Download A Book of Abstract Algebra

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