1 A Course In Commutative Algebra Robert B. Ash Preface This is a text for a basic course in commutative algebra, written in accordance with the following objectives. The course should be accessible to those who have studied algebra at the beginning graduate level. For general algebraic background, see my online text “Abstract Algebra: The Basic Graduate Year”, which can be downloaded from my web site www.math.uiuc.edu/∼ r-ash This text will be referred to as TBGY. The idea is to help the student reach an advanced level as quickly and efficiently as possible. InChapter1,thetheoryofprimarydecompositionisdevelopedsoastoapplyto modulesaswellasideals. InChapter2,integralextensionsaretreatedindetail,including the lying over, going up and going down theorems. The proof of the going down theorem does not require advanced field theory. Valuation rings are studied in Chapter 3, and the characterization theorem for discrete valuation rings is proved. Chapter 4 discusses completion,andcoverstheArtin-ReeslemmaandtheKrullintersectiontheorem. Chapter 5 begins with a brief digression into the calculus of finite differences, which clarifies some of the manipulations involving Hilbert and Hilbert-Samuel polynomials. The main result is the dimension theorem for finitely generated modules over Noetherian local rings. A corollary is Krull’s principal ideal theorem. Some connections with algebraic geometry are established via the study of affine algebras. Chapter 6 introduces the fundamental notions of depth, systems of parameters, and M-sequences. Chapter 7 develops enough homologicalalgebratoprove,underappropratehypotheses,thatallmaximalM-sequences havethesamelength. ThebriefChapter8developsenoughtheorytoprovethataregular local ring is an integral domain as well as a Cohen-Macaulay ring. After completing the course, the student should be equipped to meet the Koszul complex, the Auslander- Buchsbaumtheorems,andfurtherpropertiesofCohen-Macaulayringsinamoreadvanced course. Bibliography Atiyah,M.F.andMacdonald,I.G.,IntroductiontoCommtativeAlgebra,Addison-Wesley 1969 Balcerzyk, S. and Jozefiak, T., Commutative Noetherian and Krull Rings, Wiley 1989 Balcerzyk, S. and Jozefiak, T., Commutative Rings: Dimension, Multiplicity and Homo- logical Methods, Wiley 1989 Eisenbud, D., Commutative Algebra with a view toward algebraic geometry, Springer- Verlag 1995 Gopalakrishnan, N.S., Commutatilve Algebra, Oxonian Press (New Delhi) 1984 Kaplansky, I., Commutative Rings, Allyn and Bacon 1970 2 Kunz, E., Introduction to Commutative Algebra and Algebraic Geometry, Birkha¨user 1985 Matsumura, H., Commutatlive Ring Theory, Cambridge 1986 Raghavan, S., Singh, B., and Sridharan, S., Homological Methods in Commutative Alge- bra, Oxford 1975 Serre, J-P., Local Albegra, Springer-Verlag 2000 Sharp, R.Y., Steps in Commutative Algebra, Cambridge 2000 (cid:1)ccopyright 2003, by Robert B. Ash. Paper or electronic copies for noncommercial use maybemadefreelywithoutexplicitpermissionoftheauthor. Allotherrightsarereserved. Table of Contents Chapter 0 Ring Theory Background 0.1 Prime Avoidance 0.2 Jacobson Radicals, Local Rings, and Other Miscellaneous Results 0.3 Nakayama’s Lemma Chapter 1 Primary Decomposition and Associated Primes 1.1 Primary Submodules and Ideals 1.2 Primary Decomposition 1.3 Associated Primes 1.4 Associated Primes and Localization 1.5 The Support of a Module 1.6 Artinian Rings Chapter 2 Integral Extensions 2.1 Integral Elements 2.2 Integrality and Localization 2.3 Going Down Chapter 3 Valuation Rings 3.1 Extension Theorems 3.2 Properties of Valuation Rings 3.3 Discrete Valuation Rings Chapter 4 Completion 4.1 Graded Rings and Modules 4.2 Completion of a Module 4.3 The Krull Intersection Theorem 1 2 Chapter 5 Dimension Theory 5.1 The Calculus of Finite Differences 5.2 Hilbert and Hilbert-Samuel Polynomials 5.3 The Dimension Theorem 5.4 Consequences of the Dimension Theorem 5.5 Strengthening of Noether’s Normalization Lemma 5.6 Properties of Affine k-Algebras Chapter 6 Depth 6.1 Systems of Parameters 6.2 Regular Sequences Chapter 7 Homological Methods 7.1 Homological Dimension: Projective and Global 7.2 Injective Dimension 7.3 Tor and Dimension 7.4 Application Chapter 8 Regular Local Rings 8.1 Basic Definitions and Examples Exercises Solutions Chapter 0 Ring Theory Background We collect here some useful results that might not be covered in a basic graduate algebra course. 0.1 Prime Avoidance Let P ,P ,... ,P , s ≥ 2, be ideals in a ring R, with P and P not necessarily prime, 1 2 s 1 2 but P ,... ,P prime (if s≥3). Let I be any ideal of R. The idea is that if we can avoid 3 s the P individually, in other words, for each j we can find an element in I but not in P , j j thenwecanavoidalltheP simultaneously,thatis,wecanfindasingleelementinI that j is in none of the P . We will state and prove the contrapositive. j 0.1.1 Prime Avoidance Lemma With I and the P as above, if I ⊆∪s P , then for some i we have I ⊆P . i i=1 i i Proof. Suppose the result is false. We may assume that I is not contained in the union of any collection of s−1 of the P ’s. (If so, we can simply replace s by s−1.) Thus i for each i we can find an element ai ∈ I with ai ∈/ P1∪···∪Pi−1∪Pi+1∪···∪Ps. By hypothesis, I iscontainedintheunionofalltheP’s, soa ∈P . Firstassumes=2, with i i I (cid:7)⊆P and I (cid:7)⊆P . Then a ∈P , a ∈/ P , so a +a ∈/ P . Similarly, a ∈/ P , a ∈P , 1 2 1 1 2 1 1 2 1 1 2 2 2 so a +a ∈/ P . Thus a +a ∈/ I ⊆ P ∪P , contradicting a ,a ∈ I. Note that P 1 2 2 1 2 1 2 1 2 1 and P need not be prime for this argument to work. Now assume s > 2, and observe 2 that a1a2···as−1 ∈P1∩···∩Ps−1, but as ∈/ P1∪···∪Ps−1. Let a=(a1···as−1)+as, which does not belong to P1 ∪···∪Ps−1, else as would belong to this set. Now for all i=1,... ,s−1wehaveai ∈/ Ps,hencea1···as−1 ∈/ Ps becausePs isprime. Butas ∈Ps, so a cannot be in P . Thus a∈I and a∈/ P ∪···∪P , contradicting the hypothesis. ♣ s 1 s ItmayappearthatweonlyusedtheprimenessofP , butafterthepreliminaryreduc- s tion (see the beginning of the proof), it may very well happen that one of the other P ’s i now occupies the slot that previously housed P . s 1 2 CHAPTER 0. RING THEORY BACKGROUND 0.2 Jacobson Radicals, Local Rings, and Other Mis- cellaneous Results 0.2.1 Lemma Let J(R) be the Jacobson radical of the ring R, that is, the intersection of all maximal ideals of R. Then a∈J(R) iff 1+ax is a unit for every x∈R. Proof. Assume a∈J(R). If 1+ax is not a unit, then it generates a proper ideal, hence 1+axbelongstosomemaximalidealM. Butthena∈M,henceax∈M,andtherefore 1 ∈ M, a contradiction. Conversely, if a fails to belong to a maximal ideal M, then M+Ra = R. Thus for some b ∈ M and y ∈ R we have b+ay = 1. If x = −y, then 1+ax=b∈M, so 1+ax cannot be a unit (else 1∈M). ♣ 0.2.2 Lemma Let M be a maximal ideal of the ring R. Then R is a local ring (a ring with a unique maximal ideal, necessarily M) if and only if every element of 1+M is a unit. Proof. Suppose R is a local ring, and let a ∈ M. If 1+a is not a unit, then it must belongtoM,whichistheidealofnonunits. Butthen1∈M,acontradiction. Conversely, assumethateveryelementof1+Misaunit. WeclaimthatM⊆J(R),henceM=J(R). Ifa∈M,thenax∈Mforeveryx∈R,so1+axisaunit. By(0.2.1),a∈J(R),proving the claim. If N is another maximal ideal, then M=J(R)⊆M∩N. Thus M⊆N, and since both ideals are maximal, they must be equal. Therefore R is a local ring. ♣ 0.2.3 Lemma Let S be any subset of R, and let I be the ideal generated by S. Then I =R iff for every maximal ideal M, there is an element x∈S\M. Proof. We have I ⊂ R iff I, equivalently S, is contained in some maximal ideal M. In other words, I ⊂ R iff ∃M such that ∀x ∈ S we have x ∈ M. The contrapositive says that I =R iff ∀M ∃x∈S such that x∈/ M. ♣ 0.2.4 Lemma √ √ Let I and J be ideals of the ring R. Then I+J =R iff I+ J =R. Proof. The“onlyif”partholdsbecauseanyidealiscontainedinitsradical. Thusassume that 1=a+b with am ∈I and bn ∈J. Then (cid:3) (cid:4) (cid:2) m+n 1=(a+b)m+n = aibj. i i+j=m+n Now if i+j = m+n, then either i ≥ m or j ≥ n. Thus every term in the sum belongs either to I or to J, hence to I+J. Consequently, 1∈I+J. ♣ 0.3. NAKAYAMA’S LEMMA 3 0.3 Nakayama’s Lemma First, we give an example of the determinant trick; see (2.1.2) for another illustration. 0.3.1 Theorem Let M be a finitely generated R-module, and I an ideal of R such that IM =M. Then there exists a∈I such that (1+a)M =0. Proof(cid:5). Let x1,... ,xn generate M. Since IM = M, we have(cid:5)equations of the form x = n a x , with a ∈I. The equations may be written as n (δ −a )x =0. i j=1 ij j ij j=1 ij ij j If I is the n by n identity matrix, we have (I −A)x = 0, where A = (a ) and x is a n n ij column vector whose coefficients are the x . Premultiplying by the adjoint of (I −A), i n we obtain ∆x=0, where ∆ is the determinant of (I −A). Thus ∆x =0 for all i, hence n i ∆M =0. But if we look at the determinant of I −A, we see that it is of the form 1+a n for some element a∈I. ♣ Here is a generalization of a familiar property of linear transformations on finite- dimensional vector spaces. 0.3.2 Theorem If M is a finitely generated R-module and f : M → M is a surjective homomorphism, then f is an isomorphism. Proof. We can make M into an R[X]-module via Xx = f(x), x ∈ M. (Thus X2x = f(f(x)), etc.) Let I = (X); we claim that IM = M. For if m ∈ M, then by the hypothesisthatf issurjective, m=f(x)forsomex∈M, andthereforeXx=f(x)=m. But X ∈ I, so m ∈ IM. By (0.3.1), there exists g = g(X) ∈ I such that (1+g)M = 0. ButbydefinitionofI,g mustbeoftheformXh(X)withh(X)∈R[X]. Thus(1+g)M = [1+Xh(X)]M =0. We can now prove that f is injective. Suppose that x∈M and f(x)=0. Then 0=[1+Xh(X)]x=[1+h(X)X]x=x+h(X)f(x)=x+0=x. ♣ In (0.3.2), we cannot replace “surjective” by “injective”. For example, let f(x) = nx on the integers. If n≥2, then f is injective but not surjective. The next result is usually referred to as Nakayama’s lemma. Sometimes, Akizuki and Krull are given some credit, and as a result, a popular abbreviation for the lemma is NAK. 0.3.3 NAK (a) If M is a finitely generated R-module, I an ideal of R contained in the Jacobson radical J(R), and IM =M, then M =0. (b) If N is a submodule of the finitely generated R-module M, I an ideal of R contained in the Jacobson radical J(R), and M =N +IM, then M =N. 4 CHAPTER 0. RING THEORY BACKGROUND Proof. (a) By (0.3.1), (1+a)M = 0 for some a ∈ I. Since I ⊆ J(R), 1+a is a unit by (0.2.1). Multiplying the equation (1+a)M =0 by the inverse of 1+a, we get M =0. (b) By hypothesis, M/N =I(M/N), and the result follows from (a). ♣ Here is an application of NAK. 0.3.4 Proposition LetRbealocalringwithmaximalidealJ. LetM beafinitelygeneratedR-module,and let V =M/JM. Then (i) V is a finite-dimensional vector space over the residue field k =R/J. (ii) If {x +JM,... ,x +JM} is a basis for V over k, then {x ,... ,x } is a minimal 1 n 1 n set of generators for M. (iii) Any two minimal generating sets for M have the same cardinality. Proof. (i) Since J annihilates M/JM, V is a k-module, that is, a vector space over k. Since M is finitely gen(cid:5)erated over R, V is a finite-dimensional vector space over k. (ii)LetN = n Rx . Sincethex +JM generateV =M/JM,wehaveM =N+JM. i=1 i i By NAK, M = N, so the x generate M. If a proper subset of the x were to generate i i M, then the corresponding subset of the x +JM would generate V, contradicting the i assumption that V is n-dimensional. (iii) A generating set S for M with more than n elements determines a spanning set for V, which must contain a basis with exactly n elements. By (ii), S cannot be minimal. ♣ 0.4 Localization Let S be a subset of the ring R, and assume that S is multiplicative, in other words, 0∈/ S, 1∈S, and if a and b belong to S, so does ab. In the case of interest to us, S will be the complement of a prime ideal. We would like to divide elements of R by elements of S to form the localized ring S−1R, also called the ring of fractions of R by S. There is no difficulty when R is an integral domain, because in this case all division takes place in the fraction field of R. We will sketch the general construction for arbitrary rings R. For full details, see TBGY, Section 2.8. 0.4.1 Construction of the Localized Ring If S is a multiplicative subset of the ring R, we define an equivalence relation on R×S by (a,b)∼(c,d) iff for some s∈S we have s(ad−bc)=0. If a∈R and b∈S, we define the fraction a/b as the equivalence class of (a,b). We make the set of fractions into a ring in a natural way. The sum of a/b and c/d is defined as (ad+bc)/bd, and the product of a/bandc/disdefinedasac/bd. Theadditiveidentityis0/1,whichcoincideswith0/sfor every s∈S. The additive inverse of a/b is −(a/b)=(−a)/b. The multiplicative identity is 1/1, which coincides with s/s for every s∈S. To summarize: S−1R isaring. IfR isanintegraldomain, soisS−1R. IfR isanintegraldomainand S =R\{0}, then S−1R is a field, the fraction field of R. 0.4. LOCALIZATION 5 There is a natural ring homomorphism h : R → S−1R given by h(a) = a/1. If S has no zero-divisors, then h is a monomorphism, so R can be embedded in S−1R. In particular, a ring R can be embedded in its full ring of fractions S−1R, where S consists of all non-divisors of 0 in R. An integral domain can be embedded in its fraction field. Our goal is to study the relation between prime ideals ofR and prime ideals of S−1R. 0.4.2 Lemma If X is any subset of R, define S−1X = {x/s : x ∈ X,s ∈ S}. If I is an ideal of R, then S−1I is an ideal of S−1R. If J is another ideal of R, then (i) S−1(I+J)=S−1I+S−1J; (ii) S−1(IJ)=(S−1I)(S−1J); (iii) S−1(I∩J)=(S−1I)∩(S−1J); (iv) S−1I is a proper ideal iff S∩I =∅. Proof. The definitions of addition and multiplication in S−1R imply that S−1R is an ideal, and that in (i), (ii) and (iii), the left side is contained in the right side. The reverse inclusions in (i) and (ii) follow from a b at+bs ab ab + = , = . s t st s t st To prove (iii), let a/s = b/t, where a ∈ I, b ∈ J, s,t ∈ S. There exists u ∈ S such that u(at−bs)=0. Then a/s=uat/ust=ubs/ust∈S−1(I∩J). Finally, if s ∈ S ∩ I, then 1/1 = s/s ∈ S−1I, so S−1I = S−1R. Conversely, if S−1I = S−1R, then 1/1 = a/s for some a ∈ I, s ∈ S. There exists t ∈ S such that t(s−a)=0, so at=st∈S∩I. ♣ Ideals in S−1R must be of a special form. 0.4.3 Lemma Let h be the natural homomorphism from R to S−1R [see (0.4.1)]. If J is an ideal of S−1R and I =h−1(J), then I is an ideal of R and S−1I =J. Proof. I is an ideal by the basic properties of preimages of sets. Let a/s ∈ S−1I, with a∈I ands∈S. Thena/1=h(a)∈J, soa/s=(a/1)(1/s)∈J. Conversely, leta/s∈J, with a∈R,s∈S. Then h(a)=a/1=(a/s)(s/1)∈J, so a∈I and a/s∈S−1I. ♣ Prime ideals yield sharper results. 0.4.4 Lemma IfI isanyidealofR,thenI ⊆h−1(S−1I). TherewillbeequalityifI isprimeanddisjoint from S. Proof. If a∈I, then h(a)=a/1∈S−1I. Thus assume that I is prime and disjoint from S, and let a∈h−1(S−1I). Then h(a)=a/1∈S−1I, so a/1=b/s for some b∈I,s∈S. There exists t ∈ S such that t(as−b) = 0. Thus ast = bt ∈ I, with st ∈/ I because S∩I =∅. Since I is prime, we have a∈I. ♣ 6 CHAPTER 0. RING THEORY BACKGROUND 0.4.5 Lemma If I is a prime ideal of R disjoint from S, then S−1I is a prime ideal of S−1R. Proof. By part (iv) of (0.4.2), S−1I is a proper ideal. Let (a/s)(b/t) = ab/st ∈ S−1I, with a,b ∈ R,s,t ∈ S. Then ab/st = c/u for some c ∈ I,u ∈ S. There exists v ∈ S such that v(abu−cst) = 0. Thus abuv = cstv ∈ I, and uv ∈/ I because S∩I = ∅. Since I is prime, ab∈I, hence a∈I or b∈I. Therefore either a/s or b/t belongs to S−1I. ♣ The sequence of lemmas can be assembled to give a precise conclusion. 0.4.6 Theorem There is a one-to-one correspondence between prime ideals P of R that are disjoint from S and prime ideals Q of S−1R, given by P →S−1P and Q→h−1(Q). Proof. By (0.4.3), S−1(h−1(Q))=Q, and by (0.4.4), h−1(S−1P)=P. By (0.4.5), S−1P is a prime ideal, and h−1(Q) is a prime ideal by the basic properties of preimages of sets. If h−1(Q) meets S, then by (0.4.2) part (iv), Q=S−1(h−1(Q))=S−1R, a contradiction. Thus the maps P → S−1P and Q → h−1(Q) are inverses of each other, and the result follows. ♣ 0.4.7 Definitions and Comments If P is a prime ideal of R, then S = R\P is a multiplicative set. In this case, we write R for S−1R, and call it the localization of R at P. We are going to show that R is P P a local ring, that is, a ring with a unique maximal ideal. First, we give some conditions equivalent to the definition of a local ring. 0.4.8 Proposition For a ring R, the following conditions are equivalent. (i) R is a local ring; (ii) There is a proper ideal I of R that contains all nonunits of R; (iii) The set of nonunits of R is an ideal. Proof. (i) implies (ii): If a is a nonunit, then (a) is a proper ideal, hence is contained in the unique maximal ideal I. (ii) implies (iii): If a and b are nonunits, so are a+b and ra. If not, then I contains a unit, so I =R, contradicting the hypothesis. (iii)implies(i): IfI istheidealofnonunits,thenI ismaximal,becauseanylargeridealJ would have to contain a unit, so J =R. If H is any proper ideal, then H cannot contain a unit, so H ⊆I. Therefore I is the unique maximal ideal. ♣