Lectures on Commutative Algebra

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LecturesonCommutativeAlgebra SudhirR.Ghorpade IndianInstituteofTechnology Bombay Annual Foundation School - II (Sponsored by the National Board for Higher Mathematics) Bhaskaracharya Pratishthana, Pune and Department of Mathematics, University of Pune June 2006©SudhirR.Ghorpade DepartmentofMathematics IndianInstituteofTechnologyBombay Powai,Mumbai400076,India E-Mail: srgmath.iitb.ac.in URL:http://www.math.iitb.ac.in/∼srg/ Version1.1,April28,2008 Original Version(1.0): June1,2006 2Contents 1 RingsandModules 4 1.1 IdealsandRadicals . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 PolynomialringsandLocalizationofrings . . . . . . . . . . 9 1.3 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.4 ZariskiTopology . . . . . . . . . . . . . . . . . . . . . . . . . 13 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2 NoetherianRings 18 2.1 NoetherianRingsandModules . . . . . . . . . . . . . . . . . 18 2.2 PrimaryDecompositionofIdeals . . . . . . . . . . . . . . . . 20 2.3 ArtinianRingsandModules . . . . . . . . . . . . . . . . . . . 24 2.4 Krull’s PrincipalIdealTheorem . . . . . . . . . . . . . . . . . 28 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 3 IntegralExtensions 33 3.1 IntegralExtensions . . . . . . . . . . . . . . . . . . . . . . . . 33 3.2 NoetherNormalization . . . . . . . . . . . . . . . . . . . . . . 36 3.3 FinitenessofIntegralClosure . . . . . . . . . . . . . . . . . . 39 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4 DedekindDomains 45 4.1 DedekindDomains . . . . . . . . . . . . . . . . . . . . . . . . 46 4.2 ExtensionsofPrimes . . . . . . . . . . . . . . . . . . . . . . . 51 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 A PrimaryDecompositionofModules 56 A.1 AssociatedPrimesofModules . . . . . . . . . . . . . . . . . . 56 A.2 PrimaryDecompositionofModules . . . . . . . . . . . . . . 59 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 References 63 3Chapter1 RingsandModules Inthis chapter,we shall reviewa numberofbasic notionsand resultscon- cerning rings and modules. First, let us settle the basic terminology and notationthatweshallusethroughoutthesenotes. By a ring we mean a commutative ring with identity. Given a ring A, ∗ × we denote byA the set of all nonzero elements ofA and byA the set of all (multiplicative) units of A. For setsI,J, we writeI ⊆ J to denote that I is a subset of J and I ⊂ J to denote that I is a proper subset of J, that is, I ⊆ J and I = 6 J. We denote the set of nonnegative integers byN and n foranyn∈N,byN wedenotethesetofalln-tuplesofelementsofN. We sometimesusetheabbreviation‘iff’ tomean‘ifandonlyif’. 1.1 IdealsandRadicals Historically, the notion of an ideal arose in an attempt to prove Fermat’s Last Theorem (see Chapter 4 for more on this). From a formal viewpoint, an ideal of a ring is analogous to a normal subgroup of a group. More precisely,anidealofaringAisasubsetI ofAsatisfying(i)I isasubgroup ofAwith respectto addition,and (ii) whenevera∈ I andx∈ A, we have ax ∈ I. If A is a ring and I is an ideal of A, then we can construct a new ring, denoted by A/I and called the residue class ring or the quotient ring obtained from “moding out” A by I. The elements of A/I are the cosets x+I :=x+a :a∈IwherexvariesoverA. Additionandmultiplication inA/I isdefinedby(x+I)+(y+I)= (x+y)+I and(x+I)(y+I)=xy+I. ThefactthatI isanidealofAensuresthatthisadditionandmultiplication is well-defined andA/I is a ring with respect to theseoperations. Passing toA/I fromAhastheeffectofmakingIthenullelement. Wehaveanatural surjective homomorphismq : A → A/I given by q(x) := x+I for x∈ A. Thekernelofq ispreciselytheidealI. Conversely,ifφ :A→B isanyring homomorphism (that is, a map of rings satisfying φ(x+y) = φ(x)+φ(y) and φ(xy) = φ(x)φ(y) for every x,y ∈ A), then the kernel of φ (which, 4by definition, is the set of all a ∈ A such that φ(a) = 0) is an ideal of A; moreover, if I = kerφ denotes the kernel of φ, then A/I is isomorphic to the image of φ. In short, residue class ring and homomorphic image are identical notions. If I is an ideal of a ring A, then there is a one-to-one correspondencebetweentheidealsofAcontainingI andtheidealsofA/I ′ −1 ′ givenbyJ 7→q(J) =J/I andJ 7→q (J ). An easy way to generate examples of ideals is to look at ideals gen- erated by a bunch of elements of the ring. Given a ring A and elements a ,...,a ∈A,theset 1 n (a ,...,a ) :=a x +···+a x :x ,...,x ∈A 1 n 1 1 n n 1 n isclearlyanidealofAanditiscalledtheidealgeneratedby a ,...,a . More 1 n generally, given any ring A and a subsetE ofA, by EA we denote the set ofall finiteA-linearcombinations ofelementsofE. Clearly,EAisanideal of A and it is called the ideal generated by E. Ideals generated by a single elementare called principal. Thus,anidealI ofaringAiscalled aprincipal ideal if I = (a) for some a ∈ A. By a principal ideal ring or PIR we mean a ring in which every ideal is principal. An integral domain which is also a PIRiscalled aprincipal ideal domain orsimply,aPID. All the basic algebraic operations are applicable to ideals of a ring. Let AbearingandletI andJ beidealsofA. ThesumofI andJ isdefinedby I +J :=a+b : a∈ I, b∈ J, whereas the product ofI and J is defined P by IJ := a b : a ∈ I, b ∈ J. Clearly, I +J and IJ are ideals of A. i i i i ItmayberemarkedthattheproductIJ iscloselyrelated,butnotquitethe same as, the idealI∩J given by theintersectionofI andJ. Forexample, ifA is a PID,I = (a) andJ = (b), thenIJ = (ab) whereasI∩J = (ℓ) and I+J = (d),whereℓ = LCM(a,b)andd = GCD(a,b). Analogueofdivision is given by the colon ideal (I : J) := a ∈ A : aJ ⊆ I. Note that (I : J) ideal of A. If J equals a principal ideal (x), then (I : J) is often denoted simply by (I : x). For example, if A is a PID, I = (a) and J = (b), then (I : J) = (a/d), where d = GCD(a,b). We can also consider the radical of √ n anidealI. It is definedby I :=a∈ A : a ∈ I forsomen≥ 1 and it is readily seento be anidealofA(by Binomial Theorem). One saysthatI is √ a radical ideal if I = I. Note that the notions of sum and intersections of ideals extend easily to arbitrary families of ideals, whereas the notion of a productofidealsextendseasilytofinitefamilies ofideals. Having defined algebraic operations for ideals, it is natural to see if the basic notions of arithmetic find an analogue in the setting of ideals. It turns out that the notion of a prime number has two distinct analogues as follows. Let A be a ring and I be an ideal of A. We say that I is prime if I = 6 A and for anya,b∈ A,, wheneverab∈ I, we have a∈ I orb∈ I. We saythatI ismaximalifforanyidealJ ofAsatisfyingI⊆J,wehaveJ =I orJ = A. The set of all prime ideals ofA is denotedby Spec(A), whereas 5thesetofallmaximalidealsofAisdenotedbyMax(A).Itiseasytoseethat I isaprimeidealifandonlyifA/I isanintegraldomain,andalsothatI is amaximal idealifandonlyifA/I isafield. Usingthis(oralternatively,by a simple direct argument), we see that every maximal ideal is prime, that is,Max(A)⊆ Spec(A). Examples1.1. (i) IfAisthezeroring,thenSpec(A) =∅ = Max(A). (ii) IfAisafield,thenSpec(A) =(0) = Max(A). (iii) If A = Z, then Spec(A) = (0)∪(p) : p isaprimenumber, and Max(A) =(p) : p isaprimenumber. If A is a ring and P is a nonunit ideal of A, that is, P is an ideal of A satisfyingP 6= A, then it is evident thatP is a prime ideal if and only ifP n satisfiesthefollowingproperty: if∩ I ⊆P foranyidealsI ,...,I ofA, j 1 n j=1 thenI ⊆ P for somej. It may be interesting to note that there is also the j following counterpart where instead of an intersectionof ideals contained inaprimeideal,wehaveanidealcontainedinaunionofprimeideals. Proposition 1.2 (Prime AvoidanceLemma). Let I, P ,...,P be ideals in a 1 n n ringAsuch thatP ,...,P are prime. IfI⊆∪ P ,thenI ⊆P for somej. 1 n j j j=1 Proof. Thecasen = 1istrivial. Supposen 1. Ifthereexistx ∈I\∪ P i j6=i j for1≤i≤n,thenwehaveacontradictionsincex +x x ...x ∈I\∪ P . 1 2 3 n i i ThusI⊆∪ P ,forsomei. Thecaseofn = 1beingtrivial, theresultnow j= 6 i j followsusinginductiononn. Remark1.3. Aneasy alteration of the above proofshows that Proposition 1.2 holdsundertheweakerhypothesisthatI is a subsetofAclosedunder additionandmultiplication,andP ,...,P areidealsofAsuchthatatleast 1 n n−2 of them are prime. If A contains a field, then Proposition 1.2 can be proved, by elementary vector space arguments, without assuming any of theP ’stobeprime. i Thenotionofcongruencemoduloanintegerhasastraightforwardana- logueforideals. IfAisaringandI isanidealofA,thenforanyx,y∈Awe say thatx≡ y(modI) ifx−y ∈ I. More interestingly,Chinese Remainder Theoremforintegershasthefollowinganalogueforideals. Proposition 1.4 (Chinese Remainder Theorem). Let I ,I ,...,I be pair- 1 2 n wise comaximal ideals inaringA(i.e.,I +I =Afor alli6=j). Then: i j (i) I I ...I =I ∩I ∩···∩I . 1 2 n 1 2 n (ii) Given any x ,...,x ∈ A, there exists x ∈ A such that x ≡ x (mod I ) 1 n j j for1≤j≤n. 6(iii) Themapx+I I ···I 7→ (x+I ,...,x+I ) defines an isomorphism of 1 2 n 1 n A/I I ...I onto thedirect sumA/I ⊕A/I ⊕···⊕A/I . 1 2 n 1 2 n Proof. (i)Clearly,I I ...I ⊆I ∩I ∩···∩I . Toprovetheotherinclusion, 1 2 n 1 2 n we induct on n. The case of n = 1 is trivial. Next, if n = 2, then we can find a ∈ I and a ∈ I such that a +a = 1. Now, a ∈ I ∩I implies 1 1 2 2 1 2 1 2 that a = aa +aa , and thus a ∈ I I . Finally, if n 2, then as in (i), let 1 2 1 2 J = I ···I and note that I +J = A. Hence by induction hypothesis 1 2 n 1 1 andthecaseoftwoideals,I ∩I ∩···∩I =I ∩J =I J =I I ···I . 1 2 n 1 1 1 1 1 2 n (ii) Given any i ∈ 1,...,n, let J = I ···I I ···I . Since I + i 1 i−1 i+1 n i I = A, we can find a ∈ I such that a ≡ 1(mod I ), for all j = 6 i. Let j ij j ij i Q a = a . Then a ≡ 1(mod I ) and a ∈ J . Thus I +J = A. Now, i ij i i i i i i j6=i x =x a +···+x a satisfiesx≡x (modI )for1≤j≤n. 1 1 n n j j (iii)Themapx+I I ···I 7→ (x+I ,...,x+I )isclearlywell-defined 1 2 n 1 n andahomomorphism. By(i),itissurjectiveandby(ii),itisinjective. Of course, not every notion concerning rings is a straightforward ana- logueofanalgebraic orarithmeticnotionapplicable to integers. Thereare some basic notions, such as those defined below, which would be quite redundantoruselessintherealmofintegers. LetA be a ring. An element a of A is said to be a zerodivisor if there is ∗ b∈ A such thatab = 0. We will denotethe set of all zerodivisors inA by Z(A). Note thatZ(A) =a ∈ A : (0 : a) = 6 0. An element of A which is not a zerodivisor is called a nonzerodivisor (sometimes abbreviated as nzd). p n Elementsof (0), that is, thoseelementsa∈ A forwhicha = 0 forsome p n ∈ N, are said to be nilpotent. The set (0) of all nilpotent elements in a ringAiscalled thenilradical ofA. IfAisanonzeroring,thenclearly every p nilpotentelementofAisazerodivisor,thatis, (0)⊆Z(A). Interestingly,each ofthe above notionsis neatly connectedwith prime ideals. To begin with, the set of all nonzerodivisors in a ring enjoys prop- ertiessimilar tocomplementsofprime ideals. Moreprecisely,it is a multi- plicatively closedsetinthefollowingsense. Definition1.5. AsubsetS ofaringAissaidtobemultiplicatively closedifit satisfiesthefollowing: (i)1∈S and(ii)ifa∈S andb∈S,thenab∈S. ∗ Examples1.6. (i) If A is an integral domain, thenA = A\0 is mul- tiplicatively closed. More generally,as remarkedearlier,thesetofall nonzerodivisorsinanyringAisamultiplicativelyclosedsubsetofA. × Also, the setA of multiplicative units in any ring A is a multiplica- tivelyclosedsubsetofA. (ii) If P is a prime ideal of a ring A, then A\P multiplicatively closed. Moregenerally,ifP :α∈ Λisafamily ofprimeidealsofaringA, α thenA\∪ P is a multiplicatively closedsubsetofA. Notethat if α∈Λ α 7I is any ideal in a ring A, thenA\I is multiplicatively closed if and onlyifIisaprimeideal. n (iii) Given any element a of a ring A, the set S = a : n ∈ N of pow- ers of a is a multiplicatively closed subset of A. In particular,1 is multiplicatively closedanditisclearlythesmallestamongthemulti- plicatively closedsubsetsofA. (iv) If a ring A is a subring of a ring B and S is a multiplicatively closed subsetofA,thenS isamultiplicatively closedsubsetofB. Lemma1.7. LetAbearing. IfI isanidealofAandS isamultiplicativelyclosed subset ofA such that I∩S =∅, then there exists a prime ideal P ofA such that I ⊆P andP ∩S =∅. MoreoverP is maximal amongthe family of idealsJ ofA satisfyingI⊆J andJ∩S =∅. Proof. Considerthe familyJ : J anidealofAwithI ⊆ J andJ∩S =∅ andZornify Corollary1.8. LetI beanonunitideal ofaringA. Thenthere isamaximalideal mofAsuchthatI⊆m. Inparticular, everynonzeroringhasamaximalidealand the spectrum ofanonzero ring isnonempty. Proof. The first assertion follows from Lemma 1.7 withS =1. To prove thesecondassertion,takeI = (0). Corollary1.9. LetAbearingandI beanyideal ofA. Then √ \ p \ I = P. Inparticular, (0) = P. P∈ Spec(A) P∈ Spec(A) I⊆P √ Proof. Clearly I ⊆ P for every prime ideal P of A containing I. On the √ other hand, if a∈ P for everyP ∈ Spec(A) withI ⊆ P, buta6∈ I, then n applyingLemma1.7toS :=a :n∈N,wearriveatacontradiction. n Ifa is a nilpotent element of a ringA thena = 0 for somen∈N, and 2 n−1 hence(1−a)(1+a+a +···+a ) = 1. Inotherwords,wehaveavalid geometricseriesexpansion 1 n−1 = 1+a+···+a , 1−a which shows that 1−a is a unit in A. In fact, a similar argument shows that if a ∈ A is nilpotent, then 1− ab is a unit in A for any b ∈ A. It p follows that (0) ⊆ a ∈ A : 1− abisaunitforeveryb ∈ A. The set a ∈ A : 1− abisaunitforeveryb ∈ A will be denoted by J(A) and calledtheJacobsonradicalofA. ThefollowingresultshowsthattheJacobson 8radical of A is an an ideal of A and it is, in fact, the intersection of all the maximal ideals of A. It also gives an alternative proof of the fact that the nilradical iscontainedintheJacobsonradical. Proposition1.10. LetAbe aring. Then \ J(A) = m m∈Max(A) Proof. Let a ∈ J(A). If a ∈/ m for some m ∈ Max(A), then m + (a) is an ideal of A with m ⊂ m + (a). Since m is maximal, we have m + (a) = A. Hencethereism∈ mandb∈Asuchthatm+ab = 1.Nowmcontainsthe unitm = 1−ab, and thisis a contradiction. Ontheotherhand,ifais inm foreverym∈ Max(A), thensoisabforeveryb∈A. Nowif 1−abisnota unit inA forsomeb∈ A, thenby Corollary 1.8, thereism∈ Max(A) such that1−ab∈m,butthen1∈m,whichisacontradiction. 1.2 PolynomialringsandLocalizationofrings Formingtheresidueclassringorthequotientringisoneofthethreefunda- mental processesin Algebra for constructing new rings from a given ring. Theothertwoprocessesare forming thepolynomial ringin oneorseveral variableswithcoefficientsinthegivenringandformingthelocalization of thegivenring. Weshallfirstreviewtheformerandthendescribethelatter. Polynomial Ring: Let A be a ring and n be a nonnegative integer. We denote by AX ,...,X the ring of all polynomials in the variables 1 n X ,...,X withcoefficientsinA. ElementsofAX ,...,X looklike 1 n 1 n X i i 1 n f = a X ...X , a ∈A, i ...i i ...i 1 n 1 n 1 n n where(i ,...,i )varyoverafinitesubsetofN . Atypicalterm(excluding 1 n i i 1 n the coefficient), viz., X ...X , is called a monomial; its (usual) degree is 1 n i +···+i . Suchamonomialissaidtobesquarefree ifi ≤ 1for1≤r≤n. 1 n r Iff 6= 0,thenthe(total)degreeoff isdefinedbydegf = maxi +···+i : 1 n a 6= 0. Usual convention is that deg0 = −∞. With this in view, for i ...i 1 n every f,g ∈ AX ,...,X , we have deg(f +g) ≤ maxdegf,degg and 1 n in case A is a domain, then degfg = degf +degg. A homogeneous polyno- mialofdegreedinAX ,...,X issimply afiniteA-linear combination of 1 n monomialsofdegreed. Thesetofall homogeneouspolynomialsofdegree d is denoted by AX ,...,X . Note that any f ∈ AX ,...,X can be 1 n 1 n d uniquelywrittenasf =f +f +...,wheref ∈AX ,...,X andf = 0 0 1 i 1 n i i fori degf;wemaycallf ’stobethehomogeneouscomponentsoff. Iff 6= 0 i andd = degf,thenclearlyf = 6 0andf =f +f +···+f . AnidealI of d 0 1 d AX ,...,X is said to be a homogeneous ideal (resp: monomial ideal) if it is 1 n 9generated by homogeneous polynomials (resp: monomials). Henceforth, when we use a notation such as kX ,...,X , it will be tacitly assumed 1 n thatk denotesafieldandX ,...,X areindependentindeterminatesover 1 n k (and,ofcourse,n∈N). The process of localization described next generalizes the construction of the field of fractions of an integral domain, which in turn, is a general- izationoftheformalconstructionofrationalnumbersfromintegers. Localization: LetAbe aring andS be amultiplicatively closed subset ofA. Definearelation∼onA×Sasfollows. Givenany(a,s),(b,t)∈A×S, (a,s)∼ (b,t) ⇐⇒ u(at−bs) = 0forsomeu∈S. It is easy to see that ∼ defines an equivalence relation on A× S. Let us a −1 denotetheequivalenceclassof(a,s)∈A×S bya/s(orby ),andletS A s denotethesetofequivalenceclassesofelementsofA×S. Defineaddition −1 andmultiplicationonS Aby     a b at+bs a b ab + = and = forany(a,s),(b,t)∈A×S. s t tb s t ts It can be easily seenthat thesebinary operationsare well definedand that −1 −1 S A is a ring with respect to them. The ring S A is called the ring of fractionsorthelocalizationofAwithrespecttomultiplicativelyclosedsubset −1 S. PassingtoS AfromAhastheeffectofmakingtheelementsofS units. −1 −1 Incase 0∈ S, we seethatS A is the zero ring,and conversely,ifS A is thezeroring,then0∈S. Examples1.11. LetAbearing. −1 (i) IfAisanintegraldomainandS =A\0, thenS Aisnothingbut the quotient field or the field of fractions ofA. Note that in this case the equivalence relation∼ onA×S takes the simpler form: (a,s)∼ (b,t) ⇐⇒ (at−bs) = 0. More generally, this is the case when all theelementsofS are nonzerodivisors. Ingeneral,ifS isthesetofall −1 nonzerodivisorsinA,thenS Aiscalled thetotal quotientringofA. −1 (ii) Let S = A\ p, where p is a prime ideal of A. In this case S A is customarily denoted by A . The set pA := a/s : a ∈ p, s ∈ S is p p an ideal ofA and an element ofA that is not inpA is a unit inA . p p p p It follows that pA is the only maximal ideal of the ring A . In other p p wordsA is a local ring A local ring is a ring withonly one maximal p ideal. Ingeneral,wehavethenaturalhomomorphism a −1 φ :A→S A definedby φ(a) := fora∈A. 1 10Themapφisnotinjective,ingeneral,anditskernelisgivenby kerφ =a∈A :sa = 0forsomes∈S = (0 :s). s∈S In particular, φ is injective if S consists of nonzerodivisors; in this case A −1 may be regarded as a subring of S A. Given an ideal a of A, the ideal −1 of S A generated by φ(a) is called the extension of a, and is denoted by −1 −1 −1 −1 aS A or by S a. For an ideal b ofS A, the inverse image φ (b) is an ideal of A and is called the contraction of b to A. By abuse of language, the contraction of b is sometimes denoted by b∩ A. Basic properties of extensionandcontractionaredescribedinthefollowingresult. Proposition1.12. LetAbearing andS beamultiplicatively closed subset ofA. −1 Given anyidealaofAandanidealbofS A, wehave thefollowing: −1 −1 (i) If a = b∩ A, then b = S a = S (b∩A). In particular, b is the extension of someideal ofA. −1 −1 −1 (ii) S a∩A = (a :s). In particular, S a =S A⇔a∩S = 6 ∅. s∈S −1 (iii) a is a contraction of an ideal of S A if and only if every element of S is a −1 nonzerodivisor inA/a; inthis case a =S a∩A. −1 (iv) The prime ideals of S A are in one-to-one correspondence with the prime ideals ofAwhich donotmeetS. −1 −1 Proof. (i) Sincea = b∩A = φ (b),wehaveφ(a)⊆ bandhenceS a⊆ b. On the other hand, if a ∈ A and s ∈ S are such that a/s ∈ b, then a/1 = −1 (s/1)(a/s) ∈ b, and hence a∈ b∩A = a, and this shows thata/s∈ S a. −1 Thus,b =S a. −1 (ii) Given anyx∈ A∩S a, we havex/1 = a/t forsomea∈ a andt∈ S, and sou(tx−a) = 0 for someu∈ S. Nows = ut∈ S andsx∈ a, that is, x∈ (a :s). Ontheotherhand,ifx∈ (a :s)forsomes∈S,thensx∈a,and −1 −1 −1 sox/1 = (1/s)(sx/1)∈S a,thatis,x∈A∩S a. ThisshowsthatS a∩A −1 −1 istheunionof(a : s)assvaries overS. Inparticular, S a = S Aif and onlyif1∈ (a :s)forsomes∈S,thatis,a∩S = 6 ∅. (iii) Observethats∈S isanonzerodivisorinA/a ifandonlyif(a : s) = a. Withthisinview,(iii) isanimmediateconsequenceof(ii). −1 −1 (iv) If q is a prime ideal of S A, then q∩A = φ (q) is a prime ideal of A, being the inverse image of a prime ideal under a ring homomorphism. −1 Moreover,sinceq6=S A,weseefrom(i)abovethatq∩Aisdisjointfrom S. On the otherhand, supposep is a prime ideal ofA such thatp∩S =∅. −1 −1 Thenby (ii) above,S p6= S A. Further,ifx,y ∈ p ands,t∈ S are such 11−1 that (x/s)(y/t)∈ S p, thenuxy ∈ p for someu∈ S. Sincep is prime and −1 p∩S =∅, it follows thatx∈ p ory ∈ p, which implies thatx/s∈ S p or −1 −1 −1 y/t∈ S p. ThusS p is aprime ideal ofS A. So,in view of(i) and (iii) above, it follows thatthe processesofcontractionand extensionsetup the desiredone-to-onecorrespondence. 1.3 Modules Let A be a ring. An A-module is simply a vector space except that the scalars come from the ring A instead of a field. Apart from vector spaces over a field, basic examples of A-modules are: ideals I of A, residue class −1 rings A/I, polynomial rings AX ,...,X and localizations S A. The 1 n notionsofsubmodules,quotientmodules,directsumsofmodulesandiso- morphism of modules are defined in an obvious fashion. The concept of localization (w.r.t. multiplicatively closed subsets of A) also carries over toA-modules. A direct sum of (isomorphic) copies of A is called a free A- module. Thefinitedirectsum n A :=A⊕···⊕A z n times isreferredtoasthefreeA-moduleofrankn. LetM beanA-module. GivensubmodulesMofM,theirsum i X X M := x :x ∈M andallexceptfinitelymanyx ’sare0 i i i i i is a submodule of M. Products of submodules do not make sense but of course, intersections of submodules of M does. Further, the colon opera- tion has an interesting and important counterpart. If M and M are sub- 1 2 modulesofM,wedefine (M :M ) :=a∈A :aM ⊆M . 1 2 2 1 Note that (M : M ) is an ideal of A. The ideal (0 : M) is called the anni- 1 2 hilator ofM and is denoted by Ann(M); forx∈ M, we may write Ann(x) for the ideal (0 : x), i.e., for Ann(Ax). Note that if I is an ideal of A, then Ann(A/I) = I and if Ann(M) ⊇ I, then M may be regarded as an A/I- module. LetusalsonotethatforanysubmodulesM ,M ofM,wealways 1 2 havetheisomorphisms(M +M )/M ≃M /(M ∩M ),and,ifM ⊆M 1 2 2 1 1 2 2 1 andN isasubmoduleofM ,then(M /N)/(M /N)≃M /M . 2 1 2 1 2 We say that M is finitely generated (f.g.) or that M is a finite A-module if there exist x ,...,x ∈ M such that M = Ax +··· +Ax ; in this case 1 n 1 n x ,...,x isreferredtoasasetofgeneratorsofM. NotethatanA-module 1 r n M is finitely generated if and only if M is isomorphic to A /K for some n n∈NandasubmoduleK ofthefreeA-moduleA . 12Ingeneral,andunlikeinthecaseofvectorspaces,ifanA-moduleM is finitely generated, then a submodule ofM need not be finitely generated. Forexample,ifA =CX ,X ,...isthepolynomialringininfinitelymany 1 2 variables and I = (X ,X ,...) is the ideal of A generated by all the vari- 1 2 ables,thenA is finitelygeneratedas anA-module,but theA-submoduleI ofA is not finitely generated. We shall see in Chapter 2 that if every ideal of A is finitely generated, then every submodule of a finitely generated A-module is finitely generated. Meanwhile let us note here a very useful propertyoffinitelygeneratedA-modules. Lemma 1.13 (Nakayama’s Lemma). Let M be a finitely generated A-module andI be an ideal ofA such thatIM = M. Then (1−a)M = 0 for somea∈ I. Inparticular, ifI 6=AandifAisalocal ring, thenM = 0. P n Proof. WriteM =Ax +···+Ax . Thenx = a x ,forsomea ∈I. 1 n i ij j ij j=1 Letd = det(δ −a ). Then d = 1−a, for some a ∈ I, and, by Cramer’s ij ij rule,dx = 0forallj. j 1.4 ZariskiTopology LetAbearing. GivenanysubsetE ofA,wedefine, V(E) :=p∈ Spec(A) : E⊆p. In case E = f for some f ∈ A, then V(E) may be denoted simply by V(f). Note that ifI is the ideal ofA generatedbyE, thenV(E) = V(I) = √ V( I). Inparticular, V(E) :E⊆A =V(I) :I anidealofA =V(J) :J aradicalidealofA. Thefollowingfactsfollowdirectlyfromdefinition. 1. V(0) = Spec(A)andV(1) =∅. \ 2. V(E ) =V( E ),foranyfamilyE :α∈ ΛofsubsetsofA. α α α α∈Λ α∈Λ m m \ 3. V(E ) =V E ,foranym∈NandsubsetsE ,...E ofA. i i 1 m i=1 i=1 Thus the familyV(E) : E ⊆ A satisfies all the axioms of closed sets for a topology on Spec(A). The resulting topology on Spec(A) is called the Zariski topology. For example, if A = CX, then using the Fundamental Theoremof Algebra we seethat Spec(CX) =(X−α) : α∈C∪(0); b thus Spec(A) can be identified with the Riemann sphere C := C∪∞ 1 (or the projective lineP ). The Zariski topology on Spec(CX) coincides C 1 b with the cofinite topology on C. Reverting to the general case, for any 1 ThecofinitetopologyonanysetX isdefinedasfollows. AsubsetU ofX isopenifand onlyifU = ∅orU =X \F whereF isafinitesubsetofX. 13V ⊆ Spec(A),wedefine \ I(V) :=f ∈A :f ∈P forallP ∈V = P. P∈V Clearly,I(V)isanidealofA(infact,aradicalideal);itiscalledthevanishing ideal ofV. Forexample,ifA =CXandaisanonzeroidealofA,thena = (h(X)) for some nonzero polynomial h(X) ∈ CX. Further, if α ,...,α 1 m are the distinct roots of h(X) in C, then V = V (a) = α ,...,α and 1 m I(V) =f ∈CX : f(α ) = 0fori = 1,...m is the ideal of polynomials i whichvanishattherootsofh(X). The following result may be viewed as a version of Hilbert’s Nullstel- lensatz(seeCorollary3.19inChapter3). However,heretheproofistrivial. Weshalluseherethefollowingnotationandterminology. Proposition1.14. LetAbearing. (i) Ifaisanonunitideal ofA, thenV(a) is nonempty. √ (ii) Ifaisan ideal ofAandV =V(a) thenI(V) = a. Proof. Corollary1.8implies(i)andCorollary1.9implies(ii). The Zariski topology has a number of interesting properties. To begin with,letusnotethattheprincipal open sets D := Spec(A)\V(f) =p∈ Spec(A) :f ∈/ p, where f ∈A, f forma base forthethe ZariskitopologyonSpec(A). TheZariskitopology is almost never Hausdorff (or T ). In fact, as part (ii) of the proposition 2 below shows,a singleton set can be dense. On the other hand, part (iv) of thesamepropositionshowsthatSpec(A)withtheZariskitopologyisaT 0 topologicalspace. Proposition1.15. LetAbearingandp,q∈ Spec(A). Wehave the following: (i) p∈ Max(A) ⇐⇒ p isclosed subsetofSpec(A). (ii) p := closure ofp = V(p). Consequently, q∈p ⇐⇒ p⊆ q. In particular, ifAisan integral domain, then(0) = Spec(A). (iii) Ifp = 6 q,thenthere existsaneighbourhood ofpwhichdoesnotintersectqor a neighbourhood ofqwhich does notintersectp. Proof. (i) If p ∈ MaxA, thenp = V(p), and hence p is closed. Con- versely,if p∈ Spec(A)issuchthatpisaclosed,thenp =V(a)forsome ideal a ofA. Clearly, a 6= A and so by Corollary 1.8, there is m∈ Max(A) suchthata⊆m. Hencem∈V(a) =p andsop =mismaximal. (ii)IfV isaclosedsubsetofSpec(A)containingp,thenV =V(a)forsome 14ideal a ofA. Consequentlya ⊆ p, and hence V(p) ⊆ V(a). It follows that V(p)isthesmallestclosedsetcontainingp,andthusp =V(p). Incase A is a domain, (0) is a prime ideal and we have V((0)) = Spec(A), that is,(0) = Spec(A). (iii) If p = 6 q, then p q or q p. Supposep q. Then there is f ∈ p\q; nowD isanopensetthatcontainsqbutdoesnotcontainp. f RecallthatatopologicalspaceX is quasi-compact ifeveryopencoverof X hasafinitesubcover;X is compactifitisquasi-compactandHausdorff. Proposition1.16. Spec(A) isquasi-compact inthe Zariski topology. Proof. Given an open cover of Spec(A), we can refine it to an open cover of Spec(A) consisting of principal open sets. This implies that Spec(A) is the union of D where f vary over a subset Λ of A. Consequently, the f intersection of V(f) as f varies over Λ is empty, and so V(a) = ∅, where a is the ideal of A generated by Λ. By part (ii) of Proposition 1.14, we see thata =A,andhencetherearef ,...,f ∈ Λandg ,...,g ∈Asuchthat 1 m 1 m f g +···+f g = 1. ThisimpliesthatD :i = 1,...,mcoverSpec(A), 1 1 m m f i andhencethegivenopencoverofSpec(A)hasafinitesubcover. Exercises ThroughoutthefollowingexercisesAdenotesaring. 1. Letφ :A→Bbeahomomorphismofrings. IfJ isanidealofB,then −1 showthatφ (J)isanidealofA. Further,showthatJ ∈ Spec(B)im- −1 −1 pliesφ (J)∈ Spec(A). Isit truethatJ ∈ Max(B)impliesφ (J)∈ Max(A)? Also,isittruethatifI isanidealofA,thenφ(I)isanideal of B? What if φ is surjective? Further, if φ is surjective, then is it true thatI ∈ Spec(A) impliesφ(I)∈ Spec(B), and thatI ∈ Max(A) impliesφ(I)∈ Max(B)? Justifyyouranswers. 2. LetI beanidealofAandq :A→A/I bethenaturalhomomorphism givenbyx7→x+I. ShowthatJ 7→q(J)definesabijectivemapfrom theidealsofAcontainingI andtheidealsofA/I. Further,showthat thisbijectionpreservesinclusions,primalityandmaximality. 3. Assume that A is a PID. Given any a,b ∈ A, let d = GCD(a,b) and ℓ = LCM(a,b). IfI = (a)andJ = (b),thenshowthat IJ = (ab), I∩J = (ℓ), I +J = (d), and (I :J) = (a/d). Are these results valid if A is an arbitrary ring. What if A is a UFD? Justifyyouranswer. 154. Consideridealsa,b,cofAandthefollowingthreeequalities. ab =a∩b, (a+b)(a∩b) =ab, a∩(b+c) = (a∩b)+(a∩c). In each case, determine if the equality is valid for arbitrary a,b,c. If yes, then give a proof; otherwise give a counterexample. Also, if the answer is no, then determine if either of the inclusions ⊆ and ⊇ is valid,ingeneral. √ 5. IfI isanidealofA,thenshowthat I isanidealofA. 6. Showthatcolonscommutewithintersections,whereasradicalscom- mute with finite intersections. More precisely, if I : α ∈ Λ is a α family ofidealsofaringAandJ isanyidealofA,thenshowthat s   \ \ \ \ p (I :J) = I :J andifΛisfinite,then I = I . i i i i i∈Λ i∈Λ i∈Λ i∈Λ Give examplestoshowthattheseresultsdonothold(forfinitefami- lies)ifintersectionsarereplacedbyproducts. 7. WithI ,...,I andAasinProposition1.4,showthatthemapinpart 1 n × (iii) of Proposition 1.4 induces an isomorphism of (A/I I ...I ) 1 2 n × × × ontothedirectsum(A/I ) ⊕(A/I ) ⊕···⊕(A/I ) . Deducethat 1 2 n theEulerφ-functionismultiplicative. 8. SupposeAis notthezeroring and letN be thenilradical ofA.Show thatthefollowingareequivalent. (i) Ahasexactlyoneprimeideal. (ii) EveryelementofAiseithernilpotentoraunit. (iii) A/N isafield.  n+d−1 9. Letk beafield. Showthat dim kX ,...,X = . 1 n k d d n 10. Letf(X) =a +a X +...+a X ∈AX.Provethefollowing: 0 1 n (i) f is aunitinAXifand onlyifa isunit inAanda ,a ,...,a 0 1 2 n arenilpotentinA. (ii) f is a nilpotent inAX if and only ifa ,a ,...,a are nilpotent 0 1 n inA. ∗ (iii) f is azerodivisorinAX ifandonly ifthereexistsa∈A such thataf = 0. 11. LetSbeanymultiplicativelyclosedsubsetofA. Considertherelation onA×S definedby(a,s)∼ (b,t) ⇐⇒ (at−bs) = 0. Determineif∼ isanequivalencerelation. 16n −1 12. Givenanyf ∈A,letS =f :n∈NandA =S A. ShowthatA f f isisomorphictoAX/(Xf −1). 13. LetS and T be multiplicatively closed subsets of A with S ⊆ T and −1 let U denote the image of T under the natural map φ : A → S A. −1 −1 −1 ShowthatT AisisomorphictoU (S A). 14. Showthat localization commutes withtaking homomorphic images. More precisely, if I is an ideal of a ring A and S is a multiplica- −1 −1 tively closed subset of A, then show that S A/S I is isomorphic −1 toS (A/I),whereS denotestheimageofS inA/I. 15. LetAbeanintegraldomain. FixaquotientfieldK ofAandconsider thelocalizationA ,wherep∈ Spec(A),assubringsofK. Showthat p \ \ A = A = A . p m p∈Spec(A) m∈Max(A) 16. Consider the following ring-theoretic properties that A can have: (i) integraldomain,(ii) field,(iii) PIR,(iv) PID,and(v)UFD.Foreachof these,determineifthepropertyis preservedunderthepassagefrom Atoa(i)residueclassring,(ii)polynomialring,or(iii) localization. 17. Let M be an A-module and S be a multiplicatively closed subset of −1 A. Define carefully the localization S M ofM atS. With ideals re- placedbyA-submodules,determinewhichofthenotionsandresults inSection1.2hasananalogoueinthesettingofmodules. 18. Let (A,m) be a local ring which means that A is a local ring and m isitsuniquemaximalidealandM beafinitelygeneratedA-module. Forx∈ M, letx denotesthe image ofx in theA/m-moduleM/mM. Given any x ,...,x ∈ M, show that x ,...,x is a minimal set 1 r 1 r of generators of M if and only ifx ,...,x is a basis for the A/m- 1 r vectorspaceM/mM. Deducethatanytwominimal setofgenerators ofM havethesamecardinality,namely,dim M/mM. A/m 19. Assume thatA is not the zero ring and letm,n∈N. Use Exercise 18 m n toshowthatA andA areisomorphicasA-modulesiffm =n. 20. Given any f,g ∈ A, show that the principal open sets D and D of f g SpecAsatisfythefollowing. (i)D =∅⇐⇒f isnilpotent, (ii)D = Spec(A)⇐⇒f isaunit, f f p p (iii)D ∩D =D , and (iv)D =D ⇐⇒ (f) = (g). f g fg f g 21. Given any f ∈ A, show that the principal open set D is quasi- f compact. Further show that an open subset of Spec(A) is quasi- compactifandonlyifitisafiniteunionofprincipalopensets. 17Chapter2 NoetherianRings In this chapter we shall study a class of rings and modules named after E. Noether (1921), who first realized their importance. Basic results proved in this chapter include the Basis Theorem that goes back to the work of Hilbert (1890) onCX ,...,X and the Primary Decomposition Theorem 1 n foridealsthatgoesbacktotheworkofLasker(1905)alsoonCX ,...,X . 1 n 2.1 NoetherianRingsandModules Proposition2.1. LetAbe aring. Thefollowing conditions are equivalent. (i) (Finite Generation Condition)Every ideal ofAisfinitely generated. (ii) (Ascending Chain Condition or the a.c.c.) If I ,I ,... are ideals of A 1 2 withI ⊆I ⊆..., then there existsm≥ 1suchthatI =I forn≥m. 1 2 n m (iii) (Maximality Condition) Every nonempty set of ideals of A has a maximal element. Proof. (i) ⇒ (ii): Given a chain I ⊆ I ⊆ ... of ideals of A, the union 1 2 I =∪ I is an ideal ofA. Thus by (i), there are a ,...,a ∈ I such that n≥1 n 1 r I = (a ,...,a ). Now,ifm∈Nissuchthata ∈I fori = 1,...,r,thenwe 1 r i m clearly haveI =I forn≥m. n m (ii)⇒ (iii): LetF be a nonemptyset of ideals ofA that doesnot have a maximal element. SinceF isnonempty,thereissomeI ∈F. Moreover,I 1 1 isnotamaximal element,andsothereisI ∈F suchthatI ⊂I . Further, 2 1 2 I is not a maximal element, and so there is I ∈ F such that I ⊂ I . 2 3 2 3 Continuing in this way, we obtain a strictly ascending chainI ⊂ I ⊂··· 1 2 ofidealsinF. Thiscontradicts(ii). (iii) ⇒ (i): Let I be an ideal of A and consider the set F of ideals J ofA such thatJ ⊆ I andJ is finitely generated. By (iii),F has a maximal element,saya. Ifthereissomex∈I\a,thena+Ax∈F,whichcontradicts themaximality ofa. ThusI =a∈F,andhenceI isfinitelygenerated. 18Aringwhichsatisfieseither(andhenceall)ofthethreeequivalentcon- ditions in Proposition 2.1 is said to be noetherian. The class of noetherian ringshasaremarkablepropertythatitisclosedw.r.t. eachofthethreefun- damental processes mentioned in Section 1.2. Indeed, if A is a noetherian ring, thenit is trivial to check thatA/I is noetherianfor every idealI ofA −1 and also that S A is noetherian, for every multiplicatively closed subset S ofA. Moreover, the following basic result implies, using induction, that ifAisnoetherian,thenAX ,...,X isnoetherian. 1 n Proposition2.2(HilbertBasisTheorem). IfAisnoetherian, then soisAX. Proof. Let I be any ideal of AX. For 0 = 6 f ∈ I, let LC(f) denote the leading coefficient of f, and letJ :=0∪LC(f) : f ∈ I, f = 6 0. Then J is an ideal of A, and so we can find f ,...,f ∈ I \0 such that J = 1 r (LC(f ),...,LC(f )). Letd = maxdegf : 1 ≤ i ≤ r. For 0 ≤ i d, let 1 r i J =0∪LC(f) : f ∈ I, degf = i; thenJ is an ideal ofA, and so we i i ′ can findf ,...,f ∈ I\0 such thatJ = (LC(f ),...,LC(f )). LetI i1 ir i i1 ir i i betheidealofAXgeneratedbyf ,...,f ∪f : 0≤id, 1≤j≤r. 1 r ij i ′ ′ ′ Clearly,I ⊆I andforany0 = 6 f ∈I,weeasilyseethatthereisf ∈I such ′ ′ thatdeg(f−f ) degf. ThusaninductiveargumentyieldsI =I . AfieldaswellasaPID(e.g.,Z,theringofintegers)isclearlynoetherian. Startingfromtheseandusingiterationsofthethreefundamentalprocesses, we obtain an abundant supply of noetherian rings. Especially important amongthesearefinitelygeneratedalgebrasoverafieldor,moregenerally, overanoetherianring. Letusrecalltherelevantdefinitions. Definition 2.3. Let B be a ring and let A be a subring of B. Given any b ,...,b ∈B,wedenotebyAb ,...,b thesmallestsubringofBcontain- 1 n 1 n ing A and the elements b ,...,b . This subring consists of all polynomial 1 n expressions f(b ,...,b ) as f varies over AX ,...,X . We say that B is 1 n 1 n a finitely generated (f.g.) A-algebra or an A-algebra of finite type if there exist b ,...,b ∈ B such that B = Ab ,...,b . Finitely generated k-algebras, 1 n 1 n wherek isafield,aresometimescalled affinerings. NotethataringB isaf.g. A-algebraifandonlyifB≃AX ,...,X /I 1 n for some n ∈N and some ideal I of AX ,...,X . Hence it follows from 1 n the Hilbert basis theorem that finitely generated algebras over noetherian ringsarenoetherian. Inparticular,everyaffine ringisnoetherian. Now let us turn to modules. In the remainder of this section, we fix a ringA and considerA-modules. By submodules of a givenA-module, we alwaysmeanA-submodules. To begin with, let us note that we have the following straightforward analogueofProposition2.1. Proposition2.4. LetM beanA-module. Thefollowingconditionsareequivalent. 19(i) (Finite Generation Condition)EverysubmoduleofM isfinitelygenerated. (ii) (a.c.c.) If N ,N ,... are submodules of M with N ⊆ N ⊆ ..., then 1 2 1 2 there existsm≥ 1suchthatN =N forn≥m. n m (iii) (Maximality Condition) Every nonempty set of submodules of M has a maximal element. Proof. The proof of Proposition 2.1 carries over verbatim with ideals re- placedbyA-submodules. We define a module to be noetherian if it satisfies any (and hence all) of the equivalent conditions in Proposition 2.4. Notice that if A is any ring, thenA is noetherian as an A-module if and only if A is a noetherian ring. Thus,theterminologyformodulesisconsistentwiththatforrings. Proposition2.5. LetM beanA-module andN beasubmodule ofM. Then M is noetherian ⇐⇒ bothN andM/N are noetherian. Proof. The a.c.c. on M clearly implies the a.c.c. on N. Moreover, an as- cendingchainofA-submodulesofM/N givesrisetoanascendingchainof A-submodulesofM. Thus,ifM isnoetherian,thensoisN andM/N. Con- versely, if both N and M/N are noetherian, then for any ascending chain M ⊆ M ⊆ ... of A-submodules of M, we can find p,q ∈ N such that 0 1 M ∩N = M ∩N fori≥ p and (M +N)/N ≃ (M +N)/N fori≥ q. If i p i q m = maxp,q, thenforanyi≥ m,wehaveM ⊆ M . Further,ifx∈ M , m i i thenx = y +n forsomey∈ M andn∈ N; nown∈ M ∩N = M ∩N, m i m andhencex∈M . ThusM =M fori≥m. SoM isnoetherian. m i m n Corollary 2.6. IfM ,...,M are noetherian A-modules, then so is⊕ M . In 1 n i i=1 n particular, ifAis anoetherian ring, thenA isanoetherianA-module. n n−1 Proof. Given any n ≥ 1, letM =⊕ M and N =⊕ M . Observe that i i i=1 i=1 M ≃M/N anduseProposition2.5togetherwithinductiononn. n Proposition2.7. IfAisnoetherian andM isafinitelygeneratedA-module,then M isnoetherian. n Proof. Since M is finitely generated, M ≃ A /N for some n ∈ N and a n submoduleN ofA . NowapplyCorollary 2.6andProposition2.5. 2.2 PrimaryDecompositionofIdeals Ideals in noetherianrings admit a decompositionwhich is somewhatsim- ilar to, though much cruder than, the decomposition of positive integers intoprime–powers. 20

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