Associated Prime Submodules of a Multiplication Module

Honam Mathematical J. 39 (2017), No. 2, pp. 275–296 https://doi.org/10.5831/HMJ.2017.39.2.275

ASSOCIATED PRIME SUBMODULES OF A MULTIPLICATION MODULE

Sang Cheol Lee, Yeong Moo Song∗, and Rezvan Varmazyar

Abstract. All rings considered here are commutative rings with identity and all modules considered here are unital left modules. A submodule N of an R-module M is said to be extended to M if N = aM for some ideal a of R and it is said to be fully invariant if ϕ(L) ⊆ L for every ϕ ∈ End(M). An R-module M is called a [resp., fully invariant] multiplication module if every [resp., fully invariant] submodule is extended to M. The class of fully invariant multiplication modules is bigger than the class of multiplication modules. We deal with prime submodules and associated prime submodules of fully invariant multiplication modules. In particular, when M is a nonzero faithful multiplication module over a Noetherian ring, we characterize the zero-divisors of M in terms of the associated prime submodules, and we show that the set Aps(M) of associated prime submodules of M determines the set ZdvM (M) of zero-dvisors of M and the support Supp(M) of M.

1. Introduction

Every ring considered in this paper is a commutative ring with identity and every module considered is a unital left module. A prime number in the ring of integers is generalized to a prime ideal in a ring. Further- more a prime ideal in a ring is generalized to a prime submodule in a module. A number of authors have studied for the subject for a couple of decades and many results have been given to explore the nature of prime submodules and related concepts. In this paper, we will focus on one of the most useful and well-known types of prime submodules mentioned, the associated prime submodules.

Received April 3, 2017. Accepted May 10, 2017. 2010 Mathematics Subject Classiﬁcation. 13E05, 14A05, 13C99, 13C10. Key words and phrases. primes, associated primes, multiplication modules, fully invariant multiplication modules. The second author∗ was partially supported by Sunchon National University Research Fund in 2016. 276 Sang Cheol Lee, Yeong Moo Song ∗, and Rezvan Varmazyar

A submodule N of an R-module M is said to be extended to M if N = aM for some ideal a of R and it is said to be fully invariant if ϕ(N) ⊆ N for every ϕ ∈ End(M). An R-module M is called a multiplication module if each submodule N of M is extended. An R- module M is called a fully invariant multiplication module if each fully invariant submodule N of M is extended (see [17].) As Patrick F. Smith mentioned in the paper [17], the study of multiplication modules is due to [11]. A number of authors have studied the subject since then and many results have given to explore the nature of multiplication modules (see, for example, [16, p.180 ↑ 5].) In the paper [17], he considers the notion of fully invariant multiplication modules which is a natural generalization of multiplication modules. “The idea behind the paper was that the class of multiplication modules is quite small and much smaller than the class of fully invariant multiplication modules. However the bigger class shares some of the properties of the smaller one.” That is what he remembers. Every multiplication module is a fully invariant multiplication module. However, not every fully invariant multiplication module is a multiplication module. In section 2, we find the six conditions under which every fully invariant multiplication module is a multiplication module (see Theorem 2.3.) Assume that M is a finitely generated module over a Noetherian ring. If every prime submodule is extended and for every submodule N of M which is not prime the ideal (N :R M) is prime, then it is shown that M is a fully invariant multiplication module (see Theorem 2.6.) Now assume that M is a fully invariant multiplication module. We introduce the set X of prime submodules of M to a Zariski topology T . The resulting topological space (X, T ) is called the prime spectrum of M, denoting Spec(M). And then we study the basic properties of Spec(M) and the relative topological space (Y, TY ). Here Y = {P ∈ X | (P : M) is maximal}. If M is a finitely generated multiplication module over a ring R, then it is shown that Spec(M) is homeomorphic to Spec(R/AnnR(M)) (see Theorem 2.22.) and Y is homeomorphic to the maximal spectrum Ω(R/Ann(M)) of R/Ann(M) (see Corollary 2.23.) For any R-module M, the support of M is defined to be Supp(M) = {p ∈ Spec(R) | Mp 6= 0}. A prime ideal p of R is said to be associated to M if there exists an injective R-linear map R/p → M. The support of M and the set Ass(M) of associated primes to M are both the subsets of Spec(R) and they are both defined from Ann(M). It is shown that Ass(M) ⊆ Supp(M) (see Corollary 3.3.) A prime submodule P of an Associated Prime Submodules of a Multiplication Module 277

R-module M is said to be associated to M if the prime ideal (P :R M) of R is associated to M. If M is a finitely generated multiplication module over a ring R, then we show in Theorem 3.9 that the set Aps(M) of associated primes to M is Aps(M) = Spec(M) ∩ {AnnM (m) | m ∈ M} and in Theorem 3.12 Aps(M) ∩ Y = Spec(M) ∩ {M(m) | m ∈ Ass(M) ∩ Ω(R)}. Let R be any ring and let M be an R-module. An element r of R is called a zero-divisor on M if rm = 0 for some nonzero m of M. The set of zero-divisors on M will be written ZdvR(M). If M is a nonzero finitely generated module over a Noetherain ring, then it is shown in Theorem 4.3 that |Ass(M)| < ∞; hence [ ZdvR(M) = p. p∈Ass(M) |Ass(M)|<∞ Let R be any ring and let M be an R-module. An element m of M is called a zero-divisor of M if am = 0 for some nonzero ideal a of R. (Notice that an element m of M is called a singular element if em = 0 for some nonzero essential ideal e of R.) The set of zero-divisiors of M will be written ZdvM (M). This is not a submodule of M in general. A ring R is said to be von Neumann regular if for each a ∈ R there exists a b ∈ R such that a = aba. If M is a fully invariant multiplication module and if R is a von Neumann regular ring, then it is shown in Proposition 4.5 that ZdvR(R)M = hZdvM (M)i. Finally, we show that the set Aps(M) of associated prime submodules of an R-module M determine the set ZdvM (M) of zero-divisors of M and Supp(M). More specifically, assume that M is a nonzero faithful multiplication module over a Noetherian ring R. Then it is shown in Theorem 4.9 that |Aps(M)| < ∞; hence [ ZdvM (M) = P. P ∈Aps(M) |Aps(M)|<∞ Moreover if ψ : Spec(M) → Spec(R) defined by ψ(P ) = (P : M) is surjective, then it is shown in Corollary 4.14 that [ ψ−1(Supp(M)) = {Q}; Q∈Aps(M) |Aps(M)|<∞ hence ψ−1(Supp(M)) is closed. For undefined terms see [4], [13] and [14]. 278 Sang Cheol Lee, Yeong Moo Song ∗, and Rezvan Varmazyar

2. Prime Spectra of Fully Invariant Multiplication Modules

In this section, we will discuss that prime submodules play a critical role in fully invariant multiplication modules like prime ideals do in commutative rings. A submodule L of an R-module M is called fully invariant if ϕ(L) ⊆ L for every ϕ ∈ End(M) (see [17].) The zero submodule and M itself are fully invariant. Any submodule generated by the union of fully invariant submodules is fully invariant, and the intersection of fully invariant submodules is fully invariant. A submodule N of an R-module M is said to be extended to M if there exists an ideal a of R such that N = aM. Lemma 2.1. Let R be a ring and let M be an R-module. Then every extended submodule of M is fully invariant. Not every fully invariant submodule of an R-module is extended. For example, in the paragraph just prior to [17, Proposition 2.3], Patrick F. Smith constructed an example of a Z-module N such that the socle L of N is a fully invariant submodule of N, but N is not extended. Every multiplication module is a fully invariant multiplication module. However, not every fully invariant multiplication module is a multiplication module. The example of this is given below. Example 2.2. Let R be any ring. Consider a free R-module F of rank ≥ 2. Then by [16, Lemma 1.3 and Example 2.8] F is not a multiplication module, but by [17, Corollary 2.10] it is a fully invariant multiplication module. In view of this, it is natural for us to have a question“under what conditions is every fully invariant multiplication module a multiplication module?” To answer this question, let us consider for each element a ∈ R the map ϕa : M → M deﬁned by ϕa(m) = am, where m ∈ M. Clearly each ϕa is an endomorphism, which is called a multiplication of an R- module M by a. Let Mul(M) = {ϕa | a ∈ R} and let End(M) be the endomorphism ring of M. Then Mul(M) ⊆ End(M). The following result provides us with an answer to the question above. Theorem 2.3. Let R be any ring. Consider the following statements for any R-module M. 1. M is a multiplication module. 2. Every cyclic submodule of M can be extended to M. 3. Every ﬁnitely generated submodule of M can be extended to M. 4. Mul(M) = End(M). Associated Prime Submodules of a Multiplication Module 279

5. Every cyclic submodule of M is fully invariant. 6. Every finitely generated submodule of M is fully invariant. Then (1) ⇔ (2) ⇔ (3) ⇒ (4) ⇒ (5) ⇔ (6). If M is a fully invariant multiplication module, then the six statements are equivalent. Proof. We prove the first statement. (1) ⇔ (2) Apply [5, Proposition 1.1]. (2) ⇔ (3) Clear. (2) ⇒ (4) This follows from the fact that M is fully invariant. (4) ⇒ (5) ⇔ (6) Clear. For the second statement, if M is a fully invariant multiplication module, then clearly (5) ⇒ (2) and hence the six statements are equivalent. Remark 2.4. In the above result, (2) ⇒ (5) follows from Lemma 2.1. Lemma 2.5. Let R be any ring and let M be an R-module. Let P be any submodule of M. If P + rM and (P :M r) are extended to M, and (P :R M) is a prime ideal of R, then P is extended to M. Theorem 2.6. Assume that M is a finitely generated module over a Noetherian ring R. If every prime submodule is extended and for every submodule N of M which is not prime the ideal (N :R M) is prime, then M is a fully invariant multiplication module.

Proof. Suppose that there exists a fully invariant submodule N0 of M such that N0 is not extended. Let Σ be the set of fully invariant submodules of M which are not extended. Then N0 ∈ Σ 6= ∅. Order Σ by inclusion. Then Σ is partially ordered. Since M is a Noetherian R-module, Σ has a maximal member, say P . If P were prime, then by our assumption it could be extended to M. Hence P is not prime. Then by assumption the ideal (P :R M) is prime. By Lemma 2.5 P can be extended to M. This contradiction shows that each fully invariant submodule is extended. Therefore M is a fully invariant multiplication module. A ring R is called a multiplication ring if a ⊆ b, where a and b are ideals of R, implies that there exists an ideal c of R such that a = bc (see [8, Deﬁnition 9.12].) Theorem 2.7. Let R be a ring and let M be an R-module. If R is a multiplication ring and M is a fully invariant multiplication module, then every fully invariant submodule K of M is a fully invariant multiplication module. Proof. Let L be any fully invariant submodule of K. Then since K is a fully invariant submodule of M, it follows from [17, Lemma 2.5] that 280 Sang Cheol Lee, Yeong Moo Song ∗, and Rezvan Varmazyar

L is a fully invariant submodule of M. There exist ideals a, b of R such that L = aM and K = bM. Notice that K = (K : M)M. aM ⊆ bM implies that a ⊆ (bM : M) = (K : M). Since R is a multiplication ring, there exists an ideal c of R such that a = (K : M)c. This implies that L = aM = ((K : M)c)M = c((K : M)M) = cK. Hence L is a fully invariant multiplication module. The notion of prime submodules is a natural generalization of prime ideals. However the class of prime submodules does not share every property of prime ideals. For example, prime ideals have the following property: If the intersection of two ideals is contained in a prime ideal, then either one of these is contained in the prime ideal. However prime submodules does not share the property. In other words, we can find submodules A, B of an R-module M and a prime submodule P of M such that A ∩ B ⊆ P , but A * P and B * P . The example of this is given below. Example 2.8. Consider a polynomial ring R[x] over an integral domain R. Then R[x] and hence the external direct sum (R[x])2(= R[x] ⊕ R[x]) of two R[x]’s can be given an R-module structure. Let M be the R-module (R[x])2. Let’s write the submodules R[x] × (x) and {(f, f) | f ∈ R[x]} by A and B, respectively, and write the submodule (x)×(x) by P . Then P is a prime submodule of M. In fact, (1) P 6= M. (2) Assume that rm ∈ P , where r ∈ R and m ∈ M. Then we claim that either r ∈ (P :R M) or m ∈ P . Suppose that m∈ / P . Write m = (g, h), where g, h ∈ R[x]. Then g(0) 6= 0 or h(0) 6= 0. Assume that g(0) 6= 0. Since (rg, rh) = rm ∈ P , we have rg(0) = 0. Hence, r = 0. Similarly, if h(0) 6= 0, then r = 0 as well. In either case, r = 0 ∈ (P :R M). Therefore, P is a prime submodule of M. Moreover, A ∩ B ⊆ P , but A * P and B * P . Remark 2.9. Of course, (R[x])2 in Example 2.8 becomes a ring as well. The ideal (x) × (x) of the ring is not prime (see [6, Proposition 1.5.22].) We show that the class of prime submodules of fully invariant multiplication modules shares the property of prime ideals. We define an algebraic structure ∗ on the class S of fully invariant submodules of a fully invariant multiplication module M over a ring R. Let A, B ∈ S. Then A = IM and B = JM for some ideals I and J of R. We define Associated Prime Submodules of a Multiplication Module 281 the product A ∗ B of A and B to be A ∗ B = (IJ)M. (This definition is the same as in [1].) Then (IJ)M ∈ S. Moreover [1, Theorem 3.4] says that the product of A and B is independent of presentation ideals I,J of A, B, respectively. This independence property comes basically from the following result: Lemma 2.10. Let R be a ring and let M be an R-module. If I,I0, J, J 0 are ideals of R such that IM = I0M and JM = J 0M, then (IJ)M = (I0J 0)M. Hence ∗ is well-defined on S. It is immediate that (S, ∗) is commutative and associative. If A, B ∈ S, then A ∗ B ⊆ A ∩ B. From now on, the class of fully invariant multiplication modules will be written C. Lemma 2.11. Let M ∈ C, and let A, B ∈ S. If P is a prime submodule of M such that A ∗ B ⊆ P , then either A ⊆ P or B ⊆ P . Proof. If P is a prime submodule of M, then it is well known that the ideal (P :R M) is prime. Apply this result. Corollary 2.12. Let M ∈ C and let A, B ∈ S. If P is a prime submodule of M such that A ∩ B ⊆ P , then either A ⊆ P or B ⊆ P . Let M ∈ C and let X denote the set of all prime submodules of M. Assume that X 6= ∅. (For example, the zero submodule of a simple R- module is prime and the zero subspace of a vector space is prime.) For each fully invariant submodule N of M, let V (N) = {P ∈ X | N ⊆ P }. Then the following statements are true: 1. V (0) = X, V (M) = ∅. 2. If F is any family of fully invariant submodules of M, then X V ( N) = ∩N∈F V (N). N∈F

3. V (N1∩N2) = V (N1)∪V (N2) for fully invariant submodules N1,N2 of M. This follows from Corollary 2.12. These results show that the subfamily T = { V (N) | N is a fully invariant submodule of M} of the power set P(X) of X satisﬁes the axioms for closed sets in a topological space. The resulting topology T is called the Zariski topology. The topological space (X, T ) is called the prime spectrum of M, and is written Spec(M). This is a natural generalization of the prime spectrum Spec(R) of a ring R. If M is a simple R-module, then the resulting topology is indiscrete. Let P ∈ X. Then (P : M)M is fully invariant 282 Sang Cheol Lee, Yeong Moo Song ∗, and Rezvan Varmazyar and P ∈ V ((P : M)M). This means that every prime submodule of M is contained in a closed subset of X. Hence X = ∪P ∈X V ((P : M)M). Deﬁne a relation ∼ on X as follows. For P,Q ∈ X, P ∼ Q ⇔ (P : M) = (Q : M). Then ∼ is an equivalence relation on X. For P ∈ X, let P¯ = {Q ∈ X | P ∼ Q}.

Then for each P,Q ∈ X, either P¯ = Q¯ or P¯∩Q¯ = ∅. Also, X = ∪P ∈X P¯. Hence {P¯ | P ∈ X} is a partition of X. It is immediate that for each P ∈ X, P¯ ⊆ V ((P : M)M). Let us summarize the result: Proposition 2.13. The partition {P¯ | P ∈ X} of X is a reﬁnement of a covering {V ((P : M)M) | P ∈ X} of X. Let Y = {P ∈ X | (P : M) is maximal}. Let

TY = {(X\V (N)) ∩ Y | N is a fully invariant submodule of M}.

This is a relative topology on Y , so that (Y, TY ) is a subspace of (X, T ). For all P ∈ Y , P¯ = V ((P : M)M). Lemma 2.14. Let P,Q be proper submodules of an R-module M. Then the following are true. 1.( P : M)M ⊆ Q ⇔ (P : M) ⊆ (Q : M) 2. Let P ∈ Y,Q ∈ X. Then (P : M)M ⊆ Q ⇔ (P : M) = (Q : M). 3. Let P,Q ∈ X. Then (P : M)M * Q or (Q : M)M * P ⇔ (P : M) 6= (Q : M). 4. Let P ∈ Y . Then (P : M)M * Q ⇔ (P : M) 6= (Q : M). Theorem 2.15. For any M ∈ C, we have the following properties. 1. If the set {P } is closed in X, then P is a maximal member of X. Furthermore, if P is extended, then the converse holds. 2. The closure {P } of {P } in X is identical to V ((P : M)M). 3. Q ∈ {P } ⇔ (P : M)M ⊆ Q. In particular, for all P ∈ Y , {P } = P¯ as sets. 4. Every prime submodule of M is fully invariant, then X is T0-space. Proof. (1) (⇒) {P } = V (N) for some invariant submodule N of M. This means that the only prime submodule of M containing N is P . Hence P is a maximal member of X.(⇐) Assume that P is a maximal member of X and it is extended. Then P is fully invariant and {P } = V (P ). Associated Prime Submodules of a Multiplication Module 283

(2) We have already known that {P } ⊆ P¯ = V ((P : M)M), so that {P } ⊆ V ((P : M)M). Conversely, let Q ∈ V ((P : M)M). Let N be a fully invariant submodule of M such that {P } ⊆ V (N). Then N is extended and N ⊆ P , so that N = (N : M)M ⊆ (P : M)M ⊆ Q, so that Q ∈ V (N). Hence Q ∈ ∩{P }⊆V (N)V (N) = {P }. This shows that V ((P : M)M) ⊆ {P }. Therefore {P } = V ((P : M)M). (3) Use (2) to prove the ﬁrst part, and then use Lemma 2.14 (4) to prove the second one, which is a special case. (4) Let P,Q ∈ X with P 6= Q. Then either P * Q or Q * P . If P * Q, then by (3) X\{P } is an open neighborhood of Q and it does not contain P . Or, if Q * P , then by (3) X\{Q} is an open neighborhood of P and it does not contain Q. Remark 2.16. We hope that the reader does not confuse {P }, the closure of {P } (= the intersection of closed subsets of X containing {P }) with P¯, the equivalence class on X containing P . Corollary 2.17. Let M ∈ C. {P } is closed in X if and only if the prime submodule of M containing (P : M)M is P itself.

Proof. {P } is closed in X ⇔ {P } = {P } ⇔ V ((P : M)M) = {P } by Theorem 2.15(2). Deﬁne µ : L(M) → L(R) by µ(N) = (N : M), where N ∈ L(M) (see [12] for the notations.) Let M ∈ C and let X denote the set of all prime submodules of M. Assume that X 6= ∅. For a submodule N of a module M, deﬁne D(N) = X\V (N). Theorem 2.18. Let M ∈ C. If µ is injective and each prime submodule of M is simple, then Y is a T2-space. Proof. Assume that P 6= Q, where P,Q ∈ Y . If (Q : M)M ⊆ P , then (Q : M) ⊆ (P : M), so that (Q : M) = (P : M). Since µ is injective, we have Q = P , a contradiction. Hence (Q : M)M * P , so that P ∈ D((Q : M)M) ∩ Y . Similarly, Q ∈ D((P : M)M) ∩ Y . Moreover, if P ∩ Q 6= 0, then P = Q, because P and Q are simple. This contradiction shows that P ∩ Q = 0. Hence (D((P : M)M) ∩ Y ) ∩ (D((Q : M)M) ∩ Y ) = ∅.

This shows that Y is a T2-space. 284 Sang Cheol Lee, Yeong Moo Song ∗, and Rezvan Varmazyar

Deﬁnition 2.19. Let P be a prime submodule of an R-module M. Then P is said to be completely reducible to a prime submodule if for each proper submodule N of P there exists a prime submodule P 0 of P such that P = P 0 ⊕ N. Theorem 2.20. Let M ∈ C. If µ is injective and every prime submodule of M is completely reducible to a prime submodule, then Y is a T2-space Proof. Let P,Q ∈ Y with P 6= Q. If P ∩Q = P , then P ⊆ Q, so that (P : M) = (Q : M). Hence P = Q, a contradiction. This shows that P ∩ Q 6= P . Similarly, P ∩ Q 6= Q. By our hypothesis, there are prime submodules P 0,Q0 of P,Q , respectively, such that P = P 0 ⊕ (P ∩ Q) and Q = Q0 ⊕ (P ∩ Q). By the modular law, P ∩ Q = P ∩ (Q0 + (P ∩ Q)) = (P ∩ Q0) + (P ∩ Q) = (P 0 + (P ∩ Q)) ∩ Q0 + (P ∩ Q) ⊇ (P 0 ∩ Q0) + (P ∩ Q) ⊇ P 0 ∩ Q0. It follows that P 0 ∩ Q0 = (P 0 ∩ Q0) ∩ (P ∩ Q) = 0, so that (D((P 0 : M)M) ∩ Y ) ∩ (D((Q0 : M)M) ∩ Y ) = ∅. 0 0 0 Moreover, (P : M)M * Q and (Q : M)M * P , so that Q ∈ D((P : 0 M)M) ∩ Y and P ∈ D((Q : M)M) ∩ Y . Therefore Y is a T2-space.

Let R be any ring and let M be an R-module. It is always true that (N :R M) ∈ V (AnnR(M)) for any submodule N of M. Consider a map ϕ : Spec(M) → Spec(R/Ann(M)) by ϕ(P ) = (P :R M)/Ann(M), where P ∈ Spec(M). Then ϕ is well- deﬁned. Now, let N be a submodule of M such that (N :R M) is a prime ideal of R. Then we cannot say that N is a prime submodule of M. The example of this is given below. Example 2.21. As in Example 2.8, consider the polynomial ring R[x] over a domain R. Then R[x] can be viewed as a module over itself and R[x] × R[x] can be given an R[x]-module structure. Write M = R[x] × R[x], and N = 0 × (x).

Then N is a submodule of M.(N :R[x] M) = 0, so that (N :R[x] M) is a prime ideal of R[x] because the polynomial ring R[x] over a domain R is also a domain. However, x(0, 1) = (0, x) ∈ N, x∈ / (N :R[x] M), (0, 1) ∈/ N, so that N is not a prime submodule of M. Associated Prime Submodules of a Multiplication Module 285

Let R be any ring and let M be a nonzero R-module. Then Ann(M) $ R, so that there is a maximal ideal m in R such that Ann(M) ⊆ m. Hence Spec(R/Ann(M)) 6= ∅. Let p0 be any member of Spec(R/Ann(M)). Then p0 = p/Ann(M) for some p ∈ V (Ann(M)). If M is ﬁnitely generated, then it follows that pM 6= M and (pM : M) = p for every p ∈ V (Ann(M)). Moreover, if M is a multiplication module, then it follows from [5, Corollary 2.11] that pM is a prime submodule of M. This shows that the map ϕ : Spec(M) → Spec(R/AnnR(M)) is surjective. The following result is probably well known, but we state and prove it for our record. Theorem 2.22. If M is a nonzero ﬁnitely generated multiplication module over a ring R, then Spec(M) is homeomorphic to Spec(R/AnnR(M)). Proof. We have shown just before that the map ϕ is surjective. Since M is a multiplication module, ϕ is injective. Thus ϕ is bijective. It −1 is routine for us to show that ϕ (D(a/AnnR(M))) = D(aM) for any ideal a of R containing AnnR(M). Hence ϕ is continuous. Let ψ : Spec(R/AnnR(M)) → Spec(M) be the inverse of ϕ. Then for any sub- −1 module N of M, ψ (D(N)) = D((N :R M)/AnnR(M)). ψ is continuous. ϕ is a homeomorphism.

Let R be any ring. The set of maximal ideals of R will be written Ω(R), which is called the maximal spectrum of R. Corollary 2.23. With the same hypothesis as in Theorem 2.22, Y is homeomorphic to the maximal spectrum Ω(R/Ann(M)) of R/Ann(M).

3. Primes of ﬁnitely generated multiplication modules

As a preliminary to the next section we will find the set of associated primes of finitely generated multiplication modules and in particular the set of their associated primes P with the ideal (P :R M) being maximal. Now, we turn our attention to associated prime submodules of an R-module M. For any R-module M, Ann(M) = ∩x∈M Ann(x). Let p be a prime ideal of R. We are not sure whether p contains Ann(M) or not. Even if p contains Ann(M), we cannot say that p contains some Ann(x). (Of course, if p contains Ann(M) and M is finitely generated, then p contains some Ann(x).) In relevance to this, we can get two well known results and we can see from the results that Supp(M) and Ass(M) are very closely related to each other. 286 Sang Cheol Lee, Yeong Moo Song ∗, and Rezvan Varmazyar

Lemma 3.1. Let R be any ring and let M be an R-module. Let p be a prime ideal of R. Then

p ⊇ Ann(x) for some x ∈ M ⇐⇒ Mp 6= 0.

The support of M is deﬁned to be Supp(M) = {p ∈ Spec(R) | Mp 6= 0} (see [2, Chapter 3, Exercise 19, p.46], [9, p.25], and [13, Deﬁnition 9.14].) Then by Lemma 3.1 the support of M is the set of primes p which contain Ann(x) for some x ∈ M.

Lemma 3.2. Let R be any ring and let M be an R-module. The following are equivalent for any prime ideal p of R. 1. p = Ann(x) for some nonzero x ∈ M. 2. There exists an injective R-linear map R/p → M.

A prime ideal p of R is said to be associated to M if p satisﬁes one (and hence both) of the equivalent conditions in Lemma 3.2 (see [9, p.38] and [6, p.194].) For example, let R = k[x, y] be the polynomial ring with two indeterminates x, y over a ﬁeld k and let M = k[x, y]/(x2, xy). Then the prime ideal (x, y) of R is associated to the R-module M because 2 (x, y) = AnnR(x + (x , xy)). The set of associated primes to M is written Ass(M).

Corollary 3.3. For any R-module M, Ass(M) ⊆ Supp(M).

Proof. This follows from Lemma 3.2 and Lemma 3.1.

Deﬁnition 3.4. Let M be an R-module. A prime submodule P of M is said to be associated to M if the prime ideal (P :R M) of R is associated to M.

Example 3.5. Let R and M be the same ring and module as in the example just posterior to Lemma 3.2. Let p = (x, y). Then we have already known in there that p is an associated prime ideal of R. Let P = (x, y)/(x2, xy). Then P is a prime submodule of M and (P :R M) = p. Hence P is an associated prime submodule of M. Lemma 3.6. Let R be any ring. Then the following statements are equivalent for any R-module M. 1. A prime submodule P of M is associated to M. 2. There is an injective R-linear map R/(P :R M) → M. 3.( P :R M) = AnnR(m) for some nonzero m ∈ M. Associated Prime Submodules of a Multiplication Module 287

Deﬁnition 3.7. Let R be any ring and let M be any R-module. Then the annihilator of an element m ∈ M is deﬁned to be the submodule of M

AnnM (m) = {x ∈ M | (Rm :R M)Rx = 0}. Lemma 3.8. If M ∈ C, then for every m ∈ M

AnnR(m)M ⊆ AnnM (m) = (AnnM (m):R M)M. In particular, if M is a multiplication module, then the ﬁrst two are equal and hence the three are equal. Proof.

(Rm :R M)(AnnR(m)M) = ((Rm :R M)AnnR(m))M

= AnnR(m)((Rm :R M)M) ⊆ AnnR(m)Rm = 0, so that AnnR(m)M ⊆ AnnM (m). It is immediate that AnnM (m) is fully invariant. Since M ∈ C, AnnM (m) can be extended to M, so that AnnM (m) = (AnnM (m):R M)M. For the second statement, let x ∈ AnnM (m). Then there exist r1, ··· , rn ∈ (AnnM (m):R M) and m1, ··· , mn ∈ M such that x = r1m1 + ··· + rnmn. If M is a multiplication module, then it follows from Theorem 2.3 that for each i ∈ {1, ··· , n}, riRm = (ri(Rm :R M))M = (Rm :R M)(riM) = 0, so that rim = 0; hence ri ∈ AnnR(m). Hence x ∈ AnnR(m)M. This shows that AnnM (m) ⊆ AnnR(m)M. Therefore the second statement follows.

Let M ∈ C. The set of associated prime submodules of M will be written Aps(M). Theorem 3.9. If M is a ﬁnitely generated multiplication module over a ring R, then

Aps(M) = Spec(M) ∩ {AnnM (m) | m ∈ M}. Proof. If M is a multiplication module, then it follows from Lemma 3.6 and Lemma 3.8 that the inclusion ⊆ holds. Conversely, assume that P is a prime submodule of M and there is m ∈ M such that P = AnnM (m). Then (P :R M)M = P = AnnR(m)M. Moreover, if M is ﬁnitely generated, then it follows from [15, Corollary to Theorem 9] that

(P :R M) = (P :R M)+AnnR(M) = AnnR(m)+AnnR(M) = AnnR(m), so that P ∈ Aps(M). The reverse inclusion ⊇ holds. 288 Sang Cheol Lee, Yeong Moo Song ∗, and Rezvan Varmazyar

Let M be an R-module and p be a prime ideal of R. As in [10], and [3], we deﬁne

M(p) = {x ∈ M | ax ⊆ pM for some ideal a of R with a * p}. Then M(p) is a submodule of M such that pM ⊆ M(p). We consider under what conditions M(p) = pM. First of all, if m is a maximal ideal of R, then we can see that M(m) = mM, since if a * m, then a+m = R; hence for any x ∈ M(m), x ∈ Rx = (a + m)x = ax + mx ∈ mM. Lemma 3.10. Let R be any ring and let M be an R-module. If p ∈ Spec(M)\V (Ann(M)), then M(p) contains a cyclic submodule of M.

Proof. If p ∈ Spec(R)\V (Ann(M), then Ann(M) * p. There exists an x ∈ M such that Ann(x) * p, so that Ann(x)x = 0 ⊆ pM. Hence x ∈ M(p). Lemma 3.11. Let R be any ring and let M be an R-module. If M is a nonzero ﬁnitely generated multiplication module, then for any p ∈ V (AnnR(M)), M(p) = pM. Proof. It suﬃces to prove that M(p) ⊆ pM. This can be proved to use the fact that (pM :R M) = p and pM is a prime submodule of M (see the paragraph just prior to Theorem 2.22.)

Let R be any ring and let M be an R-module. The set of prime ideals which are associated to M will be written Ass(M), as usual. Theorem 3.12. Let M be as in Lemma 3.11 and let Y be as in the paragraph just posterior to Proposition 2.13. Then Aps(M) ∩ Y = Spec(M) ∩ {M(m) | m ∈ Ass(M) ∩ Ω(R)}. Proof. The inclusion ⊆ follows from Lemma 3.11. Conversely, assume that M(m) ∈ Spec(M) for some m ∈ Ass(M) ∩ Ω(R). Then m = AnnR(m) for some m ∈ M. Hence by the paragraph just prior to Lemma 3.10, and Lemma 3.8, M(m) = mM = AnnR(m)M = AnnM (m), so that by Theorem 3.9, M(m) ∈ Aps(M). Moreover, m ∈ V (AnnR(M)), for otherwise R = m+Ann(M), so that M = mM; hence M = M(m), a contradiction. Since M is ﬁnitely generated, it follows that (M(m):R M) = (mM :R M) = m, which is maximal. Hence M(m) ∈ Y . This shows that M(m) ∈ Aps(M) ∩ Y . We have shown that the reverse inclusion ⊇ holds. The result follows. Associated Prime Submodules of a Multiplication Module 289

4. ZdvM (M), Supp(M), and Aps(M)

In this section, we show that the set Aps(M) of associated prime submodules of an R-module M determine the set ZdvM (M) of zero- divisors of M and Supp(M). Let R be any ring. For any R-module M, set

(4.1) F := {AnnM (x) | x ∈ M, x 6= 0}. Lemma 4.1. Let M be a ﬁnitely generated multiplication module. If F has a maximal member P , then P ∈ Aps(M).

Proof. We will use Theorem 3.9 to show that P ∈ Aps(M). At ﬁrst, P ∈ F, so there is a nonzero x ∈ M such that P = AnnM (x). Next, it suﬃces to prove that P ∈ Spec(M). Suppose that P = M. Then Rx = (Rx :R M)M = (Rx :R M)P = 0, so that x = 0. This contradiction shows that P 6= M. Now, assume that rm ∈ P , where r ∈ R, m ∈ M. We claim that r ∈ (P :R M) or m ∈ P . Suppose that r∈ / (P :R M). Then rM * P , so there is u ∈ M such that ru∈ / P . Since M is a multiplication module, it follows from Theorem 2.3 that Ru = aM for some ideal a of R, so that (Rra)Rx = Rra(Rx :R M)M = (Rx :R M)Rru 6= 0. Take a nonzero v in (Rra)Rx. Then it is immediate that AnnM (x) ⊆ AnnM (v). By the maximality of P , we have AnnM (x) = AnnM (v). Rm = bM for some ideal b of R. Since rm ∈ P , it follows that (bRr)Rx = (Rx :R M)Rrm = 0. This implies

(Rv :R M)Rm = bRv ⊆ (bRra)Rx = b((Rra)Rx) ⊆ (bRr)Rx = 0; hence m ∈ AnnM (v) = AnnM (x) = P . This shows that P ∈ Spec(M), as required.

It follows from [5, Corollary 3.9] that every multiplication module over a Noetherian ring is ﬁnitely generated (in particular, every multiplication module over a Noetherian local ring is cyclic.) Hence we have the following result.

Corollary 4.2. If M is a multiplication module over a Noetherian ring R, then Aps(M) 6= ∅.

The following result is well known (see the proof of [7, Theorem 80].) For our record, another proof of this is given below. Compare the following result with [13, Exercise 9.44]. 290 Sang Cheol Lee, Yeong Moo Song ∗, and Rezvan Varmazyar

Theorem 4.3. If M is a nonzero ﬁnitely generated module over a Noetherain ring, then |Ass(M)| < ∞; hence [ ZdvR(M) = p. p∈Ass(M) |Ass(M)|<∞

Moreover, if a is an ideal of R such that a ⊆ ZdvR(M), then a is contained in some p ∈ Ass(M). Proof. At ﬁrst, since M is a module over a Noetherian ring R, it follows from [13, Corollary 9.36] that ZdvR(M) = ∪p∈Ass(M)p. Take a nonzero element m0 ∈ M and then take an element r0 ∈ Ann(m0) (notice here that Ann(m0) may not an associated prime to M.) Then r0m0 = 0, so that r0 ∈ ZdvR(M) = ∪p∈Ass(M)p. There exists a p1 ∈ Ass(M) such that r0 ∈ p1; hence p1 = Ann(m1) for some m1 ∈ M. Let ∼ ∼ M0 = 0 and M1 = Rm1. Then M1/M0 = M1 = R/Ann(m1) = R/p1. If M1 = M, then we are done. If not, M/M1 is a nonzero module over a Noetherian ring R. It has an associated prime p2 = Ann(m2 + M1). Let ∼ M2 = Rm1 + Rm2. Then M2/M1 = R/p2. Continuing in this way, we get an ascending chain of submodules Mi in M. Since M is Noetherian, we must have M = Mn for some n. Hence we get an ascending chain

0 = M0 ⊂ M1 ⊂ · · · Mn−1 ⊂ Mn = M. (In fact, we have proved [13, Exercise 9.40].) By induction, we show that Ass(Mi) is ﬁnite for each i. Ass(M0) = ∅, so it is true for i = 0. Let i ≥ 0. Assume that the result is true for i. We have an exact sequence

0 → Mi → Mi+1 → R/pi+1 → 0. By [13, Exercise 9.42] and [13, Exercise 9.41],

Ass(Mi+1) ⊆ Ass(Mi) ∪ Ass(R/pi+1) = Ass(Mi) ∪ {pi+1}, so Ass(Mi+1) is finite. This proves that each Ass(Mi) is finite. In particular, Ass(M) = Ass(Mn) is finite. In fact, |Ass(M)| ≤ n = the length of the chain < ∞. The second statement comes from the Prime Avoidence Theorem, which is well-known in [13, 3.61], and [4, Lemma 3.3]. Corollary 4.4. Let (R, m) be a Noetherian local ring, and let M be a finitely generated R-module. If every element of m is a zero-divisor on M, then m ∈ Ass(M). Associated Prime Submodules of a Multiplication Module 291

Proposition 4.5. If M is a fully invariant multiplication module, then ZdvR(R)M ⊆ hZdvM (M)i = (hZdvM (M)i :R M)M. In particular, if R is a von Neumann regular ring, then ZdvR(R)M = hZdvM (M)i.

Proof. Clearly, hZdvM (M)i is fully invariant. If M is a fully invariant multiplication module, then it can be extended to M, so that hZdvM (M)i = (hZdvM (M)i :R M)M. It is easy to prove that ZdvR(R)M ⊆ hZdvM (M)i. Conversely, let x ∈ ZdvM (M). Then ax = 0 for some nonzero ideal a of R. Take a nonzero element r ∈ a. If R is a von Neu- mann regular ring, then every cyclic ideal of R can be generated by an idempotent element. So, we may assume that r is idempotent. Set s = 1 − r. Then sr = 0. This implies that s ∈ ZdvR(R), so that x = sx ∈ ZdvR(R)M. This shows that ZdvM (M) ⊆ ZdvR(R)M, so that hZdvM (M)i ⊆ ZdvR(R)M. The assertion follows. Lemma 4.6. Let R be any ring and let M be an R-module. If M is faithful and if m ∈ ZdvM (M), then AnnM (m) 6= 0. Moreover, if M is a faithful multiplication module, then m ∈ ZdvM (M) if and only if AnnM (m) 6= 0.

Proof. Assume that M is a faithful module. Let m ∈ ZdvM (M). Then am = 0 for some nonzero ideal a of R. aM 6= 0, so that we can take a nonzero n ∈ aM. Then it is immediate that (Rm :R M)Rn = 0. Thus Rn ⊆ AnnM (m); hence AnnM (m) 6= 0. Assume now that M is a multiplication module. If AnnM (m) 6= 0, then take a nonzero n in AnnM (m). Then Rn = bM for some nonzero ideal b of R and bm = (Rm :R M)bM = (Rm :R M)Rn = 0. Hence m ∈ ZdvM (M). The results follows.

Lemma 4.7. Let R be any ring and let M be an R-module. If M is a multiplication module, then for any m, n ∈ M, (Rm :R M)Rn = (Rn :R M)Rm; hence n ∈ AnnM (m) ⇔ m ∈ AnnM (n). Corollary 4.8. Let R be any ring and let M be an R-module. As- sume that M is a faithful multiplication module. Then m ∈ ZdvM (M) ⇔ there exists a nonzero n ∈ M such that m ∈ AnnM (n). Proof. By Lemma 4.6 and Lemma 4.7.

We characterize the zero-divisors of M in terms of the associated primes. 292 Sang Cheol Lee, Yeong Moo Song ∗, and Rezvan Varmazyar

Theorem 4.9. If M is a nonzero faithful multiplication module over a Noetherian ring R, then |Aps(M)| < ∞; hence [ ZdvM (M) = P. P ∈Aps(M) |Aps(M)|<∞ Proof. (We adopt the proof of [7, Theorem 80] to prove this result.) By Corollary 4.8 ZdvM (M) = ∪m(6=0)∈M AnnM (m). Consider the set F of (4.1). Let AnnM (m) be any member of F and let

G = {AnnM (n) ∈ F | AnnM (m) ⊆ AnnM (n)}. Then G= 6 ∅. By the statement just prior to Corollary 4.2, M is ﬁnitely generated. This implies that M is a Noetherian R-module, so that G has a maximal member. This maximal member of G is also a maximal member of F. By Lemma 4.1, the maximal member is in Aps(M). Hence ZM (M) = ∪P ∈Aps(M)P . We next show that |Aps(M)| < ∞. By Theorem 3.9, for each P ∈ Aps(M) there exists xP ∈ M such that P = AnnM (xP ). Consider the submodule N of M generated by all elements xP ∈ M, where AnnM (xP ) = P ∈ Aps(M). Since M is

Noetherian, N is ﬁnitely generated, say by xP1 , ··· , xPn . If any further x0s exist we have an equation

xPn+1 = r1xP1 + ··· + rnxPn , (ri ∈ R).

From this equation we have P1 ∩ · · · Pn ⊆ Pn+1. Notice that M is a fully invariant multiplication module and each Pi is fully invariant. Then by Corollary 2.12, some Pi (i = 1, ··· , n) is contained in Pn+1. By the maximality of Pi, we have Pi = Pn+1, a contradiction. Hence 0 0 there are no further x s or P s. Hence Aps(M) = {P1, ··· ,Pn} and ZdvM (M) = P1 ∪ · · · ∪ Pn. Compare the following result with [7, Theorem 80]. Corollary 4.10. If M is a nonzero faithful multiplication module over a Noetherian ring R, then the number of the maximal members in F of (4.1) is ﬁnite, and each is the annihilator of a nonzero element of M. Proof. By Theorem 4.9 and the Prime Avoidence Theorem. Let R be any ring. If p ∈ Spec(R), then let k(p) be the ﬁeld of fractions of the integral domain R/p. Then k(p) = Rp/pRp. Let M be an R-module. Then ∼ ∼ M ⊗R k(p) = M ⊗R (Rp/pRp) = (M ⊗R Rp)/(pRp(M ⊗R Rp)). Associated Prime Submodules of a Multiplication Module 293

∼ ∼ Since M ⊗R Rp = Mp, it follows that M ⊗R k(p) = Mp/((pRp)Mp). Hence M ⊗R k(p) is a vector space over k(p). If M is finitely generated over R, then M ⊗R k(p) is a finite-dimensional vector space over k(p). Let M be a finitely generated projective module over a Noetherian ring R. Then each p ∈ Spec(R), Mp is a free module over the Noetherian local ring Rp with finite rank. But we cannot say that all p ∈ Spec(R), Mp have the same rank. In other words, there is a finitely generated projective module M over a Noetherian ring R such that p 6= q, and Mp 6= Mq. (It is not easy for us to construct the example of this, but we can do.) Now assume that M is a finitely generated projective module over a Noetherian ring R and that it has constant rank n. Then it is known that for each p ∈ Spec(R), Mp is a free module over the Noetherian local ring Rp with the same rank n, so that Mp/pMp is an n-dimensional vector space over Rp/pRp. Hence each M ⊗R k(p) is an n-dimensional vector space over k(p). Write Tp(Spec(R)) = M ⊗R k(p). Define T (Spec(R)) to be the union of the k(p)-vector spaces Tp(Spec(R)) of all primes p ∈ Spec(R): [ T (Spec(R)) = Tp(Spec(R)). p∈Spec(R)

Define a map π : T (Spec(R)) → Spec(R) by π(vp) = p, where vp ∈ Tp(Spec(R)). Then the R-module M looks like to be a vector bundle (T (Spec(R)), Spec(R), π) of rank n over the topological space Spec(R) −1 and the fiber π (p) is Tp(Spec(R)). Lemma 4.11. Let R be any ring. Let M be a finitely generated R- module. For any p ∈ Spec(R), the following statements are equivalent. 1. p ∈ Supp(M). 2. p ⊇ Ann(x) for some x ∈ M. 3. p ⊇ Ann(M). 4. M ⊗R k(p) 6= 0. Proof. By Lemma 3.1 and [9, p.26], (1) ⇔ (2) ⇔ (3). By the Nakayama lemma (1) ⇒ (4). Clearly, (4) ⇒ (1). The following lemma is probably well known. Lemma 4.12. If M is a finitely generated module over a Noetherian ring R, then Supp(M) = ∪q∈Ass(M){q}. In other words, a prime ideal of R belongs to Supp(M) if and only if it contains an associated prime. For any R-module M, consider a map ϕ : Spec(M) → Spec(R/Ann(M)) defined in the proof of Theorem 2.22. 294 Sang Cheol Lee, Yeong Moo Song ∗, and Rezvan Varmazyar

• If M is a fully invariant multiplication module, then ϕ is injective. • If ϕ is surjective, then for any p ∈ V (AnnR(M)), there exists a P ∈ Spec(M) such that (P : M) = p. Deﬁne a map f : Spec(R/Ann(M)) → Spec(R) by f(p/Ann(M)) = p. Then f is well-deﬁned. And consider the composite map ψ = f ◦ ϕ. Then we get the following commutative diagram Spec(M) ϕ ψ

v & Spec(R/Ann(M)) / Spec(R) f Now we will consider −1 ψ (Supp(M)) = {P ∈ Spec(M) | (P :R M) ∈ Supp(M)}. Theorem 4.13. Assume that M is a faithful multiplication module over a Noetherian ring R and that ψ is surjective. Then P ∈ ψ−1(Supp(M)) if and only if it contains an associated prime submodule of M.

−1 Proof. Let P ∈ ψ (Supp(M)) and let p denote (P :R M). Then p ∈ Supp(M). Then by Lemma 4.12 p contains an associated prime q, so that P (= pM) contains qM. Since ψ is surjective, there exists a Q ∈ Spec(M) such that (Q : M) = q. Moreover, since M is ﬁnitely generated and q ∈ V (AnnR(M)), it follows that (qM :R M) = q. Hence Q = (Q : M)M = qM. By Lemma 3.8 and Theorem 3.9, qM ∈ Aps(M). Hence P contains an associated prime submodule qM. Conversely, let Q be an associated prime submodule of M and P ⊇ Q. Then there is an injective map R/(Q :R M) → M, so that localization gives an injective map (R/(Q :R M))(P :RM) → M(P :RM). Moreover,

(R/(Q :R M))(P :RM) 6= 0 for otherwise there would exist s ∈ R\(P :R M) such that s(1 + (Q :R M)) = 0 + (Q :R M); hence s ∈ (Q :R M) ⊆

(P :R M), a contradiction. Hence M(P :RM) 6= 0. Corollary 4.14. Let M be as in Theorem 4.13. Assume that ψ is surjective. Then [ ψ−1(Supp(M)) = {Q}; Q∈Aps(M) |Aps(M)|<∞ hence ψ−1(Supp(M)) is closed. Associated Prime Submodules of a Multiplication Module 295

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Sang Cheol Lee Department of Mathematics Education, and Institute of Pure and Ap- plied Mathematics, Chonbuk National University Jeonju, Jeonbuk 54896, Korea. 296 Sang Cheol Lee, Yeong Moo Song ∗, and Rezvan Varmazyar

E-mail: [email protected]

Yeong Moo Song Department of Mathematics Education, Sunchon National University Sunchon, Chonnam 57922, Korea. E-mail: [email protected]

Rezvan Varmazyar Department of Mathematics, Khoy Branch, Islamic Azad University Khoy 58168-44799, Iran. E-mail: [email protected]