阿蒂亚交换代数导引习题解答

阿蒂亚《交换代数导引》部分习题解答

SolutionstoIntroductiontoCommutativeAlgebra

StevenVSam

ssam@mit.edu

June28,2008

Contents

1RingsandIdeals2Modules

3RingsandModulesofFractions

5IntegralDependenceandValuations

8Artinrings

9DiscretevaluationringsandDedekinddomains15679101RingsandIdeals

1.Exercise.LetxbeanilpotentelementofaringA.Showthat1+xisaunitofA.Deducethatthesumofanilpotentelementandaunitisaunit.

Solution.Letubeaunitandxbenilpotent.Supposeu+xisnotaunit.Thenu+x∈mforsomemaximalidealm.Butmisprime,sox∈m,whichimpliesu∈m.Thus,u+xisaunit.Lettingu=1givesthe rststatement.

4.Exercise.IntheringA[x],theJacobsonradicalisequaltothenilradical.

Solution.Wewanttoshowthat

mmaximalm= pprimep.

Sinceisprime,theinclusion isobvious.Ontheotherhand,choose everymaximalideal

nf∈mandwritef=anx+···+a0.Then1 fgisaunitforallg∈A[x],soby(Ex.2(i)),

1 a0bisaunitforallb∈A,whichmeansa0isnilpotent.Inparticular,ifg= 1,then1 +fisaunit,soaiisnilpotentforn≥i≥1.By(Ex.2(ii)),fisnilpotent,whichmeansf∈p,givingtheotherinclusion.

byM.F.AtiyahandI.G.MacDonald

1

阿蒂亚《交换代数导引》部分习题解答

1RINGSANDIDEALS2

6.Exercise.AringAissuchthateveryidealnotcontainedinthenilradicalcontainsanonzeroidempotent(thatis,anelementesuchthate2=e=0).ProvethatthenilradicalandJacobsonradicalofAareequal.

Solution.ThatthenilradicaliscontainedintheJacobsonradicalisobvious.Iffisnotcontainedinthenilradical,then(f)containsanonzeroidempotent,sayfg.Weclaim1 fgisnotaunit.Ifitwere,thenfrom(1 fg)(1+fg)=1 fg,weget1+fg=1,orfg=0,acontradiction.Thus,1 fgisnotaunit,whichmeansfisnotcontainedintheJacobson

radical.

7.Exercise.LetAbearinginwhicheveryelementxsatis esxn=xforsomen>1(dependingonx).ShowthateveryprimeidealinAismaximal.

Solution.Letpbeaprimeideal.Choosenonzerox∈A/p.Thenthereisann>1suchthatxn=x,orx(xn 1 1)=0.SinceA/pisanintegraldomain,xn 1=1.Ifn=2,thenx=1;otherwise,theinverseofxisxn 2.Thus,A/pisa eld,whichmeanspisamaximal

ideal.

8.Exercise.LetAbearing=0.ShowthatthesetofprimeidealsofAhasminimalelementswithrespecttoinclusion.

Solution.Let(P,≤)bethesetofprimeidealsofAwherep1≤p2ifp1 p2.GivenanychainCofP,weclaimthattheintersectionpofallprimeidealsinCisagainaprimeideal.Letxy∈p.Thenxy∈piforallpi∈C.Sinceeachpiisprime,eitherx∈piory∈pi.Eitherx∈piforallpi∈Corthereisaleastelementp thatcontainsx.Inthe rstcase,x∈p,andinthesecond,weconcludethaty∈psinceitmustbecontainedineveryprimeidealgreaterthanp .Thus,everychainhasamaximalelement,soPhasamaximalelement.Inourcase,thismaximalelementisminimalwithrespectto

inclusion.

9.Exercise.Letabeanideal=(1)inaringA.Showthata=r(a) aisanintersectionofprimeideals.

Solution.Ifa=r(a),thenaisanintersectionoftheprimeidealscontainingit.Conversely,ifaisanintersectionofprimeideals,thenitmustbetheintersectionofallprimeidealscontainingit,soa=r(a

).

10.Exercise.LetAbearing,Nitsnilradical.Showthatthefollowingareequivalent:

i)Ahasexactlyoneprimeideal;

ii)everyelementofAiseitheraunitornilpotent;

iii)A/Nisa eld.

Solution.i) ii)IfAhasexactlyoneprimeideal,thenitmustbemaximal,andisequaltoN.Ifx∈Aisnotnilpotent,thenx∈/N,so(x) N,whichmeans(x)=A,soxisaunit.ii) iii)IfpisanidealwithN p,thenpmustcontainaunit,soNismaximal,andA/Nisa eld.

iii) i)SinceNismaximalandistheintersectionofallprimeidealsofA,itmustbetheonlyprimeidealofA

.

11.Exercise.AringAisBooleanifx2=xforallx∈A.InaBooleanringA,showthat

i)2x=0forallx∈A;

ii)everyprimeidealpismaximal,andA/pisa eldwithtwoelements;

阿蒂亚《交换代数导引》部分习题解答

iii)every nitelygeneratedidealinAisprincipal.

Solution.

(a)Notethat(2x)2=2x=4x2=4x,so2x=4xleadsto2x=0.(b)Thefactthateveryprimeidealismaximalfollowsfromexercise7.InA/p,ifx=0,then

x2=x,andxisinvertible,whichgivesx=1,soA/phas2elements.

(c)Itisenoughtoshowthatifanidealisgeneratedby2elements,thenitisprincipalbyusing

induction.Thenweclaim(x,y)=(x xy+y).Theinclusion isclear,andtheidentitiesx(x xy+y)=xandy(x xy+y)=ygivetheother

inclusion.

12.Exercise.Alocalringcontainsnoidempotent=0,1.

Solution.LetmbethemaximalidealofalocalringA,andletebeanidempotent.Ife∈/m,

22theneisaunit,ande=e,whichmeanse=1.Ife∈m,then(1 e)(1+e)=1 e=1 e.

Since1 e∈/m(otherwise1∈m),1 eisaunit,soweget1+e=1,ore=

0.

15.Exercise.LetAbearingandletXbethesetofallprimeidealsofA.ForeachsubsetEof

A,letV(E)denotethesetofallprimeidealsofAwhichcontainE.Provethat

i)ifaistheidealgeneratedbyE,thenV(E)=V(a)=V(r(a)).

ii)V(0)=X,V(1)= .

iii)if(Ei)i∈IisanyfamilyofsubsetsofA,then

Ei=V(Ei).V

i∈Ii∈I

iv)V(a∩b)=V(ab)=V(a)∪V(b)foranyidealsa,bofA.

Solution.

(a)ThatV(E)=V(a)followsfromthefactthataisthesmallestidealcontainingE,somust

becontainedinanyprimeidealcontainingE.Sincer(a)istheintersectionofallprimeidealscontaininga,onegetsV(a) V(r(a)).Conversely,a r(a),sotheotherinclusionfollows.

(b)Thisisclearfromthefactthateveryidealcontains0andnoprimeidealcontains1. (c)Ifp ∈V(Ei),thenpcontainseachEi,whichmeansp∈V(Ei).Ontheotherhand,if p∈V(Ei),thenpcontainseachEi,sop∈V(Ei).

(d)Sincea∩b ab,wegettheinclusionV(ab) V(a∩b).Ifaprimeidealcontainsa∩b,

thenitmustcontaineitheraorbbyProp.1.11,sothisgivesV(a∩b) V(a)∪V(b).Finally,theinclusionsab aandab bgiveV(a) V(ab)andV(b) V(ab),soV(a)∪V(b) V(ab

).

17.Exercise.Foreachf∈A,letXfdenotethecomplementofV(f)inX=Spec(A).Thesets

Xfareopen.ShowthattheyformabasisofopensetsfortheZariskitopology,andthati)Xf∩Xg=Xfg;

ii)Xf= fisnilpotent;

iii)Xf=X fisaunit;

iv)Xf=Xg r((f))=r((g));

阿蒂亚《交换代数导引》部分习题解答

v)Xisquasi-compact(thatis,everyopencoveringofXhasa nitesubcovering).

vi)Moregenerally,eachXfisquas …… 此处隐藏：20175字，全部文档内容请下载后查看。喜欢就下载吧 ……