Commutative Algebra Problem Solutions

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1、SolutionstotheExercisesinChapter1,IntroductiontoCommutativeAlgebra,M.F.AtiyahandI.G.MacDonald1.Sincexisnilpotent,thereisannsuchthatxn=0.so2n1(1+x)(1x+x:::+(x))=1:Nowifaisaunit,then1+a1xisaunit,hencea+xisaunit.2.(i)SupposefisaunitinA[x].Letb+bx+:::+bxmbeitsinverse.01mClea

2、rlyab=1andab=0.Weshowbyinductionthatar+1b=0for00nmnmreach0rm.Sinceforeach0rmwehaveXaibj=0;i+j=n+mr1multiplyingar+1onbothsidesweobtainar+2b=0.Soam+1b=0,nnmr1n0thatisam+1=0.NowbyEx.1a+axnisaunit,sowemayrepeatthen0naboveargumentforan1andsoon,soa1;:::;anareallnilpotent.

3、TheotherdirectionisimmediatebyEx.1.(ii)Supposefm=0.Soam=0.Since1+fisaunit,by(i)wededucethat0a1;:::;anareallnilpotent.Conversely,sinceallcoecientsarenilpotent,itiseasytoseethatforsucientlylargeNwehavefN=0.(iii)Chooseag=b+bx+:::+bxmofleastdegreesuchthatfg=0.01mSinceanbm=0,wed

4、educethatang=0.Thenforthesamereasonweseethataig=0forany0in.Soeverynonzerocoecientofgannihilatesf.(iv)Supposeforcontradictionthatfgisprimitiveand,say,fisnotprimitive.LetI=(a;a;:::;a).SinceIisaproperideal,theringA=A=I01nA1isnontrivial.Fromthisitisreadilyseenthattheidealgene

5、ratedbythecoecientsoffgisnot(1)A.Sofgisnotprimitive,contradiction.Conversely,iffgisnotprimitivethenasaboveweconsidertheidealIgeneratedbythecoecientsoffg.SoagainA=Iisnottrivial.ConsidertheringA=I[x].LetfandgbetheimagesoffandginA=I[x]respectively.Sofandgarezerodivisors.So

6、by(iii)thereisanonzeroa2A=Isuchthataf=0.Thismeansthatthereisana2Asuchthata=2Iandaai2Iforevery0in.Sincefisprimitivewededucethata2I,contradiction.3.DoinductionandconsidertheringA[x1;:::;xr]=(x1;:::;xr1).4.ItisclearthatthenilradicaliscontainedintheJacobsonradical.Supposef2

7、J,then1+fisaunit,soby2(i)above1+a0isaunit,soa1;:::;anareallnilpotent.Since,foranyb2A,bf2J,wededuce1+ba0isaunit.Soa0isalsonilpotent.Soby2(ii)fisnilpotent.5.(i)Supposea0b0=1.Thenworkingforwardsforeachpowerofxoneeasilyseesthateachbncanbesolved.(ii)Supposefn=0.LetRbethenilradical

8、ofA.Thereisanobviousringhomomorphism:A[[x]]!A=R[[x]].Son(f)=0,whi