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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. In Chapter 1, the theory of primary decomposition is developed so as to apply to modules as well as ideals. In Chapter 2, integral extensions are treated in detail, 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, and covers the Artin-Rees lemma and the Krull intersection theorem. 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 homological algebra to prove, under approprate hypotheses, that all maximal M -sequences have the same length. The brief Chapter 8 develops enough theory to prove that a regular 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- Buchsbaum theorems, and further properties of Cohen-Macaulay rings in a more advanced course.
Bibliography
Atiyah, M.F. and Macdonald, I.G., Introduction to Commtative Algebra, 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
Kunz, E., Introduction to Commutative Algebra and Algebraic Geometry, Birkh¨auser 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
©ccopyright 2003, by Robert B. Ash. Paper or electronic copies for noncommercial use may be made freely without explicit permission of the author. All other rights are reserved.
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 1 , P 2 ,... , Ps, s ≥ 2, be ideals in a ring R, with P 1 and P 2 not necessarily prime, but P 3 ,... , Ps prime (if s ≥ 3). Let I be any ideal of R. The idea is that if we can avoid the Pj individually, in other words, for each j we can find an element in I but not in Pj , then we can avoid all the Pj simultaneously, that is, we can find a single element in I that is in none of the Pj. We will state and prove the contrapositive.
0.1.1 Prime Avoidance Lemma
With I and the Pi as above, if I ⊆ ∪si=1Pi, then for some i we have I ⊆ Pi.
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 Pi’s. (If so, we can simply replace s by s − 1.) Thus for each i we can find an element ai ∈ I with ai ∈/ P 1 ∪ · · · ∪ Pi− 1 ∪ Pi+1 ∪ · · · ∪ Ps. By hypothesis, I is contained in the union of all the P ’s, so ai ∈ Pi. First assume s = 2, with I ⊆ P 1 and I ⊆ P 2. Then a 1 ∈ P 1 , a 2 ∈/ P 1 , so a 1 + a 2 ∈/ P 1. Similarly, a 1 ∈/ P 2 , a 2 ∈ P 2 , so a 1 + a 2 ∈/ P 2. Thus a 1 + a 2 ∈/ I ⊆ P 1 ∪ P 2 , contradicting a 1 , a 2 ∈ I. Note that P 1 and P 2 need not be prime for this argument to work. Now assume s > 2, and observe that a 1 a 2 · · · as− 1 ∈ P 1 ∩ · · · ∩ Ps− 1 , but as ∈/ P 1 ∪ · · · ∪ Ps− 1. Let a = (a 1 · · · as− 1 ) + as, which does not belong to P 1 ∪ · · · ∪ Ps− 1 , else as would belong to this set. Now for all i = 1,... , s − 1 we have ai ∈/ Ps, hence a 1 · · · as− 1 ∈/ Ps because Ps is prime. But as ∈ Ps, so a cannot be in Ps. Thus a ∈ I and a /∈ P 1 ∪ · · · ∪ Ps, contradicting the hypothesis. ♣
It may appear that we only used the primeness of Ps, but after the preliminary reduc- tion (see the beginning of the proof), it may very well happen that one of the other Pi’s now occupies the slot that previously housed Ps.
0.3. NAKAYAMA’S LEMMA 3
0.3 Nakayama’sLemma
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. Let x 1 ,... , xn generate M. Since IM = M , we have equations of the form xi =
∑n j=1 aij^ xj^ , with^ aij^ ∈^ I. The equations may be written as^
∑n j=1(δij^ −^ aij^ )xj^ = 0. If In is the n by n identity matrix, we have (In − A)x = 0, where A = (aij ) and x is a column vector whose coefficients are the xi. Premultiplying by the adjoint of (In − A), we obtain ∆x = 0, where ∆ is the determinant of (In − A). Thus ∆xi = 0 for all i, hence ∆M = 0. But if we look at the determinant of In − A, we see that it is of the form 1 + a 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 X^2 x = f (f (x)), etc.) Let I = (X); we claim that IM = M. For if m ∈ M , then by the hypothesis that f is surjective, m = f (x) for some x ∈ M , and therefore Xx = 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. But by definition of I, g must be of the form Xh(X) with h(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
Let R be a local ring with maximal ideal J. Let M be a finitely generated R-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 1 + JM,... , xn + JM } is a basis for V over k, then {x 1 ,... , xn} is a minimal 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 generated over R, V is a finite-dimensional vector space over k. (ii) Let N =
∑n i=1 Rxi. Since the^ xi^ +^ JM^ generate^ V^ =^ M/JM^ , we have^ M^ =^ N^ +^ JM^. By NAK, M = N , so the xi generate M. If a proper subset of the xi were to generate M , then the corresponding subset of the xi + JM would generate V , contradicting the 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−^1 R, 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/b and c/d is defined as ac/bd. The additive identity is 0/1, which coincides with 0/s for 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−^1 R is a ring. If R is an integral domain, so is S−^1 R. If R is an integral domain and S = R \ { 0 }, then S−^1 R is a field, the fraction field of R.
6 CHAPTER 0. RING THEORY BACKGROUND
0.4.5 Lemma
If I is a prime ideal of R disjoint from S, then S−^1 I is a prime ideal of S−^1 R.
Proof. By part (iv) of (0.4.2), S−^1 I is a proper ideal. Let (a/s)(b/t) = ab/st ∈ S−^1 I, 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−^1 I. ♣
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−^1 R, given by
P → S−^1 P and Q → h−^1 (Q).
Proof. By (0.4.3), S−^1 (h−^1 (Q)) = Q, and by (0.4.4), h−^1 (S−^1 P ) = P. By (0.4.5), S−^1 P 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−^1 R, a contradiction. Thus the maps P → S−^1 P and Q → h−^1 (Q) are inverses of each other, and the result follows. ♣
0.4.7 Definitionsand Comments
If P is a prime ideal of R, then S = R \ P is a multiplicative set. In this case, we write RP for S−^1 R, and call it the localization of R at P. We are going to show that RP is 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): If I is the ideal of nonunits, then I is maximal, because any larger ideal J 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. ♣
0.4. LOCALIZATION 7
0.4.9 Theorem
RP is a local ring.
Proof. Let Q be a maximal ideal of RP. Then Q is prime, so by (0.4.6), Q = S−^1 I for some prime ideal I of R that is disjoint from S = R \ P. In other words, I ⊆ P. Consequently, Q = S−^1 I ⊆ S−^1 P. If S−^1 P = RP = S−^1 R, then by (0.4.2) part (iv), P is not disjoint from S = R \ P , which is impossible. Therefore S−^1 P is a proper ideal containing every maximal ideal, so it must be the unique maximal ideal. ♣
0.4.10 Remark
It is convenient to write the ideal S−^1 I as IRP. There is no ambiguity, because the product of an element of I and an arbitrary element of R belongs to I.
0.4.11 Localization of Modules
If M is an R-module and S a multiplicative subset of R, we can essentially repeat the construction of (0.4.1) to form the localization of M by S, and thereby divide elements of M by elements of S. If x, y ∈ M and s, t ∈ S, we call (x, s) and (y, t) equivalent if for some u ∈ S, we have u(tx − sy) = 0. The equivalence class of (x, s) is denoted by x/s, and addition is defined by
x s
y t
tx + sy st
If a/s ∈ S−^1 R and x/t ∈ S−^1 M , we define
a s
x t
ax st
In this way, S−^1 M becomes an S−^1 R-module. Exactly as in (0.4.2), if M and N are submodules of an R-module L, then
S−^1 (M + N ) = S−^1 M + S−^1 N and S−^1 (M ∩ N ) = (S−^1 M ) ∩ (S−^1 N ).
2 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES
1.1.2 Lemma
If
I is a maximal ideal M, then I is M-primary.
Proof. Suppose that ab ∈ I and b does not belong to
I = M. Then by maximality of M, it follows that M + Rb = R, so for some m ∈ M and r ∈ R we have m + rb = 1. Now m ∈ M =
I, hence mk^ ∈ I for some k ≥ 1. Thus 1 = 1k^ = (m + rb)k^ = mk^ + sb for some s ∈ R. Multiply by a to get a = amk^ + sab ∈ I. ♣
1.1.3 Corollary
If M is a maximal ideal, then Mn^ is M-primary for every n ≥ 1.
Proof. As we observed in (1.1.1),
Mn^ = M, and the result follows from (1.1.2). ♣
1.2 Primary Decomposition
1.2.1 Definitions and Comments
A primary decomposition of the submodule N of M is given by N = ∩ri=1Ni, where the Ni are Pi-primary submodules. The decomposition is reduced if the Pi are distinct and N cannot be expressed as the intersection of a proper subcollection of the Ni. We can always extract a reduced primary decomposition from an unreduced one, by discarding those Ni that contain ∩j =iNj and intersecting those Ni that are P -primary for the same P. The following result justifies this process.
1.2.2 Lemma
If N 1 ,... , Nk are P -primary, then ∩ki=1Ni is P -primary.
Proof. We may assume that k = 2; an induction argument takes care of larger values. Let N = N 1 ∩ N 2 and rM (N 1 ) = rM (N 2 ) = P. Assume for the moment that rM (N ) = P. If a ∈ R, x ∈ M, ax ∈ N , and a /∈ rM (N ), then since N 1 and N 2 are P -primary, we have x ∈ N 1 ∩ N 2 = N. It remains to show that rM (N ) = P. If a ∈ P , then there are positive integers n 1 and n 2 such that an^1 M ⊆ N 1 and an^2 M ⊆ N 2. Therefore an^1 +n^2 M ⊆ N , so a ∈ rM (N ). Conversely, if a ∈ rM (N )then a belongs to rM (Ni)for i = 1, 2, and therefore a ∈ P. ♣ We now prepare to prove that every submodule of a Noetherian module has a primary decomposition.
1.2.3 Definition
The proper submodule N of M is irreducible if N cannot be expressed as N 1 ∩ N 2 with N properly contained in the submodules Ni, i = 1, 2.
1.3. ASSOCIATED PRIMES 3
1.2.4 Proposition
If N is an irreducible submodule of the Noetherian module M , then N is primary.
Proof. If not, then for some a ∈ R, λa : M/N → M/N is neither injective nor nilpotent. The chain ker λa ⊆ ker λ^2 a ⊆ ker λ^3 a ⊆ · · · terminates by the ascending chain condition, say at ker λia. Let ϕ = λia; then ker ϕ = ker ϕ^2 and we claim that ker ϕ ∩ im ϕ = 0. Suppose x ∈ ker ϕ ∩ im ϕ, and let x = ϕ(y). Then 0 = ϕ(x) = ϕ^2 (y) , soy ∈ ker ϕ^2 = ker ϕ, so x = ϕ(y) = 0. Now λa is not injective, so ker ϕ = 0, and λa is not nilpotent, so λia can’t be 0 (because aiM ⊆ N ). Consequently, im ϕ = 0. Let p : M → M/N be the canonical epimorphism, and set N 1 = p−^1 (ker ϕ), N 2 = p−^1 (im ϕ). We will prove that N = N 1 ∩ N 2. If x ∈ N 1 ∩ N 2 , then p(x)belongs to both ker ϕ and im ϕ, so p(x)= 0, in other words, x ∈ N. Conversely, if x ∈ N , then p(x) = 0 ∈ ker ϕ ∩ im ϕ, so x ∈ N 1 ∩ N 2. Finally, we will show that N is properly contained in both N 1 and N 2 , so N is reducible, a contradiction. Choose a nonzero element y ∈ ker ϕ. Since p is surjective, there exists x ∈ M such that p(x) = y. Thus x ∈ p−^1 (ker ϕ) = N 1 (because y = p(x) ∈ ker ϕ), but x /∈ N (because p(x) = y = 0). Similarly, N ⊂ N 2 (with 0 = y ∈ im ϕ), and the result follows. ♣
1.2.5Theorem
If N is a proper submodule of the Noetherian module M , then N has a primary decom- position, hence a reduced primary decomposition.
Proof. We will show that N can be expressed as a finite intersection of irreducible sub- modules of M , so that (1.2.4)applies. Let S be the collection of all submodules of M that cannot be expressed in this form. If S is nonempty, then S has a maximal element N (because M is Noetherian). By definition of S, N must be reducible, so we can write N = N 1 ∩ N 2 , N ⊂ N 1 , N ⊂ N 2. By maximality of N , N 1 and N 2 can be expressed as finite intersections of irreducible submodules, hence so can N , contradicting N ∈ S. Thus S is empty. ♣
1.3 Associated Primes
1.3.1 Definitions and Comments
Let M be an R-module, and P a prime ideal of R. We say that P is an associated prime of M (or that P is associated to M )if P is the annihilator of some nonzero x ∈ M. The set of associated primes of M is denoted by AP(M ). (The standard notation is Ass(M). Please do not use this regrettable terminology.) Here is a useful characterization of associated primes.
1.3.2 Proposition
The prime ideal P is associated to M if and only if there is an injective R-module homo- morphism from R/P to M. Therefore if N is a submodule of M , then AP(N ) ⊆ AP(M ).
1.3. ASSOCIATED PRIMES 5
1.3.7 Proposition
If N is a submodule of M , then AP(M ) ⊆ AP(N ) ∪ AP(M/N ).
Proof. Let P ∈ AP(M ), and let h : R/P → M be a monomorphism. Set H = h(R/P ) and L = H ∩ N. Case 1: L = 0. Then the map from H to M/N given by h(r + P ) → h(r + P ) + N is a monomorphism. (If h(r + P )belongs to N , it must belong to H ∩ N = 0.)Thus H is isomorphic to a submodule of M/N , so by definition of H, there is a monomorphism from R/P to M/N. Thus P ∈ AP(M/N ). Case 2: L = 0. If L has a nonzero element x, then x must belong to both H and N , and H is isomorphic to R/P via h. Thus x ∈ N and the annihilator of x coincides with the annihilator of some nonzero element of R/P. By (1.3.5), ann x = P , so P ∈ AP(N ). ♣
1.3.8 Corollary
AP(
j∈J
Mj =
j∈J
AP(Mj ).
Proof. By (1.3.2), the right side is contained in the left side. The result follows from (1.3.7)when the index set is finite. For example,
AP(M 1 ⊕ M 2 ⊕ M 3 ) ⊆ AP(M 1 ) ∪ AP(M/M 1 ) = AP(M 1 ) ∪ AP(M 2 ⊕ M 3 ) ⊆ AP(M 1 ) ∪ AP(M 2 ) ∪ AP(M 3 ).
In general, if P is an associated prime of the direct sum, then there is a monomorphism from R/P to ⊕Mj. The image of the monomorphism is contained in the direct sum of finitely many components, as R/P is generated as an R-module by the single element 1 + P. This takes us back to the finite case. ♣ We now establish the connection between associated primes and primary decomposi- tion, and show that under wide conditions, there are only finitely many associated primes.
1.3.9 Theorem
Let M be a nonzero finitely generated module over the Noetherian ring R, so that by (1.2.5), every proper submodule of M has a reduced primary decomposition. In particular, the zero module can be expressed as ∩ri=1Ni, where Ni is Pi-primary. Then AP(M ) = {P 1 ,... , Pr }, a finite set.
Proof. Let P be an associated prime of M , so that P = ann(x), x = 0, x ∈ M. Renumber the Ni so that x /∈ Ni for 1 ≤ i ≤ j and x ∈ Ni for j + 1 ≤ i ≤ r. Since Ni is Pi-primary, we have Pi = rM (Ni) (see (1.1.1)). Since Pi is finitely generated, P (^) in iM ⊆ Ni for some ni ≥ 1. Therefore
⋂^ j
i=
P (^) in i)x ⊆
⋂^ r
i=
Ni = (0)
6 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES
so ∩ji=1P (^) in i⊆ ann(x) = P. (By our renumbering, there is a j rather than an r on the left side of the inclusion.)Since P is prime, Pi ⊆ P for some i ≤ j. We claim that Pi = P , so that every associated prime must be one of the Pi. To verify this, let a ∈ P. Then ax = 0 and x /∈ Ni, so λa is not injective and therefore must be nilpotent. Consequently, a ∈ rM (Ni) = Pi, as claimed.
Conversely, we show that each Pi is an associated prime. Without loss of generality, we may take i = 1. Since the decomposition is reduced, N 1 does not contain the intersection of the other Ni’s, so we can choose x ∈ N 2 ∩· · ·∩Nr with x /∈ N 1. Now N 1 is P 1 -primary, so as in the preceding paragraph, for some n ≥ 1 we have P 1 n x ⊆ N 1 but P 1 n −^1 x ⊆ N 1. (Take P 10 x = Rx and recall that x /∈ N 1 .)If we choose y ∈ P 1 n −^1 x \ N 1 (hence y = 0), the proof will be complete upon showing that P 1 is the annihilator of y. We have P 1 y ⊆ P 1 n x ⊆ N 1 and x ∈ ∩ri=2Ni, so P 1 n x ⊆ ∩ri=2Ni. Thus P 1 y ⊆ ∩ri=1Ni = (0), so P 1 ⊆ ann y. On the other hand, if a ∈ R and ay = 0, then ay ∈ N 1 but y /∈ N 1 , so λa : M/N 1 → M/N 1 is not injective and is therefore nilpotent. Thus a ∈ rM (N 1 ) = P 1. ♣ We can now say something about uniqueness in primary decompositions.
1.3.10 First Uniqueness Theorem
Let M be a finitely generated module over the Noetherian ring R. If N = ∩ri=1Ni is a reduced primary decomposition of the submodule N , and Ni is Pi-primary, i = 1,... , r, then (regarding M and R as fixed)the Pi are uniquely determined by N.
Proof. By the correspondence theorem, a reduced primary decomposition of (0)in M/N is given by (0)= ∩ri=1Ni/N , and Ni/N is Pi-primary, 1 ≤ i ≤ r. By (1.3.9),
AP(M/N ) = {P 1 ,... , Pr }.
But [see (1.3.1)] the associated primes of M/N are determined by N. ♣
1.3.11 Corollary
Let N be a submodule of M (finitely generated over the Noetherian ring R). Then N is P -primary iff AP(M/N ) = {P }.
Proof. The “only if” part follows from the displayed equation above. Conversely, if P is the only associated prime of M/N , then N coincides with a P -primary submodule N ′, and hence N (= N ′)is P -primary. ♣
1.3.12 Definitions and Comments
Let N = ∩ri=1Ni be a reduced primary decomposition, with associated primes P 1 ,... , Pr. We say that Ni is an isolated (or minimal )component if Pi is minimal, that is Pi does not properly contain any Pj , j = i. Otherwise, Ni is an embedded component (see Exercise 5 for an example). Embedded components arise in algebraic geometry in situations where one irreducible algebraic set is properly contained in another.
8 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES
At the beginning of the proof of (1.4.3), we have taken advantage of the isomorphism between (M/N )P ′^ and M ′/N ′. The result comes from the exactness of the localization functor. If this is unfamiliar, look ahead to the proof of (1.5.3), where the technique is spelled out. See also TBGY, Section 8.5, Problem 5.
1.4.4 Lemma
In (1.4.3), if P ⊆ P ′, then N = f −^1 (N ′), where f is the natural map from M to M ′.
Proof. As in (1.4.3), APR(M/N ) = {P }. Since P ⊆ P ′, we have R \ P ′^ ⊆ R \ P. By (1.3.6), R \ P ′^ contains no zero-divisors of M/N , because all such zero-divisors belong to P. Thus the natural map g : x → x/1 of M/N to (M/N )P ′ ∼= M ′/N ′^ is injective. (If x/1 = 0, then sx = 0 for some s ∈ S = R \ P ′, and since s is not a zero-divisor, we have x = 0.) If x ∈ N , then f (x) ∈ N ′^ by definition of f , so assume x ∈ f −^1 (N ′). Then f (x) ∈ N ′, so f (x) + N ′^ is 0 in M ′/N ′. By injectivity of g, x + N is 0 in M/N , in other words, x ∈ N , and the result follows. ♣
1.4.5Second Uniqueness Theorem
Let M be a finitely generated module over the Noetherian ring R. Suppose that N = ∩ri=1Ni is a reduced primary decomposition of the submodule N , and Ni is Pi-primary, i = 1,... , r. If Pi is minimal, then (regarding M and R as fixed) Ni is uniquely determined by N.
Proof. Suppose that P 1 is minimal, so that P 1 ⊇ Pi, i > 1. By (1.4.3)with P = Pi, P ′^ = P 1 , we have (Ni)P 1 = MP 1 for i > 1. By (1.4.4)with P = P ′^ = P 1 , we have N 1 = f −^1 [(N 1 )P 1 ], where f is the natural map from M to MP 1. Now
NP 1 = (N 1 )P 1 ∩ ∩ri=2(Ni)P 1 = (N 1 )P 1 ∩ MP 1 = (N 1 )P 1.
Thus N 1 = f −^1 [(N 1 )P 1 ] = f −^1 (NP 1 )depends only on N and P 1 , and since P 1 depends on the fixed ring R, it follows that N 1 depends only on N. ♣
1.5The Support of a Module
The support of a module M is closely related to the set of associated primes of M. We will need the following result in order to proceed.
1.5.1 Proposition
M is the zero module if and only if MP = 0 for every prime ideal P , if and only if MM = 0 for every maximal ideal M.
Proof. It suffices to show that if MM = 0 for all maximal ideals M, then M = 0. Choose a nonzero element x ∈ M , and let I be the annihilator of x. Then 1 ∈/ I (because 1 x = x = 0) , soI is a proper ideal and is therefore contained in a maximal ideal M. By hypothesis, x/1 is 0 in MM, hence there exists a /∈ M (so a /∈ I)such that ax = 0. But then by definition of I we have a ∈ I, a contradiction. ♣
1.5. THE SUPPORT OF A MODULE 9
1.5.2 Definitions and Comments
The support of an R-module M (notation Supp M )is the set of prime ideals P of R such that MP = 0. Thus Supp M = ∅ iff MP = 0 for all prime ideals P. By (1.5.1), this is equivalent to M = 0. If I is any ideal of R, we define V (I)as the set of prime ideals containing I. In algebraic geometry, the Zariski topology on Spec R has the sets V (I)as its closed sets.
1.5.3 Proposition
Supp R/I = V (I).
Proof. We apply the localization functor to the exact sequence 0 → I → R → R/I → 0 to get the exact sequence 0 → IP → RP → (R/I)P → 0. Consequently, (R/I)P ∼= RP /IP. Thus P ∈ Supp R/I iff RP ⊃ IP iff IP is contained in a maximal ideal, necessarily P RP. But this is equivalent to I ⊆ P. To see this, suppose a ∈ I, with a/ 1 ∈ IP ⊆ P RP. Then a/1 = b/s for some b ∈ P, s /∈ P. There exists c /∈ P such that c(as − b)= 0. We have cas = cb ∈ P , a prime ideal, and cs /∈ P. We conclude that a ∈ P. ♣
1.5.4 Proposition
Let 0 → M ′^ → M → M ′′^ → 0 be exact, hence 0 → M (^) P′ → MP → M (^) P′′ → 0 is exact. Then
Supp M = Supp M ′^ ∪ Supp M ′′.
Proof. Let P belong to Supp M \ Supp M ′. Then M (^) P′ = 0, so the map MP → M (^) P′′ is injective as well as surjective, hence is an isomorphism. But MP = 0 by assumption, so M (^) P′′ = 0, and therefore P ∈ Supp M ′′. On the other hand, since M (^) P′ is isomorphic to a submodule of MP , it follows that Supp M ′^ ⊆ Supp M. If MP = 0, then M (^) P′′ = 0 (because MP → M (^) P′′ is surjective). Thus Supp M ′′^ ⊆ Supp M. ♣
Supports and annihilators are connected by the following basic result.
1.5.5 Theorem
If M is a finitely generated R-module, then Supp M = V (ann M ).
Proof. Let M = Rx 1 + · · · + Rxn, so that MP = (Rx 1 )P + · · · + (Rxn)P. Then Supp M = ∪ni=1 Supp Rxi, and by the first isomorphism theorem, Rxi ∼= R/ ann xi. By (1.5.3), Supp Rxi = V (ann xi). Therefore Supp M = ∪ni=1V (ann xi) = V (ann M ). To justify the last equality, note that if P ∈ V (ann xi), then P ⊇ ann xi ⊇ ann M. Conversely, if P ⊇ ann M = ∩ni=1 ann xi, then P ⊇ ann xi for some i. ♣
And now we connect associated primes and annihilators.
