# PRIMARY DECOMPOSITION OF MODULES

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1 PRIMARY DECOMPOSITION OF MODULES SUDHIR R. GHORPADE AND JUGAL K. VERMA Department of Mathematics, Indian Institute of Technology, Mumbai , India {srg, Abstract. Basics of the theory of primary decomposition of modules are outlined here. The notions of graded rings and modules are also introduced, and then the special features of primary decomposition in the case of graded modules are discussed. Contents Introduction 2 1. Associated Primes 2 2. Primary Decomposition 5 3. Graded Rings and Modules 7 4. Primary Decomposition in Graded Modules 11 References 13 Date: November 23, This is a slightly revised version of the second chapter of the Lecture Notes of the NBHM sponsored Instructional Conference on Combinatorial Topology and Algebra, held at IIT Bombay in December As such, these notes may be viewed as a continuation of [Gh], which was a revised version of the first chapter. 1

2 2 SUDHIR R. GHORPADE AND JUGAL K. VERMA Introduction In the first two sections, we will discuss an extension of the notions of associated primes and primary decomposition to the case of modules. The classical case of ideals I in (noetherian) rings A corresponds, in this general set-up, to the case of A modules A/I; remembering this may be helpful in understanding some of the concepts and results below. Later, in Section 3, we describe the notion of graded rings and graded modules and prove some basic results concerning them. Finally, in Section 4 we discuss some special properties of primary decomposition in the case of graded modules over graded rings. 1. Associated Primes Taking into consideration the modern viewpoint that the notion of associated primes is more fundamental than primary decomposition, we shall derive in this section basic results about associated primes without mentioning primary submodules or primary decomposition. Throughout this section, we let A denote a ring and M an A module. By Spec A we shall denote the set of all prime ideals of A; Spec A is sometimes called the spectrum 1 of A. Definition: A prime ideal p of A is called an associated prime of M if p = (0 : x) for some x M. The set of all associated primes of M is denoted by Ass A (M), or simply by Ass(M). Minimal elements of Ass(M) are called the minimal primes of M, and the remaining elements of Ass(M) are called the embedded primes of M. Note that if p = (0 : x) Ass(M), then the map a ax of A M defines an embedding (i.e., an injective A module homomorphism) A/p M. Conversely, if for p Spec A, we have an embedding A/p M, then clearly p Ass(M). It may also be noted that if M is isomorphic to some A module M, then Ass(M) = Ass(M ). (1.1) Lemma. Any maximal element of {(0 : y) : y M, y 0} is a prime ideal. In particular, if A is noetherian, then Ass(M) iff M 0. Proof: Suppose (0 : x) is a maximal element of {(0 : y) : y M, y 0}. Then (0 : x) A since x 0. Moreover, if a, b A are such that ab (0 : x) and a / (0 : x), then ax 0 and b (0 : ax) (0 : x). Since (0 : x) is maximal, (0 : ax) = (0 : x). Thus b (0 : x). Thus (0 : x) is a prime ideal. The last assertion is evident. Definition: An element a A is said to be a zerodivisor of M if (0 : a) M 0, i.e., if ax = 0 for some x M with x 0. The set of all zerodivisors of M is denoted by Z(M). (1.2) Exercise: Check that the above definition of Z(M) is consistent with that in (2.4) of [Gh]. Show that if A is noetherian, then Z(M) = p. p Ass(M) 1 For an explanation of this terminology, see the article [Ta] by Taylor.

3 PRIMARY DECOMPOSITION OF MODULES 3 (1.3) Lemma. For any submodule N of M, Ass(N) Ass(M) Ass(N) Ass(M/N). More generally, if 0 = M 0 M 1 M n = M is any chain of submodules of M, then Ass(M) n i=1ass(m i /M i 1 ). Proof: The inclusion Ass(N) Ass(M) is obvious. Let x M be such that (0 : x) Ass(M). If (0 : x) / Ass(N), then we claim that (0 : x) = (0 : x) Ass(M/N), where x denotes the image of x in M/N. To see this, note that (0 : x) (0 : x) and if a A is such that a x = 0 ax, then ax N and a / (0 : x), and since (0 : x) is prime, we have b (0 : ax) ba (0 : x) b (0 : x); consequently, (0 : x) = (0 : ax) Ass(N), which is a contradiction. Thus Ass(M) Ass(N) Ass(M/N). The last assertion follows from this by induction on n. The inclusions Ass(N) Ass(M) and Ass(M) Ass(N) Ass(M/N) in the above Lemma are, in general, proper. This may be seen, for instance, when A is a domain and M = A by taking N = 0 and N = a nonzero prime ideal of A, respectively. (1.4) Corollary. Suppose M 1,..., M h are A modules such that M h i=1m i. Then Ass(M) = h i=1ass(m i ). Proof: Follows using induction on h by noting that the case of h = 2 is a consequence of the first assertion in (1.3). Example: Let G be a finite abelian group of order n. Suppose n = p e p e h h, where p 1,..., p h are distinct prime numbers and e 1,..., e h are positive integers. Then G is a Z module, and the p i Sylow subgroups P i, 1 i h, are Z submodules of G such that G P 1 P h. Clearly, Ass(P i ) = p i Z [indeed, if y is any nonzero element of P i of order p e i, then elements of (0 : y) are multiples of p e i, and if x = p e 1 i y, then (0 : x) = p i Z]. Thus by (1.4), Ass(G) = {p 1 Z,..., p h Z}. More generally, if M is a finitely generated abelian group, then M = Z r T for some r 0 and some finite abelian group T, and it follows from (1.4) that if r > 0 then Ass(M) = {l 0 Z, l 1 Z,..., l s Z}, where l 0 = 0 and l 1,... l s are the prime numbers dividing the order of T. Having discussed some properties of associated primes of quotient modules, we now describe what happens to associated prime upon localisation. Before that, let us recall that if S is m. c. subset of A, then the map p S 1 p gives a one-to-one correspondence of {p Spec A : p S = } onto Spec S 1 A. (1.5) Lemma. Suppose A is noetherian and S is a m. c. subset of A. Then Ass S 1 A(S 1 M) = {S 1 p : p Ass(M) and p S = } Proof: If p Ass(M) and S p =, then we have an embedding A/p M, which, in view of (1.2) of [Gh], induces an embedding S 1 A/S 1 p S 1 M. And S 1 p Spec S 1 A since S p =. Thus S 1 p Ass S 1 A(S 1 M). On the other hand, if p Ass S 1 A(S 1 M), then p = S 1 p for some p Spec A with p S =, and p = (0 : x ) for some x M. Write 1 p = (a 1,..., a n ). Now a i x = 0 in 1 1 S 1 M for 1 i n, and thus there exists t S such that ta i x = 0 for 1 i n. This implies that p (0 : tx). Further, if a (0 : tx), then a (0 : x) = 1 1 S 1 p so that sa p for some s S, and hence a p. Thus p Ass(M).

4 4 SUDHIR R. GHORPADE AND JUGAL K. VERMA Note that the above result implies that if p Ass(M), then pa p Ass Ap (M p ). So, in particular, M p 0. Definition: The set {p Spec A : M p 0} is called the support of M and is denoted by Supp (M). (1.6) Lemma. Supp (M) {p Spec A : p Ann(M)}. Moreover, if M is f. g., then these two sets are equal. Proof: If p Spec A and M p 0, then there is x M such that x 0 in M 1 p. Now a Ann(M) ax = 0 a / A \ p a p. Thus p Ann(M). Next, suppose M is f. g. and p Spec A contains Ann(M). Write M = Ax Ax n. If M p = 0, we can find a A \ p such that ax i = 0 for 1 i n. But then am = 0, i.e., a Ann(M), which is a contradiction. (1.7) Theorem. Suppose A is noetherian and M is finitely generated. Then there exists a chain 0 = M 0 M 1 M n = M of submodules of M such that M i /M i 1 A/p i, for some p i Spec A (1 i n). Moreover, for any such chain of submodules, we have Ass(M) {p 1,..., p n } Supp (M); furthermore, the minimal elements of these three sets coincide. Proof: The case of M = 0 is trivial. If M 0, then there exists p 1 Spec A such that A/p 1 is isomorphic to a submodule M 1 of M. If M 1 M, we apply the same argument to M/M 1 to find p 2 Spec A, and a submodule M 2 of M such that M 2 M 1 and A/p 2 M 2 /M 1. By (3.1), M has no strictly ascending chain of submodules, and therefore the above process must terminate. This yields the first assertion. Moreover, Ass(M i /M i 1 ) = Ass(A/p i ) = {p i }, and so by (1.3), we see that Ass(M) {p 1,..., p n }. Also, in view of (1.2) of [Gh], we have (M i /M i 1 ) pi A pi /p i A pi 0. Hence (M i ) pi 0. Thus {p 1,..., p n } Supp (M). Lastly, if p Supp (M), then M p 0 and so Ass Ap (M p ). Now (1.5) shows that there exists q Ass(M) with q (A \ p) =, i.e., q p. This implies the last assertion. (1.8) Corollary. If A is noetherian and M is f. g., then Ass(M) is finite. Furthermore, the minimal primes of M are precisely the minimal elements among the prime ideals of A containing Ann(M). Proof: Follows from (1.7) in view of (1.6). Remark: A chain 0 = M 0 M 1 M n = M of submodules of M is sometimes called a filtration of M. Using a filtration as in (1.7), it is often possible to reduce questions about modules to questions about integral domains. (1.9) Exercise: Show that if A is noetherian, M is f. g., and I is an ideal of A consisting only of zerodivisors of M, then there exists some x M such that x 0 and Ix = 0. (1.10) Exercise: Show that if A is noetherian and M is f. g., then Ann(M) = p = p = p Ass(M) p Supp (M) p a minimal prime of M p.

5 PRIMARY DECOMPOSITION OF MODULES 5 2. Primary Decomposition We continue to let A denote a ring and M an A module. As in the case of ideals, the primary decomposition of modules into primary submodules will be achieved using the auxiliary notion of irreducible submodules. To compare the notions and results discussed in this section to those in the classical case, you may substitute A for M. Definition: Let Q be a submodule of M. We say that Q is primary if Q M and for any a A and x M, we have ax Q and x / Q = a n M Q for some n 1. We say that Q is irreducible if Q M and for any submodules N 1 and N 2 of M we have Q = N 1 N 2 = Q = N 1 or Q = N 2. Clearly, a submodule Q of M is primary iff every zerodivisor of M/Q is nilpotent for M/Q. [An element a A is said to be nilpotent for M if a n M = 0 for some n 1. In other words, a is nilpotent for M iff a Ann(M).] If Q is a primary submodule of M and p = Ann(M/Q), we say that Q is p primary. (2.1) Exercise: Given any submodule Q of M, show that Z(M/Q) = Ann(M/Q) Q is primary = Ann(M/Q) is a primary ideal of A Use (1.2), (1.8) and (1.10) to deduce the following characterization. If A is noetherian and M is f. g., then: And also the following characterization. If A is noetherian and M is f. g., then: Q is primary Ass(M/Q) is singleton. Q is p primary Ass(M/Q) = {p}. As we shall see in the sequel, the above characterization of primary submodules [of f. g. modules over noetherian rings] is extremely useful. For this reason perhaps, it is sometimes taken as a definition of primary submodules [of arbitrary modules]. At any rate, we may tacitly use the above characterizations of primary and p primary submodules in several of the proofs below. (2.2) Lemma. Suppose A is noetherian, M is f. g., and Q 1,..., Q r are p primary submodules of M, where r is a positive integer. Then Q 1 Q r is also p primary. Proof: Clearly, Q 1 Q r M. Moreover, there is a natural injective homomorphism of M/Q 1 Q r into M/Q 1 M/Q r. Therefore, in view of (1.1), (1.3) and (1.4), we see that Ass(M/Q 1 Q r ) Ass ( r i=1m/q i ) = r i=1ass(m/q i ) = {p}. Thus it follows from (2.1) that Q 1 Q r is p primary. (2.3) Lemma. If M is noetherian, then every submodule of M is a finite intersection of irreducible submodules of M.

6 6 SUDHIR R. GHORPADE AND JUGAL K. VERMA Proof: Assume the contrary. Then we can find a maximal element, say Q, among the submodules of M which aren t finite intersections of irreducible submodules of M. Now Q can t be irreducible. Also Q M (because M is the intersection of the empty family of irreducible submodules of M). Hence Q = N 1 N 2 for some submodules N 1 and N 2 of M with N 1 Q and N 2 Q. By maximality of Q, both N 1 and N 2 are finite intersections of irreducible submodules of M. But then so is Q, which is a contradiction. (2.4) Lemma. Suppose A is noetherian, M is f. g., and Q is an irreducible submodule of M. Then Q is primary. Proof: Since Q M, Ass(M/Q). Suppose Ass(M/Q) contains two distinct prime ideals p 1 = (0 : x 1 ) and p 2 = (0 : x 2 ), where x 1, x 2 denote the images in M/Q of some elements x 1, x 2 of M. Clearly x 1 and x 2 are nonzero elements of M/Q. We claim that A x 1 A x 2 = {0} Indeed, if a x 1 = b x 2, with a, b A, is nonzero, then a / (0 : x 1 ) and b / (0 : x 2 ). Since (0 : x 1 ) is prime, we find that (0 : x 1 ) = (0 : a x 1 ) (check!). Similarly, (0 : x 2 ) = (0 : b x 2 ). This gives p 1 = p 2, which is a contradiction. Now if y (Q + Ax 1 ) (Q + Ax 2 ), then y = y 1 + ax 1 = y 2 + bx 2 for some y 1, y 2 Q and a, b A. But then a x 1 = b x 2 in M/Q and thus y Q. It follows that Q = (Q + Ax 1 ) (Q + Ax 2 ). Also since x 1 0 x 2, we have (Q + Ax 1 ) Q (Q + Ax 2 ). This contradicts the irreducibility of Q. Thus Ass(M/Q) is singleton so that Q is primary. (2.5) Lemma. Suppose A is noetherian, M is f. g., Q is a p primary submodule of M. Then the inverse image of Q p under the natural map M M p (given by x x ) is Q. 1 Proof: Suppose x M is such that x Q 1 p. Then tx Q for some t A \ p. If x / Q, then x, the image of x in M/Q, is nonzero, and thus t Z(M/Q). Hence from (2.1), we see that t p, which is a contradiction. (2.6) Remark: Given any p Spec A and a submodule Q of M p, the inverse image of Q under the natural map M M p is often denoted by Q M. Thus (2.5) can be expressed by saying that if Q is a p primary submodule of M, then Q p M = Q. Note that we have been tacitly using the fact that if Q is any submodule of M and S is any m. c. subset of A, then S 1 Q can be regarded as a submodule of S 1 M. (2.7) Theorem. Suppose A is noetherian, M is f. g., and N is any submodule of M. Then we have (i) There exist primary submodules Q 1,..., Q h of M such that N = Q 1 Q h. (ii) In (i) above, Q 1,..., Q h can be chosen such that Q i j i Q j for 1 i h, and p 1,..., p h are distinct, where p i = Ann(M/Q i ). (iii) If Q i and p i are as in (ii) above, then p 1,..., p h are unique; in fact, {p 1,..., p h } = Ass(M/N). Moreover, if p i is minimal among p 1,..., p h, i.e. p i p j for j i, then the corresponding primary submodule Q i is also unique; in fact, Q i = N pi M. Proof: Clearly, (i) is a direct consequence of (2.3) and (2.4). Given a decomposition as in (i), we can use (2.2) to reduce it by grouping together the primary submodules having the same associated prime so as to ensure that the associated primes become distinct. Then

13 PRIMARY DECOMPOSITION OF MODULES 13 (iii) If p i is minimal among p 1,..., p h, then Q i is a graded submodule of N. Proof: Applying (4.1) to M/N, we get p i = p i. So by (4.6), we find that Q i is p i primary. Since N is graded and N Q i, we have N Q i. Thus N Q 1 Q h Q 1 Q h = N, which proves (i). If N = Q 1 Q h is an irredundant primary decomposition, then p 1,..., p h are distinct, and Ass(M/N) = {p 1,..., p h }. Thus the associated primes corresponding to Q 1,..., Q h are distinct. Moreover, if the decomposition N = Q 1 Q h can be shortened, then Ass(M/N) would have less than h elements, which is a contradiction. This proves (ii). Finally, if p i is minimal among p 1,..., p h, then the corresponding primary component Q i is unique. Hence from (i), we obtain Q i = Q i, i.e., Q i is graded. Remark: As a special case of the results of this section, we obtain some useful results about ideals in graded rings. You may find it instructive to write these down explicitly. (4.8) Exercise: Show that all the results of this section remain valid if R is replaced by a Z s graded ring and M by a Z s graded module over it. Deduce from this that if I is a monomial ideal in k[x 1,..., X n ], then the associated primes of I are monomial ideals and that I has a primary decomposition such that each of the primary ideals occurring in it is a monomial ideal. The following exercise indicates a constructive method to obtain primary decompositions of monomial ideals. (4.9) Exercise: Let J be a monomial ideal of k[x 1,..., X n ] and u, v be relatively prime monomials in k[x 1,..., X n ]. Show that (J, uv) = (J, u) (J, v). Also show that if e 1,..., e n are positive integers, then the ideal (X e 1 1,..., Xn en ) is (X 1,..., X n ) primary. Use these facts to determine the associated primes and a primary decomposition of the ideal I = (X 2 Y Z, Y 2 Z, Y Z 3 ) of k[x, Y, Z]. References [Ab] S. S. Abhyankar, Algebraic Geometry for Scientists and Engineers, American Math. Society, [AM] M. Atiyah and I. G. MacDonald, Introduction to Commutative Algebra, Addison Wesley, [Bo] N. Bourbaki, Commutative Algebra, Hermann, [BH] W. Bruns and J. Herzog, Cohen-Macaulay Rings, Cambridge University Press, [Ei] D. Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry, Springer Verlag, [Gh] S. R. Ghorpade, A Quick Review of Commutative Algebra, Unpublished lecture notes, [Ka] I. Kaplansky, Commutative Rings, The University of Chicago Press, [KS] A. I. Kostrikin and I. R. Shafarevich (Eds), Algebra I : Basic Notions of Algebra, Springer Verlag, [Ku] E. Kunz, Introduction to Commutative Algebra and Algebraic Geometry, Birkhäuser, [M1] H. Matsumura, Commutative Algebra, Benjamin, [M2] H. Matsumura, Commutative Ring Theory, Cambridge University Press, [N1] D. G. Northcott, Ideal Theory, Cambridge University Press, [N2] D. G. Northcott, Lessons on Rings, Modules and Multiplicities, Cambridge University Press, [Sh] R. Y. Sharp, Steps in Commutative Algebra, Cambridge University Press, [Ta] D. E. Taylor, From rainbows to rings: a history of the idea of the spectrum, in: Fourier techniques and applications (Kensington, 1983), pp , Plenum Press, [ZS] O. Zariski and P. Samuel, Commutative Algebra, Vol. 1, Van Nostrand, 1958.

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