ContentsChapter1.TheLanguageofMathematics11.1.PropositionalCalculusand - PDF document

1 ContentsChapter1.TheLanguageofMathematic
ContentsChapter1.TheLanguageofMathematics11.1.PropositionalCalculusandLawsofInference11.2.PredicateLogicandBasicSetTheory41.3.FurtherTopicsinSetTheory81.4.Relations,FunctionsandPartialOrderings101.5.TheSubjectsofAlgebra131.6.TheNumberSystems14Chapter2.Groups232.1.Groups23Chapter3.Rings333.1.GeneralRingTheory333.2.RingHomomorphisms363.3.UniqueFactorization373.4.Polynomials423.5.AlgebraicGeometry49Chapter4.Fields534.1.FieldExtensions534.2.Somemoreconceptsfromgrouptheory574.3.GaloisTheory584.4.RadicalExtensions594.5.TheTheoremofRuniandAbel61Chapter5.VectorSpaces635.1.Fundamentals635.2.LinearTransformations655.3.Finite-dimensionalvectorspaces675.4.EigenvaluesandEigenvectors715.5.SpectralTheoryinFinite-dimensionalComplexVectorSpaces725.6.MultilinearAlgebra775.7.NormedSpacesandInnerproductSpaces79Index85iii CHAPTER1TheLanguageofMathematics1.1.PropositionalCalculusandLawsofInference1.1.1Statements.Asentence(inthegrammaticalsense)forwhichitis(inprinciple)possibletodeterminewhetheritistrueorfals

2 ewillbecalledastatement.Thus\Parisisthec
ewillbecalledastatement.Thus\ParisisthecapitalofFrance."and\Twoplusthreeissix."arestatements.\Howareyou?",\Let'ssee.",and\2+x2=11."arenot.\True",denotedbyT,and\false",denotedbyF,arecalledtruthvalues.1.1.2Connectives.Statementsmaybecombinedtoobtainotherstatements:thecom-binationisachievedusingsocalledconjunctionslike`and',`or',or`but'.Inlogicandmathematicsoneusesthefollowingconjunctionsandcallsthemconnectives:Theconjunction`and';symbol^.Thedisjunction`or'(inthesenseof`porqorboth.');symbol_.Theconditional'if-then';symbol!.Thebiconditional`ifandonlyif';symbol$.Theprecisedenitionsoftheseconnectivesaregiveninthetruthtablebelowbystatingwhatthetruthvalueofsuchacombinedstatementisgiventhetruthvaluesofitsparts.Inthefollowingpandqdenotegivenstatements.p q pandq porq porqbutnotboth Ifpthenq pifandonlyifq T T T T F T TT F F T T F FF T F T T T FF F F F F T TJustasinarithmetictheuseofparenthesesmightbenecessaryinordertoindicateprioritiesinperformingvariousoperationswhenmorecomplic

3 atedstatementsareformedsymbolically.Inst
atedstatementsareformedsymbolically.Insteadofthephrase`Ifpthenq.'oneusesalso`qifp.'and`ponlyifq.".Togetherwiththestatementp!qoneoftenconsidersalsoitsconverseq!panditscontrapositive�q!�p.1.1.3Negation.Anotherwaytomanipulateastatementistoformitsnegation.Thenegationofpiscalled`Notp.'andisdenotedby�p.Thestatement�pistrueifpisfalseandfalseifpistrue.1.1.4Tautologies.Acombinedstatementiscalledatautologyifitisalwaystruenomatterwhatthetruthvaluesofitspartsare.Themostprimitiveexamplesoftautologiesarep_(�p)whichiscalledthelawoftheexcludedmiddleand�(p^�p)whichiscalledthelawofnon-contradiction.Astatementpissaidtoimplyastatementqlogically,ifp!qisatautology.Thisisdenotedbyp)q.Forinstance,p^(p!q))q.Thiscanbeseenfromatruthtable.1 21.THELANGUAGEOFMATHEMATICSTwostatementspandqarecalledlogicallyequivalentifp$qisatautology.Wewritethenp,q.Notethatalogicalequivalenceisalogicalimplicationbothways.Thefollowingexamplesareimportant:p!q,q_�p,p!q,�q!�p,\Eitherporqbutnotboth."

4 ,(p^�q)_(�p^q).Thesecondexample,in
,(p^�q)_(�p^q).Thesecondexample,inparticular,showsthatastatementanditscontrapositivearelogicallyequivalent.Note,however,thatastatementanditsconversearenotlogicallyequivalent.Inmathematics(aswellasinmanyotheraspectsoflife)onewantstodrawconclusionsfromcertainpiecesofinformation.Theinformationisgivenintermsofstatementswhicharetakenforgranted.Theyarecalledpremises.Theconclusionisanotherstatementwhichwewillalwaysacceptprovidedweacceptthepremises.Drawingaconclusionis(bydenition)allowediftheconclusionislogicallyimplied(intheabovesense)bythepremises.Thusthetautologiesareourrulesfordrawingconclusions.Forexample,ifweknow(ortakeforgranted)thetruthsofthestatements\IfitrainsinBirminghamthenVulcanwillgetwet."and\OnMay35itrainedinBirminghamallday."thenwemayconcludethat\OnMay35Vulcangotwet.".1.1.5Lawsoflogic.Inthefollowingwelistthemostimportantlogicalimplicationsandequivalencestogetherwiththeirnamesinlogic.Note,however,thatthelistisnotcompleteandthatanytautologymaybeusedtodrawconclusions.Lawofdet

5 achmentp^(p!q))qModustollens�q^(p!q))
achmentp^(p!q))qModustollens�q^(p!q))�pLawofdisjunctivesyllogism�p^(p_q))qLawofhypotheticalsyllogism(p!q)^(q!r))p!rLawofsimplicationp^q)pLawofadditionp)p_qLawofabsurdityp!(q^�q))�pLawofconditionalizingq)p!qLawofdoublenegationp,�(�p)Lawofcontrapositionp!q,�q!�pLawofex-andimportation(p^q)!r,p!(q!r)Lawoftheconditionalp!q,q_�pLawofthebiconditionalp$q,(p!q)^(q!p),(p^q)_(�p^�q)DeMorgan'slaws�(p^q),�p_�qand�(p_q),�p^�qCommutativelawsp^q,q^pandp_q,q_pAssociativelaws(p^q)^r,p^(q^r)and(p_q)_r,p_(q_r)Distributivelawsp^(q_r),(p^q)_(p^r)andp_(q^r),(p_q)^(p_r)1.1.6Aschemeforproofs.Wewillnowintroduceaveryformalschemefortheproofofastatementfromanumberofpremises.Whilelateronwewillnotactuallyemploytheschemewewillalwaysemployitsspirit.Aproofaccordingtothisschemeconsistsofalistoflines.Eachlinehasfourentries:anumbertoenumerateit,astatement,alistofpremisesonwhichthestatementdepends,andanexplanation.Linesareaddedtothelistoflinesbyoneofthefollowingrules:Rul

6 eP:Addapremise.(Putthecurrentlinenumberi
eP:Addapremise.(PutthecurrentlinenumberinthelistofpremisesandwritePremiseasexplanation.) 1.1.PROPOSITIONALCALCULUSANDLAWSOFINFERENCE3RuleL:Addastatementwhichislogicallyimpliedbyaprecedingstatement.(Copythelistofpremisesfromthatstatementandstateasexplanationwhichlogicalimplicationwasused.)RuleC:Addtheconjunctionofpreviousstatements.(Forthenewlistofpremisesputtheunionofthelistsofpremisesoftheappropriatepreviousstatements.Asexplanationwritewhichlineshavebeencombined.)RuleCP:IfstatementSisinthelistdependingonthepremisesP1,...,PnandR,thenstatementR!SmaybeaddedasdependingonthepremisesP1,...,Pn.Considerthefollowingexampleinvolvingthestatementsn=\NapoleonexpelsSnowballfromthefarm.",w=\Therewillbeawindmill.",f=\Therewillbeahigherfoodproduction."Supposethefollowingscenario.IfNapoleonexpelsSnowballfromthefarm,thentherewillbenohigherfoodproduction.Resourcesarespentonahigherfoodproductionorawindmill.Wealsoknowforafactthatnowindmillwasbuilt.WewanttoprovethatNapoleondidnotexpelSnowballfromthefarm.Theproofc

7 anbedoneinthefollowingway:(1)n!�ff1gp
anbedoneinthefollowingway:(1)n!�ff1gpremise(2)f_wf2gpremise(3)w_ff2gfrom(2)bycommutativelaw(4)�wf4gpremise(5)�w^(w_f)f2,4gfrom(3)and(4)byRuleC(6)ff2,4gfrom(5)bylawofdisjunctivesyllogism(7)�(�f)f2,4gfrom(6)bylawofdoublenegation(8)�(�f)^(n!�f)f1,2,4gfrom(1)and(7)byRuleC(9)�nf1,2,4gfrom(8)bymodustollensUsingpremises1,2,and4wehavelogicallydeducedthatNapoleondidnotexpelSnowball.Asanotherexamplesupposewewanttodeduces!�mfromthepremises�s_`andm!�`.Wemightproceedinthefollowingway:(1)sf1gpremise(2)�(�s)f1gfrom(1)bylawofdoublenegation(3)�s_`f3gpremise(4)�(�s)^(�s_`)f1,3gfrom(2)and(3)byRuleC(5)`f1,3gfrom(4)bylawofdisjunctivesyllogism(6)�(�`)f1,3gfrom(5)bylawofdoublenegation(7)m!�`f7gpremise(8)�(�`)^(m!�`)f1,3,7gfrom(6)and(7)byRuleC(9)�mf1,3,7gfrom(8)bymodustollens(10)s!�mf3,7gfrom(9)byRuleCP1.1.7Obtaininglogicalimplicationsbyproofs.Supposealineinaproofisoftheform(n)p!sfg...wherepcan,ofcourse,beoftheformp1^:::^pk.Sincethestatem

8 entp!sdoesnotdependonanypremises(asindic
entp!sdoesnotdependonanypremises(asindicatedbyfginthethirdcolumn)itisatautologyandthereforewehaveshownthatsislogicallyimpliedbypor,symbolically,p)s.Forinstancethelawofconditionalizingmaybeprovedusingatruthtablebutalsointhefollowingway:(1)qf1gpremise(2)q_�pf1gfrom(1)bylawofaddition 101.THELANGUAGEOFMATHEMATICS1.3.9Axiomaticsettheory.Thediscussionofsettheoryinthissectioncanbelabeled\Intuitivesettheory".Thepresentation,however,hascertainpitfalls.Inparticular,wehaveconsideredthesetA=fx:P(x)gwhenPwassomeproperty.NowletPbetheproperty\isnotanelementofA"andconsiderwhetherA2A.FirstassumethatA2A.ThenP(A)istrue,i.e.,A62Awhichisacontradiction.HenceassumethatA62A.ThenP(A)mustbefalse,i.e.,A2Aagainacontradiction.ThisparadoxwasrstnoticedbyB.RussellandisknownasRussell'sparadox.In\Axiomaticsettheory"thisproblemisavoidedbyrestrictingtheuseofthewordsetinaproperway(determinedbytheaxioms).Foratreatmentofaxiomaticsettheorysee,forinstance:[1]PaulHalmos,NaiveSetTheory,Springer,NewYork,1973.[2]PatrickSuppes,A

9 xiomaticSetTheory,vanNostrand,Princeton,
xiomaticSetTheory,vanNostrand,Princeton,1960.[3]MartinZuckerman,SetsandTransniteNumbers,MacmillanPublishingCo.,NewYork,1974.1.4.Relations,FunctionsandPartialOrderings1.4.1OrderedpairsandCartesianproducts.LetAandBbetwosets(notnecessarilydistinct)andassumea2Aandb2B.Thenconsidertheorderedpair(a;b).Theadjective`ordered'emphasizesthat(a;b)and(b;a)aredierent,ingeneral.Moreprecisely,(a;b)=(c;d)ifandonlyifa=candb=d.Thus(a;b)=(b;a)ifandonlyifa=b.ThesetofallorderedpairswhichcanbeformedfromAandBiscalledtheCartesianproductofAandBandisdenotedbyAB,i.e.,AB=f(a;b):a2A^b2Bg:OnecannextdenetheCartesianproductofthreesetsA,B,andCbyABC=(AB)C.ProceedinginthismanneronecanthendenetheCartesianproductofanynitenumberofsets.1.4.2Relations.LetXandYbesets.AsetoforderedpairsinXY,i.e.,asubsetofXY,iscalledarelation.If(x;y)2RwesaythatxisinrelationRtoyanddenotethisbyxRy.GivenarelationRwedenethedomainandtheimageofRbydom(R)=fx:9y:(x;y)2Rg;im(R)=fy:9x:(x;y)2Rg:AsetRofo

10 rderedpairscanalwaysbethoughtofasasubset
rderedpairscanalwaysbethoughtofasasubsetoftheCartesianproductdom(R)im(R).Conversely,everysubsetoftheCartesianproductXYisarelation.IfRisasubsetofXYwecallRarelationfromXtoY.IfY=XwecallRarelationinX.LetRbearelationfromXtoYandAasubsetofX.ThenthesetR(A)=fy2Y:9x2A:xRygiscalledtheimageofAundertherelationR.Inparticular,im(R)=R(X).1.4.3Inverseandcompositerelations.GivenarelationRfromXtoYtheinverserelationR�1istherelationfromYtoXdenedbyR�1=f(y;x)2YX:(x;y)2Rg:IfBisasubsetofYthentheimageR�1(B)ofBundertherelationR�1isalsocalledthepreimageofBundertherelationR.Inparticular,R�1(B)=fx2X:9y2B:yR�1xg=fx2X:9y2B:xRyg:GivenarelationR1fromXtoYandarelationR2fromYtoZthecompositerelationR2R1istherelationfromXtoZdenedbyR2R1=f(x;z)2XZ:9y2Y:((x;y)2R1^(y;z)2R2)g: 121.THELANGUAGEOFMATHEMATICSLetfbeafunctionfromXtoYandAasubsetofX.Thenthefunctiong:A!Ydenedbyg(x)=f(x)forallx2AiscalledtherestrictionofftoAandisdenotedbyfjA.Conversely,thefunctionfiscalledanextension

11 ofg.Aparticularbijectivefunctionfromanon
ofg.AparticularbijectivefunctionfromanonemptysetXtoitselfistheidentityfunctionid:x7!x.1.4.7Partialorderings.ArelationRinXiscalledantisymmetricifxRyandyRxtogetherimplythatx=y.Are exive,antisymmetric,transitiverelationinXiscalledapartialorderingofX.WethensaythatXispartiallyorderedbyR.NowletRbeapartialorderingofXandAasubsetofX.Wedeneu2XisanupperboundofAifaRuforalla2A,l2XisalowerboundofAiflRaforalla2A,sup(A)2XisaleastupperboundorsupremumofAifitisanupperboundofAandalowerboundofthesetofallupperboundsofA,inf(A)2XisagreatestlowerboundorinmumofAifitisalowerboundofAandanupperboundofthesetofalllowerboundsofA,sup(A)iscalledamaximumifitisanelementofA,similarly,ifinf(A)2Aitiscalledaminimum,m2AisamaximalelementifmRaimpliesm=awhenevera2A,similarly,ifm2AandifaRmimpliesm=aforalla2Athenmiscalledaminimalelement.Thesupremumandinmumofasetareeachuniqueiftheyexist.Amaximum(orminimum)ofAisalwaysamaximal(orminimal)elementofA.Example:ConsiderthepowersetP(X)ofanonemptysetX.T

12 hentherelation\issubsetof"isapartialorde
hentherelation\issubsetof"isapartialorderingofP(X).NowletX=f1;2;3g,A=ff1g;f2g;f1;2gg,andB=ff1;2g;f2;3gg.Thenmax(A)=f1;2g,inf(A)=fg,sup(B)=f1;2;3g,andinf(B)=f2g.However,BdoesnothaveamaximumandneitherAnorBhaveaminimum.Bothf1gandf2gareminimalelementsofA.1.4.8Totalorderings.ApartialorderingRofXiscalledatotal(orlinear)orderingifxRyoryRxforanytwox;y2X,i.e.,anytwoelementsofXmaybecompared.InthiscasewecallXtotallyorderedbyR.Ifaisamaximal(minimal)elementofAXthenitisequaltomax(A)(min(A)).Inparticular,maximalandminimalelementsareuniqueiftheyexist.Example:Thesetofrealnumbersistotallyorderedbytherelation\issmallerthanorequalto".Oftenatotalorderingwillbedenotedbythesymbol.Wethenintroducealsothesymbols,,and&#x]TJ/;ø 9;&#x.962; Tf;&#x 7.7;I 0;&#x Td ;&#x[000;inthefollowingway:xyifandonlyifyx,xyifandonlyifxybutx6=y,andx&#x-278;yifandonlyifyx.1.4.9Wellorderings.AtotalorderingRofXiscalledawellorderingifeverynonemptysubsetofXcontainsaminimum.Inthiscaseweca

13 llXwell-orderedbyR.In1.6.4wewillshowthat
llXwell-orderedbyR.In1.6.4wewillshowthatthesetofnaturalnumbersiswell-orderedbytherelation\issmallerthanorequalto".1.4.10Theaxiomofchoiceandsomeofitsequivalents.IfthetwosetsXandYarenotemptythenXcontainsanelementxandYcontainsanelementy.Therefore,theCartesianproductXYcontainsatleasttheelement(x;y)andhenceisnotempty.Inanaxiomaticapproachtosettheoryitispossibletogeneralizethisresulttocollectionsofnitelymanysetsbutnottoarbitrarycollectionsofsetswithoutaddinganotheraxiom.ThisaxiomiscalledtheAxiomofChoice. 1.6.THENUMBERSYSTEMS19Euclid'salgorithmmaybeusedtocomputethegreatestcommondivisoroftwointegersa1;b1.Itworksbyrecursion:LetX=Z2andx1=(a1;b1)wherewemayassume,withoutlossofgenerality,thata1�0.Denef:Z2!Z2byf((a;b))=((r(b;a);a)ifa6=0(0;0)ifa=0wherer(b;a)2f0;:::;jaj�1gistheremainderofbafterdivisionbyaasdenedaccordingtothedivisiontheorem.BytheRecursionTheoremthereisauniquefunctionu:N!Z2suchthatu(1)=(a1;b1)andu(n+1)=f(u(n))=(an+1;bn+1).Nownotethatan0and,ifan�0,that

14 an+1=r(bn;an)an.LetM=fk2N:ak&#x-335;0g.T
an+1=r(bn;an)an.LetM=fk2N:ak&#x-335;0g.ThenMisanitesetandwedenotemax(M)bym.Thereforeum+1=(0;am).Since,bytheaboveremark,gcd(ak+1;bk+1)=gcd(ak;bk)ifak6=0weobtaingcd(a1;b1)=gcd(a2;b2)=:::=gcd(0;am)=am:1.6.11TheGCDidentity.Giventwointegersa;bwherea6=0considerthesetS=fax+by:x;y2Zg\N:Sisnotemptyandhencecontainsasmallestelementwhichwillbedenotedbyd.Thusthereexistx0;y02Zsuchthatd=ax0+by0andq;r2Zsuchthata=qd+rand0rd.Thereforer=a�qd=a(1�qx0)+b(�qy0)isinS[f0g.SincercannotbeinSwegetthatr=0andhencethatddividesa.Similarlyoneshowsthatddividesb,i.e.,disacommondivisorofaandb.Nowsupposethatc&#x-277;disalsoacommondivisorofaandb.Thenthereexistintegersn;msuchthata=ncandb=mc.Henced=c=nx0+my0isapositiveintegerstrictlysmallerthanonewhichisimpossible.Henced=gcd(a;b).WehaveprovenTheorem.Ifa;b2Zanda6=0thenthereexistx;y2Zsuchthatgcd(a;b)=ax+by.ThenumbersxandycanbecomputedbyrunningEuclid'salgorithmbackwards.1.6.12Primeandirreduciblenumbers.Anintegerpforwhichjpj&#x-277;1iscalledprimeoraprimenumberif,wh

15 eneverpdividesab,thenpdividesaorb.Aninte
eneverpdividesab,thenpdividesaorb.Anintegerpforwhichjpj&#x-277;1iscalledirreducibleifp=abimpliesthatjaj=1orjbj=1.Inotherwordspisirreducibleifitsonlydivisorsare1;�1;p,and�p.Theorem.Anintegerisprimeifandonlyifitisirreducible.Sketchofproof:Supposepisprimeandp=ab.Then(withoutlossofgenerality)pdividesa,i.e.,thereexistsn2Zsuchthata=np.Hencep=npb,i.e.,nb=1whichshowsthatpisirreducible.Nextsupposepisirreducibleandthatpdividesab.Ifpdividesanothingistobeprovenandweassumethereforethatpdoesnotdividea.Sincepisirreducibleitsonlypositivedivisorsare1andjpj.Sincepdoesnotdivideawegetthatgcd(p;a)=1.BytheGCDidentitythereexistintegersx;ysuchthat1=ax+pyandhenceb=abx+pby.Thisshowsthatpdividesb.1.6.13Uniquefactorizationtheoremforintegers.Thefollowingwell-knowntheoremisalsocalledtheFundamentalTheoremofArithmetic.Theorem.Everyintegerxotherthan0and1iseitheranirreducibleoraproductofirreducibles.Moreover,thisproductisessentiallyuniqueinthesensethat,whenx=a1:::an=b1:::bmwherea1;:::;bmareirreducibles,thenn=ma

16 ndthebjmayberearrangedsothatai=bifor
ndthebjmayberearrangedsothatai=bifori=1;:::;n. 1.6.THENUMBERSYSTEMS21AtotalorderingonRisdenedthrough:L1L2ifandonlyifL1L2.ArealnumberLiscalledpositiveifLcontainsapositiverationalnumberandnegativeifLccontainsanegativerationalnumber.Therealnumberfq2Q:q0giscalledzeroandisalsodenotedby0.Everynonzerorealnumberiseitherpositiveornegativebutnotboth.Zeroisneitherpositivenornegative.Additionofrealnumbersisdenedasfollows:L1+L2=fx1+x2:x12L1;x22L2g:Itisassociativeandcommutative.Therealnumberzeroisanadditiveidentity.EveryrealnumberLhasanadditiveinverse�L=fx2Q:(9y2Lc:x+y0)g:IfLisnegativethen�Lispositiveandviceversa.TofacilitatenotationweintroducethesetQ�0ofnonpositiverationalnumbers.Also,L+denotesthesetofpositiveelementsofL.MultiplicationoftwopositiverealnumbersisdenedbyL1
L2=Q�0[fx1x2:x12L+1;x22L+2g:NextonedenesproductsofarbitraryrealnumbersbyL1
L2=�((�L1)
L2)ifL10andL2&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;0,L1
L

17 2=�(L1
(�L2))ifL1&#x]TJ/;ø
2=�(L1
(�L2))ifL1&#x]TJ/;ø 9;&#x.962; Tf;&#x 10.;Ԗ ;� Td;&#x [00;0andL20,L1
L2=(�L1)
(�L2)ifL10andL20,andL1
L2=0ifL1=0orL2=0.Multiplicationisassociativeandcommutative.Alsoitisdistributiveoveraddition.Thenumber1=fq2Q:q1gisthemultiplicativeidentity.EverynonzerorealnumberLhasamultiplicativeinverseL�1.IfL&#x]TJ/;ø 9;&#x.962; Tf;&#x 18.;ˆ ;� Td;&#x [00;0thenL�1=fx2Q:(9y2Lc:xy1)g.IfL0thenL�1=�(�L)�1.Combiningthesefactswearriveatthefollowingtheorem.Theorem.(R;+;
)isaeld.NotethatLdoesnotcontainamaximalelement.Thusifsup(L)existsthenitisinLc,infactitistheminimumofLc,andhencearationalnumber.Wenowembed(Q;;+;
)into(R;;+;
)byidentifyingtherationalnumbersup(L),ifthisexists,withtherealnumberL.Ifsup(L)doesnotexistwecallLanirrationalnumber.Examples:LetL=fq2Q:q3=4g.ThentherealnumberLisidentiedwiththerationalnumber3=4.LetL=fq2Q:(q0_q22)g.ThenLisarealnumberbutnotarationalnumbersincesup(L)doesnotexist.Ofcourse,L

18 isusuallydenotedbyp 2.1.6.16Theleastuppe
isusuallydenotedbyp 2.1.6.16TheleastupperboundpropertyofR.EverysubsetofRwhichhasanupperboundhasinfactaleastupperbound.Thisfactaccountsfortheimportanceoftherealnumbersandsetsthemapartfromtherationalnumbers,whichdonothavealeastupperboundproperty.Sketchofproof:Letfg6=Randsupposehasanupperbound.DeneS=SL2L.ThenS2Rand,forallL2wehaveLS,i.e.,Sisanupperboundof.NowassumeTSisalsoanupperboundof.Thenthereisarationalnumberq2S�TbutalsoanL2suchthatq2L.Thisisimpossible.1.6.17Roots.DenotethesetofpositiverealnumbersbyR+.Theorem.Letnbeanaturalnumber.Theneverypositiverealnumberhasauniquen-thpositiveroot,i.e.,ify2R+thenthereisauniquex2R+suchthatxn=y. 221.THELANGUAGEOFMATHEMATICSSketchofproof:ConsiderthesetS=fs2R:s�0^snyg.BytheleastupperboundpropertyofRthenumberx=sup(S)exists.Onemayshowthatxn=y.Uniquenessfollowsfromthefactthatxn1�xn2=(x1�x2)(xn�11+xn�21x2+:::+xn�22).1.6.18Thecomplexnumbers.ThesetRRiscalledthesetofcomplexnumbers.Acomplexnum

19 berz=(x;y)isusuallydenotedbyz=x+iywherex
berz=(x;y)isusuallydenotedbyz=x+iywherexandyarerealnumbers.x=Re(z)isthencalledtherealpartandy=Im(z)theimaginarypartofthecomplexnumberz=x+iy.Wedeneadditionandmultiplication:(a+ib)+(c+id)=(a+c)+i(b+d);(a+ib)
(c+id)=(a
c�b
d)+i(b
c+a
d):Thesetofcomplexnumbersisnotequippedwithanordering.However,byidentifyingtherealnumberawiththecomplexnumbera+i0weembed(R;+;
)into(C;+;
),i.e.,additionandmultiplicationinRtransferinthenaturalwaytoC.Justasa+i0isabbreviatedbyatheexpression0+ibisabbreviatedbyiband0+i1byi.Sincei
b=(0+i1)
(b+i0)=0+ib=ibwendthatibmaybeconsideredtobeaproduct.(Itiscustomary,infact,toleaveothedotinalltheproductswehavediscussed,i.e.,a
b=ab.)Moreover,sincealsoi2=(0+i1)(0+i1)=�1+i0=�1thedenitionsofadditionandmultiplicationfollowformallyfromtherulesofadditionandmultiplicationofrealnumbers.Thenumbers0=0+i0and1=1+i0aretheadditiveandmultiplicativeidentity,respectively.Everycomplexnumbera+ibhasanadditiveinverse�(a+ib)=�a+i(�b)andev

20 erynonzerocomplexnumberhasamultiplicativ
erynonzerocomplexnumberhasamultiplicativeinverse(a+ib)�1=(a�ib)(a2+b2)�1.Toeverycomplexnumberz=x+iy,x;y2Roneassignsarealnumber,calledtheabsolutevalueofzanddenotedbyjzj,throughjzj=p x2+y2andanothercomplexnumber,calledthecomplexconjugateofzanddenotedby z,through z=x�iy.Inparticular,then,z�1= zjzj�1=2.Anytwocomplexnumbersuandv(andhencealsotworealnumbers,tworationalnumbersetc.)satisfythefollowinginequalities,calledtriangleinequalities:ju+vjjuj+jvj;ju+vjjuj�jvj:For2Rletexp(i)=cos()+isin().Fromthisoneprovesthatexp(0)=1andexp(i)exp(i)=exp(i(+)).(Thereexistsadeeprelationshipbetweentheexponentialfunctionandthetrigonometricfunctionswhichisstudiedinacourseoncomplexanalysis.)SinceCisidentiedwithRRwemayrepresentcomplexnumbersbypointsinatwo-dimensionalplane.Everycomplexnumberhasthenaso-calledpolarrepresentation:z=x+iy=rexp(i)whererdenotesthedistanceofthepoint(x;y)fromtheoriginandistheorientedanglebetweentherealaxisan

21 dthelinethrough(x;y)andtheorigin.Inparti
dthelinethrough(x;y)andtheorigin.Inparticular,x=rcos(),y=rsin(),andr=p x2+y2.1.6.19References.Forfurtherstudythefollowingbooksaresuggested:[1]PaulHalmos,NaiveSetTheory,Springer,NewYork,1973.[2]StevenG.Krantz,TheElementsofAdvancedMathematics,CRCPress,BocaRaton,1995.[3]MartinZuckerman,SetsandTransniteNumbers,MacmillanPublishingCo.,NewYork,1974. CHAPTER2Groups2.1.Groups2.1.1Groups.LetGbeasetand
abinaryoperationonG.Then(G;
)iscalledagroupiftheoperation
isassociative,ifGcontainsaleftidentityandeveryelementofGpossessesaleftinverse.Moreexplicitly,(G;
)iscalledagroupif(1)a
(b
c)=(a
b)
cforalla;b;c2G,(2)thereexistsanelement12Gsuchthat1
a=aforalla2G,and(3)foreverya2Gthereexistsb2Gsuchthatb
a=1.Thedotisusuallyomitted,i.e.,a
bissimplywrittenasab.WewillalsosaythatGisagroupundertheoperation
or,ifnoconfusioncanarisejustthatGisagroup.Agroupiscalledcommutativeorabelianifitsbinaryoperationiscommutative,i.e.,ifab=baforalla;b2G.2.1.2Basicproperties.LetGbegroupw

22 ithleftidentity1.1.Anyleftinverseofaisal
ithleftidentity1.1.Anyleftinverseofaisalsoarightinverseofa.Proof:Supposeba=1andcb=1.Thenb=(ba)bandhenceab=(cb)(ab)=c(ba)b=cb=1.2.1istheuniqueidentityinG.Proof:Becauseof1.5.2weonlyhavetoshowthat1isarightidentity.Nowleta2Gandsupposeba=ab=1.Thena=(ab)a=a(ba)=a1.3.Everyelementa2Ghasauniqueinverse,denotedbya�1.Inparticular,(a�1)�1=a.Proof:Supposeba=1andb0a=1.Thenb0=b0(ab)=(b0a)b=b.4.Theinverseoftheproductabis(ab)�1=b�1a�1.5.Cancellation:Ifca=cborac=bcthena=b.6.Foreverya;b2Gtheequationsax=bandya=bhaveunique(possibledierent)solutions,i.e.,thereexistsoneandonlyonex2Gsuchthatax=bandoneandonlyoney2Gsuchthatya=b.2.1.3Examplesofabeliangroups.Alotoffamiliarexamplesofgroupsareabeliangroups:1.(Z;+),(Q;+),(R;+),(C;+),2.(Q+;
),(R+;
),whereQ+andR+denotedthepositiverationalandreal,respectively,3.(Q�f0g;
),(R�f0g;
),(C�f0g;
),4.(f1g;
),(f0g;+),(f1;�1g;
),5.(fexp(i):2Rg;
).2.1.4Notation.IfthegroupGhasonlynitelymanyelementsthenwedenotethenu

23 mberofitselementsbyjGj.Thisnumberisalsoc
mberofitselementsbyjGj.Thisnumberisalsocalledtheorderofthegroup.IfGhasinnitelymanyelementswesaythatGisofinniteorder.Nonabeliangroupsareusuallywrittenmultiplicatively,i.e.,using
orjuxtapositiontodenotethebinaryoperation.Abeliangroups,however,areusuallywrittenadditively,i.e.,23 242.GROUPSusing+todenotethebinaryoperation.Thisaectsthenalsovariousothernotationalconventions.multiplicative additive 1 0a�1 �aQnj=1aj=a1a2:::an Pnj=1aj=a1+a2:::+anan=Qnj=1a na=Pnj=1aa0=1 0a=0a�n=Qnj=1a�1 �na=Pnj=1(�a)Ifn;m2Zanda2Gthenna+ma=(n+m)aandm(na)=(nm)aforaadditivelywrittengroupanam=an+mand(an)m=anmforamultiplicativelywrittengroup.2.1.5Groupsofbijections.LetXbeasetandconsiderthesetofallbijectionsf:X!X.Compositionisanassociativebinaryoperationinthisset.Themappingx7!x,calledtheidentitymap,isanidentity(withrespecttocomposition).Givenanybijectionf:X!Xthenthef�1isalsoabijectionfromXtoXanditistheinverseelementoff(withrespecttocomposition).Bijectiongroupsneednotbeabelian.Forexampl

24 e,letX=Randconsiderthebijectionsx7!f(x)=
e,letX=Randconsiderthebijectionsx7!f(x)=2xandx7!g(x)=x+1.Then(fg)(x)=2x+26=2x+1=(gf)(x).2.1.6Symmetrygroups.Arigidmotionisabijectionofthetwo-dimensionalplane(orthree-dimensionalspace)toitselfsuchthatdistancesbetweenpointsarepreserved.Rigidmotionsformagroupundercomposition.ArigidmotionwhichmapssomesubsetAoftheplaneontoitselfiscalledasymmetryofA.ThesetofsymmetriesofasubsetAoftheplaneisagainagroupundercomposition.Symmetrygroupsneednotbeabelian.Thisisseeninthefollowingexample.Considertheequilateraltriangletwoofwhoseverticeslieonthex-axiswhilethethirdliesonthepositivey-axis.Denotetheclockwiseandcounterclockwiserotationthrough120aboutthecenterofthetriangleby�and+,respectively.Also,denotethere ectionacrossthemedianfromtheuppervertexbyu,there ectionacrossthemedianfromthelowerleftvertexbyl,andthere ectionacrossthemedianfromthelowerrightvertexbyr.Alltheseoperationsaresymmetriesofthetriangle.Finally,denotetheidentityby.Allpossiblecompositionsofthesesymmet

25 riesaregivenbythefollowingmultiplication
riesaregivenbythefollowingmultiplicationtablewheretherstfactorofaproductistakenfromtheleftandthesecondfromthetop: �+lru  �+lru� �+rul+ +�ulrl lur+�r rlu�+u url+�Thistableshowsthatcompositionsofthesixsymmetriesofanequilateraltriangleintroducedabovedonotyieldanynewsymmetries.Arethereanyothersymmetriesoftheequilateraltriangle?Letbeanysymmetry.Thenmapsverticestovertices.Letbetherestrictionoftothesetofvertices,i.e.,=jf1;2;3g.Eachsuchfunctionisabijectionfromf1;2;3gtoitself.Sincetherearesixsuchbijectionsthereareatmostsixsymmetries.HenceD3=f;�;+;l;r;ug 2.1.GROUPS25isthesetofallsymmetriesofanequilateraltriangle.Moregenerally,Dnisthesymmetrygroupofaregularn-gon.2.1.7Perm

26 utationsandsymmetricgroups.Consideragain
utationsandsymmetricgroups.Consideragainthebijectionsof2.1.6.Anyjustrearrangesorpermutesthe\letters"1,2,and3.Thereforeiscalledapermuta-tion.Moregenerally,letM=f1;2;:::;ngandSn=fs:M!M:sisabijection)g.ThenSnisagroupundercomposition.(Sn;)(orjustSn)iscalledthegroupofpermutationsofnletters(symbols)orthesymmetricgroupofnletters.AnyelementofSniscalledapermutation.Oneusuallyusesjuxtapositiontodenotethegroupoperation,i.e.,isabbreviatedby.NotethatSnhasn!elementssincethenumberofbijectionsbetweentwosetsofnelementsisn!.Aconvenientnotationforapermutation2Snis=12:::n(1)(2):::(n).If2Snwecallthesetfk:(k)6=kgthesupportofdenotedbysupp().Twopermutationsarecalleddisjointiftheirsupportsaredisjoint.Notethat(supp())=supp().Supposethatandaredisjointpermutations.Ifkisnotinthesupportofthen(k)isalsonotinthesupportofandhence()(k)=(k)=()(k).Asimilarargumentap

27 plieswhenkisnotinthesupportof.Hence
plieswhenkisnotinthesupportof.Hencewehaveshownthattwodisjointpermutationsarecommutative.2.1.8Cycles.Apermutation2Sniscalledacycleoflengthkiftherearedistinctintegersa1;:::;ak2f1;:::ngsuchthat(a1)=a2,(a2)=a3,...,(ak)=a1,andsuchthatleavesallothern�kelementsoff1;:::;ngxed.Suchacyclewillbedenotedby(a1;a2;:::;ak).Forexamplethepermutation1234535241isacycleoflength4andcanalsobedenotedby(1;3;2;5).Theorem.Everypermutation,excepttheidentity,iseitheracycleorcanbewrittenasacomposition(product)ofdisjointcycles.Thisfactorizationisuniqueuptotheorderofthecycles.Sketchofproof:FixnandconsiderSn.LetC=f2Sn:istheidentity,acycle,oracompositionofdisjointcyclesg:andX=fk2N:82Sn:#supp()=k!2Cg:Then12Xsincethesupportofanypermutationdierentfromtheidentityhasatleasttwoelements.Supposef1;:::;kgX.Ifknthenk+12Xsincethesupportofanypermutationhasatmostnelements.Ifknconsiderapermutationwhosesupporthask+1elementsandhenceisnotempty.Picka1

28 2supp()anddenerecursivelyaj+1=
2supp()anddenerecursivelyaj+1=(aj)=j(a1).LetJ=fj:9mj:am=ajg.SinceJisnotemptyitcontainsasmallestelement`+1.Sincea`+1=(a`)=a1oneobtainsthat(a1;::::;a`)isacyclewhichwedenoteby.If`=k+1then=isacycle.If`klet =�1.Thepermutationsand aredisjointandsupp( )hasnomorethankelements.Hence,byinductionhypothesis, isinCandsois.Altogetherwehavek+12XandthesecondinductionprinciplegivesnowthateveryelementofSnexcepttheidentityisacycleoraproductofdisjointcycles. 2.1.GROUPS27hasaninversee�12G.Hencee=e�1e2=e�1e=1when1istheidentityofG.Leta;b2A.Thenb�1,theinverseofbinGisalsotheinverseinAandhenceanelementofA.Thereforea
b�1isinA,too.Suciency:SinceAisnotemptythereisanelementa2A.Then1=a
a�12A.Also,foreveryb2Awehavethatb�1=1
b�12A.Finally,a
b=a
(b�1)�12Awhenevera;b2A.Hence
isanassociativebinaryoperationonA.Examples:ForanygroupGwithidentity1thesetsf1gandGarebothsubgroupsofG.Theformeriscalledthetrivials

29 ubgroupofG.IfknthenSkisasubgroupofS
ubgroupofG.IfknthenSkisasubgroupofSn.Thegroupofintegers,(Z;+),isasubgroupof(Q;+).RigidmotionsoftheplaneformasubgroupofthegroupofallbijectionsfromR2toR2.Thesubsetsf;�;+gandf;kg,k=1;2;3,ofD3inSection2.1.6aresubgroupsofD3.ThealternatinggroupAnisasubgroupofSn.IfU1andU2aresubgroupsofagroupGthenU1\U2isalsoasubgroupofG,or,moregenerally,ifCisanynonemptycollectionofsubgroupsofGthenTU2CUisalsoasubgroupofG.2.1.12Subgroupsgeneratedbysubsetsofagroup.LetMbeanonemptysubsetofagroupGandCthecollectionofallsubgroupsofGwhichincludeM.ThenhMi=TU2CUiscalledthegroupgeneratedbyM.ThegroupgeneratedbyMistheminimumofCwhenthesetofallsubgroupsofGispartiallyorderedbyinclusion.ItisthereforealsocalledthesmallestsubgroupofGincludingM.Theorem.LetGbeagroupandManonemptysubsetofG.ThenhMi=fNYj=1anjj:N2N,aj2M,nj2Zforj=1;:::;Ng:Sketchofproof:LetAbethesetontherighthandside.Thenfg6=MA.Supposea;b2A.Thenab�12A,too.HenceAisasubgroupofGwhichincludesM,i.e.,hMiA.Nowleta2A,i.e.,a=QNj=1anjjwhereaj

30 2Mandnj2Z.Thenaj2hMiandthereforea2hMi,to
2Mandnj2Z.Thenaj2hMiandthereforea2hMi,too.2.1.13Centralizerandcenter.Let(G;
)beagroupandMasubsetofG.ThendeneC(M)=fg2G:gm=mgforeverym2Mg;thecentralizerofM.ThecentralizerC(G)ofthegroupitselfisalsocalledthecenterofG.(C(M);
)isasubgroupof(G;
)and(C(G);
)isabelian.2.1.14Orderofagroupelement.Theorderordaofanelementa2Gisdenedtobethesmallestnaturalnumberksuchthatak=1ifthisexists.Otherwiseaissaidtohaveinniteorder(orda=1).2.1.15Cyclicgroups.Leta2G.Thenhai=fan:n2ZgiscalledthecyclicsubgroupofGgeneratedbya.AgroupGsuchthathai=Gforsomea2Giscalledacyclicgroup.Acyclicgroupisalwaysabelian.Ifahasinniteorderthentheelementsanofhaiarealldistinct.Ifahasorderkthenhai=f1;a;a2;:::;ak�1g.Examples:(Z;+)iscyclic,infactZ=h1i.Thesubgroupf;�;+gofD3equalsh+i=h�iandhenceiscyclic.fexp(in):n2Zgandfexp(2in=k):n2f1;:::;kggarecyclicgroupsundermultiplicationoforderinnityandk,respectively.Theorem.Thefollowingstatementshold:1.Allsubgroupsofacyclicgroupar

31 ecyclic.2.Foraninnitecyclicgroupall
ecyclic.2.Foraninnitecyclicgroupallsubgroupsbutthetrivialonehaveinnitelymanyelements. 2.1.GROUPS292.1.20Quotientgroups.Thefollowingtheoremistrue.Theorem.IfNisanormalsubgroupofGthenG=Nisagroupunderthebinaryoperationgivenbythesubsetproduct.Sketchofproof:LetaN;bN2G=N.Then(aN)(bN)=fanbm:n;m2Ng.SinceNisnormalwendthatforalln2Nthereexistsn02Nsuchthatnb=bn0.Hence(aN)(bN)=fabn0m:n0;m2Ng=(ab)NisaleftcosetofN.ChoosingdierentrepresentativesfromaNandbNwillyieldthesameproductsetsothatmultiplicationofcosetsiswelldened.HencesubsetmultiplicationisanassociativebinaryoperationonG/N.Also,NistheidentityelementinG=Nanda�1NtheinverseelementofaN.ThegroupG=NofleftcosetsofanormalsubgroupNofGiscalledaquotientgrouporfactorgroup.IfGisnite,thenwehavejG=Nj=jGj=jNj.IfGisinniteandNisnite,thenG=Nisinnite.2.1.21Residueclasses.Letmbeaninteger.ThenthesetmZ=fmk:k2Zgisasubgroupof(Z;+).ThecosetsofmZarecalledresidueclassesmodm.Fora+(mZ)=(mZ)+awewillwriteam.Theresidueclassesmodmare

32 explicitlygivenbyam=fa+km:k2Zg:Since(mZ;
explicitlygivenbyam=fa+km:k2Zg:Since(mZ;+)isanormalsubgroupof(Z;+)wegetthatZm=Z=(mZ)isagroup(undersubsetaddition).2.1.22Directproducts.LetGandHbegroups.ThenintroduceabinaryoperationonGHby(g1;h1)(g2;h2)=(g1g2;h1h2).UnderthisbinaryoperationGHisagroupcalledthedirectproductofGandH.2.1.23Homomorphisms.LetGandHbegroups.Amapping:G!Hiscalleda(group)homomorphismif(ab)=(a)(b)foralla;b2G.Examples:TheidentitymapfromGtoGisahomomorphism.ThemapG!feg:g7!e,whereeistheidentityofanygroup,isahomomorphism.Form2NthemapZ!mZ,k7!mk,isahomomorphism.Forahomomorphism:G!Hwedenethekernelkeroftobethesetker=fg2G:(g)=1Hg.RecallthattheimageofGunderistheset(G)=f(g):g2Gg.Let:G!Hand:H!Kbegrouphomomorphisms.Thenthefollowingbasicfactshold:1.(1G)=1H.2.(a�1)=(a)�1.3.Ifg2Ghasniteorderthentheorderof(g)2Hdividestheorderofg.4.:G!Kisahomomorphism.Theorem.Foranyhomomorphism:G!Htheset(G)isasubgroupofHw

33 hichisabelianifGisabelian.Thesetker
hichisabelianifGisabelian.ThesetkerisanormalsubgroupofG.Ahomomorphism:G!Hisinjectiveifandonlyifker=f1Gg.Sketchofproof:1H=(1G)2(G).Suppose(x);(y)2(G).Then(x)(y)�1=(xy�1)2(G).IfGisabelianthen(x)(y)=(xy)=(yx)=(y)(x).1G2ker.Ifx;y2kerthen(xy�1)=(x)(y)�1=1Handhencexy�12ker.LetabeanyelementinGandxanyelementinker.Thenaxa�12ker,too.Henceker/G.Supposeker=f1Ggand(g)=(g0).Then(g�1g0)=1Handhenceg=g0.2.1.24Canonicalhomomorphisms.LetN/G.ThenthemappingG!G=N:a7!aNisahomomorphismwhichiscalledthecanonicalhomomorphismfromGtoG=N. CHAPTER3Rings3.1.GeneralRingTheory3.1.1Rings.LetRbeaset.Supposetherearetwobinaryoperations+and
onRsuchthatthefollowingpropertiesaresatised:(a)(R;+)isanabelian(commutative)groupand(b)
isassociativeand(leftandright)distributiveover+.Then(R;+;
)iscalledaring.Theringiscalledcommutativeifmultiplicatio

34 niscommutative.Theidentityelementof(R;+)
niscommutative.Theidentityelementof(R;+)isdenotedby0.Examples:Z,Q,R,andCarecommutativerings.LetRbearingandXanonemptyset,thenthesetofallR-valuedfunctionsonformsaring,whichiscommutativeifRis.Thesetofallnnmatricesisanon-commutativering.Thesetf0gcanbeconsideredaring,thezeroring.3.1.2Elementaryproperties.Let(R;+;
)bearing.Theadditiveidentityisdenotedby0andtheadditiveinverseofa2Risdenotedby�a.Thefollowingpropertiesholdtheninanyring:1.0a=a0=0foralla2R,2.(�a)b=�(ab)=a(�b).3.1.3Identity.A(multiplicative)identityorunityinaringisanonzeroelement1suchthat1a=a1=aforallelementsainthering.Ifthereissuchanelementtheringiscalledaringwithidentity(orringwithunity).Anidentity,ifitexists,isunique,for1e=1=eifboth1andeareidentities.3.1.4Directproducts.LetRandSberings.WeintroducetwobinaryoperationsonRSby(r1;s1)+(r2;s2)=(r1+r2;s1+s2)and(r1;s1)
(r2;s2)=(r1
r2;s1
s2).UnderthesebinaryoperationRSisaringcalledthedirectproductofRandS.ItiscommutativeifandonlyifRandSarecommutative.3.1.5Inte

35 gralDomains.Ifa=0orb=0thenab=0.Ifa;b6=0b
gralDomains.Ifa=0orb=0thenab=0.Ifa;b6=0butab=0thenaiscalledaleftzerodivisorandbarightzerodivisor.Examples:Everynilpotentmatrixisazerodivisorintheringofmatrices.Intheringofreal-valuedfunctionsonXanyfunctionwhichvanishesonsomenonemptysubsetofXbutnotonallofXisazerodivisor.Acommutativeringwithidentitybutwithoutzerodivisorsiscalledan(integral)do-main.Henceifab=0inanintegraldomainthena=0orb=0.Inanintegraldomainthesocalledcancellationlawholds.Thislawassertsthatab=acanda6=0implyb=c.Example:Zisanintegraldomain:givena;b2Z�f0gandab=0wemayassume(bymultiplying,perhaps,by�1)thata;b2NwhichisimpossiblesincemultiplicationisabinaryoperationinN.33 363.RINGSTheorem.AcommutativeringRwithidentityisaeldifandonlyifitsonlyidealsarethezeroandtheimproperideal.Inparticular,everyeldisaprincipalidealdomain.Sketchofproof:SupposeRisaeld.LetJbeanyidealotherthanthezeroideal.ThenJcontainsanonzeroelementaand,sinceaisaunit,R=haiJR,i.e.,J=R.NextsupposetheonlyidealsofRarethezeroandtheimproperideal.C

36 onsideranonzeroelementrinR.Thenhri=Randh
onsideranonzeroelementrinR.Thenhri=Randhence1=rsforsomes2R,i.e.,risaunit.HenceallnonzeroelementsofRareunits,i.e.,Risaeld.3.1.13Residueclassrings.LetSbeasubringoftheringR.Then(S;+)isasubgroupof(R;+).Recallthatthecosetsa+S=fa+s:s2Sg=S+aofSformapartitionofR.ThesetofallcosetsofSisdenotedbyR=S.Since(R;+)isabelian(S;+)isanormalsubgroupof(R;+).HenceR=Sisanabeliangroupwithrespecttosubsetaddition(i.e.,M+N=fm+n:m2M;n2NgforanyM;NR).Theorem.LetSbeanidealinthecommutativeringR.Deneadditionandmultiplicationofcosetsby(a+S)+(b+S)=(a+b)+Sand(a+S)(b+S)=ab+S.Then(R=S;+;
)isacommutativeringcalledthequotientringorresidueclassringofRmoduloS.Sketchofproof:Sinceadditionofcosetscoincideswithsubsetadditionweget(asexplainedabove)that(R=S;+)isanabeliangroup.Nextweprovethatmultiplicationiswelldened.Ifa+S=c+Sandb+S=d+Sthena�c;b�d2S.Henceab�cd=a(b�d)+(a�c)d2SsinceSisanideal.Thereforeab+S=cd+S,i.e.,multiplicationofcosetsiswelldened.Thevalidityoftheassociative,commutativeanddist

37 ributivelawsfollowsnowinastraightforward
ributivelawsfollowsnowinastraightforwardmanner.3.1.14Modulararithmetic.ForanynaturalnumbermthesetmZisanidealinZ.HenceZm=Z=mZistheresidueclassringofZmodulomZ.Wewilldenotetheelementofa+mZ2Zmbyamandcallittheresidueclassmodmofa.Ifam=bmwesaythataandbarecongruentmodulom.Thishappensifandonlyifa�bisanintegermultipleofm.Notethat0mand1maretheadditiveandmultiplicativeidentity,respectively,andthat(�a)mistheadditiveinverseofam.Theorem.Zmisaeldifandonlyifmisaprimenumber.Sketchofproof:Ifmisnotprimeleta;b2f2;:::;m�1gbesuchthatab=m.Thenambm=mm=0m,i.e.,Zmhaszerodivisorsandhenceisnotaeld.Ifmisprimeconsiderx2f1;:::;m�1g,i.e.,xm6=0inZm.Thengcd(x;m)=1andthusthereexistintegersk;jsuchthat1=kx+jm.Nowkmxm=(kx)m=(1�jm)m=1m,i.e.,x�1m=km.HenceeverynonzeroelementinZmisaunit,i.e.,Zmisaeld.3.1.15Primeideals.AnidealJinacommutativeringRwithidentityiscalledaprimeidealifab2Jimpliesa2Jorb2J.Theorem.JisaprimeidealinRifandonlyifR=Jisanintegraldomain.TheimproperidealRisalwaysprime,andthezer

38 oidealisprimeifandonlyRitselfisanintegra
oidealisprimeifandonlyRitselfisanintegraldomain.Propermaximalidealsareprime.InthiscaseR=Jisinfactaeld.3.2.RingHomomorphisms3.2.1Ringhomomorphisms.LetR1andR2berings.Amappingf:R1!R2iscalleda(ring)homomorphismiff(a+b)=f(a)+f(b)andf(ab)=f(a)f(b)foralla;b2R1.Itiscalleda(ring)isomorphismifitisabijectivehomomorphismanda(ring)automorphismifitisaringisomorphismfromR1toitself.Examples:f:Z!Zm,a7![a]misaringhomomorphism.Thefunctionf:R1!R2,a7!0foralla2R1isahomomorphism,calledthezerohomomorphism. 383.RINGSofaandbthensoisanyassociateofdandanytwogreatestcommondivisorofaandbareassociates.Thenotationgcd(a;b)isusedtodenoteanygreatestcommondivisorofaandb.3.3.4PrimeandirreducibleElements.Anonzeroelementpofanintegraldomainiscalledprimeifitisnotaunitandifpdividesaorbwheneverpdividesab.Anonzeroelementpofanintegraldomainiscalledirreducibleifitisnotaunitandhasonlyimproperdivisors.Inotherwordsanirreducibleelementpallowsonly\trivial"factorizationsp=up0whereuisaunit.Theorem.Aprimeelementisirreducible.Sketchofproof:

39 Assumethatp=abisprime.Thenpdividesaorb,s
Assumethatp=abisprime.Thenpdividesaorb,saya,i.e.,a=pc.Hencep=ab=pcbwhichimpliescb=1sincecancellationisallowedinanintegraldomain.Thusbisaunitandhencepisirreducible.3.3.5Divisibilityandideals.Divisibilityisstronglyrelatedtocontainmentofideals:Theorem.LetRbeanintegraldomainanda;b2R,b6=0.Then1.hai=Rifandonlyifaisaunit.2.haihbiifandonlyifbisadivisorofa.Alsohai=hbiifandonlyifbisanimproperdivisorofa,i.e.,ifandonlyifaandbareassociates.3.aisirreducibleifandonlyifhai6=f0gismaximalamongallproperprincipalideals,i.e.,thereisnoprincipalidealbutRandhaiitselfwhichcontainshai.LetJ1andJ2beidealsintheintegraldomainR.OnecallsanidealJ1amultipleofJ2orJ2afactorordivisorofJ1ifJ1J2.ThegreatestcommondivisorofJ1andJ2,denotedbygcd(J1;J2),istheidealgeneratedbyJ1[J2.TheleastcommonmultipleofJ1andJ2istheidealJ1\J2.3.3.6Uniquefactorizationdomains.LetRbeanintegraldomainandaanonzeroelementofRwhichisnotaunit.Twofactorizationsa=a1:::an=b1:::bmintoirreduciblesa1;:::;bmarecalledequivalentifn=mandifthereisapermutatio

40 n2Snsuchthatajisassociatedwithb
n2Snsuchthatajisassociatedwithbjforj=1;:::;n.Ifallfactorizationsofaintoirreduciblesareequivalentonesaysthatthefactorizationofaisessentiallyunique.AnintegraldomainRiscalledauniquefactorizationdomain(UFD)ifitsatisesthefollowingconditions:1.everynonzeroelementwhichisnotaunitisaniteproductofirreducibles,2.foreverynonzeroelementwhichisnotaunitallfactorizationsintoirreduciblesareessentiallyunique.AnytwoelementsinaUFDwhicharenotbothequaltozerohaveagreatestcommondivisor.3.3.7Primalityandirreducibility.InTheorem3.3.4wehaveshownthataprimeelementisalwaysirreducible.Iftheconverseisalsotruethenfactorizationsareessentiallyunique.AmoreprecisestatementisthefollowingTheorem.LetRbeanintegraldomaininwhicheverynonzeronon-unithasafactorizationintonitelymanyirreducibles.Thenthesefactorizationsareessentiallyunique(i.e.,Risauniquefactorizationdomain)ifandonlyifeveryirreducibleelementisprime.Sketchofproof:AssumethatRisauniquefactorizationdomain.Letp2Rbeirreducibleandassumepdividesab,i.e.,ab=

41 pc.Firstconsiderthecasethatoneofa;b,sayb
pc.Firstconsiderthecasethatoneofa;b,saybisaunit.Thena=pcb�1andhencepdividesa.Henceassumethatneitheranorbisaunit.Thenleta1:::amandb1:::bnbeafactorizationsofaandb,respectively,intoirreducibles.Sincepc=abtheproducta1:::amb1:::bnis(anessentiallyunique)factorization 3.3.UNIQUEFACTORIZATION41Sketchofproof:Assumethecontraryweretrue.Thenthereexistsanonzeronon-unita1whichisnotirreduciblenorcanitbefactoredintoaproductofnitelymanyirreducibles.Sincea1isnotirreduciblea1=a2b2whereneithera2norb2isaunit.Atleastoneofthese,saya2,isnotirreduciblenorcanitbefactoredintoaproductofnitelymanyirreduciblessinceotherwisea1couldbefactoredinthismanner.Thus(2)a1=a2b2;a2=a3b3;a3=a4b4;:::whereallaj;bjarenonzeronon-unitsandnoneoftheajisirreducibleorcanbefactoredintoaproductofnitelymanyirreducibles.Equations(2)showthatha1iha2i:::.ByTheorem3.3.10thischainterminates,i.e.,haNi=haN+1iforsomeN.ThereforeaN+1=caN=cbN+1aN+1.SinceanintegraldomainallowscancellationwehavethatcbN+1=1,i.e.,thatbN+1isaunit,thed

42 esiredcontradiction.3.3.13EveryPIDis
esiredcontradiction.3.3.13EveryPIDisaUFD.SinceaPIDisNoetherianfactorizationofnonzerononunitsisalwayspossible.Weshownowthatitisessentiallyunique.SupposepisanirreducibleelementinthePIDRandthatpdividestheproductab.Ifpdoesnotdivideathena62hpi.Sincehpiismaximalamongallproperprincipalidealsandhenceamongallproperidealsweobtainthatha;pi=R.Hence12ha;pi,i.e.,thereexistx;y2Rsuchthat1=ax+pyandb=abx+pbywhichshowsthatpdividesb.Thisimpliesthatpisprime,i.e.,inaPIDeveryirreducibleelementisprimeandhencewehavetheTheorem.EveryPIDisaUFD.3.3.14Euclideandomains.AnintegraldomainRiscalledaEuclideandomainifthereexistsafunctionN:R!N0withthefollowingproperties:1.N(r)=0ifandonlyifr=0.2.N(rs)=N(r)N(s)forallr;s2R.3.Ifa;b2Randa6=0thenthereexistqandrinRsuchthatb=aq+randN(r)N(a).NotethatN(1)=N(1)2=1.LetubeaunitintheEuclideandomainR.Then1=N(uu�1)=N(u)N(u�1)andhenceN(u)=1.Conversely,ifN(u)=1thenthereexistsaq2Rsuchthat1=qu,i.e.,qisaunit.Example:ZisaEuclideandomainwhereN(n)=jnj.EveryeldisaEuclideandomainwhenonede&

43 #12;nesN(r)=1forallr6=0.Imitatingtheproo
#12;nesN(r)=1forallr6=0.ImitatingtheproofforZyieldsthefollowingTheorem.AEuclideandomainisaPIDandhenceaUFD.WeknowthatZ[p �5]isnotaUFD.Thereforetheremustidealswhicharenotprincipal.Indeed,letJ=h2;1+p �5i.Thenx+yp �5isinJifandonlyifx�yisaneveninteger.ThereforeJisaproperideal.NowassumeJisprincipalandgeneratedbyanelementr.Thenboth2and1+p �5aremultiplesofr.ThereforeN(r)dividesbothN(2)=4andN(1+p �5)=6.SincerisnotaunitN(r)=2.However,N(z)isdierentfrom2foreveryz2Z[p �5].ThisgivesacontradictiontotheassumptionthatJisprincipal.3.3.15Gaussianintegers.TheintegraldomainZ[i]=fa+bi:a;b2ZgiscalledthesetofGaussianintegers.Sincei=p �1theGaussianintegersareaquadraticextensionoftheintegers.Itsunitsare1andi.Theorem.TheGaussianintegersformaEuclideandomain.Sketchofproof:LetN:Z[i]!N0,a+ib7!a2+b2.Weonlyhavetoprovethatdivisionwitharemainderispossible.Let;2Z[i]andsuppose=a+ib6=0.Thenthecomplexnumber=canbewrittenasm+r+i(n+s)wherem;n2Zand�1=2r;s1=2.T

44 hereforez0=�(m+in)=(r+is)(a
hereforez0=�(m+in)=(r+is)(a+ib)=(ar�bs)+i(rb+as)isaGaussianinteger.ButN(z0)=(a2+b2)(r2+s2)N()=2N(). 423.RINGS3.4.Polynomials3.4.1Polynomials.LetRbearingandP=ff:N0!R:(9N:8n�N:f(n)=0)g:Hencef2Pcanberepresentedbyasequence(a0;a1;a2;:::)ofelementsofRwhereonlynitelymanyoftheelementsajaredierentfromzero.WedeneadditionandmultiplicationinPinthefollowingway:letf;g2Pthen(f+g)(n)=f(n)+g(n);(fg)(n)=Xj+k=nf(j)g(k):Thusadditionandmultiplicationarebinaryoperations.Itiseasytocheckthattheyareassociative.Alsoadditioniscommutativeandmultiplicationisdistributiveoveraddition.Theelement(0;0;:::)isanadditiveidentityandforanyf=(a0;a1;:::)2Ptheelement�f=(�a0;�a1;:::)isanadditiveinverseoff.ThereforePisaringcalledtheringofpolynomialsoverR.AnelementofPiscalledapolynomial(withcoecientsinR).3.4.2Basicpropertiesofpolynomialrings.TheringPofpolynomialsoverRiscommutativeifandonlyifRiscommutative.IfRhasanidentity1,thensodoesPnamelytheelement(1;0;0;:::).Conv

45 ersely,if(b0;b1;:::)isanidentityofPthen,
ersely,if(b0;b1;:::)isanidentityofPthen,foranya2Rwehave(a;0;0;:::)(b0;b1;:::)=(ab0;ab1;:::)whichshowsthatb0isanidentityinR(andthatb1=b2=:::=0).Themapping:R!P,a7!(a;0;:::)isanisomorphismbetweenRand(R)P.Thereforewewillsubsequentlyabbreviatetheelement(a;0;:::)byawhennoconfusioncanarise.IfRhasanidentity1letx=(0;1;0;0;:::).Thenx2=(0;0;1;0;:::)and,ifn2N0,xn=(0;:::;0;1;0;:::)where1isinthe(n+1)stslot.Therefore(a0;a1;:::;an;0;:::)=a0+a1x+:::+anxnwhichyieldsthefamiliarexpressionforapolynomial.Itmustbestressed,however,thatapolynomialisnottobeconsideredafunctionfromR!R,i.e.,xisnottobethoughtofasavariableelementofR.Theelementx=(0;1;0;:::)2Piscalledanindeterminateand,fork=0;:::;n,theelementakiscalledacoecient(ofxk).ThesetPwillbedenotedbyR[x]andwillbecalledtheringofpolynomialsinoneindeterminateoverR.3.4.3Polynomialfunctions.LetPbearingwithidentity,letRbeaunitarysubringofP,andf=a0+:::+anxn2R[x]apolynomialoverR.Thenfgivesrisetoafunction^f:P!Pthroughthedenition^f:p7!a0+:::+anpn.Thefun

46 ction^fiscalledapolynomialfunctionoverR(
ction^fiscalledapolynomialfunctionoverR(orP).Thepolynomialfunction^f:R!Rmustnotbeconfusedwiththepolynomialf:N0!Rasthefollowingexampleshows:letR=Z2,f=x2+x+1,andg=x4+x3+x2+x+1.Thenf6=gbut^f=^g.SupposePiscommutativeandxp2P.Thenf7!^f(p)isaringhomomorphismfromR[x]toP.Thereforef7!^fisasurjectiveringhomomorphismfromR[x]tothesetofpolynomialfunctionsoverRonPwhichisthereforearing,too,calledtheringofpolynomialfunctionsoverRonP.3.4.4Thedegreeofapolynomial.Letf6=0beapolynomial.Thendeg(f)=maxfn:f(n)6=0giscalledthedegreeoff.Nodegreeisassignedtothezeropolynomial.Ifdeg(f)=nthenf(n)iscalledtheleadingcoecientoff.Thepolynomialfiscalledmonic 443.RINGS(3)(fg)0=fg0+f0gandDj(fg)=(Djf)g+f(Djg).3.4.8Thedivisionalgorithm.Wenowturntopolynomialdivision.Theorem.LetRbeacommutativeringwithidentityandf;gpolynomialsoverR.Supposethatdeg(g)=nandthataistheleadingcoecientofg.Thenthereexistpolynomialsqandrandanonnegativeintegerksuchthatakf=qg+randr=0ordeg(r)n.Sketchofproof:Iff=0chooseq=r=0.Foranynonzeropolynomialh2

47 R[x]let (h)=maxfdeg(h)�n+1;0g.Thende&
R[x]let (h)=maxfdeg(h)�n+1;0g.ThendeneP=ff2R[x]:(9q;r2R[x]:a (f)f=qg+r^(r=0_deg(r)n))gandS=fj2N:deg(f)=j�1)f2Pg:Firstweshowthat12S.Letfhavedegreezero.Ifdeg(g)=0chooser=0,q=f.Ifdeg(g)&#x-278;0chooseq=0andr=f.Hence12S.Nowassumethatf1;:::;j�1gS.Ifjnonemaychooseq=0andr=ftoshowthatj2S.Ifj&#x-278;ndenotetheleadingcoecientf(j�1)offbyb.Thenbxj�1�ngisapolynomialofdegreej�1withleadingcoecientabandhenceh=af�bxj�1�nghasdegreeatmostj�2.Byinductionhypothesish2Pandhencethereexistq;r2R[x]suchthata (h)h=qg+randr=0ordeg(r)n.Since (f)=j�n&#x-278;0and (h)maxfj�n�1;0g=j�n�1wenda (f)f=aj�n�1af=aj�n�1(h+bxj�1�ng)=aj�n�1� (h)(qg+r)+b(ax)j�n�1g=(aj�n�1� (h)q+b(ax)j�n�1)g+aj�n�1� (h)r:Hencef2Pandj2S.Thisprovesthetheorem.Corollary.IfRisaeldthenR[x]isaEuclideandomainandhenceaUFD.Sketchofproof:DeneN:R[x]!N0byN(0)=0andN(f)=2deg(f)iff6=0.3.4.9Zerosofpolynomialfunctio

48 ns.LetRbearingwithidentity,fapolynomialo
ns.LetRbearingwithidentity,fapolynomialoverR,and^ftheassociatedpolynomialfunction.Anelement2Riscalledazeroorrootof^fif^f()=0.Theorem.LetRbeacommutativeringwithidentity,f2R[x],and^ftheassociatedpolynomialfunction.Thenisazeroof^fifandonlyiffisdivisiblebyx�.Moreover,ifRisanintegraldomainthenthenumberofdistinctrootsof^fisatmostequaltothedegreeoff.Sketchofproof:Bythedivisionalgorithmthereexistpolynomialsqandrsuchthatf=q(x�)+randrisaconstant.But0=^f()=^r()sinceg7!g()isahomomorphism.Thisshowsthatr=0.Conversely,iff=q(x�)then^f()=0.NowsupposethatRisanintegraldomain,thatdeg(f)=n,andthat1;:::;n+1aredistinctzerosof^f.Then,byinduction,thereexistsapolynomialqsuchthatf=q(x�1):::(x�n+1).Thisimpliesthatdeg(f)n+1,acontradiction.Azeroof^fiscalledazeroofmultiplicitykif(x�)kdividesfbut(x�)k+1doesnot.3.4.10Polynomialsoverintegraldomains.LetRbeanintegraldomain.ThenR[x]andR[x1;:::;xn]areintegraldomains

49 alsosince,if06=f;g2R[x],thendeg(fg)=deg(
alsosince,if06=f;g2R[x],thendeg(fg)=deg(f)+deg(g)exists,i.e.,fg6=0.Thepolynomialf2R[x]isaunitifandonlyiff=ux0anduisaunitofR.Supposef=rx0hasdegreezero.ThenfisirreducibleinR[x]ifandonlyifrisirreducibleinR.Everypolynomialofdegreeoneisirreducible. 3.4.POLYNOMIALS453.4.11PrimitivepolynomialsandGauss'slemma.LetRbeaUFDandf2R[x].Thenfiscalledprimitiveifitscoecientshavenocommondivisorsotherthanunits.LetfbeanynonzeropolynomialinR[x].Thenthereexistsa2Randaprimitivepolynomialgsuchthatf=ax0g.Theelementaisdetermineduptounitmultiples.Theclassofassociatesofa(orsimplya)iscalledthecontentoffandisdenotedbyc(f).Notethatapolynomialisprimitiveifandonlyifitscontentisaunit.Inparticular,everymonicpolynomialisprimitive.TheoremLemmaofGauss.IfRisaUFDandf;g2R[x]thenc(fg)=c(f)c(g).Inparticular,theproductofprimitivepolynomialsisprimitive.Sketchofproof:Itisonlynecessarytoprovethelastassertionofthelemma.Assumethatf;gareprimitivebutthatfgisnot.Letpbeaprimefactorofc(fg).Iff=Pnk=0akxkandg=Pmj=0bjxjletsbesuchthatakisamulti

50 pleofpforksbutthatasisnotamultipleofp.Si
pleofpforksbutthatasisnotamultipleofp.Similarly,lettbethesmallestindexsuchthatbtisnotamultipleofp.Thenthecoecientofxs+tinfgisPj+k=s+takbj.Thiscoecientandeachsummandexceptforasbtisamultipleofp.Thisisimpossible.3.4.12PolynomialsoverUFDs.Theorem.IfRisauniquefactorizationdomainthensoareR[x]andR[x1;:::;xn].Sketchofproof:WeconsideronlyR[x].Werstshowthateverynonzeropolynomialisaunitoraproductofirreducibles.LetP=ff2R[x]:f2R[x]orfisirreducibleoraproductofnitelymanyirreduciblesgandS=fn2N0:deg(f)=n)f2Pg:Letf=ax0beaconstantpolynomialwhichisnotaunitandsupposea=p1:::pkisafactorizationintoirreducibles.Thenf=(p1x0):::(pkx0)isalsoaproductofirreducibles.Thereforef2Pand02S.Nextassumethatn&#x-354;0,thatf0;:::;n�1gS,andthatfisapolynomialofdegreen.Thenf=cf1forsomeconstantpolynomialcandsomeprimitivepolynomialf1.Iff1isirreduciblethenf2P.Henceassumethatf1isnotirreducibleandthatf1=gh.Thenthedegreesofbothgandharepositivebutsmallerthann.Henceg;handthereforefareinPandn2S.Thisshowsthat

51 S=N0andthateverynonzeropolynomialisaunit
S=N0andthateverynonzeropolynomialisaunit,anirreducibleoraproductofnitelymanyirreducibles.NextletJbeanontrivialidealinR[x]andhanelementofsmallestdegreeinJ.Ifistheleadingcoecientofhandf2Jthenthereexistsanonnegativeintegerksuchthathdivideskf.Leth= ~hwhere 2Rand~hisprimitive.Gauss'slemmathenimpliesthat~hdividesf.Wenowshowthateveryirreducibleisaprime.Henceletpbeirreducibleandsupposethatpdividesfgbutnotf.Ifp=^px0then^pisprimeinRandpmustdivideg.Nextassumethatdeg(p)&#x-354;0.Thenpisprimitive.LetJ=hp;fiandh=af+bp,apolynomialofsmallestdegreeinJ.Supposeh= ~hwhere 2Rand~hisprimitive.Then,accordingtowhatwasjustproved,~hdividesp.Supposep=s~h.Sincepisirreducible~horsisaunit.Ifswereaunitthenpwoulddividefwhichitdoesnot.Hence~hisaunitandthushisconstant.Nowhg=afg+bgpisdividedbyp,i.e.,hg=tp.UsingGauss'slemmaagain,hc(g)=c(t),i.e.,t=h~twhere~tisprimitiveandthisshowsg=~tp,thedesiredconclusion.3.4.13PolynomialsoverNoetheriandomains.LetRbeaPIDandJanidealinR[x].DeneLk=fr2R:(9f2J:deg(f)=k;

52 f(k)=r)g[f0g: 463.RINGSThenL0;L1;:::isan
f(k)=r)g[f0g: 463.RINGSThenL0;L1;:::isanascendingchainofidealsinR.ThereforethereexistsanindexqsuchthatLk=Lqforallkq.SinceRisaPIDwehaveLk=hakifork=0;:::;q.Foreachsuchkthereexistsapolynomialfk2Jofdegreekwithleadingcoecientak.LetP=hf0;:::;fqiJandS=fn2N0:(g2J;deg(g)=n))g2Pg:Supposeg=x02J.Then2L0andhencethereexistsr2Rsuchthat=ra0.Thereforeg=(rx0)f02Pand02S.Nextsupposek�0andf0;:::;k�1gS.LetgbeapolynomialinJandsupposedeg(g)=kandtheleadingcoecientofgis.Then2Lkandhencethereexistsanr2Rsuchthat=ramwherem=minfk;qg.Thepolynomialh=g�rxk�mfmisanelementofJwithdegreelessthank.Henceh2P.Thisshowsthatg=h+rxk�mfmisinP,too,andhencethatk2SandS=N0.ThuseverynonzeroelementofJisinP,i.e.,J=hf0;:::;fqi.Hence,ifRisaPIDthenR[x]isNoetherian.ThisproofgeneralizestothecasewhereRisNoetherianwithalittlemorenotationaleort.Itisperformed,forinstance,inZariskiandSamuel,CommutativeAlgebra,Vol.I,Springer1979,p.201f.wehavethusTheoremHilbert'sbasistheorem.If

53 RisaNoetheriandomainthensoareR[x]andR[x1
RisaNoetheriandomainthensoareR[x]andR[x1;:::;xn].3.4.14Polynomialswithrationalcoecients.Z[x]isaNoetherianUFD.Itis,how-ever,notaPIDsince,forexample,theidealh2;xiisnotprincipal.Q[x]isaEuclideandomainandhenceaPID.Letf2Z[x]Q[x]andg;h2Q[x]suchthatf=gh.Thenthereexistintegers;; ;suchthatg=(=)g1andh=( =)h1whereg1andh1areprimitivepolynomialsinZ[x].Hencef= g1h1holdsinZ[x].Gauss'slemmashowsnowthatc(f)= andhencef=c(f)g1h1,i.e.,wehavethefollowingTheorem.IfapolynomialwithintegercoecientsfactorsinQ[x]thenitfactorsalsoinZ[x].Inotherwords,ifapolynomialisirreducibleinZ[x]thenitisalsoirreducibleinQ[x].However,ifapolynomialf2Z[x]isirreducibleinQ[x]andiff=ghwithg;h2Z[x]thenoneofgandhisaconstant.Consider,forexample,thepolynomial2x.ItisirreducibleasanelementofQ[x]since2isaunitinQ[x].InZ[x],however,neither2norxisaunit.3.4.15Polynomialswithcomplexcoecients.C[x]isaEuclideandomainandhenceaPID.TheoremThefundamentaltheoremofalgebra.Iffisapol

54 ynomialinC[x]ofdegreenthen^fhasprecisely
ynomialinC[x]ofdegreenthen^fhaspreciselynroots(countingmultiplicities).Despiteitsnamethistheoremisnotapurelyalgebraicinasmuchnopurelyalgebraicproofisknown.Itcanbeprovenwithsomeknowledgeofcomplexanalysis.Corollary.ApolynomialinC[x]isirreducibleifandonlyifithasdegree1.AeldKwiththepropertythattheirreducibleelementsofK[x]arepreciselythepolynomialsofdegree1iscalledalgebraicallycomplete.3.4.16Polynomialswithrealcoecients.R[x]isaEuclideandomainandhenceaPID.Everyconstantpolynomialisaunitandeverypolynomialofdegree1isirreducible.Iff=ghhasdegreetwoandneithergnorhareunitsthenbothgandhhavedegreeone.Sinceapolynomialfunctionofdegree1hasarealroot^fhasrealrootsalso.Hence 3.5.ALGEBRAICGEOMETRY51forsomeh0;h1;:::;hk2R[y].LetN=maxfdeg(h0);:::;deg(hk)g.Thenhj=PNl=0j;lylandgN=kXj=1fjNXl=0j;lgN�l(gy)l+(gy�1)NXl=00;lgN�l(gy)l:ConsiderthisasanidentityinF[y]whereFisthefractioneldofR.Itthenholdsforally2F.Choosey=1=gtoobtaingN=kXj=1fjNXl=0j;lgN�l;i.e.,gN2Jandg2p J.Thisprove

55 sHN)1.NowassumeV(J)=fg.Thenp J=I(V(J))=I
sHN)1.NowassumeV(J)=fg.Thenp J=I(V(J))=I(fg)=RandhenceJ=R.1.)2.:LetJbeamaximalideal.ThenJ=p JandhenceI(V(J))=J.SinceI(fg)=RwehavethatV(J)isnotempty.Supposea2V(J)thenJ=I(V(J))I(fag).SinceI(fag)6=RandsinceJismaximalwegetJ=I(fag).Conversely,assumethatJisaproperidealcontainingI(fag).Thenfg6=V(J)V(I(fag))=fag.HenceJ=I(fag),i.e.,I(fag)ismaximal.2.)HN:LetJbeaproperidealinR.Then,sinceRisNoetherian,thereexistsamaximalidealMwhichcontainsJ.NotethatM=I(fag)forsomea2Cn.Sincefag=V(M)V(J)wehavethatV(J)isnotempty. CHAPTER4FieldsLetPbearingandRaunitarysubringofP.Givenapolynomialf2R[x1;:::;xn]wewilldenote,asbefore,theassociatedpolynomialfunctiononPnby^f.ThequotienteldofR[x1;:::;xn]isdenotedbyR(x1;:::;xn).4.1.FieldExtensions4.1.1Fieldsandsubelds.Acommutativeringwithidentityiscalledaeldifallitsnonzeroelementsareunits.IfFisaeldandKasubsetofFwhichiseldwithrespecttothethebinaryoperationsofF,thenKiscalledasubeldofFandFiscalledaeldextensionoverK.AsubsetKofFisasub

56 eldifandonlyifitcontainsanonzeroele
eldifandonlyifitcontainsanonzeroelementandifa;b2Kandb6=0implythata�bandab�1arealsoinK.4.1.2Primeelds.Aprimeeldisaeldwhichdoesnotcontainapropersubeld.Notethattheintersectionofallsubeldsofaeldisasubelditself.Itdoesnotproperlycontainanothersubeld,henceitisaprimeeld.Obviouslyagiveneldcontainsauniqueprimeeld.Theorem.IfisaprimeeldthenitiseitherisomorphictoQorelseitisisomorphictoZpwherepissomeprimenumber.Sketchofproof:containstheidentityelements0and1and,necessarily,theelementn1whenevernisaninteger.Notethatn1+m1=(n+m)1andthat(n1)(m1)=(nm)1,i.e.,thesetP=fn1:n2ZgisthehomomorphicimageofZunderthemap:n7!n1.BythefundamentalisomorphismtheoremforringsPisisomorphictoZ=ker().Ifker()=f0gthenPisisomorphictoZ.ButthesmallesteldcontainingZisQ.Ifker()isnontrivialthenker()=hpiforsomenaturalnumberp.Wethenhaverstlythatp�1.Secondly,sinceP(asubsetofaeld)doesnotcontainzerodivisorspmustbeprime.Nown

57 otethatZ=hpi=Zpisaeldifpisprime.
otethatZ=hpi=Zpisaeldifpisprime.4.1.3Thecharacteristicofeld.LetFbeaeldanditsprimeeld.IfisisomorphictoQwesaythatandFhavecharacteristiczero.IfisisomorphictoZpforsomepositiveprimenumberpwesaythatandFhavecharacteristicp.4.1.4Fieldextensions.LetFbeaeld,KasubeldofFandSasubsetofF.ThenwedenotebyK(S)thesmallesteldwhichcontainsK[S,i.e.,theintersectionofallsubeldsofFwhichcontainK[S.WethenhaveK(S)=f^f(s1;:::;sn) ^g(s1;:::;sn):n2N0;s1;:::;sn2S;f;g2K[x1;:::;xn];^g(s1;:::;sn)6=0g:IfShasnelementss1;:::;snwewillwriteK(S)=K(s1;:::;sn)andwewillsaythatK(S)istheeldgeneratedoverKbys1;:::;snortheeldobtainedbyadjoiningtoKtheelementss1;:::;sn.53 544.FIELDSIfS0SthenK(S0)isasubeldofK(S).IfSanysubsetofFdeneC=fTS:Tisniteg.ThenK(S)=ST2CK(T).4.1.5Fieldextensionsasvectorspaces.IfFisaeldextensionoverKthenFisalsoavectorspaceoverKwherethescalarmultiplicationisregularmultiplicationinF.ThedimensionofFasavectorspaceoverKiscall

58 edthedegreeofFoverK.Itisdenotedby[F:K].I
edthedegreeofFoverK.Itisdenotedby[F:K].If[F:K]isnitethenFiscalledaniteeldextensionoverK.4.1.6K-isomorphisms.LetFandF0betwoeldextensionsoveraeldK.Ifthereexistsanisomorphism:F!F0suchthatjKistheidentitywecallaK-isomorphismandwesaythatFandF0areK-isomorphic.ThegroupofK-automorphismsofFisdenotedbyAutK(F).ForexampletheeldsQ(p 2)andQ(�p 2)areQ-isomorphic.4.1.7Simpleeldextensions.LetFbeaeldandKasubeldofF.Ifs2FdeneS=f^f(s):f2K[x]g.ThenSisaringwhichisisomorphictoK[x]=JforsomeidealJK[x].SinceK[x]isaPIDwehavethatJ=f0gorthatJ=h'iforsomepolynomial'.Intherstcase,whereJ=f0g,theelementsiscalledtranscendentaloverKandK(s)iscalledasimpletranscendentalextensionoverK.InthiscaseK(s)isisomorphictoK(x),thequotienteldofK[x].Ontheotherhand,sinceKcanbeconsideredasubeldofK(x),K(x)itselfisasimpletranscendentalextensionoverK.Inthesecondcase,whereJ=h'i,noterstthat'isuniquelydetermineduptomultiplicationbynonzeroelementsofK.h'iisaprop

59 ermaximalidealsinceSFandhenceK[x]=h
ermaximalidealsinceSFandhenceK[x]=h'idonothavezerodivisors.ThereforeK[x]=h'iisaeldsothatK(s),K[x]=h'i,andSareallisomorphic.Since^'(s)=0thepolynomial'hasthefactorx�sandhenceitsdegreeisatleastequaltoone.WecallsalgebraicoverKandK(s)asimplealgebraicextensionoverK.Thepolynomial'iscalledtheminimalpolynomialofsoverK.NowsupposeaeldKandanirreduciblepolynomial'2K[x]aregiven.ConsidertheeldH=K[x]=h'iandthecanonicalhomomorphism :K[x]!H.Thesubset~K=fax0+h'i:a2KgistheisomorphicimageofKandhenceweconsiderKasasubsetofHbyidentifyingKand~K.NotethatH=K(s)wheres= (x)=x+h'iisanalgebraicelementofH,infact^'(s)=0.ThustheexistenceofsimplealgebraicextensionsoverKcanbeobtainedwithoutanaprioriknowledgeofaeldFcontainingtheelements.Example:LetK=Rand'=x2+1thenC=R[x]=hx2+1i=f^f(i):f2R[x]g=R(i)wherei=x+hx2+1i.4.1.8Conjugateelements.Twoelementssands0ofanextensionFoverKarecalledconjugateiftheyarealgebraicoverKandhavethesameminimalpolynomialoverK.4.1.9Thedegreeofasimpleextension.LetKbea

60 eld,'anirreduciblepolynomialinK[x]o
eld,'anirreduciblepolynomialinK[x]ofdegreen,ands=x+h'i.Then[K(s):K]=n,i.e.,thedegreeofthesimplealgebraicextensionK(s)overKisequalton.ThedegreeofasimpletranscendentalextensionoverKisinnite.Sketchofproof:Inthealgebraiccaseoneshows,usingthedivisiontheorem,thatfxj+h'i:j=0;:::;n�1gisabasisofH.Conversely,if[K(s):K]isnitethenK(s)mustbealgebraic.Forexample,ifi=x+hx2+1ithen1andiarelinearlyindependentinR[x]=hx2+1ibuti2+1=0.Hence[R[x]=hx2+1i:R]=2andR[x]=hx2+1i=fa+bi:a;b2Rg. 4.4.RADICALEXTENSIONS594.3.5ThefundamentaltheoremofGaloistheory.IfFisanormaleldextensionoverKthenalleldsEforwhichKEFareclosedandallsubgroupsofAutK(F)areclosedandhencethereisaone-to-onecorrespondencebetweentheintermediateeldsEandthesubgroups.Sketchofproof:LetEbeanintermediateeldandassumetheinclusionEEisstrict.Chooses2E�EanddenotetheminimalpolynomialofsoverEby'.Sinces62Eweknowthatdeg(')�1.SinceFisnormaloverKitisnormaloverEandhenceFcontainsa

61 lltherootsof'.Lettbearootof'dierent
lltherootsof'.Lettbearootof'dierentfroms.Then,byProposition4.1.13,theidentitymapfromEtoEextendstoanisomorphismbetweenE(s)andE(t)andthisisomorphismextendstoanelementofE.Hence(s)=t.However,sinces2Ewealsohave(s)=sandhences=t,acontradiction.HenceE=E.Nowlet=f1;:::;hgbeasubgroupofAutK(F)=Koforderh.Thereisanelementu2FsuchthatF=K(u)andhenceF=(u).Denotetheminimalpolynomialofuoverby'anddenethepolynomialf=(x�1(u)):::(x�h(u))=Phj=0ajxj.Thecoecientsajaresymmetricpolynomialsoftheroots1(u),...,h(u),inparticular,ah=1,ah�1=�Phj=0j(u),anda0=(�1)hQhj=0j(u).Thisimpliesthatk(aj)=ajwhencef2[x].Sinceuisarootoffwehavedeg(')deg(f).SinceFisnormaloverTheorem4.3.1givesthatord()=[F:]=deg(')andhenceord()ord().Thisshowsnallythat=.4.3.6TheGaloisgroupofaeldextension.IfFisanormaleldextensionoverKth

62 esetAutK(F)ofallK-automorphismofFiscalle
esetAutK(F)ofallK-automorphismofFiscalledtheGaloisgroupofFoverKanditisdenotedbyGal(F=K).4.3.7Normalextensionandnormalsubgroups.SupposeFisanormaleldextensionoverKandEisanintermediateeld.ThenEisanormalextensionoverKifandonlyifEisanormalsubgroupofK=AutK(F).Moreover,ifEisanormalextensionoverK,thentheGaloisgroupGal(E=K)isisomorphictothequotientgroupK=E.Sketchofproof:AssumethatE=K(u)andletfbetheminimalpolynomialofuoverK.SupposeEisanormalextensionoverK.If 2Kthen (u)isarootoffandhenceinE.Therefore,if2Eands2E,thens=Pn�1j=0kjujwithkj2Kandhence(  �1)(s)=n�1Xj=0kj(  �1)(u)j=n�1Xj=0kjuj=s:Thisshowsthat  �12E,i.e.,thatEisanormalsubgroupofK.NowassumethatEisanormalsubgroupofK.Then( (u))=( )(u)= (u)forall 2Kandall2E.Thisimplies (u)2E=EandhencethatallrootsoffarecontainedinE,i.e.,Ethesplittingeldoff.Finally,denethehomomorphism:K!Gal(E=K)

63 :7!jE.Thishomomor-phismissurje
:7!jE.Thishomomor-phismissurjectivebyTheorem4.1.13anditskernelisE.Thelaststatementofthetheoremfollowsnowfromthefundamentalisomorphismtheoremforcommutativerings.4.4.RadicalExtensions4.4.1Primitiverootsofunity.LetKbeaeldandletnbeanaturalnumberwhichisnotamultipleofthecharacteristicpofK(anynaturalnumberifKhascharacteristiczero).Letfbethepolynomialxn�1inK[x].If^fhasaroot2Ktheniscalledann-throotofunity.Notethat^fhasatmostnrootsandletFbethesplittingeldoff.Thederivativeoffisnxn�1andisdierentfromzero.Inthiscase^fhasthereforendistinct 604.FIELDSn-throotsofunityinF.Thesetofalln-throotsofunityformsanabeliangroupundermultiplication.Onecaninfactshowthatthisgroupiscyclic,i.e.,itisgeneratedbyoneofitselements.Anysuchelementiscalledaprimitiven-throotofunity.4.4.2Radicalextension.AsimpleeldextensionK(s)overKiscalledasimpleradicalextensionoverKifthereexistsanaturalnumbern�1suchthatsninK,or,inotherwords,iftheminimalpolynomialofsoverKisxn�rforsomer2K

64 .TheadjunctionofstoKisthencalledradical.
.TheadjunctionofstoKisthencalledradical.AeldextensionFoverKiscalledaradicalextensionoverKifthereexistelementst1,...,tminFsuchthatF=K(t1)(t2):::(tm)andeachofthesuccessiveadjunctionsisradical.Theorem.ForanyradicalextensionEoverKthereisanextensionFoverEandanascendingtowerofeldsK=F0F1:::Fk=Fwiththefollowingproperties:(1)Foreachj2f1;:::;kgthereexistsaprimenumberpjandanelementsj2Fjsuchthatrj=spjj2Fj�1andFj=Fj�1(sj).(2)FjisanormalextensionoverFj�1.Sketchofproof:BythedenitionofaradicalextensionwehaveatowerofradicalextensioneldsK=F0F1:::Fk=EwhereFj=Fj�1(sj).Ifsnj2Fj�1andnhastheprimefactorizationn=p1:::plettj;`=sn=(p1:::p`)j.ThenreplacethestringFj�1FjintheabovetowerofeldsbyFj�1Fj�1(tj;1):::Fj�1(tj;1):::(tj;)=Fj:Doingthisforalljweobtainatowerofradicalextensionshavingtherstoftheaboveproperties(withF=E).SupposeFj�1isanormalextensionoverK.IfFj�1containsj,aprimitivepj-throotofunit

65 y,thenFjisanormalextensionoverFj�1and
y,thenFjisanormalextensionoverFj�1andhenceanormalextensionoverK.IfFj�1doesnotcontainj,aprimitivepj-throotofunity,replacethestringFj�1FjinthetowerbyFj�1Fj�1(j)Fj.ThenFj�1(j)isthesplittingeldofthepolynomialxpj�1andhencenormal.AlsoFjcontains,togetherwithsj,allotherrootssjkj,k=1;:::;pjofthepolynomialxpj�rjandhenceisasplittingeld.Repeatingthisforalljresultsinatowerhavingbothoftherequiredproperties.NotethatatstepjtherootjmaynotbeanelementofEsothattheresultingeldFmaybeanextensionofE.Example:LetK=QandE=Q(3p 2).ThenF1=Q(!)where!=(�1+ip 3)=2isathirdrootofunity.AlsoF=F2=Q(!;3p 2)isthesplittingeldofx3�2.4.4.3Radicalextensionandsolvability.IfFisanormalradicalextensionKthentheGaloisgroupofFoverKissolvable.Sketchofproof:ThereexistsatowerofeldsK=F0F1:::Fk=FwiththepropertiesofTheorem4.4.2.BythefundamentaltheoremofGaloistheorythereisatowerofsubgroupsfg=FFk�1:::F1

66 K=Gal(F=K):ByTheorem4.3.7wehave
K=Gal(F=K):ByTheorem4.3.7wehavethatFjisanormalsubgroupofFj�1.ThesubgroupFjismaximalamongthenormalsubgroupsofFj�1sinceitsindexinFj�1equals[Fj:Fj�1]=pj 4.5.THETHEOREMOFRUFFINIANDABEL61whichisprime.HenceFjisamaximalnormalsubgroupofFj�1andourtheoremisprovedonceweshowthatFj�1=Fj=Gal(Fj=Fj�1)isabelian.Suppose1;22Fj�1.SinceFj=Fj�1(sj)wehavethat1and2areuniquelydeterminedrespectivelybythevalues1(sj)=k1jsjand2(sj)=k2jsjwherejisaprimitivepj-throotofunityandk1andk2areappropriateintegers.Hence(12)(sj)=1(k2jsj)=1(j)k2k1jsjand(21)(sj)=2(k1jsj)=2(j)k1k2jsj:Ifspjj=1wemayassumesj=jwhence(12)(sj)=s(k1+1)k2+k1+1j=(21)(sj):Ifspjj6=1wehavej2Fj�1andhence1(j)=2(j)=j.Thus(12)(sj)=k1+k2jsj=(21)(sj):4.5.TheTheoremof

67 RuniandAbel4.5.1Historicalfacts.The
RuniandAbel4.5.1Historicalfacts.Thequadraticformulawasknown(inasense)totheBabyloniansperhaps5000yearsago.Thecubicequationandthequarticequationweresolvedbyradicalsinthe1500sbytheItalianmathematiciansdelFerro,TartagliaandCardanandFerrari.By1799over250yearshadpassedwithoutanyonebeingabletosolvethequinticequa-tionbyradicalseventhoughattemptshadebeenmadebymanymathematiciansincludingveryfamouspeoplelikeEuler,Bezout,Vandermonde,andLagrange.Thenin1799Runiproved(perhapsnotentirelycorrectly)thatthistaskwasinfactimpossiblebuthisassertiondidnotentertheconsciousnessofthemathematicalcommunityofthetime.Thereasonwasperhapsthatnobodyreallybelievedthatitwasimpossible.AquartercenturylaterAbelgaveanotherproofofthisfactandthistimethemessagestuck.4.5.2Thegeneralpolynomialequationofdegreen.Letkbeaeldandu0,...,un�1indeterminates.Thenf=xn+un�1xn�1+:::+u0isapolynomialovertheeldK=k(u0;:::;un�1).Theequation^f(z)=0iscalledthegeneralpolynomialequationofdegreen.4.5.3Theconceptofsolvabi

68 litybyradicals.Tosolveapolynomialequatio
litybyradicals.Tosolveapolynomialequationbyradicalsmeansndingaformulaforitsrootsintermsofthecoecientssothattheformulaonlyinvolvestheoperationsofaddition,subtraction,multiplication,divisionandtakingroots,eachanitenumberoftimes.Denition.Letf2K[x].Thepolynomialequation^f(z)=0iscalledsolvablebyradicalsifthesplittingeldoffisaradicalextensionoverK.Forinstance,thegeneralquadraticequationinK=C(u0;u1)isz2+u1z+u0=0andthecorrespondingequationissolvedbytheformulaz1;2=�u1 21 2q u21�4u0intheextensioneldK(p
)where
=u21�4u0.Thepurposeoftheformulais,ofcourse,thatz1;2aretherootsofz2+u1z+u0(inK(p
))foranychoiceofu1andu0inCorevenK. 624.FIELDSHowever,ifK=C,i.e.,f2C[x],thenthesplittingeldoffisalsoCandaradicalextensionofitself.So,byourdenition,^f(z)=0issolvable,butthisdoesnotmeanthereareeectivemeanstoactuallydoit.Forthisreasonweconsideru0,...,un�1asindeterminates.4.5.4ThetheoremofRuniandAbel.LetKbeaeldofcharacteristiczeroandletf=xn+u

69 n�1xn�1+:::+u0beapolynomialinK(u0;
n�1xn�1+:::+u0beapolynomialinK(u0;:::;un�1)[x].Then^f(z)=0isnotsolvablebyradicalsifn�4.Sketchofproof:Letx1;:::;xnbenindeterminatesandletv0,...,vn�1betheelementarysymmetricpolynomialsoftheseindeterminates,i.e.,v0=(�1)nx1:::xn...vn�1=�x1�:::�xn:ThenK(x1;:::;xn)isthesplittingeldofg=xn+vn�1xn�1:::+v0overK(v0;:::;vn�1).TheGaloisgroupofK(x1;:::;xn)overK(v0;:::;vn�1)isthegroupofpermutationsoftheindeterminatesx1,...,xn.Nowsupposethat^f(z)=0issolvablebyradicalsandletFbethesplittingeldoffwhich,byassumption,isaradicalextensionofK(u0;:::;un�1).OnemayshowthattheringsK[u0;:::;un�1]andK[v0;:::;vn�1]areisomorphicandthisisomorphismextendstoanisomorphismbetweenthefractioneldsK(u0;:::;un�1)andK(v0;:::;vn�1)and,byTheorem4.1.13,toanisomorphismbetweenFandK(x1;:::;xn)andthereforetheGaloisgroupofFoverK(u0;:::;un�1)isisomorphictothepermutationgrouponnletterswhichisnotsolvablebyTheorem4.2.3. 5.2.LINEARTRANSFORMATIONS65by(v1;w1)+(v2;w2)

70 =(v1+v2;w1+w2)and:K(VW)!V&#
=(v1+v2;w1+w2)and:K(VW)!VWisdenedby(r;(v;w))7!r(v;w)=(rv;rw).Theorem.Thedimensionofadirectsumsatisesdim(VW)=dimV+dimW.Sketchofproof:LetB1beabasisofVandB2abasisofW.DeneA1=f(v;0):v2B1g,A2=f(0;w):w2B2gandB=A1[A2.ThenBisabasisofVW.Example:LetV=f(a;a):a2RgandW=f(a;b):a;b2Rg.ThenVW=f((a;a);(b;c)):a;b;c2Rg.Abasisisf((1;1);(0;0));((0;0);(1;0));((0;0);(0;1))g.5.1.9Internalsumsandinternaldirectsums.LetXandYbetwosubspacesofavectorspaceV.TheunionofXandYisnotnecessarilyasubspaceofV.WedenethesumofXandYtobethesubspacegeneratedbytheirunion,i.e.,X+Y=hX[Yi.ItturnsoutthatX+Y=fx+y:x2X;y2Yg.ThedimensionofX+YisinniteifatleastoneofXandYhasinnitedimension.Otherwiseitisgivenbydim(X+Y)=dimX+dimY�dim(X\Y):IfX\Y=f0gwedenoteX+YbyXYandcallittheinternaldirectsumofXandY.1Theorem.LetXbeasubspaceofavectorspaceV.ThenthereexistsasubspaceYsuchthatX\Y=f0gandXY=V.Sketchofproof:LetAbeabasisofX.ByTheorem5.1.7thereisabasisCofVsuchthatAC.DeneB=C�Aa

71 ndY=hBi.ThenYisasubspaceofV.Assumex2X\Y.
ndY=hBi.ThenYisasubspaceofV.Assumex2X\Y.ThenxcanbewrittenasalinearcombinationofelementsofAandasalinearcombinationofelementsofB.SubtractingthesetwoexpressionsgivesthatalinearcombinationofelementsofCequalszero.Henceallcoecientsarezeroandxisequaltozero.ThusX\Y=f0g.SinceA[BspansVwegetthatX+Y=V.5.2.LinearTransformations5.2.1Lineartransformations.LetVandWbetwovectorspacesoverthesameeldK.Afunctionf:V!Wiscalledalineartransformationoravectorspacehomomorphismiff(x+y)=f(x)+f(y)forall;2Kandallx;y2V.AbijectivelineartransformationfromVtoWiscalleda(vectorspace)isomorphism.IfW=Vitiscalleda(vectorspace)automorphism.TwovectorspacesVandWarecalledisomorphicifthereexistsanisomorphismfromVtoW.Thefollowingpropertiesholdforlineartransformationsf:V!W:|f(0)=0,f(�x)=�f(x),sincealineartransformationisagrouphomomorphismfrom(V;+)to(W;+).|Thecompositionoflineartransformationsisalineartransformation.|Therelation\isisomorphicto"isanequivalencerelationonthesetofallvectorspace

72 soverK.|f(V)isasubspaceofW.|kerf=fx2V:f(
soverK.|f(V)isasubspaceofW.|kerf=fx2V:f(x)=0gisasubspaceofV.|fisinjectiveifandonlyifkerf=f0g. 1TheinternaldirectsumofXandYisisomorphic(see5.2.1)totheir(external)directsum.Thisjustiesusingthesamenotation. 5.3.FINITE-DIMENSIONALVECTORSPACES67A.LetAbealineartransformationsfromVtoV.IfkerA2=kerAthenV=kerAA(V).Sketchofproof:NotethatA(V)\kerA=fAx:x2kerA2g.Thelatterset,however,equalsf0gsincekerA2=kerA.By5.1.9andthefundamentaltheoremoflinearalgebraweobtainthenthatdim(A(V)kerA)=dimV.SinceVisnite-dimensionalthisimpliesthatA(V)[ker(A)spansV.B.SupposeA1andA2arecommutativelineartransformationsfromVtoV,thatkerA1\kerA2istrivial,andthatkerA21=kerA1.Thenker(A1A2)=kerA1kerA2.Sketchofproof:Assumex=u+vwhereu2kerA1andv2kerA2.ThenA1A2x=A2A1u+A1A2v=0,i.e.,x2ker(A1A2).Nextsupposethatx2ker(A1A2).BypartAweknowthatx=u+A1vforsomev2Vandsomeu2kerA1.HenceA21A2v=0.SincekerA21=kerA1wehavealso0=A1A2v=A2A1v.HenceA1v2kerA2.C.AssumethatA1,...,AmarepairwisecommutativelineartransformationsfromVtoVsuch

73 thatkerA2j=kerAjforj=1;:::;mandkerAj\ker
thatkerA2j=kerAjforj=1;:::;mandkerAj\kerA`=f0gforj6=`.Thenker(A1:::Am)=kerA1:::kerAm.Sketchofproof:Firstshow(byinduction)thatkerA1\ker(Am+1�k:::Am)=f0gfork=1;:::;m�1.Hence,bypartB,ker(A1:::Am)=kerA1ker(A2:::Am).Anotherinductioncompletestheproof.5.2.4Quotientspaces.Let(V;K;+;)beavectorspaceandUasubspaceofV.Recallthat(V=U;+)isacommutativegroup.OnemaydeneascalarmultiplicationKV=U!V=Uby(;x+U)7!x+U.Thisscalarmultiplication(alsodenotedbyiswelldenedandturns(V=U;K;+;)intoavectorspace,calledthequotientspaceofVwithrespecttoU.Let':V!V=Ubethecanonicalgrouphomomorphismwhichisindeedavectorspacehomomorphism,i.e.,alinearmap.SinceU=ker(')andsince'issurjectiveweobtainfromTheorem5.2.2dim(V=U)=dimV�dimUifUisnite-dimensional.5.2.5Invariantsubspaces.Letf:V!VbealineartransformationonavectorspaceVandUasubspaceofV.Iff(U)UthenUiscalledaninvariantsubspacewithrespecttof.Forexample,thesubspaceC1(R)ofC1(R)isaninvariantsubspacewithrespecttothelinear

74 transformationy7!y0.5.3.Finite-dimension
transformationy7!y0.5.3.Finite-dimensionalvectorspaces5.3.1Existenceofabasisofanite-dimensionalvectorspace.SupposethatVisann-dimensionalvectorspaceandthatAisasubsetofVwithnelements.IfAspansV,thenitisabasisofV.Inparticular,anyn-dimensionalvectorspacehasabasisconsistingofnelements.Sketchofproof:LetVbeavectorspaceofdimensionn.Thenthereexistsasetfx1;:::;xngwhichspansV.Assumethatthissetisnotlinearlyindependent.Thenoneofitselements,sayxn,isalinearcombinationoftheothers.Thisfactisusedtoshowthatanyx2Vcanberepresentedbyalinearcombinationofx1;:::;xn�1.HenceVisinfactspannedbyfx1;:::;xn�1gbutthisisacontradictiontothefactthatVhasdimensionn.5.3.2Exchangeofbasiselements.SupposeVisavectorspaceofdimensionn�0andthatB=fb1;:::;bngspansV(andhenceisabasisofV).IfAisalinearlyindependentsubsetofVwithknelementssuchthatA\Bisempty,thenthereexistsabasisofVwhichcontainsAandn�kelementsofB. 5.3.FINITE-DIMENSIONALVECTORSPACES69withmrowsandncolumns)bylettingthej-thcolumnofMbethetupleofcoeci

75 entsofthevectorf(aj)withrespecttototheba
entsofthevectorf(aj)withrespecttotothebasisB,i.e.,iff(aj)=Pmk=1k;jbkthenM=0B@1;1:::1;n......m;1:::m;n1CA:Thencomputingf(x)isreducedtomatrixmultiplication,sincewehaveforanyx2Knf(x)=264y1...ym375B=M264x1...xn375A:Conversely,everymnmatrixrepresentsalineartransformationfromKntoKmafterchoosingorderedbasesinbothdomainandrange.Moregenerally,ifVandWarevectorspacesoverKofdimensionsnandm,respectively,thenanylineartransformationffromVtoWisrepresentedbyamatrixwithentriesinKandeverymatrixofsuitabledimensionsrepresentsalineartransformationafterorderedbases(v1;:::;vn)and(w1;:::;wm)ofVandWarechosen.Specically,giventhesebasesdeneisomorphismsg:V!Knandh:W!Kmasin5.3.4.Then,accordingtotheabove,thelineartransformationhfg�1:Kn!KmmayberepresentedbyanmnmatrixM.Thenthej-thcolumnofMish(f(g�1(ej)))=h(f(vj))=[f(vj)](w1;:::;wm),i.e.,them-tupleofcoecientsoff(vj)whenwrittenintermsoftheorderedbasis(w1;:::;wm).5.3.6Systemsoflinearequations.Considerasystemofmline

76 arequationsinnun-knownsx1,...,xn,i.e,a1;
arequationsinnun-knownsx1,...,xn,i.e,a1;1x1+:::+a1;nxn=b1;a2;1x1+:::+a2;nxn=b2;:::am;1x1+:::+am;nxn=bm:LetAbethematrixwithentriesAi;k=ai;k2K,b=(b1;:::;bm)�2Km,andx=(x1;:::;xn)�2Kn.ThentheabovesystemcanbeconciselywrittenasAx=bandAcanbeconsideredasthematrixofalineartransformationfromKntoKm(alsodenotedbyA).Wewillnowconsiderthequestionsofexistenceanduniquenessofsolutionsofsuchasystemoflinearequations.ThecolumnsofAareelementsofKm.TheyspanA(Kn),theimageofKnunderthelineartransformationA.ThenumberoflinearindependentcolumnsofAiscalledthecolumnrankof(thematrix)A.ItisequaltothedimensionofA(Kn)andhenceequaltotherankofthelineartransformationA.Let(A;b)bethem(n+1)matrixobtainedbyadjoiningbasacolumntoA.ThevectorbdependslinearlyonthecolumnsofAifandonlyifthematricesAand(A;b)havethesame(column)rank.ThisimpliesthatAx=bhasasolutionifandonlyifrank(A;b)=rankA.Sincen=rankA+dimkerAthelineartransformationAisinjectiveandhenceasolutionofAx=bisuniqueifandonlyifrankA=n.Thesolutionsofahomogeneousequation

77 Ax=0aregivenastheelementsofkerA.Ifxandxp
Ax=0aregivenastheelementsofkerA.IfxandxparebothsolutionsofAx=bthenx�xp2kerA.HencethesolutionsofAx=baregivenastheelementsofthecosetxp+kerA=fa+xp:a2kerAg.ThuswehaveproventhefollowingTheorem.LetAbeanmnmatrixandb2Km.ThenexistenceanduniquenessofsolutionsofthesystemAx=bisgivenbythefollowingtable 705.VECTORSPACES Ax=b=0 Ax=b6=0 rank(A;b)�rankA doesnothappen systemisunsolvable rank(A;b)=rankA=n x=0istheuniquesolution systemisuniquelysolvable rank(A;b)=rankAn systemhasnontrivialsolutions systemhasmanysolutionsMoreover,thesetofsolutionsofAx=0isasubspaceofKnwhosedimensionisgivenbyn�rankA.IfxpissomesolutionofAx=bthenanysolutionxofAx=bcanbeexpressedbyx=xp+xhwherexhisasolutionofAx=0.5.3.7Gaussianelimination.TwosystemsAx=bandA0x=b0ofmlinearequationsinnunknownsarecalledequivalentiftheyhavepreciselythesamesolutions.AsystemAx=bistransformedintoanequivalentsystembyanyofthefollowingelementaryrowoperations:(a)Multiplyanequationbyanonzeroscalar.(b)Addamultipleofoneequationtoanotherone.(c)Intercha

78 ngeanytwooftheequations.Totheserowoperat
ngeanytwooftheequations.Totheserowoperationsamongequationscorrespondsimilarrowoperationsamongtherowsofthematrix(A;b).Twomatricesarecalledrow-equivalentifitispossibletotransformoneintotheotherbyanitesequenceofelementaryrow-operations.Rowequivalenceis(ofcourse)anequivalencerelation.Ofcourse,thematrices(A;b)and(A0;b0)arerow-equivalentifandonlyifthesystemsAx=bandA0x=b0areequivalent.Amatrixissaidtobearow-echelonmatrixifitsatises(a)allzerorowsoccurbelowanynonzerorow,(b)therstnonzeroentryofeachnonzerorowoccurstotherightoftherstnonzeroentryofanyrowabove.Therstnonzeroentryofarowiscalledapivot.Theorem.EverymatrixAisrow-equivalenttoarow-echelonmatrixR.ThenumberoflinearlyindependentrowsofA(calledtherowrankofA)equalsboththenumberoflinearlyindependentcolumnsofA(i.e.,thecolumnrankofA)andthenumberofnonzerorowsofR.Sketchofproof:Therstclaimfollowsbyinductiononthenumberofrows.Theele-mentaryrowoperationsleavethenumberoflinearlyindependentrowsofamatrixinvariant(astheydonotchangethes

79 pacespanned).SincethesystemsAx=0andRx=0a
pacespanned).SincethesystemsAx=0andRx=0areequiv-alent,i.e.,sincekerA=kerR,weobtainrankA=n�dimkerA=n�dimkerR=rankR,assumingthatAisanmnmatrix.Hencetheelementaryrowoperationsleavealsothecolumnrankofamatrixinvariant.TheproofiscompletedbytheobservationthatcolumnrankandrowrankofRareequaltothenumberofnonzerorowsofR.ThereforeelementaryrowoperationscanbeusedtodeterminetherankofAandtherankof(A;b)andhencetoanswerthequestionsofexistenceanduniquenessofsolutionsofAx=b.Buttheyarealsousefulinactuallycomputingthesolution:Let(B;b0)bearow-echelonmatrixwhichisrowequivalentto(A;b)andassumethatrankB=rank(B;b0).ThenxsolvesAx=bifandonlyifxsolvesBx=b0.Letx=(x1;:::;xn)�.IfcolumnkofBcontainsthepivotofsomerowthenxkiscalledadeterminedandotherwiseafreevariable.Anychoiceofscalarsforthefreevariableswillleadtoasolution.Thedeterminedvariablescannowbecomputedonebyonestartingwiththeonehavingthelargestindex.Example:Consider0@1�111211111A0@x1x2x31A=0@1�5b1A: 745.VECTORSPACESwhereeachmatrixJlisaJordanbl

80 ockiscalledaJordanmatrix.5.5.6Nilpotentt
ockiscalledaJordanmatrix.5.5.6Nilpotenttransformations.AlineartransformationTfromanite-dimensionalvectorspaceVtoitselfiscallednilpotentifthereexistsanaturalnumbermsuchthatTmisthezerotransformation.SupposethatTm=0andthatisaneigenvalueofTwithassociatedeigenvectorx.ThenTmx=mxandthisshowsthat=0.Hence,anilpotenttransformationhasonlyoneeigenvalue,namelyzero.ThateigenvaluehasalgebraicmultiplicityequaltothedimensionofV.Theorem.SupposeT:V!Visanilpotentlineartransformationonann-dimensionalcomplexvectorspaceV.ThenVisadirectsumofT-cyclicsubspaces.Sketchofproof:TheproofisbyinductiononthedimensionofV.LetMbethesetofallnaturalnumbersforwhichthetheoremistrue.Then12M,becausedim(V)=1and06=x2VimplythatTx=0andhenceV=[x]T.Nextassumethatn2Mandthatdim(V)=n+1.SinceTisnilpotentdim(T(V))n.ThereforethereisasubspaceFofVofdimensionnsuchthatT(V)F.Sincedim(F)=nandFisinvariantunderTthereexistvectorsx1,...,xksuchthatF=Lkj=1[xj]Twheretheindicesarechosensuchthatm1:::mkwhenmjdenotesdim([x

81 j]T).Nowchooseg2V�F.Thenthereexistsh2
j]T).Nowchooseg2V�F.Thenthereexistsh2Fandnumbers1,...,ksuchthatTg=Th+Pkj=1jxj.Indeed,sinceTg2FwehaveTg=kXj=1mj�1X`=0j;`T`xjandcanthereforechooseh=kXj=1mj�1X`=1j;`T`�1xjandj=j;0forj=1;:::;k.Wenowdistinguishtwocases.Intherstcaseallofthenumbersjarezero.Deningxk+1=g�hgivesTxk+1=0,hxk+1i=[xk+1]T,andV=F[xk+1]T.Inthesecondcasewehaveanumberpsuchthatp6=0butj=0wheneverj�p.Inthiscasewedene~xp=(g�h)=pandF0=Lj6=p[xj]T.ThenT`+1~xp=T`xp+p�1Xj=1j pT`xj:ThisimpliesthatTmp+1~xp=0andthatf~xp;T~xp;:::;Tmp~xpgislinearlyindependent.Hence[~xp]TisaT-cyclicsubspaceofdimensionmp+1.Since[~xp]T\F0=f0gweobtainthatthesum[~xp]T+F0isdirectandhasdimensionn+1.ThusV=[~xp]TF0.Thisshowsthatn+12M.Thetheoremfollowsnowfromtheinductionprinciple.TheaboveproofisduetoGohbergandGoldberg(AsimpleproofoftheJordande-compositiontheoremformatrices,AmericanMathematicalMonthly103(1996),p.157{159).Corollary.Vhasanorderedbasiss

82 uchthatthematrixassociatedwithTisaJordan
uchthatthematrixassociatedwithTisaJordanmatrixallofwhoseJordanblockshaveeigenvaluezero.Conversely,everysuchmatrixisnilpotent. 5.6.MULTILINEARALGEBRA77thespectraltheoremimplies(zI�T)�1=X2(T)()�1Xm=0(T�I)m (z�)m+1E:ThenuseCauchy'sintegralformulaandagainthespectraltheorem.5.5.12TheminimalpolynomialandCayley'stheorem.LetT:V!Vbealineartransformationonanite-dimensionalcomplexvectorspace.Supposethat1,...,mareitsdistincteigenvalueswhoserespectiveindicesare1,...,m.Thenthepolynomial(�1)1:::(�m)miscalledtheminimalpolynomialofT.InviewofTheorem5.5.4weobtainimmediatelythevalidityofthefollowingtheoremofCayley.Theorem.Ifqis(amultipleof)theminimalpolynomialofTthenq(T)=0.Amongallpolynomialsfwithleadingcoecientonesatisfyingf(T)=0theminimalpolynomialofTistheoneoflowestdegree.5.5.13Thecharacteristicpolynomial,traceanddeterminant.LetT:V!Vbealineartransformationonanite-dimensionalcomplexvect

83 orspace.Supposethat1,...,marei
orspace.Supposethat1,...,mareitsdistincteigenvalueswhoserespectivealgebraicmultiplicitiesarek1,...,km.Thenthepolynomial(�1)k1:::(�m)km=n�a1n�1+:::+(�1)naniscalledthecharacteristicpolynomialofT.Thenumbera1=k11+:::+kmmiscalledthetraceofTandisdenotedbytrT.Thenumberan=k11:::kmmiscalledthedeterminantofTandisdenotedbydetT.SinceI�Thasaneigenvalue�jofalgebraicmultiplicitykjifandonlyifThasaneigenvaluejofthesamemultiplicityitfollowsthatthecharacteristicpolynomialofTequalsdet(I�T).Theorem.SupposeT:V!Visalineartransformationonacomplexnite-dimensionalvectorspace.ThenTisinvertibleifandonlyifdetT6=0.InthiscaseT�1isalsoalineartransformationfromVtoV.5.6.MultilinearAlgebra5.6.1Multilinearforms.SupposeVisavectorspaceovertheeldK.Toanyfunctionf:Vk!Kandanyelement(x;y)2Vj�1Vk�jwemayassociatethefunctionfx;y:V!K:t7!f(x;t;y).Thefunctionfiscalledamultilinearformora(covariant)tensorofrankk

84 overVifforallj2f1;:::;kgandall(x;y)2Vj&#
overVifforallj2f1;:::;kgandall(x;y)2Vj�1Vk�jthefunctionfx;yislinear.IfVhasdimensionnthesetTkofalltensorsofrankkoverVisavectorspaceofdimensionnk.5.6.2Antisymmetrictensors.SupposefisatensorofrankkoverthevectorspaceV.Thenfiscalledantisymmetricifforeverytranspositiontherelationshipf(x1;:::;xk)=�f(x1;:::;xk)holds.Equivalently,fiscalledantisymmetricwhenj6=landxj=xlimplythatf(x1;:::;xk)=0.TheantisymmetrictensorsofrankkformasubspaceofTk.5.6.3Determinantforms.LetVbeann-dimensionalvectorspaceandfb1;:::;bngabasisofV.Anantisymmetrictensorfofrankniscalledadeterminantform.Itisuniquely 805.VECTORSPACES(a)kfk1=maxfjf(x)j:x2[a;b]g,(b)kfkp=Rbajf(x)jpdx1=pifp1.5.7.2Innerproducts.AgainletVbeavectorspaceovereithertherealorthecomplexnumbers.Afunction(
;
):VV!Kiscalledaninnerproductorascalarproductifitsatisesthefollowingconditions:(a)(x;x)0forallx2V,(b)(x;x)=0ifandonlyifx=0,(c)(x+y;z)=(x;z)+(y;z),forall;2Kandallx;y;z2V,(d)

85 (x;y)= (y;x)forallx;y2V.Here equals
(x;y)= (y;x)forallx;y2V.Here equalsoritscomplexconjugatedependingonwhetherK=RorK=C.Notethaty=0ifandonlyif(x;y)=0forallx2V.IfK=Rtheinnerproductisbilinear(linearinbothofitsarguments).IfK=Ctheinnerproductislinearinitsrstargumentbutantilinearinitssecond:(x;y+z)= (x;y)+ (x;z).Ifthereexistsaninnerproduct(
;
)onVthen(V;(
;
))iscalledaninnerproductspace.Examples:RnandCncanbeconsideredasinnerproductspaces.AninnerproductonCnisgivenby(x;y)=Pnk=1xk yk.AlsoC0([a;b])canbeconsideredasaninnerproductspacewhen[a;b]isaclosedintervalinR:aninnerproductis(f;g)=Rbaf gdx.Theorem.1.AninnerproductsatisesSchwarz'sinequality,i.e.,j(x;y)j(x;x)1=2(y;y)1=2:2.Everyinnerproductspaceisanormedvectorspaceunderthenormx7!kxk=(x;x)1=2.Sketchofproof:1.Assume(x;y)6=0.Let=j(x;y)j=(y;x).Foranyrealr0(x�ry;x�ry)=(x;x)�2rj(x;y)j+r2(y;y):Now(y;y)6=0sinceotherwise(x;y)wouldbezero.Schwarz'sinequalityfollowsnowfromchoosingr=j(x;y)j=(y;y).2.Thetriangleinequ

86 alityfollowsfromSchwarz'sinequality.
alityfollowsfromSchwarz'sinequality.5.7.3Orthogonality.Suppose(
;
)isaninnerproductonavectorspaceV.If(x;y)=0wesaythatxandyareorthogonalanddenotethisbyx?y.LetMbeasubsetofVanddeneM?=fx2V:(8y2M:(x;y)=0)g.Ifx1;x22M?thensoisx1+x2,i.e.,M?isasubspaceofV.M?iscalledtheorthogonalcomplementofM.Theorem.LetXbeasubsetofaninnerproductspace.IftheelementsofXarenonzeroandpairwiseorthogonalthenXislinearlyindependent.Sketchofproof:Taketheinnerproductofalinearcombinationofthevectorswitheachofthevectorsthemselves.5.7.4Orthonormalsubsets.AsetXwhoseelementshavenormoneandarepairwiseorthogonaliscalledorthonormal.Theorem.LetXbealinearlyindependentcountable(e.g.,nite)subsetofaninnerprod-uctspace.ThenhXi,thespanofX,hasanorthonormalbasis.Sketchofproof:ThebasiscanbeproducedfromXusingthesocalledGram-Schmidtprocess:denez1=x1=kx1k.Thenz1hasnormoneandspanshfx1gi.Nextassumethat, 825.VECTORSPACESTisalineartransformation.ItiscalledtheadjointofT.NotethatT=Tsince(Tx;y)= (y;T&#

87 3;x)= (Ty;x)=(x;Ty):Letfx1;:::;xngbeanor
3;x)= (Ty;x)=(x;Ty):Letfx1;:::;xngbeanorthonormalbasisofV.Supposethat,withrespecttothatbasis,TisrepresentedbythematrixAwhileTisrepresentedbythematrixB,i.e.,Txj=Pnl=1Al;jxlandTxj=Pnl=1Bl;jxl.ObservethatAl;j=(Txj;xl)=(xj;Txl)=(xj;nXk=1Bk;lxk)= Bj;l:HencewefoundthatthematrixrepresentingTisthecomplexconjugateofthetransposeofthematrixrepresentingT(orjustthetransposeifthescalareldisR).WewillwriteB=A.Theorem.LetT:V!Vbealineartransformationonanite-dimensionalinnerproductspace.ThenkerT=T(V)?andkerT=T(V)?.Inparticular,TandThavethesamerank.Sketchofproof:Thefollowingsequenceofequivalencesshowstherststatement.x2kerT,Tx=0,8y:(Tx;y)=0,8y:(x;Ty)=0,x2(T(V))?:5.7.8Normallineartransformations.SupposeVisanite-dimensionalvectorspace.AlineartransformationT:V!VonaninnerproductspaceViscallednormalifitcommuteswithitsadjoint,i.e.,ifTT=TT.IfTisnormalandx2Vthen(Tx;Tx)=(Tx;Tx).ThisimpliesthatkerT=kerTforanynormallineartransformation.IfTi

88 snormalandx2kerT2,thenTx2kerT=kerT.T
snormalandx2kerT2,thenTx2kerT=kerT.Therefore0=(TTx;x)=(Tx;Tx)andhencekerT=kerT2.SinceTisnormalifandonlyifT�Iisnormal,thisimpliesthatthealgebraicandgeometricmultiplicitiesofanyeigenvalueofanormallineartransformationcoincide.Inparticular,ifV,thedomainofT,isacomplexvectorspacethenVhasabasisofeigenvectorsofT.Supposeand0aredistincteigenvaluesofanormallineartransformationTandthatxandx0aretheassociatedeigenvectors.Then(�0)(x;x0)=(Tx;x0)�(x;Tx0)=0whichimpliesthat(x;x0)=0,i.e.,thatxandx0areorthogonal.Inparticular,ifV,thedomainofT,isacomplexvectorspacethenVhasanorthonormalbasisofeigenvectorsofT.InsummarywehavetheTheoremSpectraltheorem.LetTbeanormallineartransformationonanite-dimensionalcomplexvectorspaceVwithdistincteigenvalues1;:::;m.Thenthereexistpairwiseor-thogonalsubspacesU1;:::;UmsuchthatV=U1:::UmandT=mXj=1jPjwherethePjareorthogonalprojectionsontoUj.5.7.9Self-adjointlineartransformations.Alineartransformationonaninnerproduc

89 tspaceiscalledself-adjointifT=T.Note
tspaceiscalledself-adjointifT=T.Notethateveryself-adjointlineartransformationisnormal.Example:AlineartransformationPisanorthogonalprojectionifandonlyifitisidempotentandself-adjoint.i.e.,P=P2=P. IndexK-isomorphism,54adjoint,82algebraassociative,14algebraic,54antilinear,80antisymmetric,77associative,13automorphismofvectorspaces,65axiom,7basis,64bijective,11bilinear,80binaryoperation,13cardinality,16Cartesianproduct,10center,27centralizer,27characteristic,53characteristicpolynomial,77commutative,13complementofaset,8compositionfactors,57compositionseries,57congruent,36conjugate,54connective,1constant,4contrapositive,1converse,1coordinate,68countable,17countablyinnite,17cyclic,73deMorgan'slawsinlogic,2insettheory,9denition,7determinant,77determinantform,77dierenceofsets,8dimension,64directsuminternal,65distributive,13left,13right,13domainofarelation,10domainofdiscourse,5eigenprojection,76eigenspacealgebraic,71geometric,71eigenvalue,71eigenvector,71generalized,71element,5ele

90 mentaryrowoperations,70emptyset,8equalit
mentaryrowoperations,70emptyset,8equalityofsets,8extension,12eld,14eldextension,53nite,54niteset,17xedeld,58formula,4freeabeliangroup,31freegroup,31function,11bijective,11injective,11surjective,11Galoisgroup,59greatestlowerbound,12group,14idempotent,75identity,13imageofarelation,1085 86INDEXofasetunderarelation,10indexofaneigenvalue,72inmum,12innite,17countably,17injective,11innerproduct,80innerproductspace,80intersectionofsets,8invariant,67inverseelement,13irreducible,19isomorphicvectorspaces,65isomorphismofvectorspaces,65Jordanblock,73Jordanchain,73Jordanmatrix,74Jordannormalform,75leastupperbound,12linearcombination,63linearfunctional,81linearindependence,64logicalequivalence,2logicalimplication,1lowerbound,12maximalelement,12maximum,12minimalelement,12minimalpolynomial,77minimum,12multilinearform,77multiplicityofazero,44nilpotent,74normal,82normaleldextension,56one-to-one,11onto,11orderofagroup,23ofagroupelement,27orderedpair,10orderingpartial

91 ,12total,12well,12orthogonal,80orthonorm
,12total,12well,12orthogonal,80orthonormal,80pairwisedisjoint,8parity,26partition,11permutation,25pivot,70polynomialfunction,42powerset,8predicate,4preimageofasetunderarelation,10premise,2primenumber,19proofbycontradiction,4quantierexistential,5universal,5R-module,14radicaladjunction,60radicalextension,60simple,60range,11rank,66relation,10antisymmetric,12composite,10equivalence,11inverse,10re exive,11symmetric,11transitive,11residueclass,36restriction,12ring,14root,44row-echelonmatrix,70scalarmultiplication,14scalarproduct,80scopeofaquantier,5self-adjoint,82sentence,1singular,4set,5spectrum,71splittingeld,55statement,1subeld,53subset,8proper,8subspace,63successor,14successorfunction,14supremum,12surjective,11tautology,1tensor,77theorem,7trace,77transcendental,54transformationlinear,65 INDEX87triangleinequality,22,79uncountable,17unionofsets,8unitarysubring,35upperbound,12variable,4bound,5determined,70free,5,70vectorspace,14nite-dimensional,64innite-dimensional,6