Introduction to Modern Algebra - Clark U

Once symbolic algebra was developed in the 1500s, mathematics ourished in the 1600s. Coordinates, analytic geometry, and calculus with derivatives, integrals, and series were de-veloped in that century. Algebra became more general and more abstract in the 1800s as more algebraic structures were invented.


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IntroductiontoModernAlgebraDavidJoyceClarkUniversityVersion1.2.7,5Dec201711Copyright(C)2008,2017.
iiIdedicatethisbooktomyfriendandcolleagueArthurChou.Arthurencouragedmetowritethisbook.I'msorrythathedidnotlivetoseeit nished.Arthurwasbornin1954inTaipei,Taiwan.Hereceivedhisbachelorsinmathematicsin1976fromTunghaiUniversityandhisPhDfromStonyBrookin1982.AfterayearattheInstituteforAdvancedStudyatPrinceton,hejoinedClarkUniversityin1983.HewaspromotedtoAssociateProfessorsixyearslaterandpromotedtofullprofessorin2008,theyearhedied.Besidesmathematics,hehadmanyotherinterests.Amongotherthings,hewasthegeneralmanageroftheNorthAmericaEliteYouthOrchestrawhichperformedatDallas,Beijing,andTaipei,andhewasthedeaconoftheChineseGospelChurchinSouthborough,Massachusetts.
ContentsListofFiguresviListofTablesviii1Introduction11.1Algebra.......................................11.2StructuresinModernAlgebra...........................21.2.1Operationsonsets.............................21.2.2Fields....................................41.2.3Rings....................................51.2.4Groups....................................61.2.5Otheralgebraicstructuresbesides elds,rings,andgroups.......101.3Isomorphisms,homomorphisms,etc.........................111.3.1Isomorphisms................................111.3.2Homomorphisms..............................121.3.3Monomorphismsandepimorphisms....................131.3.4Endomorphismsandautomorphisms...................141.4Alittlenumbertheory...............................151.4.1MathematicalinductiononthenaturalnumbersN...........151.4.2Divisibility..................................161.4.3Primenumbers...............................171.4.4TheEuclideanalgorithm..........................191.5Thefundamentaltheoremofarithmetic......................221.6Polynomials......................................251.6.1Divisionforpolynomials..........................261.6.2Rootsofunityandcyclotomicpolynomials................282Fields312.1Introductionto elds................................312.1.1De nitionof elds.............................312.1.2Subtraction,division,multiples,andpowers...............322.1.3Propertiesthatfollowfromtheaxioms..................332.1.4Sub elds...................................342.1.5Fieldsofrationalfunctions.........................352.1.6Vectorspacesoverarbitrary elds.....................352.2Cyclicringsand nite elds............................35iii
ivCONTENTS2.2.1ThecyclicringZn.............................362.2.2Thecyclicprime eldsZp.........................392.2.3Characteristicsof elds,andprime elds.................412.3FieldExtensions,algebraic elds,thecomplexnumbers.............412.3.1Algebraic elds...............................422.3.2The eldofcomplexnumbersC......................432.3.3Generalquadraticextensions........................442.4Realnumbersandordered elds..........................452.4.1Ordered elds................................452.4.2Archimedeanorders.............................472.4.3Completeordered elds...........................492.5Skew elds(divisionrings)andthequaternions.................502.5.1Skew elds(divisionrings).........................502.5.2ThequaternionsH.............................513Rings553.1Introductiontorings................................553.1.1De nitionandpropertiesofrings.....................553.1.2Productsofrings..............................573.1.3Integraldomains..............................573.1.4TheGaussianintegers,Z[i]........................593.1.5Finite eldsagain..............................593.2FactoringZnbytheChineseremaindertheorem.................603.2.1TheChineseremaindertheorem......................603.2.2Brahmagupta'ssolution..........................623.2.3QinJiushao'ssolution...........................623.3Booleanrings....................................633.3.1IntroductiontoBooleanrings.......................643.3.2FactoringBooleanrings..........................653.3.3ApartialorderonaBooleanring.....................663.4The eldofrationalnumbers, eldsoffractions.................673.5Categoriesandthecategoryofrings........................693.5.1Theformalde nitionofcategories.....................703.5.2ThecategoryRofrings..........................713.5.3Monomorphismsandepimorphismsinacategory............733.6Kernels,ideals,andquotientrings.........................743.6.1Kernelsofringhomomorphisms......................743.6.2Idealsofaring...............................743.6.3Quotientrings,R=I.............................763.6.4Primeandmaximalideals.........................783.7Krull'stheorem...................................793.8UFDs,PIDs,andEDs...............................803.8.1Divisibilityinanintegraldomain.....................803.8.2Uniquefactorizationdomains.......................813.8.3Principalidealdomains...........................823.8.4Euclideandomains.............................84
CONTENTSv3.9RealandcomplexpolynomialringsR[x]andC[x]................873.9.1C[x]andtheFundamentalTheoremofAlgebra.............873.9.2ThepolynomialringR[x].........................893.10Rationalandintegerpolynomialrings.......................903.10.1Rootsofpolynomials............................903.10.2Gauss'slemmaandEisenstein'scriterion.................923.10.3Primecyclotomicpolynomials.......................953.10.4PolynomialringswithcoecientsinaUFD,andpolynomialringsinseveralvariables...............................953.11Number eldsandtheirringsofintegers.....................974Groups994.1Groupsandsubgroups...............................994.1.1De nitionandbasicpropertiesofgroups.................994.1.2Subgroups..................................1004.1.3Cyclicgroupsandsubgroups........................1014.1.4Productsofgroups.............................1024.1.5CosetsandLagrange'stheorem......................1034.2SymmetricGroupsSn...............................1044.2.1Permutationsandthesymmetricgroup..................1044.2.2Evenandoddpermutations........................1064.2.3Alternatinganddihedralgroups......................1074.3Cayley'stheoremandCayleygraphs.......................1104.3.1Cayley'stheorem..............................1104.3.2Somesmall nitegroups..........................1124.4ThecategoryofgroupsG..............................1154.5Conjugacyclassesandquandles..........................1154.5.1Conjugacyclasses..............................1164.5.2Quandlesandtheoperationofconjugation................1174.6Kernels,normalsubgroups,andquotientgroups.................1204.6.1Kernelsofgrouphomomorphismsandnormalsubgroups........1204.6.2Quotientgroups,andprojections :G!G=N.............1214.6.3Isomorphismtheorems...........................1224.6.4Internaldirectproducts..........................1234.7Matrixringsandlineargroups...........................1244.7.1Lineartransformations...........................1244.7.2ThegenerallineargroupsGLn(R).....................1254.7.3Otherlineargroups.............................1264.7.4ProjectivespaceandtheprojectivelineargroupPGLn(F).......1274.8Structureof nitegroups..............................1304.8.1Simplegroups................................1314.8.2TheJordan-Holdertheorem........................1314.9Abeliangroups...................................1344.9.1ThecategoryAofAbeliangroups.....................1364.9.2FiniteAbeliangroups............................137
viCONTENTSAppendices141ABackgroundmathematics143A.1Logicandproofs..................................143A.2Sets.........................................144A.2.1Basicsettheory...............................144A.2.2Functionsandrelations...........................149A.2.3Equivalencerelations............................150A.2.4Axiomsofsettheory............................151A.3Orderedstructures.................................153A.3.1Partialordersandposets..........................153A.3.2Lattices...................................155A.3.3Booleanalgebras...............................157A.4Axiomofchoice...................................158A.4.1Zorn'slemma................................158A.4.2Well-orderingprinciple...........................159Index161
ListofFigures1.1Equilateraltrianglewithlinesofsymmetry....................81.2UnitcircleS1....................................101.3Divisorsof432...................................171.4Divisibilityupthrough12.............................172.1CyclicringsZ6;Z19;Z...............................373.1LatticeofGaussianintegersZ[i]..........................593.2FreeBooleanringontwoelements.........................663.3LatticeofEisensteinintegers............................853.4Primitive7throotsofunity.............................954.1SubgroupsofS3...................................1074.2Symmetriesofapentagon.............................1084.3Symmetriesofacubeandtetrahedron......................1094.4CayleygraphforD5................................1114.5CayleygraphforA4................................1134.6Distributivityinainvolutoryquandle.......................1184.7Aconjugacyclassinthequaterniongroup....................1194.8TheconjugacyclassoftranspositionsinS4....................1204.9TheFanoplaneZ2P2................................1284.10TheprojectiveplaneZ3P2.............................1294.11CayleygraphoftheFrobeniusgroupF21=C7oC3...............1344.12Heptahedrononatorus..............................135A.1LatticeofthePowersetof4elements.......................157vii
viiiLISTOFFIGURES
ListofTables1.1Compositiontableforsixparticularrationalfunctions..............93.1NotationsinBooleanalgebras,settheory,andBooleanrings...........654.1Listofsmallgroups.................................114A.1Standardlogicalsymbols..............................144ix
xLISTOFTABLES
Chapter1Introduction1.1AlgebraTheword\algebra"meansmanythings.Theworddatesbackabout1200yearsagotopartofthetitleofal-Khwarizm'sbookonthesubject,butthesubjectitselfgoesback4000yearsagotoancientBabyloniaandEgypt.Itwasaboutsolvingnumericalproblemsthatwewouldnowidentifyaslinearandquadraticequations.Versionsofthequadraticformulawereusedto ndsolutionstothosequadraticequations.Al-Khwarizm(ca.780{ca.850)codi edthealgorithms(\algorithm"isawordderivedfromhisname)forsolvingtheseequations.Hewroteallhisequationsoutinwordssincesymbolicalgebrahadyettobeinvented.Otherplacesintheworldalsohadalgebraanddevelopedvariousaspectsofit.TheancientChinesesolvedsystemsofsimultaneouslinearequationsandlaterdevelopedalgorithmsto ndrootsofpolynomialsofhighdegree.VariousaspectsofnumbertheorywerestudiedinChina,inIndia,andbyGreekmathematicians.Symbolicalgebrawasdevelopedinthe1500s.Symbolicalgebrahassymbolsforthearithmeticoperationsofaddition,subtraction,multiplication,division,powers,androotsaswellassymbolsforgroupingexpressions(suchasparentheses),andmostimportantly,usedlettersforvariables.Oncesymbolicalgebrawasdevelopedinthe1500s,mathematics ourishedinthe1600s.Coordinates,analyticgeometry,andcalculuswithderivatives,integrals,andserieswerede-velopedinthatcentury.Algebrabecamemoregeneralandmoreabstractinthe1800sasmorealgebraicstructureswereinvented.Hamilton(1805{1865)inventedquaternions(seesection2.5.2)andGrassmann(1809{1977)developedexterioralgebrasinthe1840s,bothofwhichledtovectorspaces.(Seesection2.1.6forvectorspaces.)Groupsweredevelopedoverthe1800s, rstasparticulargroupsofsubstitutionsorper-mutations,theninthe1850'sCayley(1821{1895)gavethegeneralde nitionforagroup.(Seechapter2forgroups.)Several eldswerestudiedinmathematicsforsometimeincludingthe eldofrealnumbersthe eldofrationalnumber,andthe eldofcomplexnumbers,buttherewasnogeneralde nitionfora elduntilthelate1800s.(Seechapter2for elds.)Ringsalsowerestudiedinthe1800s.Noether(1882{1935)gavegeneralconceptofcom-mutativeringin1921whichwaslatergeneralizedtoincludenoncommutativerings.(See1
2CHAPTER1.INTRODUCTIONchapter3forrings.)We'llintroducetheconceptsof eld,ring,andgroupintheIntroduction,thenstudyeachinturninthefollowingchapters.1.2StructuresinModernAlgebraFields,rings,andgroups.We'llbelookingatseveralkindsofalgebraicstructuresthissemester,thethreemajorkindsbeing eldsinchapter2,ringsinchapter3,andgroupsinchapter4,butalsominorvariantsofthesestructures.We'llstartbyexaminingthede nitionsandlookingatsomeexamples.Forthetimebeing,wewon'tproveanything;thatwillcomeinlaterchapterswhenwelookatthosestructuresindepth.Anoteonnotation.We'llusethestandardnotationforvariouskindsofnumbers.Thesetofnaturalnumbers,f0;1;2;:::gisdenotedN.Thesetofintegersf:::;�2;�1;0;1;2;:::gisdenotedZ(forZahlen,Germanforwholenumber).Thesetofrationalnumbers,thatis,numbersoftheformm
nwheremisanintegerandnisanonzerointeger,isdenotedQ(for\quotient").Thesetofallrealnumbers,includingallpositivenumbers,allnegativenumbers,and0,isdenotedR.Andthesetofcomplexnumbers,thatis,numbersoftheformx+iywherexandyarerealnumbersandi2=�1,isdenotedC.1.2.1OperationsonsetsForbackgroundonsets,seethesectionA.2intheappendix.We'refamiliarwithmanyoperationsontherealnumbersR|addition,subtraction,mul-tiplication,division,negation,reciprocation,powers,roots,etc.Addition,subtraction,andmultiplicationareexamplesofbinaryoperations,thatis,functionsRR!Rwhichtaketworealnumbersastheirargumentsandreturnanotherrealnumber.Divisionisalmostabinaryoperation,butsincedivisionby0isnotde ned,it'sonlyapartiallyde nedbinaryoperation.Mostofouroperationswillbede nedeverywhere,butsome,likedivison,won'tbe.Negationisaunaryoperation,thatis,afunctionR!Rwhichtakesonerealnumberasanargumentandreturnsarealnumber.Reciprocationisapartialunaryoperationsincethereciprocalofzeroisnotde ned.Theoperationswe'llconsiderareallbinaryorunary.Ternaryoperationscancertainlybede ned,butusefulternaryoperationsarerare.Someoftheseoperationssatisfyfamiliaridentities.Forexample,additionandmultipli-cationarebothcommutative;theysatisfytheidentitiesx+y=y+xandxy=yx:Abinaryoperationissaidtobecommutativewhentheorderthatthetwoargumentsareapplieddoesn'tmatter,thatis,interchangingthem,orcommutingoneacrosstheother,doesn'tchangetheresult.Subtractionanddivision,however,arenotcommutative.
1.2.STRUCTURESINMODERNALGEBRA3Additionandmultiplicationarealsoassociativebinaryoperations(x+y)+z=x+(y+z)and(xy)z=x(yz):Abinaryoperationissaidtobeassociativewhentheparenthesescanbeassociatedwitheitherthe rstpairorthesecondpairwhentheoperationisappliedtothreeargumentsandtheresultisthesame.Neithersubtractionnordivisionareassociative.Bothadditionandmultiplicationalsohaveidentityelements0+x=x=x+0and1x=x=x1:Anidentityelement,alsocalledaneutralelement,forabinaryoperationisanelementinthesetthatdoesn'tchangethevalueofotherelementswhencombinedwiththemundertheoperation.So,0istheidentityelementforaddition,and1istheidentityelementformultiplication.Subtractionanddivisiondon'thaveidentityelements.(Well,theydoontheright,sincex�0=xandx
1=x,butnotontheleft,sinceusually0�x6=xand1
x6=x.)Also,thereareadditiveinversesandmultiplicativeinverses(fornonzero)elements.Thatistosay,givenanyxthereisanotherelement,namely�x,suchthatx+(�x)=0,andgivenanynonzeroxthereisanotherelement,namely1
xsuchthatx1
x=1.Thus,abinaryoperationthathasanidentityelementissaidtohaveinversesifforeachelementthereisaninverseelementsuchthatwhencombinedbytheoperationtheyyieldtheidentityelementfortheoperation.Additionhasinverses,andmultiplicationhasinversesofnonzeroelements.Finally,thereisaparticularrelationbetweentheoperationsofadditionandmultiplica-tion,thatofdistributivity:x(y+z)=xy+xzand(y+z)x=yx+zx:Multiplicationdistributesoveraddition,thatis,whenmultiplyingasumbyxwecandis-tributethexoverthetermsofthesum.Exercise1.Onpropertiesofoperations.(a).Isthebinaryoperationxy=xy
x+yforpositivexandyacommutativeoperation?Thatis,isittruethatxy=yxforallpositivexandy?Isitassociative?Explainyouranswer.(b).Isittruethat(w�x)�(y�z)=(w�y)�(x�z)isanidentityforrealnumbers?Canyousaywhyorwhynot?(Theword\identity"isusedforanequationwhichholdswheneverbothsidesoftheequationarede nedandareequal.)(c).AlthoughmultiplicationinRdistributesoveraddition,additiondoesn'tdistributeovermultiplication.Giveanexamplewhereitdoesn't.Algebraicstructures.We'llde ne elds,rings,andgroupsasthreekindsofalgebraicstructures.Analgebraicstructurewillhaveanunderlyingset,binaryoperations,unaryoperations,andconstants,thathavesomeofthepropertiesmentionedabovelikecommuta-tivity,associativity,identityelements,inverseelements,anddistributivity.Di erentkindsofstructureswillhavedi erentoperationsandproperties.ThealgebraicstructuresareabstractionsoffamiliaroneslikethoseontherealnumbersR,butforeachkindofstructuretherewillbemorethanoneexample,aswe'llsee.
4CHAPTER1.INTRODUCTION1.2.2FieldsInformally,a eldisasetequippedwithfouroperations|addition,subtraction,multipli-cation,anddivisionthathavetheusualproperties.(Theydon'thavetohavetheotheroperationsthatRhas,likepowers,roots,logs,andthemyriadotherfunctionslikesinx.)De nition1.1(Field).A eldisasetequippedwithtwobinaryoperations,onecalledadditionandtheothercalledmultiplication,denotedintheusualmanner,whicharebothcommutativeandassociative,bothhaveidentityelements(theadditiveidentitydenoted0andthemultiplicativeidentitydenoted1),additionhasinverseelements(theinverseofxdenoted�x),multiplicationhasinversesofnonzeroelements(theinverseofxdenoted1
xorx�1),multiplicationdistributesoveraddition,and06=1.Thisde nitionwillbespelledoutindetailinchapter2.Ofcourse,oneexampleofa eldinthe eldofrealnumbersR.Whataresomeothers?Example1.2(The eldofrationalnumbers,Q).Anotherexampleisthe eldofrationalnumbers.Arationalnumberisthequotientoftwointegersa=bwherethedenominatorisnot0.ThesetofallrationalnumbersisdenotedQ.We'refamiliarwiththefactthatthesum,di erence,product,andquotient(whenthedenominatorisnotzero)ofrationalnumbersisanotherrationalnumber,soQhasalltheoperationsitneedstobea eld,andsinceit'spartofthe eldoftherealnumbersR,itsoperationshavethethepropertiesnecessarytobea eld.WesaythatQisasub eldofRandthatRisanextensionofQ.ButQisnotallofRsincethereareirrationalnumberslikep
2.Example1.3(The eldofcomplexnumbers,C).Yetanotherexampleisthe eldofcomplexnumbersC.Acomplexnumberisanumberoftheforma+biwhereaandbarerealnumbersandi2=�1.The eldofrealnumbersRisasub eldofC.We'llreviewcomplexnumbersbeforeweusethem.SeeDave'sShortCourseonComplexNumbersathttp://www.clarku.edu/~djoyce/complexInchapter2,we'llstudy eldsindetail,andwe'lllookatmanyother elds.Somewillonlyhavea nitenumberofelements.(Theywon'tbesub eldsofQ.)SomewillhaveQasasub eldbutbesub eldsthemselvesofRorC.Somewillbeevenlarger.Exercise2.On elds.Noneofthefollowingare elds.Ineachcase,theoperationsofadditionandmultiplicationaretheusualones.(a).TheintegersZdonotforma eld.Whynot?(b).Thepositiverealnumbersfx2Rjx�0gdonotforma eld.Whynot?(c).Thesetofrealnumbersbetween�10and10,thatis,(�10;10)=fx2Rj�10x10gisnota eld.Whynot?
1.2.STRUCTURESINMODERNALGEBRA51.2.3RingsRingswillhavethethreeoperationsofaddition,subtraction,andmultiplication,butdon'tnecessarilyhavedivision.Mostofourringswillhavecommutativemultiplication,butsomewon't,sowewon'trequirethatmultiplicationbecommutativeinourde nition.Alltheringswe'lllookathaveamultiplicativeidentity,1,sowe'llincludethatinthede nition.De nition1.4(Ring).Aringisasetequippedwithtwobinaryoperations,onecalledadditionandtheothercalledmultiplication,denotedintheusualmanner,whicharebothassociative,additioniscommutative,bothhaveidentityelements(theadditiveidentityde-noted0andthemultiplicativeidentitydenoted1),additionhasinverseelements(theinverseofxdenoted�x),andmultiplicationdistributesoveraddition.Ifmultiplicationisalsocom-mutative,thentheringiscalledacommutativering.Ofcourse,all eldsareautomaticallyrings,infactcommutativerings,butwhataresomeotherrings?Example1.5(Theringofintegers,Z).TheringofintegersZincludesallintegers(wholenumbers)|positive,negative,or0.Addition,subtraction,andmultiplicationsatisfytherequirementsforaring,indeed,acommutativering.Buttherearenomultiplicativeinversesforanyelementsexcept1and�1.Forinstance,1=2isnotaninteger.We'll ndthatalthoughtheringofintegerslookslikeithaslessstructurethana eld,thisverylackofstructureallowsustodiscovermoreaboutintegers.We'llbeabletotalkaboutprimenumbers,forexample.Example1.6(Polynomialrings).Awholefamilyofexamplesaretheringsofpolynomials.LetRbeanycommutativering(perhapsa eld),andletR[x]includeallpolynomialswithcoecientsinR.Weknowhowtoadd,subtract,andmultiplypolynomials,andtheseoperationshavethepropertiesrequiredtomakeR[x]acommutativering.Wehave,forinstance,theringofpolynomialswithrealcoecientsR[x],theringwithintegralcoecientsZ[x],etc.Example1.7(Matrixrings).Howaboutanexampleringthat'snotcommutative?TheringofnnmatriceswithentriesinacommutativeringRgivessuchanexample,thisringbeingdenotedMn(R).Thisring,Mn(R),won'tbecommutativewhenn2.Anexampleofamatrixringistheringof22matriceswithrealentries,M2(R).Additionandsubtractionarecomputedcoordinatewise.Theadditiveidentity,0,ofthismatrixringisthematrixwithall0entries,0=0000.Matrixmultiplicationisnotcoordinatewise,butitisassociative,andmultiplicationdoesdistributeoveraddition.Themultiplicativeidentityforthismatrixringiswhat'susuallycalledtheidentitymatrix,denotedI.Ithas1'sdownthemaindiagonaland0'selsewhere,1=1001.Sylvester(1814{1897),in1850,calledrectangulararrangementsofnumbersmatrices,andCayleywrotemuchabouttheminhispapersof1855{1858.Example1.8(Integersmodulon).Animportantfamilyofringsistheringofintegersmodulon.We'llstudythisinmoredetaillaterinsection2.2,buthere'sanincompleteoverview.Fixapositiveintegern.Thinkoftwointegersaandbasbeingthesamemodulonifndividesb�a.Inthatcase,we'llsaythataandbarecongruentmodulon,andwe'll
6CHAPTER1.INTRODUCTIONusethenotationGauss(1777{1855)developed,ab(modn),todenotethatcongruence.Congruenceiscommonlyusedinthestudyofnumbertheory.ThismeaningoftheLatinword\modulo"wasintroducedintomathematicsbyGaussin1801.Notethatthereareonlyndistinctintegersmodulon,namely0throughn�1,sincethosearetheonlyremaindersyoucangetwhenyoudivideanintegerbyn.Theseremaindersarealsocalled\residues".Wecanrepresentintegersmodulonbytheseremaindersfrom0throughn�1.Thus,we'llsay,forinstance,that5plus3equals1modulo7,bywhichwemean5+31(mod7).Thus,wecanturncongruencemodulon,whichisanequivalencerelationonZintoequalityonann-elementset.Thatn-elementsetisdenotedZ=nZ,readZmodulonZ,ormoresimplyasZn,readZmodn.So,wecantaketheelementsofZntobetheintegersfrom0throughn�1,whereweunderstandthataddition,subtraction,andmultiplicationaredonemodulon.Anditturnsoutthatthisisaring,aswe'llseewhenwestudyZnindetail.Incidentally,whennisaprimenumberp,thenZpisnotjustaring,buta eld,aswillbediscussedinsection2.2.Exercise3.Onrings.Noneofthefollowingarerings.Ineachcase,theoperationsofadditionandmultiplicationaretheusualones.(a).Thesetofnonzerointegers,fx2Zjx6=0gisnotaring.Whynot?(b).Thesetofevenintegersf2xjx2Zgisnotaring.Whynot?(c).Thesetofodddegreepolynomialswithrealcoecientsff(x)2R[x]jthedegreeoff(x)isoddgisnotaring.Whynot?(Howaboutthesetofevendegreepolynomials?)Exercise4.Onnoncommutativerings.Arethefollowingrings?(Theoperationsaretheusualmatrixoperations.)Explaininasentenceortwo,butaproofisnotnecessary.(a).Thesetofallmatriceswithrealcoecients(allsizes).(b).Thesetofall22matriceswithrealentriesoftheformab0d:(c).Thesetofall22matriceswithrealentriesoftheformab�ba:Inchapter3we'llanalyzeringsinmoredetail.1.2.4GroupsUnlike eldsandringswhichhavetwoprimarybinaryoperations,groupsonlyhaveonebinaryoperation.De nition1.9(Group).Agroupisasetequippedwithabinaryoperationthatisasso-ciative,hasanidentityelement,andhasinverseelements.If,furthermore,multiplicationis
1.2.STRUCTURESINMODERNALGEBRA7commutative,thenthegroupiscalledacommutativegrouporanAbeliangroup.Abeliangroupscanbedenotedeitheradditivelyormultiplicatively,butnonabeliangroupsareusu-allydenotedmultiplicatively.We'llusethetermorderofthegrouptoindicatehowmanyelementsagroupGhasanddenotethisorderbyjGj.Example1.10(Theunderlyingadditivegroupofaring).Ofcourse,ifyouhavea eldorring,andjustconsideraddition(andforgetaboutmultiplication)you'vegotanAbeliangroup.Sometimesthisiscalledtheunderlyingadditivegroupofthe eldorring.We'llusethesamenotationfortheunderlyingadditivegroupaswedoforthering.Thus,ZcouldmeaneithertheringofintegersortheAbeliangroupofintegersunderaddition,dependingonthecontext.Example1.11(Finitecyclicgroups).TheunderlyinggroupoftheringZniscalledacyclicgroup.It'selementsare,ofcourse,0;1;2;:::;n�1whereniscongruentto0.Cyclicgroupsarealsowrittenmultiplicatively,andthentheelementsare1;a;a2;:::;an�1wherean=1.AcommonnotationforthiscyclicgroupisCn.De nition1.12(Unitsinaring).Inordertousethemultiplicationforagroupoperation,we'llhavetoonlyincludetheunits,alsocalledinvertibleelements.AunitorinvertibleelementofaringRisanelementx2Rsuchthatthereexistsanotherelementy2Rsothatxy=yx=1.ThesubsetofunitsisdenotedR=fx2Rj9y2R;xy=1g:Youcaneasilyshowthattheunitsformagroupundermultiplication,calledthemulti-plicativegroupofunitsofR.WhenRisa eld,thenRisallofRexcept0,butforringstherewillbeotherelementsthan0thataren'tinvertible.ThegroupRwillbeAbelianwhentheringRiscommutative,butusuallyitwillbenonabelianwhenRisnotcommutative.Examples1.13.TheunitsintheringZarejust1and�1.ThegroupofunitsZisacyclicgroupoforder2.We'llseelaterthatthegroupofunitsZpwhenpisprimeisacyclicgroupoforderp�1.ItisusuallythecasethatZnwhenniscompositeisnotacyclicgroup.Example1.14(Agenerallineargroup,GL2(R)).Asaparticularexampleofamultiplicativegroupofunits,taketheinvertibleelementsofthematrixringM2(R).Theinvertible22matricesarethosematricesabcdwhosedeterminantsad�bcarenonzero.Thegroupofinvertiblennmatrices,Mn(R),isthegenerallineargroupwithcoecientsintheringR,denotedGLn(R).NotethatGLn(R)isanonabeliangroupforn2.TherealgenerallineargroupGL2(R)canbeinterpretedasthegroupofinvertiblelineartransformationsoftheplaneR2thatleavetheorigin xed.We'llstudyGL2(R)andGLn(R)inmoredetailinsection4.7.2.Exercise5.FindtwomatricesinGL2(Z)thatdon'tcommutetherebyprovingGL2(Z)isanonabeliangroup.
8CHAPTER1.INTRODUCTION
C
B
A
Figure1.1:EquilateraltrianglewithlinesofsymmetryTherearemanyexamplesof nitenonabeliangroupsfoundingeometry.We'lllookatthegroupofsymmetriesofanequilateraltriangle.Example1.15(ThedihedralgroupD3).Consideranequilateraltriangle.Placeacoordinatesystemontheplaneofthetrianglesothatitscenterisat(0;0),onevertex,A,at(1;0),andtheothertwo,BandC,at(�1
2;1
2p
3).Thistrianglehassixsymmetries.Asymmetryisatransformationoftheplanethatpreservesdistance(thatis,anisometry)thatmapsthetrianglebacktoitself.Threeofthesesymmetriesarerotationsby0,120,and240.1=1001=�1
2�1
2p
31
2p
3�1
22=�1
21
2p
3�1
2p
3�1
2Theidentitytransformation,1, xesA,B,andC;therotationby120mapsAtoB,BtoC,andCtoA;andtherotation2by240mapsAtoC,BtoA,andCtoB.Therearealsothreere ections.'=100�1'=�1
21
2p
3�1
2p
31
22'=�1
21
2p
3�1
2p
31
2Therefection' xesA,andinterchangesBandC;there ection' xesCandinterchangesAandB;andthere ection2' xesBandinterchangesAandCThisisaparticularnonabeliangroupthathas6elements.ItisasubgroupofGL2(R)mentionedabove.Example1.16(Agroupoffunctions).Manyapplicationsofgrouptheoryaretogroupsofinvertiblefunctions.Suchagroupincludesinvertiblefunctionsonsomesetsuchthatthecompositionofanytwoofthefunctionsisanotherone.Letf(x)=1=xandg(x)=1�x.Bothofthoseareinvertibleconsideredasrationalfunctions,and,infact,eachisit'sowninverse:(ff)(x)=f(1=x)=1,and(gg)(x)=
1.2.STRUCTURESINMODERNALGEBRA9g(1�x)=1�(1�x)=x.Let'sseewhatotherfunctionswecanderivefromfandgbycomposingthem.First,consider(fg)(x)=f(g(x)=f(1�x)=1
1�x;callthatcompositionhsothath(x)=1
1�x.Next,consider(gf)(x)=g(f(x)=g1
x=1�1
x=x�1
x;callthatcompositionksothatk(x)=x�1
x.Wecangetmorefunctionsifwecontinuetocomposethese.Notethat(fk)(x)=fx�1
x=x
x�1;callthat`sothat`(x)=x
x�1.Also,(gh)(x)=g1
1�x=1�1
1�x=x
x�1.Thatfunctionhasalreadybeencalled`,sogh=`.Acouplemorecomputationsshowthathh=kandkk=h.Sincefandgareeachtheirowninverses,ff=iandgg=i,whereiistheidentityfunction,i(x)=x.Alsohk=kh=i,and``=i.Also,icomposedwithanyfunction(oneitherside)isequaltothatsamefunction.Itturnsoutthatthesesixfunctionsareclosedundercomposition.Table1.1givesalloftheircompositions.
ifghk`
i
ifghk`
f
fihg`k
g
gki`fh
h
h`fkig
k
kg`ihf
`
`hkfgi
Table1.1:Compositiontableforsixparticularrationalfunctions.Notethatineachrowandeachcolumnofthetable,eachoneofthefunctionsappearsexactlyonce.ThatmakestheentriesofthetableaLatinsquare.ALatinsquareisasquarennnarray lledwithndi erentsymbols,eachoccurringexactlyonceineachrowandexactlyonceineachcolumn.Example1.17(Euler'scirclegroup).Theunitcircle,S1=fx+yi2Cjx2+y2=1g,isagroupundermultiplication.ThisissometimescalledEuler'scirclegroupsinceEuler(1707{1783)introducedtheunitcircleinthecomplexplaneforstudyinganglesandtrigonometricfunctions.Theproductoftwocomplexnumbersonthisunitcircleisanothernumberontheunitcircle.Youcandirectlyverifythatoryoucanshowitbytrigonometry.Ifx+yiisontheunitcircle,thenwecanidentifyxwithcosandywithsinwhereis,asusual,theanglebetweenthepositivex-axisandtherayfrom0tox+yi.Thentheproductoftwocomplexnumbersontheunitcirclecorrespondstoaddingtheiranglestogether.Theadditionformulasforcosinesandsinesgivethiscorrespondence.Exercise6.Computetheproductofcos+isintimescos'+isin'.Ifx+iy=(cos+isin)(cos'+isin'),thenwhatisx,therealpartoftheproduct,intermsofand'?Whatisy,theimaginarypart?
10CHAPTER1.INTRODUCTION
x
y
�1
1
�i
i
Figure1.2:UnitcircleS1Comment1.18.AlthoughthesphereS2=f(x;y;z)2R3jx2+y2+z2=1ghasnogroupstructure,the3-spherein4-spacedoes.The3-sphereisS3=f(x;y;z;w)2R4jx2+y2+z2+w2=1g:Wedon'thavetimeorspacetodiscussthatgroupstructurehere.(The2-sphereS2,infact,spheresinalldimensions,doeshavequandlestructures,whateveraquandlemightbe.Seesection4.5.2.)Inchapter4we'llstudygroupsindetail.1.2.5Otheralgebraicstructuresbesides elds,rings,andgroupsThereareanunlimitednumberofotheralgebraicstructures.Somearesimilartothoselistedabove.Forinstance,therearedivisionrings(alsocalledskew elds)thathaveallthepropertiesof eldsexceptmultiplicationdoesn'thavetobecommutative.TheprimaryexampleisthequaternionsH.We'lldiscussquaternionslaterinsection2.5.2.
Rings
Commutativerings
Skew elds
Fields
1.3.ISOMORPHISMS,HOMOMORPHISMS,ETC.11Thereareanumberofstructuresthatarejustcommutativeringsthathaveniceproperties,andwe'lllookatsomeofthemincludingintegraldomains,uniquefactorizationdomains,principalidealdomains,andEuclideandomains.Sometimesringsthatdon'thaveamultiplicativeidentityarestudied,butforus,we'llalwayshave1.You'vealreadystudiedvectorspacesovertherealnumbers.Mostofthethingsthatyou'vestudiedaboutvectorspacesoverRalsoholdforvectorspacesoverother elds.Theanalogousstructureforvectorspaceswhena eldisreplacedbyaringiscalledamoduleoverthering.Wewon'tstudymodulesoveraring,butwhenwelookatidealsinaring,theyare,infact,examplesofmodulesoverthering.Also,AbeliangroupsaremodulesovertheringZ.We'lldiscussanotheralgebraicstructure,quandles,insection4.5.2whenwediscussgroups.1.3Isomorphisms,homomorphisms,etc.Frequently,welookattwoalgebraicstructuresAandBofthesamekind,forinstance,twogroupsortworingsortwo elds,andwe'llwanttocomparethem.Forinstance,wemightthinkthey'rereallythesamething,buttheyhavedi erentnamesfortheirelements.Thatleadstotheconceptofisomorphismf:A=B,andwe'lltalkaboutthat rst.Othertimeswe'llknowthey'renotthesamething,butthereisarelationbetweenthem,andthatwillleadtothenextconcept,homomorphism,f:A!B.We'llthenlookassomespecialhomomorphismssuchasmonomorphisms.Whenwehaveahomomorphismf:A!A,we'llcallitanendomorphism,andwhenanisomorphismf:A=A,we'llcallitanautomorphism.We'lltakeeachofthesevariantsinturn.Theconceptsofinjection(one-to-onefunction),surjection(ontofunction),andbijectionaredescribedsectionA.2.2intheappendixonfunctions.We'llusethefollowingtheoremabout nitesetswhenweconsiderhomomorphismsbe-tween nitealgebraicstructures.Theorem1.19.Supposethatf:A!Bisafunctionbetweentwo nitesetsofthesamecardinality.Thenthefollowingthreeconditionsareequivalent:(1)fisabijection,(2)fisaninjection,and(3)fisasurjection.Exercise7.Provethatiff:A!Bisafunctionbetweentwo nitesetsofthesamecardinality,thenfisinjectiveifandonlyiffissurjective.1.3.1IsomorphismsWe'llsaytwoalgebraicstructuresAandBareisomorphiciftheyhaveexactlythesamestructure,buttheirelementsmaybedi erent.Forinstance,letAbetheringR[x]ofpolynomialsinthevariablexwithrealcoecientswhileBistheringR[y]ofpolynomialsiny.They'rebothjustpolynomialsinonevariable,it'sjustthatthechoiceofvariableisdi erentinthetworings.Weneedtomakethisconceptmoreprecise.
12CHAPTER1.INTRODUCTIONDe nition1.20(Ringisomorphism).TworingsAandBareisomorphicifthereisabijectionf:A!Bwhichpreservesadditionandmultiplication,thatis,forallxandyinA,f(x+y)=f(x)+f(y);andf(xy)=f(x)f(y):Thecorrespondencefiscalledaringisomorphism.Afterweintroducehomomorphisms,we'llhaveanotherwaytodescribeisomorphisms.Youcanprovevariouspropertiesofringisomorphismfromthisde nition.Exercise8.Sincethestructureofringsisde nedintermsofadditionandmultiplication,iffisaringisomorphism,itwillpreservestructurede nedintermsofthem.Verifythatfpreserves0,1,negation,andsubtraction.Exercise9.Provethatiffisaringisomorphism,thensoisitsinversefunctionf�1:B!A.Exercise10.Provethatiff:A!Bandg:B!Carebothringisomorphisms,thensoistheircomposition(gf):A!C.Sincea eldisaspecialkindofring,anditsstructureisde nedintermsofadditionandmultiplication,wedon'tneedaspecialde nitionfora eldisomorphism.A eldisomorphismisjustaringisomorphismbetween elds.Exercise11.Provethatifaringisisomorphictoa eld,thenthatringisa eld.Wedoneedadi erentde nitionforagroupisomorphismsinceagroupisde nedintermsofjustonebinaryoperationinsteadoftwo.De nition1.21(Groupisomorphism).TwogroupsAandBareisomorphicifthereisabijectionf:A!Bwhichpreservesthebinaryoperation.Ifbotharewrittenadditively,thatmeansforallxandyinA,f(x+y)=f(x)+f(y);ifmultiplicativenotationisusedinboth,thenf(xy)=f(x)f(y);ifadditiveinAbutmultiplicativeinB,thenf(x+y)=f(x)f(y);andifmultiplicativeinAandadditiveinB,thenf(xy)=f(x)+f(y):Thecorrespondencefiscalledagroupisomorphism.UsuallyAandBwillusethesamenotation,bothadditiveorbothmultiplicative,butnotalways.Exercise12.SupposethatbothAandBarewrittenmultiplicativelyandthatf:A!Bisagroupisomorphism.Provethatf(1)=1andf(x�1)=f(x)�1forallx2A.Example1.22.LetA=Zbethegroupofintegersunderaddition.LetBbetheintegralpowersof2,soB=f:::;1
4;1
2;1;2;4;:::gwithmultiplicationastheoperationinB.Provethatanisomorphismf:A!Bisde nedbyf(n)=2n.You'llneedtoshowthatf(m+n)=f(m)f(n).There'sactuallyanotherisomorphismg:A!B,too,de nedbyg(n)=2�n.1.3.2HomomorphismsWhereasisomorphismsarebijectionsthatpreservethealgebraicstructure,homomorphismsaresimplyfunctionsthatpreservethealgebraicstructure.Sincethewordhomomorphismissolong,alternatewordsareoftenusedlikemorphismandmap,especiallyinspokenmathe-matics.
1.3.ISOMORPHISMS,HOMOMORPHISMS,ETC.13De nition1.23(Ringhomomorphism).Aringhomomorphismf:A!Bbetweenringsisafunctionthatpreservesaddition,multiplication,and1.Agrouphomomorphismf:A!Bbetweengroupspreservesthebinaryoperation(ad-ditionormultiplicationdependingonthenotationusedforthegroup).Comment1.24.It'sapeculiarityofringsthatpreservingadditionandmultiplicationdoesn'timplythat1isalsopreserved,sothatconditionhastoberequiredaswell.We'llseeplentyofexamplesofhomomorphismsinthecourse,andtherearemoreexamplesinthenextsectiononmonomorphisms.Ofcourse,isomorphismsarespecialcasesofhomomorphisms.Example1.25(Aringhomomorphism).LetZ[x]betheringofpolynomialswithintegralcoecients.Evaluatingapolynomialf(x)ataparticularnumber,like3,togivef(3),isaringhomomorphism':Z[x]!Z.Itpreservesadditionsince'(f(x)+g(x))=f(3)+g(3)='(f(x))+'(g(x)),andyoucancheckthatitpreservesmultiplicationand1.Example1.26(Agrouphomomorphism).LetAbetheintegersunderaddition,andletB=f1;�1gwithmultiplicationasthebinaryoperation.Thenf:A!Bde nedbyf(n)=(�1)nisagrouphomomorphism.Youcanproveseveralpropertiesofhomomorphismsfromthede nition,butforthetimebeingI'lljustmentiontwobecausethey'llleadtotheconceptofcategorywhichwillbeintroducedinsection3.5.1.Thecompositionoftwohomomorphisms(ofthesamekind)isanotherhomomorphism.2.Theidentityfunction1A:A!A,whichmapseveryelementtoitself,isahomomor-phism,indeed,it'sanisomorphism.Whenwehaveahomomorphismf:A!B,we'llcallAthedomainoffandwe'llcallBthecodomainoff.(Sometimestheword\range"isusedforcodomain,butsomepeopleprefertouse\range"tomeanimage,whichissdi erentthing.Toavoidambiguity,we'lluse\codomain".)Amorenaturalwaytocharacterizeisomorphismisintermsofhomomorphisms.TworingsAandBareisomorphicifandonlyifthereiftherearetworinghomomorphismsf:A!Bandg:B!AsuchthatgfistheidentityonAandfgistheidentityonB.1.3.3MonomorphismsandepimorphismsTwocommonkindsofhomomorphismsaremonomorphismsandepimorphisms,oftencalledmonosandepisforshort.Whenahomomorphismf:A!Bisaninjectivefunction,it'scalledamonomorphism;andwhenitasurjectivefunction,it'sanepimorphism(but,inthecategoryofrings,we'llseetherearemoreepimorphismsthanjustthesurjectiveringhomo-morphisms).Youmightwonderwhyweneedthesewordswhenwe'vegotmorethanenoughwordsalreadytodescribeinjective(one-to-one)andsurjective(onto)aswellasothersnotmentionedhere.Themainreasonisthatthey'respecialkindsofinjectionsorsurjections|theypreservethealgebraicstructure.Anotheristhat,althoughforgrouphomomorphismsmonosandepishavetheseparticularcorrespondencestoinjectiveandsurjective,thereareothercategoriesinwhichtheydon't.Notethateveryisomorphismissimultaneouslyamonomorphismandandepimorphism.Theconverseholdsforgroups,but,surprisingly,notforrings.
14CHAPTER1.INTRODUCTIONExample1.27(Inclusion).Inclusionsaremonomorphisms.Whenonering(orgroup)Aisasubring(orsubgroup)ofanotherB,thentheinclusionfunction:A!B,whichmapsanelementtoitself,isamonomorphism.That'sanimportantexampleofamonomorphism,butthereareothers.Example1.28.Forexample,letAandBbothbetheadditivegroupofintegersZ,andletf(n)=2n.Thisfisamonomorphism,butit'snotaninclusion(whichinthiscasewouldbetheidentitymapsinceAandBarethesame).Comment1.29.Notethatiff:A!BisaringhomomorphismwhereAisa eldand06=1inB,thenfisalwaysaninjection,andsoit'samonomorphism.Youcanprovethisstatementintwostages.First,showthatiff(x)=0thenx=0.Second,showthatiff(x)=f(y),thenx=y.Thus,every eldhomomorphismisamonomorphism.Example1.30(Agroupepimorphism).We'llseeplentyofepimorphismswhenwetalkmoreabouttheintegersmodulon,butforthetimebeing,considerexample1.26ofagroupepimorphism.ThegroupAistheadditivegroupofintegersZ,andthegroupBisthetwoelementgroupf1;�1gundermultiplication.Thenf:A!Bde nedbyf(n)=(�1)nisagroupepimorphism.Evennumbersaresentto1andoddnumbersto�1.1.3.4EndomorphismsandautomorphismsAnendomorphismisjustahomomorphismf:A!Awherethedomainandcodomainarethesame,andanautomorphismisjustanisomorphismf:A!A.Theseareimportantbecausewealwayshavetheidentityautomorphism1A:A!Atocomparefto,sowehavemoreinformationwhenthedomainandcodomainarethesame.Example1.31(A eldautomorphism).LetCbethecomplex eld.Let:C!Cbecomplexconjugation,usuallydenotedbyputtingabarabovethecomplexnumber'(x+yi)=
x+yi=x�yi:Thisisclearlyabijectionsinceitisitsowninverse,
x+yi=x+yi.Also,itpreservesaddition,multiplication,and1,soit'saringisomorphism.
(x1+y1i)+(x2+y2i)=
x1+y1i+
x2+y2i
(x1+y1i)(x2+y2i)=
x1+y1i
x2+y2i
1=1Infact,it'sa eldautomorphismofC.Theexistenceofthisautomorphismsaysthatwecan'tdistinguishbetweeniand�iinthesensethatanytruestatementaboutthecomplexnumbersremainstruewhenalloccurrencesofiarereplacedby�i.Example1.32(Groupendomorphismsandautomorphisms).Therearemanygroupendo-morphismsf:Z!Zfromtheadditivegroupofintegerstoitself.Fixanyintegernandletf(x)=nx.Thisisagrouphomomorphismsincef(x+y)=n(x+y)=nx+ny=f(x)+f(y).
1.4.ALITTLENUMBERTHEORY15Forn6=0itisalsoamonomorphism.Forn=�1thisisnegation,andit'sabijection,soit'sagroupautomorphism.Thatsaysifweonlyconsideraddition,wecan'tdistinguishbetweenpositiveandnegativenumbers.Butnegationisnotaringautomorphismontheringofintegersbecause�(xy)doesnotequal(�x)(�y).Thus,withtheuseofmultiplication,wecandistinguishbetweenpositiveandnegativenumbers.1.4AlittlenumbertheoryInsciencenothingcapableofproofoughttobeacceptedwithoutproof.Thoughthisdemandseemssoreasonable,yetIcannotregarditashavingbeenmeteveninthemostrecentmethodsoflayingthefoundationsforthesimplestscience;viz.,thatpartoflogicwhichdealswiththetheoryofnumbers.Dedekind,1888Thiscourseisnotmeanttobeacourseinnumbertheory,butwewillneedalittlebitofit.We'llquicklyreviewmathematicalinductiononthenaturalnumbersN,divisibility,primenumbers,greatestcommondivisors,andtheEuclideanalgorithm.1.4.1MathematicalinductiononthenaturalnumbersNRichardDedekind(1831{1916)publishedin1888apaperentitledWassindundwassollendieZahlen?variouslytranslatedasWhatarenumbersandwhatshouldtheybe?orTheNatureofMeaningofNumbers.Inthatworkhedevelopedbasicsettheoryandcharacterizedthenaturalnumbersasasimplyin niteset.De nition1.33.(Dedekind)AsetNissaidtobesimplyin nitewhenthereexistsaone-to-onefunctionN0!Ncalledthesuccessorfunction,suchthatthereisanelement,calledtheinitialelementanddenoted1,thatisnotthesuccessorofanyelement,andifasubsetSofNcontains1andisclosedunderthesuccessorfunction,thenS=N.Suchasimplyin nitesetNmaybecalledthenaturalnumbers.Itischaracterizedbyanelement1andatransformationN0!Nsatisfyingthefollowingconditions:1.Injectivity:8n;m;n6=mimpliesn06=m0.2.Initialelement:8n;16=n0.3.Induction:IfSN,12S,and(8n;n2Simpliesn02S),thenS=N.TheDedekindaxioms,alsocalledthePeanoaxioms,arethislastcharacterizationinvolving1,thesuccessorfunction,andthethreeconditions.Amongotherthings,Peano(1858{1932)developedmuchofthenotationincommonuseinsettheory.Thelastaxiomiscalledmathematicalinduction.IfyouwanttoshowasubsetSofNisallofN, rstshowthat12S.Thenshowforeachnaturalnumbernthatn2Simpliesn+1inS.FinallyconcludethatS=N.
16CHAPTER1.INTRODUCTIONAprinciplethatislogicallyequivalenttomathematicalinductionisthewell-orderingprinciple,alsocalledtheminimizationprinciple.ItsaysthateachnonemptysubsetofNhasaleastelement.TouseittoproveasubsetSofNisallofN,assumethatitisn't,taketheleastelementninN�S,andderiveacontradiction,usuallybyshowingthere'sasmallerelementthannnotinS.AnotherprinciplelogicallyequivalenttomathematicalinductionisEuclid'sprincipleofin nitedescentwhichsaysthatthereisnoin nitedecreasingsequenceofpositiveintegers.ThisprinciplewasalsousedbyFermat(1607{1665).1.4.2DivisibilityWe'llrestrictourdiscussionnowtoN,thenaturalnumbers,thatis,thesetofpositiveintegers.Recallthatanintegermdividesanintegern,writtenm n,ifthereexistsanintegerksuchthatmk=n.Afewbasicpropertiesofdivisibilityfollowdirectlyfromthisde nition.Euclid( .ca.300B.C.E.)usessomeoftheseinBookVIIofhisElements.Youcan ndJoyce'stranslationofEuclid'sElementsonthewebathttp://aleph0.clarku.edu/~djoyce/java/elements/elements.html1.1divideseverynumber.1 n.2.Eachnumberdividesitself.n n.3.Ifonenumbermdividesanothernumbern,thenmdividesanymultipleofn,thatis,m nimpliesm kn.4.Divisibilityisatransitiverelation,thatis,m nandn kimplym k.5.Ifonenumberdividestwoothernumbers,thenitdividesboththeirsumanddi erence.m nandm kimplym (n+k)andm (n�k).6.Cancellationlaw.Onenumberdividesanotherifandonlyifanymultipleofthatonenumberdividesthesamenonzeromultipleoftheothernumber.m n()kn kn.(k6=0)Example1.34.ThedivisorsofanumbercanbedisplayedgraphicallyinwhatiscalledaHassediagramofthelatticeofdivisors.Asanexample,considerthenumber432.Itsprimefactorizationis2433,soitsdivisorsareoftheform2m3nwhere0m4and0n3.Thereare54=20ofthesedivisors.Theyare1234689121618242736485472108144216432Wecandisplaythesenumbersandemphasizewhichonesdividewhichotheronesifweputthelargenumbersatthetopofthediagram,andconnectthesmallerdivisorstothelargeroneswithlines.ThatresultsintheHassediagramin gure1.3.Sincedivisibilityistransitive,wedon'thavetoincludeallpossibleconnections.Solongasthereisapathofconnectionsfromalowernumbertoanupperone,thenwecanconcludethelowerdividestheupper.TheresultingdiagramiscalledaHassediagraminhonorofHasse(1898{1979).
1.4.ALITTLENUMBERTHEORY17
432
216
96
108
72
48
54
36
24
16
27
18
12
8
9
6
4
3
2
1
Figure1.3:Divisorsof432Exercise13.DrawHassediagramsforthedivisorsof30,32,and60.TheHassediagramforallpositiveintegersunderdivisibilityis,ofcourse,in nite.Figure1.4showsthepartofitupthrough12.
8
12
4
6
9
10
2
3
5
7
11
1
Figure1.4:Divisibilityupthrough121.4.3PrimenumbersDe nition1.35.Anaturalnumbergreaterthan1isasaidtobeaprimenumber,ormoresimplyaprime,ifitsonlydivisorsare1anditself,butifithasmoredivisors,it'scalledacompositenumber.Twopositiveintegersaresaidtoberelativelyprime,orcoprimeiftheonlypositiveintegerthatdividesthembothis1.PrimenumberswerementionedbythePythagoreansPhilolaus(470{385B.C.E.)andThymaridas(400{350B.C.E.),andbyAristotle(384{322B.C.E.)afterthem.The rstrecordedproofsaboutprimenumbersoccurinEuclid'sElements.
18CHAPTER1.INTRODUCTIONWeknowintuitivelythattherearein nitelymanyprimes,andthateverynumberisaproductofprimes.Nowlet'sprovethosestatements.We'llstartbyprovingsomethingthatwillhelpusprovethesetwostatements.Ifatheoremisnotparticularlyinteresting,butisusefulinprovinganinterestingstatement,thenit'softencalledalemma.ThisoneisfoundinEuclid'sElements.Lemma1.36(Euclid,VII.31).Everynumbergreaterthan1hasatleastoneprimedivisor.Proof.Letnbeanintegergreaterthan1.We'll ndaprimedivisorofn.Letmbethesmallestdivisorofngreaterthan1.(Notethatwe'reusingtheminimizationprinciple,alsocalledthewell-orderingprinciple,toconcludethatsuchanmexists.)We'llshowthatmisprimetherebyprovingthelemma.We'lldothatwithaproofbycontradiction,andthatmeansthat rstwe'llsupposethatmisnotprime,thenderiveacontradiction,andthatwillimplythatmmustbeprime.Supposemisnotprime,butcomposite.Themmistheproductoftwointegers,jandk,eachgreaterthan1.Now,k mandm n,sok n.Butkm.Thatgivesusadivisorofnwhichisevensmallerthanmbutstillgreaterthan1.Thatcontradictsthefactthatmisthesmallestdivisorofngreaterthan1.Thus,misprime,andit'sadivisorofn.q.e.d.Nowwecanproveoneofthetwostatements.Theorem1.37.Everynumbergreaterthan1iseitheraprimeortheproductofprimes.Proof.Thiswillbeanotherproofbycontradictionthatusesthewell-orderingprinciple.Supposethatthetheoremisfalse.Thenthereissomecompositenumbergreaterthan1thatthatisnottheproductofprimes.Letnbethesmallestsuch.Byourlemma,thisnhassomeprimedivisor,callitp.Thenm=n=pisanumbersmallerthannbutlargerthan1,so,bytheminimalityofn,miseitherprimeortheproductofprimes.Inthe rstcase,whenmisprime,thenn=pmistheproductoftwoprimes.Inthesecondcasewhenmisaproductofprimes,thenn=pmisalsoaproductofprimes.Inanycase,nistheproductofprimes,acontradiction.Thus,thetheoremistrue.q.e.d.Thislasttheoremwillformpartoftheso-calledfundamentaltheoremofarithmeticthatsayseverynumbergreaterthan1canbeuniquelyfactoredasaproductofprimes.Sofar,weonlyhavethateverynumberisaproductofprimes,butwehaven'tseentheuniqueness.We'llprovethatprettysoon.Next,let'sprovetheotherstatement,thattherearein nitelymanyprimes.ThisisEuclid'sproof.Theorem1.38(EuclidIX.20).Therearein nitelymanyprimes.Proof.Actually,Euclidprovessomethingalittlestronger.Givenany nitelistofprimes,he ndsaprimenotonthatlist.Supposethatp1;p2;:::;pkisa nitelistofprimes.Letnbetheproductoftheseprimes,n=p1p2pk:Byourlemman+1hasaprimefactor,callitp.Thisprimepcannotequalanypi,forthenpwoulddividebothnandn+1,andsowoulddividethedi erence1.Butaprimepcan'tdivide1sincep&#x-328;1.Thispisaprimenotonthelist.Itfollowsthattherearein nitelymanyprimes.q.e.d.
1.4.ALITTLENUMBERTHEORY19Thenumberofrelativelyprimeintegers.Animportantcombinatorialcountfornum-bertheoryandalgebraisthenumber'(n)ofpositiveintegerslessthanagivenintegern.Forexample,we'llshowlaterincorollary2.11thatthenumberofunitsintheringZnis'(n).We'llalsouseitinourdiscussionofcyclotomicpolynomialsinsection1.6.2.It'seasyenoughtocompute'(n)whennissmall.Forexample,'(12)=4,sincetherearefourpositiveintegerslessthan12whicharerelativelyprimeto12,namely,1,5,7,and11.De nition1.39(Euler'stotientfunction).Foragivenpositiveintegern,thenumberofpositiveintegerslessthannthatarerelativelyprimetonisdenoted'(n).Thefunction'iscalledEuler'stotientfunction.The rstfewvaluesofthetotientfunctionarelistedinthistable.
n
113456789101112131415161718
'(n)
112242646410412688166
Oneobviouspropertyofthisfunctionisthatifpisprime,then'(p)=p�1.Apropertythat'snotsoobviousisthatifmandnarerelativelyprime,then'(mn)='(m)'(n).Thatpropertyissummarizedbysayingthat'isamultiplicativefunction.ItfollowsfromtheChineseremaindertheoremdiscussedinsection3.2.1.Thatreducesthecomputationof'tocomputingitonpowerspkofprimenumbers.Thatcanbefounddirectly.Theonlypositiveintegerslessthanorequaltopkthataren'trelativelyprimetopkarethemultiplesofp,whicharep;2p;:::;pk,andtherearepk�1ofthem.Therefore,'(pk)=pk�pk�1=pk1�1
p.Theorem1.40(Euler'sproductformula).'(n)=nYp n1�1
p:Proof.Writen=pk11pkrrasaproductofpowersofdistinctprimes.Thenbythemulti-plicativityof','(n)='(pk11'(pkrr)=pk111�1
p1pk1r1�1
pr='(n)=nYp n1�1
p:q.e.d.1.4.4TheEuclideanalgorithmTheEuclideanalgorithmisanalgorithmtocomputethegreatestcommondivisoroftwonaturalnumbersmandn.EucliddescribedinBookVIIofhisElements.Euclidde nedthegreatestcommondivisoroftwonaturalnumbersmandn,oftendenotedgcd(m;n)ormoresimplyjust(m;n),asthelargestnumberdwhichisatthesametimeadivisorofmandadivisorofn.
20CHAPTER1.INTRODUCTIONAmongotherthings,greatestcommondivisorsareusedtoreducecommonfractionstolowestterms.Forexample,ifyouwantedtoreducethefraction1417
1853tolowestterms,youwouldlookforthegreatestcommondivisorofthetwonumbers1417and1853,whichis109.Thenyoucoulddivideboththenumeratorandthedenominatorbythatgreatestcommondivisortoreduce1417
1853toitslowestterms,namely,13
17.TherearetwoformsoftheEuclideanalgorithm.The rstform,asEuclidstatedit,repeatedlysubtractsthesmallernumberfromthelargerreplacingthelargerbythedi erence,untilthetwonumbersarereducedtothesamenumber,andthat'sthegreatestcommondivisor.(Notethattheprocesshastostopbythewell-orderingprinciplesinceateachstepthelargernumberisreduced.)Theotherformspeedsuptheprocess.Repeatedlydividethesmallernumberintothelargerreplacingthelargerbytheremainder.(Thisspeedsuptheprocessbecauseifthesmallernumberismuchsmallerthanthelarger,youdon'thavetosubtractitfromthelargermanytimes,justdivideonceandtaketheremainderwhichisthesameaswhatyou'dgetifrepeatedlysubtractedit.)Example1.41.Let's ndgcd(6731;5777).Since6731�5777=954,replace6731by954.We'vereducedtheproblemto ndinggcd(5777;954).Nowrepeatedlysubtract954from5777untilyougetanumbersmallerthan954andreplace5777bythatnumber.Alternatively,youcoulddivide954into5777andreplace5777bytheremainder.You'llgetthesamething,namely53.Nextto ndgcd(954;53).Ifyoukeepsubtracting53from954,eventuallyyoullget0.Orifyou'reusingdivision,whenyoudivide53into954,youllgetaremainderof0.Eitherway,youcanconclude53divides954,sotheirGCDis53itself.Thus,gcd(6731;5777)=53ThisEuclideanalgorithmworkstoproducethegcd,andtheargumentonlydependedontwopropertiesofdivisibilitymentionedabove,namelythatifonenumberdividestwoothernumbers,thenitdividesboththeirsumanddi erence.Sometimesthegcdoftwonumbersturnsouttobe1,andinthatcasewesaythetwonumbersarerelativelyprimeorthatthey'recoprime.Theorem1.42(Euclideanalgorithm).LetdbetheresultofapplyingtheEuclideanalgo-rithmtomandn.Thendisthegreatestcommondivisorgcd(m;n).Furthermore,thecommondivisorskofmandnarethedivisorsofgcd(m;n).Proof.OnestepoftheEuclideanalgorithmreplacesthepair(m;n)by(m�n;n).Itwasmentionedaboveinthepropertiesofdivisibilitythatifonenumberdividestwoothernumbers,thenitdividesboththeirsumanddi erence.Therefore,anumberkdividesbothmandnifandonlyifkdividesm�nandn.Sincethepair(m;n)havethesamesetofdivisorsasthepair(m�n;n),thereforegcd(m;n)=gcd(m�n;n).Thus,ateachstepoftheEuclideanalgorithmthegcdremainsinvariant.Eventually,thetwonumbersarethesame,butwhenthatlaststepisreached,thatnumberisthegcd.So,theendresultoftheEuclideanalgorithmisd=gcd(m;n).TheremarksaboveshowthateverydivisorkofmandnalsodividestheresultdofapplyingtheEuclideanalgorithmtomandn.Finally,ifk d,sinced mandd n,thereforek mandk n.q.e.d.
1.4.ALITTLENUMBERTHEORY21ExtendedEuclideanalgorithm.There'sstillmorethatwecangetoutofthealgorithmifweincludetheequationsimplicitinthecomputations.ThatwillleadtotheextendedEuclideanalgorithm.Example1.43.Whenwefoundgcd(6731;5777),ifwekepttrackofthequotientsaswellastheremainders,theneachstepyieldsanequation.6731�15777=9545777�6954=53954�1853=0Turningtheseequationsaround,wecan nd53asalinearcombinationof6731and5777asfollows,startingwiththenexttothelastequation.53=5777�6954=5777�6(6731�15777)=75777�66731Thus,thegcdof6731and5777isalinearcombinationofthem.Here'sthegeneralsituationto ndgcd(m;n)asalinearcombinationofmandn.Let'ssupposethatm�ntobeginwith.Wedividenintomandgetaquotientofq1andremainderofr1,thatism=q1n+r1;withr1between1andn.Thenweworkwithnandr1insteadofmandn.Divider1intontogetaquotientofq2andaremainderofr2,thatis,n=q2r1+r2:Andwekeepgoinguntileventuallywegetaremainderof0.r1=q3r2+r3r2=q4r3+r4...rs�3=qs�1rs�2+rs�1rs�2=qsrs�1+0Wehavem�n�r1�r2��rs�1andrs�1isd,thegcdwe'relookingfor.Eachequation ndsaremainderasalinearcombinationoftheprevioustworemainders.Startingwiththenexttothelastequation,wecan ndd=rs�1asalinearcombinationofrs�2andrs�3.Theequationbeforethatgivesrs�2intermsofrs�3andrs�4,sowecanalsogetdintermsofrs�3andrs�4.Workingourwaybackup,wecaneventuallygetdasalinearcombinationofmandn.Thus,we'veshownthefollowingtheorem.
22CHAPTER1.INTRODUCTIONTheorem1.44(ExtendedEuclideanalgorithm).Thegreatestcommondivisord=gcd(m;n)ofmandnisalinearcombinationofmandn.Thatis,thereexistintegersaandbsuchthatd=am+bn:Nowthatwehavethemajortheoremsongcds,thereareafewmorefairlyelementaryproprietiesofgcdsthatarestraightforwardtoprove,suchasthese.Theorem1.45.gcd(a;b+ka)=gcd(a;b).gcd(ak;bk)=kgcd(a;b).Ifd=gcd(a;b)thengcd(a=d;b=d)=1.Exercise14.Provethestatementsinthetheorem.GreatestcommondivisorsofmorethantwonumbersThegcdofmorethantwonumbersisde nedthesamewayasfortwonumbers:thegcdofasetofnumbersthelargestnumberthatdividesthemall.Forexample,gcd(14;49;91)=7.To ndagcdofthreenumbers,a,b,andc, rst ndd=gcd(a;b),then nde=gcd(d;c).Thus,gcd(a;b;c)=gcd(gcd(a;b);c);astatementthatiseasytoshow.PairwiserelativelyprimenumbersAsetofnumbersissaidtobepairwiserelativelyprimeorpairwisecoprimeifanytwoofthemarerelativelyprime.Forinstance,15,22,and49arethreepairwiserelativelyprimenumbers.Thus,a,b,andcarepairwiserelativelyprimewhengcd(a;b)=gcd(a;c)=gcd(b;c)=1:Notethatgcd(a;b;c)canbe1withouta,b,andcbeingpairwiserelativelyprime.Forinstance,gcd(6;10;15)=1,butgcd(6;10)=2,gcd(6;15)=3,andgcd(10;15)=5.LeastcommonmultiplesTheleastcommonmultipleofasetofpositiveintegersisthesmallestpositiveintegerthattheyalldivide.Itiseasytoshowthatthegreatestcommondivisoroftwointegerstimestheirleastcommonmultipleequalstheirproduct.gcd(a;b)lcm(a;b)=ab:Leastcommonmultiplescanbeusedtosumcommonfractions.Forexample,toadd5
6+4
15,notethattheleastcommonmultipleof6and15is30,soeachfractioncanbeexpressedwiththeleastcommondenominator30as25
30+8
30=25+8
30=33
30.Evenusingleastcommondenominators,itmaybethatthesumcanbesimpli edasitcaninthiscaseto11
10.1.5ThefundamentaltheoremofarithmeticWeprovedabovethateverynaturalnumbercouldbefactoredasaproductofprimes.Butwewantmorethanexistence,wewantuniqueness.Weneedtoprovethatthereisonlyonewaythatitcanbefactoredasaproductofprimes.
1.5.THEFUNDAMENTALTHEOREMOFARITHMETIC23Theuniquefactorizationtheorem,a.k.a.,thefundamentaltheoremofarithmetic.Now,inordertomakethisgeneralstatementvalidwehavetoextendalittlebitwhatwemeanbyaproduct.Forexample,howdoyouwriteaprimenumberlike7asaproductofprimes?Ithastobewrittenastheproduct7ofonlyoneprime.Sowewillhavetoacceptasinglenumberasbeingaproductofonefactor.Evenworse,whatabout1?Therearenoprimesthatdivide1.Onesolutionistoacceptaproductofnofactorsasbeingequalto1.It'sactuallyareasonablesolutiontode netheemptyproducttobe1,butuntilwe ndanotherneedforanemptyproduct,let'swaitonthatandrestrictthisuniquefactorizationtheoremtonumbersgreaterthan1.So,here'sthestatementofthetheoremwewanttoprove.Theorem1.46(Uniquefactorizationtheorem).Eachintegerngreaterthan1canbeuniquelyfactoredasaproductofprimes.Thatis,ifnequalstheproductp1p2profrprimes,anditalsoequalstheproductq1q2qsofsprimes,thenthenumberoffactorsinthetwoproductsisthesame,thatisr=s,andthetwolistsofprimesp1;p2;:::;prandq1;q2;:::;qsarethesameapartfromtheorderthelistings.We'llprovethisbyusingthestrongformofmathematicalinduction.Theformthatwe'lluseisthis:InordertoproveastatementS(n)istrueforallnumbers,provethatS(n)followsfromtheassumptionthatS(k)istrueforallkn.Thisprincipleofinductionappearstobestrongerthantheonewe'veusedbefore,but,infact,itisequivalenttoit.It'sreallythesameastheminimizationprinciple(i.e.well-orderingprinciple)appliedtothenegationofthestatement.Theadvantageinusingitisthataproofbycontradictionisnotneededmakingtheproofmoreunderstandable.Proof.We'llprovetheuniquefactorizationtheoremintwocases.Case1willbewherenisaprimenumberitself.Case2willbewhereniscomposite.Case1:Supposethatnisaprimenumber.Theonlywaythataprimenumbercanbewrittenasaproductofprimesisasitself;otherwiseitwouldnotbeprime,butcomposite.Case2:Supposethatnisacompositenumberequaltobothproductsofprimesp1p2prandq1q2qs.Notethatsinceniscomposite,bothrandsareatleast2;otherwiseitwouldnotbecomposite,butprime.Nowlookatoneoftheprimes,sayp1.Itdividesn,soitdividestheproductoftheotherprimesq1q2qs.Wesuspectthatthatimpliesithastobeoneofthoseotherprimes.Let'sputthato forabit;thatis,logicallybeforeweprovethistheorem,weneedtoproveanothertheorem,listednext,thatifaprimedividesaproductofprimes,thenitisoneofthoseprimes;butwe'llactuallydothatnext.Assumingwe'vedonethat,thenwecanconcludethatp1isoneoftheqi's.Wecanreordertheproductq1q2qstomakeitsothatq1equalsp1.Now,sincep1p2pr=q1q2qsandthe rst rstfactorsofthetwoproductsareequal,thereforep2pr=q2qs.Now,byournewinductionprinciple,thesearetwoprimefactorizationsofanumbersmallerthann,andhencearethesame,exceptfortheirorder.Therefore,theyhavethesamenumberoffactors,thatis,r=s,andallthefactorsarethesameexceptfortheirorder.Andthenumbernisthatproducttimesp1,whichequalsq1,thereforetheoriginaltwoproducts,p1p2prandq1q2qs,arethesameexceptfororder.q.e.d.
24CHAPTER1.INTRODUCTIONWell,that nishedtheproofexceptwehavetoproveanothertheorem rst,namely,thefollowingone.Theorem1.47.Ifaprimedividesaproductofprimesq1q2:::qs,thenitequalsoneoftheprimesq1;q2;:::;qs.Wecoulddothat,butwewe'llproveaslightlystrongertheorem,namely,thefollowingone.Theorem1.48.Ifaprimedividesaproductofnumbersb1b2:::bs,thenitdividesoneofthenumbersb1;b2;:::;bs.Nowthereasonthistheoremimpliestheprevioustheoremisbecauseifaprimepdividesaproductofprimesq1q2:::qs,thenitdividesoneoftheprimesq1;q2;:::;qs,buttheonlywaythatoneprimecandivideanotherisifitequalstheother.Proof.Aproductofsnumbersb1b2:::bsisactuallyaseriesofbinaryproducts.It'sb1timesb2:::bs,andb2:::bsisb2timesb3bs,etc,wherethelastproductisbs�1bsistheproductofbs�1timesbs.Thatmeansthatifweknewthefollowingtheorem,then,usingordinaryinduction,wecouldconcludethisone.q.e.d.Theorem1.49.Ifaprimedividesaproductoftwonumbers,thenitdividesoneofthenumbers.Now,wecouldprovethistheoremdirectly,butitturnsoutthatthereisaslightlystrongerversionthatwecanuseinotherplaces,solet'sproveit,theonelistednext,instead,andshowthistheoremfollowsfromit.Theorem1.50.Ifnandaarerelativelyprime,andn ab,thenn b.Proofthatthistheoremimpliesimpliesthepreviousone.Supposethataprimepdividesab.Ifpdoesn'tdividea,thenit'srelativelyprimetoa,sobythistheorem,itdividesb.Therefore,eitherp aorp b.q.e.d.Proofofthistheorem.Supposethatgcd(n;a)=1.Then,bytheextendedEuclideanal-gorithm,1isalinearcombinationofnanda,thatis,thereexistintegerstandusuchthat1=tn+ua:Multiplythatequationbybtogetb=tnb+uab:Now,ifn ab,thenndividestherighthandsideoftheequation,butthatequalsthelefthandside,son b.q.e.d.
1.6.POLYNOMIALS.25Comment1.51.Typicallyinamathematicsbookthosetheoremsthatcome rstlogicallyarepresented rst.Herewestartedwithourgoalanddiscoveredthetheoremsthatwereneededtoprovethegoal.(Actually,Imadethelistlongerthanitneededtobebystrength-eningacoupleofthembecausethestrongerversionsaremoreuseful,somethingyoucanonlytellwithhindsight.)Theadvantagetopresentingtheoremsintheirlogicalorderisthatitiseasiertofollowthelogic.Thedisadvantageisthatthemotivationforthepreliminarytheoremsisnotapparentuntilthe naltheorem,theinterestingone,isreached.Usuallywhenwewritetheprimefactorizationofanumber,we'lluseexponentsonthoseprimesthatarerepeated.Forinstance,thenumber40hadtheprimefactorization2225.Anabbreviatedformforthisfactorizationis235.Wesaythattheprime2occurswithmultiplicity3,whiletheprime5occurswithmultiplicity1.Themultiplicitiesaretheexponents.So,ingeneral,anumbernhastheprimefactorizationn=pe11pe12pekkwheretheprimesp1;p2;:::;pkarealldistinct,andtheirmultiplicitiesaretheexponentse1;e2;:::;ek,respectively.Theseexponentsarecalledtheordersoftheprimesinn.Theorderofpinnbetheexponentofpintheprimefactorizationofn,denotedordpa.Immediatecorollariestotheuniquefactorizationtheorem.Acorollaryisatheoremthatlogicallyfollowsverysimplyfromatheorem.Sometimesitfollowsfrompartoftheproofofatheoremratherthanfromthestatementofthetheorem.Inanycase,itshouldbeeasytoseewhyit'strue.Wecandrawacoupleofcorollariesfromtheuniquefactorizationtheorem.Corollary1.52.Theonlyprimesthatcandivideanumbernaretheonesthatappearinitsprimefactorizationpe11pe12pekk.Corollary1.53.Iftheprimefactorizationsofmandnarem=pe11pe12pekkandn=pf11pf12pfkk(whereheresomeoftheei'sandfi'smayequal0sowecanusethesamelistofprimesforbothnumbers),thentheirgreatestcommondivisord=gcd(m;n)hastheprimefactorizationd=pg11pg12pgkkwhereeachexponentgiistheminimumofthecorrespondingexponentseiandfi.Asanexampleofthelastcorollary,ifm=1260=22325171andn=600=233152,thentheirgcdisd=223151=60.1.6Polynomials.We'llfrequentlyusepolynomialsinourstudyof eldsandrings.We'llonlyconsiderpolyno-mialswithcoecientsin eldsandcommutativerings,notwithcoecientsinnoncommuta-tiverings.Wewon'tformallyde nepolynomials.Fornow,we'llonlylookatpolynomialsinonevariablex,butlaterinsection3.10.4we'lllookatpolynomialsintwoormorevariables.
26CHAPTER1.INTRODUCTIONInformallyapolynomialf(x)withcoecientsinacommutativeringRisanexpressionf(x)=anxn+an�1xn�1++a1x+a0whereeachcoecientai2R.We'llassumethattheleadingcoecientanisnotzerosothatdegf,thedegreeofthepolynomial,isn.Whenaniszero,thepolynomialiscalledamonicpolynomial.It'sconvenienttodenoteapolynomialeitherbyforbyf(x).Ifthevariablexisreferredtosomewherenearby,thenI'llusef(x),otherwiseI'lljustusef.Forinstance,ifIwanttomultiplytwopolynomialsfandgtogether,I'llwritefg,butifIwanttwomultiplyfbyx2�3x+2,I'llwritef(x)(x2�3x+2)orf(x)(x2�3x+2).ArootofapolynomialisanelementaofRsuchthatf(a)=0,thatis,it'sasolutionofthepolynomialequationf(x)=0.ThesetofallpolynomialswithcoecientsinacommutativeringRisdenotedR[x].Ithasaddition,subtraction,andmultiplication,andsatis estherequirementsofaring,thatis,ithasaddition,subtraction,andmultiplicationwiththeusualproperties.R[x]iscalledtheringofpolynomialswithcoecientsinR.NotethatR[x]doesn'thavereciprocalsevenwhenRisa eld,sincexhasnoinverseinR[x].Therefore,R[x]isnota eld.Nonetheless,theringRisasubringoftheringR[x]sincewecanidentifytheconstantpolynomialsastheelementsofR.1.6.1DivisionforpolynomialsAlthoughR[x]doesn'thavereciprocals,itdoeshaveadivisionalgorithm,atleastwhenthedivisorisamonicpolynomial.Theorem1.54(Thedivisionalgorithmforpolynomialsoveraring).LetRbeacommutativeringandR[x]itspolynomialringinonevariable.Letfbeapolynomial(thedividend)andgamonicpolynomial(thedivisor).Thenthereexistuniquepolynomialsq(thequotient)andr(theremainder)suchthatf=qg+rwhereeitherr=0ordegrdegg.Proofofexistence.Onecaseiswhenf=0ordegfdegg.Sincethedividendalreadyhasalowerdegree,thequotientq=0andtheremainderr=f.Thatleavesthecasewhendegfdegg.We'llproveitbyinductiononn=degfwherethebasecaseisn=0.That'sthecasewherefandgarebothconstantsintheringR,butgismonic,sog=1.Thenq=fandr=0.Nowfortheinductivestep.We'llassumetheinductivehypothesisthatthetheoremiscorrectforallpolynomialsfofdegreelessthannandshowit'strueforthoseofdegreen.Letf(x)=a0+a1x++anxnandg(x)=b0+b1x++bm�1xm�1+xmwherenm:Thepolynomialf1(x)=f(x)�anxn�mg(x)hasa0coecientforxn,soitsdegreeislessthann.Byinductivehypothesis,therearepolynomialsq1andr1suchthatf1=q1g+r1wherer1=0ordegr1degg.Equatingtherightsidesofthetwoequationsinvolvingf1,wemayconcludethatf(x)=(a1(x)+anxn�m)g(x)+f1(x):Thatgivesusthedesiredrepresentationf(x)=q(x)g(x)+r(x), nishingtheinductiveprooffortheexistencehalfoftheproof.q.e.d.
1.6.POLYNOMIALS.27Proofofuniqueness.Supposetherearealsopolynomialsq0andr0suchthatf=q0g+r0whereeitherr0=0ordegr0degg.We'llshowr=r0andq=q0.Sincef=qg+randf=q0g+r0,thereforeqg+r=qg0+r0sor�r0=g(q0�q).Supposethatr6=r0.Thenq0�q6=0,andsincegisamonicpolynomial,thereforedegg(q�q0)degg.Thereforedeg(r�r0)degg.Butdeg(r�r0)deggsincebothrandr0havedegreelessthandegg,acontradiction.Therefore,r=r0.Nowwehave0=g(q0�q),butgismonic,soq0�q=0,andq=q0.q.e.d.IfRhappenstobea eld,thereisastrongerversionofthetheoremthatdoesn'trequiregtobeamonicpolynomial.Theorem1.55(Thedivisionalgorithmforpolynomialsovera eld).LetFbea eldandF[x]itspolynomialringinonevariable.Letfbeapolynomial(thedividend)andganonzeropolynomial(thedivisor).Thenthereexistuniquepolynomialsq(thequotient)andr(theremainder)suchthatf=qg+rwhereeitherr=0ordegrdegg.Exercise15.Provetheabovetheorem.Hint:dividegbyitsleadingcoecientandusethedivisionalgorithmforpolynomialsoveraring.Therewillstillbetwoparts,oneforexistenceandoneforuniqueness.Theremaindertheoremandfactortheorem.Theremaindertheoremissomethingthat'sfrequentlycoveredinhighschoolalgebraclasses.Itsayswhenyoudivideapolynomialfbyx�a,theremainderisf(a).Itworksingeneralforpolynomialswithcoecientsinanarbitraryring.Theorem1.56(Remaindertheorem).LetRbeacommutativeringandR[x]itspolynomialring.Forf2R[x]anda2R,thereisapolynomialqsuchthatf(x)=(x�a)q(x)+f(a).Proof.Applythedivisionalgorithmforg(x)=x�a.Thenf(x)=(x�a)q(x)+rwhererisaconstant.Settingxtoa,weconcludef(a)=r.q.e.d.Thefactortheoremisacorollaryoftheremaindertheorem.|Theorem1.57(Factortheorem).Forf2R[x]anda2R,aisarootoffifandonlyif(x�a)dividesf(x).Furtherpropertiesofpolynomials.Thereareacouplemorepropertiesofpolynomialsthatapplyonlywhentheringisa eldoranintegraldomain.Asdescribedlaterinsection3.1.3,anintegraldomainisacommutativeringinwhich06=1thatsatis esoneofthetwoequivalentconditions:ithasnozero-divisors,oritsatis esthecancellationlaw.Thus, eldsarespecialcasesofintegraldomains.Onepropertyisthatapolynomialofdegreenhasatmostnroots.Theorem1.58.Thenumberofrootsofanonzeropolynomialwithcoecientsinanintegraldomainisatmostthedegreeofthepolynomial.
28CHAPTER1.INTRODUCTIONProof.We'llprovethisbyinductiononn,thedegreeofthepolynomialf.Ifn=0,thenfisaconstant,butit'snotthezeroconstant,soithasnoroots.Assumetheinductivehypothesis,namely,thetheoremholdsforallfunctionsofdegreen.We'llshowitholdsforeachfunctionfofdegreen+1.Iffhasnoroots,thenthetheoremistrue,soletrbearootoff.Bythefactortheorem,f(x)=(x�r)q(x),wherethedegreeofthequotientqequalsn.We'llshoweveryotherrootr06=roffisalsoarootofq.Sincer0isaroot,therefore0=f(r0)=(r0�r)q(r).Nowr0�risnot0,andtheringisanintegraldomainwhichhasnozero-divisors,therefore0=q(r).Thusallotherrootsoffarerootsofq.Sincedegg=n,bytheinductivehypothesis,ghasatmostnroots,thereforefhasatmostn+1roots.Thatcompletestheproofbyinduction.q.e.d.Exercise16.AnexampleofaringthatisnotanintegraldomainisZ8.Showthatthequadraticpolynomialf(x)=x2�1inZ8[x]hasmorethantworootsinZ8.Acoupleofcorollariesforpolynomialswithcoecientsinanintegraldomainfollowfromtheprevioustheorems.Corollary1.59.Ifdegf=n,anda1;a2;:::;anarendistinctrootsoff,thenf(x)=a(x�a1)(x�a2)(x�an)whereaistheleadingcoecientoff.Corollary1.60.Iftwomonicpolynomialsfandgbothofdegreenhavethesamevalueatnplaces,thentheyareequal.1.6.2RootsofunityandcyclotomicpolynomialsDe nition1.61(Rootofunity).Arootofunity,alsocalledarootof1isacomplexnumbersuchthatwhenraisedtosomepositiveintegerpoweryields1.Ifzn=1,thenziscalledannthrootofunity.Ifnisthesmallestpositiveintegerpowersuchthatzn=1,thenniscalledanthprimitiverootofunity.Amongtherealnumbers,theonlyrootsofunityare1and�1.1istheonly rstprimitiverootofunityand�1istheonlyprimitivesecondrootofunity.Thenthrootsofunityareequallyspacedaroundtheunitcircleseparatedbyanglesof2=n.See gure3.4fortheprimitiveseventhrootsofunityontheunitcircle.Annrootofunityzisarootofthepolynomialzn�1,butnotallrootsofsuchapolynomialareprimitive.Forexample,rootsofthepolynomialz2�1aresecondrootsofunity,but1,beingoneofthosetworoots,isnotaprimitivesecondrootofunity.Example1.62(Sixthrootsofunity).Thesixthrootsofunityarerootsofthepolynomialz6�1.Thispolynomialfactorsas(z3+1)(z3�1)=(z2�z+1)(z+1)(z2+z+1)(z�1).Ofcourse,twooftherootsofthispolynomialare1and�1whichaccountforthefactorsx�1andx+1.Therootsofthefactorz2�z�1arealsorootsofz3�1,soarecuberootsofunity,infact,they'rethetwoprimitivethirdrootsofunity.Thoserootsarez=1
2(�1+ip
3).Ifyoucalloneofthem!=1
2(�1+ip
3),thentheotheroneis!2=1
2(�1�ip
3).Youcansee
1.6.POLYNOMIALS.29themdisplayedinthecomplexplanein gure3.3whichillustratesthelatticeofEisensteinintegers.Therootsoftheotherfactorz2+z+1arez=1
2(1+ip
3).Theyarethetwoprimitivesixthrootsofunity.Noticethattheyare!+1and!2+1.So,altogether,therearesixsixthrootsofunity.Twoareprimitivesixthroots,twoareprimitivethirdroots,oneisaprimitivesecondroot,andoneisaprimitive rstroot.Amongthe ve fthrootsofunity,oneofthem,z=1,isnotprimitive,theotherfourare.Theyarerootsofthepolynomial5(z)=z5�1
z�1=z4+z3+z2+z+1.Ifzisaprimitiventhrootofunity,thentheentirelistofnthrootsis1;z;z2;:::;zn�1.Therootzkwon'tbeprimitiveifthereisacommondivisorofnandk.Thatleavesonly'(n)oftherootstobeprimitive,where'(n)isthenumberofpositiveintegerslessthannthatarerelativelyprimeton.Seede nition1.39forade nitionofEuler'stotientfunction'.De nition1.63(Cyclotomicpolynomial).Thepolynomialn(z)='(n)Yk=1(z�zk),wherez1;z2;:::;z(n)aretheprimitiventhrootsofunity,iscalledthenthcyclotomicpolynomial.Therearetwoprimitivethirdrootsofunityasmentionedintheexampleabove,so3(z)=z2�z�1.Therearealsotwoprimitivesixthroots,and6(z)=z2+z�1.Whenpisaprimenumber,then(p)hasdegree'(p)=p�1.Itsvalueis(p)=zp�1
z�1=zp�1++z+1.Here'sashorttableofthe rstfewcyclotomicpolynomials.n(n)
n(n)
1z�1
9z6+z3+12z+1
10z4�z3+z2�z+13z2+z+1
11z10+z9++z+14z2+1
12z4�z2+15z4+z3+z2+z+1
13z12+z9++z+16z2�z+1
14z6�z5+z4�z3+z2�z+17z6+z5+z4+z3+z2+z+1
15z8�z7+z5�z4+z3�z+18z4+1
16z8+1It'sinterestingthattheonlycoecientsthatappearinthe rstonehundredcyclotomicpolynomialsare0,1,and�1.We'llusecyclotomicpolynomialsinsection3.10.3.
30CHAPTER1.INTRODUCTION
Chapter2FieldsInformally,a eldisasetequippedwithfouroperations|addition,subtraction,multiplica-tion,anddivisionthathavetheusualproperties.We'llstudyringsinchapter3,whicharelike eldsbutneednothavedivision.2.1Introductionto eldsA eldisasetequippedwithtwobinaryoperations,onecalledadditionandtheothercalledmultiplication,denotedintheusualmanner,whicharebothcommutativeandassociative,bothhaveidentityelements(theadditiveidentitydenoted0andthemultiplicativeidentitydenoted1),additionhasinverseelements(theinverseofxbeingdenoted�x),multiplicationhasinversesofnonzeroelements(theinverseofxbeingdenoted1
x),multiplicationdistributesoveraddition,and06=1.Three eldsthatyoualreadyknowarethe eldofrealnumbersR,the eldofrationalnumbersQ,andthe eldofcomplexnumbersC.We'llseethattherearemanyother elds.Whenwehaveageneric eld,willuseacapitalFtodenoteit.2.1.1De nitionof eldsHere'samorecompletede nition.De nition2.1( eld).A eldFconsistsof1.aset,alsodenotedFandcalledtheunderlyingsetofthe eld;2.abinaryoperation+:FF!Fcalledaddition,whichmapsanorderedpair(x;y)2FFtoitssumdenotedx+y;3.anotherbinaryoperation:FF!Fcalledmultiplication,whichmapsanorderedpair(x;y)2FFtoitsproductdenotedxy,ormoresimplyjustxy;suchthat4.additioniscommutative,thatis,forallelementsxandy,x+y=y+x;31
32CHAPTER2.FIELDS5.multiplicationiscommutative,thatis,forallelementsxandy,xy=yx;6.additionisassociative,thatis,forallelementsx,y,andz,(x+y)+z=x+(y+z);7.multiplicationisassociative,thatis,forallelementsx,y,andz,(xy)z=x(yz);8.thereisanadditiveidentity,anelementofFdenoted0,suchthatforallelementsx,0+x=x;9.thereisamultiplicativeidentity,anelementofFdenoted1,suchthatforallelementsx,1x=x;10.thereareadditiveinverses,thatis,foreachelementx,thereexistsanelementysuchthatx+y=0;suchayiscalledthenegationofx;11.therearemultiplicativeinversesofnonzeroelements,thatis,foreachnonzeroelementx,thereexistsanelementysuchthatxy=1;suchayiscalledareciprocalofx;12.multiplicationdistributesoveraddition,thatis,forallelementsx,y,andz,x(y+z)=xy+xz;and13.06=1.Theconditionsfora eldareoftencallthe eldaxioms.Caveat:We'reusingtheterminologyandnotationofarithmeticthatweusefornumbers,buttheelementsofour eldsneednotbenumbers;oftentheywillbe,butsometimestheywon't.Notethatwe'llusethestandardnotationalconventionsonprecedenceforall eldssowedon'thavetofullyparenthesizeeveryexpression.Multiplicationanddivisionhaveahigherprecedencethanadditionandsubtraction,sothat,forexample,x�y=zmeansx�(y=z),not(x�y)=z.Also,operationsareexecutedfromlefttoright,sothatx�y�zmeans(x�y)�z,notx�(y�z).(Usuallyoperationsareexecutedfromlefttoright,butanexceptionisthatexponentiationisexecutedfromrighttoleft,sothatxmnmeansx(mn),not(xm)n.)Commutativityandassociativityofadditionimplythattermscanbeaddedinanyorder,soofcoursewewon'tputparentheseswhenwe'readdingmorethantwotermstogether.Likewiseformultiplication.Althoughinparts10and11ofthede nitiononlytheexistenceofanadditiveandmul-tiplicativeinversesisrequired,youcaneasilyshowuniquenessfollowsfromthede nition.Oncethatisdonewecannotethattheadditiveinverseofxiscalledthenegationofxanddenoted�x,andthemultiplicativeinverseofx,whenxisnot0,iscalledthereciprocalofxanddenoted1=x,1
x,orx�1.2.1.2Subtraction,division,multiples,andpowersWiththehelpofnegation,wecande nesubtractionasfollows.Thedi erenceoftwoelementsxandyisde nedasx�y=x+(�y).Likewise,withthehelpofreciprocation,wecande nedivision.Thequotientofanelementxandanonzeroelementyisxy�1,denotedx=yorx
y.Theexpectedpropertiesofsubtraction
2.1.INTRODUCTIONTOFIELDS33anddivisionallfollowfromthede nitionof elds.Forinstance,multiplicationdistributesoversubtraction,anddivisionbyzdistributesoveradditionandsubtraction.Likewise,wecande neintegralmultiplesofelementsina eld.First,we'llde nenonneg-ativemultiplesinductively.Forthebasecase,de ne0xas0.Thende ne(n+1)xasx+nxwhennisanonnegativeinteger.Thusnxisthesumofnx's.Forinstance,3x=x+x+x.Thenif�nisanegativeinteger,wecande ne�nxas�(nx).Theusualpropertiesofmultiples,like(m+n)x=mx+nxwill,ofcourse,hold.Furthermore,wecande neintegralpowersofx.De nex1asxforabasecase,andinductivelyfornonnegativen,de nexn+1asxxn.Thusnxistheproductofnx's.Forinstance,x3=xxx.Next,de nex0as1,solongasx6=0.(00shouldremainunde ned,butforsomepurposes,especiallyinalgebra,it'susefultode ne00tobe1.)Finally,if�nispositiveandx6=0,de nex�nas(xn)�1.Theusualpropertiesofintegralpowershold,likexm+n=xmxnand(xy)n=xnyn.2.1.3PropertiesthatfollowfromtheaxiomsTherearenumeroususefulpropertiesthatarelogicalconsequencesoftheaxioms.Generallyspeaking,thelistofaxiomsshouldbeshort,ifnotminimal,andanypropertiesthatcanbeprovedshouldbeproved.Here'salistofseveralthingsthatcanbeprovedfromtheaxioms.We'llproveafewinclass,you'llprovesomeashomework,andwe'llleavetherest.(Theymakegoodquestionsforquizzesandtests.)Inthefollowingstatements,unquanti edstatementsaremeanttobeuniversalwiththeexceptionthatwheneveravariableappearsinadenominator,thatvariableisnottobe0.Exercise17.Provethat0isunique.Thatis,thereisonlyoneelementxofa eldthathasthepropertythatforally,x+y=y.(Theproofthat1isuniqueissimilar.)Exercise18.Provethateachnumberhasonlyonenegation.Thatis,foreachxthereisonlyoneysuchthatx+y=0.(Theproofthatreciprocalsofnonzeroelementsareuniqueissimilar.)Exercise19.Provethattheinversesoftheidentityelementsarethemselves,thatis,�0=0,and1�1=1.Exercise20.Provethatmultiplicationdistributesoversubtraction:x(y�z)=xy�xz.Exercise21.Provethat0timesanyelementina eldis0:0x=0.Exercise22.Provethefollowingpropertiesconcerningmultiplicationbynegatives:(�1)x=�x,�(�x)=x,(�x)y=�(xy)=x(�y),and(�x)(�y)=xy.Exercise23.Provethefollowingpropertiesconcerningreciprocals:(x�1)�1=x,and(xy)�1=x�1y�1.Exercise24.Provethatwhenyandzarebothnonzerothatx
y=w
zifandonlyifxz=yw.Exercise25.Provethefollowingpropertiesconcerningdivision:x
yw
z=xzyw
yz,x
yw
z=xw
yz,andx
y.w
z=xz
yw.
34CHAPTER2.FIELDSAssumethatanytimeatermappearsinadenominatorthatitdoesnotequal0.Exercise26.Provethatifxy=0,theneitherx=0ory=0.2.1.4Sub eldsFrequentlywe'll ndone eldcontainedinanother eld.Forinstance,the eldofrationalnumbersQispartofthe eldofrealnumbersR,andRispartofthe eldofcomplexnumbersC.They'renotjustsubsets,QRC,buttheyhavethesameoperations.Here'stheprecisede nitionofsub eld.De nition2.2(sub eld).A eldEisasub eldofa eldFif1.theunderlyingsetofEisasubsetoftheunderlyingsetofF;2.theadditionoperation+EonEistherestrictionoftheadditionoperation+FonF,thatis,forallxandyinE,x+Ey=x+Fy;and3.themultiplicationoperationEonEistherestrictionofthemultiplicationoperationFonF,thatis,forallxandyinE,xEy=xFy.WhenEisasub eldofF,we'llalsosaythatFisanextensionofE.Whenyouknowone eldisasub eldofanother,there'snoneedtosubscripttheopera-tionssincetheyarethesame.Thereisanalternatecharacterizationofsub eld.Theproofofthefollowingtheoremisstraightforward,buttherearemanysteps.Theorem2.3.IfasubsetEofa eldFhas0,1,andisclosedunderaddition,multiplication,negation,andreciprocationofnonzeroelements,thenEisasub eldofF.The eldofrationalnumbersQ.Whenwe'retryingto ndthesmallestexampleofa eld,itlookslikeitwillhavetobeQ.Laterinsection2.2we'llseethatit'snotthesmallest!Buthere'sanargument(whichmusthavea awinit)whichsaysweneedalltherationalnumberstobeinany eldF.Tobeginwith,0and1havetobeinF.Butwealsohavetohave1+1inFandwe'lldenotethat2,ofcourse.Andwe'llneed1+1+1=2+1whichwe'lldenote3.Andsoforth,sowe'vegot0andallthepositiveintegersinF.Wealsoneednegationsofthem,soallthenegativeintegersareinF,too.Butarationalnumberm=nisjustanintegermdividedbyapositiveintegern,sowe'llhavetohaveallrationalnumbersinF.ThatshowsthatQisasub eldofF.Thus,itlookslikeevery eldFincludesthesmallest eldQ,the eldofrationalnumbers.There'soneminor awintheargumentabove,butlet'snotpickitapartrightnow.Prettysoonwe'lllookat eldsthatdon'tcontainQ.
2.2.CYCLICRINGSANDFINITEFIELDS352.1.5FieldsofrationalfunctionsArationalfunctionwithcoecientsinFisaquotientoftwopolynomialsf(x)
g(x).Rationalfunctionsdoforma eld,the eldF(x)ofrationalfunctionswithcoecientsinF.NoticethattheringofpolynomialsF[x]isdenotedwithsquarebracketswhilethe eldofrationalfunctionsF(x)isdenotedwithroundparentheses.Forexample,onerationalfunctioninQ(x)is5x2�3x+1=2
x3+27.Notethatthe eldFisasub eldoftheF(x).Again,wecanidentifytheconstantrationalfunctionsastheelementsofF.Forexample,Qisasub eldofQ(x),andbothRandQ(x)aresub eldsofR(x).Also,thetheringofpolynomialswithcoecientsisasubringofthe eldofrationalfunctions.ThatisFF[x]F(x).2.1.6Vectorspacesoverarbitrary eldsWhenyoustudiedvectorspaces,youmayhavestudiedonlyvectorspacesovertherealnumbers,althoughvectorspacesoverother eldsmighthavebeenmentioned.Infact,vectorspacesoveranarbitrary eldFhavethesamebasicpropertiesasvectorspacesoverR.Then-dimensionalstandardvectorspaceFnisde nedthesamewayasRnexceptthen-tupleshavecoordinatesinF.Additionandscalarmultiplicationarede nedthesamewayforFnastheyareforRn.Furthermore,matricesinMmn(F)arede nedthesamewayasmatricesinMmn(R)excepttheentriesareinFinsteadofR.Thematrixoperationsarethesame.YoucanusethesamemethodsofeliminationtosolveasystemoflinearequationswithcoecientsinFor ndtheinverseofamatrixinMnn(F)ifitsdeterminantisnonzero.Determinantshavethesameproperties.YoucanusemethodsoflinearalgebratostudygeometryinFnjustasyoucanforRn(althoughitmaynotbepossibletovisualizewhatFnissupposedtolooklike,andthingslikeareasoftriangleshavevaluesinF).Theabstracttheoryof nitedimensionalvectorspacesoverFisthesame,too.Linearindependence,span,basis,dimensionareallthesame.Rankandnullityofamatrixarethesame.Changeofbasisisthesame.Eigenvalues,eigenvectors,andeigenspacesmayhaveproblemsoversome elds.Infact,whenyoustudiedtransformationsRn!Rn,sometimesyouhadcomplexeigenvalues,andtheironlyeigenvectorswereinCn.LikewisewhenlookingattransformationsFn!Fnandtheeigenvaluesaren'tinF,you'llmayhavetogotosome eldextensionF0ofFto ndthemandtoF0nto ndtheeigenvectors.Likewise,canonicalformsformatriceswilldependonF.2.2Cyclicringsand nite eldsInthissectionwe'lllookat eldsthatare nite,andwe'lldiscoverthatQactuallyisn'tthesmallest eld.Althoughthey'resmaller elds|they're nite|theywon'tbesub eldsofQ.Firstwe'lllookabitattheconceptofcongruencemodulon,wherenisapositiveinteger.Thenwe'lllookattheringofintegersmodulon,denotedZ=nZormoresimplyZn.We'll
36CHAPTER2.FIELDSseewhythey'recalledcyclicrings.Finally,we'lllookatthecasewherenisprime,andwe'lldenoteitpthen,whereZpturnsouttobea eld,andwe'llexaminesomeofthecyclic elds.De nition2.4(Congruencemodulon).Fixn,apositiveinteger.Wesaythattwointegersxandyarecongruentmodulonifnevenlydividesthedi erencex�y.We'llusethestandardnotationfromnumbertheoryxy(modn)toindicatethatxiscongruenttoymodulon,andthenotationn mtoindicatethattheintegerndividestheintegerm(withnoremainder).Thenxy(modn)i n (x�y):Whenndoesn'tdividethedi erencex�y,wesayaisnotcongruenttob,denotedx6y(modn).You'refamiliarwithcongruencemodulo12;it'swhat12-hourclocksuse.ThegeneraltheoryofequivalencerelationsinsectionA.2.3.Theorem2.5.Congruencemodulonisanequivalencerelation.Proof.Forre exivity,xx(modn)holdssincen (x�x).Forsymmetry,weneedtoshowthatxy(modn)impliesyx(modn).Butifn (x�y),thenn (y�x).Fortransitivity,supposethatxy(modn)andyz(modn).Thenn (x�y)andn (y�z),sothereexistkandmsuchthatnk=x�yandnm=y�z.Thereforen(k+m)=x�z,showingthatn (x�z).Hencexz(modn).q.e.d.2.2.1ThecyclicringZnDe nition2.6(Integersmodulon).Theintegersmodulon,Znisthesetofequivalenceclassesofintegersundertheequivalencerelationwhichiscongruencemodulon.We'lldenotetheseequivalenceclasseswithsquarebracketssubscriptedbyn.Thus,forinstance,theelement0inZ6isreally[0]6,whichwe'lldenote[0]whenmodulo6isunderstood.Thisequivalenceclassisthesetofallxsuchthatx0(mod6).This[0]6=f:::;�18;�12;�6;0;6;12;18;:::g.Likewisetheelement1inZ6isreallytheequivalenceclassof1,whichistheset[1]6=fx2Zjx1(mod6)g=f:::;�17;�11;�5;1;7;13;19;:::g:Notethat[1]6=[7]6=[13]6allnamethesameequivalenceclass.Anequationinequivalenceclasses,suchas[x]6+[3]6=[5]6,isthesamethingasancongruence,x+35(mod6).Thecongruencenotationisusuallymoreconvenient.Whenthemodulusnisknownbycontext,we'lldispencewiththesubscriptn,andabusingnotation,we'llfrequentlydropthesquarebrackets.Therearetwowaysyoucanthinkaboutintegersmodulon.Oneistothinkofthemasregularintegersfrom0throughn�1,dothearithmeticmodulon,andadjustyouranswersoit'sinthesamerange.Forexample,wecantakeZ6=f0;1;2;3;4;5g.Then,todosome
2.2.CYCLICRINGSANDFINITEFIELDS37computation,say5(2�4)(mod6), rstcompute5(2�4)asintegerstoget�10,andthen,since�102(mod6),saytheansweris2.Thatworksverywellforcomputation,butit'sprettymessywhenyou'retryingtodoanythingwithvariablesortryingtoproveanythingingeneral.AbetterwayistosaythatanelementofZnisnamedbyaninteger,buttwointegersnamethesameelementofZnifthey'recongruentmodulon.Thus,xandynamethesameelementofZnifxy(modn).Thiswillworkbecausecongruencemodulonisanequivalencerelationaswesawearlier.Inanycase,ithelpsconceptuallytothinkoftheelementsofZnasbeingarrangedonacirclelikeweimaginetheelementsofZbeingarrangedonaline.See gure2.1ofacoupleofcyclicringsZntoseewheretheword\ring"camefrom.Z
r0r1r2r�1r�2Z19r0
r1
r2
r3
r�1
r�2
r�3
Z6r0
r1
r2
r�33
r�2
r�1
Figure2.1:CyclicringsZ6;Z19;ZTheoperationsonZn.Ourequivalencerelationiscongruencemodulon,soourequiva-lenceclassesarealsocalledcongruenceclasses.Congruencemodulonismorethanjustanequivalencerelation;itworkswellwithaddi-tion,subtraction,andmultiplication,asyoucaneasilyshow.Theorem2.7.Ifxy(modn),anduv(modn),thenx+uy+v(modn),x�uy�v(modn),andxuyv(modn).Thesepropertieswillallowustode nearingstructureonZn,asdonebelow.Butcongruencemodulondoesn'tworksowellwithdivision.Although60(mod6),itisnotthecasethat6=20=2(mod6).Thus,wecan'texpectthatZnwillbea eld,atleastwhenn=6.
38CHAPTER2.FIELDSOurjobistode neaddition,subtraction,andmultiplicationonZn.Wheneverasetisde nedasaquotientset,thatis,onequivalenceclasses,asZnis,anextrastepisrequiredwhende ninganoperationonit,aswe'llsee.Wewouldliketode neadditiononZnbysaying[x]+[u]=[x+u],thatis,thesumoftheequivalenceclassofxandtheequivalenceclassofushouldbetheequivalenceclassofx+u.Butwhatifwenamedtheequivalenceclassxbysomeotherinteger,sayy,andtheequivalenceofofubysomeotherintegerv?Howdoweknowwethat[y+v]isthesameequivalenceclassas[x+u]?Wecanstatethisquestioninacoupleofotherways.Howdoweknow[x]=[y]and[u]=[v]implies[x+u]=[y+v]?Thatasksthequestion:howdoweknowxy(modn)anduv(modn)impliesx+uy+v(modn)?That'soneofthepropertiesofcongruencementionedabove.ThatpropertysaysadditiononZnis\well-de ned".Likewise,sincemultiplicationworkswellwithcongruence,xy(modn)anduv(modn)implyxuyv(modn);wecande nemultiplicationonZnby[x][u]=[xu].Furthermore,alltheringaxiomswillbesatis edinZnsincethey'resatis edinZ.Thus,Znisaring,andit'scalledacyclicring.Theprojection :Z!Zn.Thefunction :Z!Znde nedby (k)=[k]mapsanintegertoitsequivalenceclassmodulon.Wede nedadditionandmultiplicationinZn[x+u]=[x]+[u]and[xu]=[x][u]so preservesadditionandmultiplication.Furthermore,since (1)=[1],itpreserves1.Therefore isaringhomomorphism.Itis,ofcourse,onto,soitisaringepimorphism.It'scalledaprojectionoracanonicalhomomorphismtothequotientring.Insection3.6,we'llgeneralizethisconstructiontoringsbesidesZandtheirquotients,andwe'llhaveprojectionsforthegeneralizations,too.Thecharacteristicofaring.What'sweirdaboutthesecyclicringsisthatifyoustartwith1andadd1overandover,you'llreachzero.Forinstance,inZ5,wehave1+1+1+1+1=50(mod5).Thiscorrespondstothegeometricinterpretationofthesecyclicringsbeingshapedlikerings.De nition2.8.Ifsomemultipleof1equals0inaring,thenthecharacteristicoftheringisthesmallestsuchmultiple.Ifnomultipleof1equals0,thenthecharacteristicissaidtobe0.We'reprimarilyinterestedincharacteristicswhenwe'retalkingabout elds,andwe'llseesoonthatthecharacteristicofa eldiseither0oraprimenumber.Example2.9.ThecharacteristicofZ5is5,and,ingeneral,thecharacteristicofa nitecyclicringZnisn.
2.2.CYCLICRINGSANDFINITEFIELDS392.2.2Thecyclicprime eldsZpSincedivisiondoesn'tworkwellwithcongruence,wecan'texpectZntoalwayshaverecipro-cals,sowedon'texpectittobea eld.Let's rstseewhenanelementinZnisaunit.Thetermunitinaringreferstoanelementxoftheringthatdoeshaveareciprocal.1isalwaysaunitinaring,andeverynonzeroelementina eldisaunit.Theorem2.10.AnelementkinZnisaunitifandonlyifkisrelativelyprimeton.Proof.First,supposethatkisaunitinZn.Thatmeansthereexistslsuchthatkl1(modn).Thennj(kl�1),andhencenisrelativelyprimetok.Second,supposethatkisrelativelyprimeton.Then,bytheextendedEuclideanalgo-rithm,theirgreatestcommondivisor,1,isalinearcombinationofkandn.Thus,thereareintegersxandysothat1=xk+yn.Then1xk(modn),andkdoeshaveareciprocal,namelyx,inZn.ThuskisaunitinZn.q.e.d.Recallfromde nition1.39thatthetotientfunction'(n)denotesthenumberofpositiveintegerslessthannthatarerelativelyprimeton.Corollary2.11(UnitsinZn).ThenumberofunitsinZnis(n).Theorem2.12.ThecyclicringZnisa eldifandonlyifnisprime.Proof.Partofthistheoremisadirectcorollaryofthepreviousone.Supposenisprime.TheneverynonzeroelementofZnisrelativelyprimeton.Therefore,Znisa eld.Nextwe'llshowthatifniscomposite,theringisnota eld.Letnbetheproductoftwointegersmandk,bothgreaterthan1.ThenneithermnorkcanhaveareciprocalinZn.Whynot?Supposethatm�1didexistinZn.Then(m�1m)k1kk(modn)m�1(mk)m�1n0(modn)Butk60(modn),acontradiction.Som�1doesn'texist.Therefore,Znisnota eld.q.e.d.Corollary2.13.Thecharacteristicofa eldis0oraprimenumber.Proof.We'llshowthatifthecharacteristicnis nite,itmustbeaprimenumber.Supposen=st.Then0=n1=(st)1=(s1)(t1).Therefore,eithers1=0ort1=0.Butnisthesmallestpositiveintegersuchthatn1=0,soeithers=nort=n.Thereforenisprime.q.e.d.Thisproofworksaswellinintegraldomainsintroducedinsection3.1.3.Thistheoremwillbementionedagainatthattime.Example2.14.Z2.Notethatthereisonlyonenonzeroelement,namely1,anditisitsowninverse.TheadditionandmultiplicationtablesforZ2areparticularlysimple.+
01
0
01
1
10

01
0
00
1
01
NotethatsubtractionisthesameasadditioninZ2sincex�yx+y(mod2).
40CHAPTER2.FIELDSExample2.15.Z3.Here,therearetwononzeroelements,namely1and2,but,forsym-metry'ssake,we'llcallthetwononzeroelements1and�1.Notethateachofthesetwoaretheirowninverses.Theadditionandmultiplicationtablesarestillprettysimple.+
�101
�1
1�10
0
�101
1
01�1

�101
�1
10�1
0
000
1
�101
Example2.16.Z13.Whatarethereciprocalsofthe12nonzeroelements?Wecannamethenonzeroelementsas1;2;:::;6.Youcanverifythatthistablegivestheirinverses.
x
1
2
3
4
5
6
x�1
1
6
4
3
5
2
Forinstance,thereciprocalof2is�6since2(�6)�121(mod13).These elds,Zpwherepisprime,arethe niteprime elds.Butthereareother nite elds.Example2.17.A eldoforder9.We'llmakeanextensionofZ3togeta eldoforder9.Notethat�1isnotasquaremodulo3.Wecanappendp
�1toZ3togeta eldalgebraicoveritinexactlythesamewaywegotCfromR.Let'suseiasanabbreviationforp
�1,asusual.ThenZ3(i)=fx+yijx;y2Z3gAddition,subtraction,andmultiplicationgiveusnoproblems.Wejusthavetocheckthatnonzeroelementshaveinverses.That'sexactlyasbefore.1
x+yi=x�yi
(x+yi)(x�yi)=x�yi
x2+y2=x
x2+y2+�y
x2+y2iThus,ifx+yiisnot0inZ3(i),thatis,notbothofxandyarearecongruentto0modulo3,thenx2+y260(mod3),andtheexpressionontherightgives(x+yi)�1.Notethatthecharacteristicofthis eldis3since1+1+1is0inthis eld.Exercise27.Youcanconstructa eldoforder25fromZ5,butithastobedonesomewhatdi erentlybecausep
�1alreadyexistsinZ5since(2)2=4�1inZ5.ThesquaresofthenonzeroelementsinZ5include1=(12and4=(2)2,but2isnotamongthesquares.ShowthattheringZ5[p
2]isa eldby ndinganinverseofanonzeroelementx+yp
2wherexandyareelementsofZ5butnotbothare0.Hint:(x�yp
2)(x+yp
2)=x2�2y2cannotbe0.Infact,thereare nite eldsoforderpnforeachpowerofaprimep.ThesearecalledtheGalois eldsGF(pn).Notethatcyclicprime eldarethesimplestGalois elds;ZpisGF(p).TheexampleconstructedGF(32)andtheexerciseGF(52).Theproofthata nite eldofcharacteristicphastohavepnelementsfollowsfromthetheoremsinsection4.9.2onAbeliangroups.It'sonlydependentonadditioninthe nite eld,notonmultiplication.
2.3.FIELDEXTENSIONS,ALGEBRAICFIELDS,THECOMPLEXNUMBERS412.2.3Characteristicsof elds,andprime eldsThecharacteristicofaringwasde nedabove,sowealreadyhavethede nitionforthecharacteristicofa eld.Those eldsthathavecharacteristic0allhaveQasasub eld.The awedproofwesawearlierincludedthemistakenassumptionthatalltheelements0;1;2;:::weredistinct,which,aswe'veseenwiththese nite elds,isn'talwaysthecase.Butwecancorrectthe awedprooftovalidatethefollowingtheorem.First,ade nition.De nition2.18.Aprime eldisa eldthatcontainsnopropersub eld.Equivalently,everyelementinitisamultipleof1.Theorem2.19.Each eldFhasexactlyoneoftheprime eldsasasub eld.ItwillhaveZpwhenithascharacteristicp,butitwillhaveQifithascharacteristic0.TheFrobeniusendomorphism.Exponentiationtothepowerphasaninterestingprop-ertywhenacommutativeringRhasprimecharacteristicp:(x+y)p=xp+ypTherearevariouswaystoprovethis.Forinstance,youcanshowthatthebinomialcoecientpkisdivisiblebypwhen1kp.Sincepk=p!
2)inGF(52).2.3FieldExtensions,algebraic elds,thecomplexnumbersAlotof eldsarefoundbyextendingknown elds.Forinstance,the eldofcomplexnumbersCisextendedfromthe eldofrealnumbersRandGF(32)isextendedfromZ3=GF(3).We'lllookatthegeneralcaseofextending eldsbyaddingsquarerootstoknown elds,thesmallestkindofextension,calledaquadraticextension.
42CHAPTER2.FIELDS2.3.1Algebraic eldsWe'velookedatsomequadraticextensionsof elds.Nowwe'lllookatalgebraicextensionsinmoredetail.De nition2.21(Algebraicandtranscendentalnumbers).Analgebraicnumberisanumberthatisarootofapolynomialwithrationalcoecients.Ifthepolynomialismonic,thenthealgebraicnumberisanalgebraicinteger.Arealnumberoracomplexnumberthatisnotalgebraiciscalledatranscendentalnumber.Forinstance,x=p
2isanalgebraicnumbersinceitistherootofthepolynomialx2�2;infact,it'sanalgebraicinteger.Ontheotherhand,x=p
1=2isarootofthepolynomial2x2�1,soit'sanalgebraicnumber,butnotanalgebraicinteger.Therearemanyrealnumbersusedinanalysisthataretranscendental.In1873CharlesHermite(1882{1901)provedthatthenumbereistranscendental.Itfollowsthatmanyrelatednumbersaretranscendentalsuchase2andp
e.De nition2.22(Algebraicandtranscendental eldextensions).Moregenerally,ifxsatis esapolynomialequationf(x)=0wherethepolynomialfhascoecientsina eldF,thenwesayxisalgebraicoverF.A eldextensionF0ofF,allofwhoseelementsarealgebraicoverFissaidtobeanalgebraicextensionofF.Fieldextensionsthatarenotalgebraicarecalledtranscendentalextensions.AnalgebraicextensionofQisalsocalledanalgebraicnumber eld,ormoresimplyanumber eld.In1882LindemannextendedHermite'sresulttoshowthateaistranscendentalforallnonzeroalgebraicnumbersa.Thusep
2istranscendental.Moreimportantly,Lindemann'stheoremshowsthat=eiistranscendental,forifitwerealgebraic,thenei=�1wouldbetranscendental,whichitisn't.Weierstrassprovedanevenmoregeneraltheoremin1885.Ifa1;:::;anaredistinctnonzeroalgebraicnumbers,thenthenumbersea1;:::;eanarealgebraicallyindependentmeaningeacheaiistranscendentaloverthe eldQ(ea1;:::;^eai;ean).Thehatovereaimeansthatisomittedfromthelist.Example2.23.Weknowthatthesquarerootof2,p
2isnotarationalnumber.The eldQ(p
2)isthesmallest eldthatcontainsp
2.Infact,itselementsarealloftheformx+yp
2wherex2Qandy2Q:It'sprettyobviousthatmostofthe eldaxiomshold.Theonlyonethat'snotobviousistheexistenceofreciprocalsofnonzeroelements,thatistosay,thestatement\(x+yp
2)�1isoftheformx0+y0p
2wherex0andy0arerationalandnotboth0"isnotsoobvious.Butthetrickof\rationalizingthedenominator"showsushow.1
x+yp
2=x�yp
2
(x+yp
2)(x�yp
2)=x�yp
2
x2�2y2=x
x2�2y2+�2y
x2�2y2p
2Notethatx2�2y2cannotbe0whenxandyarerationalandnotboth0.Forifx2�2y2=0,then2=(x=y)2,andthenp
2wouldbearationalnumber,whichitisn't.Thus,Q(p
2)isa eld.
2.3.FIELDEXTENSIONS,ALGEBRAICFIELDS,THECOMPLEXNUMBERS43Thetrickwastomultiplyanddividebytheconjugate.Let'sgiveanotationtothisconjugate:
x+yp
2=x�yp
2.Conjugationhassomeniceproperties.Itpreservesalltheelementsofthebase eldQ,thatis,ifx2Q,then
x=x.Itpreservesadditionandmultiplication,thatis,if and areelementsofQ(p
2),then
+ =
+
and
=

.Finally,theoperationofconjugation,
:Q(p
2)!Q(p
2),isitsowninverse,
= .Thus,conjugationisa eldautomorphism.Furthermore,theelements it xes,
= ,arejusttheelementsofthebase eldQ.2.3.2The eldofcomplexnumbersCInthesamewaywejustadjoinedp
2toQtogetQ(p
2),wecanadjoinp
�1toRtogetR(p
�1),whichisC.Algebraically,theprocessisidentical,butconceptuallyit'salittledi erentbecausewethoughtthatp
2,beingarealnumber,existedbeforeweappendedittoQ,whileitmaynotbesoclearthatp
�1existsbeforeweappendittoR.Butp
�1,usuallydenotedi,hasthepropertyi2=�1,soitisanalgebraicnumbersinceit'stherootofthepolynomialx2+1.Infact,R(i)consistsofelementsoftheformx+yiwithx;y2RasdescribedbyEuler.Additionandsubtractionare\coordinatewise"(x1+y1i)(x2+y2i)=(x1+x2)+(y1+y2)iwhilemultiplicationisonlyslightlymorecomplicated(x1+y1i)(x2+y2i)=x1x2+x1y2i+x2y1i+y1y2i2=(x1x2�y1y2)+(x1y2+x2y1)iWecan ndreciprocalsbyrationalizingthedenominatoraswedidabove.1
x+yi=x�yi
(x+yi)(x�yi)=x�yi
x2+y2=x
x2+y2+�y
x2+y2iWecande necomplexconjugationby
x+yi=x�yi.It'sa eldautomorphismofC,andits xedsub eldisR.Wecanalsode neanormonConcewehaveconjugation.Forz=x+yi2Q,letjzj2=z
z=(x+yi)(x�yi)=x2+y2:Sincejzj2isanonnegativerealnumber,ithasasquarerootjzj.AmatrixrepresentationofC.ConsiderthesubsetCofthematrixringM2(R)con-sistingofmatricesoftheformxy�yxwherex;y2R:
44CHAPTER2.FIELDSYoucaneasilyshowthatthisisasubringofM2(R)sincethe0matrixandtheidentitymatrixareofthisform,thesumanddi erenceofmatricesofthisformareofthisform,andsoistheproductasyoucanseeherexy�yxuv�vu=xu�yvxv+yu�yu�vx�yv+xu:Thus,CisasubringofM2(R).Furthermore,it'sacommutativesubringeventhoughM2(R)isnotacommutativeringsincethesameproductresultswhenthetwofactorsareinter-changed:uv�vuxy�yx=ux�vyuy+vx�vx�uy�vy+ux:FurthermoreCisa eldbecausenonzeromatricesinithaveinverses.Forsupposenotbothxandyare0.Thenxy�yxx
x2+y2�y
x2+y2y
x2+y2x
x2+y2=1001:Infact,Cisisomorphictothecomplex eldCasdescribedabove.Theisomorphismisdescribedbytheone-to-onecorrespondencexy�yx$x+yi:Notethatarealnumberxcorrespondstothematrixx00xwhileapurelyimaginarynumberyicorrespondstothematrix0y�y0.Notethatcomplexconjugationinthisrepresentationisjustmatrixtransposition.ThisalternaterepresentationofthecomplexnumbersasmatricesdirectlyexplainshowacomplexnumberactsasalineartransformationontherealplaneR2.Thecomplexnumberx+yimapsapoint(a;b)ofR2tothepoint(ax+by;�ay+bx)sincexy�yxab=ax+by�ay+bx:Matrixrepresentationsofvarious elds,rings,andgroupsareusefulfortworeasons.Oneisthattheygiveusgeometricinterpretationsfortheelementsasillustratedabove.Theotheristhatallthetoolsoflinearalgebraareavailabletousoncewehavethematrixrepresentation.2.3.3GeneralquadraticextensionsNowthatwe'veseenacoupleofquadraticextensions,let'sseehowitworksingeneral.LetFbea eldandeanelementofFthatisnotasquare.Inotherwords,thepolynomialx2�ehasnorootsinF.We'llconsiderorderedpairs(a1;a2)2FF,butwe'llwritethemasa1+a2p
e.We'llde neadditioncoordinatewise(a1+a2p
e)+(b1+b2p
e)=(a1+b1)+(a2+b2)p
e
2.4.REALNUMBERSANDORDEREDFIELDS45andde nemultiplicationby(a1+a2p
e)(b1+b2p
e)=(a1b1+ea2b2)+(a1b2+a2b2)p
e:Youcancheckthatthesede nitionsgiveusaring.But,doesitgiveusa eld?Aswedidbefore,we'll ndareciprocalofanonzeroelementa1+a2p
e1
a1+a2p
e=a1�a2p
e
(a1+a2p
e)(a1�a2p
e)=a1�a2p
e
a21�ea22Inorderforthistobethereciprocal,allwehavetodoisshowthedenominatora21�ea22isnot0.Inthecasethata2=0weknowa16=0sincenotbothare0,sointhatcasea21�ea22isnot0.Thatleavesusthecasethata26=0.Supposethata21�ea22=0.Thenea22=a21,anddividingbya22,weconcludee=(a1=a2)2.ButeisnotasquareinF.Thusa21�ea22isnot0inthiscase,too.Therefore,we'vefoundthereciprocal.Thus,wehavea eld,F(p
e).Whenwelookatmoregeneral eldextensions,we'llhavealotmoretheory,andwewon'thavedetailstocheckaswedidhere.Thattheorywillinvolvetheconceptof\ideals"inaringasdiscussedinsection3.6.2.4Realnumbersandordered eldsWe'lllooknowatR,the eldofrealnumbers.What'ssospecialabouttherealnumber eld?Foronething,it'sgotanorderonit;wecancomparetworealnumbersxandyandsaywhichissmallerorifthey'reequal.That'sanextrastructureona eld.We'llstartbylookingatthisconceptofordered eld.Beforewegettoofar,youshouldknowthatthatisn'tenoughtodistinguishRfromother elds.Thereareplentyofotherordered elds,suchasQandallthe eldsbetweenQandR.2.4.1Ordered eldsTheeasiestwaytode neanordered eldisbysayingit'spartitionedintopositiveelements,negativeelements,and0,andrequiringacouplepropertiesontheseparts.De nition2.24(Ordered eld).Anordered eldconsistsofa eldFalongwithasubsetPwhoseelementsarecalledpositivesuchthat1.Fispartitionedintothreeparts:P,f0g,andNwhereN=fx2Fj�x2PgtheelementsofNarecallednegative;2.thesumoftwopositiveelementsispositive;and3.theproductoftwopositiveelementsispositive.
46CHAPTER2.FIELDSPropertiesofordered elds.Youcanshowfromthisde nitionthat1.thesumofnegativeelementsisnegative2.theproductofanegativeelementandapositiveelementisnegative3.theproductoftwonegativeelementsispositive4.1ispositive,and�1isnegativeExercise29.Provethefourpropertiesabove.Examples.R,Q,andall eldsbetweenthemareordered eldswheretheusualpositivenumbersinthe eldformP.AlthoughQandRareordered elds, nite eldsandChavenoordering.Exercise30.ShowthatCisnotanordered eld.Hint:showwhyican'tbepositive,zero,ornegative.Thebinaryorderrelations.FromPwecande nethebinaryorderrelations,,&#x]TJ/;༗ ;.9;Ւ ;&#xTf 9;&#x.104;&#x 0 T; [0;,and.Forinstance,xymeansy�xiszeroorpositive,whilexymeansy�xispositive.Thatcanbestatedformallyasfollows:xyi y�x2P[f0gxyi y�x2P:Alltheexpectedpropertiesoftheseorderrelationsfollow.Hereareafew.1.Trichotomy:Foreachpairx;y,exactlyoneofthethreerelationsxy,x=y,orx&#x-277;yholds.2.Transitivity:xyandyzimplyxz.3.Ifxispositiveandyz,thenxyxz.4.Ifxisnegativeandyz,thenxy&#x-277;xz.5.Ifxispositive,thensois1=x.6.Forpositivexandy,ifxy,then1=y1=x.Exercise31.Provethesixpropertiesabove.Theorem2.25.Thecharacteristicofanordered eldis0.Proof.SupposeFisanordered eldofcharacteristicp6=0.Since1ispositive,thenanysumof1swillbepositive.Thenpispositive.Butpequals0whichisnotpositive.Acontradiction.Thereforeanordered eldcannothavenonzerocharacteristic.q.e.d.ItfollowsthatQisasub eldofeveryordered eld.
2.4.REALNUMBERSANDORDEREDFIELDS47Example2.26.Anorderedextensionoftherealnumberswithin niteelementsandin- nitesimalelements.Wecangivethe eldofrationalfunctionsR(x)anorderasfollows.First,we'llde newhenapolynomialf(x)=anxn+an�1xn�1++a1x+a0ispositive,andthatwillbewhenitsleadingcoecientanisapositiverealnumber.Next,we'llde newhenarationalfunctionf(x)=g(x)ispositive,andthatwillbewhenfandgarebothpositivepolynomialsorbothnegativepolynomials.Itfollowsthatf(x)=g(x)isnegativeoneoffandgispositiveandtheotherisnegative.Only0=g(x),whichequals0,won'tbepositiveornegative.Youcaneasilyshowthatthesumandproductofpositiverationalfunctionsispositive.TherealnumbersRisanorderedsub eldofR(x),meaningthatit'sasub eldanditselementshavethesameorderwhethertheorderonRisusedortheorderonR(x)isused.Withthisorder,thereareelementsthatarelargerthananyrealnumbera,forexample,x�asincex�aispositive.Inotherwords,xisanin niteelement.Likewise,therearepositiveelementsthataresmallerthananypositiverealnumber,1=x,forexample,so1=xisanin nitesimalnumber.2.4.2ArchimedeanordersThelastexampleisanexampleofanordered eldwithin niteelementsandin nitesimals.Everyordered eldFisanextensionofQ,sowecande neanin niteelementofFtobeanelementx2Fgreaterthaneveryrationalnumber,andwecande neapositivein nitesimalelementasapositivex2Fsmallerthaneverypositiverationalnumber.Notethatthereciprocalofanin niteelementisanin nitesimal,andviceversa.De nition2.27.AnArchimedeanordered eldor,moresimply,anArchimedean eld,issimplyanordered eldFwithoutin niteelementsorin nitesimals.BeforeArchimedes,EuclidusedthispropertyinhisElementsinBookVandfollowingbooks.ThecontentofBookVisduetoEudoxus,soabetternamefortheArchimedeanpropertywouldhavebeenEudoxus'property.Thereareequivalentcharacteristicsthatcouldbeusedforthede nition.Herearetwo.EachelementofFislessthansomeinteger.EachpositiveelementofFisgreaterthanthereciprocalofsomepositiveinteger.Ofcourse,theprecedingexampleisanon-Archimedean eld.Anotherinterestingnon-Archimedean eldisthatofsurrealnumberscreatedbyJohnConway.Surrealnumbersincludeallrealnumbers,allordinalnumbersandmore.Sinceordinalnumbersformaproperclass,sodosurrealnumbers.Foraniceintroductiononsurrealnumbers,seeDonaldKnuth'sbookSurrealNumbers.Still,thereareloadsofArchimedean elds,namelyQ,R,andalltheintermediate elds.Westillhaven'tansweredthequestionaboutwhatmakesRspecial.Beforewegoon,however,let'sseehowelementsinanArchimedean eldaredeterminedbyhowtheycomparetorationalnumbers.ForanArchimedean eldF,sinceFisordered,ithascharacteristic0,soithasasasub eld,indeed,anorderedsub eld,the eldofrationalnumbersQ.Theorem2.28(Density).BetweenanytwodistinctelementsofanArchimedean eld,thereliesarationalnumber.
48CHAPTER2.FIELDSProof.LetxyinanArchimedean eld.We'relookingforarationalnumberm
nbetweenxandy.Ifxisnegativewhileyispositive,thentherationalnumber0liesbetweenthem.Wecanreducethecasewherethey'rebothnegativetothecasewherethey'rebothpositivebynotingthatifm
nliesbetween�xand�y,then�m
nliesbetweenxandy.Sowemayassumethatbothxandyarepositive.Ifwecan ndsomemultiplenofthemsothatny�nx�1,thensomeintegermliesbetweennyandnx,butnxmnygivesxm
ny.Andwecan ndsuchamultiplesincey�xisgreaterthanthereciprocal1
nofsomepositiveintegersincethe eldisArchimedean.q.e.d.AnelementaofFpartitionsQintotwoparts(La;Ra)La=fx2QjxagandRa=fx2Qjxag:Thesetwopartshaveaspecialproperty.De nition2.29.ADedekindcutoftherationalnumbersisapartitionofQintotwononemptyparts(L;R)|aleftpartLandarightpartR|suchthateveryelementofLislessthaneveryelementofR.Furthermore,theleftpartdoesnothaveagreatestelement.Theorem2.30.AnelementaofanArchimedean eldFisdeterminedbyitsDedekindcut(La;Ra).Thatis,if(La;Ra)=(Lb;Rb),thena=b.Proof.Ifa6=b,thenthereisarationalnumberbetweenthem,sothatrationalnumberwillbeinoneleftpartbuttheotherrightpart.q.e.d.InanArchimedean eldFnoteveryDedekindcuthastodetermineanelement.Forexample,inQ,thecut(L;R)whereL=fxjx0orx22gandR=fxjx&#x]TJ/;༗ ;.9;Ւ ;&#xTf 2;.74; 0 ;&#xTd [;0andx2&#x]TJ/;༗ ;.9;Ւ ;&#xTf 2;.74; 0 ;&#xTd [;2gisnotthecutofanyrationalnumber.Butthatsamecutwithp
2includedinRisthecutofp
2.Therealnumbersarespecialinthateverycutisthecutofsomerealnumber.AlthoughtheremightnotbeaelementofFforeverycut,thecutsareenoughtodeter-mine,alongwiththeorderonFandthe eldstructureofQ,the eldstructureofF.IthelpsinproofstocutinhalftheinformationofaDedekindcutfrom(L;R)tojustL.Itissucienttode neaDedekindcutjustintermsofoftheleftpart.Youcanprovethefollowinglemmatosimplifythestatementandtheproofofthefollowingtheorem.Lemma2.31.If(L;R)isaDedekindcut,thenLhasthefollowingthreepropertiesi.Lisanonempty,propersubsetofQ;ii.ify2Landx2Qsuchthatxy,thenx2L;andiii.foreachx2C,thereexistsy2CsuchthatxyConversely,ifLhasthesethreeproperties,then(L;R)isacutwhereRisthecomplementofL.Theorem2.32.InanArchimedean eldF,additionandmultiplicationaredeterminedbyDedekindcutsinthesensethatIfaandbaretwoelementsofF,thentheleftpartoftheirsuma+bisdeterminedbytheirleftpartsLa+b=fx+yjx2Laandy2Lbg:IfaandbaretwopositiveelementsofF,thentheleftpartoftheirproductisdeterminedbytheirleftpartsLab=fxyjx2La;x&#x-278;0;y2Lbandy&#x-278;0g[fxjx0g:
2.4.REALNUMBERSANDORDEREDFIELDS492.4.3Completeordered eldsTherearevariousde nitionsgivenforcompleteordered elds,alllogicallyequivalent.Here'sone.De nition2.33.Acompleteordered eldisanArchimedean eldthatcannotbeextendedtoalargerArchimedean eld.Equivalently,everyDedekindcutdeterminesanelementofthe eld.Completenessisthe nalpropertythatcharacterizesR.Actually,rightnowwehaven'tprovedthatthereisatleastonecompleteordered eld,andwehaven'tprovedthatthereisatmostonecompleteordered eld.Oncewedo,wecan nallyproperlyde neR.Existenceofacompleteordered eldWe'llstartbystatingthetheoremwhichgivesthecomponentsforonewayofconstructingacompleteordered eldF.Tomakeitcomplete,wejusthavetomakesurethateveryDedekindcutdeterminesanelementofthe eld.Thewaytodothat,ofcourse,tode nethe eldtobethecuts,andthede nitionoftheoperationsofadditionandmultiplicationaredeterminedbythecutsasseeninthelasttheorem.Theorem2.34.Thereisacompleteordered eldF.ItselementsareDedekindcutsofQ.IfL1andL2areleftpartsoftwocuts,thentheleftpartofthesumisdeterminedbytheleftpartL+=fx+yjx2L1andy2L2g:IfListheleftpartapositivecut(onethatcontainsatleastonepositiverationalnumber),thenitsnegationisdeterminedbytheleftpartL�=f�xjx=2Lgexcept,ifthisL�hasalargestelement,thatlargestelementisremoved.IfL1andL2areleftpartsoftwopositivecuts,thentheleftpartoftheproductisdeterminedbytheleftpartL=fxyjx2L1;x�0;y2L2andy�0g[fxjx0g:TherearemanydetailstoshowtoverifythatRisacompleteordered eld.First,thatthesetsL+,L�,andLareleftparts.thenthe eldaxiomsneedtobeveri ed,thentheorderaxioms,thenthat'sit'sanArchimedean eld.Thelaststep,thatit'scompleteisalmostobviousfromtheconstruction.Nooneofthesestepsisdicult,buttherearemanydetailstocheck.Therearealternatewaystoconstructcompleteordered elds.OneisbymeansofCauchysequences.Thespiritisdi erent,buttheresultisthesame,since,aswe'reabouttosee,thereisonlyonecompleteordered eld.Uniquenessofthecompleteordered eldWehavetosomehowexcludethepossibilitythattherearetwodi erentArchimedean eldsthatcan'tbeextendedtolargerArchimedean elds.Wedon'twanttocounttwoisomorphic eldsasbeingdi erent,since,inessence,they'rethesame eldbutthenamesoftheelementsarejustdi erent.So,whatwewantisthefollowingtheorem.
50CHAPTER2.FIELDSTheorem2.35.Anytwocompleteordered eldsareisomorphicasordered elds.Further-more,thereisonlyoneisomorphismbetweenthem.Proof.Wemaytreatthe eldQasasub eldofthetwocompleteordered eldsF1andF2.ThenasaDedekindcutdeterminesanelementa12F1andanelementa2inF2,wehaveabijectionF1!F2.Youonlyneedtoverifythatpreservesadditionandmultiplication,whichitdoes,sinceinanArchimedeanring,additionandmultiplicationaredeterminedbyDedekindcuts.q.e.d.Risthecompleteordered eldWenowknowthatthereisonlyonecompleteordered elduptoisomorphism.Anysuchcompleteordered eldmaybetakenastherealnumbers.2.5Skew elds(divisionrings)andthequaternionsSirWilliamRowanHamilton,whoearlyfoundthathisroad[tosuccesswithvec-tors]wasobstructed|heknewnotbywhatobstacle|sothatmanypointswhichseemedwithinhisreachwerereallyinaccessible.Hehaddoneaconsiderableamountofgoodwork,obstructedashewas,when,abouttheyear1843,heper-ceivedclearlytheobstructiontohisprogressintheshapeofanoldlawwhich,priortothattime,hadappearedlikealawofcommonsense.Thelawinquestionisknownasthecommutativelawofmultiplication.KellandandTait,18732.5.1Skew elds(divisionrings)Skew elds,alsocalleddivisionrings,haveallthepropertiesof eldsexceptthatmultipli-cationneednotbecommutative.Whenmultiplicationisnotassumedtobecommutative,acoupleofthe eldaxiomshavehavetobestatedintwoforms,aleftformandarightform.Inparticular,werequire1.thereisamultiplicativeidentity,anelementofFdenoted1,suchthat8x;1x=x=x1;2.therearemultiplicativeinversesofnonzeroelements,thatis,8x6=0;9y;xy=1=yx;and3.multiplicationdistributesoveraddition,thatis,8x;8y;8z;x(y+z)=xy+xzand8x;8y;8z;(y+z)x=yx+zx.Alltheotheraxiomsremainthesame,exceptwenolongerrequirecommutativemultiplica-tion.Thevariouspropertiesof eldsthatfollowfromthe eldaxiomsalsofollowfromtheskew eldaxioms,althoughsomehavetostatedintwoforms.Themostimportantskew eldisthequaternions,mentionednext.Waringshowedthattherewereno niteskew eldsthatweren't elds(adicultproof).
2.5.SKEWFIELDS(DIVISIONRINGS)ANDTHEQUATERNIONS512.5.2ThequaternionsHWe'renotgoingtostudyskew elds,butoneisofparticularimportance,thequaternions,denotedH.TheletterHisinhonorofHamilton,theirinventor.Wecande neaquaternionaasanexpressiona=a0+a1i+a2j+a3kwherea0;a1;a2,anda3arerealnumbersandi;j;andkareformalsymbolssatisfyingthepropertiesi2=j2=k2=�1andij=k;jk=i;ki=j:Thei,j,andkareallsquarerootsof�1,buttheydon'tcommuteasyoucanshowfromthede nitionthatji=�k;kj=�i;ik=�j:Thisdoesn'tleadtoacommutativemultiplication,butnotethatifaisreal(i.e.,itspurequaternionpartsa1;a2,anda3areall0),thenawillcommutewithanyquaternionb.AdditionandsubtractionarecoordinatewisejustlikeinC.Here'smultiplication.(a0+a1i+a2j+a3k)(b0+b1i+b2j+b3k)=(a0b0�a1b1�a2b2�a3b3)+(a0b1+a1b0+a2b3�a3b2)i+(a0b2�a1b3+a2b0+a3b1)j+(a0b3+a1b2�a2b1�a3b0)kIt'seasytocheckthatalltheaxiomsforanoncommutativeringaresatis ed.TheonlythinglefttoinordertoshowthatHisaskew eldisthatreciprocalsexist.Wecanuseavariantofrationalizingthedenominatorto ndthereciprocalofaquaternion.1
a0+a1i+a2j+a3k=a0�a1i�a2j�a3k
(a0�a1i�a2j�a3k)(a0+a1i+a2j+a3k)=a0�a1i�a2j�a3k
a20+a21+a22+a23Thus,anonzeroquaterniona0+a1i+a2j+a3k,thatis,onewherenotalloftherealnumbersa0;a1;a2,anda3are0,hasaninverse,sincethedenominatora20+a21+a22+a23isanonzerorealnumber.Theexpressiona0�a1i�a2j�a3kusedtorationalizethedenominatoristheconjugateoftheoriginalquaterniona0+a1i+a2j+a3k.It'sworthwhiletohaveanotationforit.
a0+a1i+a2j+a3k=a0�a1i�a2j�a3k;aswedoforC.We'llalsode nethenormofaquaternionabyjaj2=a
a.It'sanonnegativerealnumber,soithasasquarerootjaj.Notethatjaj2=a20+a21+a22+a23.Thus,ifaisanonzeroquaternion,thenitsinverseis1
a=
a
jaj2.
52CHAPTER2.FIELDSForC,the eldofcomplexnumbers,conjugationwasa eldautomorphism,butforH,it'snotquiteanautomorphism.Ithasallofthepropertiesofanautomorphismexceptone.Itpreserves0,1,additionandsubtraction
ab=
a
b,andreciprocation
1=a=1=
a,butitreversestheorderofmultiplication
ab=
b
a.We'llcallsuchathinganantiautomorphism.NotethatHextendsCinmanyways.Theassignmentx+iy2Ctox+iy2Hisone,butx+iy2Ctox+jy2Hisanother.Thereare,infact,in nitelymanywaysthattheskew eldHextendsthe eldC.Theorem2.36.Thenormofaproductistheproductofthenorms.Proof.jabj2=ab
ab=ab
b
a=ajbj2
a=a
ajbj2=jaj2jbj2.q.e.d.Ifweunpacktheequationjaj2jbj2=jabj2,we'llgetasacorollaryLagrange'sidentityonrealnumberswhichshowshowtoexpresstheproductoftwosumsoffoursquaresasthesumoffoursquares.Corollary2.37(Lagrange).Theproductofthesumoffoursquaresofintegersisasumoffoursquaresofintegers(a20+a21+a22+a23)(b20+b21+b22+b23)=(a0b0�a1b1�a2b2�a3b3)2+(a0b1+a1b0+a2b3�a3b2)2+(a1b2+a2b1+a3b1�a1b3)2+(a2b3+a3b2+a1b2�a2b1)2Notethatthisequationnotonlyworksforrealnumbers,butalsoforintegers,indeedwhenthecoecientslieinanycommutativering.Lagrangeusedthisidentitytoshowthateverynonnegativeintegernisthesumoffoursquares.Theidentityaboveisusedtoreducethegeneralcasetothecasewhennisprime.Lagrangestillhadworktodototakecareoftheprimecase.Frobenius'stheoremandtheoctonions.Thequaternionsareveryspecialinthesensethatthey'retheonly nite-dimensionaldivisionalgebraoverRotherthanRitselfandC.ThistheoremwasprovedbyFrobeniusin1877.AdivisionalgebraovertherealnumbersRisadivisionring(skew eld)thathastherealsasasub eld.ItsdimensionisthedimensionithasasavectorspaceoverR.Thereisalsoaneight-dimensionalnon-associativealgebraovertherealnumberscalledtheoctonions,O.OctonionswerediscoveredbyJohnT.Gravesin1843.AlthoughOisnotassociative,itdoessatisfyweakerassociativityidentitieswhentwoofthethreevariablesarethesame:x(xy)=(xx)y,x(yy)=(xy)y,and(xy)x=x(yx).Italsosatis estheMoufangidenties:z(x(zy))=((zx)z)y,x(z(yz))=((xz)y)z,(zx)(yz)=(z(xy))z,and(zx)(yz)=z((xy)z).FurthermoreOhasanorm.OctonionsoverRareaspecialcaseofaCayleyalgebraovera eld.
2.5.SKEWFIELDS(DIVISIONRINGS)ANDTHEQUATERNIONS53AmatrixrepresentationforH.TherearevariousmatrixrepresentationsforH.ThisonewillmakeHasubringoftherealmatrixringM4(R).We'llrepresent1bytheidentitymatrix,andi,j,andkbythreeothermatriceswhich,youcanverify,satisfyi2=j2=k2=�1andij=k;jk=i;ki=j:1$266410000100001000013775i$26640�1001000000�100103775j$266400�10000110000�1003775k$2664000�100�10010010003775Thenagenericquaterniona+bi+cj+dkcorrespondstothematrix2664a�b�c�dba�dccda�bd�cba3775Quaternionsandgeometry.Eachquaternionaisthesumofarealparta0andapurequaternionparta1i+a2j+a3k.Hamiltoncalledtherealpartascalarandpurequaternionpartavector.Wecaninterpreta1i+a2j+a3kasavectora=(a1;a2;a3)inR3.Additionandsubtractionofpurequaternionsthenarejustordinaryvectoradditionandsubtraction.Hamiltonrecognizedthattheproductoftwovectors(purequaternions)hadbothavectorcomponentandascalarcomponent(therealpart).ThevectorcomponentoftheproductaboftwopurequaternionsHamiltoncalledthevectorproduct,nowoftendenotedabora_b,andcalledthecrossproductortheouterproduct.ThenegationofthescalarcomponentHamiltoncalledthescalarproduct,nowoftendenotedab,(a;b),ha;bi,orhajbiandcalledthedotproductortheinnerproduct.Thusab=ab�ab:Hamilton'squaternionswereverysuccessfulinthe19thcenturyinthestudyofthree-dimensionalgeometry.Here'satypicalproblemfromKellandandTait's1873IntroductiontoQuaternions.Ifthreemutuallyperpendicularvectorsbedrawnfromapointtoaplane,thesumofthereciprocalsofthesquaresoftheirlengthsisindependentoftheirdirections.Matriceswereinventedlaterinthe19thcentury.(Butdeterminantswereinventedearlier!)Matrixalgebrasupplantedquaternionalgebraintheearly20thcenturybecause(1)theydescribedlineartransformations,and(2)theyweren'trestrictedtothreedimensions.Exercise32.ShowthatHcanberepresentedasasubringofthecomplexmatrixringM2(C)where1$1001i$i00�ij$01�10k$0ii0
54CHAPTER2.FIELDSsothatagenericquaterniona+bi+cj+dkcorrespondstothematrixa+bic+di�c+dia�biUnitquaternionsandS3asagroup.Thequaternionsa=a0+a1i+a2j+a3kwithnorm1arecalledunitquaternions.Examplesofunitquaternionsare1;i;j;k,buttherearemanymore.Unitquaternionsarethequaternionsforwhicha20+a21+a22+a23=1.Thatequationispreciselytheequationthatde nestheunitthreesphereS3in4-spaceR4,althoughS3isusuallydescribedwithdi erentvariables:S3=f(w;x;y;z)2R4jw2+x2+y2+z2=1g:Aswesawabove,theproductofthenormsoftwoquaternionsisthenormoftheproduct,thereforemultiplicationisclosedonthis3-sphere.Furthermore,1isaunitquaternion,andthereciprocalofaunitquaternionisanotherone,and,multiplicationisassociative,somultiplicationofquaternionsmakesthe3-sphereS3intoagroup.
Chapter3RingsRingsarethingslikeZthathavethethreeoperationsofaddition,subtraction,andmulti-plication,buttheydon'tneeddivision.Thelackofadivisionoperationmakesthemmorecomplicatedandmoreinteresting.Theconceptofprime,forexample,isuninterestingfor elds,butveryinterestingforZandotherrings.Mostofourringswillhavecommutativemultiplication,butsomewon't,sowewon'trequirethatmultiplicationbecommutativeinourde nition.Wewillrequirethateveryringhave1.Theformalde nitionforringsisverysimilartothatfor elds,butweleaveoutacoupleoftherequirements.Inthischapterwe'llconcentratemainlyoncommutativeringsandtheirproperties.We'llconsidercommutativeringswithvariousniceproperties.Thoseringswithnicepropertieswe'llgivespecialnamesinincreasingnicenesssuchasintegraldomain,uniquefactorizationdomain,principalidealdomain,andEuclideandomain3.1IntroductiontoringsAringisasetequippedwithtwobinaryoperations,onecalledadditionandtheothercalledmultiplication,denotedintheusualmanner,whicharebothassociative,additioniscommu-tative,bothhaveidentityelements(theadditiveidentitydenoted0andthemultiplicativeidentitydenoted1),additionhasinverseelements(theinverseofxdenoted�x),andmulti-plicationdistributesoveraddition.If,furthermore,multiplicationiscommutative,thentheringiscalledacommutativering.3.1.1De nitionandpropertiesofringsHereisamorecompletede nition.De nition3.1.AringRconsistsof1.aset,alsodenotedRandcalledtheunderlyingsetofthering;2.abinaryoperation+:RR!Rcalledaddition,whichmapsanorderedpair(x;y)2RRtoitssumdenotedx+y;55
56CHAPTER3.RINGS3.anotherbinaryoperation:RR!Rcalledmultiplication,whichmapsanorderedpair(x;y)2RRtoitsproductdenotedxy,ormoresimplyjustxy;suchthat4.additioniscommutative,thatis,8x;8y;x+y=y+x;5.additionisassociative,thatis,8x;8y;(x+y)+z=x+(y+z);6.multiplicationisassociative,thatis,8x;8y;(xy)z=x(yz);7.thereisanadditiveidentity,anelementofFdenoted0,suchthat8x;0+x=x;8.thereisamultiplicativeidentity,anelementofFdenoted1,suchthat8x;1x=x;9.thereareadditiveinverses,thatis,8x;9y;x+y=0;and10.multiplicationdistributesoveraddition,thatis,8x;8y;8z;x(y+z)=xy+xz.Whenmultiplicationisalsocommutative,thatis,8x;8y;xy=yx,theringiscalledacom-mutativering.Theconditionsforaringareoftencalltheringaxioms.Subtraction,multiples,andpowers.Aswedidwith elds,wecande nesubtraction,integralmultiples,andnonnegativeintegralpowers.Wewon'thavedivisionornegativeintegralpowerssincewedon'thavereciprocals.Asbefore,wede nesubtractionintermsofnegation.Thedi erenceoftwoelementsxandyisx�y=x+(�y).Theexpectedpropertiesofsubtractionallfollowfromtheringaxioms.Forinstance,multiplicationdistributesoversubtraction.Likewise,wecande neintegralmultiplesofelementsinaring.De ne0xas0,theninductivelyde ne(n+1)x=x+nxwhenn0.Thenif�nisanegativeinteger,de ne�nxas�(nx).Theusualpropertiesofmultiples,like(m+n)x=mx+nxstillhold.Furthermore,wecande nepositiveintegralpowersofx.De nex1asxforabasecase,andinductively,xn+1=xxn.Thusnxistheproductofnx's.Forinstance,x3=xxx.Sinceringsneedn'thavereciprocals,wecan'tde nenegativeintegralpowersofx.Examples3.2(rings).Ofcourse,all eldsareautomaticallyrings,butwhataresomeotherrings?We'vetalkedaboutsomeothersalready,including1.theringofintegersZwhichincludesallintegers(wholenumbers)|positive,negative,or0.2.theringofpolynomialsR[x]withcoecientsinacommutativeringR.3.thematrixringMn(R)ofnnmatriceswithentriesinacommutativeringR.Thisexampleisanoncommutativeringwhenn2.4.theringofuppertriangularmatricesisasubringofMn(R).5.thecyclicringZn,theringofintegersmodulon,wherenisaparticularinteger.6.thepowerset}(S)consistingofsubsetsofasetSbecomesaring,calledaBooleanring,whereA+Bisthesymmetricdi erenceandABistheintersectionoftwosubsetsAandB.
3.1.INTRODUCTIONTORINGS57Propertiesthatfollowfromtheringaxioms.Therearenumeroususefulpropertiesthatfromtheaxioms,butnotsomanyasfollowfromthe eldaxioms.Here'salistofseveralofthem.1.0isunique.Thatis,thereisonlyoneelementxofaringthathasthepropertythat8y;x+y=y.Likewise,1isunique.2.Multiplicationdistributesoversubtraction.x(y�z)=xy�xzand(y�z)x=yx�zx.3.�0=0.4.0x=0.5.(�1)x=�x,(�x)y=�(xy)=x(�y),and(�x)(�y)=xy.Therearesomeexpectedpropertiesthatarenotincludedhere.I'llshowwhynotusingexamplesfromZ6.1.Iftheproductoftwoelementsis0,xy=0,itdoesnotfollowthateitherx=0ory=0.Forexample,inZ6theproductof2and3is0.2.Cancellationdoesnotalwayswork.Thatis,ifxy=xzandx6=0,itdoesn'tfollowthaty=z.Forexample,inZ6,32=34,but26=4.3.1.2ProductsofringsIfR1andR2aretworings,youcanconstructtheirproductringR.TheunderlyingsetofRistheproductR1R2oftheunderlyingsetsofthetworings,andaddition,subtraction,andmultiplicationarecoordinatewise.Thus,(x1;x2)(y1;y2)=(x1y1;x2y2)and(x1;x2)(y1;y2)=(x1y1;x2y2):TheadditiveidentityinR1R2is0=(0;0),andthemultiplicativeidentityis1=(1;1).Sincealltheoperationsareperformedcoordinatewise,theringaxiomsaresatis edinR1R2,soit'saring.Theprojectionfunctions1:R1R2!R1and2:R1R2!R2de nedby1(x1;x2)=x1and2(x1;x2)=x2arebothringhomomorphisms.Theypreserveaddition,multiplication,and1.Productsofmorethan2ringscanbede nedanalogously,evenproductsofin nitelymanyrings.Wedidn'tdiscussproductsof eldsinthechapteron eldbecausetheproductoftwo eldsisnotanother eld.Itisatleastaring,however.3.1.3IntegraldomainsMuchofthetimewewillwantthecancellationpropertythatwasmentionedabovetohold,sowe'llgiveaspecialnametocommutativeringsthatsatisfythem.Itwillhelpifwemakeacoupleofde nitions.
58CHAPTER3.RINGSDe nition3.3.Anonzeroelementxinacommutativeringisazero-divisorifthereexistsanonzeroysuchthatxy=0.Ofcourse,0isalwaysazero-divisor.We'llsayacommutativeringhasnozerodivisorsif0istheonlyzero-divisor.De nition3.4.We'llsayacommutativeringsatis esthecancellationlawif8x6=0;8y;8z;xy=xzimpliesy=z:Wefoundintheexampleabovethat2and3arezero-divisorsinZ6,andthatZ6didnotsatisfythecancellationlaw.YoucanexamineZntodeterminewhichnonzeroelementsarezero-divisorsandwhichhavereciprocals.There'saconnectionbetweenzero-divisorsandthecancellationlaw.Theorem3.5.Acommutativeringsatis esthecancellationlawifandonlyifithasnozero-divisors.Proof.Supposetheringsatis esthecancellationlaw.Letxbeanonzeroelementinthering.Ifxy=0,thenxy=x0,sobythatcancellationlaw,y=0.Thenxcan'tbeazero-divisor.Thustheringhasnozero-divisors.Nextsupposethattheringhasnozero-divisors.We'llshowitsatis esthecancellationlaw.Ifx6=0andxy=xz,thenx(y�z)=0,andsincexisnotazerodivisor,thereforey�z=0,soy=z.Thustheringsatis esthecancellationlaw.q.e.d.De nition3.6(integraldomain).AnintegraldomainisacommutativeringDinwhich06=1thatsatis esoneofthetwoequivalentconditions:ithasnozero-divisors,oritsatis esthecancellationlaw.Allthe eldsandmostoftheexamplesofcommutativeringswe'velookedatareintegraldomains,butZnisnotanintegraldomainifnisnotaprimenumber.Notethatanysubringofa eldoranintegraldomainwillanintegraldomainsincethesubringstillwon'thaveanyzero-divisors.Notethatproductsof(nontrivial)ringsareneverintegraldomainssincetheyalwayshavethezerodivisors(1;0)and(0;1)whoseproductis0.Corollary2.13statedthatthecharacteristicofa eldiseither0oraprimenumber.Theproofthereworksaswellforintegraldomains.Thecharacteristicofanintegraldomainiseither0oraprimenumber.GroupringsYoucanformaringZGoutofagroupGasfollows.AssumethatGiswrittenmultiplicatively.The niteformalsumsofelementsofGaretheelementsofZG.Thus,ifnisanonnegativeintegeranda1;:::;an2G,thentheformalsumx1a1++xnannamesanelementofthegroupringZG.Additioniscoordinatewise.Multiplicationusesthegroupoperation.Thisde nitioncanbegeneralizedsothatgroupringshavetheircoordinatesinanycom-mutativeringR,notjustZ.ThisresultsinagroupringRG.Exercise33.LetGbethetwoelementcyclicgroupG=f1;agwherea2=1.AtypicalelementofZGisx+yawherex;y2Z.Multiplicationisde nedby(x1+y1a)(x2+y2a)=(x1x2+y1y2)+(x1y2+x2y1)a.ShowthatthesquareofanynonzeroelementinZGisnotzero,butshowthatZGdoeshavezero-divisorsby ndingapairofnonzeroelementswhoseproductis0.
3.1.INTRODUCTIONTORINGS593.1.4TheGaussianintegers,Z[i]OneimportantexampleofanintegraldomainisthatoftheGaussianintegersZ[i].Itselementsareoftheformx+yiwherex;y2Z,sotheycanbeviewedasalatticeofpointsinthecomplexplaneasin gure3.1.YoucancheckthatZ[i]isclosedunderaddition,subtraction,multiplication,andincludes1,soitisasubringofthe eldC.Therefore,it'sanintegraldomain.We'llseelaterthatZ[i]isaparticularlyniceintegraldomaincalledaEuclideandomain.
1
i
Figure3.1:LatticeofGaussianintegersZ[i]Therearefourunits(elementshavingreciprocals)intheGaussianintegers.Besides1and�1,iand�iarealsounits.Notethat(1+i)(1�i)=2,so2isnotprimeinZ[i]eventhoughitisprimeinZ.We'llcomebacktoZ[i]whenwestudyEuclideandomainsinsection3.8.4.AlsoZ[i]isanexampleofa\ringofintegers"tobede nedinsection3.11.Eisensteinintegers.TheEisensteinintegersaresimilartotheGaussianintegers,butinsteadofconsistingofasquarelatticesofcomplexnumbers,theyconsistofatriangularlatticeofcomplexnumbers.Theyincludecomplexnumbersoftheformz=x+y!where!isthecuberootof1,!=1
2(�1+ip
3)=e2i=3.See gure3.3forthelatticeofEisensteinintegers.3.1.5Finite eldsagainWewon't ndanyexamplesof niteintegraldomainsthataren't eldsbecausetherearen'tany.Theorem3.7(Wedderburn).IfRisa niteintegraldomain,thenRisa eld.
60CHAPTER3.RINGSProof.LetxbeanonzeroelementofR.Considerthepositivepowersofx:x;x2;x3;:::;xn::::Sincetherearein nitelymanypowers,butonly nitelymanyelementsinR,thereforeatleasttwodistinctpowersareequal.Let,then,xm=xnwithmn.Cancelxmfromeachsideoftheequation(whichispossiblebecauseRisanintegraldomain)toconcludexn�m=1.Therefore,thereciprocalofxisxn�m�1.Therefore,everynonzeroelementhasaninverse.q.e.d.ThistheoremcanbeusedtogiveashortproofthatZpisa eldwhenpisaprime,sinceit'seasytoshowthatZpisanintegraldomain.We'llshowithasnozero-divisors.Supposethatxy0(modp).Thenp xy.Butifaprimedividesaproduct,itdividesoneofthefactors,soeitherp xorp y,inotherwords,eitherx0(modp)ory0(modp).Thus,Zpisanintegraldomain,andhence,bytheabovetheorem,it'sa eld.Ourearlier,morecomplicatedproofusedtheextendedEuclideanalgorithmto ndaninverseforx.That'sactuallyamuchmoreecientwayto ndtheinversethantolookthroughthepowersofx.3.2FactoringZnbytheChineseremaindertheoremWe'lllookatthestructureofthecyclicringZnwhenniscompositeinmoredetail.Inparticular,whennisnotapowerofaprimenumber,thenZnisaproductofsmallercyclicrings.3.2.1TheChineseremaindertheoremThistheoremsaysthatifmandkarerelativelyprimeandn=mk,thenZn=ZmZk.Let'sillustratethatwithm=7andk=12toshowhowZ84=Z7Z12.Startingwithanumberxmodulo84,we'llgetapairofnumbers,onebeingxmodulo7,theotherxmodulo12.Wecandisplaythisina712tablewhereeachrowisanumbermodulo7,eachcolumnanumbermodulo12,andtheentryatrowiandcolumnjisthatnumberwhichisimodulo7andjmodulo12.It'seasytoconstructthetable.Start llingthediagonal.Afteryoureachthelastrow,gonexttothetoprow,andafteryoureachtherightcolumn,gonexttotheleftcolumn.
01234567891011
0
0491463287742756217035
1
3615015642978438572271
2
7237251166530794495823
3
24733835217663180451059
4
60257439453186732814611
5
12612675405541968338247
6
48136227764165520693483
Allthenumbersinthe rstrowarecongruentto0modulo7,sothey'redivisibleby7,butlookingatthem,theyseemtoberatherrandomlyarranged.Likewise,allthenumbersinthe rstcolumnaredivisibleby12.
3.2.FACTORINGZNBYTHECHINESEREMAINDERTHEOREM61Thepairoflinearcongruencesxi(mod7)andxj(mod12)canbeeasilysolvedforxinbylookinginrowiandcolumnj.Forexample,takethisChineseremainderproblem.Findanumbersuchthatwhenyoudivideitby7yougetaremainderof3,butwhenyoudivideitby12yougetaremainderof8.Theanswer,80,isrightinthetable.Theorem3.8(Chineseremaindertheorem).Supposethatn=kmwherekandmarerelativelyprime.ThenZn=ZkZm:Moregenerally,ifnistheproductk1krwherethefactorsarepairwiserelativelyprime,thenZn=Zk1Zkr=rYi=1Zki:Inparticular,iftheprimefactorizationofnisn=pe11perr.ThenthecyclicringZnfactorsastheproductofthecyclicringsZpeii,thatis,Zn=rYi=1Zpeii:Proof.Thethirdstatementisaspecialcaseofthesecond.Thesecondfollowsfromthe rstbyinductiononr.Thatleavesuswiththe rststatement.Inonedirection,Zn!ZkZm,thefunctiongivingtheisomorphismisfairlyobvious;it'sbuiltofthetwofunctionsZn!ZkandZn!Zmthatareeasytodescribe.ThereisanobviouscandidateforaringfunctionZn!Zk,namely[x]n7![x]kbywhichismeanttheequivalenceclassofxmodulonissenttotheequivalenceclassofxmodulok.First,wehavetocheckthatthisfunctioniswellde ned.Suppose[x]n=[y]n.Thenxy(modn),son (x�y).Butk n,thereforek (x�y).Hence,xy(modk),and[x]k=[y]k.Sothefunctioniswell-de ned.Youcanchecktherest,thatthisfunctionpreservestheringoperationsothatit'saringhomomorphism.PuttingtogetherthetworinghomomorphismsZn!ZkandZn!ZmwehavearinghomomorphismZn!ZkZm[x]n7!([x]k;[x]m)Inordertoshowthatthisisanisomorphism,allweneedtodoistoshowthatit'sabijection,andforthat,allweneedtodoistoshowthatit'saninjectionsincethesetsZnandZkZmhavethesamecardinality.Supposethat[x]nand[y]naresenttothesameelementinZkZm.Then[x]k=[y]kand[x]m=[y]m,thatis,k (x�y)andm (x�y).Sincetheybothdividex�y,sodoestheirleastcommonmultiple.Butthey'rerelativelyprime,sotheirLCMistheirproduct,n.Thusn (x�y),so[x]n=[y]n.Therefore,thisisaone-to-onefunction,henceaone-to-onecorrespondence.Thus,theringhomomorphismisanisomorphism.q.e.d.
62CHAPTER3.RINGSTheinverse.Well,sinceit'sabijection,itshouldn'tbetoohardto nditsinverseZkZm!Zn.Inotherwords,solveforx(modn)thepairofsimultaneouscongruencesxa(modk)xb(modm)It'stoomuchworktoconstructtheentirekmtableaswasdoneforthe712.There'sabetterway.Wecan ndasolutionwiththeextendedEuclideanalgorithm.Sincegcd(m;k)=1,therefore1isalinearcombinationofmandk,thatis,thereareintegerssandtsothatsm+tk=1.Multiplybyb�atoconcludes(b�a)m+t(b�a)k=b�a.Therefore,t(b�a)k+a=b�s(b�a)m.Letthatbex.Thenxa(modk)andxb(modm)asrequired.ProblemslikethisinindeterminateanalysisweresolvedinancientChinaandinancientIndia.TheearliestappearedinSunzisuanjing(MasterSun'sMathematicalManual)intheaboutthefourthcenturyC.E.inChina.In1247QinJiushaogaveageneralmethodforsolvinglinearcongruencesinhisShushujiuzhang(MathematicalTreatiseinNineSections).3.2.2Brahmagupta'ssolutionInIndiaintheseventhcenturyC.E.,Brahmaguptaalsogaveageneralalgorithmforsolv-ingtheselinearcongruencesinhisBrahmasphut.asiddhanta(CorrectAstronomicalSystemofBrahma).Ifmorethantwocongruencesweregiven,he rstreducedtheproblemtosolvingpairsofcongruencesaswedidabove.Hissolutionistheonedescribedabove.Asanexample, ndx(mod210)ifx11(mod45)x4(mod56)Here'showhediditinmodernnotation,explainedwiththenumericalexampleabove.We'relookingforavalueofxsothatx=45s+11=56t+4forsomeintegerssandt.Soweneedsandtsothat45s+7=56t.Thatreducesto45(s�t)+7=11t.Lets0=s�t.Tosolve45s0+7=11t,since45=411+1,reduceittos0+7=11(t�4s0).Lett0=t�4s0.Wecansolves0+7=11t0bysettings0=4andt0=1.Substitutingtheseinthede ningequations,we ndt=t0+4s0=17,ands=s0+t=21.Therefore,x=45s+11=956,theanswer.Ofcourse,Brahmaguptadidnotusevariables.Hisissolutionwasdescribedasafairlysimplealgorithmthatjustusedthefourarithmeticoperations.3.2.3QinJiushao'ssolutionThealgorithmthatQinJiushaodescribedwasfairlydi erentandapplieddirectlytomanylinearcongruencessolongasthemoduliwerepairwiserelativelyprime.Let'sillustrateitwiththesystemofthreecongruencesx45(mod121)x31(mod63)x30(mod100)
3.3.BOOLEANRINGS63Sincethemoduliarepairwiserelativelyprime,wecan ndauniquesolutiontothissystemmodulo762300,theproductofthemoduli.Step1.Foreachmodulus, ndareciprocaloftheproductoftheremainingmodulimodulothegivenmodulus.Forthe rstmodulus,121,thatmeansweneedthereciprocalof6300modulo121,thatis,weneedtosolve6300y1(mod121):That'sthesameas8y1(mod121).TheextendedEuclideanalgorithmgivesus1=(�15)8+1121,soy=�15isasolution.Forthesecondmodulus,63,weneedthereciprocalof12100modulo63.That'sthesameasthereciprocalof4modulo63,whichis16.Forthethirdmodulus,100,weneedthereciprocalof7623modulo100.That'sthesameasthereciprocalof23modulo100.TheChinesemathematicianscalled ndingareciprocalmodulonbytheterm\ ndingone".BytheextendedEuclideanalgorithm,(�13)23+38=1,so�13isthereciprocalof23modulo100.Step2.Togetxsumthreeproductsabc,oneforeachcongruence,whereaistheconstantinthecongruence,bistheproductoftheothermoduli,andcisthereciprocalfoundinthepreviousstep.Thatgivesus456300(�15)+311210016+307623(�13)=�283515+6001600�2972970=2745115andthenreducethisnumbermodulotheproduct762300ofallthreemoduli.Thatgivesa nalanswerofx458215(mod762300).Exercise34.Solvethefollowingsystemofsimultaneouslinearcongruences.YoucanuseeitherBrahmagupta'salgorithm,QinJiushao'salgorithm,orsomethingofyourowndevising.x4(mod33)x22(mod35)x41(mod53)Besuretoshowhowyouderivedthesolution.3.3BooleanringsRepresentingbyxtheclass\men,"andbyy\Asiatics,"letzrepresentthead-jective\white"tothecollectionofmenexpressedbythephrase\MenexceptAsiatics,"isthesameastosay\WhitemenexceptwhiteAsiatics."Hencewehavez(x�y)+zx�zy:Thisisalsoinaccordancewiththelawsofordinaryalgebra.GeorgeBoole,1854.AnInvestigationoftheLawsofThoughtonwhicharefoundedthemathematicaltheoriesoflogicandprobabilities.
64CHAPTER3.RINGSGeorgeBoole(1815-1864).Boolewantedtobringlogicintotherealmofmathematics,whichhedidbyalgebrizingit.We'llincorporatehisinvestigationsinourstudyofringtheory,butchangehisnotationslightly.Booledidnotallowasumoftwothingsunlesstheyweredisjoint,sox+xhadnomeaningforhim.We'lljusttake+tobeanexclusiveor(symmetricdi erence),sox+xwillbe0forus.3.3.1IntroductiontoBooleanringsWesawbeforethatpowerset}(S)ofasetSbecomesaringwhenwede neA+Btobethesymmetricdi erenceandABtobetheintersectionoftwosubsetsAandB.The0elementoftheringistheemptyset?,whilethe1elementisS.ThecomplementofasubsetAis1�A(whichequals1+A).We'llde newhataBooleanringisintermsofidempotents.De nition3.9.Anelementeofaringissaidtobeidempotentwhene2=e.Noticethat0and1arealwaysidempotentinanyring.Otherexamplesofidempotentelementsinringsareprojections.Thetransformationf:R2!R2whichprojectsapointintheplanetothex-axis,de nedbyf(x;y)=(x;0),isidempotentasisanyprojectionfromaspacetoasubspaceofitself.De nition3.10.ABooleanringisaringinwhicheveryelementisidempotent.Thering}(S)isevidentlyanexampleofaBooleanring.Twopropertiesthatfollowfromthede nitionare(1)thattheaBooleanringhaschar-acteristic2,(2)Booleanringsarecommutative.Theorem3.11.AnontrivialBooleanringhascharacteristic2.Proof.Since1+1isidempotent,(1+1)2=1+1.Therefore,1+1+1+1=1+1,andso1+1=0.q.e.d.Asinanyringofcharacteristic2,negationdoesnothing,�x=x,andsubtractionisthesameasaddition,x�y=x+y.Theorem3.12.Booleanringsarecommutative.Proof.LetxandybetwoelementsofaBooleanring.Sincex+yisidempotent,(x+y)2=x+y.Expandingthatequationusingthefactthatmultiplicationdistributesoveradditionineveryring,commutativeornot,itfollowsthatx2+xy+yx+y2=x+y.Butx2=xandy2=y,sothatlastequationsimpli estoxy+yx=0.Therefore,xy=�yx,and�yx=yx,soxy=yx.q.e.d.BooleanringsarethesamethingassomethingcalledBooleanalgebras,buttheapproachesaredi erent.ABooleanringisthoughtofasaspecialkindofring,whileaBooleanalgebraisaspecialkindofpartiallyorderedsetwhoseelementsaretruthvalues.BooleanalgebrasarereviewedintheappendixsectionA.3.3.Table3.1comparescommonnotationsinBooleanalgebras,settheory,andBooleanrings.Here,PandQarepropositionsorpredicates,AandBaresubsetsofaset
,andxandy
3.3.BOOLEANRINGS65
Boolean
Set
Boolean
algebras
theory
rings
T(true)


1
F(false)
?
0
P^Q(and)
A\B
xy
P_Q(inclusiveor)
A[B
x+y+xy
PQ(exclusiveor)
AB
x+y
:P(not)
Ac
1+x
P()Q
A=B
x=y
P)Q
AB
xy=x
T_Q()T

[B=

1+y+1y=1
F_Q()Q
?[B=B
0+y+0y=y
T^Q()Q

\B=B
1y=y
F^Q()F
?\B=?
0y=0
P^Q()Q^P
P\Q=Q\P
xy=yx
P_Q()Q_P
P[Q=Q[P
x+y+xy=y+x+yx
:(P^Q)():P_:Q
(A\B)c=Ac[Bc
1+xy=(1+x)+(1+y)+(1+x)(1+y)
:(P_Q)():P^:Q
(A[B)c=Ac\Bc
1+(x+y+xy)=(1+x)(1+y)
Table3.1:NotationsinBooleanalgebras,settheory,andBooleanrings.areelementsofaBooleanring.Thesearejustafewcorrespondences.Youcanaddmanymore.FormoreonBooleanalgebras,seesectionA.3.3intheappendix.FreeBooleanringsSomeBooleanrings,calledfreeBooleanrings,havespecialproperties.GivenasetSwhoseelementsarecallegenerators,thefreeBooleanringonSistheBooleanringB(S)whichcomesequippedwithafunction:S!B(S)thatsatis esthefollowinguniversalproperty:foreachBooleanringRandfunctionf:S!R,thereexistsauniqueringhomomorphism^f:B(S)!Rsuchthatf=^f.Examples3.13(FreeBooleanrings).IfS=istheemptyset,thenB(?)consistsofonlytwoelements,0and1,whichwhenidenti edwithtruthvaluesare?and�,respectively.IfS=fpgisasingletonset,thenB(fpg)hasfourelements,0,p,1�p,and1,whichwhenidenti edwithtruthvaluesare?,p,:p,and�.IfS=fp;qghastwoelements,thenB(fp;qg)has16elements.Theyaredisplayedin gure3.2astruthvalues.ItisisomorphictotheBooleanringwhichisthepowersetofasetoffourelementsin gureA.1.3.3.2FactoringBooleanringsSupposethatasetSispartitionedintosubsetsS1;S2;:::;Sn.ThatmeansSistheunionofallthesesubsets,andtheyarepairwisedisjoint.Thenthering}(S)isisomorphictoaproductoftherings}(Si).Thefunction}(S)=}(S1)}(S2)}(Sn)A7!(A\S1;A\S2;:::;A\Sn)
66CHAPTER3.RINGS

p_q
q!p
p!q
:p_:q
p
q
p$q
p$:q
:q
:p
p^q
p^:q
:p^q
:p^:q
?
Figure3.2:FreeBooleanringontwoelementsgivestheringhomomorphisminonedirection,andit'sabijectionsinceAisthedisjointunionofthetermsontheright.Infact,thisworksevenwhenSispartitionedintoarbitrarilymanysubsets.SinceSisthedisjointunionofitssingletonsS=[x2Sfxg,therefore}=Qx2S}(fxg).Inotherwords,}isapowerofthe2-elementring.FactoringworksinageneralBooleanringaswellasthoseoftheform}(S).LetRbeaBooleanring,andeanyidempotentinitotherthan0or1.Let
e=1�e,sothat1=e+
efromwhichitfollowsthatx=xe+x
eforallx2R.LetRe=fxejx2Rg,andletR
e=fxejx2Rg.YoucancheckthatbothReandR
eareBooleanrings,wherethemultiplicativeidentitiesareeand
e,respectively.Furthermore,R=ReR
ex7!(xe;x
e)3.3.3ApartialorderonaBooleanringIfwede nexytomeanxy=y,thenourBooleanringwillhaveapartialordering.Recallthatapartialorderingonasetisare exive,antisymmetric,andtransitiverelation.1.Re exive:xx,sincex2=x.2.Antisymmetric:xyandyximplyx=y,sincexy=xandyx=yimplyx=y.3.Transitive:xyandyzimplyxz,sincexy=xandyz=yimplyxz=x.(Proof:xz=(xy)z=x(yz)=xy=x.)Inthispartialorder,theproductxyisthemeetx^yofxandy,thatis,it'sthelargestelementzsuchthatzxandzy.Likewise,x+y+xyisthejoinx_yofxandy,thatis,it'sthesmallestelementzsuchthatxzandyz.Apartialorderthathasmeetsand
3.4.THEFIELDOFRATIONALNUMBERS,FIELDSOFFRACTIONS67joinsofpairsofelementsiscalledalattice.Notalllatticeshavethedistributivepropertieswheremeetandjoindistributeovereachother(x_y)^z=(x^z)_(y^z)and(x^y)_z=(x_z)^(y_z)butBooleanringsdo,soBooleanringsareexamplesofdistributivelatticesAminimalelementofaBooleanringisanonzeroelementsuchthatthereisnosmallernonzeroelement.Everyelementofa niteBooleanringisasumoftheminimalelementslessthanorequaltoit.Sincetherearenoelementslessthan0,0hastobetreatedastheemptysum.Theorem3.14.IfRisa niteBooleanring,thenR=}(S)whereS=fx2Rjxisminimalg:Exercise35.Provetheprecedingtheorem.Hint:seethesectionabove3.3.2onfactoringBooleanrings.Inductionmayhelp.3.4The eldofrationalnumbers, eldsoffractionsSupposethatwealreadyhaveconstructedtheintegraldomainofintegersZ,butforsomereasondonothavethe eldofrationalnumbersQ.ThenwecouldconstructQfromZsinceeachrationalnumbercanbenamedbyapairofintegers.We'lldothat.ThestepsweuseonlydependonZbeinganintegraldomain.Thatmeansthattheconstructionweusecanalsobeusedtocreatea eldoffractionsFfromanyintegraldomainR.Inthefollowing,thinkoftheintegraldomainRasbingZandthe eldFasbeingQ.Anequivalencerelationonpairsofintegers.Firstofall,arationalnumberm
ncanbenamedbyapairofintegers(m;n)wherethesecondintegerndoesnotequal0.Butdi erentpairs(m;n)and(k;l)cannamethesameintegerm
n=k
lifml=nk.Thatsuggestsifwewanttocreaterationalnumbersfromintegers,we'llneedanequivalencerelationonpairsofelementsoftheintegraldomainR.We'llstartwiththesetRR6=0oforderedpairs(m;n)ofelementsofanintegraldomainRwithn6=0.De nearelationonthissetby(m;n)(k;l)i ml=nk:Youcaneasilyverifythatthisrelationisanequivalencerelation.Re exivity:(m;n)(m;n).That'svalidsincemn=mn.Symmetry:(m;n)(k;l)implies(k;l)(m;n).That'svalidsinceml=nkimplieskn=lm.Transitivity:(m;n)(k;l)and(k;l)(s;t)imply(m;n)(s;t).Weneedtoshowthatml=nkandkt=lsimplymt=ns.Multiplythe rstequationbytandthesecondbyn.Thenmlt=nktandnkt=nls,somlt=nls.ButRisanintegraldomain,socancellationisvalidwhenl6=0,somt=ns.
68CHAPTER3.RINGSThus,isanequivalencerelationonRR6=0.LetFbethequotientsetF,anddenoteanelement[(m;n)]ofFbym
n.Sofar,we'vegottheunderlyingsetforourproposed eldF,butwedon'thavetheoperationsfora eld.Beforewede nethem(andshowthey'rewell-de ned),let'sverifythatthefunctionR!RR6=0!FwhichsendsanelementmofR rstto(m;1)thentom
1isaone-to-onefunction.Supposethatm
1=n
1.Thatmeansm1=1n,som=n.ThuswemayinterpretR!FasmakingRasubsetofFbyidentifyingmwithm
1.AdditiononF.We'dliketode nethesumm
n+k
lasml+nk
nl;butasourfractionsarereallyequivalenceclasses,weneedtoshowthat'swellde ned.Indetail,weneedtoshowthatm
n=m0
n0andk
l=k0
l0implyml+nk
nl=m0l0+n0k0
n0l0:Thatreducestoshowingthatmn0=nm0andkl0=lk0imply(ml+nk)n0l0=nl(m0l0+n0k0):Butthatcanbeshownbymultiplyingthe rstequationbyll0,thesecondbynn0andaddingthetworesultingequations.Thus,thisadditiononFiswell-de ned.MultiplicationonF.We'dliketode netheproductm
nk
lasmk
nl;Weneedtoshowthat'swellde ned.You'll ndthattheproofiseasierthantheoneaboveforaddition.Next,weneedtoverifythatwiththesede nitionsFsatis esthe eldaxioms.Aproofisneededforeach eldaxiom.Commutativityofaddition.m
n+k
l=k
l+m
n.That'seasilyveri edsinceml+nk
nl=kn+lm
ln.(Thatdependsoncommutativityofadditionandmulti-plicationinR.)Commutativityofmultiplication.m
nk
l=k
lm
n.That'seasilyveri edsincemk
nl=km
ln.Associativityofaddition.Youcaneasilyshowit,butit'sabigmess.Associativityofmultiplication.Prettyeasy.Additiveidentity.0
1+k
l=k
l.Easy.Multiplicativeidentity1
1k
l=k
l.Easy.
3.5.CATEGORIESANDTHECATEGORYOFRINGS69Negation.m
n+�m
n=0
1.Prettyeasy.Reciprocation.Form
n6=0
1,m
nn
m=1
1.Prettyeasy.Multiplicationdistributesoveraddition.Easybutmessy.06=1.Weneedtoshowthat0
16=1
1inF.Butthat'sthesameas016=11intheintegraldomainR,andpartofthede nitionofintegraldomainrequires06=1.Thus,Fisa eld.Exercise36.Selectfouroftheaxiomsabove,andprovethem.Asalways,yourproofsshouldincludejusti cations.We'llsummarizethisresultasatheorem.Theorem3.15.AnintegraldomainRisasubringofa eldF,calledthe eldoffractions,whereeachelementofFcanberepresentedasm
nwheremandnareelementsofRandn6=0.Thisgivesusanotherproofthatthecharacteristicofanintegraldomainiseither0oraprimenumbersinceithasthesamecharacteristicofits eldoffractions.Examples3.16.TheprimaryexampleofthisistheconstructionofQfromZ.Foranotherexample,taketheGaussianintegersZ[i]fortheintegraldomainR.Thenthe eldoffractionsFisthe eldQ(i).TheelementsofQ(i)areoftheformx+yiwherexandyarerationalnumbers.Yetforanotherexample,takethepolynomialringF[x]withcoecientsina eldF.It'sanintegraldomain,andits eldoffractionsistherationalfunction eldF(x)withcoecientsinF.Stoppingshortofinvertingallelements.Sometimesyoumaywanttocreatereciprocalsforsomeelementsofanintegraldomain,butnotforallelements.Thiscanbedonebyaminormodi cationoftheaboveprocess.Suppose,forinstance,thatyouwanttoextendZtoincludethereciprocalof2butnotofanyotherprimenumber.ThatwouldleadtothedomainofdyadicrationalsZ[1
2]wherethedenominatorsarepowersof2.Ontheotherhand,ifyouwanttoextendZtoincludethereciprocalsofalltheprimesexcept2,justincludeodddenominators.ThisiscalledlocalizingZat2.Theseotherconstructionsareusefulinmathematics,butwewon'tusethemourselves.3.5CategoriesandthecategoryofringsCategoriesarehigher-orderalgebraicstructures.We'lllookatthecategoryofringsinwhichtheobjectsofthecategoryarealltherings.Thepurposeofacategoryistostudytheinterrelationsofitsobjects,andtodothatthecategoryincludesmorphismsbetweentheobjects.Inthecaseofthecategoryofrings,themorphismsaretheringhomomorphisms.We'llstartwiththeformalde nitionofcategories.We'llusethecategoryofringsbothtoillustratecategoricalconceptsandtostudyrings.CategorytheorywasdevelopedbyEilenbergandMacLaneinthe1940s.
70CHAPTER3.RINGS3.5.1Theformalde nitionofcategoriesUnlike elds,rings,andgroups,wewon'trequirethatcategoriesbuildonsets.Inacategorythecollectionofallitsobjectswon'tbeasetbecausethecollectionislargerthananyset.That'snotaproblemsincetheoriesdon'thavetobebuiltonsettheory.Indeed,settheoryitselfisnotbuiltonsettheory.De nition3.17.AcategoryCconsistsof1.objectsoftendenotedwithuppercaseletters,and2.morphisms(alsocalledmapsorarrows)oftendenotedwithlowercaseletters.3.Eachmorphismfhasadomainwhichisanobjectandacodomainwhichisalsoanobject.IfthedomainoffisAandthecodomainisB,thenwewritef:A!BorAf!B.ThecollectionofallmorphismsfromAtoBisdenotedHom(A;B).4.ForeachobjectAthereisamorphism1A:A!AcalledtheidentitymorphismonA.(WhenAcanbedeterminedbycontext,itsdenotedsimply1.)5.GiventwomorphismsAf!BandBg!CwherethecodomainofoneisthesameasthedomainoftheotherthereisanothermorphismAgf�!Ccalledthecompositionofthetwomorphisms.ThiscompositionisillustratedbythecommutativediagramABC
-f
[email protected]@@@@Rgf(Sometimesgfisdenotedfg.)Adiagramofobjectsandmorphismsinacategoryissaidtocommute,orbeacommu-tativediagramifanytwopathsofmorphisms(inthedirectionofthearrows)betweenanytwoobjectsyieldequalcompositions.6.ForallAf!B,f1A=fand1Bf=f.ThesecompositionsareillustratedbythetwocommutativediagramsAAB
?1A
[email protected]@@@@Rf1AABB
-f
[email protected]@@@@R1Bf7.ForallAf!B,Bg!C,andCh!D,(hg)f=h(gf).Inthediagrambelow,ifthetwotrianglesinthediagrameachcommute,thentheparallelogramcommutes.
3.5.CATEGORIESANDTHECATEGORYOFRINGS71ABCD
-f
[email protected]@@@@Rgf
[email protected]@@@@RhgIsomorphismsinacategoryC.Althoughonlymorphismsarede nedinacategory,it'seasytodeterminewhichonesareisomorphisms.Amorphismf:A!Bisanisomorphismifthereexistsanothermorphismg:B!A,calleditsinverse,suchthatfg=1Aandgf=1B.Indeed,themainreasonidentitymorphismsareincludedinthede nitionofcategoriesistobeabletode neisomorphisms.Examples3.18(Thecategoriesofsets,groups,rings,and elds).Althoughwe'remoreinterestedinthecategoryofringsrightnow,thecategorySofsetsisalsorelevant.AnobjectinSisaset,andamorphisminSisafunction.Thedomainandcodomainofamorphismarejustthedomainandcodomainofthefunction,andcompositioniscomposition.Isomorphismsarebijections.TheobjectsofthecategoryGofgroupsaregroups,andthemorphismsofGaregrouphomomorphisms.TheobjectsofthecategoryRofringsarerings,andthemorphismsofRareringhomo-morphisms.Theobjectsofthecategoryof eldsare elds,anditsmorphismsare eldhomomorphisms,whicharejustringhomomorphisms.Thecategoryof eldsisasubcategoryofthecategoryofrings.Ineachoftheseotherthreecategories|groups,rings, elds|isomorphismsinthecategoryarewhatwehavecalledisomorphisms.3.5.2ThecategoryRofringsRecallthataringhomomorphismf:A!Bbetweenringsisafunctionthatpreservesaddition,multiplication,and1.Thecategoryofringshasasitsobjectsallringsandasitsmorphismsallringhomomorphisms.Theidentitymorphism1Aonaringistheidentityhomomorphism,andcompositionistheusualcompositionofhomomorphisms.Thus,wehaveacategoryRofrings.Ifthiswerealltherewastocategorytheory,therewouldn'tbemuchpointtoit.Butbyemphasizingthemorphismsanddeemphasizingelementsinringswecanidentifywhat'simportantaboutcertainringsandcertainringconstructions.We'lllookatproductsofrings rsttoseewhatcharacterizesthem.We'llalsolookatacoupleofspecialrings,namelyZandZ[x],forcharacterizingpropertiesofthem.We'llalsoseehowtocharacterizemonomor-phisms.Theuniversalpropertyofproducts.RecallthattheproductR1R2oftworingsisconsistsoforderedpairs(x1;x2)withx12R1andx22R2,andtheringoperationsforR1R2areperformedcoordinatewise.Furthermore,wehavetheprojectionringhomomorphismsR1R21!R1andR1R22!R2whichpickoutthetwocoordinates.
72CHAPTER3.RINGSThisproducthastheuniversalpropertythatforeachringSandringhomomorphismsSf1!R1andSf2!R1,thereexistsauniqueringhomomorphismS!R1R2,whichwewilldenote(f1;f2),suchthatf1=1(f1;f2)andf2=2(f1;f2),asillustratedbythediagrambelow.R1R2R1R2S����[email protected]@@@R2
-(f1;f2)1f1PPPPPPPPPPPPPqf2Infact,theproductischaracterizedbythisuniversalpropertyinthesensethatifanotherringRhasthisuniversalproperty,thenthereisaringisomorphismR!R1R2.Inmoredetail,ifRp1!R1andRp2!R2havethisproductproperty(namely,thatforeachringSandringhomomorphismsSf1!R1andSf2!R1,thereexistsauniqueringhomomorphismSf!Rsuchthatf1=p1fandf2=p2f),thenthereexistsauniqueringisomorphismRh!R1R2suchthat1h=p1and2h=p2.Althoughthischaracterizationofproductswasdescribedforthecategoryofrings,itisthede nitionfortheproductoftwoobjectsinanycategory.AproductR1R2ischaracterizedbythepropertythatamorphismtotheproductcorrespondtoapairofmorphismstothefactors.TheproductoftwosetsinthecategorySofsetshasthissameuniversalpropertyasdoestheproductoftwogroupsinthecategoryGofgroups.Thereare,however,noproductsinthecategoryof elds.Zistheinitialobjectinthecategoryofrings.Wecanalsousecategorytheorytopindownwhat'ssospecialabouttheringZ.IthasthepropertythatgivenanyringR,thereisauniqueringhomomorphismZf!R,andit'sde nedbyf(n)=n.Anobjectinacategorywiththatpropertyiscalledtheinitialobjectinthecategory.Anytwoinitialobjectsinacategoryareisomorphic.Thetrivialringisthe nalobjectinthecategoryofrings.Dualtotheinitialobjectisa nalobject,whichinthecategoryofringsisthetrivialordegeneratering0.Thisringhasonlyoneelementinwhich0=1.Infact,it'stheonlyringinwhich0=1(since0=0x=1x=x).The nalobjectinacategoryhasthepropertythatthere'sauniquemorphismtoitfromeachobjectinthecategory.Thetrivialringhasthatpropertyinthecategoryofrings.Exercise37.Determinetheinitialobjectandthe nalobjectinthecategorySofsets.
3.5.CATEGORIESANDTHECATEGORYOFRINGS73TheuniversalpropertyofthepolynomialringZ[x].GivenanyringRandanyelementa2R,thereisauniqueringhomomorphismZ[x]!Rthatmapsxtoa.Thishomomorphismisjustevaluationata,andapolynomialf(x)ismappedtotheelementf(a)inR.3.5.3MonomorphismsandepimorphismsinacategoryAlthoughwede nedamonomorphismf:A!Basaone-to-onehomomorphism,wecancharacterizemonomorphismsentirelyintermsofcategorytheory.De nition3.19.Amorphismf:A!Bismonic,oramonomorphism,whenifgandhareanytwomorphismsfromanyanotherobjectCtoAsuchthatfg=fh,theng=h.ABC
-f
-g
-hAmonomorphisminthecategorySofsetsisaninjection.Thisde nitionagreeswithourpreviousde nitionforringmonomorphismintermsofele-ments,andonewaytoseethecorrespondenceistoletCbeZ[x].Likewise,amonomorphisminthecategoryGofgroupsagreeswithourpreviousde nitionofgroupmonomorphism.Epimorphisms.Theconceptofepimorphismisdualtothatofmonomorphism.Ifwechangethedirectionofallthearrowsinthede nitionofmonomorphism,we'llgetthede -nitionofepimorphism.De nition3.20.Amorphismf:A!Bisepic,oranepimorphism,whenifgandhareanytwomorphismsfromBtoanyanotherobjectCsuchthatgf=hf,theng=h.BCA
-f
-g
-hInthecategorySofsets,anepimorphismisasurjection.Likewise,itturnsoutthatinthecategoryGofgroups,anepimorphismisasurjection.InthecategoryRofrings,it'seasyenoughtoshowthatiffisasurjectiveringhomomor-phism,thenfisanepimorphism,butthereareotherepimorphismsthataren'tsurjections.Example3.21.Considertheinclusionfunction:Z!Q.We'llshowthatit'sanepimor-phism.LetgandhbeanytwomorphismsfromQtoanyanotherringCsuchthatg=h.Theng(n)=h(n)foranyintegern.Letm
nbearationalnumberwithn6=0.Theng(m)=h(m)andg(n)=h(n).So,gm
ng(n)=gm
nn=g(m)=h(m)=hm
nn=hm
nh(n)=hm
ng(n):Sincen6=0,thereforeg(n)6=0aswell.Canceltheg(n)attheendsofthecontinuedequationtoconcludegm
n=hm
n.Thus,g=h.Therefore,theringhomomorphism:Z!QisanepimorphisminR,thecategoryofrings.Itisalsoamonomorphism.Butitisnotanisomorphism.
74CHAPTER3.RINGSInmanycategories,ifamorphismisbothmonicandepic,thenit'salsoanisomorphism.That'strueinthecategorySofsetsandinthecategoryGofgroups,butnotinthecategoryRofrings.ThisexampleshowsthatRisasomewhatunusualcategory.3.6Kernels,ideals,andquotientringsThesethreeconceptsarecloselyrelated.Foraringhomomorphismf:R!S,theinverseimageof0isasubsetofRcalledthekerneloffanddenotedKerf.Itcan'tbejustanysubset,aswe'llsee,sinceit'sclosedunderadditionandmultiplicationbyelementsofR.Asubsetwiththosepropertieswe'llcallanidealofR.EveryidealIofRisthekernelofsomeringhomomorphismf:R!S.We'lluseanidealIofaringRtode neaquotientringR=Iandaprojection :R!R=I.TheseprojectionswillbegeneralizationsoftheprojectionsZ!Znthatwestudiedearlier.3.6.1KernelsofringhomomorphismsDe nition3.22.Letf:R!Sbearinghomomorphism.ThoseelementsofRthataresentto0inSformthekerneloff.Kerf=f�1(0)=fx2Rjf(x)=0g:We'lllookatpropertiesofthiskernelandseewhatittellsusaboutthefunctionf.Example3.23.It'sagoodideatohaveinmindanexampleortwowheneveranewconceptisde ned.Thede nitionofthekernelofaringhomomorphismisgivenabove,andagoodexampleforitistheringhomomorphismf:Z!Znwherenisa xedinteger.That'sanespeciallygoodexamplewecanuseitthroughoutthisdiscussionofrings,ideals,andquotientrings.Forthatf:Z!Zn,anelementx2ZisinKerfifitissentto[0]n,the0elementintheringZn,thatis,if[x]n=[0]n,or,moresimply,ifn x.Therefore,thekerneloffconsistsofthemultiplesofn.AstandardnotationforthemultiplesofanintegernisnZ.Thus,forthisfunctionf,Kerf=nZ.Kernelsaren'tjustanysubsetsofR;theyhavesomespecialproperties.Wehave,ofcourse,02Kerf,sincef(0)=0,Also,ifxandyarebothinKerf,thenf(x+y)=f(x)+f(y)=0+0=0,sotheirsumx+yisalsoinKerf.Furthermore,ifx2KerfandyisanyelementofR,thenf(xy)=f(x)f(y)=0f(y)=0,soxy2Kerf,andlikewiseyx2Kerf.Besidestellinguswhatelementsaresentto0byf,thekerneloffalsotellsuswhentwoelementsaresenttothesameelement.Sincef(x)=f(y)ifandonlyiff(x�y)=0,therefore,fwillsendxandytothesameelementofSifandonlyifx�y2Kerf.3.6.2IdealsofaringThepropertiesofkernelsofhomomorphismsthatwejustfoundwe'llusetode neidealsofrings.Historically,idealshadadi erentpurpose,butwe'llgettothatpurposelater.Theword\ideal"isshortforidealnumberoridealelement.
3.6.KERNELS,IDEALS,ANDQUOTIENTRINGS75De nition3.24.AnidealIofaringRisasubsetthat(1)includes0,(2)isclosedunderaddition,and(3)isclosedundermultiplicationbyelementsofR.Wecansummarizetheserequirementssymbolicallyby02I,I+II,RII,andIRI.Bothofthelasttworequirements,RIIandIRI,areneededwhenRisanon-commutativering.Mostofthetimewe'llbedealingwithcommutativeringssoonewilldo.Notethatf0gisalwaysanidealinaringR.It'scalledthetrivialideal.We'llusuallyjustdenoteit0.Also,theentireringRisanideal,butnotaproperideal.AproperidealisanyidealI6=R.Theorem3.25.Theintersectionofidealsisanideal.Proof.Here'stheprooffortwoidealsI1andI2ofaringR.Thisproofcanbegeneralizedtoanynumber,includinganin nitenumber,ofideals.WeneedtoshowthatI1\I2(1)includes0,(2)isclosedunderaddition,and(3)isclosedundermultiplicationbyelementsofR.First,since02I1and02I2,therefore02I1\I2.Second,giventwoelementsx;y2I1\I2,toshowx+y2I1\I2.Sincex;y2I1\I2,thereforex;y2I1andx;y2I2.Thereforex+y2I1andx+y2I2,andsox+y2I1\I2.Third,givenx2I1\I2andy2R,toshowxy2I1\I2.Sincex2I1\I2,thereforex2I1andx2I2.Therefore,xy2I1andxy2I2,andsoxy2I1\I2.q.e.d.Principalidealsandidealsgeneratedbyaset.Thesimplestexamplesofidealsarewhatarecalledprincipalideals.LetabeanelementofacommutativeringR.Thesetofallmultiplesofa,(a)=fxajx2Rg;isanidealofR,asyoucaneasilycheck.Theseidealsarecalledprincipalidealsbecausetheyaregeneratedbyoneelement.AnalternatenotationfortheprincipalidealgeneratedbytheelementaisRaoraR.Notethat(0),theidealgeneratedby0,isjustthe0ideal,while(1),theidealgeneratedby1,isallofR.Sometimesittakesmorethanoneelementtogenerateanideal.LetAbeasubsetofacommutativeringR.ThesmallestidealthatcontainsAiscalledtheidealgeneratedbyA.ItmustcontainalllinearcombinationsofelementsofAsinceanidealisclosedunderadditionandclosedundermultiplicationbyelementsofR,butthat'senough.Usually,we'reonlyinterestedingeneratinganidealfroma nitenumberofelementsA=fa1;a2;:::;akg.ThentheidealgeneratedbyAis(a1;a2;:::;ak)=fx1a1++xkakjeachxi2Rg:Anexampleofanidealgeneratedbytwoelementsbutnotprincipal(notbyoneelement)is(5;x2)inZ[k],thepolynomialringwithintegralcoecients.Exercise38.Asyouknow,ifn2Z,thennZ,alsowritten(n),isanidealoftheringZ.ConsiderthetwoidealsI=6ZandJ=10ZoftheZ.(a).DeterminetheirintersectionI\JasaprincipalidealofZ.(b).ProvethattheunionI[JisnotanidealofZ.
76CHAPTER3.RINGS3.6.3Quotientrings,R=IAsmentionedabovethekernelofaringhomomorphismftellsuswhentwoelementsaresenttothesameelement:f(x)=f(y)ifandonlyifx�y2Kerf.WecanuseKerftoconstructa\quotientring"R=KerfbyidentifyingtwoelementsxandyinRiftheirdi erenceliesinKerf.Infact,wecandothisnotjustforkernelsofhomomorphisms,butforanyidealI.Thatis,wecanuseanidealIofRtodeterminewhentwoelementsxandyaretobeidenti ed,xy,andwe'llendupwitharingR=I.Theidenti cationiscalledacongruence.ThisconceptofcongruencegeneralizescongruencemodulononZ.De nition3.26.AcongruenceonaringRisanequivalencerelationsuchthatforallx;x0;y;y02R,xx0andyy0implyx+yx0+y0andxyx0y0:Sincewe'redealingwithringswith1,we'llusuallyinsistthat061.Theequivalenceclassesforacongruencearecalledcongruenceclasses.Theorem3.27.IfisacongruenceonaringR,thenthequotientsetR=,thatis,thesetofcongruenceclasses,isaringwhereadditionisde nedby[x]+[y]=[x+y]andmultiplicationby[x][y]=[xy].Proof.Firstweneedtoshowthattheproposedde nitionsareactuallywellde ned.Thatis,ifadi erentrepresentativex0ischosenfromthecongruenceclass[x]andy0from[y],thenthesameclasses[x0+y0]and[x0y0]result.Thatis[x]=[x0]and[y]=[y0]imply[x+y]=[x0+y0]and[xy=xy0]:That'sthesameastherequirementsmetinthede nitionofcongruence(whichexplainswhytheyareinthede nition).Also,eachoftheaxiomsforaringneedtobeveri ed,butthey'reallautomatic.Here'scommutativityofaddition,forexample.[x]+[y]=[x+y]=[y+x]=[y]+[x]:Wecouldsaythatthequotientringinheritsthepropertiesfromthering.q.e.d.Inthenexttheoremwe'llseethatanidealIdeterminesacongruence.We'llwritethecongruencexy(modI)ratherthanjustxywhenwewanttoemphasizetheroleofI.Thecongruenceclassesmaybewritten[x]or[x]I,orx+I.Thelastnotationisagoodonesince[x]=fx+yjy2Ig.Theorem3.28(Congruencemoduloanideal).LetIbeanidealofaringR.Acongruence,calledcongruencemoduloI,isde nedbyxy(modI)ifandonlyifx�y2I:Thequotientring,R=,isdenotedR=I.
3.6.KERNELS,IDEALS,ANDQUOTIENTRINGS77Proof.First,weneedtoshowthatit'sanequivalencerelation.Re exivity.xx(modI).That'sokaysincex�x=02I.Symmetry.xy(modI)impliesyx(modI).That'sokaybecauseifx�y2I,theny�x=�(x�y)2I.Transitivity.xy(modI)andyz(modI)implyxz(modI).That'sokay,too.Ifx�y2Iandy�z2I,thensoistheirsumx�z2I.Thus,it'sanequivalencerelation.Nexttoshowthatxx0(modI)andyy0(modI)implyx+yx0+y0(modI)andxyx0y0(modI):Thatrequirementreducestothestatementx�x02Iandy�y02Iimply(x+y)�(x0+y0)2Iand(xy�x0y0)2I;which,youcancheck,followfromthede nitionofideal.q.e.d.Exercise39.Provethelaststatementabove:ifx�x02Iandy�y02I,then(x+y)�(x0+y0)2Iand(xy�x0y0)2I.Example3.29(Cyclicrings).Aswesawabove,I=nZisanidealofZ.Thecongruencede nedhereisthesameonewehadbefore.Thus,xy(modI)meansxy(modn).ThequotientringisZ=nZ,whichwehavestudiedbeforeanddenotedZnforshort.Comment3.30.TheringstructureonthequotientR=Iwasde nedfromtheringstructureonR,sotheprojection :R!R=Iisaringhomomorphism.ThisringR=IiscalledaquotientringofR.(Itisalsosometimescalledafactorring,butthattermshouldberestrictedtothecasewhenRfactorsasaproductofrings,oneofwhichisR=I.AnexampleofthatistheChineseremaindertheorem.)Examples3.31(Quadratic eldextensions.).We'velookedatQ(p
2),C=R(i),andotherquadratic eldextensions.Wecaninterpretthemasquotientrings.Let'stakeQ(p
2) rst.ConsidertheringR=Q[x]ofpolynomialswithrationalcoe-cients.AnidealinRistheprincipalidealI=(x2�2)generatedbythepolynomialx2�2.InthequotientringR=I=Q[x]=(x2�2),wehavex2�20(modI),thatis,x22(modI),soinR=I,we ndthat2doeshaveasquareroot,namelyx.SinceinR=Ieverypolynomialanxn++a1x+a0iscongruenttoapolynomialofdegree1(becausex22),butnotwolinearpolynomialsarecongruentmodI(becausea1x+a0b1x+b0(modI)implies(a1�b1)x+(a0�b0)2Isoa1=b1anda0=b0),thereforeeveryelementinR=Iisuniquelyrepresentedasalinearpolynomiala1x+a0.Ifwedenotexbythesymbolp
2,thenwe ndQ[x]=(x2�2)isthesame eldasQ(p
2)thatwedescribedbefore.Likewise,R[x]=(x2+1)isC.We'll ndthisconstructionofnewringsasquotientringsisveryuseful,especiallywhenwetakequotientsringsofpolynomialringslikewedidhere.
78CHAPTER3.RINGSTheimageofaringhomomorphismisisomorphictotheringmoduloitskernel.Letf:R!Sbearinghomomorphism.Theimageoff,denotedf(R),isthesetf(R)=ff(x)2Sjx2Rg:ItisasubringofS,asyoucaneasilyverify.Youcanalsoshowthefollowingisomorphismtheorem,calledthe rstisomorphismforrings.Theorem3.32.Iff:R!SisaringhomomorphismthenthequotientringR=Kerfisisomorphictotheimageringf(R),theisomorphismbeinggivenbyR=Kerf!f(R)x+Kerf7!f(x)Exercise40.Provetheprecedingtheorem.(a).Firstshowthattheassignmentx+Kerftof(x)iswellde ned.Thatmeansthatifx+Kerf=x0+Kerf,thenf(x)=f(x0).Callthatfunction(x).(b).Showthatassignmentisaringhomomorphism.Show(1)(1)=1,(2)(x+y)=(x)+(y),and(3)(xy)=(x)(y).Thisgivesustwowaystolookattheimage,eitherasaquotientringofthedomainRorasasubringofthecodomainS.Furthermore,wecannowtreataringhomomorphismf:R!Sasacompositionofthreeringhomomorphisms.R!R=Kerf=f(R)!SThe rstistheprojectionfromRontoitsquotientringR=Kerf,thesecondistheisomor-phismR=Kerf=f(R),andthethirdistheinclusionoftheimagef(R)asasubringofS.3.6.4PrimeandmaximalidealsSometimesitoccursthatR=Iisnotjustaring,buteitheranintegraldomainorevena eld.ThoseresultsoccurwhentheidealIisaprimeidealoramaximalideal,respectively,aswe'llde nenow.De nition3.33.AnidealIinacommutativeringRissaidtobeaprimeidealifR=Iisanintegraldomain.Equivalently,Iisaprimeidealif(1)I6=R,and(2)8x;y2R,ifxy2I,theneitherx2Iory2I.AnidealIissaidtobemaximalit'saproperideal,butitisnotcontainedinanylargerproperideal.Exercise41.ProvethatR=IisanintegraldomainifandonlyifR=Isatis esbothconditions(1)I6=R,and(2)8x;y2R,ifxy2I,theneitherx2Iory2I.Example3.34.TheidealsofZthatareprimearethoseoftheformpZwherepisaprimenumber,andthe0ideal.Infact,pZaremaximalideals,but0isnotmaximal.Ina eldFthereisonlyoneproperideal,namely0.Inanintegraldomain,the0idealisaprimeideal,andconversely,if0isanidealinacommutativering,thentheringisanintegraldomain.
3.7.KRULL'STHEOREM79Theorem3.35.Everymaximalidealisprime.Proof.LetIbeamaximalidealofacommutativeringR,andletxy2I.Supposex=2I.ThenxR+I=fxu+vju2R;v2IgisanidealcontainingI.SinceIisanmaximalideal,thereforexR+IisnotaproperidealbutallofR.Therefore1=xu+vforsomeu2R,v2I.Hencey=yxu+yv2Iu+I=I.Thus,Isatis estheconditionstobeaprimeideal.q.e.d.Wewon'tshowitrightnow,butwe'llprovelaterKrull'stheoremwhichsaysthateveryidealiscontainedinamaximalideal.We'llneedtodiscusstheaxiomofchoiceandZorn'slemmabeforewecanproveit.Theorem3.36(Maximalidealtheorem).LetIbeanidealofacommutativeringR.ThenIisamaximalidealifandonlyifR=Iisa eld.Proof.We'llusethenotation[x]forx+Itostressthatwe'rethinkingofitasanelementofR=I.SupposethatIisamaximalideal,andlet[x]beanynonzeroelementofR=I,thatisx=2I.Asinthelastproof,xR+I=R.Therefore1=xu+vforsomeu2R,v2I.Then,inR=Iwehave[1]=[x][u]+[v]=[x][u]+[0]=[x][u].Therefore[x]hasareciprocal,andR=Iisa eld.NowsupposethatR=Iisa eld.Letx=2I.We'llshowthatxR+I=RwhichwillshowthatIisamaximalideal.InR=I,[x]6=[0],so[x]hasaninverse[y],[x][y]=[1],so1�xy2I,so12xR+I,henceR=xR+I.q.e.d.3.7Krull'stheoremWe'dliketoproveKrull'stheoremthateveryidealinacommutativeringiscontainedinamaximalideal,butinordertodothatingeneralwe'llneedsomethingcalledZorn'slemma.It'sastatementthat'slogicallyequivalenttothebetterknownaxiomofchoice.SeesectionA.4intheappendixforareviewoftheaxiomofchoiceandZorn'slemma.Theorem3.37(Krull).LetIbeaproperidealofacommutativeringR.ThenthereisamaximalidealJsuchthatIJ.Proof.ConsiderthecollectionMofproperidealsofRthatcontainI.NotethatMisnonemptysinceI2M.We'llshowthateverychainCinMhasanupperboundinM.LetB=[A2CA.CertainlyBisanupperboundforCsinceBisjusttheunionofelementsofC.WestillhavetoshowBisanideal,whichrequiresRBBandB+BB.Forthe rst,RB=R [A2CA!=[A2CRA=[A2CA=B.Nowletx;y2B.Thenx2A1forsomeA12Candy2A2forsomeA22C.ButCisachain,soeitherA1A2orA2A1.Inthe rstcase,x;y2A2,sox+y2A2B,andinthesecondx;y2A1,sox+y2A1B.Thus,B+BB.
80CHAPTER3.RINGSNowwecanapplyZorn'slemma.ItimpliesMhasamaximalelementJ.Clearly,IJ,andJisaproperidealofR,buttherearenolargerproperidealsofRthatcontainJ,soJisamaximalideal.q.e.d.NotehowwehavenotactuallyfoundJ.Theremaybemanydi erentmaximalidealsthatcontainI,andonewasselectedbyachoicefunction,butwedon'tevenknowwhatthechoicefunctionissowecan'tevendetermineJinprinciple.It'sactuallythecasethatKrull'stheoremislogicallyequivalenttotheAxiomofChoice.Thatis,ifKrull'stheoremistakenasanaxiom,thentheAxiomofChoicecanbeprovedfromit.TherearemanyotherapplicationsofZorn'slemma.Forinstance,youcanprovethateveryvectorspacehasabasis,evenwhenthevectorspaceisin nitedimensional.3.8Uniquefactorizationdomains,principalidealdo-mains,andEuclideandomainsNoteveryintegraldomainisasniceastheringofintegers.Theringofintegershasthreeniceproperties.Oneisuniquefactorization|everyintegerisuniquelyaproductofprimenumbers.Asecondisthateveryidealisaprincipalideal.AthirdisthatthereisadivisionalgorithmthatisthebasisoftheEuclideanalgorithm.Therearen'tmanyringsthathavealltheseproperties,andsomeringshavenoneofthem.We'llinvestigatethesepropertiesandtheirinterrelations.We'llusethesethreepropertiestode nethreespecialkindsofintegraldomains:uniquefactorizationdomains(UFDs),principalidealdomains(PIDs),andEuclideandomains(EDs).Whenwedowe'll ndeveryEuclideandomainisaprincipalidealdomain,everyprincipalidealdomainisauniquefactorizationdomain,everyuniquefactorizationdomainisanintegraldomain;andeveryintegraldomainisaring.EDsPIDsUFDsIntegraldomainsCommutativerings3.8.1DivisibilityinanintegraldomainWe'llborrowtheconceptsofdivisibilityandgreatestcommondivisorfromZandapplythemtointegraldomains.We'llseparatetheconceptofprimenumberinZintotwoconceptssinceinsomeoftheintegraldomainswe'lllookatthey'reactuallydi erent.De nition3.38.Thefollowingde nitionsapplytoelementsofanintegraldomain.Letaandbbenonzeroelements.We'llsayadividesb,writtena b,ifthereexistscsuchthatac=b.We'llsaythatdisagreatestcommondivisorofaandb,ifddividesbothaandb,andwheneveranotherelementedividesbothaandb,thenedividesd.Anelementxthatisnotzeroandnotaunitisirreducibleifwheneverx=yz,eitheryorzisaunit,otherwiseitisreducible
3.8.UFDS,PIDS,ANDEDS81Anelementxthatisnotzeroandnotaunitisprimeifwheneverx yz,thenx yorx z.Notethatwewon'tusethenotationd=gcd(a;b)whendisagreatestcommondivisorsincetherewillbeothergreatestcommondivisors,thatis,thegreatestcommondivisorisonlyuniqueuptoaunit.Later,whenwelookatprincipalidealdomains,wecanusethenotation(c)=(a;b)forgreatestcommondivisorswhichsaystheprincipalideal(c)isthesameastheidealgeneratedbyaandb.Exercise42.Severalpropertiesofdivisibilityfollowdirectlyfromthede nitionjustliketheydowiththeintegraldomainisZ.Provethefollowingpropertiesfromtheabovede nitions.(a).1divideseveryelement.(b).Eachelementdividesitself.(c).Ifa bthena bc.(d).Divisibilityistransitive.(e).Ifoneelementdividestwootherelements,thenitdividesboththeirsumanddi er-ence.(f).Cancellation:Whenc6=0,a bifandonlyifac bc.Theorem3.39.Ifanelementinanintegraldomainisprime,thenitirreducible.Proof.Letxbeprime.Supposethatx=yz.Thenx yz,soeitherx yorx z.Inthe rstcase,xw=yforsomew.Thereforexwz=yz=x.Cancelthextoconcludewz=1.Thenzisaunit.Likewise,inthesecondcaseyisaunit.Thereforexisirreducible.q.e.d.Theconverseofthistheoremdoesnothold.Thatis,thereareintegraldomainswherenotallirreducibleelementsareprime.We'llseethatinthisnextexample.Butthenalittlelater,we'llseethatinprincipalidealdomains(abouttobede ned),irreducibleelementsareprime.Example3.40(anonUFD).We'll ndanumberofotherUFDs,but,it'simportanttoknowthatnoteveryintegraldomainhasuniquefactorization.ConsidertheintegraldomainR=Z[p
10].Anelementofitisoftheformx+yp
10wherexandyareintegers.Inthisintegraldomain9canbefactoredintwoways.9=32=(p
10+1)(p
10�1);but3,p
10+1,andp
10�1areallirreducible.Thisintegraldomain,andmanyothers,arenotUFDs.Althoughthethreeelements3,p
10+1,andp
10�1areirreducible,nonedividesanyother,sononeofthemisprime,asyoucanseebytheequationinvolving9,above.3.8.2UniquefactorizationdomainsUniquefactorizationisapropertythatwemightexpect,butitturnsoutitdoesn'tholdineveryintegraldomain.GivenanyelementxinaringD,weexpectthatwecanfactoritinto`atoms,'thingsthatcan'tbecutfurther,andthatthere'sonlyonewaytodothat.Ofcourse,withourexperiencewiththeintegers,weknowthatthere'sabitofdicultyinstatingtheuniquenesspartoftheclaim.Foronething,theorderofthefactorsisvariable,and,foranother,thereareunits,like1and�1thatcanbeinsertedtochangetheformallistingofthefactors.Still,thesearesmallthingsthatwecandealwith.
82CHAPTER3.RINGSDe nition3.41.Anintegraldomainisauniquefactorizationdomain(UFD)ifeveryelementinitisaproductofirreducibleelementsanditisaproductofirreducibleelementsinonlyonewayapartfromtheorderoftheproductandfactorsofunits.TheringZofintegersis,ofcourse,auniquefactorizationdomain.Aninteger,suchas6canbewritteninmorethanonewayasaproductofirreducibleelements(primes,inthecaseofintegers)6=23=(�3)(�2),buttheonlydi erenceistheorderoftheprimesandtheinsertionsofunitsinthefactorization.RecallthatanidealIinacommutativeringRisaprimeidealwhenR=Iisanintegraldomain.Equivalently,Iisaprimeidealif(1)I6=R,and(2)forallx;y2R,ifxy2I,theneitherx2Iory2I.Theorem3.42.AnnonzeroelementxisanintegraldomainDisprimeifandonlyiftheprincipalideal(x)isaprimeideal.Exercise43.Provetheprecedingtheorem.Notethattherearetwothingstoproveinanif-and-only-ifstatement.3.8.3PrincipalidealdomainsAsecondnicepropertythattheringofintegershasisthateveryidealinZisgeneratedbyasingleelement.IfIisanidealinZ,thenthegcdofallit'snonzeroelementsisanelementofIandallotherelementsaremultiplesofthisgcd.Thiswillbeourde nitionofaprincipalidealdomain(PID),andwe'llshowthateveryPIDisaUFD.ThereareUFDsthataren'tPIDs,forinstance,Z[x],theringofpolynomialswithintegercoecientsisone;onenonprincipalidealisgeneratedby2andx.De nition3.43.Anintegraldomainisaprincipalidealdomain(PID)ifeveryidealinthedomainisprincipal,thatis,generatedbyoneelement.BesidesZ,otherprominentPIDsareF[x]whereFisa eld.We'llprovethisinsection3.8.4onEuclideandomainswhicharespecialkindsofPIDs.We'llshowinacoupleofstepsthateveryPIDisaUFD.The rstonemakesaconnectionbetweengreatestcommondivisorsandideals.Theorem3.44.LetDbeaprincipalidealdomainwithnonzeroelementsaandb.Theideal(a;b)isprincipal,soitisequalto(c)forsomeelementc.Thencisagreatestcommondivisorofaandb.Proof.Sincea2c,thereforec a.Likewise,c b.Wealsoknowthatc2(a;b),soc=xa+ybforsomeelementsxandy.Toshowthatcisagreatestcommondivisor,supposedissomeothercommondivisorofaandb.Thena=udandb=vdforsomeelementsuandv.Now,c=xa+yb=xud+yvd=(xu+yv)d:Therefore,d c.Thuscisagreatestcommondivisorofaandb.q.e.d.Theorem3.45.Inaprincipalidealdomain,irreducibleelementsareprime.
3.8.UFDS,PIDS,ANDEDS83Proof.Supposethatpisirreducibleandp ab.We'llshoweitherp aorp b.We'lldothatbyshowingthatifpdoesn'tdividea,thenitdoesdivideb.Supposepdoesnotdividea.Thentheideal(p;a)is(1)sincepisirreducible.Since12(p;a),1=xp+yaforsomeelementsxandy.Therefore,b=bxp+aby.Sincep ab,thereforep bxp+aby,sop b.Thus,theirreducibleelementpisalsoprime.q.e.d.Next,we'llusethefollowinglemmatoshowthatelementshavefactorizationsinPIDs.We'llstillhavetoshowthey'reuniqueafterthat.Theconditioninthelemmaiscalledtheascendingchaincondition(ACC)onideals,andringsthatsatisfyitarecalledNoetherianringsinhonorofNoetherwhostudiedsuchrings.Lemma3.46.Inaprincipalidealdomain,therearenoin nitelyascendingchainsofideals.Thatis,(a1)$(a2)$(a3)$doesnotexist.Proof.Supposethereweresuchanin nitelyascendingchainofideals.ThentheunionI=1[i=1(ai)isanideal,asyoucaneasilycheck.Itmustbeprincipal,soI=(a)forsomeelementa.Butaisintheunion,soit'sinoneoftheideals(ai).Then(a)(ai)$(ai+1)(a);acontradiction.q.e.d.Therearerings,infactUFDs,thatarenotNoetherian.Anexampleisapolynomialringwithin nitelymanyvariablessuchasQ[x1;x2;x3;:::].Anin nitelyascendingchainofidealsinthatringis(x1)$(x1;x2)$(x1;x2;x3)$.Theorem3.47.Inaprincipalidealdomain,everyelementthatisnotzeroandnotaunithasafactorizationintoirreducibleelements.Proof.Supposethatanonzeroelementa1hasnofactorizationintoirreducibleelements.We'llderiveacontradiction,butwe'llneedanelementwithnofactorizationwithanextraproperty.We'llgetthatelement,denotedanbelow,asfollows.Startingwiththeideal(a1),formanyascendingchainofidealsgeneratedbyotherelementswithnofactorizations,andextendthechainasfaraspossible.Bythelemma,itstopssomewhere,sayat(an).(a1)$(a2)$$(an):Wenowhaveanelementanwhichhasnofactorizationintoirreducibleelementswithanextraproperty,namely,anyidealstrictlycontaining(an)isgeneratedbyanelementthatdoeshavesuchafactorization.Now,anisnotirreducibleitself,forthatwouldbeafactorization,soan=bcwhereneitherbnorcisaunit.Sinceb an,therefore(an)(b).But(an)6=(b),forotherwiseb=andforsomed,andthenandc=bc=an,sodc=1makingcaunit,whichitisnot.So(an)$(b)andlikewise(an)$(c),thereforebothbandchavefactorizations,andtheproductofthosefactorizationsgivesafactorizationforan,acontradiction.q.e.d.
84CHAPTER3.RINGSTheorem3.48.Everyprincipalidealdomainisauniquefactorizationdomain.Proof.Thelasttheoremgavetheexistenceofatleastonefactorizationforanelementa.Westillhavetoshowthatthere'satmostonefactorization.Supposethatahastwofactorizationsasproductsofirreducibleelements.a=p1pn=q1qmSincetheirreducibleelementp1isprime(inaPID),p1dividesoneoftheqi's,whichwecanrenumberasq1.Thenp1=u1q1whereu1isaunit.Substituteu1q1forp1,andcancelq1togettheequationu1p2pn=q2qm:Thatcompletestheinductivestepofmathematicalinductiononn.Thebasecase,whenn=1,islefttothereader.q.e.d.3.8.4EuclideandomainsThethirdnicepropertythatZhasisthatthereisadivisionalgorithmthatisthebasisoftheEuclideanalgorithm.SomeexampleEuclideandomainsbesidesZthatwe'lldiscussinthissectionincludetheGaussianintegersZ[i],theEisensteinintegersZ[!]where!isaprimitivecuberootof1,andpolynomialringsF[x]overa eldF.Fortheintegers,thedivisionalgorithmstartswithanintegera(thedividend)andanonzerointegerb(thedivisor)anddeliversq(thequotient)andr(theremainder)suchthata=qb+rand0rb:ThispropertyallowedustoconstructtheEuclideanalgorithmfor ndingGCDsaswellastheextendedEuclideanalgorithmtoshowthatthegreatestcommondivisoroftwonumbersisalinearcombinationofthem.Thereareafewotherintegraldomainsthathavethesamekindofdivisionalgorithmwheretheremainderissomehow\smaller"thanthedivisor,buttheconceptofsmallerandhowto ndqandrdi ersfromdomaintodomain.De nition3.49.AEuclideanvaluationonanintegraldomainDisafunctionv:D�0!Z0thatsatis estheconditions1.fornonzeroelementsaandb,v(a)v(ab),and2.foreachelementa(thedividend)andnonzeroelementb(thedivisor),thereareelementsq(thequotient)andr(theremainder)suchthata=qb+rwhereeitherr=0orv(r)v(b):AnintegraldomainthatadmitsaEuclideanvaluationiscalledEuclideandomain.
3.8.UFDS,PIDS,ANDEDS85Ofcourse,ZisaEuclideandomainwiththevaluationbeingtheabsolutevaluev(a)=jaj.AnotherclassofEuclideandomainsaretheringsofpolynomials(inonevariable)withcoecientsinagiven eld.Thefollowingtheoremisessentiallyjustlongdivisionforpoly-nomials.We'llmakeitsimplebymakingthedivisorg(x)amonicpolynomial,thatis,apolynomialwhoseleadingcoecientis1.Itdirectlyfollowsfromthedivisionalgorithmforpolynomialsovera eld,theorem1.55,thata eld'spolynomialringisaEuclideandomain.Corollary3.50.ThepolynomialringF[x]withcoecientsina eldFisaEuclideandomainwherethevaluationvassignstoapolynomialf(x)thedegreeoff.Soonwe'llstudypolynomialringsinmoredetail.ThereareotherEuclideandomainsincludingtheGaussianintegersandtheEisensteinintegers.TheGaussianintegersZ[i]isaEuclideandomain.TheringofGaussianintegersisZ[i]=fa1+a2ija1;a22Zg.Itsvaluationfunction,alsocalledthenorm,isv(a1+a2i)=a21+a22,thesquareofthedistancetotheorigin.InordertodivideoneGaussianintegerb1+b2iintoanothera1+a2itogetaquotientq1+q2iandremainderr1+r2i,youcanperformthecomplexdivisiona1+a2i
b1+b2itogetanexactquotient,andchooseq1+q2itobetheclosestGaussianintegertothatexactquotient.Theremainderisthendetermined.
1
!
!2
Figure3.3:LatticeofEisensteinintegers
86CHAPTER3.RINGSEisensteinintegersWhereasabasisfortheGaussianintegersconsistsof1andi,abasisfortheEisensteinintegersconsistsof1and!where!=1
2(�1+ip
3)isaprimitivecuberootofunity.Aprimitivecuberootofunitysatis estheequationx3�1
x�1=0whichsimpli estox2+x+1=0.ThelatticeofEisensteinintegersisatriangularlatticesince1+!+!2=0.Thelatticeisshownin gure3.3.Thedottedlinesshowcoordinatesrelativetothebasisconsistingof1and!.TherearesixunitsintheEisensteinintegers.Theyarethesixsixthrootsofunity:1itself,aprimitivesixthroot!�1,aprimitivecuberoot!,theprimitivesquareroot�1,andaprimitivesixthroot�!.Theyareequallyspacedat60aroundtheunitcircle.LiketheGaussianintegers,theEisensteinintegersalsoareaEuclideandomain.Thevaluationisv(a+b!)=a2�ab+b2.TheEuclideanalgorithminEuclideandomains.First,we'llshowthatEuclideandomainsareprincipalidealdomains,andsincePIDsarealsoUFDs,thereforeEuclideando-mainsarealsouniquefactorizationdomains.Thenwe'lllookatanexampleoftheEuclideanalgorithminaEuclideandomainotherthanZ.Theorem3.51.AEuclideandomainisaprincipalidealdomain.Proof.LetIbeanidealinaEuclideandomainDwithvaluationv.We'llshowIisaprincipalideal.IfIisthezeroideal(0),thenit'sprincipalofcourse.AssumenowthatIhasanonzeroelement,andletS=fv(x)j06=x2Ig.Thisisanonemptysubsetofthenonnegativeintegers,soithasaleastelement,andletthatbev(a).Thus,aisanonzeroelementofI,so(a)I.LetxbeanyothernonzeroelementinI.Thenv(a)v(x).Furthermore,thereareelementsqandrinDsuchthatx=aq+randeitherr=0orv(f)v(a).Butr=x�aq2I,soifr6=0,thenv(r)&#x-340;v(a)contradictsv(a)v(r).Therefore,r=0,andhencex=aq,soa x.Therefore,I=(a).Thus,DisaPID.q.e.d.TheEuclideanalgorithmworksinanyEuclideandomainthesamewayitdoesforintegers.Itwillcomputethegreatestcommondivisor(uptoaunit),andtheextendedEuclideanalgorithmwillconstructthegreatestcommondivisorasalinearcombinationoftheoriginaltwoelements.Example3.52.Let'stakeanexamplefromthepolynomialringQ[x].Let's ndthegreatestcommondivisoroff1(x)=x4+2x3�x�2andf2(x)=x4�x3�4x2�5x�3.Theyhavethesamedegree,sowecantakeeitheroneofthemasthedivisor;let'stakef2(x).Dividef2intof1togetaquotientof1andremainderoff3(x)=3x3+4x2+4x+1.Thendividef3intof2togetaquotientandaremainderf4,andcontinueuntiltheremainderis0(whichoccursonthenextiteration.f1(x)=x4+2x3�x�2f1(x)=1f2(x)+f3(x)f2(x)=x4�x3�4x2�5x�3f2(x)=(1
3x�7
9)f3(x)+f4(x)f3(x)=3x3+4x2+4x+1f3(x)=(27
20x�9
20)f4(x)f4(x)=�20
9x2�20
9x�20
9
3.9.REALANDCOMPLEXPOLYNOMIALRINGSR[X]ANDC[X]87Thus,agreatestcommondivisorisf4(x),whichdi ersbyaunitfactorfromthesimplergreatestcommondivisorx2+x+1.Wecanreadtheequationsontherightinreversetogetf4asalinearcombinationoff1andf2.f4(x)=f2(x)�(1
3x�7
9)f3(x)=f2(x)�(1
3x�7
9)(f1(x)�f2(x))=(1
3x+2
9)f2(x)�(1
3x�7
9)f1(x)3.9RealandcomplexpolynomialringsR[x]andC[x]WeknowafairamountaboutF[x],theringofpolynomialsovera eldF.Ithasadivisionalgorithm,soit'saEuclideandomainwheretheEuclideanvaluationisthedegreeofapoly-nomial,soithasdivisionandEuclideanalgorithms.Sinceit'sEuclidean,it'salsoaprincipalidealdomain,andthatmeansirreducibleelementsareprime.Andsinceit'saPID,it'salsoauniquefactorizationdomain,thatis,everypolynomialuniquelyfactorsasaproductofirreduciblepolynomials.Ratherthancallingirreduciblepolynomialsprimepolynomials,we'llusetheterm\irre-duciblepolynomial".That'sthecommonpractice.ThenonzeroprimeidealsofF[x]arejusttheprincipalideals(f)generatedbyirreduciblepolynomialsf2F[x],and,furthermore,they'remaximalideals,soF[x]=(f)isa eld.We'veseenexamplesofthis,forinstance,R[x]=(x2+1)=R[i]=C,Q[x]=(x2�2)=Q(p
2),andZ3[x]=(x2+1)=Z3(i).So,irreduciblepolynomialsinF[x]give eldextensionsofF.ThemainquestionforF[x]is:whataretheirreduciblepolynomials?We'llstartwithC[x]andR[x]followedbyQ[x]andZ[x].3.9.1C[x]andtheFundamentalTheoremofAlgebraInthe16thcenturyCardano(1501{1576)andTartaglia(1500{1557)andothersfoundformu-lasforrootsofcubicandquarticequationsintermsofsquarerootsandcuberoots.Atthetime,onlypositivenumberswerecompletelylegitimate,negativenumberswerestillsome-whatmysterious,andthe rstinklingofacomplexnumberappeared.Incidentally,atthistimesymbolicalgebrawasstillbeingdeveloped,andtheywrotetheirequationsinwordsinsteadofsymbols!Here'sanillustrationofhowcomplexnumbersarose.OneofCardano'scubicformulasgivesthesolutiontotheequationx3=cx+dasx=3q
d=2+p
e+3q
d=2�p
ewheree=(d=2)2�(c=3)3.Bombelliusedthistosolvetheequationx3=15x+4,whichwasknowntohave4asasolution,togetthesolutionx=3q
2+p
�121+3q
2�p
�121:
88CHAPTER3.RINGSNow,p
�121isnotarealnumber;it'sneitherpositive,negative,norzero.Bombellicontin-uedtoworkwiththisexpressionuntilhefoundequationsthatleadhimtothesolution4.Assumingthattheusualoperationsofarithmeticheldforthese\numbers,"hedeterminedthat3q
2+p
�121=2+p
�1and3q
2�p
�121=2�p
�1and,therefore,thesolutionx=4.Cardanohadnotedthatthesumofthethreesolutionsofacubicequationx3+bx2+cx+d=0is�b,thenegationofthecoecientofx2.Bythe17thcenturythetheoryofequationshaddevelopedsofarastoallowGirard(1595{1632)tostateaprincipleofalgebra,whatwecallnow\thefundamentaltheoremofalgebra."Hisformulation,whichhedidn'tprove,alsogivesageneralrelationbetweenthensolutionstoannthdegreeequationanditsncoecients.Foragenericequationxn+an�1xn�1++a1x+a0=0Girardrecognizedthattherecouldbensolutions,ifyouallowallrootsandcountrootswithmultiplicity.So,forexample,theequationx2+1=0hasthetwosolutionsp
�1and�p
�1,andtheequationx2�2x+1=0hasthetwosolutions1and1.Girardwasn'tparticularlyclearwhatformhissolutionsweretohave,justthattherewerenofthem:x1;x2;:::;xn.Girardgavetherelationbetweenthenrootsx1;x2;:::;xnandthencoecientsa1;:::;anthatextendedCardano'sremark.First,thesumoftherootsx1+x2++xnis�a1(Cardano'sremark).Next,thesumofallproductsofpairsofsolutionsisa2.Next,thesumofallproductsoftriplesofsolutionsis�a3.Andsoonuntiltheproductofallnsolutionsiseitheran(whenniseven)or�an(whennisodd).He guredthisoutbyusingaversionofoneofthepropertiesofpolynomialsmentionedabove,namely,ifa1;a2;:::;anarerootsofamonicpolynomialf(x)ofdegreen,thenf(x)=(x�a1)(x�a2)(x�an):Ifyouexpandtherightsideoftheequation,you'llderivehisresult.Here'sanexample.The4thdegreeequationx4�6x3+3x2+26x�24=0hasthefoursolutions�2,1,3,and4.Thesumofthesolutionsequals6,thatis�2+1+3+4=6.Thesumofallproductsofpairs(sixofthem)is(�2)(1)+(�2)(3)+(�2)(4)+(1)(3)+(1)(4)+(3)(4)whichis3.Thesumofallproductsoftriples(fourofthem)is(�2)(1)(3)+(�2)(1)(4)+(�2)(3)(4)+(1)(3)(4)whichis�26.Andtheproductofallfoursolutionsis�24.Overtheremainderofthe17thcentury,negativenumbersroseinstatustobefull- edgednumbers.Butcomplexnumbersremainedsuspectthroughmuchofthe18thcentury.They
3.9.REALANDCOMPLEXPOLYNOMIALRINGSR[X]ANDC[X]89weren'tconsideredtoberealnumbers,buttheywereusefulinthetheoryofequationsandbecomingmoreandmoreusefulinanalysis.Itwasn'tevenclearwhatformthesolutionstoequationsmighttake.Certainly\numbers"oftheforma+bp
�1weresucienttosolvequadraticequations,evencubicandquarticequations.Eulerdidaprettygoodjobofstudyingcomplexnumbers.Forinstance,hestudiedtheunitcircleassigningthevaluecos+isintothepointontheunitcircleatanangleclockwisefromthepositiverealaxis.Hemeasuredtheanglebythelengthofthearccutoftheunitcirclecuto bytheangle.Wecallthatmeasurementradiansnow,buttheword\radian"wasn'tcoineduntillater.InhisstudyofthiscirclehedevelopedwhatwecallEuler'sidentityei=cos+isin:Thiswasanespeciallyusefulobservationinthesolutionofdi erentialequations.Becauseofthisandotherusesofi,itbecamequiteacceptableforuseinmathematics.Bytheendofthe18thcenturynumbersoftheformx+iywereinfairlycommonusebyresearchmathematicians,anditbecamecommontorepresentthemaspointsintheplane.Yetmaybesomeotherformof\number"wasneededforhigher-degreeequations.ThepartoftheFundamentalTheoremofAlgebrawhichstatedthereactuallyarensolutionsofannthdegreeequationwasyettobeproved,pending,ofcourse,somedescriptionofthepossibleformsthatthesolutionsmighttake.Still,atnearlytheendofthe18thcentury,itwasn'tyetcertainwhatformallthesolutionsofapolynomialequationmighttake.Leibniz,forexample,statedin1702thatx4�a4didn'thaverootsoftheformx+yp
�1,butEulershoweditdidin1742.D'Alembert,Euler,deFoncenex,Lagrange,andLaplacedevelopedpartialproofs.Finally,in1799,Gauss(1777{1855)publishedhis rstproofoftheFundamentalTheoremofAlgebra.Wewon'tlookathisoranyotherproofofthetheorem.That'susuallyprovedinacourseincomplexanalysis.Wewill,however,usethetheorem.De nition3.53.A eldFisalgebraicallyclosedifeverypolynomialf(x)2F[x]factorsasaproductoflinearfactors.Equivalently,apolynomialf(x)ofdegreenhasnrootsinFcountingmultiplicities.Aweakerde nitioncouldbemade,andthat'sthateverypolynomialofdegreeatleast1hasatleastonerootinF.Byinduction,theremainingrootscanbeshowntoexist.Thus,theFundamentalTheoremofAlgebraisastatementthatCisanalgebraicallyclosed eld.Therefore,thealgebraofC[x]isparticularlysimple.Theirreduciblepolynomialsarethelinearpolynomials.3.9.2ThepolynomialringR[x]Let'sturnourattentionnowtopolynomialswithrealcoecients.MuchofwhatwecansayaboutR[x]comesfromtherelationofRasasub eldC,andconsequently=fromtherelationofR[x]asasubringofC[x].Thatistosay,wecaninterpretapolynomialf(x)withrealcoecientsasapolynomialwithcomplexcoecients.Theorem3.54.Ifapolynomialf(x)withrealcoecientshasacomplexrootz,thenitscomplexconjugate
zisalsoaroot.
90CHAPTER3.RINGSProof.Letf(x)=anxn++a1x+a0whereeachai2R.Ifzisarootoff,thenf(z)=anzn++a1z+a0=0.Takethecomplexconjugateoftheequation,andnotethat
ai=ai.Thenf(
z)=an
zn++a1
z+a0=0.Thus,
zisalsoaroot.q.e.d.Thistheoremtellsusforapolynomialf(x)withrealcoecients,itsrootseithercomeinkpairsofacomplexnumberorsinglyasrealnumbers.Wecannamethe2kcomplexrootsasz1;
z1;z2;
z2;:::;zk;
zk:Writingz1=x1+y1i;:::;zk=xi+yki,thecomplexrootsarex1+y1i;x1�y1i;x2+y2i;x2�y2i;:::;xk+yki;xk�ykiandthen�2krealrootsasr2k+1;:::;rn:UsingthefactthatCisalgebraicallyclosed,wecanwritef(x)asf(x)=an(x�z1)(x�
z1)(x�zk)(x�
zk)(x�r2k+1)(x�rn)=an(x2�2x1x+x21+y21)(x2�2xkx+x2k+y2k)(x�r2k+1)(x�rn)Thislastexpressionhasfactoredf(x)asaproductofirreduciblequadraticandlinearpoly-nomialswithrealcoecients.Theorem3.55.TheirreduciblepolynomialsinR[x]arethelinearpolynomialsandthequadraticpolynomialswithnegativediscriminant.Proof.Theremarksaboveshowthatonlylinearandquadraticpolynomialscanbeirreducible.Linearpolynomialsarealwaysirreducible.Aquadraticpolynomialwillhavenorealrootswhenitsdiscriminantisnegative.q.e.d.3.10RationalandintegerpolynomialringsWe'vestudiedtheirreduciblepolynomialsinC[x]andR[x]withthehelpoftheFundamentalTheoremofAlgebraandfoundthemtobeeasilyclassi ed.TheirreduciblepolynomialsinC[x]arethelinearpolynomials,andirreduciblepolynomialsinR[x]arethelinearpolynomialsandquadraticpolynomialswithnegativediscriminant.DeterminingwhichpolynomialsinQ[x]areirreducibleismuchharder.Ofcourse,allthelinearonesare,andwe'llbeabletotellwhichquadraticandcubiconesareirreduciblefairlyeasily.Afterthatitbecomesdicult.3.10.1RootsofpolynomialsThequadraticcase.Consideraquadraticpolynomialf(x)=ax2+bx+cwithcoecientsinQ.Itsrootsaregivenbythequadraticformula�bp
b2�4ac
2awhichcanbeshownbytheprocessknownascompletingthesquare.Thediscriminantofaquadraticpolynomialis
3.10.RATIONALANDINTEGERPOLYNOMIALRINGS91=b2�4ac.Whenispositive,therearetworealroots;when0,thereisonedoubleroot;andwhennegative,therootsareapairofcomplexconjugatenumbers.Whenisaperfectrationalsquare,thatis,thesquareofarationalnumber,thenf(x)factors,thatis,it'sreducible.Otherwise,it'sirreducible.Thus,f(x)isirreducibleifandonlyifthediscriminantisnotaperfectsquare.Thecubiccase.Itismorediculttodeterminewhenacubicpolynomialf(x)=ax3+bx2+cx+dwithrationalcoecientsisirreducible,butnottoodicult.Notethatiff(x)factors,thenoneofthefactorshastobelinear,sothequestionofreducibilityreducestotheexistenceofarationalrootoff(x).Varioussolutionsofacubicequationax3+bx2+cx+d=0havebeendeveloped.Here'sone.First,wemayassumethatfismonicbydividingbytheleadingcoecient.Ourequationnowhastheformx3+bx2+cx+d=0.Second,wecaneliminatethequadratictermbyreplacingxbyy�1
3b.Thenewpolynomialinywillhavedi erentroots,butthey'reonlytranslationsby1
3b.Wenowhavethecubicequationy3+(c�1
3b2)y+(2
27b3�1
3bc+d)=0whichwe'llwriteasy3+py+q=0:Bytheway,thissubstitutionwhichresultsinapolynomialwhosetermaftertheleadingtermis0hasaname.ItiscalledaTschirnhaussubstitution.Therootsofthenewpolynomialwillsumto0.We'llfollowViete'smethodandperformanothersubstitution.Replaceybyz�p
3z.Aftersimplifyingandclearingthedenominatorswe'llhavetheequationz6+qz3�p3
27z=0whichisaquadraticequationinz3.Itscomplexsolutionsarez3=�qp
q2+4p3=27
2=�q
2r
q
22+p
33:Takingcomplexcuberootstogetthreevaluesforz,thenusingy=z�p
3ztodetermineyandx=y�1
3btodeterminex,wehavetheallthreecomplexsolutionstotheoriginalequation.Atleastoneofthesethreecomplexsolutionsisreal,andperhapsallthree.Wehaveawayofdeterminingwhetheracubicpolynomialisreducible.Firstq2+4p3=27needstobeaperfectrationalsquarer2,thenoneof�q+rand�q�rneedstobeaperfectrationalcube.Thereisanotherwaytodetermineifthereisarationalroot.Rationalrootsofapolynomial.Ifwe'relookingfortherootsofapolynomialwithrationalcoecients,wecansimplifythejobalittlebitbyclearingthedenominatorssothatallthecoecientsareintegers.Thefollowingtheoremhelpsin ndingroots.
92CHAPTER3.RINGSTheorem3.56(Rationalroottheorem).Letf(x)=anxn++a1x+a0beapolynomialwithintegralcoecients.Ifr=sisarationalrootoffwithr=sinlowestterms,thenrdividestheconstanta0andsdividestheleadingcoecientan.Proof.Sincer=sisaroot,thereforef(x)=an(r=s)n+an(r=s)n�1++a1(r=s)+a0=0;andso,clearingthedenominators,wehaveanrn+anrn�1s++a1rsn�1+a0sn=0:Wecanrewritethisequationas(anrn�1+anrn�2s++a1sn�1)r=�a0sn:Now,sincerdivides�a0sn,andrisrelativelyprimetos,andhencetosn,thereforerdividesa0.Inlikemanner,youcanshowsdividesan.q.e.d.Forexample,to ndtherationalrootsr=soff(x)=27x4+30x3+26x2�x�4,rwillhavetodivide4,sothepossibilitiesforrare1;2;4,andswillhavetodivide27,sothepossibilitiesforsare1;3;9;27(sincewemayassumesispositive).Thatgives24rationalnumberstocheck,andamongthemwillbefoundthetworationalroots1
3and�4
9.Afterone,r
s,isfoundf(x)canbedividedbyx�r
stolowerthedegreeofthepolynomialto ndtherestoftheroots.Ifapolynomialdoeshavearationalroot,thenit'sclearlyreduciblesincethatrationalrootdeterminesalinearfactorofthepolynomial.Thatgivesusanotherwaytodetermineifacubicpolynomialisreducible.Forpolynomialsofdegree4orhigher,knowingthattherearenorationalrootsisin-sucienttoconcludethepolynomialisirreducible.Itstillmayfactorasquadraticandhigherdegreeterms.Forexample,x4+x2+1hasnorationalroots,butitfactorsas(x2+x+1)(x2�x+1),soitisreducible.3.10.2Gauss'slemmaandEisenstein'scriterionFurtherstudyofQ[x]willrequirelookingatZ[x].Inotherwords,inordertostudypolynomi-alswithrationalcoecients,we'llhavetolookatpolynomialswithintegralcoecients.Wecantakeapolynomialwithrationalcoecientsandmultiplyitbytheleastcommonmultipleofthedenominatorsofitscoecientstogetanotherpolynomialwiththesamerootsbutwithintegralcoecients.Wecanalsodividebythegreatestcommondivisoroftheresultingcoecientstogetyetanotherpolynomialwiththesameroots,withintegralcoecients,andthegreatestcommondivisorofallitscoecientsis1.Suchapolynomialiscalledprimitive.Afterthat,we'llbeabletoproveGauss'slemmawhichsaysthataprimitivepolynomialf(x)2Z[x]isreducibleinQ[x]ifandonlyifit'sreducibleinZ[x].Wecanmakemoreuseoftheseresultsif,insteadofconsideringjustthecaseofthedomainZandits eldoffractionsQ,wegeneralizetoanyuniquefactorizationdomainDandits eldoffractionsF.So,forthefollowingdiscussion, xaUFDD,andletFdenoteits eld
3.10.RATIONALANDINTEGERPOLYNOMIALRINGS93offractions.Though,keepinmindthebasiccasewhenD=Z,F=Q,D=(p)=Zp,andD=(p)[x]=Zp[x]togetabetterideaofwhat'sgoingon.WhenwehaveaprimepinD,theprojection :D!D=(p)inducesaringepimorphismD[x]!D=(p)[x]betweenpolynomialringswherethecoecientsoffarereducedmodulopgivingapolynomialinD=(p)[x].We'lldenotetheresultingpolynomialinD=(p)[x]byfp.De nition3.57.ThecontentofapolynomialinD[x]isthegreatestcommondivisorofallofitscoecients.Ifthecontentis1,thepolynomialiscalledprimitive.Forexample,iff(x)=3x2�9x+6thenthecontentoffis3.Also,thecontentofamonicpolynomialis1,soallmonicpolynomialsareprimitive.Thecontentofapolynomialisonlyde neduptoaunit.Evidently,everypolynomialinD[x]equalsaconstanttimesaprimitivepolynomial,theconstantbeingitscontent.Lemma3.58(Gauss).TheproductoftwoprimitivepolynomialsinD[x]isprimitive,andthecontentoftheproductofanytwopolynomialsinD[x]istheproductoftheircontents(uptoaunit).Proof.Inordertoshowthe rststatement,we'llshowiftheproductisnotprimitive,thenoneofthetwopolynomialsisnotprimitive.Letfandgbeprimitivepolynomialsandsupposethattheirproductfgisnotprimitive.ThensomeprimepofDdividesthecontentoffg,sopdivideseverycoecientoffg.Therefore,inD=(p)[x],(fg)p=0,sofpgp=0.ButD=(p)[x]isanintegraldomain(infact,aUFD),soeitherfp=0orgp=0.Therefore,peitherdividesallthecoecientsofforallthecoecientsofg,henceoneortheotherisnotprimitive.Thesecondstatementfollowsfromthe rstjustbyusingthefactthatapolynomialequalsitscontenttimesaprimitivepolynomial.q.e.d.Theorem3.59(Gauss'slemma).IfaprimitivepolynomialinD[x]canbefactoredastheproductoftwopolynomialsinF[x],thenitcanbefactoredastheproductoftwopolynomialsinD[x]ofthesamedegrees.Proof.Givenf2D[x]asaproductghwithg;h2F[x].Wecanwritegh=p
quvwhereuandvareprimitivepolynomialsinD[x],andpandqarerelativelyprimeintegers.Thenqf=puv.Sincefisprimitive,thecontentofqfequalsthecontentofq.Sinceuandvareprimitive,soisuv,andthereforethecontentpfpuvequalsthecontentofp.Thusp=q,andthey'reboth1,andsof=uv.Notethatthedegreesofuandvarethesameasthedegreesofgandh,respectively.q.e.d.ThefollowingcorollaryissometimescalledGauss'slemma.Itfollowsdirectlyfromtheabovesincemonipolynomialsareprimitive.Corollary3.60.AmonicpolynomialinD[x]isreducibleoverF[x]ifandonlyifit'sreducibleoverD[x].Thereareirreducibilitytestsforpolynomialswithintegercoecients,sobythiscorollary,we'llbeabletotestirreducibilityforpolynomialswithrationalcoecients.OnetestforirreducibilityofpolynomialswithintegercoecientsistomovetoaquotientringZp.ThatalsogeneralizestoanyUFDD.IfyoucanfactoritinD,youcanfactoritinaquotientring,atleastiftheleadingtermdoesn'tdisappearinthequotient.
94CHAPTER3.RINGSTheorem3.61(Modulopirreducibilitytest.).Letpbeaprimeinteger,andletfbeapolynomialwhoseleadingcoecientisnotdivisiblebyp.IffisreducibleinF[x],thenfpisreducibleinD=(p)[x].IffpisirreducibleinD=(p)[x],thenfisirreducibleinF[x].Proof.SupposefisreducibleinF[x].Thenthereexistg;h2D[x]suchthatf=ghwherethedegreesofgandhareatleast1.Sincef=gh,therefore,fp=gphp.Sincepdoesnotdividetheleadingcoecientoff,neitherdoesitdividetheleadingcoecientsofgorh.Thereforedeggp=degg1anddeghp=degh1.Thus,fpisreducible.Thelaststatementofthetheoremisthecontrapositiveofthe rststatement.q.e.d.Example3.62.ConsideranycubicpolynomialfinQ[x]withanoddleadingcoecient,anoddconstant,andoneoftheothertwocoecientsodd,forinstance,f(x)=77x3+15x2+8x+105.ByGauss'slemma,it'sreducibleinQ[x]ifandonlyifit'sreducibleinZ[x].Todeterminethat,usethemodulo2irreducibletest.Forf(x)=77x3+15x2+8x+105,you'llgetf2(x)=x3+x2+1.Theresultingf2willhavenorootsinZ2sinceithasthreenonzeroterms.Acubicpolynomialwithnorootsisirreducible,sof2isirreducibleinZ2[x].Hence,fisirreducibleinQ[x].Theconverseofthemodpirreducibilitytestisnotvalid.ApolynomialcanbereduciblemodpbutirreducibleinZ[x].Takef(x)=77x3+15x2+8x+105,forexample,whichweknowisirreducibleinZ[x].Modulop=5,however,itfactorsintolinearfactors:f5(x)2x3�2x=2(x+1)x(x�1),soisreducible.Exercise44.Showthepolynomialf(x)=x4+x3+2x2+2x+1isirreducibleinQ[x].Hint:Consideritmodulo2.Firstcheckforroots,thenseeifit'sdivisiblebyairreduciblequadratic.Therearen'tmanyirreduciblequadraticsmodulo2;x2+1isn'tsinceitfactorsas(x+1)2modulo2.Neitherarex2orx2+xsincethey'rebothdivisiblebyx.AnotherusefulirreducibilitytestisEisenstein'scriterion.Theorem3.63(Eisenstein'scriterion).Letf2D[x].Ifaprimepdoesnotdividetheleadingcoecientoff,butitdoesdividealltheothercoecients,andp2doesnotdividetheconstantoff,thenfisirreducibleinF[x].Proof.Supposefisreducible.Asintheprevioustheorem,thereexistg;h2D[x]suchthatf=ghwherethedegreesofgandhareatleast1.Reduceeverythingmodulop.Thenanxn=fp(x)=gp(x)hp(x)whereanistheleadingcoecientoff.NowZp[x]isaUFD,andsincefp(x)istheunitantimestheirreduciblexraisedtothenthpower,thereforexdividesbothgp(x)andhp(x).Thereforegp(0)=hp(0)=0.Thatmeansthatpdividestheconstanttermsofbothgandh,whichimpliesp2dividestheconstanttermoff,contrarytotheassumption.q.e.d.Example3.64.Considerthepolynomialf(x)=xn�a.Aslongasahasaprimefactorthatappearstothe rstpower,thenEisenstein'scriterionimpliesfisirreducible.Exercise45.Showthatthepolynomialf(x)=xn+10x+15isirreducibleinQ[x]
3.10.RATIONALANDINTEGERPOLYNOMIALRINGS953.10.3PrimecyclotomicpolynomialsCyclotomicpolynomialswereintroducedinde ntion1.63.Foraprimep,thepthcyclotomicpolynomialisp(x)=xp�1
x�1=xp�1++x+1:We'lluseEisenstein'scriteriontoshowpisirreducible,butnotdirectly.First,we'lluseatranslation.Letf(x)=p(x+1)=(x+1)p�1
x=xp�1+pp�1xp�2++p2x+p1:ThenEisenstein'scriterionappliestof.Sincefisirreducible,soisp.
1
!1
!2
!3
!4
!5
!6

Figure3.4:Primitive7throotsofunityTherootsofparethep�1primitivepthrootsofunity.Inthecasethatp=7,therearesixprimitive7throotsofunity.One,labelled!in gure3.4,islocatedatanangleof =2=7,andtheothersarepowersofit.Thereisonemore7throotofunity,namely1.1isnotprimitive7throotofunitybutinsteadaprimitive rstrootofunitysinceit'sarootofthepolynomialx�1ofdegree1.3.10.4PolynomialringswithcoecientsinaUFD,andpolyno-mialringsinseveralvariables.Gauss'slemmahasmoreusesthanwe'veuseditfor.WecanuseittoshowthatifDisaUFD,thensoisthepolynomialringD[x].AndwecanapplythatstatementtoconcludeapolynomialringD[x;y]intwoorD[x1;:::;xn]morevariablesisalsoaUFD.AlthoughtheseringsareUFDs,they'renotPIDs.Theorem3.65.LetDbeauniquefactorizationdomainandFitsringoffractions.ThenD[x]isalsoaUFD.TheirreduciblepolynomialsinD[x]areeitherirreducibleelementsofDorhavecontent1andareirreduciblepolynomialsinF[x].
96CHAPTER3.RINGSProof.LetfbeanonzeropolynomialinD[x].Itisequaltoitscontenttimesaprimitivepolynomial.ItscontentisanelementofD,and,sinceDisaUFD,itscontentuniquelyfactors(uptoaunit)asaproductofirreducibleelementsofD.We'rereducedtheprooftoshowingthatthataprimitivepolynomialfinD[x]ofdegreeatleast1uniquelyfactorsasaproductofirreduciblepolynomials.SincefisapolynomialinD[x],it'salsoapolynomialinF[x],andweknowF[x]isaUFDbeingapolynomialringwithcoecientsina eldF.Thus,funiquelyfactorsinF:f(x)=f1(x)f2(x)fk(x)whereeachfi(x)isirreducibleinF[x].WeonlyneedtoshowthatthisfactorizationcanbecarriedoutinD[x].Eachpolynomialfi(x)isaelementaiofFtimesaprimitivepolynomialf0i(x)inD[x],sof(x)=a1akf01(x)f0k(x):Sincef(x)isprimitiveandtheproductf01(x)f0k(x)isalsoprimitive,thereforea1akisaunitinD.Thus,f(x)factorsinD[x].YoucanalsoshowthatitcanfactorinonlyonewayinD[x]sinceitonlyfactorsinonewayinF[x].q.e.d.Corollary3.66.IfDisaUFD,thenapolynomialringinseveralvariablesD[x1;x2;:::;xr]withcoecientsinDisalsoaUFD.Ingeneral,thesearen'tPIDs.Forexample,(2;x)isnotaprincipalidealinZ[x],and(x;y)isnotaprincipalidealinQ[x;y].Irreduciblepolynomialsand eldextensions.We'llseeirreduciblepolynomialsina eldF[x]correspondtomaximalideals.Quotientringsbyidealsare eldsifandonlyiftheidealisamaximalideal,asshowninsection3.6.4.Thereforeirreduciblepolynomialscorrespondto eldextensions.Theorem3.67.Theidealgeneratedbyapolynomialwithcoecientsina eldismaximalifandonlyifthepolynomialisirreducibleoverthe eld.Proof.Letf2F[x].Suppose rstthat(f)isamaximalidealofF[x].We'llshowthatfcan'tbereducible.SupposethatffactorsasghinF[x]wheregandhhavelowerdegreesthanf,butneithergnorhisaunit.Then(f)(g)F[x].Since(f)ismaximal,therefore(g)isequaltoeither(f)orF[x].Neitherofthesecanoccur,forif(g)=(f),thentheyhavethesamedegree;butif(g)=F[x],thengisaunit.Thereforefisnotareduciblepolynomial.NextsupposethatfisanirreduciblepolynomialinF[x].LetIbeanyidealsuchthat(f)IF[x].SinceF[x]isaEuclideandomain(seesection3.8.4),itisalsoaprincipalidealdomain,soI=(g)forsomepolynomialg.Thereforef=ghforsomeh2F[x].Butfisirreducible,soeithergorhisaunit.Butgisnotaunitsince(g)=I6=F[x],sohisaunit.Therefore(f)=(g).SoanyproperidealIthatcontains(f)is(f)itself.Thus(f)isamaximalideal.q.e.d.Corollary3.68.ThequotientringF[x]=(f)ofapolynomialringovera eldFbytheidealgeneratedbyanirreduciblepolynomialisa eldextensionofF.
3.11.NUMBERFIELDSANDTHEIRRINGSOFINTEGERS97Thiscorollaryfollowsdirectlytheprecedingtheoremandthemaximalidealtheorem,theorem3.36,.Example3.69.WesawearlierthatanycubicpolynomialfinQ[x]withanoddleadingcoecient,anoddconstant,andoneoftheothertwocoecientsoddisirreducible.So,forexample,f(x)=x3�x�1isirreducibleoverQ.ThatmeansK=Q[x]=(f)isa eld.This eldKcanalsobedenotedK=Q[x]=(x3=x+1)sinceinthequotient,x3�x�1is0.Ratherthanusingthesymbolxinthequotient,itwouldbebettertohaveadi erentsymbolsothatxcanstillbeusedasourvariable.Let'susew.Theneveryelementinitisoftheformaw2+bw+c.Additionisdoneasusualforpolynomials,andmultiplicationisasusualexceptwheneverw3appears,itisreplacedbyw+1.Forexample,theproduct(w2+3w�3)(2w2�5)=2w3+w2�9w+15w2�7w+17.Asthisisa eld,therearereciprocalsofnonzeroelementsanddivisionbynonzeroelements.Findingreciprocalsisnotsoeasy.Forexample,to ndthereciprocalofw,weneedanelementaw2+bw+csuchthatitsproductwithwequals1.Now,(aw2+bw+c)w=aw3+bw2+cw=bw2+(a+c)w+a,soforthattoequal1,a=1,b=0,andc=�1.Sothereciprocalisw�1=w2�1.Exercise46.FindanirreduciblecubicpolynomialinZ2[x]toconstructa eldwitheightelements.Writedownamultiplicationtableforthat eld.Youcanleaveout0and1fromthetablesinceit'sobvioushowtheymultiply,buthavesixrowsandcolumnslabeleda=x,b=x+1,c=x2,d=x2+1,e=x2+x,andf=x2+x+1.3.11Number eldsandtheirringsofintegersInsection2.3anumber eldKwasde neda nite eldextensionoftherationalnumber eldQ.SomeexampleswereQ[p
2]andQ[p
�1]=Q[i].TheGaussianintegersZ[i]isanexampleofwhatiscalledaringofintegers.We'llseeinthissectionwhataringofintegersisandstudysomeoftheirproperties.Anumber eldwasde nedtobeanalgebraicextensionofQ,andanalgebraicintegerwasde nedinsection2.3.1tobearootofamonicpolynomialwithcoecientsintheintegers.Our rstgoalistoshowthatthesetofallalgebraicintegersinnumber eldK,thatsetbeingdenotedOKisasubringofK.Thatwilltakeafewsteps.De netheminimalpolynomialofanalgebraicintegeraisbeingthatmonicpolynomialfinQofminimaldegreesuchthatf(a)=0.Notethatwe'renotrequiringftohavecoecientsinZ,butwe'llprovethatbelow.Lemma3.70.TheminimalpolynomialofanalgebraicintegerdivideseverypolynomialinQ[x]ofwhichitisaroot.Proof.Letabeanalgebraicintegerwithminimalpolynomialf.Bythedivisionalgorithm,therearepolynomialsqandrinQ[x]suchthatg=qf+r,whereeitherr=0ordegrdegf.Thenr(a)=g(a)�q(a)f(a)=0,soaisarootofr.Sincefisthepolynomialofleastpositivedegreewithroot ,sor=0.q.e.d.Lemma3.71.TheminimalpolynomialofanalgebraicintegerhascoecientsinZ.
98CHAPTER3.RINGSProof.Letfbetheminimalpolynomialofa,andletgbeamonicpolynomialinZ[x]suchthatg(a)=0.Bythepreviouslemma,g=fhforsomeh2Q[x].Supposethatf=2Z[x],thensomeprimenumberpdividesthedenominatorofsomecoecientoff.Letpibethelargestpowerofpdividingthatdenominator,soi1.Letpjbethelargestpowerofpthatdividessomedenominatorofacoecientofh,withj0.Thenpi+jg=(pif)(pjh).Nowtakethatequationmodulop.Modulop,theleftsideis0,butneitherpolynomialontherightsideis0,acontradictionsinceZp[x]isanintegraldomain.q.e.d.***Theorem3.72.Thesetofallalgebraicintegers,OK,inanumber eldKisasubringofthat eld.ThisringOKiscalledtheringofintegersinthenumber eldK.Exercise47.ProvethattheGaussianintegersZ[i]istheringofintegersinthenumber eldQ[i].Sinceiistherootofthemonicpolynomialx1�1,allthat'sneededtoproveisthattherearenootherintegralelementsinQ[i]otherthanthoseinZ[i].
Chapter4GroupsRecallthatagroupisasetequippedwithonebinaryoperationthatisassociative,hasanidentityelement,andhasinverseelements.Ifthatbinaryoperationiscommutative,thenthegroupiscalledanAbeliangroup.4.1Groupsandsubgroups4.1.1De nitionandbasicpropertiesofgroupsWe'lllookatbasicpropertiesofgroups,andsincewe'lldiscussgroupsingeneral,we'lluseamultiplicativenotationeventhoughsomeoftheexamplegroupsareAbelian.De nition4.1.Theaxiomsforagroupareveryfew.AgroupGhasanunderlyingset,alsodenotedG,andabinaryoperationGG!Gthatsatis esthreeproperties.1.Associativity.(xy)z=x(yz).2.Identity.Thereisanelement1suchthat1x=x=x1.3.Inverses.Foreachelementxthereisanelementx�1suchthatxx�1=x�1x=1.Theorem4.2.Fromthesefewaxiomsseveralpropertiesofgroupsimmediatelyfollow.1.Uniquenessoftheidentity.Thereisonlyoneelementesuchthatex=x=xe,anditise=1.Outlineofproof.Thede nitionsaysthatthereisatleastonesuchelement.Toshowthatit'stheonlyone,supposeealsohasthepropertyofanidentityandprovee=1.2.Uniquenessofinverses.Foreachelementxthereisonlyoneelementysuchthatxy=yx=1.Outlineofproof.Thede nitionsaysthatthereisatleastonesuchelement.Toshowthatit'stheonlyone,supposethatyalsohasthepropertyofaninverseofxandprovey=x�1.3.Inverseofaninverse.(x�1)�1=x.Outlineofproof.Showthatxhasthepropertyofaninverseofx�1andusethepreviousresult.99
100CHAPTER4.GROUPS4.Inverseofaproduct.(xy)�1=y�1x�1.Outlineofproof.Showthaty�1x�1hasthepropertyofaninverseofxy.5.Cancellation.Ifxy=xz,theny=z,andifxz=yz,thenx=y.6.Solutionstoequations.Givenelementsaandbthereareuniquesolutionstoeachoftheequationsax=bandya=b,namely,x=a�1bandy=ba�1.7.Generalizedassociativity.Thevalueofaproductx1x2xnisnota ectedbytheplacementofparentheses.Outlineofproof.Theassociativityinthede nitionofgroupsisforn=3.Inductionisneededforn�3.8.Powersofanelement.Youcande nexnfornonnegativevaluesofninductively.Forthebasecase,de nex0=1,andfortheinductivestep,de nexn+1=xxn.Fornegativevaluesofn,de nexn=(x�n)�1.9.Propertiesofpowers.Usingthede nitionabove,youcanproveusinginductionthefollowingpropertiesofpowerswheremandnareanyintegers:xmxn=xm+n,(xm)n=xmn.Notethat(xy)ndoesnotequalxnyningeneral,althoughitdoesforAbeliangroups.4.1.2SubgroupsAsubgroupHofGisagroupwhoseunderlyingsetisasubsetoftheunderlyingsetofGandhasthesamebinaryoperation,thatis,forx;y2H,xHy=xGywhereHdenotesisthemultiplicationinHwhileGdenotesisthemultiplicationinG.Sincetheyarethesame,wewon'thavetosubscriptthemultiplicationoperation.AnalternatedescriptionofasubgroupHisthatitisasubsetofGthatisclosedundermultiplication,has1,andisclosedunderinverses.Ofcourse,Gisasubgroupofitself.AllothersubgroupsofG,thatis,thosesubgroupsthatdon'thaveeveryelementofGinthem,arecalledpropersubgroups.Also,f1gisasubgroupofG,usuallysimplydenoted1.It'scalledthetrivialsubgroupofG.Example4.3.ConsiderthecyclicgroupofsixelementsG=f1;a;a2;a3;a4;a5gwherea6=1.Besidesthetrivialsubgroup1andtheentiresubgroupG,therearetwoothersubgroupsofG.Oneisthe3-elementsubgroupf1;a2;a4gandtheotheristhe2-elementsubgroupf1;a3g.TheintersectionH\KoftwosubgroupsHandKisalsoasubgroup,asyoucaneasilyshow.Indeed,theintersectionofanynumberofsubgroupsisasubgroup.Theunionoftwosubgroupsisneverasubgroupunlessoneofthetwosubgroupsiscontainedintheother.Exercise48.Aboutintersectionsandunionsofsubgroups.(a).Showthattheintersectionoftwosubgroupsisalsoasubgroup.(b).Giveacounterexamplewheretheunionoftwosubgroupsisnotasubgroup.
4.1.GROUPSANDSUBGROUPS101Example4.4(SubgroupsofZ).ConsiderthegroupZunderaddition.AsubgroupofZhastobeclosedunderaddition,include0,andbeclosedundernegation.Besides0andZitself,whatarethesubgroupsofZ?Ifthesubgroupisnontrivial,thenithasasmallestpositiveelement,n.Butifnliesinasubgroup,thenallmultiples,bothpositiveandnegative,ofnalsomustbeinthesubgroup.Thus,nZisthatsubgroupofZ.Usefulsubgroupsofagroup.Thereareanumberofothersubgroupsofagroupthatareimportantinstudyingnonabeliangroupssuchasthecenterofagroupandthecentralizerofanelementofagroup.De nition4.5.Centerandcentralizer.ThecenterofagroupGisZ(G)=fx2Gjax=xaforalla2Gg,Fora2G,thecentralizerofaisZa(G)=fx2Gjax=xag.Exercise49.Showthefollowingpropertiesaboutcentersandcentralizers.(a).ProvethatZ(G)isasubgroupofG.(b).ProvethatthecenterofGistheintersectionofallthecentralizersubgroupsofG.(c).ProvethatZa(G)isasubgroupofG.De nition4.6(Commutatorsubgroup).ThecommutatoroftwoelementsxandyinagroupGistheelementx�1y�1xy.Itisdenoted[x;y].ThesubgroupofGgeneratedbyallthecommutatorsofitselementsiscalledthecom-mutatorsubgroupofG,denotedG0.NotethatforanAbeliangroup,allthecommutatorsare1,andthethecommutatorsubgroupistrivial.IfSisasubsetofG,thenthereisasmallestsubgrouphSiofGcontainingS.ItcanbedescribedastheintersectionofallsubgroupsHcontainingS,hSi=\SHH:Alternatively,itcanbedescribedasthesubsetofGofallproductsofpowersofelementsofS,hSi=fxe11xe22xennjn0;eachxi2S;andeachei2Zg:4.1.3CyclicgroupsandsubgroupsIfaisanelementofagroupG,thenthesubsetofGgeneratedbyahai=fanjn2ZgisasubgroupofG.ThissubgroupgeneratedbyaiscalledacyclicsubgroupofG.IfGitselfisgeneratedbysomeelementa,thenGiscalledacyclicgroup.De nition4.7(Orderandinvolution).TheorderofagroupGisthenumberofelementsinit,thatis,thecardinalityofitsunderlyingset.It'susuallydenotedjGj.
102CHAPTER4.GROUPSTheorderofanelementainagroupisthesmallestpositiveintegernsuchthatan=1.It'sdenotedorda.Ifeverypositivepoweran6=1,thentheorderofnisde nedtobe1.So,forexample,theorderof1is1since11=1.Aninvolutionaisanelementofagroupwhichisitsowninverse,a�1=a.Clearly,theorderofaninvolutionais2unlessa=1,inwhichcasetheorderofais1.Exercise50.Provethattheorderofaisalsoequaltotheorderofthecyclicgroup(a)generatedbya.Thatis,orda=jhaij.AnabstractcyclicgroupofordernisoftendenotedCn=f1;a;a2;:::;an�1gwhentheoperationiswrittenmultiplicatively.ItisisomorphictotheunderlyingadditivegroupoftheringZnwhereanisomorphismisf:Zn!Cnisde nedbyf(k)=ak.Exercise51.Provethatanysubgroupofacyclicgroupisitselfcyclic.Exercise52.LetGbeacyclicgroupofordernandaanelementofG.ProvethatageneratesG,thatis,hai=G,ifandonlyiforda=n.CyclicgroupsareallAbelian,sinceanam=am+n=aman.TheintegersZunderadditionisanin nitecyclicgroup,whileZn,theintegersmodulon,isa nitecyclicgroupofordern.Exercise53.ProvethateverycyclicgroupisisomorphiceithertoZortoZnforsomen.Exercise54.Provethatifkisrelativelyprimeton,thenkgeneratesZn.4.1.4ProductsofgroupsJustasproductsofringsarede nedcoordinatewise,soareproductsofgroups.Usingmul-tiplicativenotation,ifGandHaretwogroupsthenGHisagroupwheretheproduct(x1;y1)(x2;y2)isde nedby(x1x2;y1y2).TheidentityelementinGHis(1;1),andtheinverse(x;y)�1is(x�1;y�1).Theprojections1:GH!Gand2:GH!Haregroupepimorphismswhere1(x;y)=xand2(x;y)=y.Also,1:G!GHand2:H!GHaregroupmonomorphismswhere1(x)=(x;1)and2(y)=(1;y).Thus,wecaninterpretGandHassubgroupsofGH.NotethatGandHarebothAbeliangroupsifandonlyifGHisanAbeliangroup.TheproductoftwoAbeliangroupsisalsocalledtheirdirectsum,denotedGH.TheunderlyingadditivegroupofaringisanAbeliangroup,andsomeoftheresultswehaveforringsgiveustheoremsforAbeliangroups.Inparticular,theChineseremaindertheoremforcyclicringsZngivesusatheoremforcyclicgroupsCn.Theorem4.8(Chineseremaindertheoremforgroups).Supposethatn=kmwherekandmarerelativelyprime.ThenthecyclicgroupCnisisomorphictoCkCn.Moregenerally,ifnistheproductk1krwherethefactorsarepairwiserelativelyprime,thenCn=Ck1Ckr=rYi=1Cki:Inparticular,iftheprimefactorizationofnisn=pe11perr.ThenthecyclicgroupCnfactorsastheproductofthecyclicgroupsCpeii,thatis,Cn=rYi=1Cpeii:
4.1.GROUPSANDSUBGROUPS1034.1.5CosetsandLagrange'stheoremCosetsareusefulindevelopingthecombinatoricsof nitegroups,thatis,forcountingsub-groupsandotherthingsrelatedtoa nitegroup.Theycomeinbothleftandrightformsasyou'llseeinthede nitionbelow,butwe'llonlyuseleftcosets.Our rstcombinatorialtheoremiscalledLagrange'stheoremwhichsaysthattheorderofasubgroupdividestheorderofagroup.Sincethesubgrouphaigeneratedbyasingleelementhasanorderthatdividestheorderofthegroup,thereforetheorderofanelementdividestheorderofthegroup,too.We'llhaveour rstclassi cationtheoremasacorollary,andthatisthatagroupwhoseorderisaprimenumberiscyclic.Thus,uptoisomorphism,thereisonlyonegroupofthatorder.De nition4.9.LetHbeasubgroupofG.AleftcosetofHisasetoftheformaH=fahjh2HgwhilearightcosetisoftheformHa=fhajh2Hg.Theorem4.10.Severalpropertiesofcosetsfollowfromthisde nition.1.Thecoset1HisjustthesubgroupHitself.Infact,ifh2HthenhH=H.2.Moregenerally,aH=bHifandonlyifab�12H.Thus,thesamecosetcanbenamedinmanydi erentways.3.Cosetsaredisjoint.IfaH6=bH,thenaH\bH=?.Outlineofproof.It'sprobablyeasiertoshowthecontrapositive:ifaH\bH6=?thenaH6=bH.Supposeanelementisintheintersection.Thenitcanbewrittenasahorasbh0wherebothhandh0areelementsofH.Therestreliesonthepreviousstatement.4.CosetsofHallhavethesamecardinality.Outlineofproof.Checkthatthefunctionf(ah)=bhisabijectionaH!bH.5.Thus,thecosetsofHpartitionGintosubsetsallhavingthesamecardinality.6.Lagrange'stheorem.IfGisa nitegroup,andHasubgroupofG,thenjHjdividesjGj.Moreover,jGj=jHjisthenumberofcosetsofH.Outlineofproof.Followsfromtheprecedingstatement.De nition4.11.TheindexofasubgroupHofagroupGisthenumberofcosetsofH.Theindexisdenoted[G:H].ByLagrange'stheorem,[G:H]=jGj=jHjwhenGisa nitegroup.Corollary4.12.Iftheorderofagroupisaprimenumber,thenthegroupiscyclic.Proof.LetjGj=p,aprime.Sincephasnodivisorsexcept1andp,therefore,byLagrange'stheorem,Gonlyhasitselfandthetrivialsubgroupasitssubgroups.Leta6=1beanelementofG.Itgeneratesacyclicsubgrouphaiwhichisn'ttrivial,sohai=G.ThusGiscyclic.q.e.d.
104CHAPTER4.GROUPSCorollary4.13.Ifagroupis nite,thentheorderofeveryelementdividestheorderofthegroup.Proof.Letabeanelementofa nitegroupG.ThentheorderofthesubgrouphaidividesjGj.Butordaistheorderofhai.ThereforeordadividesjGj.q.e.d.Productsofsubsetsinagroup.Occasionallywe'llwanttolookatproductsHKofsubsetsHandK,especiallywhenHandKaresubgroupsofagroupG.Thisproductisde nedbyHK=fxyjx2H;y2Kg:EvenwhenHandKaresubgroups,itisn'tnecessarythatHKisasubgroup,butthereisasimplecriteriontotestifitis.Abeliangroupsareoftenwrittenadditively.Inthatcase,ratherthanusingthenotationHK,thenotationH+Kispreferred:H+K=fx+yjx2H;y2Kg.Theorem4.14.LetHandKbesubgroupsofG.ThenHKisalsoasubgroupofGifandonlyifHK=KH.Proof.=):SupposethatHKisasubgroup.We'llshowthatKHHK.Letxy2KHwithx2Kandy2H.Sincex=1x2HKandy=y12HK,thereforetheirproductxyisalsoinHK.Thus,KHHK.LikewiseHKKH.ThereforeHK=KH.(=:SupposethatHK=KH.Toshowit'sasubgroup, rstnote12HKsince12Hand12K.Second,we'llshowthatHKisclosedundermultiplication.Letx1y1andx2y2beelementsofHKwithx1;x22Handy1;y22K.Theny1x22KH=HK,soy1x2=x3y3wherex32Handy32K.Therefore,(x1y1)(x2y2)=(x1x3)(y3y2)2HK.Third,we'llshowthatHKisclosedunderinverses.Letxy2HKwithx2Handy2K.Then(xy)�1=y�1x�12KH=HK.q.e.d.Corollary4.15.IfHandKaresubgroupsofanAbeliangroupG,thenH+KisalsoasubgroupofG.4.2SymmetricGroupsSnWe'velookedatseveralexamplesofgroupsalready.It'stimetoexaminesomeinmoredetail.4.2.1PermutationsandthesymmetricgroupDe nition4.16.ApermutationofasetXisjustabijection:X!Xonthatset.ThepermutationsonXformagroupcalledthesymmetricgroup.We'reprimarilyinterestedinpermutationsona niteset.We'llcalltheelementsofthe nitesetletters,butwe'lldenotethemwithnumbers.Thesymmetricgrouponnelements1;2;:::;nisdenotedSn.NotethattheorderofthesymmetricgrouponnlettersisjSnj=n!.
4.2.SYMMETRICGROUPSSN105Example4.17.ConsiderthepermutationonsetX=f1;2;3;4;5;6gthatexchanges2with4,sends1to3,3to5,and5to1,andleaves6 xed.Youcandescribeinatablelikethis:
n
123456
(n)
345216
Thattablehasalotofredundantinformation.The rstrowisjustthenamesoftheelements.TodescribeonanorderedsetlikeX,it'senoughtolisttheelementsinthesecondrow:3,4,5,2,1,6.Unfortunately,thatmakesitharderto gureoutwheresendsanelement.Thecyclenotation,mentionednext,iscompactandmakesiteasiertoseehowacts.For,thisnotationwilllooklike(135)(24).Thethreeelementsforma3-cycle17!37!57!1ofdenoted(135).Alsonote27!47!2,so(24)isa2-cycleof.Anothernamefora2-cycleistransposition.Since(6)=6,therefore(6)byitselfisa1-cycle,alsocalleda xedpoint,of.Thecyclenotationforthispermutationis=(135)(24).Notethat xedpointsarenotdenotedinthisnotation.Alternatively,thispermutationcouldbedenoted(24)(135),or(531)(42),orseveralothervariants.Since xedpointsaren'tdenotedincyclenotation,we'llneedaspecialnotationfortheidentitypermutationsinceit xesallpoints.We'lluse1todenotetheidentitysincewe'reusing1todenotetheidentityinagroupwrittenmultiplicatively.Inmanytextbookstheidentityisdenotede.There'sabitofexperienceneededtoquicklymultiplytwopermutationstogetherwhenthey'reincyclenotation.Let=(146)(23)and=(15)(2643).Bymean rstperformthepermutationthenperform(inotherwords,thecompositionifwethinkofthesepermutationsasfunctions).Thenweneedsimplifythecyclenotation=(146)(23)(15)(2643):Notethat rstsends1to4,thensends4to3,thereforesends1to3.Next37!27!6,so37!6,likewise67!17!5,so67!5,and57!57!1,so57!1.Thus,wehaveacycleof,namely,(1365).Youcancheckthat(2)and(4)are xedpointsof.Thus,wefoundtheproduct.(146)(23)(15)(2643)=(1365).Incidentally, ndingtheinverseofapermutationincyclenotationisveryeasy|justreverseallthecycles.Theinverseof=(146)(23)is�1=(641)(32).SmallsymmetricgroupsWhenn=0orn=1,there'snothinginthesymmetricgroupexcepttheidentity.Thesymmetricgroupontwoletters,S2,hasonenontrivialelement,namely,thetranspo-sition(12).Thisisthesmallestnontrivialgroup,andit'sisomorphictoanygroupoforder2.Itis,ofcourse,anAbeliangroup.Thesymmetricgrouponthreeletters,S3,hasorder6.Wecannameitselementsusingthecyclenotation.1;(12);(13);(23);(123);(132)Besidestheidentity,therearethreetranspositionsandtwo3-cycles.ThisisnotanAbeliangroup.Forinstance(12)(13)=(123),but(13)(12)=(132).
106CHAPTER4.GROUPSThesymmetricgrouponfourletters,S4,hasorder24.Besidestheidentity,thereare�42=6transpositions,�432=83-cycles,64-cycles,and3productsoftwo2-cycles,like(12)(34).Exercise55.Completethefollowingtablelistingall24oftheelementsofS4.
theidentity
1
transpositions
(12);(13);(14);(23);(24);(34)
3-cycles
4-cycles
productsof2transpositions
4.2.2EvenandoddpermutationsFirstwe'llnotethateverycycle,andthereforeeverypermutation,canbeexpressedasaproductoftranspositions.We'llsoonseeafterthatthatapermutationcaneitherbeexpressedasaproductofanevennumberoftranspositionsorasaproductofanoddnumberoftranspositions,butnotboth.Thatwilljustifythede nitionofevenandoddpermutations.Theorem4.18.Anycyclecanbeexpressedasaproductoftranspositions.Proof.Thecycle(a1a2a3ak)istheproduct(a1a2)(a1a3):::(a1ak).q.e.d.We'lllookataninvariantthatwillhelpusdistinguishevenfromoddpermutations.ItisPn,theproductofalldi erencesoftheformi�jwhere0ijn.Pn=Y0n(i�j)=(1�2)(1�3)(1�n)(2�3)(2�n)((n�1)�n)Lemma4.19.Thee ectofapplyingatranspositiontotheintegersthatmakeupPnistochangethesignofPn.Proof.Letthetranspositionbe(ab)where0abn.TheproductPnismadeofthreefactorsPn=P0P00P000whereP0=(a�b),P00istheproductoffactorsthathaveeitheraorbbutnotboth,andP000istheproductoffactorsthatdon'thaveeitheraorb.Nowthetransposition(ab)hasnoe ectatallonP000butnegatesP0.Itse ectonP00ismorecomplicated.Supposecisanotherletter.Case1.cab.Thefactors(c�a)and(c�b)ofP00areinterchangedbythetransposition(ab).Case2.acb.Thefactors(a�c)and(c�b)areinterchangedandbothnegated.Case3.abc.Likecase1.ThusP00doesnotchangeitsvalue.SinceonlyP0isnegated,Pnisnegated.q.e.d.Theorem4.20.Apermutationiseithertheproductofanevennumberoftranspositionsortheproductofanoddnumberoftranspostions,butitcan'tbeboth.
4.2.SYMMETRICGROUPSSN107Proof.SinceeachtranspositionnegatesPn,theproductofanevennumberoftranspositionsleavesPnalone,buttheproductofanoddnumberoftranspositionsnegatesPn.Itcan'tbebothsincePnisnot0.q.e.d.De nition4.21.Apermutationisevenifit'stheproductofanevennumberoftranspo-sitions,it'soddifit'stheproductofanoddnumberoftranspositions.Theidentity1isanevenpermutation.Notethatacycleisanevenpermutationifithasanoddlength,butit'sanoddpermu-tationifithasanevenlength.Also,theproductoftwoevenpermutationsiseven,theproductoftwooddsiseven,andtheproductofanevenandanoddisodd.Examples4.22.ThesymmetricgroupS3hasorder6.It'selementsare1,(12),(13),(23),(123),and(132):Threeofthem,namely1,(123),and(132)areevenwhiletheotherthree(12),(13),and(23)areodd.ThesymmetricgroupS4has12evenpermutations(theidentity,eight3-cycles,andthreeproductsoftwo2-cycles)and12oddpermutations(sixtranspositionsandsix4-cycles).4.2.3AlternatinganddihedralgroupsDe nition4.23(ThealternatinggroupAn).Sincetheproductofevenpermutationsiseven,andtheinverseofanevenpermutationiseven,thereforethesetofevenpermutationsinthesymmetricgroupSnisasubgroupofSn.Itiscalledthealternatinggrouponnletters,denotedAn.Forn2,thenumberofevenpermutationsinSnisthesameasthenumberofoddpermutations,sincemultiplyingbythetransposition(12)setsupthebijection.Therefore,theorderofAnishalftheorderofSn.SojAnj=1
nn!.Example4.24(SubgroupsofS3).ThesymmetricgroupS3onlyhassixelements,soitdoesn'thavemanyelements.There'sthetrivialsubgroup1oforder1.Therearethreecyclicsubgroupsoforder2eachisomorphictoC2;besides1,theotherelementisoneofthetranspo-sitions(12),(13)or(23).There'sonesubgroupoforderthree,namely,D3=f1;(123);(132)g.(NotethatA3isthesamegroupasD3.TheHassediagramforthesubgroupsisfairlysimple.
S3
A3
C2
C2
C2
1
Figure4.1:SubgroupsofS3
108CHAPTER4.GROUPSExample4.25(ThedihedralgroupD5).ThedihedralgroupsDnarethesymmetrygroupsofregularn-gons.Wealreadylookedatthecasen=3ofanequilateraltriangle.Consideraregularpolygonwithn=5vertices.1
r1r2r3r4r5
r5r1r2r3r4'
r1r5r4r3r2Figure4.2:SymmetriesofapentagonWecanlabeltheverticesinorderfrom1ton.Asymmetryofaplane gureisatrans-formationoftheplanethatmapsthe guretoitself.We'reonlyinterestedinisometries,transformationsthatpreservedistance,rightnow,butothertransformationshavetheirap-plications,too.Figure4.2showsshowsapentagon.(Thepentagonshownhereisinthehyperbolicplane,butthatdoesn'tmatter.)Oneofitssymmetriesistheonethatrotatesthepentagon72counterclockwise.Itmapsthevertexlabelled1to2,maps2to3,andsoforth.Knowingwheretheverticesaremappedisenoughtodeterminethetransformation,sowecanidentifywiththepermutationitdescribesonthesetofvertices.Thisisthepermutation(12345).Anotherofthesymmetriesofthepentagonisare ectionlike'shownabove,are ectionacrossahorizontalaxis.Incyclenotation'=(25)(34).Infact,thereare10symmetriesoftheregularpentagon,sojD5j=10.IngeneraljDnj=2n.InD5,besidestheidentity,therearefourrotationsand vere ections.identity=1=(12345)2=(13524)3=(14253)4=(15432)'=(25)(34)'=(12)(35)'2=(13)(45)'3=(14)(23)'4=(15)(24)Therearenomoresymmetriesalthoughwecanwritemoreexpressionsintermsof'and,forinstance'.But'=(15)(24)whichis'4.Thus,wecanseenowhowtorepresentthedihedralgroup,D5,asasubgroupofthesymmetricgroupS5.Infact,it'srepresentedasasubgroupofthealternatinggroup,A5aswell,sinceallthepermutationsareevenpermutations.Example4.26(Symmetriesofacubeandtetrahedron).Consideracubewithvertices120340102304andtheinscribedregulartetrahedron1234shownin gure4.3.Thefourdiagonalsofthecube,110,220,330,and440,aredrawningreen.Therearemanysymmetriesofatetrahedron.Theypermutethevertices1234.Therearerotationsof120and240aboutanyofthefourdiagonals.Thoserotationsabouttheline110arethepermutations(234)and(243).Therotationsabouttheotherthreediagonalsare(123),(132),(124),(142),(134),and(143).Besidestheserotations,therearethree180rotationsaboutthethreelinesjoiningthemidpointsoftheoppositeedgesofthetetrahedron.Alongwiththeidentity,thatmakes12permutations,allofwhichpreserveorientation,that
4.2.SYMMETRICGROUPSSN109is,they'rerigidmotions.ThegroupoforientationpreservingsymmetriesofthetetrahedronformthegroupS4.Besidesthese,therearesymmetriesofthetetrahedronwhicharere ectionsacrossplanes.Theyareorientationreversingsymmetries.Forexample,there ectionacrosstheplanepassingthroughvertices1and2andthemidpointofedge34leavesvertices1and2 xedbutitexchangesvertices3and4;it'sthetransposition(34).Thegroupofallthesymmetriesofthetetrahedron,includingboththeorientationpreservingsymmetriesandtheorientationreversingones,formthegroupS4.Eachofthesymmetriesofthetetrahedrongivesasymmetryoftheenclosingcube.Thesymmetriesofacubepermuteitseightvertices123410203040.Forexample,thesymmetry(123)ofthetetrahedrongivesthesymmetry(123)(102030)ofthecube.Butthereareothersymmetriesofacubesincethetetrahedron1234doesn'thavetobepreservedunderasymmetryofthecube;itcouldbesenttotheoppositetetrahedron10203040.Otherorientationpreservingsymmetriesthatsendthetetrahedrontotheoppositetetrahedronincludethesix90and270rotationsaboutthecentersofthefacesandthesix180rotationsaboutthelinejoiningmidpointsofoppositesides.Thatmakes24orientationpreservingsymmetriesforthecube.Eachonepermutesthefourdiagonals,andnotwoofthempermutethefourdiagonalsinthesameway,sothissymmetrygroupisS4.Notethatthesymmetry(110)(220)(330)(440)thatexchangesavertexwithitsoppositevertexreversesorientation.Theentiregroupofsymmetriesofthecubeincludesthe24orientationpreservingsym-metriesandeachofthosetimes(110)(220)(330)(440).Thatmakes48symmetriesofthecube.
1
20
3
40
10
2
30
4
Figure4.3:SymmetriesofacubeandtetrahedronExercise56.Verifythestatementsmadeintheexample.(a).TheorientationpreservingsymmetriesofatetrahedronformthegroupA4.(b).ThegroupofallthesymmetriesofthetetrahedronformthegroupS4.
110CHAPTER4.GROUPS(c).TheorientationpreservingsymmetriesofatetrahedronformthegroupS4.(c).Explainwhythepermutation(110)(220)(330)(440)ofthecubereversesorientation.Presentationsbygeneratorsandrelations.Althoughit'snicetohaveagrouprepre-sentedinasymmetricgroup,sometimesit'smoreconvenienttodescribeitmorealgebraicallyintermsofgeneratorsandrelations.ForD5wecanseethatand'aresucienttogen-eratethewholegroupinthesensethateveryelementinthegroupcanbewrittenassomeexpressioninvolvingand'.Buttherearecertainrelations,actuallyequations,thatand'satisfyinthisgroup,namely5=1,2=1,and'='�1.Thus,wecanpresentthegroupasD5=h;':5=1;2=1;'='�1i:Thedicultywithapresentationofthistypeisknowingwhenyouhaveenoughgeneratorsandrelations.Ifyoudon'thaveenoughgenerators,youwon'tgeneratethewholegroup.Ifyoudon'thaveenoughrelations,you'llgeneratealargergroup,butnottheoneyouwant.Aproofneedstobesuppliedtobeassuredthatthisistherightpresentation.Frequently,adiagramofsomesort llsthebill.4.3Cayley'stheoremandCayleygraphsOneofthereasonssymmetricgroupsaresoimportantisthateverygroupisisomorphictoasubgroupofasymmetricgroup,aresultofCayley.Thisgivesusanotherwaytolookatgroups,especiallysmall niteones.We'llproveCayley'stheorem,thenlookatafewCayleygraphswhichdependonCayley'stheorem.4.3.1Cayley'stheoremRecallthatapermutationofasetXisjustabijection:X!Xonthatsetandpermu-tationsonXformagroupcalledthesymmetricgroupS(X).Whenthesetis nite,wecanwriteitasf1;2;:::;ng,andSndenotestheitssymmetricgroup.Cayley'stheoremcanbestatedforin nitegroupsaswellas nitegroups.Theorem4.27(Cayley).LetGbeagroup,andletS(G)bethesymmetricgrouponG,thatis,thegroupofpermutationsontheunderlyingsetofG.Thefunction':G!S(G)de nedby'(a)(x)=axisagroupmonomorphism.Therefore,GisisomorphictoasubgroupofS(G).Proof.'(a)isthepermutationonGthatmapsxtoax.It'sabijectionsinceitsinversesendsxtoa�1x.Toshowthatit'sagrouphomomorphism,itisonlynecessarytoshowthat'(ab)='(a)'(b)foraandbinG.But'(ab)(x)=abx,and('(a)'(b))(x)='(a)('(b)(x))='(a)(bx)=abx.Finally,':G!S(G)isamonomorphismsinceif'(a)='(b),thenevaluatingthetwopermutationsat1givesa1=b1,soa=b.q.e.d.Althoughthisrepresentationtheoremdoesshowthateverygroupisasubgroupofasymmetricgroup(uptoisomorphism),it'spracticallynotallthatusefulsinceifthegroupGhasordern,it'sbeingrepresentedinagroupofordern!,whichismuchtoolargetodealwithifnisatalllarge.Still,it'sausefulrepresentationfortheoreticalpurposes.
4.3.CAYLEY'STHEOREMANDCAYLEYGRAPHS111Cayleygraphs.WithaCayleygraphwecanrepresentagroupGbyagraphwithverticesandlabeled,directededges.EachelementofGisavertexofthegraph,andforeachelementa,wealsohaveadirectededgelabeledafromavertexxtothevertexax.Inotherwords,theCayleygraphisarepresentationofGbytheCayleytheoremtoS(G).Forasmallexample,letGbethecyclicgroupG=f1;a;bgwherea2=banda3=1.TheCayleygraphforGhasthreevertexes,labeled1,a,andb.Eachnodehasalooponitlabeled1since1x=x.Therearethreeedgeslabelleda,1a!aa!ba!1,andthreeedgeslabelledb,1b!bb!ab!1.Thisisprobablymostconvenientlydrawninatriangular gure.There'salotofredundancyinthegraphinthesensethatyoudon'tneedalltheinfor-mationtoreconstructthegroup.Theloopslabelled1mightjustaswellbedroppedsinceforanygroup1x=x.Ifweknowtheedgeslabelleda,thenwecandeterminetheedgeslabelledbsinceyoujusttraveltwoa-edgestogetab-edge.Thatleavesjustthetriangle1a!aa!ba!1.Moregenerally,ifyouknowtheedgesforgeneratorsofagroup,thenalltheotheredgesaredetermined.Example4.28(D5).RecallthatthedihedralgroupD5has10elementsandthepresentationD5=h;':5='2=(')2=1i:The rstrelation,5=1givesusa vecycle1!!2!3!4!1whichwecandrawasapentagon,thecenterpentagoninthegraphbelow.Thesecondrelation,'2=1,meanswehavethe2-cycle1'!''!1,and,moregenerally,foranyelementa,wehavea2-cyclea'!a''!a.We'lldraw2-cyclesasundirectededgesa'�a.Weget veoftheseedges,oneateachvertexofthecenterpentagon.Thethirdrelation,(')2=1,describesasquarea'!a'!a''!a''!a:Startingateachofthenewouterverticesofthegraph,followthreeedgestoreachanotheroutervertex,anddrawa-edgebacktowhereyoustarted.Whenyou nish,youhavetheCayleygraphforD5in gure4.4 
 
-r1rr2r3r4
r'r'r2'r3'r4'
'
'
'
'
'Figure4.4:CayleygraphforD5
112CHAPTER4.GROUPSNoticethatthegraphiscompletelysymmetric.Youcouldlabelanyvertex1and llinthenamesoftherestoftheverticesbyfollowingthelabelledarcs.Forthatreason,theverticesofaCayleygraphneedn'tbelabelled.ThereisanotherpresentationforD5thatgivesadi erentlookingCayleygraph.Let =.ThenD5=h'; :'= 2=(' )5i:TheCayleygraphhasthesametenvertices,buttheedgesareallundirectedandtheyformacycleoflength10withlabelsalternatingbetween'and .Example4.29(A4).Recallthatthealternatinggrouponf1;2;3;4ghas12elements.It'snotcyclic,soatleasttwogeneratorsarerequiredtogenerateit.Infact,twowilldo.Considerthethreeelementsa=(123)b=(124)c=ab=(14)(23)ThetwoelementsaandbaresucienttogenerateA4asarethetwoelementsaandcandmanyotherpairsofelements(butnotallpairswilldo).Infact,A4canberepresentedineitherofthefollowingtwoways:ha;b:a3=b3=(ab)2=1iha;c:a3=c2=(ac)2=1iSo,ifwehavetheCayleygraphwithonlya-andb-edges,thenwehaveenoughinformationtodetermineA4,orifwehavethegraphwithonlya-andc-edges,thenthat'senough.Althoughthesetwographsbothhave12vertices(sincejA4j=12),theydon'tlookverymuchalike.Let'slookattheCayleygraphwithallthreekindsofedges,a-edgesandb-edgesandc-edges.It'sdisplayedin gure4.5asaplanargraph,butmoreofthesymmetrywouldbeappar-entifitweredisplayedinthreedimensionswheretheverticesandedgeswerethoseofanicosahedron.Someofthetrianglesinthe gureareblue.Theirthreesidesofthetrianglearea-edges.Likewise,sometrianglesaregreenwithb-edges.Notethatallthea-andb-trianglesareorientedcounterclockwiseexcepttheouterb-triangle.Theremainingedgesaretheredc-edges,andtosavespace,sincecisaninvolution,ratherthanputtingintwoedges,onepointingonewayandtheotherpointingtheotherway,justasinglethickundirectededgeisincluded.Eachvertexinthegraphhasana-edgecominginandonecomingout,ab-edgecominginandonecomingout,andanundirectedc-edgemeaningthatitgoesbothinandout.SinceitonlytakestwoofthesethreeelementstogenerateA4,thisgraphhassuper uousinformation.Alltheedgeslabelledbyoneoftheletterscanberemovedmakingthegraphsimpler.Exercise57.FindaCayleygraphforthesymmetricgroupS4.Therearevariouspairsortriplesofgeneratorsyoucanuse.Oneisthepaira=(1234);b=(12).4.3.2Somesmall nitegroupsWe'veseenafewfamiliesof nitegroupsincludingCnthecyclicgroupofordern,Dnthedihedralgroupoforder2n,Snthesymmetricgroupofordern!,andAnthealternatinggroupofordern!=2.
4.3.CAYLEY'STHEOREMANDCAYLEYGRAPHS113
Figure4.5:CayleygraphforA4Theclassi cationof nitegroups(uptoisomorphism,ofcourse)isextremelydicult.DanielGorenstein(1923{1992)wasaleaderofmathematicianswhoeventuallyclassi ed nitesimplegroups.HewasfacultymemberatClarkUniversityfor13years.We'lllookatafewmoresmall nitegroups.Later,we'lllookattheclassi cationof niteAbeliangroups,and ndthatthey'reallproductsofcyclicgroups.Table4.1liststhesmallgroupsuptoisomorphismoforderupthrough24.order123456789101112131415numberofgroups111212152215121Wewon'tprovethattheseareallofthem,butwewilllookatthemall.Therearecombina-torialtheorems,themostimportantbeingtheSylowtheorems,thathelpinclassifying nitegroups.Weknownearlyallofthese27groups.ThecyclicgroupsCnaccountfor15ofthem.Thereare12others.Someofthemareproductsofsmallerones,forinstance,theothergroupoforder4isC2C2,sometimescalledtheKlein4-group.Thesecondgroupoforder6isD3,whichisthesameasS3.Twoofthegroupsoforder8areproducts,namely,C4C2andC2C2C2.AnotherisD4andtheremainingoneiscalledthequaterniongroup.Example4.30(Thequaterniongroup).ThisgroupconsistsofeightoftheunitsofthedivisionringH,thequaternions.LetQ=f1;i;j;kg.Recallthatthemultiplicationofquaternionshasi2=j2=k2=�1,ij=k,jk=i,andki=j,sothissetofunitsisclosedundermultiplicationandformsagroup,calledthequaterniongroup.Exercise58.ConstructaCayleygraphforthequaterniongroup.Thesecondgroupoforder9istheAbeliangroupC3C3,andthesecondgroupoforder10isD5.Wealreadyknowtheothergroupsoforder12:D6,C2C6,D3C2,andA4,and
114CHAPTER4.GROUPS
Order
1234
5678
Abeliangroups
C1C2C3C4
C5C6C7C8
C2C2
C4C2
C2C2C2
Non-Abelian
D3D4
Q
Order
9101112
13141516
Abeliangroups
C9C10C11C12
C13C14C15C16
C3C3C2C6
C2C8
C2C2C4
C2C2C2C2
C4C4
Non-Abelian
D5D6
D7D8
A4
8others
C3C4
Order
17181920
21222324
Abeliangroups
C17C18C19C20
C21C22C23C24
C3C6C2C10
C2C12
C2C2C6
Non-Abelian
D9D10
D11D12
S3C3Dic5
C7oC311others
C23oC2C5oC4
Table4.1:Listofsmallgroups
4.4.THECATEGORYOFGROUPSG115theothergroupoforder14isD7.4.4ThecategoryofgroupsGThecategoryGofgroupswasmentionedbrie yinsection3.5whenthecategoryofringswasintroduced.Theobjectsinthiscategoryaregroups,andthemorphismsaregrouphomomorphisms.Productsinacategoryarede nedbyauniversalpropertyratherthanbyorderedpairs,butinG,theproductoftwogroupsiswhatwascalledthetheproductofgroupsinsection4.1.4:GH=f(x;y)jx2G;y2HgOthercategoricalconceptsincludetheinitialand nalobject.InGthesearethesamegroup,namelythetrivialgroupthathasonlyoneelement1.Thereisauniquegrouphomomorphismfromeachgroupto1,andthere'sauniquegrouphomomorphismfrom1toeachgroup.Theuniversalpropertyofanin nitecyclicgroupinthecategoryofgroups.TheadditionoperationonZmakesitanin nitecyclicgroupsinceeachelementinitisamultipleof1.YoucanalsowriteitmultiplicativelyasC1=hai=f:::;a�2;a�1;1;a;a2;:::g.Thisin nitecyclicgrouphasthefollowinguniversalproperty.GivenanygroupGandanyelementc2G,thereisauniquegrouphomomorphismhai!Gthatmapsatoc.Ingeneral,itmapsantocn.TheimageofthishomomorphismisthesubgroupofGgeneratedbyc.Freegroups.Thein nitecyclicgrouphaiisaspecialcaseofafreegroup.It'safreegroupononeelement.Therearefreegroupsonmorethanoneelementwithanalogousuniversalproperties.We'lllookatthefreegroupontwoelements.Letaandbbetwosymbols.Formthegroupha;biasfollows.Anelementinitisnamedbyastringofa'aandb'sraisedtovariousintegralpowers,suchasba�3b2b4a2.Di erentnamesaretobeconsideredtonamethesameelementifadjacentsymbolsarethesame,inwhichcase,theycanbecombinedbytheusualpowerrule.Forexample,ba�3b2b44a2=ba�3b6a2.Also,anysymboltothepower0istobetreatedastheidentityelement1,and1timesanystringsimpli estothatstring.Aformalproofthatha;biis,infact,agrouprequiresinduction.We'llomitthatproof.Thisgroupha;bihasthefollowinguniversalproperty.GivenanygroupGandanyele-mentsc;d2G,thereisauniquegrouphomomorphismha;bi!Gthatmapsatocandbtod.4.5ConjugacyclassesandquandlesWe'llconsideranotherwaytoexaminethestructureofgroups.Thatdependsonanalyzingtheoperationofconjugationinagroup.De nition4.31(Conjugateelementinagroup).IfxandyareelementsofagroupG,thenyxy�1iscalledconjugatesofx.Inthatcase,elementsyandyxy�1aresaidtobeconjugates.
116CHAPTER4.GROUPSExercise59.Showthatbeinganconjugateinagroupisanequivalencerelationbyprovingthat(a)anyelementisconjugatetoitself,(b)ifoneelementisconjugatetosecond,thenthesecondisconjugatetothe rst,and(c)ifoneelementisconjugatetoasecondandthesecondconjugatetoathird,thenthe rstisconjugatetothethird.4.5.1ConjugacyclassesSincebeingconjugatesinagroupisanequivalencerelation,thecorrespondingequivalenceclassescansayalotaboutthegroup.De nition4.32.Eachoftheequivalenceclassesofagroupunderconjugacyiscalledaconjugacyclass,andthesetofallconjugatesofaparticularelementxiscalledtheconjugacyclassofx.Exercise60.Ifxisanelementoforderninagroup,showeveryconjugateofxalsohasordern.Example4.33(Conjugacyclassesinsymmetricgroups).Conjugationandconjugacyclassesinsymmetricgroupsareparticularlyeasytoidentifyusingcyclenotation.Letx=(13)(245)andy=(142)betwoelementsinSn.Theny�1xy=(124)(13)(245)(142)=(43)(125).Notehowyconjugatesthecycle(13)tothecycle(43),anditconjugatesthecycle(245)to(125).Thecyclestructuresforxandy�1xyarethesame,buttheelementsinthecyclesarepermutedbyy.Thisisgenerallythecaseforsymmetricgroups.ItfollowsthataconjugacyclassinSnconsistsofalltheelementsinSnwithagivenstructure.Thus,forexample,theconjugacyclassof(13)(235)consistsofallelementsoftheform(ab)(cde)wherea;b;c;d,andeare5distinctintegersbetween1andn.ForS5thesizeofthatconjugacyclassis�522=20.Exercise61.DeterminealltheconjugacyclassesofS5andtheirsizes.(Thesumoftheirsizeswillequal120,ofcourse.)Theorem4.34.IfHisasubgroupofG,andx2G,thenxHx�1isalsoasubgroupofG,calledasubgroupconjugatetoH.Proof.First,12xHx�1sincex1x�1=1.Next,ifxyx�1;xzx�12xHx�1withy;z2H,thentheirproductxyx�1xzx�1=x(yz)x�12xHx�1.Finally,givenxyx�12xHx�1withy2H,thentheinverse(xyx�1)�1=xy�1x�12xHx�1.Therefore,xHx�1isasubgroupofG.q.e.d.Similarlytotheargumentintheexerciseabove,beingconjugatesubgroupsofagivengroupisanequivalencerelation.Theorem4.35.IfnoothersubgroupofGhasthesameorderasH,thenHisnormal.Proof.SinceanyconjugatesubgroupxHx�1isinone-to-onecorrespondencewithH,ithasthesamenumberofelements,somustequalH.q.e.d.Exercise62.IfHisasubgroupofGandNisanormalsubgroupofG,provethatH\NisanormalsubgroupofH.
4.5.CONJUGACYCLASSESANDQUANDLES117Exercise63.IfHisasubgroupofGandNisanormalsubgroupofG,provethatHNisasubgroupofG.(Hint:showHN=NH.)Exercise64.Provethattheintersectionoftwonormalsubgroupsisalsoanormalsubgroup.Exercise65.ProvethatifHandNarenormalsubgroupsofG,thentheirproductisalsoanormalsubgroupofG,infact,it'sthesubgroupgeneratedbyH[N.4.5.2QuandlesandtheoperationofconjugationTheoperationsofconjugationhavecertainproperties.Ifwethinkofy�1xyasabinaryoperationx.y,andyxy�1asanotheroperationx.-1y,thenthesetwooperationssatisfythepropertiesstatedinthenextde nition.De nition4.36.Aquandleisasetequippedwithtwooperations,.and.-1satisfyingthefollowingthreeconditionsforallelementsx,y,andz.Q1:x.x=x:Q2:(x.y).-1y=x=(x.-1y).y:Q3:(x.y).z=(x.z).(y.z):Thesymbol.ispronounced\through",and.-1\backthrough".Exercise66.ProvethatifQisaconjugacyclassinagroupGthenQisaquandlewheretheoperationx.yis�1xy,andx.-1yisyxy�1.Involutoryquandles.Aquandlesatisfyingtheidentityx.y.y=x,equivalentlyx.y=x.-1y,iscalledaninvolutoryquandleora2-quandle.Thetwooperationsofaquandlearethesameinaninvolutoryquandle.Thereisananalogousde nitionforann-quandle.Firstde nex.nyasx.y..ywhere.yoccursntimes.Ann-quandleisaquandlethatsatis estheidentityx.ny=x.Aconjugacyclassofinvolutionsinagroupisaninvolutoryquandle,whiletheconjugacyclassofanelementofordernisann-quandle.Conjugacyclassesofinvolutionsareusefulinthestudyofgroups.Besidesconjugacyclassesofgroups,involutoryquandlesappearascoresofagroup.ThecoreofagroupGhasthesameelementsasGbutwiththeoperationx.y=yx�1y.Exercise67.Provethatthecoreofagroupisaninvolutoryquandle.Involutoryquandleswithgeodesics.Involutoryquandleshaveanicegeometricinter-pretationwheretheelementsarepointsandthelinesaredeterminedbytheoperation.Example4.37(Theplaneasaquandle).ConsidertheEuclideanplaneR2withtheoperationwhichsendsapointpthroughapointqtoyieldthepointp.qonthelinethatpassesthroughpandqandontheoppositesideofqthatpliesonbutequallyfarawayfromq.Ifphappenstoequalq,thende nep.qtobeq.Algebraically,p.q=2q�p.ThisoperationmakesR2andinvolutoryquandle.TheselfdistributivityaxiomQ3,whichsays(p.q).r=(p.r).(q.r),isillustratedin gure4.6.
118CHAPTER4.GROUPS
p
q
r
p.q
p.r
q.r
(p.q).r
Figure4.6:DistributivityinainvolutoryquandleSymmetricspaces.Asymmetricspaceisaparticularkindofmanifold.Ateachpointinthespace,thereisanisometry(thatis,atranslationthatpreservesdistance)forwhichthatpointisanisolatedsingularity.OttmarLoosdiscoveredin1967thattheintrinsicalgebraicstructureofasymmetricisaninvolutoryquandle.Thus,asymmetricspaceasadi erentiableinvolutoryquandleinwhicheverypointisanisolated xedpointofthesymmetrythroughit.BesidestheplaneR2,everyvectorspacesVoverany eldisasymmetricspaces.Theoperationthatmakesitaninvolutoryquandleisgivenbev.w=2w�v.That'sthesameoperationasdescribedaboveforR2.Therearelotsofothersymmetricspaces.TheordinarysphereS2aswellashigherdimensionalspheresSnareallsymmetricspaces.Soareothergeometricspacesincludingprojectivespaces,hyperbolicspaces,andinversivespaces.Theycanallbeusedtorepresentquandlesgeometricallyassubspaces.Geodesics.Ageodesicinamanifoldisacurvewhichforpointsclosetogetheristhecurveofshortestlengththatjoinsthem.InEuclideanspace,ageodesicisastraightline.OnthesphereS2,ageodesicisagreatcircle,thatis,theintersectionofaplanepassingthroughthecenterofthespherewiththesphere.Geodesicsonmanifoldshavemetrics,thatis,there'sadistancebetweenanytwopointsonthegeodesic.Giventwopointspandq,theentireinvolutivequandlegeneratedbythemliesononegeodesic.Thatmeansthatanyotherexpressionthatcanbemadefrompandqlieonageodesic.Inparticular,thepointsp.(q.p),q.p,p,q,p.q,andq.(p.q)lieonageodesic,andthey'reequallyspacedonit.
p.(q.p)
q.p
p
q
p.q
q.(p.q)
Sincegeodesicshavealltheinformationneededtodescribethequandlestructure,wecandrawinvolutoryquandles,atleastthesmallones,toseewhattheylooklike.Somewillbe
4.5.CONJUGACYCLASSESANDQUANDLES119generatedbytwoelements,sothey'lleitherlooklikethelineabove,orbequotientsofit.Example4.38(Re ectionsofapentagon).ThegroupD5ofsymmetriesofthepentagonwasillustratedin gure4.2.There ectionsofapentagonareinvolutions,andtheyformaconjugacyclassinD5.Thereare vere ections:p=(25)(34),q=(13)(25),p.q=(15)(24),q.p=(14)(23),andp.(q.p)=q.(p.q)=(12)(35).Sincep.(q.p)=q.(p.q),the veofthemlieonacircleasillustratedtotheright.
(25)(34)
(13)(45)
(15)(24)
(12)(35)
(14)(23)
Exercise68.Letthesevenverticesofaregularheptagonbedenoted1,2,3,4,5,6,and7.Describethehowthesymmetry(27)(36)(45)actsontheheptagoninwords.Determinetheconjugacyclassof(27)(36)(45)inD7andillustrateitaspointsonacircle.SomeconjugacyclassesofinvolutionsarecyclicliketheonesinD5above,butmostaren't.Herearetwoexamplesof6-elementconjugacyclassesinsmallgroups.Example4.39(Aconjugacyclassinthequaterniongroup).Thequaterniongroupwasintroducedinsection4.30.Ithaseightelements,namely1,�1,i,�i,j,�j,k,and�k.Sixofthem,allthoseexcept1,areinvolutionsandtheyformaconjugacyclass.It'sillustratedin gure4.7.Notethati.j=jij=�i,soi,j,�i,and�jareequallyspacedaroundacircle.Likewisefori,k,�i,and�kandfork,j,�k,and�k.Althoughthespacingdoesn'tappearequalontheEuclideanplaneasshown,itiswhenrepresentedonasphere.
i
k
j
�i
�j
�k
Figure4.7:AconjugacyclassinthequaterniongroupExample4.40(TheconjugacyclassoftranspositionsinS4).Thetranspositionsinasym-metricgroupformaconjugacyclass.ThethesymmetricgroupS4therearesixtranspositions,
120CHAPTER4.GROUPS
(12)
(13)
(23)
(34)
(14)
(24)
Figure4.8:TheconjugacyclassoftranspositionsinS4namely,(12),(13),(14),(23),(24),and(34).Theinvolutoryquandletheyformisshownin gure4.8.Therearefourcircles,eachwiththreetranspositions.Onecircle,forexample,includes(12),(13),and(23)since(12).(13)=(23),(13).(23)=(12),and(23).(12)=(13).Alsonotethat(12).(34)=(12),butnogeodesicisshownhavingthosetwotranspositions.Itreducestheclutterinthediagramtosuppressgeodesicswithonlytwoelements.4.6Kernels,normalsubgroups,andquotientgroupsThekernelKerfofagrouphomomorphismf:G!Hplaysthesameroleasthekernelofaringhomomorphism.It'sde nedasthetheinverseimageoftheidentity.ItisasubgroupofthedomainG,butaparticularkindofsubgroupcalledanormalsubgroup.We'llseethateverynormalsubgroupNofGisthekernelofsomegrouphomomorphism,infact,ofaprojectionG!G=NwhereG=NisaquotientgroupofG.4.6.1KernelsofgrouphomomorphismsandnormalsubgroupsWe'llusemultiplicativenotation.De nition4.41.Letf:G!Hbeagrouphomomorphism.ThoseelementsofGthataresenttotheidentity1inHformthekerneloff.Kerf=f�1(1)=fx2Gjf(x)=1g:Example4.42.LetGbethesymmetricgroupSnandf:G!f1;�1gmapevenpermuta-tionsto1andoddpermutationsto�1.Thenfisagrouphomomorphism,andKerf=An,thealternatingsubgroupofSn.Theorem4.43.Thekernelofagrouphomomorphismf:G!HisasubgroupN=KerfofGsuchthatforeachx2G,xNx�1N.
4.6.KERNELS,NORMALSUBGROUPS,ANDQUOTIENTGROUPS121Proof.ToshowthatNisasubgroupofG,notethat(1)it'sclosedundermultiplication,(2)itincludes1,and(3)it'sclosedunderinverses.For(1),ifx;y2N,thenf(x)=f(y)=1,sof(xy)=f(x)f(y)=1,thereforexy2N.(2)isobvious.For(3),ifx2N,thenf(x)=1,sof(x�1)=f(x)�1=1�1=1,thereforex�12N.ThusNisasubgroupofG.Nowtoshowthatforx2G,xNx�1N.Considerxyx�1wherey2N.Thenf(y)=1,sof(xyx�1)=f(x)f(y)f(x)�1=f(x)1f(x)�1=f(x)f(x)�1=1.Therefore,xyx�12N.Thus,xNx�1N.q.e.d.Besidestellinguswhatelementsaresentto1byf,thekerneloffalsotellsuswhentwoelementsaresenttothesameelement.Sincef(x)=f(y)ifandonlyiff(xy�1)=1,therefore,fwillsendxandytothesameelementofSifandonlyifxy�12Kerf.Thepropertiesofkernelsofgrouphomomorphismsthatwejustfounddeterminethefollowingde nition.De nition4.44.AsubgroupNofagroupGissaidtobeanormalsubgroupifforeachx2G,xNx�1N.NotethatsinceanormalsubgroupagroupGisclosedunderconjugation,thereforeanormalsubgroupofGistheunionofsomeoftheconjugacyclassesinG.Exercise69.ShowthatasubgroupNisnormalinGifandonlyifforeachx2G,xNx�1=N.Exercise70.ShowthatasubgroupNisnormalinGifandonlyifforeachx2G,xN=Nx.BoththetrivialsubgroupofGandGitselfarealwaysnormalsubgroups.IfGisanAbeliangroup,theneverysubgroupofGisanormalsubgroup.Theorem4.45.Anysubgroupofindex2isanormalsubgroup.Proof.LetNbeasubgroupofagroupGofindex2.We'llshowthatxN=Nxforeachx2G.Incasex2N,thenxN=N=Nx.Nowconsiderthecasex=2N.ThentherearetwoleftcosetsofN,namelyNitselfandxN,andtherearetworightcosets,NandNx.ThatgivesustwopartitionsofG,butsinceNisapartofeachpartition,theotherparts,namelyxNandNxmustbeequal.q.e.d.4.6.2Quotientgroups,andprojections :G!G=NAsmentionedabovethekernelofagrouphomomorphismftellsuswhentwoelementsaresenttothesameelement:f(x)=f(y)ifandonlyifxy�12Kerf.WecanuseKerftoconstructa\quotientgroup"G=KerfbyidentifyingtwoelementsxandyinGifxy�1liesinKerf.Infact,wecandothisnotjustforkernelsofhomomorphisms,butforanynormalsubgroupN.Thatis,wecanuseanormalsubgroupNofGtodeterminewhentwoelementsxandyaretobeidenti ed,xy,andwe'llendupwithagroupG=N.De nition4.46.AcongruenceonagroupGisanequivalencerelationsuchthatforallx;x0;y;y02G,xx0andyy0implyxyx0y0:Theequivalenceclassesforacongruencearecalledcongruenceclasses.Theorem4.47.IfisacongruenceonagroupG,thenthequotientsetG=,thatis,thesetofcongruenceclasses,isagroupwherethebinaryoperationisde nedby[x][y]=[xy].
122CHAPTER4.GROUPSProof.Firstweneedtoshowthattheproposedde nitionsareactuallywellde ned.Thatis,ifadi erentrepresentativex0ischosenfromthecongruenceclass[x]andy0from[y],thenthesameclass[x0y0]results.Thatis[x]=[x0]and[y]=[y0]imply[xy=xy0]:Butthatistherequirementinthede nitionofcongruence.Also,eachoftheaxiomsforagroupneedtobeveri ed,butthey'reallautomaticasthey'reinheritedfromthegroupG.q.e.d.Justasanidealinaringdeterminesacongruenceonthering,anormalsubgroupofagroupdeterminesacongruenceonagroup,andtheproofissimilar.Theorem4.48(Congruencemoduloanormalsubgroup).LetNbeanormalsubgroupofagroupG.Acongruence,calledcongruencemoduloN,isde nedbyxy(modN)ifandonlyifxy�12N:Thequotientgroup,G=,isdenotedG=N.ThecongruenceclassesarecosetsofN,thatis[x]=xN.Thefunction :G!G=Nde nedby (x)=[x]=xNisagrouphomomorphism,infact,anepimorphism.It'scalledaprojectionoracanonicalhomomorphismtothequotientgroup.It'skernelisN.Exercise71.IfisacongruenceonagroupG,showthatthecongruenceclassoftheidentity,[1]=N,isanormalsubgroupofG,andthecongruencedeterminedbyNistheoriginalcongruence.4.6.3IsomorphismtheoremsTheimageofagrouphomomorphismisisomorphictothegroupmoduloitskernel.Letf:G!Hbearinghomomorphism.Theimageoff,denotedf(G),isthesetf(G)=ff(x)2Hjx2Gg:Exercise72.Verifythattheimagef(G)isasubgroupofH.Exercise73.Provethefollowingtheorem.You'llneedtoshowthattheproposedfunctioniswell-de ned,thatitisagrouphomomorphism,andthenthatit'sanisomorphism.Theorem4.49(Firstisomorphismtheorem,Jordan,1870).Iff:G!HisagrouphomomorphismthenthequotientgroupG=Kerfisisomorphictotheimageringf(G),theisomorphismbeinggivenbyG=Kerf!f(G)xKerf7!f(x)Thisgivesustwowaystolookattheimage,eitherasaquotientgroupofthedomainGorasasubgroupofthecodomainH.Furthermore,wecannowtreatagrouphomomorphismf:G!Hasacompositionofthreegrouphomomorphisms.G !G=Kerf=f(G)!H
4.6.KERNELS,NORMALSUBGROUPS,ANDQUOTIENTGROUPS123The rstistheprojectionfromGontoitsquotientringG=Kerf,thesecondistheisomor-phismG=Kerf=f(G),andthethirdistheinclusionoftheimagef(G)asasubgroupofH.Theorem4.50(Secondisomorphismtheorem).IfHisasubgroupofGandNisanormalsubgroupofG,thenH=(H\N)=(HN)=N:Proof.Letf:H!(HN)=Nbede nedbyf(x)=xN.Thisfisagrouphomomorphismsincef(xy)=xyN=xNyN=f(x)f(y).Next,we'llshowthatfisanepimorphism.LetxN2(HN)=Nwherex2HN.Thenx=yzforsomey2Handz2N.SoxN=yzN=yN=f(y).Thus,fisanepimorphism,thatis,f(H)=(HN)=N.bythe rstisomorphismtheorem,wehaveH=Kerf=(HN)=N:Finally,we'llshowthatKerf=H\KwhichwillimplyH=(H\N)=(HN)=N.LetxbeanelementofHwhichliesinKerf.ThenxNistheidentityelementNin(HN)=N,sox2N.Butx2Halso,sox2H\N.Conversely,x2H\Nimpliesx2Kerf.q.e.d.Theorem4.51(Thirdisomorphismtheorem).IfHandKarebothnormalsubgroupsofGwithHK,then(G=H)=(K=H)=G=K:Exercise74.Provethethirdisomorphismtheorem.De nef:G=H!G=Kbyf(aH)=aK.Checkthatthisisawell-de nedhomomorphism.ShowKerf=H.ShowtheimageoffisallofG=K.Applythe rstisomorphismtheoremto nishtheproof.Theorem4.52(Correspondencetheorem).LetNbeanormalsubgroupofG.Thesub-groupsofGcontainingNareinone-to-onecorrespondencewiththesubgroupsofG=N.Thus,ifHisasubgroupofGcontainingN,thenH=NisasubgroupofG=N,andeverysubgroupofG=Nsoarises.Furthermore,HisnormalinGifandonlyifH=NisnormalinG=N.Exercise75.Provethecorrespondencetheorem.ShowthatforHNthatH=Nis,indeed,asubgroupofG=N.Showthatif
HisanysubgroupofG=NthatthesetH=fx2Gjx=N2
HgisasubgroupofGcontainingN.Verifythatthesetwooperationsareinversetoeachother.Finally,verifythelaststatement.4.6.4InternaldirectproductsWecanrecognizewhenagroupGisisomorphictoaproductoftwoormoregroups.RecallthatifG=MN,thenwecaninterpretMandNassubgroupsofG.AssuchtheyarenormalsubgroupsofGandtheirintersectionistrivial.Furthermore,G=MN.De nition4.53.AgroupGissaidtobeaninternaldirectproductoftwosubgroupsMandNifM\N=1,MN=G,andbothMandNarenormalsubgroupsofG.We'llshowinamomentthatifGistheinternaldirectproductofMandN,thenGisisomorphictotheproductgroupMN.But rst,alemma.
124CHAPTER4.GROUPSLemma4.54.IfMandNaretwonormalsubgroupsofGwhoseintersectionistrivial,thenelementsofMcommutewithelementsofN.Proof.Letm2Mandn2N.Inordertoshowthatmn=nm,we'llshowtheequivalentmnm�1n�1=1.Letx=mnm�1n�1.Sincex=(mnm�1)n�1,andbothmnm�1andn�1areelementsofthenormalsubgroupN,thereforex2N.Butsincex=m(nm�1n�1),andbothmandnm�1n�1areelementsofthenormalsubgroupM,thereforex2M.Sincex2M\N=1,thereforex=1.q.e.d.Theorem4.55.IfGistheinternaldirectproductofMandN,thenMN=Gwheretheisomorphismisgivenby(m;n)7!mn.Proof.Outline.Usethelemmatoverifythattheproposedisomorphismisahomomorphism.It'sevidentlyasurjectionsinceMN=G.Toshowthatit'saninjection,showthatthekernelistrivial.Suppose(m;n)7!mn=1.Thenm=n�1liesinbothMandN,soit'strivial,thatis,m=n=1.q.e.d.Exercise76.ProvethatGisaninternaldirectproductoftwonormalsubgroupsMandNifandonlyifeveryelementx2Gcanbeuniquelyrepresentedasaproductmnwithm2Mandn2N.Althoughwe'veonlylookedatinternaldirectproductsoftwosubgroups,thede nitioncanbegeneralizedtomorethantwosubgroups.We'llsaythatGistheinternaldirectproductofrnormalsubgroupsN1;N2;:::;Nrif(1)theyjointlygenerateG,thatis,N1N2Nr=G,and(2)theintersectionofanyoneNiwiththesubgroupgeneratedbytherestistrivial.ItfollowsthatN1N2Nr=G.Furthermore,anequivalentconditiontobeingainternaldirectproductofthenormalsubgroupsN1;N2;:::;Nristhateveryelementx2Gcanbeuniquelyrepresentedasaproductn1n2nrwitheachni2Ni.4.7MatrixringsandlineargroupsTherepresentationofringsandgroupsassubringsorsubgroupsofmatrixringsisveryhelpfulforacoupleofreasons.Oneisthatmatricesdescribelineartransformations.Thatmeansthattheelementsoftheringorgroupcanbeinterpretedasgeometrictransformations.Asecondisthatmatrixnotationissoveryconvenient.Usuallythecoecientsaretakentobeelementsofafamiliar eldlikeC,R,orQ,butforspecialpurposesthecoecientsmaybetakeninsomeotherintegraldomainsuchasZ.Forexample,the eldcomplexnumbersCcanberepresentedasacertainsubringofM2(R),theringof22matriceswithcoecientsinR,andthedivisionringofquaternionsHcanberepresentedasacertainsubringofM4(R).Mostofourexampleshavenequalto2or3andthecoecientsarereal.4.7.1LineartransformationsTheringofnnmatriceswithrealcoecients,M2(R),isanoncommutativeringwhenn2.WecaninterpreteachmatrixA2M2(R)asalineartransformationA:Rn!Rn
4.7.MATRIXRINGSANDLINEARGROUPS125wherea(column)n-vectorx2Rnismappedtoanothern-vectorAx=26664a11a12:::a1na21a22:::a2n............an1an2:::ann3777526664x1x2...xn37775=26664a11x1+a12x2++a1nxna21x1+a22x2++a2nxn...an1x1+an2x2++annxn37775TheidentitymatrixI=2666410:::001:::0............00:::137775correspondstotheidentitytransformationI:Rn!RnwhereIx=x.Alineartransformationfromavectorspacetoitselfisalsocalledalinearoperator.4.7.2ThegenerallineargroupsGLn(R)TheinvertiblennmatricesinMn(R),thatis,theunitsintheringMn(R),formthegenerallineargroupwithcoecientsinthecommutativeringR,denotedGLn(R).TheydescribenonsingulartransformationsRn!Rn.RecallthatamatrixAhasaninverseifandonlyifitsdeterminantjAjisaunitinR.Let'sinterpretsomeoftheseinthecasewhenn=2.ThedeterminantofA=abcdisjAj=ad�bc,andwhenthat'saunitinR,theinverseofAisA=1
jAjd�b�ca.NotethatthedeterminantisagrouphomomorphismGLn(R)!RfromthegenerallineargrouptotheinvertibleelementsofR.Thedeterminantoftheidentitymatrixis1,thedeterminantoftheproductoftwomatricesistheproductoftheirdeterminants,andthedeterminantoftheinverseofamatrixisthereciprocalofthedeterminantofthematrix.Let'sletRbethe eldofrealnumbersR.TherealgenerallineargroupGL2(R)canbeinterpretedasthegroupofinvertiblelineartransformationsoftheplaneR2thatleavetheorigin xed.Hereareafewlineartransformationsoftheplane.Rotationbyanangleabouttheoriginisdescribedbythematrixcos�sinsincossinceapointxyinR2issenttothepointxcos�ysinxsin+ycosThedeterminantofarotationmatrixis1.Re ectionacrossalinethroughtheoriginatanangletothex-axisisdescribedbythematrixcos2sin2sin2�cos2:Thedeterminantis�1.
126CHAPTER4.GROUPSExpansionsandcontractionsaredescribedbyscalarmatricesr00rwhereristheratio.Ifr�1,thenit'sanexpansion(alsocalleddilation),butif0r1,thenit'sacontraction.Therearenumerousotherkindsoftransformations.Here'sjustonemoreexample1101,anexampleofashearparalleltothex-axis.Pointsabovethex-axisaremovedright,pointsbelowleft,andpointsonthex-axisare xed.Inthreedimensionsyoucandescriberotations,re ections,andsoforth,aswell.4.7.3OtherlineargroupsThereareanumberofinterestingsubgroupsofGLn(R).ThespeciallineargroupsSLn(R).ThereareseveralsubgroupsofGLn(R),oneofwhichisthespeciallineargroupSLn(R)whichconsistsofmatriceswhosedeterminantsequal1,alsocalledunimodularmatrices.(Thereareotherlineargroupscalled\special"andineachcaseitmeansthedeterminantis1.)AmongtheexamplesinGL2(R)mentionedabove,therotationsandshearsaremem-bersofSL2(R),butre ectionshavedeterminant�1andexpansionsandcontractionshavedeterminantsgreaterorlessthan1,sononeofthembelongtothespeciallineargroup.SincetheabsolutevalueofthedeterminantistheJacobianofthetransformationRn!Rn,thereforetransformationsinSL2(R)preservearea.Sincethedeterminantispositive,thesetransformationspreserveorientation.Thus,transformationsinSL2(R)arethelineartransformationsthatpreserveorientationandarea.MoregenerallythoseinSLn(R)preserveorientationandn-dimensionalcontent.Rotationsandshears,andtheirproducts,arealwaysinSLn(R).Exercise77.Showthatthematrix2001=2liesinSL2(R).Describeinwordshowthistransformationactsontheplane.TheorthogonalgroupsO(n).ThesearesubgroupsofGLn(R).Anorthogonaltransfor-mationisonethatpreservesinnerproducts(alsocalleddotproductsorscalarproducts).I'llusethenotationhajbi=a1b1+a2b2++anbnfortheinnerproductofthevectorsa=(a1;a2;:::;an)andb=(b1;b2;:::;bn).Othercommonnotationsare(a;b)orab.ForthetransformationdescribedbythematrixAtopreserveinnerproductsmeansthathAajAbi=hajbi.Sincethelengthofavectorjajisdeterminedbytheinnerproduct,jaj2=hajai,thereforeanorthogonaltransformationpreservesdistance,too:jAaj=jaj.Conversely,ifApreservesdistance,itpreservesinnerproducts.Notethatsincedistanceispreserved,soisareaindimension2orn-dimensionalcontentindimensionn.It'satheoremfromlinearalgebrathatamatrixAdescribesanorthogonaltransformationifandonlyifitsinverseequalsitstransform:A�1=AT;equivalently,AAT=1.Thesema-
4.7.MATRIXRINGSANDLINEARGROUPS127trices,ofcourse,arecalledorthogonalmatrices.Notethatthedeterminantofanorthogonalmatrixis1.TheorthogonalgroupO(n)isthesubgroupofGLn(R)oforthogonalmatrices.It'snotasubgroupofSLn(R)sincehalftheorthogonalmatriceshavedeterminant�1,meaningtheyreverseorientation.ThespecialorthogonalgroupSO(n)isthesubgroupofO(n)ofmatriceswithdeterminant1.IntwodimensionsO(2)consistsofrotationsandre ectionswhileSO(n)consistsofonlytherotations.InthreedimensionsO(3)consistsofrotations(bysomeanglearoundsomelinethrough0)andre ections(acrosssomeplanethrough0).Again,SO(3)onlyhastherotations.TheunitarygroupsU(n).Formatriceswithcomplexcoecients,themostusefulanal-ogousgroupcorrespondingtotheorthogonalgroupforrealcoecientsissomethingcalledaunitarygroup.Theinnerproduct,alsocalledtheHermitian,forthecomplexvectorspaceCnisde nedashajbi=a1
b1+a2
b2+an
bnforthecomplexvectorsa=(a1;a2;:::;an)andb=(b1;b2;:::;bn)wherethebarindicatescomplexconjugation.AmatrixA,andthetransformationCn!Cnthatitdescribes,arecalledunitaryifitpreservestheHermitian.ThecollectionofallunitarymatricesinGLn(C)iscalledtheunitarygroupU(n).AnothertheoremfromlinearalgebraisthatamatrixAisunitaryifandonlyifitsinverseisthetransformofitsconjugate,A�1=
AT,equivalently,A
AT=I.Therearemanypropertiesofcomplexunitarymatricesthatcorrespondtopropertiesofrealorthogonalmatrices.4.7.4ProjectivespaceandtheprojectivelineargroupPGLn(F)Projectiveplanesandprojectivespace.Projectivegeometrydi ersfromEuclideangeometryinacoupleofways:alllinesinaplaneintersect,anddistanceandanglesarenotconsidered.Let'sstartwithEuclideanplanegeometry,thendropdistanceandangles,thenaddpointsatin nitytogettheprojectiveplane.WhendistanceandanglesarenotconsideredinEuclideangeometry,what'sleftiscalledanegeometry.Pointsandlinesstillremain.TheaneplaneisstillmodelledbyR2,butanetransformationsdon'thavetopreservedistanceorangles.So,forinstance,thelineartransformationsknownasexpansions,contractions,andsheartransformationsareallanetransformations.InfacteveryelementinGL2(R)describesananeplanartransforma-tion.Thesearetheanetransformationstht xtheorigin.Also,translations,whicharenotlineartransformations,areanetranstransformations.Similarly,indimensionn,anetransformationsarecomposedoftranslationsandelementsofGLn(R).AnespacesFncanbesimilarlyde nedforother eldsFbesidestherealsR.Sofar,we'vedroppeddistanceandangles,butparallellinesremaininanegeometry.Thenextstepistoallenoughpoints,calledpointsatin nity,sothatparallellinesmeetatthem.
128CHAPTER4.GROUPSParallelismisanequivalencerelationonlines.We'llassumethatalineisparalleltoitself,soparallelismisre exive.It'salsosymmetric,andit'stransitive:ifonelineisparalleltoanother,andtheothertoathird,thenthe rstisparalleltothethird.Foreachparallelismequivalenceclass,addonepoint,apointatin nitytoanespaceandspecifythateverylineinthatequivalenceclasspassesthroughthatpoint.Addonemoreline,thelineatin nity,andspecifythateverypointatin nitypassesthroughit.Theresultingspaceistheprojectivespacecorrespondingtotheanespace.Projectivespaceandprojectivecoordinates.LetFbea eld,suchasthe eldofrealnumbers.TheprojectivelineargroupPGLn(F)isusedtostudyprojectivespace.Amoreformalwaytode neprojectiven-spaceovera eldFisbymodellingpointsoftheprojectiveplanebylinesinanen+1-space,Fn+1,throughtheoriginbymeansananalgebraicequivalencerelation.Twopointsa=(a0;a1;:::;an)andb=(b0;b1;:::;bn)ofFn+1namethesamepointofFPniftheircoordinatesareproportional,thatis,ifthereexistsanonzeroelement2Fsuchthatbi=ai=fori=0;1;:::;n.We'lllet[a0;a1;:::;an]denotethepointinFPnnamedby(a0;a1;:::;an)2Fn+1.Thus,[a0;a1;:::;an]=[a0;a1;:::;an].Thenotation[a0;a1;:::;an]iscalledprojectivecoordinates.Geometrically,thisconstructionaddspointsatin nitytotheaneplane,onepointforeachsetofparallellines.Linescanalsobenamedwithprojectivecoordinatesb=[b0;b1;:::;bn].Ifyoudothat,thenapointa=[a0;a1;:::;an]liesonthelinebiftheirinnerproducthajbiis0.Example4.56(TheFanoplaneZ2P2).TheprojectiveplaneZ2P2hasaname,theFanoplane,namedafterGinoFano(1871{1952),afounderof nitegeometries.Figure4.9showsarepresentationofZ2P2.Thereare7pointsand7lines,eachlinewith3points,andeachpointon3lines.
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4.7.MATRIXRINGSANDLINEARGROUPS129Example4.57(TheprojectiveplaneZ3P2).Figure4.10showsarepresentationofthe niteprojectiveplaneZ3P2.Thereare13pointsand13lines,eachlinewith4points,andeachpointon4lines.Wecannamethe9pointsintheaneplaneZ23withthirdcoordinate1,andthe4pointsatin nitywiththirdcoordinate0.Thefourpointsatin nitylineonalineatin nity.Eachofthesepointsatin nitylieonallthoselinewithaparticularslope.Forinstance,thepoint[1;�1;0]liesonthethreelineswithslope�1(anditliesonthelineatin nity,too).Z3P2
r[�1;�1;1]r[�1;0;1]r[�1;1;1]r[0;�1;1]r[0;0;1]r[0;1;1]r[1;�1;1]r[1;0;1]r[1;1;1]r[1;0;0]r[0;1;0]r[1;1;0]r[1;�1;0]
Figure4.10:TheprojectiveplaneZ3P2Finiteprojectiveplanes.There'sasimplerde nitionofaprojectiveplanethatcanbemadeaxiomatically.Itstatesthattwopointsdeterminealine,andtwolinesdetermineapoint.Anondegeneracyaxiomisalsorequiredthatthereareatleastthreepointswhichdon'talllieonthesameline(fromwhichitfollowsthatthereareatleastthreelineswhichdon'tallmeetatonepoint).Itturnsoutthatthisaxiomaticde nitionadmitsprojectiveplanesthatdon'tderivefrom elds.We'lllookattheonesthatdo.Finiteprojectiveplanesexistforeach nite eld.LetGF(pn)beaGalois eldofq=pnelements.Therewillbeq2pointsontheaneplaneGF(pn)2withthirdcoordinate1,andq+1pointsonthelineatin nitywiththirdcoordinate0.Sothe niteprojectiveplaneGF(pn)P2hasq2+q+1pointsaltogether.Ithasthesamenumberoflines.Theseprojectiveplanesallhaveacoupleofniceproperties.TheyareallDesarguesianandPappian,thatis,Desargue'stheoremandPappas'stheorembothholdfortheseprojectiveplanes.Thesetwotheoremsstatethatcertaincon gurationsofpointsandlinesholdforthe
130CHAPTER4.GROUPSprojectiveplane.Desarguesdevelopedprojectivegeometryinthe1600s,andoneofPappus'stheoremsapplytoprojectivegeometry.Thereareotherprojectiveplanesthataren'tbasedon nite eldsthataren'tDesarguesianandPappian.ProjectivelineargroupPGLn(F).Aswede nedprojectiven�1-spaceovera eldFasaquotientofnonzeroelementsofn-space,sotoowecande neaquotientofGLn(F)togettheprojectivelineargroupPGLn(F)actingonprojectiven�1-space.TwomatricesAandBinGLn(F)namethesameelementofPGLn(F)ifeachisamultipleoftheother,thatis,thereexists6=02FsuchthatB=A.ThenPGLn(F)actsonFPn�1,sinceAaandAanamethesameelementofFPn�1.IfFisa nite eldwithqelements,thentheorderofthegroupPGLn(F)istheorderofGL(n;F)dividedbyq�1,sojPGLn(F)j=(qn�1)(qn�q)(qn�q2)(qn�qn�1)
q�1.ProjectivespeciallineargroupPSLn(F).Theprojectivespeciallineargroup,PSLn(F),isthesubgroupofPGLn(F)namedbyunimodularmatrices.It'sSLn(F)moduloscalarmatrices!Iwhere!isannthrootofunity.TheorderofPSLn(F)isequaltotheorderofPGLn(F)dividedbygcd(n;q�1)whereqisthenumberofelementsofF.Exceptforsmallvaluesofntheprojectivespeciallineargroupsareallsimple.Simplicityisde nedinthenextsection.ThegroupsPSL3(Z3)isactuallythesameasPGL3(Z3)since3and2arerelativelyprime.Example4.58.TheprojectivelineargroupPGL3(Z2)=PSL3(Z2)whichactsontheFanoplaneZ2P2has764
1=168elements.It'ssmallenoughsothatitsconjugacyclassescanbedeterminedwithoutresortingtoadvancedmethods.Therearesixconjugacyclassesofsizes1,21,56,42,24,and24.Asalways,theidentityformsaconjugacyclassof1element.Referto gure4.9tonameelements.Theconjugacyclassoftheinvolution(12)(56)hassize21.Theconjugacyclassof(124)(365)has56elements.Theconjugacyclassof(0124)(36)has42elements.Theconjugacyclassof(0125463)has24elements,andtheconjugacyclassofitsinversealsohas24elements.Also,PGL3(Z3)=PSL3(Z3),actingontheprojectiveplaneZ3P2,hasorder262418
2=5616.4.8Structureof nitegroupsTheclassi cationof nitegroupsisextremelydicult,butthereareatoolswecanusetoseehowthatclassi cationbegins.Inthenextsectionwe'llclassify niteAbeliangroupsandseethatthey'reisomorphictoproductsofcyclicgroups,butthesituationforgeneralgroupsmuchmorecomplicated.
4.8.STRUCTUREOFFINITEGROUPS1314.8.1SimplegroupsThewaywe'llanalyzegroupsisbytheirnormalsubgroupsandquotients.Inparticular,ifNisamaximal,propernormalsubgroupofG,thenG=Nhasnosubgroups,forifitdid,bythecorrespondencetheorem,therewouldbeanormalsubgroupbetweenNandG.De nition4.59.Anontrivialgroupissaidtobesimpleifithasnoproper,nontrivial,normalsubgroups.Exercise78.ProvethattheonlyAbeliansimplegroupsarecyclicofprimeorder.Therearemanynonabeliansimplegroups.Thereareseveralin nitefamiliesofthem,andafewthataren'tinin nitefamilies,calledsporadicsimplegroups.Onein nitefamilyofsimplegroupsconsistsofalternatinggroupsAnwithn5.Indeed,A5isthesmallestnonabeliansimplegroup.Theprojectivespeciallineargroupsmentionedinthesectionaboveformanotherfamilyof nitesimplegroups.Exercise79(NonsimplicityofA4).Verifythatthereare veconjugacyclassesinA4asshowninthefollowingtable.GeneratorSizeOrder111(12)(34)32(123)43(132)43AnormalsubgroupofA4wouldbeaunionofsomeoftheseconjugacyclassesincludingtheidentityconjugacyclassofsize1,butitsorderwouldhavetodivide12.FindallthepropernontrivialnormalsubgroupsofA4.Exercise80(SimplicityofA5).Verifythatthereare veconjugacyclassesinA5asshowninthefollowingtable.GeneratorSizeOrder111(12)(34)152(123)203(12345)125(12354)125AnormalsubgroupofA5wouldbeaunionofsomeoftheseconjugacyclassesincludingtheidentityconjugacyclassofsize1,butitsorderwouldhavetodivide60.Verifythatnocombinationofthenumbers1,15,12,12,and20,where1isincludedinthethecombination,yieldsasumthatdivides60(thosenumbersbeing2,3,4,6,10,12,15,20,and30)exceptjust1itselfandthesumofall venumbers.Thus,thereisnopropernontrivialnormalsubgroupofA5.4.8.2TheJordan-HoldertheoremDe nition4.60.AcompositionseriesforagroupGisa nitechainofsubgroups1=NnNn�1N1N0=G
132CHAPTER4.GROUPSsuchthateachNi�1isamaximalpropernormalsubgroupofNi.Thenumberniscalledthelengthofthecompositionseries,andthenquotientgroupsNn�1=1;:::;N1=N2;G=N1whichareallasimplegroups,arecalledcompositionfactorsdeterminedbythecompositionseries.Itisevidentthatany nitegroupGhasatleastonecompositionseries.JusttakeN1tobeamaximalpropernormalsubgroupofG,N2tobeamaximalpropernormalsubgroupofN1,etc.In nitegroupsmayalsohavecompositionseries,butnotallin nitegroupsdo.Exercise81.FindacompositionseriesforthesymmetricgroupS4.Exercise82.Provethatanin nitecyclicgrouphasno( nite)compositionseries.Althougha nitegroupmayhavemorethanonecompositionseries,thelengthoftheseriesisdeterminedbythegroupasarecompositionfactorsatleastuptoisomorphismaswe'llseeinamoment.Thus,theseareinvariantsofthegroup.Theydonot,however,completelydeterminethegroup.Exercise83.ShowthatthedihedralgroupD5andthecyclicgroupC10havecompositionserieswiththesamelengthandsamefactors.Theorem4.61(Jordan-Holder).Anytwocompositionseriesfora nitegrouphavethesamelengthandthereisaone-to-onecorrespondencebetweenthecompositionfactorsofthetwocompositionseriesforwhichthecorrespondingcompositionfactorsareisomorphic.Proof.We'llprovethisbyinductionontheorderofthegroupunderquestion.Thebasecaseisforthetrivialgroupwhichhasonlythetrivialcompositionseries.AssumenowthatagroupGhastwocompositionseries1=NmMm�1M1M0=G;and1=NnNn�1N1N0=GIfM1=N1,thenbyinductionweconcludethatthelengthsoftherestofthecompositionareequalandthecompositionfactorstherestoftherestoftheseriesarethesame,andofcourse,thefactorsG=M1andG=N1areequal,sothecaseM1=N1is nished.ConsidernowthecaseM16=N1.SincebothM1andN1arenormalsubgroupsofG,soistheirintersectionK2=M1\N1.Let1=KkKk�1K3K2beacompositionseriesfortheirintersection.ThesesubgroupsofGareillustratedinthefollowingdiagram.1Mm�1��M2M1Kk�1��K2Nn�1��N2N1G
Bythesecondisomorphismtheorem,wehaveM1=(M1\N1)=G=N1.Therefore,K2isamaximalnormalsubgroupofM1.Thus,wehavetwocompositionseriesforM1,andbytheinductivehypothesis,theyhavethesamelength,som=k,andtheyhavethesamefactors
4.8.STRUCTUREOFFINITEGROUPS133uptoisomorphisminsomeorder.LikewisewehavetwocompositionseriesforN1,andtheyhavethesamelength,sok=n,andthesamefactorsuptoisomorphisminsomeorder.WenowhavefourcompositionseriesforG,twoincludingM1andtwoincludingN1.Theyallhavethesamelength,andsinceG=M1=N1=K2andG=N1=M1=K2,theyallhavethesamefactorsuptoisomorphisminsomeorder.q.e.d.Thereisageneralizationofthistheoremthatappliestoin nitegroupsthathavecompo-sitionseriesbutitsproofisconsiderablylonger.Thelistofcompositionfactorsisnotenoughtocharacterizethegroup.Thatistosay,therearenon-isomorphicgroupsthathavethesamecompositionfactors.ThesmallestpairofsuchgroupsareA3andC6oforder6.Asporadicgroup.Most nitesimplegroupscomeinin niteparameterizedfamiliessuchasthecyclicgroupsCpforprimep,andthealternatinggroupsAnforn5.Thereareseveralofthesein nitefamiliesofsimplegroups.Therearealsoafewsimplegroupsthatdon'tbelongtoanyofthesein nitefamilies.We'lllookatoneofthem,theMathieugroupM11.MathieudiscoveredM11in1861.It'sthesmallestsporadicgroup,andithasorder7920=891011.ItcanbedescribedasasubgroupofthesymmetricgroupS11generatedbythepairofpermutations(123456789te)and(37e8)(4t56).(Heretisusedfor10andefor11.)M11haselementsoforder1,2,3,4,5,6,8,and11.Ithas165=3511elementsoforder2,thatis,involutions.Theyareallconjugatesof(3t)(49)(56)(8e).AsS11actsonasetof11elements,sodoesM11.Infact,theactionissharply4-transitive.Transitivemeansthatforanypairxandyofelementsintheset,thereisagroupelementthatmapsthextoy.Doublytransitivemeansthatforx1;x2andy1;y2,distinctpairs,there'sagroupelementthatsendsx1toy1atthesametimeassendingx2toy2.Moregenerally,andn-transitiveactionisonesuchthatforallpairwisedistinctn-tuplesx1;:::;xnandpairwisedistincty1;:::;ynthereisagroupelementthatmapseachxitoyi.Whenthereisexactlyonegroupelementforpairofn-tuples,thegroupissaidtoactsharply.SolvablegroupsOneoftheapplicationsofgrouptheoryisGaloistheoryforalgebraic elds.Thegroupsofautomorphismsofthese eldsarecloselyrelatedtothesolutionsofalgebraicequations.Inparticular,thesegroupscantellyouiftheequationshavesolutionsthatcanbeexpressedintermsofradicals,thatissquareroots,cuberoots,andhigherroots.Theconditionforsuchsolvabilityisnonethefactorsinacompositionseriesforagrouparenonabeliansimplegroups,equivalently,thatallthefactorsarecyclicgroupsofprimeorder.De nition4.62.Agroupissaidtobesolvableifithasacompositionseriesallofwhosefactorsarecyclic.Exercise84.Provethatiftheorderofagroupisapowerofaprimenumber,thenthatgroupissolvable.Example4.63(TheFrobeniusgroupF21=C7oC3).Thisgroupwillhave21elements.ItiswhatiscalledasemidirectproductofthecyclicgroupC7=f1;a;a2;:::;a6gof7elementswiththecyclicgroupC3=f1;b;b2gof3elements.Eachelementcanbewrittenintheformbnamwith0b2and0a6,butaandbdon'tcommute.Forthisgroup,bab�1=a2.
134CHAPTER4.GROUPSThegroupisdenotedC7oC3.ThegroupC7isanormalsubgroup,butC3isnotanormalsubgroup.
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Figure4.11:CayleygraphoftheFrobeniusgroupF21=C7oC3Thisgroupcanbepresentedasha;b:a7=1;b3=1;ba=a2bi.ItsCayleygraphisshownin gure4.11withtheunderstandingthatthetoplineisidenti edwiththebottomline,andtheleftlineisidenti edwiththerightline.Abluearrowindicatesmultiplicationbyawhilearedoneismultiplicationbyb.ThegroupC7oC3isagroupofsymmetriesofaheptahedrononatorus.Aheptahedronhas7hexagonalfaceswhichmeetthreeatatimeatavertex,14vertices,and27edges.Itisatilingofthetoruswhichisillustratedin gure4.12.Eachofthesevenhexagonsislabelled1through7andcoloredadi erentcolor.Theouteredgesaretobeidenti edsothattheedgesABCDareidenti edontheupperleftandlowerright,theedgesDEFAareidenti edontheupperleftandlowerright,andtheedgesAGHDareidenti edontheleftandtheright.Theresultingtopologicalspaceisatorus.Youcanalsointerpretthisasacoloringofthetilingoftheplanebyhexagonswherethelabelsofsomenearbyhexagonsareshowninthe gure.ThegroupC7oC3isasubgroupofthegroupofsymmetriesofthisheptahedron.Theelementaoforder7describesthepermutationofthefaces(1234567)whichmovesthehexagonstotheupperright.Theelementboforder3describesthepermutation(142)(356)whichisarotationabouthexagon7by120.It'seasilyveri edthatbaanda2bbothdescribethepermutation(157)(364)whichisarotationabouthexagon2.Exercise85.Verifythattherotationc=(154623)abouthexagon7by60isasymmetryoftheheptahedron.Evidentlyc2=b.(a).Determinetherelationbetweenaandcoftheformca=akc,thatis, ndk.(b).ThisgroupisasemidirectproductC7oC6.DrawitsCayleygraph.Muchmorecanbesaidaboutsolvablegroupsthanwehavetimefor.4.9AbeliangroupsCommutativegroupsarecalledAbeliangroupsinhonorofNeilsHenrikAbel(1802{1829)whoworkedwithgroupsofsubstitutionsinordertounderstandsolutionsofpolynomialequations.
4.9.ABELIANGROUPS135
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136CHAPTER4.GROUPSWe'lluseadditivenotationthroughoutthissectiononAbeliangroups.Also,we'llcalltheproductoftwoAbeliangroupsAandBadirectsumanddenoteitABratherthanAB.EverysubgroupofanAbeliangroupisnormal,sowe'lljustrefertothemassubgroupsandleaveo theadjective\normal."WealreadyknowafairamountaboutAbeliangroups.WeknowaboutcyclicgroupsandtheChineseremaindertheorem.Forexample,weknowZ12=Z3Z4whereanelementnmodulo12correspondstothepairnmodulo3andnmodulo4.Likewise,Z6=Z2Z3.ThisgivesusthreewaystotreatthegroupZ2Z3Z4sinceitisisomorphictobothZ2Z12andZ6Z4.Ourcharacterizationofinternaldirectproductlooksalittledi erentwhenthegroupiswrittenadditively.Hereitis,rewrittenforAbeliangroups.AnAbeliangroupGistheinternaldirectsumofsubgroupsMandNif(1)theyjointlygenerateG,thatis,M+N=G,and(2)theintersectionM\N=0.IfGistheinternaldirectsumofMandN,thenMN=G.Furthermore,anequivalentconditiontobeingainternaldirectsumisthateveryelementx2Gcanbeuniquelyrepresentedasasumm+nwithm2Mandn2N.FortheexampleZ2Z3Z4above,itistheinternaldirectsumofZ2and0Z3Z4aswellastheinternaldirectsumofZ2Z30andZ4.4.9.1ThecategoryAofAbeliangroupsThecategoryofAbeliangroupsisaparticularlynicecategory.Notonlydoesithaveproducts,butitalsohascoproducts,tobede nednext,andtheproductsarecoproducts,andthat'swhywe'recallingthemdirectsums.It'snottheonlycategorywithdirectsums.Thecategoryofvectorspacesovera xed eldhasthemtoo.CoproductsinacategoryandtheiruniversalpropertyWhenallthearrowsinadiagramarereversed,asimilardiagram,calledthedualresults.Recallthatproductsinacategoryarecharacterizedbyadiagram.TheproductABinacategoryalongwiththetwoprojectionsAB1!AandAB2!BhastheuniversalpropertythatforeachobjectXandmorphismsX!AandX!B,thereisauniquemorphismX!AB,suchthatthediagrambelowcommutes.ABABX����[email protected]@@@R2
-1PPPPPPPPPPPPPq
4.9.ABELIANGROUPS137Ifweturnaroundallthearrows,we'llgetthecharacterizingpropertyforcoproducts.ThecoproductA`BinacategoryalongwiththetwoinjectionsA 1!A`BandB 1!A`BhastheuniversalpropertythatforeachobjectXandmorphismsA!XandB!X,thereisauniquemorphismA`B!X,suchthatthediagrambelowcommutes.ABA`[email protected]@@@R 1���� 2
-PPPPPPPPPPPPPq1InthecategorySofsetscoproductsaredisjointunions.ThedisjointunionoftwosetsSandThasoneelementforeachelementofSandadi erentelementforeachelementofT.SothecardinalityoftheirdisjointunionisjSj+jTj.Exercise86.InthecategoryofAbeliangroups,thecoproductobjectA`Biswhatwe'vecalledthedirectsumAB,whichisthesameastheproductAB.TheinjectionsA 1!A`BandB 1!A`BforAbeliangroupsarede nedby 1(x)=(x;0)and 1(y)=(0;y).Verifythattheuniversalpropertyholds.4.9.2FiniteAbeliangroupsTheclassi cationof nitegroupsisverydicult,buttheclassi cationof niteAbelianisnotsodicult.Itturnsout,aswe'llsee,thata niteAbeliangroupisisomorphictoaproductofcyclicgroups,andthere'sacertainuniquenesstothisrepresentation.Thisclassi cationissometimescalledthefundamentaltheoremof niteAbeliangroups.Thetheoremaboveoninternaldirectsumsisessentialinthisclassi cation.Theorem4.64.LetGbea niteAbeliangroupofordermnwheremandnarerelativelyprime,bothgreaterthan1.LetM=fx2Gjmx=0gandN=fx2Gjnx=0g.ThenMandNaresubgroupsofG,andGistheinternaldirectsumofMandN.Furthermore,jMj=mandjNj=n.Proof.Outline.ThatMandNaresubgroupsisquicklyveri ed.Sincemandnarerelativelyprime,therefore1isalinearcombinationofthem,thatis,thereareintegerssandtsuchthat1=sm+tn.TheirintersectionM\Nistrivialsinceifx2M\N,thenmx=nx=0,hencex=1x=(sm+tn)x=smx+tnx=0.TogetherMandNgenerateG,sinceforx2G,x=smx+tnx,butsmx2Nsincensmx=(nm)sx=0,likewisetnx2M.ThusM+N=G.Therefore,GistheinternaldirectsumofMandN.q.e.d.
138CHAPTER4.GROUPSp-primarygroups.LetGbeaAbeliangroupandpaprimenumber.ThesetG(p)=fxjpkx=0forsomek0gisasubgroupofG.Itiscalledthep-primarycomponentofG.AsacorollarytotheabovetheoremconsiderthecasewhenjGjisfactoredasapowerofprimes.Corollary4.65(Primarydecompositiontheorem).LetGbea niteAbeliangroupwhoseorderhasprimefactorizationpe11pe22perr.ThenGisadirectsumofthepi-primarycompo-nentsG=G(p1)G(p2)G(pr)andjG(pi)j=peiiforeachi.We'vereducedtheproblemofclassifying niteAbeliangroupstoclassifyingthosewhoseordersarepowersofaprimep.Suchgroupsarecalledp-primarygroupsorsimplyp-groups.Ifthepowerisgreaterthan1,thentherearedi erentgroupsofthatorder.Forexample,therearethreedistinctAbeliangroupsoforder125,namely,Z125,Z25Z5andZ5Z5Z5.The rsthasanelementoforder125,buttheothertwodon't,whilethesecondhasanelementoforder25,butthethirddoesn't.Hence,theyarenotisomorphic.We'llseesoonthateveryp-groupGisaisomorphictouniquedirectsumofcyclicp-groupsG=Ze1pZe2p2Zerprwherethesume1+2e2++rerisequalton,wherejGj=pn.Example4.66.We'll ndallthe2-groupsoforder32uptoisomorphism.Since32=25,We'llneede1+2e2++rer=5.Eachsolutionwillgiveusawayofmakingasumofpositiveintegersequalto5.Apartitionofnisalistofpositiveintegersthatsumton.Here'satablewhichgivesallthepartitionsof5andtheassociated2-groups.5Z321+4Z2Z162+3Z4Z81+1+3Z22Z41+2+2Z2Z241+1+1+2Z32Z41+1+1+1+1Z52Exercise87.Completethistableofofthenumberofpartitionsofnupthroughn=10.Workitoutyourself.n
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4.9.ABELIANGROUPS139Fundamentaltheoremof niteAbeliangroups.Ourstrategyforap-primarygroupwillbetopicko directsummandscontainingelementsofmaximalorders,oneatatime.Thatwillshowthatap-primarygroupisadirectsumofcyclicgroupswhoseordersarenonincreasingpowersofp.We'llthenshowthosepowersofparedeterminedbythep-primarygroup.Adicultyintheproofisthattherearemanychoicestobemaderesultingindi erentdirectsums,butwe'llseethattheordersofthecyclicsubgroupsturnsouttobethesamenomatterhowwemakethechoices.Theproofofthetheoremisparticularlytechnical,sowe'llseparatepartsoftheproofaslemmas.Lemma4.67.LetGbeanoncyclicp-primarygroupandaanelementofGofmaximalorder.Thenthereisanelementbinthecomplementofhaioforderp.Proof.Letcbeanelementinthecomplementofhaiofsmallestorder.Sincetheorderofpcis1=ptimestheorderofc,whichisasmallerorderthantheorderofc,thereforepcliesinhai.Sopc=kaforsomeintegerk.Letpmdenotetheorda,thelargestorderofanyelementinG.Thenord(ka)pm�1sincepm�1(ka)=pm�1pc=pmc=0.Therefore,kaisnotageneratorofthecyclicgrouphaisincethatgrouphaspmelements.Hence,gcd(pm;k)6=1,andsopdividesk.Letk=pj.Thenpb=ka=pji.Letb=c�ja.Thenpb=0,butb=2haiasc=b+ka=2hai.q.e.d.Proof.LetjGj=pnandorda=pmwithmn.We'llprovethelemmabyinduction.Assumeitisvalidforallgroupsoforderlessthanpn.Letbbeanelementinthecomplementofhaioforderpshowntoexistinthepreviouslemma.Sinceordb=pand=2hai,therefore=2hai\=2hbi=0.We'llreducemodulohbitoasmallerp-primarygroupG=hbiwherewecanusetheinductivehypothesis,thenbringtheresultsbackuptoG.First,we'llshowthata+hbi,whichistheimageofainG=hbi,hasthesameorderthatadoesinG,namelypm,whichimpliesthata+hbiisanelementofmaximalorderinthegroupG=hbi.Supposeord(a+hbi)pm.Thenpm�1(a+hbi)isthe0elementofG=hbi,inotherwords,pm�1a2hbi.Butpm�1a2hai,andtheintersectionofhaiandhbiistrivial.Therefore,pm�1a=0whichcontradictsorda=pm.Wenowknowa+hbiisanelementofmaximalorderinthegroupG=hbi,sowecanapplytheinductivehypothesistoconcludethatG=hbiisthedirectsumofthecyclicsubgroupgeneratedbya+hbiandanothersubgroupK=hbi.Notethatbythecorrespondencetheorem,everysubgroupofaquotientgroupG=hbiistheimageofagroupinG,sowemaytakeKtobeasubgroupofG.We'llshowthatG=haiKbyshowingthat(1)hai\K=0,and(2)haiK=G.(1).Ifx2hai\K,thenitsimagex+hbiinthequotientgroupG=hbiliesinboththecyclicsubgroupgeneratedbya+hbiandK=hbi.Buttheirintersectionisthe0elementinG=hbi,thereforex2hbi.Sincex2haialso,andx2hai\hbiistrivial,thereforex=0.(2).WecanshowhaiKisallofGbyacountingargument.WeknowthattheorderofG=hbiistheproductoftheorderofthecyclicsubgroupgeneratedbya+hbiandtheorderofK=hbi,theorderofGisptimestheorderofG=hbi,theorderofhaiisthesameastheorderofthecyclicsubgroupgeneratedbya+hbi,andtheorderofKisptimestheorderof
140CHAPTER4.GROUPSKhbi.Therefore,theorderofGequalstheproductoftheorderofhaiandtheorderofK.ThushaiK=G.q.e.d.Youcanprovethe rststatementoffollowingtheorembyinductionusingthelemmawejustproved,thenapplytheprimarydecompositiontheoremforthesecondstatement.Thisistheexistencehalfofthetheoremwewant.We'llstillneedsomekindofuniquenessofthetermsinthedirectsum.Theorem4.68.Ap-primarygroupisadirectsumofcyclicgroupswhoseordersarepowersofp.A niteAbeliangroupisthedirectsumofcyclicgroups.Thereareacoupleofwaystodescribetheuniquenessoftheterms.Sincewe'vebeenusingcyclicgroupswhoseordersareprimepowers,let'ssticktothat.There'saconceptwe'llneedinthefollowinglemma.IfGisanAbeliangroupandpaninteger,thenthesubsetGp=fxjpx=0gisasubgroupofG.Infact,it'sjustthekernelofthegrouphomomorphismG!Gthatmapsxtopx.Exercise88.Showthatitis,indeed,agrouphomomorphism.Lemma4.69.SupposethatGisap-primarygroupthatcanbewrittenasadirectsumofnontrivialcyclicsubgroupsintwowaysG=H1H2Hm=K1K2KnwherejH1jjH1jjHmjandjK1jjK1jjKnj.Thenm=nandforeachi,jHij=jKij.Proof.Outline.ByinductionontheorderofG.FirstverifythatGp=Hp1Hp2Hpm=Kp1Kp2Kpn:IfanyofthegroupsHpiorKpjaretrivial,thendropthemtogetGp=Hp1Hp2Hpm0=Kp1Kp2Kpn0togettwodirectsumsofnontrivialcyclicsubgroups.Byinduction,m0=n0andforeachim0,jHpij=jKpij.SincejHij=pjHpijandjKij=pjKpij,thereforejHij=jKijforeachim0.FinishwithacountingargumenttoshowthatthenumberoftrivialgroupsthatweredroppedisthesamefortheH'sasfortheK's.They'rethesubgroupsHiandKiofordern.q.e.d.Puttingthelasttheoremandlemmatogether,wehavethefollowingtheorem.Theorem4.70(Fundamentaltheoremof niteAbeliangroups).A niteAbeliangroupisthedirectsumofcyclicgroupswhoseordersareprimepowers.Thenumberoftermsinthedirectsumandtheordersofthecyclicgroupsaredeterminedbythegroup.
Appendices141
AppendixABackgroundmathematicsA.1LogicandproofsTheorems.Logicandproofsareattheheartofmathematics.Astatementwillnotbeacceptedbyamathematicianifthere'snoproofofit.Atheoremisastatementthathasanaccompanyingproof.Ifastatementissuspectedtobetrue,butthere'snoproofyet,thenitwillbecalledaconjecture.Sometimessomeonewillsupplytheconjecturewithaproof,thenitbecomesatheorem;sometimesacounterexampletotheconjectureisdiscoveredsoitfailstobeatheorem.Atypicaltheorembeginswiththeword\Theorem"followedbythestatementofthetheorem.Thatstatementusuallydoesn'thavemuchmathematicalsymbolismorvariables,butit'swrittenasmuchaspossibleinEnglishsentences.Aftertheproofiscomplete,it'sendedwithQ.E.D.(\QuodEratDemonstrandum",Latinfor\thatwhichwastobeshown")orsomespecialsymbollikeabox.Corollariesandlemmasarealsotheorems.Acorollaryisatheoremthatfollowsquiteeasilyfromtheprecedingtheorem.Sometimestheproofofacorollaryisomittedandlefttothereadertoprovide.Alemmaisatheoremthatprecedesanothertheorem.Lemmasareoftentechnicalandoflittleinterestinthemselves,butarenecessaryforthetheoremswhichfollowsthem.Some-timesacomplicatedproofwillbesplitupandpartsdeclaredaslemmas.Thatmakesiteasiertounderstandthelogical owoftheproof.Somestandardsymbolsseeninproofs.Thereareawholelotofsymbolsandabbrevia-tionsthatareusedinproofs.SomearelistedintableA.1.Althoughthesearefairlystandard,sometimesothersymbolsareusedinstead.Theyareusedalotwhenwritingmathematicsonablackboardtosavetime.They'renotascommonintextbooks.Besidesthesesymbols,thesymbol*standsfor\since",andthesymbol)standsfor\therefore".Theyrarelyappearintextbooks,butoftenonblackboards.Anexampleofuniversalquanti cationistheexpression8x;(x�2)x2�4)whichmeansforallx,ifxisgreaterthan2,thenx2isgreaterthan4.Typicallyconditionslikex�2afteruniversalquanti ersareincludedinthequanti ersothattheimplicationdoesn'thavetobeexpressedseparately.Thatlastexpressioncanbeabbreviatedas8x�2;x2�4.143
144APPENDIXA.BACKGROUNDMATHEMATICSOperation,symbolReadAsExplanation
Conjunction,^andThestatementA^BistrueifAandBarebothtrue;elseitisfalse.Disjunction,_(inclusive)orThestatementA_BistrueifAorB(orboth)aretrue;ifbotharefalse,thestatementisfalse.Negation,:notThestatement:AistruejustwhenAisfalseImplication,)implies;if...thenA)BmeansifAistrue,thenBisalsotrue;ifAisfalse,thennothingissaidaboutB.Bi-implication,()\i ",ifandonlyifA()BmeansbothA)BandB)A.Universalquanti cation,8forall;forany;foreachwhenit'strueuniversallyExistentialquanti cation,9thereexists;thereisanwhenthere'satleastoneUniqueexistentialquanti -cation,9!thereexistsauniquewhenthereisexactlyoneTableA.1:StandardlogicalsymbolsAnexampleofexistentialquanti cationistheexpression9x;(x�1^x2=4)whichmeansthereisanxsuchthatxisgreaterthan1andx2=4.Typicallyconditionslikex�1afterexistentialquanti ersareincludedinthequanti ersothattheconjunctiondoesn'thavetobeexpressedseparately.Thatlastexpressioncanbeabbreviatedas9x�1;x2=4.A.2SetsJustalittlebitaboutsets.We'llusethelanguageofsetsthroughoutthecourse,butwe'renotusingmuchofsettheory.ThisnotejustcollectsthebackgroundthatyouneedtoknowaboutsetsinoneplaceA.2.1BasicsettheoryAsetitselfisjustsupposedtobesomethingthathaselements.Itdoesn'thavetohaveanystructurebutjusthaveelements.Theelementscanbeanything,butusuallythey'llbethingsofthesamekind.Ifyou'veonlygotoneset,however,there'snoneedtoevenmentionsets.It'swhenseveralsetsareunderconsiderationthatthelanguageofsetsbecomesuseful.Therearewaystoconstructnewsets,too,andtheseconstructionsareimportant.Themostimportantoftheseisawaytoselectsomeoftheelementsinasettoformanotherset,asubsetofthe rst.Examples.Let'sstartwithsetsofnumbers.Therearewaysofconstructingthesesets,butlet'snotdealwiththatnow.Let'sassumethatwealreadyhavethesesets.
A.2.SETS145Thenaturalnumbers.Thesearethecountingnumbers,thatis,wholenonnegativenum-bers.Thatmeanswe'llinclude0asanaturalnumber.(Sometimes0isn'tincluded.)ThereisastructureonN,namelythereareoperationsofaddition,subtraction,etc.,butasaset,it'sjustthenumbers.You'lloftenseeNde nedasN=f0;1;2;3;:::gwhichisreadas\Nisthesetwhoseelementsare0,1,2,3,andsoforth."That'sjustaninformalwayofdescribingwhatNis.Acompletedescriptioncouldn'tgetawaywith\andsoforth."Ifyouwanttoseeallofwhat\andsoforth"entails,youcanreadDedekind's1888paperWassindundwassollendieZahlen?andJoyce'scommentsonit.InthatarticleDedekindstartso developingsettheoryandendsupwiththenaturalnumbers.Therealnumbers.Theseincludeallpositivenumbers,negativenumbers,and0.Besidesthenaturalnumbers,theirnegationsand0areincluded,fractionslike22
7,algebraicnumberslikep
5,andtranscendentalnumberslikeande.Ifanumbercanbenameddecimallywithin nitelymanydigits,thenit'sarealnumber.We'lluseRtodenotethesetofallrealnumbers.LikeN,Rhaslotsofoperationsandfunctionsassociatedwithit,buttreatedasaset,allithasisitselements,therealnumbers.NotethatNisasubsetofRsinceeverynaturalnumberisarealnumber.Elementsandmembership.ThestandardnotationtosayanelementxisamemberofasetSisx2S.The2symbolvariesabit.Sometimesitappearsasanepsilonor"orE.Readx2Sas\xisanelementofS,"oras\xbelongstoS,ormoresimply\xisinS."It'snegationisthesymbol=2.So,forexamplep
52R,butp
5=2N.Asmentionedabove,setsarecompletelydeterminedbytheirelements,sotwosetsareequaliftheyhaveexactlythesameelements.S=Tifandonlyif(1)forallx2S;x2T;and(2)forallx2T;x2S:Thetwohalvesoftheconditionontherightleadtotheconceptofsubset.Subsets.Ifyouhaveasetandalanguagetotalkaboutelementsinthatset,thenyoucanformsubsetsofthatsetbypropertiesofelementsinthatlanguage.Forinstance,wehavearithmeticonR,sosolutionstoequationsaresubsetsofR.Thesolutionstotheequationx3=xare0,1,and�1.WecandescribeitssolutionsetusingthenotationS=fx2Rjx3=xgwhichisreadas\SisthesetofxinRsuchthatx3=x."Wecouldalsodescribethatsetbylistingitselements,S=f0;1;�1g.Whenyounameasetbylistingitselements,theorderthatyounamethemdoesn'tmatter.WecouldhavealsowrittenS=f�1;0;1gforthesameset.ThissetSisasubsetofR.AsetSisasubsetofasetTifeveryelementofSisalsoanelementofT,thatisSTifandonlyifforallx2S;x2T:ReadSTas\SisasubsetofT."
146APPENDIXA.BACKGROUNDMATHEMATICSNotethatS=TifandonlyifSTandTS.Thereareacoupleofnotationsforsubsets.We'llusethenotationAStosaythatAisasubsetofS.WeallowSS,thatis,weconsiderasetStobeasubsetofitself.IfasubsetAdoesn'tincludealltheelementsofS,thenAiscalledapropersubsetofS.TheonlysubsetofSthat'snotapropersubsetisSitself.We'llusethenotationAStoindicatethatAisapropersubsetofS.(Warning.There'sanalternatenotationalconventionforsubsets.InthatnotationASmeansAisanysubsetofS,whileA(SmeansAisapropersubsetofS.Ipreferthethenotationwe'reusingbecauseit'sanalogoustothenotationsforlessthanorequal,andforlessthan.)Operationsonsubsets.Frequentlyyoudealwithseveralsubsetsofaset,andthereareoperationsofintersection,union,anddi erencethatdescribenewsubsetsintermsofpreviouslyknownsubsets.TheintersectionA\BoftwosubsetsAandBofagivensetSisthesubsetofSthatincludesalltheelementsthatareinbothAandB,asshownintheVenndiagrambelow.(It'sinterestingthatVenncalledthemEulercirclesasEulerhadusedthemearlier,butLeibnizhadalsousedthem,andRamonLlull(RaymondLully)inthe13thcentury.)ReadA\Bas\theintersectionofAandB"oras\AintersectB."Notethattheoperationofinter-sectionisassociativeandcommutative.
A
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A\B=fx2Sjx2Aandx2Bg:TwosetsAandBaresaidtobedisjointiftheirunionisempty,A\B=?.Severalsetsaresaidtobepairwisedisjointifeachpairofthosesetsaredisjoint.
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A[B=fx2Sjx2Aorx2Bg:TheunionA[BoftwosubsetsAandBofagivensetSisthesubsetofSthatincludesalltheele-mentsthatareinAorinBorinboth.ReadA[Bas\theunionofAandB"oras\AunionB."Likeintersection,theoperationofunionisalsoassocia-tiveandcommutative.Itisusualinmathematicstotaketheword\or"tomeananinclusiveor.Itimplicitlyincludes\orboth."Intersectionandunioneachdistributeovertheother:(A\B)[C=(A[C)\(B[C)(A[B)\C=(A\C)[(B\C)
A.2.SETS147Thedi erenceA�BoftwosubsetsAandBofagivensetSisthesubsetofSthatincludesalltheelementsthatareinAbutnotinB.
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A�B=fx2Sjx2Aandx=2BgThere'salsothecomplementofasubsetAofasetS.ThecomplementisjustS�A,alltheelementsofSthataren'tinA.WhenthesetSisunderstood,thecomplementofAoftenisdenotedmoresimplyaseitherAc,
A,orA0ratherthanS�A.IpreferthenotationAc.Theseoperationssatisfylotsofidentities.I'lljustnameacoupleofimportantones.DeMorgan'slawsdescribeadualitybetweenintersectionandunion.Theycanbewrittenas(A\B)c=Ac[Bcand(A[B)c=Ac\BcUnionsandintersectionssometimesaretakenofmanysubsets,evenin nitelymany.SupposethatA1;A2;:::;AnaresubsetsofS.Theintersectionofallofthemcanbewritteninanindexednotationasn\i=1Ai=A1\A2\\Anandtheirunionasn[i=1Ai=A1[A2[[An:Andwhentherearein nitelymany,A1;A2;:::;An;:::,as1\i=1Ai=fx2Sjx2Aiforalligandtheirunionas1[i=1Ai=fx2Sjx2Aiforatleastoneig:DeMorgan'slawsandthedistributivitylawsalsoapplytoindexedintersectionsandunions. n\i=1Ai!c=n[i=1Aic n[i=1Ai!c=n\i=1Aic n\i=1Ai![C=n\i=1(Ai[C) n[i=1Ai!\C=n[i=1(Ai\C)
148APPENDIXA.BACKGROUNDMATHEMATICSPartitions.AsetSissaidtobepartitionedintosubsetsA1;A2;:::;AnwheneachelementofSbelongstoexactlyoneofthesubsetsA1;A2;:::;An.That'slogicallyequivalenttosayingthatSisthedisjointunionoftheA1;A2;:::;An.WhenyouhaveapartitionA1;A2;:::;AnofasetSlikethat,itinducesapartitionE\A1;E\A2;:::;E\AnoneachsubsetEofS.EachelementofEbelongstoexactlyoneofitssubsetsE\A1;E\A2;:::;E\An.Productsofsets.Sofarwe'velookedatcreatingsetswithinset.Therearesomeoper-ationsonsetsthatcreatebiggersets,themostimportantbeingcreatingproductsofsets.Thesedependontheconceptoforderedpairsofelements.Thenotationfororderedpair(a;b)oftwoelementsextendstheusualnotationweuseforcoordinatesinthexy-plane.Theimportantpropertyoforderedpairsisthattwoorderedpairsareequalifandonlyiftheyhavethesame rstandsecondcoordinates:(a;b)=(c;d)i a=candb=d:TheproductoftwosetsSandTconsistsofalltheorderedpairswherethe rstelementcomesfromSandthesecondelementcomesfromT:ST=f(a;b)ja2Sandb2Tg:Thus,theusualxy-planeisRR,usuallydenotedR2.BesidesbinaryproductsST,youcananalogouslyde neternaryproductsSTUintermsoftriples(a;b;c)wherea2S,b2T,andc2U,andhigherproducts,too.Setsofsubsets;powersets.Anotherwaytocreatebiggersetsistoformsetsofsubsets.IfyoucollectallthesubsetsofagivensetSintoaset,thenthesetofallthosesubsetsiscalledthepowersetofS,denoted}(S)orsometimes2S.Forexample,letSbeasetwith3elements,S=fa;b;cg.ThenShaseightsubsets.Therearethreesingletonsubsets,thatis,subsetshavingexactlyoneelement,namelyfag,fbg,andfcg.Therearethreesubsetshavingexactlytwoelements,namelyfa;bg,fa;cg,andfb;cg.There'sonesubsethavingallthreeelements,namelySitself.Andthere'sonesubsetthathasnoelements.Youcoulddenoteitfg,butit'salwaysdenoted?andcalledtheemptysetornullset.Thus,thepowersetofShaseightelements}(S)=f?;fag;fbg;fcg;fa;bg;fa;cg;fb;cg;Sg:Cardinality,countableversusuncountablesets.ThecardinalityofasetSisthenumberofelementsinit,denotedjSj.So,forexample,ifSfa;b;cg,thenjSj=3,andj}(S)j=23=8.Somesetsarein nite,sotheircardinalityisnota nitenumber.Amorecarefulde nitionisneeded.TwosetsSandTaresaidtohavethesamecardinalityifthereisaone-to-onecorrespondenceoftheirelements.Thatmeansthatthereissomefunctionf:S!Twhichisinjective(alsocalledone-to-one)andsurjective(alsocalledonto).Afunctionwhichisbothinjectiveandsurjectiveiscalledabijection.Forabijectionf:S!T,theinversefunctionf�1:T!Sisalsoabijection.ThenotationjSj=jTjindicatesSandThavethesamecardinality.
A.2.SETS149IfthereisaninjectionS!T,thenthecardinalityofSislessthanorequaltothatofT,writtenjSjjTj.Itisevidentthatisatransitiverelationoncardinalities.TheSchroder-BernsteintheoremstatesthatifthereareinjectionsbothwaysbetweenSandT,thentheyhavethesamecardinality.Thus,isapartialorderoncardinalities.ThenotationjSjjTjmeansjSjjTjbutnotjSj=jTj.AsGeorgCantor(1845{1918)discovered,notallin nitesetshavethesamecardinality.Somein nitesetsarebiggerthanothers.Usinghisfamousdiagonalproof,heprovedthatforanyset,evenifit'sin nite,jSjj}(S)j.Thesmallestsizeanin nitesetcanbeisthatofthenaturalnumbersN.AsetthathasthesamecardinalityasNiscalledacountablyin niteset.Anin nitesetthatdoesn'thavethesamecardinalityasNiscalledanuncountableset.ThesetofrealnumbersRisuncountable.Finitesetsarealsosaidtobecountable.Thus,asetiscountableifit'seither niteorcountablyin nite.A.2.2FunctionsandrelationsAfunctionfisassociatedtoapairofsets,adomainSandacodomainT.Theusualnotationsforthataref:S!TandSf!T.Inordertobeafunction,eachelementx2SmustbeassociatedtoaparticularelementofT,denotedf(x).ThegraphofafunctionfisasubsetoftheproductST,namely,thesetF=f(x;y)2STjy=f(x)g.Twofunctionsaresaidtobethesameiftheyhavethesamegraph,sothegraphcharac-terizesthefunction.Frequently,textbooksde neafunctionf:S!Tasitsgraph,thatis,asubsetFofSTsuchthatforallx2X,thereisauniquey2Tsuchthat(x;y)2F.Whenf:S!T,itissaidthatfmapsStoT,andthatfmapsxtof(x).Thiselementf(x)iscalledtheimageofxunderf.Themappingofxtof(x)isdenotedx7!f(x).Theconceptofimageisextendedtosubsetsofthedomain.IfAS,thenfmapsAtothesetf(A)=ff(x)jx2Ag,calledtheimageofAunderf.Anotherrelatedconceptisthatofpreimage,alsocalledinverseimage.IfBisasubsetofthecodomainT,thenthepreimageofBunderfisthesetf�1(B)=fx2Ajf(x)2Bg.Composition.Iff:S!Tandg:T!U,thenthecompositiongf:S!Uisde nedby(gf)(x)=g(f(x)).Compositionisassociative.(hg)f=h(gf).Sincecompositionisassociative,parenthesesarenotnecessarywhencomposingthreeormorefunctions.ForeachsetSthereisanidentityfunction1S:S!SwhichmapseveryelementinStoitself,1S(x)=x.Theidentityfunctionsactasunitsforcomposition.Iff:S!T,then1Tf=fandf=f1S.Injections,surjections,andbijections.Thesearewordsthatdescribecertainfunctionsf:S!Tfromonesettoanother.Aninjection,alsocalledaone-to-onefunctionisafunctionthatmapsdistinctelementstodistinctelements,thatis,ifx6=y,thenf(x)6=f(y).Equivalently,iff(x)=f(y)thenx=y.IfSisasubsetofT,thenthereisanaturalinjection:S!T,calledtheinclusionfunction,de nedby(x)=x.
150APPENDIXA.BACKGROUNDMATHEMATICSAsurjection,alsocalledanontofunctionisonethatincludesallofTinitsimage,thatis,ify2T,thenthereisanx2Ssuchthatf(x)=y.Abijection,alsocalledaone-to-onecorrespondence,isafunctionthatissimultaneouslyinjectiveandbijective.Anotherwaytodescribeabijectionistosaythatthereisaninversefunctiong:T!Ssothatthecompositiongf:S!SistheidentityfunctiononSwhilefg:T!TistheidentityfunctiononT.Theusualnotationforthefunctioninversetofisf�1.Inthissituationfandgareinversetoeachother,thatis,ifgisf�1,thenfisg�1.Thus,(f�1)�1=f.Relations.Relationsincludefunctions,butaremoregeneral.AbinaryrelationR:S!Tdoesn'thavetoassociateeachelementofStoexactlyoneelementofT.ItcanassociateanelementofStoanynumberofelementsinTincludingthepossibilityofnoelementsinTatall.Inotherwords,arelationR:S!TisdeterminedbyanarbitrarysubsetofST.Themostusefulrelationsarethosethathavespecialproperties.Thenextsectiondiscussesequivalencerelations.Atypicalequivalencerelationiscongruencemodulon.OrderrelationsarediscussedinsectionA.3.Atypicalorderrelationisonnumbers.A.2.3EquivalencerelationsTherearevarioussymbolsusedforequivalencerelations,suchas=,,,,',,andsoforth.We'lluseforagenericequivalencerelation.De nitionA.1(Equivalencerelation).AnequivalencerelationonasetSisarelationthatisre exive,symmetric,andtransitive.ArelationonasetSmaybeidenti edwithasubsetoftheproductSS.Foranequivalencerelation,thismeansxycorrespondstothestatementthattheorderedpair(x;y)isanelementofthatsubset.Re exivity:Forallx,xx:Symmetry:Forallxandy,xyimpliesyx:Transitivity:Forallx,y,andz,xyandyzimpliesxz:Equivalenceclassesandpartitionsofsets.Anequivalencerelationonasetdeterminesapartitiononthatset,andconversely,aswe'llseepresently.De nitionA.2(Equivalenceclass).Givenanequivalencerelationonaset,anequivalenceclassofanelementx,denoted[x],isthesetofallelementsequivalenttox,[x]=fyjyxg:Youcaneasilyshowtheseveralpropertiesofequivalenceclasses.TheoremA.3.IfisanequivalencerelationonasetS,thenthefollowingfourstatementsareequivalent1.xy.2.[x]=[y].
A.2.SETS1513.x2[y].4.[x]\[y]6=?.Furthermore,foreachx2S,thereisexactlyoneequivalenceclasscontainingx,namely,[x].De nitionA.4(Partitionofaset).ApartitionofasetSisacollectionofnonemptysubsets,calledparts,ofSwhicharepairwisedisjointandwhoseunionisallofS.Thus,eachelementofSbelongstoexactlyoneoftheparts.Theabovetheoremshowsthattheequivalenceclassesformapartition.Theconverseisalsotrueasyoucaneasilyshow.TheoremA.5.Foreachequivalenceclassonaset,theequivalenceclassespartitiontheset.Conversely,apartitionofasetdeterminesanequivalencerelationwheretwoelementsareequivalentifthey'reinthesamepart.ThesetofequivalenceclassesissometimesdenotedS=,andit'ssometimescalledaquotientset.Usingequivalenceclassestoconstructnewsetsofthingsisacommonpracticeinmathematicsandespeciallyinalgebra.KeepinmindthatyoucanalwaysnameanelementofS=bynaminganelementofS,buttwoelementsxandyofSwillnamethesameelementofS=,thatis,[x]=[y],ifxy.Thefunction :S!S=de nedby (x)=[x]iscalledaprojection,orthecanonicalfunction,fromthesettoitsquotientset.A.2.4AxiomsofsettheoryAlthoughtheaxiomsofsettheorydon'tplayanimportantroleinanintroductorycourseinmodernalgebra,occasionallytheymaybeuseful.HereisasummaryofaxiomsofZermelo-Fraenkelsettheory,abbreviatedZFsettheory.Axiomofextensionality.Thisistheaxiomthatsaystwosetsarethesameiftheyhavethesameelements.8A;8B;(8x;(x2A,x2B)()A=B):Axiomofseparation.Thisaxiomisalsocalledtheaxiomofspeci cation.Itsaysifyouhaveapredicate'onsetsandagivensetA,thenthereisasubsetBofAonwhichthatpredicateholds.8A;9B;8x;(x2B()x2S^'(x)):It'sanaxiomschemaratherthanasingleaxiombecauseadi erentaxiomisneededforeachpredicate'.Thisaxiomallowsthecreationofsmallersetsfromagivenset.Forexample,ifA=Risthesetofrealnumbers,bytheaxiomofseparationthereisasetBsuchthattheelementsofBaretherealnumbersthatsatisfytheequationx3�3x=1.Here,thepredicate'atx,writtenaboveas'(x)isthatequation.Theaxiomofseparationisthejusti cationforthe\setbuilding"notationB=fx2Rjx3�3x=1g.
152APPENDIXA.BACKGROUNDMATHEMATICSAxiomofpairing.Theaxiomofpairingallowsthecreationofasetcontainingtwoele-ments(oroneifthey'rethesameelement).8x;8y;9A;(z2A()z=x_z=y):ThesetAisusuallydenotedfx;yg.Ifithappensthatx=y,thenAonlyhasoneelementinsteadoftwosincefx;xg=fxg.Ansetwithonlyoneelementiscalledasingletonset,orjuatasingleton.Axiomofunion.GivenasetAofsets,thissaystheunionCofthesetsinAisaset.8A;9C;8x;(x2C()9B;(x2B^B2A):TheusualnotationforCisSA,orSB2AB.WhenAisthepairfD;EgthenSB2ABisthepairwiseunionD[E.Theredoesn'tneedtobeanaxiomforintersectionsorforrelativecomplimentsbecauseintersectionsandrelativecomplementscanbeprovedfromtheaxiomofseparation.Axiomofpowersets.ItsaysgivenasetA,thereisasetwhichcontainsallthesubsetsofA.8A;9B;8C;(C2B()CA):OnecommonnotationforthepowersetBofAis}(A).Axiomofin nity.Sofar,therearenoaxiomsthatsaythereareanysetsatall.Thisaxiomsaysthatthereisanin nitesetwhichcontainstheemptyset?,soamongotherthings,itsaystheemptysetexists.Whenstudyingthetheoryof nitesets,theaxiomofin nityisnotincluded,butanexplicitaxiomisneededtosaytheemptysetexists.De neS(A)todenoteA[fAgwhereAisaset.S(A)iscalledthesuccessorofA.TheaxiomofpairingsaysthatifAisaset,thensoisfAg,andtheaxiomofunionthenimpliesthatA[fAgisaset.Theaxiomofin nitysaysthatthereisasetBthathas?asanelementandisclosedunderS.9B;(?2B^8y2B;S(y)2B):Alongwiththeaxiomofregularity,thisaxiomimpliesthatthereisatleastonein niteset.Withtheotheraxioms,itcanbeshownthatthereisasmallestsuchset.ThatsmallestsetisamodelforthesetofnaturalnumbersN.Inthatmodel,theemptyset?actsas0,it'ssuccessorS(?)actsas1,S(S(?))actsas2,andsoforth.Axiomofregularity.Thisaxiomsisalsocalledtheaxiomoffoundation.ItisatechnicalaxiomthatsaysthatgivenanonemptysetA,thereisanelementofAwhichisdisjointfromit.8A6=?;9x2A;8y2x;y=2A:Theaxiomofregularityimpliesthatnosetisanelementofitself,noristherea nitecycleofmembershipswhereA12A222An2A1.Furthermore,thereisnoin nitedescendingmemberships2An2A22A1.Oneofthemainreasonsfortheaxiomofregularityistodevelopthetheoryofordinals.
A.3.ORDEREDSTRUCTURES153Axiomofreplacement.Liketheaxiomofseparation,thisisanotheraxiomschema.Thistechnicalaxiomcreatesimagesoffunctionsdescribedbypredicates.Apredicate'describesafunctionifforallx,thereexistsauniqueysuchthat'(x;y).Inthatcase,afunctionsymbollikefisusedsothatf(x)=yexpresses'(x;y).(Forthisaxiom,thepredicatecanhaveotherargumentsthatwon'tbementionedexplicitly.)8A;((8x2A;9!y;'(x;y)))9B;8x2A;9y2B;'(x;y):TheBintheaxiomisusuallydenotedf(A),theimageofAunderf.Axiomofchoice.TheaxiomofchoiceisnotpartofZFsettheory,butwhenit'sincluded,thesettheoryisdenotedZFCsettheory,Zermelo-Fraenkelsettheorywiththeaxiomofchoice.ThisaxiomisdiscussedinmoredetailinthesectionA.4.VonNeumann{Bernays{Godelsettheory(NBG).ThisisanextensionofZFthatincludesproperclasses.Whereassetscanbeelementsofothersetsandproperclasses,properclassescannotbeelements.NBGisaconservativeextensionofZFinthatsensethatanytheoremnotmentioningclassesandprovableinonetheorycanbeprovedintheother.NBGmakesitpossibletotalkaboutthingsliketheclassofallsets,ortheclassofallgroups,etc.A.3OrderedstructuresSeveralmathematicalstructuresarede nedintermsofanorderrelation.Theseorderrela-tionshavesomethingincommonwiththeorderrelation\lessthanorequal"ontherealnumbers.Manyofthemarenottotalorderslike,butonlypartialorders.Havingfewernicepropertiesthan,however,canmakethemmoreinteresting.Inparticular,we'lllookatpartialorders,lattices,andBooleanalgebras.A.3.1Partialordersandposets.You'refamiliarwiththeorderonrealnumbers.It'sabinaryrelationwiththefollowingfourproperties.1.Re exivity:forallx,xx.2.Anti-symmetry:forallxandy,ifxyandyx,thenx=y.3.Transitivity:forallx,y,andz,ifxyandyz,thenxz.4.Totality:forallxandy,eitherxyoryx(orbothinwhichcasex=y.Thereareotherusefulbinaryrelationsinmathematicswitheitherthosefourpropertiesoratleastthe rstthree.Althoughsometimessuchbinaryrelationsaredenotedwiththesamesign,frequentlyasimilarbutvisuallydistinctsignsuchasisused.Bothareread\lessthanorequalto".Ofcourse,there'salsoagreaterthanorequalto,writtenandde nedbyxyifandonlyifyx.
154APPENDIXA.BACKGROUNDMATHEMATICSDe nitionA.6(Totalorder).Atotalorder,alsocalledalinearorder,onasetisabinaryrelationhavingthefourproperties:re exivity,anti-symmetry,transitivity,andtotality.Asetwithaspeci edtotalorderiscalledatotallyorderedsetorachain.Thestrictformofatotalorder(orapartialorderde nedbelow)isde nedbyxyifanadonlyofxyandx6=y:Ausefulweakeningoftotalordersiswhatiscalledapartialorder.De nitionA.7(Partialorder).Apartialorderonasetisabinaryrelationhavingthe rstthreeofthoseproperties:re exivity,anti-symmetry,andtransitivity.Asetwithaspeci edpartialorderiscalledapartiallyorderedsetorposetforshort.Twoelementsxandyinapartiallyorderedsetaresaidtobecomparableifeitherxyoryx.Otherwisethey'reincomparable.ExampleA.8.Thepositiveintegersarepartiallyorderedbydivisibility.Divisibilityisre exivesincex x;it'santi-symmetricsinceifx yandy x,thenx=y,andit'stransitivesinceifx yandy z,thenx z.Divisibilityisnotapartialorderonallintegerssince2 �2and�2 2but26=�2.Itis,however,apre-order.Apre-orderisre exiveandtransitivebutneednotbeanti-symmetric.ExampleA.9.AnycollectionTofsubsetsofasetSisapartiallyorderedsetwherethebinaryrelationis.Inparticular,thepowerset}(S)consistingofallthesubsetsofSispartiallyordered.Hassediagrams.ApartiallyorderedsetcanbedescribedwithakindofagraphcalledaHassediagram.Theelementsofthesetaretheverticesofthegraph,andtheedgesindicatewhichelementsarelessthanorequaltowhichotherelements.Ifab,thenanedgeisdrawnfromatobwiththelargerelementabovethesmallerone.Transitivityoftheorderrelationisassumedsothatifabc,thananedgedoesn'thavetobedrawnbetweenaandc.De nitionA.10.AnupperboundofasubsetSinaposetisanyelementintheposetwhichisgreaterthanorequaltoallelementsinS.Thatelementneedn'tbeanelementofthesubsetS.Likewise,alowerboundofSisanelementthatislessthanorequaltoalltheelementsinS.Aleastupperbound,alsocalledsupremumofSisanupperboundofSwhichislessthanorequaltoallotherupperboundsofS.ItisdenotedlubSorsupS.Likewise,agreatestlowerbound,alsocalledinfemumofSisanlowerboundofSwhichisgreaterthanorequaltoallotherlowerboundsofS.ItisdenotedglbSorinfS.Leastupperboundsandgreatestlowerboundsofsubsetsneednotalwaysexist.ExampleA.11.WiththeusualorderingontherealnumbersR,boththeopeninterval(2;3)andtheclosedinterval[2;3]havethesameleastupperbound3andthesamegreatestlowerbound2.WiththeusualorderingontherationalnumbersQ,thesubsetS=fxjx2=2ghasneitheraleastupperboundnoragreatestlowerboundsincep
2and�p
2arenotrationalnumbers.
A.3.ORDEREDSTRUCTURES155De nitionA.12(Maximalandminimalelements).Amaximalelementinapartiallyorderedsetisanelementwhichisnotlessthanorequaltoanyotherelement.Aminimalelementinapartiallyorderedsetisanelementwhichisnotgreaterthanorequaltoanyotherelement.Maximalandminimalelementsdon'thavetobeunique.Apartiallyorderedsetcanhavemorethanoneofeachnononeatall.De nitionA.13(Meetandjoin).Themeetoftwoelementsaandbisthegreatestlowerboundofthesetfa;bg.Thatis,itisanelementxlessthanorequaltobothaandbandgreaterthanorequaltoallotherelementsgreaterthanorequaltobothaandb.Ifthatmeetexists,itisdenoteda^b.Thejoinoftwoelementsaandbinapartiallyorderedsetistheleastupperboundofthesetfa;bg.Thatis,itisanelementxgreaterthanorequaltobothaandbandlessthanorequaltoallotherelementsgreaterthanorequaltobothaandb.Ifthatjoinexists,itisdenoteda_b.Meetsandjoinsaren'tparticularlyinterestingintotallyorderedsets.Inatotallyorderedset,themeetoftwoelementsistheminimumofthetwowhilethejoinoftwoelementsisthemaximumofthetwo.ExampleA.14.Considerthepositiveintegerspartiallyorderedbydivisibility.Themeetoftwointegersmandnisthatnumberdwhichdividesthembothforwhichanyotherdivisorofbothdividesd.Inotherwords,ameetinthispartiallyorderedsetisthegreatestcommondivisor.Likewise,ajoinistheleastcommonmultiple.ExampleA.15.Sometimesmeetsandjoinsdon'texistinapartiallyorderedset.Considertheposetwithfourelements,fa;b;c;dgwherebothaandbarelessthanbothcandd.
a
b
c
d
Thejoinc_ddoesn'texistsincethereisnoupperboundforcandd.Thejoina_bdoesn'texistbecausetherearetwoupperboundsforaandb,butnoleastupperbound.Likewise,thetwomeetsc^danda^bdon'texist.A.3.2LatticesLatticesarepartiallyorderedsetsthathavemeetsandjoinsDe nitionA.16(Lattice).Alatticeisapartiallyorderedsetinwhichallmeetsandjoinsoftwoelementsexist,hasasmallestelement(oftendenoted0or?)andalargestelement(oftendenoted1or�),inwhichthefollowingidentitieshold.Idempotency:x=x^x=x_x.Commutativity:x^y=y^xandx_y=y_x.Associativity:(x^y)^z=x^(y^z)and(x_y)_z=x_(y_z).Absorption:a^(a_b)=aanda_(a^b)=a.Identity:a^1=aanda_0=0.Latticescanbede nedwithoutreferencetoapartialorderastherelationabcanbecharacterizedintermsofmeetsandjoinsasinthefollowingtheorem.
156APPENDIXA.BACKGROUNDMATHEMATICSTheoremA.17.Thefollowingthreeconditionsareequivalent:ab,a^b=aanda_b=b.Proof.First,supposeab,thenbyde nition,themeetofaandbisawhilethejoinofaandbisb.Thus,the rstconditioninthestatementofthetheoremimpliestheothertwo.Nowsupposea^b=a,sincea^bb,thereforeab.Thusthesecondconditionimpliesthe rst.Similarly,thethirdconditionimpliesthe rst.q.e.d.Sincecanbecharacterizedintermsof^and_,thereisanalternatede nitionoflattice.De nitionA.18(Alternatede nitionoflattice).Alatticeisasetequippedwithtwobinaryoperations^and_andtwoconstants0and1whichsatisfytheidentitiesinthepreviousde nition.Itfollowsfromthatde nitionthata=a^bifandonlyifb=b_a.(Proof:a=a^bimpliesb=b_(b^a)=(a^b)_b=a_bbycommutativityandabsorption.)Thepartialordercanthenberecoveredbyde ningabifandonlyifa^b=aanda_b=b.Thereareacoupleotheridentitiesthatfollowfromthede nition,namely,0^a=0and1_a=1.ExampleA.19.Thepowerset}(S)ofasetSisalattice.(It'sactuallyaBooleanring,discussedlater.)Here'stheHassediagramfor}(fa;b;cg).
fa;b;cg
fb;cg
fa;cg
fa;bg
fcg
fbg
fag
?
Thepowersetofasetwithfourelementshas16elements.It'salittlehardertodrawasaHassediagramwhichisdisplayedin gureA.1.Thenamesofthesubsetsareabbreviatedsothat,forexample,thesubsetfa;b;cgisdisplayedasabc.Modularanddistributivelattices.Notethatdistributivityisnotlistedamongtheidentitiesabove.That'sbecauseitdoesn'tholdinalllattices.Anotheridentitythatdoesn'tholdinalllatticesismodularity.De nitionA.20.Alatticeissaidtobemodularifforallelementsa,b,andcforwhichac,itisthecasethata_(b^c)=(a_b)^c.Alatticeissaidtobedistributiveifforallelementsa,b,andc,itisthecasethata^(b_c)=(a^b)_(a^c)anda_(b^c)=(a_b)^(a_c).
A.3.ORDEREDSTRUCTURES157
abcd
abc
abd
acd
bcd
ab
ac
ad
abc
bd
cd
a
b
c
d
?
FigureA.1:LatticeofthePowersetof4elementsItcanbeshownthateverydistributivelatticeisalsomodular,buttherearemodularlatticesthatarenotdistributive.Itcanalsobeshownthateitheroneofthedistributiveidentitiesimplytheother.Thepowerset}(S)latticeisadistributivelattice.Thesubgroupsofagroupwithinclusionasapartialorderalwaysformamodularlattice,butnotalwaysadistributivelattice.ExampleA.21.ThesymmetricgroupS3hasfoursubgroupsbesidesitselfandthetrivialsubgroup.Thesubgroupgen-eratedbythepermutation(123)hasorder3whilethethreesubgroupsgeneratedbythethreetrans-postions(12),(13),and(23)eachhaveorder2.Thelatticeofsubgroupsismodular,butit'snotdis-tributive.
S3
(123)
(12)
(13)
(23)
1
A.3.3Booleanalgebras.ABooleanalgebraisadistributivelatticewithonemoreoperation.De nitionA.22(Booleanalgebra).ABooleanalgebraisadistributivelatticewithaunaryoperation,calledcomplementationornegation,denoted:satisfyingtheidentitiesa_:a=1anda^:a=0.Actually,notalltheidentitiesfrombooleanlatticesarenecessaryforthede nitionsinceabsorptioncanbeshownfromtherest.Otheridentitiesthatfollowfromthede nitioninclude:0=1,:1=0,and::a=a.Asdescribedinsection3.3,BooleanalgebrasarethesamethingasBooleanrings.Theonlydi erenceisnotational.Truthvalues.Thetwo-elementBooleanalgebrathatconsistsonlyof0and1isisusedinlogic.0,or?isthetruthvalue\false"while1,or�isthetruthvalue\true".
158APPENDIXA.BACKGROUNDMATHEMATICSA.4AxiomofchoiceGivenacollectionofnonemptysets,theaxiomofchoicesaysthatthereisafunctionthatchoosesoneelementfromeachset.Thisisanaxiomofsettheory.Therearemanyaxiomsofsettheory,mostofwhicharefairlyobviousanduncontroversial.Moreprecisely,theaxiomofchoicesaysthatgivenanysetS,thereexistsa\choicefunction" :}(S)�?!SwhichchoosesfromanynonemptysetTSanelement (T)2T.Insomesense,anytheoremthatreliesontheaxiomofchoiceis awedsincetheaxiomofchoiceisnotconstructive.So,forinstance,afterprovinganidealisasubidealofamaximalideal,wewon'thaveanywaytoidentifythatmaximalideal.Here'sasimpletheoremthatreliesontheaxiomofchoice.TheoremA.23.Letf:A!BbeasurjectivefunctionbetweensetsAandB.Thenthereexistsg:B!AsuchthatfgistheidentityfunctiononB.Proof.Let beachoicefunctionforA.Thengisthefunctiong(y)= (f�1(y))= (fxjf(x)=yg):Sincefissurjective,f�1(y)isnottheemptyset,sothechoicefunction willchoosesomeelementxoutoff�1(y)withf(x)=y.q.e.d.Thattheoremisactuallylogicallyequivalenttotheaxiomofchoice,thatis,theaxiomofchoicefollowsfromit.Independenceoftheaxiomofchoice.Theaxiomofchoiceisindependentoftherestoftheaxiomsofsettheory.Godelprovedin1938thatsettheorywiththeaxiomsofchoiceaddedisasconsistentassettheory,whileCohenin1963provedthatsettheorywiththenegationoftheaxiomofchoiceaddedisasconsistentassettheory.Inotherwords,theaxiomofchoiceisindependentoftherestoftheaxioms.A.4.1Zorn'slemmaAlthoughtheaxiomofchoiceiseasytostate,it'snotusuallyeasytouse.Zorn'slemma,whichislogicallyequivalenttotheaxiomofchoiceishardtostate,buteasytouse.Anotheristhewell-orderingprinciple.ThislemmaisappliedtoanonemptycollectionMofsubsetsofasetS.SectionA.3.1onpartially-orderedsetsde nedachain,upperbound,andmaximalele-ment.AchaininMisacollectionCofelementsofMlinearlyorderedbysubsetinclusion.Inotherwords,ifAandBareelementsofC,eitherABorBA.AnupperboundofCisasubsetBofSwhichcontainsallelementsofC.AmaximalelementBofMisonenotcontainedinanylargerelementofM.Zorn'slemma.IfeverychaininMhasanupperboundinM,thenMhasamaximalelement.Wewon'tprovethattheAxiomofChoiceisequivalenttoZorn'slemmabecauseitwouldtaketoolong.
A.4.AXIOMOFCHOICE159A.4.2Well-orderingprincipleThemostcommonformoftheaxiomsofchoiceusedinalgebraisZorn'slemma.Anotherthat'ssometimesusedisthewell-orderingprinciple.De nitionA.24(Well-ordering).Apartiallyorderedsetiswellorderedifeverynonemptysubsetofithasaleastelement.Itfollowsfromthede nitionthateverywell-orderingistotallyordered.Giventwoele-mentsxandy,thesubsetfx;yghasasmallestelement,eitherxinwhichcasexy,oryinwhichcaseyx.ExampleA.25.ThenaturalnumbersNiswell-orderedbyitsusualordering.TheintegersZisnotwellorderedbyitsusualorderingbecausetheentiresetdoesn'thaveasmallestelement.ForthesamereasonRisnotwell-ordered.Thenon-negativerealnumbersaren'twell-orderedbyitsusualorderingbecauseanyopeninterval(a;b)doesn'thavealeastelement.Anysubsetofawell-orderedsetiswell-orderedbythesameordering.Lexicographicordering.TheproductNNhasawell-orderingcalledtheemlexico-graphicordering.Theorderedpair(a;b)isde nedtobelessthanorequaltotheorderedpair(c;d)ifeithera=candbdorac.Thus,theelementsofNNlistedinincreasingorderare(0;0);(0;1);(0;2);:::;(1;0);(1;1);(1;2);:::;(2;0);(2;1);(2;2);:::;:::Moregenerally,ifAandBarebothwellordered,thenthelexicographicorderonABisawell-ordering.Furthermore, niteproductsA0A2A2ofwell-orderedsetsarewellorderedbyalexicographicordering.Thewell-orderingprinciple.Thisprinciplestatesthateverysethasawell-ordering,thatis,foreachset,thereissomewell-orderingofthatset.Theaxiomofchoice,Zorn'slemma,andthewell-orderingprinciplecaneachbeprovedfromtheeachother.Here'saproofthatthewell-orderingprinciplefollowsfromZorn'slemma.TheoremA.26.Thewell-orderingprinciplefollowsfromZorn'slemma.Proof.LetSbeaset.LetWbethesetofwell-orderingsofsubsetsofS.PartiallyorderWsogivensubsetsAandBbothwithwill-orderings,de neABifABandthetwoorderingsagreeonA.Inotherwords,theorderingonAextendstothatonB.TouseZorn'slemma,weneedtoshowthateverychainCinWhasanupperbound.AchainCconsistsofsubsetsAandBwhereifAB,theorderingofAisextendedtoB.TheunionofallthesesubsetsisasetCwhich,whengiventheextendedordering,soliesinW,isitselfawell-orderedsetthatcontainseverysubsetA2C.Thus,everychaininWhasanupperbound.ByZorn'slemma,WhasamaximalelementM.ThisMisawell-orderedsubsetofSwhichcannotbeextended(sinceit'smaximal).IftherewereanelementofS�M,the
160APPENDIXA.BACKGROUNDMATHEMATICSorderingonMcouldbeextendedtoawell-orderingbymakingthatelementlessthaneveryelementinM.Therefore,therearenoelementsinS�M.Thus,S=M,andsoShasawell-ordering.q.e.d.Thisprincipleimpliesthatthereissomewell-orderingoftherealnumbersR.It'snottheusualorder,ofcourse,sincetheusualorderdoesnotwellorderR.Infact,noparticularwell-orderingofRcaneverbedescribed.
IndexA4,108,112An,seeAlternatinggroupD3,8D5,108,111,119Dn,seeDihedralgroupF21=C7oC3,133GF(pn),seeGalois eldGL2(R),7GLn(R),seeGenerallineargroupPGLn(F),seeProjectivelineargroupS1,seeUnitcircleS2,seeSphereS3,see3-sphere,see3-sphereS4,119S5,108Sn,seeSymmetricgroupC,seeComplexnumbersC[x],seeComplexpolynomialsN,seeNaturalnumbersp,seePrimecyclotomicpolynomialsQ,seeRationalnumbersR,seeRealnumbersR[x],seeRealpolynomials),seeImplicationZ,seeIntegersZ[i],seeGaussianintegersZ3P2,128Zn,seeIntegersmodulon.-1backthrough,117*,since,143=,seeIsomorphism ,seeDivisibility,seeEquivalencerelation(modn),seeCongruencemodulon9,seeExistentialquanti cation8,seeUniversalquanti cation(),seeBi-implication^,seeConjunction_,seeDisjunctionH,seeQuaternionsC,seeCategoryG,seeCategoryofgroupsR,seeCategoryofringsS,seeCategoryofsets:,seeNegation(logical),seeDirectsum},seePowerset,seePartialorder),therefore,143.through,117p-group,138p-primarycomponent,138p-primarygroup,1383-Sphere,54Abel,NeilsHenrik(1802{1829),134Abeliangroup,7,99 nite,137{140ACC(ascendingchaincondition),83Anegeometry,127al-Khwarizm(ca.780{ca.850),1Algebra,1Boolean,157Cayley,52division,52Algebraic eldextension,42Algebraic elds,42Algebraicinteger,42Algebraicnumber,42Algebraicstructure,2{11Algebraicallyclosed eld,89AlgorithmBrahmagupta's,62division,84Euclidean,19{20,86extendedEuclidean,22QinJiushao's,62AlternatinggroupAn,107,112Antiautomorphism,52Antisymmetry,66Archimedeanordered eld,47{48ArchimedesofSyracuse(ca.287{212B.C.E.),47161
162INDEXAristotle(384{322B.C.E.),17Arrow,seemorphismAscendingchaincondition,83Associativity,2Automorphism,14 eld,43Axiomofchoice,153,158{160ofextensionality,151ofin nity,152ofpairing,151ofpowersets,152ofregularity,152ofreplacement,152ofseparation,151ofunion,152AxiomsDedekind/Peano,15 eld,31group,99ofsettheory,151{153ring,55Backthrough.-1,117Bernays,PaulIsaak(1888{1977),153Bernstein,Felix(1878{1956),148Bi-implication(),144Bijection,150Binaryoperation,2Binaryorderrelation,46Bombelli,Rafael(1526{1572),87Boole,George(1815{1864),56,63Booleanalgebra,157Booleanring,56,63{67Boundgreatestlower,154leastupper,154lower,154upper,154Brahmagupta(598{670),62Brahmagupta'salgorithm,62Cancellation,58Canonicalfunction,151Canonicalhomomorphism,38Cardano,Gerolamo(1501{1576),87Cardinality,101,148Category,69{74coproduct,136 nalobject,72,115generic,C,13initialobject,72,115ofAbeliangroupsA,136{137of elds,71ofgroupsG,71,115ofringsR,71{74ofsetsS,71Cauchysequence,49Cauchy,AugustinLouis(1789{1857),49Cayleyalgebra,52Cayley'stheorem,110{112Cayley,Arthur(1821{1895),1,5,52,110Centerofagroup,101Centralizer,101Chain,158Characteristicofa eld,41ofaring,38ofanintegraldomain,58,69CharlesHermite(1882{1901),42Chineseremaindertheorem,60{63,77,102Circleunit,9,89Codomain,70,149Cohen,Paul(1934{2007),158Commutativediagram,70Commutativegroup,seeAbeliangroupCommutativering,55Commutativity,2Commutatoroftwogroupelements,101subgroup,101Comparableelements,154Completeordered eld,49{50Complexconjugation,14,43,89ComplexnumbersC,2,4,43{44,87{89ComplexpolynomialsC[x],87Compositenumber,17Compositionoffunctions,149ofhomomorphisms,13ofmorphisms,70Compositionfactor,131Compositionseries,131Congruencegroup,121ring,76Congruenceclass,37,76,121
INDEX163Congruencemodulon,5,36Conjecture,143Conjugacyclass,116Conjugateelementinagroup,115subgroup,116Conjugationcomplex,14,43,89foraquadraticextension eld,43quaternion,51Conjunction^,144Contentofapolynomial,93Contraction,125Conway,JohnH.,47Coprime,17Coproductinacategory,136Coreofagroup,117Corollary,143Correspondencetheoremforgroups,123Coset,103{104Countableset,149Crossproduct,53CRT,seeChineseremaindertheoremCubicequation,87Cubicpolynomial,91Cyclenotation,105Cyclic eld,39{40Cyclicgroup,101{102Cyclotomicpolynomial,29d'Alembert,JeanleRond(1717{1783),89Dave'sShortCourseonComplexNumbers,4deFoncenex,FrancoisDaviet(1734{1799),89DeMorgan'slaws,147DeMorgan,Augustus(1806{1871),147Dedekindcut,48Dedekind,Richard(1831{1916),15Dedekind/Peanoaxioms,15DesarguesGirard(1591{1661),129Determinantsasgrouphomomorphisms,125Diagramcommutative,70DihedralgroupDn,108,111Dilation,125Directsum,102Disjointpairwise,146sets,146union,137Disjunction_,144Distributivity,3,33,67Divisibility=,80{81Divisibility ,16{17Divisionalgorithm,84forpolynomials,26Divisionring,10,50{54Domain,70,149Euclidean,84{87integral,57{60,78,80principalideal,82{84,86uniquefactorization,81{82,84Dotproduct,53,seeinnerproductDyadicrational,69ED,seeEuclideandomainEilenberg,Samuel(1913{1998),69Eisensteinintegers,59,86Eisenstein'scriterion,92{95Eisenstein,Gotthold(1823{1852),59,86,92Elementidentity,3initial,15inverse,3irreducible,80{82maximal,158orderof,101positiveandnegative,45prime,81,82Elements,145ElementsofEuclid,16{18Endomorphism,14Epimorphism,13,73Equivalenceclass,150{151Equivalencerelation,67,77,150{151EuclidofAlexandria( .ca.300B.C.E.),16{19,47Euclideanalgorithm,19{20,86Euclideandomain,84{87Euclideangeometry,117,127Euclideanvaluation,84Eudoxus( .350B.C.E),47Euler'scirclegroup,9Euler'sidentity,89Euler,Leonhard(1707{1783),9,19,43,89,146Evenpermutation,106{107
164INDEXExistentialquanti cation9,144Expansion,125ExtendedEuclideanalgorithm,22Extension eld,41{43Factortheorem,27Fanoplane,128Fano,Gino(1871{1952),128Fermat,Pierrede(1607{1665),16Field,2,4,31{54,79algebraic,42algebraicnumber,42algebraicallyclosed,89Archimedean,47{48axioms,31category,71completeordered,49{50de nition,4,31extension,41{43homomorphism,14isomorphism,12number,42,97{98ofcomplexnumbers,43{44ofrationalfunctions,35,69ofrationalnumbers,34,67{69ordered,45{50prime,39{41skew,10,50{54Fieldextensionalgebraic,42quadratic,41{45,77transcendental,42Finalobject,72,115Findingone,63FiniteAbeliangroup,137{140Finitegroup,103,104,112{115Firstisomorphismtheoremforgroups,122Firstisomorphismtheoremforrings,78Fixedpoint,104,105Foursquaresidentity,52Fraenkel,Abraham(1891{1965),151FreeBooleanring,65Freegroup,115Frobeniusendomorphism,41Frobenius,FerdinandGeorg(1849{1917),41,52,133FTA,seeFundamentaltheoremofalgebraFunction,149canonical,151choice,158codomain,149composition,149domain,149graph,149identity,13,149image,149inclusion,149injective,13,73,149inverse,150preimage,149projection,151rational,35,69successor,15surjective,13,73,149Fundamentaltheoremofalgebra,88{89ofarithmetic,22{24of niteAbeliangroups,140Godel,Kurt(1906{1978),153,158Galois eldGF(pn),40,129Galois,Evariste(1811-1832),40,133Gauss'slemma,92{95Gauss,CarlFriedrich(1777{1855),5,59,69,85,89,93Gaussianintegers,59GaussianintegersZ[i],69,85GCD,seegreatestcommondivisorGenerallineargroupGLn(R)GLn(R),7,125{126Geodesic,118Geodesics,117Geometryane,127Eucldean,117,127GeorgCantor(1845{1918),149Girard,Albert(1595{1632),88Gorenstein,Daniel(1923{1992),112Grassmann,Hermann(1809{1977),1Graves,JohnT.(1806{1870),52Greatestcommondivisor,19,22,80Greatestlowerbound,154Group,2,6{10,99{140Abelian,7,99,134{140alternating,107,112axioms,99category,71,115center,101
INDEX165circle,9core,117cyclic,7,101{102,111de nition,6dihedral,108,111 nite,103,104,112{115 niteAbelian,137{140free,115Frobenius,133generallinear,7,125{126homomorphism,13isomorphism,12Klein4-group,113linear,124{130ofunitsinaring,7order,7,101,103orthogonal,126presentation,110primary,138projectivelinear,127,130projectivespeciallinear,130quaternion,113,119quotient,121{123simple,131{134solvable,133speciallinear,126sporadic,133symmetric,104{107,110,112,116,119unitary,127Groupactiontransitive,133Groupring,58Holder,Otto(1859{1937),131Hamilton,WilliamRowan(1805{1865),1,50Hassediagram,16,154Hasse,Helmut(1898{1979),16,154Heptahedron,134Hermite,Charles(1822{1901),127Hermitian,127Homset,70Homomorphism,12{13 eld,14group,13ring,13,71Hyperbolicspace,118Ideal,74{79generatedbyaset,75maximal,78{79prime,78{79,82principal,75proper,75trivial,75Idempotentelement,64Identityelement,3Identitymorphism,70Image,149Implication),144Inclusion,14Inclusionfunction,149Indexofasubgroup,103Infemum,seeGreatestlowerboundInitialelement,15Initialobject,72,115Injection,13,73,149Innerproduct,53,126Integeralgebraic,42IntegersEisenstein,59,86Gaussian,59,69,85IntegersZ,2,4,5,69,72Integersmodulon,Zn,5,7,35{39,56,60{63,74,77,102de nition,36Integraldomain,57{60,78,80Internaldirectproduct,123{124Internaldirectsum,136Intersection,146,147ofsubgroups,100Inverseelement,3Inversefunction,150Inversivespace,118Invertibleelement,seeunitInvolution,101Involutoryquandle,117IrreducibilitytestEisenstein'scriterion,94modulop,93Irreducibleelement,80{82Irreduciblepolynomial,87,89{97Isomorphism=,11{12,71,74 eld,12group,12ring,11Isomorphismtheorem rstforgroups,122
166INDEX rstforrings,78secondforgroups,123thirdforgroups,123Join,66Jordan,Camille(1838{1922),122,131Jordan-Holdertheorem,131{133Joyce,David,4,16,145Kelland,Philip(1808{1879),50Kernelofagrouphomomorphism,120{124ofaringhomomorphism,74Klein,Felix(1849{1925),113Knuth,Donald,47Krull'stheorem,79Krull,Wolfgang(1899{1971),79Lagrange'stheorem,103{104Lagrange,Joseph-Louis(1736{1813),52,89,103Laplace,Pierre-Simon(1749{1827),89Latinsquare,9Lattice,67,155distributive,67,156modular,156Leastcommondenominator,22Leastcommonmultiple,22Leastupperbound,154Leibniz,GottfriedWilhelm(1646{1716),89,146Lemma,143Lexicographicordering,159Lindemann,Ferdinandvon(1852{1939),42Lineargroup,124{130Linearorder,153Lineartransformation,7,44,53,124{127Llull,Ramon(ca.1232{ca.1315),146Localization,69Logicalsymbols,143Loos,Ottmar,117Lowerbound,154MacLane,Saunders(1909{2005),69Map,12,seemorphism,149Mathematicalinduction,15strongform,23Mathieu,EmileLeonard(1835{1890),133Matrixunimodular,126unitary,127MatrixrepresentationofC,43ofH,53Matrixring,5,35,43,124{130Maximalideal,78{79Meet,66Membership,145Minimizationprinciple,15Module,11Modulopirreducibilitytest,93Monomorphism,13,73Morphism,12,70Moufang,Ruth(1905{1977),52Multiplicativefunction,19Multiplicativegroupofunits,7NaturalnumbersN,2,15Negation(logical):,144Neutralelement,seeidentityelementNoether,Emmy(1882{1935),1Noether,EmmyAmalie(1882{1935),83Noetherianring,83Normofacomplexnumber,43ofaquaternion,51Normalsubgroup,120{124Numberalgebraic,42complex,2,4,43{44,87{89composite,17greatestcommondivisor,22integer,2,4,5,69,72natural,2,15,144prime,17{19,22{39rational,2,4,34,67{69real,2,45{50,89{90,145relativelyprime,17,19{20,24,39,61,102,137surreal,47transcendental,42whole,seeintegersNumber eld,42,97{98Numbertheory,15{25Object,70Octonions,52Oddpermutation,106{107One-to-onecorrespondence,seebijectionOne-to-onefunction,seeinjectionOntofunction,seesurjection
INDEX167Operation,2{3associative,2binary,2commutative,2unary,2Orderlexicographic,159linear,153ofagroup,7,101,103ofaprimeinanumber,25ofanelementinagroup,101partial,66,153{155total,153Ordered eld,45{50Archimedean,47{48complete,49{50Orthogonalgroup,126Orthogonaltransformation,126Outerproduct,53Pairwiserelativelyprimenumbers,22PappusofAlexandria(ca.290{ca.350),129Partialorder,66,153{155Partition,151Partitionofanumber,138Partitionofaset,148,150{151Peano,Giuseppe(1858{1932),15Permutation,104evenandodd,106{107Philolaus(470{385B.C.E.),17PID,seeprincipalidealdomainPolynomial,25{29complex,87content,93cubic,91cyclotomic,29irreducible,87,89{97monic,25primecyclotomic,95primitive,92quadratic,90rationalroottheorem,91real,89root,26Polynomialevaluation,13,73Polynomialring,5,26,73,85{97Poset,153{155Powerset},56,64,148Pre-order,154Preimage,149Presentationbygeneratorsandrelations,110Primarycomponent,138Primarydecompositiontheorem,138Primarygroup,138Primecyclotomicpolynomialsp,95Primeelement,81,82Prime eld,39{41Primeideal,78{79,82Primenumber,17{19,22{39in nitelymany,18Primitivepolynomial,92Primitiverootofunity,28Primitiverootsofunity,95Principalideal,75Principalidealdomain,82{84,86Principleofin nitedescent,16Productinacategory,71internaldirect,123{124ofgroups,102ofrings,57,71ofsets,148semidirect,133Productsofsubsetsinagroup,104Projection,38,151ProjectivelineargroupPGLn(F),127,130ProjectiveplaneDesarguesian,129 nite,129Pappian,129Projectivespace,118,127ProjectivespeciallineargroupPSLn(F),130Q.E.D.,143QinJiushao(1202{1261),62QinJiushao'salgorithm,62Quadratic eldextension,41{45,77Quadraticpolynomial,90Quandle,10,11,117involutory,117withgeodesics,117Quaterniongroup,113,119QuaternionsH,10,50{54unit,54Quotientgroup,121{123Quotientring,76{79Quotientset,37,68,76,151
168INDEXRadian,89Rationalfunction,35,69Rationalnumbers,2,4,34,67{69Rationalroottheorem,91Realnumbers,45{50,89{90RealnumbersR,2RealpolynomialsR[x],89Reducible,80Re ection,125Re exivity,66,150Relation,150antisymmetric,66binaryorder,46equivalence,67,77,150{151partialorder,seePartialorderre exive,66,150symmetric,150transitive,16,66,150Relativelyprime,17,19{20,24,39,61,102,137pairwise,22Remaindertheorem,27Residue,6Ring,2,5{6,55{98algebraicintegers,97{98axioms,55Boolean,56,63{67category,71{74commutative,55cyclic,35{38,56,77de nition,5division,10,50{54freeBoolean,65homomorphism,13,71isomorphism,11matrix,5,35,124{130Noetherian,83ofintegers,seeintegersofpolynomials,5,26,73,85{97quotient,76{79trivial,72Rootofunity,28{29,95primitive,28,95Rotation,125Scalar,53Scalarproduct,53Schroder,Ernst(1841{1902),148Secondisomorphismtheoremforgroups,123Semidirectproduct,133Set,15,144{149category71complement,147countable,149di erence,147element,145 nite,11in nite,15intersection,146,147membership,145operationon,2{3partiallyordered,153{155partition,148,151permutation,104power,56,64,148productof,148quotient,37,68,76,151singleton,152subset,145uncountable,149underlying,3,31,55,99union,146,147Settheoryaxioms,151{153Shear,126Simplegroup,131{134Simplyin nite,15Singletonset,152Skew eld,10,50{54Solvablegroup,133Spacehyperbolic,118inversive,118projective,118,127Speciallineargroup,126Sphere,118SphereS2,10Structurealgebraic,2{11Sub eld,34de nition,34Subgroup,14,100{104commutator,101conjugate,116generatedbyaset,101generatedbyanelement,101index,103normal,120{124ofZ,101
INDEX169ofS3,107proper,100trivial,100Subring,14,58Subset,145SubstitutionTschirnhaus,91Successorfunction,15SunZi( .400),62Supremum,seeLeastupperboundSurjection,13,73,149Surrealnumbers,47Sylvester,JamesJoseph(1814{1897),5SymmetricgroupSn,104{107,110,112,116,119Symmetricspace,117Symmetriesofacube,108ofapentagon,108,111,119ofatetrahedron,108ofatriangle,8Symmetry,150Tait,PeterGuthrie(1831{1901),50Tartaglia,NicoloFontana(1500{1557),87Theorem,143Thirdisomorphismtheoremforgroups,123Through.,117Thymaridas(400{350B.C.E.),17Torus,134Totalorder,153Totientfunction,19,29,39Transcendenceof,42ofe,42Transcendental eldextensions,42Transcendentalnumber,42Transformationlinear,7,44,53,124{127Transitivegroupaction,133Transitivity,16,46,66,150Transposition,105,106Trichotomy,46Trivialring,72Tschirnhaussubstitution,91Tschirnhaus,EhrenfriedWalthervon(1651{1708),91UFD,seeuniquefactorizationdomainUnaryoperation,2Uncountableset,149Underlyingset,3Unimodularmatrix,126Union,146,147disjoint,137Uniquefactorizationdomain,81{82,84,95Uniquefactorizationtheorem,22{24Unitcircle,9inZn,19inaring,7UnitcircleS1,89Unitarygroup,127Unitarymatrix,127Unitarytransformation,127Unityrootof,28{29,95Universalpropertyofanin nitecyclicgroup,115ofcoproducts,136of nalobjects,72,115offreegroups,115ofinitialobjects,72,115ofproducts,71oftheringZ,72,73Universalquanti cation8,144Upperbound,154ValuationEuclidean,84Vector,53Vectorproduct,53Vectorspace,35Venndiagram,146Venn,John(1834{1923),146Viete,Francois(1540{1603),91vonNeumann,John(1903{1957),153Waring,Edward(1736{1798),50Wedderburn,Joseph(1882{1948),59Weierstrass,Karl(1815{1897),42Well-orderingprinciple,15,159Zermelo,Ernst(1871{1953),151Zero-divisor,58Zorn'slemma,79,158{160Zorn,MaxAugust(1906{1993),158

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