An Introduction to Algebraic Number Theory, and the ClassNumber FormulaSamChowDepartmentofMathematicsandStatisticsUniversityofMelbourneParkvilleVIC3010Australiasam.chow42@gmail.comSUPERVISOR:DRCRAIGWESTERLANDHonoursthesis,October2011.AbstractAlgebraicnumber theorystudies algebraicproperties of theringof algebraicintegers inanumber eld.Wedescribevariousalgebraicinvariantsofnumberelds, aswell astheirapplications. Theseapplicationsrelate to prime ramication, the niteness of the class number, cyclotomic extensions, and the unit theorem.Finally, we present an exposition of the class number formula, which generalizes a result about the Riemannzeta function. The class number formula is a deep connection between algebraic number theory and analyticnumbertheory,anditsproofemploysmanyofthetechniquesdevelopedinthemainbodyofthethesis.iAcknowledgmentsThankstomysupervisor,CraigWesterland,foraveryinterestingproject. Theyearhaspresentedmewithmany challenges, and Craig has always been there to provide emotional support and sage advice. Thanks toJohnGrovesforteachingmetherelevantcommutativealgebraoverthesummer. IamextremelygratefultoTrithangTran, whogenerouslygaveuphistimetoreadthroughmostofthisdocument, ndmistakes,andsuggestwaystoimproveit. ThankstoKimRamchenfordrawingmyattentiontoafewtypographicalerrors. Thankstothefellowstudentswholistenedtometalkaboutnumbertheoryeveryweek;theyreallymotivatedmetomakesteadyprogress. Finally, I wouldliketothankmypartner, Julia, as well as myparentsandfriends,forbeingsosupportiveofmycareerandlifechoices,aswellasformakingmehappyingeneral.iiContents1 Background(algebra) 11.1 Ringsandmodulesoffractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Tensorproducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Chineseremaindertheorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Fieldtheory. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.5 Noetherianringsandmodules. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 Algebraicnumbertheory 122.1 Ringsofintegers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Norms,tracesanddiscriminants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 Dedekinddomains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.4 Theidealclassgroup. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.5 Discretevaluations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.6 Factorizationinextensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.7 Normsofideals,andnitenessoftheclassnumber . . . . . . . . . . . . . . . . . . . . . . . . 522.8 Cyclotomicextensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 602.9 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 682.10 Unitsintheringofintegers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 742.11 Fundamentalunitsandregulators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803 Theclassnumberformula 81iii1 Background(algebra)Thischapterdenesnotationsandpresentsbasicresultsfromabstractalgebra. SeeAtiyah-Macdonald[1],chapters1-3inparticular.1.1 RingsandmodulesoffractionsLetaandbbeidealsinaringA(denoteda, b A). Theirproduct isab :=_n

i=1aibi: ai a, bi b, n Z0_,whichisanidealinA.LetAbeacommutativeunitalring,andletX A. Theideal generatedbyXis(X) :=_n

i=1aixi: ai A, xi X, n Z0_.Letf: A Bbearinghomomorphism. Leta Aandb B. Theextensionaeofaisae:= Bf(a) :=_n

i=1bif(ai) : bi B, ai a, n Z0_, (1.1)whichisanidealinB. Thecontractionbcofbisbc:= f1(b), (1.2)whichisanidealinA.Lemma1.1. Letf:A Bbeahomomorphismbetweencommutativeunital rings, andletbbeaprimeideal inB. Thenbcisaprimeideal inA.Proof. (Properideal:) Letx, y bcanda a. Thenf(x), f(y) b, sof(x + y)=f(x) + f(y) bandf(x)= f(x) b, sox + y, x bc. Moreover, f(ax)=f(a)f(x) b(asbB), soax bc. HencebcA. Further,b ,= B 1/ b 1/ bc. HencebcisaproperidealinA.(Prime:) Letx, y bc. Thenf(x)f(y)=f(xy) b. Asbisaprimeideal,f(x) borf(y) b,sox bcory bc. HencebcisaprimeidealinA.However,ifaisaprimeideal,thenaeneednotbeaprimeideal,forinstancef: Z Q,x xhas(2Z)e= Q,whichisnotaprimeidealin Q.LetAbeacommutativeunitalring. Amultiplicativelyclosedsubset ofAisasubsetS Asuchthat:1 1 S. Ifa, b S,thenab S.TheringoffractionsisS1A :=_as: a A, s S__ =,whereas=btifthereexistsu Ssuchthatu(at bs) = 0. If0 SthenS1Aistrivial,sohenceforthweassumethat0/ S.Let A be a commutative unital ring, let S be a multiplicatively closed subset of A, and let Mbe an A-module.ThemoduleoffractionsisS1M:=_ms: m M, s S__ =,wherems=ntifthereexistsu Ssuchthatu(mt ns) = 0.Wecandeneextendedandcontractedidealsinringsoffractionsbyconsideringf: A S1Af(x) =x1.Ifa A,thenae:= a(S1A) = S1a.Proof. (:)Letn Z0. Fori=1, 2, . . . , n, letati aandaisiS1A. Then

ni=1atiaisiisanarbitraryelementofa(S1A).n

i=1atiaisi=n

i=1atiaitis S1A, (1.3)wheres =

nj=1sjandti=

j,=isj.(:)Leta a, s S. ThenasisanarbitraryelementofS1A,andas= a1s a(S1A). (1.4)Lemma1.2. LetAbeacommutativeunital ring, andletSbeamultiplicativelyclosedsubsetofA. Theneveryideal inS1Aisanextendedideal.Proof. Letb S1A. Leta := bc= x A :x1 b. Thenb = S1a:(:)Letxs b. Thenx1= s xs b,sox a. Hencexs S1a.(:)Letas S1a. Thena a,soa1 b,soas=a11s b.Thusb = aeisanextendedideal,soeveryidealinS1Aisanextendedideal.2Proposition1.3. Let Abeacommutativeunital ring, andlet Sbeamultiplicativelyclosedsubset of A.Thenp S1pisabijectionfromtheset of primeidealsinAwhichdonot meet Stotheset of primeidealsinS1A.Proof. (Well dened:)Let p be a prime ideal in A which does not meet S. As pA and Sis multiplicativelyclosed,S1p S1A. As p ,= A,1/ p,so11/ S1p,so S1p ,= S1A. Letxsyt S1p,wherexs,yt S1A.Thenxy p,sox pory p. Thusxs S1poryt S1p. HenceS1pisaprimeidealinS1A.(Injective:) LetI andJbeprimeidealsinAwhichdonotmeetS, andassumethatS1I=S1J. WeshowthatI J,andthereverseinclusionwillfollowbysymmetry. Letp I. Thenp1 S1I= S1J,sothere exist q Jand s Ssuch thatp1=qs. Now there exists u Ssuch that u(ps q) = 0. ups = uq J,butus S, sous/ J. AsJisaprimeideal, us/ J, and(us)p J, itfollowsthatp J. HenceI J,andbysymmetryI= J.(Surjective:) LetJbeaprimeidealinS1A. (Jcis a prime ideal in A that does not meet S:)As 1/ J, 1/ Jc, so Jcis a proper ideal in A. Supposex, y Aandxy Jc. Thenx1y1 J. AsJisaprimeidealinS1A,x1 Jory1 J. Thusx Jcory Jc. HenceJcisaprimeidealinA. ItremainstoprovethatJc S= . Proofbycontradiction:assumethatthereexistss S Jc. Thens1 Jands1= 1 S1A,soJ= S1A,contradictingthefactthatJisaprimeidealinS1A. HenceJcisaprimeidealinAthatdoesnotmeetS. (Jce=J:) Bydenition, JceJ. Theideal Jce S1Aisgeneratedby x1: x Jc J, andJ S1A,thereforeJce= J.Lemma1.4. LetSbeamultiplicativelyclosedsubsetofacommutativeunital ringA. LetI, J A. ThenS1(IJ) = (S1I)(S1J).Proof. (:)Letn Z0,a1, . . . , an I, b1, . . . , bn J, s S. Then

ni=1 aibis S1(IJ)isarbitraryand

ni=1aibis=n

i=1aisbi1 (S1I)(S1J). (1.5)(:) Letn Z0,a1, . . . , an I, b1, . . . , bn J, s1, . . . , sn, t1, . . . , tn S. Then

ni=1aisibiti (S1I)(S1J)isarbitraryandn

i=1aisibiti=

ni=1aibixiX, (1.6)whereX=

nj=1sjtjandxi=

j,=isjtj(fori=1, 2, . . . , n). AsSismultiplicativelyclosed, X S. AsIJ A,aibixi IJ(fori = 1, 2, . . . , n). Hencen

i=1aisibiti=

ni=1aibixiX S1(IJ). (1.7)3Aringislocalifithaspreciselyonemaximalideal.Lemma1.5. Let pbeaprimeideal inacommutativeunital ringA, andlet S=A p. ThenSis amultiplicativelyclosedsubsetofA,and(thelocalizationofAatp)Ap:= S1Aisalocal integral domainwhosemaximal ideal ispAp.Proof. (Sismultiplicativelyclosed:) 1/ ptherefore1 S. Leta, b S=A p. Weneedtoshowthatab S. Proof bycontradiction: assumethatab / S. Thenab p, soa porb p, pisaprimeideal.Thusa/ Sorb/ S,contradiction. Henceab S,soSismultiplicativelyclosed.(ApisalocalringwhosemaximalidealispAp:) WeneedtoshowthatpAp:=_ps: p p, s/ p_is the only maximal ideal in Ap. As pA and Sis multiplicatively closed, pApAp. As11/ pAp, pAp ,= Ap.It now suces to prove that pApcontains all non-units in Ap. Ifus Ap pAp, then u A p, sousis a unit(withinversesu)inAp. HencepApcontainsallnon-unitsinAp,andthereforecontainsanyproperidealinAp.(Apisanintegral domain:) Apisasubringof theeldof fractionsof A, and11 Ap, thereforeApisanintegraldomain.1.2 TensorproductsSomeoftheproofsareomitted;Atiyah-Macdonald[1]isagoodreference.Let A be a commutative unital ring,and let Mand Nbe A-modules. The tensorproduct of Mand Nis anA-module Twith an A-bilinear mapping g: MN Tsuch that: if Pis an A-module and f: MN PisanA-bilinearmapping,thenthereexistsauniqueA-linearmappingft: T Psuchthatf= ft g.Remark(i) ThetensorproductM N:= M A N:= Texists,andisuniqueuptoisomorphism.(ii) If xiiIgenerateMand yjjJgenerateN,then xiyjiI,jJgenerateM N. Inparticular,thetensorproductoftwonitelygeneratedmodulesisnitelygenerated. Moreover, ifMandNareA-moduleswithrespectivebases aiand bj,then aibjisabasisforM N.(iii) Tensorproductscommutewithdirectsumsviathecanonicalisomorphism(iMi) A (jNj) i,j(MiA Nj)(

imi) (

jnj)

i,j(minj).4Lemma1.6. LetAbeacommutativeunital ring,andletPbeanA-module. LetMt M Mtt 0beanexactsequenceofA-modules. ThenMtA P M A P MttA P 0isanexactsequenceofA-modules.Lemma1.7. LetAbeacommutativeunital ring,andletM, N, KbeA-modulessuchthatM = N. ThenM A K = N A K.Proof. Letf: M Nandg: N Mbeinversemodulehomomorphisms. Thenid f andid gareinversemodulehomomorphismsbetweenM A KandN A K.Corollary1.8. LetAbeacommutativeunital ring. LetMbeanA-module,andleta A. ThenMaM=Aa A M. (1.8)Proof. Taketheexactsequence0 a A Aa 0,andtensorwithMoverA,togivetheexactsequencea A M AA M Aa A M 0. (1.9)Bytherstisomorphismtheorem,Aa A M =AA Ma A M.ItremainstoshowthatAA Ma A M=MaM[a m] [am] (1.10)isanisomorphism.(Welldened:) Assumethatai A, xi Mand[

aixi] = [

biyi]. Then1

(aixibiyi) =

aixi

biyi a A M,so

(aixibiyi) aM,so[aixi] = [biyi].(Injective:) Ifai A, xi Mand[

aixi] = [0],then

aixi aM,so

aixi= 1

aixi a M.(Surjective:) Letm M. Then[1 m] [m].Hence,MaM=AA Ma A M=Aa A M.51.3 ChineseremaindertheoremLet A be a commutative unital ring. Ideals I, J A are relativelyprimeif I +J= A. If I A, then elementsp, q AarecongruentmoduloI(denotedp q modI)ifp q I.Theorem1.9(Chineseremaindertheorem). Let Abeacommutativeunital ring. Let n Z>0andleta1, . . . , an Aberelativelyprimeinpairs.1Ifx1, . . . , xn A,thenthecongruencesx ximodai, i = 1, . . . , n (1.11)haveasimultaneoussolutionx A. Moreover,ifxisonesolution,theny Aisasolutionifandonlyifx y ni=1ai=n

i=1ai.Moreover,A

ni=1 ai=n

i=1Aai. (1.12)Proof. (Existence:)First consider the situation with two ideals. Let I, Jbe relatively prime ideals in A, andletx, y A. Weneedtondanelementk Asuchthatk x modIandk y modJ. AsI + J= A,thereexista Iandb Jsuchthata +b = 1. Ifk := ay +bxthenk = ay +bx bx (1 a)x x modIk = ay +bx ay (1 b)y y modJ.Nowconsiderexistenceinthegeneralcase. Letn Z>2andleta1, . . . , an Aberelativelyprimeinpairs.Letx1, . . . , xn A. Weneedtoshowthatthereexistsk Asuchthatk ximodaifori=1, 2, . . . , n.Fori = 1, 2, . . . , n,wewillndyisuchthatyi 1 modai,yi 0modaj, j ,= i.Thenk =

ni=1xiyiwillsatisfythenecessarycondition.Bysymmetry, itsucestondy Asuchthaty 1 moda1andy 0 modajforj=2, 3, . . . , n. Fori = 2, 3, . . . , n,thereexistai a1andbi aisuchthatai +bi= 1. Then

ni=2(ai +bi) = 1andn

i=2(ai +bi) n

i=2bimoda1=n

i=2(ai +bi) a1 +n

i=2ai, (1.13)soa1 +

ni=2 ai= A.Weshowedexistencefortworelativelyprimeideals,thereforethereexistsy Asuchthaty 1 moda1andy 0 mod

ni=2 ai. Forj =2, 3, . . . , n, y ni=2 ai aj, i.e. y 0 modaj.Hencethereexistsk Asuchthatk ximodaifori = 1, 2, . . . , n.1Thatis,ifi = jthenai + aj= A.6(Uniquenessmodulo ni=1ai:) Letn Z>0andletx, y Asatisfyx y modaifori=1, 2, . . . , n. Thenx y aifori = 1, 2, . . . , n,thereforex y ni=1ai.Thus,thesolutionstothecongruences(1.11)arecongruentmodulo ni=1ai.Wealsowanttoshowthatni=1ai=n

i=1ai.(:)Letx

ni=1 ai. Thenx n

i=1ai aj, j= 1, 2, . . . , n x ni=1ai.Henceni=1ai n

i=1ai.(:)Proofbyinduction. Assume that I, J A and I +J= A. Then there exist a Iand b Jsuch that a+b = 1. If x I Jthenx = (a +b)x = ax +bx IJ(asax IJandbx IJ). HenceI J IJ. Now let n Z>2and let a1, . . . , anA be relatively prime in pairs, and assume that ni=2ai

ni=2 ai.Weshowedintheproofofexistencethata1 +

ni=2 ai= A,thereforeni=1ai= a1 (ni=2ai) a1n

i=2ai a1

n

i=2ai=n

i=1ai.Byinduction,weconcludethatifn Z>0anda1, . . . , an Aarerelativelyprimeinpairs,thenni=1ai n

i=1ai.Henceni=1 ai=n

i=1ai. (1.14)Finally,thefollowingisanisomorphism: :A

ni=1 ain

i=1Aai[a] ([a], . . . , [a]).(Welldened:) Ifx, y Aand[x] = [y] A

ni=1 aithenx y n

i=1ai= ni=1ai=([x], . . . , [x]) = ([y], . . . , [y]) n

i=1Aai.7Henceiswelldened.(Injective:) Assumethata Aand([a], . . . , [a]) = 0

ni=1Aai. Thena ni=1ai=n

i=1ai=[a] = 0 A

ni=1 ai.Henceisinjective.(Surjective:) Letx1, . . . , xn A. Then([x1], . . . , [xn])isanarbitraryelementof

ni=1Aai, andwewanttoshowthat[x1, . . . , xn] isintheimageof. Bytherstpartofthetheorem, thereexistsx Asuchthatx ximodaifori = 1, 2, . . . , n. Then[x] ([x1], . . . , [xn]) n

i=1Aai,soissurjective.Thus,isanisomorphism,soA

ni=1 ai=n

i=1Aai.1.4 FieldtheoryWebeginwithsomebasicresults. SeeMilne[7],oranyintroductorytextonabstractalgebra.E/Fisaeldextension(alsodenotedE F)ifEandFareeldsandE F. Thedegree(ordimension)oftheextension,[E: F],isthedegreeofthevectorspaceEoverF. TheextensionE/Fisniteif[E: F]isnite.Let E/Fbe a eld extension, and let E. is algebraicif there exists g F[X] 0 such that g() = 0.TheextensionE/Fisalgebraic(orEisalgebraicoverF)ifeveryelementofEisalgebraicoverF.LetE/Fbeaeldextension, andlet EbealgebraicoverF. Theminimal polynomial of overF,denotedirr(, F), istheuniquemonicirreduciblepolynomial f F[X] suchthatf()=0. Thesetofpolynomialsg F[X] suchthatg()=0formsanonzeroprincipal ideal inF[X], whichisgeneratedbyirr(, F).LetE/Fbeaeldextension, andlet E. Letf F[X] beamonic, irreduciblepolynomial. F[] isastemeldforfiff() = 0.Lemma1.10. LetE/Fbeaeldextension, andlet EbealgebraicoverF. Letf F[X] beamonicpolynomial suchthatf() = 0. Thenthefollowingareequivalent:(a) fisirreducible.(b) Ifg F[X]andg() = 0thenfdividesg.8(c) Ifg F[X]andg() = 0thendeg(f) deg(g).Moreover,F[X](f)F[][X] isanF-isomorphism.2Corollary1.11. Let F beaeld, andlet F[] andF[] bestemelds for F. Thefollowingis anF-isomorphismF[] F[] .Lemma1.12. AeldextensionE/FisniteifandonlyifEisalgebraicandnitelygenerated(asaeld)overF.Lemma1.13. LetFbeaeld,andletf F[X]beamonicirreduciblepolynomial ofdegreen. ThenF[X](f)isaeldextensionofFofdegreen. AbasisforF[X](f)overFis [1], [X], . . . , [Xn1]. Consequently(from1.10),astemeldF[]forFisofdegreenoverF,withbasis 1, , . . . , n1.Lemma1.14. Let Fbeaeld, andlet bealgebraicoverF, andlet f betheminimal polynomial of overF. LetbeaeldcontainingF. Thenthefollowingisabijection:F-homomorphisms : F() rootsoffin ().Proof. (Welldened:) Let : F() beanF-homomorphism. Thenf(()) = (f()) = (0) = 0, (1.15)so()isarootoff. Hencethefunctioniswelldened.(Injective:) LetandbeF-homomorphismsF() suchthat()=(). AsandagreeonFand,theyagreeonF(). Hence = ,sothefunctionisinjective.(Surjective:) Let bearootoff. TheF-homomorphism :F() ismappedto. Hencethefunctionissurjective.2Thatis,anisomorphismwhichxeselementsofF. FembedsnaturallyintoF[X](f)andF[]viac [c]andc c.9LetE/Fbeaeldextension,andletf F[X]. Esplitsf,orfsplitsinE[X],ifthereexistm Z0and1, . . . , m Esuchthatf(X) =m

i=1(X i),andEisasplittingeldforfif,further,E= F[1, . . . , m].LetFbeaeld. Apolynomialf F[X]isseparableoverFifnoneofitsirreduciblefactorshasamultipleroot (in a splitting eld). An algebraic extension E/Fis separableif the minimal polynomial of every elementofEisseparable(otherwiseitisinseparable). Fisperfect ifeveryniteextensionofFisseparable.Weshallalsousesomehigher-levelresultswithoutproof, astheywillmakelaterproofsmoreconcise. SeeLang[4]fordetails.LetE/Fbeanalgebraicextension, andlet:FLbeanembeddingofFintoanalgebraicallyclosedeldL. LetSbethesetofextensionsoftoanembeddingofEintoL. TheseparabledegreeofEoverFis[E: F]s= [S[.Then[E: F]s [E: F],withequalityifandonlyifE/Fisseparable. NowassumethatE/Fisnite. TheinseparabledegreeofEoverFis[E: F]i=[E: F][E: F]s.Aniteextension[E: F]isseparableifandonlyif[E: F]i= 1.Lemma1.15. LetE/kbeaniteextension,andletE F kbeatowerofelds. Then(a) [E: k]s= [E: F]s [F: k]s.(b) [E: k]i= [E: F]i [F: k]i.(c) E/kisseparableifandonlyifE/FandF/kareseparable.3AnalgebraicextensionE/Fisnormal if theminimal polynomial (overF)of everyelementof EsplitsinE[X].Lemma1.16.(a) LetE F kbeatowerofelds. IfE/kisanormal extension,thenE/Fisanormal extension.(b) LetEandFbeanormal extensionsofaeldk. ThenEFandE Farenormal extensionsofk.Theorem1.17(Primitive element theorem). LetEbeanite,separableextensionofaeldk. Thenthereexistsaprimitiveelement,i.e. anelement EsuchthatE= k().3Separableextensionsformadistinguishedclassof extensions, sinceE/kisseparableifandonlyifeachstepofthetowerisseparable.10ExampleQ[2,3] = Q[2 +3]:(:)2 +3 Q[2,3] Q[2 +3] Q[2,3].(:)3 2 =12 +3 Q[2 +3]2 =(2 +3) (3 2)2 Q[2 +3]3 =(2 +3) + (3 2)2 Q[2 +3]HenceQ[2,3] Q[2 +3].Weconcludethat Q[2,3] = Q[2 +3]. Itisnotobvioushowtondaprimitiveelementingeneral.Lemma1.18. AnalgebraicextensionE/F is normal andseparableif andonlyif, foreach E, theminimal polynomial ofhas[F[ : F]]distinctrootsinE.Theorem1.19. Let E/Fbeaeldextension, andlet Aut(E/F)bethegroupof F-automorphismsof E.Thefollowingareequivalent:(a) Eisthesplittingeldofaseparablepolynomial f F[X].(b) F= EGforsomenitegroupGofautomorphismsofE.(c) Eisnormal andseparable,andofnitedegree,overF.(d) F= EAut(E/F).Ifthesestatementshold,theextensionE/FisGalois,4andwedenetheGaloisgroupGal(E/F) := Aut(E/F).IfE/FisaGaloisextension,then[Gal(E/F)[ = [E: F],sotheGaloisgroupisnite.LetL/Kbeanalgebraicextension. Let L, andletf:=irr(, K). Theconjugates of inLaretherootsoffinL.Lemma1.20. LetL/Kbeaniteseparableextension. ThenthereexistsaeldE LsuchthatE/KisGalois.4Someotherdenitionsdier.11Proof. Bytheprimitiveelementtheorem(1.17),thereexists LsuchthatL = K(). Letf K[X]betheminimal polynomial of , andlet=1, . . . , mbetheconjugatesof . ThenE=L[1, . . . , m] isthesplittingeldoff K[X], andfisseparable(beingtheminimal polynomial of L, whereL/Kisseparable),soE/KisGalois.Let L/Kbeaniteseparableextension. TheGalois closure of L/Kistheminimal (byinclusion) eldE LsuchthatE/KisGalois.1.5 NoetherianringsandmodulesSeeAtiyah-Macdonald[1]fortheproofs.AcommutativeunitalringAisNoetherianifitsatisesthefollowingequivalentconditions: Everynon-emptysetofidealsinAhasamaximalelement,i.e. if

isanon-emptysetofidealsinA,thenthereexistsI

suchthatifJ

andJ IthenJ= I. If (Ik)isasequenceof idealsinAsuchthatI1 I2 . . ., thenthereexistsN Z>0suchthatifk NthenIk= IN.5 EveryidealinAisnitelygenerated.AmoduleMisNoetherianifitsatisesthefollowingequivalentconditions: Everynon-emptyset of submodules of Mhas amaximal element, i.e. if

is anon-emptyset ofsubmodulesofM,thenthereexistsI

suchthatifJ

andJ IthenJ= I. If(Mk)isasequenceofsubmodulesofMsuchthatM1 M2 . . .,thenthereexistsN Z>0suchthatifk NthenMk= MN.6 EverysubmoduleofMisnitelygenerated.Lemma1.21. Let AbeaNoetherianring, andlet MbeanitelygeneratedA-module. ThenMis aNoetherianA-module.2 AlgebraicnumbertheoryMost of theideas inthis chapter canbeattributedtoDedekind, Kummer andNoether. TheapproachpresentedfollowsMilne[8]fairlyclosely.5Inwords,everyascendingchainofidealsisstationary.6Inwords,everyascendingchainofsubmodulesofMisstationary.122.1 RingsofintegersAlgebraicintegersareageneralizationofintegers. Specically,thealgebraicintegersinthenumbereld Qare the elements of Z, but larger number elds contain more algebraic integers. We will see that the algebraicintegersinanumbereldformaringwithsomeniceproperties.LetAbeanintegraldomain,andletL Abeaeld. Anelement Lisintegral overAifitistherootof a monic polynomial with coecients in A. We want to prove that the elements of L that are integral overAformaring. Theproofwewillgive,whichisduetoDedekind,reliesonthefollowinglemma.Lemma2.1. LetAbeanintegraldomain,andletL Abeaeld. Anelement LisintegraloverAifandonlyifthereexistsanonzeronitelygeneratedA-submoduleMofLsuchthatM M.Proof. (:) Let LbeintegraloverA. Thenthereexistn Z>0anda1, . . . , an Asuchthatn+a1n1+. . . +an1 +an= 0.Let Mbe the A-module generated by 1, , . . . , n1. Then Mis a nitely generated A-submodule of L suchthatM M.(:) AssumethatthereexistsanonzeronitelygeneratedA-submoduleMofLsuchthatM M. Letn Z>0andlet MbetheA-modulegeneratedby(nonzero) v1, . . . , vn. For i =1, 2, . . . , n, thereexistai1, . . . , ain Asuchthatvi=

nj=1aijvj. ThusCv= 0, (2.1)whereC=______ a11a12. . . a1na21 a22. . . a2n............an1. . . an,n1 ann______, v=______v1v2...vn______.In order to show that is integral over A,it suces to prove that det(C) = 0. As Cv= 0,multiplying bothsidesbytheadjugatematrixofCyields0 = adj(C)Cv= det(C)Inv= det(C)v.Asv1 ,= 0,weconcludethatdet(C) = 0. HenceisintegraloverA.Theorem2.2. Let Abeanintegral domain, andlet L Abeaeld. Thentheelementsof Lthat areintegral overAformaring.Proof. Let , L be integral over A. Then 1 and are integral over A, so it remains to show that +andareintegraloverA.By 2.1, there exist nonzero nitely generated A-submodules Mand Nof B such that M Mand N N.ConsiderMN:=_n

i=1aibi: n Z0, a1, . . . , an M, b1, . . . , bn N_.13 MN ,= 0,sinceM ,= 0, N ,= 0. (MNisanA-submoduleofL:) Let

ni=1aibi,

mj=1cidi MNanda A. Thenn

i=1aibi +m

j=1cidi MN, an

i=1aibi=n

i=1(aai)bi MN. (MNis anitelygeneratedA-module:) Let m, n Z>0. Let Mbe the A-module generatedbyu1, . . . , um,andletNbetheA-modulegeneratedbyv1, . . . , vn. ThenMNisgeneratedby_uivj: i 1, 2, . . . , m, j 1, 2, . . . , n_. ((+)MN MNand ()MN MN:)Let n Z>0and let a1, . . . , an M, b1, . . . , bn N. Then

ni=1aibiisanarbitraryelementofMN.( +)n

i=1aibi=n

i=1_(ai)bi +ai(bi)_, n

i=1aibi=n

i=1(ai)(bi).AsM MandN N,wehaveai Mandbi Nfori = 1, 2, . . . , n. Thus( +)n

i=1aibi MN, n

i=1aibi MN.Hence, by2.1, + andareintegral overA. Thus, thesetof elementsof Lthatareintegral overAformaring.Let A be an integral domain, and let L A be a eld. The integral closure of A in L is the set Bof elementsofLthatareintegraloverA. Bisaring,by2.2. Bisanintegraldomain,since1 BandBisasubringofaeld(namelyL). AringC Aisintegral overAifeachelementofCisintegraloverA. LetK Qbeaeld. Anelement Kisanalgebraicintegerifisintegralover Z.An algebraicnumbereld(or numbereld) is a nite (and hence algebraic) extension of Q. If L is a numbereld,theringofintegersinL,denoted OL,istheintegralclosureof ZinL.Anyeldofcharacteristic0isperfect. Inparticular, Qisperfect,soanynumbereldisanite,separableextensionof Q. By1.17,thisimpliesthatanynumbereldLhasaprimitiveelementover Q,i.e. L = Q()forsome L.Proposition2.3. Let Abeanintegral domain, andlet Kbetheeldof fractionsof A. Let LbeaeldcontainingK,andlet LbealgebraicoverK. Thenthereexistsd Asuchthatdisintegral overA.Proof. Letm Z>0andletk1, . . . , km Kbesuchthatm+k1m1+. . . +km= 0. (2.2)Letd Abetheproductofthedenominatorsofk1, . . . , km. Then(d)m+k1d(d)m1+. . . +kmdm= 0, k1d, . . . , kmdm A. (2.3)HencedisintegraloverA.14Anintegraldomainisintegrallyclosedifitisitsownintegralclosureinitseldoffractions.Proposition2.4. LetAbeauniquefactorizationdomain. ThenAisintegrallyclosed.Proof. SinceanyelementinAisintegraloverA,itremainstoshowthatifisintheeldoffractionsofAandisintegral overAthen A. Leta, b A, b ,=0. Wecanwriteabinsimplestterms bywritingaandbasproductsofprimes,andthencancelling. ThuswemayassumethatnoprimeelementofAdividesbothaandb. SupposethatabisintegraloverA. Toshow: bisaunit. Proofbycontradiction: assumethatbisnotaunit. Thenthereexistsaprime Asuchthat [ bbut ,[ a. Let_ab_n+an1_ab_n1+. . . +a0= 0, (2.4)wheren Z0anda0, . . . , an1 A. Thenan+an1an1b +. . . +a0bn= 0. (2.5)Now[an,so[a,sinceisprime. Thiscontradictstheassumptionthataandbarerelativelyprime,sobisaunitandab A. HenceAisintegrallyclosed.Consequently,theringofintegersin Qis Z.Proposition2.5. LetAbeanintegrallyclosedintegral domainwitheldoffractionsK,andletL/Kbeaniteextension. Thenanelement Lisintegral overAifanonlyifitsminimal polynomial overKhascoecientsinA.Proof. Let L.(:)Assumethattheminimalpolynomialf K[X]ofoverKhascoecientsinA. Asfismonic,isintegraloverA.(:)Conversely, assumethatisintegraloverA. Letf K[X]betheminimalpolynomialofoverK.We need to show that the coecients of flie in A. As the coecients of flie in K, and A is integrally closed,itsucestoprovethatthecoecientsoffareintegraloverA. BytheVietaformulae,thecoecientsoffaresumsandproductsoftherootsoff. AstheintegralclosureofAinLisaring,itsucestoprovethattherootsoffareintegraloverA. Let Lbeanyrootoff.Letn Z>0anda1, . . . , an Asatisfyg() = n+a1n1+. . . +an1 +an= 0, (2.6)whereg A[X]isdenedbyg(X) = Xn+a1Xn1+. . . +an1X +an. Themap: K[] K[] (2.7)isaK-isomorphismby1.11. Nowg() = g(()) = (g()) = (0) = 0, (2.8)soisintegraloverA. ThusallrootsoffareintegraloverA,andweconcludefromthatf A[X].15ExampleThis example demonstrates that rings of integers are not necessarily unique factorization domains.Weshallseethattheringofintegersof Q[5]is Z[5]. Consider2 3 = (1 +5)(1 5).Wewillshowthattheelements2, 3, 1 +5and1 5areirreduciblein Z[5]. Considerthemap7Nm : Z[5] Z0a +b5 a2+ 5b2.Weprovesomepropertiesofthismap.(ThemapNmismultiplicative:) Leta, b, c, d Z. ThenNm((a +b5)(c +d5)) = Nm((ac 5bd) + (ad +bc)5) = (ac 5bd)2+ 5(ad +bc)2= a2c2+ 5a2d2+ 5b2c2+ 25b2d2= (a2+ 5b2)(c2+ 5d2) = Nm(a +b5)Nm(c +d5).Hence,themapNm : Z[5] Z0ismultiplicative.Let Z[5]. Thefollowingareequivalent:(a) Nm() = 1.(b) = 1.(c) isaunitin Z[5].Proof. Let = a +b5, a, b Z. ((a) =(b):)Nm() = (a +b5)(a b5) = .Thus,ifNm() = 1,then = 1. ((b) =(c):) Assumethat = 1. Thenisaunitin Z[5]withinverse Z[5]. ((c) =(a):) AssumethatisaunitinZ[5]withinverse Z[5]. ThenNm()Nm() = Nm() = Nm(1) = 1,thereforeNm() = 1. AsNm() 0,wemusthaveNm() = 1.If 2=xywithx, y Z[5], thenNm(x)Nm(y) =Nm(2) =4. Without loss of generalityNm(x) Nm(y). As 2 / Image(Nm), it follows that Nm(x) =1, so x is a unit. Hence 2 is irreducible inZ[5]. Similarly, 3 / Image(Nm), so3is irreducible inZ[5]. Similarly, as 2 / Image(Nm) andNm(1 +5) = Nm(1 5) = 6,itfollowsthat1 +5and1 5areirreduciblein Z[5]. Thus2 3 = (1 +5)(1 5)7Thisisthenormmapthatwewillsoonencounter,whichistheproductofthe(notnecessarilydistinct)Galoisconjugates.16constitutestwodistinctfactorizationsof6 Z[5]intoirreduciblefactors.8Though rings of integers are not necessarily unique factorization domains, they do have many nice properties.Wenowaimtoestablishthattheringof integerinanumbereldisintegrallyclosed, whichessentiallyinvolvesprovingthatintegralclosuresareintegrallyclosed.Lemma2.6. LetA, B, CberingssuchthatA B C. IfBisnitelygeneratedasanA-moduleandCisnitelygeneratedasaB-module,thenCisnitelygeneratedasanA-module.Proof. Assume that Bis nitely generated as an A-module,and that Cis nitely generated as a B-module.LetBbetheA-modulegeneratedby 1, . . . , m, andletCbetheB-modulegeneratedby 1, . . . , n,wherem, n Z>0.LetDbetheA-modulegeneratedby_ij: i 1, . . . , m, j 1, . . . , n_. InordertoshowthatCisnitely-generatedasanA-module,itsucestoprovethatC= D. (C D :)AsCisaB-module,Ccontainsijfori 1, 2, . . . , mandj 1, 2, . . . , n. AsCisanA-module(beingaB-module),C D. (C D:)Let C. Thenthereexistb1, . . . , bn Bsuchthat=

nj=1bjj. Forj=1, 2, . . . , n,thereexista1j, . . . , amj Asuchthatbj=

mi=1aiji. Now=

i,jaij(ij), aij A,soC D.ThusC= D,soCisnitelygeneratedasanA-module.Lemma2.7. Let Abeanintegral domain, andlet B Abearingwhichisintegral overAandnitelygeneratedasanA-algebra. ThenBisnitelygeneratedasanA-module.Proof. Proofbyinductiononr,forringsoftheformA[1, . . . , r]. LetC=A[], andassumethatCisintegral overA. Thenthereexistn Z>0anda1, . . . , an Asuchthatn+a1n1+. . . +an1 +an= 0.Letx C= A[]. ThenthereexistN Z>0andc0, c1, . . . , cN Asuchthatx = c0 +c1 +. . . +cNN. (2.9)Substitutingn= (a1n1+. . . +an1 +an)intoequation(2.9)yieldsanexpressionoftheformx = d0 +d1 +. . . +dn1n1, d0, d1, . . . , dn1 A.HenceCistheA-modulegeneratedby 1, , . . . , n1.8Thesefactorsdonotdierbymultiplicationbyaunit(considertheirnorms).17 AssumethatD AisaringthatisnitelygeneratedasanA-module,andletE= D[]beintegraloverA. Toshow: EisnitelygeneratedasanA-module.EisintegraloverA,andD A,thereforeEisintegraloverD. Fromthebasestepofourinduction,EisthereforenitelygeneratedasaD-module. Additionally,DisnitelygeneratedasanA-module,so,by2.6,EisnitelygeneratedasanA-module.Hence,byinduction,BisnitelygeneratedasanA-module.Lemma2.8. LetA, BandCbeintegral domainssuchthatBisintegral overAandCisintegral overB.ThenCisintegral overA.Proof. Let C. Toshow: isintegral overA. By2.1, itsucestondanonzero, nitelygeneratedA-submoduleEof CsuchthatE E. Asisintegral overB, thereexistn Z>0andb1, . . . , bn Bsuchthatn+b1n1+. . . +bn1 +bn= 0.LetD = A[b1, . . . , bn],andletE= D[]. ThenEisanonzeroA-submoduleofC(as, b1, . . . , bn C),andE E(as E). ItremainstoshowthatEisnitelygeneratedasanA-module.Disintegral overAandnitelygeneratedasanA-algebra. Thus, by2.7, DisnitelygeneratedasanA-module. By2.6,itthereforesucestoprovethatE= D[]isnitelygeneratedasaD-module. Asn+b1n1+. . . +bn1 +bn= 0, b1, . . . , bn D,Eis the D-module generated by 1, , . . . , n1. Hence Eis nitely generated as a D-module. We concludethatisintegraloverA. Thus,CisintegraloverA.Theorem2.9. LetAbeanintegral domainwitheldof fractionsK. LetL/Kbeanalgebraicextension.Theintegral closureofAinLisintegrallyclosed.Proof. LetBbetheintegral closureof AinL, andletCbetheintegral closureof BinL. Toshow: Bisintegrallyclosed. AsLisaeldcontainingB, itsucestoprovethatC=B. WeknowthatC B,therefore it suces to prove that C B. Thus, it suces to prove that Cis integral over A. As Bis integraloverAandCisintegraloverB,CisintegraloverAby2.8. WeconcludethatBisintegrallyclosed.By2.9(withA = ZandLasanynumbereld),theringofintegersinanynumbereldisintegrallyclosed.2.2 Norms,tracesanddiscriminantsInvariants are at the heart of algebraic number theory. We will encounter many powerful theorems involvingalgebraicinvariants,aswellasafewsurprisingrelationshipsbetweenthem. Themaingoalsofthissectionaretoprovidedenitionsandtointroduceformulaeforlateruse. TheseformulaehavebeentakenfromMilne[8], whichiswhereanyof theomittedproofsinthissectioncanbefound. Wewill alsoshowthatringsofintegersarenitelygeneratedas Z-modules, allowingustodeneanintegral basis. Finally, using18the discriminant, we will completely describe the rings of integers for number elds of the form Q[d], wheredisasquare-freeinteger.LetA BberingssuchthatBisafreeA-moduleofrankn. Thenany BdenesanA-linearmapB Bx x,which,ifwechooseabasisforBasanA-module,canberepresentedbyamatrix. Thetrace(TrB/A)andnorm(NmB/A)ofintheextensionB/Aarerespectivelythetraceanddeterminantofthismatrix. Thetraceandnormarewelldened,aschoosingadierentbasiswouldgiverisetoasimilarmatrix.Ifa Aand, t Bthen Tr( +t) = Tr() + Tr(t) Tr(a) = aTr() Tr(a) = na Nm(t) = Nm()Nm(t) Nm(a) = an.Lemma2.10. LetE/MandM/Fbeniteeldextensions. Then(a) TrE/F= TrM/F TrE/M(b) NmE/F= NmM/F NmE/M.Proof. SeeLang[4],Theorem5.1. ThisisalsodiscussedlessformallyinMilne[7],Proposition5.44.Proposition2.11. LetL/Kbeaeldextensionofdegreen,andlet L. Letf K[X] betheminimalpolynomial ofoverKandlet1= , 2, . . . , mbetherootsoffinthealgebraicclosureofK. ThenTrL/K() = r(1 +. . . +m), NmL/K() = (1. . . m)r,wherer := [L : K[]] =nm.Proof. Thematrixofx x : K[] K[]withrespecttothebasis 1, , . . . , m1forK[]overKis____________0 0 0 . . . 0 (1)m

mi=1i1 0 0 . . . 0 ?0 1 0 . . . 0......... ?0 0 . . . 0 1

mi=1i____________.19Thetraceofthismatrixis

mi=1iandthedeterminantis

mi=1i. HenceTrK[]/K() =m

i=1i, NmK[]/K() =m

i=1i.By2.10,TrL/K() = TrK[]/K(TrL/K[]()) = TrK[]/K(r), as K[]= rTrK[]/K() = r(1 +. . . +m)andNmL/K() = NmK[]/K(NmL/K[]()) = NmK[]/K(r), as K[]= (NmK[]/K())r= (1. . . m)rCorollary2.12. LetL/Kbeaniteeldextension,andlet L. ThenTrL/K(), NmL/K() K.Proof. Let fK[X] betheminimal polynomial of overK. Uptosign, thesumandproduct of theconjugatesofarecoecientsoff,andthereforelieinK. By2.11,thetraceandnormlieinK.Corollary2.13. LetAbeanintegrallyclosedintegral domainwitheldoffractionsK, andletL/Kbeaniteextension. If Lisintegral overA,thenTrL/K(), NmL/K() A.Proof. Let L be integral over A. By 2.12, TrL/K(), NmL/K() K. As A is integrally closed, it sucestoprovethatTrL/K()andNmL/K()areintegraloverA.Letf K[X]betheminimalpolynomialofoverK. By2.5, f A[X]. As, inaddition, fismonic, theroots of fare integraloverA. By2.11,TrL/K()and NmL/K()are sums andproducts of these roots,andarethereforeintegraloverA(sincetheintegralclosureofAinLisaring).ThusTrL/K(), NmL/K() A.Corollary2.14. LetL/Kbeaseparableextensionof degreen Z>0, andlet L. Let 1, . . . , nbethesetofdistinctK-homomorphismsL ,where/KisaGaloisextensionand L. ThenTrL/K() = 1 +. . . +n, NmL/K() = 1n. (2.10)Proof. Letf K[X]betheminimalpolynomialofoverK, andlet1=, 2, . . . , mbetherootsoff(distinctbecauseL/Kisaseparableextension). By2.11, weneedtoshowthateachioccursrtimesinthemultiset 1(), . . . , n(),wherer = [L : K[]] =nm.By 1.14,each ioccurs once in the set 1, . . . , m,where 1, . . . , mare the distinct K-homomorphismsK[] . Thesearedenedby9j: K[] j.9See1.14.20Beinginjective,eachjisanembeddingofK[]into. TheextensionL/K[]isseparable,by1.15,so[L : K[]]s= [L : K[]] = r.Hence each jhas rdistinct extensions to a K-homomorphism L . As rm = n, every K-homomorphismL canbeobtainedinthisway(astheremustbe[L:K]s=[L:K]=noftheseK-homomorphismsL ). We therefore conclude that each ioccurs rtimes in the multiset i, where r = [L : K[]] =nm,andtherefore2.11yieldsTrL/K() = 1 +. . . +n, NmL/K() = 1n.Lemma2.15. LetKbeanumbereld. Anelement OKisaunitifandonlyifNmK/Q() = 1.Proof. (:) Assume that OKis a unit. Thenthere exists OKsuchthat =1. ThusNm()Nm() =1. Theminimal polynomials of andoverQhavecoecients inZ, by2.5. Thesehave constant coecients equal to Nm() and Nm() respectively, so Nm(), Nm() Z. SinceNm()Nm() = 1,itnowfollowsthatNm() = 1.(:)Assumethat OKandNm() = 1. Denei : KC, x x.Now1 = Nm() =

:KC =

,=i. (2.11)Thenhasinverseequalto

,=i OK,so OKisaunit.LetL/Kbeaniteeldextension,andlet e1, . . . , embeabasisforLoverK. ThediscriminantofL/KisthediscriminantofthesymmetricbilinearformL L K(, ) TrL/K(),i.e. thedeterminantofthematrix(TrL/K(eiej)).Let A and B be rings such that B A and B is a free A-module of rank m. The discriminantof 1, . . . , m BisD(1, . . . , m) = det(TrB/A(ij)). (2.12)Lemma2.16. LetA, BberingssuchthatB AandBisafreeA-moduleofrankm. Let1, . . . , m B.Forj= 1, 2, . . . , m,letj=

mi=1ajiiwhereaij Afori, j 1, 2, . . . , m. ThenD(1, . . . , m) = det(aij)2 D(1, . . . , m). (2.13)Proof. Fork, l 1, 2, . . . , m,TrB/A(kl) =

i,jTrB/A(akiialjj) =

i,jakiTrB/A(ij)alj.21Hence(TrB/A(kl)) = MTrB/A(ij)MT, M= (aij).Thusdet(TrB/A(kl)) = det(M)2 det(TrB/A(ij))D(1, . . . , m) = det(aij)2 D(1, . . . , m).If 1, . . . , mand 1, . . . , marebasesforBoverA,thendet(aij)isaunit. ThusthediscriminantofabasisiswelldeneduptoamultiplicationbythesquareofaunitinA. Inparticular,theidealinAthatitgeneratesisindependentofthechoiceofbasis. Thediscriminant ofBoverAisdisc(B/A) = [D(1, . . . , m)] A(A)2,where 1, . . . , m is a basis for B over A. We often regard disc(B/A) as a representative D(1, . . . , m) Aofdisc(B/A)inA(A)2.Corollary2.17. LetAandBbeintegral domainssuchthatB A. AssumethatBisafreeA-moduleofrankm,andthatdisc(B/A) ,= 0. Then1, . . . , n BformabasisforBasanA-moduleifandonlyif(D(1, . . . , m)) = (disc(B/A)) A.Proposition2.18. LetBbeanintegraldomainsuchthatB Z,andassumethatBisafree Z-moduleofrankm. Then(a) Elements1, . . . , m Bgeneratea Z-submodule10NofniteindexinBifandonlyifD(1, . . . , m) ,= 0.(b) If1, . . . , m BandD(1, . . . , m) ,= 0then11D(1, . . . , m) = (B: N)2 disc(B/Z). (2.14)Proposition2.19. LetL/Kbeaniteseparableextensionofdegreem,andlet1, . . . , mbethedistinctK-homomorphismsL, whereisaGaloisextensionof KcontainingL.12If 1, . . . , misabasisforLoverK,thenD(1, . . . , m) = det(ij)2,= 0. (2.15)Proposition2.20. Let Kbeaeldof characteristic0, let bealgebraicoverK, andlet L=K[]. Letf(X)betheminimal polynomial ofoverK,andletf(X) =

mi=1(X i)intheGaloisclosureofL/K.ThenD(1, , . . . , m1) =

1i0,andletV beanm-dimensional vectorspaceoveraeldK. Let e1, . . . , embeabasisforV ,andlet: V V Kbeanondegenerate13bilinearform. Thenthereexistsabasis et1, . . . , etmforV ,suchthat(ei, etj) = ij, i, j 1, 2, . . . , m,whereistheKroneckerdelta.Proposition2.22. Let Abeanintegrallyclosedintegral domainwitheldof fractionsK. Let L/Kbeaseparableextensionofdegreem Z>0,andletBbetheintegral closureofAinL. Then(a) ThereexistfreeA-submodulesMandMtofL,ofrankm,suchthatM B Mt.(b) IfAisaNoetherianring,thenBisanitelygeneratedA-module.(c) IfAisaprincipal ideal domain,thenBisafreeA-moduleofrankm.Proof.(a) Let 1, . . . , mbeabasisforLoverK. By2.3, thereexistd1, . . . , dm Asuchthatdii Bfori=1, 2, . . . , m. Letd=

mi=1di. Thend A, i:=di Bfori=1, 2, . . . , m, and 1, . . . , misabasisforLoverK. Thebilinearform: L L K(x, y) TrL/K(xy)isnondegenerate,sinceifx Lthen(x,1x) = 1 ,= 0. Thus,by2.21,thereexistsabasis t1, . . . , tmforLoverKsuchthatTrL/K(itj) = ij, i, j 1, 2, . . . , m,whereistheKroneckerdelta.It remains to show that the free A-submodules M:= A1 +. . . +Amand Mt:= At1 +. . . +Atmof LsatisfyM B Mt. As1, . . . , m B,itfollowsimmediatelythatM B. NowitremainstoshowthatB Mt. Let B,andwrite=m

j=1bjtj, b1, . . . , bj K.13Thatis,theredoesnotexistv V \ {0}suchthatifx V then(v, x) = 0.23Itremainstoprovethatbi Afori=1, 2, . . . , m. Leti 1, 2, . . . , m. By2.13, TrL/K(i) A.Moreover,TrL/K(i) = TrL/K(m

j=1bjtji) =m

j=1bjTrL/K(tji) =m

j=1bjij= bi,sobi A. Hence=

mj=1bjtj Mt,soweconcludethatB Mt.(b) AssumethatAisaNoetherianring. From(a), thereexistsafreeA-moduleMtof rankmsuchthatB Mt. AsMtisnitelygeneratedA-module,1.21impliesthatMtisaNoetherianA-module. HenceBisnitelygeneratedasanA-module,beingasubmoduleoftheNoetherianA-moduleMt.(c) Assume that A is a principal ideal domain. Then A is a Noetherian ring, since every ideal in A is nitelygenerated. Hence, thereexistfreeA-submodulesMandMtofL, ofrankm, suchthatM B Mt.Then B is a free A-module, being a nitely generated submodule of the free module Mt over the principalidealdomainA. Moreover,m = rank(M) rank(B) rank(Mt) = m,sorank(B) = m.In particular, applying 2.22(c) with A = Z reveals that rings of integers are nitely generated free Z-modules.LetKbeanumbereld. Anintegral basisforKisabasisfor OKasa Z-module.LetKbeanumbereld,[K: Q] = m. Then OKisfreeofrankmover Z,sodisc(OK/Z)isawelldenedinteger(choosingtwointegralbasesforK,theirdiscriminantdiersbymultiplicationbyasquareofaunitin Z,i.e. multiplicationby1). Thisintegeriscalledthediscriminant ofK,andissometimesdenotedDKorK.Lemma2.23. Any Z-independentsetofsizem:=[K: Q] (suchasanintegral basisforK)isabasisforK/Q.Proof. Let1, . . . , mbe Z-independentelementsinK. Itsucestoprovethattheyare Q-independent(sincetherearem = [K: Q]ofthem).Letp1q1, . . . ,pmqm Q(sop1, . . . , pm Zandq1, . . . , qm Z 0)satisfy

piqii= 0.Multiplyingbothsides by

qiyields

(piSi)i=0, where Si:=

j,=iqj. EachpiSiZ, thereforepiSi=0fori=1, . . . , m(since1, . . . , mare Z-independent). Fori=1, . . . , m, Si ,=0 pi=0. Hence1, . . . , misanindependentsetofvectorsinK(asavectorspaceover Q). As [1, . . . , m[ = [K: Q],1, . . . , misabasisforKasavectorspaceover Q.Consequently,TrcK/Z= TrK/QcK, NmcK/Z= NmK/QcK. (2.17)Lemma2.24. Let Kbeanumbereld. Thenthesignof disc(K/Q)is(1)s, where2sisthenumberofhomomorphismsKCwithimagenotcontainedin R.24Theorem2.25(Stickelbergers theorem). LetKbeanalgebraicnumbereld. Thendisc(OK/Z)iscongru-entto0or1modulo4.Wecanusethediscriminanttodeterminesomeringsofintegers:Theorem2.26. Let mbeasquare-freeinteger. Thentheringof integers of thealgebraicnumber eldQ[m]is(a)Z[m]ifm 2, 3mod4,(b)Z[1+m2]ifm 1mod4.Proof. LetK= Q[m].(a) Assumethatmiscongruentto2or3modulo4. D(1,m) = disc(X2m) = 4m. As m OK,2.18yields4m = D(1,m) = disc(OK/Z)(OK: Z[m])2. (2.18) Case: (OK: Z[m)iseven. Thenm = disc(OK/Z) _(OK: Z[m])2_2. (2.19)As(cK:Z[m])2 Z, we know that_(cK:Z[m])2_2is congruent to 0 or 1 modulo 4. By Stickelbergerstheorem, 2.25, disc(OK/Z) is congruent to 0 or 1 modulo 4. Hence m is congruent to 0 or 1 modulo4,contradictingtheassumptionthatmiscongruentto2or3modulo4. Case: (OK: Z[m]) is odd. Then gcd(4, (OK: Z[m])2) = 1, so, by Euclids lemma, (OK: Z[m])2dividesm. Asmissquarefree,(OK: Z[m])2= 1,so OK= Z[m].Weconcludethat OK= Z[m].(b) Assume that m is congruent to 1 modulo 4. Note that1+m2 OK, as the minimal polynomial of1+m2isX2X +1m4 Z[X]. Now2.18yieldsm = disc_X2X +1 m4_ = D(1,m) = disc(OK/Z) _OK: Z_1 +m2__2. (2.20)Asmissquarefreeand_OK: Z_1+m2__2dividesm,_OK: Z_1 +m2__2= 1, (2.21)so OK= Z_1+m2_.252.3 DedekinddomainsInthissection,weprovethatringsofintegersinnumbereldsareDedekinddomains. WestudyDedekinddomainsmoregenerallyand, inparticular, provethatnonzeroproperidealsinDedekinddomainsfactoruniquelyintoproductsofnonzeroprimeideals.ExampleBy2.26,theringofintegersin Q[5]is Z[5]. In Z[5],(2, 1 +5)(2, 1 5) = (4, 2 25, 2 + 25, 6) = (2).Infact(2, 1 +5)and(2, 1 5)areprimeidealsin Z[5].Forinstance,toshowthat(2, 1 +5)isprime,weneedtoshowthatZ[5](2,1+5)isanintegraldomain. Infactitisaeld,sincethefollowingmapisaringisomorphism: :Z2Z Z[5](2, 1 +5)[x] [x]. (2.22) (Welldened:) Assumethatx, y Zand[x] = [y] Z2Z. Thenx y 2Z (2, 1 +5)Z[5],so[x] = [y] Z[5](2,1+5).Henceiswelldened. (Injective:) Assumethatx Zand[x] [0] Z[5](2,1+5). Thenx (2, 1 +5)Z[5]. Thusthereexistp, q, r, s Zsuchthatx = 2(p +q5) + (1 +5)(r +s5) = (2p +r 5s) + (2q +r +s)5. (2.23)x Z,therefore2q +r +s = 0. Thus,x = 2p +r 5s = 2p +r 5(2q r) = 2p + 10q + 6r 2Z. (2.24)Hence[x] = 0 Z2Z,andweconcludethatisinjective. (Surjective:) Leta, b Z[5]. Weneedtoshowthat[a +b5]isintheimageofthemap.[a b] [a b] = [a b +b(1 +5)] = [a +b5].Henceissurjective.ThusZ[5](2,1+5)is isomorphic toZ2Z, which is a eld. Hence (2, 1 +5) is a maximal (and therefore prime)idealin Z[5]. Similarly,(2, 1 5)isaprimeidealin Z[5].Thisexampleshowshowtofactorize(2) Z[5]astheproductofnonzeroprimeidealsin Z[5]. Thisfactorizationisunique,thoughwehavenotshownit.Proposition2.27. LetAbeaprincipal ideal domain. Thenthefollowingareequivalent.(a) Ahasexactlyonenonzeroprimeideal.26(b) Uptoassociates,Ahasexactlyoneprimeelement.(c) Aislocal andnotaeld.Proof. ((a) (b):) AssumethatAhasexactlyonenonzeroprimeideal, (y), wherey A. Theny Aisprime. Thus,weneedtoshowthatifx Aisprimethenx y. Letx AbeprimeinA. Then(x)isanonzeroprimeidealinA,so(x) = (y). Hencex y. ThusAhasexactlyoneprimeelement,uptoassociates. ((b) (c):) Assumethatuptoassociates,Ahasexactlyoneprimeelement. Fieldshavenononzeroproperideals, andthereforenononzeroprimeideals, andthereforenoprimeelements, soAisnotaeld. ItremainstoshowthatAislocal.Let(x)and(y)bemaximalidealsinA. Thenx, y Aareprime, sox y, i.e. (x)=(y). HenceAhasatmostonemaximalideal. ItremainstoshowthatAcontainsamaximalideal.AsAisnotaeld,thereexistsz Asuchthatz ,= 0andzisnotaunitinA. Now(z)isanonzeroproperidealinA. Hence(z)iscontainedwithina(nonzero)maximalidealinA.14 ((c) (a):) SinceAislocal,Ahaspreciselyonemaximalideal(m). m ,= 0,sinceAisnotaeld15.Hence(m)isanonzeroprimeidealinA. Itremainstoshowthat(m)isuniqueinthissense.Let(x)beaprimeideal inA. Then(x)iscontainedwithinamaximal ideal, and(m)istheonlymaximal ideal inA, so(x) (m). Thusm[x. Letx=am, wherea A. Sincexisprime, itisirreducible, soamustbeaunit(notethatmisnotaunit, since(m)isaproperideal inA). Hencex m,so(x) = (m).Adiscretevaluationringisaprincipalidealdomainsatisfyingthethreeequivalentconditionsin2.27. IfAisadiscretevaluationringwithprimeelement, thenanynonzeroelementcanbeuniquelyexpressedasum, wherem Z0andu Aisaunit. Theonlynonzeroprimeideal inAisp=(), soanynonzeroidealinAhastheformpmforawelldenedintegerm 0.Proposition2.28. Anintegral domainAis adiscretevaluationringif andonlyif thefollowingthreeconditionshold:(a) AisaNoetherianring,(b) Aisintegrallyclosed,and(c) Ahasexactlyonenonzeroprimeideal.Proof.(:)AssumethatAisadiscretevaluationring.14ThisisadirectapplicationofZornslemma,orderingbyinclusionthe(non-empty)setofidealsinAcontaining(z). Seecorollary1.4ofAtiyah-Macdonald[1].15Lety A\ {0}beanon-unit. Then(0) (y) = A,so(0)isnotamaximalidealinA.27 (A is a Noetherianring:) Letpbe the unique nonzeroprimeideal,andletI1 I2 . . .be a sequenceofidealsinA. IfI1= I2= . . . = 0,thenthesequence(Ik)isstationary.Otherwise,letn = mink Z>0: Ik ,= 0,andletIn= pm,wherem Z>0. ThenIk: k Zn p, p2, . . . , pm =: S,whichisaniteset. Thereforesomeelementof Sappearsinnitelyoftenin(Ik), sotheascendingchain(Ik)iseventuallystationary. HenceAisaNoetherianring. (Aisintegrallyclosed:) Aisaprincipal ideal domain, thereforeAisauniquefactorizationdomain.By2.4,Aisintegrallyclosed. Aisadiscretevaluationring,thereforeAhasexactlyonenonzeroprimeideal.(:) Let A be an integral domain which is Noetherian and integrally closed,and assume that A has exactlyonenonzeroprimeideal. Toshow: Aisadiscretevaluationring. AsAisanintegraldomainwithexactlyonenonzeroprimeideal,itremainstoshowthateveryidealinAisprincipal. AsA = (1)and 0 = (0),itremainstoshowthateverynonzeroidealinAisprincipal.FirstwelldeterminethenonzeroprimeidealinA. Letc A 0beanon-unit. Foreachm A 0,Im:= a A : c [ amisaproperidealinA. Let

:= Im: m A (c). AsAisaNoetherianring,

hasamaximalelementp := Ib= a A : c [ ab,whereb A (c). Thenc p,sop ,= (0).(pisaprimeidealinA:) p ,thereforepisaproperidealinA. Assume,forthesakeofcontradiction,that p is not a prime ideal in A. Then there exist x, y A p such that xy p. Then yb ,= 0 (as y ,= 0, b ,= 0andAisanintegraldomain). NowA ,= Iyb p + (x)p,asx Iyb c[xyb xy Ib=pandif a pthenc[ab c[ayb a Iyb. Wehavecontradictedthemaximalityofpin

,thereforepisaprimeidealinA. Thus,pistheuniquenonzeroprimeidealinA,asAhaspreciselyonenonzeroprimeideal.LetKbetheeldoffractionsofA. Notethatb A (c) c ,[b bc/ A.(p=Acb, andthereforecb A:) Inordertoshowthatp=Acb, itsucestoshowthatp bc=A(thenp=Acb). Bydenition, pb (c), sop bc A. Moreover, p bcisanideal inA(beginanA-submoduleofA). Assume, forthesakeofcontradiction, thatp bc ,=A. ByZornslemma, thereexistsamaximal idealcontainingp bc. However,Ahasexactlyonenonzeroprimeideal(namelyp),andp bc ,= (0),thereforep bc p. (2.25)However, p is a nonzero A-submodule of K(as pA), and is nitely generated (as pA and A is a Noetherianring),andbc/ A,so(2.25)contradicts2.1. Hencep bc= A,sop = Acb. Finally,cb p A.28LetabeanonzeroidealinA,andwewanttoshowthataisprincipal. Let=cb,sothatp = (), A,and1 K A. Considerthesequencea a1 a2 . . . (2.26)ofA-submodulesofK.(The sequence (2.26) is strictly increasing:) Proof by contradiction: assume that the sequence is not strictlyincreasing. Thenthereexistsr Z0suchthatar=ar1. Now1a=a. However, aisanonzero,nitelygenerated(asaAandAisaNoetherianring)A-submoduleofK, so1isintegral overA(by2.1). As KandAisintegrallyclosed, 1A, contradiction. Hencethesequence(2.26)isstrictlyincreasing.AsAisaNoetherianring,thisimpliesthatthesequence(2.26)isnotasequenceofidealsinA. Letm := minn Z0: an A.Nowam A. Moreover,am1 A,soam () = p. Asam,= (0)isnotcontainedinp(whichistheonlymaximalidealinA),Zornslemmaimpliesthatam= A,soa = Am= (m)isprincipal.Hence A is a principal ideal domain with exactly one nonzero prime ideal, and is therefore a discrete valuationring.A Dedekinddomainis an integral domain A which is not a eld, and which satises the following conditions:(a) AisNoetherian.(b) Aisintegrallyclosed.(c) EverynonzeroprimeidealinAismaximal.Lemma2.29. LetAbealocalintegraldomain. ThenAisaDedekinddomainifandonlyifAisadiscretevaluationring.Proof. (:)AssumethatAisaDedekinddomain. By2.28, itremainstoshowthatAhasexactlyonenonzeroprimeideal. Aislocal,soletMbetheuniquemaximalidealinA. LetIbeanonzeroprimeidealinA. ThenI isamaximal ideal, sinceAisaDedekinddomain. SinceI ,=0, thisimpliesthatI =M.HenceMistheonlynonzeroprimeidealinA,soAisadiscretevaluationring.(:)AssumethatAbeadiscretevaluationring. By2.28, itremainstoshowthateverynonzeroprimeideal in A is maximal. By 2.28, A has exactly one nonzero prime ideal, I. To show: Iis maximal. Note thatthere exists a maximal ideal Jwhich contains I(see footnote 14). Being a maximal ideal, J ,= (0) must alsobeaprimeideal,soJ= I. HenceIisamaximalideal,soAisaDedekinddomain.Thus,alocalintegraldomainisaDedekinddomainifandonlyifitisadiscretevaluationring.WenowaimtoshowthatlocalizingaDedekinddomainatanonzeroprimeidealyieldsadiscretevaluationring. Themotivationforthisisasfollows. Laterinthissectionwewill studythefactorizationof prime29idealsinextensions,whichisofgreatimportanceinalgebraicnumbertheory. Ausefultoolwillbetotakeanequationinvolvingideals,andtolocalizebothsidesoftheequationataprimeideal.16TheideaisthatdiscretevaluationringshavestrongerpropertiesthangeneralDedekinddomains.Lemma2.30. LetAbeanintegral domain,andletSbeamultiplicativelyclosedsubsetofA.(a) IfAisaNoetherianring,thenS1AisaNoetherianring.(b) IfAisintegrallyclosed,thenS1Aisintegrallyclosed.Proof.(a) AssumethatAisaNoetherianring. LetS1I(whereIA)beanarbitraryideal inS1A(by1.2).SinceAisNoetherian, I isnitelygeneratedinA, sayI=(a1, . . . , an), wherea1, . . . , an A. ThenS1I=(a11, . . . ,an1)isnitelygeneratedinS1A. Thuseveryideal inS1Aisnitelygenerated, soS1AisaNoetherianring.(b) AssumethatAisintegrallyclosed. AandS1Ahavethesameeldoffractions,sayK. Let KbeintegraloverS1A. Wewanttoshowthat S1A.m+am1m1+. . . +a1 +a0= 0,wherem Z>0anda0, . . . , am1 S1A. Thus, thereexists1, . . . , sm Ssuchthatsiai Afori = 1, 2, . . . , m. Multiplyingbothsidesbysm,wheres = s0s1. . . sm1,yields(s)m+sam1(s)m1+. . . +sm1a1(s) +sma0= 0. (2.27)In(2.27),allofthecoecientslieinA. HencesisintegraloverA,sos A,i.e. S1A.Proposition2.31. Let AbeaDedekinddomain, andlet pbeanonzeroprimeideal inA. ThenApisadiscretevaluationring.Proof. Apisalocalintegraldomain(by1.5), soby2.29itsucestoshowthatApisaDedekinddomain.By2.30, ApisNoetherianandintegrallyclosed. Apisnotaeldbecauseithasanonzeromaximal idealpAp. ItremainstoshowthateverynonzeroprimeidealinApismaximal.LetIApbeanarbitrarynonzeroprimeidealinAp, whereIisanonzeroprimeidealinAsuchthatI p(using1.3withS= A p). ThenImustbeamaximalidealinA,sinceAisaDedekinddomain. However,I p, andpisaproperideal inA, thereforeI=p. NowIAp=pAp, whichisthemaximal ideal inAp.Henceeverynonzeroprimeideal inApisamaximal ideal. Weconcludethat Apisadiscretevaluationring.Wenowmovetowardsprovingthefollowingimportanttheorem:16Localizationisanimportantconceptinabstractalgebragenerally, becausetheresultisalocalring. Inourcasewegetadiscretevaluationring,whichisevennicer !30Theorem2.32. Let AbeaDedekinddomain, andlet abeanonzeroproperideal inA. Thenacanbewrittenintheforma = pr11. . . prmm, (2.28)wherem Z>0,p1, , pmarepairwisedistinctnonzeroprimeidealsandr1, . . . , rm Z>0. Moreover,thepiandriareuniquelydetermined,uptorearrangement.Lemma2.33. Let AbeaNoetherianring. Theneverynonzeroideal inAcontainsaproduct of nonzeroprimeideals.Proof. Proof bycontradiction: assumethat thereexists anonzeroideal inAwhichdoes not containaproductofnonzeroprimeideals. Let

bethesetofnonzeroidealsinAwhichdonotcontainaproductof nonzeroprimeideals. As

isnon-emptyandAisaNoetherianring, thereexistsamaximal elementI

.17ThenIisnotaprimeideal,andI ,= A,thereforethereexistx, y A Isuchthatxy I. SinceI +(x)Iand I +(y)I,the maximality of Iimplies that I +(x) and I +(y) contain a product of primeideals. NowI=(I+ (x))(I+ (y))containsaproductofprimeideals, contradiction. HenceeverynonzeroidealinAcontainsaproductofnonzeroprimeideals.Lemma2.34. Let Abeacommutativeunital ring, andlet aandbberelativelyprimeidealsinA. Letm, n Z>0. Thenamandbnarerelativelyprime.Proof. Asa + b = A,thereexista aandb bsuchthata +b = 1. Now1 = (a +b)m+n=m+n

i=1_m+ni_aibm+ni=m+n

i=m+1_m+ni_aibm+ni+m

i=1_m+ni_aibm+ni am+ bn am+ bn= A,soamandbnarerelativelyprime.Lemma2.35. LetAbeacommutativeunital ring, andletpbeamaximal ideal inA. Letm Z>0, andletq = pApAp. Thenthemap :Apm Apqm[a] _a1_isanisomorphism.Proof. LetS:= A p. By1.4,qm:= (pAp)m= (S1p)m= S1(pm).(Welldened:) Leta, b A,andassumethat[a] = [b] Apm. Thena b pm,soa1 b1=a b1 S1pm= qm,17Thatis,ifJ andJ IthenJ= I.31as1 S. Thus_a1 =_b1Apqm,soiswelldened.(Injective:) Assumethatx Aand[x1]=[0] Apqm. Thenx1 qm=S1(pm), sothereexistb pmandt Ssuchthatx1=bt S1A. Thusthereexistsu Ssuchthatu(xt b) = 0. Withs := ut S(sinceSismultiplicative),sx = utx = ub pm.Wewanttoshowthat[x] = [0] Apm,i.e. thatx pm.As s/ p, the ideal (s)+p properly contains p. As p is a maximal ideal in A, it follows that (s)+p = A, so (s)and p are relatively prime ideals in A. By 2.34, (s) and pmare relatively prime ideals in A, so (s) +pm= A.Thus,thereexistsa Asuchthatas 1 modpm.Nowsx pm=asx pm=x pm,as0 asx x modpm.Thus[x] = [0] Apm,soisinjective.(Surjective:) Let a A and s S. Nowasis an arbitrary element of Ap, and we want to show that [as] Apqmisintheimageof. As(s) + pm= A,18thereexistsx Asuchthatxs 1 modpm. Nowasx a modpm asx a pm.Thusax1as=asx as S1pm,so_as_ =_ax1_ = ([ax]) Apqm.Henceissurjective.Weconcludethatisanisomorphism.Nowweprove2.32. Weusethestandardmotifoflocalizingandworkingwithdiscretevaluationrings.Proofof2.32. (Existence:) By 2.33, there exist mZ>0, r1, . . . , rmZ, and distinct nonzero prime idealsp1, . . . , pm Asuchthata b := pr11. . . prmm.By 2.34, the ideals pr11, . . . , prmmare relatively prime, so by the Chinese remainder theorem (1.9), the mapAb m

i=1Aprii[x] ([x], . . . , [x])18Ass/ p, theideal (s) + pproperlycontainsp. Aspisamaximal ideal inA, itfollowsthat(s) + p=A, so(s)andparerelativelyprimeidealsinA. By2.34,(s)andpmarerelativelyprimeidealsinA,so(s) + pm= A.32isanisomorphism. Fori=1, 2, . . . , m, deneqi:=piApi, whichistheuniquemaximal ideal inthediscretevaluationringApi(by1.5). By2.35,themapm

i=1Apriim

i=1Apiqrii([x1], . . . , [xm]) __x11_, . . . ,_xm1__(2.29)isanisomorphism. Thus,themap :Ab m

i=1Apiqrii[x] __x1_, . . . ,_x1__(2.30)isanisomorphism,beingacompositionofisomorphisms.Fori =1, 2, . . . , m, theringsApiarediscretevaluationringswithcorrespondingmaximal ideal qi.ThereforetheimageofainthemapA

mi=1Apiis

mi=1 qsii,forsomes1, . . . , sm Zwithsi rifori = 1, 2, . . . , m. Thus,_ab_ =m

i=1qsiiqrii= _ps11 psmmb_, (2.31)whereab:= x + b : x a,qsiiqrii:= x + qrii: x qsii , i = 1, 2, . . . , m,andps11 psmmb:= x + b : x ps11 psmm .Asisanisomorphism,equation(2.31)yieldsab=ps11 psmmb. (2.32)Asa bandps11 psmm b,a = ps11 psmm, (2.33)sothefactorizationexists. (Uniqueness:) Assumethatps11 psmm= pt11ptmm ,wherem Z>0,s1, . . . , sm, t1, . . . , tm Z0,andp1, . . . , pmaredistinctnonzeroprimeidealsinA.19Fori = 1, 2, . . . , m,localizingatpiyields_m

j=1psjj_Api=_m

j=1ptjj_Apim

j=1(pjApi)sj=m

j=1(pjApi)tj(2.34)19Weaddfactorsp0iifnecessary.33(If i, j 1, 2, . . . , mandi ,=j thenpi+ pj=A:) Leti, j 1, 2, . . . , m, withi ,=j. Toshow:pi+ pj=A. Proof bycontradiction: assumethat pi+ pj ,=A. AsAisaDedekinddomain, thenonzeroprimeidealspiandpjaremaximal. Hencepi=pi + pj=pj, soi=j, contradiction. Thus,pi + pj= A.For i, j 1, 2, . . . , m withi ,=j, there therefore exists xpjsuchthat x 1pi. Thus,1 pjApi Api,sopjApi= Api. Nowequation(2.34)becomesqsii= qtii,where qi:= piApiis the unique maximal ideal in the discrete valuation ring Api(by 1.5). Hence si= tifori = 1, 2, . . . , m,sothefactorizationisunique.Let A be a Dedekind domain. Let Ibe a nonzero proper ideal in A, and let p be a nonzero prime ideal in A.ThenpdividesIifpoccursintheprimefactorizationofI. Thefollowingresultprovidesusefulequivalentdenitions.Lemma2.36. Let AbeaDedekinddomain, let I beanonzeroproperideal inA, andlet pbeanonzeroprimeideal inA. Thenthefollowingareequivalent:(a) pdividesI.(b) IAp ,= Ap.(c) I p.Proof. LetI= pe11. . . pegg, (2.35)whereg, e1, . . . , eg Z>0,andwherep1, . . . , pgareprimeidealsinA. NowIAp=g

i=1(piAp)ei. (2.36) ((a) (c):) AssumethatpoccursintheprimefactorizationofI. Thenthereexistsi 1, 2, . . . , gsuchthatp = pi. Thus,p pe11 pegg= I. ((c) (b):) AssumethatI p. ThenIAp pApAp,soIAp ,= Ap. ((b) (a):) AssumethatIAp ,= Ap. Theng

i=1(piAp)ei,= Ap,sothereexistsi 1, 2, . . . , nsuchthatpiAp ,= Ap.Claim: pi p. Proofbycontradiction: assumethatthereexistsx pi p. Thenx1 piApisaunitinAp,sopiAp= Ap,contradiction. Thuspi p.However piis a prime ideal in A, and is therefore a maximal ideal in A, since A is a Dedekind domain.Asp ,= A,thisimpliesthatp = pi,sopdividesI.34Manymoretechnicalresultsfollowfrom2.32. Wediscussthemhere,butthereadermaywishtoskipthemfornow,andreturntothemwhentheyareneeded.Lemma2.37. LetAbeaDedekinddomain,andleta, b A. Thena bifandonlyifaAp bApforallnonzeroprimeidealspinA. Thus,a = bifandonlyifaAp= bApforall nonzeroidealspinA.Proof. Thecasesa = (0), a = A, b = (0)andb = Aareeasilychecked,soweexcludethemhenceforth. (:)Assumethata b,andletpbeanonzeroprimeidealinA. ThenaAp bAp. (:)AssumethataAp bApforallnonzeroprimeidealspinA. Leta = pr11. . . prmm, b = ps11. . . psmm, (2.37)where m Z>0, r1, . . . , rm, s1, . . . , sm Z0,and where p1, . . . , pmare nonzero prime ideals in A (addfactorsp0iifnecessary). Fori = 1, 2, . . . , n,localizebothsidesoftheequationsin(2.37)atpi,andtheassumptionaAp bApbecomes(piApi)ri (piApi)si,sori sifori = 1, 2, . . . , n. Hencea b.Therefore, a bifandonlyifaAp bApforall nonzeroprimeidealspinA. Consequently, a=bifandonlyifaAp= bApforallnonzeroidealspinA.Corollary2.38. Let AbeaDedekinddomain, andlet a, bbenonzeroidealsinAsuchthat a b. Thenthereexistsa Asuchthata = b + (a).Proof. Letb =m

i=1prii, a =m

i=1psii, (2.38)wherem Z>0, r1, . . . , rm, s1, . . . , sm Z0, andwherep1, . . . , pmarenonzeroprimeideals inA(addfactorsp0iifnecessary). Thenri sifori = 1, 2, . . . , m(asa b).For i = 1, 2, . . . , m, choose xi A such that xi psiips1+1i. By the Chinese remainder theorem, there existsa Asuchthata ximodprii(fori = 1, 2, . . . , m). Toshow: a = b + (a).By2.37,itsucestoprovethataAp + bAp= (a)Ap(2.39)forallnonzeroprimeidealspinA. By2.36(b),bothsidesofequation(2.39)areequaltoApifpdoesnotoccurintheprimefactorizationof a. Henceitsucestoshowthatif i 1, . . . , m, r=ri, s=siandp = pithenpsAp= prAp + (a)Ap.Leti 1, . . . , m, r = ri, s = siandp = pi. ItremainstoshowthatpsAp= prAp + (a)Ap. (2.40)35 (:)Notethata ps,so(a)Ap psAp. Also,r s,thereforeprAp psAp. Hence,prAp + (a)Ap psAp. (:)Notethatr s.Case: r = s. ThenpsAp= prAp prAp + (a)Ap.Case: r s + 1. Thena ximodpr=a xi pr ps+1=a ximodps+1i,soa ps ps+1. Hence(a)Ap psApand(a)Apps+1Ap.AsApisadiscretevaluationringwithuniqueprimeidealpAp,weknowthatanynonzeroidealinAphastheformpnAp, forsomen Z>0. Moreover, (a)Ap psApand(a)Apps+1Ap,therefore(a)Ap= psAp. NowprAp + (a)Ap= prAp + psAp psAp.HencepsAp= prAp + (a)Ap,andweconcludethata = b + (a).Corollary2.39. LetAbeaDedekinddomain,letabeanonzeroidealinA,andleta a0. Thenthereexistsb asuchthata = (a, b).Proof. Apply2.38witha (a) ,= 0.Lemma2.40. LetAbeaDedekinddomain,andletabeanonzeroideal inA.(a) Leta a. Thenthereexistsanideal ainAsuchthataa= (a).(b) Letcbeanonzeroideal inA. Thenthereexistsanideal ainAsuchthataaisaprincipal ideal inAandc + a= A.Proof.(a) If a=0, thena=(0)satisesaa=(a). If a=A, thena=(a)sayisesaa=(a). Henceforth,assumethata = 0andthata ,= A. Now(a) = pr11. . . prmm, a = ps11. . . psmm, (2.41)wherem Z>0, r1, . . . , rm, s1, . . . , sm Z0, andwherep1, . . . , pmarenonzeroprimeidealsinA(addfactorsp0iifnecessary). Now(a) aimpliesthatri sifori = 1, 2, . . . , m.20Nowa=n

i=1pirisi(2.42)satisesaa= (a).20Seetheproofof2.37ifthisisnotclear.36(b) a ac ,= (0), so, by 2.38, there exists a A such that a = ac +(a). Moreover, from part (a), there existsanidealainAsuchthataa= (a). Thus,a = ac + aa= a(c + a), (2.43)thereforec + a= A.21Importantly,weareabletoreconcileourdenitionofrelativelyprimeidealswiththeintuitivenotion:Lemma2.41. Let AbeaDedekinddomain, andlet aandbbenonzeroproper ideals inA. Thenthefollowingareequivalent:(a) a + b = A.(b) Theredoesnotexistanonzeroprimeideal p Asuchthatpisintheprimefactorizationofaandb.Proof. By2.36,aprimeidealp Adividesaandbifandonlyifitcontainsaandb. Ifa + b = Aandpcontainsaandb,thenp = p + b a + b = A,contradiction. Ifa + b ,= A,thenthereexistsamaximalidealp a + b. Thenpisprimeandpcontainsaandb.WewillprovethatringsofintegersinnumbereldsareDedekinddomains,butrstweneedthefollowinglemma:Lemma2.42. Letkbeaeld,andletB kbeanintegraldomainwhichisalgebraicoverk. ThenBisaeld.Proof. Itremainstoshowthateverynonzeroelementof BhasaninverseinB. Let0 ,= B. Asisalgebraicoverk,theringk[]isanite-dimensionalvectorspaceoverk. Themapk[] k[]x xisaninjectivelineartransformationbetweenvectorspacesofthesamedimension,andthereforesurjective.Hencethereexistsx k[] Bsuchthatx = 1.Theorem2.43. Let Abe aDedekinddomainwitheldof fractions K, andlet Lbe anite separableextensionofK. LetBbetheintegral closureofAinL. ThenBisaDedekinddomain.Proof. By2.9,Bisanintegrallyclosedintegraldomain.21Ifc + a = A,thenfactorizingaandc + aandsubstitutingintoequation(2.43)willyieldacontradiction.37 (BisaNoetherianring:) AsAisaNoetherianring, andBisanitelygeneratedA-module, BisaNoetherianA-module(by1.21). AnyIBisthereforenitelygeneratedasanA-module, andthesame nite set of generators generates Ias a B-module, i.e. as an ideal in B. Hence Bis a Noetherianring. (EverynonzeroprimeidealinBismaximal:) LetqbeanonzeroprimeidealinB. Inordertoshowthatqismaximal,weneedtoshowthatBqisaeld.Firstweestablishthatq AisamaximalidealinA. AsAisaDedekinddomain,itsucestoprovethatq AisanonzeroprimeidealinA.(q A ,= 0:) Let q, ,= 0. Thenthereexistn Z>0anda1, . . . , an Asuchthatn+a1n1+. . . +an= 0, (2.44)andan ,= 0(otherwisethepolynomialn+a1n1+. . . +anisreducible). Then0 ,= an= (n+a1n1+. . . +an1) q A. (2.45)Henceq A ,= (0).(q A ,= A:) 1/ q q A,thereforeq A ,= A.(q AisaprimeidealinA:) Assumethatx, y Aandxy q A. AsqisaprimeidealinB,weknowthatx qory q. Thusx q Aory q A. Thereforeq AisaprimeidealinA.Henceq AisanonzeroprimeidealinA. AsAisaDedekinddomain,q AisamaximalidealinA.Thus,Apisaeld,wherep := q A. TheimageoftheinjectivemapAp Bq[a] [a]is aisomorphic toAp, bythe rst isomorphismtheorem, andis therefore aeldcontainedinBq .Moreover,Bqisanintegral domain, sinceqisaprimeideal inB. By2.42, weconcludethatBqisaeld. Thus,qisamaximalidealinB.HenceBisaDedekinddomain.Corollary2.44. LetKbeanumbereld. Then OKisaDedekinddomain.Proof. In2.43,letA = Z,andletLbeanynumbereld.2.4 TheidealclassgroupLetAbeaDedekinddomainwitheldoffractionsK. Afractional ideal ofAisanonzeroA-submoduleaofKsuchthatda := da : a a A38forsomed A 0. Inthiscase,da AisanA-module,soda A.FractionalidealsofAarenotbonadeidealsunlesstheyarecontainedinA,soweshallrefertoidealsasintegral idealsifthereisanyconfusion. Afractionalidealisprincipal ifittakestheform(b) := bA := ba : a A,forsomeb K.Weshallseethattheproductoftwofractionalidealsisafractionalideal:ab =_n

i=1aibi: ai a, bi b, n Z0_. (2.46)Forprincipalfractionalideals,(a)(b) = (ab).Thefollowingtheoremresultsinuniqueprimefactorizationof fractional ideals, allowingnegativepowers,andneatlydescribeswhatall of thefractional idealsare. Thefactthatfractional idealshaveinversesisextremelyuseful,andnontrivial.22Theorem2.45. LetAabeaDedekinddomainwitheldoffractionsK. Then,withthegroupoperationasinequation(2.46),thesetId(A)offractional idealsofAisthefreeabeliangrouponthesetofprimeidealsinA.Proof. First we show that Id(A) is a group. Let a, b, c Id(A) be such that da, eb, fc A, where d, e, f A. (ab Id(A):) Notethat(de)ab A. Moreover, abisanA-submoduleofK, sinceitisclosedunderaddition and scalar multiplication, and that it contains additive inverses. Hence ab is a fractional idealofA. TheoperationiscommutativebecauseKiscommutative:ab =_n

i=1aibi: ai a, bi b, n Z0_ =_n

i=1biai: ai a, bi b, n Z0_ = ba. Aistheidentity. (Inversesexist:) Werstprovethisforintegralideals. Let(0) ,= a Aand0 ,= a a. By2.40,thereexistsa Asuchthataa= (a). Nowa(a1a) = A,soa1a Id(A)isaninversefora.Nowleta Id(A). Thenthereexistsd A 0suchthatda A. Nowd(da)1isaninversefora.HenceId(A) is agroup. Tocompletetheproof, weneedtoshowthat anyfractional ideal of Acanbeexpresseduniquelyintheformpt11ptmm ,wherem Z>0, t1, . . . , tm Z,andp1, . . . , pmareprimeidealsinA.Leta Id(A),andletd Abesuchthatda A.22Thatis,withoutthepowerfultechnicalframeworkthatwehavedeveloped.39(Existence:)da = pr11. . . prmm, (d) = ps11. . . psmm, ri, si 0.Nowda = (d)a =a = (da)(d)1=n

i=1prisii.(Uniqueness:) Assumethatn Z>0andn

i=1prii=n

i=1psii, ri, si Z.Thenn

i=1pc+rii=n

i=1pc+sii,wherec = [ri[ +[si[. Thenc +ri, c +si Z0,so2.32yieldsri= siforalli.LetAbeaDedekinddomain. Theideal classgroupofAisthequotientCl(A) =Id(A)P(A) ,whereP(A)isthesubgroupofprincipalfractionalidealsinId(A). IfCl(A)isnite,theclassnumber ofAistheorderofCl(A).LetKbeanumbereld. Theideal classgroupofKisCl(OK),andtheclassnumber ofKistheorderoftheidealclassgroupofK. Weshallseethatthisisalwaysnite.Theidealclassgroupandtheclassnumberofanumbereldareimportantinvariantsinalgebraicnumbertheory. Intuitively, they measure the failure of prime factorization of elements in the ring of integers.23Thisismotivatedbythefollowingresult:Proposition2.46. LetAbeaDedekinddomain. ThenAisaprincipal ideal domainifandonlyifAisauniquefactorizationdomain.Proof. (:)AssumethatAisaprincipal ideal domain. ThenAisauniquefactorizationdomain, asanyprincipalidealdomainisauniquefactorizationdomain.(:)AssumethatAisauniquefactorizationdomain. 0=(0)andA=(1), soitremainstoshowthateverynonzeroproperidealinAisprincipal. Byprimefactorizationofideals(2.32),itsucestoprovethateverynonzeroprimeidealinAisprincipal.Let pbeanonzeroprimeideal inA. Let 0 ,=a p. Thenais not aunit inA. SinceAis auniquefactorizationdomain,afactorsasaproductofprimeelementsinA:a = 1 n, n Z>0, 1, . . . , nprime.23Thetaskofndingprecisewaystodescribethisisthesubjectofcurrentresearch. SeeMartin[6],2011.40Aspis prime,there existsi 1, . . . , n suchthati p. Nowp (i).Moreoever,(i)isa maximalidealinA(sinceitisanonzeroprimeidealinAandAisaDedekinddomain),thereforep = (i). HencepisaprincipalidealinA.WeconcludethatAisaprincipalidealdomain.2.5 DiscretevaluationsPreviously, wedeneddiscretevaluationringwithoutdeningdiscretevaluations. Weshall seethatthisterminologyisnocoincidence, butmoreimportantlywegiveexamplesofdiscretevaluationswhichweuseinlatersections,andwhichareusefulmorebroadlyinnumbertheory.LetKbeaeld. AdiscretevaluationonKisanonzerogrouphomomorphismv: K (Z, +)suchthatifa, b Kthenv(a +b) minv(a), v(b).A discrete valuation is normalizedif it is surjective. Note that Image(v) is a nonzero subgroup of Z, so thereexistsm Z>0suchthatImage(v)=mZ. Ifvisadiscretevaluation, thenx m1v(x)isanormalizeddiscretevaluation.LetKbeaeld,andletv: K Zbeadiscretevaluation. Byinduction,ifm Z2anda1, . . . , am Kthenv(a1 +. . . +am) minv(ai) : i 1, . . . , m.Notethatv(1) = 0,asv(1) = v(1) +v(1). Also,v(1) = 0,asv(1) +v(1) = v(1) = 0. Thus,ifx Kthenv(x) = v(1) +v(x) = v(x). Itissometimesconvenienttodenev(0) := .ExampleLetAbeaDedekinddomainwitheldoffractionsK,andletpbeanonzeroprimeidealinA.Forx K,letordp(x)betheexponentofpinthefactorizationof(x) Id(A). Thenordp: K Zisanormalizeddiscretevaluation.Proof. Werstshowthatordp: K Zisadiscretevaluation: Thefunctionordp: K Ziswelldened,sinceId(A)isthefreeabeliangrouponthesetofprimeidealsinA(from2.45). Letx, y K. Factorizing(x)and(y)asin2.45yieldsordp(xy) = ordp(x) + ordp(y), (2.47)soordp: K Zisagrouphomomorphism. Weshowthatordp(x +y) minordp(x), ordp(y). Proofbycontradiction: assumethatordp(x +y) < ordp(x), ordp(x +y) < ordp(y). (2.48)41Letm := ordp(x +y). Thenordp(x) m+ 1andordp(y) m+ 1. Nowx, y pm+1=x +y pm+1=ordp(x +y) m+ 1, (2.49)contradiction. Henceordp(x +y) minordp(x), ordp(y).Thereforeordp: K Zisadiscretevaluation.It remains to show that ordpis normalized, i.e. that 1 Image(ordp). By 2.40, there exists pA such thatppisaprincipalidealinAandp + p= A. Thep ,= (0),soletx A 0 Kbesuchthatpp= (x). (2.50)Nowordp(x) = 1,sincepdoesnotoccurintheprimefactorizationofp(by2.41). Henceordp: K Zisanormalizeddiscretevaluation.Proposition2.47. LetKbeaeld,andletv: K ZbeadiscretevaluationonK. ThenA := 0 a K: v(a) 0 (2.51)isadiscretevaluationringwithuniquenonzeroprimeidealm := 0 a K: v(a) > 0. (2.52)Moreover,ifv(K) = mZ,m > 0,andv() = m,thenm = ().Proof. We want to establish that A is a discrete valuation ring. By 2.27, it suces to prove that A is a localprincipalidealdomainwhichisnotaeld.(AisasubringofK,andisthereforeanintegraldomain:) Letx, y A 0.24 0 A,bydenition. 1 A,asv(1) = 0. v(x +y) minv(x), v(y) 0 x +y A. v(x) = v(x) 0 x A. v(xy) = v(x) +v(y) 0 xy A.HenceAisasubringoftheeldK,andisthereforeanintegraldomain.(mA:) Letx, y m 0anda A 0.25 0 mbydenition. v(x) = v(x) > 0 x m.24Thecasex = 0andthecasey= 0aretrivial.25Again,theremainingcasesaretrivial.42 v(x +y) minv(x), v(y) > 0 x +y m. v(ax) = v(a) +v(x) v(x) > 0 ax m.HencemA.(Aisalocalringwithuniquemaximalidealm:) 1 A m,som ,= A. Itsucestoprovethatmcontainsallnon-unitsinA. Letxbeanon-unitinA. Then1x K A,sov(1x) < 0. Now0 > v_1x_ = v(1) v(x) = v(x) v(x) > 0,sox m. Hencemcontainsallnon-unitsinA. WeconcludethatAisalocalringwithmaximalidealm.(Aisaprincipalidealdomain:) LetIbeanonzeroproperidealinA. Thereexistsx Isuchthatify Ithenv(y) v(x). ThenI=(x): ify Ithenv(yx)=v(y) v(x) 0 yx A y (x). HenceAisaprincipalidealdomain.Fromtheaboveparagraph, weconcludethatifv(K)=mZ, m>0, thenm=()forany Ksuchthatv() = m. Finally,sinceAisalocalprincipalidealdomainwhichisnotaeld,26weconcludethatAisadiscretevaluationring.2.6 FactorizationinextensionsLetAbeaDedekinddomainwitheldoffractionsK. LetL/Kbeaniteseparableextension,andletBbetheintegralclosureofAinL. LetpbeanonzeroprimeidealinA. BisaDedekinddomain,by2.43,sopBfactorizesuniquelyintoaproductofnonzeroprimeidealsinB(by2.32):pB= Pe11 Pegg, ei 1. (2.53)Thenonzeroprimeideal pAramies inB(orL)if thereexistsi 1, 2, . . . , gsuchthatei>1, andramiestotallyifg= 1ande1> 1. LetPbeaprimeidealinB. TheprimeidealPdividespifPdividespB. The ramication index, e(P/p), is the exponent of P in the prime factorization of pB. The residue classdegreeisgivenbyf(P/p) :=_BP:Ap_.27TheprimeidealpsplitsinB(orL)ifei= fi= 1foralli,andisinert inB(orL)ifpBisaprimeidealinB.Lemma 2.48. Let A be a Dedekind domain with eld of fractions K. Let L/Kbe a nite separable extension,andlet Bbetheintegral closureof AinL. Let pbeanonzeroprimeideal inA, andlet Pbeanonzeroprimeideal inB. ThenPdividespifandonlyifp = P K.Proof. By2.36,PdividespifandonlyifpB P.(:)AssumethatPdividesp. p pB Pandp A K. Hencep P K.26Asvisnonzero, thereexistsx Ksuchthatv(x)>0(ify Kandv(y)0. Thenx A \ {0}butv(1x) < 0 1x/ A.27pandPare(nonzero)primeandhencemaximalidealsintherespectiveDedekinddomainsAandB,thereforeBPandApareelds. Themap[x] [x] :ApBPisaninclusion,sowecanregardBPasavectorspaceoverAp .43 AsAisintegrallyclosed,P K B K= A. KisanA-module(beingaK-module)andPisanA-module(beingaB-module),thereforePKisanA-module. SinceP K A,thismeansthatP K A. 1/ P P K,soP K ,= A.ThusP Kisaproperideal inAwhichcontainsp. Sincepisamaximal ideal (aspisanonzeroprimeidealintheDedekinddomainA),weconcludethatp = P K.(:) Assumethatp = P K. ThenpB= (P K)B PB= P,soPdividesp.Lemma2.49. Let0 ABC 0beanexactsequenceofvectorspacesoveraeld. ThendimB= dimA+ dimC.Proof. Applytherank-nullitytheoremtoand,notingexactness:dimB= dim(Image) + dim(ker) = dimC + dim(Image)= dimC + dimAdim(ker) = dimC + dimA.The following formula links the ramication indices and residue class degrees to the degree of the extension:Theorem2.50. Let Abe aDedekinddomainwitheldof fractions K. Let L/Kbe anite separableextensionofdegreem,andletBbetheintegral closureofAinL. Letpbeaprimeideal inA,andletpBprimefactorizeinBasfollows:pB= Pe11 Pegg, ei 1.Then(a)g

i=1eifi=_BpB:Ap_ = m. (2.54)(b) IfL/KisaGaloisextension,thenall oftheramicationindicesareequal,andall oftheresidueclassdegressareequal,soefg= m. (2.55)Wenowaimtoprove2.50.44Lemma 2.51. Let A be a Dedekind domain with eld of fractions K. Let L/Kbe a nite separable extensionofdegreem,andletBbetheintegral closureofAinL. Letpbeanonzeroprimeideal inA,andletPbeanonzeroprimeideal inBwhichdividesp. Letf= f(P/p)andletr Z>0. ThenBPrisavectorspaceofdimesionrfoverAp.Proof.Apisaeldbecausepisa(nonzero)prime(andthereforemaximal)ideal intheDedekinddomainA. Theinclusion[x] [x] :ApBPrmakesBPravectorspaceoverAp. Weprovebyinductionthatitsdimensionisrf. Bydenition,_BP:Ap=f. Letn Z>0,andassumethat_BPn:Ap=nf.28Toshow:_BPn+1:Ap=(n + 1)f.Applying2.49totheshortexactsequence(ofvectorspacesoverAp)0 PnPn+1 BPn+1 BPn 0 (2.56)yieldsdim_BPn+1_ = dim_PnPn+1_+ dim_BPn_ =_PnPn+1:Ap_+nf. (2.57)Thus,itsucestoprovethat_PnPn+1:Ap = f. By2.38,thereexistsy PnsuchthatPn= Pn+1+ (y) B. (2.58)whichimpliesthatordP(y) = n,asy Pn Pn+1. Themap :BP PnPn+1[x] [xy]isanisomorphism:(Welldened:) Ifa, b Band[a] = [b] BP,thena b P=ay by= (a b)y Pn+1=[ay] = [by] PnPn+1.Henceiswelldened.(Injective:) Ifx Bandxy Pn+1,thenordP(x) = ordP(xy) ordP(y) (n + 1) n = 1 x P.Henceisinjective.(Surjective:) Lett Pn. Fromequation(2.58),thereexistsx Bsuchthatt xy Pn+1.Now[x] [xy] = [t] PnPn+1,soissurjective.28ThisnotationdoesnotmeantoimplythatBPnisaeld; itmerelydescribesthedimensionofBPnasavectorspaceovertheeldAp .45ThuswecanregardPnPn+1asavectorspaceofdimension1overBP. Now_PnPn+1:Ap_ =_PnPn+1:BP_

_BP:Ap_ = f. (2.59)Weconcludethat_BPn+1:Ap=(n + 1)f.Byinduction,_BPr:Ap = rf.Lemma2.52. LetAbeaDedekinddomainwitheldoffractionsK,andletpbeanonzeroprimeideal inA. LetL/KbeaGaloisextensionwithG := Gal(L/K),andletBbetheintegral closureofAinL. Thenanyelement of GrestrictstoabijectionB B, andGactstransitivelyontheset of primeidealsinBwhichdividep.Proof. Werstshowthattheactioniswelldened. Let G.(B= B:) (:) Letb B. Thenf(b)=0forsomemonicpolynomial f A[x], sof((b))=(f(b))=0, so(b) B. (:) Letb B. Thenb = (1(b)) B.HenceB= B. As[B: B Bisabijectivehomomorphism,[B: B Bisanisomorphism,soittakesprimeidealstoprimeideals: ifPisaprimeidealinB,thenPisaprimeidealinB.(IfPdividesp,thenPdividesp:) AssumethatPdividesp. ThenP K= P K= (P K), asisinjective= P K, as Gal(L/K)= p, by2.48, (2.60)soPdividesp(by2.48). Hencetheactionof Gonthesetof primeidealsinBwhichdividepiswelldened.(Theactionistransitive:) Proof bycontradiction: assumethat, areprimeidealsinBdividingp, andassumethatand arenotconjugate. Fori=1, 2, . . . , m, thereexistsci i.29BytheChineseremainder theorem (by 2.41, distinct nonzero prime ideals are relatively prime), there exists x Bsuch thatx 0 modx c1mod1...x cmmodm.29By2.43,BisaDedekinddomain,soeverynonzeroprimeidealinBismaximal. Inparticular,isamaximalidealinB,soitcannotbestrictlycontainedini.46Nowx ,butx/ ifori = 1, 2, . . . , n. Deneb :=

Gx = NmL/K(x) K, (2.61)by2.14and2.12. Asx ,b = x

G\id]x . (2.62)Thus,b K= p(by2.48). Let G. Thenx/ 1,sox/ . However,

Gx = b p pB ,contradictingthefactthatisaprimeidealinB. Hencetheactionistransitive.Proofof2.50. (a)_

gi=1eifi=_BpB:Ap_:_BytheChineseremaindertheorem,BpB=B

Peii=

BPeii.Using2.51,_BpB:Ap_ =g

i=1_BPeii:Ap_ =g

i=1eifi.__BpB:Ap_ = m:_WerstestablishthatifAisaprincipalidealdomain,then_BpB:Ap_ = m:AssumethatAisaprincipal ideal domain. By2.22, BisafreeA-moduleof rankm. ThenthereisanisomorphismAm = BofA-modules,so(using1.7)Ap A Am =Ap A B. (2.63)By1.8,thisbecomesBpB=AmpAm=_Ap_m. (2.64)Thus,ifAisaPID,then_BpB:Ap_ = m.NowrelaxtheassumptionthatAisaPID.LetS:= A p. ThenSisamultiplicativelyclosedsubsetofAandofB,and0/ S. LetAt:= S1A = ApandBt:= S1B. ThenKistheeldoffractionsofAtandListheeldoffractionsofBt.(BtistheintegralclosureofAtinL:) (:)If Bt, thenthereexistss Ssuchthats B. AsBisintegral overA, thereexistn Z>0anda1, . . . , an Asuchthat(s)n+a1(s)n1+. . . +an= 0 (2.65) n+a1sn1+. . . +ansn= 0, (2.66)soisintegraloverAt.47 (:)Assumethat LisintegraloverAt. Thenthereexistn Z>0,a1, . . . , an Aands1, . . . , sn Ssuchthatn+a1s1n1+. . . +ansn= 0 (2.67) (s)n+a1ss1(s)n1+. . . +snsnan= 0, s :=n

i=1si. (2.68)ThecoecientsofslieinA,sos B,so Bt.HenceBtistheintegralclosureofAtinL.Moreover,Atisadiscretevaluationring(by2.31). NowS1(pB) = S1g

i=1Peii(2.69)pBt=g

i=1(PiBt)ei. (2.70)Henceg

i=1eifi=_BtpBt:AtpAt_, (2.71)as_B

PiB

:A

pA

_=_BP:Ap_=fi.30Moreover, Atis aprincipal ideal domainwitheldoffractionsK,andListheeldoffractionsofBt,so_BtpBt:AtpAt_ = m. (2.72)Fromequations(2.71)and(2.72),g

i=1eifi=_BtpBt:AtpAt_ = m, (2.73)andwehadalreadyshownthat_BpB:Ap_ = m. Hence,g

i=1eifi=_BpB:Ap_ = m.(b) AssumethatL/KisaGaloisextension,andletG = Gal(L/K) = 1, . . . , m. Let G,andletPbeaprimeidealinBwhichdividesp. By2.52,restrictstoanisomorphismB B. (e(P/p) = e(P/p) :)pB=g

i=1Peii. (2.74)Applyingtobothsidesyields31pB=g

i=1(Pi)ei. (2.75)30Letn = fi. If {[x1], . . . , [xn]}isabasisforBPioverAp ,then {

x11

, . . . ,

xn1

}isabasisforB

PiB

overA

pA

).31(pB)isanideal inBwhichcontains(p)=p(notethatxesK p), so(pB) pB. Similarly, pB 1(pB) (pB) pB. Hence(pB) = pB.48Hencee(P/p) = e(P/p). (f(P/p) = f(P/p) :)Abasis1, . . . , f (2.76)forBPcorrespondstoabasis1, . . . , f (2.77)forB. Hencef(P/p) = f(P/p).Leti, j 1, . . . , g. By2.52,thereexists GsuchthatPj= Pi. Nowej= e(Pj/p) = e(Pi/p) = e(Pi/p) = ei(2.78)andfj= f(Pj/p) = f(Pi/p) = f(Pi/p) = fi, (2.79)soalloftheramicationindicesandresidueclassdegreesareequal. Hence,efg= m.Thefollowingisanimportantcriteriontodecidewhetherornotaprimeramies:Theorem 2.53. Let Kbe a number eld, and let L be a nite separable extension of K. Let A be a Dedekinddomainwitheldof fractionsK,32andlet Bbetheintegral closureof AinL. Assumethat BisafreeA-module. Let pbeanonzeroprimeideal inA. Thenpramies inLif andonlyif p[disc(B/A).33Inparticular,onlynitelyprimeidealsramify.Weneedsomepreliminaryresultstoprove2.53.Lemma2.54. LetAbeacommutativeunital ringandletBbeacommutativeringcontainingA. AssumethatBadmitsanitebasis e1, . . . , emasanA-module,andleta A. Then [e1], . . . , [em]isabasisfortheAa -moduleBaB,andD([e1], . . . , [em]) = [D(e1, . . . , em)] Aa . (2.80)Proof. TheisomorphismAmB(a1, . . . , am) m

i=1aieigives,whentensored(overA)withAa,anisomorphism_Aa_mAmAa B Aa BaB32e.g. A = OK33Thismeansthatpcontainsanyrepresentativeofdisc(B/A),orthatpdividestheidealgeneratedbydisc(B/A).49([a1], . . . , [am]) (a1, . . . , am) [1] m

i=1aiei[1] m

i=1ai[ei]. (2.81)Hence [e1], . . . , [em]isabasisfortheAa-moduleBaB.D([e1], . . . , [em]) = det_TrBaB/Aa[eiej]_i,j1,2,...,m]=_det_TrB/A(eiej)__= [D(e1, . . . , em)] Aa . (2.82)Lemma2.55. LetAbeacommutativeunital ring, andletB1, . . . , BgbecommutativeringscontainingAthatarenitelygeneratedfreeA-modules. Thendisc__g

i=1Bi__A_ =g

i=1disc(Bi/A).Proof. Fori 1, 2, . . . , g,let ei1, . . . , einibeabasisforBi. Thentheirunion, e11, . . . , egng,isabasisfortheA-module

gi=1Bi.D(e11, . . . , egng) = det M,whereMisablockmatrix,withablock__Tr(e2i1) Tr(ei1ei2) . . . Tr(ei1eini)Tr(ei1ei2) Tr(e222) . . . Tr(ei2eini)............Tr(einiei1) Tr(einiei2) . . . Tr(e2ini).__(2.83)foreachi 1, 2, . . . , g.34NowD(e11, . . . , egng) =g

i=1D(ei1, . . . , eini), (2.84)sodisc__g

i=1Bi__A_ =g

i=1disc(Bi/A).Let Abeacommutativeunital ring. Anelement Ais nilpotent if thereexists m Z>0suchthatm= 0. Acommutativeunitalringisreducedifithasnononzeronilpotentelements.Lemma2.56. Letkbeaperfecteld,andletBbeacommutativek-algebrawhichhasnitedimensionasavectorspaceoverk. ThenBisreducedifandonlyifdisc(B/k) ,= 0.34Thisfollowsfromthefactthatifi = jtheneiaejb= 0 gi=1 Bi.50Proof. (:) We prove the contrapositive statement. Assume that Bis not reduced. Then Bhas a nilpotentelement ,=0. Wecanextend toabasis =e1, . . . , emforB(asavectorspaceoverk). Thus, ifj 1, 2, . . . , mthenejisnilpotent,sothemapB Bx ejxisnilpotent. Inparticular,itsmatrixisnilpotent,andisthereforetraceless.35Henceeveryentryintherstrowofthematrix(Tr(eiej))iszero,soD(e1, . . . , em) = det(Tr(eiej)) = 0.(:) Assume that Bis reduced. Bis a commutative unital ring,so the intersection of the prime ideals in B(thenilradical)isthesetofnilpotentelementsinB,whichis 0.36(EveryprimeidealinBismaximal:) LetpbeaprimeidealinB. ThenBpisanintegraldomaincontainingk.37Toshow:Bpisaeld. By2.42, itremainstoshowthatBpisalgebraicoverk. Ifb B, thenbistheroot of some f k[X].38Then [b] Bpis algebraic. HenceBpis algebraic over k. By 2.42,Bpis a eld. ThuspisamaximalidealinB. Hence,everyprimeidealinBismaximal.Let p1, . . . , prbedisinct primeideals inB. Sincetheyaremaximal, theyarepairwiserelativelyprime.Hence,bytheChineseremaindertheorem,Bri=1pi=r

i=1Bpi. (2.85)Now[B: k] _Bri=1pi: k_ =_r

i=1Bpi: k_ =r

i=1_Bpi: k_ r. (2.86)ThereforeBonlyhasnitelymanyprimeideals,sayp1, . . . , pg,where gi=1pi= 0. NowB=Bgi=1pi=g

i=1Bpi. (2.87)EachBpiisanite(sinceBisanite-dimensional vectorspaceoverk)separable(askisaperfecteld)extensionofk. By2.19,disc_Bpi_k_,= 0. (2.88)Using2.55,disc(B/k) = disc_g

i=1Bpi_k_ =g

i=1disc_Bpi_k_,= 0. (2.89)35LetMbethematrixofthemap,andletMa=0. TheminimalpolynomialofMdividesXa,andthereforehastheformXb. ThecharacteristicpolynomialofMisdivisiblebyXb, andthereforehastheformXc. ThetraceofMisthesumoftheeigenvaluesofM,whichisthesumoftherootsofthecharacteristicpolynomialofM,whichis0(bytheVietaformulae).36SeeAtiyah-Macdonald[1],chapter1.37ToshowthatBp k,weneedtoshowthatp k= {0}. Ifx p kthenx pisnotaunitinB,soxisnotaunitink,sox = 0. HenceBp k.38LetBhavedimensionmasak-vectorspace. Then1, b, b2, . . . , bm+1aredependentvectorsoverk.51Proofof2.53. Wewillregarddisc(B/A)asanelementinArepresentingdisc(B/A) A(A)2. By2.54,[disc(B/A)] = disc_BpB_Ap_Ap . (2.90)NotethatBpBisanalgebraoverAp,andthatBpBisnite-dimensionalasavectorspaceoverAp(by2.50,itsdimensionis[L : K]).(Apisaperfecteld:) The(equivalent)denitionof aperfecteldgiveninLang[4], p252, is: aeldkofcharacteristicpisperfect if kp=k, i.e. if everyelementof kisapthpower. Inourcase, KandApbothcontain1, sotheyhavecommoncharacteristict.39Letx A K. Thenthereexistsy Ksuchthatx = yt. Nowyt x = 0,x A, y K,andKisintegrallyclosed,soy A. Now[x] = [y]tAp,soApisaperfecteld.Nowby2.56,[disc(B/A)] = [0] ApifandonlyifBpBisnotreduced. Inotherwords,pdividesdisc(B/A)ifandonlyifBpBisnotreduced. LetpB=g

i=1Peii.SinceAisaDedekinddomain,thePiaremaximal,andthereforerelativelyprimeinpairs. BytheChineseremaindertheorem,BpB=g

i=1BPeii. (2.91) If p ramies, then there exists i 1, 2, . . . , g such that ei> 1. Let x PiP2i. Then [x] is a nonzeronilpotent inBPeii, so we can lift [x] to a nonzero nilpotent ([0], . . . , [0], [x], [0], . . . , [0]) in

gi=1Beii. ThusBpBisnotreduced,sop[disc(B/A). Ifpdoesnotramify, thenBpB=

gi=1Beiiisadirectsumofelds, soitisaeld. Inparticular, itisreduced,40sop ,[ disc(B/A).HencepramiesinLifandonlyifp [ disc(B/A). Thisimpliesthatonlynitelymanyprimeidealsramify,sincewecanfactorize(disc(B/A))asaproductofnonzeroprimeidealsintheDedekinddomainA.2.7 Normsofideals,andnitenessoftheclassnumberInthissection, weintroducenormsofideals, whichappearintheclassnumberformula. Weusenormsofideals to establish that the class number of any number eld is nite, as well as some other important results.Lemma 2.57. Let A be a Dedekind domain with eld of fractions K. Let L/Kbe a nite separable extension,andletBbetheintegral closureofAinL. LetPbeanonzeroprimeideal inB,andletp := P A. Then(a) pisaprimeideal inA.39Everyeldofcharacteristic0isperfect,soassumethatt > 0.40Letkbeaeld. If0 =x kandt Z2andxt=0, thenx=xt(x1)t1=0, contradiction. Hencekhasnononzeronilpotents. Weconcludethatanyeldisreduced.52(b) p ,= (0).(c) Pdividesp.Proof.(a) Ifx, y pthen x, x +y p. Ifx p,a A,thenax P(asPB)andax A,soax p. Hencep A. 1/ P p p ,= A. HencepisaproperidealinA. Ifx, y Aandxy p,thenx Pory P(asPisaprimeidealinB),sox pory p.HencepisaprimeidealinA.(b) Let0 ,=x P. Letn:=[L:K], andletG:= 1, . . . , nbethesetofdistinctK-homomorphismsL ,where LisaGaloisextensionofK. By2.14,NmL/K(x) = 1(x)n(x). (2.92)(EachGaloisconjugateof xliesinB:) Let G, andletf A[X] betheminimal polynomial of xoverK(by2.5). Thenf((x)) = (f(x)) = 0,so(x) B(by2.5). Thus,eachGaloisconjugateofxliesinB.Deneid : L , x x. Fromequation(2.92),NmL/K(x) = 1(x)n(x) = x

G\id](x) Bx. (2.93)ThenNm(x) Bx PandNm(x) A(by2.13),soNm(x) P A = p. Moreover,Nm(x) ,= 0,fromequation(2.92). Hencep ,= (0).(c) By2.36,P pBimpliesthatPdividesp.Recall that Id(B) is the free abelian group on the set of nonzero prime ideals in B. Dene a homomorphismA= AL/K: Id(B) Id(A)P pf, fornonzeroprimeidealsPB,wherep = P Aandf= f(P/p) =_BP:Ap_.This generalised denition coincides with a much simpler denition in the case of rings of integers in numberelds. LetKbeanumbereld, andlet(0) ,=IOK. Thenumerical normof I isN(I):=(OK: I),41whichweshallprovetobenite.41ThisistheindexofthesubgroupIoftheadditivegroup OK.53Proposition2.58. LetKbeanumbereld.(a) Let(0) ,= I OK. Then AK/Q(I) = N(I)Z.(b) LetK R0. ThentherearenitelymanynonzeroidealsI OKsuchthatN(I) K.(c) Leta, b Id(OK),andassumethata b. Then(a : b) = N(a1b).Proof. (a) LetI=g

i=1prii(2.94)bethefactorizationofIintoprimeidealsintheDedekinddomain OK,anddenepi Z>0byZpi= Z pi. (2.95)BytheChineseremaindertheorem,OKI=g

i=1OKprii, (2.96)soN(I) =g

i=1(OK: prii). (2.97)By2.51,cKpriiisavectorspaceofdimensionrifioverZZpi(wherefi= f(pi/Zpi)),therefore(OK: prii) = prifii. (2.98)Thus,N(I) =g

i=1(OK: prii) =g

i=1prifii(2.99)N(I)Z =g

i=1(piZ)rifi= A(I). (2.100)(b) AssumethatI OKisanonzeroidealsuchthatN(I) K. LetI=g

i=1prii(2.101)bethefactorizationofIintoprimeidealsintheDedekinddomain OK,anddenepi Z>0byZpi= Z pi. (2.102)ThenK N(I) =g

i=1prifii. (2.103)Thus, therearenitelymanypossibilitiesforgandtheri. Moreover, therearenitelymanypossiblepi, andthereforenitelymanypossiblepi(primeidealsin OKwhichdividepiOK). Hence, therearenitelymanypossibilitiesforI=

gi=1 prii.Therefore,therearenitelymanynonzeroidealsI OKsuchthatN(I) K.54(c) Thereexistsd OK 0suchthatda, db A. ThemapK Kx dxisanisomorphismof additivegroups, therefore(da: db)=(a: b). DeneI :=daandJ:=db, soI, J AandI J ,= (0). Now(a : b) = (I: J) =(OK: J)(OK: I)=N(J)N(I)= N(I1J) = N((da)1(db)) = N(a1b). (2.104)Themap Aistransitive:Lemma2.59. LetAbeaDedekinddomainwitheldoffractionsK. LetE L Kbeatowerofeldssuchthateachstepofthetowerisaniteseparableextension. LetBbetheintegral closureofAinL,andletCbetheintegral closureofBinE. ThenAL/K AE/L= AE/K: Id(E) Id(K). (2.105)Proof. Note that B and Care Dedekind domains, by 2.43. Both AL/KAE/L and AE/Kare homomorphismsId(E) Id(K),andId(E)isthefreeabeliangroupgeneratedbythesetofnonzeroprimeidealsinC,soitsucestoprovethatifPisanonzeroprimeidealinCthen AL/K(AE/L(P)) = AE/K(P).LetPisanonzeroprimeidealinC. LetP = P B,andletp = P A. Thenp = P A,andf(P/p) =_CP:Ap_ =_CP:BP_

_BP:Ap_ = f(P/P)f(P/p), (2.106)soAL/K(AE/L(P)) = AL/K(Pf(P/P)) = AL/K(P)f(P/P)= pf(P/p)f(P/P)= pf(P/p)= AE/K(P). (2.107)ThusAL/K AE/L= AE/K: Id(E) Id(K),i.e. Aistransitive.Proposition2.60. Let AbeaDedekinddomainwitheldof fractionsK. Let L/Kbeaniteseparableextension,andletBbetheintegral closureofAinL.(a) Let(0) ,= a Aandm := [L : K]. ThenA(aB) = am. (2.108)(b) AssumethatL/KisGalois. LetPbeanonzeroprimeideal inB, andletp=P A. From2.50, weknowthattheramicationindicesofpinBareequal,soletpB= (P1 Pg)e(2.109)55betheprimefactorizationof pBinB. Moreover, 2.50alsotellsusthat theresidueclassdegreesareequal,soletfbethecommonresidueclassdegree. ThenA(P)B= (P1 Pg)ef=

Gal(L/K)P. (2.110)(c) Let0 ,= B. ThenNm()A = A(B). (2.111)(d) Thefollowingdiagramcommutes:Lb(b)Id(B)Nm__AKa(a)Id(A).(2.112)Proof.(a) (IfpisanonzeroprimeidealinA,then A(pB) = pm:) LetpbeanonzeroprimeidealinA,andprimefactorizepBaspB=

gi=1 Peii.ThenA(pB) = A_g

i=1Peii_ = p

gi=1 eifi= pm, (2.113)sincep = Pi Afori = 1, . . . , g,42andusing2.50.NowprimefactorizeaintheDedekinddomainAasa =

hi=1 prii.A(aB) = A_h

i=1(piB)ri_ =h

i=1A(piB)ri=h

i=1pmrii=_h

i=1prii_m= am. (2.114)(b) As A(P) = pf,A(P)B= (pB)f= (P1 Pg)ef. (2.115)LetG:=Gal(L/K). Inordertoshowthat(P1 Pg)ef=

GP, itsucestoshowthateachPioccurseftimesinthefamily P : G. As [G[ = mandef=mg ,itsucestoshowthateachPioccursthesamenumberoftimesinthefamily P : G.Leti, j 1, . . . , g, i ,=j. By2.52, Gactstransitivelyontheset P1, . . . , Pg. Hencethereexists GsuchthatPi=Pj. If G, thenP=Piif andonlyif P=Pj. HencethenumberoftimesPioccursinthefamily P : GisthesameasthenumberoftimesPjoccursinthefamilyP : G. SinceGisagroup,G = G,so P : G = P : Gasfamilies. HencePiandPjoccurthesamenumberoftimesinthefamily P : G. WeconcludethatA(P)B= (P1 Pg)ef=

Gal(L/K)P.42p pB Piandp A,sop Pi A. Moreover,Pi A Pi K= p(by2.48).56(c) LetEbetheGaloisclosureofL/K,43letd:=[E:L], andletG:=Gal(E/K). LetCbetheintegralclosureofBinE,andletc := C. Dene : Id(A) Id(C)a aC. (isinjective:) AssumethatI, J Id(A)andIC= J