atiyah

AdamAllanSolutionstoAtiyahMacdonald1Chapter1:RingsandIdeals1.1.Showthatthesumofanilpotentelementandaunitisaunit.Ifxisnilpotent,then1¡xisaunitwithinverseP1i=0xi.Soifuisaunitandxisnilpotent,thenv=1¡(¡u¡1x)isaunitsince¡u¡1xisnilpotent.Hence,u+x=uvisaunitaswell.1.2.LetAbearingwithf=a0+a1x+¢¢¢+anxninA[x].a.Showthatfisauniti®a0isaunitanda1;:::;anarenilpotent.Ifa1;:::;anarenilpotentinA,thena1x;:::;anxnarenilpotentinA[x].Sincethesumofnilpotentelementsisnilpotent,a1x+¢¢¢+anxnisnilpotent.Sof=a0+(a1x+¢¢¢+anxn)isaunitwhena0isaunitbyexercise1.1.NowsupposethatfisaunitinA[x]andletg=b0+b1x+¢¢¢+bmxmsatisfyfg=1.Thena0b0=1,andsoa0isaunitinA[x].Noticethatanbm=0,andsupposethat0·r·m¡1satis¯esar+1nbm¡r=arnbm¡r¡1=¢¢¢=anbm=0Noticethat0=fg=m+nXi=00@iXj=0ajbi¡j1Axi=m+nXi=0cixiwherewede¯neaj=0forjnandbj=0forjm.Thismeansthateachci=0,andso0=ar+1ncm+n¡r¡1=nXj=0ajar+1nbm+n¡r¡1¡j=ar+2nbm¡r¡1sincem+n¡r¡1¡j¸m¡rforj·n¡1.Sobyinductionam+1nb0=0.Sinceb0isaunit,weconcludethatanisnilpotent.Thismeansthatf¡anxnisaunitsinceanxnisnilpotentandfisaunit.Byinduction,a1;:::;anareallnilpotent.b.Showthatfisnilpotenti®a0;:::;anarenilpotent.Clearlyf=a0+a1x+:::+anxnisnilpotentifa0;:::;anarenilpotent.Assumefisnilpotentandthatfm=0form2N.Theninparticular(anxn)m=0,andsoanxnisnilpotent.Thus,f¡anxnisnilpotent.Byinduction,akxkisnilpotentforallk.Thismeansthata0;:::;anarenilpotent.c.Showthatfiszero-divisori®bf=0forsomeb6=0.Ifthereisb6=0forwhichbf=0,thenfisclearlyazero-divisor.Sosupposefisazero-divisorandchooseanonzerog=b0+b1x+¢¢¢+bmxmofminimaldegreeforwhichfg=0.Theninparticular,anbm=0.Sinceang¢f=0andang=anb0+¢¢¢+anbm¡1xm¡1,weconcludethatang=0byminimality.Hence,anbk=0forallk.Supposethatan¡rbk=an¡r+1bk=¢¢¢=anbk=0forallkThenasinpartaweobtaintheequation0=m+n¡r¡1Xj=0am+n¡r¡1¡jbj=an¡r¡1bm2Againweconcludethatan¡r¡1g=0.Hence,byinductionajbk=0forallj;k.Chooseksothatb=bk6=0.Thenbf=0withb6=0.d.Provethatf;gareprimitivei®fgisprimitive.LethbeanypolynomialinA[x].IfhisnotprimitivethenthereisamaximalminAcontainingthecoe±cientsofh.Letkbetheresidue¯eldofmandconsiderthenaturalmap¼:A[x]!k[x].Then¼(h)=0.Thisconditionisalsosu±cientforshowingthathisnotaprimitivepolynomial.Soiffgisnotprimitive,then¼(fg)=0asaboveforsomemaximalm.But¼(fg)=¼(f)¼(g)andk[x]isanintegraldomainsothat¼(f)=0or¼(g)=0.Inotherwords,eitherfisnotprimitiveorgisnotprimitive.Theconversefollowssimilarly.1.3.Generalizetheresultsofexercise2toA[x1;:::;xr]wherer¸2.Letf2A[x1;:::;xr].Usemulti-indexnotationtowritef=XI2Nr®IxIwherexI=xI11¢¢¢xIrrWecanalsowritef=nXi=0gxirwhereg2A[x1;:::;xr¡1]b.Showthatfisnilpotenti®each®Iisnilpotent.Supposethatfisnilpotent.Theng0;:::;gnarenilpotentpolynomialsinA[x1;:::;xr¡1]byexercise1.2.Sobyinductioneacha®isnilpotent.Ifeach®Iisnilpotenttheneach®IxIisnilpotent,sothatfisnilpotent.a.Showthatfisauniti®theconstantcoe±cientisaunitandeach®IisnilpotentforjIj0.Supposethatfisaunit.TheninA[x1;:::;xr¡1]weknowthatg0isaunitandg1;:::;gnarenilpotent.Sobypartbweseethat®IisnilpotentwheneverI(r)0.Bysymmetry®IisnilpotentwheneverjIj0.Theconstantcoe±cientisclearlyaunit.Ontheotherhand,iftheconstantcoe±cientisaunitandallothercoe±cientsarenilpotent,thenfisclearlyaunit.c.Showthatfisazero-divisori®bf=0forsomeb6=0.LetabeanyidealinA[x1;:::;xn]andsupposega=0forsomenon-zerog2A[x1;:::;xn].SinceA[x1;:::;xn]=A[x1;:::;xn¡1][xn],exercise1.2allowsustoassumethatg2A[x1;:::;xn¡1].Nowgivenf2awecanwritef=Pfixinwhereeachfi2A[x1;:::;xn¡1].LetbbethesubsetofA[x1;:::;xn¡1]consistingofallsuchfi,asfrangesacrossa.Thenbisanidealsinceaisanideal,andgb=0sinceg2A[x1;:::;xn¡1]byhypothesis.Sobyinduction,thereisb6=0satisfyingbb=0,andhenceba=0.Nowweapplythisresulttoa=(f)togetthedesiredconclusion.d.Showthatfandgareprimitivei®fgisprimitive.LethbeanypolynomialinA[x1;:::;xr].IfhisnotprimitivethenthereisamaximalminAcontainingthecoe±cientsofh.Letkbetheresidue¯eldofmandconsiderthenaturalmap¼:A[x1;:::;xr]!3k[x1;:::;xr].Then¼(h)=0ink[x1;:::;xr].Thisconditionisalsosu±cientforshowingthathisnotaprimitivepolynomial.Soiffgisnotprimitive,then¼(fg)=0asaboveforsomemaximalm.But¼(fg)=¼(f)¼(g)andk[x1;:::;xr]isanintegraldomainsothat¼(f)=0or¼(g)=0.Inotherwords,eitherfisnotprimitiveorgisnotprimitive.Theconverseisobvious.1.4.ShowthatR(A[x])=N(A[x])foreveryringA.AswithanyringN(A[x])µR(A[x]).Sosupposethatf2R(A[x]).Then1¡fxisaunit.Iff=a0+:::+anxnthismeansthat1¡a0x¡:::¡anxn+1isaunit,sothata0;:::;anarenilpotentbyexercise1.2.Byexercise1.2thismeansthatfisnilpotent,andsof2N(A[x]).HenceR(A[x])µN(A[x]),givingthedesiredresult.1.5.LetAbearingwithf=P10anxninA[[x]].a.Showthatfisauniti®a0isaunit.Supposefisaunit.Thenthereisg(x)=P10bnxnsatisfyingfg=1.Inparticular,a0b0=1,implyingthata0isaunit.Conversely,supposethata0isaunit.Wewishto¯ndbnforwhichfg=1.Thisisequivalentto¯ndingbnsatisfyinga0b0=1anda0bn+n¡1Xi=0an¡ibi=0forn0Sowede¯neb0=a¡10andbn=¡a¡10n¡1Xi=0an¡ibiforn0Thisconstructivelyshowsthatfisaunit.b.Showthateachaiisnilpotentiffisnilpotent,andthattheconverseisfalse.Supposethatfisnilpotentandchoosen0forwhichfn=0.Thenan0=0.Hencea0isnilpotent,asisf¡a0.Nowbyinductionweseethateveryanisnilpotent.Theconver