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Kyungpook Mathematical Journal 2018; 58(3): 463-472

Published online September 30, 2018

Copyright © Kyungpook Mathematical Journal.

Some Properties of Dedekind Modules and Q-modules

Shahram Motmaen, Ahmad Yousefian Darani*, and Mahdi Rahmatinia

Department of Mathematics and Applications, Univeristy of Mohaghegh Ardabili, P.O. Box 179, Ardabil, Iran
e-mail : sh.motmaen@uma.ac.ir, yousefian@uma.ac.ir and m.rahmati@uma.ac.ir

Received: March 23, 2016; Revised: August 26, 2016; Accepted: February 4, 2017

Abstract

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A Q-module is a module in which every nonzero submodule of M is a finite product of primary submodules of M. This paper is devoted to study some properties of Dedekind modules and Q-modules.

Keywords: Dedekind module, multiplication module, invertible submodule, Q-module.

1. Introduction

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Throughout this paper all rings are considered commutative rings with identiry and all modules are considered unitary. Let R be a ring and M an R-module. A proper submodule N of M is a prime submodule if for each r ∈ R and for each m ∈ M with rm ∈ N, we have m ∈ N or r ∈ (N:R M), where (N:R M) = Ann(M/N) = {r ∈ R|rM ⊆ N}. Also N is called a primary submodule of M if for each r ∈ R and for each m ∈ M with rm ∈ N, we have m ∈ N or rn ∈ (N:R M) for a positive integer n. We say that a submodule N of M is a radical submodule of M if N=N, where N=(N:RM)M.

The R-module M is said to be a multiplication R-module if every submodule N of M has the form IM for some ideal I of R. If M be a multiplication R-module and N a submodule of M, then N = IM for some ideal I of R. Hence I ⊆ (N:R M) and so N = IM ⊆ (N:R M)M ⊆ N. Therefore N = (N:R M)M [8]. Let M be a multiplication R-module, N = IM and L = JM be submodules of M for ideals I and J of R. Then, the product of N and L is denoted by N.L or NL and is defined by IJM [5]. An R-module M is called a cancellation module if IM = JM for two ideals I and J of R implies I = J [1]. By [13, Corollary 1 to Theorem 9], finitely generated faithful multiplication modules are cancellation modules. It follows that if M is a finitely generated faithful multiplication R-module, then (IN:R M) = I(N:R M) for all ideals I of R and all submodules N of M. If R is an integral domain and M a faithful multiplication R-module, then M is a finitely generated R-module [9]. Let R be a ring, Z(R) the set of zero-divisors of R and S = R Z(R). Then T(R) denotes the total quotient ring of R. A non-zero-divisor of a ring R is called a regular element and an ideal of R is said to be regular if it contains a regular element. For a non-zero ideal I of R, Let

I-1={x∈T(R):xI⊆R}.

In this case II−1 ⊆ R. I is called an invertible ideal of R if II−1 = R. An integral domain R is called a Dedekind domain if every nonzero ideal of R is invertible.

Let M be an R-module. An element r ∈ R is said to be a zero-divisor on M if rm = 0 for some nonzero element m ∈ M. We denote by Z(M) the set of all zero-divisors of M. It is easy to see that Z(M) is not necessarily an ideal of R, but it has the property that if a, b ∈ R with ab ∈ Z(M), then either a ∈ Z(M) or b ∈ Z(M). Let M be an R-module and set

T={t∈S:for all m∈M,tm=0 implies m=0}=(RZ(M))∩(RZ(R)).

Then T is a multiplicatively closed subset of R with T ⊆ S, and if M is torsion-free then T = S. In particular, T = S if M is a faithful multiplication R-module [9, Lemma 4.1]. Let N be a nonzero submodule of M. Then we write N−1 = (M:RTN) = {x ∈ RT: xN ⊆ M}. Then N−1 is an R-submodule of RT, R ⊆ N−1 and NN−1 ⊆ M. We say that N is invertible in M if NN−1 = M. Clearly 0 ≠ M is invertible in M. An R-module M is called a Dedekind module if every nonzero submodule of M is invertible. In Section 2, we investigate some properties of Dedekind modules. It is proved that if M is a faithful multiplication R-module over an integral domain R, then M is Dedekind R-module if and only if every proper submodule of M is a finite product of prime submodules of M. In Section 3 we prove some results on Q-modules. Let R be a ring and M a finitely generated faitful multiplication R-module. We show that if M is a Noetherian module with dim(M) = 1, then M is a Q-module. Finally we prove that if M a Noetherian finitely generated multiplication module over R, then M is a Q-module if and only if every prime submodule which is not a maximal submodule of M is a multiplication submodule.

Here we list some preliminaries and results used throughout the paper.

Lemma 1.1([9])

Let M be multiplication module and let N be a submodule of M. Then N = Ann(M/N)M

Lemma 1.2

([9, Theorem 2.5]) Let M be a nonzero multiplication R-module. Then,

  • every proper submodule of M is contained in a maximal submodule of M;

  • K is a maximal submodule of M if and only if there exists a maximal ideal P of R such that K = PM ≠ M.

Theorem 1.3

([9, Corollary 2.11]) Let R be ring and M an R-module. The following statements are equivalent for a proper submodule N of M:

  • N is a prime submodule of M;

  • Ann(M/N) is a prime ideal of R;

  • N = PM for some prime ideal P of R whit Ann(M) ⊆ P.

Theorem 1.4

([9, Theorem 3.1]) Let R be a ring and M a faithful multiplication R-module. Then the following statements are equivalent:

  • is finitely generated;

  • AM ⊆ BM if and only if A ⊆ B;

  • for each submodule N of M, there exists a unique ideal I of R such that N = IM;

  • M ≠ AM for any proper ideal A of R;

  • M ≠ PM for any maximal ideal P of R.

Definition 1.5

Let R be a ring and M be an R-module and let N be a submodule of M such that N = IM for same ideal I of R. Then, we say that I is a presentation ideal of N.

Theorem 1.6

([5, Theorem 3.4]) Let N = IM and K = JM be submodules of a multiplication R-module M. Then, the product of N and K is independent of presentations of N and K.

Definition 1.7

Let R be a ring, M an R-module and N a submodule of M. Then N is called decomposable if it has a primary decomposition N = Q1 ∩…∩Qn where for each 1 ≤ i ≤ n, Qi is Pi-primary. Such a primary decomposition of N is said to be a minimal primary decomposition if

  • P1, …, Pn are distinict prime ideal of R.

  • ∩i=1,i≠jn⊈Qj for all j = 1, …, n.

It is proved that every decomposable submodule of M has a minimal primary decomposition.

Theorem 1.8([12])

Let R be a ring and M a Noetherian R-module. Then every proper submodule of M is decomposable.

A commutative ring R is called a Q-ring if every ideal in R is a finite product of primary ideals in R. First, the class of Noetherian Q-rings have been studied and characterized by D. D. Anderson in [6]. Then Anderson and Mahaney in [7] have studied Q-rings in general.

2. Dedekind Modules

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Proposition 2.1

Let R be a ring and M a multiplication R-module. If N, K, L are submodules of M such that NK = NL and N is invertible, then K = L.

Proof

Let N, K, L are submodules of M such that NK = NL and N is invertible. Then K = MK = N−1NK = N−1NL = ML = L.

Lemma 2.2

Let R be a ring, M a multiplication R-module and N1, ···, Nn submodules of M. Then the submodule N1 ··· Nn is invertible if and only if for each 1 ≤ i ≤ n, Ni is invertible.

Proof

Let I1, I2, ···, In be ideals of R such that N1 = I1M, N2 = I2M, ···, Nn = InM. Suppose N1N2 ···Nn is invertible submodule. If K is a fractional ideal of R such that KN1N2 ···Nn = M, then for each i = 1, 2, ···, n, we have,

(KI1I2⋯Ii-1Ii+1⋯In)Ni=(KI1I2⋯Ii-1Ii+1⋯In)IiM=(KI1I2⋯In)M=K(I1I2⋯In)M=KN1N2⋯Nn=M.

So Ni is invertible. Conversely, suppose for each 1 ≤ i ≤ n, Ni is invertible. Then

(N1-1N2-1⋯Nn-1)(N1N2⋯Nn)=(N1-1N1)(N2-1N2)⋯(Nn-1Nn)=M.

So N1N2 ···Nn is invertible submodule of M.

Lemma 2.3

Let R be an integral domain and M a faithful multiplication R-module. If K1K2 ··· Kn = N = L1L2 ··· Lm where Ki, Li are prime submodules of M and Ki is invertible then n = m and Ki = Li for each i = 1, 2, ···, n.

Proof

The proof is by induction on n. Suppose n = 1 and K1 = N = L1L2 ··· Lm and J1, J2, ···, Jn, I1, I2, ···, Im are prime ideals of R such that Kj = JjM and Li = IiM. So we have JM = I1I2 ··· ImM, since M is cancelative R-module J = I1I2 ··· Im. So after reindexing J = I1, thus K1 = L1. If n > 1, choose one of the Ki, say K1, such that K1 does not properly contain Ki, for i = 2, 3, ···, n. Since

I1I2⋯ImM=L1L2⋯Lm=K1K2⋯Kn=J1J2⋯JnM⊂K1

and M is cancelative, we have

I1I2⋯Im=J1J2⋯Jn⊂J1

and J1 is prime so by prime avoidenc Theorem there exists some Ii, say I1, is contained in J1. Similarly since

J1J2⋯Jn=I1I2⋯Im⊂I1

so Ji ⊆ I1. Hence Ji ⊆ I1 ⊆ J1 and so Ki ⊆ L1 ⊆ K1. By the minimality of K1 we must have Ki = L1 = K1. Since K1 = L1 is invertible, Proposition 2.1 implies that K2K3 ··· Kn = L2L3 ··· Lm. Therefore by the induction hypothesis n = m and after reindexing Ki = Li for i = 1, 2, ···, n.

Proposition 2.4

Let R be a ring and M be a finitely generated faithful multiplication R-module in which every proper submodule is a finite product of prime submodules. Then every proper ideal of R is a finite product of prime ideals of R.

Proof

Let I be a proper ideal of R. Then IM is a proper submodule of M, so IM = K1K2 ··· Kn where Ki, i = 1, 2, ···, n is prime submodule of M, and there exist prime ideals P1, P2, ···, Pn of R such that Ki = PiM for each i = 1, 2, ···, n. So we have

IM=K1K2⋯Kn=(P1M)(P2M)⋯(PnM)=P1P2⋯Pn.

Since M is a finitely generated faithful multiplication R-module, M is a cancelative module, hence we must have I = P1P2 ··· Pn.

Theorem 2.5

Let R be a ring and M be a finitely generated faithful multiplication R-module in which every proper submodule is the product of a finite number of prime submodules. Then every invertible prime submodule of M is maximal.

Proof

Let N be an invertible prime submodule of M. So there exists prime ideal P of R such that N = PM. Since N is invertible, P is an invertible ideal of R. Hence, by [10, Theorem 6.5], P is a maximal ideal of R. Therefore N is maximal, because M is cancelation R-module.

Proposition 2.6

Every faithful multiplication module over an integral domain is a D1module.

Proof

See [11, Remark 3.8].

Theorem 2.7

Let R be an integral domain and M be a faithful multiplication R-module in which every proper submodule is the product of a finite number of prime submodules. Then every prime submodule of M is invertible.

Proof

Suppose N is a nonzero prime submodule of M and 0 ≠ a ∈ N. Then Ra = K1K2 ··· Kn where Pi is a prime submodule of M for all i = 1, 2, ···, n. There exist prime ideals P, P1, P2, ···, Pn such that N = PM and for each 1 ≤ i ≤≤ n, Ki = PiM. Since

(P1P2⋯Pn)M=(P1M)(P2M)⋯(PnM)=K1K2⋯Kn=Ra⊆N=PM

and M is a cancelation R-module, P1P2 ··· Pn ⊆ P. Therefore for some k, Pk ⊆ P and hence Kk ⊆ N. Since by Proposition 2.6, Ra is invertible, Kk is invertible, by Lemma 2.2. Hence Kk is invertible prime submodule. So Kk is maximal by Theorem 2.5, whence N = Kk. Therefore N is maximal and invertible.

Theorem 2.8

Let R be an integral domain and M be a faithful multiplication R-module. Then M is Dedekind R-module if and only if every proper submodule of M is a finite product of prime submodules of M.

Proof

Let N be anonzero submodule of M. Choose maximal submodule KN such that N ⊆ KN ⊊ M. If N = M, let KM = R. Now we have

KN-1N⊆KN-1KN⊆M

therefore KN-1N is a submodule of M and contains N. If N is proper submodule of M, then N⊊KN-1N, because, if not

M=RM=RMRM=(N-1N)(KN-1KN)=N-1(NKN-1)KN=N-1NKN=MKN=KN

is a contradiction. Let S be the set of all submodules of M and define a function f: S → S by N↦KN-1N. Given a proper submodule N, there exists a function φ: N → S such that φ(0) = N and φ(n + 1) = f(φ(n)). If we denote φ(n) by Nn and KNn by Kn, then we have an ascending chain of submodules

N=N0⊂N1⊂N2⊂⋯

such that Nn+1=f(Nn)=Kn-1Nn. Since M is Dedekind, M is Notherian R-module and N is proper submodule of M, there is a least integer l such that

N=N0⊊N1⊊⋯⊊Nl-1⊊Nl=Nl+1.

Thus Nl=Nl+1=f(Nl)=Kl-1Nl. So we must have Nl = M. Consequently,

M=Nl=f(Nl-1)=Kl-1-1Nl-1

whence

Nl-1=Nl-1M=Nl-1Kl-1-1Kl-1=MKl-1=Kl-1.

Since Kl−1 = Nl−1 ⊊ Nl = M, Kl−1 is a maximal submodule of M. The minimality of l insures that each of K0, ···, Kl−2 is also maximal, because, if not we have Ki = M, whence

Ni+1=Ki-1Ni=M-1Ni=RNi=Ni

is a contradiction. Now we have

Kl-1=Nl-1=Kl-2-1Nl-2=Kl-2-1Kl-3-1Nl-3=⋯=Kl-2-1⋯K1-1K0-1N.

Consequently, since each Ki is invertible,

(K0K1⋯Kl-2)Kl-1=(K0K1⋯Kl-2)Kl-2-1⋯K1-1K0-1N=N.

Conversely, by Lemma 2.2 and Theorem 2.7, M is a Dedekind R-module.

3. Q-modules

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Definition 3.1

Let R be a ring and M an R-module. Then M is called a Q-module if every submodule of M is a finite product of primary submodules of M.

It is clear that a Q-module is a Dedekind module.

Theorem 3.2

Let R be a ring and M a finitely generated faithful multiplication R-module. If M is a Q-module, then

  • MS is a Q-module for multiplicative subset S of R.

  • M/N is a Q-module for each submodule N of M.

Proof

(1) Let j be a submodule of MS. Then j ∩ M is a submodule of M. So j ∩M = P1…Pn where for each 1 ≤ i ≤ n, Pi is a primary submodule of M. Hence j = S−1(j∩M) = S−1(P1…Pn) = (S−1P1)…(S−1Pn) which is a product of primary submodules of MS. Therefore MS is a Q-module.

(2) Let K/N be a submodule of M/N where K is a submodule of M. Then K = P1…Pn where for each 1 ≤ i ≤ n, Pi is a primary submodule of M. Hence K/N = P1…Pn/N = (P1/N)…(Pn/N) which is a product of primary submodules of M/N. Therefore M/N is a Q-module.

Remark 3.3

Let R be a ring, M a multiplication R-module, I an ideal of R and N a submodule of M. Then (N:R M)Mn = (N:R M)MMn−1 = NMn−1 = NMMn−2 = … = NM = N and IMn = I(RM…RM) = IM.

Lemma 3.4

Let R be a ring, M a finitely generated multiplication R-module, I an ideal of R and N a submodule of M. Then

  • N is a product of primary submodules of M if and only if (N:R M) is a product of primary ideals of R.

  • I is a product of primary ideals of R if and only if IM is a product of primary submodules of M.

Proof

(1) Let N = P1…Pn where for each 1 ≤ i ≤ n, Pi is a primary submodule of M. Then (N:R M) = (P1…Pn:R M) = (P1:R M)…(Pn:R M) where for each 1 ≤ i ≤ n, (Pi:R M) is a primary ideal of R, by [3, Lemma 4]. Conversely, let (N:R M) = P1…Pn where for each 1 ≤ i ≤ n, Pi is a primary ideals of R. Hence, by [3, Lemma 4], N = (N:R M)Mn = (P1…Pn)Mn = (P1M)…(PnM) where for each 1 ≤ i ≤ n, PiM is a primary submodule of M.

(2) let I = P1…Pn where for each 1 ≤ i ≤ n, Pi is a primary ideals of R. Hence, by [3, Lemma 4], IM = IMn = (P1…Pn)Mn = (P1M)…(PnM) where for each 1 ≤ i ≤ n, PiM is a primary submodule of M. Conversely, let IM = P1…Pn where for each 1 ≤ i ≤ n, Pi is a primary submodule of M. Then I = (IM:R M) = (P1…Pn:R M) = (P1:R M)…(Pn:R M) where for each 1 ≤ i ≤ n, (Pi:R M) is a primary ideal of R, by [3, Lemma 4].

Now we have the following Corollary.

Corollary 3.5

Let R be a ring and M be a finitely generated multiplication R-module. Then R is a Q-ring if and only if M is a Q-module.

Theorem 3.6

Let R be a ring and M be a finitely generated multiplication R-module. If a submodule N of M is a finite product of primary submodules, then there are only finitely many prime submodules of M which are minimal over N.

Proof

Let N be a product of primary submodules of M. Then, by Lemma 3.4, (N:R M) is a product of primary ideals. Hence, by [6, Lemma 4], there are only finitely many minimal prime submodules over (N:R M).

Therefore, by [3, Lemma 4], there are only finitely many prime submodules of M which are minimal over N.

Corollary 3.7

Let R be a ring and M be a finitely generated multiplication R-module. If M is a Q-module, then there are only finitley many minimal prime submodules over any submodule of M.

Lemma 3.8

Let R be a ring, M a multiplication R-module and N, K submodules of M. IfN+K=M, then N + K = M. Moreover, NK = N ∩ K.

Proof

Let N+K=M. Then ((N:RM)+(K:RM))=(N:RM)M+(K:RM)M=M. So (N:RM)+(K:RM)=R. Hence (N:R M)+(K:R M) = R and thus (N:R M)(K:R M) = (N:R M) ∩ (K:R M). Therefore N + K = (N:R M)M + (K:R M)M = M. Moreover, KN = (N:R M)M(K:R M)M = (N:R M)M ∩ (K:R M)M = N ∩ K.

Theorem 3.9

Let R be a ring and M a finitely generated faitful multiplication R-module. Let M be a Noetherian module with dim(M) = 1. Then M is a Q-module.

Proof

Let N be a submodule of M. Then N has a minimal primary decomposition, say, N = Q1 ∩ Q2… ∩ Qn where for each 1 ≤ i ≤ n, Qi is a Pi-primary submodule of M. Since dim(M) = 1, each nonzero prime submodule of M is maximal. So for each 1 ≤ i ≤ n, Pi is a maximal submodule of M. Hence, Pi+Pj = M for all i ≠ j.

Thus, by Lemma 3.8, Qi + Qj = M for all i ≠ j. Then N = Q1 ∩ Q2… ∩ Qn = Q1Q2…Qn. Therefore M is a Q-module.

Note that if R is a ring, M is a multiplication R-module and N a submodule of M, then N is a multiplication R-submodule of M if and only if (N:R M) is a multiplication ideal of R. For this, let N be a multiplication R-submodule of M and I an ideal of R such that I ⊆ (N:R M). Then IM ⊆ N. So IM = KN for an ideal K of R. Hence I = (IM:R M) = K(N:R M). Therefore (N:R M) is a multiplication ideal of R. Conversely, let (N:R M) be a multiplication ideal of R and K a submodule of N. Then (K:R M) ⊆ (N:R M). So (K:R M) = I(N:R M) for an ideal I of R. Hence K = IN. Therefore N is a multiplication submodule of M.

Proposition 3.10

Let R be a ring, M a multiplication R-module and N be a multiplication submodule of M. If P is a prime submodule of M with P ⊊ N, thenP⊆∩n=1∞Nn.

Proof

Let N be a multiplication submodule of M. Then (N:R M) is a multiplication ideal of R. Let P be a prime submodule of M with P ⊊ N. Then (P:R M) is a prime ideal of R with (P:R M) ⊊ (N:R M). Hence, by [6], (P:RM)⊆∩n=1∞(N:RM)n. Therefore

P=(P:RM)Mn⊆∩n=1∞(N:RM)nMn=∩n=1∞Nn.

It is shown that if R is a ring and M is a finitely generated faithful multiplication R-module, then M is a Noetherian R-module if and only if R is a Noetherian ring.

Theorem 3.11

Let R be a ring and M a Noetherian finitely generated multiplication R-module. Then M is a Q-module if and only if every prime submodule which is not a maximal submodule of M is a multiplication submodule.

Proof

It is obvious that M is a Noetherian module if and only if R is a Noetherian ring. Then M is a Q-module if and only if R is a Q-ring by Theorem 3.5, if and only if every prime ideal which is not maximal in R is multiplication by [6, Theorem 10], if and only if every prime submodule which is not a maximal submodule in M is a multiplication submodule.

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