *.Theset *withtheoperationofmultiplication,isthegroup〈 *,·〉,orsimply *. Finally, posdenotesthegroupofallthepositiverationalnumbers,withmultiplication. posdenotes thegroupofallthepositiverealnumbers,withmultiplication. Groupsoccurabundantlyinnature.Thisstatementmeansthatagreatmanyofthealgebraicstructures which can be discerned in natural phenomena turn out to be groups. Typical examples, which we shall examine later, come up in connection with the structure of crystals, patterns of symmetry, and various kinds of geometric transformations. Groups are also important because they happen to be one of the fundamentalbuildingblocksoutofwhichmorecomplexalgebraicstructuresaremade. Especially important in scientific applications are the finite groups, that is, groups with a finite numberofelements.Itisnotsurprisingthatsuchgroupsoccurofteninapplications,forinmostsituations oftherealworldwedealwithonlyafinitenumberofobjects. Theeasiestfinitegroupstostudyarethosecalledthegroupsofintegersmodulon(wherenisany positive integer greater than 1). These groups will be described in a casual way here, and a rigorous treatmentdeferreduntillater. Letusbeginwithaspecificexample,say,thegroupofintegersmodulo6.Thisgroupconsistsofa setofsixelements, {0,1,2,3,4,5} andanoperationcalledadditionmodulo6,whichmaybedescribedasfollows:Imaginethenumbers0 through5asbeingevenlydistributedonthecircumferenceofacircle.Toaddtwonumbershandk,start withhandmoveclockwisekadditionalunitsaroundthecircle:h+kiswhereyouendup.Forexample, 3 + 3 = 0, 3 + 5 = 2, and so on. The set {0, 1, 2, 3, 4, 5} with this operation is called the group of integersmodulo6,andisrepresentedbythesymbol 6. Ingeneral,thegroupofintegersmodulonconsistsoftheset {0,1,2,…,n−1} with the operation of addition modulo n, which can be described exactly as previously. Imagine the numbers 0 through n − 1 to be points on the unit circle, each one separated from the next by an arc of length2π/n. To add h and k, start with h and go clockwise through an arc of k times 2π/n. The sum h + k will, of course,beoneofthenumbers0throughn−1.Fromgeometricalconsiderationsitisclearthatthiskindof addition (by successive rotations on the unit circle) is associative. Zero is the neutral element of this group,andn−hisobviouslytheinverseofh[forh+(n−h)=n,whichcoincideswith0].Thisgroup, thegroupofintegersmodulon,isrepresentedbythesymbol n. Oftenwhenworkingwithfinitegroups,itisusefultodrawupan“operationtable.”Forexample,the operationtableof 6is Thebasicformatofthistableisasfollows: withonerowforeachelementofthegroupandonecolumnforeachelementofthegroup.Then3+4,for example,islocatedintherowof3andthecolumnof4.Ingeneral,anyfinitegroup〈G,*〉hasatable Theentryintherowofxandthecolumnofyisx*y. Let us remember that the commutative law is not one of the axioms of group theory; hence the identitya*b=b*aisnottrueineverygroup.IfthecommutativelawholdsinagroupG,suchagroupis calledacommutativegroupor,morecommonly,anabeliangroup. Abelian groups are named after the mathematician Niels Abel, who was mentioned in Chapter 1 and who was a pioneer in the study of groups.Alltheexamplesofgroupsmentioneduptonowareabeliangroups,buthereisanexamplewhich isnot. LetGbethegroupwhichconsistsofthesixmatrices withtheoperationofmatrixmultiplicationwhichwasexplainedonpage8.Thisgrouphasthefollowing operationtable,whichshouldbechecked: Inlinearalgebraitisshownthatthemultiplicationofmatricesisassociative.(Thedetailsaresimple.)It isclearthatIistheidentityelementofthisgroup,andbylookingatthetableonecanseethateachofthe sixmatricesin{I,A,B,C,D,K}hasaninversein{I,A,B,C,D,K}.(Forexample,Bistheinverseof D,AistheinverseofA,andsoon.)Thus,Gisagroup!NowweobservethatAB=CandBA=K,soG isnotcommutative. EXERCISES A.ExamplesofAbelianGroups Provethateachofthefollowingsets,withtheindicatedoperation,isanabeliangroup. InstructionsProceedasinChapter2,ExerciseB. 1x*y=x+y+k (kafixedconstant),ontheset oftherealnumbers. 2 ,ontheset{x∈ :x≠0}. 3x*y=x+y+xy,ontheset{x∈ :x≠−1}. 4 ,theset{x∈ :−1<x<1}. B.GroupsontheSet × Thesymbol × representsthesetofallorderedpairs(x,y)ofrealnumbers. × may therefore be identified with the set of all the points in the plane. Which of the following subsets of × , with the indicatedoperation,isagroup?Whichisanabeliangroup? InstructionsProceedasintheprecedingexercise.Tofindtheidentityelement,whichintheseproblems isanorderedpair(e1,e2)ofrealnumbers,solvetheequation(a,b)*(e1,e2)=(a,b)fore1ande2. To findtheinverse(a′,b′)of(a,b),solvetheequation(a,b)*(a′,b′)=(e1,e2)fora′andb′.[Rememberthat (x,y)=(x′,y′)ifandonlyifx=x′andy=y′.] 1(a,b)*(c,d)=(ad+bc,bd),ontheset{(x,y)∈ × :y≠0}. #2(a,b)*(c,d)=(ac,bc+d),ontheset{(x,y)∈ × :x≠0}. 3Sameoperationasinpart2,butontheset × . 4(a,b)*(c,d)=(ac−bd,ad+bc),ontheset × withtheorigindeleted. 5Considertheoperationoftheprecedingproblemontheset × .Isthisagroup?Explain. C.GroupsofSubsetsofaSet IfAandBareanytwosets,theirsymmetricdifferenceisthesetA+Bdefinedasfollows: A+B=(A−B)∪(B−A) NOTE:A−BrepresentsthesetobtainedbyremovingfromAalltheelementswhichareinB. ItisperfectlyclearthatA+B=B+A;hencethisoperationiscommutative.Itisalsoassociative,as the accompanying pictorial representation suggests: Let the union of A, B, and C be divided into seven regionsasillustrated. A+Bconsistsoftheregions1,4,3,and6. B+Cconsistsoftheregions2,3,4,and7. A+(B+C)consistsoftheregions1,3,5,and7. (A+B)+Cconsistsoftheregions1,3,5,and7. Thus,A+(B+C)=(A+B)+C. IfDisaset,thenthepowersetofDisthesetPDofallthesubsetsofD.Thatis, PD={A:A⊆D} Theoperation+istoberegardedasanoperationonPD. 1Provethatthereisanidentityelementwithrespecttotheoperation+,whichis_________. 2ProveeverysubsetAofDhasaninversewithrespectto+,whichis_________.Thus,〈PD, +〉 is a group! 3LetDbethethree-elementsetD={a,b,c}.ListtheelementsofPD.(Forexample,oneelementis{a}, anotheris{a,b},andsoon.DonotforgettheemptysetandthewholesetD.)Thenwritetheoperation tablefor〈PD,+〉. D.ACheckerboardGame Ourcheckerboardhasonlyfoursquares,numbered1,2,3,and4.Thereisasinglecheckerontheboard, andithasfourpossiblemoves: V: H: D: I: Movevertically;thatis,movefrom1to3,orfrom3to1,orfrom2to4,orfrom4to2. Movehorizontally;thatis,movefrom1to2orviceversa,orfrom3to4orviceversa. Movediagonally;thatis,movefrom2to3orviceversa,ormovefrom1to4orviceversa. Stayput. We may consider an operation on the set of these four moves, which consists of performing moves successively.Forexample,ifwemovehorizontallyandthenvertically,weendupwiththesameresultas ifwehadmoveddiagonally: H*V=D Ifweperformtwohorizontalmovesinsuccession,weendupwherewestarted:H*H=I.Andsoon.If G={V,H,D,I},and*istheoperationwehavejustdescribed,writethetableofG. Grantingassociativity,explainwhy〈G,*〉isagroup. E.ACoinGame Imaginetwocoinsonatable,atpositionsAandB.Inthisgamethereareeightpossiblemoves: M1: FlipoverthecoinatA. M2: FlipoverthecoinatB. M3: Flipoverbothcoins. M4: Switchthecoins. M5: FlipcoinatA;thenswitch. M6: FlipcoinatB;thenswitch. M7: Flipbothcoins;thenswitch. I: Donotchangeanything. Wemayconsideranoperationontheset{I,M1,…,M7},whichconsistsofperforminganytwomovesin succession.Forexample,ifweswitchcoins,thenflipoverthecoinatA,thisisthesameasfirstflipping overthecoinatBthenswitching: M4*M1=M2*M4=M6 IfG={I,M1,…,M7}and*istheoperationwehavejustdescribed,writethetableof〈G,*〉. Grantingassociativity,explainwhy〈G,*〉isagroup.Isitcommutative?Ifnot,showwhynot. F.GroupsinBinaryCodes ThemostbasicwayoftransmittinginformationistocodeitintostringsofOsandIs,suchas0010111, 1010011,etc.Suchstringsarecalledbinarywords, and the number of 0s and Is in any binary word is calleditslength.Allinformationmaybecodedinthisfashion. When information is transmitted, it is sometimes received incorrectly. One of the most important purposesofcodingtheoryistofindwaysofdetectingerrors,andcorrectingerrorsoftransmission. Ifaworda=a1a2…anissent,butawordb=b1b2…bnisreceived(wheretheaiandthebj are0s or1s),thentheerrorpatternistheworde=e1e2…enwhere Withthismotivation,wedefineanoperationofaddingwords,asfollows:Ifaandbarebothoflength1, weaddthemaccordingtotherules 0+0=0 1+1=0 0+1=1 1+0=1 Ifa and b are both of length n, we add them by adding corresponding digits. That is (let us introduce commasforconvenience), (a1,a2,…,an)+(b1,b2,…,bn)=(a1+b1a2+b2,…,an+bn Thus,thesumofaandbistheerrorpatterne. Forexample, The symbol will designate the set of all the binary words of length n. We will prove that the operationofwordadditionhasthefollowingpropertieson : 1. 2. 3. 4. Itiscommutative. Itisassociative. Thereisanidentityelementforwordaddition. Everywordhasaninverseunderwordaddition. First,weverifythecommutativelawforwordsoflength1: 0+1=1=1+0 1Showthat(a1,a2,…,an)+(b1,b2,…,bn)=(b1,b2,…,bn)+(a1,a2,…,an). 2Toverifytheassociativelaw,wefirstverifyitforwordsoflength1: 1+(1+1)=1+0=1=0+1=(1+1)+1 1+(1+0)=1+1=0=0+0=(l+l)+0 Checktheremainingsixcases. 3Showthat(a1,…,an)+[(b1,…,bn)+(c1,…,cn)]=[(a1,…,an)+(b1,…,bn)]+(c1,…,cn). 4Theidentityelementof ,thatis,theidentityelementforaddingwordsoflengthn,is__________. 5Theinverse,withrespecttowordaddition,ofanyword(a1,…,an)is__________. 6Showthata+b=a−b[wherea−b=a+(−b*)]. 7Ifa+b=c,showthata=b+c. G.TheoryofCoding:Maximum-LikelihoodDecoding WecontinuethediscussionstartedinExerciseF:Recallthat designatesthesetofallbinarywordsof lengthn.Byacodewemeanasubsetof .Forexample,belowisacodein 5. The code, which we shallcallC1,consistsofthefollowingbinarywordsoflength5: 00000 00111 01001 01110 10011 10100 11010 11101 Notethatthereare32possiblewordsoflength5,butonlyeightofthemareinthecodeC,.Theseeight words are called codewords; the remaining words of B5 are not codewords. Only codewords are transmitted. If a word is received which is not a codeword, it is clear that there has been an error of transmission.Inawell-designedcode,itisunlikelythatanerrorintransmittingacodewordwillproduce anothercodeword(ifthatweretohappen,theerrorwouldnotbedetected).Moreover,inagoodcodeit shouldbefairlyeasytolocateerrorsandcorrectthem.Theseideasaremadepreciseinthediscussion whichfollows. TheweightofabinarywordisthenumberofIsintheword:forexample,11011hasweight4.The distance between two binary words is the number of positions in which the words differ. Thus, the distancebetween11011and01001is2(sincethesewordsdifferonlyintheirfirstandfourthpositions). The minimum distance in a code is the smallest distance among all the distances between pairs of codewords.ForthecodeC1,above,pairwisecomparisonofthewordsshowsthattheminimumdistance is 2. What this means is that at least two errors of transmission are needed in order to transform a codeword into another codeword; single errors will change a codeword into a noncodeword, and the errorwillthereforebedetected.Inmoredesirablecodes(forexample,theso-calledHammingcode),the minimumdistanceis3,soanyoneortwoerrorsarealwaysdetected,andonlythreeerrorsinasingle word(averyunlikelyoccurrence)mightgoundetected. Inpractice,acodeisconstructedasfollows:ineverycodeword,certainpositionsareinformation positions,andtheremainingpositionsareredundancypositions. For instance, in our code C1, the first threepositionsofeverycodewordaretheinformationpositions:ifyoulookattheeightcodewords(and confine your attention only to the first three digits in each word), you will see that every three-digit sequenceof0sandIsistherenamely, 000, 001, 010, 011, 100, 101, 110, 111 Thenumbersinthefourthandfifthpositionsofeverycodewordsatisfyparity-checkequations. #1Verifythateverycodeworda1a2a3a4a5inC1satisfiesthefollowingtwoparity-checkequations:a4= a1+a3;a5=a1+a2+a3. 2LetC2bethefollowingcodein . The first three positions are the information positions, and every codeworda1a2a3a4a5a6satisfiestheparity-checkequationsa4=a2,a5=a1+a2,anda6=a1+a2+a3. #(a) ListthecodewordsofC2. (b) FindtheminimumdistanceofthecodeC2. (c) HowmanyerrorsinanycodewordofC2aresuretothedetected?Explain. 3 Design a code in where the first two positions are information positions. Give the parity-check equations,listthecodewords,andfindtheminimumdistance. Ifaandbareanytwowords,letd(a,b)denotethedistancebetweenaandb.Todecodeareceived wordx(whichmaycontainerrorsoftransmission)meanstofindthecodewordclosesttox,thatis,the codewordasuchthatd(a,x)isaminimum.Thisiscalledmaximum-likelihooddecoding. 4DecodethefollowingwordsinC1:11111,00101,11000,10011,10001,and10111. You may have noticed that the last two words in part 4 had ambiguous decodings: for example, 10111 may be decoded as either 10011 or 00111. This situation is clearly unsatisfactory. We shall see nextwhatconditionswillensurethateverywordcanbedecodedintoonlyonepossiblecodeword. Intheremainingexercises,letCbeacodein ,letmdenotetheminimumdistanceinC,andleta andbdenotecodewordsinC. 5Provethatitispossibletodetectuptom−1errors.(Thatis,ifthereareerrorsoftransmissioninm− 1orfewerpositionsofacodeword,itcanalwaysbedeterminedthatthereceivedwordisincorrect.) #6Bythesphereofradiuskaboutacodewordawemeanthesetofallwordsin whosedistancefrom aisnogreaterthank.ThissetisdenotedbySk (a);hence Sk (a)={x:d(a,x)≤k} If ,provethatanytwospheresofradiust,saySt(a)andSt(b),havenoelementsincommon. [HINT:Assumethereisawordxsuchthatx∈St(a)andx∈St,(b).Usingthedefinitionsoftandm,show thatthisisimpossible.] 7Deducefrompart6thatiftherearetorfewererrorsoftransmissioninacodeword,thereceivedword willbedecodedcorrectly. 8LetC2bethecodedescribedinpart2.(IfyouhavenotyetfoundtheminimumdistanceinC2, do so now.)Usingtheresultsofparts5and7,explainwhytwoerrorsinanycodewordcanalwaysbedetected, andwhyoneerrorinanycodewordcanalwaysbecorrected. CHAPTER FOUR ELEMENTARYPROPERTIESOFGROUPS Is it possible for a group to have two different identity elements? Well, suppose e1 and e2 are identity elementsofsomegroupG.Then e1*e2=e2 becausee1isanidentityelement,and e1*e2=e1 becausee2isanidentityelement Therefore e1=e2 Thisshowsthatineverygroupthereisexactlyoneidentityelement. Cananelementainagrouphavetwodifferentinverses!Well,ifa1anda2arebothinversesofa, then a1*(a*a2)=a1*e=a1 and (a1*a)*a2=e*a2=a2 Bytheassociativelaw,a1*(a*a2)=(a1*a)*a)*a2;hencea1=a2.Thisshowsthatineverygroup, eachelementhasexactlyoneinverse. Uptonowwehaveusedthesymbol*todesignatethegroupoperation.Other,morecommonlyused symbolsare+and·(“plus”and“multiply”).When+isusedtodenotethegroupoperation,wesayweare usingadditivenotation,andwerefertoa+basthesumofaandb.(Rememberthataandbdonothave to be numbers and therefore “sum” does not, in general, refer to adding numbers.) When · is used to denotethegroupoperation,wesayweareusingmultiplicativenotation’,weusuallywriteabinsteadof a-b,andcallabtheproductofaandb.(Onceagain,rememberthat“product”doesnot,ingeneral,refer tomultiplyingnumbers.)Multiplicativenotationisthemostpopularbecauseitissimpleandsavesspace. Intheremainderofthisbookmultiplicativenotationwillbeusedexceptwhereotherwiseindicated.In particular,whenwerepresentagroupbyalettersuchasGorH,itwillbeunderstoodthatthegroup’s operationiswrittenasmultiplication. There is common agreement that in additive notation the identity element is denoted by 0, and the inverseofaiswrittenas−a.(Itiscalledthenegativeofa.)Inmultiplicativenotationtheidentityelement iseandtheinverseofaiswrittenasa−1(“ainverse”).Itisalsoatraditionthat+istobeusedonlyfor commutativeoperations. Themostbasicruleofcalculationingroupsisthecancellationlaw,whichallowsustocancelthe factoraintheequationsab=acandab=ca.Thiswillbeourfirsttheoremaboutgroups. Theorem1IfGisagroupanda,b,careelementsofG,then (i) ab=ac implies b=c and (ii) ba=ca implies b=c Itiseasytoseewhythisistrue:ifwemultiply(ontheleft)bothsidesoftheequationab=acbya −1 ,wegetb=c.Inthecaseofba=ca,wemultiplyontherightbya−1.Thisistheideaoftheproof;now hereistheproof: Part(ii)isprovedanalogously. Ingeneral,wecannotcancelaintheequationab=ca.(Whynot?) Theorem2IfGisagroupanda,bareelementsofG,then ab=e implies a=b−1 and b=a−l Theproofisverysimple:ifab=e,thenab=aa−1sobythecancellationlaw,b=a−1.Analogously, a=b−l. Thistheoremtellsusthatiftheproductoftwoelementsisequaltoe,theseelementsareinversesof eachother.Inparticular,ifaistheinverseofb,thenbistheinverseofa. Thenexttheoremgivesusimportantinformationaboutcomputinginverses. Theorem3IfGisagroupanda,bareelementsofG,then (i) (ab−1=b−1a−1 and (ii)(a−1)−1=a Thefirstformulatellsusthattheinverseofaproductistheproductoftheinversesinreverseorder. Thenextformulatellsusthataistheinverseoftheinverseofa.Theproofof(i)isasfollows: Since the product of ab and b−1a−1 is equal to e, it follows by Theorem 2 that they are each other’s inverses.Thus,(ab)−1=b−1a−1.Theproofof(ii)isanalogousbutsimpler:aa−l=e,sobyTheorem2a istheinverseofa−1,thatis,a=(a−1)−1. The associative law states that the two products a(bc) and (ab)c are equal; for this reason, no confusioncanresultifwedenoteeitheroftheseproductsbywritingabc(withoutanyparentheses),and callabctheproductofthesethreeelementsinthisorder. Wemaynextdefinetheproductofanyfourelementsa,b,c,anddinGby abcd=a(bcd) Bysuccessiveusesoftheassociativelawwefindthat a(bc)d=ab(cd)=(ab)(cd)=(ab)cd Hence the product abed (without parentheses, but without changing the order of its factors) is defined withoutambiguity. Ingeneral,anytwoproducts,eachinvolvingthesamefactorsinthesameorder,areequal.Thenet effectoftheassociativelawisthatparenthesesareredundant. Havingmadethisobservation,wemayfeelfreetouseproductsofseveralfactors,suchasa1a2 ··· an, without parentheses, whenever it is convenient. Incidentally, by using the identity (ab)−l = b−la−l repeatedly,wefindthat IfGisafinitegroup,thenumberofelementsinGiscalledtheorderofG.Itiscustomarytodenote theorderofGbythesymbol |G| EXERCISES RemarkonnotationIntheexercisesbelow,theexponentialnotationanisusedinthefollowingsense:if aisanyelementofagroupG, then a2 means aa, a3 means aaa, and, in general, a1 is the product of n factorsofa,foranypositiveintegern. A.SolvingEquationsinGroups Leta,b,c,andxbeelementsofagroupG.Ineachofthefollowing,solveforxintermsofa,b,andc. ExampleSolvesimultaneously: x2=b and x5=e Fromthefirstequation, b=x2 Squaring, b2=x4 Multiplyingontheleftbyx,xb2=xx4=x5=e.(Note:x5=ewasgiven.) Multiplyingby(b2)−1,xb2(b2)−1.Therefore,x=(b2)−1. Solve: 1axb=c 2x2b=xa−1c Solvesimultaneously: 3x2a=bxc−1 and acx=xac 4ax2=b and x3=e 5x2=a2 and x5=e 6(xax)3=bx and x2a=(xa)−l B.RulesofAlgebrainGroups Foreachofthefollowingrules,eitherprovethatitistrueineverygroupG,orgiveacounterexampleto show that it is false in some groups. (All the counterexamples you need may be found in the group of matrices{I,A,B,C,D,K}describedonpage28.) 1Ifx2=e,thenx=e. 2Ifx2=a2,thenx=a. 3(ab)2=a2b2 4Ifx2=x,thenx=e. 5Foreveryx∈G,thereissomey∈Gsuchthatx=y2.(Thisisthesameassayingthateveryelementof Ghasa“squareroot.”) 6ForanytwoelementsxandyinG,thereisanelementzinGsuchthaty=xz. C.ElementsThatCommute IfaandbareinGandab=ba,wesaythataandbcommute.Assumingthataandbcommute,provethe following: #1a−1andb−1commute. 2aandb−1commute.(HINT:Firstshowthata=b−1ab.) 3acommuteswithab. 4a2commuteswithb2. 5xax−1commuteswithxbx−1,foranyx∈G. 6ab=ba iff aba∈1=b. (Theabbreviationiffstandsfor“ifandonlyif.”Thus,firstprovethatifab = ba, then aba−1 = b. Next,provethatifaba−1=b,thenab=ba.ProceedroughlyasinExerciseA.Thus,assumingab=ba, solveforb.Next,assumingaba−1=b,solveforab.) 7ab=ba iff aba−1b−1=e. †D.GroupElementsandTheirInverses1 LetGbeagroup.Leta,b,cdenoteelementsofG,andletebetheneutralelementofG. 1Provethatifab=e,thenba=e.(HINT:SeeTheorem2.) 2Provethatifabc=e,thencab=eandbca=e. 3Stateageneralizationofparts1and2 Provethefollowing: 4Ifxay=a−1,thenyax=a−1. 5Leta,b,andceachbeequaltoitsowninverse.Ifab=c,thenbc=aandca=b. 6Ifabcisitsowninverse,thenbcaisitsowninverse,andcabisitsowninverse. 7Letaandbeachbeequaltoitsowninverse.Thenbaistheinverseofab. 8a=a−1 iff aa=e.(Thatis,aisitsowninverseiffa2=e.) 9Letc=c−1.Then ab=c iff xy2abc=e. †E.CountingElementsandTheirInverses Let G be a finite group, and let S be the set of all the elements of G which are not equal to their own inverse.Thatis,S={x∈G:x≠x−1}.ThesetScanbedividedupintopairssothateachelementis pairedoffwithitsowninverse.(Seediagramonthenextpage.)Provethefollowing: 1InanyfinitegroupG,thenumberofelementsnotequaltotheirowninverseisanevennumber. 2 The number of elements of G equal to their own inverse is odd or even, depending on whether the numberofelementsinGisoddoreven. 3IftheorderofGiseven,thereisatleastoneelementxinGsuchthatx≠eandx=x−1. Inparts4to6,letGbeafiniteabeliangroup,say,G={e,a1,a2,…,an}.Provethefollowing: 4(a1a2⋯an)2=e 5Ifthereisnoelementx≠einGsuchthatx=x−1,thena1a2⋯an=e. 6Ifthereisexactlyonex≠einGsuchthatx=x−1,thena1a2⋯an=x. †F.ConstructingSmallGroups Ineachofthefollowing,letGbeanygroup.LetedenotetheneutralelementofG. 1Ifa,bareanyelementsofG,proveeachofthefollowing: (a)Ifa2=a,thena=e. (b)Ifab=a,thenb=e. (c)Ifab=b,thena=e. 2Explainwhyeveryrowofagrouptablemustcontaineachelementofthegroupexactlyonce.(HINT: Supposejcappearstwiceintherowofa: Nowusethecancellationlawforgroups.) 3Thereisexactlyonegrouponanysetofthreedistinctelements,saytheset{e,a,b}.Indeed,keepingin mindparts1and2above,thereisonlyonewayofcompletingthefollowingtable.Doso!Youneednot proveassociativity. 4ThereisexactlyonegroupGoffourelements,sayG={e,a,b,c}, satisfying the additional property thatxx=eforeveryx∈G.Usingonlypart1above,completethefollowinggrouptableofG: 5ThereisexactlyonegroupGoffourelements,sayG={e,a,b,c},suchthatxx=eforsomex≠einG, andyy≠eforsomey∈G(say,aa=eandbb≠e).CompletethegrouptableofG,asinthepreceding exercise. 6 Use Exercise E3 to explain why the groups in parts 4 and 5 are the only possible groups of four elements(exceptforrenamingtheelementswithdifferentsymbols). G.DirectProductsofGroups IfGandHareanytwogroups,theirdirectproductisanewgroup,denotedbyG × H, and defined as follows:G×Hconsistsofalltheorderedpairs(x,y)wherexisinGandyisinH.Thatis, G×H={(x,y):x∈G and y∈H} TheoperationofG×Hconsistsofmultiplyingcorrespondingcomponents: (x,y)·(x′y′)=(xx′,yy′) IfGandHaredenotedadditively,itiscustomarytodenoteG×Hadditively: (x,y)+(x′y′)=(x+x′,y+y′) 1ProvethatG×Hisagroupbycheckingthethreegroupaxioms,(Gl)to(G3): (G1) (x1,y1)[(x2,y2)(x3,y3)]=( , ) [(x1,y1)(x2,y2)](x3,y3)=( , ) (G2) LeteGbetheidentityelementofG,andeHtheidentityelementofH. TheidentityelementofG×His(,).Check (G3)Foreach(a,b)∈G×H,theinverseof(a,b)is(,).Check. 2Listtheelementsof 2× 3,andwriteitsoperationtable.(NOTE:Therearesixelements,eachofwhich isanorderedpair.Thenotationisadditive.) #3IfGandHareabelian,provethatG×Hisabelian. 4SupposethegroupsGandHbothhavethefollowingproperty: Everyelementofthegroupisitsowninverse. ProvethatG×Halsohasthisproperty. H.PowersandRootsofGroupElements LetGbeagroup,anda,b∈G.Foranypositiveintegernwedefineanby Ifthereisanelementx∈Gsuchthata=x2,wesaythatahasasquarerootinG.Similarly,ifa=y3for somey∈G,wesayahasacuberootinG.Ingeneral,ahasannthrootinGifa=znforsomez∈G. Provethefollowing: 1 (bab−l)n = banb−l, for every positive integer Prove by induction. (Remember that to prove a formula suchasthisonebyinduction,youfirstproveitforn=l;nextyouprovethatifitistrueforn=k,thenit must be true for n = k + 1. You may conclude that it is true for every positive integer n. Induction is explainedmorefullyinAppendixC.) 2Ifab=ba,then(ab)n=anbnforeverypositiveintegern.Provebyinduction. 3Ifxax=e,then(xa)2n=an. 4Ifa3=e,thenahasasquareroot. 5Ifa2=e,thenahasacuberoot. 6Ifa∈1hasacuberoot,sodoesa. 7Ifx2ax=a−1,thenahasacuberoot.(HINT:Showthatxaxisacuberootofa−1.) 8Ifxax=b,then06hasasquareroot. 1Whentheexercisesinasetarerelated,withsomeexercisesbuildingonprecedingonessothattheymustbedoneinsequence,thisis indicatedwithasymboltinthemargintotheleftoftheheading. CHAPTER FIVE SUBGROUPS LetGbeagroup,andSanonemptysubsetofG.Itmayhappen(thoughitdoesn’thaveto)thattheproduct ofeverypairofelementsofSisinS.Ifithappens,wesaythatSisclosedwithrespecttomultiplication. Then,itmayhappenthattheinverseofeveryelementofSisinS.Inthatcase,wesaythatSisclosedwith respecttoinverses.Ifboththesethingshappen,wecallSasubgroupofG. When the operation of G is denoted by the symbol +, the wording of these definitions must be adjusted: if the sum of every pair of elements of S is in S, we say that S is closed with respect to addition.IfthenegativeofeveryelementofSisinS,wesaythatSisclosedwithrespecttonegatives.If boththesethingshappen,SisasubgroupofG. Forexample,thesetofalltheevenintegersisasubgroupoftheadditivegroup of the integers. Indeed,thesumofanytwoevenintegersisaneveninteger,andthenegativeofanyevenintegerisaneven integer. As another example, * (the group of the nonzero rational numbers, under multiplication) is a subgroupof *(thegroupofthenonzerorealnumbers,undermultiplication).Indeed, *⊆ * because everyrationalnumberisarealnumber.Furthermore,theproductofanytworationalnumbersisrational, andtheinverse(thatis,thereciprocal)ofanyrationalnumberisarationalnumber. Animportantpointtobenotedisthis:ifSisasubgroupofG,theoperationofSisthesameasthe operationofG.Inotherwords,ifaandbareelementsofS,theproductabcomputedinSispreciselythe productabcomputedinG. Forexample,itwouldbemeaninglesstosaythat〈 *,·〉isasubgroupof〈 ,+〉;foralthoughitistrue that *isasubsetof ,theoperationsonthesetwogroupsaredifferent. Theimportanceofthenotionofsubgroupstemsfromthefollowingfact:ifGisagroupandSisa subgroupofG,thenSitselfisagroup. Itiseasytoseewhythisistrue.Tobeginwith,theoperationofG, restricted to elements of S, is certainlyanoperationonS.Itisassociative:forifa,b,andcareinS,theyareinG(becauseS⊆G);but Gisagroup,soa(bc)=(ab)c.Next,theidentityelementeofGisinS(andcontinuestobeanidentity elementinS)forSisnonempty,soScontainsanelementa;butSisclosedwithrespecttoinverses,soS alsocontainsa–1;thus,Scontainsaa–1 = e, because S is closed with respect to multiplication. Finally, everyelementofShasaninverseinSbecauseSisclosedwithrespecttoinverses.Thus,Sisagroup! Onereasonwhythenotionofsubgroupisusefulisthatitprovidesuswithaneasywayofshowing thatcertainthingsaregroups.Indeed,ifGisalreadyknowntobeagroup,andSisasubgroupofG,we may conclude that S is a group without having to check all the items in the definition of “group.” This conclusionisillustratedbythenextexample. Many of the groups we use in mathematics are groups whose elements are functions. In fact, historically,thefirstgroupseverstudiedassuchweregroupsoffunctions. ( )representsthesetofallfunctionsfrom to ,thatis,thesetofallreal-valuedfunctionsofa realvariable.Incalculuswelearnedhowtoaddfunctions:iffandgarefunctionsfrom to ,theirsum isthefunctionf+ggivenby [f+g](x)=f(x)+g(x) foreveryrealnumberx Clearly,f+gisagainafunctionfrom to ,andisuniquelydeterminedbyfandg. ( ),withtheoperation+foraddingfunctions,isthegroup〈 ( ),+〉,orsimply ( ).Thedetailsare simple,butfirst,letusrememberwhatitmeansfortwofunctionstobeequal.Iffandgarefunctionsfrom to ,thenfandgareequal(thatis,f=g)ifandonlyiff(x)=g(x)foreveryrealnumberx. In other words,tobeequal,fandgmustyieldthesamevaluewhenappliedtoeveryrealnumberx. Tocheckthat+isassociative,wemustshowthatf+[g+h]=[f+g]+h,foreverythreefunctions, f,g,andhin ( ).Thismeansthatforanyrealnumberx,{f+[g+h]}(x)={[f+g]+h}(x).Well, {f+[g+h]}(x)=f(x)+[g+h](x)=f(x)+g(x)+h(x) and{[f+g]+h}(x)hasthesamevalue. Theneutralelementof ( )isthefunction givenby (x)=0 foreveryrealnumberx Toshowthat +f=f,onemustshowthat[ +f](x)=f(x)foreveryrealnumberx.Thisistruebecause[ +f](x)= (x)+f(x)=0+f(x)=f(x). Finally,theinverseofanyfunctionfisthefunction–fgivenby [–f](x)=–f(x) foreveryrealnumberx Oneperceivesimmediatelythatf+[–f]= ,foreveryfunctionf. ( )representsthesetofallcontinuousfunctionsfrom to .Now, ( ),withtheoperation+,isa subgroup of ( ), because we know from calculus that the sum of any two continuous functions is a continuousfunction,andthenegative–fofanycontinuousfunctionfisacontinuousfunction.Becauseany subgroupofagroupisitselfagroup,wemayconcludethat ( ),withtheoperation+,isagroup.Itis denotedby〈 ( ),+〉,orsimply ( ). ( ) represents the set of all the differentiable functions from to . It is a subgroup of ( ) becausethesumofanytwodifferentiablefunctionsisdifferentiable,andthenegativeofanydifferentiable functionisdifferentiable.Thus, ( ),withtheoperationofaddingfunctions,isagroupdenotedby〈 ( ), +〉,orsimply ( ). By the way, in any group G the one-element subset {e}, containing only the neutral element, is a subgroup.Itisclosedwithrespecttomultiplicationbecauseee=e,andclosedwithrespecttoinverses because e–1 = e. At the other extreme, the whole group G is obviously a subgroup of itself. These two examplesare,respectively,thesmallestandlargestpossiblesubgroupsofG.Theyarecalledthetrivial subgroupsofG.AlltheothersubgroupsofGarecalledpropersubgroups. Suppose G is a group and a, b, and c are elements of G. Define S to be the subset of G which containsallthepossibleproductsofa,b,c,andtheirinverses,inanyorder,withrepetitionoffactors permitted.Thus,typicalelementsofSwouldbe abac–1 c–1a–1bbc andsoon.ItiseasytoseethatSisasubgroupofG:foriftwoelementsofSaremultipliedtogether,they yieldanelementofS,andtheinverseofanyelementofSisanelementofS.Forexample,theproductof abaandcb–1acis abacb–1ac andtheinverseofab–1c–1ais a–1cba–1 SiscalledthesubgroupofGgeneratedbya,b,andc. Ifa1,…,anareanyfinitenumberofelementsofG,wemaydefinethesubgroupgeneratedbya1, …, an in the same way. In particular, if a is a single element of G, we may consider the subgroup generatedbya.Thissubgroupisdesignatedbythesymbol〈a〉,andiscalledacyclicsubgroupofG;ais calleditsgenerator.Notethat〈a〉consistsofallthepossibleproductsofaanda−1,forexample,a–1aaa– 1andaaa–1aa–1.However,sincefactorsofa–1cancelfactorsofa,thereisnoneedtoconsiderproducts involvingbothaanda–1sidebyside.Thus,〈a〉contains a,aa,aaa,..., a–1,a–1a–1,a–1a–1a1,…, aswellasaa–1=e. If the operation of G is denoted by +, the same definitions can be given with “sums” instead of “products.” Inthegroupofmatriceswhosetableappearsonpage28,thesubgroupgeneratedbyDis〈D〉={I,B, D}andthesubgroupgeneratedbyAis〈A〉={I,A}.(Thestudentshouldcheckthetabletoverifythis.)In fact,theentiregroupGofthatexampleisgeneratedbythetwoelementsAandB. IfagroupGisgeneratedbyasingleelementa,wecallGacyclicgroup,andwriteG − 〈a〉. For example,theadditivegroup 6iscyclic.(Whatisitsgenerator?) Every finite group G is generated by one or more of its elements (obviously). A set of equations, involving only the generators and their inverses, is called a set of defining equations for G if these equationscompletelydeterminethemultiplicationtableofG. Forexample,letGbethegroup{e,a,b,b2,ab,ab2}whosegeneratorsaandbsatisfytheequations a2=e b3=e ba=ab2 (1) These three equations do indeed determine the multiplication table of G. To see this, note first that the equationba=ab2allowsustoswitchpowersofawithpowersofb,bringingpowersofatotheleft,and powersofbtotheright.Forexample,tofindtheproductofabandab2,wecomputeasfollows: ButbyEquations(1),a2=eandb4=b3b=b;sofinally,(ab)(ab2)=b.AlltheentriesinthetableofG maybecomputedinthesamefashion. When a group is determined by a set of generators and defining equations, its structure can be efficientlyrepresentedinadiagramcalledaCayleydiagram.ThesediagramsareexplainedinExercise G. EXERCISES A.RecognizingSubgroups Inparts1–6below,determinewhetherornotHisasubgroupofG.(AssumethattheoperationofHisthe sameasthatofG.) InstructionsIfHisasubgroupofG,showthatbothconditionsinthedefinitionof“subgroup”are satisfied.IfHisnotasubgroupofG,explainwhichconditionfails. ExampleG= *,themultiplicativegroupoftherealnumbers. H={2n:n∈ } His isnot□ asubgroupofG. (i)If2n,2m∈H,then2n2m=2n+m.Butn+m∈ ,so2n+m∈H. (ii)If2n∈H,then1/2n=2–n.But–n∈ ,so2n+m∈H. (Note that in this example the operation of G and H is multiplication. In the next problem, it is addition.) 1G=〈 ,+〉,H={loga:a∈ ,a>0}. His□ isnot□ asubgroupofG. 2G=〈 ,+〉,H={logn:n∈ ,n>0}. His□ isnot□ asubgroupofG. 3G=〈 ,+〉,H={x∈ :tanx∈ }. His□ isnot□ asubgroupofG. HINT:Usethefollowingformulafromtrigonometry: 4G=〈 *,·〉,H={2n3m:m,n∈ }. His□ isnot□ asubgroupofG. 5G=〈 × ,+〉,H={(x,y):y=2x}. His□ isnot□ asubgroupofG. 6G=〈 × ,+〈,H={(x,y):x2+y2>0}. His□ isnot□ asubgroupofG. 7LetCandDbesets,withC⊆D.ProvethatPCisasubgroupofPD.(SeeChapter3,ExerciseC.) B.SubgroupsofFunctions Ineachofthefollowing,showthatHisasubgroupofG. ExampleG=〈 ( ),+〉,H={f∈ ( ):f(0)=0} (i)Supposef,g∈H;thenf(0)=0andg(0)=0,so[f+g](0)=f(0)+g(0)=0+0=0.Thus,f+g∈ H. (ii)Iff∈H,thenf(0)=0.Thus,[–f](0)=–f(0)=–0=0,so–f∈H. 1G=〈 ( ),+〉,H={f∈ ( ):f(x)=0foreveryx∈[0,1]} 2G=〈 ( ),+〉,H={f∈ ( ):f(–x)=–f(x)} 3G=〈 ( ),+〉,H={f∈ ( ):fisperiodicofperiodπ} REMARK:Afunctionfissaidtobeperiodicofperiodaifthereisanumbera,calledtheperiodoff,such thatf(x)=f(x+na)foreveryx∈ andn∈ . 4G=〈 ( ),+〉,H={f∈ ( ): 5G=〈 ( ),+〉,H={f∈ ( ):df/dxisconstant} 6G=〈 ( ),+〉,H={f∈ ( ):f(x)∈ foreveryx∈ } C.SubgroupsofAbelianGroups Inthefollowingexercises,letGbeanabeliangroup. 1IfH={x∈G:x=x–1},thatis,HconsistsofalltheelementsofGwhicharetheirowninverses,prove thatHisasubgroupofG. 2Letnbeafixedinteger,andletH={x∈G:xn=e}.ProvethatHisasubgroupofG. 3LetH={x∈G:x=y2forsomey∈G};thatis,letHbethesetofalltheelementsofGwhichhavea squareroot.ProvethatHisasubgroupofG. 4LetHbeasubgroupofG,andletK={x∈G:x2∈H}.ProvethatKisasubgroupofG. #5.LetHbeasubgroupofG,andletKconsistofalltheelementsxinGsuchthatsomepowerofxisin H.Thatis,K={x∈G:forsomeintegern>0,xn∈H).ProvethatKisasubgroupofG. 6SupposeHandKaresubgroupsofG,anddefineHKasfollows: HK={xy:x∈Handy∈K} ProvethatHKisasubgroupofG. 7Explainwhyparts4–6arenottrueifGisnotabelian. D.SubgroupsofanArbitraryGroup LetGbeagroup. 1IfHandKaresubgroupsofagroupG,provethatH∩KisasubgroupofG.(Rememberthatx∈H∩K iffx∈Handx∈K.) 2LetHandKbesubgroupsofG.ProvethatifH⊆K,thenHisasubgroupofK. 3BythecenterofagroupGwemeanthesetofalltheelementsofGwhichcommutewitheveryelement ofG,thatis, C={a∈G:ax=xaforeveryx∈G) ProvethatCisasubgroupofG. 4LetC′={a∈G:(ax)2=(xa)2foreveryx∈G).ProvethatC′isasubgroupofG. # 5 Let G be a finite group, and let S be a nonempty subset of G. Suppose S is closed with respect to multiplication. Prove that S is a subgroup of G. (HINT: It remains to prove that S contains e and is closedwithrespecttoinverses.LetS={a1,…,an}.Ifai∈S,considerthedistinctelementsaia1,aia2, …,aian.) 6LetGbeagroupandf:G→Gafunction.AperiodoffisanyelementainGsuchthatf(x)=f(ax)for everyx∈G.Prove:ThesetofalltheperiodsoffisasubgroupofG. #7LetHbeasubgroupofG,andletK={x∈G:xax−1∈Hiffa∈H}.Prove: (a)KisasubgroupofG. (b)HisasubgroupofK. 8LetGandHbegroups,andG×Htheirdirectproduct. (a)Provethat{(x,e):x∈G}isasubgroupofG×H. (b)Provethat{(x,x):x∈G}isasubgroupofG×G. E.GeneratorsofGroups 1Listallthecyclicsubgroupsof〈 2Showthat 10,+〉. 10isgeneratedby2and5. 3Describethesubgroupof 12generatedby6and9. 4Describethesubgroupof generatedby10and15. 5Showthat isgeneratedby5and7. 6Showthat 2× 3isacyclicgroup.Showthat 3× 4isacyclicgroup. #7Showthat 2× 4isnotacyclicgroup,butisgeneratedby(1,1)and(1,2). 8SupposeagroupGisgeneratedbytwoelementsaandb.Ifab=ba,provethatGisabelian. F.GroupsDeterminedbyGeneratorsandDefiningEquations #1LetGbethegroup{e,a,b,b2,ab,ab2}whosegeneratorssatisfya2=e,b3=e,ba=ab2.Writethe tableofG. 2LetGbethegroup{e,a,b,b2,b3,ab,ab2,ab3}whosegeneratorssatisfya2=e,b4=e,ba=ab3.Write thetableofG.(GiscalledthedihedralgroupD4.) 3LetGbethegroup{e,a,b,b2,b3,ab,ab2,ab3}whose,generatorssatisfya4 = e, a2 = b2, ba = ab3. WritethetableofG.(Giscalledthequaterniongroup.) 4LetGbethecommutativegroup{e,a,b,c,ab,be,ac,abc}whosegeneratorssatisfya2=b2=c2=e. WritethetableofG. G.CayleyDiagrams Every finite group may be represented by a diagram known as a Cayley diagram. A Cayley diagram consistsofpointsjoinedbyarrows. Thereisonepointforeveryelementofthegroup. Thearrowsrepresenttheresultofmultiplyingbyagenerator. For example, if G has only one generator a (that is, G is the cyclic group 〈a〉), then the arrow → representstheoperation“multiplybya”: e→a→a2→a3··· Ifthegrouphastwogenerators,sayaandb,weneedtwokindsofarrows,say and→,where means “multiplybya,”and→means“multiplybyb.” Forexample,thegroupG={e,a,b,b2,ab,ab2}wherea2=e,b3=e,andba=ab2(seepage47) hasthefollowingCayleydiagram: Movingintheforwarddirectionofthearrow→meansmultiplyingbyb, whereasmovinginthebackwarddirectionofthearrowmeansmultiplyingbyb–1: (Notethat“multiplyingxbyb”isunderstoodtomeanmultiplyingontherightbyb:itmeansxb,notbx.) Itisalsoaconventionthatifa2=e(hencea=a–1),thennoarrowheadisused: forifa=a–1,thenmultiplyingbyaisthesameasmultiplyingbya–1. TheCayleydiagramofagroupcontainsthesameinformationasthegroup’stable.Forinstance,to findtheproduct(ab)(ab2)inthefigureonpage51,westartatabandfollowthepathcorrespondingto ab2(multiplyingbya,thenbyb,thenagainbyb),whichis Thispathleadstob;hence(ab)(ab2)=b. Asanotherexample,theinverseofab2isthepathwhichleadsfromab2backtoe.Wenoteinstantly thatthisisba. Apoint-and-arrowdiagramistheCayleydiagramofagroupiffithasthefollowingtwoproperties: (a)Foreachpointxandgeneratora,thereisexactlyonea-arrowstartingatx,andexactlyonea-arrow ending at x; furthermore, at most one arrow goes from x to another point y. (b) If two different paths startingatx lead to the same destination, then these two paths, starting at any point y, lead to the same destination. Cayleydiagramsareausefulwayoffindingnewgroups. Write the table of the groups having the following Cayley diagrams: (REMARK: You may take any point to represent e, because there is perfect symmetry in a Cayley diagram. Choose e, then label the diagramandproceed.) H.CodingTheory:GeneratorMatrixandParity-CheckMatrixofaCode Forthereaderwhodoesnotknowthesubject,linearalgebrawillbedevelopedinChapter28.However, somerudimentsofvectorandmatrixmultiplicationwillbeneededinthisexercise;theyaregivenhere: Avectorwithn components is a sequence of n numbers: (a1, a2, …, an). The dot product of two vectorswithncomponents,saythevectorsa=(a1,a2,…,an)andb=(b1,b2,…,bn),isdefinedby a·b=(a1,a2,…,an)·(b1,b2,…,bn)=a1b1+a2b2+···+anbn thatis,youmultiplycorrespondingcomponentsandadd.Forexample, (1,4,–2,3)·(6,2,4,–2)=1(6)+4(2)+(–2)4+3(–2)=0 Whena·b=0,asinthelastexample,wesaythataandbareorthogonal. Amatrixisarectangulararrayofnumbers.An“mbynmatrix”(m×n matrix) has m rows and n columns.Forexample, is a 3 × 4 matrix: It has three rows and four columns. Notice that each row of B is a vector with four components,andeachcolumnofBisavectorwiththreecomponents. IfAisanym×nmatrix,leta1,a2,…,anbethecolumnsofA.(EachcolumnofAisavectorwithm components.)Ifxisanyvectorwithmcomponents,xAdenotesthevector xA=(x·a1,x·a2,…,x·an) That is, the components of xA are obtained by dot multiplying x by the successive columns of A. For example,ifBisthematrixofthepreviousparagraphandx=(3,1,–2),thenthecomponentsofxBare thatis,xB=(−7,3,−15,8). IfAisanm×nmatrix,leta(1),a(2),…,a(m)betherowsofA.Ifyisanyvectorwithncomponents, Aydenotesthevector Ay=(y·a(1),y·a(2),…,y·a(m)) Thatis,thecomponentsofAyareobtainedbydotmultiplyingywiththesuccessiverowsofA.(Clearly, AyisnotthesameasyA.)Fromlinearalgebra,A(x+y)=Ax+Ayand(A+B)x=Ax+Bx. WeshallnowcontinuethediscussionofcodesbeguninExercisesFandGofChapter3.Recallthat isthesetofallvectorsoflengthnwhoseentriesare0sand1s.InExerciseF,page32,itwasshown that isagroup.AcodeisdefinedtobeanysubsetCof .Acodeiscalledagroup code if C is a subgroupof .ThecodesdescribedinChapter3,aswellasallthosetobementionedinthisexercise, aregroupcodes. Anm×nmatrixGisageneratormatrixforthecodeCifCisthegroupgeneratedbytherowsof G.Forexample,ifC1isthecodegivenonpage34,itsgeneratormatrixis YoumaycheckthatalleightcodewordsofC1aresumsoftherowsofG1. Recallthateverycodewordconsistsofinformationdigitsandparity-checkdigits.InthecodeC1the firstthreedigitsofeverycodewordareinformationdigits,andmakeupthemessage;thelasttwodigits areparity-checkdigits.Encodingamessageistheprocessofaddingtheparity-checkdigitstotheendof themessage.Ifxisamessage,thenE(x) denotes the encoded word. For example, recall that in C1 the parity-check equations are a4 = a1 + a3 and a5 = a1 + a2 + a3. Thus, a three-digit message a1a2a3 is encodedasfollows: E(a1,a2,a3)=(a1,a2,a3,a1+a3,a1+a2+a3) Thetwodigitsaddedattheendofawordarethosedictatedbytheparitycheckequations.Youmayverify that Thisistrueinallcases:IfGisthegeneratormatrixofacodeandxisamessage,thenE(x)isequaltothe productxG.Thus,encodingusingthegeneratormatrixisveryeasy:yousimplymultiplythemessagexby thegeneratormatrixG. Now,theparity-checkequationsofC1(namely,a4=a1+a3anda5=a1+a2+a3)canbewrittenin theform a1+a3+a4=0 and a1+a2+a3+a5=0 whichisequivalentto (a1,a2,a3,a4,a5)·(1,0,1,1,0)=0 and (a1,a2,a3,a4,a5)·(1,1,1,0,1)=0 The last two equations show that a word a1a2a3a4a5 is a codeword (that is, satisfies the parity-check equations)ifandonlyif(a1,a2,a3,a4,a5)isorthogonaltobothrowsofthematrix: Hiscalledtheparity-checkmatrixofthecodeC1.Thisconclusionmaybestatedasatheorem: Theorem1LetHbetheparity-checkmatrixofacodeCin .Aword x in is a codeword if andonlyifHx=0. (RememberthatHxisobtainedbydotmultiplyingxbytherowsofH.) 1FindthegeneratormatrixG2andtheparity-checkmatrixH2ofthecodeC2describedinExerciseG2of Chapter3. 2 Let C3 be the following code in : the first four positions are information positions, and the paritycheckequationsarea5=a2+a3+a4,a6=a1+a3+a4,anda7=a1+a2+a4.(C3iscalledtheHamming code.)FindthegeneratormatrixG3andparity-checkmatrixH3ofC3. The weight of a word x is the number of Is in the word and is denoted by w(x). For example, w(11011)=4.TheminimumweightofacodeCistheweightofthenonzerocodewordofsmallestweight inthecode.(Seethedefinitionsof“distance”and“minimumdistance”onpage34.)Provethefollowing: #3d(x,y)=w(x+y). 4w(x)=d(x,0),where0isthewordwhosedigitsareall0s. 5TheminimumdistanceofagroupcodeCisequaltotheminimumweightofC. 6(a)Ifxandyhaveevenweight,sodoesx+y. (b)Ifxandyhaveoddweight,x+yhasevenweight. (c)Ifxhasoddandyhasevenweight,thenx+yhasoddweight. 7Inanygroupcode,eitherallthewordshaveevenweight,orhalfthewordshaveevenweightandhalf thewordshaveoddweight.(Usepart6inyourproof.) 8H(x+y)=0ifandonlyifHx=Hy,whereHdenotestheparity-checkmatrixofacodein andxand yareanytwowordsin ). CHAPTER SIX FUNCTIONS The concept of a function is one of the most basic mathematical ideas and enters into almost every mathematicaldiscussion.Afunctionisgenerallydefinedasfollows:IfAandBaresets,thenafunction from A to B is a rule which to every element x in A assigns a unique element y in B. To indicate this connectionbetweenxandyweusuallywritey=f(x),andwecallytheimageofxunderthefunctionf. Thereisnothinginherentlymathematicalaboutthisnotionoffunction.Forexample,imagineAtobe asetofmarriedmenandBtobethesetoftheirwives.Letfbetherulewhichtoeachmanassignshis wife.ThenfisaperfectlygoodfunctionfromAtoB;underthisfunction,eachwifeistheimageofher husband.(Nopunisintended.) Takecare,however,tonotethatiftherewereabachelorinAthenfwouldnotqualifyasafunction fromAtoB;forafunctionfromAtoBmustassignavalueinBtoeveryelementofA,withoutexception. Now,supposethemembersofAandBareAshanti,amongwhompolygamyiscommon;inthelandofthe Ashanti, f does not necessarily qualify as a function, for it may assign to a given member of A several wives.If/isafunctionfromAtoB,itmustassignexactlyoneimagetoeachelementofA. IffisafunctionfromAtoBitiscustomarytodescribeitbywriting f:A→B ThesetAiscalledthedomainoff.TherangeoffisthesubsetofBwhichconsistsofalltheimagesof elementsofA.Inthecaseofthefunctionillustratedhere,{a,b,c}isthedomainoff,and{x,y}isthe range of f(z is not in the range of f). Incidentally, this function f may be represented in the simplified notation ThisnotationisusefulwheneverAisafiniteset:theelementsofAarelistedinthetoprow,andbeneath eachelementofAisitsimage. Itmayperfectlywellhappen,iffisafunctionfromAtoB,thattwoormoreelementsofAhavethe sameimage.Forexample,ifwelookatthefunctionimmediatelyabove,weobservethata and b both havethesameimagex.Iffisafunctionforwhichthiskindofsituationdoesnotoccur,thenfiscalledan injectivefunction.Thus, Definition1Afunctionf:A→BiscalledinjectiveifeachelementofBistheimageofnomore thanoneelementofA. Theintendedmeaning,ofcourse,isthateachelementyinBistheimageofnotwodistinctelements ofA.Soif that is, xl and x2 have the same image y, we must require that xl be equal to x2. Thus, a convenient definitionof“injective”isthis:afunctionf:A→Bisinjectiveifandonlyif f(x1)=f(x2) implies x1=x2 IffisafunctionfromAtoB,theremaybeelementsinBwhicharenotimagesofelementsofA. If thisdoesnothappen,thatis,ifeveryelementofBistheimageofsomeelementofA, we say that f is surjective. Definition2Afunctionf:A→BiscalledsurjectiveifeachelementofBistheimageofatleast oneelementofA. ThisisthesameassayingthatBistherangeoff. Now,supposethatfisbothinjectiveandsurjective.ByDefinitions1and2,eachelementofBisthe imageofatleastoneelementofA,andnomorethanoneelementofA.SoeachelementofBistheimage ofexactlyoneelementofA.Inthiscase,fiscalledabijectivefunction,oraone-to-onecorrespondence. Definition3Afunctionf:A→Biscalledbijectiveifitisbothinjectiveandsurjective. Itisobviousthatunderabijectivefunction,eachelementofAhasexactlyone“partner”inB and each elementofBhasexactlyonepartnerinA. The most natural way of combining two functions is to form their “composite.” The idea is this: supposefisafunctionfromAtoB,andgisafunctionfromBtoC.WeapplyftoanelementxinAand get an element y in B; then we apply g to y and get an element z in C. Thus, z is obtained from x by applyingfandginsuccession.Thefunctionwhich consistsofapplyingfandginsuccessionisafunctionfromAtoC,andiscalledthecompositeoffand g.Moreprecisely, Letf:A→Bandg:B→Cbefunctions.Thecompositefunctiondenotedbyg∘fisafunction fromAtoCdefinedasfollows: [g∘f](x)=g(f(x)) foreveryx∊A Forexample,consideronceagainasetAofmarriedmenandthesetBoftheirwives.LetCbethesetof allthemothersofmembersofB.Letf:A→BbetherulewhichtoeachmemberofAassignshiswife, andg:B→CtherulewhichtoeachwomaninBassignshermother.Theng ∘f is the “mother-in-law function,”whichassignstoeachmemberofAhiswife’smother: Foranother,moreconventional,example,letfandgbethefollowingfunctionsfrom to :f(x) = 2x;g(x)=x+1.(Inotherwords,fistherule“multiplyby2”andgistherule“add1.”)Theircomposites arethefunctionsg∘fandf∘ggivenby [f∘g](x)=f(g(x))=2(x+1) and [g∘f](x)=g(f(x))=2x+1 f∘gandg∘faredifferent:f∘gistherule“add1,thenmultiplyby2,”whereasg∘fistherule“multiply by2andthenadd1.” Itisanimportantfactthatthecompositeoftwoinjectivefunctionsisinjective,thecompositeoftwo surjective functions is surjective, and the composite of two bijective functions is bijective. In other words,iff:A→Bandg:B→Carefunctions,thenthefollowingaretrue: Iffandgareinjective,theng∘fisinjective. Iffandgaresurjective,theng∘fissurjective. Iffandgarebijective,theng∘fisbijective. Letustackleeachoftheseclaimsinturn.Wewillsupposethatfandgareinjective,andprovethatg∘fis injective.(Thatis,wewillprovethatif[g∘f](x)=[g∘f](y),thenx=y.) Suppose[g∘f](x)=[g∘f](y),thatis, g(f(x))=g(f(y)) Becausegisinjective,weget f(x)=f(y) andbecausefisinjective, x=y Next,letussupposethatfandgaresurjective,andprovethatg ∘fissurjective.Whatweneedto showhereisthateveryelementofC is g ∘ f of some element of A. Well, if z ∈ C, then (because g is surjective)x=g(y)forsomey∈B;butfissurjective,soy=f(x)forsomex∈A.Thus, z=g(y)=g(f(x))=[g∘f](x) Finally,iffandgarebijective,theyarebothinjectiveandsurjective.Bywhatwehavealreadyproved,g ∘fisinjectiveandsurjective,hencebijective. AfunctionffromAtoBmayhaveaninverse,butitdoesnothaveto.Theinverseoff,ifitexists,isa functionf−1(“finverse”)fromBtoAsuchthat x=f−1(y) ifandonlyif y=f(x) Roughlyspeaking,iffcarriesxtoythenf−lcarriesytox.Forinstance,returning(forthelasttime)tothe exampleofasetAofhusbandsandthesetBoftheirwives,iff:A→Bistherulewhichtoeachhusband assignshiswife,thenf−l:B→Aistherulewhichtoeachwifeassignsherhusband: Ifwethinkoffunctionsasrules,thenf−1istherulewhichundoeswhateverfdoes.Forinstance,iffis the real-valued function f(x) = 2x, then f−1 is the function f−1(x) = x/2 [or, if preferred, f−1(y) = y/2]. Indeed,therule“divideby2”undoeswhattherule“multiplyby2”does. Whichfunctionshaveinverses,andwhichothersdonot?Iff,afunctionfromAtoB,isnotinjective, itcannothaveaninverse;for“notinjective”meansthereareatleasttwodistinctelementsx1andx2with thesameimagey; Clearly,x1=f−1(y) and x2 = f−1(y) so f−l(y) is ambiguous (it has two different values), and this is not allowedforafunction. Iff,afunctionfromAtoBisnotsurjective,thereisanelementyinBwhichisnotanimageofany elementofA;thusf−l(y)doesnotexist.Sof−1cannotbeafunctionfromB(thatis,withdomainB)toA. Itisthereforeobviousthatiff−lexists,fmustbeinjectiveandsurjective,thatis,bijective.Onthe otherhand,iffisabijectivefunctionfromAtoB,itsinverseclearlyexistsandisdeterminedbytherule thatify=f(x)thenf−1(y)=x. Furthermore,itiseasytoseethattheinverseoffisalsoabijectivefunction.Tosumup: Afunctionf:A→Bhasaninverseifandonlyifitisbijective. Inthatcase,theinversef−lisabijectivefunctionfromBtoA. EXERCISES A.ExamplesofInjectiveandSurjectiveFunctions Eachofthefollowingisafunctionf: → .Determine (a)whetherornotfisinjective,and (b)whetherornotfissurjective. Proveyouranswerineithercase. Example1f(x)=2x fisinjective. PROOFSupposef(a)=f(b),thatis, 2a=2b Then a=b Thereforefisinjective.■ fissurjective. PROOFTakeanyelementy∈ .Theny=2(y/2)=f(y/2). Thus,everyy∈ isequaltof(x)forx=y/2. Thereforefissurjective.■ Example2f(x)=x2 fisnotinjective. PROOFByexhibitingacounterexample:f(2)=4f(−2),although2≠−2.■ fisnotsurjective. PROOFByexhibitingacounterexample:−1isnotequaltof(x)foranyx∈ .■ 1 f(x)=3x+4 2 f(x)=x3+1 3 f(x)=|x| #4 f(x)=x3−3x 5 #6 7 Determinetherangeofeachofthefunctionsinparts1to6. B.Functionson and Determinewhethereachofthefunctionslistedinparts1−4isorisnot(a)injectiveand(b) surjective. ProceedasinExerciseA. 1 f: →(0,∞),definedbyf(x)=ex. 2 f:(0,1)→,definedbyf(x)=tanx. 3 f: → ,definedbyf(x)=theleastintegergreaterthanorequaltox. 4 f: → ,definedby 5 Findabijectivefunctionffromtheset oftheintegerstothesetEoftheevenintegers. C.FunctionsonArbitrarySetsandGroups Determinewhethereachofthefollowingfunctionsisorisnot(a)injectiveand(b)surjective.Proceedas inExerciseA. Inparts1to3,AandBaresets,andA×Bdenotesthesetofalltheorderedpairs(x,y)asxranges overAandyoverB. 1 f:A×B→A,definedbyf(x,y)=x. 2 f:A×B→B×A,definedbyf(x,y)=(y,x). 3 f:A×B,definedbyf(x)=(x,b),wherebisafixedelementofB. 4 Gisagroup,a∈G,andf:G→Gisdefinedbyf(x)=ax. 5 Gisagroupandf:G→Gisdefinedbyf(x)=x−1. 6 Gisagroupandf:G→Gisdefinedbyf(x)=x2. D.CompositeFunctions Inparts1−3findthecompositefunction,asindicated. 1 f: → isdefinedbyf(x)=sinx. g: → isdefinedbyg(x)=ex. Findf∘gandg∘f. 2 AandBaresets;f:A×B→B×Aisgivenbyf(x,y)=(y,x). g:B×A→Bisgivenbyg(y,x)=y. Findg∘f. 3 f:(0,1)→ isdefinedbyf(x)=1/x. g: → isdefinedbyg(x)=Inx. Findg∘f.Explainwhyf∘gisundefined. 4 Inschool,JackandSamexchangednotesinacodefwhichconsistsofspellingeverywordbackwards andinterchangingeveryletterswitht.Alternatively,theyuseacodegwhichinterchangesthelettersa witho,iwithu,ewithy,andswitht.Describethecodesf∘gandg∘f.Aretheythesame? 5 A={a,b,c,d};fandgarefunctionsfromAtoA;inthetabularformdescribedonpage57,theyare givenby Givef∘gandg∘finthesametabularform. 6 Gisagroup,andaandbareelementsofG. f:G→Gisdefinedbyf(x)=ax. g:G→Gisdefinedbyg(x)=bx. Findf∘gandg∘f. 7 Indicatethedomainandrangeofeachofthecompositefunctionsyoufoundinparts1to6. E.InversesofFunctions Eachofthefollowingfunctionsfisbijective.Describeitsinverse. 1f:(0,∞)→(0,∞),definedbyf(x)=1/x. 2f: →(0,∞),definedbyf(x)=ex. 3f: → ,definedbyf(x)=x3+1. 4f: → ,definedby 5A={a,b,c,d},B={1,2,3,4}andf:A→Bisgivenby 6Gisagroup,a∈G,andf:G→Gisdefinedbyf(x)=ax. F.FunctionsonFiniteSets 1ThemembersoftheU.N.PeaceCommitteemustchoose,fromamongthemselves,apresidingofficerof their committee. For each member x, let f(x) designate that member’s choice for officer. If no two membersvotealike,whatistherangeoff? 2LetAbeafiniteset.Explainwhyanyinjectivefunctionf:A→A is necessarily surjective. (Look at part1.) 3IfAisafiniteset,explainwhyanysurjectivefunctionf:A→Aisnecessarilyinjective. 4Arethestatementsinparts2and3truewhenAisaninfiniteset?Ifnot,giveacounterexample. #5IfAhasnelements,howmanyfunctionsaretherefromAtoA?Howmanybijectivefunctionsarethere fromAtoA? G.SomeGeneralPropertiesofFunctions Inparts1to3,letA,B,andCbesets,andletf:A→Bandg:B→Cbefunctions. 1Provethatifg∘fisinjective,thenfisinjective. 2Provethatifg∘fissurjective,thengissurjective. 3 Parts 1 and 2, together, tell us that if g ∘ f is bijective, then f is injective and g is surjective. Is the converseofthisstatementtrue:Ifxy0isinjectiveandgsurjective,isg∘fbijective?(If“yes,”proveit;if “no,”giveacounterexample.) 4Letf:A→Bandg:B→Abefunctions.Supposethaty=f(x)iffx=g(y).Provethatfisbijective,and g=f−1 H.TheoryofAutomata Digitalcomputersandotherelectronicsystemsaremadeupofcertainbasiccontrolcircuits.Underlying suchcircuitsisafundamentalmathematicalnotion,thenotionoffiniteautomata, also known as finitestatemachines. Afiniteautomatonreceivesinformationwhichconsistsofsequencesofsymbolsfromsomealphabet A.Atypicalinputsequenceisawordx=x1x2…xn,wherexlx2,...aresymbolsinthealphabetA. The machine has a set of internal components whose combined state is called the internal state of the machine. At each time interval the machine “reads” one symbol of the incoming input sequence and respondsbygoingintoanewinternalstate:theresponsedependsbothonthesymbolbeingreadandon themachine’spresentinternalstate.LetSdenotethesetofinternalstates;wemaydescribeaparticular machinebyspecifyingafunctionα:S×A→S. If si is an internal state and aj is the symbol currently beingread,thenα(siaj )=sk isthemachine’snextstate.(Thatis,themachine,whileinstatesirespondsto the symbol aj by going into the new state sk .) The function α is called the next-state function of the machine. ExampleLetM1bethemachinewhosealphabetisA={0,1},whosesetofinternalstatesisS={s0, s1},andwhosenext-statefunctionisgivenbythetable (Thetableasserts:Wheninstates0andreading0,remainins0.Whenins0andreading1,gotostates1. Whenins1andreading0,remainins1Whenins1andreading1,gotostates0.) A possible use of M1 is as a parity-check machine, which may be used in decoding information arriving on a communication channel. Assume the incoming information consists of sequences of five symbols, 0s and Is, such as 10111. The machine starts off in state s0. It reads the first digit, 1, and as dictatedbythetableabove,goesintostates1.Thenitreadstheseconddigit,0,andremainsins1.Whenit readsthethirddigit,1,itgoesintostates0,andwhenitreadsthefourthdigit,1,itgoesintos1.Finally,it reachesthelastdigit,whichistheparity-checkdigit:ifthesumofthefirstfourdigitsiseven,theparitycheckdigitis0;otherwiseitis1.Whentheparity-checkdigit,1,isread,themachinegoesintostates0. Endinginstates0indicatesthattheparity-checkdigitiscorrect.Ifthemachineendsins1,thisindicates thattheparity-checkdigitisincorrectandtherehasbeenanerroroftransmission. A machine can also be described with the aid of a state diagram, which consists of circles interconnectedbyarrows:thenotation meansthatifthemachineisinstatesiwhenxisread,itgoesintostatesj .ThediagramofthemachineM1 ofthepreviousexampleis In parts 1−4 describe the machines which are able to carry out the indicated functions. For each machine,givethealphabetA,thesetofstatesS, and the table of the next-state function. Then draw the statediagramofthemachine. #1Theinputalphabetconsistsoffourletters,a,b,c, and d. For each incoming sequence, the machine determineswhetherthatsequencecontainsexactlythreea’s. 2Thesameconditionspertainasinpart1,butthemachinedetermineswhetherthesequencecontainsat leastthreea’s. 3Theinputalphabetconsistsofthedigits0,1,2,3,4.Themachineaddsthedigitsofaninputsequence; theadditionismodulo5(seepage27).Thesumistobegivenbythemachine’sstateafterthelastdigitis read. 4Themachinetellswhetherornotanincomingsequenceof0sandIsendswith111. 5IfMisamachinewhosenext-statefunctionisα,define ᾱasfollows:Ifxisaninputsequenceandthe machine(instatesi.)beginsreadingx,thenᾱ(si,x)isthestateofthemachineafterthelastsymbolofxis read.Forinstance,ifM1isthemachineoftheexamplegivenabove,then ᾱ(s0,11010)=s1.(Themachine isins0beforethefirstsymbolisread;each1altersthestate,butthe0sdonot.Thus,afterthelast0is read,themachineisinstates1.) (a)ForthemachineM1,giveᾱ(s0,x)forallthree-digitsequencesx. (b)Forthemachineofpart1,giveᾱ(si,x)foreachstatesiandeverytwo-lettersequencex. 6WitheachinputsequencexweassociateafunctionTx:S→Scalledastatetransitionfunction,defined asfollows: Tx(si)=ᾱ(si,x) ForthemachineM1oftheexample,ifx=11010,Txisgivenby Tx(s0)=s1 and Tx(s1)=s0 (a) Describe the transition function Tx for the machine M1 and the following sequences: x = 01001, x =10011,x=01010. (b)ExplainwhyM1hasonlytwodistincttransitionfunctions.[Note:Twofunctionsfandg are equal if f(x)=g(x)foreveryx;otherwisetheyaredistinct.] (c) For the machine of part 1, describe the transition function T1 for the following x: x = abbca, x = babac,x=ccbaa. (d)Howmanydistincttransitionfunctionsarethereforthemachineofpart3? I.Automata,Semigroups,andGroups ByasemigroupwemeanasetAwithanassociativeoperation.(Theredoesnotneedtobeanidentity element,nordoelementsnecessarilyhaveinverses.)Everygroupisasemigroup,thoughtheconverseis clearlyfalse.WitheverysemigroupAweassociateanautomatonM=M(A)calledtheautomatonofthe semigroupA.ThealphabetofMisA,thesetofstatesalsoisA,andthenext-statefunctionisα(s,a)=sa [orα(s,a)=s+aiftheoperationofthesemigroupisdenotedadditively]. 1DescribeM( 4).Thatis,givethetableofitsnext-statefunction,aswellasitsstatediagram. 2DescribeM(S3). If M is a machine and S is the set of states of M, the state transition functions of M (defined in ExerciseH6ofthischapter)arefunctionsfromStoS.Inthenextexerciseyouwillbeaskedtoshowthat Ty ∘ Tx = Txy; that is, the composite of two transition functions is a transition function. Since the compositionoffunctionsisassociative[f ∘(g ∘h)=(f ∘g) ∘h],itfollowsthatthesetofalltransition functions, with the operation °, is a semigroup. It is denoted by (M) and called the semigroup of the machineM. 3ProvethatTxy=Ty∘Tx. 4LetM1bethemachineoftheexampleinExerciseHabove.Givethetableofthesemigroup (M1.Does (M1)haveanidentityelement?Is (M1)agroup? 5LetM2bethemachineofExerciseH3.Howmanydistinctfunctionsaretherein (M3)?Givethetable of (M3).Is (M3)agroup?(Why?) 6Findthetableof (M)ifMisthemachinewhosestatediagramis CHAPTER SEVEN GROUPSOFPERMUTATIONS Inthischapterwecontinueourdiscussionoffunctions,butweconfineourdiscussionstofunctionsfroma settoitself.Inotherwords,weconsideronlyfunctionsf:A→AwhosedomainisasetA and whose rangeisinthesamesetA. Tobeginwith,wenotethatanytwofunctionsfandg(fromAtoA)areequalifandonlyiff(x) = g(x)foreveryelementxinA. IffandgarefunctionsfromAtoA,theircompositef∘gisalsoafunctionfromAtoA.Werecallthat itisthefunctiondefinedby [f∘g](x)=f(g(x)) foreveryxinA (1) Itisaveryimportantfactthatthecompositionoffunctionsisassociative.Thus,iff,g,andh are three functionsfromAtoA,then f∘(g∘h)=(f∘g)∘h Toprovethatthefunctionsf∘(g∘h)and(f∘g)∘hareequal,onemustshowthatforeveryelementxin A, {f∘[g∘h]}(x)={[f∘g]∘h}(x) WegetthisbyrepeateduseofEquation(1): By a permutationof a set A we mean a bijective function from A to A, that is, a one-to-one correspondencebetweenAanditself.Inelementary algebrawelearnedtothinkofapermutationasarearrangementoftheelementsofaset.Thus,fortheset {1,2,3,4,5}, we may consider the rearrangement which changes (1,2,3,4,5) to (3,2,1,5,4); this rearrangementmaybeidentifiedwiththefunction which is obviously a one-to-one correspondence between the set {1,2,3,4,5} and itself. It is clear, therefore,thatthereisnorealdifferencebetweenthenewdefinitionofpermutationandtheold.Thenew definition,however,ismoregeneralinaveryusefulwaysinceitallowsustospeakofpermutationsof setsAevenwhenAhasinfinitelymanyelements. InChapter6wesawthatthecompositeofanytwobijectivefunctionsisabijectivefunction.Thus, the composite of any two permutations of A is a permutation of A. It follows that we may regard the operation ∘ofcompositionasanoperationonthesetofallthepermutationsofA.Wehavejustseen thatcompositionisanassociativeoperation.Isthereaneutralelementforcomposition? ForanysetA,theidentityfunctiononA,symbolizedbyεAorsimplyε,isthefunctionx→xwhich carrieseveryelementofAtoitself.Thatis,itisdefinedby ε(x)=x foreveryelement x∈A ItiseasytoseethatεisapermutationofA(itisaone-to-onecorrespondencebetweenAanditself);and iffisanyotherpermutationofA,then f∘ε=f and ε∘f=f Thefirstoftheseequationsassertsthat[f∘ε](x)=f(x)foreveryelementxinA,whichisquiteobvious, since[f∘ε](x)=f(ε(x))=f(x).Thesecondequationisprovedanalogously. WesawinChapter6thattheinverseofanybijectivefunctionexistsandisabijectivefunction.Thus, theinverseofanypermutationofAisapermutationofA.Furthermore,iffisanypermutationofAandf −1 isitsinverse,then f−1∘f=ε and f∘f−1=ε ThefirstoftheseequationsassertsthatforanyelementxinA, [f−1∘f](x)=ε(x) thatis,f−1(f(x))=x: This is obviously true, by the definition of the inverse of a function. The second equation is proved analogously. Letusrecapitulate:Theoperation ∘ofcompositionoffunctionsqualifiesasanoperationontheset ofallthepermutationsofA.Thisoperationisassociative.Thereisapermutationεsuchthatε∘f=fandf ∘ε=fforanypermutationfofA.Finally,foreverypermutationfofAthereisanotherpermutationf−1of Asuchthatf∘f−1=εandf−1∘f=ε.Thus,thesetofallthepermutationsofA,withtheoperation ∘of composition,isagroup. ForanysetA,thegroupofallthepermutationsofAiscalledthesymmetricgrouponA, and it is representedbythesymbolSA.Foranypositiveintegern,thesymmetricgroupontheset{1,2,3,...,n}is calledthesymmetricgrouponηelements,andisdenotedbySn. LetustakealookatS3.First,welistallthepermutationsoftheset{1,2,3}: Thisnotationforfunctionswasexplainedonpage57;forexample, isthefunctionsuchthatβ(1)=3,β(2)=1,andβ(3)=2.Amoregraphicwayofrepresentingthesame functionwouldbe TheoperationonelementsofS3iscomposition.Tofindα∘β,wenotethat Thus Notethatinα∘β,βisappliedfirstandαnext.Agraphicwayofrepresentingthisis TheothercombinationsofelementsofS3maybecomputedinthesamefashion.Thestudentshould checkthefollowingtable,whichisthetableofthegroupS3: ByagroupofpermutationswemeananygroupSAorSn,oranysubgroupofoneofthesegroups. Amongthemostinterestinggroupsofpermutationsarethegroupsofsymmetriesofgeometricfigures.We willseehowsuchgroupsarisebyconsideringthegroupofsymmetriesofthesquare. Wemaythinkofasymmetryofthesquareasanywayofmovingasquaretomakeitcoincidewithits formerposition.Everytimewedothis,verticeswillcoincidewithvertices,soasymmetryiscompletely describedbyitseffectonthevertices. Letusnumbertheverticesasinthefollowingdiagram: ThemostobvioussymmetriesareobtainedbyrotatingthesquareclockwiseaboutitscenterP, through anglesof90°,180°,and270°,respectively.Weindicateeachsymmetryasapermutationofthevertices; thusaclockwiserotationof90°yieldsthesymmetry for this rotation carries vertex 1 to 2,2 to 3, 3 to 4, and 4 to 1. Rotations of 180° and 270° yield the followingsymmetries,respectively: TheremainingsymmetriesareflipsofthesquareaboutitsaxesA,B,C,andD: For example, when we flip the square about the axis A, vertices 1 and 3 stay put, but 2 and 4 change places;sowegetthesymmetry Inthesameway,theotherflipsare and Onelastsymmetryistheidentity whichleavesthesquareasitwas. Theoperationonsymmetriesiscomposition:Ri∘Rj istheresultoffirstperformingRj ,andthenRi. Forexample,R1 ∘R4istheresultoffirstflippingthesquareaboutitsaxisA,thenrotatingitclockwise 90°: Thus,theneteffectisthesameasifthesquarehadbeenflippedaboutitsaxisC. The eight symmetries of the square form a group under the operation ∘ of composition, called the groupofsymmetriesofthesquare. For every positive integer n ≥ 3, the regular polygon with n sides has a group of symmetries, symbolizedbyDn,whichmaybefoundaswedidhere.Thesegroupsarecalledthedihedralgroups.For example,thegroupofthesquareisD4,thegroupofthepentagonisD5,andsoon. Everyplanefigurewhichexhibitsregularitieshasagroupofsymmetries.Forexample,thefollowing figure, has a group of symmetries consisting of two rotations (180° and 360°) and two flips about the indicatedaxes.Artificialaswellasnaturalobjectsoftenhaveasurprisingnumberofsymmetries. Farmorecomplicatedthantheplanesymmetriesarethesymmetriesofobjectsinspace.Modern-day crystallography and crystal physics, for example, rely very heavily on knowledge about groups of symmetriesofthree-dimensionalshapes. Groupsofsymmetryarewidelyemployedalsointhetheoryofelectronstructureandofmolecular vibrations.Inelementaryparticlephysics,suchgroupshavebeenusedtopredicttheexistenceofcertain elementaryparticlesbeforetheywerefoundexperimentally! Symmetries and their groups arise everywhere in nature: in quantum physics, flower petals, cell division,theworkhabitsofbeesinthehive,snowflakes,music,andRomanesquecathedrals. EXERCISES A.ComputingElementsofS6 1Considerthefollowingpermutationsf,g,andhinS6: Computethefollowing: 2f°(g∘h)= 3g∘h−1= 4h∘g−1°f−1= 5g∘g∘g= B.ExamplesofGroupsofPermutations 1LetGbethesubsetofS4consistingofthepermutations ShowthatGisagroupofpermutations,andwriteitstable: 2ListtheelementsofthecyclicsubgroupofS6generatedby 3Findafour-elementabeliansubgroupofS5.Writeitstable. 4ThesubgroupofS5generatedby hassixelements.Listthem,thenwritethetableofthisgroup: C.GroupsofPermutationsof Ineachofthefollowing,Aisasubsetof andGisasetofpermutationsofA.ShowthatGisasubgroup ofSA,andwritethetableofG. 1Aisthesetofallx∈ suchthatx≠0,1.G={ε,f,g},wheref(x)=1/(1−x)andg(x)=(x−1)/x. 2Aisthesetofallthenonzerorealnumbers.G={ε,f,g,h},wheref(x)=1/x,g(x)= −x,andh(x)=– 1/x. 3Aisthesetofalltherealnumbersx≠0,1.G={ε,f,g,h,k},wheref(x)=1 −x,g(x)=1/x, h(x) = 1/(1−x),j(x)=(x−1)/x,andk(x)=x/(x−1). 4Aisthesetofalltherealnumbersx≠0,1,2.GisthesubgroupofSAgeneratedbyf(x)=2 −xandg(x) =2/x.(Ghaseightelements.Listthem,andwritethetableofG.) †D.ACyclicGroupofPermutations Foreachintegern,definefnbyfn(x)=x+n. 1Prove:Foreachintegern,fnisapermutationof ,thatis,fn∈SR. 2Provethatfn∘fm=fn+mand . 3LetG={fn:n∈ }.ProvethatGisasubgroupofSR. 4ProvethatGiscyclic.(IndicateageneratorofG.) †E.ASubgroupofS Foranypairofrealnumbersa≠0andb,defineafunctionfa,basfollows: fa,b(x)=ax+b 1Provethatfa,bisapermutationof ,thatis,fa,b∈S . 2Provethatfa,b∘fc,d=fac,ad+b. 3Provethat . 4LetG={fa,b:a∈ ,b∈ ,a≠0}.ShowthatGisasubgroupofSR. F.SymmetriesofGeometricFigures 1LetGbethegroupofsymmetriesoftheregularhexagon.ListtheelementsofG(thereare12ofthem), thenwritethetableofG. 2LetGbethegroupofsymmetriesoftherectangle.ListtheelementsofG(therearefourofthem),and writethetableofG. 3ListthesymmetriesoftheletterZandgivethetableofthisgroupofsymmetries.Dothesameforthe lettersVandH. 4Listthesymmetriesofthefollowingshape,andgivethetableoftheirgroup. (Assumethatthethreearmsareofequallength,andthethreecentralanglesareequal.) G.SymmetriesofPolynomials Considerthepolynomialp=(x1−x2)2+(x3−x4)2.Itisunalteredwhenthesubscriptsundergoanyofthe followingpermutations: Forexample,thefirstofthesepermutationsreplacespby (x2−x1)2+(x3−x4)2 thesecondpermutationreplacespby(x1−x2)2+(x4−x3)2;andsoon.Thesymmetriesofapolynomial ρareallthepermutationsofthesubscriptswhichleaveρunchanged.Theyformagroupofpermutations. Listthesymmetriesofeachofthefollowingpolynomials,andwritetheirgrouptable. 1 p=x1x2+x2x3 2 p=(x1−x2)(x2−x3)(x1−x3) 3 p=x1x2+x2x3+x1x3 4 p=(x1−x2)(x3−x4) H.PropertiesofPermutationsofaSetA 1LetAbeasetanda∈A.LetGbethesubsetofSAconsistingofallthepermutationsfofAsuchthatf(a) =a.ProvethatGisasubgroupofSA. #2IffisapermutationofAanda∈A,wesaythatfmovesaiff(a)≠a.LetAbeaninfiniteset,andlet GbethesubsetofSAwhichconsistsofallthepermutationsfofAwhichmoveonlyαfinitenumberof elementsofA.ProvethatGisasubgroupofSA. 3LetAbeafiniteset,andBasubsetofA.LetGbethesubsetofSAconsistingofallthepermutationsfof Asuchthatf(x)∈Bforeveryx∈B.ProvethatGisasubgroupofSA. 4Giveanexampletoshowthattheconclusionofpart3isnotnecessarilytrueifAisaninfiniteset. I.AlgebraofKinshipStructures(Anthropology) Anthropologistshaveusedgroupsofpermutationstodescribekinshipsystemsinprimitivesocieties.The algebraicmodelforkinshipstructuresdescribedhereisadaptedfromAnAnatomyofKinshipbyH.C. White.Themodelisbasedonthefollowingassumptions,whicharewidelysupportedbyanthropological research: (i) Theentirepopulationofthesocietyisdividedintoclans.Everypersonbelongstöone,andonlyone, clan.Letuscalltheclansk1,k2,...,kn. (ii) Ineveryclanki,allthemenmustchoosetheirwivesfromamongthewomenofaspecifiedclankj . Wesymbolizethisbywritingw(ki)=kj . (iii) Menfromtwodifferentclanscannotmarrywomenfromthesameclan.Thatis,ifki≠kj ,thenw(ki) ≠w(kj ). (iv) Allthechildrenofacoupleareassignedtosomefixedclan.Soifamanbelongstoclanki,allhis childrenbelongtoaclanwhichwesymbolizebyc(ki). (v) Childrenwhosefathersbelongtodifferentclansmustthemselvesbeindifferentclans.Thatis,ifki≠ kj ,thenc(ki)≠c(kj ). (vi) Amancannotmarryawomanofhisownclan.Thatis,w(ki)≠ki. NowletK={k1,k2,...,kn}bethesetofallthedistinctclans.By(ii),wisafunctionfromKtoK, andby(iv),cisafunctionfromKtoK. By (iii), w is an injective function; hence (see Exercise F2 of Chapter6)wisapermutationofK.Likewise,by(v),cisapermutationofK. LetGbethegroupofpermutationsgeneratedbycandw;thatis,Gconsistsofc,w,c−1,w−1,andall possiblecompositeswhichcanbeformedfromthese—forexample,c ∘w ∘w ∘c−1 ∘w−1. Clearly the identityfunctionεisinGsince,forexample,ε=c∘c−1.Herearetwofinalassumptions: (vii) Everyperson,inanyclan,hasarelationineveryotherclan.Thismeansthatforanykiandkj inK, thereisapermutationαinGsuchthatα(ki)=kj (viii) Rulesofkinshipapplyuniformlytoallclans.Thus,foranyαandβinG,ifα(kj )=β(kj )forsome specificclankj ,itnecessarilyfollowsthatα(ki)=β(ki)foreveryclanki Proveparts1–3: 1Letα∈G.Ifα(ki)=kiforanygivenki,thenα=ε. 2LetaαG.Thereisapositiveintegerm≤nsuchthatαm=ε. [αm=α∘α∘···∘α(mfactorsofα).HINT:Considerα(k1),α2(k1),etc.] 3ThegroupGconsistsofexactlynpermutations. Explainparts4–9. 4Ifapersonbelongstoclanki,thatperson’sfatherbelongstoclanc−1(ki).Ifawomanbelongstoclankj , herhusbandbelongstoclanw−1(kj ). 5Ifanymanisinthesameclanashisson,thenc=ε.Ifanywomanisinthesameclanasherson,thenc= w. 6 If a person belongs to clan ki, the son of his mother’s sister belongs to clan c ∘ w−1 ∘ w ∘ c−1(ki). Concludethatmarriagebetweenmatrilateralparallelcousins(marriagebetweenawomanandthesonof hermother’ssister)isprohibited. 7Marriagebetweenamanandthedaughterofhisfather’ssisterisprohibited. 8Ifmatrilateralcross-cousinsmaymarry(thatis,awomanmaymarrythesonofhermother’sbrother), thenc∘w=w−1∘c. 9Ifpatrilateralcross-cousinsmaymarry(awomanmaymarrythesonofherfather’ssister),thencandw −1 commute. CHAPTER EIGHT PERMUTATIONSOFAFINITESET Permutations of finite sets are used in every branch of mathematics—for example, in geometry, in statistics, in elementary algebra—and they have a myriad of applications in science and technology. Because of their practical importance, this chapter will be devoted to the study of a few special propertiesofpermutationsoffinitesets. Ifnisapositiveinteger,considerasetofnelements.Itmakesnodifferencewhichspecificsetwe consider,justaslongasithasnelements;soletustaketheset{1,2,…,n).Wehavealreadyseenthatthe groupofallthepermutationsofthissetiscalledthesymmetricgrouponηelementsandisdenotedbySn. Intheremainderofthischapter,whenwesay“permutation”wewillinvariablymeanapermutationofthe set{1,2,…,n}foranarbitrarypositiveintegern. One of the most characteristic activities of science (any kind of science) is to try to separate complex things into their simplest component parts. This intellectual “divide and conquer” helps us to understand complicated processes and solve difficult problems. The savvy mathematician never misses thechanceofdoingthiswhenevertheopportunitypresentsitself.Wewillseenowthateverypermutation canbedecomposedintosimplepartscalled“cycles,”andthesecyclesare,inasense,themostbasickind ofpermutations. Webeginwithanexample:take,forinstance,thepermutation andlookathowfmovestheelementsinitsdomain: Noticehowfdecomposesitsdomainintothreeseparatesubsets,sothat,ineachsubset,theelementsare permuted cyclically so as to form a closed chain. These closed chains may be considered to be the componentpartsofthepermutation;theyarecalled“cycles.”(Thiswordwillbecarefullydefinedina moment.)Everypermutationbreaksdown,justasthisonedid,intoseparatecycles. Leta1,a2,…,asbedistinctelementsoftheset{1,2,…,n}.Bythecycle(a1a2…as)wemeanthe permutation of{1,2,…,n}whichcarriesa1toa2,a2toa3,…,as−1toas,andastoa1,whileleavingalltheremaining elementsof{1,2,…,n}fixed. Forinstance,ins6,thecycle(1426)isthepermutation InS5,thecycle(254)isthepermutation Becausecyclesarepermutations,wemayformthecompositeoftwocyclesintheusualmanner.The composite of cycles is generally called their productand it is customary to omit the symbol °. For example,inS5, Actually,itisveryeasytocomputetheproductoftwocyclesbyreasoninginthefollowingmanner:Letus continuewiththesameexample, Rememberthatthepermutationontherightisappliedfirst,andthepermutationontheleftisappliednext. Now, βcarries1to2,andαcarries2to4;henceαβcarries1to4. βcarries2to4,andαcarries4to5;henceαβcarries2to5. βleaves3fixedandsodoesα;henceαβleaves3fixed. βcarries4to1andαleaves1fixed,soαβcarries4to1. βleaves5fixedandαcarries5to2;henceαβcarries5to2. If(a1a2…as)isacycle,theintegersiscalleditslength;thus,(a1a2…as)isacycleoflengths.For example,(1532)isacycleoflength4. Iftwocycleshavenoelementsincommontheyaresaidtobedisjoint.Forexample,(132)and(465) aredisjointcycles,but(132)and(453)arenotdisjoint.Disjointcyclescommute:thatis,if(a1…ar)and (bl…bs)aredisjoint,then Itiseasytoseewhythisistrue:αmovesthea’sbutnotthefc’s,whileβmovestheft’sbutnotthea’s. Thus,ifβcarriesbitobj ,thenαβdoesthesame,andsodoesβαSimilarly,ifacarriesahtoak thenβα doesthesame,andsodoesαβ. Wearenowreadytoprovewhatwasassertedatthebeginningofthischapter:Everypermutation canbedecomposedintocycles—infact,intodisjointcycles.Moreprecisely,wecanstatethefollowing: Theorem1Everypermutationiseithertheidentity,asinglecycle,oraproductofdisjointcycles. Webeginwithanexample,becausetheproofusesthesametechniqueastheexample.Considerthe permutation andletuswritefasaproductofdisjointcycles.Webeginwith1andnotethat Wehavecomeacompletecircleandfoundourfirstcycle,whichis(135).Next,wetakethefirstnumber whichhasn’tyetbeenused,namely,2.Weseethat Again we have come a complete circle and found another cycle, which is (24). The only remaining numberis6,which/leavesfixed.Wearedone: f=(135)(24) Theproofforanypermutationffollowsthesamepatternastheexample.Leta1bethefirstnumber in{1,…,n)suchthatf(a1)≠a1.Leta2=f(a1),a3=f(a2),andsooninsuccessionuntilwecometoour firstrepetition,thatis,untilf(ak )isequaltooneofthenumbersa1,a2,…,ak −1.Sayf(ak )=aiIfaiisnot a1,wehave soaiistheimageoftwoelements,ak andai−1,whichisimpossiblebecausefisbijective.Thus,ai=a1, andthereforef(ak )=a1.Wehavecomeacompletecircleandfoundourfirstcycle,namely,(a1a2···ak ). Next,letb1bethefirstnumberwhichhasnotyetbeenexaminedandsuchthatf(b1)≠b1.Weletb2= f(b1), b3 = f(b2), and proceed as before to obtain the next cycle, say (b1 ··· bt). Obviously (b1 ··· bt) is disjointfrom(a1…,ak ).Wecontinuethisprocessuntilallthenumbersin{1,…,n)havebeenexhausted. Thisconcludestheproof. Incidentally,itiseasytoseethatthisproductofcyclesisunique,exceptfortheorderofthefactors. Nowourcuriositymayprodustoask:onceapermutationhasbeenwrittenasaproductofdisjoint cycles,hasitbeensimplifiedasmuchaspossible?Oristheresomewayofsimplifyingitfurther? Acycleoflength2iscalledatransposition.Inotherwords,atranspositionisacycle(ai,aj )which interchanges the two numbers ai and aj .It is a fact both remarkable and trivial that every cycle can be expressedasaproductofoneormoretranspositions.Infact, (a1a2…ar)=(arar−1)(arar−2)…(ara3)(ara2ara1) whichmaybeverifiedbydirectcomputation.Forexample, (12345)=(54)(53)(52)(51) However, there is more than one way to write a given permutation as a product of transpositions. For example,(12345)mayalsobeexpressedasaproductoftranspositionsinthefollowingways: aswellasinmanyotherways. Thus, every permutation, after it has been decomposed into disjoint cycles, may be broken down further and expressed as a product of transpositions. However, the expression as a product of transpositionsisnotunique,andeventhenumberoftranspositionsinvolvedisnotunique. Nevertheless, when a permutation π is written as a product of transpositions, one property of this expressionisunique:thenumberoftranspositionsinvolvediseitheralwayseven or always odd. (This fact will be proved in a moment.) For example, we have just seen that (12345) can be written as a productoffourtranspositionsandalsoasaproductofsixtranspositions;itcanbewritteninmanyother ways,butalwaysasaproductofanevennumberoftranspositions.Likewise,(1234)canbedecomposed inmanywaysintotranspositions,butalwaysanoddnumberoftranspositions. Apermutationiscalledevenifitisaproductofanevennumberoftranspositions,andoddifitisa productofanoddnumberoftranspositions.Whatweareasserting,therefore,isthateverypermutationis unambiguouslyeitheroddoreven. Thismayseemlikeaprettyuselessfact—butactuallytheveryoppositeistrue.Anumberofgreat theorems of mathematics depend for their proof (at that crucial step when the razor of logic makes its decisivecut)onnoneotherbutthedistinctionbetweenevenandoddpermutations. Webeginbyshowingthattheidentitypermutation,ε,isanevenpermutation. Theorem 2 No matter how ε is written as a product of transpositions, the number of transpositionsiseven. PROOF:Lett1,t2,…,tmbemtranspositions,andsupposethat ε=t1t2…tm (1) Weaimtoprovethatεcanberewrittenasaproductofm−2transpositions.Wewillthenbedone:for ifεwereequaltoaproductofanoddnumberoftranspositions,andwewereabletorewritethisproduct repeatedly, each time with two fewer transpositions, then eventually we would get ε equal to a single transposition(ab),andthisisimpossible. Letxbeanynumeralappearinginoneofthetranspositionst2,…,tm.Lettk =(xa),andsupposetk is thelasttranspositioninEquation(1)(readingfromlefttoright)inwhichxappears: Now, tk −1 is a transposition which is either equal to (xa), or else one or both of its components are differentfromχanda.Thisgivesfourpossibilities,whichwenowtreatasfourseparatecases. CaseI tk −1=(xa). Then tk −1tk = (xa)(xa), which is equal to the identity permutation. Thus, tk −1tk may be removed withoutchangingEquation(1).Asaresult,εisaproductofm−2transpositions,asrequired. CaseII tk −1=(xb)whereb≠x,a. Then tk −1)tk =(xb)(xa) But (xb)(xa)=(xa)(ab) Wereplacetk −1tk by(xa)(ab)inEquation(1).Asaresult,thelastoccurrenceofxisonepositionfurther leftthanitwasatthestart. CaseIIItk −1=(ca),wherec≠x,a. Then tk −1tk =(ca)(xa) But (ca)(xa)=(xc)(ca) Wereplacetk −1tk by(xa)(bc)inEquation(1),asinCaseII. CaseIVtk −1=(bc),whereb≠x,aandc≠x,a Then tk −tk =(bc)(xa) But (bc)(xa)=(xa)(bc) Wereplacetk −1tk by(xa)(bc)inEquation(1),asinCasesIIandIII. InCaseI,wearedone.InCasesII,III,andIV,werepeattheargumentoneormoretimes.Eachtime, thelastappearanceofχisonepositionfurtherleftthanthetimebefore.ThismusteventuallyleadtoCase I.Forotherwise,weendupwiththelast(hencetheonly)appearanceofxbeingint1.Thiscannotbe:for ift1=(xa)andxdoesnotappearint2,…,tm,thenε(x)=a,whichisimpossible!■ (Thebox■isusedtomarktheendingofaproof.) Ourconclusioniscontainedinthenexttheorem. Theorem3Ifπ∈Sn,thenπcannotbebothanoddpermutationandanevenpermutation. Supposeπcanbewrittenastheproductofanevennumberoftranspositions,anddifferentlyasthe productofanoddnumberoftranspositions.Thenthesamewouldbetrueforπ−1Butε = π° π−1: thus, writing π−1 as a product of an even number of transpositions and π as a product of an odd number of transpositions, we get an expression for ε as a product of an odd number of transpositions. This is impossiblebyTheorem2. ThesetofalltheevenpermutationsinSnisasubgroupofSn.ItisdenotedbyAn,andiscalledthe alternatinggroupontheset{1,2,…,n}. EXERCISES A.PracticeinMultiplyingandFactoringPermutations 1ComputeeachofthefollowingproductsinS9.(Writeyouranswerasasinglepermutation.) (a)(145)(37)(682) (b)(17)(628)(9354) (c)(71825)(36)(49) (d)(12)(347) #(e)(147)(1678)(74132) (f)(6148)(2345)(12493) 2Writeeachofthefollowingpermutationsins9asaproductofdisjointcycles: (a) (b) (c) (d) 3ExpresseachofthefollowingasaproductoftranspositionsinS8: (a)(137428) (b)(416)(8235) (c)(123)(456)(1574) (d) 4Ifα=(3714),β=(123),andγ=(24135)ins7,expresseachofthefollowingasaproductofdisjoint cycles: (a)α−1β (b)γ−lα (c)α2β (d)β2αγ (e)γ4 #(f)γ3α−1 (g)β−1γ (h)α−1γ2α (NOTE:α2=α∘α,γ3=γ∘γ∘γ,etc.) 5 In S5, write (12345) in five different ways as a cycle, and in five different ways as a product of transpositions. 6InS5,expresseachofthefollowingasthesquareofacycle(thatis,expressasα2whereαisacycle): (a)(132) (b)(12345) (c)(13)(24) B.Powersof If π is any permutation, we write π ∘ π = π2, π ∘ π ∘ π = π3, etc. The convenience of this notation is evident. 1Computeα−1,α2,α3,α4,α5where (a)α=(123) (b)α=(1234) (c)α=(123456). Inthefollowingproblems,letαbeacycleoflengths,sayα=(α1α2…αs). # 2 Describe all the distinct powers of α. How many are there? Note carefully the connection with additionofintegersmodulos(page27). 3Findtheinverseofa,andshowthatα−1=αs Proveeachofthefollowing: #4α2isacycleiffsisodd. 5Ifsisodd,αisthesquareofsomecycleoflengthα.(Findit.HINT:Showα=αs+1.) 6Ifsiseven,says=2t,thenα2istheproductoftwocyclesoflengtht.(Findthem.) 7Ifsisamultipleofk,says=kt,thenαk istheproductofkcyclesoflengtht. 8Ifsisaprimenumber,everypowerofαisacycle. C.EvenandOddPermutations 1Determinewhichofthefollowingpermutationsiseven,andwhichisodd. (a) (b)(71864) (c)(12)(76)(345) (d)(1276)(3241)(7812) (e)(123)(2345)(1357) Proveeachofthefollowing: 2(a)Theproductoftwoevenpermutationsiseven. (b)Theproductoftwooddpermutationsiseven. (c)Theproductofanevenpermutationandanoddpermutationisodd. 3(a)Acycleoflengthliseveniflisodd. (b)Acycleoflengthlisoddifliseven. 4(a)Ifαandβarecyclesoflength/andra,respectively,thenaßisevenorodddependingonwhetherl +m−2isevenorodd. (b)Ifπ=β1…βrwhereeachβiisacycleoflengthli,thenπisevenorodddependingonwhetherl1+ l2+…+lr−risevenorodd. D.DisjointCycles Ineachofthefollowing,letαandβbedisjointcycles,say α=(a1a2…as) and β=(b1b2…br) Proveparts1−3: 1Foreverypositiveintegern,(αβ)n=αnβn. 2Ifαβ=ε,thenα=εandβ=ε. 3If(αβ)t=ε,thenαt=εandβt=ε(wheretisanypositiveinteger).(Usepart2inyourproof.) 4Findatranspositionγsuchthatαβγisacycle. 5Letγbethesametranspositionasintheprecedingexercise.Showthatayβandγαβarecycles. 6Letαandβbecyclesofoddlength(notdisjoint).Provethatifa2=β2,thenα=β. †E.ConjugateCycles ProveeachofthefollowinginSn: 1 Let α = (a1, …, as) be a cycle and let π be a permutation in Sn. Then παπ−1 is the cycle (π(a1), …, π(as)). Ifαisanycycleandπ any permutation, παπ−1 is called a conjugate of α. In the following parts, let π denoteanypermutationinSn. #2Concludefrompart1:Anytwocyclesofthesamelengthareconjugatesofeachother. 3Ifαandβaredisjointcycles,thenπαπ−1andπβπ−1aredisjointcycles. 4Letσbeaproductα1…αtoftdisjointcyclesoflengthsl1…,lt,respectively.Thenπσπ−1isalsoa productoftdisjointcyclesoflengthsl1,…,lt 5Letα1andα2becyclesofthesamelength.Letβ1andβ2becyclesofthesamelength.Letα1andβ1be disjoint,andletα2andβ2bedisjoint.Thereisapermutationπ∈Snsuchthatα1β1=πα2β2π−1 †F.OrderofCycles 1ProveinSn:Ifα=(a1…as)isacycleoflengths,thenαs=ε,α2s=ε,andα3s=ε.Isαk =εforany positiveintegerk<s?(Explain.) Ifαisanypermutation,theleastpositiveintegernsuchthatαn=εiscalledtheorderofα. 2ProveinSn:Ifα=(α1…as)isanycycleoflengths,theorderofαiss. 3Findtheorderofeachofthefollowingpermutations: (a)(12)(345) (b)(12)(3456) (c)(1234)(56789) 4Whatistheorderofαβ,ifaandβaredisjointcyclesoflengths4and6,respectively?(Explainwhy. Usethefactthatdisjointcyclescommute.) 5Whatistheorderofαβifαandβaredisjointcyclesoflengthsrands,respectively?(Ventureaguess, explain,butdonotattemptarigorousproof.) †G.Even/OddPermutationsinSubgroupsofSn ProveeachofthefollowinginSn: 1Letα1,…,αrbedistinctevenpermutations,andβanoddpermutation.Thenα1β,…,αrβarerdistinct oddpermutations.(SeeExerciseC2.) 2Ifβ1,…,βraredistinctoddpermutations,thenβ1βl,β1β2,…,βlβrarerdistinctevenpermutations. 3InSn,therearethesamenumberofoddpermutationsasevenpermutations.(HINT:Usepart1toprove thatthenumberofevenpermutations≤isthenumberofoddpermutations.Usepart2toprovethereverse ofthatinequality.) 4 The set of all the even permutations is a subgroup of Sn. (It is denoted by An and is called the alternatinggrouponnsymbols.) 5LetHbeanysubgroupofSn.Heithercontainsonlyevenpermutations,orHcontainsthesamenumber ofoddasevenpermutations.(Useparts1and2.) †H.GeneratorsofAnandSn RememberthatinanygroupG,asetSofelementsofGissaidtogenerateGifeveryelementofGcanbe expressedasaproductofelementsinSandinversesofelementsinS.(Seepage47.) 1ProvethatthesetTofallthetranspositionsinSngeneratesSn. #2ProvethatthesetT1={(12),(13),…,(1n)}generatesSn. 3 Prove that every even permutation is a product of one or more cycles of length 3. [HINT: (13)(12) = (123);(12)(34)=(321)(134).]ConcludethatthesetUofallcyclesoflength3generatesAn. 4ProvethatthesetU1={(123),(124),…,(12n)}generatesAn.[HINT:(abc)=(1ca)(1ab),(1ab)=(1b2) (12a)(12b),and(1b2)=(12b)2.] 5Thepairofcycles(12)and(12···n)generatesSn.[HINT:(1···n)(12)(1…n)−1=(23);(12)(23)(12)= (13).] CHAPTER NINE ISOMORPHISM Humanperception,aswellasthe“perception”ofso-calledintelligentmachines,isbasedontheabilityto recognize the same structure in different guises. It is the faculty for discerning, in different objects, the samerelationshipsbetweentheirparts. Thedictionarytellsusthattwothingsare“isomorphic”iftheyhavethesamestructure.Thenotion ofisomorphism—ofhavingthesamestructure—iscentraltoeverybranchofmathematicsandpermeates allofabstractreasoning.Itisanexpressionofthesimplefactthatobjectsmaybedifferentinsubstance butidenticalinform. In geometry there are several kinds of isomorphism, the simplest being congruence and similarity. Twogeometricfiguresarecongruentifthereexistsaplanemotionwhichmakesonefigurecoincidewith theother;theyaresimilarifthereexistsatransformationoftheplane,magnifyingorshrinkinglengthsina fixedratio,which(again)makesonefigurecoincidewiththeother. Wedonotevenneedtoventureintomathematicstomeetsomesimpleexamplesofisomorphism.For instance,thetwopalindromes aredifferent,butobviouslyisomorphic;indeed,thefirstonecoincideswiththesecondifwereplaceM byR,AbyO,andDbyT. Hereisanexamplefromappliedmathematics:Aflownetworkisasetofpoints,witharrowsjoining some of the points. Such networks are used to represent flows of cash or goods, channels of communication,electriccircuits,andsoon.Theflownetworks(A)and(B),below,aredifferent,butcan beshowntobeisomorphic.Indeed,(A)canbemadetocoincidewith(B)ifwesuperimposepoint1on point6,point2onpoint5,point3onpoint8,andpoint4onpoint7.(A)and(B) then coincide in the senseofhavingthesamepointsjoinedbyarrowsinthesamedirection.Thus,network(A)istransformed intonetwork(B)ifwereplacepoints1by6,2by5,3by8,and4by7.Theone-to-onecorrespondence whichcarriesoutthistransformation,namely, iscalledanisomorphismfromnetwork(A)tonetwork(B),forittransforms(A)into(B). Incidentally,theone-to-onecorrespondence is an isomorphism between the two palindromes of the preceding example, for it transforms the first palindromeintothesecond. Ournextandfinalexampleisfromalgebra.ConsiderthetwogroupsG1andG2describedbelow: TableofG1 TableofG2 G1andG2aredifferent,butisomorphic.Indeed,ifinG1wereplace0bye,1bya,and2byb,thenG1 coincideswithG2,thetableofG1beingtransformedintothetableofG2.Inotherwords,theone-to-one correspondence transforms G1 to G2. It is called an isomorphism from G1 to G2. Finally, because there exists an isomorphismfromG1toG2,G1andG2areisomorphictoeachother. Ingeneral,byanisomorphismbetweentwogroupswemeanaone-to-onecorrespondencebetween them which transforms one of the groups into the other. If there exists an isomorphism from one of the groupstotheother,wesaytheyareisomorphic.Letusbemorespecific: IfG1andG2areanygroups,anisomorphismfromG1toG2isaone-to-onecorrespondencef from G1toG2withthefollowingproperty:ForeverypairofelementsaandbinG1, Iff(a)=a′andf(b)=b′thenf(ab)=a′b′ (1) Inotherwords,if/matchesawitha′andbwithb′itmustmatchabwitha′b′. Itiseasytoseethatif/hasthispropertyittransformsthetableofG1intothetableofG2: Thereisanother,equivalentwayoflookingatthissituation:IftwogroupsG1andG2areisomorphic, wecansaythetwogroupsareactuallythesame,exceptthattheelementsofG1havedifferentnamesfrom the elements of G2. G1 becomes exactly G2 if we rename its elements. The function which does the renamingisanisomorphismfromG1toG2.Thus,inourlastexample,if0isrenamede,1isrenameda, and2isrenamed6,G1becomesexactlyG2, with the same table. (Note that we have also renamed the operation:itwascalled+inG1and·inG2.) Bytheway,property(1)maybewrittenmoreconciselyasfollows: f(ab)=f(a)f(b) (2) Sowemaysumupourdefinitionofisomorphisminthefollowingway: DefinitionLetG1andG2be groups. A bijective function f : G1 → G2 with the property that for anytwoelementsaandbinG1, f(ab)=f(a)f(b) (2) iscalledanisomorphismfromG1toG2. IfthereexistsanisomorphismfromG1toG2,wesaythatG1isisomorphictoG2. If there exists an isomorphism f from G1 to G2, in other words, if G1 is isomorphic to G2, we symbolizethisfactbywriting G1≅G2 toberead,“G1isisomorphictoG2.” Howdoesonerecognizeiftwogroupsareisomorphic?Thisisanimportantquestion,andnotquiteso easytoanswerasitmayappear.ThereisnowayofspontaneouslyrecognizingwhethertwogroupsG1 andG2areisomorphic.Rather,thegroupsmustbecarefullytestedaccordingtotheabovedefinition. G1andG2 are isomorphic if there exists an isomorphism from G1 to G2. Therefore, the burden of proof is upon us to find an isomorphism from G1 to G2, and show that it is an isomorphism. In other words,wemustgothroughthefollowingsteps: 1. Makeaneducatedguess,andcomeupwithafunctionf:G1 G2whichlooksasthoughitmightbean isomorphism. 2. Checkthatfisinjectiveandsurjective(hencebijective). 3. Checkthatfsatisfiestheidentity f(ab)=f(a)f(b) Here’s an example: is the group of the real numbers with the operation of addition. pos is the groupofthepositiverealnumberswiththeoperationofmultiplication.Itisaninterestingfactthat and posareisomorphic.Toseethis,letusgothroughthestepsoutlinedabove: 1. Theeducatedguess:Theexponentialfunctionf(x)=exisafunctionfrom to poswhich,ifwerecall itsproperties,mightdothetrick. 2. fisinjective:Indeed,iff(a)=f(b),thatis,ea=eb,then,takingthenaturallogonbothsides,wegeta =b. fissurjective:Indeed,ify∈ pos,thatis,ifyisanypositiverealnumber,theny=elny=f(lny); thus,y=f(x)forx=lny. 3. Itiswellknownthatea+b=ea·eb,thatis, f(a+b)=f(a)·f(b) Incidentally, note carefully that the operation of is +, whereas the operation of pos is ·. That is the reasonwehavetouse+ontheleftsideoftheprecedingequation,and·ontherightsideoftheequation. Howdoesonerecognizewhentwogroupsarenotisomorphic?Inpracticeitisusuallyeasiertoshow thattwogroupsarenotisomorphicthantoshowtheyare.Rememberthatiftwogroupsareisomorphic theyarereplicasofeachother;theirelements(andtheiroperation)maybenameddifferently,butinall otherrespectstheyarethesameandsharethesameproperties.Thus,ifagroupG1hasapropertywhich groupG2doesnothave(orviceversa),theyarenotisomorphic!Herearesomeexamplesofpropertiesto lookoutfor: 1. PerhapsG1iscommutative,andG2isnot. 2. PerhapsG1hasanelementwhichisitsowninverse,andG2doesnot. 3. Perhaps G1 is generated by two elements, whereas G2 is not generated by any choice of two of its elements. 4. Perhaps every element of G1 is the square of an element of G1, whereas G2 does not have this property. This list is by no means exhaustive; it merely illustrates the kind of things to be on the lookout for. Incidentally,thekindofpropertiestowatchforarepropertieswhichdonotdependmerelyonthenames assignedtoindividualelements;forinstance,inourlastexample,0∈G1and0∉G2,butneverthelessG1 andG2areisomorphic. Finally,letusstatetheobvious:ifG1andG2cannotbeputinone-to-onecorrespondence(say,G1 hasmoreelementsthatG2),clearlytheycannotbeisomorphic. Intheearlydaysofmodernalgebratheword“group”hadadifferentmeaningfromthemeaningit hastoday.Inthosedaysagroupalwaysmeantagroupofpermutations.Theonlygroupsmathematicians used were groups whose elements were permutations of some fixed set and whose operation was composition. There is something comforting about working with tangible, concrete things, such as groups of permutationsofaset.Atalltimeswehaveaclearpictureofwhatitisweareworkingwith.Later,asthe axiomaticmethodreshapedalgebra,agroupcametomeananysetwithanyassociativeoperationhaving aneutralelementandallowingeachelementaninverse.Thenewnotionofgrouppleasesmathematicans becauseitissimplerandmoreleanandsparingthantheoldnotionofgroupsofpermutations;itisalso more general because it allows many new things to be groups which are not groups of permutations. However,itishardertovisualize,preciselybecausesomanydifferentthingscanbegroups. Itwasthereforeagreatrevelationwhen,about100yearsago,ArthurCayleydiscoveredthatevery groupisisomorphictoagroupofpermutations.Roughly,thismeansthatthegroupsofpermutationsare actuallyallthegroupsthereare!Everygroupis(orisacarboncopyof)agroupofpermutations.This greatresultisaclassictheoremofmodernalgebra.Asabonanza,itsproofisnotverydifficult. Cayley’sTheoremEverygroupisisomorphictoagroupofpermutations. PROOF: Let G be a group; we wish to show that G is isomorphic to a group of permutations. The first questiontoaskis,“Whatgroupofpermutations?Permutationsofwhatset?”(Afterall,everypermutation mustbeapermutationofsomefixedset.)Well,theonesetwehaveathandisthesetG,sowehadbetter fixourattentiononpermutationsofG.ThewaywematchupelementsofG with permutations of G is quiteinteresting: WitheachelementainGweassociateafunctionπa:G→Gdefinedby πa(x)=ax Inotherwords,πaisthefunctionwhoserulemaybedescribedbythewords“multiplyontheleftbya,” WewillnowshowthatπaisapermutationofG: 1. πaisinjective:Indeed,ifπa(x1)=πa(x2),thenax1=ax2,sobythecancellationlaw,x1=x2. 2. πaissurjective:Forify∈G,theny=a(a−1y)=πa(a−1y).Thus,eachyinGisequaltoπa(x)forx =a−1y. 3. SinceπaisaninjectiveandsurjectivefunctionfromGtoG,πaisapermutationofG. LetusrememberthatwehaveapermutationπaforeachelementainG;forexample,ifbandcareother elementsinG,πbisthepermutation“multiplyontheleftbyb,”πcisthepermutation“multiplyontheleft by c,” and so on. In general, let G* denote the set of all the permutations πa as a ranges over all the elementsofG: G*={πa:a∈G} Observe now that G* is a set consisting of permutations of G—but not necessarily all the permutationsofG.InChapter7weusedthesymbolSGtodesignatethegroupofallthepermutationsof G.WemustshownowthatG*isasubgroupofSG,forthatwillprovethatG*isagroupofpermutations. ToprovethatG*isasubgroupofSG,wemustshowthatG*isclosedwithrespecttocomposition, andclosedwithrespecttoinverses.Thatis,wemustshowthatifπaandπbareanyelementsofG*,their compositeπa∘πbisalsoinG*;andifπaisanyelementofG*,itsinverseisinG*. First,weclaimthatifaandbareanyelementsofG,then πa∘πb=πab (3) Toshowthatπa∘πbandπabarethesame,wemustshowthattheyhavethesameeffectoneveryelement x:thatis,wemustprovetheidentity[πa∘πb](x)=πab(x).Well,[πa∘πb](x)=πa(πb(x))=πa(bx)=a(bx) =(ab)x=πab(x).Thus,πa∘πb=πab;thisprovesthatthecompositeofanytwomembersZaandπbofG* isanothermemberπabofG*.Thus,G*isclosedwithrespecttocomposition. Itiseachtoseethatπeistheidentityfunction:indeed, πe(x)=ex=x Inotherwords,πeistheidentityelementofSG. Finally,byEquation(3), πa∘πa−1=πaa−1=πe SobyTheorem2ofChapter4,theinverseofπaisπa 1.Thisprovesthattheinverseofanymemberπaof G*isanothermemberπa 1ofG*.Thus,G*isclosedwithrespecttoinverses. − − SinceG*isclosedwithrespecttocompositionandinverses,G*isasubgroupofSG. Wearenowinthefinallapofourproof.WehaveagroupofpermutationsG*,anditremainsonlyto show that G is isomorphic to G*. To do this, we must find an isomorphism f : G → G*. Let f be the function f(a)=πa Inotherwords,fmatcheseachelementainGwiththepermutationπainG*.Wecanquicklyshowthatf isanisomorphism: 1. fisinjective:Indeed,iff(a)=f(b)thenπa=πb.Thus,πa(e)=πb(e),thatis,ae=be,so,finally,a=b. 2. fissurjective:Indeed,everyelementofG*issomeπa,andπa=f(a). 3. Lastly,f(ab)=πab=πa∘πb=f(a)∘f(b). Thus,fisanisomorphism,andsoG≅G*.■ EXERCISES A.IsomorphismIsanEquivalenceRelationamongGroups Thefollowingthreefactsaboutisomorphismaretrueforallgroups: (i)Everygroupisisomorphictoitself. (ii)IfG1≅G2,thenG2≅G1. (iii)IfG1≅G2andG2≅G3,thenG1≅G3. Fact(i)assertsthatforanygroupG,thereexistsanisomorphismfromGtoG. Fact(ii)assertsthat,ifthereisanisomorphismffromG1toG2, there must be some isomorphism fromG2toG1.Well,theinverseoffissuchanisomorphism. Fact (iii) asserts that, if there are isomorphisms f : G1 → G2 and g : G2 → G3, there must be an isomorphismfromG1toG3.Onecaneasilyguessthatg ∘fissuchanisomorphism.Thedetailsoffacts (i),(ii),and(iii)areleftasexercises. 1LetGbeanygroup.Ifε:G→Gistheidentityfunction,ε(x)=x,showthatεisanisomorphism. 2LetG1andG2begroups,andf:G1→G2anisomorphism.Showthatf−1:G2→G1isanisomorphism. [HINT:ReviewthediscussionofinversefunctionsattheendofChapter6.Then,forarbitraryelementsc, d∈G2,thereexista,b∈G1,suchthatc=f(a)andd=f(b).Notethata=f−1(c)andb=f−1(d). Show thatf−1(cd)=f−1(c)f−1(d).] 3LetG1,G2,andG3begroups,andletf:G1→G2andg:G2→G3beisomorphisms.Provethatg ∘f: G1→G3isanisomorphism. B.ElementsWhichCorrespondunderanIsomorphism RecallthatanisomorphismffromG1toG2isaone-to-onecorrespondencebetweenG1andG2satisfying f(ab)=f(a)f(b).fmatcheseveryelementofGlwithacorrespondingelementofG2.Itisimportanttonote that: (i) fmatchestheneutralelementofG1withtheneutralelementofG2. (ii) IffmatchesanelementxinG1withyinG2,then,necessarily,fmatchesx−lwithy−1Thatis,ifx↔ y,thenx−l↔y−1. (iii) fmatchesageneratorofG1withageneratorofG2. Thedetailsofthesestatementsarenowleftasanexercise.LetG1andG2begroups,andletf:G1→G2 beanisomorphism. 1Ife1denotestheneutralelementofG1ande2denotestheneutralelementofG2,provethatf(e1) = e2. [HINT:Inanygroup,thereisexactlyoneneutralelement;showthatf(e1)istheneutralelementofG2.] 2ProvethatforeachelementainG1f(a−l)=[f(a)]−1.(HINT:YoumayuseTheorem2ofChapter4.) 3IfG1isacyclicgroupwithgeneratora,provethatG2isalsoacyclicgroup,withgeneratorf(a). C.IsomorphismofSomeFiniteGroups Ineachofthefollowing,GandHarefinitegroups.DeterminewhetherornotG≅H.Proveyouranswer ineithercase. To find an isomorphism from G to H will require a little ingenuity. For example, if G and H are cyclicgroups,itisclearthatwemustmatchageneratoraofGwithageneratorbofH;thatis,f(a)=b. Thenf(aa)=bb,f(aaa)=bbb,andsoon.IfGandHarenotcyclic,wehaveotherways:forexample,if G has an element which is its own inverse, it must be matched with an element of H having the same property.Often,thespecificsofaproblemwillsuggestanisomorphism,ifwekeepoureyesopen. Toprovethataspecificone-to-onecorrespondencef:G→Hisanisomorphism,wemaycheckthat ittransformsthetableofGintothetableofH. #1GisthecheckerboardgamegroupofChapter3,ExerciseD.Histhegroupofthecomplexnumbers {i,−i,1,−1}undermultiplication. 2Gisthesameasinpart1.H= 4. 3GisthegroupP2ofsubsetsofatwo-elementset.(SeeChapter3,ExerciseC.)Hisasinpart1. #4GisS3,Histhegroupofmatricesdescribedonpage28ofthetext. 5GisthecoingamegroupofChapter3,ExerciseE.HisD4,thegroupofsymmetriesofthesquare. 6Gisthegroupofsymmetriesoftherectangle.Hisasinpart1. D.SeparatingGroupsintoIsomorphismClasses Eachofthefollowingisasetoffourgroups.Ineachset,determinewhichgroupsareisomorphictowhich others.Proveyouranswers,anduseExerciseA3whereconvenient. 1 4 2× 2 P2 V [P2denotesthegroupofsubsetsofatwo-elementset.(SeeChapter3,ExerciseC.)Vdenotesthegroupof thefourcomplexnumbers{i,−i,1,−1}withrespecttomultiplication.] 2S3 6 3× 2 ( denotesthegroup{1,2,3,4,5,6}withmultiplicationmodulo7.Theproductmodulo7ofaandbisthe remainderofabafterdivisionby7.) 3 8 P3 2× 2× 2 D4 (D4isthegroupofsymmetriesofthesquare.) 4ThegroupshavingthefollowingCayleydiagrams: E.IsomorphismofInfiniteGroups #1LetEdesignatethegroupofalltheevenintegers,withrespecttoaddition.Provethat ≅E. 2LetGbethegroup{10n:n∈ }withrespecttomultiplication.ProvethatG≅ .(Rememberthatthe operationof isaddition.) 3Provethat = × . 4 We have seen in the text that is isomorphic to pos. Prove that is not isomorphic to * (the multiplicativegroupofthenonzerorealnumbers).(HINT:Considerthepropertiesofthenumber−1in *. Does haveanyelementwiththoseproperties?) 5Provethat isnotisomorphicto . 6Wehaveseenthat ≅ pos.However,provethat isnotisomorphicto pos.( posisthemultiplicative groupofpositiverationalnumbers.) F.IsomorphismofGroupsGivenbyGeneratorsandDefiningEquations IfagroupGisgenerated,say,bya,b,andc,thenasetofequationsinvolvinga,b,andciscalledasetof definingequationsforGiftheseequationscompletelydeterminethetableofG.(SeeendofChapter5.) IfG′isanothergroup,generatedbyelementsa′,b′,andc′satisfyingthesamedefiningequationsasa, b, andc,thenG′hasthesametableasG(becausethetablesofGandG′arecompletelydeterminedbythe definingequations,whicharethesameforGasforG′). Consequently,ifweknowgeneratorsanddefiningequationsfortwogroupsGandG′,andifweare abletomatchthegeneratorsofG with those of G′ so that the defining equations are the same, we may concludethatG≅G′. ProvethatthefollowingpairsofgroupsG,G′areisomorphic. #1G=thesubgroupofS4generatedby(24)and(1234);G′={e,a,b,b2,b3,ab,ab2,ab3}wherea2=e, b4=e,andba=ab3. 2G=S3;G′={e,a,b,ab,aba,abab}wherea2=e,b2=e,andbab=aba. 3G=D4;G′={e,a,b,ab,aba,(ab)2,ba,bab}wherea2=b2=eand(ab)4=e. 4G= 2× 2× 2;G′={e,a,b,c,ab,ac,bc,abc}wherea2=b2=c2=eand(ab)2=(bc)2=(ac)2=e. G.IsomorphicGroupsontheSet 1 G is the set {x ∈ : x ≠ −1} with the operation x * y = x + y + xy. Show that f(x) = x − 1 is an isomorphismfrom *toG.Thus, *≅G. 2Gisthesetoftherealnumberswiththeoperationx*y=x+y+l.Findanisomorphismf: →Gand showthatitisanisomorphism. 3Gisthesetofthenonzerorealnumberswiththeoperationx*y=xy/2.Findanisomorphismfrom *to G. 4Showthatf(x,y)=(−1)yxisanisomorphismfrom pos× 2to *.(REMARK:Tocombineelementsof pos× ,onemultipliesfirstcomponents,addssecondcomponents.)Concludethat *≅ pos× . 2 2 H.SomeGeneralPropertiesofIsomorphism 1LetGandHbegroups.ProvethatG×H≅H×G. #2IfG1≅G2andH1≅H2,thenG1×H1≅G2×H2. 3LetGbeanygroup.ProvethatGisabelianiffthefunctionf(x)=x−1isanisomorphismfromGtoG. 4LetGbeanygroup,withitsoperationdenotedmultiplicatively.LetHbeagroupwiththesamesetasG andletitsoperationbedefinedbyx*y=y·x(where·istheoperationofG).ProvethatG≅H. 5LetcbeafixedelementofG.LetHbeagroupwiththesamesetasG,andwiththeoperationx*y= xcy.Provethatthefunctionf(x)=c−lxisanisomorphismfromGtoH. I.GroupAutomorphisms IfGisagroup,anautomorphismofGisanisomorphismfromGtoG.Wehaveseen(ExerciseA1)that theidentityfunctionε(x)=xisanautomorphismofG.However,manygroupshaveotherautomorphisms besidesthisobviousone. 1Verifythat isanautomorphismof 6. 2Verifythat and areallautomorphismsof 5. 3IfGisanygroup,andaisanyelementofG,provethatf(x)=axa’−1isanautomorphismofG. 4SinceeachautomorphismofGisabijectivefunctionfromGtoG,itisapermutationofG.Provethe set Aut(G) ofalltheautomorphismsofGisasubgroupofSG.(Rememberthattheoperationiscomposition.) J.RegularRepresentationofGroups ByCayley’stheorem,everygroupGisisomorphictoagroupG*ofpermutationsofG. Recall that we matcheachelementainGwiththepermutationπadefinedbyπa=ax,thatis,therule“multiplyontheleft bya.”WeletG*={πa:a∈G};withtheoperation∘ofcompositionitisagroupofpermutations,called theleftregularrepresentationofG.(Itiscalleda“representation”ofGbecauseitisisomorphictoG.) Insteadofusingthepermutationsπa,wecouldjustaswellhaveusedthepermutationsρadefinedby ρa(x)=xa,thatis,“multiplyontherightbya.”ThegroupGρ={ρa:a∈G}iscalledtheright regular representationofG. IfGiscommutative,thereisnodifferencebetweenrightandleftmultiplication,soG*andGρarethe same,andaresimplycalledtheregularrepresentationofG.Also,iftheoperationofGisdenotedby+, thepermutationcorrespondingtoais“adda”insteadof“multiplybya.” ExampleTheregularrepresentationof 3consistsofthefollowingpermutations: Theregularrepresentationof 3hasthefollowingtable: Thefunction iseasilyseentobeanisomorphismfrom 3toitsregularrepresentation. Findtherightandleftregularrepresentationofeachofthefollowinggroups,andcomputetheirtables.(If thegroupisabelian,finditsregularrepresentation.) 1P2,thegroupofsubsetsofatwo-elementset.(SeeChapter3,ExerciseC.) 2 4. 3ThegroupGofmatricesdescribedonpage28ofthetext. CHAPTER TEN ORDEROFGROUPELEMENTS Let G be an arbitrary group, with its operation denoted multiplicadvely. Exponential notation is a convenientshorthand:foranypositiveintegern,wewillagreetolet and a0=e Take care to observe that we are considering only integer exponents, not rational or real exponents. Raising a to a positive power means multiplying a by itself the given number of times. Raising a to a negativepowermeansmultiplyinga−1byitselfthegivennumberoftimes.Raisingatothepower0yields thegroup’sidentityelement. Thesearethesameconventionsusedinelementaryalgebra,andtheyleadtothesamefamiliar“laws ofexponents.” Theorem1:Lawsofexponents if G is a group and a ∈G, the following identities hold for all integersmandn: (i)aman=am+n (ii)(am)n=amn (iii)a−n=(a−l)n=(an)−l Theselawsareveryeasytoprovewhenmandnarepositiveintegers.Indeed, (i) (ii) Next,bydefinitiona−n=a−1···a−1=(a−1)n.Finally,sincetheinverseofaproductistheproductofthe inversesinreverseorder, (an)−1=(aa···a)−1=a−1a−1···a−1=a−n ToproveTheorem1completely,weneedtochecktheothercases,whereeachoftheintegersmand n is allowed to be zero or negative. This routine case-by-case verification is left to the student as ExerciseAattheendofthischapter. In order to delve more deeply into the behavior of exponents we must use an elementary but very importantpropertyoftheintegers:Fromelementaryarithmeticweknowthatwecandivideanyintegerby anypositiveintegertogetanintegerquotientandanintegerremainder.Theremainderisnonnegativeand less than the dividend. For example, 25 may be divided by 8, giving a quotient of 3, and leaving a remainderof1: Similarly,−25maybedividedby8,withaquotientof−4andaremainderof7: Thisimportantprincipleiscalledthedivisionalgorithm.Initspreciseform,itmaybestatedasfollows: Theorem 2: Division algorithm if m and n are integers and n is positive, there exist unique integersqandrsuchthat m=nq+r and 0≥r<n Wecallqthequotient,andrtheremainder,inthedivisionofmbynx. Atthisstagewewilltakethedivisionalgorithmtobeapostulateofthesystemoftheintegers.Later, in Chapter 19, we will turn things around, starting with a simpler premise and proving the division algorithmfromit. LetGbeagroup,andaanelementofG.Letusobservethat Ifthereexistsanonzerointegermsuchthatam=e,thenthereexistsapositiveintegernsuchthatan= e. Indeed,ifam=ewheremisnegative,thena−m=(am)−1=e−1=e.Thus,a−m=ewhere −mispositive. Thissimpleobservationiscrucialinournextdefinition.LetGbeanarbitrarygroup,andaanelementof G: DefinitionIfthereexistsanonzerointegermsuchthatam–e,thentheorderoftheelementais definedtobetheleastpositiveintegernsuchthatan=e. Iftheredoesnotexistanynonzerointegermsuchthatam=e,wesaythatahasorderinfinity. Thus,inanygroupG,everyelementhasanorderwhichiseitherapositiveintegerorinfinity.Ifthe order of a is a positive integer, we say that a has finite order; otherwise, a has infinite order. For example,letusglanceatS3,whosetableappearsonpage72.(Itiscustomarytouseexponentialnotation forcomposition:forinstance,β ∘β=β2,β ∘β ∘β=β3,andsoon.)Theorderofαis2,becauseα2=ε and2isthesmallestpositiveintegerwhichsatisfiesthatequation.Theorderofβis3,becauseβ3=ε,and 3isthelowestpositivepowerofβequaltoε.Itisclear,similarly,thatγhasorder2,δhasorder3,andκ hasorder2.Whatistheorderofε? It is important to note that one speaks of powers of a only when the group’s operation is called multiplication.Whenweuseadditivenotation,wespeakofmultiplesofa instead of powers of a. The positivemultiplesofaarea,a+a,a+a+a,andsoon,whilethenegativemultiplesofaare−a,(−a)+ (−a), (−a) + (−a) + (−a), and so on. In 6, the number 2 has order 3, because 2 + 2 + 2 = 0, and no smallermultipleof2isequalto0.Similarly,theorderof3is2,theorderof4is3,andtheorderof5is 6. In ,thenumber2hasinfiniteorder,becausenononzeromultipleof2isequalto0.Asamatterof fact,in ,everynonzeronumberhasinfiniteorder. Themainfactabouttheorderofelementsisgiveninthenexttwotheorems.Ineachofthefollowing theorems,GisanarbitrarygroupandaisanyelementofG. Theorem3:Powersofa,ifahasfiniteorderiftheorderofaisn,thereareexactlyndifferent powersofa,namely, a0,a,a2,a3,…,an−1 Whatthistheoremassertsisthateverypositiveornegativepowerofaisequaltooneoftheabove, andtheabovearealldifferentfromoneanother. Beforegoingon,rememberthattheorderofainn,hence an=e andnisthesmallestpositiveintegerwhichsatisfiesthisequation. PROOFOFTHEOREM3:Letusbeginbyprovingthateverypowerofaisequaltooneofthepowers a0,a1,a2,…,an−1.Letambeanypowerofa.Usethedivisionalgorithmtodividembyn: m=nq+r 0≤r<n Then am=anq+r=(anqar=(an)qar=eqar=ar Thus,am=ar,andrisoneoftheintegers0,1,2,…,n−1. Next,wewillprovethata0,a1,a2,…,an−1arealldifferent,Supposenot;supposear=as,wherer andsaredistinctintegersbetween0andn −1.Eitherr<sors<r,says<r.Thus0≤s<r<n, and consequently, 0<r−s<n (1) However,thisisimpossible,becausebyEquation(1),r −sisapositiveintegerlessthann,whereasn (theorderofa)isthesmallestpositiveintegersuchthatan=e. Thisprovesthatwecannothavear=aswherer≠s.Thus,a0,a1,a2,…,an−1arealldifferent!■ Theorem4:Powersofa,ifahasinfiniteorderIfahasorderinfinity,thenallthepowersofa aredifferent.Thatis,ifrandsaredistinctintegers,thenar≠as. PROOF:Letrandsbeintegers,andsupposear=as. Butahasorderinfinity,andthismeansthatamisnotequaltoeforanyintegermexpect0.Thus,r−s=0, sor=s.■ Thischapterconcludeswithatechnicalpropertyofexponents,whichisoftremendousimportancein applications. Ifaisanelementofagroup,theorderofaistheleastpositiveintegernsuchthatan=e.Butthere areotherintegers,sayt,suchthata=e.Howaretheyrelatedton?Theanswerisverysimple: Theorem5Supposeanelementainagrouphasordern.Thenat=eifftisamultipleofn(“tisa multipleofn”meansthatt=nqforsomeintegerq). PROOF:Ift=nq,thenat=anq=(an)q=eq=e.Conversely,supposeat=e.Dividetbynusingthe divisionalgorithm: t=nq+r 0≤r<n Then e=at=anq+r=(an)qar=eqar=ar Thus,ar=e,where0≤r<n.Ifr≠0,thenrisapositiveintegerlessthann,whereasn is the smallest positiveintegersuchthatan=e.Thusr=0,andthereforet=nq.■ Ifaisanyelementofagroup,wewilldenotetheorderofaby ord(a) EXERCISES A.LawsofExponents LetGbeagroupanda∈G. 1Provethataman=am+ninthefollowingcases: (a) m=0 (b) m<0andn>0 (c) m<0andn<0 2 Provethat(am)n=amninthefollowingcases: (a) m=0 (b) n=0 (c) m<0andn>0 (d) m>0andn<0 (e) m<0andn<0 3 Provethat(an)−1=a−ninthefollowingcases: (a) n=0 (b) n<0 B.ExamplesofOrdersofElements 1 Whatistheorderof10in 25? 2 Whatistheorderof6in 16? 3 Whatistheorderof inS6 4 Whatistheorderof1in *?Whatistheorderof1in ? 5 IfAisthesetofalltherealnumbersx≠0,1,2,whatistheorderof inSA? 6 Cananelementofaninfinitegrouphavefiniteorder?Explain. 7 In 24,listalltheelements(a)oforder2;(b)oforder3;(c)oforder4;(d)oforder6. C.ElementaryPropertiesofOrder Leta,b,andcbeelementsofagroupG.Provethefollowing: 1 Ord(a)=1 iff a=e. 2 Iford(a)=n,thenan−r=(ar)−1. 3 Ifak =ewherekisodd,thentheorderofaisodd. 4Ord(a)=ord(bab−1). 5 Theorderofa−listhesameastheorderofa. 6 Theorderofabisthesameastheorderofba.[HINT:if thenaistheinverseofx.Thus,ax=e.] 7 Ord(abc)=ord(cab)=ord(bca). 8 Letx=a1a2···an,andletybeaproductofthesamefactors,permutedcyclically.(Thatis,y=ak ak+1 ···ana1···ak−1.)Thenord(x)=ord(y). D.FurtherPropertiesofOrder LetabeanyelementoffiniteorderofagroupG.Provethefollowing: 1 Ifap=ewherepisaprimenumber,thenahasorderp.(a≠e.) 2 Theorderofak isadivisor(factor)oftheorderofa. 3 Iford(a)=km,thenord(ak )=m. 4 Iford(a)=nwherenisodd,thenord(a2)=n. 5Ifahasordern,andar=as,thennisafactorofr−s. 6 IfaistheonlyelementoforderkinG,thenaisinthecenterofG.(HINT:UseExerciseC4.Also,see Chapter4,ExerciseC6.) 7 Iftheorderofaisnotamultipleofra,thentheorderofak isnotamultipleofm.(HINT:Usepart2.) 8 Iford(a)=mkandark =e,thenrisamultipleofm. †E.Relationshipbetweenord(ab),ord(a),andord(b) Let a and b be elements of a group G. Let ord(a) = m and ord(b) = n; let lcm(m, n) denote the least commonmultipleofmandn.Proveparts1–5: 1Ifaandbcommute,thenord(ab)isadivisoroflcm(m,n). 2 If m and n are relatively prime, then no power of a can be equal to any power of b (except for e). (REMARK:Twointegersaresaidtoberelativelyprimeiftheyhavenocommonfactorsexcept±1.)(HINT: UseExerciseD2.) 3Ifmandnarerelativelyprime,thentheproductsaibj (0≤i<m,0≤j<n)arealldistinct. 4Letaandbcommute.Ifmandnarerelativelyprime,thenord(ab)=mn.(HINT:Usepart2.) 5Letaandbcommute.ThereisanelementcinGwhoseorderislcm(m,n).(HINT:Usepart4,above, togetherwithExerciseD3.Letc=aibwhereaiisacertainpowerofa.) 6Giveanexampletoshowthatpart1isnottrueifaandbdonotcommute. Thus,thereisnosimplerelationshipbetweenord(ab),ord(a),andord(b)ifaandbfailtocommute. †F.OrdersofPowersofElements Letabeanelementoforder12inagroupG. 1Whatisthesmallestpositiveintegerksuchthata8k =e?(HINT:UseTheorem5andexplaincarefully!) #2Whatistheorderofa8? 3Whataretheordersofa9,a10,a5? 4Whichpowersofahavethesameorderasa?[Thatis,forwhatvaluesofkisord(ak )=12?] 5 Let a be an element of order m in any group G. What is the order of ak ? (Look at the preceding examples,andgeneralize.Donotprove.) 6 Let a be an element of order m in any group G. For what values of k is ord(ak ) = m? (Look at the precedingexamples.Donotprove.) †G.Relationshipbetweenord(a)andord(ak) From elementary arithmetic we know that every integer may be written uniquely as a product of prime numbers.Twointegersmandnaresaidtoberelativelyprimeiftheyhavenoprimefactorsincommon. (Forexample,15and8arerelativelyprime.)Hereisausefulfact:Ifmandnarerelativelyprime,andm isafactorofnk,thenmisafactorofk.(Indeed,alltheprimefactorsofmarefactorsofnkbutnotofn, hencearefactorsofk.) LetabeanelementoforderninagroupG. 1Provethatifmandnarerelativelyprime,thenamhasordern.(HINT:Ifamk =e, use Theorem 5 and explainwhynmustbeafactorofk.) 2Provethatifamhasordern,thenmandnarerelativelyprime.[HINT:Assumemandnhaveacommon factorq>1,hencewecanwritem=m′qandn=n′q.Explainwhy(am)n=e,andproceedfromthere.] 3Letlbetheleastcommonmultipleofmandn.Letl/m=k.Explainwhy(am)k =e. 4Prove:If(am)t=e,thennisafactorofmt.(Thus,mtisacommonmultipleofmandn.)Concludethat 5Useparts3and4toprovethattheorderofamis[lcm(m,n)]/m. †H.RelationshipbetweentheOrderofaandtheOrderofanykthRootofa LetadenoteanelementofagroupG. 1Letahaveorder12.Provethatifahasacuberoot,saya=b3forsomeb∈G,thenb>hasorder36. {HINT:Showthatb36=e;thenshowthatforeachfactorkof36,bk =eisimpossible.[Example:Ifb12= e,thenb12=(b3)4=a4=e.]Deriveyourconclusionfromthesefacts.} #2Letahaveorder6.IfahasafourthrootinG,saya=b4,whatistheorderofb? 3Letahaveorder10.IfahasasixthrootinG,saya=b6,whatistheorderofb? 4Letahaveordern,andsupposeahasasixthrootinG,saya = bk . Explain why the order of b is a factorofnk. Let 5Provethatnandlarerelativelyprime.[HINT:Supposenandlhaveacommonfactorq>1.Thenn= qn′andl=ql′,so Thusbn,k =e.(Why?)Drawyourconclusionfromthesefacts.] Thus, if a has order n and a has a kth root b, then b has order nk/l, where n and l are relatively prime. 6Letahaveordern.Letkbeanintegersuchthateveryprimefactorofkisafactorofn.Prove:Ifahasa kthrootb,thenord(b)=nk. CHAPTER ELEVEN CYCLICGROUPS IfGisagroupanda∈G,itmayhappenthateveryelementofGisapowerofa.Inotherwords,Gmay consistofallthepowersofa,andnothingelse: G={an:n∈ } Inthatcase,Giscalledacyclicgroup,andaiscalleditsgenerator.Wewrite G=〈a〉 andsaythatGisthecyclicgroupgeneratedbya. IfG=〈a〉isthecyclicgroupgeneratedbya,andahasordern,wesaythatGisacyclicgroupof ordern.Wewillseeinamomentthat,inthatcase,Ghasexactlynelements.IfthegeneratorofG has order infinity, we say that G is a cyclic group of order infinity. In that case, we will see that G has infinitelymanyelements. Thesimplestexampleofacyclicgroupis ,whichconsistsofallthemultiplesof1.(Rememberthat inadditivenotationwespeakof“multiples”insteadof“powers.”) isacyclicgroupoforderinfinity;its generatoris1.Anotherexampleofacyclicgroupis 6,whichconsistsofallthemultiplesof1,added modulo6. 6isacyclicgroupoforder6;1isageneratorof 6,but 6hasanothergeneratortoo.Whatis it? Suppose〈a〉isacyclicgroupwhosegeneratorahasordern.Since〈a〉isthesetofallthepowersof a,itfollowsfromTheorem3ofChapter10that 〈a〉={e,a,a2,...,an−1} Ifwecomparethisgroupwith n,wenoticearemarkableresemblance!Foronething,theyareobviously inone-to-onecorrespondence: Inotherwords,thefunction f(i)=ai isaone-to-onecorrespondencefrom nto〈a〉.Butthisfunctionhasanadditionalproperty,namely, f(i+j)=ai+j =aiaj =f(i)f(j) Thus,fisanisomorphismfrom nto〈a〉.Inconclusion, n ≅〈a〉 Let us review the situation in the case where a has order infinity. In this case, by Theorem 4 of Chapter10, 〈a〉={...,a−2,a−1,e,a,a2,...} Thereisobviouslyaone-to-onecorrespondencebetweenthisgroupand : Inotherwords,thefunction f(i)=ai isaone-to-onecorrespondencefrom to〈a〉.Asbefore,fisanisomorphism,andtherefore ≅〈a〉 Whatwehavejustprovedisaveryimportantfactaboutcyclicgroups;letusstateitasatheorem. Theorem1:Isomorphismofcyclicgroups (i) For every positive integer n, every cyclic group of order n is isomorphic to n. Thus, any two cyclicgroupsofordernareisomorphictoeachother. (ii) Every cyclic group of order infinity is isomorphic to , and therefore any two cyclic groups of orderinfinityareisomorphictoeachother. IfGisanygroupanda∈G,itiseasytoseethat 1.Theproductofanytwopowersofaisapowerofa;foraman=am+n. 2.Furthermore,theinverseofanypowerofaisapowerofa,because(an)−1=a−n. 3.ItthereforefollowsthatthesetofallthepowersofaisasubgroupofG. This subgroup is called the cyclic subgroup of G generated by a. It is obviously a cyclic group, and therefore we denote it by 〈a〉. If the element a has order n, then, as we have seen, 〈a〉 contains the n elements {e, a, a2,. . ., an −1}. If a has order infinity, then 〈a〉 = {. . ., a−2, a−1, e, a, a2,...} and has infinitelymanyelements. Forexample,in ,〈2〉isthecyclicsubgroupof whichconsistsofallthemultiplesof2.In 15,〈3〉 is the cyclic subgroup {0,3,6,9,12} which contains all the multiples of 3. In S3, 〈β〉 = {ε, β, δ}, and containsallthepowersofβ. Can a cyclic group, such as , have a subgroup which is not cyclic? This question is of great importance in our understanding of cyclic groups. Its answer is not obvious, and certainly not selfevident: Theorem2Everysubgroupofacyclicgroupiscyclic. LetG=〈a〉beacyclicgroup,andletHbeanysubgroupofG.WewishtoprovethatHiscyclic. Now, G has a generator a; and when we say that H is cyclic, what we mean is that H too has a generator(callitb),andHconsistsofallthepowersofb.Thegistofthisproof,therefore,istofind a generatorofH,andthencheckthateveryelementofHisapowerofthisgenerator. Hereistheidea:Gisthecyclicgroupgeneratedbya,andHisasubgroupofG.EveryelementofH isthereforeinG,whichmeansthateveryelementofHissomepowerofa.ThegeneratorofHwhichwe aresearchingforisthereforeoneofthepowersofa—oneofthepowersofawhichhappenstobeinH; butwhichone?Obviouslythelowestone!Moreaccurately,thelowestpositivepowerofainH. Andnow,carefully,hereistheproof: PROOF:Letmbethesmallestpositiveintegersuchthatam∈H.Wewillshowthateveryelementof Hisapowerofam,henceamisageneratorofH. LetatbeanyelementofH.Dividetbymusingthedivisionalgorithm: t=mq+r Then Solvingforar, 0≤r<m at=amq+r=amqar ar=(amq)−1at=(am)−qat Butam∈Handat∈H;thus(am)−q∈H. Itfollowsthatar∈H.Butr<mandmisthesmallestpositiveiritegersuchthatam∈H.Sor=0, andthereforet=mq. Weconcludethateveryelementat∈Hisoftheformat=(am)q,thatis,apowerofam.Thus,Histhe cyclicgroupgeneratedbyam.■ Thischapterendswithafinalcommentregardingthedifferentusesoftheword“order”inalgebra. LetGbeagroup;aswehaveseen,theorderofanelementainGistheleastpositiveintegernsuch that (Theorderofaisinfinityifthereisnosuchn.) Earlier,wedefinedtheorderofthegroupGtobethenumberofelementsinG.Rememberthatthe orderofGisdenotedby|G|. Thesearetwoseparateanddistinctdefinitions,nottobeconfusedwithoneanother.Nevertheless, thereisaconnectionbetweenthem:Letabeanelementoforderninagroup.ByChapter10,Theorem3, thereareexactlyndifferentpowersofa,hence〈a〉hasnelements.Thus, If ord(a)=n then |〈a〉|=n Thatis,theorderofacyclicgroupisthesameastheorderofitsgenerator. EXERCISES A.ExamplesofCyclicGroups 1Listtheelementsof〈6〉in 16. 2Listtheelementsof〈f〉inS6,where 3Describethecyclicsubgroup in *.Describethecyclicgroup 4Iff(x)=x+1,describethecyclicsubgroup〈f〉ofSR. in . 5Iff(x)=x+1,describethecyclicsubgroup〈f〉of ( ). 6Showthat −1,aswellas1,isageneratorof .Arethereanyothergeneratorsof ?Explain!Whatare thegeneratorsofanarbitraryinfinitecyclicgroup〈a〉? 7Is *cyclic?Trytoproveyouranswer.HINT:Supposekisageneratorof *: Ifk<1,thenk>k2>k3>⋯ Ifk>1,thenk<k2<k3<⋯ B.ElementaryPropertiesofCyclicGroups Proveeachofthefollowing: #1IfGisagroupofordern,GiscycliciffGhasanelementofordern. 2Everycyclicgroupisabelian. 3IfG=〈a〉andb∈G,theorderofbisafactoroftheorderofa. 4Inanycyclicgroupofordern,thereareelementsoforderkforeveryintegerkwhichdividesn. 5LetGbeanabeliangroupofordermn,wheremandnarerelativelyprime.IfGhasanelementoforder mandanelementofordern,Giscyclic.(SeeChapter10,ExerciseE4.) 6Let〈a〉beacyclicgroupofordern.Ifnandmarerelativelyprime,thenthefunctionf(x)=xm is an automorphismof〈a〉.(HINT:UseExerciseB3andChapter10,Theorem5.) C.GeneratorsofCyclicGroups Foranypositiveintegern,letϕ(n)denotethenumberofpositiveintegerslessthannandrelativelyprime ton.Forexample,1,2,4,5,7,and8arerelativelyprimeto9,soϕ(9)=6.Letahaveordern,andprove thefollowing: 1arisageneratorof〈a〉iffrandnarerelativelyprime.(HINT:SeeChapter10,ExerciseG2.) 2〈a〉hasϕ(n)differentgenerators.(Usepart1inyourproof.) 3Foranyfactormofn,letCm={x∈〈a〉:xm=e).Cmisasubgroupof〈a〈. #4Cmhasexactlymelements.(HINT:UseExerciseB4.) 5Anelementxin〈a〉hasordermiffxisageneratorofCm. 6Thereareϕ(m)elementsofordermin〈a〉.(Useparts1and5.) # 7 Let n = mk. ar has order m iff r = kl where l and m are relatively prime. (HINT: See Chapter 10, ExerciseG1,2.) 8Ifcisanygeneratorof〈a〉,then{cr:risrelativelyprimeton}isthesetofallthegeneratorsof〈a〉. D.ElementaryPropertiesofCyclicSubgroupsofGroups LetGbeagroupandleta,b∈G.Provethefollowing: 1Ifaisapowerofb,saya=bk ,then〈a〉⊆〈b〉. 2Supposeaisapowerofb,saya=bk .Thenbisequaltoapowerofaiff〈a〉=〈b〉. 3Supposea∈〈b〉.Then〈a〉=〈b〉iffaandbhavethesameorder. 4Letord(a)=n,andb=ak .Then〈a〉=〈b〉iffnandkarerelativelyprime. 5Letord(a)=n,andsupposeahasakthroot,saya=bk .Then〈a〉=〈b〉iffkandnarerelativelyprime. 6Anycyclicgroupofordermnhasauniquesubgroupofordern. E.DirectProductsofCyclicGroups LetGandHbegroups,witha∈Gandb∈H.Proveparts1-4: 1If(a,b)isageneratorofG×H,thenaisageneratorofGandbisageneratorofH. 2IfG×Hisacyclicgroup,thenGandHarebothcyclic. 3Theconverseofpart2isfalse.(Giveanexampletodemonstratethis.) 4Letord(a)=mandord(b)=n.Theorderof(a,b)inG×Histheleastcommonmultipleofmandn. (HINT:UseChapter10,Theorem5.Nothingelseisneeded!) 5Concludefrompart4thatifmandnarerelativelyprime,then(a,b)hasordermn. 6Suppose(c,d)∈G×H,wherechasordermanddhasordern.Prove:Ifmandnarenot relatively prime(hencehaveacommonfactorq>1),thentheorderof(c,d)islessthanmn. 7Concludefromparts5and6that〈a〉×〈b〉iscyclicifford(a)andord(b)arerelativelyprime. 8LetGbeanabeliangroupofordermn,wheremandnarerelativelyprime.Prove:IfGhasanelement aofordermandanelementbofordern,thenG≅〈a〉×〈b〉.(HINT:SeeChapter10,ExerciseE1,2.) 9Let〈a〉beacyclicgroupofordermn,wheremandnarerelativelyprime,andprovethat〈a〉≅〈am〉× 〈an〉. †F.kthRootsofElementsinaCyclicGroup Let〈a〉beacyclicgroupofordern.Foranyintegerk,wemayask:whichelementsin〈a〉haveakthroot? Theexerciseswhichfollowwillanswerthisquestion. 1Letahaveorder10.Forwhatintegersk(0≤k≤12),doesahaveakthroot?Forwhatintegersk(0≤k ≤12),doesa6havefcthroot? Letkandnbeanyintegers,andletgcd(k,n)denotethegreatestcommondivisorofkandn.Alinear combinationofkandnisanyexpressionck+dnwherecanddareintegers.Itisasimplefactofnumber theory(theproofisgivenonpage219),thatanintegermisequaltoalinearcombinationofkandniffm isamultipleofgcd(k,n).Usethisfacttoprovethefollowing,whereaisanelementoforderninagroup G. 2Ifmisamultipleofgcd(k,n),thenamhasakth root in 〈a〉. [HINT: Compute am, and show that am = (ac)k forsomeac∈〈a〉.] #3Ifamhasakthrootin〈a〉,thenmisamultipleofgcd(k,n).Thus,amhasakthrootin〈a〉iffgcd(k,n) isafactorofm. 4ahasakthrootin〈a〉iffkandnarerelativelyprime. 5Letpbeaprimenumber. (a)Ifnisnotamultipleofp,theneveryelementin〈a〉hasapthroot. (b)Ifnisamultipleofp,andamhasapthroot,thenmisamultipleofp. (Thus,theonlyelementsin〈a〉whichhavepthrootsaree,ap,a2p,etc.) 6Thesetofalltheelementsin〈a〉havingakthrootisasubgroupof〈a〉.Explainwhythissubgroupis cyclic,say〈am〉.Whatisthevalueofm?(Usepart3.) CHAPTER TWELVE PARTITIONSANDEQUIVALENCERELATIONS Imagineemptyingajarofcoinsontoatableandsortingthemintoseparatepiles,onewiththepennies, onewiththenickels,onewiththedimes,onewiththequarters,andonewiththehalf-dollars.Thisisa simple example of partitioning a set. Each separate pile is called a class of the partition; the jarful of coinshasbeenpartitionedintofiveclasses. Herearesomeotherexamplesofpartitions:Thedistributoroffarm-fresheggsusuallysortsthedaily supplyaccordingtosize,andseparatestheeggsintothreeclassescalled“large,”“medium,”and“small.” ThedelegatestotheDemocraticnationalconventionmaybeclassifiedaccordingtotheirhomestate, thusfallinginto50separateclasses,oneforeachstate. A student files class notes according to subject matter; the notebook pages are separated into four distinctcategories,marked(letussay)“algebra,”“psychology,”“English,”and“Americanhistory.” Every time we file, sort, or classify, we are performing the simple act of partitioning a set. To partition a set A is to separate the elements of A into nonempty subsets, say A1, A2, A3 etc., which are calledtheclassesofthepartition.Anytwodistinctclasses,sayAiandAj aredisjoint,whichmeansthey havenoelementsincommon.AndtheunionoftheclassesisallofA. Instead of dealing with the process of partitioning a set (which is awkward mathematically), it is more convenient to deal with the result of partitioning a set. Thus, {A1, A2, A3, A4}, in the illustration above,iscalledapartitionofA.Wethereforehavethefollowingdefinition: ApartitionofasetAisafamily{Ai:i∈I}ofnonemptysubsetsofAwhicharemutuallydisjoint andwhoseunionisallofA. The notation {Ai: i ∈ I} is the customary way of representing a family of sets {Ai, Aj , Ak , …} consistingofonesetAiforeachindexiinI.(TheelementsofIarecalledindices’,thenotation{Ai:i∈ I}mayberead:thefamilyofsetsAi,asirangesoverI.) Let{Ai:i∈I}beapartitionofthesetA.Wemaythinkoftheindicesi,j,k,…aslabelsfornaming theclassesAi,Aj ,Ak ,.…Now,inpracticalproblems,itisveryinconvenienttoinsistthateachclassbe namedonceandonlyonce.Itissimplertoallowrepetitionofindexingwheneverconveniencedictates. Forexample,thepartitionillustratedpreviouslymightalsoberepresentedlikethis,whereA1isthesame classasA5,A2isthesameasA6,andA3isthesameasA7. Aswehaveseen,anytwodistinctclassesofapartitionaredisjoint;thisisthesameassayingthatif twoclassesarenotdisjoint,theymustbeequal.TheotherconditionfortheclassesofapartitionofAis thattheirunionmustbeequaltoA;thisisthesameassayingthateveryelementofAliesinoneofthe classes.Thus,wehavethefollowing,moreexplicitdefinitionofpartition: ByapartitionofasetAwemeanafamily{Ai:i∈I}ofnonemptysubsetsofAsuchthat (i) Ifanytwoclasses,sayAiandAj ,haveacommonelementx(thatis,arenotdisjoint),thenAi= Aj and (ii) EveryelementxofAliesinoneoftheclasses. Wenowturntoanotherelementaryconceptofmathematics.Arelation on a set A is any statement which is either true or false for each ordered pair (x, y) of elements of A. Examples of relations, on appropriate sets, are “x = y,” “x < y,” “x is parallel to y,” “x is the offspring of y,” and so on. An especiallyimportantkindofrelationonsetsisanequivalencerelation.Sucharelation,willusuallybe representedbythesymbol∼,sothatx∼ymayberead“xisequivalenttoy.”Hereishowequivalence relationsaredefined: ByanequivalencerelationonasetAwemeanarelation∼whichis Reflexive:thatis,x∼xforeveryxinA; Symmetric:thatis,ifx∼y,theny∼x;and Transitive:thatis,ifx∼yandy∼z,thenx∼z. Themostobviousexampleofanequivalencerelationisequality,buttherearemanyotherexamples,as weshallbeseeingsoon.Someexamplesfromoureverydayexperienceare“xweighsthesameasy,”“x isthesamecolorasy,”“xissynonymouswithy,”andsoon. Equivalencerelationsalsoariseinanaturalwayoutofpartitions.Indeed,if{Ai:i∈I}isapartition ofA,wemaydefineanequivalencerelation∼onAbylettingx∼yiffxandyareinthesameclassof thepartition. Inotherwords,wecalltwoelements“equivalent”iftheyaremembersofthesameclass.Itiseasytosee thattherelation∼isanequivalencerelationonA.Indeed,x∼xbecausexisinthesameclassasx;next, ifxandyareinthesameclass,thenyandxareinthesameclass;finally,ifxandyareinthesameclass, andyandzareinthesameclass,thenxandzareinthesameclass.Thus,∼isanequivalencerelationon A;itiscalledtheequivalencerelationdeterminedbythepartition{Ai:i∈I}. Let∼beanequivalencerelationonAandxanelementofA.Thesetofalltheelementsequivalentto xiscalledtheequivalenceclassofx,andisdenotedby[x].Insymbols, [x]={y∈A:y∼x} Ausefulpropertyofequivalenceclassesisthis: LemmaIfx∼y,then[x]=[y]. Inotherwords,iftwoelementsareequivalent,theyhavethesameequivalenceclass.Theproofofthis lemma is fairly obvious, for if x ∼ y, then the elements equivalent to x are the same as the elements equivalenttoy. Forexample,letusreturntothejarfulofcoinswediscussedearlier.IfAisthesetofcoinsinthejar, call any two coins “equivalent” if they have the same value: thus, pennies are equivalent to pennies, nickelsareequivalenttonickels,andsoon.IfxisaparticularnickelinA,then[x],theequivalenceclass ofx,istheclassofallthenickelsinA.Ifyisaparticulardime,then[y]isthepileofallthedimes;andso forth.Thereareexactlyfivedistinctequivalenceclasses.Ifweapplythelemmatothisexample,itstates simplythatiftwocoinsareequivalent(thatis,havethesamevalue),theyareinthesamepile.Bythe way, the five equivalence classes obviously form a partition of A; this observation is expressed in the nexttheorem. TheoremIf∼isanequivalencerelationonA,thefamilyofalltheequivalenceclasses, that is, {[x]:x∈A),isapartitionofA. This theorem states that if ∼ is an equivalence relation on A and we sort the elements of A into distinctclassesbyplacingeachelementwiththeonesequivalenttoit,wegetapartitionofA. Toprovethetheorem,weobservefirstthateachequivalenceclassisanonemptysubsetofA.(Itis nonemptybecausex∼x,sox∈[x].)Next,weneedtoshowthatanytwodistinctclassesaredisjoint— or,equivalently,thatiftwoclasses[x]and[y]haveacommonelement,theyareequal.Well,if[x]and[y] haveacommonelementu,thenu∼xandu∼y.Bythesymmetricandtransitivelaws,x∼y.Thus,[x]= [y]bythelemma. Finally,wemustshowthateveryelementofAliesinsomeequivalenceclass.Thisistruebecausex ∈[x].Thus,thefamilyofalltheequivalenceclassesisapartitionofA. When∼isanequivalencerelationonAandA is partitioned into its equivalence classes, we call thispartitionthepartitiondeterminedbytheequivalencerelation∼. Thestudentmayhavenoticedbynowthatthetwoconceptsofpartitionandequivalence relation, while superficially different, are actually twin aspects of the same structure on sets. Starting with an equivalencerelationonA,wemaypartitionAintoequivalenceclasses,thusgettingapartitionofA. But fromthispartitionwemayretrievetheequivalencerelation,foranytwoelementsxandyareequivalent ifftheylieinthesameclassofthepartition. Goingtheotherway,wemaybeginwithapartitionofAanddefineanequivalencerelationbyletting any two elements be equivalent iff they lie in the same class. We may then retrieve the partition by partitioningAintoequivalenceclasses. Asafinalexample,letAbeasetofpokerchipsofvariouscolors,sayred,blue,green,andwhite. Callanytwochips“equivalent”iftheyarethesamecolor.Thisequivalencerelationhasfourequivalence classes:thesetofalltheredchips,thesetofbluechips,thesetofgreenchips,andthesetofwhitechips. ThesefourequivalenceclassesareapartitionofA. Conversely,ifwebeginbypartitioningthesetAofpokerchipsintofourclassesaccordingtotheir color,thispartitiondeterminesanequivalencerelationwherebychipsareequivalentifftheybelongtothe sameclass.Thisispreciselytheequivalencerelationwehadpreviously. A final comment is in order. In general, there are many ways of partitioning a given set A; each partitiondetermines(andisdeterminedby)exactlyonespecificequivalencerelationonA.Thus,ifAisa set of three elements, say a, b, and c, there are five ways of partitioning A, as indicated by the accompanying illustration. Under each partition is written the equivalence relation determined by that partition. Onceagain,eachpartitionofAdetermines,andisdeterminedby,exactlyoneequivalencerelationonA. EXERCISES A.ExamplesofPartitions Provethateachofthefollowingisapartitionoftheindicatedset.Thendescribetheequivalencerelation associatedwiththatpartition. 1Foreachintegerr∈{0,1,2,3,4},letArbethesetofalltheintegerswhichleavearemainderofr whendividedby5.(Thatis,x∈Ariffx=5q+rforsomeintegerq.)Prove:{A0, A1, A2, A3, A4} is a partitionof . #2Foreachintegern,letAn={x∈ :n≤x<n+1}.Prove{An:n∈ }isapartitionof . 3Foreachrationalnumberr,letAr={(m,n)∈ × *:m/n=r}.Provethat{Ar:r∈ }isapartitionof × *,where *isthesetofnonzerointegers. 4Forr∈{0,1,2,…,9},letArbethesetofalltheintegerswhoseunitsdigit(indecimalnotation)is equaltor.Prove:{A0,A1,A2,…,A9}isa.partitionof . 5Foranyrationalnumberx,wecanwritex=q+n/mwhereqisanintegerand0≤n/m<1.Calln/mthe fractionalpartofx.Foreachrationalr∈{x:0≤x<1},letAr={x∈ :thefractionalpartofxisequal tor}.Prove:{Ar:0≤r<1}isapartitionof . 6Foreachr∈ ,letAr={(x,y)∈ × :x−y=r}.Prove:{Ar:r∈ }isapartitionof × . B.ExamplesofEquivalenceRelations Prove that each of the following is an equivalence relation on the indicated set. Then describe the partitionassociatedwiththatequivalencerelation. 1In ,m∼niff|m|=|n|. 2In ,r∼siffr−s∈ . 3Let⌈x⌉denotethegreatestinteger≤x.In ,leta∼biff⌈a⌉=⌈b⌉. 4In ,letm∼niffm−nisamultipleof10. 5In ,leta∼biffa−b∈ . 6In ( ),letf∼gifff(0)=g(0). 7In ( ),letf∼gifff(x)=g(x)forallx>c,wherecissomefixedrealnumber. 8IfCisanyset,PCdenotesthesetofallthesubsetsofC.LetD⊆C.InPC,letA∼BiffA∩D=B∩ D. 9In × ,let(a,b)∼(c,d)iffa2+b2=c2+d2. 10In *,leta∼biffa/b∈ ,where *isthesetofnonzerorealnumbers. C.EquivalenceRelationsandPartitionsof × Inparts1to3,{Ar:r∈ }isafamilyofsubsetsof × .Proveitisapartition,describethepartition geometrically,andgivethecorrespondingequivalencerelation. 1 Foreachr∈ ,Ar={(x,y):y=2x+r}. 2 Foreachr∈ ,Ar={(x,y):x2+y2=r2}. 3 Foreachr∈ ,Ar={(x,y):y=|x|+r}. Inparts4to6,anequivalencerelationon × isgiven.Proveitisanequivalencerelation,describeit geometrically,andgivethecorrespondingpartition. 4 (x,y)∼(u,υ)iffax2+by2=au2+bυ2(wherea,b>0). 5 (x,y)∼(u,υ)iffx+y=u+υ 6 (x,y)∼(u,υ)iffx2−y=u2−υ. D.EquivalenceRelationsonGroups LetGbeagroup.Ineachofthefollowing,arelationonGisdefined.Proveitisanequivalencerelation. Thendescribetheequivalenceclassofe. 1IfHisasubgroupofG,leta∼biffab−1∈H 2 If H is a subgroup of G, let a ∼ b iff a−xb ∈ H. Is this the same equivalence relation as in part 1? Prove,orfindacounterexample. 3Leta∼biffthereisanx∈Gsuchthata=xbx−1. 4Leta∼biffthereisanintegerksuchthatak =bk #5Leta∼biffab−1commuteswitheveryx∈G. 6Leta∼biffab−1isapowerofc(wherecisafixedelementofG). E.GeneralPropertiesofEquivalenceRelationsandPartitions 1Let{Ai:i∈I}beapartitionofA.Let{Bj :j∈J}beapartitionofB.Provethat{A,×Bj ,(i,j)∈I×J} isapartitionofA×B. 2 Let ∼I be the equivalence relation corresponding to the above partition of A, and let ∼J be the equivalencerelationcorrespondingtothepartitionofB.Describetheequivalencerelationcorresponding totheabovepartitionofA×B. 3Letf:A→Bbeafunction.Define∼bya∼bifff(a)=f(b).Provethat∼isanequivalencerelationon A.Describeitsequivalenceclasses. 4 Let f: A → B be a function, and let {Bi: i ∈ I} be a partition of B. Prove that {f−1(Bi): i ∈ I} is a partition of A. If ∼I is the equivalence relation corresponding to the partition of B, describe the equivalencerelationcorrespondingtothepartitionofA.[REMARK:ForanyC⊆B,f−1C)={x∈A:f(x) ∈C}.] 5 Let ∼1 and ∼2 be distinct equivalence relations on A. Define ∼3 by a ∼3 b iff a ∼1 b and a ∼2 b. Provethat∼3isanequivalencerelationonA.If[x],denotestheequivalenceclassofxfor∼i(i=1,2, 3),provethat[x]3=[x]1∩[x]2. CHAPTER THIRTEEN COUNTINGCOSETS Justastherearegreatworksinartandmusic,therearealsogreatcreationsofmathematics.“Greatness,” in mathematics as in art, is hard to define, but the basic ingredients are clear: a great theorem should contributesubstantialnewinformation,anditshouldbeunexpected!.Thatis,itshouldrevealsomething which common sense would not naturally lead us to expect. The most celebrated theorems of plane geometry,asmayberecalled,comeasacompletesurprise;astheproofunfoldsinsimple,surestepsand wereachtheconclusion—aconclusionwemayhavebeenskepticalabout,butwhichisnowestablished beyondadoubt—wefeelacertainsenseofawenotunlikeourreactiontotheironicortragictwistofa greatstory. In this chapter we will consider a result of modern algebra which, by all standards, is a great theorem.Itissomethingwewouldnotlikelyhaveforeseen,andwhichbringsneworderandsimplicityto therelationshipbetweenagroupanditssubgroups. We begin by adding to our algebraic tool kit a new notion—a conceptual tool of great versatility whichwillserveuswellinalltheremainingchaptersofthisbook.Itistheconceptofacoset. LetGbeagroup,andHasubgroupofG.ForanyelementainG,thesymbol aH denotesthesetofallproductsah,asaremainsfixedandhrangesoverH.aHiscalledaleft cosetofHinG. Insimilarfashion Ha denotesthesetofallproductsha,asaremainsfixedandhrangesoverH.Haiscalledaright cosetofHinG. Inpractice,itwillmakenodifferencewhetherweuseleftcosetsorrightcosets,justaslongaswe remainconsistent.Thus,fromhereon,wheneverweusecosetswewilluserightcosets.Tosimplifyour sentences,wewillsaycosetwhenwemean“rightcoset.” WhenwedealwithcosetsinagroupG,wemustkeepinmindthateverycosetinGisasubsetofG. Thus,whenweneedtoprovethattwocosetsHaandHbareequal,wemustshowthattheyareequalsets. Whatthismeans,ofcourse,isthateveryelementx∈HaisinHb,andconversely,everyelementy∈Hb isinHa.Forexample,letusprovethefollowingelementaryfact: Ifa∈Hb,thenHa=Hb (1) Wearegiventhata∈Hb,whichmeansthata=h1bforsomeh1∈H.WeneedtoprovethatHa=Hb. Letx∈Ha;thismeansthatx=h2aforsomeh2∈H.Buta=h1b,sox=h2a=(h2h1)b,andthelatter isclearlyinHb.Thisprovesthateveryx∈HaisinHb;analogously,wemayshowthateveryy∈Hbis inHa,andthereforeHa=Hb. Thefirstmajorfactaboutcosetsnowfollows.LetGbeagroupandletHbeafixedsubgroupofG: Theorem1ThefamilyofallthecosetsHa,asarangesoverG,isapartitionofG. PROOF:First,wemustshowthatanytwocosets,sayHaandHb,areeitherdisjointorequal.Ifthey aredisjoint,wearedone.Ifnot,letx∈Ha∩Hb.Becausex∈Ha,x=hxaforsomeh1∈H.Becausex ∈Hb,x=h2bforsomeh2∈H.Thus,h1a=h2b,andsolvingfora,wehave Thus, a∈Hb ItfollowsfromProperty(1)abovethatHa=Hb. Next, we must show that every element c ∈ G is in one of the cosets of H. But this is obvious, becausec=ecande∈H;therefore, c=ec∈Hc Thus,thefamilyofallthecosetsofHisapartitionofG.■ Beforegoingon,itisworthmakingasmallcomment:Agivencoset,sayHb,maybewritteninmore thanoneway.ByProperty(1)ifaisanyelementinHb,thenHbisthesameasHa.Thus,forexample,if acosetofHcontainsndifferentelementsa1, a2, …, an, it may be written in n different ways, namely, Ha1,Ha2,…,Han. The next important fact about cosets concerns finite groups. Let G be a finite group, and H a subgroupofG.WewillshowthatallthecosetsofHhavethesamenumberofelements!Thisfactisa consequenceofthenexttheorem. Theorem2IfHaisanycosetofH,thereisaone-to-onecorrespondencefromHtoHa. PROOF:ThemostobviousfunctionfromHtoHaistheonewhich,foreachh∈H,matchesh with ha.Thus,letf:H→Habedefinedby f(h)=ha Rememberthataremainsfixedwhereashvaries,andcheckthatfisinjectiveandsurjective. fisinjective:Indeed,iff(h1)=f(h2),thenh1a=h2a,andthereforeh1=h2. fissurjective,becauseeveryelementofHaisoftheformhaforsomeh∈H,andha=f(h). Thus,fisaone-to-onecorrespondencefromHtoHa,asclaimed.■ ByTheorem2,anycosetHahasthesamenumberofelementsasH,andthereforeallthecosetshave thesamenumberofelements! LetustakeacarefullookatwhatwehaveprovedinTheorems1and2.LetGbeafinitegroupand H any subgroup of G. G has been partitioned into cosets of H, and all the cosets of H have the same numberofelements(whichisthesameasthenumberofelementsinH).Thus,thenumberofelementsin G is equal to the number of elements in H, multiplied by the number of distinct cosets of H. This statementisknownasLagrange’stheorem.(Rememberthatthenumberofelementsinagroupiscalledthe group’sorder.) Theorem3:Lagrange’stheoremLetGbeafinitegroup,andHanysubgroupofG.Theorderof GisamultipleoftheorderofH. Inotherwords,theorderofanysubgroupofagroupGisadivisoroftheorderofG. Forexample,ifGhas15elements,itspropersubgroupsmayhaveeither3or5elements.IfGhas7 elements,ithasnopropersubgroups,for7hasnofactorsotherthan1and7.Thislastexamplemaybe generalized: LetGbeagroupwithaprimenumberpofelements.Ifa∈Gwherea≠e,thentheorderofa is someintegerm≠1.Butthenthecyclicgroup〈a〉hasmelements.ByLagrange’stheorem,m must be a factor of p. But p is a prime number, and therefore m = p. It follows that 〈a〉 has p elements, and is thereforeallofG!Conclusion: Theorem 4 If G is a group with a prime number p of elements, then G is a cyclic group. Furthermore,anyelementa≠einGisageneratorofG. Theorem 4, which is merely a consequence of Lagrange’s theorem, is quite remarkable in itself. What it says is that there is (up to isomorphism) only one group of any given prime order p. For example,theonlygroup(uptoisomorphism)oforder7is 7,theonlygroupoforder11is 11,andsoon! Sowenowhavecompleteinformationaboutallthegroupswhoseorderisaprimenumber. Bytheway,ifaisanyelementofagroupG,theorderofaisthesameastheorderofthecyclic subgroup〈a〉,andbyLagrange’stheoremthisnumberisadivisoroftheorderofG.Thus, Theorem5Theorderofanyelementofafinitegroupdividestheorderofthegroup. Finally,ifGisagroupandHisasubgroupofG,theindexofHinGisthenumberofcosetsofHin G.Wedenoteitby(G:H). Since the number of elements in G is equal to the number of elements in H, multipliedbythenumberofcosetsofHinG, EXERCISES A.ExamplesofCosetsinFiniteGroups Ineachofthefollowing,HisasubgroupofG.Inparts1–5listthecosetsofH.Foreachcoset,listthe elementsofthecoset. ExampleG= 4,H={0,2}. (REMARK:IftheoperationofGisdenotedby+,itiscustomarytowriteH+xforacoset,ratherthan Hx.)ThecosetsofHinthisexampleare H=H+0=H+2={0,2} 1 2 3 4 5 6 and H+1=H+3={1,3} G=S3,H={ε,β,δ}. G=S3,H={ε,α}. G= 15,H=〈5〉. G=D4,H={R0,R4}.(ForD4,seepage73.) G=S4,H=A4.(ForA4,seepage86.) Indicatetheorderandindexofeachofthesubgroupsinparts1to5. B.ExamplesofCosetsinInfiniteGroups Describethecosetsofthesubgroupsdescribedinparts1–5: 1 Thesubgroup〈3〉of . 2 Thesubgroup of . 3 ThesubgroupH={2n:n∈ }of *. 4 Thesubgroup〈 〉ofR*;thesubgroup〈 〉of . 5 ThesubgroupH={(x,y):x=y}of( × . 6 Foranypositiveintegerm,whatistheindexof〈m〉in ? 7 Findasubgroupof *whoseindexisequalto2. C.ElementaryConsequencesofLagrange’sTheorem LetGbeafinitegroup.Provethefollowing: 1IfGhasordern,thenxn=eforeveryxinG. 2LetGhaveorderpq, where p and q are primes. Either G is cyclic, or every element x ≠ e in G has orderporq. 3LetGhaveorder4.EitherGiscyclic,oreveryelementofGisitsowninverse.Concludethatevery groupoforder4isabelian. 4IfGhasanelementoforderpandanelementoforderq,wherepandqaredistinctprimes,thenthe orderofGisamultipleofpq. 5 If G has an element of order k and an element of order m, then |G| is a multiple of lcm(k, m), where lcm(k,m)istheleastcommonmultipleofkandm. #6Letpbeaprimenumber.Inanyfinitegroup,thenumberofelementsoforderpisamultipleofp−1. D.FurtherElementaryConsequencesofLagrange’sTheorem LetGbeafinitegroup,andletHandKbesubgroupsofG.Provethefollowing: 1SupposeH⊆K(thereforeHisasubgroupofK).Then(G:H)=(G:K)(K:H). 2TheorderofH∩KisacommondivisoroftheorderofHandtheorderofK. 3LetHhaveordermandKhaveordern,wheremandnarerelativelyprime.ThenH∩K={e}. 4SupposeHandKarenotequal,andbothhaveorderthesameprimenumberp.ThenH∩K={e}. 5SupposeHhasindexpandKhasindexp,wherepandgaredistinctprimes.ThentheindexofH∩Kis amultipleofpq. #6IfGisanabeliangroupofordern,andmisanintegersuchthatmandnarerelativelyprime,thenthe functionf(x)=xmisanautomorphismofG. E.ElementaryPropertiesofCosets LetGbeagroup,andHasubgroupofG.LetaandbdenoteelementsofG.Provethefollowing: 1Ha=Hbiffab−1∈H. 2Ha=Hiffa∈H. 3IfaH=HaandbH=Hb,then(ab)H=H(ab). #4IfaH=Ha,thena−1H=Ha−1. 5If(ab)H=(ac)H,thenbH=cH. 6ThenumberofrightcosetsofHisequaltothenumberofleftcosetsofH. 7IfJisasubgroupofGsuchthatJ=H∩K,thenforanya∈G,Ja=Ha∩Ka.ConcludethatifHandK areoffiniteindexinG,thentheirintersectionH∩KisalsooffiniteindexinG. Theorem5ofthischapterhasausefulconverse,whichisthefollowing: Cauchy’stheoremIfGisafinitegroup,andpisaprimedivisorof|G|,thenGhasanelementof orderp. Forexample,agroupoforder30musthaveelementsoforders2,3,and5.Cauchy’stheoremhasan elementaryproof,whichmaybefoundonpage340. Inthenextfewexercisesets,wewillsurveyallpossiblegroupswhoseorderis≤10.ByTheorem4 of this chapter, if G is a group with a prime number p of elements, then G ≅ p. This takes care of all groupsoforders2,35,and7.InExerciseG6ofChapter15,itwillbeshownthatifGisagroupwithp2 elements(wherepisaprime),thenG≅ p2orG≅ p× p.Thiswilltakecareofallgroupsoforders4 and9.Theremainingcasesareexaminedinthenextthreeexercisesets. †F.SurveyofAllSix-ElementGroups LetGbeanygroupoforder6.ByCauchy’stheorem,Ghasanelementaoforder2andanelementb of order3.ByChapter10,ExerciseE3,theelements e,a,b,b2,ab,ab2 arealldistinct;andsinceGhasonlysixelements,thesearealltheelementsinG.Thus,baisoneofthe elementse,a,b,b2,ab,orab2. 1Provethatbacannotbeequaltoeithere,a,b,orb2.Thus,ba=aborba=ab2. EitherofthesetwoequationscompletelydeterminesthetableofG.(Seethediscussionattheendof Chapter5.) 2Ifba=ab,provethatG≅ 6. 3Ifba=ab2,provethatG≅S3. Itfollowsthat 6andS3are(uptoisomorphism),theonlypossiblegroupsoforder6. †G.SurveyofAll10-EIementGroups LetGbeanygroupoforder10. 1ReasonasinExerciseFtoshowthatG={e,a,b,b2,b3,b4,ab,ab2,ab3,ab4}, where a has order 2 andbhasorder5. 2Provethatbacannotbeequaltoe,a,b,b2,b3,orb4. 3Provethatifba=ab,thenG≅ 10. 4Ifba=ab2,provethatba2=a2b4,andconcludethatb=b4.Thisisimpossiblebecausebhasorder5; henceba≠ab2.(HINT:Theequationba=ab2tellsusthatwemaymoveafactorafromtherighttothe leftofafactorb,butinsodoing,wemustsquareb.Toproveanequationsuchastheprecedingone,move allfactorsatotheleftofallfactorsb.) 5Ifba=ab3,provethatba2=a2b9=a2b4,andconcludethatb=b4.Thisisimpossible(why?);henceba ≠ab3. 6Provethatifba=ab4,thenG≅D5(whereD5isthegroupofsymmetriesofthepentagon). Thus,theonlypossiblegroupsoforder10(uptoisomorphism),are 10andD5. †H.SurveyofAllEight-ElementGroups LetGbeanygroupoforder8.IfGhasanelementoforder8,thenG≅ 8.LetusassumenowthatGhas noelementoforder8;hencealltheelements≠einGhaveorder2or4. 1Ifeveryx≠einGhasorder2,leta,b,cbethreesuchelements.ProvethatG={e,a,b,c,ab,bc,ac, abc}.ConcludethatG≅ 2× 2× 2. Intheremainderofthisexerciseset,assumeGhasanelementaoforder4.LetH=〈a〉={e,a,a2, a3}.Ifb∈GisnotinH,thenthecosetHb={b,ab,a2b,a3b}.ByLagrange’stheorem,Gistheunionof He=HandHb;hence G={e,a,a2,a3,b,ab,a2b,a3b} 2AssumethereisinHbanelementoforder2.(Letbbethiselement.)Ifba=a2b,provethatb2a=a4b2, hencea=a4,whichisimpossible.(Why?)Concludethateitherba=aborba=a3b. 3Letbbeasinpart2.Provethatifba=ab,thenG≅ 4× 2. 4Letbbeasinpart2.Provethatifba=a3b,thenG≅D4. 5Nowassumethehypothesisinpart2isfalse.Thenb,ab,a2b,anda3ballhaveorder4.Provethatb2= a2.(HINT:Whatistheorderofb2?WhatelementinGhasthesameorder?) 6Prove:Ifba=ab,then(a3b)2=e,contrarytotheassumptionthatord(a3b)=4.Ifba=a2b,thena=b4a =e,whichisimpossible.Thus,ba=a3b. 7Theequationsa4=b4=e,a2=b2,andba=a3bcompletelydeterminethetableofG.Writethistable. (GisknownasthequarterniongroupQ.) Thus,theonlygroupsoforder8(uptoisomorphism)are 8, 2× 2× 2, 4× 2,D4,andQ. †I.ConjugateElements Ifa∈G,aconjugateofaisanyelementoftheformxax−1,wherex∈G.(Roughlyspeaking,aconjugate ofaisanyproductconsistingofasandwichedbetweenanyelementanditsinverse.)Proveeachofthe following: 1Therelation“aisequaltoaconjugateofb” is an equivalence relation in G. (Write a ∼ b for “a is equaltoaconjugateofb.”) Thisrelation∼partitionsanygroupGintoclassescalledconjugacyclasses.(Theconjugacyclass ofais[a]={xax−1:x∈G}.) Foranyelementa∈G,thecentralizerofa,denotedbyCa,isthesetofalltheelementsinGwhich commutewitha.Thatis, Ca={x∈G:xa=ax}={x∈G:xax−1=a} Provethefollowing: 2Foranya∈G,CaisasubgroupofG. 3x−1ax=y−1ayiffxy−1commuteswithaiffxy−1∈Ca. 4x−1ax=y−1ayiffCax=Cay.(HINT:UseExerciseEl.) 5Thereisaone-to-onecorrespondencebetweenthesetofalltheconjugatesofaandthesetofallthe cosetsofCa.(HINT:Usepart4.) 6Thenumberofdistinctconjugatesofaisequalto(G:Ca),theindexofCainG.Thus,thesizeofevery conjugacyclassisafactorof|G. †J.GroupActingonaSet LetAbeaset,andletGbeanysubgroupofSA.GisagroupofpermutationsofA;wesayitisagroup actingonthesetA.AssumeherethatGisafinitegroup.Ifu∈A,theorbitofu(withrespecttoG)isthe set O(u)={g(u):g∈G} 1Definearelation∼onAbyu∼υiffg(w)=υforsomeg∈G.Provethat∼isanequivalencerelation onA,andthattheorbitsareitsequivalenceclasses. Ifu∈A,thestabilizerofuisthesetGu={g∈G:g(u)=u},thatis,thesetofallthepermutations inGwhichleaveufixed. 2ProvethatGuisasubgroupofG. #3Letα=(12)(34)(56)andβ=(23)inS6.LetGbethefollowingsubgroupofS6:G={ε,α,β,αβ, βα,αβα,βαβ,(αβ)2}.FindO(1),O(2),O(5),G1,G2,G4,G5. 4Letf,g∈G.ProvethatfandgareinthesameleftcosetofGuifff(u)=g(u).(HINT:UseExerciseEl modifiedforleftcosets.) 5Usepart4toshowthatthenumberofelementsinO(u)isequaltotheindexofGuinG.[HINT:Iff(u)= υ,matchthecosetoffwithυ.] 6Concludefrompart5thatthesizeofeveryorbit(withrespecttoG)isafactoroftheorderofG. In particular,iff∈SA,thelengthofeachcycleoffisafactoroftheorderoffinSA. K.CodingTheory:CosetDecoding In order to undertake this exercise set, the reader should be familiar with the introductory paragraphs (precedingtheexercises)ofExercisesFandGofChapter3andExerciseHofChapter5. Recallthat isthegroupofallbinarywordsoflengthn.AgroupcodeCisasubgroupof . To decodeareceivedwordxmeanstofindthecodewordaclosesttox,thatis,thecodewordasuchthatthe distanced(a,x)isaminimum. Butd(a,x)=w(a+x),theweight(numberofIs)ofa+x.Thus,todecodeareceivedwordxisto findthecodewordasuchthattheweightw(a+x)isaminimum.Now,thecosetC+xconsistsofallthe sums c + x as c ranges over all the codewords; so by the previous sentence, if a + x is the word of minimumweightinthecosetC+x,thenaisthewordtowhichxmustbedecoded. Nowa=(a+x)+x;soaisfoundbyaddingxtothewordofminimumweightinthecosetC+x.To recapitulate: In order to decode a received word x you examine the coset C + x, find the word e of minimumweightinthecoset(itiscalledthecosetleader),andaddetox.Thene+xisthecodeword closesttox,andhencethewordtowhichxmustbedecoded. 1LetC1bethecodedescribedinExerciseGofChapter3. (a) ListtheelementsineachofthecosetsofC1. (b) Findacosetleaderineachcoset.(Theremaybemorethanonewordofminimumweightinacoset; chooseoneofthemascosetleader.) (c) Usetheproceduredescribedabovetodecodethefollowingwordsx:11100,01101,11011,00011. 2LetC3betheHammingcodedescribedinExerciseH2ofChapter5. List the elements in each of the cosetsofC3andfindaleaderineachcoset.Thenusecosetdecodingtodecodethefollowingwordsx: 1100001,0111011,1001011. 3LetCbeacodeandletHbetheparity-checkmatrixofC.ProvethatxandyareinthesamecosetofC ifandonlyifHx=Hy.(HINT:UseExerciseH8,Chapter5.) Ifxisanyword,Hxiscalledthesyndromeofx.Itfollowsfrompart3thatallthewordsinthesame cosethavethesamesyndrome,andwordsindifferentcosetshavedifferentsyndromes.Thesyndromeofa wordxisdenotedbysyn(x). 4LetacodeChaveqcosets,andletthecosetleadersbee 1,e 2,…,e q.Explainwhythefollowingistrue: Todecodeareceivedwordx,comparesyn(x)withsyn(e 1,…,syn(e q)andfindthecosetleadere i such thatsyn(x)=syn(e i).Thenxistobedecodedtox+e i. 5Findthesyndromesofthecosetleadersinpart2.Thenusethemethodofpart4todecodethewordsx= 1100001andx=1001011. CHAPTER FOURTEEN HOMOMORPHISMS Wehaveseenthatiftwogroupsareisomorphic,thismeansthereisaone-to-onecorrespondencebetween themwhichtransformsoneofthegroupsintotheother.NowifGandHareanygroups,itmayhappenthat thereisafunctionwhichtransformsGintoH,althoughthisfunctionisnotaone-to-onecorrespondence. Forexample, 6istransformedinto 3by aswemayverifybycomparingtheirtables: IfGandHareanygroups,andthereisafunctionfwhichtransformsGintoH,wesaythatifisa homomorphic image of G. The function f is called a homomorphism from G to f. This notion of homomorphism is one of the skeleton keys of algebra, and this chapter is devoted to explaining it and definingitprecisely. First,letusexaminecarefullywhatwemeanbysayingthat“ftransformsGintoH.”Tobeginwith,f mustbeafunctionfromGontoH;butthatisnotall,becausefmustalsotransformthetableofGintothe tableofH.Toaccomplishthis,fmusthavethefollowingproperty:foranytwoelementsaandbinG, if f(a)=a′ and f(b)=b′, then f(ab)=a′b′ (1) Graphically, Condition(1)maybewrittenmoresuccinctlyasfollows: f(ab)=f(a)f(b) (2) Thus, DefinitionifGandHaregroups,ahomomorphismfromGtoHisafunctionf:G→Hsuchthat foranytwoelementsaandbinG, f(ab)=f(a)f(b) IfthereexistsahomomorphismfromGontoH,wesaythatHisahomomorphicimageofG. Groups have a very important and privileged relationship with their homomorphic images, as the nextfewexampleswillshow. LetPdenotethegroupconsistingoftwoelements,eando,withthetable Wecallthisgrouptheparitygroupofevenandoddnumbers.Weshouldthinkofeas“even”ando as “odd,”andthetableasdescribingtheruleforaddingevenandoddnumbers.Forexample,even+odd= odd,odd+odd=even,andsoon. Thefunctionf: →Pwhichcarrieseveryevenintegertoeandeveryoddintegertooisclearlya homomorphismfrom toP.Thisiseasytocheckbecausethereareonlyfourdifferentcases:forarbitrary integersrands,randsareeitherbotheven,bothodd,ormixed.Forexample,ifrandsarebothodd, theirsumiseven,sof(r)=0,f(s)=o,andf(r+s)=e.Sincee=o+o, f(r+s)=f(r)+f(s) Thisequationholdsanalogouslyintheremainingthreecases;hencefisahomomorphism.(Notethatthe symbol + is used on both sides of the above equation because the operation, in as well as in P, is denotedby+.) ItfollowsthatPisahomomorphicimageof ! Now, what do P and have in common? P is a much smaller group than , therefore it is not surprisingthatveryfewpropertiesoftheintegersaretobefoundinP.Nevertheless,one aspect of the structureof isretainedabsolutelyintactinP,namely,thestructureoftheoddandevennumbers.(The fact of being odd or even is called the parity of integers.) In other words, as we pass from to P we deliberatelyloseeveryaspectoftheintegersexcepttheirparity;theirparityalone(withitsarithmetic)is retained,andfaithfullypreserved. Anotherexamplewillmakethispointclearer.RememberthatD4isthegroupofthesymmetriesof thesquare.Now,everysymmetryofthe squareeitherinterchangesthetwodiagonalsherelabeled1and2,orleavesthemastheywere.Inother words,everysymmetryofthesquarebringsaboutoneofthepermutations ofthediagonals. ForeachRi∈D4,letf(Ri) be the permutation of the diagonals produced by Ri Then f is clearly a homomorphism from D4 onto S2. Indeed, it is clear on geometrical grounds that when we perform the motionRifollowedbythemotionRj onthesquare,weare,atthesametime,carryingoutthemotionsf(Ri) followedbyf(Rj )onthediagonals.Thus, f(Rj ºRi)=f(Rj )ºf(Ri) Itfollowsthat52isahomomorphicimageofD4.NowS2isasmallergroupthanD4,andtherefore very few of the features of D4 are to be found in S2. Nevertheless, one aspect of the structure of D4 is retained absolutely intact in S2, namely, the diagonal motions. Thus, as we pass from D4 to 52, we deliberately lose every aspect of plane motions except the motions of the diagonals; these alone are retainedandfaithfullypreserved. Afinalexamplemaybeofsomehelp;itrelatestothegroup describedinChapter3,ExerciseE. Here,briefly,isthecontextinwhichthisgrouparises:Themostbasicwayoftransmittinginformationis tocodeitintostringsofOsandIs,suchas0010111,1010011,etc.Suchstringsarecalledbinarywords, andthenumberof0sandIsinanybinarywordiscalleditslength.Thesymbol designatesthegroup consistingofallbinarywordsoflengthn,withanoperationofadditiondescribedinChapter3,Exercise E. Considerthefunctionf: → whichconsistsofdroppingthelasttwodigitsofeveryseven-digit word.Thiskindoffunctionarisesinmanypracticalsituations:forexample,itfrequentlyhappensthatthe firstfivedigitsofawordcarrythemessagewhilethelasttwodigitsareanerrorcheck.Thus,fseparates themessagefromtheerrorcheck. Itiseasytoverifythatfisahomomorphism;hence isahomomorphicimageof .Aswepass from to , the message component of words in is exactly preserved while the error check is deliberatelylost. Theseexamplesillustratethebasicideainherentintheconceptofahomomorphicimage.Thecases whichariseinpracticearenotalwayssoclear-cutasthese,buttheunderlyingideaisstillthesame:Ina homomorphic image of G, some aspect of G is isolated and faithfully preserved while all else is deliberatelylost. Thenexttheorempresentstwoeleirientarypropertiesofhomomorphisms. Theorem1LetGandHbegroups,andf:G→Hahomomorphism.Then (i) f(e)=e,and (ii) f(a−l)=[f(a)]−lforeveryelementa∈G. Intheequationf(e)=e,thelettereontheleftreferstotheneutralelementinG,whereasthelettereonthe rightreferstotheneutralelementinH. Toprove(i),wenotethatinanygroup, if yy=y then y=e (Usethecancellationlawontheequationyy=ye.)Now,f(e)f(e)=f(ee)=f(e);hencef(e)=e. To prove (ii), note that f(a)f(a−1) = f(aa−1) = f(e). But f(e) = e, so f(a)f(a−1) = e. It follows by Theorem2ofChapter4thatf(a−1)istheinverseoff(a),thatis,f(a−l)=[f(a)]−1. Before going on with our study of homomorphisms, we must be introduced to an important new concept.IfaisanelementofagroupG,aconjugateofαisanyelementoftheformxax−l,wherex∈G. Forexample,theconjugatesofαinS3are aswellasaitself,whichmaybewrittenintwoways,asε∘α∘ε−1orasα∘α∘α−1.ifHisanysubset ofagroupG,wesaythatHisclosedwithrespecttoconjugatesifeveryconjugateofeveryelementofH isinH.Finally, DefinitionLetHbeasubgroupofagroupG.HiscalledanormalsubgroupofGifitisclosed withrespecttoconjugates,thatis,if forany a∈H and x∈G xax−l∈H (Note that according to this definition, a normal subgroup of G is any nonempty subset of G which is closedwithrespecttoproducts,withrespecttoinverses,andwithrespecttoconjugates.) Wenowreturntoourdiscussionofhomomorphisms. DefinitionLetf:G→Hbeahomomorphism.ThekerneloffisthesetKofalltheelementsofG whicharecarriedbyfontotheneutralelementofH.Thatis, K={x∈G:f(x)=e} Theorem2Letf:G→Hbeahomomorphism. (i) ThekerneloffisanormalsubgroupofG,and (ii) TherangeoffisasubgroupofH. PROOF:LetKdenotethekerneloff.Ifa,b∈K,thismeansthatf(a)=eandf(b)=e.Thus,f(ab)= f(a)f(b)=ee=e;henceab∈k. Ifa∈k,thenf(a)=e.Thus,f(a−l)=[f(a)]−1=e−1=e,soa−1∈K. Finally,ifa−Kandx∈G,thenf(xax−l)=f(x)f(a)f(x−l)=f(x)f(a)[f(x)]−1=e,whichshowsthatxax −l ∈K.Thus,KisanormalsubgroupofG. Nowwemustprovepart(ii).Iff(a)andf(b)areintherangeoff,thentheirproductf(a)f(b)=f(ab) isalsointherangeoff. Iff(a)isintherangeoff,itsinverseis[f(a)]−l=f(a−l),whichisalsointherangeoff. Thus, the rangeoffisasubgroupofH.■ Iffisahomomorphism,werepresentthekerneloffandtherangeoffwiththesymbols ker(f) and ran(f) EXERCISES A.ExamplesofHomomorphismsofFiniteGroups 1Considerthefunctionf: 8→ 4givenby Verifythatfisahomomorphism,finditskernelK,andlistthecosetsofK.[REMARK:Toverifythatfisa homomorphism,youmustshowthatf(a+b)=f(a)+f(b) for all choices of a and b in 8; there are 64 choices.Thismaybeaccomplishedbycheckingthatftransformsthetableof 8tothetableof 4, as on page136.] 2Considerthefunctionf:S3→ 2givenby Verifythatfisahomomorphism,finditskernelK,andlistthecosetsofK. 3 Find a homomorphism f: 15 → 5, and indicate its kernel. (Do not actually verify that f is a homomorphism.) 4Imagineasquareasapieceofpaperlyingonatable.Thesidefacingyouisside A.ThesidehiddenfromviewissideB.Everymotionofthesquareeitherinterchangesthetwosides(that is, side B becomes visible and side A hidden) or leaves the sides as they were. In other words, every motionRiofthesquarebringsaboutoneofthepermutations ofthesides;callitg(Ri).Verifythatg:D4 S2isahomomorphism,andgiveitskernel. 5Everymotionoftheregularhexagonbringsaboutapermutationofitsdiagonals,labeled1,2,and3.For eachRi∈D6,letf(Ri)bethepermutationof thediagonalsproducedbyRiArgueinformally(appealingtogeometricintuition)toexplainwhyf: D6∈ S3isahomomorphism.Thencompletethefollowing: (Thatis,findthevalueoffonall12elementsofD6.) # 6 Let B ⊂ A. Let h : PA → PB be defined by h(C) = C = C B. For A = {1,2,3} and B = {1, 2}, completethefollowing: ForanyAandB⊂A,showthathisahomomorphism. B.ExamplesofHomomorphismsofInfiniteGroups Provethateachofthefollowingisahomomorphism,anddescribeitskernel. 1Thefunctionϕ: ( )→ givenbyϕ(f)=f(0). 2Thefunctionϕ: ( )→ ( )givenbyϕ(f)=f′. (R)isthegroupofdifferentiablefunctionsfrom to f′isthederivativeoff. 3Thefunctionf: × → givenbyf(x,y)=x+y. 4Thefunctionf: *→ posdefinedbyf(x)=|x|. 5Thefunctionf: *→ posdefinedbyf(a+bi)= 6LetGbethemultiplicativegroupofall2×2matrices . satisfyingad−bc≠0.Letf:G→ *begivenbyf(A)=determinantofA=ad−bc. C.ElementaryPropertiesofHomomorphisms LetG,H,andKbegroups.Provethefollowing: 1Iff:G→Gandg:H→Karehomomorphisms,thentheircompositegºf:G→Kisahomomorphism. #2Iff:G→HisahomomorphismwithkernelK,thenfisinjectiveiffK={e}. 3Iff:G→HisahomomorphismandisanysubgroupofG,thenf(K)={f(x):x∈K}isasubgroupof H. 4Iff:G→HisahomomorphismandjisanysubgroupofH,then f−1(J)={x∈G:f(x)∈J} isasubgroupofG.Furthermore,kerf⊆f−1(J). 5Iff:G→HisahomomorphismwithkernelK,andJisasubgroupofG,letfJdesignatetherestriction offtoJ.(InotherwordsfJisthesamefunctionasf,exceptthatitsdomainisrestrictedtoJ.)ThenkerfJ =J K. 6ForanygroupG,thefunctionf:G→Gdefinedbyf(x)=eisahomomorphism. 7ForanygroupG,{e}andGarehomomorphicimagesofG. 8Thefunctionf:G→Gdefinedbyf(x)=x2isahomomorphismiffGisabelian. 9Thefunctionsf1(x,y)=xandf2(x,y)=y,fromG×HtoGandH,respectively,arehomomorphisms. D.BasicPropertiesofNormalSubgroups Inthefollowing,letGdenoteanarbitrarygroup. 1Findallthenormalsubgroups(a)ofS3and(b)ofD4. Provethefollowing: 2Everysubgroupofanabeliangroupisnormal. 3ThecenterofanygroupGisanormalsubgroupofG. 4LetHbeasubgroupofG.Hisnormaliffithasthefollowingproperty:ForallaandbinG,ab∈Hiff ba∈H. 5LetHbeasubgroupofG.HisnormaliffaH=Haforeverya∈G. 6AnyintersectionofnormalsubgroupsofGisanormalsubgroupofG. E.FurtherPropertiesofNormalSubgroups LetGdenoteagroup,andHasubgroupofG.Provethefollowing: #1IfHhasindex2inG,thenHisnormal.(HINT:UseExerciseD5.) 2Supposeanelementa∈Ghasorder2.Then(〈a〉)isanormalsubgroupofGiffaisinthecenterofG. 3IfaisanyelementofG,(〈a〉)isanormalsubgroupofGiffahasthefollowingproperty:Foranyx∈ G,thereisapositiveintegerksuchthatxa=ak x. 4InagroupG,acommutatorisanyproductoftheformaba−1b−1,whereaandbareanyelementsofG. IfasubgroupHofGcontainsallthecommutatorsofG,thenHisnormal. 5IfHandKaresubgroupsofG,andKisnormal,thenHKisasubgroupofG.(HKdenotesthesetofall productshkashrangesoverHandkrangesoverK.) # 6 Let S be the union of all the cosets Ha such that Ha = aH. Then S is a subgroup of G, and H is a normalsubgroupofS. F.HomomorphismandtheOrderofElements Iff:G→Hisahomomorphism,proveeachofthefollowing: 1Foreachelementa∈G,theorderoff(a)isadivisoroftheorderofa. 2Theorderofanyelementb≠eintherangeoffisacommondivisorof|G|and|H|.(Usepart1.) 3Iftherangeoffhasnelements,thenxn∈kerfforeveryx∈G. 4Letmbeanintegersuchthatmand|H|arerelativelyprime.Foranyx∈G,ifxm∈kerf,thenx∈kerf. 5Lettherangeoffhavemelements.Ifa∈Ghasordern,wheremandnarerelativelyprime,thenais inthekerneloff.(Usepart1.) 6Letpbeaprime.Ifranfhasanelementoforderp,thenGhasanelementoforderp. G.PropertiesPreservedunderHomomorphism A property of groups is said to be “preserved under homomorphism” if, whenever a group G has that property, every homomorphic image of G does also. In this exercise set, we will survey a few typical propertiespreservedunderhomomorphism.Iff:G→HisahomomorphismofGontoH,proveeachof thefollowing: 1IfGisabelian,thenHisabelian. 2IfGiscyclic,thenHiscyclic. 3IfeveryelementofGhasfiniteorder,theneveryelementofHhasfiniteorder. 4IfeveryelementofGisitsowninverse,everyelementofHisitsowninverse. 5IfeveryelementofGhasasquareroot,theneveryelementofHhasasquareroot. 6IfGisfinitelygenerated,thenHisfinitelygenerated.(Agroupissaidtobe“finitelygenerated”ifitis generatedbyfinitelymanyofitselements.) †H.InnerDirectProducts IfGisanygroup,letHandKbenormalsubgroupsofGsuchthatH K={e}.Provethefollowing: 1Leth1andh2beanytwoelementsofH,andk1andk2anytwoelementsofK. h1k1=h2k (HINT:Ifh1k1=h2k2,then implies h1∈H Kandk2 h1=h2 and k1=k2 ∈H K.Explainwhy.) 2 For any h ∈ H and k ∈ K, hk = kh. (HINT: hk = kh iff hkh−lk−l = e. Use the fact that H and K are normal.) 3Now,maketheadditionalassumptionthatG=HKthatis,everyxinGcanbewrittenasx=hkforsome h∈Handk∈K.Thenthefunctionϕ(h,k)=hkisanisomorphismfromH×KontoG. Wehavethusprovedthefollowing:IfHandKarenormalsubgroupsofG,suchthatH K= {e} andG=HK,thenG≅H×K.GissometimescalledtheinnerdirectproductofHandK. †I.ConjugateSubgroups LetHbeasubgroupofG.Foranya∈G,letaHa−l={axa−l:x∈H};aHa−liscalledaconjugateofH. Provethefollowing: 1Foreacha∈G,aHa−lisasubgroupofG. 2Foreacha∈G,H≅aHa−1. 3HisanormalsubgroupofGiffH=aHa−1foreverya∈G. Intheremainingexercisesofthisset,letGbeafinitegroup.BythenormalizerofHwemeanthesetN(H) ={a∈G:axa−1∈Hforeveryx∈H}. 4Ifa∈N(H),thenaHa−1=H.(RememberthatGisnowafinitegroup.) 5N(H)isasubgroupofG. 6H⊆N(H).Furthermore,HisanormalsubgroupofN(H). Inparts7–10,letN=N(H). 7Foranya,b∈G,aHa−1=bHb−1iffb−1a∈N(H). #8Thereisaone-to-onecorrespondencebetweenthesetofconjugatesofHandthesetofcosetsofN. (Thus,thereareasmanyconjugatesofHascosetsofN.) 9Hhasexactly(G:N)conjugates.Inparticular,thenumberofdistinctconjugatesofifHisadivisorof |G|. 10LetKbeanysubgroupofG,letK*={Na:a∈K},andlet XK={aHa−l:a∈K} Argueasinpart8toprovethatXKisinone-to-onecorrespondencewithK*.Concludethatthenumberof elementsinXKisadivisorof|K|. CHAPTER FIFTEEN QUOTIENTGROUPS InChapter14welearnedtorecognizewhenagroupHisahomomorphicimageofagroupG.Nowwe will make a great leap forward by learning a method for actually constructing all the homomorphic imagesofanygroup.Thisisaremarkableprocedure,ofgreatimportanceinalgebra.Inmanycasesthis constructionwillallowustodeliberatelyselectwhichpropertiesofagroupGwewishtopreserveina homomorphicimage,andwhichotherpropertieswewishtodiscard. The most important instrument to be used in this construction is the notion of a normal subgroup. RememberthatanormalsubgroupofGisanysubgroupofGwhichisclosedwithrespecttoconjugates. Webeginbygivinganelementarypropertyofnormalsubgroups. Theorem1IfHisanormalsubgroupofG,thenaH=Haforeverya∈G. (Inotherwords,thereisnodistinctionbetweenleftandrightcosetsforanormalsubgroup.) PROOF:Indeed,ifxisanyelementofaH,thenx=ahforsomeh∈H.ButHisclosedwithrespect to conjugates; hence aha−1 ∈ H. Thus, x = ah = (aha−1)a is an element of Ha. This shows that every elementofaHisinHa;analogously,everyelementofHaisinaH.Thus,aH=Ha.■ LetGbeagroupandletHbeasubgroupofG.Thereisawayofcombiningcosets,calledcoset multiplication,whichworksasfollows:thecosetofa,multipliedbythecosetofb,isdefinedtobethe cosetofab.Insymbols, Ha·Hb=H(ab) Thisdefinitionisdeceptivelysimple,foritconcealsafundamentaldifficulty.Indeed,itisnotatallclear that the product of two cosets Ha and Hb, multiplied together in this fashion, is uniquely defined. RememberthatHamaybethesamecosetasHc(thishappensiffcisinHa),and,similarly,Hbmaybe thesamecosetasHd.Therefore,theproductHa·HbisthesameastheproductHe·Hd.Yetitmayeasily happenthatH(ab)isnotthesamecosetasH(cd).Graphically, Forexample,ifG=S3andH={ε,α},then andyet H(β∘δ)=Hε≠Hβ=H(γ∘κ) Thus,cosetmultiplicationdoesnotworkasanoperationonthecosetsofH={ε,α}inS3.Thereasonis that, although H is a subgroup of S3, H is not a normal subgroup of S3. If H were a normal subgroup, cosetmultiplicationwouldwork.Thenexttheoremstatesexactlythat! Theorem2LetHbeanormalsubgroupofG.IfHa=HeandHb=Hd,thenH(ab)=H(cd). PROOF:IfHa=Hc,thena∈Hc;hencea=h1cforsomeh1∈H.IfHb=Hd,thenb∈Hd;henceb= h2dfromsomeh2∈H.Thus, ab=h1ch2d=h1(ch2)d Butch2∈cH=Hc(thelastequalityistruebyTheorem1).Thus,ch2=h3cforsomeh3∈H.Returningto ab, ab=h1(ch2)d=h1(h3c)d=(h1h3)(cd) andthislastelementisclearlyinH(cd). Wehaveshownthatab∈H(cd).Thus,byProperty(1)inChapter13,H(ab)=H(cd).■ Wearenowreadytoproceedwiththeconstructionpromisedatthebeginningofthechapter.LetG beagroupandletHbeanormalsubgroupofG.ThinkofthesetwhichconsistsofallthecosetsofH. ThissetisconventionallydenotedbythesymbolG/H.Thus,ifHa,Hb,He,...arecosetsofH,then G/H={Ha,Hb,Hc,...} Wehavejustseenthatcosetmultiplicationisavalidoperationonthisset.Infact, Theorem3G/Hwithcosetmultiplicationisagroup. PROOF:Cosetmultiplicationisassociative,because TheidentityelementofG/HisH=He,forHa·He=HaandHe·Ha=HaforeverycosetHa. Finally,theinverseofanycosetHaisthecosetHa−1,becauseHa·Ha−1=Haa−1=HeandHa−1· Ha=Ha−1aHe. ThegroupG/Hiscalledthefactorgroup,orquotientgroupofGbyH. Andnow,thepiècederésistance: Theorem4G/HisahomomorphicimageofG. PROOF:ThemostobviousfunctionfromGtoG/Histhefunctionfwhichcarrieseveryelementtoits owncoset,thatis,thefunctiongivenby f(x)=Hx Thisfunctionisahomomorphism,because f(xy)=Hxy=Hx·Hy=f(x)f(y) f is called the natural homomorphism from G onto G/H. Since there is a homomorphism from G onto G/H,G/HisahomomorphicimageofG.■ Thus,whenweconstructquotientgroupsofG,weare,infact,constructinghomomorphicimagesof G.Thequotientgroupconstructionisusefulbecauseitisawayofactuallymanufacturinghomomorphic imagesofanygroupG.Infact,aswewillsoonsee,itisawayofmanufacturingall the homomorphic imagesofG. Ourfirstexampleisintendedtoclarifythedetailsofquotientgroupconstruction.Let bethegroup oftheintegers,andlet〈6〉bethecyclicsubgroupof whichconsistsofallthemultiplesof6.Since is abelian,andeverysubgroupofanabeliangroupisnormal,〈6〉isanormalsubgroupof .Therefore,we mayformthequotientgroup /〈6〉.Theelementsofthisquotientgroupareallthecosetsofthesubgroup 〈6〉,namely: Theseareallthedifferentcosetsof〈6〉,foritiseasytoseethat〈6〉+6=〈6〉+0,〈6〉+7=〈6〉+1,〈6〉+ 8=〈6〉+2,andsoon. Now,theoperationon isdenotedby+,andthereforewewillcalltheoperationonthecosetscoset additionratherthancosetmultiplication.Butnothingischangedexceptthename;forexample,thecoset 〈6〉+1addedtothecoset〈6〉+2isthecoset〈6〉+3.Thecoset〈6〉+3addedtothecoset〈6〉+4isthe coset〈6〉+7,whichisthesameas〈6〉+1.Tosimplifyournotation,letusagreetowritethecosetsinthe followingshorterform: Then /〈6〉 consists of the six elements followingtable: and , and its operation is summarized in the Thereaderwillperceiveimmediatelythesimilaritybetweenthisgroupand 6. As a matter of fact, the quotientgroupconstructionof /〈6〉isconsideredtobetherigorouswayofconstructing 6.Sofromnow on,wewillconsider 6tobethesameas /〈6〉;and,ingeneral,wewillconsider ntobethesameas / 〈n〉.Inparticular,wecanseethatforanyn, nisahomomorphicimageof . Let us repeat: The motive for the quotient group construction is that it gives us a way of actually producingallthehomomorphicimagesofanygroupG.However,whatisevenmorefascinatingaboutthe quotient group construction is that, in practical instances, we can often choose H so as to “factor out” unwanted properties of G, and preserve in G/H only “desirable” traits. (By “desirable” we mean desirablewithinthecontextofsomespecificapplicationoruse.)Letuslookatafewexamples. First,wewillneedtwosimplepropertiesofcosets,whicharegiveninthenexttheorem. Theorem5LetGbeagroupandHasubgroupofG.Then (i) Ha=Hb iff ab−1∈H and (ii) Ha=H iff a∈H PROOF:IfHa=Hb,thena∈Hb,soa=hbforsomeh∈H.Thus, ab−1=h∈H If ab−1 ∈ H, then ab−1 = h for h ∈ H, and therefore a = hb ∈ Hb. It follows by Property (1) of Chapter13thatHa=Hb. Thisproves(i).ItfollowsthatHa=Heiffae−1=a∈H,whichproves(ii).■ Forourfirstexample,letGbeanabeliangroupandletHconsistofall the elements of G which have finite order. It is easy to show that H is a subgroup of G. (The details may be supplied by the reader.)Rememberthatinanabeliangroupeverysubgroupisnormal;henceHisanormalsubgroupofG, andthereforewemayformthequotientgroupG/H.WewillshownextthatinG/H,noelementexceptthe neutralelementhasfiniteorder. ForsupposeG/HhasanelementHxoffiniteorder.SincetheneutralelementofG/HisH,thismeans thereisanintegerm≠0suchthat(Hx)m=H,thatis,Hxm=H.Therefore,byTheorem5(ii),xm∈H,so xmhasfiniteorder,sayt: (xm)t=xmt=e Butthenxhasfiniteorder,sox∈H.Thus,byTheorem5(ii),Hx=H.ThisprovesthatinG/H,theonly elementHxoffiniteorderistheneutralelementH. Let us recapitulate: If H is the subgroup of G which consists of all the elements of G which have finiteorder,theninG/H, no element (except the neutral element) has finite order. Thus, in a sense, we have“factoredout”alltheelementsoffiniteorder(theyareallinH)andproducedaquotient group GIHwhoseelementsallhaveinfiniteorder(exceptfortheneutralelement,whichnecessarilyhasorder 1). Our next example may bring out this idea even more clearly. Let G be an arbitrary group; by a commutatorofGwemeananyelementoftheformaba−1b−1whereaandbareinG.Thereasonsucha productiscalledacommutatoristhat aba−1b−1=e iff ab=ba In other words, aba−1b−1 reduces to the neutral element whenever a and b commute—and only in that case!Thus,inanabeliangroupallthecommutatorsareequaltoe.Inagroupwhichisnotabelian,the numberofdistinctcommutatorsmayberegardedasameasureoftheextenttowhichGdepartsfrombeing commutative.(Thefewerthecommutators,thecloserthegroupistobeinganabeliangroup.) WewillseeinamomentthatifHisasubgroupofGwhichcontainsallthecommutatorsofG,then G/Hisabelian!Whatthismeans,inafairlyaccuratesense,isthatwhenwefactoroutthecommutators of G we get a quotient group which has no commutators (except, trivially, the neutral element) and whichisthereforeabelian. TosaythatG/HisabelianistosaythatforanytwoelementsHxandHyinG/H,HxHy=HyHx;that is,Hxy=Hyx.ButbyTheorem5(ii), Hxy=Hyx iff xy(yx)−1∈H Nowxy(yx)−1isthecommutatorxyx−1y−1;soifallcommutatorsareinH,thenG/Hisabelian. EXERCISES A.ExamplesofFiniteQuotientGroups Ineachofthefollowing,GisagroupandHisanormalsubgroupofG.ListtheelementsofG/Handthen writethetableofG/H. ExampleG= 6 and H={0,3} TheelementsofG/HarethethreecosetsH=H+0={0,3},H+1={1,4},andH+2={2,5}.(Note thatH+3isthesameasH+0,H+4isthesameasH+1,andH+5isthesameasH+2.)Thetableof G/His 1G= 10,H={0,5}.(ExplainwhyG/H≅Z5.) 2G=S3,H={ε,β,δ}. 3G=D4,H={R0,R2}.(Seepage73.) 4G=D4,H={R0,R2,R4,R5}. 5G= 4× 2,H=〈(0,1)〉=thesubgroupof 4× 2generatedby(0,1). 6G=P3,H={ø,{1}}.(P3isthegroupofsubsetsof{1,2,3}.) B.ExamplesofQuotientGroupsof × Ineachofthefollowing,Hisasubsetof × . (a)ProvethatHisanormalsubgroupof × .(Rememberthateverysubgroupofanabeliangroupis normal.) (b)Ingeometricalterms,describetheelementsofthequotientgroupG/H. (c)Ingeometricaltermsorotherwise,describetheoperationofG/H. 1H={(x,0):x∈ } 2H={(x,y):y=−x} 3H={(x,y):y=2x} C.RelatingPropertiesofHtoPropertiesofG/H Inparts1-5below,GisagroupandifisanormalsubgroupofG.Provethefollowing(Theorem5 will playacrucialrole): 1Ifx2∈Hforeveryx∈G,theneveryelementofG/Hisitsowninverse.Conversely,ifeveryelementof G/Hisitsowninverse,thenx2∈Hforallx∈G. 2Letmbeafixedinteger.Ifxm∈Hforeveryx∈G,thentheorderofeveryelementinG/Hisadivisor ofm.Conversely,iftheorderofeveryelementinG/Hisadivisorofm,thenxm∈Hforeveryx∈G. 3Supposethatforeveryx∈G, there is an integer n such that xn ∈ H; then every element of G/H has finiteorder.Conversely,ifeveryelementofG/Hhasfiniteorder,thenforeveryx∈Gthereisaninteger nsuchthatxn∈H. #4EveryelementofG/Hhasasquarerootiffforeveryx∈G,thereissomey∈Gsuchthatxy2∈H. 5G/Hiscycliciffthereisanelementa∈Gwiththefollowingproperty:foreveryx∈G,thereissome integernsuchthatxan∈H. 6IfGisanabeliangroup,letHpbethesetofallx∈Hwhoseorderisapowerofp.ProvethatHpisa subgroupofG.ProvethatG/Hphasnoelementswhoseorderisanonzeropowerofp 7 (a)IfG/Hisabelian,provethatHcontainsallthecommutatorsofG. (b)LetKbeanormalsubgroupofG,andHanormalsubgroupofK.IfG/Hisabelian,provethatG/K andK/Harebothabelian. D.PropertiesofGDeterminedbyPropertiesofG/HandH There are some group properties which, if they are true in G/H and in H, must be true in G. Here is a sampling.LetGbeagroup,andHanormalsubgroupofG.Provethefollowing: 1IfeveryelementofG/Hhasfiniteorder,andeveryelementofHhasfiniteorder,theneveryelementof Ghasfiniteorder. 2IfeveryelementofG/Hhasasquareroot,andeveryelementofHhasasquareroot,theneveryelement ofGhasasquareroot.(AssumeGisabelian.) 3Letpbeaprimenumber.IfG/HandHarep-groups,thenGisap-group.AgroupGiscalledap-group iftheordereveryelementxinGisapowerofp. # 4 If G/H and H are finitely generated, then G is finitely generated. (A group is said to be finitely generatedifitisgeneratedbyafinitesubsetofitselements.) E.OrderofElementsinQuotientGroups LetGbeagroup,andHanormalsubgroupofG.Provethefollowing: 1Foreachelementa∈G,theorderoftheelementHainG/HisadivisoroftheorderofainG.(HINT: UseChapter14,ExerciseF1.) 2If(G:H)=m,theorderofeveryelementofG/Hisadivisorofm. 3If(G:H)=p,wherepisaprime,thentheorderofeveryelementa∉HinGisamultipleofp.(Use part1.) 4IfGhasanormalsubgroupofindexp,wherepisaprime,thenGhasatleastoneelementoforderp. 5If(G:H)=m,thenam∈Gforeverya∈G. #6In / ,everyelementhasfiniteorder. †F.QuotientofaGroupbyItsCenter The center of a group G is the normal subgroup C of G consisting of all those elements of G which commutewitheveryelementofG.SupposethequotientgroupG/Cisacyclicgroup;sayitisgenerated bytheelementCaofG/C.Proveparts1-3: 1Foreveryx∈G,thereissomeintegermsuchthatCx=Cam. 2Foreveryx∈G,thereissomeintegermsuchthatx=cam,wherec∈C. 3ForanytwoelementsxandyinG,xy=yx.(HINT:Usepart2towritex=cam,y=c′an,andremember thatc,c′∈C.) 4ConcludethatifG/Ciscyclic,thenGisabelian. †G.UsingtheClassEquationtoDeterminetheSizeoftheCenter {Prerequisite:Chapter13,ExerciseI.) LetGbeafinitegroup.ElementsaandbinGarecalledconjugatesofoneanother(insymbols,a~ b)iffa=xbx−1forsomex∈G(thisisthesameasb−x−1ax).Therelation~isanequivalencerelation inG; the equivalence class of any element a is called its conjugacy class. Hence G is partitioned into conjugacyclasses(asshowninthediagram);thesizeofeachconjugacyclassdividestheorderofG.(For thesefacts,seeChapter13,ExerciseI.) “EachelementofthecenterCisaloneinitsconjugacyclass.” LetS1,S2,...,StbethedistinctconjugacyclassesofG,andletk1,k2,...,ktbetheirsizes.Then|G|= k1+k2+…+kt(ThisiscalledtheclassequationofG.) LetGbeagroupwhoseorderisapowerofaprimep,say|G|=pk .LetCdenotethecenterofG. Proveparts1-3: 1Theconjugacyclassofacontainsa(andnootherelement)iffa∈C. 2LetcbetheorderofC.Then|G|=c+ks+ks+1+···+kt,whereks,...,ktarethesizesofallthedistinct conjugacyclassesofelementsx∉C. 3Foreachi∈{s,s+1,...,t},kiisequaltoapowerofp.(SeeChapter13,ExerciseI6.) 4Solvingtheequation|G|=c+ks+···+ktforc,explainwhycisamultipleofp Wemayconcludefrompart4thatCmustcontainmorethanjusttheoneelemente;infact,|C|isamultiple ofp. 5Prove:If|G|=p2,Gmustbeabelian.(UsetheprecedingExerciseF.) #6Prove:If|G|=p2,theneitherG≅ p 2orG≅∈p ×∈p . †H.Inductionon|G|:AnExample Manytheoremsofmathematicsareoftheform“P(n)istrueforeverypositiveintegern.”[Here,P(n)is used as a symbol to denote some statement involving n.] Such theorems can be proved by induction as follows: (a)ShowthatP(n)istrueforn=1. (b)Foranyfixedpositiveintegerk,showthat,ifP(n)istrueforeveryn<k,thenP(n)mustalsobetrue forn=k. Ifwecanshow(a)and(b),wemaysafelyconcludethatP(n)istrueforallpositiveintegersn. Sometheoremsofalgebracanbeprovedbyinductionontheordernofagroup.Hereisaclassical example:LetGbeafiniteabeliangroup.WewillshowthatGmustcontainatleastoneelementoforder p,foreveryprimefactorpof|G|.If|G|=1,thisistruebydefault,sincenoprimepcanbeafactorof1. Next,let|G|=k,andsupposeourclaimistrueforeveryabeliangroupwhoseorderislessthank.Letp beaprimefactorofk. Takeanyelementa≠einG.Iford(a)=poramultipleofp,wearedone! 1Iford(a)=tp(forsomepositiveintegert),whatelementofGhasorderp? 2Supposeord(a)isnotequaltoamultipleofp.ThenG/〈a〉 is a group having fewer than k elements. (Explainwhy.)TheorderofG/〈a〉isamultipleofp.(Explainwhy.) 3WhymustG/(a)haveanelementoforderp? 4ConcludethatGhasanelementoforderp.(HINT:UseExerciseEl.) CHAPTER SIXTEEN THEFUNDAMENTALHOMOMORPHISMTHEOREM LetGbeanygroup.InChapter15wesawthateveryquotientgroupofGisahomomorphicimageofG. Nowwewillseethat,conversely,everyhomomorphicimageofGisaquotientgroupofG.Moreexactly, everyhomomorphicimageofGisisomorphictoaquotientgroupofG. It will follow that, for any groups G and H, H is a homomorphic image of G iff H is (or is isomorphic to) a quotient group of G. Therefore, the notions of homomorphic image and of quotient groupareinterchangeable. Thethreadofourreasoningbeginswithasimpletheorem. Theorem1Letf:G→HbeahomomorphismwithkernelK.Then f(a)=f(b) iff Ka=Kb (Inotherwords,anytwoelementsaandbinGhavethesameimageunderfifftheyareinthesamecoset ofK.) Indeed, Whatdoesthistheoremreallytellus?ItsaysthatiffisahomomorphismfromGtoHwithkernelK, thenalltheelementsinanyfixedcosetofKhavethesameimage,and,conversely,elementswhichhave thesameimageareinthesamecosetofK. It is therefore clear already that there is a one-to-one correspondence matching cosets of K with elementsinH.Itremainsonlytoshowthatthiscorrespondenceisanisomorphism.Butfirst,howexactly doesthiscorrespondencematchupspecificcosetsofKwithspecificelementsofH?Clearly,foreachx, thecosetKxismatchedwiththeelementf(x).Oncethisisunderstood,thenexttheoremiseasy. Theorem2Letf:G→HbeahomomorphismofGontoH.IfKisthekerneloff,then H≅G/K PROOF:ToshowthatG/KisisomorphictoH,wemustlookforanisomorphismfromG/KtoH.We havejustseenthatthereisafunctionfromG/KtoHwhichmatcheseachcosetKxwiththeelementf(x); callthisfunctionϕ.Thus,ϕisdefinedbytheidentity ϕ(Kx)=f(x) This definition does not make it obvious that ϕ(Kx) is uniquely defined. (If it is not, then we cannot properlycallϕafunction.)WemustmakesurethatifKaisthesamecosetasKb,thenϕ(Ka)isthesame asϕ(Kb):thatis, if Ka=Kb then f(a)=f(b) Asamatteroffact,thisistruebyTheorem1. Now,letusshowthatϕisanisomorphism: ϕisinjective:Ifϕ(Ka)=ϕ(Kbthenf(a)=f(b);sobyTheorem1,Ka=Kb. ϕissurjective,becauseeveryelementofHisoftheformf(x)=ϕ(Kx).Finally,ϕ(Ka·Kb)=ϕ(Kab) = f(ab)=f(a)f(b)=ϕ(Ka)ϕ(Kb). Thus,ϕisanisomorphismfromG/KontoH.■ Theorem 2 is often called the fundamental homomorphism theorem. It asserts that every homomorphic image of G is isomorphic to a quotient group of G. Which specific quotient group of G? Well,iffisahomomorphismfromGontoH,thenHisisomorphictothequotientgroupofGbythekernel off ThefactthatfisahomomorphismfromGontoHmaybesymbolizedbywriting Furthermore,thefactthatKisthekernelofthishomomorphismmaybeindicatedbywriting Thus,incapsuleform,thefundamentalhomomorphismtheoremsaysthat Letusseeafewexamples: WesawintheopeningparagraphofChapter14that is a homomorphism from 6 onto 3. Visibly, the kernel of f is {0, 3}, which is the subgroup of generatedby〈3〉,thatis,thesubgroup(3).Thissituationmaybesymbolizedbywriting 6 WeconcludebyTheorem2that 3≅ 6/〈3〉 For another kind of example, let G and H be any groups and consider their direct product G × H.RememberthatG×Hconsistsofalltheorderedpairs(x,y)asxrangesoverGandyrangesoverH. Youmultiplyorderedpairsbymultiplyingcorrespondingcomponents;thatis,theoperationonG×H is givenby (a,b)·(c,d)=(ac,bd) Now,letfbethefunctionfromG×HontoHgivenby f(x,y)=y Itiseasytocheckthatfisahomomorphism.Furthermore,(x,y)isinthekerneloffifff(x,y)=y = e. Thismeansthatthekerneloffconsistsofalltheorderedpairswhosesecondcomponentise. Call this kernelG*;then G*={(x,e):x∈G} Wesymbolizeallthisbywriting Bythefundamentalhomomorphismtheorem,wededucethatH≅(G×H)/G*.[ItiseasytoseethatG*is anisomorphiccopyofG;thus,identifyingG*withG,wehaveshownthat,roughlyspeaking,(G×H)/G≅ H.] Otherusesofthefundamentalhomomorphismtheoremaregivenintheexercises. EXERCISES In the exercises which follow, FHT will be used as an abbreviation for fundamental homomorphism theorem. A.ExamplesoftheFHTAppliedtoFiniteGroups Ineachofthefollowing,usethefundamentalhomomorphismtheoremtoprovethatthetwogivengroups areisomorphic.Thendisplaytheirtables. Example 2and 6/〈2〉. isahomomorphismfrom 6onto 2.(Donotprovethatfisahomomorphism.)Thekerneloffis{0,2,4} =〈2〉.Thus, ItfollowsbytheFHTthat 2≅ 6/〈2〉. 1 5and 20/〈5〉. 2 3and 6/〈3〉. 3 2andS3/{ε,β,δ}. 4P2andP3/K,whereK={0,{c}}.[HINT:Considerthefunctionf(C)=C∩{a,b}.P3isthegroupof subsetsof{a,b,c},andP2of{a,b}.] 5 3and( 3× 3)/K,whereK={(0,0),(1,1),(2,2)}.[HINT:Considerthefunctionf(a,b)=a−bfrom × 3to 3.] 3 B.ExampleoftheFHTAppliedto ( ) Letα: ( )→ bedefinedbyα(f)=f(l)andletβ: ( )→ bedefinedbyβ(f)=f(2). 1Provethatαandβarehomomorphismsfrom ( )onto . 2LetJbethesetofallthefunctionsfrom to whosegraphpassesthroughthepoint(1,0)andletKbe thesetofallthefunctionswhosegraphpassesthrough(2,0).UsetheFHTtoprovethat ≅ ( )/Jand ≅ ( )/K 3Concludethat ( )/J≅ ( )/K C.ExampleoftheFHTAppliedtoAbelianGroups LetGbeanabeliangroup.LetH={x2:x∈G}andK={x∈G:x2=e}. 1Provethatf(x)=x2isahomomorphismofGontoH. 2Findthekerneloff. 3UsetheFHTtoconcludethatH≅G/K †D.GroupofInnerAutomorphismsofaGroupG LetGbeagroup.ByanautomorphismofGwemeananisomorphismf:G→G. #1ThesymbolAut(G)isusedtodesignatethesetofalltheautomorphismsofG.ProvethatthesetAut (G),withtheoperation∘ofcomposition,isagroupbyprovingthatAut(G)isasubgroupofSG. 2ByaninnerautomorphismofGwemeananyfunctionϕaofthefollowingform: foreveryx∈G ϕa(x)=axa−1 ProvethateveryinnerautomorphismofGisanautomorphismofG. 3Provethat,forarbitrarya,b∈G. ϕa∘ϕb=ϕab and (ϕa)−1=ϕa−1 4LetI(G)designatethesetofalltheinnerautomorphismsofG.Thatis,I(G)={ϕa:a∈G}.Usepart3to provethatI(G)isasubgroupofAut(G).ExplainwhyI(G)isagroup. 5BythecenterofGwemeanthesetofallthoseelementsofGwhichcommutewitheveryelementofG, thatis,thesetCdefinedby C={a∈G:ax=xaforeveryx∈G} Provethata∈Cifandonlyifaxa−1=xforeveryx∈G. 6Leth:G→I(G)bethefunctiondefinedbyh(a)=ϕa.ProvethathisahomomorphismfromGontoI(G) andthatCisitskernel. 7UsetheFHTtoconcludethatI(G)isisomorphicwithG/C. †E.TheFHTAppliedtoDirectProductsofGroups LetGandHbegroups.SupposeJisanormalsubgroupofGandKisanormalsubgroupofH. 1Showthatthefunctionf(x,y)=(Jx,Ky)isahomomorphismfromG×Honto(G/J)×(H/K). 2Findthekerneloff. 3UsetheFHTtoconcludethat(G×H)/(J×K)≅(G/J)×(H/K). †F.FirstIsomorphismTheorem LetGbeagroup;letHandKbesubgroupsofG,withHanormalsubgroupofG.Provethefollowing: 1H KisanormalsubgroupofK #2IfHK={xy:x∈Handy∈K},thenHKisasubgroupofG. 3HisanormalsubgroupofHK. 4EverymemberofthequotientgroupHK/HmaybewrittenintheformHkforsomek∈K. 5Thefunctionf(k)=HkisahomomorphismfromKontoHK/H,anditskernelisH K 6BytheFHT,K/(H K)≅HK/H.(Thisisreferredtoasthefirstisomorphismtheorem.) †G.ASharperCayleyTheorem IfHisasubgroupofagroupG,letXdesignatethesetofalltheleftcosetsofHinG.Foreachelementa ∈G,definepa:X→Xasfollows: pa(xH)=(ax)H 1ProvethateachpaisapermutationofX. 2Provethath:G→SXdefinedbyh(a)=paisahomomorphism. #3Provethattheset{a∈H:xax−1∈Hforeveryx∈G},thatis,thesetofalltheelementsofHwhose conjugatesareallinH,isthekernelofh. 4ProvethatifHcontainsnonormalsubgroupofGexcept{e},thenGisisomorphictoasubgroupofSX. †H.QuotientGroupsIsomorphictotheCircleGroup Everycomplexnumbera+bimayberepresentedasapointinthecomplexplane. Theunitcircleinthecomplexplaneconsistsofallthecomplexnumberswhosedistancefromtheorigin is1;thus,clearly,theunitcircleconsistsofallthecomplexnumberswhichcanbewrittenintheform cosx+isinx forsomerealnumberx. #1Foreachx∈ ,itisconventionaltowritecisx=cosx+isinx.Provethateis(x+y)=(cisx)(cis y). 2LetTdesignatetheset{cisx:x∈ },thatis,thesetofallthecomplexnumberslyingontheunitcircle, withtheoperationofmultiplication.Usepart1toprovethatTisagroup.(Tiscalledthecirclegroup.) 3Provethatf(x)=cisxisahomomorphismfrom ontoT. 4Provethatkerf={2nπ:n∈ }=〈2π〉. 5UsetheFHTtoconcludethatT≅ /〈2〉. 6Provethatg(x)=cis2πxisahomomorphismfrom ontoT,withkernel . 7ConcludethatT≅ / . †I.TheSecondIsomorphismTheorem LetHandKbenormalsubgroupsofagroupG,withH kDefineϕ:G/H→G/Kbyϕ(Ha)=Ka. Proveparts1–4: 1ϕisawell-definedfunction.[Thatis,ifHa=Hb,thenϕ(Ha)=ϕ(Hb).] 2ϕisahomomorphism. 3ϕissurjective. 4kerϕK/H 5Conclude(usingtheFHT)that(G/H)K/H)≅G/K. †J.TheCorrespondenceTheorem LetfbeahomomorphismfromGontoHwithkernelK: IfSisanysubgroupofH,letS*={x∈G:f(x)∈S}.Prove: 1S*isasubgroupofG. 2K S*. 3LetgbetherestrictionofftoS.*[Thatis,g(x)=f(x)foreveryx∈S*,andS*isthedomainofg.]Then gisahomomorphismfromS*ontoS,andK=kerg. 4S≅S*/K. †K.Cauchy’sTheorem Prerequisites:Chapter13,ExerciseI,andChapter15,ExercisesGandH. IfGisagroupandpisanyprimedivisorof|G|,itwillbeshownherethatGhasatleastoneelement of order p. This has already been shown for abelian groups in Chapter 15, Exercise H4. Thus, assume herethatGisnotabelian.Theargumentwillproceedbyinduction;thus,let|G|=k,andassumeourclaim istrueforanygroupoforderlessthank.LetCbethecenterofG,letCabethecentralizerofaforeacha ∈G,andletk=c+ks+⋯+ktbetheclassequationofG,asinChapter15,ExerciseG2. 1Prove:Ifpisafactorof|Ca|foranya∈G,wherea∉C,wearedone.(Explainwhy.) 2Provethatforanya∉CinG,ifpisnotafactorof|Ca|,thenpisafactorof(G:Ca). 3Solvingtheequationk=c+ks+⋯+ktforc,explainwhypisafactorofc.Wearenowdone.(Explain why.) †L.Subgroupsofp-Groups(PreludetoSylow) Prerequisites:ExerciseJ;Chapter15,ExercisesGandH. Letpbeaprimenumber.Ap-groupisanygroupwhoseorderisapowerofp.Itwillbeshownhere thatif|G|=pk thenGhasanormalsubgroupoforderpmforeverymbetween1andk.Theproofisby inductionon|G|;wethereforeassumeourresultistrueforall/^-groupssmallerthanG.Proveparts1and 2: 1ThereisanelementainthecenterofGsuchthatord(a)=p.(SeeChapter15,ExercisesGandH.) 2〈a〉isanormalsubgroupofG. 3ExplainwhyitmaybeassumedthatG/〈a〉hasanormalsubgroupoforderpm−1. #4UseExerciseJ4toprovethatGhasanormalsubgroupoforderpm. SUPPLEMENTARYEXERCISES ExercisesetsMthroughQareincludedasachallengefortheambitiousreader.Twoimportantresultsof grouptheoryareprovedintheseexercises:oneiscalledSylow’stheorem,theotheriscalledthebasis theoremoffiniteabeliangroups. †M.p-SylowSubgroups Prerequisites:ExercisesJandKofthisChapter,ExerciseI1ofChapter14,andExerciseD3ofChapter 15. Letpbeaprimenumber.AfinitegroupGiscalledap-groupiftheorderofeveryelementxinGis a power p. (The orders of different elements may be different powers of p.) If H is a subgroup of any finitegroupG,andHisap-group,wecallHap-subgroipofG.Finally,ifKisap-subgroupofG,andK ismaximal(inthesensethatKisnotcontainedinanylargerp-subgroupofG),thenKiscalledap-Sylow subgroupofG. 1Provethattheorderofanyp-groupisapowerofp.(HINT:UseExerciseK.) 2Provethateveryconjugateofap-SylowsubgroupofGisap-SylowsubgroupofG. LetKbeap-SylowsubgroupofG,andN=N(K)thenormalizerofK. 3Leta∈N,andsupposetheorderofKainN/Kisapowerofp.LetS=〈Ka〉bethecyclicsubgroupof N/KgeneratedbyKa.ProvethatNhasasubgroupS*suchthatS*/Kisap-group. (HINT: See Exercise J4.) 4ProvethatS*isap-subgroupofG(useExerciseD3,Chapter15).ThenexplainwhyS*=K,andwhyit followsthatKa=K. 5Useparts3and4toprove:noelementofN/K has order a power of p (except, trivially, the identity element). 6Ifa∈Nandtheorderofpisapowerofp,thentheorderofKa(inN/K)isalsoapowerofp.(Why?) Thus,Ka=K.(Why?) 7Usepart6toprove:ifaKa−l=Kandtheorderofaisapowerofp,thena∈K. †N.Sylow’sTheorem Prerequisites:ExercisesKandMofthisChapterandExerciseIofChapter14. LetGbeafinitegroup,andKap-SylowsubgroupofG.LetXbethesetofalltheconjugatesofK. SeeExerciseM2.IfC1,C2∈X,letC1∼C2iffC1=aC2a−lforsomeα∈K 1Provethat∼isanequivalencerelationonX. Thus,∼partitionsXintoequivalenceclasses.IfC∈,XlettheequivalenceclassofCbedenotedby [C]. 2ForeachC∈X,provethatthenumberofelementsin[C]isadivisorof|K|.(HINT:UseExerciseI10of Chapter14.)ConcludethatforeachC∈X,thenumberofelementsin[C]iseither1orapowerofp. 3UseExerciseM7toprovethattheonlyclasswithasingleelementis[K], 4Useparts2and3toprovethatthenumberofelementsinXiskp+1,forsomeintegerk. 5Usepart4toprovethat(G:N)isnotamultipleofp. 6Provethat(N:K)isnotamultipleofp.(UseExercisesKandM5.) 7Useparts5and6toprovethat(G:K)isnotamultipleofp. 8Conclude:LetGbeafinitegroupoforderpk m,wherepisnotafactorofm.Everyp-Sylowsubgroup KofGhasorderpk . Combiningpart8withExerciseLgives Let G be a finite group and let p be a prime number. For each n such that pn divides |G|, G has a subgroupoforderpn. ThisisknownasSylow’stheorem. †O.LiftingElementsfromCosets ThepurposeofthisexerciseistoproveapropertyofcosetswhichisneededinExerciseQ.LetGbea finiteabeliangroup,andletabeanelementofGsuchthatord(a)isamultipleoford(x)foreveryx∈G. LetH=〈a〉.Wewillprove: Foreveryx∈G,thereissomey∈GsuchthatHx=Hyandord(y)=ord(Hy). ThismeansthateverycosetofHcontainsanelementywhoseorderisthesameasthecoset’sorder. LetxbeanyelementinG,andletord(a)=t,ord(x)=s,andord(Hx)=r. 1Explainwhyristheleastpositiveintegersuchthatxrequalssomepowerofa,sayxr=am. 2Deducefromourhypothesesthatrdividess,andsdividest. Thus,wemaywrites=ruandt=sυ,soinparticular,t=ruυ. 3Explainwhyamu=e,andwhyitfollowsthatmu=tzforsomeintegerz.Thenexplainwhym=rυz. 4Settingy=xa−υz,provethatHx=Hyandord(y)=r,asrequired. †P.DecompositionofaFiniteAbelianGroupintop-Groups LetGbeanabeliangroupoforderpk m,wherepk andmarerelativelyprime(thatis,pk andmhaveno commonfactorsexcept±1).(REMARK:Iftwointegersjandkarerelativelyprime,thenthereareintegers sandtsuchthatsj+tk=1.Thisisprovedonpage220.) Let Gpk be the subgroup of G consisting of all elements whose order divides pk . Let Gm be the subgroupofGconsistingofallelementswhoseorderdividesra.Prove: k 1Foranyx∈Gandintegerssandt,xsp ∈Gmandxtm∈Gpk . 2Foreveryx∈G,therearey∈Gpk andz∈Gmsuchthatx=yz. 3Gpk Gm={e}. 4G≅Gpk ×Gm.(SeeExerciseH,Chapter14.) 5Suppose|G|hasthefollowingfactorizationintoprimes: Gnwhereforeachi=1,…,n,Giisapi-group. .ThenG≅G1×G2× ⋯ × Q.BasisTheoremforFiniteAbelianGroups Prerequisite:ExerciseP. Asaprovisionaldefinition,letuscallafiniteabeliangroup“decomposable”ifthereareelements al,…,an∈Gsuchthat: (Dl)Foreveryx∈G,thereareintegersk1,…,knsuchthat …,lnsuchthat then . .(D2)Ifthereareintegersl1, If(Dl)and(D2)hold,wewillwriteG=[a1,a2,…,an].Assumethisinparts1and2. 1LetG′bethesetofallproducts andG′=[a2,…,an]. rangeover .ProvethatG′isasubgroupofG, 2Prove:G≅〈a1〉×G′.ConcludethatG≅〈a1〉×〈a2〉×⋯×〈an Intheremainingexercisesofthisset,letpbeaprimenumber,andassumeGisafiniteabeliangroup suchthattheorderofeveryelementinGissomepowerofp.Leta∈Gbeanelementwhoseorderisthe highestpossibleinG.WewillarguebyinductiontoprovethatGis“decomposable.”LetH=〈a〉. 3ExplainwhywemayassumethatG/H=[Hb1,…,Hbn]forsomebl,…,bn∈G. ByExerciseO,wemayassumethatforeachi=1,…,n,ord(bi)=(Hbi).WewillshowthatG=[a, b1,…,bn]. 4Provethatforeveryx∈G,thereareintegersk0,k1,…,knsuchthat 5Provethatif ,then .ConcludethatG=[a,b,1,…,bn]. 6UseExerciseP5,togetherwithparts2and5above,toprove:EveryfiniteabeliangroupGisadirect productofcyclicgroupsofprimepowerorder.(Thisiscalledthebasistheoremoffiniteabeliangroups.) It can be proved that the above decomposition of a finite abelian group into cyclic p-groups is unique, except for the order of the factors. We leave it to the ambitious reader to supply the proof of uniqueness. CHAPTER SEVENTEEN RINGS:DEFINITIONSANDELEMENTARYPROPERTIES Inpresentingscientificknowledgeitiselegantaswellasenlighteningtobeginwiththesimpleandmove towardthemorecomplex.Ifwebuilduponaknowledgeofthesimplestthings,itiseasiertounderstand the more complex ones. In the first part of this book we dedicated ourselves to the study of groups— surely one of the simplest and most fundamental of all algebraic systems. We will now move on, and, using the knowledge and insights gained in the study of groups, we will begin to examine algebraic systemswhichhavetwooperationsinsteadofjustone. Themostbasicofthetwo-operationalsystemsiscalledaring:itwillbedefinedinamoment.The surprising fact about rings is that, despite their having two operations and being more complex than groups, their fundamental properties follow exactly the pattern already laid out for groups. With remarkable,almostcompellingease,wewillfindtwo-operationalanalogsofthenotionsofsubgroupand quotient group, homomorphism and isomorphism—as well as other algebraic notions— and we will discoverthatringsbehavejustlikegroupswithrespecttothesenotions. Thetwooperationsofaringaretraditionallycalledadditionandmultiplication,andaredenotedas usualby+and·,respectively.Wemustremember,however,thattheelementsofaringarenotnecessarily numbers(forexample,thereareringsoffunctions,ringsofswitchingcircuits,andsoon);andtherefore “addition” does not necessarily refer to the conventional addition of numbers, nor does multiplication necessarilyrefertotheconventionaloperationofmultiplyingnumbers.Infact,+and·arenothingmore thansymbolsdenotingthetwooperationsofaring. ByaringwemeanasetAwithoperationscalledadditionandmultiplicationwhichsatisfythe followingaxioms: (i) Awithadditionaloneisanabeliangroup. (ii) Multiplicationisassociative. (iii) Multiplicationisdistributiveoveraddition.Thatis,foralla,b,andcinA, a(b+c)=ab+ac and (b+c)a=ba+ca SinceAwithadditionaloneisanabeliangroup,thereisinAaneutralelementforaddition:itiscalled the zero element and is written 0. Also, every element has an additive inverse called its negative; the negativeofaisdenotedby−a.Subtractionisdefinedby a−b=a+(−b) The easiest examples of rings are the traditional number systems. The set of the integers, with conventionaladditionandmultiplication,isaringcalledtheringoftheintegers.Wedesignatethisring simply with the letter . (The context will make it clear whether we are referring to the ring of the integersortheadditivegroupoftheintegers.) Similarly, istheringoftherationalnumbers, theringoftherealnumbers,and theringofthe complexnumbers.Ineachcase,theoperationsareconventionaladditionandmultiplication. Remember that ( ) represents the set of all the functions from to ; that is, the set of all realvaluedfunctionsofarealvariable.Incalculuswelearnedtoaddandmultiplyfunctions:iffandgareany twofunctionsfrom to ,theirsumf+gandtheirproductfgaredefinedasfollows: [f+g](x)=f(x)+g(x) foreveryrealnumberx and [fg](x)=f(x)g(x) foreveryrealnumberx ( )withtheseoperationsforaddingandmultiplyingfunctionsisaringcalledtheringofrealfunctions. Itiswrittensimplyas ( ).Onpage46wesawthat ( )withonlyadditionoffunctionsisanabelian group. It is left as an exercise for you to verify that multiplication of functions is associative and distributiveoveradditionoffunctions. Therings , , , ,and ( )areallinfiniterings,thatis,ringswithinfinitelymanyelements.There arealsofiniterings:ringswithafinitenumberofelements.Asanimportantexample,considerthegroup n ,anddefineanoperationofmultiplicationon n byallowingtheproductabtobetheremainderofthe usualproductofintegersaandbafterdivisionbyn.(Forexample,in 5,2·4=3,3·3=4,and4·3=2.) Thisoperationiscalledmultiplicationmodulon. nwithadditionandmultiplicationmoduloηisaring: thedetailsaregiveninChapter19. LetAbeanyring.SinceAwithadditionaloneisanabeliangroup,everythingweknowaboutabelian groups applies to it. However, it is important to remember that A with addition is an abelian group in additive notation and, therefore, before applying theorems about groups to A, these theorems must be translatedintoadditivenotation.Forexample,Theorems1,2,and3ofChapter4readasfollowswhen thenotationisadditiveandthegroupisabelian: a+b=a+c a+b=0 implies impliesa=−b −(a+b)=−(a)+−(b) and b=c b=−a and−(−a)=a (1) (2) (3) ThereforeConditions(1),(2),and(3)aretrueineveryring. What happens in a ring when we multiply elements by zero? What happens when we multiply elementsbythenegativesofotherelements?Thenexttheoremanswersthesequestions. Theorem1LetaandbbeanyelementsofaringA. (i) a0=0 and 0a=0 (ii) a(−b)=−(ab) and (−a)b=−(ab) (iii) (−a)(−b)=ab Part (i) asserts that multiplication by zero always yields zero, and parts (ii) and (iii) state the familiar rulesofsigns. PROOF:Toprove(i)wenotethat Thus, aa + 0 = aa + a0. By Condition (1) above we may eliminate the term aa on both sides of this equation,andtherefore0=a0. Toprove(ii),wehave Thus,a(−b)+ab=0.ByCondition(2)abovewededucethata(−b)=−(ab).Thetwinformula(−a)b= −(−ab)isdeducedanalogously. Weprovepart(iii)byusingpart(ii)twice: (−a)(−b)=−[a(−b)]=−[−(ab)]=ab■ The general definition of a ring is sparse and simple. However, particular rings may also have “optionalfeatures”whichmakethemmoreversatileandinteresting.Someoftheseoptionsaredescribed next. By definition, addition is commutative in every ring but mutiplication is not. When multiplication alsoiscommutativeinaring,wecallthatringacommutativering. A ring A does not necessarily have a neutral element for multiplication. If there is in A a neutral elementformultiplication,itiscalledtheunityofA,andisdenotedbythesymbol1.Thus,a.1=aand1 .a=aforeveryainA.IfAhasaunity,wecallAaringwithunity.Therings , , , ,and ( )areall examplesofcommutativeringswithunity. Incidentally, a ring whose only element is 0 is called a trivial ring; a ring with more than one elementisnontrivial.Inanontrivialringwithunity,necessarily1≠0.Thisistruebecauseif1=0andx isanyelementofthering,then x=x1=x0=0 Inotherwords,if1=0theneveryelementoftheringisequalto0;hence0istheonlyelementofthering. If A is a ring with unity, there may be elements in A which have a multiplicative inverse. Such elementsaresaidtobeinvertible.Thus,anelementaisinvertibleinaringifthereissomexinthering suchthat ax=xa=1 Forexample,in everynonzeroelementisinvertible:itsmultiplicativeinverseisitsreciprocal.Onthe otherhand,in theonlyinvertibleelementsare1and−1. Zero is never an invertible element of a ring except if the ring is trivial; for if zero had a multiplicativeinversex,wewouldhave0x=1,thatis,0=1. IfAisacommutativeringwithunityinwhicheverynonzeroelementisinvertible,A is called a field.Fieldsareoftheutmostimportanceinmathematics;forexample, , ,and arefields.Thereare also finite fields, such as 5 (it is easy to check that every nonzero element of 5 is invertible). Finite fieldshavebeautifulpropertiesandfascinatingapplications,whichwillbeexaminedlaterinthisbook. Inelementarymathematicswelearnedthecommandmentthatiftheproductoftwonumbersisequal tozero,say ab=0 then one of the two factors, either a or b (or both) must be equal to zero. This is certainly true if the numbersarereal(orevencomplex)numbers,buttheruleisnotinviolableineveryring.Forexample,in 6, 2·3=0 eventhoughthefactors2and3arebothnonzero.Suchnumbers,whentheyexist,arecalleddivisors of zero. Inanyring,anonzeroelementaiscalledadivisorofzeroifthereisanonzeroelementbinthe ringsuchthattheproductaborbaisequaltozero. (Notecarefullythatbothfactorshavetobenonzero.)Thus,2and3aredivisorsofzeroin 6;4isalsoa divisor of zero in 6, because 4·3 = 0. For another example, let 2( ) designate the set of all 2 × 2 matricesofrealnumbers,withadditionandmultiplicationofmatricesasdescribedonpage8.Thesimple taskofcheckingthat 2( )satisfiestheringaxiomsisassignedasExerciseC1attheendofthischapter. 2( )isrampantwithexamplesofdivisorsofzero.Forinstance, hence arebothdivisorsofzeroin 2( ). Ofcourse,thereareringswhichhavenodivisorsofzeroatall!Forexample, , , ,and donot haveanydivisorsofzero.Itisimportanttonotecarefullywhatitmeansforaringtohavenodivisorsof zero:itmeansthatiftheproductoftwoelementsintheringisequaltozero,atleastoneofthefactors iszero.(Ourcommandmentfromelementarymathematics!) Itisalsodecreedinelementaryalgebrathatanonzeronumberamaybecanceledintheequationax =aytoyieldx=y.Whileundeniablytrueinthenumbersystemsofmathematics,thisruleisnottruein everyring.Forexample,in 6, 2.5=2.2 yet we cannot cancel the common factor 2. A similar example involving 2×2 matrices may be seen on page9.Whencancellationispossible,wesaytheringhasthe“cancellationproperty.” Aringissaidtohavethecancellationpropertyif ab=ac or ba=ca implies b=c foranyelementsa,b,andcintheringifa≠0. Thereisasurprisingandunexpectedconnectionbetweenthecancellationpropertyanddivisorsofzero: Theorem2Aringhasthecancellationpropertyiffithasnodivisorsofzero. PROOF: The proof is very straightforward. Let A be a ring, and suppose first that A has the cancellationproperty.ToprovethatAhasnodivisorsofzerowebeginbylettingab=0,andshowthata orbisequalto0.Ifa=0,wearedone.Otherwise,wehave ab=0=a0 sobythecancellationproperty(cancellinga),b=0. Conversely,assumeAhasnodivisorsofzero.ToprovethatAhasthecancellationproperty,suppose ab=acwherea≠0.Then ab−ac=a(b−c)=0 Remember,therearenodivisorsofzero!Sincea≠0,necessarilyb−c=0,sob=c.■ Anintegraldomainisdefinedtobeacommutativeringwithunityhavingthecancellationproperty. By Theorem 2, an integral domain may also be defined as a commutative ring with unity having no divisors of zero. It is easy to see that every field is an integral domain. The converse, however, is not true:forexample, isanintegraldomainbutnotafield.Wewillhavealottosayaboutintegraldomains inthefollowingchapters. EXERCISES A.ExamplesofRings In each of the following, a set A with operations of addition and multiplication is given. Prove that A satisfiesalltheaxiomstobeacommutativeringwithunity.Indicatethezeroelement,theunity,and thenegativeofanarbitrarya. 1Aistheset oftheintegers,withthefollowing“addition”⊕and“multiplication” : a⊕b=a+b−1 a b=ab−(a+b)+2 2Aistheset oftherationalnumbers,andtheoperationsare⊕and definedasfollows: a⊕b=a+b+1 a b=ab+a+b #3Aistheset × oforderedpairsofrationalnumbers,andtheoperationsarethefollowingaddition ⊕andmultiplication : 4 withconventionaladditionandmultiplication. 5Provethattheringinpart1isanintegraldomain. 6Provethattheringinpart2isafield,andindicatethemultiplicativeinverseofanarbitrarynonzero element. 7Dothesamefortheringinpart3. B.RingofRealFunctions 1Verifythat ( )satisfiesalltheaxiomsforbeingacommutativeringwithunity.Indicatethezeroand unity,anddescribethenegativeofanyf. #2Describethedivisorsofzeroin ( ). 3Describetheinvertibleelementsin ( ). 4Explainwhy ( )isneitherafieldnoranintegraldomain. C.Ringof2×2Matrices Let 2( )designatethesetofall2×2matrices whoseentriesarerealnumbersa,b,c,andd,withthefollowingadditionandmultiplication: and 1 Verifythat 2( )satisfiestheringaxioms. 2 Showthat 2( )isnotcommutativeandhasaunity. 3 Explainwhy 2( )isnotanintegraldomainorafield. D.RingsofSubsetsofaSet IfDisaset,thenthepowersetofDisthesetPDofallthesubsetsofD.Additionandmultiplicationare definedasfollows:IfAandBareelementsofPD(thatis,subsetsofD),then A+B=(A−B)∪(B−A) and AB=A∩B ItwasshowninChapter3,ExerciseC,thatPDwithadditionaloneisanabeliangroup. #1Prove:PDisacommutativeringwithunity.(Youmayassume∩isassociative;forthedistributive law,usethesamediagramandapproachaswasusedtoprovethatadditionisassociativeinChapter3, ExerciseC.) 2DescribethedivisorsofzeroinPD. 3DescribetheinvertibleelementsinPD. 4ExplainwhyPDisneitherafieldnoranintegraldomain.(AssumeDhasmorethanoneelement.) 5GivethetablesofP3,thatis,PDwhereD={a,b,c}. E.RingofQuaternions Aquaternion(inmatrixform)isa2×2matrixofcomplexnumbersoftheform 1Provethatthesetofallthequaternions,withthematrixadditionandmultiplicationexplainedonpages 7and8,isaringwithunity.Thisringisdenotedbythesymbol .Findanexampletoshowthat isnot commutative. (You may assume matrix addition and multiplication are associative and obey the distributivelaw.) 2Let Showthatthequaterniona,definedpreviously,maybewrittenintheform α=al+bi+cj+dk (Thisisthestandardnotationforquaternions.) #3Provethefollowingformulas: i2=j2=k2=−l ij=−ji=k jk=−kj=i ki=−ik=j 4Theconjugateofαis Thenormofαisa2+b2+c2+d2,andiswritten∥α∥.Showdirectly(bymatrixmultiplication)that Concludethatthemultiplicativeinverseofαis(1/t)ᾱ. 5Askewfieldisa(notnecessarilycommutative)ringwithunityinwhicheverynonzeroelementhasa multiplicativeinverse.Concludefromparts1and4that isaskewfield. F.RingofEndomorphisms LetGbeanabeliangroupinadditivenotation.AnendomorphismofGisahomomorphismfromGtoG. Let End(G) denote the set of all the endomorphisms of G, and define addition and multiplication of endomorphismsasfollows: 1ProvethatEnd(G)withtheseoperationsisaringwithunity. 2ListtheelementsofEnd( 4),thengivetheadditionandmultiplicationtablesforEnd( 4). REMARK:Theendomorphismsof 4areeasytofind.Anyendomorphismsof 4willcarry1toeither 0,1,2,or3.Forexample,takethelastcase:if thennecessarily hencefiscompletelydeterminedbythefactthat G.DirectProductofRings IfAandBarerings,theirdirectproductisanewring,denotedbyA×B,anddefinedasfollows:A×B consistsofalltheorderedpairs(x,y)wherexisinAandyisinB.AdditioninA×Bconsistsofadding correspondingcomponents: (x1,y1)+(x2,y2)=(x1+x2,y1+y2) MultiplicationinA×Bconsistsofmultiplyingcorrespondingcomponents: (x1,y1)·(x2,y2)=(x1x2,y1y2) 1IfAandBarerings,verifythatA×Bisaring. 2IfAandBarecommutative,showthatA×Biscommutative.IfAandBeachhasaunity,showthatA×B hasaunity. 3DescribecarefullythedivisorsofzeroinA×B. #4DescribetheinvertibleelementsinA×B. 5ExplainwhyA×Bcanneverbeanintegraldomainorafield.(AssumeAandBeachhavemorethan oneelement.) H.ElementaryPropertiesofRings Proveparts1−4: 1Inanyring,a(b−c)=ab−acand(b−c)a=ba−ca. 2Inanyring,ifab=−ba,then(a+b)2=(a−b)2=a2+b2. 3Inanyintegraldomain,ifa2=b2,thena=±b. 4Inanyintegraldomain,only1and −1aretheirownmultiplicativeinverses.(Notethatx=x−1iffx2= 1.) 5Showthatthecommutativelawforadditionneednotbeassumedindefiningaringwithunity:itmaybe provedfromtheotheraxioms.[HINT:Usethedistributivelawtoexpand(a+b)(1+1)intwodifferent ways.] #6LetAbeanyring.ProvethatiftheadditivegroupofAiscyclic,thenAisacommutativering. 7Prove:Inanyintegraldomain,ifan=0forsomeintegern,thena=0. I.PropertiesofInvertibleElements Provethatparts1−5aretrueinanontrivialringwithunity. 1Ifaisinvertibleandab=ac,thenb=c. 2Anelementacanhavenomorethanonemultiplicativeinverse. 3Ifa2=0thena+1anda−1areinvertible. 4Ifaandbareinvertible,theirproductabisinvertible. 5ThesetSofalltheinvertibleelementsinaringisamultiplicativegroup. 6Bypart5,thesetofallthenonzeroelementsinafieldisamultiplicativegroup.NowuseLagrange’s theoremtoprovethatinafinitefieldwithmelements,xm−1=1foreveryx≠0. 7Ifax=1,xisarightinverseofa;ifya=1,yisaleftinverseofa.Provethatifahasarightinversey andaleftinversey,thenaisinvertible,anditsinverseisequaltoxandtoy.(Firstshowthatyaxa=1.) 8Prove:Inacommutativering,ifabisinvertible,thenaandbarebothinvertible. J.PropertiesofDivisorsofZero Provethateachofthefollowingistrueinanontrivialring. 1Ifa≠±1anda2=1,thena+1anda−1aredivisorsofzero. #2Ifabisadivisorofzero,thenaorbisadivisorofzero. 3Inacommutativeringwithunity,adivisorofzerocannotbeinvertible. 4Supposeab≠0inacommutativering.Ifeitherαorisadivisorofzero,soisab. 5Supposeaisneither0noradivisorofzero.Ifab=ac,thenb=c. 6A×Balwayshasdivisorsofzero. K.BooleanRings AringAisabooleanringifa2=aforeverya∈A.Provethatparts1and2aretrueinanybooleanring A. 1 Foreverya∈A,a=−a.[HINT:Expand(a+a)2.] 2 Usepart1toprovethatAisacommutativering.[HINT:Expand(a+b)2.] Inparts3and4,assumeAhasaunityandprove: 3 Everyelementexcept0and1isadivisorofzero.[Considerx(x−1).] 4 1istheonlyinvertibleelementinA. 5 Lettinga∨b=a+b+abwehavethefollowinginA: a∨bc=(a∨b)(a∨c) a∨(1+a)=1 a∨a=a a(a∨b)=a L.TheBinomialFormula Animportantformulainelementaryalgebraisthebinomialexpansionformulaforanexpression(a+b)n. Theformulaisasfollows: wherethebinomialcoefficient Thistheoremistrueineverycommutativering.(IfKisanypositiveintegerandaisanelementofaring, ka refers to the sum a + a + ⋯ + a with k terms, as in elementary algebra.) The proof of the binomial theorem in a commutative ring is no different from the proof in elementary algebra. We shall review it here. Theproofofthebinomialformulaisbyinductionontheexponentn.Theformulaistriviallytruefor n=1.Intheinductionstep,weassumetheexpansionfor(a+b)nisasabove,andwemustprovethat Now, Collectingterms,wefindthatthecoefficientofan+1−kbk is Bydirectcomputation,showthat Itwillfollowthat(a+b)n+1isasclaimed,andtheproofiscomplete. M.NilpotentandUnipotentElements Anelementaofaringisnilpotentifan=0forsomepositiveintegern. 1Inaringwithunity,provethatifaisnilpotent,thena+1anda −1arebothinvertible.[HINT:Usethe factorization 1−an=(1−a)(1+a+a2+⋯+an−1) for1−a,andasimilarformulafor1+a.] 2Inacommutativering,provethatanyproductxaofanilpotentelementabyanyelementxisnilpotent. #3Inacommutativering,provethatthesumoftwonilpotentelementsisnilpotent.(HINT:Youmustuse thebinomialformula;seeExerciseL.) Anelementaofaringisunipotentiff1−aisnilpotent. 4Inacommutativering,provethattheproductoftwounipotentelementsaandbisunipotent.[HINT:Use thebinomialformulatoexpand1−ab=(1−a)+a(1−b)topowern+m.] 5Inaringwithunity,provethateveryunipotentelementisinvertible.(HINT:UsePart1.) CHAPTER EIGHTEEN IDEALSANDHOMOMORPHISMS Wehavealreadyseenseveralexamplesofsmallerringscontainedwithinlargerrings.Forexample, isa ring inside the larger ring , and itself is a ring inside the larger ring . When a ring B is part of a largerringA,wecallBasubringofA.Thenotionofsubringisthepreciseanalogforringsofthenotion ofsubgroupforgroups.Herearetherelevantdefinitions: LetAbearing,andBanonemptysubsetofA.IfthesumofanytwoelementsofBisagaininB,then Bisclosedwithrespecttoaddition.IfthenegativeofeveryelementofBisinB,thenBisclosed with respecttonegatives.Finally,iftheproductofanytwoelementsofBisagaininB,thenBisclosedwith respect to multiplication. B is called a subring of A if B is closed with respect to addition, multiplication,andnegatives.WhyisBthencalledasubringofA?Quiteelementary: IfanonemptysubsetB⊆Aisclosedwithrespecttoaddition,multiplication,andnegatives,then BwiththeoperationsofAisaring. Thisfactiseasytocheck:Ifa,b,andcareanythreeelementsofB,thena,b,andcarealsoelementsofA becauseB⊆A.ButAisaring,so Thus,inBadditionandmultiplicationareassociativeandthedistributivelawissatisfied.Now,B was assumedtobenonempty,sothereisanelementb∈BbutBisclosedwithrespecttonegatives,so −bis alsoinB.Finally,Bisclosedwithrespecttoaddition;henceb+(−b)∈B.Thatis,0isinB.Thus,B satisfiesalltherequirementsforbeingaring. Forexample, isasubringof becausethesumoftworationalnumbersisrational,theproductof tworationalnumbersisrational,andthenegativeofeveryrationalnumberisrational. By the way, if B is a nonempty subset of A, there is a more compact way of checking that B is a subringofA: BisasubringofAifandonlyifBisclosedwithrespecttosubtractionandmultiplication. ThereasonisthatBisclosedwithrespecttosubtractioniffBisclosedwithrespecttobothaddition andnegatives.Thislastfactiseasytocheck,andisgivenasanexercise. Awhile back, in our study of groups, we singled out certain special subgroups called normal subgroups. We will now describe certain special subrings called ideals which are the counterpart of normalsubgroups:thatis,idealsareinringsasnormalsubgroupsareingroups. LetAbearing,andBanonemptysubsetofA.WewillsaythatBabsorbsproductsinA(or,simply, Babsorbsproducts)if,wheneverwemultiplyanelementinBbyanelementinA(regardlessofwhether thelatterisinsideBoroutsideB),theirproductisalwaysinB.Inotherwords, forallb∈Bandx∈A,xbandbxareinB. AnonemptysubsetBofaringAiscalledanidealofAifBisclosedwithrespecttoadditionand negatives,andBabsorbsproductsinA. Asimpleexampleofanidealistheset oftheevenintegers. isanidealof becausethesumof twoevenintegersiseven,thenegativeofanyevenintegeriseven,and,finally,theproductofaneven integerbyanyintegerisalwayseven. Inacommutativeringwithunity,thesimplestexampleofanidealisthesetofallthemultiplesofa fixedelementabyalltheelementsinthering.Inotherwords,thesetofalltheproducts xa asaremainsfixedandxrangesoveralltheelementsofthering.Thissetisobviouslyanidealbecause and y(xa)=(yx)a Thisidealiscalledtheprincipalidealgeneratedbya,andisdenotedby 〈a〉 Asinthecaseofsubrings,ifBisanonemptysubsetofA,thereisamorecompactwayofchecking thatBisanidealofA: BisanidealofAifandonlyifBisclosedwithrespecttosubtractionandBabsorbsproductsin A. Weshallseepresentlythatidealsplayanimportantroleinconnectionwithhomomorphism Homomorphismsarealmostthesameforringsasforgroups. AhomomorphismfromaringAtoaringBisafunctionf:A→Bsatisfyingtheidentities f(xl+x2)=f(x1)+f(x2) and f(x1x2)=f(x1)f(x2) Thereisalongerbutmoreinformativewayofwritingthesetwoidentities: 1.Iff(x1)=y1andf(x2)=y2thenf(x1+x2)=y1+y2. 2.Iff(x1)=y1andf(x2)=y2thenf(x1x2)=y1y2 Inotherwords,iffhappenstocarryxltoy1andx2toy2,then,necessarily,itmustcarryx1+x2toy1+y2 andx1x2toy1y2.Symbolically, If and ,thennecessarily One can easily confirm for oneself that a function f with this property will transform the addition and multiplication tables of its domain into the addition and multiplication tables of its range. (We may imagineinfiniteringstohave“nonterminating”tables.)Thus,ahomomorphismfromaringAontoaringB isafunctionwhichtransformsAintoB. Forexample,thering 6istransformedintothering 3by aswemayverifybycomparingtheirtables.Theadditiontablesarecomparedonpage136,andwemay dothesamewiththeirmultiplicationtables: If there is a homomorphism from A onto B, we call B a homomorphic image of A. If f is a homomorphismfromaringAtoaringB,notnecessarilyonto,therangeof/isasubringofB.(Thisfactis routinetoverify.)Thus,therangeofaringhomomorphismisalwaysaring.Andobviously,therangeofa homomorphismisalwaysahomomorphicimageofitsdomain. Intuitively, if B is a homomorphic image of A, this means that certain features of A are faithfully preserved in B while others are deliberately lost. This may be illustrated by developing further an exampledescribedinChapter14.TheparityringPconsistsoftwoelements,eando,withadditionand multiplicationgivenbythetables We should think of e as “even” and o as “odd,” and the tables as describing the rules for adding and multiplyingoddandevenintegers.Forexample,even+odd=odd,eventimesodd=even,andsoon. Thefunctionf: →Pwhichcarrieseveryevenintegertoeandeveryoddintegertooiseasilyseen tobeahomomorphismfrom toPthisismadeclearonpage137.Thus,Pisahomomorphicimageof . Although the ring P is very much smaller than the ring , and therefore few of the features of can be expectedtoreappearinP,neverthelessoneaspectofthestructureof isretainedabsolutelyintactinP, namely,thestructureofoddandevennumbers.Aswepassfrom toP,theparityoftheintegers(their beingevenorodd),withitsarithmetic,isfaithfullypreservedwhileallelseislost.Otherexampleswill begivenintheexercises. IffisahomomorphismfromaringAtoaringB,thekerneloffisthesetofalltheelementsofA whicharecarriedbyfontothezeroelementofB.Insymbols,thekerneloffistheset K={x∈A:f(x)=0} ItisaveryimportantfactthatthekerneloffisanidealofA.(Thesimpleverificationofthisfactisleftas anexercise.) If A and B are rings, an isomorphism from A to B is a homomorphism which is a one-to-one correspondencefromAtoB.Inotherwords,itisaninjectiveandsurjectivehomomorphism.Ifthereisan isomorphismfromAtoBwesaythatAisisomorphictoB,andthisfactisexpressedbywriting A≅B EXERCISES A.ExamplesofSubrings Provethateachofthefollowingisasubringoftheindicatedring: 1{x+ y:x,y≅ }isasubringof . 2{x+21/3y+22/3z:x,y,z∈ }isasubringof . 3{x2y:x,y∈ }isasubringof . #4Let ( )bethesetofallthefunctionsfrom to whicharecontinuouson(−∞,∞)andlet ( )be the set of all the functions from to which are differentiable on (−∞, ∞). Then ( ) and ( ) are subringsof ( ). 5Let ( )bethesetofallfunctionsfrom to whicharecontinuousontheinterval[0,1].Then ( )isa subringof ( ),and ( )isasubringof ( ). 6Thesubsetof 2( )consistingofallmatricesoftheform isasubringof 2( ). B.ExamplesofIdeals 1Identifywhichofthefollowingareidealsof × ,andexplain:{(n,n):n∈ };{(5n,0):n∈ };{(n, m):n+miseven};{(n,m):nmiseven};{(2n,3m):n,m∈ }. 2Listalltheidealsof 12. #3Explainwhyeverysubringof nisnecessarilyanideal. 4ExplainwhythesubringofExerciseA6isnotanideal. 5Explainwhy ( )isnotanidealof ( ). 6Provethateachofthefollowingisanidealof ( ): (a)Thesetofallfsuchthatf(x)=0foreveryrationalx. (b)Thesetofallfsuchthatf(0)=0. 7ListalltheidealsofP3.(P3isdefinedinChapter17,ExerciseD.) 8GiveanexampleofasubringofP3whichisnotanideal. 9Giveanexampleofasubringof 3× 3whichisnotanideal. C.ElementaryPropertiesofSubrings Proveparts1–6: 1AnonemptysubsetBofaringAisclosedwithrespecttoadditionandnegativesiffB is closed with respecttosubtraction. 2 Conclude from part 1 that B is a subring of A iff B is closed with respect to subtraction and multiplication. 3IfAisafiniteringandBisasubringofA,thentheorderofBisadivisoroftheorderofA. #4 If a subring B of an integral domain A contains 1, then B is an integral domain. (B is then called a subdomainofA.) #5Everysubringcontainingtheunityofafieldisanintegraldomain. 6IfasubringBofafieldFisclosedwithrespecttomultiplicativeinverses,thenBisafield.(Bisthen calledasubfieldofF.) 7Findsubringsof 18whichillustrateeachofthefollowing: (a)Aisaringwithunity,BisasubringofA,butBisnotaringwithunity. (b)AandBareringswithunity,BisasubringofA,buttheunityofBisnotthesameastheunityofA. 8LetAbearing,f:A→Aahomomorphism,andB={x∈A:f(x)=x}.ProvethatBisasubringofA. 9ThecenterofaringAisthesetofalltheelementsa∈Asuchthatax=xaforeveryx∈A.Provethat thecenterofAisasubringofA. D.ElementaryPropertiesofIdeals LetAbearingandJanonemptysubsetofA. 1 Using Exercise C1, explain why J is an ideal of A iff J is closed with respect to subtraction and J absorbsproductsinA. 2 If A is a ring with unity, prove that J is an ideal of A iff J is closed with respect to addition and J absorbsproductsinA. 3ProvethattheintersectionofanytwoidealsofAisanidealofA. 4ProvethatifJisanidealofAand1∈J,thenJ=A. 5ProvethatifJisanidealofAandJcontainsaninvertibleelementaofA,thenJ=A. 6ExplainwhyafieldFcanhavenonontrivialideals(thatis,noidealsexcept{0}andF). E.ExamplesofHomomorphisms Provethateachofthefunctionsinparts1–6isahomomorphism.Thendescribeitskernelanditsrange. 1ϕ: ( )→ givenbyϕ(f)=f(0). 2h: × → givenbyh(x,y)=x. 3h: → 2( )givenby 4h: × → 2( )givenby #5LetAbetheset × withtheusualadditionandthefollowing“multiplication”: (a,b) (c,d)=(ac,bc) GrantingthatAisaring,letf:A→ 2( )begivenby 6h:Pc→Pcgivenbyh(A)=A D,whereDisafixedsubsetofC. 7Listallthehomomorphismsfrom 2to 4;from 3to 6. F.ElementaryPropertiesofHomomorphisms LetAandBberings,andf:A→Bahomomorphism.Proveeachofthefollowing: 1f(A)={f(x):x∈A}isasubringofB. 2ThekerneloffisanidealofA. 3f(0)=0,andforeverya∈A,f(−a)=−f(a). 4fisinjectiveiffitskernelisequalto{0}. 5IfBisanintegraldomain,theneitherf(l)=1orf(l)=0.Iff(l)=0,thenf(x)=0foreveryx∈A.Iff(1) =1,theimageofeveryinvertibleelementofAisaninvertibleelementofB. 6 Any homomorphic image of a commutative ring is a commutative ring. Any homomorphic image of a fieldisafield. 7IfthedomainAofthehomomorphismfisafield,andiftherangeoffhasmorethanoneelement,thenf isinjective.(HINT:UseExerciseD6.) G.ExamplesofIsomorphisms 1LetAbetheringofExerciseA2inChapter17.Showthatthefunctionf(x)=x −1isanisomorphism from toAhence ≅A. 2Let bethefollowingsubsetof 2( ): Provethatthefunction isanisomorphismfrom to .[REMARK:Youmustbeginbycheckingthatfisawell-definedfunction; thatis,ifa+bi=c+di,thenf(a+bi)=f(c+di).Todothis,notethatifa+bi=c+dithena−c=(d − b)i;thislastequationisimpossibleunlessbothsidesareequaltozero,forotherwiseitwouldassertthat agivenrealnumberisequaltoanimaginarynumber.] 3Provethat{(x,x):x∈ }isasubringof × ,andshow{(x,x):x∈ }≅ . 4Showthatthesetofall2×2matricesoftheform isasubringof 2( ),thenprovethissubringisisomorphicto . Foranyintegerk,letk designatethesubringof whichconsistsofallthemultiplesofk. 5Provethat ∉2 thenprovethat2 ∉3 .Finally,explainwhyifk≠l,thenk ∉l .(REMEMBER:How doyoushowthattworings,orgroups,arenotisomorphic?) H.FurtherPropertiesofIdeals LetAbearing,andletJandKbeidealsofA. Proveparts1-4.(Inparts2-4,assumeAisacommutativering.) 1IfJ K={0},thenjk=0foreveryj∈Jandk∈K. 2Foranya∈A,Ia={ax+j+k:x∈A,j∈J,k∈K}isanidealofA. #3TheradicalofJisthesetradJ={a∈A:an∈Jforsomen∈ }.ForanyidealJ,radJisanideal ofA. 4Foranya∈A,{x∈A:ax=0}isanideal(calledtheannihilatorofa). Furthermore,{x∈A:ax=0foreverya∈A}isanideal(calledtheannihilatingidealofA).IfAisa ringwithunity,itsannihilatingidealisequalto{0}. 5Showthat{0}andAareidealsofA.(Theyaretrivialideals;everyotheridealofAisaproperideal.) AproperidealJofAiscalledmaximalifitisnotstrictlycontainedinanystrictlylargerproperideal: thatis,ifJ⊆K,whereKisanidealcontainingsomeelementnotinJ,thennecessarilyK=A.Showthat thefollowingisanexampleofamaximalideal:In ( ),theidealJ={f:f(0)=0}.[HINT:UseExercise D5.Notethatifg∈Kandg(0)≠0(thatis,g∉J),thenthefunctionh(x)=g(x)−g(0)isinJhenceh(x) −g(x)∈K.Explainwhythislastfunctionisaninvertibleelementof ( ).] I.FurtherPropertiesofHomomorphisms LetAandBberings.Proveeachofthefollowing: 1Iff:A→BisahomomorphismfromAontoBwithkernelK,andJisanidealofAsuchthatK Jthen f(J)isanidealofB. 2Iff:A→BisahomomorphismfromAontoB,andBisafield,thenthekerneloffisamaximalideal. (HINT:Usepart1,withExerciseD6.MaximalidealsaredefinedinExerciseH5.) 3Therearenonontrivialhomomorphismsfrom to .[Thetrivialhomomorphismsaref(x)=0andf(x)= x.] 4Ifnisamultipleofm,then misahomomorphicimageof n. 5Ifnisodd,thereisaninjectivehomomorphismfrom 2into 2n . †J.ARingofEndomorphisms LetAbeacommutativering.Proveeachofthefollowing: 1ForeachelementainA,thefunctionπadefinedbyπa(x)=axsatisfiestheidentityπa(x+y)=πa(x)+ πa(y).(Inotherwords,πaisanendomorphismoftheadditivegroupofA.) 2πaisinjectiveiffaisnotadivisorofzero.(Assumea≠0.) 3πaissurjectiveiffaisinvertible.(AssumeAhasaunity.) 4Let denotetheset{πa:a∈A}withthetwooperations [πa+πb](x)=πa(x)+πb(x) and πaπb=πa∘πb Verifythat isaring. 5Ifϕ:A→ isgivenbyϕ(a)=πa,thenϕisahomomorphism. 6 If A has a unity, then ϕ is an isomorphism. Similarly, if A has no divisors of zero then ϕ is an isomorphism. CHAPTER NINETEEN QUOTIENTRINGS Wecontinueourjourneyintotheelementarytheoryofrings,travelingaroadwhichrunsparalleltothe familiarlandscapeofgroups.Inourstudyofgroupswediscoveredawayofactuallyconstructingallthe homomorphic images of any group G. We constructed quotient groups of G, and showed that every quotient group of G is a homomorphic image of G. We will now imitate this procedure and construct quotientrings. Webeginbydefiningcosetsofrings: LetAbearing,andJanidealofA.Foranyelementa∈A,thesymbolJ+adenotesthesetofall sumsj+a,asaremainsfixedandjrangesoverJ.Thatis, J+a={j+a:j∈J} J+aiscalledacosetofJinA. It is important to note that, if we provisionally ignore multiplication, A with addition alone is an abelian group and J is a subgroup of A. Thus, the cosets we have just defined are (if we ignore multiplication)preciselythecosetsofthesubgroupJinthegroupA,withthenotationbeingadditive. Consequently, everything we already know about group cosets continues to apply in the present case— only,caremustbetakentotranslateknownfactsaboutgroupcosetsintoadditivenotation.Forexample, Property(1)ofChapter13,withTheorem5ofChapter15,readsasfollowsinadditivenotation: Wealsoknow,bythereasoningwhichleadsuptoLagrange’stheorem,thatthefamilyofallthecosetsJ+ a,asarangesoverA,isapartitionofA. Thereisawayofaddingandmultiplyingcosetswhichworksasfollows: Inotherwords,thesumofthecosetofaandthecosetofbisthecosetofa+b;theproductofthecosetof aandthecosetofbisthecosetofab. It is important to know that the sum and product of cosets, defined in this fashion, are determined withoutambiguity.RememberthatJ+amaybethesamecosetasJ+c[byCondition(1)thishappensiff cisanelementofJ+a],and,likewise,J+bmaybethesamecosetasJ+d. Therefore, we have the equations ObviouslywemustbeabsolutelycertainthatJ+(a + b) = J + (c + d) and J + ab = J + cd. The next theoremprovidesuswiththisimportantguarantee. Theorem1LetJbeanidealofA.IfJ+a=J+candJ+b=J+d,then (i) J+(a+b)=J+(c+d),and (ii) J+ab=J+cd. PROOF:WearegiventhatJ+a=J+candJ+b=J+d;hencebyCondition(2), a–c∈J and b–d∈J SinceJisclosedwithrespecttoaddition,(a −c)+(b − d) = (a + b) − (c + d) is in J. It follows by Condition(2)thatJ+(a+b)=J+(c+d),whichproves(i).Ontheotherhand,sinceJabsorbsproducts inA, andtherefore(ab −cb)+(cb −cd)=ab −cdisinJ.ItfollowsbyCondition(2)thatJ+ab=J+cd. Thisproves(ii).■ Now,thinkofthesetwhichconsistsofallthecosetsofJinA.Thissetisconventionallydenotedby thesymbolA/J.Forexample,ifJ+a,J+b,J+c,…arecosetsofJ,then A/J={J+a,J+b,J+c,…} Wehavejustseenthatcosetadditionandmultiplicationarevalidoperationsonthisset.Infact, Theorem2A/Jwithcosetadditionandmultiplicationisaring. PROOF: Coset addition and multiplication are associative, and multiplication is distributive over addition.(Thesefactsmayberoutinelychecked.)ThezeroelementofA/JisthecosetJ=J+0,forifJ+ aisanycoset, (J+a)+(J+0)=J+(a+0)=J+a Finally,thenegativeofJ+aisJ+(–a),because (J+a)+(J+(–a))=J+(a+(–a))=J+0■ TheringA/JiscalledthequotientringofAbyJ. Andnow,thecrucialconnectionbetweenquotientringsandhomomorphisms: Theorem3A/JisahomomorphicimageofA. Followingtheplanalreadylaidoutforgroups,thenatural homomorphism from A onto A/J is the functionfwhichcarrieseveryelementtoitsowncoset,thatis,thefunctionfgivenby f(x)=J+x Thisfunctionisveryeasilyseentobeahomomorphism. Thus,whenweconstructquotientringsofA,weare,infact,constructinghomomorphicimagesofA. The quotient ring construction is useful because it is a way of actually manufacturing homomorphic imagesofanyringA. Thequotientringconstructionisnowillustratedwithanimportantexample.Let betheringofthe integers,andlet〈6〉betheidealof whichconsistsofallthemultiplesofthenumber6.Theelementsof thequotientring /〈6〉areallthecosetsoftheideal(6),namely: Wewillrepresentthesecosetsbymeansofthesimplifiednotation andmultiplyingcosetsgiveusthefollowingtables: .Therulesforadding Onecannotfailtonoticetheanalogybetweenthequotientring /〈6〉andthering 6.Infact,wewill regardthemasoneandthesame.Moregenerally,foreverypositiveintegern,weconsider ntobethe sameas /〈n〉.Inparticular,thismakesitclearthat nisahomomorphicimageof . By Theorem 3, any quotient ring A/J is a homomorphic image of A. Therefore the quotient ring constructionisawayofactuallyproducinghomomorphicimagesofanyringA.Infact,aswewillnow see,itisawayofproducingallthehomomorphicimagesofA. Theorem4Letf:A→BbeahomomorphismfromaringAontoaringB,andletKbethekernel off.ThenB≅A/B. PROOF: To show that A/K is isomorphic with B, we must look for an isomorphism from A/K to B. Mimickingtheprocedurewhichworkedsuccessfullyforgroups,weletϕbethefunctionfromA/K to B whichmatcheseachcosetK+xwiththeelementf(x);thatis, ϕ(K+x)=f(x) Remember that if we ignore multiplication for just a moment, A and B are groups and f is a group homomorphismfromAontoB,withkernelK.ThereforewemayapplyTheorem2ofChapter16:ϕ is a well-defined,bijectivefunctionfromA/KtoB.Finally, Thus,ϕisanisomorphismfromA/KontoB.■ Theorem4iscalledthefundamentalhomomorphismtheoremforrings.Theorems3and4together assert that every quotient ring of A is a homomorphic image of A, and, conversely, every homomorphic image of A is isomorphic to a quotient ring of A. Thus, for all practical purposes, quotients and homomorphicimagesofaringarethesame. Asinthecaseofgroups,therearemanypracticalinstancesinwhichitispossibletoselectanideal JofAsoasto“factorout”unwantedtraitsofA,andobtainaquotientringA/Jwith“desirable”features. Asasimpleexample,letAbearing,notnecessarilycommutative,andletJbeanidealofAwhich containsallthedifferences ab–ba asaandbrangeoverA.ItisquiteeasytoshowthatthequotientringA/Jisthencommutative.Indeed,to saythatA/JiscommutativeistosaythatforanytwocosetsJ+aandJ+b, (J+a)(J+b)=(J+b)(J+a) thatis J+ab=J+ba ByCondition(2)thislastequationistrueiffab−ba∈J.Thus,ifeverydifferenceab −baisinJ,then anytwocosetscommute. Anumberofimportantquotientringconstructions,similarinprincipletothisone,aregiveninthe exercises. AnidealJofacommutativeringissaidtobeaprimeidealifforanytwoelementsaandbinthe ring, If ab∈J then a∈J or b∈J Whenever J is a prime ideal of a commutative ring with unity A, the quotient ring A/J is an integral domain.(Thedetailsareleftasanexercise.) Anidealofaringiscalledproperifitisnotequaltothewholering.AproperidealJofaringAis calledamaximalidealifthereexistsnoproperidealKofAsuchthatJ⊆KwithJ≠K(inotherwords, Jisnotcontainedinanystrictlylargerproperideal).ItisanimportantfactthatifAisacommutativering withunity,thenJisamaximalidealofAiffA/Jisafield. Toprovethisassertion,letJbeamaximalidealofA.IfAisacommutativeringwithunity,itiseasy toseethatA/Jisonealso.Infact,itshouldbenotedthattheunityofA/JisthecosetJ+1,becauseifJ+ aisanycoset,(J+a)(J+1)=J+a1=J+a.Thus,toprovethatA/Jisafield,itremainsonlytoshow thatifJ+aisanynonzerocoset,thereisacosetJ+xsuchthat(J+a)(J+x)=J+1. ThezerocosetisJ.Thus,byCondition(3),tosaythatJ+aisnotzero,istosaythata∉J.Now,let Kbethesetofallthesums xa+j asxrangesoverAandjrangesoverJ.ItiseasytocheckthatKisanideal.Furthermore,K contains a becausea=1a+0,andKcontainseveryelementj∈Jbecausejcanbewrittenas0a+j.Thus,Kisan ideal which contains J and is strictly larger than J (for remember that a ∈ K but a ∉ J). But J is a maximalideal!Thus,KmustbethewholeringA. Itfollowsthat1∈K,so1=xa+jforsomex∈Aandj∈J.Thus,1–xa=j∈J,sobyCondition (2),J+1=J+xa=(J+x)(J+a).InthequotientringA/J,J+xisthereforethemultiplicativeinverseof J+a. The converse proof consists, essentially, of “unraveling” the preceding argument; it is left as an entertainingexercise. EXERCISES A.ExamplesofQuotientRings Ineachofthefollowing,AisaringandJisanidealofA.ListtheelementsofA/J, and then write the additionandmultiplicationtablesofA/J. ExampleA= 6,J={0,3}. TheelementsofA/JarethethreecosetsJ=J+0={0,3},J+1={1,4},andJ+2={2,5}.The tablesforA/Jareasfollows: 1A= 10,J={0,5}. 2A=P3,J={0,{a}}.(P3isdefinedinChapter17,ExerciseD.) 3A= 2× 6;J={(0,0),(0,2),(0,4)}. B.ExamplesoftheUseoftheFHT Ineachofthefollowing,usetheFHT(fundamentalhomomorphismtheorem)toprovethatthetwogiven groupsareisomorphic.Thendisplaytheirtables. Example 2and 6/〈2〉. Thefollowingfunctionisahomomorphismfrom 6onto 2: (DonotprovethatJisahomomorphism.) Thekerneloffis{0,2,4}=(2).Thus: ItfollowsbytheFHTthat 2≅ 6/〈2〉. 1 5and 20/〈5〉. 2 3and 6/〈3〉. 3P2andP3/K,whereK={0,{c}}.[HINT:SeeChapter18,ExerciseE6.Considerthefunctionf(X)=X ∩{a,b}.] 4 2and 2× 2/K,whereK={(0,0),(0,1)}. C.QuotientRingsandHomomorphicImagesin ( ) 1Letϕbethefunctionfrom ( )to × definedbyϕ(f)=(f(0),f(1)).Provethatϕisahomomorphism from ( )onto × ,anddescribeitskernel. 2 Let J be the subset of ( ) consisting of all f whose graph passes through the points (0,0) and (1,0). Referringtopart1,explainwhyJisanidealof ( ),and ( )/J≅ × . 3Letϕbethefunctionfrom ( )to ( , )definedasfollows: ϕ(f)=f =therestrictionofJto (NOTE: The domain of f is and on this domain f is the same function as f.) Prove that ϕ is a homomorphismfrom ( )onto ( , ),anddescribethekernelofϕ.[ ( , )istheringoffunctionsfrom to .] 4LetJbethesubsetof ( )consistingofallfsuchthatf(x)=0foreveryrationalx.Referringtopart3, explainwhyJisanidealof ( )and ( )/J≅ ( ). D.ElementaryApplicationsoftheFundamentalHomomorphismTheorem IneachofthefollowingletAbeacommutativering.Ifa∈Aandnisapositiveinteger,thenotationna willstandfor a+a+⋯+a (nterms) 1Suppose2x=0foreveryx∈A.Provethat(x+y)2=x2+y2forallxandy in A. Conclude that the functionh(x)=x2isahomomorphismfromAtoA.IfJ={x∈A:x2=0}andB={x2:x∈A),explain whyJisanidealofA,BisasubringofA,andA/J≅B. 2Suppose6x=0foreveryx∈A.Provethatthefunctionh(x)=3xisahomomorphismfromAtoA.IfJ= {x:3x=0}andB={3x:x∈A},explainwhyJisanidealofA,BisasubringofA,andA/J≅B. 3IfaisanidempotentelementofA(thatis,a2=a),provethatthefunctionπa(x)=axisahomomorphism fromAintoA.ShowthatthekernelofπaisIa,theannihilatorofa(definedinExerciseH4ofChapter18). Showthattherangeofπais〈a〉.ConcludebytheFHTthatA/Ia=〈a〉. 4 For each a ∈ A, let πa be the function given by πa(x) = ax. Define the following addition and multiplicationon ={πa:a∈A}: πa+πb=πa+bandπaπb=πab ( isaring;however,donotprovethis.)Showthatthefunctionϕ(a)=πaisahomomorphismfromAonto . Let I designate the annihilating ideal of A (defined in Exercise H4 of Chapter 18). Use the FHT to showthatA/I≅ . E.PropertiesofQuotientRingsA/JinRelationtoPropertiesofJ LetA be a ring and J an ideal of A. Use Conditions (1), (2), and (3) of this chapter. Prove each of the following: #1EveryelementofA/Jhasasquarerootiffforeveryx∈A,thereissomey∈Asuchthatx–y2∈J. 2EveryelementofA/Jisitsownnegativeiffx+x∈Jforeveryx∈A. 3A/Jisabooleanringiffx2–x∈Jforeveryx∈A.(AringSiscalledabooleanringiffs2=sforevery s∈S.) 4IfJistheidealofallthenilpotentelementsofacommutativeringA,thenA/Jhasnonilpotentelements (exceptzero).(NilpotentelementsaredefinedinChapter17, Exercise M; by M2 and M3 they form an ideal.) 5EveryelementofA/JisnilpotentiffJhasthefollowingproperty:foreveryx∈A,thereisapositive integernsuchthatxn∈J. #6A/Jhasaunityelementiffthereexistsanelementa∈Asuchthatax–x∈Jandxa–x∈Jforevery x∈A. F.PrimeandMaximalIdeals LetAbeacommutativeringwithunity,andJanidealofA.Proveeachofthefollowing: 1A/Jisacommutativeringwithunity. 2JisaprimeidealiffA/Jisanintegraldomain. 3 Every maximal ideal of A is a prime ideal. (HINT: Use the fact, proved in this chapter, that if J is a maximalidealthenA/Jisafield.) 4IfA/Jisafield,thenJisamaximalideal.(HINT:SeeExerciseI2ofChapter18.) G.FurtherPropertiesofQuotientRingsinRelationtoTheirIdeals LetAbearingandJanidealofA.(Inparts1–3and5assumethatAisacommutativeringwithunity.) #1ProvethatA/Jisafieldiffforeveryelementa∈A,wherea∉J,thereissomeb∈Asuchthatab–1 ∈J. 2 Prove that every nonzero element of A/J is either invertible or a divisor of zero iff the following propertyholds,wherea,x∈A:Foreverya∉J,thereissomex∉Jsuchthateitherax∈Jorax–1∈ J. 3AnidealJofaringAiscalledprimaryiffforalla,b∈A,ifab∈J,theneithera∈Jorbn∈Jfor somepositiveintegern.ProvethateveryzerodivisorinA/JisnilpotentiffJisprimary. 4AnidealJofaringAiscalledsemiprimeiffithasthefollowingproperty:Foreverya∈A,ifan∈J for some positive integer n, then necessarily a ∈ J. Prove that J is semiprime iff A/J has no nilpotent elements(exceptzero). 5Provethatanintegraldomaincanhavenononzeronilpotentelements.Thenusepart4,togetherwith ExerciseF2,toprovethateveryprimeidealinacommutativeringissemiprime. H.ZnasaHomomorphicImageofZ Recallthatthefunction f(a)=ā isthenaturalhomomorphismfrom onto n.Ifapolynomialequationp=0issatisfiedin , necessarily f(p)=f(0)istruein n.Letustakeaspecificexample;thereareintegersxandysatisfying11x2 −8y2+ 29=0(wemaytakex=3andy=4).Itfollowsthattheremustbeelements and in 6whichsatisfy in 6, that is, . (We take and .) The problems which followarebasedonthisobservation. 1Provethattheequationx2–7y2 −24=0hasnointegersolutions.(HINT:Ifthereareintegersxandy satisfyingthisequation,whatequationwill and satisfyin 7?) 2Provethatx2+(x+1)2+(x+2)2=y2hasnointegersolutions. 3Provethatx2+10y2=n(wherenisaninteger)hasnointegersolutionsifthelastdigitofnis2,3,7,or 8. 4Provethatthesequence3,8,13,18,23,…doesnotincludethesquareofanyinteger.(HINT:Theimage ofeachnumberonthislist,underthenaturalhomomorphismfrom to 5,is3.) 5Provethatthesequence2,10,18,26,…doesnotincludethecubeofanyinteger. 6Provethatthesequence3,11,19,27,…doesnotincludethesumoftwosquaresofintegers. 7Provethatifnisaproductoftwoconsecutiveintegers,itsunitsdigitmustbe0,2,or6. 8Provethatifnistheproductofthreeconsecutiveintegers,itsunitsdigitmustbe0,4,or6. CHAPTER TWENTY INTEGRALDOMAINS Letusrecallthatanintegraldomainisacommutativeringwithunityhavingthecancellationproperty,that is, if a≠0and ab=acthen b=c (1) AttheendofChapter17wesawthatanintegraldomainmayalsobedefinedasacommutativeringwith unityhavingnodivisorsofzero,whichistosaythat if ab=0 then a=0 or b=0 (2) foraswesaw,(1)and(2)areequivalentpropertiesinanycommutativering. The system of the integers is the exemplar and prototype of integral domains. In fact, the term “integraldomain”meansasystemofalgebra(“domain”)havingintegerlikeproperties.However, isnot theonlyintegraldomain:thereareagreatmanyintegraldomainsdifferentfrom . Our first few comments will apply to rings generally. To begin with, we introduce a convenient notationformultiples,whichparallelstheexponentnotationforpowers.Additively,thesum a+a+···+a ofnequaltermsiswrittenasn·a.Wealsodefine0·atobe0,andlet(-n)·a=−(n·a)forallpositive integersn.Then m·a+n·a=(m+n)·a and m·(n·a)=(mn)·a foreveryelementaofaringandallintegersmandn.Theseformulasarethetranslationsintoadditive notationofthelawsofexponentsgiveninChapter10. IfAisaring,Awithadditionaloneisagroup.Rememberthatinadditivenotationtheorder of an elementainAistheleastpositiveintegernsuchthatn·a=0.Ifthereisnosuchpositiveintegern,thena is said to have order infinity. To emphasize the fact that we are referring to the order of a in terms of addition,wewillcallittheadditiveorderofa. Inaringwithunity,if1hasadditiveordern,wesaytheringhas“characteristicn.”Inotherwords, ifAisaringwithunity, thecharacteristicofAistheleastpositiveintegernsuchthat Ifthereisnosuchpositiveintegern,Ahascharacteristic0. Theseconceptsareespeciallysimpleinanintegraldomain.Indeed, Theorem1Allthenonzeroelementsinanintegraldomainhavethesameadditiveorder. PROOF:Thatis,everya≠0hasthesameadditiveorderastheadditiveorderof1.Thetruthofthis statementbecomestransparentlyclearassoonasweobservethat n·a=a+a+···+a=1a+···+1a=(1+···+1)a=(n·1)a hencen·a=0iffn·1=0.(Rememberthatinanintegraldomain,iftheproductoftwofactorsisequalto 0,atleastonefactormustbe0.)■ Itfollows,inparticular,thatifthecharacteristicofanintegraldomainisapositiveintegern,then n·x=0 foreveryelementxinthedomain. Furthermore, Theorem 2 In an integral domain with nonzero characteristic, the characteristic is a prime number. PROOF:Ifthecharacteristicwereacompositenumbermn,thenbythedistributivelaw, Thus,eitherm·1=0orn·1=0,whichisimpossiblebecausemnwaschosentobetheleast positive integersuchthat(mn)·1=0.■ Averyinterestingruleofarithmeticisvalidinintegraldomainswhosecharacteristicisnotzero. Theorem3Inanyintegraldomainofcharacteristicp, (a+b)p=ap+bp forallelementsaandb PROOF:Thisformulabecomesclearwhenwelookatthebinomialexpansionof(a+b)p.Remember thatbythebinomialformula, wherethebinomialcoefficient ItisdemonstratedinExerciseLofChapter17thatthebinomialformulaiscorrectineverycommutative ring. Notethatifpisaprimenumberand0<k<p,then becauseeveryfactorofthedenominatorislessthanp,hencepdoesnotcancelout.Thus,eachtermofthe binomial expansion above, except for the first and last terms, is of the form px, which is equal to 0 becausethedomainhascharacteristicp.Thus,(a+b)p=ap+bp.■ It is obvious that every field is an integral domain: for if a ≠ 0 and ax = ay in a field, we can multiply both sides of this equation by the multiplicative inverse of a to cancel a. However, not every integraldomainisafield:forexample, isnotafield.Nevertheless, Theorem4Everyfiniteintegraldomainisafield. Listtheelementsoftheintegraldomaininthefollowingmanner: 0,1,a1,a2,…,an Inthismanneroflisting,therearen+2elementsinthedomain.Takeanyai,andshowthatitisinvertible: tobeginwith,notethattheproducts ai0,ai1,aia1,aia2,…,aian arealldistinct:forifaix=aiy,thenx=y.Thus,therearen+2distinctproductsaix;butthereareexactly n + 2 elements in the domain, so every element in the domain is equal to one of these products. In particular,1=aixforsomex;henceaiisinvertible. OPTIONAL Theintegraldomain isnotafieldbecauseitdoesnotcontainthequotientsm/nofintegers.However, canbeenlargedtoafieldbyaddingtoitallthequotientsofintegers;theresultingfield,ofcourse,is , the field of the rational numbers. consists of all quotients of integers, and it contains (or rather, an isomorphiccopyof )whenweidentifyeachintegernwiththequotientn/1.Wesaythat isthefieldof quotientsof . It is a fascinating fact that the method for constructing from can be applied to any integral domain.StartingfromanyintegraldomainA,itispossibletoconstructafieldwhichcontainsA:afield ofquotientsofA.Thisisnotmerelyamathematicalcuriosity,butavaluableadditiontoourknowledge.In applicationsitoftenhappensthatasystemofalgebrawearedealingwithlacksaneededproperty,butis containedinalargersystemwhichhasthatproperty—andthatisalmostasgood!Inthepresentcase,Ais notafieldbutmaybeenlargedtoone. Thus, if A is any integral domain, we will proceed to construct a field A* consisting of all the quotientsofelementsinA;andA*willcontainA,orratheranisomorphiccopyofA, when we identify eachelementaofAwiththequotienta/1.Theconstructionwillbecarefullyoutlinedandthebusywork leftasanexercise. GivenA,letSdenotethesetofallorderedpairs(a,b)ofelementsofA,whereb≠0.Thatis, S={(a,b):a,b∈Aandb≠0} In order to understand the next step, we should think of (a, b) as a/b. [It is too early in the proof to introduce fractional notation; nevertheless each ordered pair (a, b) should be thought of as a fraction a/b.}Nowaproblemofrepresentationariseshere,becauseitisobviousthatthequotientxa/xbisequal tothequotienta/b;toputthesamefactdifferently,thequotientsa/bandc/dareequalwheneverad=bc. Thatis,ifad=bc,thena/bandc/daretwodifferentwaysofwritingthesamequotient.Motivatedbythis observation,wedefine(a,b)~(c,d) to mean that ad = bc, and easily verify that ~ is an equivalence relationontheset5.(EquivalencerelationsareexplainedinChapter12.)Thenwelet[a,b] denote the equivalenceclassof(a,b),thatis, [a,b]={(c,d)∈S:(c,d)~(a,b)} Intuitively,allthepairswhichrepresentagivenquotientarelumpedtogetherinoneequivalenceclass; thus,eachquotientisrepresentedbyexactlyoneequivalenceclass. Letusrecapitulatetheformaldetailsofourconstructionuptothispoint:GiventhesetSofordered pairsofelementsinA,wedefineanequivalencerelation—inSbyletting(a,b)~(c,d)iffad=bc.We let [a, b] designate the equivalence class of (a, b), and finally, we let A* denote the set of all the equivalenceclasses[a,b].TheelementsofA*willbecalledquotients. Beforegoingon,observecarefullythat [a,b]=[r,s] iff (a,b)~(r,s) iff as=br (3) Asournextstep,wedefineoperationsofadditionandmultiplicationinA*: [a,b]+[c,d]=[ad+bc,bd] and [a,b]·[c,d]=[ac,bd] Tounderstandthesedefinitions,simplyremembertheformulas Wemustmakecertainthesedefinitionsareunambiguous;thatis,if[a,b]=[r,s]and[c,d]=[t, u], we haveequations andwemustthereforeverifythat[ad+bc,bd]=[ru+st,su]and[ac,bd]=[rt,su].Thisisleftasan exercise. It is also left for the student to verify that addition and multiplication are associative and commutativeandthedistributivelawissatisfied. Thezeroelementis[0,1],because[a,b]+[0,1]=[a,b].Thenegativeof[a,b]is[-a,b],for[a,b] +[-a,b]=[0,b2]=[0,1].[ThelastequationistruebecauseofEquation(3).]Theunityis[1,1],andthe multiplicativeinverseof[a,b]is[b,a],for[a,b]·[b,a]=[ab,ab]=[1,1].Thus,A*isafield! Finally,ifA′isthesubsetofA*whichcontainsevery[a,1],weletϕbethefunctionfromA to A′ definedbyϕ(a)=[a,1].Thisfunctionisinjectivebecause,byEquation(3),if[a,1]=[b,1]thena=b.It isobviouslysurjectiveandiseasilyshowntobeahomomorphism.Thus,ϕisanisomorphismfromAtoA ′soA*containsanisomorphiccopyA′ofA. EXERCISES A.CharacteristicofanIntegralDomain LetAbeafiniteintegraldomain.Proveeachofthefollowing: 1LetabeanynonzeroelementofA.Ifn·a=0,wheren≠0,thennisamultipleofthecharacteristicof A. 2IfAhascharacteristiczero,n≠0,andn·a=0,thena=0. 3IfAhascharacteristic3,and5·a=0,thena=0. 4IfthereisanonzeroelementainAsuchthat256·a=0,thenAhascharacteristic2. 5IftherearedistinctnonzeroelementsaandbinAsuchthat125·a=125·b,thenAhascharacteristic5. 6IftherearenonzeroelementsaandinAsuchthat(a+b)2=a2+b2,thenAhascharacteristic2. 7IftherearenonzeroelementsaandbinAsuchthat10a=0and14b=0,thenAhascharacteristic2. B.CharacteristicofaFiniteIntegralDomain LetAbeanintegraldomain.Proveeachofthefollowing: 1IfAhascharacteristicq,thenqisadivisoroftheorderofA. 2IftheorderofAisaprimenumberp,thenthecharacteristicofAmustbeequaltop. 3IftheorderofAispm,wherepisaprime,thecharacteristicofAmustbeequaltop. 4IfAhas81elements,itscharacteristicis3. 5IfA,withadditionalone,isacyclicgroup,theorderofAisaprimenumber. C.FiniteRings LetAbeafinitecommutativeringwithunity. 1 Prove: Every nonzero element of A is either a divisor of zero or invertible. (HINT: Use an argument analogoustotheproofofTheorem4.) 2Prove:Ifa≠0isnotadivisorofzero,thensomepositivepowerofaisequalto1.(HINT:Considera, a2,a3,....SinceAisfinite,theremustbepositiveintegersn<msuchthatan=am.) 3Usepart2toprove:Ifaisinvertible,thena−1isequaltoapositivepowerofa. D.FieldofQuotientsofanIntegralDomain ThefollowingquestionsrefertotheconstructionofafieldofquotientsofA,asoutlinedonpages203to 205. 1If[a,b]=[r,s]and[c,d]=[t,u],provethat[a,b]+[c,d]=[r,s]+[t,u]. 2If[a,b]=[r,s]and[c,d]=[t,u],provethat[a,b][c,d]=[r,s][t,u]. 3If(a,b)~(c,d)meansad=bc,provethat~isanequivalencerelationonS. 4ProvethatadditioninA*isassociativeandcommutative. 5ProvethatmultiplicationinA*isassociativeandcommutative. 6ProvethedistributivelawinA*. 7Verifythatϕ:A→A′isahomomorphism. E.FurtherPropertiesoftheCharacteristicofanIntegralDomain LetAbeanintegraldomain.Proveparts1-4: 1Leta∈A.IfAhascharacteristicp,andn·a=0wherenisnotamultipleofp,thena=0. 2Ifpisaprime,andthereisanonzeroelementa∈Asuchthatp·a=0,thenAhascharacteristicp. 3Ifpisaprime,andthereisanonzeroelementa∈Asuchthatpm·a=0forsomeintegerm,thenAhas characteristicp. 4IfAhascharacteristicp,thenthefunctionf(a)=apisahomomorphismfromAtoA. #5LetAhaveorderp,wherepisaprime.Explainwhy A={0,1,2·1,3·1,...,(p−1).1} ProvethatA≅ p. #6IfAhascharacteristicp,provethatforanypositiveintegern, n n n (a)(a+b)p =ap +bp (b) 7LetA⊆BwhereAandBareintegraldomains.Prove:AhascharacteristicpiffBhascharacteristicp. F.FiniteFields ByTheorem4,“finiteintegraldomain”and“finitefield”arethesame. 1Prove:Everyfinitefieldhasnonzerocharacteristic. 2ProvethatifAisafinitefieldofcharacteristicp,thefunctionf(a)=apisanautomorphismofA;thatis, anisomorphismfromAtoA.(HINT:UseExerciseE4aboveandExerciseF7ofChapter18.Toshowthat fissurjective,comparethenumberofelementsinthedomainandintherangeoff.) Thefunctionf(a)=apiscalledtheFroebeniusautomorphism. 3Usepart2toprove:Inafinitefieldofcharacteristicp,everyelementhasap-throot. CHAPTER TWENTY-ONE THEINTEGERS Therearetwopossiblewaysofdescribingthesystemoftheintegers. Ontheonehand,wemayattempttodescribeitconcretely. Ontheotherhand,wemayfindalistofaxiomsfromwhichitispossibletodeducealltheproperties oftheintegers,sotheonlysystemwhichhasallthesepropertiesisthesystemoftheintegers. Thesecondofthesetwowaysisthewayofmathematics.Itiselegant,economical,andsimple.We selectasaxiomsonlythoseparticularpropertiesoftheintegerswhichareabsolutelynecessaryinorder toprovefurtherpropertiesoftheintegers.Andweselectasufficientlycompletelistofaxiomssothat, usingthem,onecanproveallthepropertiesoftheintegersneededinmathematics. We have already seen that the integers are an integral domain. However, there are numerous examplesofintegraldomainswhichbearlittleresemblancetothesetoftheintegers.Forexample,there arefiniteintegraldomainssuchas 5,fields(rememberthateveryfieldisanintegraldomain)suchas and ,andothers.Thus,inordertopindowntheintegers—thatis,inordertofindalistofaxiomswhich appliestotheintegersandonlytheintegers—wemustselectsomeadditionalaxiomsandaddthemtothe axiomsofintegraldomains.Thiswewillnowproceedtodo. Most of the traditional number systems have two aspects. One aspect is their algebraic structure: theyareintegraldomainsorfields.Theotheraspect—whichwehavenotyettouchedupon—isthattheir elementscanbeordered.Thatis,ifaandbaredistinctelements,wecansaythataislessthanborbis lessthana.Thissecondaspect—theorderingofelements—willnowbeformalized. AnorderedintegraldomainisanintegraldomainAwitharelation,symbolizedby<,havingthe followingproperties: 1. ForanyaandbinA,exactlyoneofthefollowingistrue: a=b a<b or b<a Furthermore,foranya,b,andcinA, 2. Ifa<bandb<c,thena<c. 3. Ifa<b,thena+c<b+c. 4. Ifa<b,thenac<bcontheconditionthat0<c. Therelation<iscalledanorderrelationonA.Thefourconditionswhichanorderrelationmustfulfill arefamiliartoeveryone.Properties1and2requirenocomment.Property3assertsthatweareallowed toaddanygivenctobothsidesofaninequality.Property4assertsthatwemaymultiplybothsidesofan inequalitybyanyc,ontheconditionthatcisgreaterthanzero. Asusual,a>bhasthesamemeaningasb<a.Furthermore,a≤bmeans“a<bora=b,”andb≥a meansthesameasa≤b. InanorderedintegraldomainA,anelementaiscalledpositiveifa>0.Ifa<0wecallanegative. Note that if a is positive then −a is negative. (Proof: Add −a to both sides of the inequality a > 0.) Similarly,ifaisnegative,then−aispositive. In any ordered integral domain, the square of every nonzero element is positive. Indeed, if c is nonzero,theneitherc>0ofc<0.Ifc>0,then,multiplyingbothsidesoftheinequalityc>0byc, cc>c0=0 soc2>0.Ontheotherhand,ifc<0,then (−c)>0 hence (−c)(−c)>0(−c)=0 But(−c)(−c)=c2,soonceagain,c2>0. Inparticular,since1=l2,1isalwayspositive. From the fact that 1 >0, we immediately deduce that 1 + 1 > 1, 1 + 1 + 1 > 1 + 1, and so on. In general,foranypositiveintegern, (n+1)·1>n·1 wheren·1designatestheunityelementoftheringAaddedtoitselfntimes.Thus,inanyordered integraldomainA,thesetofallthemultiplesof1isorderedasin :namely ⋯<(−2)·1<(−1)·1<0<1<2·1<3·1<⋯ ThesetofallthepositiveelementsofAisdenotedbyA+.AnorderedintegraldomainAiscalledan integralsystemifeverynonemptysubsetofA+hasaleastelement.Inotherwords,ifeverynonemptyset ofpositiveelementsofAhasaleastelement.Thispropertyiscalledthewell-orderingpropertyforA+. Itisobviousthat isanintegralsystem,foreverynonemptysetofpositiveintegerscontainsaleast number.Forexample,thesmallestelementofthesetofallthepositiveevenintegersis2.Notethat and arenotintegralsystems.Foralthoughbothareorderedintegraldomains,theycontainsetsofpositive numbers,suchas{x:0<x<l},whichhavenoleastelement. Inanyintegralsystem,thereisnoelementbetween0and1.ForsupposeAisanintegralsystemin whichthereareelementsχbetween0and1.Thentheset{x ∊A:0<x<1}isanonemptysetofpositive membersofA,sobythewell-orderingpropertyithasaleastelementc.Thatis, 0<c<1 andcistheleastelementofAwiththisproperty.Butthen(multiplyingbyc), 0<c2<c Thus,c2isbetween0and1andislessthanc,whichisimpossible. Thus,thereisnoelementofAbetween0and1. Finally,inanyintegralsystem,everyelementisamultipleof1.Ifthatwerenotthecase,wecould usethewell-orderingprincipletopicktheleastpositiveelementofAwhichisnotamultipleof1:callit b.Now,b>0andtherearenoelementsofAbetween0and1,sob>l.(Rememberthatbcannotbeequalto 1becausebisnotamultipleof1.)Sinceb>1,itfollowsthatb−1>0.Butb−1<bandbistheleast positiveelementwhichisnotamultipleof1,sob−1isamultipleof1.Say b−1=n·1 Butthenb=n·1+1=(n+1)·1,whichisimpossible. Thus,inanyintegralsystem,alltheelementsaremultiplesof1andtheseareorderedexactlyasin . Itisnowamereformalitytoprovethateveryintegralsystemisisomorphicto ThisisleftasExercise Dattheendofthischapter. Since every integral system is isomorphic to , any two integral systems are isomorphic to each other. Thus is, up to isomorphism, the only integral system. We have therefore succeeded in giving a completeaxiomaticcharacterizationof Henceforwardweconsider tobedefinedbythefactthatitisanintegralsystem. Thetheoremwhichfollowsisthebasisofproofsbymathematicalinduction.Itisintuitivelyclear andeasytoprove. Theorem1LetKrepresentasetofpositiveintegers.Considerthefollowingtwoconditions’. (i)1isinK. (ii)Foranypositiveintegerk,ifk∈K,thenalsok+1∈K. IfKisanysetofpositiveintegerssatisfyingthesetwoconditions,thenKconsistsofallthepositive integers. PROOF:Indeed,ifKdoesnotcontainallthepositiveintegers,thenbythewell-orderingprinciple, thesetofallthepositiveintegerswhicharenotinKhasaleastelement.Callitb;bistheleastpositive integernotinK.ByCondition(i),b≠1,henceb>1. Thus,b−1>0,andb−1∈K.Butthen,byCondition(ii),b∈K,whichisimpossible.■ LetthesymbolSnrepresentanystatementaboutthepositiveintegern.Forexample,Snmightstand for“nisodd,”or“nisaprime,”oritmightrepresentanequationsuchas(n −1)(n+1)=n2 −1oran inequalitysuchasn≤n2.If,letussay,Snstandsforn≤n2,thenS1assertsthat1≤12,S2assertsthat2≤ 22,S3assertsthat3≤32,andsoon. Theorem2:PrincipleofmathematicalinductionConsiderthefollowingconditions: (i) S1istrue. (ii) Foranypositiveintegerk,ifSk istrue,thenalsoSk+1istrue. IfConditions(i)and(ii)aresatisfied,thenSnistrueforeverypositiveintegern. PROOF:Indeed,ifKisthesetofallthepositiveintegersksuchthatSkistrue,thenKcomplieswith the conditions of Theorem 1. Thus, K contains all the positive integers. This means that Sn is true for everyn.■ As a simple illustration of how the principle of mathematical induetion is applied, let Sn be the statementthat thatis,thesumofthefirstnpositiveintegersisequalton(n+1)/2.ThenS1issimply whichisclearlytrue.Suppose,next,thatkisanypositiveintegerandthatSkistrue.Inotherwords, Then,byaddingk+1tobothsidesofthisequation,weobtain thatis, However,thislastequationisexactlySk+1.WehavethereforeverifiedthatwheneverSkistrue,Sk+1also istrue.Now,theprincipleofmathematicalinductionallowsustoconcludethat foreverypositiveintegern. Avariantoftheprincipleofmathematicalinduction,calledtheprincipleofstronginduction,asserts thatSnistrueforeverypositiveintegernontheconditionsthat (i) S1istrue,and (ii) Foranypositiveintegerk,ifSiistrueforeveryi<k,thenSkistrue. ThedetailsareoutlinedinExerciseHattheendofthischapter. One of the most important facts about the integers is that any integer m may be divided by any positiveintegerntoyieldaquotientqandapositiveremainderr.(Theremainderislessthanthedivisor n.)Forexample,25maybedividedby8togiveaquotientof3andaremainderof1: Thisprocessisknownasthedivisionalgorithm.Itisstatedinaprecisemannerasfollows: Theorem 3: Division algorithm If m and n are integers and n is positive, there exist unique integersqandrsuchthat m=nq+r and 0≤r<n Wecallqthequotient,andrtheremainder,inthedivisionofmbyn. PROOF:Webeginbyshowingasimplefact: Thereexistsanintegerxsuchthatxn≤m. (*) Rememberthatnispositive;hencen≥1.Asform,eitherm≥0orm<0.Weconsiderthesetwocases separately: Supposem≥0.Then Supposem<0.Wemaymultiplybothsidesofn≥1bythepositiveinteger −mtoget(−m)n≥ −m. Addingmn+mtobothsidesyields Thus,regardlessofwhethermispositiveornegative,thereissomeintegerxsuchthatxn≤m. LetWbethesubsetof consistingofallthenonnegativeintegerswhichareexpressibleintheform m−xn,wherexisanyinteger.By(*)Wisnotempty;hencebythewell-orderingproperty,Wcontainsa leastintegerr.Becauser∈W,risnonnegativeandisexpressibleintheformm−nqforsomeintegerq. Thatis, r≥0 and r=m−nq Thus,wealreadyhavem=nq+rand0≤r.Itremainsonlytoverifythatr<n.Supposenot:suppose n≤r,thatis,r−n≥0.But r−n=(m−nq)−n=m−n(q+1) andclearlyr −n<r.Thismeansthatm −n(q+1)isanelementofWlessthanr,whichisimpossible becauseristheleastelementofW.Weconcludethatn≤risimpossible;hencer<n. Theverificationthatqandrareuniqueisleftasanexercise.■ EXERCISES A.PropertiesofOrderRelationsinIntegralDomains LetAbeanorderedintegraldomain.Provethefollowing,foralla,b,andcinA: 1 2 3 4 5 6 7 8 Ifa≤bandb≤c,thena≤c. Ifa≤b,thena+c≤b+c. Ifa≤bandc≥0,thenac≤bc. Ifa<bandc<0,thenbc<ac. a<biff−b<−a. Ifa+c<b+c,thena<b. Ifac<bcandc>0,thena<b. Ifa<bandc<d,thena+c<b+d. B.FurtherPropertiesofOrderedIntegralDomains LetAbeanorderedintegraldomain.Provethefollowing,foralla,b,andcinA: 1 2 3 4 5 6 a2+b2≥2ab a2+b2≥abanda2+b2≥−ab a2+b2+c2≥ab+bc+ac a2+b2>,ifa2+b2≠0 a+b<ab+1,ifa,b>1 ab+ac+bc+1<a+b+c+abc,ifa,b,c>1 C.UsesofInduction Proveparts1−7,usingtheprincipleofmathematicalinduction.(Assumenisapositiveinteger.) 1 1+3+5+⋯+(2n−1)=n2(Thesumofthefirstnoddintegersisn2.) 2 13+23+⋯+n3=(1+2+⋯+n)2 3 12+22+⋯+(n−1)2< <12+22+⋯+n2 4 13+23+⋯+(n−1)3< <13+23+⋯+n3 5 12+22+⋯+n2= n(n+1)(2n+1) 6 13+23+⋯+n3= n2(n+1)2 7 8TheFibonaccisequenceisthesequenceofintegersF1,F2, F3,…defined as follows: F1 = 1; F2 = 1; Fn+2=Fn+1+Fnforallpositiveintegersn.(Thatis,everynumber,afterthesecondone,isthesumof thetwoprecedingones.)Useinductiontoprovethatforalln>0, Fn+1Fn+2−FnFn+3=(−l)n D.EveryIntegralSystemIsIsomorphicto Let A be an integral system. Let h: →A be defined by: h(n) = n · 1. The purpose of this exercise is to provethathisanisomorphism,fromwhichitfollowsthatA≅ 1 Prove:Foreverypositiveintegern,n·1>0.WhatisthecharacteristicofA 2 Provethathisinjectiveandsurjective. 3 Provethathisanisomorphism. E.AbsoluteValues Inanyorderedintegraldomain,define|a|by Usingthisdefinition,provethefollowing: 1 2 3 4 5 6 7 8 9 |−a|=|a| a≤|a| a≤−|a| Ifb>0,|a|≤biff−b≤a≤b |a+b|≤|a|+|b| |a−b|≤|a|+|b| |ab|=|a|·|b| |a|−|b|≤|a−b| ||a|−|b||≤|a−b| F.ProblemsontheDivisionAlgorithm Proveparts1−3,wherek,m,n,q,andrdesignateintegers. 1 Letn>0andk>0.Ifqisthequotientandristheremainderwhenmisdividedbyn,thenqisthe quotientandkristheremainderwhenkmisdividedbykn. # 2 Let n > 0 and k > 0. If q is the quotient when m is divided by n, and q1 is the quotient when q is dividedbyk,thenq1isthequotientwhenmisdividedbynk. 3Ifn≠0,thereexistqandrsuchthatm=nq+rand0≤r<|n|.(UseTheorem3,andconsiderthecase whenn<0.) 4 InTheorem3,supposem=nq1+r1=nq2+r2where0≤r1,r2 < n. Prove that r1 − r2 = 0. [HINT: Considerthedifference(nq1+r1−(nq2+r2).] 5 Use part 4 to prove that q1 − q2 = 0. Conclude that the quotient and remainder, in the division algorithm,areunique. 6 Ifristheremainderwhenmisdividedbyn,provethatm=rin n. G.LawsofMultiples Thepurposeofthisexerciseistogiverigorousproofs(usinginduction)ofthebasicidentitiesinvolvedin the use of exponents or multiples. If A is a ring and a ∈ A, we define n · a (where n is any positive integer)bythepairofconditions: (i)1·a=a, and (ii)(n+1)·a=n·a+a Usemathematicalinduction(withtheabovedefinition)toprovethatthefollowingaretrueforallpositive integersnandallelementsa,b∈A: 1 2 3 4 5 6 n·(a+b)=n·a+n·b (n+m)·a=n·a+m·a (n·a)b=a(n·b)=n·(ab) m·(n·a)=(mn)·a n·a=(n·1)a where1istheunityelementofA (n·a)(m·b)=(nm)·ab (Useparts3and4.) H.PrincipleofStrongInduction Provethefollowingin : 1 LetKdenoteasetofpositiveintegers.Considerthefollowingconditions: (i)I∈K. (ii)Foranypositiveintegerk,ifeverypositiveintegerlessthankisinK,thenk∈K. IfKsatisfiesthesetwoconditions,provethatKcontainsallthepositiveintegers. 2 LetSnrepresentanystatementaboutthepositiveintegern.Considerthefollowingconditions: (i)S1istrue. (ii)Foranypositiveintegerk,ifSiistrueforeveryi<k,Skistrue. IfConditions(i)and(ii)aresatisfied,provethatSnistrueforeverypositiveintegern. CHAPTER TWENTY-TWO FACTORINGINTOPRIMES It has been said that the two events most decisive in shaping the course of our development were the invention of the wheel and the discovery of numbers. From the time—ages ago—when humans first learnedtheuseofnumbersforcounting,theyhavebeenasourceofendlessfascination.Alchemistsand astrologersextolledthevirtuesof“mystic”numbersandfoundinthemthekeytopotentmagic.Others, more down to earth, found delight in observing the many regularities and unexpected properties of numbers.Theintegershavebeenaseeminglyinexhaustiblesourceofproblemsgreatandsmallonwhich mathematicshasfedandcontinuestodrawinourday. Thepropertiesofprimenumbersalonehaveoccupiedtheenergiesofmathematiciansfromthetime ofEuclid.Newquestionsrelatingtotheprimescontinuetoappear,andmanycontinuetoresistthebest efforts to solve them. Most importantly, a large part of number theory starts out from a few basic facts aboutprimenumbers.Theywillbeoutlinedinthischapter. Modernnumbertheoryistheoldestaswellasoneofthenewestpartsofmathematics.Itrestsupon somebasicdataregardingthestructureofthedomain oftheintegers.Anunderstandingofthisstructure isafundamentalpartofanymathematicaleducation. Animportantfeatureofanyringisthestructureofitsideals.Wethereforebeginbyasking:Whatare theidealsof ?Wehavealreadymadeuseofprincipalidealsof ,suchastheideal 〈6〉={…,−18,−12,−6,0,6,12,18,…} which consist of all the multiples of one fixed integer. It is natural to inquire whether has any ideals whicharenotprincipal,andwhattheymightlooklike.Theanswerisfarfromself-evident. Theorem1Everyidealof isprincipal. PROOF:LetJbeanyidealof .If0istheonlyintegerinJ,thenJ=〈0〉,theprincipalidealgenerated by0.IftherearenonzerointegersinJ,thenforeachxinJ,−xisalsoinJ;thustherearepositiveintegers inJ.Bythewell-orderingpropertywemaypicktheleastpositiveintegerinJ,andcallitn. WewillprovethatJ=〈n〉,whichistosaythateveryelementofJissomemultipleofn.Well,letm beanyelementofJ.Bythedivisionalgorithmwemaywritem=nq+rwhere0≤r<n.Nowm was choseninJ,andn∈J,hencenqisinJ.Thus, r=m−nq∈J Rememberthatriseither0orelsepositiveandlessthann.Thesecondcaseisimpossible,becausenis theleastpositiveintegerinJ.Thus,r=0,andthereforem=nq,whichisamultipleofn. WehaveprovedthateveryelementofJisamultipleofn,whichistosaythatJ=〈n〉.■ Itisusefultonotethatbytheprecedingproof,anyidealJisgeneratedbytheleastpositiveinteger inJ. Ifrandsareintegers,wesaythatsisamultipleofrifthereisanintegerksuchthat s=rk Ifthisisthecase,wealsosaythatrisafactorofs,orrdividess,andwesymbolizethisbywriting r|s Notethat1and−1divideeveryinteger.Ontheotherhand,anintegerrdivides1iffrisinvertible. In thereareveryfewinvertibleelements.Asamatteroffact, Theorem2Theonlyinvertibleelementsof are1and−1. PROOF:Ifsisinvertible,thismeansthereisanintegerrsuchthat rs=1 Clearlyr≠0ands≠0(otherwisetheirproductwouldbe0).Furthermore,randsareeitherbothpositive orbothnegative(otherwisetheirproductwouldbenegative). Ifrandsarebothpositive,thenr=1orr>1.Incaser>1wemaymultiplybothsidesof1<rbys togets<rs=1;thisisimpossiblebecausescannotbepositiveand<1.Thus,itmustbethatr = 1; hence1=rs=1s=s,soalsos=1. Ifrandsarebothnegative,then−rand−sarepositive.Thus, 1=rs=(−r)(−s) andbytheprecedingcase,−r=−s=1.Thus,r=s=−1.■ Apairofintegersrandsarecalledassociatesiftheydivideeachother,thatis,ifr|sands|r.Ifrand sareassociates,thismeansthereareintegerskandlsuchthatr=ksands=lr.Thus,r=ks=klr,hence kl=1.ByTheorem2,kandlare±1,andthereforer=±s.Thus,wehaveshownthat Ifrandsareassociatesin ,thenr=±s. (1) Anintegertiscalledacommondivisorofintegersrandsift|randt|s.Agreatestcommondivisor ofrandsisanintegertsuchthat (i) t|randt|sand (ii) Foranyintegeru,ifu|randthenu|t. Inotherwords,tisagreatestcommondivisorofrandsiftisacommondivisorofrands,andevery othercommondivisorofrandsdividest.Notethattheadjective“greatest”inthisdefinitiondoesnot mean primarily that t is greater in magnitude than any other common divisor, but, rather, that it is a multipleofanyothercommondivisor. Thewords“greatestcommondivisor”arefamiliarlyabbreviatedbygcd.Asanexample,2isagcd of8and10;but−2alsoisagcdof8and10.Accordingtothedefinition,twodifferentgcd’smustdivide eachother;hencebyProperty(1)above,theydifferonlyinsign.Ofthetwopossiblegcd’s±tforrands, weselectthepositiveone,callitthegcdofrands,anddenoteitby gcd(r,s) Doeseverypairr,sofintegershaveagcd?Ourexperiencewiththeintegerstellsusthattheanswer is“yes.”Wecaneasilyprovethis,andmore: Theorem3Anytwononzerointegersrandshaveagreatestcommondivisort.Furthermore,tis equaltoa“linearcombination”ofrands.Thatis, t=kr+ls forsomeintegerskandl. PROOF:LetJbethesetofallthelinearcombinationsofrands,thatis,thesetofallur+υsasuand υrangeover .Jisclosedwithrespecttoadditionandnegativesandabsorbsproductsbecause and w(ur+υs)=(wu)r+(wυ)s Thus, J is an ideal of . By Theorem 1, J is a principal ideal of , say J = 〈t〉. (J consists of all the multiplesoft.) NowtisinJ,whichmeansthattisalinearcombinationofrands: t=kr+ls Furthermore,r=1r+0sands=0r+1s,sorandsarelinearcombinationsofrands;thusrandsarein J.ButalltheelementsofJaremultiplesoft,sorandsaremultiplesoft.Thatis, t|r and t|s Now,ifuisanycommondivisorofrands,thismeansthatr=xuands=yuforsomeintegersxandy. Thus, t=kr+ls=kxu+lyu=u(kx+ly) Itfollowsthatu|t.Thisconfirmsthattisthegcdofrands.■ Awordofwarning:thefactthatanintegermisalinearcombinationofrandsdoesnotnecessarily implythatmisthegcdofrands.Forexample,3=(1)15+(−2)6,and3isthegcdof15and6.Onthe otherhand,27=(1)15+(2)6,yet27isnotagcdof15and6. Apairofintegersrandsaresaidtoberelativelyprimeiftheyhavenocommondivisorsexcept±1. Forexample,4and15arerelativelyprime.Ifrandsarerelativelyprime,theirgcdisequalto1;soby Theorem3,thereareintegerskandlsuchthatkr+ls=1.Actually,theconverseofthisstatementistrue too:ifsomelinearcombinationofrandsisequalto1(thatis,ifthereareintegerskandlsuchthatkr+ ls=1),thenrandsarerelativelyprime.Thesimpleproofofthisfactisleftasanexercise. Ifmisanyinteger,itisobviousthat±1and±marefactorsofm.Wecallthesethetrivialfactorsof m.Ifmhasanyotherfactors,wecallthemproperfactorsofm.Forexample,±1and±6arethetrivial factorsof6,whereas±2and±3areproperfactorsof6. Ifanintegermhasproperfactors,miscalledcomposite.Ifanintegerp≠0,1hasnoproperfactors (thatis,ifallitsfactorsaretrivial),thenwecallpaprime.Forexample,6iscomposite,whereas7isa prime. CompositenumberlemmaIfapositiveintegermiscomposite,thenm=rswhere 1<r<m and 1<s<m PROOF:Ifmiscomposite,thismeansthatm=rsforintegersrandswhicharenotequaleitherto1 ortom.Wemaytakerandstobepositive;hence1<rand1<s.Multiplyingbothsidesof1<rbys givess<rs=m.Analogously,wegetr<m.■ Whathappenswhenacompositenumberisdividedbyaprime?Thenextlemmaprovidesananswer tothatquestion. Euclid’slemmaLetmandnbeintegers,andletpbeaprime. Ifp|(mn),theneitherp|morp|n. PROOF:Ifp|mwearedone.Soletusassumethatp does not divide m. What integers are common divisorsofpandm? Well,theonlydivisorsofpare±1and±p.Sinceweassumethatpdoesnotdividem,pand −pare ruledoutascommondivisorsofpandm,hencetheironlycommondivisorsare1and−1. Itfollowsthatgcd(p,m)=1,sobyTheorem3, kp+lm=1 forsomeintegercoefficientskandl.Thus, kpn+lmn=n Butp|(mn),sothereisanintegerhsuchthatmn=ph.Therefore, kpn+lph=n thatis,p(kn+lh)=n.Thus,p|n.▪ Corollary1Letm1,…,mt,beintegers,andletpbeaprime.Ifp|(m1⋯mt),thenp|miforoneofthe factorsmiamongm1,…,mt. Wemaywritetheproductm1⋯mtasm1(m2 ⋯mt),sobyEuclid’slemma,p|m1orp|m2 ⋯mt.Inthe firstcasewearedone,andinthesecondcasewemayapplyEuclid’slemmaonceagain,andrepeatthis uptottimes. Corollary 2 Let q1, …, qt and p be positive primes. If p|(q1 ⋯ qt), then p is equal to one of the factorsq1,…,qt. PROOF:ByCorollary1,pdividesoneofthefactorsq1,…,qt,sayp|qiButqiisaprime,soitsonly divisorsare±1and±qi;pispositiveandnotequalto1,soifp|qi,necessarilyp=qi.■ Theorem 4: Factorization into primes Every integer n > 1 can be expressed as a product of positiveprimes.Thatis,thereareoneormoreprimesp1,…,prsuchthat n=p1p2⋯pr PROOF:LetKrepresentthesetofallthepositiveintegersgreaterthan1whichcannotbewrittenasa productofoneormoreprimes.Wewillassumetherearesuchintegers,andderiveacontradiction. Bythewell-orderingprinciple,Kcontainsaleastintegerm;mcannotbeaprime,becauseifitwere aprimeitwouldnotbeinK.Thus,miscomposite;sobythecompositenumberlemma, m=rs forpositiveintegersrandslessthanmandgreaterthan1;randsarenotinKbecausem is the least integerinK.Thismeansthatrandscanbeexpressedasproductsofprimes,hencesocanm=rs. This contradictionprovesthatKisempty;henceeveryn>1canbeexpressedasaproductofprimes.■ Theorem 5: Unique factorization Suppose n can be factored into positive primes in two ways, namely, n=p1⋯pr=q1⋯qt Then r = t, and the pi are the same numbers as the qj except, possibly, for the order in which they appear. PROOF:Intheequationp1 ⋯pr=q1 ⋯qt,letuscancelcommonfactorsfromeachside,onebyone, untilwecandonomorecanceling.Ifallthefactorsarecanceledonbothsides,thisprovesthetheorem. Otherwise,weareleftwithsomefactorsoneachside,say pi⋯pk =qj ⋯qm Now,piisafactorofpi⋯pk ,sopi|qj ⋯qm.Thus,byCorollary2toEuclid’slemma,p1isequaltooneof thefactorsqj ,…,qm,whichisimpossiblebecauseweassumedwecandonomorecanceling.■ ItfollowsfromTheorems4and5thateveryintegermcanbefactoredintoprimes,andthattheprime factorsofmareunique(exceptfortheorderinwhichwehappentolistthem). EXERCISES A.PropertiesoftheRelation“aDividesa” Provethefollowing,foranyintegersa,b,andc: 1Ifa|bandb|c,thena|c. 2a|biffa|(−b)iff(−a)|b. 31|aand(−1)|a. 4a|0. 5Ifc|aandc|b,thenc|(ax+by)forallx,y∈ . 6Ifa>0andb>0anda|b,thena≤b. 7a|biffac|bc,whenc≠0. 8Ifa|bandc|d,thenac|bd. 9Letpbeaprime.Ifp|anforsomen>0,thenp|a. B.Propertiesofthegcd Prove the following, for any integers a, b, and c. For each of these problems, you will need only the definitionofthegcd. #1Ifa>0anda|b,thengcd(a,b)=a. 2gcd(a,0)=a,ifa>0. 3gcd(a,b)=gcd(a,b+xa)foranyx∈ . 4Letpbeaprime.Thengcd(a,p)=1orp.(Explain.) 5Supposeeverycommondivisorofaandbisacommondivisorofcandd,andviceversa.Thengcd(a, b)=gcd(c,d). 6Ifgcd(ab,c)=1,thengcd(a,c)=1andgcd(b,c)=1. 7Letgcd(a,b)=c.Writea=ca′andb=cb′.Thengcd(a′,b′)=1. C.PropertiesofRelativelyPrimeIntegers Provethefollowing,forallintegersa,b,c,d,r,ands.(Theorem3willbehelpful.) 1Ifthereareintegersrandssuchthatra+sb=1,thenaandbarerelativelyprime. 2Ifgcd(a,c)=1andc|ab,thenc|b.(ReasonasintheproofofEuclid’slemma.) 3Ifa|dandc|dandgcd(a,c)=1,thenac|d. 4Ifd|abandd|cb,wheregcd(a,c)=1,thend|b. 5Ifd=gcd(a,b)wherea=drandb=ds,thengcd(r,s)=1. 6Ifgcd(a,c)=1andgcd(b,c)=1,thengcd(ab,c)=1. D.FurtherPropertiesofgcd’sandRelativelyPrimeIntegers Provethefollowing,forallintegersa,b,c,d,r,ands: 1Supposea|bandc|bandgcd(a,c)=d.Thenac|bd. 2Ifac|bandad|bandgcd(c,d)=1,thenacd|b. #3Letd=gcd(a,b).Foranyintegerx,d|xiffxisalinearcombinationofaandb. 4Supposethatforallintegersx,x|aandx|biffx|c.Thenc=gcd(a,b). 5Foralln>0,ifgcd(a,b)=1,thengcd(a,bn)=1.(Provebyinduction.) 6Supposegcd(a,b)=1andc|ab.Thenthereexistintegersrandssuchthatc=rs,r|a,s|b,andgcd(r,s)= 1. E.APropertyofthegcd Letaandbbeintegers. Proveparts1and2: #1Supposeaisoddandbiseven,orviceversa.Thengcd(a,b)=gcd(a+b,a−b). 2Supposeaandbarebothodd.Then2gcd(a,b)=gcd(a+b,a−b). 3Ifaandbarebotheven,explainwhyeitherofthetwopreviousconclusionsarepossible. F.LeastCommonMultiples Aleastcommonmultipleoftwointegersaandbisapositiveintegercsuchthat(i)a|candb|c;(ii)ifa|x andb|x,thenc|x. 1Prove:Thesetofallthecommonmultiplesofaandbisanidealof 2Prove:Everypairofintegersaandbhasaleastcommonmultiple.(HINT:Usepart1.) Thepositiveleastcommonmultipleofaandbisdenotedbylcm(a,b).Provethefollowingforall positiveintegersa,b,andc: #3a·lcm(b,c)=lcm(ab,ac). 4Ifa=a1candb=b1cwherec=gcd(a,b),thenlcm(a,b)=a1b1c. 5lcm(a,ab)=ab. 6Ifgcd(a,b)=1,thenlcm(a,b)=ab. 7Iflcm(a,b)=ab,thengcd(a,b)=1. 8Letgcd(a,b)=c.Thenlcm(a,b)=ab/c. 9Letgcd(a,b)=candlcm(a,b)=d.Thencd=ab. G.Idealsin Provethefollowing: 1〈n〉isaprimeidealiffnisaprimenumber. 2Everyprimeidealof isamaximalideal.[HINT:If〈p〉⊆〈a〉,but〈p〉≠〈a〉,explainwhygcd(p,a)=1 andconcludethat1∈〈a〉.]Assumetheidealisnot{0}. 3Foreveryprimenumberp, pisafield.(HINT:Remember p= /〈p〉.UseExercise4,Chapter19.) 4Ifc=lcm(a,b),then〈a〉∩〈b〉=〈c〉. 5Everyhomomorphicimageof isisomorphicto nforsomen. 6LetGbeagroupandleta,b∈G.ThenS={n∈ :abn=bna}isanidealof . 7LetGbeagroup,HasubgroupofG,anda∈G.Then S={n∈ :an∈H} isanidealof . 8Ifgcd(a,b)=d,then〈a〉+〈b〉=〈d〉.(NOTE:IfJandKareidealsofaringA,thenJ+K={x+y:x∈J andy∈K}.) H.Thegcdandthe1cmasOperationson Foranytwointegersaandb,leta*b=gcd(a,b)anda∘b?=lcm(a,b).Provethefollowingproperties oftheseoperations: 1*and∘areassociative. 2Thereisanidentityelementforº,butnotfor*(onthesetofpositiveintegers). 3Whichintegershaveinverseswithrespecttoº? 4Prove:a*(b∘c)=(a*b)∘(a*c). CHAPTER TWENTY-THREE ELEMENTSOFNUMBERTHEORY(OPTIONAL) Almost as soon as children are able to count, they learn to distinguish between even numbers and odd numbers.Thedistinctionbetweenevenandoddisthemostelementalofallconceptsrelatingtonumbers. Itisalsothestartingpointofthemodernscienceofnumbertheory. Fromasophisticatedstandpoint,anumberiseveniftheremainder,afterdividingthenumberby2,is 0.Thenumberisoddifthatremainderis1. Thisnotionmaybegeneralizedinanobviousway.Letnbeanypositiveinteger:anumberissaidto becongruentto0,moduloniftheremainder,whenthenumberisdividedbyft,is0.Thenumberissaid to be congruent to 1, modulo n if the remainder, when the number is divided by is 1. Similarly, the numberiscongruentto2,moduloniftheremainderafterdivisionbyftis2;andsoon.Thisisthenatural wayofgeneralizingthedistinctionbetweenoddandeven. Notethat“even”isthesameas“congruentto0,modulo2”;and“odd”isthesameas“congruentto 1,modulo2.” Inshort,thedistinctionbetweenoddandevenisonlyonespecialcaseofamoregeneralnotion.We shallnowdefinethisnotionformally: Letnbeanypositiveinteger.Ifaandbareanytwointegers,weshallsaythataiscongruenttob, modulo n if a and b, when they are divided by n, leave the same remainder r. That is, if we use the divisionalgorithmtodivideaandbbyn,then a=nq1+r b=nq2+r and wheretheremainderristhesameinbothequations. Subtractingthesetwoequations,weseethat a−b=(nq1+r)−(nq2+r)=n(q1−q2) Thereforewegetthefollowingimportantfact: aiscongruenttob,modulon iff ndividesa−b (1) Ifaiscongruenttob,modulon,weexpressthisfactinsymbolsbywriting a≡b(modn) whichshouldberead“aiscongruenttob,modulon.”Werefertothisrelationascongruencemodulon. ByusingCondition(1),itiseasilyverifiedthatcongruencemodulonisareflexive,symmetric,and transitiverelationon .Itisalsoeasytocheckthatforanyn>0andanyintegersa,b,andc, a≡b(modn) implies a+c≡b+c(modn) and a≡b(modn) implies ac≡bc(modn) (Theproofs,whichareexceedinglyeasy,areassignedasExerciseCattheendofthischapter.) Recallthat 〈n〉={…,−3n,−2n,−n,0,n,2n,3n,…} istheidealof whichconsistsofallthemultiplesofn.Thequotientring /〈n〉isusuallydenotedby n, anditselementsaredenotedby .Theseelementsarecosets: andsoon.Itisclearbyinspectionthatdifferentintegersareinthesamecosetifftheydifferfromeach otherbyamultipleofn.Thatis, Ifaisanyinteger,thecoset(in n)whichcontainsawillbedenotedby .Forexample,in 6, Inparticular, meansthataandbareinthesamecoset.ItfollowsbyCondition(2)that On account of this fundamental connection between congruence modulo n and equality in n, most facts about congruence can be discovered by examining the rings n. These rings have very simple properties,whicharepresentednext.Fromthesepropertieswewillthenbeabletodeduceallweneedto knowaboutcongruences. Letnbeapositiveinteger.Itisanimportantfactthatforanyintegera, Indeed,ifaandnarerelativelyprime,theirgcdisequalto1.Therefore,byTheorem3 of Chapter 22, thereareintegerssandtsuchthatsa+tn=1.Itfollowsthat 1−sa=tn∈〈n〉 sobyCondition(2)ofChapter19,1andsabelongtothesamecosetin /〈n〉.Thisisthesameassaying that ;hence isthemultiplicativeinverseof in n.Theconverseisprovedbyreversingthe stepsofthisargument. It follows from Condition (4) above, that if n is a prime number, every nonzero element of n is invertible!Thus, p isafieldforeveryprimenumberp. (5) In any field, the set of all the nonzero elements, with multiplication as the only operation (ignore addition),isagroup.Indeed,theproductofanytwononzeroelementsisnonzero,andthemultiplicative inverseofanynonzeroelementisnonzero.Thus,in p,theset withmultiplicationasitsonlyoperation,isagroupoforderp−1. RememberthatifGisagroupwhoseorderis,letussay,m,thenxm=eforeveryxinG. (This is truebyTheorem5ofChapter13.)Now, hasorderp −1anditsidentityelementis ,so for every . If we use Condition (3) to translate this equality into a congruence, we get a classicalresultofnumbertheory: LittletheoremofFermatLetpbeaprime.Then, ap−1≡1(modp) forevery a 0(modp) Corollaryap≡a(modp)foreveryintegera. Actually,aversionofthistheoremistruein nevenwherenisnotaprimenumber.Inthiscase,let Vndenotethesetofalltheinvertibleelementsin n.Clearly,Vnisagroupwithrespecttomultiplication. (Reason:Theproductoftwoinvertibleelementsisinvertible,and,ifaisinvertible,soisitsinverse.) For any positive integer n, let ϕ(n) denote the number of positive integers, less than n, which are relativelyprimeton.Forexample,1,3,5,and7arerelativelyprimeto8;henceϕ(8)=4.ϕ is called Eulef ’sphi−function. ItfollowsimmediatelyfromCondition(4)thatthenumberofelementsinVnisϕ(n). Thus, Vn is a group of order ϕ(n), and its identity element is . Consequently, for any in .IfweuseCondition(3)totranslatethisequationintoacongruence,weget: Euler’stheoremIfaandnarerelativelyprime,αϕ(n)≡1(modn). FURTHERTOPICSINNUMBERTHEORY Congruencesaremoreimportantinnumbertheorythanwemightexpect.Thisisbecauseavastrangeof problemsinnumbertheory—problemswhichhavenothingtodowithcongruencesatfirstsight—canbe transformedintoproblemsinvolvingcongruences,andaremosteasilysolvedinthatform.Anexampleis givennext: ADiophantineequationisanypolynomialequation(inoneormoreunknowns)whosecoefficients areintegers.TosolveaDiophantineequationistofindintegervaluesoftheunknownswhichsatisfythe equation. We might be inclined to think that the restriction to integer values makes it easier to solve equations;infact,theveryoppositeistrue.Forinstance,eveninthecaseofanequationassimpleas4x+ 2y=5,itisnotobviouswhetherwecanfindanintegersolutionconsistingofxandyin .(Asamatterof fact,thereisnointegersolution;trytoexplainwhynot.) SolvingDiophantineequationsisoneoftheoldestandmostimportantproblemsinnumbertheory. Even the problem of solving Diophantine linear equations is difficult and has many applications. Therefore,itisaveryimportantfactthatsolvinglinearDiophantineequationsisequivalenttosolving linearcongruences.Indeed, ax+by=c iff by=c−ax iff ax≡c(modb) Thus,anysolutionofax≡c(modb)yieldsasolutioninintegersofax+by=c. Finding solutions of linear congruences is therefore an important matter, and we will turn our attentiontoitnow. A congruence such as ax ≡ b (mod n) may look very easy to solve, but appearances can be deceptive.Infact,manysuchcongruenceshavenosolutionsatall!Forexample,4x≡5(mod2)cannot have a solution, because 4x is always even [hence, congruent to 0 (mod 2)], whereas 5 is odd [hence congruentto1(mod2)].Ourfirstitemofbusiness,therefore,istofindawayofrecognizingwhetheror notalinearcongruencehasasolution. Theorem1Thecongruenceax≡b(modn)hasasolutioniffgcd(a,n)|b. Indeed, Next,bytheproofofTheorem3inChapter22,ifJistheidealofallthelinearcombinationsofaandn, thengcd(a,n)istheleastpositiveintegerinJ.Furthermore,everyintegerinJisamultipleofgcd(a,n). Thus,bisalinearcombinationofaandniffb∈Jiffbisamultipleofgcd(a, n). This completes the proofofourtheorem. Nowthatweareabletorecognizewhenacongruencehasasolution,letusseewhatsuchasolution lookslike. Consider the congruence ax ≡ b (mod n). By a solution modulo n of this congruence, we mean a congruence x≡c(modn) suchthatanyintegerxsatisfiesx≡c(modn)iffitsatisfiesax≡b(modn).[Thatis,thesolutionsofax≡ b (mod n) are all the integers congruent to c, modulo n.] Does every congruence ax ≡ b (mod n) (supposingthatithasasolution)haveasolutionmodulon?Unfortunatelynot!Nevertheless,asastarter, wehavethefollowing: LemmaIfgcd(a,n)=1,thenax≡b(modn)hasasolutionmodulon. PROOF:Indeed,by(3),ax≡b(modn)isequivalenttotheequality in n.ButbyCondition (4), hasamultiplicativeinversein n;hencefrom weget .Setting ,weget in n,thatis,x≡c(modn).■ Thus,ifa and n are relatively prime, ax ≡ b (mod n) has a solution modulo n. If a and n are not relativelyprime,wehavenosolutionmodulon;nevertheless,wehaveaveryniceresult: Theorem 2 If the congruence ax ≡ b (mod n) has a solution, then it has a solution modulo m, where Thismeansthatthesolutionofax≡b(modn)isoftheformx ≡ c (mod m); it consists of all the integerswhicharecongruenttoc,modulom. PROOF.Toprovethis,letgcd(a,n)=d,andnotethefollowing: Buta/dandn/darerelativelyprime(becausewearedividingaandnbyd,whichistheirgcd);henceby thelemma, hasasolutionxmodn/d.ByCondition(6),thisisalsoasolutionofax≡b(modn).■ As an example, let us solve 6x ≡ 4 (mod 10). Gcd(6,10) = 2 and 2|4, so by Theorem 1, this congruencehasasolution.ByCondition(6),thissolutionisthesameasthesolutionof Thisisequivalenttotheequation in 5,anditssolutionis .Sofinally,thesolutionof6x≡4 (mod10)isx≡4(mod5). How do we go about solving several linear congruences simultaneously? Well, suppose we are givenkcongruences, a1x≡b1(modn1), a2x≡b2(modn2),…,ak x=bk (modnk ) Ifeachofthesecongruenceshasasolution,wesolveeachoneindividuallybymeansofTheorem2.This givesus x≡c1(modm1), x≡c2(modm2), …, x≡ck (modmk ) Weareleftwiththeproblemofsolvingthislastsetofcongruencessimultaneously. Isthereanyintegerxwhichsatisfiesallkofthesecongruences?Theanswerfortwo simultaneous congruencesisasfollows: Theorem 3 Consider x ≡ a (mod n) and x ≡ b (mod m). There is an integer x satisfying both simultaneouslyiffa≡b(modd),whered=gcd(m,n). PROOF:Ifxisasimultaneoussolution,thenn|(x−a)andm|(x−b).Thus, x−a=nq1 and x−b=mq2 Subtractingthefirstequationfromthesecondgives a−b=mq2−nq1 Butd|mandd|n,sod|(a−b);thus,a≡b(modd). Conversely,ifa≡b(modd),thend|(a−b),soa−b=dq.ByTheorem3ofChapter22,d=rn+ tmforsomeintegersrandt.Thus,a−b=rqn+tqm.Fromthisequation,weget a−rqn=b+tqm Setx=a−rqn=b+tqm;thenx−a=−rqnandx−b=tqm;hencen|(x−a)andm|(x−b),so x≡a(modn) and x≡b(modm)■ Nowthatweareabletodeterminewhetherornotapairofcongruenceshasasimultaneoussolution, letusseewhatsuchasolutionlookslike. Theorem4Ifapairofcongruencesx≡a(modn)andx≡b(modm)hasasimultaneoussolution, thenithasasimultaneoussolutionoftheform x≡c(modt) wheretistheleastcommonmultipleofmandn. Beforeprovingthetheorem,letusobservethattheleastcommonmultiple(1cm)ofanytwointegers ra and n has the following property: let t be the least common multiple of m and n. Every common multipleofmandnisamultipleoft,andconversely.Thatis,forallintegersx, m|x and n|x iff t|x (SeeExerciseFattheendofChapter22.)Inparticular, m|(x−c) n|(x−c) iff t|(x−c) x≡c(modn) iff x≡c(modt)(7) and hence x≡c(modm) and Returningtoourtheorem,letcbeanysolutionofthegivenpairofcongruences(remember,weare assumingthereisasimultaneoussolution).Thenc≡a(modn)andc≡b(modm).Anyotherintegerxisa simultaneoussolutioniffx≡c(modn)andx≡c(modm).ButbyCondition(7),thisistrueiffx≡c(mod t).Theproofiscomplete. AspecialcaseofTheorems3and4isveryimportantinpractice:itisthecasewheremandn are relativelyprime.Notethat,inthiscase,gcd(m,n)=1andlcm(m,n)=mn.Thus,byTheorems3and4, Ifmandnarerelativelyprime,thepairofcongruencesx≡a(modn)andx≡b(modm)always hasasolution.Thissolutionisoftheformx≡c(modmn). Thisstatementcaneasilybeextendedtothecaseofmorethantwolinearcongruences.Theresultis knownasthe Chinese remainder theorem Let m1,m2,…,mk be pairwise relatively prime. Then the system of simultaneouslinearcongruences x≡c1(modm1), x≡c2(modm2),…,x=ck (modmk ) alwayshasasolution,whichisoftheformx≡c(modm1m2…mk ). UseTheorem4tosolvex≡c1(modm1)andx≡c2(modm2)simultaneously.Thesolutionisofthe form x ≡ d (mod m1m2). Solve the latter simultaneously with x ≡ c3 (mod m3), to get a solution mod m1m2m3.Repeatthisprocessk−1times. EXERCISES A.SolvingSingleCongruences 1Foreachofthefollowingcongruences,findmsuchthatthecongruencehasauniquesolutionmodulom. Ifthereisnosolution,write“none.” (a)60x≡12(mod24) (b)42x≡24(mod30) (c)49x≡30(mod25) (d)39x≡14(mod52) (e)147x≡47(mod98) (f)39x≡26(mod52) 2Solvethefollowinglinearcongruences: (a)12x≡7(mod25) (b)35x≡8(mod12) (c)15x≡9(mod6) (d)42x≡12(mod30) (e)147x≡49(mod98) (f)39x≡26(mod52) 3(a)Explainwhy2x2≡8(mod10)hasthesamesolutionsasx2≡4(mod5).(b)Explainwhyx≡2(mod 5)andx≡3(mod5)areallthesolutions4of2x2≡8(mod10). 4Solvethefollowingquadraticcongruences.(Ifthereisnosolution,write“none.”) (a)6x2≡9(mod15) (b)60x2≡18(mod24) (c)30x2≡18(mod24) (d)4(x+1)2≡14(mod10) (e)4x2−2x+2≡0(mod6) #(f)3x2−6x+6≡0(mod15) 5Solvethefollowingcongruences: (a)x4≡4(mod6) (b)2(x−1)4≡0(mod8) (c)x3+3x2+3x+1=0(mod8) (d)x4+2x2+1≡4(mod5) 6SolvethefollowingDiophantineequations.(Ifthereisnosolution,write“none.”) (a)14x+15y=11 (b)4x+5y=1 (c)21x+10y=9 #(d)30x2+24y=18 B.SolvingSetsofCongruences ExampleSolvethepairofsimultaneouscongruencesx≡5(mod6),x≡7(mod10). ByTheorems3and4,thispairofcongruenceshasasolutionmodulo30.Fromx≡5(mod6),weget x≡6q+5.Introducingthisintox≡7(mod10)yields6q+5≡7(mod10).Thus,successively,6q≡2 (mod10),3q≡1(mod5),q≡2(mod5),q=5r+2.Introducingthisintox=6q+5givesx=6(5r+2)+ 5=30r+17.Thus,x≡17(mod30).Thisisoursolution. 1Solveeachofthefollowingpairsofsimultaneouscongruences: (a)x≡7(mod8);x≡11(mod12) (b)x≡12(mod18);x≡30(mod45) (c)x≡8(mod15);x=11(mod14) 2Solveeachofthefollowingpairsofsimultaneouscongruences: (a)10x≡2(mod12);6x≡14(mod20) (b)4x≡2(mod6);9x≡3(mod12) (c)6x≡2(mod8);10x≡2(mod12) #3UseTheorems3and4toprovethefollowing:Supposewearegivenkcongruences x=c1(modm1), x≡c2(modm2) …, x≡ck (modmk ) There is an x satisfying all k congruences simultaneously if for all i, j ∈ {1, …, k}, ci ≡ cj (mod dij ), wheredij =gcd(mi,mj ).Moreover,thesimultaneoussolutionisoftheformx≡c(modt),wheret=lcm (m1,m2,…,mk ). 4Solvingeachofthefollowingsystemsofsimultaneouslinearcongruences;ifthereisnosolution,write “none.” (a)x≡2(mod3);x≡3(mod4);x=1(mod5);x≡4(mod7) (b)6x≡4(mod8);10x≡4(mod12);3x≡8(mod10) (c)5x≡3(mod6);4x≡2(mod6);6x≡6(mod8) 5SolvethefollowingsystemsofsimultaneousDiophantineequations: #(a)4x+6y=2;9x+12y=3 (b)3x+4y=2;5x+6y=2;3x+10y=8. C.ElementaryPropertiesofCongruence Provethefollowingforallintegersa,b,c,dandallpositiveintegersmandn: 1Ifa≡b(modn)andb≡c(modn),thena≡c(modn). 2Ifa≡b(modn),thena+c≡b+c(modn). 3Ifa≡b(modn),thenac≡bc(modn). 4a≡b(mod1). 5Ifab≡0(modp),wherepisaprime,thena≡0(modp)orb≡0(modp). 6Ifa2≡b2(modp),wherepisaprime,thena≡±b(modp). 7Ifa≡b(modm),thena+km≡b(modm),foranyintegerk.Inparticular,a+km≡a(modm). 8Ifac≡bc(modn)andgcd(c,n)≡1,thena≡b(modn). 9Ifa≡b(modn),thena≡b(modm)foranymwhichisafactorofn. D.FurtherPropertiesofCongruence Provethefollowingforanintegersa,b,candallpositiveintegersmandn: 1Ifac≡bc(modn),andgcd(c,n)≡d,thena≡b(modn/d). 2Ifa≡b(modn),thengcd(a,n)=gcd(b,n). 3Ifa≡b(modp)foreveryprimep,thena=b. 4Ifa≡b(modn),thenam≡bm(modn)foreverypositiveintegerm. 5Ifa≡b(modm)anda≡b(modn)wheregcd(m,n)=1,thena≡b(modmn). #6Ifab≡1(modc),ac≡1(modb)andbc≡1(moda),thenab+bc+ac≡1(modabc).(Assumea,b, c>0.) 7Ifa2≡1(mod2),thena2≡1(mod4). 8Ifa≡b(modn),thena2+b2≡2ab(modn2),andconversely. 9Ifa≡1(modm),thenaandmarerelativelyprime. E.ConsequencesofFermat’sTheorem 1Ifpisaprime,findϕ(p).UsethistodeduceFermat’stheoremfromEuler’stheorem. Proveparts2-6: 2Ifp>2isaprimeanda 0(modp),then a(p−1)/2≡±1(modp) 3 (a)Letpaprime>2.Ifp≡3(mod4),then(p−1)/2isodd. (b)Letp>2beaprimesuchthatp≡3(mod4).Thenthereisnosolutiontothecongruencex2+1≡0 (mod p). [HINT: Raise both sides of x2 ≡ −1 (mod p) to the power (p − 1)/2, and use Fermat’s little theorem.] #4Letpandqbedistinctprimes.Thenpq−1+qp−1≡1(modpq). 5Letpbeaprime. (a)If,(p−1)|m,thenam≡1(modp)providedthatp a. (b)If,(p−1)|m,thenam+1≡a(modpq)forallintegersa. #6Letpandqbedistinctprimes. (a)If(p−1)|mand(q−1)|m,thenam≡1(modpq)foranyasuchthatp aandq a. (b)If(p−1)|mand(q−1)|m,thenam+1≡a(modpq)forintegersa. 7Generalizetheresultofpart6tondistinctprimes,p1…,pn.(Stateyourresult,butdonotproveit.) 8Usepart6toexplainwhythefollowingaretrue: #(a)a19≡a(mod133). (b)a10≡1(mod66),providedaisnotamultipleof2,3,or11. (c)a13≡a(mod105). (d)a49≡a(mod1547).(HINT:1547=7×13×17.) 9Findthefollowingintegersx: (a)x≡838(mod210) (b)x≡757(mod133) (c)x≡573(mod66) F.ConsequencesofEuler’sTheorem Proveparts1-6: 1Ifgcd(a,n)=1,thesolutionmodulonofax≡b(modn)isx≡aϕ(n)-1b(modn). 2Ifgcd(a,n)=1,thenamϕ(n)≡1(modn)forallvaluesofm. 3Ifgcd(m,n)=gcd(a,mn)=1,thenaϕ(m)ϕ(n)≡1(modmn). 4Ifpisaprime,ϕ(pn)=pn −pn −1=pn −1(p −1).(HINT:Foranyintegera,aandpnhaveacommon divisor≠±1iffaisamultipleofp.Thereareexactlypn−1multiplesofpbetween1andpn.) n 5Foreverya 0(modp),ap (p−1)),wherepisaprime. 6 Under the conditions of part 3, if t is a common multiple of ϕ (m) and ϕ (n), then at ≡ 1 (mod mn). Generalizetothreeintegersl,m,andn. 7Useparts4and6toexplainwhythefollowingaretrue: (a)a12≡1(mod180)foreveryasuchthatgcd(a,180)=1. (b)a42≡1(mod1764)ifgcd(a,1764)=1.(REMARK:1764=4×9×49.) (c)a60≡1(mod1800)ifgcd(a,1800)=1. #8Ifgcd(m,n)=l,provethatnϕ(m)+mϕ(n)≡1(modmn). 9Ifl,m,narerelativelyprimeinpairs,provethat(mn)ϕ(l)+(ln)ϕ(m)+(lm)ϕ(n)≡1(modlmn). G.Wilson’sTheorem,andSomeConsequences Inanyintegraldomain,ifx2=1,thenx2−1=(x+1)(x−1)=0;hencex=±1.Thus,anelementx≠±1 cannot be its own multiplicative inverse. As a consequence, in p the integers may be arrangedinpairs,eachonebeingpairedoffwithitsmultiplicativeinverse. Provethefollowing: 1In . 2(p−2)!≡1(modp)foranyprimenumberp. 3(p−1)!+1≡0(modp)foranyprimenumberpThisisknownasWilson’stheorem. 4Foranycompositenumbern≠4,(n−1)!≡0(modn).[HINT:Ifpisanyprimefactorofn,thenpisa factorof(n−1)!Why?] Before going on to the remaining exercises, we make the following observations: Let p > 2 be a prime.Then Consequently, REASON:p−1≡−1(modp),p−2≡−2(modp),···,(p+1)/2≡−(p−1)/2(modp). Withthisresult,provethefollowing: 5[(p−l)/2]!2≡(-1)(p+1)/2(modp),foranyprimep>2.(HINT:UseWilson’stheorem.) 6Ifp≡1(mod4),then(p+1)/2isodd.(Why?)Concludethat 7Ifp≡3(mod4),then(p+1)/2iseven.(Why?)Concludethat 8Whenp>2isaprime,thecongruencex2+1≡0(modp)hasasolutionifp≡1(mod4). 9Foranyprimep>2,x2≡−1(modp)hasasolutioniffp 3(mod4).(HINT:Usepart8andExercise E3.) H.QuadraticResidues Anintegeraiscalledaquadraticresiduemodulomifthereisanintegerxsuchthatx2≡a(modm).This isthesameassayingthatāisasquarein m.Ifaisnotaquadraticresiduemodulom,thenaiscalleda quadraticnonresiduemodulom.Quadraticresiduesareimportantforsolvingquadraticcongruences,for studyingsumsofsquares,etc.Here,wewillexaminequadraticresiduesmoduloanarbitraryprimep>2. Leth: bedefinedby . 1Provehisahomomorphism.Itskernelis . #2Therangeofhhas(p −1)/2elements.Prove:Ifranh=R,Risasubgroupof havingtwocosets. Onecontainsalltheresidues,theotherallthenonresidues. TheLegendresymbolisdefinedasfollows: 3Referringtopart2,letthetwocosetsofRbecalled1and-1.Then foreveryintegerαwhichisnotamultipleofp. .Explainwhy . #4Evaluate 5Prove:ifa≡b(modp),then .Inparticular, 6Prove: (HINT:UseExercisesG6and7.) 7 Themostimportantruleforcomputing isthelawofquadraticreciprocity,whichassertsthatfordistinctprimesp,q>2, (The proof may be found in any textbook on number theory, for example, Fundamentals of Number TheorybyW.J.LeVeque.) 8Useparts5to7andthelawofquadraticreciprocitytofind: Is14aquadraticresidue,modulo59? 9Whichofthefollowingcongruencesissolvable? (a)x2=30(mod101) (b)x2≡6(mod103) (c)2x2≡70(mod106) NOTE:x2≡a(modp)issolvableiffaisaquadraticresiduemodulopiff I.PrimitiveRoots RecallthatVnisthemultiplicativegroupofalltheinvertibleelementsin n.IfVnhappenstobecyclic, sayVn=〈m〉,thenanyintegera≡m(modn)iscalledaprimitiverootofn. 1ProvethataisaprimitiverootofnifftheorderofāinVnisϕ(n). 2Provethateveryprimenumberphasaprimitiveroot.(HINT:Foreveryprime isacyclicgroup. ThesimpleproofofthisfactisgivenasTheorem1inChapter33.) 3Findprimitiverootsofthefollowingintegers(iftherearenone,sayso):6,10,12,14,15. 4Supposeaisaprimitiverootofm.Prove:Ifbisanyintegerwhichisrelativelyprimetom,thenb≡ak (modm)forsomek 1. 5Supposemhasaprimitiveroot,andletnberelativelyprimetoϕ(m).(Supposen>0.)Provethatifa isrelativelyprimetom,thenxn≡a(modm)hasasolution. 6Letp>2beaprime.Provethateveryprimitiverootofpisaquadraticnonresidue,modulop. (HINT: Supposeaprimitiverootαisaresidue;theneverypowerofaisaresidue.) 7Aprimepoftheformp=2m+1iscalledaFermatprime.LetpbeaFermâtprime.Provethatevery quadraticnonresiduemodpisaprimitiverootofp.(HINT: How many primitive roots are there? How manyresidues?Compare.) CHAPTER TWENTY-FOUR RINGSOFPOLYNOMIALS Inelementaryalgebraanimportantroleisplayedbypolynomialsinanunknownx.Theseareexpressions suchas whose terms are grouped in powers of x. The exponents, of course, are positive integers and the coefficientsarerealorcomplexnumbers. Polynomialsareinvolvedincountlessapplications—applicationsofeverykindanddescription.For example,polynomialfunctionsaretheeasiestfunctionstocompute,andthereforeonecommonlyattempts toapproximatearbitraryfunctionsbypolynomialfunctions.Agreatdealofefforthasbeenexpendedby mathematicianstofindwaysofachievingthis. Asidefromtheirusesinscienceandcomputation,polynomialscomeupverynaturallyinthegeneral studyofrings,asthefollowingexamplewillshow: Supposewewishtoenlargethering byaddingtoitthenumberπ.Itiseasytoseethatwewillhave to adjoin to other new numbers besides just π; for the enlarged ring (containing π as well as all the integers)willalsocontainsuchthingsas−π,π+7,6π2−11,andsoon. Asamatteroffact,anyringwhichcontains asasubringandwhichalsocontainsthenumberπwill havetocontaineverynumberoftheform aπn+bπn”1+⋯+kπ+l wherea,b,…,k,lareintegers.Inotherwords,itwillcontainallthepolynomialexpressionsinπwith integercoefficients. Butthesetofallthepolynomialexpressionsinπwithintegercoefficientsisaring.(Itisasubringof becauseitisobviousthatthesumandproductofanytwopolynomialsinπisagainapolynomialinπ.) Thisringcontains becauseeveryintegeraisapolynomialwithaconstanttermonly,anditalsocontains π. Thus,ifwewishtoenlargethering byadjoiningtoitthenewnumberπ,itturnsoutthatthe“next largest”ringafter whichcontains asasubringandincludesπ,isexactlytheringofallthepolynomials inπwithcoefficientsin . Asthisexampleshows,asidefromtheirpracticalapplications,polynomialsplayanimportantrole intheschemeofringtheorybecausetheyarepreciselywhatweneedwhenwewishtoenlargearingby addingnewelementstoit. In elementary algebra one considers polynomials whose coefficients are real numbers, or in some cases,complexnumbers.Asamatteroffact,thepropertiesofpolynomialsareprettymuchindependentof theexactnatureoftheircoefficients.Allweneedtoknowisthatthecoefficientsarecontainedinsome ring.Forconvenience,wewillassumethisringisacommutativeringwithunity. LetAbeacommutativeringwithunity.Uptonowwehaveusedletterstodenoteelementsorsets, but now we will use the letter x in a different way. In a polynomial expression such as ax2 + bx + c, wherea,b,c∈A,wedonotconsiderxtobeanelementofA,butratherxisasymbolwhichweuseinan entirelyformalway.Laterwewillallowthesubstitutionofotherthingsforx,butatpresentxissimplya placeholder. Notationally,thetermsofapolynomialmaybelistedineitherascendingordescendingorder.For example, 4x3 – 3x2 + x + 1 and 1 + x – 3x2 + 4x3 denote the same polynomial. In elementary algebra descendingorderispreferred,butforourpurposesascendingorderismoreconvenient. LetAbeacommutativeringwithunity,andxanarbitrarysymbol.Everyexpressionoftheform a0+a1x+a2x2+⋯+anxn iscalledapolynomialinxwithcoefficientsinA,ormoresimply,apolynomialinxoverA.The expressionsak xk ,fork∈{1,…,n},arecalledthetermsofthepolynomial. Polynomialsinxaredesignatedbysymbolssuchasa(x),b(x),q(x),andsoon.Ifa(x)=a0+a1x+ ⋯ + an xn is any polynomial and ak xk is any one of its terms, ak is called the coefficient of xk . By the degreeofapolynomiala(x)wemeanthegreatestnsuchthatthecoefficientofxnisnotzero.Inother words, if a(x) has degree n, this means that an ≠ 0 but am = 0 for every m > n. The degree of a(x) is symbolizedby dega(x) Forexample,1+2x−3x2+x3isapolynomialdegree3. The polynomial 0 + 0x + 0x2 + ⋯ all of whose coefficients are equal to zero is called the zero polynomial,andissymbolizedby0.Itistheonlypolynomialwhosedegreeisnotdefined(becauseithas nononzerocoefficient). If a nonzero polynomial a(x) = a0 + a1x + ⋯ + anxn has degree n, then an is called its leading coefficient:itisthelastnonzerocoefficientofa(x).Thetermanxnisthencalleditsleadingterm,while a0iscalleditsconstantterm. Ifapolynomiala(x) has degree zero, this means that its constant term a0 is its only nonzero term: a(x) is a constant polynomial. Beware of confusing a polynomial of degree zero with the zero polynomial. Two polynomials a(x) and b(x) are equal if they have the same degree and corresponding coefficientsareequal.Thus,ifa(x)=a0+⋯+anxnisofdegreen,andb(x)=b0+⋯+bmxmisofdegree m,thena(x)=b(x)iffn=mandak =bk foreachkfrom0ton. Thefamiliarsigmanotationforsumsisusefulforpolynomials.Thus, withtheunderstandingthatx0=1. Additionandmultiplicationofpolynomialsisfamiliarfromelementaryalgebra.Wewillnowdefine these operations formally. Throughout these definitions we let a(x) and b(x) stand for the following polynomials: Here we do not assume that a(x) and b(x) have the same degree, but allow ourselves to insert zero coefficientsifnecessarytoachieveuniformityofappearance. Weaddpolynomialsbyaddingcorrespondingcoefficients.Thus, a(x)+b(x)=(a0+b0)+(a1,+b1)x+⋯+(an+bn)xn Notethatthedegreeofa(x)+b(x)islessthanorequaltothehigherofthetwodegrees,dega(x)anddeg b(x). Multiplicationismoredifficult,butquitefamiliar: a(x)b(x) =a0b0+(a0b1+b0a1)x+(a0b2+a1b1+a2b0)x2+⋯+anbnx2n Inotherwords,theproductofa(x)andb(x)isthepolynomial c(x)=c0+c1x+⋯+c2nx2n whosekthcoefficient(foranykfrom0to2n)is Thisisthesumofalltheaibj forwhichi+j=k.Notethatdeg[a(x)b(x)]≤dega(x)+degb(x). IfAisanyring,thesymbol A[x] designatesthesetofallthepolynomialsinxwhosecoefficientsareinA,withadditionandmultiplication ofpolynomialsaswehavejustdefinedthem. Theorem1LetAbeacommutativeringwithunity.ThenA[x]isacommutativeringwithunity. PROOF:Toprovethistheorem,wemustshowsystematicallythatA[x] satisfies all the axioms of a commutative ring with unity. Throughout the proof, let a(x), b(x), and c(x) stand for the following polynomials: The axioms which involve only addition are easy to check: for example, addition is commutative because Theassociativelawofadditionisprovedsimilarly,andisleftasanexercise.Thezeropolynomialhas alreadybeendescribed,andthenegativeofa(x)is –a(x)=(–a0)+(–a1)x+⋯+(–an)xn Toprovethatmultiplicationisassociativerequiressomecare.Letb(x)c(x)=d(x),whered(x)=d0 +d1x+⋯+d2nx2n.Bythedefinitionofpolynomialmultiplication,thekthcoefficientofb(x)c(x)is Then a(x)[b(x)c(x)] = a(x)d(x) = e(x), where e(x) = e0 + e1x + ⋯ + e3nx3n. Now, the lth coefficient of a(x)d(x)is Itiseasytoseethatthesumontherightconsistsofallthetermsahbicj suchthath+i+j=lThus, Foreachlfrom0to3n,elisthelthcoefficientofa(x)[b(x)c(x)]. Ifwerepeatthisprocesstofindthelthcoefficientof[a(x)b(x)]c(x),wediscoverthatit,too,isel Thus, a(x)[b(x)c(x)]=[a(x)b(x)]c(x) Toprovethedistributivelaw,leta(x)[b(x)+c(x)]=d{x)whered(x)=d0+d1x+ ⋯+d2nx2n. By thedefinitionsofpolynomialadditionandmultiplication,thekthcoefficienta(x)[b(x)+c(x)]is But Σi+j = k aibj . is exactly the kth coefficient of a(x) b(x), and Σi + j = k aicj is the kth coefficient of a(x)c(x),hencedk isequaltothekthcoefficientofa(x)b(x)+a(x)c(x).Thisprovesthat a(x)[b(x)+c(x)]=a(x)b(x)+a(x)c(x) Thecommutativelawofmultiplicationissimpletoverifyandislefttothestudent.Finally,theunity polynomialistheconstantpolynomial1.■ Theorem2IfAisanintegraldomain,thenA[x]isanintegraldomain. PROOF:Ifa(x)andb(x)arenonzeropolynomials,wemustshowthattheirproducta(x) b(x) is not zero.Letanbetheleadingcoefficientofa(x),andbmtheleadingcoefficientofb(x).Bydefinition,an≠0, and bm ≠ 0. Thus anbm ≠ 0 because A is an integral domain. It follows that a(x) b(x) has a nonzero coefficient(namely,anbm),soitisnotthezeropolynomial.■ IfAisanintegraldomain,werefertoA[x]asadomainofpolynomials,becauseA[x]isanintegral domain.Notethatbytheprecedingproof,ifanandbmaretheleadingcoefficientsofa(x)andb(x), then anbmistheleadingcoefficientofa(x)b(x).Thus,dega(x)b(x)=n+m:InadomainofpolynomialsA[x], whereAisanintegraldomain, deg[a(x)·b(x)]=dega(x)+degb(x) In the remainder of this chapter we will look at a property of polynomials which is of special interestwhenallthecoefficientslieinafield.Thus,fromthispointforward,letFbeafield,andletus considerpolynomialsbelongingtoF[x]. ItwouldbetemptingtobelievethatifFisafieldthenF[x]alsoisafield.However,thisisnotso, for one can easily see that the multiplicative inverse of a polynomial is not generally a polynomial. Nevertheless,byTheorem2,F[x]isanintegraldomain. Domains of polynomials over a field do, however, have a very special property: any polynomial a(x)maybedividedbyanynonzeropolynomialb(x)toyieldaquotientq(x)andaremainderr(x). The remainderiseither0,orifnot,itsdegreeislessthanthedegreeofthedivisorb(x).Forexample,x2may bedividedbyx–2togiveaquotientofx+2andaremainderof4: Thiskindofpolynomialdivisionisfamiliartoeverystudentofelementaryalgebra.Itiscustomarilyset upasfollows: Theprocessofpolynomialdivisionisformalizedinthenexttheorem. Theorem3:DivisionalgorithmforpolynomialsIfa(x)andb(x) are polynomials over a field F, andb(x)≠0,thereexistpolynomialsq(x)andr(x)overFsuchthat a(x)=b(x)q(x)+r(x) and r(x)=0 or degr(x)<degb(x) PROOF: Let b(x) remain fixed, and let us show that every polynomial a(x) satisfies the following condition: Thereexistpolynomialsq(x)andr(x)overFsuchthata(x)=b(x)q(x)+r(x),andr(x)=0ordeg r(x)<degb(x). Wewillassumetherearepolynomialsa(x)whichdonotfulfillthecondition,andfromthisassumption we will derive a contradiction. Let a(x) be a polynomial of lowest degree which fails to satisfy the conditions.Notethata(x)cannotbezero,becausewecanexpress0as0=b(x) · 0 + 0, whereby a(x) wouldsatisfytheconditions.Furthermore,dega(x)≥degb(x),forifdega(x)<degb(x)thenwecould writea(x)=b(x)·0+a(x),soagaina(x)wouldsatisfythegivenconditions. Leta(x)=a0+⋯+anxnandb(x)=b0+⋯+bmxm.Defineanewpolynomial Thisexpressionisthedifferenceoftwopolynomialsbothofdegreenandbothhavingthesameleading termanxn.Becauseanxncancelsinthesubtraction,A(x)hasdegreelessthann. Rememberthata(x)isapolynomialofleastdegreewhichfailstosatisfythegivencondition;hence A(x)doessatisfyit.Thismeanstherearepolynomialsp(x)andr(x)suchthat A(x)=b(x)p(x)+r(x) wherer(x)=0ordegr(x)<degb(x).Butthen If we let p(x) + (an/bm)xn – m be renamed q(x), then a(x) = b(x)q(x) + r(x), so a(x) fulfills the given condition.Thisisacontradiction,asrequired.■ EXERCISES A.ElementaryComputationinDomainsofPolynomials REMARKONNOTATION:Insomeoftheproblemswhichfollow,weconsiderpolynomialswithcoefficients in nforvariousn.Tosimplifynotation,wedenotetheelementsof nby1,2,…,n–1ratherthanthe morecorrect . #1Leta(x)=2x2+3x+1andb(x)=x3+5x2+x.Computea(x)+b(x),a(x)–b(x)anda(x)b(x)in [x], 5[x], 6[x],and 7[x]. 2Findthequotientandremainderwhenx3+x2+x+1isdividedbyx2+3x+2in [x]andin 5[x]. 3Findthequotientandremainderwhenx3+2isdividedby2x2+3x+4in [x],in 3[x],andin 5[x]. Wecallb(x)afactorofa(x)ifa(x)=b(x)q(x)forsomeq(x),thatis,iftheremainderwhena(x)is dividedbyb(x)isequaltozero. 4ShowthatthefollowingistrueinA[x]foranyringA:Foranyoddn, (a)x+1isafactorofxn+1. (b)x+1isafactorofxn+xn–1+⋯+x+1. 5Provethefollowing:In 3[x],x+2isafactorofxm+2,forallm.In n[x],x+(x–1)isafactorofxm+ (n−1),forallmandn. 6Provethatthereisnointegermsuchthat3x2+4x+misafactorof6x4+50in [x]. 7Forwhatvaluesofnisx2+1afactorofx5+5x+6in n[x]? B.ProblemsInvolvingConceptsandDefinitions 1Isx8+1=x3+1in 5[x]?Explainyouranswer. 2IsthereanyringAsuchthatinA[x],somepolynomialofdegree2isequaltoapolynomialofdegree4? Explain. #3Writeallthequadraticpolynomialsin 5[x].Howmanyarethere?Howmanycubicpolynomialsare therein 5[x]?Moregenerally,howmanypolynomialsofdegreemaretherein n[x]? 4LetAbeanintegraldomain;provethefollowing: If(x+1)2=x2+1inA[x],thenAmusthavecharacteristic2. If(x+1)4=x4+1inA[x],thenAmusthavecharacteristic2. If(x+1)6=x6+2x3+1inA[x],thenAmusthavecharacteristic3. 5 Find an example of each of the following in 8[x]: a divisor of zero, an invertible element. (Find nonconstantexamples.) 6ExplainwhyxcannotbeinvertibleinanyA[x],hencenodomainofpolynomialscaneverbeafield. 7ThereareringssuchasP3inwhicheveryelement≠0,1isadivisorofzero.Explainwhythiscannot happeninanyringofpolynomialsA[x],evenwhenAisnotanintegraldomain. 8ShowthatineveryA[x],thereareelements≠0,1whicharenotidempotent,andelements≠0,1whichare notnilpotent. C.RingsA[x]WhereAIsNotanIntegralDomain 1Prove:IfAisnotanintegraldomain,neitherisA[x]. 2Giveexamplesofdivisorsofzero,ofdegrees0,1,and2,in 4[x]. 3In 10[x],(2x+2)(2x+2)=(2x+2)(5x3+2x+2),yet(2x+2)cannotbecanceledinthisequation. Explainwhythisispossiblein 10[x],butnotin 5[x]. 4Giveexamplesin 4[x],in 6[x],andinZ9[x]ofpolynomialsa(x)andb(x)suchthatdega(x)b(x)<deg a(x)+degb(x). 5IfAisanintegraldomain,wehaveseenthatinA[x], dega(x)b(x)=dega(x)+degb(x) Show that if A is not an integral domain, we can always find polynomials a(x) and b(x) such that deg a(x)b(x)<dega(x)+degb(x). 6ShowthatifAisanintegraldomain,theonlyinvertibleelementsinA[x]aretheconstantpolynomials withinversesinA.Thenshowthatin 4[x]thereareinvertiblepolynomialsofalldegrees. #7Giveallthewaysoffactoringx2intopolynomialsofdegree1in 9[x];in 5[x].Explainthedifference inbehavior. 8Findallthesquarerootsofx2+x+4in 5[x].Showthatin 8[x],thereareinfinitelymanysquareroots of1. D.DomainsA[x]WhereAHasFiniteCharacteristic Ineachofthefollowing,letAbeanintegraldomain: 1ProvethatifAhascharacteristicp,thenA[x]hascharacteristicp. 2Usepart1togiveanexampleofaninfiniteintegraldomainwithfinitecharacteristic. 3 Prove: If A has characteristic 3, then x + 2 is a factor of xm + 2 for all m. More generally, if A has characteristicp,thenx+(p–1)isafactorofxm+(p−1)forallm. 4ProvethatifAhascharacteristicp,theninA[x],(x+c)p=xp+cp.(Youmayuseessentiallythesame argumentasintheproofofTheorem3,Chapter20.) 5Explainwhythefollowing“proofofpart4isnotvalid:(x+c)p=xp+cpinA[x]because(a+c)p=ap +cpforalla,c∈A.(Notethefollowingexample:in 2,a2+1=a4+1foreverya,yetx2+1≠x4+1in 2[x].) #6Usethesameargumentasinpart4toprovethatifAhascharacteristicp,then[a(x)+b(x)]p=a(x)p+ b(x)pforanya(x),b(x)∈A[x].Usethistoprove: E.SubringsandIdealsinA[x] 1ShowthatifBisasubringofA,thenB[x]isasubringofA[x]. 2IfBisanidealofA,B[x]isanidealofA[x]. 3LetSbethesetofallthepolynomialsa(x)inA[x]forwhicheverycoefficientaiforoddiisequalto zero.ShowthatSisasubringofA[x].Whyisthesamenottruewhen“odd”isreplacedby“even”? 4LetJconsistofalltheelementsinA[x]whoseconstantcoefficientisequaltozero.ProvethatJisan idealofA[x]. #5LetJconsistofallthepolynomialsa0+a1x+⋯+anxninA[x]suchthata0+a1+⋯+an=0.Prove thatJisanidealofA[x]. 6Provethattheidealsinbothparts4and5areprimeideals.(AssumeAisanintegraldomain.) F.HomomorphismsofDomainsofPolynomials LetAbeanintegraldomain. 1Leth:A[x]→Amapeverypolynomialtoitsconstantcoefficient;thatis, h(a0+a1x+⋯+anxn)=a0 ProvethathisahomomorphismfromA[x]ontoA,anddescribeitskernel. 2Explainwhythekernelofhinpart1consistsofalltheproductsxa(x),foralla(x)∈A[x].Whyisthis thesameastheprincipalideal〈x〉inA[x]? 3Usingparts1and2,explainwhyA[x]/〈x〉≅A. 4 Let g : A[x] → A send every polynomial to the sum of its coefficients. Prove that g is a surjective homomorphism,anddescribeitskernel. 5Ifc∈A,leth:A[x]→A[x]bedefinedbyh(a(x))=a(cx),thatis, h(a0+a1x+⋯+anxn)=a0+a1cx+a2c2x2+⋯+ancnxn Provethathisahomomorphismanddescribeitskernel. 6Ifhisthehomomorphismofpart5,provethathisanautomorphism(isomorphismfromA[x]toitself) iffcisinvertible. G.HomomorphismsofPolynomialDomainsInducedbyaHomomorphismofthe RingofCoefficients LetAandBberingsandleth:A→BbeahomomorphismwithkernelK.Define :A[x]→B[x]by (a0+a1x+⋯+anxn)=h(a0)+h(a1)x+⋯+h(an)xn (Wesaythat isinducedbyh.) 1Provethat isahomomorphismfromA[x]toB[x]. 2Describethekernel of . #3Provethat issurjectiveiffhissurjective. 4Provethat isinjectiveiffhisinjective. 5Provethatifa(x)isafactorofb(x),then (a(x))isafactorof (b{x)). 6Ifh : → n is the natural homomorphism, let : [x] → n[x] be the homomorphism induced by h. Provethat (a(x))=0iffndivideseverycoefficientofa(x). 7Let beasinpart6,andletnbeaprime.Provethatifa(x)b(x)∈ker ,theneithera(x)orb(x)isin ker .(HINT:UseExerciseF2ofChapter19.) H.PolynomialsinSeveralVariables A[x1, x2] denotes the ring of all the polynomials in two letters x1 and x2 with coefficients in A. For example,x2–2xy+y2+x −5isaquadraticpolynomialin [x,y].Moregenerally,A[x1,…,xn]isthe ringofthepolynomialsinnlettersx1,…,xnwithcoefficientsinA.Formallyitisdefinedasfollows:Let A[x1]bedenotedbyA1;thenA1[x2]isA[x1,x2].Continuinginthisfashion,wemayadjoinonenewletterxi atatime,togetA[x1,…,xn]. 1ProvethatifAisanintegraldomain,thenA[x1,…,xn]isanintegraldomain. 2 Give a reasonable definition of the degree of any polynomial p(x, y) in A[x, y] and then list all the polynomialsofdegree≤3inZ3[x,y]. Letusdenoteanarbitrarypolynomialp(x,y)inA[x,y]byΣaij xiyj whereΣrangesoversomepairsi,jof nonnegativeintegers. 3ImitatingthedefinitionsofsumandproductofpolynomialsinA[x],giveadefinitionofsumandproduct ofpolynomialsinA[x,y]. 4Provethatdega(x,y)b(x,y)=dega(x,y)+degb(x,y)ifAisanintegraldomain. I.FieldsofPolynomialQuotients LetAbeanintegraldomain.BytheclosingpartofChapter20,everyintegraldomaincanbeextendedtoa “fieldofquotients.”Thus,A[x]canbeextendedtoafieldofpolynomialquotients,whichisdenotedby A(x).NotethatA(x)consistsofallthefractionsa(x)/b(x)fora(x)andb(x)≠0inA[x],andthesefractions areadded,subtracted,multiplied,anddividedinthecustomaryway. 1ShowthatA(x)hasthesamecharacteristicasA. 2Usingpart1,explainwhythereisaninfinitefieldofcharacteristicp,foreveryprimep. 3 If A and B are integral domains and h : A → B is an isomorphism, prove that h determines an isomorphism :A(x)→B(x). J.DivisionAlgorithm:UniquenessofQuotientandRemainder In the division algorithm, prove that q(x) and r(x) are uniquely determined. [HINT: Suppose a(x) = b(x)q1(x)+r1(x)=b(x)q2(x)+r2(x),andsubtractthesetwoexpressions,whicharebothequaltoa(x).] CHAPTER TWENTY-FIVE FACTORINGPOLYNOMIALS Justaseveryintegercanbefactoredintoprimes,soeverypolynomialcanbefactoredinto“irreducible” polynomials which cannot be factored further. As a matter of fact, polynomials behave very much like integers when it comes to factoring them. This is especially true when the polynomials have all their coefficientsinafield. Throughoutthischapter,weletFrepresentsomefieldandweconsiderpolynomialsoverF.Itwill be found that F[x] has a considerable number of properties in common with . To begin with, all the idealsofF[x]areprincipalideals,whichwasalsothecasefortheidealsof . Note carefully that in F[x], the principal ideal generated by a polynomial a(x) consists of all the productsa(x)s(x)asa(x)remainsfixedands(x)rangesoverallthemembersofF[x]. Theorem1EveryidealofF[x]isprincipal. PROOF:LetJbeanyidealofF[x].IfJcontainsnothingbutthezeropolynomial,J is the principal idealgeneratedby0.IftherearenonzeropolynomialsinJ,letb(x)beanypolynomialoflowestdegreein J.WewillshowthatJ=〈(b(x)〉,whichistosaythateveryelementofJisapolynomialmultipleb(x)q(x) ofb(x). Indeed,ifa(x)isanyelementofJ,wemayusethedivisionalgorithmtowritea(x)=b(x)q(x)+r(x), wherer(x)=0ordegr(x)<degb(x).Now,r(x)=a{x)−b(x)q(x);buta(x)waschoseninJ,andb(x)∈ J;henceb(x)q(x)∈J.Itfollowsthatr(x)isinJ. If r(x) ≠ 0, its degree is less than the degree of b(x). But this is impossible because b(x) is a polynomialoflowestdegreeinJ.Therefore,ofnecessity,r(x)=0. Thus,finally,a(x)=b(x)q(x);soeverymemberofJisamultipleofb(x),asclaimed.■ ItfollowsthateveryidealJofF[x]isprincipal.Infact,astheproofaboveindicates,Jisgenerated byanyoneofitsmembersoflowestdegree. Throughoutthediscussionwhichfollows,rememberthatweareconsideringpolynomialsinafixed domainF[x]whereFisafield. Leta(x)andb(x)beinF[x].Wesaythatb(x)isamultipleofa(x)if b(x)=a(x)s(x) forsomepolynomials(x)inF[x].Ifb(x)isamultipleofa(x),wealsosaythata(x)isafactorofb(x),or thata(x)dividesb(x).Insymbols,wewrite a(x)∣b(x) Everynonzeroconstantpolynomialdivideseverypolynomial.Forifc≠0isconstantanda(x)=a0+ …+anxn,then hencec∣a(x).Apolynomiala(x)isinvertibleiffitisadivisoroftheunitypolynomial1.Butifa(x)b(x) =1,thismeansthata(x)andb(x)bothhavedegree0,thatis,areconstantpolynomials:a(x)=a,b(x)=b, andab=1.Thus, theinvertibleelementsofF[x]areallthenonzeroconstantpolynomials. A pair of nonzero polynomials a(x) and b(x) are called associates if they divide one another: a(x)∣b(x)andb(x)∣a(x).Thatistosay, a(x)=b(x)c(x) and b(x)=a(x)d(x) forsomec(x)andd(x).Ifthishappenstobethecase,then a(x)=b(x)c(x)=a(x)d(x)c(x) henced(x)c(x)=1becauseF[x]isanintegraldomain.Butthenc(x)andd(x)areconstantpolynomials, andthereforea(x)andb(x)areconstantmultiplesofeachother.Thus,inF[x], a(x)andb(x)areassociatesifftheyareconstantmultiplesofeachother. Ifa(x)=a0+ ⋯+anxn,theassociatesofa(x) are all its nonzero constant multiples. Among these multiplesisthepolynomial whichisequalto(1/an)a(x),andwhichhas1asitsleadingcoefficient.Anypolynomialwhoseleading coefficient is equal to 1 is called monk. Thus, every nonzero polynomial a(x) has a unique monic associate.Forexample,themonicassociateof3+4x+2x3is . A polynomial d(x) is called a greatest common divisor of a(x) and b(x) if d(x) divides a(x) and b(x),andisamultipleofanyothercommondivisorofa(x)andb(x);inotherwords, (i) d(x)∣a(x)andd(x)∣b(x),and (ii) For any u(x) in F[x], if u(x)∣a(x) and u(x)∣b(x), then u(x)∣d(x). According to this definition, two differentgcd’sofa(x)andb(x)divideeachother,thatis,areassociates.Ofallthepossiblegcd’sof a(x)andb(x),weselectthemonicone,callitthegcdofa(x)andb(x),anddenoteitbygcd[a(x),b(x)]. Itisimportanttoknowthatanypairofpolynomialsalwayshasagreatestcommondivisor. Theorem2Anytwononzeropolynomialsa(x)andb(x)inF[x]haveagcdd(x).Furthermore,d(x) canbeexpressedasa“linearcombination” d(x)=r(x)a(x)+s(x)b(x) wherer(x)ands(x)areinF[x]. PROOF:Theproofisanalogoustotheproofofthecorrespondingtheoremforintegers.IfJistheset ofallthelinearcombinations u(x)a(x)+υ(x)b(x) asu(x)andυ(x)rangeoverF[x],thenJisanidealofF[x],saytheideal〈d(x)〉generatedbyd(x). Now a(x) = la(x) + 0b(x) and b(x) = 0a(x) + 1b(x), so a(x) and b(x) are in J. But every element of 7 is a multipleofd(x),so d(x)∣a(x) and d(x)∣b(x) Ifk(x)isanycommondivisorofa(x)andb(x),thismeanstherearepolynomialsf(x)andg(x) such thata(x)=k(x)f(x)andb(x)=k(x)g(x).Now,d(x)∈J,sod(x)canbewrittenasalinearcombination hencek(x)∣d(x).Thisconfirmsthatd(x)isthegcdofa(x)andb(x).■ Polynomialsa(x)andb(x)inF[x]aresaidtoberelativelyprimeiftheirgcdisequalto1.(Thisis equivalenttosayingthattheironlycommonfactorsareconstantsinF.) Apolynomiala(x)ofpositivedegreeissaidtobereducible over F if there are polynomials b(x) andc(x)inF[x],bothofpositivedegree,suchthat a(x)=b(x)c(x) Because b(x) and c(x) both have positive degrees, and the sum of their degrees is deg a(x), each has degreelessthandega(x). A polynomial p(x) of positive degree in F[x] is said to be irreducible over F if it cannot be expressedastheproductoftwopolynomialsofpositivedegreeinF[x].Thus,p(x)isirreducibleiffitis notreducible. Whenwesaythatapolynomialp(x)isirreducible,itisimportantthatwespecifyirreducible over thefieldF.ApolynomialmaybeirreducibleoverF,yetreducibleoveralargerfieldE. For example, p(x)=x2+1isirreducibleover ;butover ithasfactors(x+i)(x−i). We next state the analogs for polynomials of Euclid’s lemma and its corollaries. The proofs are almostidenticaltotheircounterpartsin ;thereforetheyareleftasexercises. Euclid’s lemma for polynomials Let p(x) be irreducible. If p(x)∣ a(x)b(x), then p(x)∣a(x) or p(x)∣b(x). Corollary 1 Let p(x) be irreducible. If p(x) ∣a1(x)a2(x) ⋯ an(x), then p(x) ∣ ai(x) for one of the factorsai(x)amonga1(x),...,an(x). Corollary 2 Let q1(x), …, qr(x) and p(x) be monic irreducible polynomials. If p(x) ∣ q1 (x) … qr(x),thenp(x)isequaltooneofthefactorsq1(x),...,qr(x). Theorem3:FactorizationintoirreduciblepolynomialsEverypolynomiala(x)ofpositivedegree inF[x]canbewrittenasaproduct a(x)=kp1(x)p2(x)…pr(x) wherekisaconstantinFandp1(x),…,pr(x)aremonicirreduciblepolynomialsofF[x]. Ifthiswerenottrue,wecouldchooseapolynomiala(x)oflowestdegreeamongthosewhichcannot befactoredintoirreducibles.Thena(x)isreducible,soa(x)=b(x)c(x)whereb(x)andc(x)havelower degreethana(x).Butthismeansthatb(x)andc(x)canbefactoredintoirreducibles,andthereforea(x) canalso. Theorem 4: Unique factorization If a(x) can be written in two ways as a product of monic irreducibles,say a(x)=kp1(x)⋯pr(x)=lq1(x)⋯qs(x) thenk=l,r=s,andeachpi(x)isequaltoaqJ(x). The proof is the same, in all major respects, as the corresponding proof for ; it is left as an exercise. Inthenextchapterwewillbeabletoimprovesomewhatonthelasttworesultsinthespecialcases of [x]and [x].Also,wewilllearnmoreaboutfactoringpolynomialsintoirreducibles. EXERCISES A.ExamplesofFactoringintoIrreducibleFactors 1Factorx4−4intoirreduciblefactorsover ,over ,andover . 2Factorx6−16intoirreduciblefactorsover ,over ,andover . 3Findalltheirreduciblepolynomialsofdegree≤4in 2[x]. # 4 Show that x2 + 2 is irreducible in 5[x]. Then factor x4 − 4 into irreducible factors in 5[x]. (By Theorem3,itissufficienttosearchformonicfactors.) 5Factor2x3+4x+1in 5[x].(FactoritasinTheorem3.) 6In 6[x],factoreachofthefollowingintotwopolynomialsofdegree1:x, x + 2, x + 3. Why is this possible? B.ShortQuestionsRelatingtoIrreduciblePolynomials LetFbeafield.ExplainwhyeachofthefollowingistrueinF[x]: 1Everypolynomialofdegree1isirreducible. 2Ifa(x)andb(x)aredistinctmonicpolynomials,theycannotbeassociates. 3Anytwodistinctirreduciblepolynomialsarerelativelyprime. 4Ifa(x)isirreducible,anyassociateofa(x)isirreducible. 5Ifa(x)≠,a(x)cannotbeanassociateof0. 6In p[x],everynonzeropolynomialhasexactlyp−1associates. 7x2+1isreduciblein p[x]iffp=a+bwhereab≡1(modp). C.NumberofIrreducibleQuadraticsoveraFiniteField 1Withoutfindingthem,determinehowmanyreduciblemonicquadraticstherearein 5[x].[HINT:Every reduciblemonicquadraticcanbeuniquelyfactoredas(x+a)(x+b).] 2Howmanyreduciblequadraticsaretherein 5[x]?Howmanyirreduciblequadratics? 3Generalize:Howmanyirreduciblequadraticsarethereoverafinitefieldofnelements? 4Howmanyirreduciblecubicsarethereoverafieldofnelements? D.IdealsinDomainsofPolynomials LetFbeafield,andletJdesignateanyidealofF[x].Proveparts1−4. 1AnytwogeneratorsofJareassociates. 2Jhasauniquemonicgeneratorm(x).Anarbitrarypolynomiala(x)∈F[x]isinJiffm(x)∣a(x). 3Jisaprimeidealiffithasanirreduciblegenerator. #4Ifp(x)isirreducible,then〈p(x)〉isamaximalidealofF[x].(SeeChapter18,ExerciseH5.) 5LetSbethesetofallpolynomialsa0+a1x+⋯+anxninF[x]whichsatisfya0+a1+⋯+an=0.Ithas beenshown(Chapter24,ExerciseE5)thatSisanidealofF[x].Provethatx−1∈S,andexplainwhyit followsthatS=〈−1〉. 6Concludefrompart5thatF[x]/〈x−1〉≅F.(SeeChapter24,ExerciseF4.) 7LetF[x,y]denotethedomainofallthepolynomialsΣaij xiyj intwolettersxandy,withcoefficientsin F. Let J be the ideal of F[x, y] which contains all the polynomials whose constant coefficient in zero. ProvethatJisnotaprincipalideal.ConcludethatTheorem1isnottrueinF[x,y]. E.ProofoftheUniqueFactorizationTheorem 1ProveEuclid’slemmaforpolynomials. 2ProvethetwocorollariesofEuclid’slemma. 3Provetheuniquefactorizationtheoremforpolynomials. F.AMethodforComputingthegcd Leta(x)andb(x)bepolynomialsofpositivedegree.Bythedivisionalgorithm,wemaydividea(x) by b(x): a(x)=b(x)ql(x)+r1,(x) 1Provethateverycommondivisorofa(x)andb(x)isacommondivisorofb(x)andr1(x). It follows from part 1 that the gcd of a(x) and b(x) is the same as the gcd of b(x) and r1(x). This procedurecannowberepeatedonb(x)andr1(x);divideb(x)byr1(x): b(x)=r1(x)q2(x)r2(x) Next r1(x)=r2(x)q3(x)+r3(x) Finally, rn−1(x)=rn(x)qn+1(x)+0 Inotherwords,wecontinuetodivideeachremainderbythesucceedingremainder.Sincetheremainders continuallydecreaseindegree,theremustultimatelybeazeroremainder.Butwehaveseenthat gcd[a(x),b(x)]=gcd[b(x),r1(x)]=⋯=gcd[rn−1(x),rn(x)] Sincern(x)isadivisorofrn−1(x),itmustbethegcdofrn(x)andrn−.1.Thus, rn(x)=gcd[a(x),b(x)] Thismethodiscalledtheeuclideanalgorithmforfindingthegcd. #2Findthegcdofx3+1andx4+x3+2x2+x −1.Expressthisgcdasalinearcombinationofthetwo polynomials. 3Dothesameforx24—1andx15−1. 4Findthegcdofx3+x2+x+1andx4+x3+2x2+2xin 3[x]. G.ATransformationofF[x] LetGbethesubsetofF[x]consistingofallpolynomialswhoseconstanttermisnonzero.Leth:G→G bedefinedby h(a0+a1x+⋯+anxn)=an+an−1x+⋯+a0xn Proveparts1−3: 1hpreservesmultiplication,thatis,h[a(x)b(x)]=h[a(x)]h[b(x)]. 2hisinjectiveandsurjectiveandh∘h=ε. 3a0+a1x+⋯+anxnisirreducibleiffan+an−lx+⋯+a0xnisirreducible. 4Leta0+a1x+⋯+anxn=(b0+⋯+bmxm)(c0+⋯+cqxq).Factor an+an−1x+⋯+a0xn 5Leta(x)=a0+a1x+⋯+anxnandâ(x)=an+an−1x+⋯+a0xn.Ifc∈F,provethata(c)=0iffâ(1/c) =0. CHAPTER TWENTY-SIX SUBSTITUTIONINPOLYNOMIALS Uptonowwehavetreatedpolynomialsasformalexpressions.Ifa(x)isapolynomialoverafieldF,say a(x)=a0+a1x+⋯+anxn thismeansthatthecoefficientsa0,a1,…,anareelementsofthefieldF,whiletheletterxisaplaceholder whichplaysnootherrolethantooccupyagivenposition. When we dealt with polynomials in elementary algebra, it was quite different. The letter x was calledanunknownandwasallowedtoassumenumericalvalues.Thismadea(x)intoafunctionhavingx asitsindependentvariable.Suchafunctioniscalledapolynomialfunction. Thischapterisdevotedtothestudyofpolynomialfunctions.Webeginwithafewcarefuldefinitions. Leta(x)=a0+a1x+⋯+anxnbeapolynomialoverF.IfcisanyelementofF,then a0+a1c+⋯+ancn isalsoanelementofF,obtainedbysubstitutingcforxinthepolynomiala(x).Thiselementisdenoted bya(c).Thus, a(c)=a0+a1c+⋯+ancn SincewemaysubstituteanyelementofFforx,wemayregarda(x)asafunctionfromFtoF.Assuch,it iscalledapolynomialfunctiononF. Thedifferencebetweenapolynomialandapolynomialfunctionismainlyadifferenceofviewpoint. Givena(x)withcoefficientsinF:ifxisregardedmerelyasaplaceholder,thena(x)isapolynomial;ifx isallowedtoassumevaluesinF,thena(x)isapolynomialfunction.Thedifferenceisasmallone,and wewillnotmakeanissueofit. Ifa(x)isapolynomialwithcoefficientsinF,andcisanelementofFsuchthat a(c)=0 thenwecallcarootofa(x).Forexample,2isarootofthepolynomial3x2+x−14∈ [x],because3· 22+2−14=0. There is an absolutely fundamental connection between roots of a polynomial and factors of that polynomial.Thisconnectionisexploredinthefollowingpages,beginningwiththenexttheorem: Leta(x)beapolynomialoverafieldF. Theorem1cisarootofa(x)iffx−cisafactorofa(x). PROOF:Ifx−cisafactorofa(x),thismeansthata(x)=(x−c)q(x)forsomeq(x).Thus,a(c)=(c − c)q(c)=0,socisarootofa(x).Conversely,ifcisarootofa(x),wemayusethedivisionalgorithmto dividea(x)byx −c:a(x)=(x −c)q(x)+r(x).Theremainderr(x)iseither0orapolynomialoflower degreethanx −c;butlowerdegreethanx—cmeansthatr(x)isaconstantpolynomial:r(x)=r ≥ 0. Then 0=a(c)=(c−c)q(c)+r=0+r=r Thus,r=0,andthereforex−cisafactorofa(x).■ Theorem1tellsusthatifcisarootofa(x),thenx −cisafactorofa(x)(andviceversa).Thisis easily extended: if c1 and c2 are two roots ofa(x), then x − c1 and x − c2 are two factors of a(x). Similarly, three roots give rise to three factors, four roots to four factors, and so on. This is stated conciselyinthenexttheorem. Theorem2Ifa(x)hasdistinctrootscl,…,cminF,then(x −c1)(x −c2) ⋯(x −cm)isafactorof a(x). PROOF:Toprovethis,letusfirstmakeasimpleobservation:ifapolynomiala(x)canbefactored, anyrootofa(x)mustbearootofoneofitsfactors.Indeed,ifa(x)=s(x)t(x)anda(c)=0,thens(c)t(c) =0,andthereforeeithers(c)=0ort(c)−0. Letcl,…,cmbedistinctrootsofa(x).ByTheorem1, a(x)=(x−c1)q1(x) Byourobservationintheprecedingparagraph,c2mustbearootofx−c1orofq1(x).Itcannotbearoot ofx−c1becausec2−c1≠0;soc2isarootofq1(x).Thus,q1(x)=(x−c2)q2(x),andtherefore a(x)=(x−c1)(x−c2)q2(x) Repeatingthisargumentforeachoftheremainingrootsgivesusourresult.■ Animmediateconsequenceisthefollowingimportantfact: Theorem3Ifa(x)hasdegreen,ithasatmostnroots. PROOF:Ifa(x)hadn+1rootsc1,…,cn+1,thenbyTheorem2,(x −c1) ⋯(x −cn + 1) would be a factorofa(x),andthedegreeofa(x)wouldthereforebeatleastx+1.■ Itwasstatedearlierinthischapterthatthedifferencebetweenpolynomialsandpolynomialfunctions ismainlyadifferenceofviewpoint.Mainly,butnotentirely!Rememberthattwopolynomialsa(x) and b(x)areequaliffcorrespondingcoefficientsareequal,whereastwofunctionsa(x)andb(x)areequaliff a(x)=b(x)foreveryxintheirdomain.Thesetwonotionsofequalitydonotalwayscoincide! Forexample,considerthefollowingtwopolynomialsin 5[x]: Youmaycheckthata(0)=b(0),a(1)=b(l),…,a(4)=b(4);hencea(x)andb(x)areequalfunctionsfrom 5to 5.Butaspolynomials,a(x)andb(x)arequitedistinct!(Theydonotevenhavethesamedegree.) ItisreassuringtoknowthatthiscannothappenwhenthefieldFisinfinite.Supposea(x) and b(x) arepolynomialsoverafieldFwhichhasinfinitelymanyelements.Ifa(x)andb(x)areequalasfunctions, thismeansthata(c)=b(c)foreveryc∈F.Definethepolynomiald(x)tobethedifferenceofa(x) and b(x): d(x) = a(x) − b(x). Then d(c) = 0 for everyc ∈ F. Now, if d(x) were not the zero polynomial, it wouldbeapolynomial(withsomefinitedegreen)havinginfinitelymanyroots,andbyTheorem3thisis impossible!Thus,d(x)isthezeropolynomial(allitscoefficientsareequaltozero),andthereforea(x)is thesamepolynomialasb(x).(Theyhavethesamecoefficients.) ThistellsusthatifFisafieldwithinfinitelymanyelements(suchas , ,or ),thereisnoneedto distinguishbetweenpolynomialsandpolynomialfunctions.Thedifferenceis,indeed,justadifferenceof viewpoint. POLYNOMIALSOVER AND In scientific computation a great many functions can be approximated by polynomials, usually polynomialswhosecoefficientsareintegersorrationalnumbers.Suchpolynomialsarethereforeofgreat practical interest. It is easy to find the rational roots of such polynomials, and to determine if a polynomialover isirreducibleover .Wewilldothesethingsnext. First,letusmakeanimportantobservation: Leta(x)beapolynomialwithrationalcoefficients,say Wemaynowfactoroutsfromallbutthefirsttermtoget Thepolynomialb(x)hasintegercoefficients;andsinceitdiffersfroma(x)onlybyaconstantfactor,it hasthesamerootsasa(x).Thus,foreverypolynomialwithrationalcoefficients,thereisapolynomial withintegercoefficientshavingthesameroots.Therefore,forthepresentwewillconfineourattention topolynomialswithintegercoefficients.Thenexttheoremmakesiteasytofindalltherationalrootsof suchpolynomials: Let s/t be a rational number in simplest form (that is, the integers s and t do not have a common factorgreaterthan1).Leta(x)=a0+⋯+anxnbeapolynomialwithintegercoefficients. Theorem4Ifs/tisarootofa(x),thens|a0andt|an. PROOF:Ifs/tisarootofa(x),thismeansthat a0+a1(s/t)+⋯+an(sn/tn)=0 Multiplyingbothsidesofthisequationbytnweget a0tn+a1stn−1+⋯+ansn=0 (1) Wemaynowfactoroutsfromallbutthefirsttermtoget −a0tn =s(altn−1+⋯+an sn−1 Thus,s/a0tn;andsincesandthavenocommonfactors,s|a0.Similarly,inEquation(1), we may factor outtfromallbutthelasttermtoget t(a0tn−1+⋯+an−1sn−l)=−ansn Thus,t\ansn;andsincesandthavenocommonfactors,t\an.■ AsanexampleofthewayTheorem4maybeused,letusfindtherationalrootsofa(x)=2x4+7x3+ 5x2+7x+3.Anyrationalrootmustbeafractionsitwheresisafactorof3andtisafactorof2.The possiblerootsaretherefore±1,±3,± and± .Testingeachofthesenumbersbydirectsubstitutioninto theequationa(x)=0,wefindthat− and−3areroots. Beforegoingtothenextstepinourdiscussionwenoteasimplebutfairlysurprisingfact. LemmaLeta(x)=b(x)c(x),wherea(x),b(x),andc(x)haveintegercoefficients.Ifaprimenumber pdivideseverycoefficientofa(x),iteitherdivideseverycoefficientofb(x)oreverycoefficientofc(x). PROOF:Ifthisisnotthecase,letbrbethefirstcoefficientofb(x)notdivisiblebyp,andletctbethe firstcoefficientofc(x)notdivisiblebyp.Now,a(x)=b(x)c(x),so ar+t=b0cr+t+⋯+brct+⋯+br+tc0 Eachtermontheright,exceptbrctisaproductbiCj whereeitheri>rorj>t.Byourchoiceofbrandct, ifi>rthenp | bi, and j > t then p | cj . Thus, p is a factor of every term on the right with the possible exceptionofbrcnbutpisalsoafactorofar+t.Thus,pmustbeafactorofbrcnhenceofeitherbrorcn andthisisimpossible.■ Wesaw(inthediscussionimmediatelyprecedingTheorem4)thatanypolynomiala{x)withrational coefficientshasaconstantmultipleka(x),withintegercoefficients,whichhasthesamerootsasa(x).We cangoonebetter;leta(x)∈ [x]: Theorem 5 Suppose a(x) can be factored as a(x) = b(x)c(x), where b(x) and c(x) have rational coefficients. Then there are polynomials B(x) and C(x) with integer coefficients, which are constant multiplesofb(x)andc(x),respectively,suchthata(x)=B(x)C(x). PROOF. Let k and l be integers such that kb(x) and lc(x) have integer coefficients. Then kla(x) = [kb(x)][lc(x)].Bythelemma,eachprimefactorofklmaynowbecanceledwithafactorofeitherkb(x)or lc(x).■ Remember that a polynomial a(x) of positive degree is said to be reducible over F if there are polynomialsb(x)andc(x)inF[x],bothofpositivedegree,suchthata(x)=b(x)c(x).Iftherearenosuch polynomials,thena(x)isirreducibleoverF. Ifweusethisterminology,Theorem5statesthatanypolynomialwithintegercoefficientswhichis reducibleover isreduciblealreadyover . InChapter25wesawthateverypolynomialcanbefactoredintoirreduciblepolynomials.Inorder tofactorapolynomialcompletely(thatis,intoirreducibles),wemustbeabletorecognizeanirreducible polynomial when we see one! This is not always an easy matter. But there is a method which works remarkablywellforrecognizingwhenapolynomialisirreducibleover : Theorem6:Eisenstein’sirreducibilitycriterionLet a(x)=a0+a1x+⋯+anxn be a polynomial with integer coefficients. Suppose there is a prime number p which divides every coefficientofa(x)excepttheleadingcofficientan;supposepdoesnotdivideanandp2doesnotdivide a0.Thena(x)isirreducibleover . PROOF:Ifa(x)canbefactoredover asa(x)=b(x)c(x),thenbyTheorem5 we may assume b(x) andc(x)haveintegercoefficients:say b(x)=b0+⋯+bk xk and c(x)=c0+⋯+cmxm Now,a0=b0c0;pdividesa0butp2doesnot,soonlyoneofb0,c0isdivisiblebyp.Sayp|c0andp ∤b0. Next,an=bk cmandp∤an,sop∤cm. Letsbethesmallestintegersuchthatp∤cs.Wehave a s=b 0cs+b1cs−1+⋯+bsc0 andbyourchoiceofcs,everytermontherightexceptb0csisdivisiblebyp.Butasalsoisdivisiblebyp, andthereforeb0csmustbedivisiblebyp.Thisisimpossiblebecausep∤b0andp ∤cs.Thus,a(x)cannot befactored.■ For example, x3 + 2x2 + 4x + 2 is irreducible over because p = 2 satisfies the conditions of Eisenstein’scriterion. POLYNOMIALSOVER AND One of the most far-reaching theorems of classical mathematics concerns polynomials with complex coefficients.Itissoimportantintheframeworkoftraditionalalgebrathatitiscalledthefundamental theoremofalgebra.Itstatesthefollowing: Everynonconstantpolynomialwithcomplexcoefficientshasacomplexroot. (Theproofofthistheoremisbasedupontechniquesofcalculusandcanbefoundinmostbookson complexanalysis.Itisomittedhere.) It follows immediately that the irreducible polynomials in [x] are exactly the polynomials of degree1.Forifa(x)isapolynomialofdegreegreaterthan1in [x],thenbythefundamentaltheoremof algebraithasarootcandthereforeafactorx−c. Now,everypolynomialin [x]canbefactoredintoirreducibles.Sincetheirreduciblepolynomials areallofdegree1,itfollowsthatifa(x)isapolynomialofdegreenover ,itcanbefactoredinto a(x)=k(x−c1)(x−c2)⋯(x−cn) Inparticular,ifa(x)hasdegreenithasn(notnecessarilydistinct)complexrootsc1…cn. Since every real number a is a complex number (a = a + 0i), what has just been stated applies equallytopolynomialswithrealcoefficients.Specifically,ifa(x)isapolynomialofdegreenwithreal coefficients, it can be factored into a(x) = k(x − c1) ⋯ (x − cn), where cl, …,cn are complex numbers (someofwhichmaybereal). Forourclosingcomments,weneedthefollowinglemma: LemmaSupposea(x)∈ [x].Ifa+biisarootofa(x),soisa−bi. PROOF:Rememberthata−biiscalledtheconjugateofa+bi.Ifrisanycomplexnumber,wewrite rforitsconjugate.Itiseasytoseethatthefunctionf(r)= isahomomorphismfrom to (infact,itis anisomorphism).Foreveryrealnumbera,f(a)=a.Thus,ifa(x)hasrealcoefficients,thenf(a0+a1r+ ⋯+an rn )=a0+a1 +⋯+an n .Sincef(0)=0,itfollowsthatifrisarootofa(x),sois .■ Nowleta(x)beanypolynomialwithrealcoefficients,andletr=a+bibeacomplexrootofa(x). Then isalsoarootofa(x),so (x−r)(x− )=x2−2ax+(a2+b2) andthisisaquadraticpolynomialwithrealcoefficients!Wehavethusshownthatanypolynomialwith realcoefficientscanbefactoredintopolynomialsofdegree1or1in [x].Inparticular,theirreducible polynomialsofthelinearpolynomialsand[x]arethelinearpolynomialsandtheirreduciblequadratics (thatis,theax2+bx+cwhereb2−4ac<0). EXERCISES A.FindingRootsofPolynomialsoverFiniteFields Inordertofindarootofa(x)inafinitefieldF,thesimplestmethod(ifFissmall)istotesteveryelement ofFbysubstitutionintotheequationa(x)=0. 1Findalltherootsofthefollowingpolynomialsin 5[x],andfactorthepolynomials: x3+x2+x+1; 3x4+x2+1; x5+1; x4+1; x4+4 #2UseFermat’stheoremtofindalltherootsofthefollowingpolynomialsin 7[x]: x100−1; 3x98+x19+3; 2x74−x55+2x+6 3 Using Fermat ’s theorem, find polynomials of degree ≤6 which determine the same functions as the followingpolynomialsin 7[x]: 3x75−5x54+2x13−x2; 4x108+6x101−2x81; 3x103−x73+3x55−x25 4Explainwhyeverypolynomialin p[x]hasthesamerootsasapolynomialofdegree<p. B.FindingRootsofPolynomialsover #1Findalltherationalrootsofthefollowingpolynomials,andfactorthemintoirreduciblepolynomials in [x]: 2Factoreachoftheprecedingpolynomialsin [x]andin [x]. 3Findassociateswithintegercoefficientsforeachofthefollowingpolynomials: 4Findalltherationalrootsofthepolynomialsinpart3andfactorthemover . 5Does2x4+3x2−2haveanyrationalroots?Canitbefactoredintotwopolynomialsoflowerdegreein [x]?Explain. C.ShortQuestionsRelatingtoRoots LetFbeafield. Provethatparts1−3aretrueinF[x]. 1Theremainderofp(x),whendividedbyx−c,isp(c). 2(x−c)|(p(x)−p(c)). 3Everypolynomialhasthesamerootsasanyofitsassociates. 4Ifa(x)andb(x)havethesamerootsinF,aretheynecessarilyassociates?Explain. 5Prove:Ifa(x)isamonicpolynomialofdegreen,anda(x)hasnrootsc1,…,cn∈F,thena(x)=(x − c1)⋯(x−cn). 6Supposea(x)andb(x)havedegree<n.Ifa(c)=b(c)fornvaluesofc,provethata(x)=b(x). 7Thereareinfinitelymanyirreduciblepolynomialsin 5[x]. #8Howmanyrootsdoesx2−xhavein 10?In 11?Explainthedifference. D.IrreduciblePolynomialsin [x]byEisenstein’sCriterion(andVariationsonthe Theme) 1Showthateachofthefollowingpolynomialsisirreducibleover : 2 It often happens that a polynomial a(y), as it stands, does not satisfy the conditions of Eisenstein’s criterion,butwithasimplechangeofvariabley=x+c,itdoes.Itisimportanttonotethatifa(x)canbe factored into p(x)q(x), then certainly a(x + c) can be factored into p(x + c)q(x + c). Thus, the irreducibilityofa(x+c)impliestheirreducibilityofa(x). (a) Use the change of variable y = x + l to show that x4 + 4x + l is irreducible in [x]. [In other words,test(x+l)4+4(x+1)+1byEisenstein’scriterion.] (b)Findanappropriatechangeofvariabletoprovethatthefollowingareirreduciblein [x]: x4+2x2−l; x3−3x+1; x4+1; x4−10x2+1 #3Provethatforanyprimep,xp −l+xp −2+ ⋯ + x + 1 is irreducible in [x].[HINT: By elementary algebra, (x−1)(xp−1+xp−2+⋯+x+1)=xp−1 hence Usethechangeofvariabley=x+1,andexpandbythebinomialtheorem. 4ByExerciseG3ofChapter25,thefunction h(a0+a1x+⋯+anxn)=an+an−1x+⋯+a0xn restricted to polynomials with nonzero constant term matches irreducible polynomials with irreducible polynomials.UsethisfacttostateadualversionofEisenstein’sirreducibilitycriterion. 5Usepart4toshowthateachofthefollowingpolynomialsisirreduciblein [x]: E.IrreducibilityofPolynomialsofDegree≤4 1LetFbeanyfield.Explainwhy,ifa(x)isaquadraticorcubicpolynomialinF[x],a(x)isirreduciblein F[x]iffa(x)hasnorootsinF. 2Provethatthefollowingpolynomialsareirreduciblein [x]: 3Supposeamonicpolynomiala(x)ofdegree4inF[x]hasnorootsinF.Thena(x)isreducibleiffitisa productoftwoquadraticsx2+ax+bandx2+cx+d,thatis,iff a(x)=x4+(a+c)x3+(ac+b+d)x2+(bc+ad)x+bd If the coefficients of a(x) cannot be so expressed (in terms of any a, b,c, d ∈ F) then a(x) must be irreducible. Examplea(x)=x4+2x3+x+1;thenbd=1,sob=d=±1;thus,bc+ad=±(a+c),buta+c=2 andbc+ad=1;whichisimpossible. Provethatthefollowingpolynomialsareirreduciblein [x](useTheorem5,searchingonlyforinteger valuesofa,b,c,d): x4−5x2+1; 3x4−x2−2; x4+x3+3x+1 4Provethatthefollowingpolynomialsareirreduciblein 5[x]: 2x3+x2+4x+l; x4+2; x4+4x2+2; x4+1 F.Mappingonto ntoDetermineIrreducibilityover Ifh: → nisthenaturalhomomorphism,let : [x]→ n[x]bedefinedby (a0+a1x+⋯+anxn)=h(a0)+h(al)x+⋯+h(an)xn InChapter24,ExerciseG,itisprovedthat isahomomorphism.Assumethisfactandprove: #1If (a(x))isirreduciblein n[x]anda(x)ismonic,thena(x)isirreduciblein [x]. 2x4+10x3+7isirreduciblein [x]byusingthenaturalhomomorphismfrom to 5. 3Thefollowingareirreduciblein [x](findtherightvalueofnandusethenaturalhomomorphismfrom to n): x4−10x2+1; x4+7x3+14x2+3; x5+1 G.RootsandFactorsinA[x]WhenAIsanIntegralDomain It is a useful fact that Theorems 1, 2, and 3 are still true in A[x] when A is not a field, but merely an integraldomain.TheproofofTheorem1mustbealteredabittoavoidusingthedivisionalgorithm.We proceedasfollows: Ifa(x)=a0+a1x+⋯+anxnandcisarootofa(x),consider a(x)−a(c)=a1(x−c)+a2(x2−c2)+⋯+an(xn−cn) 1Provethatfork=1,…,n: ak (xk −ck )=ak (x−c)(xk−l+xk−2c+⋯+ck−l) 2Concludefrompart1thata(x)−a(c)=(x−c)q(x)forsomeq(x). 3CompletetheproofofTheorem1,explainingwhythisparticularproofisvalidwhenA is an integral domain,notnecessarilyafield. 4CheckthatTheorems2and3aretrueinA[x]whenAisanintegraldomain. H.PolynomialInterpolation Oneofthemostimportantapplicationsofpolynomialsistoproblemswherewearegivenseveralvalues ofx(say,x=a0,a1,…,an)andcorrespondingvaluesofy(say,y=b0,b,…,bn),andweneedtofinda function y = f(x) such that f(a0) = b0, f(a1) = b1,…, f(an) = bn. The simplest and most useful kind of functionforthispurposeisapolynomialfunctionofthelowestpossibledegree. Wenowconsideracommonlyusedtechniqueforconstructingapolynomialp(x)ofdegreen which assumesgivenvaluesb0,b1,…,bnaregivenpointsa0,a1,…,an,.Thatis, p(a0)=b0,p(a1)=b,…,p(an)=bn First,foreachi=0,1,…,n,let qi(x)=(x−a0)⋯(x−ai−1)(x−ai+1)⋯(x−an) 1Showthatqi(aj )=0forj≠i,andqi(ai)≠0. Letqi(ai)=ci,anddefinep(x)asfollows: (ThisiscalledtheLagrangeinterpolationformula.) 2Explainwhyp(a0)=b0,p(ax)=,…,p(an)=bn. #3Provethatthereisoneandonlyonepolynomialp(x)ofdegree≤nsuchthatp(a0)=b0,…,p(an)=bn. 4UsetheLagrangeinterpolationformulatoprovethatifFisafinitefield,everyfunctionfromFtoFis equaltoapolynomialfunction.(Infact,thedegreeofthispolynomialislessthanthenumberofelements inF.) 5Ift(x)isanypolynomialinF[x],anda0,…,an∈F,theuniquepolynomialp(x)ofdegree≤nsuchthat p(a0)=t(a0),…,p(an)=t(an)iscalledtheLagrangeinterpolatorfort(x)anda0,…,an.Provethatthe remainder,whent(x)isdividedby(x−a0)(x−a1)⋯(x−an),istheLagrangeinterpolator. I.PolynomialFunctionsoveraFiniteField 1Findthreepolynomialsin 5[x]whichdeterminethesamefunctionas x2−x+1 2Provethatxp−xhasprootsin p[x],foranyprimep.Drawtheconclusionthatin p[x],xp −xcanbe factoredas xp−x=x(x−1)(x−2)⋯[x−(p−1)] 3Provethatifa(x)andb(x)determinethesamefunctionin p[x],then (xp−x)|(a(x)−b(x)) Inthenextfourparts,letFbeanyfinitefield. #4Leta(x)andb(x)beinF[x].Provethatifa(x)andb(x)determinethesamefunction,andifthenumber ofelementsinFexceedsthedegreeofa(x)aswellasthedegreeofb(x),thena(x)=b(x). 5Prove:Thesetofalla(x)whichdeterminethezerofunctionisanidealofF[x].Whatitsgenerator? 6Let (F)betheringofallfunctionsfromFtoF,definedinthesamewayas ( ).Leth:F[x]→ (F) send every polynomial a(x) to the polynomial function which it determines. Show that h is a homomorphismfromF[x]onto (F).(NOTE:Toshowthathisonto,useExerciseH4.) 7LetF={C1,…,cn}andp(x)=(x−c1)⋯(x−cn).Provethat F[x]/〈p(x)〉≅ (F) CHAPTER TWENTY-SEVEN EXTENSIONSOFFIELDS Inthefirst26chaptersofthisbookweintroducedthecastandsetthesceneonavastandcomplexstage. Nowitistimefortheactiontobegin.Wewillbesurprisedtodiscoverthatnoneofourefforthasbeen wasted; for every notion which was defined with such meticulous care, every subtlety, every fine distinctionwillhaveitsuseandplayitsprescribedroleinthestorywhichisabouttounfold. Wewillseemodernalgebrareachingoutandmergingwithotherdisciplinesofmathematics;wewill see its machinery put to use for solving a wide range of problems which, on the surface, have nothing whatever to do with modern algebra. Some of these problems—ancient problems of geometry, riddles about numbers, questions concerning the solutions of equations—reach back to the very beginnings of mathematics. Great masters of the art of mathematics puzzled over them in every age and left them unsolved,forthemachinerytosolvethemwasnotthere.Now,withalighttouchmodernalgebrauncovers theanswers. Modern algebra was not built in an ivory tower but was created part and parcel with the rest of mathematics—tied to it, drawing from it, and offering it solutions. Clearly it did not develop as methodicallyasithasbeenpresentedhere.Itwouldbepointless,inafirstcourseinabstractalgebra,to replicate all the currents and crosscurrents, all the hits and misses and false starts. Instead, we are provided with a finished product in which the agonies and efforts that went into creating it cannot be discerned. There is a disadvantage to this: without knowing the origin of a given concept, without knowingthespecificproblemswhichgaveitbirth,thestudentoftenwonderswhatitmeansandwhyit waseverinvented. Wehope,beginningnow,toshedlightonthatkindofquestion,tojustifywhatwehavealreadydone, and to demonstrate that the concepts introduced in earlier chapters are correctly designed for their intendedpurposes. Mostofclassicalmathematicsissetinaframeworkconsistingoffields,especially , ,and .The theoryofequationsdealswithpolynomialsover and ,calculusisconcernedwithfunctionsover ,and planegeometryissetin × .Itisnotsurprising,therefore,thatmoderneffortstogeneralizeandunify thesesubjectsshouldalsocenteraroundthestudyoffields.Itturnsoutthatagreatvarietyofproblems, ranging from geometry to practical computation, can be translated into the language of fields and formulated entirely in terms of the theory of fields. The study of fields will therefore be our central concernintheremainingchapters,thoughwewillseeotherthemesmergingandflowingintoitlikethe tributariesofagreatriver. If F is a field, then a subfield of F is any nonempty subset of F which is closed with respect to additionandsubtraction,multiplicationanddivision.(Itwouldbeequivalenttosay:closedwithrespect to addition and negatives, multiplication and multiplicative inverses.) As we already know, if K is a subfieldofF,thenKisafieldinitsownright. IfKisasubfieldofF,wesayalsothatFisanextensionfieldofK.Whenitisclearincontextthat bothFandKarefields,wesaysimplythatFisanextensionofK. GivenafieldF,wemaylookinwardfromFatallthesubfieldsofF. On the other hand, we may lookoutwardfromFatalltheextensionsofF.JustastherearerelationshipsbetweenFanditssubfields, there are also interesting relationships between F and its extensions. One of these relationships, as we shallseelater,ishighlyreminiscentofLagrange’stheorem—aninside-outversionofit. Whyshouldwebeinterestedinlookingattheextensionsoffields?Thereareseveralreasons,but one is very special. If F is an arbitrary field, there are, in general, polynomials over F which have no roots in F. For example, x2 + 1 has no roots in . This situation is unfortunate but, it turns out, not hopeless.For,asweshallsoonsee,everypolynomialoveranyfieldFhasroots.Iftheserootsarenot alreadyinF,theyareinasuitableextensionofF.Forexample,x2+1=0hassolutionsin . Inthematteroffactoringpolynomialsandextractingtheirroots, isutopia!In everypolynomial a(x)ofdegreenhasexactlynrootsc1,…,cnandcanthereforebefactoredasa(x)=k(x−c1)(x −c2) ⋯ (x−cn).Thisidealsituationisnotenjoyedbyallfields—farfromit!InanarbitraryfieldF,apolynomial of degree n may have any number of roots, from no roots to n roots, and there may be irreducible polynomials of any degree whatever. This is a messy situation, which does not hold the promise of an eleganttheoryofsolutionstopolynomialequations.However,itturnsoutthatF always has a suitable extensionEsuchthatanypolynomiala{x)ofdegreenoverF has exactly n solutions in E. Therefore, a(x)canbefactoredinE[x]as a(x)=k(x−c1)(x−c2)⋯(x−cn) Thus,paradiseisregainedbytheexpedientofenlargingthefieldF.Thisisoneofthestrongestreasons forourinterestinfieldextensions.Theywillgiveusatrimandeleganttheoryofsolutionstopolynomial equations. Now,letusgettowork!LetEbeafield,FasubfieldofE,andcany elementofE.Wedefinethesubstitutionfunctionσcasfollows: Foreverypolynomiala(x)inF[x], σc(a(x))=a(c) Thus, σc is the function “substitute c for x.” It is a function from F[x] into E. In fact, σc is a homomorphism.Thisistruebecause and Thekernelofthehomomorphismσcisthesetofallthepolynomialsa(x)suchthata(c)=σc(a(x))= 0.Thatis,thekernelofσcconsistsofallthepolynomialsa(x)inF[x]suchthatcisarootofa(x). LetJcdenotethekernelofac;sincethekernelofanyhomomorphismisanideal,Jcisanidealof F[x). Anelementc in E is called algebraic over F if it is the root of some nonzero polynomial a(x) in F[x]. Otherwise, c is called transcendental over F. Obviously c is algebraic over F iff Jc contains nonzeropolynomials,andtranscendentaloverFiffJc={0}. Wewillconfineourattentionnowtothecasewherecisalgebraic.Thetranscendentalcasewillbe examinedinExerciseGattheendofthischapter. Thus,letcbealgebraicoverF,andletJcbethekernelofσc(whereσcisthefunction“substitutec forx”).RememberthatinF[x]everyidealisaprincipalideal;henceJc=〈p(x)〉=thesetofallmultiples ofp(x),forsomepolynomialp(x).SinceeverypolynomialinJcisamultipleofp(x),p(x)isapolynomial of lowest degree among all the nonzero polynomials in Jc. It is easy to see that p(x) is irreducible; otherwisewecouldfactoritintopolynomialsoflowerdegree,sayp(x)=f(x)g(x).Butthen0=p(c) = f(c)g(c),sof(c)=0org(c)=0,andthereforeeitherf(x)org(x)isinJc.Thisisimpossible,becausewe havejustseenthatp{x)hasthelowestdegreeamongallthepolynomialsinJc,whereasf(x)andg(x)both havelowerdegreethanp(x). Sinceeveryconstantmultipleofp(x)isinJc,wemaytakep(x)tobemonic,thatis,tohaveleading coefficient1.Thenp(x)istheuniquemonicpolynomialoflowestdegreeinJc.(Also,itistheonlymonic irreduciblepolynomialinJc.)Thispolynomialp(x)iscalledtheminimumpolynomialofcoverF, and willbeofconsiderableimportanceinourdiscussionsinalaterchapter. Letuslookatanexample: isanextensionfieldof ,and containstheirrationalnumber .The function isthefunction“substitute forx”;forexample (x4−3x2+1)= −3 +1=−1. By our discussion above, : [x] → . is a homomorphism and its kernel consists of all the polynomialsin [x]whichhave asoneoftheirroots.Themonicpolynomialofleastdegreein [x] having asarootisp(x)=x2−2;hencex2−2istheminimumpolynomialof over . Now,letusturnourattentiontotherangeofσc.Sinceσcisahomomorphism,itsrangeisobviously closedwithrespecttoaddition,multiplication,andnegatives,butitisnotobviouslyclosedwithrespect tomultiplicativeinverses.Notobviously,butinfactitisclosedformultiplicativeinverses,whichisfar fromself-evident,andquitearemarkablefact.Inordertoprovethis,letf(c)beanynonzeroelementinthe rangeofσc.Sincef(c)≠0,f(x)isnotinthekernelofσc.Thus,f(x)isnotamultipleofp(x), and since p(x)isirreducible,itfollowsthatf(x)andp(x)arerelativelyprime.Thereforetherearepolynomialss(x) andt(x)suchthats(x)f(x)+t(x)p(x)=1.Butthen andtherefores(c)isthemultiplicativeinverseoff(c). WehavejustshownthattherangeofσcisasubfieldofE.Now,therangeofσcisthesetofallthe elementsa(c),foralla(x)inF[x]: Rangeσc={a(c):a(x)∈F[x]} Wehavejustseenthatrangeσcisafield.Infact,itisthesmallestfieldcontainingFandc:indeed,any otherfieldcontainingFandcwouldinevitablycontaineveryelementoftheform a0+a1c+⋯+ancn (a0,…,an∈F) inotherwords,wouldcontaineveryelementintherangeofσc. BythesmallestfieldcontainingFandcwemeanthefieldwhichcontainsFandcandiscontained inanyotherfieldcontainingFandc.ItiscalledthefieldgeneratedbyF and c, and is denoted by the importantsymbol F(c) Now,hereiswhatwehave,inanutshell:σcisahomomorphismwithdomainF[x],rangeF(c),and kernelJc=〈p(x)〉.Thus,bythefundamentalhomomorphismtheorem, Finally,hereisaninterestingsidelight:ifcanddarebothrootsofp(x),wherecanddareinE,then, bywhatwehavejustproved,F(c)andF(d)arebothisomorphictoF[x]/〈p(x)〉,andthereforeisomorphic toeachother: Ifcanddarerootsofthesameirreduciblepolynomialp(x)inF[x],thenF(c)≅F(d) Inparticular,thisshowsthat,givenFandc,F(c)isuniqueuptoisomorphism. Itistimenowtorecallourmainobjective:ifa(x)isapolynomialinF[x]whichhasnorootsinF, wewishtoenlargeFtoafieldEwhichcontainsarootofa(x).Howcanwemanagethis? An observation is in order: finding extensions of F is not as easy as finding subfields of F. A subfieldofFisasubsetofanexistingset:itistherelButanextensionofF is not yet there. We must somehowbuilditaroundF. Letp(x)beanirreduciblepolynomialinF[x].WehavejustseenthatifFcanbeenlargedtoafieldE containingarootcofp(x),thenF(c)isalreadywhatwearelookingfor:itisanextensionofFcontaining arootofp(x).Furthermore,F(c)isisomorphictoF[x]/〈p(x)〉.Thus,thefieldextensionwearesearching forispreciselyF[x]/〈p(x)〉.Ourresultissummarizedinthenexttheorem. Basic theorem of field extensions Let F be a field and a(x) a nonconstant polynomial in F[x]. ThereexistsanextensionfieldEofFandanelementcinEsuchthatcisarootofa(x). PROOF: To begin with, a(x) can be factored into irreducible polynomials in F[x]. If p(x) is any nonconstantirreduciblefactorofa(x),itisclearlysufficienttofindanextensionofFcontainingarootof p(x),sincesucharootwillalsobearootofa(x). In Exercise D4 of Chapter 25, the reader was asked to supply the simple proof that, if p(x) is irreducible in F[x], then 〈p(x)〉 is a maximal ideal of F[x]. Furthermore, by the argument at the end of Chapter19,if〈p(x)〉isamaximalidealofF[x],thenthequotientringF[x]/〈p(x)〉isafield. It remains only to prove that F[x]/ 〈p(x)〉 is the desired field extension of F. When we write J = 〈p(x)〉, let us remember that every element of F[x]/J is a coset of J. We will prove that F[x]/J is an extensionofFbyidentifyingeachelementainFwithitscosetJ+a. Tobeprecise,defineh:F→F[x]/Jbyh(a)=J+a.Notethathisthefunctionwhichmatchesevery ainFwithitscosetJ+ainF[x]/J.Wewillnowshowthathisanisomorphism. By the familiar rules of coset addition and multiplication, h is a homomorphism. Now, every homomorphism between fields is injective. (This is true because the kernel of a homomorphism is an ideal,andafieldhasnonontrivialideals.)Thus,hisanisomorphismbetweenitsdomainanditsrange. What is the range of h? It consists of all the cosets J + a where a ∈ F, that is, all the cosets of constantpolynomials.(IfaisinF,thenaisaconstantpolynomial.)Thus,Fisisomorphictothesubfield of F[x]/J containing all the cosets of constant polynomials. This subfield is therefore an isomorphic copyofF,whichmaybeidentifiedwithF,soF[x]/JisanextensionofF. Finally,ifp(x)=a0+a1x+⋯+anxn,letusshowthatthecosetJ+xisarootofp(x)inF[x]/J.Of course,inF[x]/J,thecoefficientsarenotactuallya0,a1,…,an,buttheircosetsJ+a0,J+a1,…,J+an. Writing J+a0=ā0,…,J+an=ān and J+x= wemustprovethat ā0+ā1 +⋯+ān n=J (Jisthezerocoset) Well, This completes the proof of the basic theorem of field extensions. Observe that we may use this theoremseveraltimesinsuccessiontogetthefollowing: Leta(x)beapolynomialofdegreeninF[x],Thereisanextensionfield∈ofFwhichcontainsall nrootsofa(x). EXERCISES A.RecognizingAlgebraicElements ExampleToshowthat that isarootofp(x). isalgebraicover ,onemustfindapolynomialp(x)∈ [x]such Leta= ;thena2=1+ ,a2−1= ,andfinally,(a2−l)2=2.Thus,asatisfiesp(x)=x4 −2x2−1=0. 1Provethateachofthefollowingnumbersisalgebraicover : (a) i (b) (c) 2+3i (d) #(e) (f) + (g) 2Provethateachofthefollowingnumbersisalgebraicoverthegivenfield: (a) over (π) (b) over (π2) (c) π2−1over (π3) NOTE:Recognizingatranscendentalelementismuchmoredifficult,sinceitrequiresprovingthatthe elementcannotbearootofanypolynomialoverthegivenfield.Inrecenttimesithasbeenproved,using sophisticatedmathematicalmachinery,thatπandearetranscendentalover . B.FindingtheMinimumPolynomial 1 Find the minimum polynomial of each of the following numbers over . (Where appropriate, use the methodsofChapter26,ExercisesD,E,andFtoensurethatyourpolynomialisirreducible.) (a) 1+2i (b) 1+ (c) 1+ #(d) (e) + (f) 2Showthattheminimumpolynomialof +iis (a) x2−2 x+3over (b) x4−2x2+9over (c) x2−2ix−3over (i) 3Findtheminimumpolynomialofthefollowingnumbersovertheindicatedfields: +i over ;over :over (i);over ( over ;over (i);over ( ) );over 4Foreachofthefollowingpolynomialsp(x),findanumberasuchthatp(x)istheminimumpolynomial ofaover : (a) x2+2x−1 (b) x4+2x2−1 (c) x4−10x2+1 5Findamonicirreduciblepolynomialp(x)suchthat [x]/〈p(x)〉isisomorphicto (a) ( ) (b) (1+ ) (c) C.TheStructureofFieldsF[x]/〈p(x)〉 Letp(x)beanirreduciblepolynomialofdegreenoverF.Letcdenotearootofp(x)insomeextensionof F(asinthebasictheoremonfieldextensions). 1Prove:EveryelementinF(c)canbewrittenasr(c),forsomer(x)ofdegree<ninF[x].[HINT:Given anyelementt(c)∈.F(c),usethedivisionalgorithmtodividet(x)byp(x).] 2Ifs(c)=t(c)inF(c),wheres(x)andt(x)havedegree<n,provethats(x)=t(x). 3Concludefromparts1and2thateveryelementinF(c)canbewrittenuniquelyasr(c),withdegr(x)< n. # 4 Using part 3, explain why there are exactly four elements in 2[x]/〈x2 + x + 1〉. List these four elements,andgivetheiradditionandmultiplicationtables.{HINT:Identify 2[x]/〈x2+x+1〉with 2(c), where c is a root of x2 + x + 1. Write the elements of 2(c) as in part 3. When computing the multiplicationtable,usethefactthatc2+c+1=0.} 5Describe 2[x]/〈x3+x+1〉,asinpart4. 6Describe 3[x]/〈x3+x2+2〉,asinpart4. D.ShortQuestionsRelatingofFieldExtensions LetFbeanyfield. Proveparts1–5: #1IfcisalgebraicoverF,soarec+1andkc(wherek∈F). 2Ifc≠0andcisalgebraicoverF,sois1/c. 3IfcdisalgebraicoverF,thencisalgebraicoverF(d).Ifc+disalgebraicoverF,thencisalgebraic overF(d)(Assumec≠0andd≠0.) 4IftheminimumpolynomialofaoverFisofdegree1,thena∈F,andconversely. 5SupposeF⊆Kanda∈K.Ifp(x)isamonicirreduciblepolynomialinF[x],andp(a)=0,thenp(x)is theminimumpolynomialofaoverF. 6Nameafield(≠ or )whichcontainsarootofx5+2x3+4x2+6. #7Prove: (1+i)≅ (1−i).However, ( )≅ ( ). 8Ifp(x)isirreducibleandhasdegree2,provethatF[x]/〈p(x)〉containsbothrootsofp(x). E.SimpleExtensions RecallthedefinitionofF(a).Itisafieldsuchthat(i)F⊆F(a);(ii)a∈F(a);(iii)anyfieldcontainingF andacontainsF(a). Usethisdefinitiontoproveparts1–5,whereF⊆K,c∈F,anda∈K: 1F(a)=F(a+c)andF(a)=F(ca).(Assumec≠0.) 2F(a2)⊆F(a)andF(a+b)⊆F(a,b).[F(a,b)isthefieldcontainingF,a,andb,andcontainedinany otherfieldcontainingF,aandb.]Whyarethereverseinclusionsnotnecessarilytrue? 3a+cisarootofp(x)iffaisarootofp(x+c);caisarootofp(x)iffaisarootofp(cx). 4Letp(x)beirreducible,andletabearootofp{x+c).Then F[x]/〈p(x+c)〉≅F(a) and F[x]/〈p(x)〉≅F(a+c) ConcludethatF[x]/〈p(x+c)〉≅F[x]/〈p(x)〉. 5Letp(x)beirreducible,andletabearootofp(cx).ThenF[x]/〈p(cx)〉≅F(a)andF[x]/〈p(x)〉≅F(ca). ConcludethatF[x]/〈p(cx)〉≅F[x]/〈p(x)〉. 6Useparts4and5toprovethefollowing: (a) 11[x]/〈x2+1〉≅ 11[x]/〈x2+x+4〉. (b) Ifaisarootofx2−2andbisarootofx2−4x+2,then (a)≅ (b). (c) Ifaisarootofx2−2andbisarootofx2− ,then (a)≅ (b). †F.QuadraticExtensions IftheminimumpolynomialofaoverFhasdegree2,wecallF(a)aquadraticextensionofF. 1Provethat,ifFisafieldwhosecharacteristicis≠2,anyquadraticextensionofFisoftheformF( forsomea∈F(HINT:Completethesquare,anduseExerciseE4.) ), LetFbeafinitefield,andF*themultiplicativegroupofnonzeroelementsofF.ObviouslyH={x2: x∈F*}isasubgroupofF*; since every square x2 in F* is the square of only two different elements, namely±x,exactlyhalftheelementsofF*areinH.Thus,Hhasexactlytwocosets:Hitself,containing allthesquares,andaH(wherea∉H),containingallthenonsquares.Ifaandbarenonsquares,thenby Chapter15,Theorem5(i), Thus:ifaandbarenonsquares,a/bisasquare.Usetheseremarksinthefollowing: 2LetFbeafinitefield.Ifa,b∈F,letp(x)=x2−aandq{x)=x2−bbeirreducibleinF[x],andlet and denoterootsofp(x)andq(x)inanextensionofF.Explainwhya/bisasquare,saya/b=c2for somec∈F.Provethat isarootofp(cx). 3Usepart2toprovethatF[x]/〈p(cx)〉≅F( );thenuseExerciseE5toconcludethatF( )≅F( ). 4Usepart3toprove:Anytwoquadraticextensionsofafinitefieldareisomorphic. 5Ifaandbarenonsquaresin ,a/bisasquare(why?).Usethesameargumentasinpart4toprovethat anytwosimpleextensionsof areisomorphic(henceisomorphicto ). G.QuestionsRelatingtoTranscendentalElements LetFbeafield,andletcbetranscendentaloverF.Provethefollowing: 1{a(c):a(x)∈F[x]}isanintegraldomainisomorphictoF[x]. #2F(c)isthefieldofquotientsof{a(c):a(x)∈F[x]},andisisomorphictoF(x),thefieldofquotients ofF[x]. 3IfcistranscendentaloverF,soarec+1,kc(wherek∈Fandk≠0),c2. 4IfcistranscendentaloverF,everyelementinF(c)butnotinFistranscendentaloverF. †H.CommonFactorsofTwoPolynomials:OverFandoverExtensionsofF LetFbeafield,andleta(x),b(x)∈F[x].Provethefollowing: 1Ifa{x)andb{x)haveacommonrootcinsomeextensionofF,theyhaveacommonfactorofpositive degreeinF[x].[Usethefactthata(x),b(x)∈kerσc.] 2Ifa(x)andb(x)arerelativelyprimeinF[x],theyarerelativelyprimeinK[x],foranyextensionKofF. Conversely,iftheyarerelativelyprimeinK[x],thentheyarerelativelyprimeinF[x]. †I.DerivativesandTheirProperties Leta(x)=a0+a1x+⋯+anxn∈F[x].Thederivativeofa(x)isthefollowingpolynomiala′(x)∈F[x]: a′(x)=a1+2a2x+…+nanxn−1 (Thisisthesameasthederivativeofapolynomialincalculus.)Wenowprovetheanalogsoftheformal rulesofdifferentiation,familiarfromcalculus. Leta(x),b(x)∈F[x],andletk∈F. Proveparts1–4: 1[a(x)+b(x)]′=a′(x)+b′(x) 2[a(x)b(x)]′=a′(x)b(x)+a(x)b′(x) 3[ka(x)]′=ka′(x) 4IfFhascharacteristic0anda′(x)=0,thena(x)isaconstantpolynomial.Whyisthisconclusionnot necessarilytrueifFhascharacteristicp≠0? 5Findthederivativeofthefollowingpolynomialsin 5[x]: x6+2x3+x+1 x5+3x2+1 x15+3x10+4x5+1 6IfFhascharacteristicp≠0,anda′(x)=0,provethattheonlynonzerotermsofa(x) are of the form ampxmpforsomem.[Thatis,a(x)isapolynomialinpowersofxp.] †J.MultipleRoots Supposea(x)≅F[x],andKisanextensionofF.Anelementc∈Kiscalledamultiplerootofa(x)if(x −c)m|a(x)forsomem>1.Itisoftenimportanttoknowifalltherootsofapolynomialaredifferent,or not.Wenowconsideramethodfordeterminingwhetheranarbitrarypolynomiala(x)≅F[x]hasmultiple rootsinanyextensionofF. LetKbeanyfieldcontainingalltherootsofa(x).Supposea(x)hasamultiplerootc. 1Provethata(x)=(x−c)2q(x)∈K[x]. 2Computea′(x),usingpart1. 3Showthatx −cisacommonfactorofa(x)anda′(x).Use Exercise hi to conclude that a(x) and a′(x) haveacommonfactorofdegree>1inF[x]. Thus, if a(x) has a multiple root, then a(x) and a′(x) have a common factor in F[x]. To prove the converse,supposea(x)hasnomultipleroots.Thena(x)canbefactoredasa(x) = (x − c1) ⋯ (x − cn) wherec1,…,cnarealldifferent. 4Explainwhya′(x)isasumoftermsoftheform (x−c1)⋯(x−ci−1)(x−ci+1)⋯(x−cn) 5Usingpart4,explainwhynoneoftherootsc1,…,cnofa(x)arerootsofa′(x). 6Concludethata(x)anda′(x)havenocommonfactorofdegree>1inF[x]. Thisimportantresultisstatedasfollows:Apolynomiala(x)inF[x]hasamultiplerootiffa(x)and a′(x)haveacommonfactorofdegree>1inF[x]. 7 Show that each of the following polynomials has no multiple roots in any extension of its field of coefficients: x3−7x2+8∈ [x] x2+x+1∈ 5[x] x100−1∈ 7[x] Theprecedingexampleismostinteresting:itshowsthatthereare100differenthundredthrootsof1 over 7. (The roots ±1 are in 7, while the remaining 98 roots are in extensions of 7.) Corresponding resultsholdformostotherfields. CHAPTER TWENTY-EIGHT VECTORSPACES Many physical quantities, such as length, area, weight, and temperature, are completely described by a single real number. On the other hand, many other quantities arising in scientific measurement and everyday reckoning are best described by a combination of several numbers. For example, a point in spaceisspecifiedbygivingitsthreecoordinateswithrespecttoanxyzcoordinatesystem. Here is an example of a different kind: A store handles 100 items; its monthly inventory is a sequenceof100numbers(a1,a2,…,a100)specifyingthequantitiesofeachofthe100itemscurrentlyin stock. Such a sequence of numbers is usually called a vector. When the store is restocked, a vector is addedtothecurrentinventoryvector.Attheendofagoodmonthofsales,avectorissubtracted. As this example shows, it is natural to add vectors by adding corresponding components, and subtractvectorsbysubtractingcorrespondingcomponents.Ifthestoremanagerintheprecedingexample decided to double inventory, each component of the inventory vector would be multiplied by 2. This showsthatanaturalwayofmultiplyingavectorbyarealnumberkistomultiplyeachcomponentbyk. Thiskindofmultiplicationiscommonlycalledscalarmultiplication. Historically,astheuseofvectorsbecamewidespreadandtheycametobeanindispensabletoolof science,vectoralgebragrewtobeoneofthemajorbranchesofmathematics.Todayitformsthebasisfor muchofadvancedcalculus,thetheoryandpracticeofdifferentialequations,statistics,andvastareasof appliedmathematics.Scientificcomputationisenormouslysimplifiedbyvectormethods;forexample,3, or300,or3000individualreadingsofscientificinstrumentscanbeexpressedasasinglevector. Inanybranchofmathematicsitiselegantanddesirable(butnotalwayspossible)tofindasimple listofaxiomsfromwhichalltherequiredtheoremsmaybeproved.Inthespecificcaseofvectoralgebra, wewishtoselectasaxiomsonlythoseparticularpropertiesofvectorswhichareabsolutelynecessary for proving further properties of vectors. And we must select a sufficiently complete list of axioms so that,byusingthemandthemalone,wecanproveallthepropertiesofvectorsneededinmathematics. Adelightfullysimplelistofaxiomsisavailableforvectoralgebra.Theremarkablefactaboutthis axiom system is that, although we conceive of vectors as finite sequences (a1, a2, …, an) of numbers, nothingintheaxiomsactuallyrequiresthemtobesuchsequences!Instead,vectorsaretreatedsimplyas elementsinaset,satisfyingcertainequations.Hereisourbasicdefinition: A vector space over a field F is a set V, with two operations + and · called vector addition and scalarmultiplication,suchthat 1.Vwithvectoradditionisanabeliangroup. 2.Foranyk∊Fanda∊V,thescalarproductkaisanelementofV,subjecttothefollowingconditions: forallk,l∊Fanda,b∊V, (a)k(a)+b)=ka+kb, (b)(k+l)a=ka+la, (c)k(la)=(kl)a, (d)1a=a. TheelementsofVarecalledvectorsandtheelementsofthefieldFarecalledscalars, InthefollowingexpositionthefieldFwillnotbespecificallyreferredtounlessthecontextrequires it.Fornotationalclarity,vectorswillbewritteninboldtypeandscalarsinitalics. Thetraditionalexampleofavectorspaceistheset nofalln-tuplesofrealnumbers,(a1, a2, …, an),withtheoperations (a1,a2,…,an)+(b1,b2,…,bn)=(a1+b1,a2+b2,…,an+bn) and k(a1,a2,…,an)=(ka1,ka2,…,kan) Forexample, 2isthesetofalltwo-dimensionalvectors(a,b),while 3isthesetofallvectors(a,b,c) ineuclideanspace.(Seethefigureonthenextpage.) However,thesearenottheonlyvectorspaces!Ourdefinitionofvectorspaceissoverysimplethat manyotherthings,quitedifferentinappearancefromthetraditionalvectorspaces,satisfytheconditions ofourdefinitionandaretherefore,legitimately,vectorspaces. Forexample, ,youmayrecall,isthesetofallfunctionsfrom to .Wedefinethesumf+g of twofunctionsbytherule [f+g](x)=f(x)+g(x) andwedefinetheproductaf,ofarealnumberaandafunctionf,by [af](x)=af(x) Itisveryeasytoverifythat ,withtheseoperations,satisfiesalltheconditionsneededinordertobea vectorspaceoverthefield . Asanotherexample,let denotethesetofallpolynomialswithrealcoefficients.Polynomialsare addedasusual,andscalarmultiplicationisdefinedby k(a0+a1x+⋯+anxn)=(ka0)+(ka1)x+⋯+(kan)xn Again,itisnothardtoseethat isavectorspaceover . LetVbeavectorspace.SinceVwithadditionaloneisanabeliangroup,thereisazeroelementinV calledthezerovector,writtenas0.EveryvectorainVhasanegative,writtenas −a.Finally,sinceV withvectoradditionisanabeliangroup,itsatisfiesthefollowingconditionswhicharetrueinallabelian groups: a+b=a+c a+b=0 implies a=−b −(a+b)=(−a)+(−b) implies b=c (1) b=−a (2) −(−a)=a (3) and and There are simple, obvious rules for multiplication by zero and by negative scalars. They are containedinthenexttheorem. Theorem1IfVisavectorspace,then: (i) 0a=0,foreverya∈V (ii) k0=0,foreveryscalark. (iii) Ifka=0,thenk=0ora=0. (iv) (−1)a=−aforeverya∈V. ToproveRule(i),weobservethat 0a=(0+0)a=0a+0a hence0+0a=0a+0a.ItfollowsbyCondition(1)that0=0a. Rule(ii)isprovedsimilarly.AsforRule(iii),ifk=0,wearedone.Ifk≠0,wemaymultiplyka=0 by1/ktogeta=0.Finally,forRule(iv),wehave a+(−1)a=1a+(−1)a=(1+(−1))a=0a=0 sobyCondition(2),(−1)a=−a. LetVbeavectorspace,andU⊆V.WesaythatUisclosedwithrespecttoscalarmultiplicationif ka∈Uforeveryscalarkandeverya∈U.WecallUasubspaceofV if U is closed with respect to additionandscalarmultiplication.ItiseasytoseethatifVisavectorspaceoverthefieldF,andUisa subspaceofV,thenUisavectorspaceoverthesamefieldF. Ifa1,a2,…,anareinVandk1,k2,…,knarescalars,thenthevector k1a1+k2a2+⋯+knan iscalledalinearcombinationofa1,a2,…,an.Thesetofallthelinearcombinationsofa1,a2,…,anisa subspaceofV.(Thisfactisexceedinglyeasytoverify.) IfUisthesubspaceconsistingofallthelinearcombinationsofa1,a2,…,an,wecallUthesubspace spannedbya1,a2,…,an.Anequivalentwayofsayingthesamethingisasfollows:aspace(orsubspace) Uisspannedbya1,a2,…,aniffeveryvectorinUisalinearcombinationofa1,a2,…,an. IfUisspannedbya1,a2,…,an,wealsosaythata1,a2,…,anspanU. LetS={a1,a2,…,an}beasetofdistinctvectorsinavectorspaceV.ThenSissaidtobelinearly dependentiftherearescalarsk1,…,kn,notallzero,suchthat k1a1+k2a2+⋯+knan=0 (4) Obviously this is the same as saying that at least one of the vectors in S is a linear combination of the remainingones.[Solveforanyvectorai,inEquation(4)havinganonzerocoefficient.] IfS={a1,a2,…,an}isnotlinearlydependent,thenitislinearlyindependent.Thatis,Sislinearly independentiff k1a1+k2a2+⋯+knan=0 implies k1=k2=⋯=kn=0 ThisisthesameassayingthatnovectorinSisequaltoalinearcombinationoftheothervectorsinS. It is obvious from these definitions that any set of vectors containing the zero vector is linearly dependent.Furthermore,theset{a},containingasinglenonzerovectora,islinearlyindependent. Thenexttwolemmas,althoughveryeasyandatfirstglancerathertrite,areusedtoprovethemost fundamentaltheoremsofthissubject. Lemma 1 If {a1, a2, …, an} is linearly dependent, then some ai, is a linear combination of the precedingones,a1,a2,…,ai−1. PROOF:Indeed,if{a1,a2,…,an}islinearlydependent,thenk1a1 + ⋯ + knan = 0 for coefficients k1,k2,…,knwhicharenotallzero.Ifkiisthelastnonzerocoefficientamongthem,thenk1ai+⋯+kiai=0, andthisequationcanbeusedtosolveforaiintermsofa1,⋯,ai−1.■ Let{a1a2,…, ,…,an}denotetheset{a1,a2,…,an}afterremovalofai. Lemma2If{a1,a2,…,an}spansV,andaiisalinearcombinationofprecedingvectors,then{a1, …, ,…,an}stillspansV. PROOF:Ourassumptionisthatai,=k1a1+ ⋯+ki−1ai−1 for some scalars k1, …, ki−1. Since every vectorb∈Visalinearcombination b=l1a1+⋯+liai+⋯+lnan itcanalsobewrittenasalinearcombination b=l1a1+⋯+li(k1a1+⋯+ki−1ai−1)+⋯lnan inwhichaidoesnotfigure.■ Asetofvectors{a1,…,an}inViscalledabasisofVifitislinearlyindependentandspansV. Forexample,thevectorsε1=(1,0,0),ε2=(0,1,0),andε3=(0,0,1)formabasisof( 3. They are linearly independent because, obviously, no vector in {ε1, ε2, ε3} is equal to a linear combination of precedingones.[Anylinearcombinationofε1andε2isoftheformaε1+bε2=(a,b,0),whereasε3isnot ofthisform;similarly,anylinearcombinationofε1aloneisoftheformaε1=(a,0,0),andε2isnotofthat form.]Thevectorsε1,ε2,ε3span 3becauseanyvector(a,b,c)in canbewrittenas(a,b,c)=aε1+bε2 +cε3. Actually,{ε1,ε2,ε3}isnottheonlybasisof 3.Anotherbasisof 3consistsofthevectors(1,2,3), (1,0,2),and(3,2,1);infact,thereareinfinitelymanydifferentbasesof 3.Nevertheless,allbasesof 3 haveonethingincommon:theycontainexactlythreevectors!Thisisaconsequenceofournexttheorem: Theorem2AnytwobasesofavectorspaceVhavethesamenumberofelements. PROOF:Suppose,onthecontrary,thatVhasabasisA={a1,…,an}andabasisB={b1, …, bm} wherem≠n.Tobespecific,supposen<m.Fromthisassumptionwewillderiveacontradiction. PutthevectorblinthesetA,soAnowcontains{b1,a1a2, …, an}. This set is linearly dependent becauseb1isalinearcombinationofa1,…,an.Butthen,byLemma1,someaiisalinearcombinationof precedingvectors.ByLemma2wemayexpelthisai,andtheremainingset{b1,al,…, ,…,an} still spansV. Repeatthisargumentasecondtimebyputtingb2inA,soAnowcontains{b2,b1,a1,a2, …, , …, an}.Thissetislinearlydependentbecause{b1,a1,…, , …, an} spans V and therefore b2 is a linear combinationofb1,a1,…, ,…,an.ByLemma1,someaj isalinearcombinationofprecedingvectors inA,sobyLemma2wemayremoveaj ,and{b2,b1,a1,a2,…, ,…, ,an}stillspansV. This argument is repeated n times. Each time, a vector from B is put into A and a vector ak is removed.Attheendofthenthrepetition,Acontainsonlyb1,…,bn,and{b1,…,bn}stillspansV.Butthis isimpossiblebecauseitimpliesthatbn+1isalinearcombinationofb1,…,bn,whereasinfact,B={bl, …,bn,…,bm}islinearlyindependent! ThiscontradictionprovesthatanytwobasesofVmustcontainthesamenumberofelements!■ If V has a basis {a1, …, an}, we call V a finite-dimensional vector space and say that V is of dimensionn.Inthatcase,byTheorem2everybasisofVhasexactlynelements. In the sequel we consider only finite-dimensional vector spaces. The next two lemmas are quite interesting.Thefirstonestatesthatif{a1…,am}spansV,thereisawayofremovingvectorsfromthis set,onebyone,untilweareleftwithanindependentsetwhichstillspansV. Lemma3Iftheset{a1,…,am}spansV,itcontainsabasisofV. PROOF:If{a1,…,am}isanindependentset,itisabasis,andwearedone.Ifnot,somea1isalinear combinationofprecedingones,so{a1,…, ,…,am}stillspansV.Repeatingthisprocess,wediscard vectorsonebyonefrom{a1…,am}and,eachtime,theremainingvectorsstillspanV.Wekeepdoingthis untiltheremainingsetisindependent.(Intheworstcase,thiswillhappenwhenonlyonevectorisleft.)■ Thenextlemmaassertsthatif{a1,…,as}isanindependentsetofvectorsinV,thereisawayof addingvectorstothissetsoastogetabasisofV. Lemma4Iftheset{a1,…,as}islinearlyindependent,itcanbeextendedtoabasisofV. PROOF:If{b1,…,bn}isanybasisofV,then{a1,…,as,b1,…,bn}spansV.BytheproofofLemma 3, we may discard vectors from this set until we get a basis of V. Note that we never discard any ai, because,byhypothesis,ai,isnotalinearcombinationofprecedingvectors.■ ThenexttheoremisanimmediateconsequenceofLemmas3and4. Theorem3LetVhavedimensionn.If{a1,…,an}isanindependentset,itisalreadyabasisof V.If{bl,…,bn}spansV,itisalreadyabasisofV. If{a1,…,an}isabasisofV,theneveryvectorcinVhasauniqueexpressionc=k1a1+ ⋯+knan asalinearcombinationofa1,…,an.Indeed,if c=k1a1+⋯+knan=l1a1+⋯+lnan then (k1−l1)a1+⋯+(kn−ln)an=0 hence k1−l1=⋯=kn−ln=0 sok1=l1,…kn=ln. If c=k1al+⋯+knan, the coefficientsk1,…,knarecalledthecoordinatesof cwithrespecttothebasis{al,…,an}.Itisthenconvenienttorepresentcasthen-tupie c=(k1,…,kn) IfUandVarevectorspacesoverafieldF,afunctionh:U→Visahomomorphismifitsatisfies thefollowingtwoconditions: h(a+b)=h(a)+h(b) and h(ka)=kh(a) Ahomomorphismofvectorspacesisalsocalledalineartransformation. Ifh:U→Visalineartransformation,itskernel[thatis,thesetofalla∈Usuchthath(a)=0]isa subspace of U, called the null space of h. Homomorphisms of vector spaces behave very much like homomorphismsofgroupsandrings.Theirpropertiesarepresentedintheexercises. EXERCISES A.ExamplesofVectorSpaces 1Provethat n,asdefinedonpage283,satisfiesalltheconditionsforbeingavectorspaceover . 2Provethat ( ),asdefinedonpage284,isavectorspaceover . 3Provethat ,asdefinedonpage284,isavectorspaceover . 4 Prove that 2( ), the set of all 2 × 2 matrices of real numbers, with matrix addition and the scalar multiplication isavectorspaceover . B.ExamplesofSubspaces #1 Provethat{(a,b,c):2a−3b+c=0}isasubspaceof 3. 2 Provethatthesetofall(x,y,z)∈ 3whichsatisfythepairofequationsax+by+c=0,dx+ey+f= 0isasubspaceof 3. 3 Provethat{f:f(l)=0}isasubspaceof ( ). 4 Provethat{f:fisaconstantontheinterval[0,1]}isasubspaceof ( ). 5 Provethatthesetofallevenfunctions[thatis,functionsfsuchthatf(x)=f(−x)]isasubspaceof ( ). Isthesametrueforthesetofalltheoddfunctions[thatis,functionsfsuchthatf(−x)=−f(x)]? 6 Provethatthesetofallpolynomialsofdegree≤nisasubspaceof C.ExamplesofLinearIndependenceandBases 1 Provethat{(0,0,0,1),(0,0,1,1),(0,1,1,1),(1,1,1,1)}isabasisof 4. 2 Ifa=(1,2,3,4)andb=(4,3,2,1),explainwhy{a,b}maybeextendedtoabasisof 4.Thenfinda basisof 4whichincludesaandb. 3 LetAbethesetofeightvectors(x,y,z)wherex,y,z=1,2.ProvethatAspans 3,andfindasubset ofAwhichisabasisof 3. 4 If isthesubspaceof consistingofallpolynomialsofdegree≤n,provethat{1,x,x2,…,xn}is abasisof .Thenfindanotherbasisof . 5 Findabasisforeachofthefollowingsubspacesof 3: #(a)S1={(x,y,z):3x−2y+z=0} (b)S2={(x,y,z):x+y−z=0and2x−y+z=0} 6 Findabasisforthesubspaceof 3spannedbythesetofvectors(x,y,z)suchthatx2+y2+z2=1. 7 LetUbethesubspaceof ( )spannedby{cos2x,sin2x,cos2x}.FindthedimensionofU,andthen findabasisofU. 8 Findabasisforthesubspaceof spannedby {x3+x2+x+1,x2+1,x3−x2+x−1,x2−1} D.PropertiesofSubspacesandBases Let V be a finite-dimensional vector space. Let dim V designate the dimension of V. Prove each of the following: 1 IfUisasubspaceofV,thendimU≤dimV. 2 IfUisasubspaceofV,anddimU=dimV,thenU=V. 3 Anysetofvectorscontaining0islinearlydependent. 4 Theset{a},containingonlyonenonzerovectora,islinearlyindependent. 5 Any subset of an independent set is independent. Any set of vectors containing a dependent set is dependent. #6 If{a,b,c}islinearlyindependent,sois{a+b,b+c,a+c}. 7 If{a1,…,an}isabasisofV,sois{k1a1,…,knan}foranynonzeroscalars 8 Thespacespannedby{a1,…,an}isthesameasthespacespannedby{b1,…,bm}iffeachai is a linearcombinationofb1,…,bm,andeachbj isalinearcombinationofa1…,an. E.PropertiesofLinearTransformations Let U and V be finite-dimensional vector spaces over a field F, and let h : U → V be a linear transformation.Proveparts1–3: 1 ThekernelofhisasubspaceofU.(Itiscalledthenullspaceofh.) 2 TherangeofhisasubspaceofV.(Itiscalledtherangespaceofh.) 3 hisinjectiveiffthenullspaceofhisequalto{0}. LetNbethenullspaceofh,and therangespaceofh.Let{a1,…,ar} be a basis of N. Extend it to a basis{al,…,ar,…,an}ofU. Proveparts4–6: 4 Everyvectorb∈ isalinearcombinationofh(ar+1),…,h(an). #5 {h(ar+1),…,h(an)}islinearlyindependent. 6 Thedimensionof isn−r. 7 Concludeasfollows:foranylineartransformationh,dim(domainh)=dim(nullspaceofh)+dim (rangespaceofh). 8 LetUandVhavethesamedimensionn.Usepart7toprovethathisinjectiveiffhissurjective. F.IsomorphismofVectorSpaces Let U and V be vector spaces over the field F, with dim U = n and dim V = m. Let h : U → V be a homomorphism. Provethefollowing: 1 Let h be injective. If {a1.. ., ar} is a linearly independent subset of U, then {h(a1), …, h(ar)} is a linearlyindependentsubsetofV. #2 hisinjectiveiffdimU=dimh(U). 3 SupposedimU=dimV;hisanisomorphism(thatis,abijectivehomomorphism)iffhisinjectiveiff hissurjective. 4 Anyn-dimensionalvectorspaceVoverFisisomorphictothespaceFnofalln-tupiesofelementsof F. †G.SumsofVectorSpaces LetTandUbesubspacesofV.ThesumofTandU,denotedbyT+U,isthesetofallvectorsa + b, wherea∈Tandb∈U. 1 ProvethatT+UandT∩UaresubspacesofV. VissaidtobethedirectsumofTandUifV=T+UandT∩U={0}.Inthatcase,wewriteV=T ⊕U. #2 Prove:V=T⊕Uiffeveryvectorc∈Vcanbewritten,inauniquemanner,asasumc=a + b wherea∈Uandb∈U. 3 LetTbeak-dimensionalsubspaceofann-dimensionalspaceV.Provethatan(n − k)-dimensional subspaceUexistssuchthatV=T⊕U. 4 IfTandUarearbitrarysubspacesofV,provethat dim(T+U)=dimT+dimU−dim(T∩U) CHAPTER TWENTY-NINE DEGREESOFFIELDEXTENSIONS In this chapter we will see how the machinery of vector spaces can be applied to the study of field extensions. LetFandKbefields.IfKisanextensionofF,wemayregardKasbeingavectorspaceoverF.We maytreattheelementsinKas“vectors” andtheelementsinFas“scalars.”Thatis,whenweaddelementsinK,wethinkofitasvectoraddition; whenweaddandmultiplyelementsinF,wethinkofthisasadditionandmultiplicationofscalars;and finally,whenwemultiplyanelementofFbyanelementofK,wethinkofitasscalarmultiplication. Wewillbeespeciallyinterestedinthecasewheretheresultingvectorspaceisoffinitedimension. IfK,asavectorspaceoverF,isoffinitedimension,wecallKafiniteextensionofF.Ifthedimension of the vector space K is n, we say that K is an extension of degree n over F. This is symbolized by writing [K:F]=n whichshouldberead,“thedegreeofKoverFisequalton.” Let us recall that F(c) denotes the smallest field which contains F and c. This means that F(c) containsFandc,andthatanyotherfieldKcontainingFandcmustcontainF(c).WesawinChapter 27 thatifcisalgebraicoverF,thenF(c)consistsofalltheelementsoftheforma(c),foralla(x) in F[x]. SinceF(c)isanextensionofF,wemayregarditasavectorspaceoverF.IsF(c)afiniteextensionofF? Well,letcbealgebraicoverF,andletp(x)betheminimumpolynomialofcoverF.[Thatis,p(x)is themonicpolynomialoflowestdegreehavingcasaroot.]Letthedegreeofthepolynomialp(x)beequal ton.Itturnsout,then,thatthenelements 1,c,c2,…,cn−1 are linearly independent and span F(c). We will prove this fact in a moment, but meanwhile let us recordwhatitmeans.Itmeansthatthesetofn“vectors”{1,c,c2,…,cn−1}isabasisofF(c);henceF(c) isavectorspaceofdimensionnoverthefieldF.Thismaybesummedupconciselyasfollows: Theorem1ThedegreeofF(c)overFisequaltothedegreeoftheminimumpolynomialofcover F. PROOF: It remains only to show that the n elements 1, c, …, cn − l span F(c) and are linearly independent.Well,ifa(c)isanyelementofF(c),usethedivisionalgorithmtodividea(x)byp(x): a(x)=p(x)q(x)+r(x) wheredegr(x)≤n−1 Therefore, ThisshowsthateveryelementofF(c)isoftheformr(c)wherer(x)hasdegreen−lorless.Thus,every elementofF(c)canbewrittenintheform a0+a1c+⋯+an−1cn−1 whichisalinearcombinationof1,c,c2,…,cn−1. Finally,toprovethat1,c,c2,…,cn−larelinearlyindependent,supposethata0+a1c+⋯+an−lcn − 1=0.Ifthecoefficientsa ,a ,…,a 0 1 n−1werenotallzero,cwouldbetherootofanonzeropolynomialof degree n − 1 or less, which is impossible because the minimum polynomial of c over F has degree n. Thus,a0=a1=⋯=an−1=0.■ Forexample,letuslookat ( ):thenumber isnotarootofanymonicpolynomialofdegree1 over .Forsuchapolynomialwould havetobe ,andthelatterisnotin [x]because isirrational.However, isarootofx2−2, whichisthereforetheminimumpolynomialof over ,andwhichhasdegree2.Thus, Inparticular,everyelementin ( )isthereforealinearcombinationof1and ,thatis,anumberof theform wherea,b∈ . Asanotherexample,iisarootoftheirreduciblepolynomialofoverx2+1in [x].Thereforex2+1 istheminimumpolynomialofiover x2+1hasdegree2,so[ (i): ]=2.Thus, (i)consistsofallthe linearcombinationsof1andiwithrealcoefficients,thatis,allthea+biwherea,b∈ .Clearlythen, (i)= ,sothedegreeof over isequalto2. Inthesequelwewilloftenencounterthefollowingsituation:EisafiniteextensionofK,whereKis afiniteextensionofF.Ifweknowthe degree of E over K and the degree of K over F, can we determine the degree of E over F This is a questionofmajorimportance!Fortunately,ithasaneasyanswer,basedonthefollowinglemma: LemmaLeta1a2,…,ambeabasisofthevectorspaceKoverF,andletb1,b2,…,bnbeabasisof thevectorspaceEoverK.Thenthesetofmnproducts{aibj }isabasisofthevectorspaceEoverthe fieldF. PROOF:Toprovethattheset{aibj }spansE,notethateachelementcinEcanbewrittenasalinear combination c = k1b1 + ⋯ + knbn with coefficients ki in K. But each k,i because it is in K, is a linear combination ki=li1a1+⋯+limam withcoefficientslij inF.Substituting, andthisisalinearcombinationoftheproductsaibj withcoefficientlij inF. Toprovethat{aibj }islinearlyindependent,suppose∑lij aibj =0.Thiscanbewrittenas (l11a1+⋯+l1mam)b1+⋯+(ln1a1+⋯+lnmam)bn=0 andsinceb1,…,bnareindependent,lila1+···+limam=0foreachi.Buta1,…,amarealsoindependent, soeverylij =0.■ Withthisresultwecannowconcludethefollowing: Theorem2SupposeF⊆K⊆EwhereEisafiniteextensionofK and K is a finite extension of F.Then∈isafiniteextensionofF,and [E:F]=[E:K][K:F] Thistheoremisapowerfultoolinourstudyoffields.Itplaysaroleinfieldtheoryanalogoustothe roleofLagrange’stheoremingrouptheory.Seewhatitsaysaboutanytwoextensions,KandEofafixed “basefield”F:IfKisasubfieldofE9thenthedegreeofK(overF)dividesthedegreeof∈(overF). IfcisalgebraicoverF,wesaythatF(c)isobtainedbyadjoiningctoF.Ifcandd are algebraic overF,wemayfindadjoinctoF,therebyobtainingF(c),andthenadjoindtoF(c).Theresultingfieldis denotedF(c,d)andisthesmallestfieldcontainingF,candd.[Indeed,anyfieldcontainingF, c and d must contain F(c), hence also F(c,d).] It does not matter whether we first adjoin c and then d or vice versa. Ifc1,…,cnarealgebraicoverF,weletF(c1,…,cn)bethesmallestfieldcontainingFandc1,…, cn. We call it the field obtained by adjoining c1,…,cn to F. We may form F(c1, …, cn) step by step, adjoiningoneciatatime,andtheorderofadjoiningtheciisirrelevant. AnextensionF(c)formedbyadjoiningasingleelementtoFiscalledasimpleextensionofF. An extension F(c1, …, cn)formed by adjoining a finite number of elements c1, …, cnis called an iterated extension.Itiscalled“iterated”becauseitcanbeformedstepbystep,onesimpleextensionatatime: F⊆F(c1)⊆F(c1,c2)⊆F(cl,c2,c3)⊆…⊆F(c1,…,cn (1) Ifc1,…,cnarealgebraicoverF,thenbyTheorem1,eachextensioninCondition(1)isafiniteextension. By Theorem 2, F(c1 c2) is a finite extension of F; applying Theorem 2 again, F(c1,c2,c3) is a finite extension of F; and so on. So finally, if c1, …, cn are algebraic over F, then F(c1, …, cn) is a finite extensionofF. Actually,theconverseistruetoo:everyfiniteextensionisaniteratedextension.Thisisobvious: forifKisafiniteextensionofF,sayanextensionofdegreen,thenKhasabasis{a1,…,an}overF.This meansthateveryelementinKisalinearcombinationofa1,…,anwithcoefficientsinF; but any field containingFanda1, …, an obviously contains all the linear combinations of a1, …, an; hence K is the smallestfieldcontainingFanda1,…,an.Thatis,K=F(a1,…,an). Infact,ifKisafiniteextensionofFandK=F(a1,…,an),thena1,…,anhavetobealgebraicover F.Thisisaconsequenceofasimplebutimportantlittletheorem: Theorem3IfKisafiniteextensionofF,everyelementofKisalgebraicoverF PROOF:Indeed,supposeKisofdegreen;overF,andletcbeanyelementofK.Thentheset{1,c,c2, …, cn) is linearly dependent, because it has n + 1 elements in a vector space K of dimension n. Consequently,therearescalarsa0,…,an∈F,notallzero,suchthata0+a1c+⋯+ancn=0.Thereforec isarootofthepolynomiala(x)=a0+a1x+⋯+anxninF[x].■ Let us sum up: Every iterated extension F(c1, …, cn), where c1, …, cn are algebraic over F, is finiteextensionofextensionofF.Conversely,everyfiniteextensionofFisaniteratedextensionF(c1, …,cn),wherec1,…,cnarealgebraicoverF. Hereisanexampleoftheconceptspresentedinthischapter.Wehavealreadyseenthat ( )isof degree2over ,andtherefore ( )consistsofallthenumbersa+b wherea,b∈ .Observethat cannotbein ( );forifitwere,wewouldhave =a+b forrationalaandb;squaringboth sidesandsolvingfor wouldgiveus =arationalnumber,whichisimpossible. Since is not in ( ), cannot be a root of a polynomial of degree 1 over ( ) (such a polynomial would have to be x − ). But is a root of x2 − 3, which is therefore the minimum polynomialof over ( ).Thus, ( , )isofdegree2over ( ),andthereforebyTheorem2, ( , )isofdegree4over . BythecommentsprecedingTheorem1,{1, }isabasisof ( )over ,and{1, }isabasisof ( , )over ( ).Thus,bythelemmaofthischapter,{1, , , }isabasisof ( , )over .Thismeansthat ( , )consistsofallthenumbersa+b +c +d ,forallab,c,anddin . Forlaterreference.Thetechnicalobservationwhichfollowswillbeneededlater. BythecommentsimmediatelyprecedingTheorem1,everyelementofF(c1)isalinearcombination ofpowersofc1,withcoefficientsinF.Thatis,everyelementofF(c1)isoftheform wherethekiareinF.Forthesamereason,everyelementofF(c1c2)isoftheform wherethecoefficientslj areinF(c1).Thus,eachcoefficientlj isequaltoasumoftheform(2).Butthen, clearingbrackets,itfollowsthateveryelementofF(c1c2)isoftheform wherethecoefficientskij areinF. Ifwecontinuethisprocess,itiseasytoseethateveryelementofF(c1,c2,…,cn)isasumoftermsof theform wherethecoefficientkofeachtermisinF. EXERCISES A.ExamplesofFiniteExtensions 1 Find a basis for (i ) over , and describe the elements of (i ).(See the two examples immediatelyfollowingTheorem1.) 2Showthateveryelementof (2+3i)canbewrittenasa+bi,wherea,b∈ Concludethat (2+3i)= . # 3 If , show that {1, 21/3, 22/3, a,21/3a, 22/3a} is a basis of (a) over . Describe the elementsof (a). #4Findabasisof ( )over ,anddescribetheelementsof . 5Findabasisof over anddescribetheelementsof ( ).(Seetheexampleatthe endofthischapter.) 6Findabasisof ( , , )over ,anddescribetheelementsof ( , , . 7Nameanextensionof overwhichπisalgebraicofdegree3. †B.FurtherExamplesofFiniteExtensions LetFbeafieldofcharacteristic≠2.Leta≠bbeinF. 1ProvethatanyfieldFcontaining + alsocontains and .[HINT:Compute( + )2 and showthat ∈F.Thencompute ( + ),whichisalsoinF]ConcludethatF( + )=F( , ). 2Provethatifb≠x2aforanyx∈F,then ∉F( ).ConcludethatF( , )isofdegree4overF. 3Showthatx= + satisfiesx4 −2(a+b)x2+(a − b)2 = 0. Show that x = also satisfiesthisequation.Concludethat 4Usingparts1to3,findanuncomplicatedbasisfor (d)over ,wheredisarootofx4 − 14x2 + 9. Thenfindabasisfor over . C.FiniteExtensionsofFiniteFields Bytheproofofthebasictheoremoffieldextensions,ifp(x)isanirreduciblepolynomialofdegreen in F[x],thenF[x]/〈p(x)〉≅F(c)wherecisarootofp(x).ByTheorem1inthischapter,F(c)isofdegreeη overF.UsingtheparagraphprecedingTheorem1: 1ProvethateveryelementofF(c)canbewrittenuniquelyasa0+a1c+⋯+an −lcn −1,forsomea0,…, an−1∈F. #2Constructafieldoffourelements.(Itistobeanextensionof 2.)Describeitselements,andsupply itsadditionandmultiplicationtables. 3Constructafieldofeightelements.(Itistobeanextensionof 2). 4ProvethatifFhasqelements,andaisalgebraicoverFofdegreen,thenF(a)hasqnelements. 5Provethatforeveryprimenumberp,thereisanirreduciblequadraticin p[x].Concludethatforevery primep,thereisafieldwithp2elements. D.DegreesofExtensions(ApplicationsofTheorem2) LetFbeafield,andKafieldextensionofF.Provethefollowing: 1[K:F]=1iffK=F. #2If[K:F]isaprimenumber,thereisnofieldproperlybetweenFandK(thatis,thereisnofieldL suchthatF L K). 3If[K:F]isaprime,thenK=F(a)foreverya∈K−F. 4Supposea,b∈KarealgebraicoverFwithdegreesmandη,wheremand«arerelativelyprime.Then: (a) F(a,b)isdegreemnoverF. (b) F(a) F(b)=F. 5IfthedegreeofF(a)overFisaprime,thenF(a)=F(an)foranyn(ontheconditionthatan∉F). 6Ifanirreduciblepolynomialp(x)∈F[x]hasarootinK,thendegp(x)|[K:F]. E.ShortQuestionsRelatingtoDegreesofExtensions LetFbeafield. Proveparts1−3: 1ThedegreeofaoverFisthesameasthedegreeof1/aoverF.Itisalsothesameasthedegreesofa+ candacoverF,foranyc∈F. 2aisofdegree1overFiffa∈F. 3Ifarealnumbercisarootofanirreduciblepolynomialofdegree>1in [x],thencisirrational. 4Usepart3andEisentein’sirreducibilitycriteriontoprovethat (wherera,mn∈ )isirrational ifthereisaprimenumberwhichdividesrabutnotn,andwhosesquaredoesnotdividera. 5Showthatpart4remainstruefor whereq>1. 6IfaandbarealgebraicoverF,provethatF(a,b)isafiniteextensionofF. †F.FurtherPropertiesofDegreesofExtensions LetFbeafield,andKafiniteextensionofF.Proveeachofthefollowing: 1AnyelementalgebraicoverKisalgebraicoverF,andconversely. 2IfbisalgebraicoverK,then[F(b:F]|[K(b):F]. 3IffeisalgebraicoverK,then[K(b):K]≤[F(b):F].(HINT:TheminimumpolynomialofboverFmay factorinK[x],and6willthenbearootofoneofitsirreduciblefactors.) #4IfbisalgebraicoverK,then[K(b):F(b)]≤[K:F].[HINT:NotethatF⊆K⊆K(b)andF⊆F(b)⊆ K(b).Relatethedegreesofthefourextensionsinvolvedhere,usingpart3.] #5Letp(x)beirreducibleinF[x].If[K:F]anddegp(x)arerelativelyprime,thenp(x)isirreduciblein K[x]. †G.FieldsofAlgebraicElements:AlgebraicNumbers LetF⊆Kanda,b∈K.Wehaveseenonpage295thatifaandbarealgebraicoverF,thenF(a,b)isa finiteextensionofF. Usetheabovetoproveparts1and2. 1IfaandbarealgebraicoverF,thena+b,a −b,ab,anda/barealgebraicoverF.(Inthelastcase, assumeb≠0.) 2Theset{x∈K:xisalgebraicoverF}isasubfieldofK,containingF. Anycomplexnumberwhichisalgebraicover iscalledanalgebraicnumber.Bypart2,thesetof allthealgebraicnumbersisafield,whichweshalldesignateby . Leta(x)=a0+a1x+⋯+anxnbein [x],andletcbeanyrootofa(x).Wewillprovethatc∈ .To beginwith,allthecoefficientsofa(x)arein (a0,a1,…,an). 3Prove: (a0,a1,…,an)isafiniteextensionof . Let (a0,…,an)= 4 1Sincea(x)∈ 1[x],cisalgebraicover 1Proveparts4and5: 1(c)isafiniteextensionof 1henceafiniteextensionof 5c∈ . .(Why?) Conclusion: The roots of any polynomial whose coefficients are algebraic numbers are themselves algebraicnumbers. AfieldFiscalledalgebraicallyclosediftherootsofeverypolynomialinF[x]areinF.Wehave thusprovedthat isalgebraicallyclosed. CHAPTER THIRTY RULERANDCOMPASS TheancientGreekgeometersconsideredthecircleandstraightlinetobethemostbasicofallgeometric figures,otherfiguresbeingmerelyvariantsandcombinationsofthesebasicones.Tounderstandthisview wemustrememberthatconstructionplayedaveryimportantroleinGreekgeometry:whenafigurewas defined, a method was also given for constructing it. Certainly the circle and the straight line are the easiestfigurestoconstruct,fortheyrequireonlythemostrudimentaryofallgeometricinstruments:the rulerandthecompass.Furthermore,theruler,inthiscase,isasimple,unmarkedstraightedge. Rudimentary as these instruments may be, they can be used to carry out a surprising variety of geometricconstructions.Linescanbedividedintoanynumberofequalsegments,andanyanglecanbe bisected.Fromanypolygonitispossibletoconstructasquarehavingthesamearea,ortwiceorthree times the area. With amazing ingenuity, Greek geometers devised ways to cleverly use the ruler and compass, unaided by any other instrument, to perform all kinds of intricate and beautiful constructions. Theyweresosuccessfulthatitwashardtobelievetheywereunabletoperformthreelittletaskswhich, atfirstsight,appeartobeverysimple:doublingthecube,trisectinganyangle,andsquaringthecircle. Thefirsttaskdemandsthatacubebeconstructedhavingtwicethevolumeofagivencube.Thesecond asksthatanyanglebedividedintothreeequalparts.Thethirdrequirestheconstructionofasquarewhose areaisequaltothatofagivencircle.Remember,onlyarulerandcompassaretobeused! Mathematicians,inGreekantiquityandthroughouttheRenaissance,devotedagreatdealofattention totheseproblems,andcameupwithmanybrilliantideas.Buttheyneverfoundwaysofperformingthe above three constructions. This is not surprising, for these constructions are impossible! Of course, the Greeks had no way of knowing that fact, for the mathematical machinery needed to prove that these constructions are impossible—in fact, the very notion that one could prove a construction to be impossible—wasstilltwomillenniaaway. The final resolution of these problems, by proving that the required constructions are impossible, camefromamostunlikelysource:itwasaby-productofthearcanestudyoffieldextensions,intheupper reachesofmodernalgebra. Tounderstandhowallthisworks,wewillseehowtheprocessofruler-and-compassconstructions canbeplacedintheframeworkoffieldtheory.Clearly,wewillbemakinguseofanalyticgeometry. If isanysetofpointsintheplane,consideroperationsofthefollowingtwokinds: 1. Ruleroperation:Throughanytwopointsin ,drawastraightline. 2. Compassoperation:GiventhreepointsA,B,andCin ,drawacirclewithcenterCandradiusequal inlengthtothesegmentAB. Thepointsofintersectionofanytwoofthesefigures(line-line,line-circle,orcircle-circle)aresaid tobeconstructibleinonestepfrom .ApointPiscalledconstructiblefrom iftherearepointsP1, P2, …, Pn = P such that P1 is constructible in one step from , P2 is constructible in one step from ,andsoon,sothatPiisconstructibleinonestepfrom . Asasimpleexample,letusseethatthemidpointofalinesegmentABisconstructiblefromthetwo pointsAandBintheabovesense.Well,givenAandB,firstdrawthelineAB.Then,drawthecirclewith center A and radius and the circle with center A and radius ; let C and D be the points of intersection of these circles. C and D are constructible in one step from {A, B}. Finally, draw the line throughCandD;theintersectionofthislinewithABistherequiredmidpoint.Itisconstructiblefrom{A, B}. Asthisexampleshows,thenotionofconstructiblepointsisthecorrectformalizationoftheintuitive ideaofruler-and-compassconstructions. Wecallapointintheplaneconstructibleifitisconstructiblefrom × ,thatis,fromthesetofall pointsintheplanewithrationalcoefficients. How does field theory fit into this scheme? Obviously by associating with every point its coordinates.Moreexactly,witheveryconstructiblepointPweassociateacertainfieldextensionof , obtainedasfollows: SupposePhascoordinates(a,b)andisconstructedfrom × inonestep.WeassociatewithPthe field (a,b),obtainedbyadjoiningto thecoordinatesofP.Moregenerally,supposePisconstructible from × innsteps:therearethennpointsP1,P2,…,Pn=PsuchthateachPiisconstructibleinone stepfrom × {P1,…,Pi−1}.LetthecoordinatesofP1,…,Pnbe(a1,b1),…,(an,bn),respectively. WiththepointsP1,…,PnweassociatefieldsK1,…,KnwhereK1= (a1,b1),andforeachi>1, Ki=Ki−1(ai,bi) Thus,K1= (a1,b1),K2=K1(a2,b2),andsoon:beginningwith ,weadjoinfirstthecoordinatesofP1, thenthecoordinatesofP2,andsoonsuccessively,yieldingthesequenceofextensions ⊆K1⊆K2⊆···⊆Kn=K WecallKthefieldextensionassociatedwiththepointP. Everythingwewillhavetosayinthesequelfollowseasilyfromthenextlemma. LemmaIfK1,…,Knareasdefinedpreviously,then[Ki:Ki−1]=1,2,or4. PROOF:RememberthatKi−1alreadycontainsthecoordinatesofP1,...,Pi−1,andKiisobtainedby adjoiningtoKi−1thecoordinatesxi,yiofPi.ButPiisconstructibleinonestepfrom × {P1,...,Pi −1},sowemustconsiderthreecases,correspondingtothethreekindsofintersectionwhichmayproduce Pi,namely:lineintersectsline,lineintersectscircle,andcircleintersectscircle. Lineintersectsline:Supposeonelinepassesthroughthepoints(a1,a2)and(b1,b2),andtheother linepassesthrough(c1,c2)and(d1,d2).Wemaywriteequationsfortheselinesintermsoftheconstants a1,a2,b1,b2,c1,c2andd1,d2(allofwhichareinKi−1),andthensolvetheseequationssimultaneouslyto give the coordinates x, y of the point of intersection. Clearly, these values of x and y are expressed in termsofa1,a2,b1,b2,c1,c2,d1,d2,hencearestillinKi−1.Thus,Ki,=Ki−1. Line intersects circle: Consider the line AB and the circle with center C and radius equal to the distance .LetA,B,Chavecoordinates(a1,a2),(b1,b2),and(c1,c2),respectively.Byhypothesis, Ki − 1 contains the numbers a1, a2, b1, b2, c1, c2, as well as k2 = the square of the distance . (To understandthelastassertion,rememberthatKi−1containsthecoordinatesofDandE;seethefigureand usethePythagoreantheorem.) Now,thelineABhasequation andthecirclehasequation (x−c1)2+(y−c2)2=k2 (2) Solvingforxin(1)andsubstitutinginto(2)gives This is obviously a quadratic equation, and its roots are the x coordinates of S and T. Thus, the x coordinatesofbothpointsofintersectionarerootsofaquadraticpolynomialwithcoefficientsinKi −1. The same is true of the y coordinates. Thus, if Ki = Ki − 1(xi, yi) where (xi, yi) is one of the points of intersection,then {Thisassumesthatxi,yi,∉Ki−1.Ifeitherxioryi,orbotharealreadyinKi−1,then[Ki−1,(xi,yi):Ki−1] =1or2.} Circleintersectscircle:Supposethetwocircleshaveequations x2+y2+ax+by+c=0 (3) x2+y2+dx+ey+f=0 (4) and Thenbothpointsofintersectionsatisfy (a−d)x+(b−e)y+(c−f)=0 (5) obtained simply by subtracting (4) from (3). Thus, x and y may be found by solving (4) and (5) simultaneously,whichisexactlytheprecedingcase.■ Wearenowinapositiontoprovethemainresultofthischapter: Theorem 1: Basic theorem on constructible points If the point with coordinates (a, b) is constructible,thenthedegreeof (a)over isapowerof2,andlikewiseforthedegreeof (b)over . PROOF:LetPbeaconstructiblepoint;bydefinition,therearepointsP1...,Pnwithcoordinates(a1, b1),…,(an,bn)suchthateachPiisconstructibleinonestepfrom × {P1,...,Pi−1},andPn=P.Let thefieldsassociatedwithP1,...,PnbeK1,...,Kn.Then [Kn: ]=[Kn:Kn−1][Kn−1:Kn−2]⋯[K1: ] andbytheprecedinglemmathisisapowerof2,say2m.But [Kn: ]=[Kn: (a)][ (a): ] hence[ (a): ]isafactorof2m,hencealsoapowerof2.■ We will now use this theorem to prove that ruler-and-compass constructions cannot possibly exist forthethreeclassicalproblemsdescribedintheopeningtothischapter. Theorem2“Doublingthecube”isimpossiblebyrulerandcompass. PROOF:Letusplacethecubeonacoordinatesystemsothatoneedgeofthecubecoincideswiththe unitintervalonthexaxis.Thatis,itsendpointsare(0,0)and(1,0).Ifwewereabletodoublethecubeby rulerandcompass,thismeanswecouldconstructapoint(c,0)suchthatc3=2.However,byTheorem1, [ (c): ]wouldhavetobeapowerof2,whereasinfactitisobviously3.Thiscontradictionproves thatitisimpossibletodoublethecubeusingonlyarulerandcompass.■ Theorem3“Trisectingtheangle”byrulerandcompassisimpossible.Thatis,thereexistangles whichcannotbetrisectedusingarulerandcompass. PROOF: We will show specifically that an angle of 60° cannot be trisected. If we could trisect an angleof60°,wewouldbeabletoconstructapoint(c,0)(seefigure)wherec=cos20°;hencecertainly wecouldconstruct(b,0)whereb=2cos20°. Butfromelementarytrigonometry cos3θ=4cos3θ−3cosθ hence Thus,b=2cos20°satisfiesb3−3b−1=0.Thepolynomial p(x)=x3−3x−1 isirreducibleover becausep(x+1)=x3+3x2 −3isirreduciblebyEisenstein’scriterion.Itfollows that (b)hasdegree3over ,contradictingtherequirement(inTheorem1)thatthisdegreehastobea powerof2.■ Theorem4“Squaringthecircle”byrulerandcompassisimpossible. PROOF.Ifwewereabletosquarethecirclebyrulerandcompass,itwouldbepossibletoconstruct thepoint ;hencebyTheorem1, wouldbeapowerof2.Butitiswellknownthatπ istranscendentalover .ByTheorem3ofChapter29,thesquareofanalgebraicelementisalgebraic; hence is transcendental. It follows that is not even a finite extension of , much less an extensionofsomedegree2masrequired.■ EXERCISES †A.ConstructibleNumbers IfOandIareanytwopointsintheplane,consideracoordinatesystemsuchthat theintervalOIcoincideswiththeunitintervalonthexaxis.Let bethesetofrealnumberssuchthat iffthepoint(a,0)isconstructiblefrom{O,I}.Provethefollowing: 1If ,then and . 2If ,then .(HINT:Usesimilartriangles.Seetheaccompanyingfigure.) 3If ,then .(Usethesamefigureasinpart2.) 4Ifa>0and ,then .(HINT:Intheaccompanyingfigure,ABisthediameterofacircle.Use anelementarypropertyofchordsofacircletoshowthat .) Itfollowsfromparts1to4that isafield,closedwithrespecttotakingsquarerootsofpositive numbers. iscalledthefieldofconstructiblenumbers. 5 . 6Ifaisarealrootofanyquadraticpolynomialwithcoefficientsin ,then .(HINT: Complete the squareandusepart4.) †B.ConstructiblePointsandConstructibleNumbers Proveeachofthefollowing: 1 Let be any set of points in the plane; (a, b) is constructible from iff (a, 0) and (0, b) are constructiblefrom . 2IfapointPisconstructiblefrom{O,I}[thatis,from(0,0)and(1,0)],thenPisconstructiblefrom × . #3Everypointin × isconstructiblefrom{O,I}.(UseExerciseA5andthedefinitionof .) 4IfapointPisconstructiblefrom × ,itisconstructiblefrom{O,I}. Bycombiningparts2and4,wegetthefollowingimportantfact:AnypointPisconstructiblefrom × iff P is constructible from {O,I}. Thus, we may define a point to be constructible iff it is constructiblefrom{O,I}. 5ApointPisconstructibleiffbothitscoordinatesareconstructiblenumbers. †C.ConstructibleAngles AnangleαiscalledconstructibleiffthereexistconstructiblepointsA,B,andC such that ∠LABC= α. Provethefollowing: 1Theangleαisconstructibleiffsinαandcosαareconstructiblenumbers. 2cos iffsin . 3Ifcosα,cos ,thencos(α+ß),cos . 4cos iffcos . 5Ifαandβareconstructibleangles,soare ,andnαforanypositiveintegern. #6Thefollowinganglesareconstructible: . 7Thefollowinganglesarenotconstructible:20°;40°,140°.(HINT:UsetheproofofTheorem3.) D.ConstructiblePolygons Apolygoniscalledconstructibleiffitsverticesareconstructiblepoints.Provethefollowing: #1Theregularn-gonisconstructibleifftheangle2π/nisconstructible. 2Theregularhexagonisconstructible. 3Theregularpolygonofninesidesisnotconstructible. †E.AConstructiblePolygon We will show that 2π/5 is a constructible angle, and it will follow that the regular pentagon is constructible. 1Ifr=cosk+isinkisacomplexnumber,provethat1/r=cosk−isink.Concludethatr+1/r=2cos k. BydeMoivre’stheorem, isacomplexfifthrootofunity.Since x5−1=(x−1)(x4+x3+x2+x+1) ωisarootofp(x)=x4+x3+x2+x+1. 2Provethatω2+ω+1+ω−1+ω−2=0. 3Provethat (HINT:Useparts1and2.)Concludethatcos(2π/5)isarootofthequadratic4x2−2x−1. 4Usepart3andA6toprovethatcos(2π/5)isaconstructiblenumber. 5Provethat2π/5isaconstructibleangle. 6Provethattheregularpentagonisconstructible. †F.ANonconstructiblePolygon BydeMoivre’stheorem, isacomplexseventhrootofunity.Since x7−1=(x−1)(x6+x5+x4+x3+x2+x+1) ωisarootofx6+x5+x4+x3+x2+x+1. 1Provethatω3+ω2+ω+1+ω−1+ω−2+ω−3=0. 2Provethat (Usepart1andExerciseEl.)Concludethatcos(2π/7)isarootof8x3+4x2−4x−1. 3Provethat8x3+4x2−4x−1hasnorationalroots.Concludethatitisirreducibleover . 4Concludefrompart3thatcos(2π/7)isnotaconstructiblenumber. 5Provethat2π/7isnotaconstructibleangle. 6Provethattheregularpolygonofsevensidesisnotconstructible. G.FurtherPropertiesofConstructibleNumbersandFigures Proveeachofthefollowing: 1Ifthenumberaisarootofanirreduciblepolynomialp(x)∈ [x]whosedegreeisnotapowerof2, thenaisnotaconstructiblenumber. #2 Any constructible number can be obtained from rational numbers by repeated addition, subtraction, multiplication,division,andtakingsquarerootsofpositivenumbers. 3 isthesmallestfieldextensionof closedwithrespecttosquarerootsofpositivenumbers(thatis, anyfieldextensionof closedwithrespecttosquarerootscontains ).(Usepart2andExerciseA.) 4Alltherootsofthepolynomialx4−3x2+1areconstructiblenumbers. A line is called constructible if it passes through two constructible points. A circle is called constructibleifitscenterandradiusareconstructible. 5Thelineax+by+c=0isconstructibleif . 6Thecirclex2+y2+ax+by+c=0isconstructibleif . CHAPTER THIRTY-ONE GALOISTHEORY:PREAMBLE Field extensions were used in Chapter 30 to settle some of the most puzzling questions of classical geometry.Nowtheywillbeusedtosolveaproblemequallyancientandimportant:theywillgiveusa definiteandeleganttheoryofsolutionsofpolynomialequations. Wewillbeconcernednotsomuchwithfinding solutions (which is a problem of computation) as withthenatureandpropertiesofthesesolutions.Asweshalldiscover,thesepropertiesturnouttodepend lessonthepolynomialsthemselvesthanonthefieldswhichcontaintheirsolutions.Thisfactshouldbe keptinmindifwewanttoclearlyunderstandthediscussionsinthischapterandChapter32.Wewillbe speakingoffieldextensions,butpolynomialswillalwaysbelurkinginthebackground.Everyextension will be generated by roots of a polynomial, and every theorem about these extensions will actually be sayingsomethingaboutthepolynomials. Letusquicklyreviewwhatwealreadyknowoffieldextensions,fillinginagaportwoaswego along. Let F be a field; an element a (in an extension of F) is algebraic over F if a is a root of some polynomialwithitscoefficientsinF.TheminimumpolynomialofaoverFisthemonicpolynomialof lowestdegreeinF[x]havingaasaroot;everyotherpolynomialinF[x]havingaasarootisamultipleof theminimumpolynomial. ThebasictheoremoffieldextensionstellsusthatanypolynomialofdegreeninF[x]hasexactlyn rootsinasuitableextensionofF.However,thisdoesnotnecessarilymeanndistinctroots.Forexample, in [x]thepolynomial(x −2)5hasfiverootsallequalto2.Suchrootsarecalledmultiple roots. It is perfectlyobviousthatwecancomeupwithpolynomialssuchas(x −2)5havingmultipleroots;butare thereanyirreduciblepolynomialswithmultipleroots?Certainlytheanswerisnotobvious.Hereitis: Theorem 1 If F has characteristic 0, irreducible polynomials over F can never have multiple roots. PROOF:Toprovethis,wemustdefinethederivativeofthepolynomiala(x)=a0+a1x+⋯+anxn.It is a′(x) = a1 + 2a2x + ⋯ + nanxn−l. As in elementary calculus, it is easily checked that for any two polynomialsf(x)andg(x), (f+g)′=f′+g′ and (fg)′=fg′+f′g Nowsupposea(x)isirreducibleinF[x]andhasamultiplerootc: then in a suitable extension we can factora(x)asa(x)=(x−c)2q(x),andthereforea′(x)=2(x−c)q(x)+(x −c)2q′(x).Sox −cisafactor ofa′(x),andthereforecisarootofa′(x).Letp(x)betheminimumpolynomialofcoverF;sincebotha(x) anda′(x)havecasaroot,theyarebothmultiplesofp(x). But a(x) is irreducible: its only nonconstant divisor is itself; so p(x) must be a(x). However, a(x) cannotdividea′(x)unlessa′(x)=0becausea′(x)isoflowerdegreethana(x).Soa′(x)=0andtherefore its coefficient nan is 0. Here is where characteristic 0 comes in: if nan = 0 then an = 0, and this is impossiblebecauseanistheleadingcoefficientofa(x).■ Intheremainingthreechapterswewillconfineourattentiontofieldsofcharacteristic0.Thus,by Theorem1,anyirreduciblepolynomialofdegreenhasndistinctroots. Letusmoveonwithourreview.LetEbeanextensionofF.WecallEafiniteextensionofFifE,as avectorspacewithscalarsinF,hasfinitedimension.Specifically,ifEhasdimensionn,wesaythatthe degreeofEoverFisequalton,andwesymbolizethisbywriting[E:F]=n.IfcisalgebraicoverF,the degreeofF(c)overFturnsouttobeequaltothedegreeofp(x),theminimumpolynomialofcoverF. F(c),obtainedbyadjoininganalgebraicelementctoF,iscalledasimpleextensionofF.F(cl,…, cn),obtainedbyadjoiningnalgebraicelementsinsuccessiontoF,iscalledaniteratedextensionofF. Any iterated extension of F is finite, and, conversely, any finite extension of F is an iterated extension F(c1,…,cn).Infact,evenmoreistrue;letFbeofcharacteristic0. Theorem2EveryfiniteextensionofFisasimpleextensionF(c). PROOF:Wealreadyknowthateveryfiniteextensionisaniteratedextension.Wewillnowshowthat anyextensionF(a,b)isequaltoF(c)forsomec.Usingthisresultseveraltimesinsuccessionyieldsour theorem.(Ateachstage,wereduceby1thenumberofelementsthatmustbeadjoinedtoFinordertoget thedesiredextension.) Well,givenF(a,b),letA(x)betheminimumpolynomialofaoverF,andletB(x)betheminimum polynomialofboverF.LetKdenoteanyextensionofFwhichcontainsalltherootsa1,…,anofA(x)as wellasalltherootsb1,…,bmofB(x).Leta1beaandletb1beb. LettbeanynonzeroelementofFsuchthat Crossmultiplyingandsettingc=a+tb,itfollowsthatc≠ai+tbj ,thatis, c−tbj ≠ai whileforevery;j≠1, foralli≠1andj≠1 Thus,bistheonlycommonrootofh(x)andB(x). Wewillprovethatb∈F(c),hencealsoa=c −tb∈F(c),andthereforeF(a,b)⊆F(c). But c ∈ F(a,b),soF(c)⊆F(a,b).ThusF(a,b)=F(c). So,itremainsonlytoprovethatb∈F(c).Letp(x)betheminimumpolynomialofboverF(c).Ifthe degreeofp(x)is1,thenp(x)isx −b,sob∈F(c),andwearedone.Letussupposep(x)≥2andgeta contradiction:observethath(x)andB(x)mustbothbemultiplesofp(x)becausebothhaveb as a root, andp(x)istheminimumpolynomialofb.Butifh(x)andB(x)haveacommonfactorofdegree≥2, they musthavetwoormorerootsincommon,contrarytothefactthatbistheironlycommonroot.Ourproofis complete.■ Forexample,wemayapplythistheoremdirectlyto ( , ).Takingt=1,wegetc= + , hence ( , )= ( + ). Ifa(x)isapolynomialofdegreeninF[x],letitsrootsbec1,…,cn.ThenF(c1,…,cn)isclearlythe smallestextensionofFcontainingalltherootsofa(x).F(c1,…,cn)iscalledtherootfieldofa(x)over F.Wewillhaveagreatdealtosayaboutrootfieldsinthisandsubsequentchapters. Isomorphisms were important when we were dealing with groups, and they are important also for fields. You will remember that if F1 and F2 are fields, an isomorphism from F1 to F2 is a bijective functionh:F1→F2satisfying h(a+b)=h(a)+h(b) and h(ab)=h(a)h(b) Fromtheseequationsitfollowsthath(0)=0,h(l)=1,h(−a)=−h(a),andh(a−l)=(h(a))−1 SupposeF1andF2arefields,andh:F1→F2isanisomorphism.LetK1andK2beextensionsofF1 andF2,andlet :K1→K2alsobeanisomorphism.Wecall andextensionof if (x)= (x)foreveryx inFlthatis,ifhand arethesameonF1.( isanextensionofhintheplainsensethatitisformedby “addingon”toh.) Asanexample,givenanyisomorphismh:F1→F2,wecanextendhtoanisomorphism : F1[x]→ F2[x].(NotethatF[x]isanextensionofFwhenwethinkoftheelementsofFasconstantpolynomials;of course,F[x]isnotafield,simplyanintegraldomain,butinthepresentexamplethisfactisunimportant.) Nowweask:Whatisanobviousandnaturalwayofextendingh?Theanswer,quiteclearly,istolet sendthepolynomialwithcoefficientsa0,a1,…,antothepolynomialwithcoefficientsh(a0),h(a1), …, h(an): (a0+a1x+⋯+anxn)=h(a0)+h(a1)x+⋯+h(an)xn It is child’s play to verify formally that is an isomorphism from F1[x] to F2[x]. In the sequel, the polynomial (a(x)), obtained in this fashion, will be denoted simply by ha(x). Because is an isomorphism,a(x)isirreducibleiffha(x)isirreducible. Averysimilarisomorphismextensionisgiveninthenexttheorem. Theorem3Leth:F1→F2beanisomorphism,andletp(x)beirreducibleinF1[x].Supposeaisa rootofp(x),andbarootofhp(x).Thenhcanbeextendedtoanisomorphism :F1,(a)→F2(b) Furthermore, (a)=b. PROOF:RememberthateveryelementofF1(a)isoftheform c0+c1a+⋯+cnan wherec0,…,cnareinF1,andeveryelementofF2(b)isoftheformd0+d1b+⋯+dnbnwhered0,…,dn areinF2.Imitatingwhatwedidsuccessfullyintheprecedingexample,welet sendtheexpressionwith coefficients c0,…,cn to the expression with coefficientsh(c0),…,h(cn): (c0+c1a+⋯+cnan)=h(c0)+h(c1)b+⋯+h(cn)bn Again,itisroutinetoverifythat isanisomorphism.DetailsarelaidoutinExerciseHattheendofthe chapter.■ MostoftenweyseTheorem3inthespecialcasewhereF1andF2arethesamefield—letuscallitF —andhistheidentityfunctionε:F→F.[Rememberthattheidentityfunctionisε(x)=x.]Whenweapply Theorem3totheidentityfunctionε:F→F,weget Theorem4Supposeaandbarerootsofthesameirreduciblepolynomialp(x)inF[x].Thenthere isanisomorphismg:F(a)→F(b)suchthatg(x)=xforeveryxinF,andg(a)=b. Nowletusconsiderthefollowingsituation:KandK′arefiniteextensionsofF,andKandK′havea commonextensionE.Ifh:K→K′isanisomorphismsuchthath(x)=xforeveryxinF,wesaythath fixesF.LetcbeanelementofK;ifhfixesF,andcisarootofsomepolynomiala(x)=a0+⋯+anxnin F[x],h(c)alsoisarootofa(x).Itiseasytoseewhy:thecoefficientsofa(x)areinFandaretherefore notchangedbyh.Soifa(c)=0,then Whatwehavejustshownmaybeexpressedasfollows: (*)Leta(x)beanypolynomialinF[x].AnyisomorphismwhichfixesFsendsrootsofa(x)toroots ofa(x). IfKhappenstobetherootfieldofa(x)overF,thesituationbecomesevenmoreinteresting.SayK= F(cl,c2,…,cn),wherecl,c2,…,cnaretherootsofa(x).Ifh:K→K′isanyisomorphismwhichfixesF, thenby(*),hpermutescl,c2,…,cn.Now,bythebriefdiscussionheaded“Forlaterreference”onpage 296,everyelementofF(c1,…,cn)isasumoftermsoftheform wherethecoefficientkisinF.BecausehfixesF,h(k)=k.Furthermore,c1, c2, …, cn are the roots of a(x),soby(*),theproduct istransformedbyhintoanotherproductofthesameform.Thus,h sendseveryelementofF(c1,c2,…,cn)toanotherelementofF(c1,c2,…,cn). Theabovecommentsaresummarizedinthenexttheorem. Theorem5LetKandK′befiniteextensionsofF.AssumeKistherootfieldofsomepolynomial overF.Ifh:K→K′isanisomorphismwhichfixesF,thenK=K′. PROOF:FromTheorem2,KandK′aresimpleextensionsofF,sayK=F(a)andK′=F(b).ThenE= F(a, b) is a common extension of K and K′. By the comments preceding this theorem, h maps every elementofKtoanelementofK′;henceK′⊆K.Sincethesameargumentmaybecarriedoutforh−1, we alsohaveK⊆K′.■ Theorem5isoftenusedintandemwiththefollowing(seethefigureonthenextpage): Theorem6LetLandL′befiniteextensionsofF.LetKbeanextensionofLsuchthatKisaroot fieldoverF.Anyisomorphismh:L→L′whichfixesFcanbeextendedtoanisomorphism :K→K. PROOF:FromTheorem2,KisasimpleextensionofL,sayK=L(c).NowwecanuseTheorem3to extendtheisomorphismh:L→L′toanisomorphism ByTheorem5appliedto ,K=K′.■ REMARK:ItfollowsfromthetheoremthatL′⊆,Ksinceranh⊆ran =K. For later reference. The following results, which are of a somewhat technical nature, will be needed later.Thefirstpresentsasurprisinglystrongpropertyofrootfields. Theorem7LetKbetherootfieldofsomepolynomialoverF.Forevery irreducible polynomial p(x)inF[x],ifp(x)hasonerootinK,thenp(x)musthaveallofitsrootsinK. PROOF:Indeed,supposep(x)hasarootainK,andletbbeanyotherrootofp(x).FromTheorem4, there is an isomorphism h : F(a)→ F(b) fixing F. But F(a) ⊆ K; so from Theorem 6 and the remark followingitF(b)⊆K;henceb∈K.■ Theorem8SupposeI⊆E⊆K,whereEisafiniteextensionofIandKisafiniteextensionofE.If KistherootfieldofsomepolynomialoverI,thenKisalsotherootfieldofsomepolynomialoverE. PROOF: Suppose K is a root field of some polynomial over I. Then K is a root field of the same polynomialoverE.■ EXERCISES A.ExamplesofRootFieldsover ExampleFindtherootfieldofa(x)=(x2−3)(x3−1)over . SOLUTIONThecomplexrootsofa(x)are± ,1, (−1± i),sotherootfieldis (± i)).Thesamefieldcanbewrittenmoresimplyas ( ,i). 1Showthat ( ,1, (−1± ,i)istherootfieldof(x2−2x−2)(x2+1)over . Comparingpart1withtheexample,wenotethatdifferentpolynomialsmayhavethesamerootfield. Thisistrueevenifthepolynomialsareirreducible. 2Provethatx2 −3andx2 −2x −2arebothirreducibleover .Thenfindtheirrootfieldsover and showtheyarethesame. 3Findtherootfieldofx4−2,firstover ,thenover . 4Explain: (i, )istherootfieldofx4−2x2+9over ,andistherootfieldofx2−2 x+3over ( ). 5Findirreduciblepolynomialsa(x)over ,andb(x) over (i), such that (i, ) is the root field of a(x)over ,andistherootfieldofb(x)over (i).Thendothesamefor ( , ). #6Whichofthefollowingextensionsarerootfieldsover ?Justifyyouranswer: (i); ( ); ( ), where istherealcuberootof2; (2+ ); (i+ ); (i, , ). B.ExamplesofRootFieldsover p ExampleFindtherootfieldofx2+1over 3. SOLUTIONBythebasictheoremoffieldextensions, whereuisarootofx2+1.In 3(u),x2+1=(x+u)(x−u),becauseu2+1=0.Since 3(u)contains±u, itistherootfieldofx2+1over 3.Notethat 3(u)hasnineelements,anditsadditionandmultiplication tablesareeasytoconstruct.(SeeChapter27,ExerciseC4). 1Showthat,inanyextensionof 3whichcontainsarootuof a(x)=x3+2x+1∈ 3[x] ithappensthatu+1andu+2aretheremainingtworootsofa(x).Usethisfacttofindtherootfieldofx3 +2x+1over 3.Listtheelementsoftherootfield. 2Findtherootfieldofx2+x+2over 3,andwriteitsadditionandmultiplicationtables. 3Findtherootfieldofx3+x2+1∈ 2[x]over 2.Writeitsadditionandmultiplicationtables. 4Findtherootfieldover 2ofx3+x+1⊆ 2[x].(CAUTION:Thiswillprovetobealittlemoredifficult thanpart3.) #5Findtherootfieldofx3+x2+x+2over 3.Findabasisforthisrootfieldover 3. C.ShortQuestionsRelatingtoRootField Proveeachofthefollowing 1Everyextensionofdegree2isarootfield. 2IfF⊆I⊆KandKisarootfieldofa(x)overF,thenKisarootfieldofa(x)overI. 3Therootfieldover ofanypolynomialin [x]is or . 4Ifcisacomplexrootofacubica(x)∈ [x],then (c)istherootfieldofa(x)over . #5Ifp(x)=x4+ax2+bisirreducibleinF[x],thenF[x]/〈p(x)〉istherootfieldofp(x)overF. 6IfK=F(a)andKistherootfieldofsomepolynomialoverF,thenKistherootfieldoftheminimum polynomialofaoverF. 7EveryrootfieldoverFistherootfieldofsomeirreduciblepolynomialoverF.(HINT:Usepart6and Theorem2.) 8Suppose[K:F]=n,whereKisarootfieldoverF.ThenKistherootfieldoverFofeveryirreducible polynomialofdegreeninF[x]havingarootinK. 9Ifa(x)isapolynomialofdegreeninF[x],andKistherootfieldofa(x)overF,then[K:F]dividesn! D.ReducingIteratedExtensionstoSimpleExtensions 1Findcsuchthat ( , )= (c).Dothesamefor ( , ) 2Letabearootofx3 −x+1,andbarootofx2 −2x −1.Findcsuchthat (a,b) = (c). (HINT: Use calculustoshowthatx3−x+1hasonerealandtwocomplexroots,andexplainwhynotwoofthesemay differbyarealnumber.) #3Findcsuchthat ( , , )= (c). 4Findanirreduciblepolynomialp(x)suchthat ( , )istherootfieldofp(x)over .(HINT: Use ExerciseC6.) 5Dothesameasinpart4for ( , , ). DeMoivre’stheoremprovidesanexplicitformulatowritethencomplexnthrootsof1.(SeeChapter16, ExerciseH.)BydeMoivre’sformula,thenthrootsofunityconsistofω=cos(2π/n)+isin(2π/n)andits firstnpowers,namely,1,ω,ω2,…,ωn−1Wecallωaprimitiventhrootofunity,becausealltheothernth rootsofunityarepowersofω.Clearly,everynthrootofunity(except1)isarootof †E.RootsofUnityandRadicalExtensions Thispolynomialisirreducibleifnisaprime(seeChapter26,ExerciseD3).Proveparts1−3,whereω denotesaprimitiventhrootofunity. 1 (ω)istherootfieldofxn−1over . 2Ifnisaprime,[ (ω): ]=n−1. 3Ifnisaprime,ωn−1isequaltoalinearcombinationof1,ω,…,ωn−2withrationalcoefficients. 4Find[ (ω): ],whereωisaprimitiventhrootofunity,forn=6,7,and8. 5Provethatforanyr∈{1,2,…,n−1}, ωrisannthrootofa.Concludethat , ω,…, ωn−1 arethencomplexnthrootsofa. 6Provethat (ω, )istherootfieldofxn−aover . 7Findthedegreeof (ω, )over ,whereωisaprimitivecuberootof1.Alsoshowthat (ω, )= ( ,iV3)(HINT:Computeω.) 8ProvethatifKistherootfieldofanypolynomialover ,andKcontainsannthrootofanynumbera, thenKcontainsallthenthrootsofunity. †F.SeparableandInseparablePolynomials Let F be a field. An irreducible polynomial p(x) in F[x] is said to be separable over F if it has no multiplerootsinanyextensionofF.Ifp(x)doeshaveamultiplerootinsomeextension,itisinseparable overF. 1ProvethatifFhascharacteristic0,everyirreduciblepolynomialinF[x]isseparable. Thus,forcharacteristic0,thereisnoquestionwhetheranirreduciblepolynomialisseparableornot. However,forcharacteristicp≠0,itisdifferent.Thiscaseistreatednext.Inthefollowingproblems,let Fbeafieldofcharacteristicp≠0. 2 If a′(x) = 0, prove that the only nonzero terms of a(x) are of the form ampxmp for some m. [In other words,a(x)isapolynomialinpowersofxp.] 3Provethatifanirreduciblepolynomiala(x)isinseparableoverF,thena(x)isapolynomialinpowers ofxp.(HINT:Usepart2,andreasonasintheproofofTheorem1.) 4UseChapter27,ExerciseJ(especiallytheconclusionfollowingJ6)toprovetheconverseofpart3. Thus,ifFisafieldofcharacteristicp≠0,anirreduciblepolynomiala(x)∈F[x]isinseparableiff a(x)isapolynomialinpowersofxp.Forfinitefields,wecansayevenmore: 5ProvethatifFisanyfieldofcharacteristicp≠0,theninF[x], (HINT:SeeChapter24,ExerciseD6.) 6 If F is a finite field of characteristic p ≠ 0, prove that, in F[x], every polynomial a(xp) is equal to [b(x)]pforsomeb(x).[HINT:Usepart5andthefactthatinafinitefieldofcharacteristicp,everyelement hasapthroot(seeChapter20,ExerciseF).] 7Useparts3and6toprove:Inanyfinitefield,everyirreduciblepolynomialisseparable. Thus,fieldsofcharacteristic0andfinitefieldssharethepropertythatirreduciblepolynomialshave nomultipleroots.Theonlyremainingcaseisthatofinfinitefieldswithfinitecharacteristic.Itistreated inthenextexerciseset. †G.MultipleRootsoverInfiniteFieldsofNonzeroCharacteristic If p[y]isthedomainofpolynomials(inthelettery)over p,letE= p(y)bethefieldofquotientsof p p [y].LetKdenotethesubfield p (y )of p (y). 1Explainwhy p(y)and p(yp)areinfinitefieldsofcharacteristicp. 2Provethata(x)=xp − yp has the factorization xp − yp = (x − y)p in E[x], but is irreducible in K[x]. Concludethatthereisanirreduciblepolynomiala(x)inK[x]witharootwhosemultiplicityisp. Thus,overaninfinitefieldofnonzerocharacteristic,anirreduciblepolynomialmayhavemultiple roots. Even these fields, however, have a remarkable property: all the roots of any irreducible polynomialhavethesamemultiplicity.Thedetailsfollow:LetFbeanyfield,p(x)irreducibleinF[x],a and b two distinct roots of p(x), and K the root field of p(x) over F. Let i: K → i(K) = K′ be the isomorphism of Theorem 4, and : K[x] → K′[x] the isomorphism described immediately preceding Theorem3. 3Provethat leavesp(x)fixed. 4Provethat ((x−a)m)=(x−b)m. 5Provethataandbhavethesamemultiplicity. †H.AnIsomorphismExtensionTheorem(ProofofTheorem3) LetF1,F2,h,p(x),a,b,and beasinthestatementofTheorem3.Toprovethat isanisomorphism,it mustfirstbeshownthatitisproperlydefined:thatis,ifc(a)=d(a)inF1(a),then (c(a))= (d(a)). 1 If c(a) = d(a), prove that c(x) −d(x) is a multiple of p(x). Deduce from this that hc(x) − hd(x) is a multipleofhp(x). #2Usepart1toprovethat (c(a))= (d(a)). 3Reversingthestepsoftheprecedingargument,showthat isinjective. 4Showthat issurjective. 5Showthat isahomomorphism. †I.UniquenessoftheRootField Leth:F1→F2beanisomorphism.Ifa(x)∈F1[x],letKlbetherootfieldofa(x)overF1,andK2theroot fieldofha(x)overF2. 1Prove:Ifp(x)isanirreduciblefactorofa(x),u∈K1isarootofp(x),andυ∈K2isarootofhp(x), thenF1(u)≅F2(υ). 2F1(u)=KliffF2(υ)=K2. #3Useparts1and2toformaninductiveproofthatK1≅K2. 4 Draw the following conclusion: The root field of a polynomial a(x) over a field F is unique up to isomorphism. †J.ExtendingIsomorphism In the following, let F be a subfield of . An injective homomorphism h: F→ is called a monomorphism;itisobviouslyanisomorphismF→h(F). 1 Let ω be a complex pth root of unity (where p is a prime), and let h: (ω)→ be a monomorphism fixing .Explainwhyhiscompletelydeterminedbythevalueofh(ω).Thenprovethatthereexistexactly p−1monomorphisms (ω)→ whichfix . #2Letp(x)beirreducibleinF[x],andcacomplexrootofp(x).Leth:F→ beamonomorphism.Ifdeg p(x)=n,provethatthereareexactlynmonomorphismsF(c)→ whichareextensionsofh. 3 Let F ⊆ K⊆ , with [K: F] = n. If h: F→ is a monomorphism, prove that there are exactly n monomorphismsK→ whichareextensionsofh. #4Prove:Theonlypossiblemonomorphismh: → ish(x)=x.Thus,anymonomorphismh: (a)→ necessarilyfixes . 5Prove:Thereareexactlythreemonomorphisms ( )→ ,andtheyaredeterminedbytheconditions: → ; → ω; → ω2,whereωisaprimitivecuberootofunity. K.NormalExtensions IfKistherootfieldofsomepolynomiala(x)overF,KisalsocalledanormalextensionofF.Thereare otherpossiblewaysofdefiningnormalextensions,whichareequivalenttotheabove.Weconsiderthe twomostcommononeshere:theyarepreciselythepropertiesexpressedintheorems7and6.LetKbea finiteextensionofF. 1 Suppose that for every irreducible polynomial p(x) in F[x], if p(x) has one root in K, then p(x) must haveallitsrootsinK.ProvethatKisanormalextensionofF. 2Supposethat,ifhisanyisomorphismwithdomainKwhichfixesF,thenh(K)⊆K.ProvethatK is a normalextensionofF. CHAPTER THIRTY-TWO GALOISTHEORY:THEHEARTOFTHEMATTER IfKisafieldandhisanisomorphismfromKtoK,wecallhanautomorphismofK (automorphism = “self-isomorphism”). WebeginthischapterbyrestatingTheorems5and6ofChapter31: LetKbetherootfieldofsomepolynomialoverF;supposea∈K: (i) AnyisomorphismwithdomainKwhichfixesFisanautomorphismofK. (ii) Ifaandbarerootsofanirreduciblepolynomialp(x)inF[x],thereisanautomorphismofKfixing Fandsendingatob. Rule(i)ismerelyarestatementofTheorem5ofChapter31,usingthenotionofautomorphism.Rule (ii) is a result of combining Theorem 4 of Chapter 31 [which asserts that there exists an F-fixing isomorphismfromL=F(a)toL′=F(b)]withTheorem6ofthesamechapter. LetKbetherootfieldofapolynomiala(x)inF[x],Ifc1,c2,…,cnaretherootsofa(x),thenK = F(c1,c2,…,cn),and,by(*)onpage316,anyautomorphismhofKwhichfixesFpermutesc1,c2,…,cn. Ontheotherhand,rememberthateveryelementainF(c1,c2,…,cn)isasumoftermsoftheform wherethecoefficientkofeachtermisinF.IfhisanautomorphismwhichfixesF,hdoesnotchangethe coefficients,soh(a)iscompletelydeterminedonceweknowh(c1),…,h(cn).Thus,everyautomorphism ofKfixingFiscompletelydeterminedbyapermutationoftherootsofa(x). Thisisveryimportant! WhatitmeansisthatwemayidentifytheautomorphismsofKwhichfixFwithpermutationsofthe rootsofa(x). Itmustbepointedoutherethat,justasthesymmetriesofgeometricfiguresdeterminetheirgeometric properties, so the symmetries of equations (that is, permutations of their roots) give us all the vital informationneededtoanalyzetheirsolutions.Thus,ifKistherootfieldofourpolynomiala(x)overF, wewillnowpayverycloseattentiontotheautomorphismsofKwhichfixF. To begin with, how many such automorphisms are there? The answer is a classic example of mathematicaleleganceandsimplicity. Theorem1LetKbetherootfieldofsomepolynomialoverF.ThenumberofautomorphismsofK fixingFisequaltothedegreeofKoverF. PROOF:Let[K:F]=n,andletusshowthatKhasexactlynautomorphismsfixingF.ByTheorem2 ofChapter31,K=F(a)forsomea∈K.Letp(x)betheminimumpolynomialofaoverF;ifbisanyroot ofp(x),thenby(ii)onthepreviouspage,thereisanautomorphismofKfixingFandsendingatob.Since p(x)hasnroots,thereareexactlynchoicesofb,andthereforenautomorphismsofKfixingF. [RememberthateveryautomorphismhwhichfixesFpermutestherootsofp(x)andthereforesends atosomerootofp(x);andhiscompletelydeterminedoncewehavechosenh(a).]■ Forexample,wehavealreadyseenthat ( )isofdegree2over . ( )istherootfieldofx2– 2over because ( )containsbothrootsofx2–2,namely± .ByTheorem1,thereareexactlytwo automorphismsof ( )fixing :onesends to ;itistheidentityfunction.Theothersends to – ,andisthereforethefunctiona+b →a−b . Similarly,wesawthat = (i),and isofdegree2over .Thetwoautomorphismsof whichfix are the identity function and the function a + bi → a − bi which sends every complex number to its complexconjugate. Asafinalexample,wehaveseenthat ( , )isanextensionofdegree4over ,sobyTheorem 1,therearefourautomorphismsof ( , )whichfix :Now, ( , )istherootfieldof(x2–2) (x2−3)over foritcontainstherootsofthispolynomial,andanyextensionof containingtherootsof (x2–2)(x2−3)certainlycontains and .Thus,by(*)onpage316,eachofthefourautomorphisms whichfix sendsrootsofx2–2torootsofx2 −2,androotsofx2 −3torootsofx2–3.Butthereare onlyfourpossiblewaysofdoingthis,namely, Sinceeveryelementof ( , )isoftheform shallcallthemε,α,β,andγ)arethefollowing: ,thesefourautomorphisms(we If K is an extension of F, the automorphisms of K which fix F form a group. (The operation, of course,iscomposition.)Thisisperfectlyobvious:forifgandhfixF,thenforeveryxinF, thatis,g∘hfixesF.Furthermore,if thatis,ifhfixesFsodoesh–1 Thisfactisperfectlyobvious,butnonethelessofgreatimportance,foritmeansthatwecannowuse allofouraccumulatedknowledgeaboutgroupstohelpusanalyzethesolutionsofpolynomialequations. AndthatispreciselywhatGaloistheoryisallabout. IfKistherootfieldofapolynomiala(x)inF[x],thegroupofalltheautomorphismsofKwhichfix FiscalledtheGaloisgroupofa(x).WealsocallittheGaloisgroupofKoverF,anddesignateitbythe symbol Gal(K:F) Inourlastexamplewesawthattherearefourautomorphismsof ( , ε,α,β,andγ.Thus,theGaloisgroupof ( , )over isGal( ( operationiscomposition,givingusthetable )whichfix .Wecalledthem , ) : ) = {ε, α, β, γ}; the Asonecansee,thisisanabeliangroupinwhicheveryelementisitsowninverse;almostataglanceone canverifythatitisisomorphicto 2× 2. LetKbetherootfieldofa(x),wherea(x) is in F[x]. In our earlier discussion we saw that every automorphismofKfixingF[thatis,everymemberoftheGaloisgroupofa(x)]maybeidentifiedwitha permutationoftherootsofa(x).However,itisimportanttonotethatnoteverypermutationoftheroots ofa(x)needbeintheGaloisgroupofa(x),evenwhena(x)isirreducible.Forexample,wesawthat ( , )= ( + ),where + isarootoftheirreduciblepolynomialx4–10x2+1over . Sincex4–10x2+1hasfourroots,thereare4!=24permutationsofitsroots,onlyfourofwhichareinits Galoisgroup.Thisisbecauseonlyfourofthepermutationsaregenuinesymmetriesofx4–10x2+1,in thesensethattheydetermineautomorphismsoftherootfield. Inthediscussionthroughouttheremainderofthischapter,letFandKremainfixed.Fisanarbitrary fieldandKistherootfieldofsomepolynomiala(x)inF[x].Thethreadofourreasoningwillleadusto speakaboutfieldsIwhereF⊆I⊆K,thatis,fields“between”FandK.Wewill refertothemasintermediatefields.SinceKistherootfieldofa(x)overF,itisalsotherootfieldof a(x)overIforeveryintermediatefieldI. TheletterGwilldenotetheGaloisgroupofKoverF.WitheachintermediatefieldI,weassociate thegroup I*=Gal(K:I) thatis,thegroupofalltheautomorphismsofKwhichfixI.ItisobviouslyasubgroupofG.Wewillcall I*thefixerofI. Conversely, with each subgroup H of G we associate the subfield of K containing all the a in K whicharenotchangedbyanyπ∈H.Thatis, {a∈K:π(a)=aforeveryπ∈H} One verifies in a trice that this is a subfield of K. It obviously contains F, and is therefore one of the intermediatefields.ItiscalledthefixedfieldofH.ForbrevityandeuphonywecallitthefixfieldofH. Let us recapitulate: Every subgroup H of G fixes an intermediate field I, called the fixfield of H. EveryintermediatefieldIisfixedbyasubgroupHofG,calledthefixerofI.Thissuggestsverystrongly thatthereisaone-to-onecorrespondencebetweenthesubgroupsofGandthefieldsintermediatebetween F and K. Indeed, this is correct. This one-to-one correspondence is at the very heart of Galois theory, becauseitprovidesthetie-inbetweenpropertiesoffieldextensionsandpropertiesofsubgroups. Just as, in Chapter 29, we were able to use vector algebra to prove new things about field extensions, now we will be able to use group theory to explore field extensions. The vector-space connectionwasarelativelightweight.Theconnectionwithgrouptheory,ontheotherhand,givesusatool oftremendouspowertostudyfieldextensions. WehavenotyetprovedthattheconnectionbetweensubgroupsofGandintermediatefieldsisaoneto-onecorrespondence.Thenexttwotheoremswilldothat. Theorem2IfHisthefixerofI,thenIisthefixfieldofH. PROOF:LetHbethefixerofI,andI′bethefixfieldofH.Itfollowsfromthedefinitionsoffixerand fixfieldthatI⊆I′,sowemustnowshowthatI′⊆I.Wewilldothisbyprovingthata∉Iimpliesa∉I′. Well,ifaisanelementofKwhichisnotinI,theminimumpolynomialp(x)ofaoverImusthavedegree ≥2 (for otherwise, a ∈ I). Thus, p(x) has another root b. By Rule (ii) given at the beginning of this chapter,thereisanautomorphismofKfixingIandsendingatob.Thisautomorphismmovesa,soa∉I′. ■ LemmaLetHbeasubgroupofG,andIthefixfieldofH.ThenumberofelementsinHisequalto [K:I]. PROOF: Let H have r elements, namely, h1, …, hr. Let K = I(a). Much of our proof will revolve aroundthefollowingpolynomial: b(x)=[x–h1(a)][x−h2(a)]⋯[x−hr(a)] Sinceoneofthehiistheidentityfunction,onefactorofb(x)is(x−a),andthereforeaisarootofb(x). Inthenextparagraphwewillseethatallthecoefficientsofb(x)areinI,sob(x)∈I[x].Itfollowsthat b(x)isamultipleoftheminimumpolynomialofaoverI,whosedegreeisexactly[K:I].Sinceb(x)isof degreer,thismeansthatr≥[K:I],whichishalfourtheorem. Well, let us show that all the coefficients of b(x) are in I. We saw on page 314 that every isomorphism hi : K → K can be extended to an isomorphism i, : K[x] → K[x]. Because i is an isomorphismofpolynomials,weget Buthi∘h1,hi ∘h2,…,hi ∘hrarerdistinctelementsofH,andHhasexactlyrelements,sotheyareall theelementsofH(thatis,theyareh1,…,hr,possiblyinadifferentorder).Sothefactorsof i(b(x))are the same as the factors of b(x), merely in a different order, and therefore i(b(x)) = b(x). Since equal polynomialshaveequalcoefficients,hileavesthecoefficientsofb(x)invariant.Thus,everycoefficientof b(x)isinthefixfieldofH,thatis,inI. Wehavejustshownthat[K:I]≤r.Fortheoppositeinequality,rememberthatbyTheorem1,[K:I] isequaltothenumberofI-fixingautomorphismsofK.Butthereareatleastrsuchautomorphisms,namely h1,…,hr.Thus,[K:I]≥r,andwearedone.■ Theorem3IfIisthefixfieldofH,thenHisthefixerofI. PROOF:LetIbethefixfieldofH,andI*thefixerofI. It follows from the definitions of fixer and fixfieldthatH⊆I*.WewillproveequalitybyshowingthatthereareasmanyelementsinHasinI*.By thelemma,theorderofHisequalto[K:I].ByTheorem2,IisthefixfieldofI*,sobythelemmaagain, theorderofI*isalsoequalto[K:I].■ ItfollowsimmediatelyfromTheorems2and3thatthereisaone-to-onecorrespondencebetween thesubgroupsofGal(K:F)andtheintermediatefieldsbetweenKandF.Thiscorrespondence,which matcheseverysubgroupwithitsfixfield(or,equivalently,matcheseveryintermediatefieldwithitsfixer) is called a Galois correspondence. It is worth observing that larger subfields correspond to smaller subgroups;thatis, Asanexample,wehaveseenthattheGaloisgroupof ( , )over isG={ε,α,β,γ}withthe tablegivenonpage325.Thisgrouphasexactlyfivesubgroups—namely,{ε},{ε,α},{ε,β},{ε,γ},and thewholegroupG.Theymayberepresentedinthe“inclusiondiagram”: Ontheotherhand,thereareexactlyfivefieldsintermediatebetween and ( berepresentedintheinclusiondiagram: , ),whichmay IfHisasubgroupofanygaloisgroup,letH°designatethefixfieldofH.ThesubgroupsofGinour examplehavethefollowingfixfields: (This is obvious by inspection of the way ε, α, β, and γ were defined on page 325.) The Galois correspondence,forthisexample,maythereforeberepresentedasfollows: InordertoeffectivelytieinsubgroupsofGwithextensionsofthefieldF,weneedonemorefact,to bepresentednext. SupposeE⊆I⊆K,whereKisarootfieldoverEandIisarootfieldoverE.(HencebyTheorem 8ofChapter31,KisarootfieldoverI.)Ifh∈Gal(K:E),hisanautomorphismofKfixingE.Consider therestrictionofhtoI,thatis,hrestrictedtothesmallerdomainI.ItisanisomorphismwithdomainI fixingE,sobyRule(i)givenatthebeginningofthischapter,itisanautomorphismofI,stillfixingE.We havejustshownthatifh∈Gal(K:E),thentherestrictionofhtoIisinGal(I:E).Thispermitsusto defineafunctionμ:Gal(K:E)→Gal(I:E)bytherule μ(h)=therestrictionofhtoI Itisveryeasytocheckthatμisahomomorphism.μissurjective,becauseeveryE-fixing automorphism ofIcanbeextendedtoanE-fixingautomorphismofK,byTheorem6inChapter31. Finally,ifh∈Gal(K:E),therestrictionofhtoIistheidentityfunctioniffh(x)=xforeveryx∈I, thatis,iffhfixesI.ThisprovesthatthekernelofμisGal(K:I). Torecapitulate:μisahomomorphismfromGal(K:E)ontoGal(I:E)withkernelGal(K:I).Bythe FHT,weimmediatelyconcludeasfollows: Theorem4SupposeE⊆I⊆K,whereIisarootfieldoverEandKisarootfieldoverE.Then Itfollows,inparticular,thatGal(K:I)isanormalsubgroupofGal(K:E). EXERCISES †A.ComputingaGaloisGroup 1 Showthat (i, )istherootfieldof(x2+1)(x2−2)over . #2 Findthedegreeof (i, )over . 3 ListtheelementsofGal( (i, ): )andexhibititstable. 4 WritetheinclusiondiagramforthesubgroupsofGal( (i, ): ),andtheinclusiondiagramforthe fieldsintermediatebetween and (i, ).IndicatetheGaloiscorrespondence. †B.ComputingaGaloisGroupofEightElements 1 Showthat ( , , )istherootfieldof(x2–2)(x2–3)(x2−5)over . 2 Showthatthedegreeof ( , , )over is8. 3 ListtheeightelementsofG=Gal( ( , , ): )andwriteitstable. 4 List the subgroups of G. (By Lagrange’s theorem, any proper subgroup of G has either two or four elements.) 5 ForeachsubgroupofG,finditsfixfield. 6 IndicatetheGaloiscorrespondencebymeansofadiagramliketheoneonpage329. †C.AGaloisGroupEqualtoS3 1 Showthat ( ,i )istherootfieldofx3–2over ,where designatestherealcuberootof2. (HINT:Computethecomplexcuberootsofunity.) 2 Showthat[ ( ): ]=3. 3 Explainwhyx2+3isirreducibleover ( ),thenshowthat[ ( ,i ): ( )]=2.Concludethat [ ( ,i ): ]=6. 4 Usepart3toexplainwhyGal( ( , ) : ) has six elements. Then use the discussion following Rule (ii) on page 323 to explain why every element of Gal( ( , i ) : ) may be identifed with a permutationofthethreecuberootsof2. 5 Usepart4toprovethatGal( ( ,i ): )≅S3. †D.AGaloisGroupEqualtoD4 If α = is a real fourth root of 2, then the four fourth roots of 2 are ±α and ±iα. Explain parts 1–6, brieflybutcarefully: #1 (α,i)istherootfieldofx4−2over . 2[ (α): ]=4. 3i∉ (α);hence[ (α,i): (α)]=2. 4[ (α,i): ]=8. 5{1,α,α2,α3,i,iα,iα2,iα3}isabasisfor (α,i)over . 6Any -fixingautomorphismhof (α,i)isdeterminedbyitseffectontheelementsinthebasis.These, inturn,aredeterminedbyh(α)andh(i). 7Explain:h(α)mustbeafourthrootof2andh(i)mustbeequalto±i.Combiningthefourpossibilities forh(α)withthetwopossibilitiesforh(i)giveseightpossibleautomorphisms.Listthemintheformat 8 Compute the table of the group Gal( (α, i) : ) and show that it is isomorphic to D4, the group of symmetriesofthesquare. †E.ACyclicGaloisGroup #1DescribetherootfieldKofx7−1over .Explainwhy[K: ]=6. 2Explain:Ifαisaprimitiveseventhrootofunity,anyh∈Gal(K: )mustsendαtoaseventhrootof unity.Infact,hisdeterminedbyh(α). 3Usepart2tolistexplicitlythesixelementsofGal(K: ).ThenwritethetableofGal(K: )andshow thatitiscyclic. 4ListallthesubgroupsofGal(K: ),withtheirfixfields.ExhibittheGaloiscorrespondence. 5DescribetherootfieldLofx6−1over ,andshowthat[L: ]=2.Explainwhyitfollowsthatthere arenointermediatefieldsbetween andL(exceptfor andLthemselves). #6LetLbetherootfieldofx6–2over .ListtheelementsofGal(L: )andwriteitstable. †F.AGaloisGroupIsomorphictoS5 Leta(x)=x5−4x4+2x+2∈ [x],andletr1,…,r5betherootsofa(x)in .LetK (r1,…,r5)bethe rootfieldofa(x)over . Prove:parts1–3: 1a(x)isirreduciblein ⌈x⌉. 2a(x)hasthreerealandtwocomplexroots.[HINT:Usecalculustosketchthegraphofy=a(x),andshow thatitcrossesthexaxisthreetimes.] 3Ifr1denotesarealrootofa(x),[ (r1): ]=5.Usethistoprovethat[K: ]isamultipleof5. 4Usepart3andCauchy’stheorem(Chapter13,ExerciseE)toprovethatthereisanelementαoforder5 inGal(K: ).Sinceαmaybeidentifiedwithapermutationof{r1,…,r5},explainwhyitmustbeacycle oflength5.(HINT:Anyproductofdisjointcycleson{r1,…,r5}hasorder≠5.) 5ExplainwhythereisatranspositioninGal(K: ).[Itpermutestheconjugatepairofcomplexrootsof a(x).] 6 Prove: Any subgroup of S5 which contains a cycle of length 5 and a transposition must contain all possibletranspositionsinS5,henceallofS5.Thus,Gal(K: )=S5. G.ShorterQuestionsRelatingtoAutomorphismsandGaloisGroups LetFbeafield,andKafiniteextensionofF.Supposea,b∈K.Proveparts1–3: 1IfanautomorphismhofKfixesFanda,thenhfixesF(a). 2F(a,b)*=F(a)* F(b)*. 3Asidefromtheidentityfunction,thereareno -fixingautomorphismsof ( containsonlyrealnumbers.] 4Explainwhytheconclusionofpart3doesnotcontradictTheorem1. ).[HINT:Notethat ( ) Inthenextthreeparts,letωbeaprimitivepthrootofunity,wherepisaprime. 5Prove:Ifh∈Gal( (ω): ),thenh(ω)=ωk forsomekwhere1≤k≤p−1. 6Usepart5toprovethatGal( (ω): )isanabeliangroup. 7Usepart5toprovethatGal( (ω): )isacyclicgroup. †H.TheGroupofAutomorphismsofC 1Prove:Theonlyautomorphismof istheidentityfunction.[HINT:Ifhisanautomorphism,h(1)=1; henceh(2)=2,andsoon.] 2Prove:Anyautomorphismof sendssquaresofnumberstosquaresofnumbers,hencepositivenumbers topositivenumbers. 3Usingpart2,provethatifhisanyautomorphismof ,a<bimpliesh(a)<h(b). #4Useparts1and3toprovethattheonlyautomorphismof istheidentityfunction. 5ListtheelementsofGal( : ). 6Provethattheidentityfunctionandthefunctiona+bi→a−biaretheonlyautomorphismsof which fix . I.FurtherQuestionsRelatingtoGaloisGroups Throughout this set of questions, let K be a root field over F, let G = Gal(K : F), and let I be any intermediatefield.Provethefollowing: 1I*=Gal(K:I)isasubgroupofG. 2IfHisasubgroupofGandH°={a∈K:π(a)=aforeveryπ∈H},thenH°isasubfieldofK,andF ⊆H°. 3LetHbethefixerofI,andI′thefixfieldofH.ThenI⊆I′.LetIbethefixfieldofH,andI*thefixerof I.ThenH⊆I*. #4LetIbeanormalextensionofF(thatis,arootfieldofsomepolynomialoverF).IfGisabelian,then Gal(K:I)andGal(I:F)areabelian.(HINT:UseTheorem4.) 5LetIbeanormalextensionofF.IfGisacyclicgroup,thenGal(K:I)andGal(I:F)arecyclicgroups. 6 If G is a cyclic group, there exists exactly one intermediate field I of degree k, for each integer k dividing[K:F]. †J.NormalExtensionsandNormalSubgroups SupposeF⊆K,whereKisanormalextensionofF.(ThismeanssimplythatKistherootfieldofsome polynomialinF[x]:seeChapter31,ExerciseK.)LetI1⊆I2beintermediatefields. 1DeducefromTheorem4that,ifI2isanormalextensionofI1,then isanormalsubgroupof . 2ProvethefollowingforanyintermediatefieldI:Leth∈Gal(K:F),g∈I*,a∈I,andb=h(a).Then [h∘g∘h−1](b)=b.Concludethat hI*h−1⊆h(I)* 3Usepart2toprovethathI*h−1=h(I)*. TwointermediatefieldsI1andI2arecalledconjugateiffthereisanautomorphism[i.e.,anelementi ∈Gal(K:F)]suchthati(I1)=I2. 4Usepart3toprovethatI1andI2areconjugateiff and areconjugatesubgroupsintheGaloisgroup. 5Usepart4toprovethatforanyintermediatefieldsI1andI2:iff isanormalsubgroupof ,thenI2is anormalextensionofI1. Combiningparts1and5wehave:I2isanormalextensionofI1iff isanormalsubgroupof . (Historically,thisresultistheoriginoftheword“normal”intheterm“normalsubgroup.”) CHAPTER THIRTY-THREE SOLVINGEQUATIONSBYRADICALS In this final chapter, Galois theory will be used explicitly to answer practical questions about solving equations. Intheintroductiontothisbookwesawthatclassicalalgebrawasdevotedlargelytofindingmethods forsolvingpolynomialequations.Thequadraticformulayieldsthesolutionsofeveryequationofdegree 2, and similar formulas have been found for polynomials of degrees 3 and 4. But every attempt to find explicit formulas, of the same kind as the quadratic formula, which would solve a general equation of degree5orhigherendedinfailure.ThereasonforthiswasfinallydiscoveredbytheyoungGalois,who showed that an equation is solvable by the kind of explicit formula we have in mind if and only if its group of symmetries has certain properties. The group of symmetries is, of course, the Galois group which we have already defined, and the required group properties will be formulated in the next few pages. Galois showed that the groups of symmetries of all equations of degree ≤4 have the properties neededforsolvability,whereasequationsofdegree5ormoredonotalwayshavethem.Thus,notonlyis the classical quest for radical formulas to solve all equations of degree > 4 shown to be futile, but a criterion is made available to test any equation and determine if it has solutions given by a radical formula.Allthiswillbemadeclearinthefollowingpages. Everyquadraticequationax2+bx+c=0hasitsrootsgivenbytheformula Equationsofdegree3and4canbesolvedbysimilarformulas.Forexample,thecubicequationx3+ax+ b=0hasasolutiongivenby Such expressions are built up from the coefficients of the given polynomials by repeated addition, subtraction,multiplication,division,and taking roots. Because of their use of radicals, they are called radicalexpressionsorradicalformulas.Apolynomiala(x)issolvablebyradicalsifthereisaradical expressiongivingitsrootsintermsofitscoefficients. Letusreturntotheexampleofx3+ax+b=0,whereaandbarerational,andlookagainatFormula (1). We may interpret this formula to assert that if we start with the field of coefficients , adjoin the squareroot ,thenadjointhecuberoots ,wereachafieldinwhichx3+ax+b=0has itsroots. Ingeneral,tosaythattherootsofa(x)aregivenbyaradicalexpressionisthesameassayingthat wecanextendthefieldofcoefficientsofa(x)bysuccessivelyadjoiningnthroots(forvariousn),andin thiswayobtainafieldwhichcontainstherootsofa(x).Wewillexpressthisnotionformallynow,inthe languageoffieldtheory. F(c1,…,cn)iscalledaradicalextensionofFif,foreachi,somepowerofciisinF(c1,…,ci−1). Inotherwords,F(c1,…,cn)isaniteratedextensionofFobtainedbysuccessivelyadjoiningnthroots,for variousn.Wesaythatapolynomiala(x)inF[x]issolvablebyradicalsifthereisaradicalextensionof Fcontainingalltherootsofa(x),thatis,containingtherootfieldofa(x). Todealeffectivelywithnthrootswemustknowalittleaboutthem.Tobeginwith,thenthrootsof1, callednthrootsofunity,are,ofcourse,thesolutionsofxn −1=0.Thus,foreachn,thereareexactlyn nthrootsofunity.Asweshallsee,everythingweneedtoknowaboutrootswillfollowfrompropertiesof therootsofunity. In thenthrootsofunityareobtainedbydeMoivre’stheorem.Theyconsistofanumberωandits first n powers: 1 = ω0, ω, ω2, …, ωn − 1. We will not review de Moivre’s theorem here because, remarkably,themainfactsaboutrootsofunityaretrueineveryfieldofcharacteristiczero.Everythingwe needtoknowemergesfromthefollowingtheorem: Theorem1Anyfinitegroupofnonzeroelementsinafieldisacyclicgroup.(Theoperationinthe groupisthefield’smultiplication.) PROOF:IfF* denotes the set of nonzero elements of F, suppose that G ⊆ F*, and that G, with the field’s“multiply”operation,isagroupofnelements.WewillcompareGwith nandshowthatG,like n ,hasanelementofordernandisthereforecyclic. Foranyintegerk,letg(k)bethenumberofelementsoforderkinG,andletz(k)bethenumberof elementsoforderkin n.Foreverypositiveintegerkwhichisafactorofn,theequationxk =1hasat mostksolutionsinF;thus, (*)Gcontainsatmostkelementswhoseorderisafactorofk. IfGhasanelementaoforderk,then〈a〉={e,a,a2,…,ak −1)areallthedistinctelementsofGwhose orderisafactorofk.[By(*),therecannotbeanyothers.]In n,thesubgroup containsalltheelementsof nwhoseorderisafactorofk. Since〈a〉and〈n/k〉arecyclicgroupswiththesamenumberofelements,theyareisomorphic;thus, thenumberofelementsoforderkin〈a〉isthesameasthenumberofelementsoforderkin〈n/k〉. Thus, g(k)=z(k). Letusrecapitulate:ifGhasanelementoforderk,theng(k)=z(k);butifGhasno such elements, theng(k)=0.Thus,foreachpositiveintegerkwhichisafactorofn,thenumberofelementsoforderkin Gislessthan(orequalto)thenumberofelementsoforderkin n. Now, every element of G (as well as every element of n) has a well-defined order, which is a divisorofn.Imaginetheelementsofbothgroupstobepartitionedintoclassesaccordingtotheirorder, andcomparetheclassesinGwiththecorrespondingclassesin n.Foreachk,Ghasasmanyor fewer elementsoforderkthan ndoes.SoifGhadnoelementsofordern(while ndoeshaveone),thiswould mean that G has fewer elements than n, which is false. Thus, G must have an element of order n, and thereforeGiscyclic.■ Thenthrootsofunity(whicharecontainedinForasuitableextensionofF)obviouslyformagroup withrespecttomultiplication.ByTheorem1,itisacyclicgroup.Anygeneratorofthisgroupiscalleda primitiventhrootofunity.Thus,ifωisaprimitiventhrootofunity,thesetofallthenthrootsofunityis 1,ω,ω2,…,ωn−1 Ifωisaprimitiventhrootofunity,F(ω)isanabelianextensionofFinthesensethatg∘h=h ∘g foranytwoF-fixingautomorphismsgandhofF(ω)).Indeed,anyautomorphismmustobviouslysendnth roots of unity to nth roots of unity. So if g(ω) = ωr and h(ω) = ωs, then g ∘ h(ω) = g(ωs) = ωrs and analogously,h∘g(ω)=ωrs.Thus,g∘h(ω)=h∘g(ω).SincegandhfixF,andeveryelementofF(ω)is alinearcombinationofpowersofωwithcoefficientsinF,g∘h=h∘g. Now,letFcontainaprimitiventhrootofunity,andthereforeallthenthrootsofunity.Supposea∈ F,andahasannthrootbinF.Itfollows,then,thatallthenthrootsofaareinF,fortheyareb, bω, bω2,…,bωn−1.Indeed,ifcisanyothernthrootofa,thenclearlyc/bisannthrootof1,sayωr;hencec =bω4.WemayinferfromtheabovethatifFcontainsaprimitiventhrootofunity,andbisannthroot ofa,thenF(b)istherootfieldofxn−aoverF. In particular, F(b) is an abelian extension of F. Indeed, any F-fixing automorphism of F(b) must sendnthrootsofatonthrootsofa:forifcisanynthrootofaandgisanF-fixingautomorphism,then g(c)n=g(cn)=g(a)=a;henceg(c)isannthrootofa.Soifg(b)=bωrandh(b)=bωs,then g∘h(b)=g(bωs)=bωrωs=bωr+s and h∘g(b)=h(bωr)=bωsωr=bwr+s hence g ∘ h(b) = h ∘ g(b). Since g and h fix F, and every element in F(b) is a linear combination of powersofbwithcoefficientsinF,itfollowsthatg∘h=h∘g. Ifa(x)isinF[x],rememberthata(x)issolvablebyradicalsjustaslongasthereexistssomeradical extensionofFcontainingtherootsofa(x).[AnyradicalextensionofFcontainingtherootsofa(x) will do.] Thus, we may as well assume that any radical extension used here begins by adjoining to F the appropriate roots of unity; henceforth we will make this assumption. Thus, if K = F(c1, …, cn) is a radicalextensionofF,then is a sequence of simple abelian extensions. (The extensions are all abelian by the comments in the precedingthreeparagraphs.) Still,thisisnotquiteenoughforourpurposes:Inordertousethemachinerywhichwassetupinthe previous chapter, we must be able to say that each field in (2) is a root field over F. This may be accomplishedasfollows:SupposewehavealreadyconstructedtheextensionsI0⊆I1⊆⋯⊆Iqin(2)so thatIqisarootfieldoverF.WemustextendIqtoIq+1,soIq+1isarootfieldoverF.Also,Iq+1must includetheelementcq+1,whichisthenthrootofsomeelementa∈Iq. Let H = {h1, …, hr) be the group of all the F-fixing automorphisms of Iq, and consider the polynomial b(x)=[xn−h1(a)][xn−h2(a)]⋯[xn−hr(a)] By the proof of the lemma on page 327, one factor of b(x) is (xn − a); hence cq + 1 is a root of b(x). Moreover,bythesamelemma,everycoefficientofb(x)isinthefixfieldofH,thatis,inF.Wenowdefine Iq + 1 to be the root field of b(x) over F. Since all the roots of b(x) are nth roots of elements in Iq, it follows that Iq + 1 is a radical extension of Iq. The roots may be adjoined one by one, yielding a successionofabelianextensions,asdiscussedpreviously.Toconclude,wemayassumein(2)thatKisa rootfieldoverF. IfGdenotestheGaloisgroupofKoverF,eachofthesefieldsIk hasafixerwhichisasubgroupof G.Thesefixersformasequence Foreachk,byTheorem4ofChapter32, isanormalsubgroupof ,and ≅ Gal(Ik + 1: Ik whichisabelianbecauseisanabelianextensionofIk .Thefollowingdefinitionwasinventedpreciselyto accountforthissituation. AgroupGiscalledsolvableifithasasequenceofsubgroups{e}=H0⊆H1⊆⋯⊆Hm=Gsuch thatforeachk,Gk isanormalsubgroupofGk+1andGk+1/Gk isabelian. WehaveshownthatifKisaradicalextensionofF,thenGal(K:F)isasolvablegroup.Wewishto go further and prove that if a(x) is any polynomial which is solvable by radicals, its Galois group is solvable.Todoso,wemustfirstprovethatanyhomomorphicimageofasolvablegroupissolvable.A key ingredient of our proof is the following simple fact, which was explained on page 152: G/H is abelianiffHcontainsalltheproductsxyx−1y−1forallxandyinG.(Theproductsxyx−1y−1arecalled “commutators”ofG.) Theorem2Anyhomomorphicimageofasolvablegroupisasolvablegroup. PROOF:LetGbeasolvablegroup,withasequenceofsubgroups {e)⊆H1⊆⋯⊆Hm=G asspecifiedinthedefinition.Letf:G→XbeahomomorphismfromGontoagroupX.Thenf(H0),f(H1), …,f(Hm)aresubgroupsofX,andclearly{e}⊆f(H0)⊆f(H1)⊆⋯⊆f(Hm)=X.Foreachiwehavethe following:iff(a)∈f(Hi)andf(x)∈f(Hi+1),thena∈Hiandx∈Hi+1;hencexax−1∈Hiandtherefore f(x)f(a)f(x)−1∈f(Hi).Sof(Hi)isanormalsubgroupoff(Hi+1).Finally,sinceHi+1/Hiisabelian,every commutatorxyx−1y−1(forallxandyinHi+1)isinHi;henceeveryf(x)f(y)f(x)−1f(y)−1isinf(Hi).Thus, f(Hi+1)/f(Hi)isabelian.■ Nowwecanprovethemainresultofthischapter: Theorem 3 Let a(x) be a polynomial over a field F. If a(x) is solvable by radicals, its Galois groupisasolvablegroup. PROOF: By definition, if K is the root field of a(x), there is an extension by radicals F(c1, …, cn) such that F ⊆ K ⊆ F (c1, …, cn). It follows by Theorem 4 of Chapter 32 that Gal(F(c1, …, cn): F)/Gal(F(c1, …, cn): K) ≅ Gal(K: F); hence by that theorem, Gal(K: F) is a homomorphic image of Gal(F(c1,…,cn):F)whichweknowtobesolvable.Thus,byTheorem2Gal(K:F)issolvable.■ Actually,theconverseofTheorem3istruealso.Allweneedtoshowisthat,ifKisanextensionof FwhoseGaloisgroupoverFissolvable,thenKmaybefurtherextendedtoaradicalextensionofF.The detailsarenottoodifficultandareassignedasExerciseEattheendofthischapter. Theorem 3 together with its converse say that a polynomial a(x) is solvable by radicals iff its Galoisgroupissolvable. Webringthischaptertoaclosebyshowingthatthereexistgroupswhicharenotsolvable,andthere exist polynomials having such groups as their Galois group. In other words, there are unsolvable polynomials.First,hereisanunsolvablegroup: Theorem4ThesymmetricgroupS5isnotasolvablegroup. PROOF:SupposeS5hasasequenceofsubgroups {e}=H0⊆H1⊆…⊆Hm=S5 asinthedefinitionofsolvablegroup.ConsiderthesubsetofS5containingallthecycles(ijk)oflength3. WewillshowthatifHicontainsallthecyclesoflength3,sodoesthenextsmallergroupHi−1.Itwould followinmstepsthatH0={e}containsallthecyclesoflength3,whichisabsurd. So let Hi contain all the cycles of length 3 in S5. Remember that if α and β are in Hi, then their commutatorαβα−1β−1isinHi−1.Butanycycle(ijk)isequaltothecommutator (ilj)(jkm)(ilj)−1(jkm)−1=(ilj)(jkm)(jli)(mkj)=(ijk) henceevery(ijk)isinHi−1,asclaimed.■ Before drawing our argument toward a close, we need to know one more fact about groups; it is containedinthefollowingclassicalresultofgrouptheory: Cauchy’stheoremLetGbeafinitegroupofnelements.Ifpisanyprimenumberwhichdivides n,thenGhasanelementoforderp. Forexample,ifGisagroupof30elements,ithaselementsoforders2,3,and5.Togiveourproof atrimmerappearance,wewillproveCauchy’stheoremspecificallyforp=5(theonlycasewewilluse here,anyway).However,thesameargumentworksforanyvalueofp. PROOF:Considerallpossible5-tuples(a,b,c,d,k)ofelementsofGwhoseproductabcdk=e.How manydistinct5-tuplesofthiskindarethere?Well,ifweselecta,b,c,anddatrandom,thereisauniquek =d−1c−1b−1a−1inGmakingabcdk=e.Thus,therearen4such5-tuples. Calltwo5-tuplesequivalentifoneismerelyacyclicpermutationoftheother.Thus,(a,b,c,d,k)is equivalenttoexactlyfivedistinct5-tuples,namely,(a,b,c,d,k),(b,c,d,k,a),(c,d,k,a,b),(d,k,a,b, c)and(k,a,b,c,d).Theonlyexceptionoccurswhena5-tupleisoftheform(a,a,a,a,a) with all its componentsequal;itisequivalentonlytoitself.Thus,theequivalenceclassofany5-tupleoftheform(a, a,a,a,a)hasasinglemember,whilealltheotherequivalenceclasseshavefivemembers. Arethereanyequivalenceclasses,otherthan{(e,e,e,e,e)},withasinglemember?Ifnot,then5| 4 (n −1)[fortherearen45-tuplesunderconsideration,less(e,e,e,e,e)];hencen4≡1(mod5).Butwe areassumingthat5|n;hencen4≡0(mod5),whichisacontradiction. Thiscontradictionshowsthattheremustbea5-tuple(a,a,a,a,a)≠(e,e,e,e,e)suchthataaaaa= a5=e.Thus,thereisanelementa∈Goforder5.■ Wewillnowexhibitapolynomialin [x]havingS5asitsGaloisgroup(rememberthatS5isnot a solvablegroup). Leta(x)=x5 −5x −2.ByEisenstein’scriterion,a(x+2)isirreducibleover ;hencea(x)alsois irreducibleover .Byelementarycalculus,a{x)hasasinglemaximumat(−1,2),asingleminimumat (1,−6),andasinglepointofinflectionat(0,−2).Thus(seefigure),itsgraphintersectsthexaxisexactly threetimes.Thismeansthata(x)hasthreerealroots,r1,r2,andr3,andthereforetwocomplexroots,r4 andr5,whichmustbecomplexconjugatesofeachother. Let K denote the root field of a(x) over , and G the Galois group of a(x). As we have already noted,everyelementofGmaybeidentifiedwithapermutationoftherootsr1,r2,r3,r4,r5ofa(x),soG maybeviewedasasubgroupofS5.WewillshowthatGisallofS5. Now,[ (r1): ]=5becauser1isarootofanirreduciblepolynomialofdegree5over .Since[K: ]=[K: (r1)][ (r1): ],itfollowsthat5isafactorof[K: ].Then,byCauchy’stheorem,Gcontains anelementoforder5.Thiselementmustbeacycleoflength5:forbyChapter8,everyotherelementof S5isaproductoftwoormoredisjointcyclesoflength<5,andsuchproductscannothaveorder5.(Try thetypicalcases.)Thus,Gcontainsacycleoflength5. Furthermore,Gcontainsatranspositionbecausecomplexconjugationa+bi→a−biisobviously a -fixing automorphism of K; it interchanges the complex roots r4 and r5 while leaving r1, r2, and r3 fixed. AnysubgroupG⊆S5whichcontainsatranspositionτandacycleσoflength5necessarilycontains all the transpositions. (They are στσ−1, σ2τσ−2, σ3τσ−3, σ4τσ−4, and their products; check this by direct computation!)Finally,ifGcontainsallthetranspositions,itcontainseverythingelse:foreverymemberof S5isaproductoftranspositions.Thus,G=S5. Wehavejustgivenanexampleofapolynomiala(x)ofdegree5over whoseGaloisgroupS5 is notsolvable.Thus,a(x)isanexampleofapolynomialofdegree5whichcannotbesolvedbyradicals.In particular, there cannot be a radical formula (on the model of the quadratic formula) to solve all polynomialequationsofdegree5,sincethiswouldimplythateverypolynomialequationofdegree5has aradicalsolution,andwehavejustexhibitedonewhichdoesnothavesuchasolution. InExerciseBitisshownthatS3,S4,andalltheirsubgroupsaresolvable;henceeverypolynomialof degree≤4issolvablebyradicals. EXERCISES A.FindingRadicalExtensions 1Findradicalextensionsof containingthefollowingcomplexnumbers: (a) (b) (c) 2Showthatthefollowingpolynomialsin [x]arenotsolvablebyradicals: #(a) 2x5−5x4+5 (b) x5−4x2+2 (c) x5−4x4+2x+2 3Showthata(x)=x5−10x4+40x3−80x2+79x −30issolvablebyradicalsover ,andgiveitsroot field.[HINT:Compute(x−2)5−(x−2).] 4Showthatax8+bx6+cx4+dx2+eissolvablebyradicalsoveranyfield.(HINT:Lety=x;usethefact thateveryfourth-degreepolynomialissolvablebyradicals.) 5 Explain why parts 3 and 4 do not contradict the principal finding of this chapter: that polynomial equationsofdegreen≥5donothaveageneralsolutionbyradicals. †B.SolvableGroups LetGbeagroup.ThesymbolH Giscommonlyusedasanabbreviationof“Hisanormalsubgroupof G.”AnormalseriesofGisafinitesequenceH0,H1,…,HnofsubgroupsofGsuchthat {e}=H0 H1 ⋯ Hn=G SuchaseriesiscalledasolvableseriesifeachquotientgroupHi+1/Hiisabelian.Giscalledasolvable groupifithasasolvableseries. 1Explainwhyeveryabeliangroupis,trivially,asolvablegroup. 2LetGbeasolvablegroup,withasolvableseriesH0,…,Hn.LetKbeasubgroupofG.ShowthatJ0= K∩H0,…,Jn=K∩HnisanormalseriesofK. #3UsetheremarkimmediatelyprecedingTheorem2toprovethatJ0,…,JnisasolvableseriesofK. 4Useparts2and3toprove:Everysubgroupofasolvablegroupissolvable. 5Verifythat{ε}⊆{ε,β,δ}⊆S3isasolvableseriesforS3.ConcludethatS3,andallofitssubgroups, aresolvable. 6InS4,letA4bethegroupofalltheevenpermutations,andlet B={ε,(12)(34),(13)(24),(14)(23)} Show that {ε} ⊆ B ⊆ A4 ⊆ S4 is a solvable series for S4. Conclude that S4 and all its subgroups are solvable. The next three sets of exercises are devoted to proving the converse of Theorem 3: If the Galois groupofa(x)issolvable,thena(x)issolvablebyradicals. †C.pthRootsofElementsinaField Letpbeaprimenumber,andωaprimitivepthrootofunityinthefieldF. 1Ifdisanyrootofxp −a∈F[x],showthatF(ω, d) is a root field of xp − a. Suppose xp − a is not irreducibleinF[x]. 2Explainwhyxp−afactorsinF[x]asxp−a=p(x)f(x),wherebothfactorshavedegree≤2. #3Ifdegp(x)=m,explainwhytheconstanttermofp(x)(letuscallitb)isequaltotheproductofmpth rootsofa.Concludethatb=ωk dmforsomek. 4Usepart3toprovethatbp=am. 5Explainwhymandparerelativelyprime.Explainwhyitfollowsthatthereareintegerssandt such thatsm+tp=1. 6Explainwhybsp=asm.Usethistoshowthat(bsat)p=a. 7Conclude:Ifxp−aisnotirreducibleinF[x],ithasaroot(namely,bsat)inF. Wehaveproved:xp−aeitherhasarootin.ForisirreducibleoverF. †D.AnotherWayofDefiningSolvableGroups LetGbeagroup.ThesymbolH Gshouldberead,“HisanormalsubgroupofG.”Amaximalnormal subgroupofGisanH Gsuchthat,ifH J G,thennecessarilyJ=HorJ=G.Provethefollowing: 1IfGisafinitegroup,everynormalsubgroupofGiscontainedinamaximalnormalsubgroup. 2Letf:G→Hbeahomomorphism.IfJ H,thenf−1(J)<G. #3LetK G.If isasubgroupofG/K,let denotetheunionofallthecosetswhicharemembersof . If G/K,then G.(Usepart2.) 4IfKisamaximalnormalsubgroupofG,thenG/Khasnonontrivialnormalsubgroups.(Usepart3.) 5IfanabeliangroupGhasnonontrivialsubgroups,Gmustbeacyclicgroupofprimeorder.(Otherwise, choosesomea∈Gsuchthat〈a〉isapropersubgroupofG.) 6IfH K G,thenG/KisahomomorphicimageofG/H. #7LetH G,whereG/Hisabelian.ThenGhassubgroupsH0,…,HqsuchthatH=H0 H1 ⋯ Hq= G,whereeachquotientgroupHi+1/Hiiscyclicofprimeorder. Itfollowsfrompart7thatifGisasolvablegroup,then,by“fillingingaps,”Ghasanormalseries in which all the quotient groups are cyclic of prime order. Solvable groups are often defined in this fashion. E.IfGal(K:F)IsSolvable,KIsaRadicalExtensionofF LetKbeafiniteextensionofF,whereKisarootfieldoverF,withG=Gal(K:F)asolvablegroup.As remarkedinthetext,wewillassumethatFcontainstherequiredrootsofunity.ByExerciseD,letH0,…, HnbeasolvableseriesforGinwhicheveryquotientHi+1/Hiiscyclicofprimeorder.Foranyi=1,…, n,letFi.andFi+1bethefixfieldsofHiandHi+1. 1Prove:FiisanormalextensionofFi+1,and[Fi:Fi+1]isaprimep. 2LetπbeageneratorofGal(Fi:Fi+1),ωapthrootofunityinFi+1,andb∈Fi.Set c=b+ωπ−1(b)+ω2π−2(b)+⋯+ωp−1π−(p−1)(b) Showthatπ(c)=ωc. 3Usepart2toprovethatπk (cp)=cpforeveryk,anddeducefromthisthatcp∈Fi+1 4ProvethatFiistherootfieldofxp−cpoverFi+1. 5ConcludethatKisaradicalextensionofF. APPENDIX A REVIEWOFSETTHEORY Asetisacollectionofobjects.Theobjectsinthesetarecalledtheelementsoftheset.Ifxisanelement ofasetA,wedenotethisfactbywritingx∈A(tobereadas“xisanelementofA”).Ontheotherhand,if xisnotanelementofA,wewritex∉A.(Thisshouldbereadas“xisnotanelementofA.”) SupposethatAandBaresetsandthatAisapartofB.Inotherwords,supposeeveryelementofAis anelementofB.Inthatcase,wecallAasubsetofBandwriteA⊆B.WesaythatAandBareequal(and wewrite“A=B”)ifAandBhavethesameelements.ItiseasytoseethatifA⊆BandB⊆A,thenA= B.Andconversely,ifA=B,thenA⊆5andB⊆A. Setsareoftenrepresentedpictoriallybycircularorovalshapes.ThefactthatA⊆Bisrepresented asinthefollowingfigure: Manysetscaneasilybedescribedinwords;forexample,“thesetofallthepositiverealnumbers,”“the setofalltheintegers,”or“thesetofalltherationalnumbersbetween1and2.”However,itisoften moreconvenienttodescribeasetinsymbols.Wedothisbywriting {x:________________} Followingthecolon,wenamethepropertywhichalltheelementsintheset,andnootherelements,have incommon.Forexample, {x:x∈ andx≤2} isthesetofalltherealnumbers(elementsof )whicharelessthanorequalto2.Wemayalsosymbolize thissetby {x∈ :x≤2} tobereadas“thesetofallxin satisfyingx≤2.” IfAandBaresets,thereareseveralwaysofformingnewsetsfromAandB: 1. TheunionofAandB(denotedbyA B)isthesetofallthoseelementswhichareelementsofA or elementsofB.Here,theword“or”isusedinthe“inclusive”sense:AnelementxisinA Bifxisin A,orinB,orinbothAandB.Thus,A Bistheshadedareainthefigure: 2. TheintersectionofAandB(denotedbyA B)isthesetofallthoseelementswhichareinAandinB. Thus,xisinA Bifx∈Aandx∈B.ThesetA Bisrepresentedbytheshadedareainthefigure: 3. Thedifferenceoftwosetsisdefinedasfollows:A−BisthesetofallthoseelementswhichareinA, butnotinB.Thus,xisinA−Bifx∈Aandx∉B.ThesetA−Bistheshadedareainthefigure: Inshort, Wedefinetheemptysettobethesetwithnoelementsatall.Theemptysetisdenotedbythesymbol 0. Thereareseveralbasicidentitiesinvolvingthesetoperationswhichhavejustbeenintroduced.As anexampleofhowtoprovethem,wegiveaproofoftheidentity A B=B A PROOF:WeneedtoshowthateveryelementinA BisinB A,andconversely,thateveryelement inB AisinA B.Well,supposex∈A B;thismeansthatx∈A or x ∈ B. But this is the same as saying:x∈Borx∈A.Thus,x∈B A. Thesamereasoningshowsthatifx∈B A,thenx∈A B.■ Next,weprovetheidentity A (B C)=(A B) (A C) PROOF:First,wewillshowthateveryelementofA (B C)isanelementof(A B) (A C). Well,supposex∈A (B C):Thismeansthat (i) x∈A,and (ii) x∈B C. From(ii),x∈Borx∈C.Ifx∈B,thenfrom(i),x∈A B.Ifx∈C,thenfrom(i),x∈A C.Ineither case,xisin(A B) (A C). Conversely,supposexisin(A B) (A C).Then (i) x∈A B,or (ii) x∈A C. Ineithercase,x∈Borx∈C,sothatx∈B C.Alsoineithercase,x∈A.Thus,x∈A (B C).We’re done.■ Weconcludethissectionwithonemoredefinition:IfAandBaresets,thenA×Bdenotesthesetof allorderedpairs(a,b),wherea∈Aandb∈B.Thatis, A×B={(a,b):a∈Aandb∈B} WecallA×BthecartesianproductofAandB.Note,bytheway,that(a,b)isnotthesameas(b, a). Moreover,(a,b)=(c,d)ifa=candb=d.Thatis,twoorderedpairsareequaliftheyhavethesame firstcomponentandthesamesecondcomponent. EXERCISES Provethefollowing: 1IfA⊆BandB⊆C,thenA⊆C. 2IfA=BandB=C,thenA=C. 3A⊆A BandB⊆A B. 4A B⊆AandA B⊆B. 5A B=B A. 6A A=AandA A=A. 7A (B C)=(A B) (A C). 8A 0=AandA 0=0. 9A (A B)=A. 10.A (A B)=A. 11A (B−C)=(A B)−C. 12(A B)−C=(A−C) (B−C). 13IfA⊆B,thenA B=B.Conversely,ifA B=B,thenA⊆B. 14.IfA⊆B,thenA B=A.Conversely,ifA B=A,thenA⊆B. IfSisaset,andAisasubsetofS,thenthecomplementofAinSisthesetofalltheelementsofS whicharenotinA.ThecomplementofAisdenotedbyA′: Provethefollowing’. 15(A B)′=A′ B′. 16(A B)′=A′ B′. 17(A′)′=A. 18A A′=0. 19IfA⊆B,thenA B′=0,andconversely. 20IfA⊆BandC=B−A,thenA=B−C. Provethefollowingidentitiesinvolvingcartesianproducts: 21A×(B C)=(A×B) (A×C). 22A×(B C)=(A×B) (A×C). 23(A×B) (C×D)=(A C)×(B D). 24A×(B−D)=(A×B)−(A×D). APPENDIX B REVIEWOFTHEINTEGERS One of the most important facts about the integers is that any integer m can be divided by any positive integerntoyieldaquotientqandapositiveremainderr.(Theremainderislessthanthedivisorn.)For example,25maybedividedby8togiveaquotientof3andaremainderof1: Thisprocessisknownasthedivisionalgorithm.Itmaybestatedpreciselyasfollows: Theorem 1: Division algorithm If m and η are integers and η is positive, there exist unique integersqandrsuchthat m=nq+r and 0≤r<n Wecallqthequotientandrtheremainderwhenmisdividedbyn. Hereweshalltakethedivisionalgorithmasapostulateofthesystemoftheintegers.(InChapter21 westartedwithamorefundamentalpremiseandprovedthedivisionalgorithmfromit.) Ifrandsareintegers,wesaythatsisamultipleofrifthereisanintegerksuchthat s=rk In this case, we also say that r is a factor of s, or r divides s, and we symbolize this relationship by writing r|s Forexample,3isafactorof12,sowewrite:3|12.Someoftheelementarypropertiesofdivisibilityare statedinthenexttheorem. Theorem2:Thefollowingaretrueforallintegersa,b,andc: (i) Ifa|bandb|c,thena|c. (ii) 1|a. (iii) a|0. (iv) Ifc|aandc|b,thenc|(ax+by)forallintegersxandy. (v) Ifa|bandc|d,thenac|bd. The proofs of these relationships follow from the definition of divisibility. For instance, we give heretheproofof(iv):Ifc|aandc|b,thismeansthata=kcandb=lcforsomekandl.Then ax+by=kcx+lcy=c(kx+ly) Visibly,cisafactorofc(kx+ly),andhenceafactorofax+by.Insymbols,c|(ax+by),andwearedone. Anintegertiscalledacommondivisorofintegersrandsift|randt|s.Agreatestcommondivisor ofrandsisanintegertsuchthat (i) t|randt|s,and (ii) Foranyintegeru,ifu|randu|s,thenu|t. In other words, t is a greatest common divisor of r and s if ί is a common divisor of r and s and everyothercommondivisorofrandsdividest. Itisanimportantfactthatanytwononzerointegersrandsalwayshaveapositivegreatestcommon divisor: Theorem3:Anytwononzerointegersrandshaveauniquepositivegreatestcommondivisort. Moreover,tisequaltoa“linearcombination”ofrands.Thatis, t=kr+ls forsomeintegerskandl. Theuniquepositivegreatestcommondivisorofrandsisdenotedbythesymbolgcd(r,s). Apairofintegersrandsaresaidtoberelativelyprimeiftheyhavenocommondivisorsexcept±1. Forexample,4and15arerelativelyprime.Ifrandsarerelativelyprime,theirgcdisequalto1.Soby Theorem3,thereareintegerskandlsuchthat kr+ls=1 Actually,theconverseofthisstatementistruetoo,andisstatedinthenexttheorem: Theorem4:Twointegersrandsarerelativelyprimeifandonlyifthereareintegerskandlsuch thatkr+ls=1. Theproofofthistheoremisleftasanexercise.FromTheorem4,wededucethefollowing: Theorem5Ifrandsarerelativelyprime,andr|st,thenr|t. PROOFFromTheorem4weknowthereareintegerskandlsuchthatkr+ls=1.Multiplyingthrough byt,weget krt+lst=t (1) Butwearegiventhefactthatr|st;thatis,stisamultipleofr,sayst=mr.SubstitutionintoEquation(1) giveskrt+lmr=t,thatis,r(kt+lm)=t.Thisshowsthatrisafactoroft,asrequired.■ Ifanintegermhasfactorsnotequalto1or-1,wesaythatmiscomposite.Ifapositiveintegerm≠1 isnotcomposite,wecallitaprime.Forexample,6iscomposite(ithasfactors±2and±3),while7is prime. If p is a prime, then for any integer n, p and n are relatively prime. Thus, Theorem 5 has the followingcorollary: CorollaryLetmandnbeintegers,andletpbeaprime:Ifp|mn,theneitherp|morp|n. It is a major fact about integers that every positive integer m > 1 can be written, uniquely, as a productofprimes.(TheproofisgiveninChapter22.) Byaleastcommonmultipleoftwointegersrandswemeanapositiveintegermsuchthat (i) r|mands|m,and (ii) Ifr|xands|x,thenm|x. Inotherwords,misacommonmultipleofrandsandmisafactorofeveryothercommonmultipleofr ands.InChapter22itisshownthateverypairofintegersrandshasauniqueleastcommonmultiple. The least common multiple of r and s is denoted by lcm(r, s). The least common multiple has the followingproperties: Theorem6Foranyintegersa,b,andc, (i) Ifgcd(a,b)=1,thenlcm(a,b)=ab. (ii) Conversely,iflcm(a,b)=ab,thengcd(a,b)=1. (iii) Ifgcd(a,b)=c,thenlcm(a,b)=ab|c. (iv) lcm{a,ab)=ab. Theproofsareleftasexercises. EXERCISES Provethatthefollowingaretrueforanyintegersa,b,andc: 1Ifa|bandb|c,thena|c. 2Ifa|b,thena|(−b)and(−a)|b. 31|aand(−1)|a. 4a|0. 5Ifa|b,thenac|bc. 6Ifa>0,thengcd(a,0)=a. 7Ifgcd(ab,c)=1,thengcd(a,c)=1andgcd(b,c)=1. 8Ifthereareintegerskandlsuchthatka+lb=1,thenaandarerelativelyprime. 9Ifa|dandc|dandgcd(a,c)=1,thenac|d. 10Ifd|abandd|cbandgcd(a,c)=1,thend|b. 11Ifgcd(a,b)=1,thenlcm(a,b)=ab. 12Iflcm(a,b)=c,thengcd(a,b)=1. 13Ifgcd(a,b)=c,thenlcm(a,b)=ab/c. 14lcm(a,ab)=ab. 15a·lcm(b,c)=lcm(ab,ac). APPENDIX C REVIEWOFMATHEMATICALINDUCTION Thebasicassumptiononemakesabouttheorderingoftheintegersisthefollowing: Well-orderingprinciple.Everynonemptysetofpositiveintegershasaleastelement. Fromthisassumption,itiseasytoprovethefollowingtheorem,whichunderliesthemethodofproof byinduction: Theorem 1: Principle of mathematical induction Let A represent a set of positive integers. Considerthefollowingtwoconditions: (i) 1isinA. (ii) Foranypositiveintegerk,ifkisinA,thenk+1isinA. If A is any set of positive integers satisfying these two conditions, then A consists of all the positiveintegers. PROOF:IfAdoesnotcontainallthepositiveintegers,thenbythewell-orderingprinciple(above), thesetofallthepositiveintegerswhicharenotinAhasaleastelement;callitb.FromCondition(i),b≠ 1;henceb>1. Thus, b − 1 > 0, and b — 1 ∈ A because b is the least positive integer not in A. But then, from Condition(ii),b∈A,whichisimpossible.■ Here is an example of how the principle of mathematical induction is used: We shall prove the identity thatis,thesumofthefirstnpositiveintegersisequalton(n+1)/2. LetAconsistofallthepositiveintegersnforwhichEquation(1)istrue.Then1isinAbecause Next,supposethatkisanypositiveintegerinA;wemustshowthat,inthatcase,k+1alsoisinA.Tosay thatkisinAmeansthat Byaddingk+1tobothsidesofthisequation,weget thatis, Fromthislastequation,k+1∈A. We have shown that 1 ∈ A, and moreover, that if k ∈ A, then k + 1 ∈ A. So by the principle of mathematicalinduction,allthepositiveintegersareinA.ThismeansthatEquation(1)istrueforevery positiveinteger,asclaimed. EXERCISES Usemathematicalinductiontoprovethefollowing: 11+3+5+⋯+(2n−1)=n2 (Thatis,thesumofthefirstnoddintegersisequalton2.) 213+23+⋯+n3=(1+2+⋯+n)2 312+22+⋯+n2= n(n+1)(2n+1) 413+23+⋯+n3= n(n+1)2 5 612+22+⋯+(n−1)2< <12+22+⋯+n2 ANSWERSTOSELECTEDEXERCISES CHAPTER2 A3 Thisisnotanoperationon ,sincea*bisnotuniquelydefinedforanya,b∈ ,a≠0,andb≠0.If a≠0andb≠0,thentheequationx2−a2b2=0hastworoots—namely,x=a*b=±ab. B7 (ii) Associativelaw: (iii) Identityelement:Tofindanidentityelement(ifoneexists),wetrytosolvetheequationx*e= xfore: Crossmultiplyinggivesxe=x2+xe+x,andthus0=x(x+1).Consequently,thisequationcanbe solvedonlyforx=0andx=–1.Forpositivevaluesofxthereisnosolution. C3 Weshallexamineonlyoneoftheoperations—namely,theonewhosetableis Tosaythattheoperationisassociativeistosaythattheequation x*(y*z)=(x*y)*z istrueforallpossiblechoicesofx,ofy,andofz.Eachofthevariablesmaybeequaltoeitheraorb, yieldingeightcasestobechecked: 1. a*(a*a)=a*b=a, (a*a)*a=b*a=a 2. a*(a*b)=a*a=b, (a*a)*b=b*b=a 3. a*(b*a)=a*a=b, (a*b)*a=a*a=b 4. a*(b*b)=a*a=b, (a*b)*b=a*b=a 5. b*(a*a)=b*b=a, (b*a)*a=a*a=b 6. b*(a*b)=b*a=a, (b*a)*b=a*b=a 7. b*(b*a)=b*a=a, (b*b)*a=a*a=b 8. b*(b*b)=b*a=a, (b*b)*b=a*b=a Sincea*(a*b)≠(a*a)*b,*isnotassociative. Bytheway,thisoperationiscommutative,sincea*b=a=b*a.Usingcommutativity,weneed notcheckalleightcasesabove;infact,equalityincases1,3,6,and8followsusingcommutativity. Forexample,incase3,itfollowsfromcommutativitythata*(b*a)=(b*a)*a=(a*b)*a. Thus,forcommutativeoperations,onlyfouroftheeightcasesneedtobechecked.Ifyouareableto showthatitissufficienttocheckonlycases2and4forcommutativeoperations,yourworkwillbe furtherreduced. Wehavecheckedonlyoneofthe16operations.Checktheremaining15. CHAPTER3 A4 (ii)Associativelaw: (iii)Identityelement:Solvex*e=xfore. If thene(1−x2)=0,soe=0.Nowcheckthatx*0=0*x=x. (iv)Inverses:Solvex*x′=0forx′.(Youshouldgetx′=−x.)Thencheckthatx*(−x)=0=(−x) *x. B2 (ii)Associativelaw: (a,b)*[(c,d)*(e,f)]=(a,b)*(ce,de+f)=(ace,bce+de+f) [(a,b)*(c,d)]*f)=(ac,bc+d)*(e,f)=(ace,bce+de+f) (iii)Identityelement:Solve(a,b)*(e1,e2)=(a,b)for(e1,e2).Suppose (a,b)*(e1,e2)=(ae,be1+e2)=(a,b) Thisimpliesthatae1=aandbe1+e2=b,soe1=1ande2=0.Thus,(e1,e2)=(l,0).Nowcheckthat (a,b)*(1,0)=(1,0)*(a,b)=(a,b). (iv)Inverses:Solve(a,b)*(a′,b′)=(1,0)for(x′,b′).Youshouldgeta′=1/aandb′= −b/a.Then check. G1Theequationa4=a1+a3meansthatthefourthdigitofeverycodewordisequaltothesumofthefirst andthirddigits.WecheckthisfactfortheeightcodewordsofC1inturn:0=0+0,1=0+1,0=0+ 0,1=0+1,1=1+0,0=1+1,1=1+0,and0=1+1. G2 (a) As stated, the first three positions are the information positions. Therefore there are eight codewords; omitting the numbers in the last three positions (the redundancy positions), the codewordsare000,001,010,011,100,101,110,and111.Thenumbersintheredundancypositions arespecifiedbytheparity-checkequationsgivenintheexercise.Thus,thecompletecodewordsare 000000,001001,….(Completethelist.) G6 Letak ,bk ,xk denotethedigitsinthekthpositionofa,b,andx,respectively.Notethatifxk ≠ak and xk ≠bk ,thenak =bk .Butifxk ≠ak andxk =bk ,thenak ≠bk .Andifxk =ak andxk ≠bk ,thenak ≠ bk .Finally,ifxk =ak andxk =bk ,thenak =bk .Thus,adiffersfrombinonlythosepositionswherea differsfromxorbdiffersfromx.Sinceadiffersfromxintorfewerpositions,andbdiffersfromx intorfewerpositions,acannotdifferfrombinmorethan2tpositions. CHAPTER4 A3 Takethefirstequation,x2a=bxc−1,andmultiplyontherightbyc: x2ac=(bxc−1)c=bx(c−1c)=bxe=bx Fromthesecondequation,x2ae=x(xac)=xacx.Thus, xaca=bx Bythecancellationlaw(cancelxontheright), xac=b Nowmultiplyontherightby(ac)−1tocompletetheproblem. C1 FromTheorem3,a−1b−1=(ba)−1andb−1a−1=(ab)−1. D2 FromTheorem2,if(ab)c=e,thencistheinverseofab. G3 Let G and H be abelian groups. To prove that G × H is abelian, let (a, b) and (c, d) be any two elementsofG×H(sothata∈G,c∈G,b∈H,andd∈H)andshowthat(a,b)·(c,d)=(c,d)· (a,b). PROOF: CHAPTER5 B5 Supposef,g∈H.Thendf/dx=aanddg/dx=bforconstantsaandb.Fromcalculus,d(f+g)dx = df/dx+dg/dx=a+b,whichisconstant.Thus,f+g∈H. C5 PROOF:(i)Kisclosedunderproducts.Letx,y∈K.Thenthereareintegersn,m>0suchthatxn∈H and ym ∈ H. Since H is a subgroup of G, it is closed under products—and hence under exponentiation(whichisrepeatedmultiplicationofanelementbyitself).Thus(xn)m∈Hand(xm)n ∈H.Setq=mn.SinceGisabelian,(xy)q=xqyq=xmnymn=(xn)m(ym)n∈Hsinceboth(xn)mand (ym)nareinH.Completetheproblem. D5 S={a1…,an}hasnelements.Thenproductsa1a1,a1a2,…,a1anareelementsofS(why?)andno two of them can be equal (why?). Hence every element of S is equal to one of these products. In particular, a1 = a1ak for some k. Thus, a1e = a1ak , and hence e = ak . This shows that e ∈S. Now completetheproblem. D7 (a)Supposex∈K.Then (i)ifa∈H,thenxax−1∈H,and (ii)ifxbx−1∈H,thenb∈H. Weshallprovethatx−1∈K:wemustfirstshowthatifa∈H,thenx−1a(x−1)−1=x−1ax∈H. Well,a=x(x−1ax)x−1andifx(x−1ax)x−1∈H,thenx−1ax∈Hby(ii)above.[Use(ii)withx−1ax replacingb.]Conversely,wemustshowthatifx−1ax∈H,thena∈H.Well,ifx−1ax∈H,thenby (i)above,x(x−1ax)x−1=a∈H. E7 Webeginbylistingalltheelementsof 2× 4obtainedbyadding(1,1)toitselfrepeatedly:(1,1); (1,1)+(1,1)=(0,2);(1,1)+(1,1)+(1,1)=(1,3);(1,1)+(1,1)+(1,1)+(1,1)=(0,0).Ifwe continue adding (1, 1) to multiples of itself, we simply obtain the above four pairs over and over again.Thus,(1,1)isnotageneratorofallof 2× 4. Thisprocessisrepeatedforeveryelementof 2× 4.Noneisageneratorof 2× 4;hence 2× isnotcyclic. F1 ThetableofGisasfollows: 4 Usingthedefiningequationsa2=e,b3=e,andba=ab2,wecomputetheproductofabandab2 in thisway: (ab)(ab2)=a(ba)b2=a(ab2)b2=a2b3b=eeb=b Completetheproblembyexhibitingthecomputationofallthetableentries. H3 Recallthedefinitionoftheoperation+inChapter3,ExerciseF:x+yhassinthosepositionswhere xandydiffer,and0selsewhere. CHAPTER6 A4 Fromcalculus,thefunctionf(x)=x3–3xiscontinuousandunbounded.Itsgraphisshownbelow.[f isunboundedbecausef(x)=x(x2 −3)isanarbitrarilylargepositivenumberforsufficientlylarge positive values of x, and an arbitrarily large negative number (large in absolute value) for sufficientlylargenegativevaluesofx.]Becausefiscontinuousandunbounded,therangeoffis . Thus,fissurjective.Nowdeterminewhetherfisinjective,andproveyouranswer. Graphoff(x)=x3−3x A6fisinjective:Toprovethis,notefirstthatifxisanintegerthenf(x)isaninteger,andifxisnotan integer,thenf(x)isnotaninteger.Thus,iff(x)=f(y),thenxandyareeitherbothintegersorboth nonintegers.Case 1, both integers: then f(x) = 2x, f(y) = 2y, and 2x = 2y; so x = y. Case 2, both nonintegers:⋯(Completetheproblem.Determinewhetherfissurjective.) F5 LetA={a1,a2,…,an).IffisanyfunctionfromAtoA,therearenpossiblevaluesforf(ax),namely, a1,a2,…,an.Similarlytherearenpossiblevaluesforf(a2).Thustherearen2pairsconsistingofa valueoff(a1)togetherwithavalueoff(a2).Similarlytherearen3 triples consisting of a value of f(a1),avalueoff(a2),andavalueoff(a3).Wemaycontinueinthisfashionandconcludeasfollows: Sinceafunctionfisspecifiedbygivingavalueforf(a1),avalueforf(a2),andsoonuptoavalue for f(an), there are nn functions from A to A. Now, by reasoning in a similar fashion, how many bijectivefunctionsarethere? H1 Thefollowingisoneexample(thoughnottheonlypossibleone)ofamachinecapableofcarrying outtheprescribedtask: A={a,b,c,d} S={s0,s1,s2,s3,s4} Thenext-statefunctionisdescribedbythefollowingtable: Toexplainwhythemachinecarriesouttheprescribedfunction,notefirstthatthelettersb,c,and dnevercausethemachinetochangeitsstate—onlyadoes.Ifthemachinebeginsreadingasequence instates0,itwillbeinstates3afterexactlythreea’shavebeenread.Anysubsequenta’swillleave themachineinstates4.Thus,ifthemachineendsins3,thesequencehasreadexactlythreea’s.The machine’sstatediagramisillustratedbelow: I4 M1hasonlytwodistincttransitionfunctions,whichweshalldenotebyToandTe(whereomaybe anysequencewithonoddnumberof1s,andeanysequencewithanevennumberof1s).ToandTe maybedescribedasfollows: To(s0),=s1, To(s1)=s0 Te(s0),=s0, Te(s1)=s1 Now,Te ∘To=Toebypart3.Sinceeisasequencewithanevennumberof1sandoisasequence withanoddnumberof1s,oeisasequencewithanoddnumberof1s;henceToe=To.Thus,Te ∘To =To.Similarly Te∘Te=Te, To∘Te=To, Inbrief,thetableof (M1)isasfollows: andTo∘To=Te The table shows that (M1) is a two-element group. The identity element is Te, and To is its own inverse. CHAPTER7 D2 fn+misthefunctiondefinedbytheformula fn+m(x)=x+n+m Usethisfacttoshowthatfn∘fm=fn+m. Inordertoshowthatf−nistheinverseoffn,wemustverifythatfn∘f−n=εandf−n∘fn=ε.We verifythefirstequationasfollows:f−n(x)=x+(−n);hence [fn∘f−n](x)=fn(f−n(x))=fn(x−n)=x−n+n=x=ε(x) Since[fn∘f−n](x)=ε(x)foreveryxin ,itfollowsthatfn∘f−n=ε. E3 Toprovethatf1/a,−b/aistheinverseoffa,b,wemustverifythat fa,b∘f1/a,−b/a=ε and f1/a,−b/a∘fa,b=ε Toverifythefirstequation,webeginwiththefactthatf1/a, −b/a(x) = x/a − b/a. Now complete the problem. H2 Iff,g∈G,thenfmovesafinitenumberofelementsofA,saya1,…,an,andgmovesafinitenumber ofelementsofA,sayb1,…,bm.NowifxisanyelementofAwhichisnotoneoftheelementsa1,…, an,b1,…,bm,then [f∘g](x)=f(g(x))=f(x)=x Thus,f∘gmovesatmostthefinitesetofelements{a1,…,an,b1,…,bm}. CHAPTER8 A1(e) A4(f) γ3=γ∘γ∘γ=(12345);α−1=(4173);thus;γ3α−1=γ3∘α−1=1(12345)∘(4173)=(1 74235) B2 andsoon.Notethatα2(a1)=a3,α2(a2)=a4,…,α2(as −2)=as.Finally,α2(as −1)=alandα2(as)= a2.Thus,α2(ai)=a?Completetheproblemusingadditionmodulos,page27. B4 Letα=(a1a2⋯as)wheresisodd.Thenα2=(a1a3a5⋯asa2a4⋯as−1).Ifsiseven,thenα2=? Completetheproblem. E2 If α and β are cycles of the same length, α = (a1 ⋯ as) and β = (b1 ⋯ bs, let π be the following permutation:π(k)=bifori=1,…,sandπ(k)=kfork≠a1, …, as, b1 …, bs. Finally, let π map distinctelementsof{b1…,bs)−{a1,…,as)todistinctelementsof{a1,…,as}–{b1,…,bs}.Now completetheproblem,supplyingdetails. F1 whenkisapositiveinteger,k<s Forwhatvaluesofkcanyouhaveαk =ε? H2 UseExerciseH1andthefactthat(ij)=(1i)(1j)(1i). CHAPTER9 C1 ThegrouptablesforGandHareasfollows: GandHarenotisomorphicbecauseinGeveryelementisitsowninverse(VV=I,HH=I,andDD =I),whereasinHthereareelementsnotequaltotheirinverse;forexample,(-i)(-i)= −1≠1.Find atleastoneotherdifferencewhichshowsthatG H. C4ThegrouptablesforGandHareasfollows: GandHareisomorphic.Indeed,letthefunctionf:G→Hbedefinedby Byinspection,ftransformsthetableofGintothetableofH,Thus,fisanisomorphismfromGtoH. El Showthatthefunctionf: →Egivenbyf(n)=2nisbijectiveandthatf(n+m)=f(n)+f(m). F1 Checkthat(24)2=ε,(1234)4=ε,and(1234)(24)=(24)(1234)3.NowexplainwhyG≅G′. H2 LetfG:G1→G2andfH:H1→H2beisomorphisms,andfindanisomorphismf:G1×H1→G2× H2. CHAPTER10 A1(c) Ifm<0andn<0,letm=−kandn=l,wherek,l>0.Thenm+n=−(k+l).Now, am=a−k =(a−1)k and an=a−1=(a−1)l Hence aman=(a−1)k (a−1)l=(a−1)k+l=a−(k+l)=am+n B3 Theorderoffis4.Explainwhy. C4 Foranypositiveintegerk,ifak =e,then Conversely,if(bab−1k =bak b−1=e,thenak =e.(Why?)Thus,foranypositiveintegerk,ak =eiff (bab−1)k =e.Nowcompletetheproblem. D2Lettheorderofabeequalton.Then(ak )n=ank =(an)k =ek =e.NowuseTheorem5. F2 Theorderofa8is3.Explainwhy. H2 Theorderis24.Explainwhy. CHAPTER11 A6 Ifkisageneratorof ,thismeansthat consistsofallthemultiplesofk;thatis,k,2k, 3k, etc., as wellas0and−k,−2k,−3k,etc.: B1 LetGbeagroupofordern,andsupposeGiscyclic,sayG=〈a〉.Thena,thegeneratorofG,isan element of order n. (This follows from the discussion on the first two pages of this chapter.) Conversely, let G be a group of order n (that is, a group with n elements), and suppose G has an elementaofordern.ProvethatGiscyclic. C4 ByExerciseB4,thereisanelementbofordermin〈a〉,andb∈Cm.SinceCmisasubgroupof〈a〉, whichiscyclic,weknowfromTheorem2thatCmiscyclic.SinceeveryelementxinCmsatisfiesxm =e,noelementinCmcanhaveordergreaterthanm.Nowcompletetheargument. C7 First,assumethatord(ar)=m.Then(ar)m=arm=e.UseTheorem5ofChapter10toshowthatr= klforsomeintegerl.Toshowthatlandmarerelativelyprime,assumeonthecontrarythatlandm haveacommonfactorq;thatis, l=jq m=hq h<m Nowraiseartothepowerhandderiveacontradiction.Conversely,supposer=lkwherelandm arerelativelyprime.Completetheproblem. D6 Let(a)beacyclicgroupofordermn.If〈a〉hasasubgroupofordern,thissubgroupmustbecyclic, andgeneratedbyoneoftheelementsin〈a〉.Say〈ak 〉isasubgroupofordern.Thenak hasordern, so(ak )n=akn=e.ByTheorem5ofChapter10,kn=mnqforsomeintegerq(explainwhy);hencek =mqandthereforeak =(am)q∈〈am〉. Usethisinformationtoshowthat(i)〈a〉hasasubgroupofordernand(ii)〈a〉hasonlyone subgroupofordern. F3 Ifthereissomejsuchthatam=(aj )k =ajk ,thene=am(ajk )−1=am−jk ;sobyTheorem5ofChapter 10,m −jk=nqforsomeintegerq.(Explainwhy.)Thus,m=jk+nq.Nowcompletetheproblem, supplyingalldetails. CHAPTER12 A2 Letxbeanyrationalnumber(anyelementof ),andsupposex∈Ai.andx∈Aj ,whereiandj are integers.Theni≤x<i+1andj≤x<j+1.Nowifi<j,theni+1≥j;sox<i+1impliesthatx< j,whichisacontradiction.Thuswecannothavei<j,norcanwehavej<i;hencei=j.Wehave shown that if Ai. and Aj have a common element x, then Ai = Ai: this is the first condition in the definitionofpartition.Forthesecondcondition,notethateveryrationalnumberisanintegerorlies betweentwointegers:ifi≤x<i+1,thenxisinAi C1 Ar consists of all the points (x, y) satisfying y = 2x + r; that is, Ar is the line with slope 2 and y interceptequaltor.Thus,theclassesofthepartitionareallthelineswithslope2. Forthecorrespondingequivalencerelation,wesaythattwopoints(x,y)and(u,υ)are “equivalent,”thatis, (x,y)~(u,υ) ifthetwopointslieonthesamelinewithslope2: Completethesolution,supplyingdetails. D5 Ifab−1commuteswitheveryxinG,thenwecanshowthatba−1commuteswitheveryxinG: ba−1x=(x−1ab−1)’1=(ab−1x−1)−1 (why?) CHAPTER13 B1 Notefirstthattheoperationinthecaseofthegroup isaddition.Thesubgroup(3)consistsofallthe multiplesof3,thatis, 〈3〉={...,−9,−6,−3,0,3,6,9,…} Thecosetsof(3)are(3)+0=(3),aswellas 〈3〉+1={...,−8,−5,−2,1,4,7,10,…} 〈3〉+2={...,−7,−4,−1,2,5,8,11,…} Notethat〈3〉+3=〈3〉,〈3〉+4=〈3〉+1,andsoon;hencethereareonlythreecosetsof〈3〉,namely, 〈3〉=〈3〉+0 〈3〉+1〈3〉+2 C6Everyelementaoforderpbelongsinasubgroup〈a〉.Thesubgroup〈a〉hasp–1elements(why?), andeachoftheseelementshasorderp(why?).Completethesolution. D6 Foronepartoftheproblem,useLagrange’stheorem.Foranotherpart,usetheresultofExerciseF4, Chapter11. E4 TosaythataH=HaistosaythateveryelementinaHisinHaandconversely.Thatis,foranyh∈ H,thereissomek∈Hsuchthatah=kaandthereissomel∈Hsuchthatha=al.(Explainwhythis is equivalent to aH = Ha.) Now, an arbitrary element of a− 1H is of the form a− 1h = (h− 1a)− 1. Completethesolution. J3 O(1)=O(2)={1,2,3,4};G1={ε,β};G2={ε,αβα}.Completetheproblem,supplyingdetails. CHAPTER14 A6 Weusethefollowingpropertiesofsets:ForanythreesetsX,YandZ, (i)(X∪Y)∩Z=(X∩Z)∪(Y∩Z) (ii)(X−Y)∩Z=(X∩Z)−(Y∩Z) Nowhereistheproofthathisahomomorphism:LetCandDbeanysubsetsofA;then Nowcomplete,using(i)and(ii). C2 Letfbeinjective.ToshowthatK={e},takeanyarbitraryelementx∈Kandshowthatnecessarily x=e.Well,sincex∈K,f(x)=e=f(e).Nowcomplete,andprovetheconverse:AssumeK={e} …. D6 ConsiderthefollowingfamilyofsubsetsofG:{Hi:i∈I},whereeachHiisanormalsubgroupof G.ShowthatH= Hi is a normal subgroup of G. First, show that H is closed under the group operation:well,ifa,b∈H,thena∈Hiandb∈Hiforeveryi∈I.SinceeachHiisasubgroupof G,ab∈Hiforeveryi∈I;henceab∈ Hi.Nowcomplete. E1 IfHhasindex2,thengispartitionedintoexactlytworightcosetsofH;alsoG is partitioned into exactlytwoleftcosetsofH.OneofthecosetsineachcaseisH. E6 First,showthatifx∈Sandy∈S,thenxy∈S.Well,ifx∈S,thenx∈Ha=aHforsomea∈G. Andify∈S,theny∈Hb=bHforsomeb∈G.Showthatxy∈H(ab)andthatH(ab)=(ab)Hand thencompletetheproblem. I4 It is easy to show that aHa− 1 ⊆ H. Show it. What does Exercise I2 tell you about the number of elementsinaHa−1 I8 LetX={aHa−1:a∈G}bethesetofalltheconjugatesofH,andletY={aN:a∈G}bethesetof allthecosetsofN.Findafunctionf:X→Yandshowthatfisbijective. CHAPTER15 C4 EveryelementofG/HisacosetHx.AssumeeveryelementofG/Hhasasquareroot:thismeansthat foreveryx∈G, Hx=(Hy)2 forsomey∈G.AvailyourselfofTheorem5inthischapter. D4 Let H be generated by {h1, …, hn and let G/H be generated by {Ha1, …, Ham). Show that G is generatedby {a1,…,am,h1,…,hn} thatis,everyxinGisaproductoftheaboveelementsandtheirinverses. E6 Everyelementof / isacoset +(m/n). G6 IfGiscyclic,thennecessarilyG≅ p2.(Why?) IfGisnotcyclic,theneveryelementx≠einGhasorderp.(Why?)Takeanytwoelementsa≠eand b≠einGwherebisnotapowerofa.Completetheproblem. CHAPTER16 D1 Letf∈Aut(G);thatis,letfbeanisomorphismfromGontoG.Weshallprovethatf−1 ∈ Aut(G); thatis,f−1isanisomorphismfromGontoG.Tobeginwith,itfollowsfromthelastparagraphof Chapter 6 that f− 1 is a bijective function from G onto G. It remains to show that f− 1 is a homomorphism.Letf−(c)=aandf− 1(d) = b, so that c = f(a) and d = f(b). Then cd = f(a)f(b) = f(ab),whencef−1(cd=ab.Thus, f−1(cd)=ab=f−1(c)f−1(d) whichshowsthatf−1isahomomorphism. F2 Ifa,b∈HK,thena=h1k1andb=h2k2,whereh1,h2∈Handk1,k2 ∈ K. Then ab = h1k1h2k2 = . G3 Notethattherangeofhisagroupoffunctions.Whatisitsidentityelement? H1 Fromcalculus,cos(x+y)=cosxcosy−sinxsiny,andsin(x+y)=sinxcosy+cosxsiny. L4Thenaturalhomomorphism(Theorem4,Chapter15)isahomomorphismf:G→G/〈a〉 with kernel 〈a〉.LetSbethenormalsubgroupofG/〈a〉whoseorderispm −1.(TheexistenceofSisassuredby part3ofthisexerciseset.)ReferringtoExerciseJ,showthatS*isanormalsubgroupofG,andthat theorderofS*ispm. CHAPTER17 A3 Weprovethat isassociative: Thus,(a,b) [(c,d) (p,q)]=[(a,b) (c,d)] (p,q). B2 A nonzero function f is a divisor of zero if there exists some nonzero function g such that fg = 0, where0isthezerofunction(page46).Theequationfg=0meansthatf(x)g(x)=0(x)foreveryx∈ .Veryprecisely,whatfunctionsfhavethisproperty? D1 Forthedistributivelaw,refertothediagramonpage30,andshowthatA∩(B+C)=(A∩B)+(A ∩C): B + C consists of the regions 2, 3, 4, and 7; A ∩ (B + C) consists of the regions 2 and 4. Now completetheproblem. E3 G4 (a,b)isaninvertibleelementofA×Biffthereisanorderedpair(c,d)inA×Bsatisfying(a,b)· (c,d)=(1,1).Nowcomplete. H6 IfAisaring,then,aswehaveseen,Awithadditionastheonlyoperationisanabeliangroup:this groupiscalledtheadditivegroupoftheringA.Now,supposetheadditivegroupofA is a cyclic group,anditsgeneratorisc.IfaandbareanytwoelementsofA,then a=c+c+⋯+c (mterms) and b=c+c+⋯+c (nterms) forsomepositiveintegersmandn. J2 Ifabisadivisorof0,thismeansthatab≠0andthereissomex≠0suchthatabx=0.Moreover,a≠ 0andb≠0,forotherwiseab=0. M3 Supposeam=0andbn=0.Showthat(a+b)m+n=0.Explainwhy,ineverytermofthebinomial expansionof(a+b)m+n,eitheraisraisedtoapower≥m,orbisraisedtoapower≥n. CHAPTER18 A4 Fromcalculus,thesumandproductofcontinuousfunctionsarecontinuous. B3 Theproofhingesonthefactthatifkandaareanytwoelementsof n,then ka=a+a+·+a (kterms) C4 If the cancellation law holds in A, it must hold in B. (Why?) Why is it necessary to include the conditionthatBcontains1? C5 Let B be a subring of a field F. If b is any nonzero element of B, then b− 1 is in F, though not necessarilyinB.(Why?)Completetheargument. E5 f(x,y)f(u,υ)= =f[x,y) (u,υ)]= Completetheproblem. H3 Ifan∈Jandbm ∈ J, show that (a + b)n + m ∈ J. (See the solution of Exercise M3, Chapter 17.) Completethesolution. CHAPTER19 E1 TosaythatthecosetJ+xhasasquarerootistosaythatforsomeelementyinA,J+x=(J+y)(J+ y)=J+y2. E6 AunityelementofA/JisacosetJ+asuchthatforanyx∈A, (J+a)(J+x)=J+x and (J+x)(J+a)=J+a G1 Tosaythata∉JisequivalenttosayingthatJ+a≠J;thatis,J+aisnotthezeroelementofA/J. Explainandcomplete. CHAPTER20 E5 RestrictyourattentiontoAwithadditionalone,andseeChapter13,Theorem4. E6 Forn=2youhave Provetherequiredformulabyreasoningsimilarlyandusinginduction:assumetheformulaistruefor n=k,andproveforn=k+1. CHAPTER21 B5 Usetheproduct(a−1)(b−1). C8 Intheinductionstep,youassumetheformulaistrueforn=kandproveitistrueforn=k+1.That is,youassume Fk+1Fk+2−FkFk+3=(–1)k andprove Fk+2Fk+3−Fk+1Fk+4=(−1)k+1 RecallthatbythedefinitionoftheFibonaccisequence,Fn+2=Fn+1+Fnforeveryn>2.Thus, Fk+2=Fk+1+FkandFk+4=Fk+3+Fk+2.Substitutetheseinthesecondoftheequationsabove. E5 Anelegantwaytoprovethisinequalityisbyuseofpart4,witha+bintheplaceofa,and|a|+|b|in theplaceofb. E8 Thiscanbeprovedeasilyusingpart5. F2 Youaregiventhatm=nq+randq=kq1+r1where0≤r<nand0≤r1<k.(Explain.)Thus, m=n(kq1+r1+r=(nk)q1+(nr1+r) Youmustshowthatnr1+r<nk,(Why?)Beginbynotingthatk−r1>0;hencek–r1≥1,son(k–r1) ≥n. G5 Fortheinductionstep,assumek·a=(k·1)a,andprove(k+1)·a=[(k+1)·1]a.From(ii)inthis exercise,(k+1)·1=k·1+1. CHAPTER22 B1 Assumea>0anda|b.Tosolvetheproblem,showthataisthegreatestcommondivisorofaandb. First,aisacommondivisorofaandb:a|aanda|b.Next,supposetisanycommondivisorofaand b:t|aandt|b.Explainwhyyouarenowdone. D3 FromtheproofofTheorem3,disthegeneratoroftheidealconsistingofallthelinearcombinations ofaandb. E1 Supposeaisoddandbiseven.Thena+bisoddanda −bisodd.Moreover,iftisanycommon divisorofa–banda+b,thentisodd.(Why?)Notealsothatifrisacommondivisorofa−band a+b,thentdividesthesumanddifferenceofa+banda−b. F3Ifl=lcm(ab,ac),thenl=abx=acyforsomeintegersxandy.Fromtheseequationsyoucanseethat aisafactorofl,sayl=am.Thus,theequationsbecomeam=abx=acy.Cancela,thenshowthatm =lcm(b,c). G8 LookattheproofofTheorem3. CHAPTER23 A4(f) 3x2−6x+6=3(x2−2x+1)+3=3(x−1)2+3.Thus,wearetosolvethecongruence3(x−1)2 ≡−3(mod15).Webeginbysolving3y≡−3(mod15),thenwewillsety=(x−1)2. WenotefirstthatbyCondition(6),inacongruenceax=b(modn),ifthethreenumbersa,b,and nhaveacommonfactord,then (That is, all three numbers a, b, and n may be divided by the common factor d.) Applying this observationtoourcongruence3y≡−3(mod15),weget 3y≡−3(mod15) isequivalentto y≡−1(mod5) Thisisthesameasy≡4(mod5),becausein 5thenegativeof1is4. Thus,ourquadraticcongruenceisequivalentto (x−1)2≡4(mod5) In 5,22=4and32=4;hencethesolutionsarex−1=2(mod5)andx−1≡3(mod5),orfinally, x≡3(mod5) and x≡4(mod5) A6(d) Webeginbyfindingallthesolutionsof30z+24y=18,thensetz=x2.Now, 30z+24y=18 iff 24y=18−30z iff 30z≡18(mod24) Bycommentsintheprevioussolution,thisisequivalentto5z≡3(mod4).Butin 4, = ;hence5z =3(mod4)isthesameasz≡3(mod4). Nowsetz=x2.Thenthesolutionisx2≡3(mod4).Butthislastcongruencehasnosolution, becausein 4, isnotthesquareofanynumber.Thus,theDiophantineequationhasnosolutions. B3 Hereistheidea:ByTheorem3,thefirsttwoequationshaveasimultaneoussolution;byTheorem4, itisoftheformx≡c(modt),wheret=lcm(m1,m2).Tosolvethelattersimultaneouslywithx≡c3 (mod m3), you will need to know that c3 ≡ c [mod gcd(t, m3)]. But gcd(t, m3) = lcm(d13, d23). (Explainthiscarefully,usingtheresultofExerciseH4inChapter22.)FromTheorem4(especially the last paragraph of its proof), it is clear that since c3 ≡ c1 (mod d13) and c3 ≡ c2 (mod d23), thereforec3≡c[modlcm(d13,d23)]. Thus,c3≡c[modgcd(t,m3)],andthereforebyTheorem3thereisasimultaneoussolutionofx≡ c(modt)andx≡c3(modm3).Thisisasimultaneoussolutionofx≡c1(modm1),x≡c2(modm2), andx≡c3(modm3).Repeatthisprocessktimes. Anelegant(andmathematicallycorrect)formofthisproofisbyinductiononk,wherekisthe numberofcongruences. B5(a) SolvingthepairofDiophantineequationsisequivalenttosolvingthepairoflinearcongruences 4x≡2(mod6)and9x=3(mod12).NowseetheexampleatthebeginningofExerciseB. D6 Usethedefinitions,andformtheproduct(ab−1)(ac−1)(bc−1). E4 Usetheproduct(pq−1−1)(qp−1−1). E6 UseExerciseD5. E8(a) Notethefollowing: (i)133=7×19. (ii)18isamultipleofboth7-1and19-1. Nowusepart6(b). F8 Consider(nϕ(m)−1)(mϕ(n)−1). H2 Inanycommutativering,ifa2=b2,then(a+b)(a −b)=0;hencea=±b.Thus,foranya≠0,the twoelementsaand−ahavethesamesquare;andnox≠±acanhavethesamesquareas±a. H4 ( )=−1because17isnotaquadraticresiduemod23. CHAPTER24 A1 Wecomputea(x)b(x)in 6[x]: a(x)b(x)=(2x2+3x+1)(x3+5x2+x)=2x5+13x4+18x3+8x2+x Butthecoefficientsarein 6,andin 6 13=1 18=0 and 8=2 Thus,a(x)b(x)=2x5+x4+2x2+xin 6[x]. Notethatwhilethecoefficientsarein 6,theexponentsarepositiveintegersandarenotin 6. Thus,forexample,in 6[x]thetermsx8andx2arenotthesame. B3 In 5[x],thereare5×5×5=125differentpolynomialsofdegree2orless.Explainwhy.Thereare 5×5=25polynomialsofdegree1or0;hencethereare125 −25=100differentpolynomialsof degree2.Explain,thencompletetheproblem. C7 In 9[x],x2=(x+3)(x+6)=(2x+3)(5x+6)=etc.Systematicallylistallthewaysoffactoringx2 intofactorsofdegree1,andcompletetheproblem. D6Afterprovingthefirstpartoftheproblem,youcanprovethesecondpartbyinductiononthenumbern oftermsinthepolynomial.Forn=2,itfollowsfromthefirstpartoftheproblem.Next,assumethe formulaforkterms, andprovefork+1terms: [(a0+a1x+⋯ak xk )+ak+1xk+1]p= (Complete.) E5 Ifa(x)=a0+a1x+⋯anxn∈J,thena0+a1+⋯+an=0.Ifb(x)isanyelementinA[x],sayb(x)= b0+b1x+⋯+bmxm,letB=b0+b1+⋯+bm.Explainwhythesumofallthecoefficientsina(x)b(x) isequalto(a0+a1⋯+an)B.Supplyallremainingdetails. G3 Ifhissurjective,theneveryelementofBisoftheformh(a)forsomeainA.Thus,anypolynomial withcoefficientsinBisoftheformh(a0)+h(a1)x+⋯+h(an)xn. CHAPTER25 A4 Assumetherearea,b∈ 5suchthat(x+a)(x+b)=x2+2.Notethatin 5,only1and4aresquares. D4 Let〈p(x)〉⊂JwhereJisanidealofF[x],andassumethereisapolynomiala(x)∈Jwherea(x)∉ 〈p(x)〉.Thena(x)andp(x)arerelativelyprime.(Why?)Completethesolution. F2 Thegcdofthegivenpolynomialsisx+1. CHAPTER26 A2 Tofindtherootsofx100−1in 7[x],reasonasfollows:FromFermat’stheorem,ifa∈ 7,thenin 7, a69=1foreveryintegerq.Inparticular,a96=1;hencea100=a96a4=a4.Thusanyroot(in 7)ofx100 −1isarootofx4−1. B1 Anyrationalrootof9x3+18x2−4x−8isafractions/twhere s=±1,±8,±2,or±4 and t=±1,±9,or±3 Thus,thepossiblerootsare±1,±2,±4,±8,±1/9,±1/3,±8/9,±8/3,±2/9,±2/3,±4/9,and ±4/3.Onceasingleroothasbeenfoundbysubstitutionintothepolynomial,theproblemmaybe simplified.Forexample,−2isonerootoftheabove.Youmaynowdividethegivenpolynomialbyx +2.Theremainingrootsarerootsofthequotient,whichisaquadratic. C5Provebyinduction:Forn=1,weneedtoshowthatifa(x)isamonicpolynomialofdegree1,say a(x)=x–c,anda(x)hasonerootc1,thena(x)=x–c1.Well,ifc1isarootofa(x)=x–c, then a(c1)=c1–c=0,soc=c1;thus,a(x)=x−c1. Fortheinductionstep,assumethetheoremistruefordegreekandkroots,andproveitistrue fordegreek+1andk+1roots. C8 Ifx2−x=x(x−1)=0,thenx=0orx−1=0.Doesthisruleholdinboth 10and 11?Why? D3 Notethatifpisaprimenumber,and0<k<p,thenthebinomialcoefficient isamultipleofp. (Seepage202.) F1 Thefactthata(x)ismonicisessential.Explainwhythedegreeof (a(x))isthesameasthedegreeof a(x). H3 Assumetherearetwopolynomialsp(x)andq(x),eachofdegree≤n,satisfyingthegivencondition. Showthatp(x)=q(x). I4 Ifa(x)andb(x)determinethesamefunction,thena(x)−b(x)isthezerofunction;thatis,a(c)–b(c)= 0foreveryc∈F. CHAPTER27 A1(e) Let ;thena2=i− anda4=−1− i+2=1−2 i.Thus,a4−1= −2 i,and(a4−1)2=−8.Thereforeaisarootofthepolynomial(x4–1)2+8,thatis, x8−2x4+9 B1(d) Ofthesixparts,(a)−(f),ofthisexercise,only(d)involvesapolynomialofdegreehigherthan4; henceitisthemostpainstaking.Leta= ;thena2=2+31/3anda2−2=31/3.Thus,(a2− 2)3=3,soaisarootofthepolynomial p(x)=(x2−2)3−3=x6−x6−12x2−11 The bulk of the work is to ensure that this polynomial is irreducible. We will use the methods of Chapter26,ExercisesFandE.Bythefirstofthesemethods,wetransformp(x)intoapolynomialin Z3[x]: x6−2=x6+1 Sincenoneofthethreeelements0,1,2in 3isarootofthepolynomial,thepolynomialhasnofactor ofdegree1in 3[x].Sotheonlypossiblefactoringsintononconstantpolynomialsare x6+1=(x3+ax2+bx+c)(x3+dx2+ex+f) or x6+1=(x4+ax3+bx2+cx+d)(x2+ex+f) Fromthefirstequation,sincecorrespondingcoefficientsareequal,wehave: From(1),c=f=±1,andfrom(5),a+d=0.Consequently,af+cd=c(a+d)=0,andby(3), eb=0.Butfrom(2)(sincec=f),b+e=0,andthereforeb=e=0.Itfollowsfrom(4)thatc+f= 0,whichisimpossiblesincec=f=±1.Wehavejustshownthatx6+1cannotbefactoredintotwo polynomialseachofdegree3.Completethesolution. C4 Frompart3,theelementsof 2(c)are0,1,c,andc+1.(Explain.)Whenaddingelements,notethat0 and1areaddedasin 2,since 2(c)isanextensionof 2.Moreover,c+c=c(1+1)=c0=0.When multiplyingelements,notethefollowingexample:(c+1)2=c2+c+c+1=c(becausec2+c+1= 0). D1 SeeExerciseE3,thischapter. D7 In ( ),1+1isasquare;soif ( )≅ ( ),then1+1isasquarein ( ),thatis, ∈ ( ).Butalltheelementsof ( )areoftheforma +bfora,b∈∈.(Explainwhy.)Sowe wouldhave =a +b.Squaringbothsidesandsolvingfor (supplydetails)showsthat isarationalnumber.Butitisknownthat isirrational. G2 LetQdenotethefieldofquotientsof{a(c):a(x)∈F[x]}.SinceF(c)isafieldwhichcontainsF andc,itfollowsthatF(c)containseverya(c)=a0+a1c+⋯+ancnwherea0,…,an∈F.Thus,Q ⊆F(c).Conversely,bydefinitionF(c)iscontainedinanyfieldcontainingFandc;henceF(c)⊆Q. Completethesolution. CHAPTER28 B1 LetU={(a,b,c):2a−3b+c=0}.ToshowthatUisclosedunderaddition,letu,v∈U.Thenu= (a1,b1,c1)where2a1−3b1+c1=0,andv=(a2,b2,c2)where2a2−3b2+c2=0.Adding, u+v=(a1+a2,b1+b2,c1+c2) and 2(a1+a2)−3(b1+b2)+(c1+c2)=0 C5(a) S1={(x,y,z):z=2y −3x}.AsuggestedbasisforS1wouldconsistofthetwovectors(1,0,?) and(0,1,?)wherethethirdcomponentsatisfiesz=2y−3x,thatis, (1,0,−3) and (0,1,2) NowshowthatthesetofthesetwovectorsislinearlyindependentandspansS1. D6 Supposetherearescalarsk,l,andmsuchthat k(a+b)+l(b+c)+m(a+c)=0 Usethefactthat{a,b,c}isindependenttoshowthatk=l=m=0. E5 Letkr+1,kr+2,…,knbescalarssuchthat kr+1h(ar+1)+⋯+knh(an)=b Hence h(kr+1ar+1+⋯+knan)=0 Thenkr+1ar+1+⋯+knan∈ .Recallthat{a1,…,ar}isabasisfor ,andcompletetheproof. F2 UsetheresultofExerciseE7. G2 Assumethateveryc∈Vcanbewritten,inauniquemanner,asc=a+bwherea∈Tandb∈U. Thus,everyvectorinVisanelementofT+U,andconversely,sinceTandUaresubspacesofV, everyvectorinT+UisanelementofV.Thus, V=T+U InordertoshowthatV=T⊕U,itremainstoshowthatT U={0};thatis,ifc∈T Uthenc=0. Completethesolution. CHAPTER29 A3 Notefirstthata2−= ;hence ∈ (a)andtherefore (a)= ( ,a).(Explain.)Nowx3−2 (whichisirreducibleover byEisenstein’scriterion)istheminimumpolynomialof ;hence[ ( ): ]=3.Next,in ( ),asatisfiesa2−1− =0,soaisarootofthepolynomialx2−(1 + ). This quadratic is irreducible over ( )[x] (explain why); hence x2 − (1 + ) is the minimum polynomial of a over ( )[x]. Thus, [ ( , a) : ( ] = 2. Use Theorem 2 to complete. A4 Thisissimilartopart3;adjoinfirst ,thena. C2 Notethatx2+x+1isirreduciblein 2[x]. D2 SupposethereisafieldLproperlybetweenFandK,thatis,F L K.Asvectorspacesoverthe fieldF,LisasubspaceofK.(Why?)UseChapter28,ExerciseD1. F4 Therelationshipbetweenthefieldsmaybesketchedasfollows: [K(b):F]=[K(b):F(b)]·[F(b):F]=[K(b):K]·[K:F] F5 Reasoningasinpart4(andusingthesamesketch),showthat[K(b):K]=[F(b):F],Here,b is a rootofp(x). CHAPTER30 B3 If(a,b)∈ × ,thenaandarerationalnumbers.ByExerciseA5andthedefinitionof ,(a,0)and (b,0)areconstructiblefrom{O,I}.Withtheaidofacompass(markoffthedistancefromtheorigin to b along the y axis), we can construct the point (0, b). From elementary geometry there exists a ruler-and-compass construction of a perpendicular to a given line through a specified point on the line.Constructaperpendiculartothexaxisthrough(a,0)andaperpendiculartotheyaxisthrough (0,b).Theselinesintersectat(a,b);hence(a,b)isconstructiblefrom{O,I}. C6 Describearuler-and-compassconstructionofthe30-60-90°triangle. D1 Fromgeometry,theangleateachvertexofaregularn-gonisequaltothesupplementof2π/n,thatis, π−(2π/n). G2 Anumberisconstructibleiffitisacoordinateofaconstructiblepoint.(Explain,usingExerciseA.) IfPisaconstructiblepoint,thismeanstherearesomepoints,saynpointsP1,P2,…,Pn=P, such thateachPiisconstructibleinonestepfrom × {P1,…,Pi−1.Intheproofofthelemmaofthis chapteritisshownthatthecoordinatesofPicanbeexpressedintermsofthecoordinatesofPx,…, Pi −1 using addition, subtraction, multiplication, division, and taking of square roots. Thus, the coordinates of P are obtained by starting with rational numbers and sucessively applying these operations. Usingtheseideas,writeaproofbyinductionofthestatementofthisexercise. CHAPTER31 A6 Let bearealcuberootof2. ( )isnotarootfieldover becauseitdoesnotcontainthe complexcuberootsof2. B5 First,p(x)=x3+x2+x+2isirreducibleover 3because,bysuccessivelysubstitutingx=0,1,2, oneverifiesthatp(x)hasnorootsin 3.[Explainwhythisfactprovesthatp(x)isirreducibleover 3.]Ifuisarootofp(x)insomeextensionof 3,then 3(u)isanextensionof 3ofdegree3.(Why?) Thus, 3(u)consistsofallelementsau2+bu+c,wherea,b,c∈ 3.Hence 3(u)has27elements. Dividingp(x)=x3+x2+x+2byx−ugives p(x)=(x−u)[x2+(u+1)x+(u2+u+1)] whereq(x)=x2+(u+l)x+(u2+u+1)isinZ3(u)[x].(Why?) In 3(u)[x],q(x)isirreducible:thiscanbeshownbydirectsubstitutionofeachofthe27 elementsof 3(u),successively,intoq(x).Soifwdenotesarootofq(x)insomeextensionof 3(u), thenZ3(u,w)includesuandbothrootsofq(x).(Explain.)Thus, 3(u,w)istherootfieldofp(x)over 3.Itisofdegree6over 3.(Why?) C5 WemayidentifyF[x]/〈p(x)〉withF(c),wherecisarootofp(x).ThenF(c)isanextensionofdegree 4overF.(Why?)Nowcompletethesquare: D3 D5 E4 H2 FormtheiteratedextensionF(d1,d2),whered1isarootofx2−[(a2/4) −b]andd2isarootofx2 − [(a/2)+d1].Explainwhy(a)F(d1,d2)=F(c)and(b)F(d1,d2)containsalltherootsofp(x). Frompage313, ( , )= ( + );hence ( , , )= ( + , ). AsintheillustrationforTheorem2,takingt=1givesc= + + .(Providedetails.) Usetheresultofpart3.Thedegreeof ( , , )over is8(explainwhy).Nowfinda polynomialp(x)ofdegree8having + + asaroot:ifp(x)ismonicandofdegree8, p(x)mustbetheminimumpolynomialof + + over .(Why?)Hencep(x) must be irreducible.Explainwhytherootsofp(x)are± ± ± ,allofwhicharein ( , , ). Forn=6,wehavex6−1=(x −1)(x+1)(x2 −x+1)(x2+x+1).Fromthequadraticformula,the rootsofthelasttwofactorsarein ( ). Notethatifc(x)=c0+c1x+⋯+cnxnthen and h(c(a))=h(c0)+h(c1)b+⋯+h(cn)bn Ford(x)=d0+d1x+⋯+dnxn,wehavesimilarformulas.Showthath(c(a))=h(d(a))iffbisaroot ofhc(x)−hd(x). I4 Theproofisbyinductiononthedegreeofa(x).Ifdega(x)=1,thenK1=F1 and K2 = F2 (explain why),andthereforeK1≅K2.Nowletdega(x)=n,andsupposetheresultistrueforanypolynomial ofdegreen−1. Letp(x)beanirreduciblefactorofa(x);letubearootofp(x)andυarootofhp(x).IfF1(u)= K1,thenbyparts1and2wearedone.Ifnot,let =F1(u)and =F2(υ);hcanbeextendedtoan isomorphismh: → ,withh(u)=υ.In [x],a(x)=(x−u)a1(x),andin [x],ha(x)= a(x)= (x−υ)ha1(x),wheredega1(x)=degha1(x)=n−1.Completethesolution. J2 Explainwhyanymonomorphism :F(c)→ ,whichisanextensionofh,isfullydeterminedbythe valueof (c),whichisarootofhp(x). J4 Sinceh(1)=1,beginbyprovingbyinductionthatforanypositiveintegern,h(n)=n.Thencomplete thesolution. CHAPTER32 A2 Inthefirstplace, ( )isofdegree2over .Next,x2+1isirreducibleover ( why.)Completethesolution. D1 Thecomplexfourthrootsof1are±1and±i.Thus,thecomplexfourthrootsof2are 1), (i),and (−i),thatis: E1 E6 H4 I4 ). (Explain (1), (− Explainwhyanyfieldcontainingthefourthrootsof2containsαandi,andconversely. SeeChapter31,ExerciseE. Letaα bearealsixthrootof2.Then[ (α): ]=6.Explainwhyx2+3isirreducibleover (α)andwhy[ (a, i): (α)]=2.Ifω=(1/2)+( /2)i(whichisaprimitivesixthrootof1), thenthecomplexsixthrootsof2areα,αω,αω2,αω3,αω4,andαω5.Anyautomorphismof (α, i)fixing mapssixthrootsof2tosixthrootsof2,atthesametimemapping i to ± i (and hencemappingωtoω5).Providedetailsandcompletethesolution. Supposea≠h(a),saya<h(a).Letrbearationalnumbersuchthata<r<h(a). Every subgroup of an abelian group is abelian; every homomorphic image of an abelian group is abelian. CHAPTER33 A2(a) Thepolynomialisirreducibleover byEisenstein’scriterion,andithasthreerealrootsandtwo complexroots.(Explainwhy.)Argueasintheexampleofpage340andshowthattheGaloisgroup isS5. B3 Itmustbeshownthatforeachi,i=0,1,…,n–1,thequotientgroupJi+1/Jiisabelian.Itmustbe shownthatJicontainsxyx−1y−1forallxandyinJi+1.Providedetailsandcomplete. C3 FhasanextensionKwhichcontainsalltherootsd1,d2,…,dpofxp −a.InK,xp – a factors into linearfactors: xp−a=(x−d1)(x−d2)⋯(x−dp) By uniqueness of factorization, p(x) is equal to the product of m of these factors, while f(x) is the productoftheremainingfactors. D3 Iff:G→G/Kisthenaturalhomomorphism,then =f−1( )={x∈G:f(x)∈ }. D7 ProvethisstatementbyinductionontheorderofG/H.Let|G/H|=n,andassumethestatementistrue forallgroupsH′ G′,where|G′/H′|<n.IfGhasnonormalsubgroupJsuchthatH⊂J⊂G(H≠J ≠G),thenHisamaximalnormalsubgroupofG;sowearedonebyparts4and5.Otherwise,|G/J|< nand|J/H|<n.Completethesolution. INDEX Abelianextension,336 Abeliangroup,28 Absorbsproducts,182 Additionmodulon,27 Additivegroupofintegersmodulon,27 Algebraicclosure,300 Algebraicelements,273 Algebraicnumbers,299 Algebraicstructures,10 Alternatinggroup,86 Annihilatingideal,189 Annihilator,188 Associates,219,252 Associativeoperation,21 Automata,24,65 Automorphism: ofafield,323 Froebenius,207 ofagroup,101,161 inner,161 Basisforavectorspace,286 Basistheoremforfiniteabeliangroups,167 Bijectivefunction,58 Binarycodes,33 Binomialformula,179 Booleanalgebra,9 Booleanrings,179 Cancellationlaw,37 Cancellationpropertyinrings,173 Cauchy’stheorem,131,164,339 Cayleydiagram,51 Cayley’stheorem,95 Center: ofagroup,50,154 ofaring,186 Centralizer,134 Characteristicofanintegraldomain,201 Chineseremaindertheorem,233 Circlegroup,163 Classequation,154 Closure:algebraic,300 withrespecttoaddition,44 withrespecttoinverse,44 withrespecttomultiplication,44 Code,33 Commutativeoperation,20 Commutativering,172 Commutator,144,152 Commutingelements,40 Compositefunction,59 Congruence,modulon,227 linear,230 Conjugacyclass,134 Conjugatecycles,88 Conjugateelements,133,140 Conjugatesubgroups,145 Constructibleangles,308 Constructiblenumbers,307 Constructiblepoints,303 Correspondencetheorem,164 Coset,126-134,190 left,127 right,127 Cosetaddition,150 Cosetdecoding,134 Cosetmultiplication,149 Cycles,81-82 conjugate,88 disjoint,82 lengthof,82 orderof,88 Cyclicgroup,112-118 generatorof,112 Cyclicsubgroup,47,114 Definingequations,50 Degree: ofanextension,292 ofapolynomial,241 Derivative,279 Dihedralgroup,74 Dimensionofavectorspace,287 Diophantineequation,230 Directproduct: ofgroups,42 inner,145 ofrings,177 Disjointcycles,82 Distributivelaw,15,170 Divisionalgorithm,213,245 Divisorofzero,173 Domainofafunction,57 Eisenstein’sirreducibilitycriterion,263 Endomorphism,177 Equivalenceclass,121 Equivalencerelation,121-125 Euclideanalgorithm,257 Euclid’slemma,221,254 Euler’sphi-function,229 Euler’stheorem,229 Evenpermutation,84-86 Extension: abelian,336 algebraic,296 degreeof,292 finite,292,312 iterated,295 normal,322,333 quadratic,278 radical,319,335 simple,278,295,312 Extensionfield,270-281 degreeof,292-300 Fermat’slittletheorem,229 Field,172 Fieldextension,270-281 degreeof,292-300 Fieldofquotients,203 Finiteextension,292,312 Firstisomorphismtheorem,162 Fixedfield,326 Fixer,326 Fixfield,326 Flownetwork,91 Froebeniusautomorphism,207 Functions,56-68 bijective,58 composite,59 domainof,57 identity,70 injective,57 inverse,60 rangeof,57 surjective,58 Fundamentalhomomorphismtheorem,157-168 forgroups,158 forrings,193 Galois,Évariste,17 Galoiscorrespondence,328 Galoisgroup,325 Galoistheory,311-333 Generatormatrix,54 Greatestcommondivisor,219,253 Group(s),25-43,69-79,103-118,147-156 abelian,28 actingonaset,134 alternating,86 circle,163 cyclic,112-118 dihedral,74 finite,26 Galois,325 ofintegersmodulon,26 orderof,39 parity,137 ofpermutations,71 quaternion,133 quotient,147-156 solvable,338,342 symmetric,71 ofsymmetriesofasquare,72 Groupcode,53 Homomorphisms,136-146,157-168,183-184,187-189 Ideal(s),182-189 annihilating,189 annihilatorof,188 maximal,189,195 primary,198 prime,194 principal,183,251 proper,189,195 radicalof,188 semiprime,198 Identityelement,21 Identityfunction,70 Indexofasubgroup,129 Induction: mathematical,211,214 strong,212 Injectivefunction,57 Innerautomorphism,161 Innerdirectproduct,145 Input/outputsequence,24 Integers,208-217 Integraldomain,174,200-207 characteristicof,201 ordered,209 Integralsystem,210 Inversefunction,60 Invertibleelementinaring,172 Irreduciblepolynomial,254 Isomorphism,90-103 Iteratedextension,295 Kernel: ofagrouphomomorphism,140 ofaringhomomorphism,185 Lagrange’stheorem,129 Leadingcoefficient,242 Leastcommonmultiple,224 Legendresymbol,238 Linearcombination,285 Lineardependence,285 Linearindependence,285 Lineartransformation,288 Matrices,7-9 Matrixmultiplication,8 Maximalideal,189,195 Minimumpolynomial,273 Modulon: addition,27 congruence,227,230 multiplication,171 Monicpolynomial,253 Monomorphism,322 Multiplicationmodulon,171 Nilpotentelement,180 Normalextension,322,333 Normalseries,342 Normalsubgroup,140 Nullspace,290 Oddpermutation,84-86 Operations,10,19-24 Orbit,134 Order: ofanelement,105 ofagroup,39 infinite,105 p-group,165 p-subgroup,165 p-Sylowsubgroup,165 Parity-checkequations,34 Parity-checkmatrix,54 Partition,119-125 Permutation,69-86 even,84-86 odd,84-86 Polynomial(s),240-270 inseparable,320 irreducible,254 minimum,273 monic,253 reducible,254 root(s)of,259 multiple,280,312 separable,320 Polynomialinterpolation,268 Primeideal,194 Primenumbers,217-225 Primitiveroots,239,337 Principalideal,183,251 Properideal,189,195 Quadraticreciprocity,239 Quadraticresidue,238 Quaternion,176 ringof,176 Quaterniongroup,133 Quotientgroup,147 Quotientring,190-199 Radical(s),335-343 Radicalextension,319,335 Radicalofanideal,188 Rangeofafunction,57 Redundancy,34 Regularrepresentationofgroups,102 Relativelyprime,220,254 Ring(s),169-181,190-199,240-251 commutative,172 ofendomorphisms,177 ofpolynomials,240-251 ofquaternions,176 quotient,190-199 trivial,172 withunity,172 Rootfield,313 Root(s)ofapolynomial,259 multiple,280,312 Secondisomorphismtheorem,163 Solvablebyradicals,335 Solvablegroup,338,342 Solvableseries,342 Stabilizer,134 State,65 diagram,66 internal,65 next,65 Subgroup(s),44-52 cyclic,47,114 generatorsof,47 index,of,129 normal,140 p-,165 p-Sylow,165 proper,46 trivial,46 Subring,181 Surjectivefunction,58 Sylowgroups,165 Sylowsubgroup,165 Sylow’stheorem,165 Symmetricdifference,30 Symmetricgroup,71 Transcendentalelements,273 Transposition,83 Uniquefactorization,222,255 Unity,ringwith,172 Vectorspace,282-291 basisof,286 dimensionof,287 Weight,55 Well-orderingproperty,210 Wilson’stheorem,237