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### Article

Kyungpook Mathematical Journal 2022; 62(3): 407-423

Published online September 30, 2022

### On Representable Rings and Modules

Seyed Ali Mousavi, Fatemeh Mirzaei and Reza Nekooei∗

Department of Pure Mathematics, Mahani Mathematical Research Center, Shahid Bahonar University of Kerman, Kerman, Iran
e-mail : s110m@yahoo.com, mirzaee0269@yahoo.com and rnekooei@uk.ac.ir

Received: November 11, 2020; Revised: October 8, 2021; Accepted: October 10, 2021

In this paper, we determine the structure of rings that have secondary representation (called representable rings) and give some characterizations of these rings. Also, we characterize Artinian rings in terms of representable rings. Then we introduce completely representable modules, modules every non-zero submodule of which have secondary representation, and give some necessary and sufficient conditions for a module to be completely representable. Finally, we define strongly representable modules and give some conditions under which representable module is strongly representable.

Keywords: Secondary modules, Secondary representation, Representable rings and modules, Laskerian rings and modules, Semiperfect rings, Artinian rings, Noetherian spectrum, Compactly packed rings, Q-rings, AB5* modules, Linearly compact modules.

### 1. Introduction

Throughout this paper, R will denote a commutative ring with a non-zero identity and every module will be unitary. Given an R-module M, we shall denote the annihilator of M (in R) by AnnR(M) or Ann(M). We shall follow Macdonald's terminology in [18] concerning secondary representation. Thus, an R-module N is secondary if N ≠ 0 and for each r ∈ R, either rN=N or there exists some positive integer n, such that rnN=0. If N is a secondary module then, Ann(N) is a primary ideal and hence P=Ann(N) is prime and we say that N is P-secondary. A secondary representation of M is an expression for M as a sum M=N1+N2++Nt of finitely many secondary submodules of M, such that Ni is Pi-secondary for i=1,,t. If such a representation exists, we shall say M is representable. Such a representation is said to be minimal if P1,,Pt are all different and none of the summands Ni are redundant. Every representable module has a minimal secondary representation. As for the primary decomposition of submodules, we have two uniqueness theorems for secondary representation of modules. The first uniqueness theorem (see [18, 2.2]) says that if M=N1++Nt, with Ni being Pi-secondary, is a minimal secondary representation of M, then the set {P1,...,Pt} is independent of the choice of minimal secondary representation of M. This set is called the set of attached prime ideals of M and is denoted by Att(M). Every Artinian module and every injective module over a Noetherian ring is representable. These and other propositions about representable modules can be found in [5, 18, 22].

An R-module M is said to be Laskerian if every proper submodule of M is an intersection of a finite number of primary submodules, i.e. has a primary decomposition. A ring R is Laskerian if it is Laskerian as an R-module over itself.

We denote the set of all prime ideals and the set of all maximal ideals of a ring R by Spec(R) and Max(R), respectively. The Jacobson radical J(R) of a ring R is defined to be the intersection of all the maximal ideals of R. The set of all nilpotent elements of R is called the nilradical of R and denoted by N(R). The Krull dimension of R is denoted by dim(R). If I is a proper ideal of R, then I and Min(I) denote the radical ideal of I and the set of prime ideals of R minimal over I, respectively.

A topological space X is said to be irreducible if X, and whenever X=Z1Z2 with Zi closed, we have X=Z1 or X=Z2. The maximal irreducible subsets of X are called irreducible components of X. A topological space X is said to be Noetherian if the ascending chain condition holds for open subsets of X. If X=Spec(R) with the Zariski topology, then X is Noetherian if and only if R satisfies the ascending chain condition for radical ideals.

In Section 2, we investigate representable rings and show that these rings and Artinian rings have similar properties in some cases. We then show that representable rings are strictly between Artinian and semiperfect rings. Therefore, we determine the structure of these rings and characterize them. Next we characterize Artinian rings in terms of representable rings.

In Section 3, we consider modules for which all non-zero submodules are representable and called these modules completely representable. Then we give some necessary and sufficient conditions for a module to be completely representable.

In Section 4, we define strongly representable modules as modules that can be written as a direct sum of finitely many secondary submodules and then give some conditions such that a representable module is strongly representable.

### 2. Representable Rings

A ring R is representable if it is representable as a module over itself. In this section, we first show that representable rings are strictly between Artinian and semiperfect rings. Then we determine the structure of these rings. Finally we give some characterization of these rings and characterize Artinian rings in terms of representable rings. By [18, 5.2], any Artinian ring is representable. But the converse is not true.

Example 2.1. The ring R=k[X1,X2,]/(X1,X2,)2, where k is a field, is representable (in fact it is m=(x1,x2,)-secondary, where xi denotes the equivalence class of Xi in R) but it is not Artinian, because it is not Noetherian.

Representable rings have similar properties to Artinian rings. Artinian rings are Noetherian. We show that representable rings are Laskerian.

Proposition 2.2. Let R be a representable ring. We have the following statements.

• (i) If R is a domain, then R is a field.

• (ii) dim(R)=0.

• (iii) J(R)=N(R), and hence J(R) is a nil ideal.

• (v) Spec(R) is Noetherian.

Proof. (i) Let R be a representable domain and R=I1+I2++It be a minimal secondary representation of R, where Ii is Pi-secondary (i=1,,t). Let rPi. Then there exists n1 such that rnIi=0. But Ii0, hence r=0. Thus, Pi=0 for all i, and hence R=I1. So, R is 0-secondary and hence is a field.

(ii) Let P be a prime ideal of R. Then R/P is a representable R-module and also a representable domain. Hence by part (i), R/P is a field. Thus, P is a maximal ideal.

(iii) It follows by part (ii).

(iv) Let I be a proper ideal of a representable ring R.

Then R/I is a representable R-module. Let R/I=J1++Jt be a minimal secondary representation of R/I. Then I=Ann(R/I)= i=1tAnn(Ji) is a primary decomposition for I.

(v) It follows from (iv) and [10, Theorem 4].

Every representable ring has finitely many prime ideals. Indeed, prime ideals are maximal, and we have the following.

Proposition 2.3. Let R be a Laskerian ring. If dim(R)=0, then it has finitely many maximal ideals. In particular, every representable ring has finitely many maximal ideals.

Proof. Since R is Laskerian, it follows from [10][Theorem 4] that R has Noetherian spectrum. Hence by [20, Theorem 3.A.16], the Spec(R) has finitely many irreducible components and by [20, Corollary 3.A.14], these components are of the forms V(P)={QSpec(R)PQ}, where P is a minimal prime ideal of R. Since dim(R)=0, Min(R)=Max(R). Thus, V(P)={P}, for every PMin(R). Therefore, Spec(R) is finite. The last statement follows from Proposition 2.2 (ii), (iv).

In the following, we show that every representable ring is semiperfect. By [27, 42.6], a ring R is semiperfect if and only if R/J(R) is semisimple and idempotents lift modulo J(R).

Proposition 2.4. Every representable ring is semiperfect.

Proof. Let R be a representable ring and let Max(R)={m1,m2,,mt}=Spec(R). By the Chinese Remainder Theorem, R/J(R)i=1tR/mi. Hence R/J(R) is semisimple. By Proposition 2.2 (iii), J(R) is a nil ideal. Thus by [27, 42.7], idempotents lift modulo J(R). Therefore, R is a semiperfect ring.

The converse of Proposition 2.4, is not true in general. For example, the ring of formal power series R=k[[X]] where k is a field, is semiperfect (because it is local) but it is not representable (because dim(R)0). Thus, representable rings are strictly between Artinian and semiperfect rings.

In the following, we will determine the structure of representable rings. To this purpose we need the following.

Proposition 2.5. Let R be a ring and e ∈ R be an idempotent element. Then we have the following statements.

• (i) Re is a ring with 1Re=e.

• (ii) Every ideal of Re is of the form Ie, where I is an ideal of R.

• (iii) If m is a maximal (prime) ideal of R and e∉ m, then me is a maximal (prime) ideal of Re.

• (iv) If I is an P-secondary (as R-module) ideal of R, then Ie is either zero or Pe-secondary (as Re-module) ideal of Re.

Proof. (i), (ii) and (iv) are easily proven by definitions. For (iii), one can consider ring isomorphism Re/Ie(R/I)(e+I).

Lemma 2.6. A ring R is representable and local if and only if it is secondary as an R-module.

Proof. Let (R,m) be a representable and local ring. Then m=N(R), and hence every elements of R is a unit or nilpotent. Thus, R is m-secondary. Conversely, if R is m-secondary, then RU(R)=m, where U(R) is the set of unit elements of R. Thus, R is local with unique maximal ideal m.

Lemma 2.7. Let R be a representable ring with Max(R)={m1,,mt} and e∈ R be an idempotent element such that e1m1 and e i=2tmi. Then Re is a representable local ring with unique maximal ideal m1e.

Proof. According to Lemma 2.6, it is sufficient to show that the ring Re is m1e-secondary. Let R=I1+I2++Is be a minimal secondary representation for R, where Ii be mi-secondary (i=1,,s). In fact, s=t. Because, if s<t then for rmt i=1smi, we have rR=rI1++rIs=I1++Is=R. So r is unit, which is a contradiction. Since em1 and emi(2is), we have, Re=I1e(=I1). Therefore, Re is m1e-secondary.

Theorem 2.8. (Structure theorem for representable rings) A representable ring is uniquely (up to isomorphism) a finite direct product of representable local rings.

Proof. Let Max(R)={m1,,mt}. Since J(R) is a nil ideal, hence by [27, 42.7], every idempotent in R/J(R) can be lifted to an idempotent in R. But by Chines Reminder Theorem, R/J(R) i=1tR/mi. Thus, there exists a set {e1,,et} of orthogonal idempotents (i.e. ei2=ei for all i and eiej=0 for ij) in R such that ei1mi and ei jitmj, for i=1,,t. Thus, we have (e11)(e21)(et1)J(R)=N(R). So there exists n1, such that (e11)n(e21)n(et1)n=0. Hence e1+e2++et=1. By [13, Exercise 24, page 135], we have RRe1××Ret. But by Lemma 2.7, each Rei is representable and local. Therefore, R is a finite direct product of representable local rings.

For uniqueness, suppose R i=1sRi, where the (Ri,mi) are representable local (=mi-secondary) rings. For each i, we have a natural injective homomorphism φi:RiR. Let Ii=Im(ϕi). Then one can simply see that R=I1++Is is a minimal secondary representation of R. On the other hand by proof of Lemma 2.7, R=Re1++Ret is a minimal secondary representation of R and all the secondary components Rei are isolated. Hence by 2nd uniqueness theorem for secondary representation (see [18, 3.2]), we have s=t and RiIiRei for all i, and the proof is complete.

Corollary 2.9. Let R be a representable ring. Then every non-zero ideal of R is representable (as an R-module).

Proof. By Theorem 2.8, R=R1××Rn, where Ri is a representable local ring (or equivalently secondary ring by Lemma 2.6). Let I be a non-zero ideal of R. Then I=I1××In, where Ii is an ideal of Ri for all i. Since every non-zero ideal of a secondary ring is secondary, Ii is secondary for all i with Ii0. For simplicity, we can assume that all of Ii's are non-zero. Then I=(I1×0××0)++(0××0×In), is a secondary representation of I and the proof is complete.

Proposition 2.10. Let R be a ring. Then R is a representable ring if and only if Spec(R) is Noetherian and for every non-zero ideal I of R and every minimal prime ideal P of Ann(I), there exists rRP such that rI is P-secondary.

Proof. Let R be a representable ring. Then by Proposition 2.2 (v), Spec(R) is Noetherian. Let I be a non-zero ideal of R and P be a minimal prime ideal of Ann(I). By Corollary 2.9, I has a minimal secondary representation, say, I=I1++It. Thus, Ann(I)= i=1tPiP. So, there exists i such that PiP, and hence Pi=P because dim(R)=0. By rearranging Ii's (if necessary), we can assume that i=1. If t=1, then I=I1 is P-secondary and the proof is complete for r=1. Otherwise, suppose s( i=2tPi)P. Then there exists k1 such that skAnn(Ii) for all i, 2it. Let r=sk. Then rI=rI1++rIt=rI1=I1 is P-secondary.

Conversely, let P0 be a minimal prime ideal of R=I0 . Then by assumption, there exists r0RP0 such that r0R is P0-secondary. Let I1=(0:Rr0)=AnnR(Rr0). Since r02I0=r0I0, so we have R=r0I0+I1. If I1=0, then R=r0I0 is a representation of R. Otherwise, let P1 be a minimal prime ideal of Ann(I1). Again by assumption, there exists r1RP1 such that r1I1 is P1-secondary and hence I1=r1I1+I2 , where I2=(0:Rr1). Thus, R=r0I0+r1I1+I2. If I2=0, then R=r0I0+r1I1 is representation of R. Otherwise we continue this process and claim that there exists some n1 such that In=0. If not, then since I3I2I1I0=R and hence Ann(R)Ann(I1)Ann(I2) . Since Spec(R) is Noetherian, there exists n1 such that Ann(In)=Ann(In+1). Thus, there exists t1 such that rntIn=0. But rntIn=rnIn. So rnAnn(In)Pn, which is a contradiction. Therefore, there exists n0 such that R=r0I0+r1I1++rnIn, a representation of R and the proof is complete.

In the next theorem we give some characterization of representable rings. A ring R is called von Neumann regular ring if for each a ∈ R, there exists b ∈ R such that a=a2b. A ring R is said to be a Q-ring, if every ideal of R is a finite product of primary ideals. A ring R is called a completely packed ring if whenever I αΓPα, where I is an ideal and Pα's are prime ideals of R then IPβ, for some βΓ. It is well known that R is a compactly packed ring if and only if every prime ideal is the radical of a principle ideal [24, Theorem]. We need the following lemma.

Lemma 2.11. Let I⊆ P be ideals of a ring R with P a prime ideal. Then the following statements are equivalent.

• (i) P is a minimal prime ideal of I.

• (ii) For each x ∈ P, there is yRP and a positive integer n such that yxnI.

Proof. [12, Theorem 2.1, Page 2].

Theorem 2.12. Let R be a ring. The following statements are equivalent.

• (i) R is a representable ring.

• (ii) R/J(R) is a semisimple ring and J(R) is a nil ideal.

• (iii) R/N(R) is a Noetherian and von Neumann regular ring.

• (iv) R is a zero dimensional compactly packed ring.

• (v) R has Noetherian spectrum and dim(R)=0.

• (vi) R is a zero dimensional Q-ring.

Proof. (i)(ii) It follows from Propositions 2.2 and 2.4.

(ii)(iii) A ring is semisimple if and only if it is Noetherian Von Neumann regular ring [17, Theorem 4.25]. Since J(R) is a nil ideal, J(R)=N(R). Hence the result follows.

(iii)(iv)(v)(vi) [15, Theorem 2.12].

Now by Proposition 2.10, we show that the equivalent conditions (iv) and (v) imply (i).

Let I be a non-zero ideal of R and P a minimal prime ideal of Ann(I). By [24][Theorem], there exists a ∈ R such that P=(a). Now by Lemma 2.11, there exists rRP and n1 such that ranAnn(I). We claim that rI is P-secondary. Since r∉ P, rI ≠ 0. Let s ∈ R. If s ∈ P, then there exists m1 such that sm(a). Thus sm=ta, for some t ∈ R. We have snm(rI)=tnan(rI)=tn(ranI)=0. If s ∉ P, since P is maximal, then R=Rs+P. Therefore, 1=ys+x, for some y ∈ R and s ∈ P. Since P=(a), sk=za for some k1 and z ∈ R. Now we have 1=(ys+x)kn=αs+xkn=αs+znan, for some αR. Thus, rI=(αsr+znanr)I=αsrI. So, s(rI)rI=s(αrI)s(rI). Therefore, s(rI)=rI and the proof is complete.

The following result is the expected generalization of [3, Theorem 8.5].

Corollary 2.13. A ring R is representable if and only if R is Laskerian and dim(R)=0.

Proof. It follows from Proposition 2.2, [10, Theorem 4] and Theorem 2.12 (v).

Corollary 2.14. Let R be a local ring. Then R is representable if and only if dim(R)=0.

Proof. It follows form Proposition 2.2 (ii) and Theorem 2.12 (iii).

Corollary 2.15. A ring R is reduced and representable if and only if R is a Noetherian von Neumann regular ring (or equivalently a semisimple ring).

Proof. Let R be reduced (i.e. N(R)=0) and representable. Then by Theorem 2.12 (iii), R is a Noetherian von Neumann regular ring. Conversely, if R is a Noetherian von Neumann regular ring, then it is semisimple [17, Theorem 4.25]. So it is a finite direct product of fields and hence is representable and reduced.

Representability is not a local property. However, we have the following theorem.

Theorem 2.16. Let R be a ring. Then Rm is a representable ring for all maximal ideal m of R if and only if dim(R)=0.

Proof. If dim(R)=0 then by Corollary 2.14, Rm is a representable ring, for all maximal ideals m of R. Now let Rm be a representable ring, for all maximal ideal m of R. Let q be a prime ideal and m be a maximal ideal of R such that q ⊆ m. Then by Proposition 2.2 (ii), dim(Rm)=0. So qm=mm, and hence q=m. This shows that dim(R)=0.

Corollary 2.17. Let R be a ring. Then R is reduced and Rm is a representable ring, for all maximal ideal m of R if and only if R is a von Neumann regular ring.

Proof. It follows from Theorem 2.16 and [12, Remark, Page 5].

Now we characterize Artinian rings in terms of representable rings.

Theorem 2.18. Let R be a ring. The following statements are equivalent.

• (i) R is an Artinian ring.

• (ii) R is a representable ring and locally Noetherian.

• (iii) R is a Noetherian ring and Rm is a representable ring, for all maximal ideals m of R.

Proof. (i)(ii) Follows from [18, 5.2] and [3, Theorem 8.5].

(ii)(iii) By Theorem 2.12 (vi), R is a Q-ring with dim(R)=0. Now by [14, Theorem 3] and Theorem 2.16, (iii) holds.

(iii)(i) Follows from Theorem 2.16 and [3, Theorem 8.5].

### 3. Completely Representable Modules

In this section, we consider modules that all non-zero submodules are representable. This is similar to definition of Laskerian modules and, in a sense, a dual of that notion. We call these modules "completely representable". Artinian modules (see [18]), representable modules over regular rings ([8, Theorem 2.3]) and modules over local rings that their maximal ideals are nilpotent, are examples of such modules.

Definition 3.1. An R-module M is said to be completely representable, if M ≠ 0 and every non-zero submodule of M is representable. A ring R is completely representable, if it is completely representable as a module over itself.

Remark 3.2. Both representable and completely representable modules are generalizations of Artinian modules. But the second seems to be a better generalization, because we know that every submodule of an Artinian module is again Artinian, and we may want this feature to be preserved in generalization and this holds in definition of completely representable modules (but not in representable modules).

For the case of rings, representable and completely representable are the same.

Proposition 3.3. A ring R is completely representable if and only if it is representable.

Proof. If R is completely representable, then R is representable by definition. The converse is Corollary 2.9.

By definition, every completely representable module is representable. But the converse is not true.

Example 3.4. The -module is (0)-secondary and hence representable. But it is not completely representable, because the submodule is not representable.

In the next, we give some necessary and some sufficient conditions for a module to be completely representable. An R-module M is said to satisfy (dccr) if the descending chain INI2N terminates, for every submodule N of M and every finitely generated ideal I of R. The following result mentioned in [25, Proposition 3] without complete proof. We give its proof.

Proposition 3.5. Let M be a completely representable R-module. Then M satisfies (dccr).

Proof. By [25, Theorem, Page 2 ], M satisfies (dccr) if and only if for any submodule N of M and a ∈ R, N=akN+(0:Nak), for all large k. Let N=N1++Nt be a minimal secondary representation of N with Ni, Pi-secondary. If a i=1tPi, then aNi=Ni, for all i. Hence N=aN and the proof is complete. Otherwise, let (by rearranging if necessary) a i=1lPi and a i=l+1tPi. Then there exists an integer k such that akNi=0, for all i, 1il. So, akN=Nl+1++Nt and N1++Nl(0:Nak). Hence N=akN+(0:Nak) and the proof is complete.

The following examples show that the converse of Proposition 3.5, is not true in general.

Example 3.6. Let R be a non Noetherian von Neumann regular ring (e.g. R= i=12). Then dim(R)=0. So, by [26, Proposition 1.2], R satisfies (dccr). By Theorem 2.12, R is not representable. (Note that any von Neumann regular ring is reduced).

Example 3.7. Let M=pp, where is the set of all prime numbers. Then by [26, Remark 1.10], M satisfies (dccr) condition as a -module. But this module is not representable (and therefore, is not completely representable). Because if N is a p-secondary submodule of M, for some p, then every component of every element of N is zero except probably the component that belongs to p. Obviously, the finite sum of this submodules can not be equal to M.

Therefore, the class of completely representable modules is strictly between modules that satisfies (dcc) (i.e. Artinian modules) and modules that satisfies (dccr) condition. Also, by Proposition 3.3 and [26, Proposition 1.2], representable rings are strictly between Artinian rings and rings with zero dimension.

Although, the converse of Proposition 3.5, is true under some additional conditions. A module M over a Noetherian ring R is said to have finite Goldie dimension, if M does not contain an infinite direct sum of non-zero submodules.

Proposition 3.8. Let M be a module of finite Goldie dimension over a commutative Noetherian ring R. Then M is completely representable if and only if M satisfies (dccr).

Proof. [25, Proposition 3].

Bourbaki in [4, Chap. IV, Sect. 2, Exercise 23, Page 295], give a necessary and sufficient conditions for a finitely generated module to be Laskerian. In the following we dualize these conditions for a module to be completely representable.

Let N be an R-module and S be a multiplicatively closed subset of R. We consider S(N)= rSrN. If P be a prime ideal of R and S=RP, we denote S(N) by SP(N).

Proposition 3.9. Let M be a completely representable R-module. Then we have

• (i) For every non-zero submodule N of M and every multiplicatively closed subset S of R, there exists r ∈ S such that S(N)=rN.

• (ii) For every non-zero submodule N of M, every increasing sequence (Sn(N))n1 is stationary, where (Sn)n1 is any decreasing sequence of multiplicatively closed subset of R.

Proof. (i) Let N= i=1tNi be a minimal secondary representation with Ann(Ni)=Pi, (1it). If SPi= for all i, then for every s ∈ S, sNi=Ni for all i, and hence sN=N. Thus, S(N)=N and (i) holds, for r=1∈ S. Otherwise, there exists l, 1l<t, such that PiS=, (1il) and PiS (l+1it). Thus, for every s ∈ S, sN=N1++Nl+sNl+1++sNt. On the other hand, there exists r ∈ S such that rNi=0, (l+1it). So rN=N1++NlS(N)rN=N1++Nl. Thus, S(N)=rN and (i) holds.

(ii) Suppose the contrary is true; i.e., there exists a decreasing sequence (Sn)n1 such that S1(N)S2(N). Let PiS1=(i=1,,l) and PiS1(i=l+1,,t). Since S2S1 and S1(N)S2(N), we have S2Pk=, for some k, (l+1kt). If similarly continue this, then there exists some i, such that SiPj=(j=1,2,,t), and so N=Si(N)=Si+1(N)=, which is a contradiction.

Corollary 3.10. Let M be a completely representable R-module. Then for every non-zero submodule N of M and every minimal prime ideal P over Ann(N), there exists rRP such that rN is P-secondary.

Proof. Let S=RP. Then by Proposition 3.9 (i), there exists rRP such that SP(N)=rN. Let s ∈ R. If s ∉ P, rN=SP(N)srNrN. Thus, s(rN)=rN. If s∈ P, by Lemma 2.11, there exists tRP and n1 such that tsnAnn(N). So, tsnN=0, and hence tsnrN=0. But trN=rN. Thus, sn(rN)=0. Therefore, rN is P-secondary.

Conditions of Proposition 3.9 are sufficient for a finitely cogenerated AB5* module to be completely representable.

An R-module M is said to be finitely cogenerated if for every family {Mi}iI of submodules of M with iIMi=0, there is a finite subset FI with iFMi=0. It is clear that every submodule of a finitely cogenerated module is finitely cogenerated. Also if N and M/N are finitely cogenerated then so is M. Hence every direct sum of finitely cogenerated modules is finitely cogenerated.

A family {Mi}iI of submodules of a module M is called inverse (direct) if, for all i,j ∈ I there exists k ∈ I such that MkMiMj (Mi+MjMk). For example, every chain of submodules is an inverse and direct family. The module M is said to be satisfy the AB5* (AB5) condition (and is called an AB5* (AB5) module) if, for every submodule K of M and every inverse (direct) family {Mi}iI of submodules of M, K+ iIMi= iI(K+Mi) (K( iIMi)= iIKMi). Every Artinian module is finitely cogenerated and AB5*. For more information about this class of modules one can see [6, 9, 19, 27].

Theorem 3.11. Let M be a finitely cogenerated AB5* module. Then M is completely representable if and only if

• (i) For every non-zero submodule N of M and every minimal prime P over Ann(N), there exists r∈ R∖ P such that SP(N)=rN (or equivalently rN is P-secondary).

• (ii) For every non-zero submodule N of M, every increasing sequence (Sn(N))n1 is stationary, where (Sn)n1 is any decreasing sequence of multiplicatively closed subset of R.

Proof. () It follows by Proposition 3.9. () Let conditions (i) and (ii) holds and let N be a non-zero submodule of M. Let P1 be a minimal prime of Ann(N). Then by (i), and by proof of Corollary 3.10, there exists r1RP1 such that Q1=r1N is P1-secondary. Let N1=(0:Nr1). Since r12N=r1N, so we have N=Q1+N1. If N1=0, then N=Q1 is representable. Otherwise, let Σ={0GMN=Q1+G,GN1}. N1Σ, and hence Σ. Since M is finitely cogenerated and AB5* module, so every chain in Σ has a lower bound. Hence by Zorn's lemma, Σ has a minimal element with respect to inclusion, N1 say. Let P2 be a minimal prime of Ann(N1) and r2RP2 such that Q2=r2N1 is P2-secondary. Then N1=Q2+N2 where N2=(0:N1r2). Thus, N=Q1+Q2+N2. If N2=0, N=Q1+Q2 is representable. Otherwise, let Σ={0GMN=Q1+Q2+G,GN2}. Again by Zorn's lemma, Σ has a minimal element N2 with respect to inclusion. We continue this process and claim that there exists n1 such that Nn=0. Suppose on the contrary that Nn0 for all n. Let Sn=R i=1 nPi (n=1,2,3,). We show that SnAnn(Nn). If SnAnn(Nn)=, then Ann(Nn) i=1 nPi and by Prime Avoidance Theorem, Ann(Nn)Pi for some i, 1in. But, N2N2N1N1N. So, NnNiNi, and hence, Ann(Ni)Ann(Ni)Ann(Nn)Pi. Thus, riPi, a contradiction. Therefore, SnAnn(Nn). Let sSnAnn(Nn). Then Q1++QnSn(N)sN=Q1++Qn, so; Sn(N)=Q1++Qn (n=1,2,). Now we have a decreasing sequence (Sn)n1 of multiplicatively closed subsets of R, such that the sequence (Sn(N))n1 is strictly increasing. Because if Sn(N)=Sn+1(N) for some n; then Qn+1Q1++Qn. Hence, N=Q1++Qn+Nn+1. But, Nn+1Nn, and hence by minimality of Nn, Nn+1=Nn. Since, rn+1Nn+1=0, thus rn+1Nn+1=0. So, rn+1Nn=0, and hence rn+1Ann(Nn)Pn+1. Thus, rn+1Pn+1, which is a contradiction. Therefore, the sequence (Sn(N))n1 is not stationary, which contradicts condition (ii). Thus, there exists n1 such that Nn=0, and hence N=Q1++Qn is representable.

Remark 3.12. We note that the notion of "completely representable" is the dual of the notion "primary decomposition". In some basic theorems for primary decomposition, the authors assume that the condition "finitely generated" (see [4, Chap. IV, Sect. 2, Exercise 23]). Also the AB5 condition is true for all modules (see [19, Lemma 6.22]). Since the "cofinitely generated" and "AB5*" are the dual notions of "finitely generated" and "AB5" conditions respectively, so it is natural to use these conditions in the proof or results about "completely representable".

Proposition 3.13. Let M be a finitely cogenerated AB5* module and N be a non-zero submodule of M. Then M is completely representable if and only if both N and M/N are completely representable.

Proof. If M is completely representable, then it is straightforward to show that N and M/N are completely representable. Conversely, Let N and M/N be completely representable. Let K be a non-zero submodule of M and P be a minimal prime of Ann(K). By condition (i) of Proposition 3.9, there exist r1, r2RP such that SP(KN)=r1(KN) and SP(K/KN)=r2(K/KN). Let r=r1r2. We show that SP(K)=rK. If αrK, then there exists xK such that α=rx. We have

r2(x+KN)r2(K/KN)=SP(K/KN)\$.

So,

r2x+KN sRPs(K/KN)= sRP((sK+KN)/KN)= sRP(sK+KN)/KN.

Thus, r2x sRP(sK+KN). Hence,

α=r1r2xr1( sRP(sK+KN)) sRP(r1sK+r1(KN)).

On the other hand, it is obvious that for every sRP, r1(KN)=r1s(KN). So

α sRP(sK+r1sK)= sRPsK=SP(K).

Thus, αSP(K). Now we check condition (ii) of Theorem 3.11. Let (Sn)n1 be a decreasing sequence of multiplicatively closed subset of R. Then there exists n1 and r1,r2Sn+1 such that Sn+1(KN)=Sn(KN)=r1(KN) and Sn+1(K/KN)=Sn(K/KN)=r2(K/KN). Let r=r1r2. Then Sn(K)=rK=Sn+1(K). Thus, conditions (i) and (ii) of Theorem 3.11 are hold. So M is completely representable.

Proposition 3.14. Let M1, M2,,Mn be R-modules such that are finitely cogenerated, AB5* and completely representable. If M=M1M2Mn is AB5*, then M is completely representable.

Proof. We prove this by induction on n. For n=1, M=M1 is completely representable. Now assume the assertion is true for n=k. For n=k+1, M=M1M2MkMk+1. Since M is AB5*, M=M1M2Mk is AB5*. Hence by induction hypothesis, it is completely representable. Since MM/Mk+1 and Mk+1 are completely representable, by Proposition 3.13, M is completely representable and the proof is complete by induction.

Note that finite direct sum of AB5* modules need not to be AB5* ([6, Lemma 2.5]).

A family of sets has the finite intersection property if every finite subfamily has a nonempty intersection. A module M is linearly compact (with respect to the discrete topology) if every collection of cosets of submodules of M which has the finite intersection property has non-empty intersection. Linearly compact modules are AB5* (see [9, Lemma 7.1]). Also every finite direct sum of linearly compact modules is linearly compact (see [27, 29.8(2)]). So, we have the following Corollary.

Corollary 3.15. Let M1, M2,,Mn be linearly compact, finitely cogenerated and completely representable R-modules. Then M=M1M2Mn is completely representable.

For relation between linearly compact and AB5* modules, see ([1],[2],[6]).

Heinzer, in [11, Proposition 2.1], gives a new restatement of Bourbaki's conditions. Inspired by this, we have the following results.

Theorem 3.16. Let M be an R-module such that Spec(R/Ann(M)) is Noetherian and for all non-zero submodule N of M, there exists a minimal prime P of Ann(N) and rRP such that rN is P-secondary. Then M is completely representable.

Proof. Let N be a non-zero submodule of M. Then by assumption, there exists a minimal prime P1 of Ann(N) and r1RP1 such that r1N is P1-secondary. Let N1=(0:Nr1). Since r12N=r1N, so we have N=r1N+N1. If N1=0, then N=r1N is representation of N. Otherwise, there exists a minimal prime P2 of Ann(N1) and r2RP2 such that r2N1 is P2-secondary and N1=r2N1+N2, where N2=(0:N1r2). Thus, N=r1N+r2N1+N2. If N2=0, then N=r1N+r2N1 is representation of N. Otherwise, we continue this process and claim that there exists some n1 such that Nn=0. If not, then since N3N2N1NM and hence Ann(M)Ann(M)Ann(N)Ann(N1)Ann(N2). Since Spec(R/Ann(M)) is Noetherian so there exists n1 such that Ann(Nn)=Ann(Nn+1). Thus, there exists t1 such that rn+1tNn=0. But rn+1tNn=rn+1Nn. So rn+1Ann(Nn)Pn+1, rn+1Pn+1, which is a contradiction. Therefore, there exists n1 such that N=r1N+r2N1++rnNn+1, a representation of N and the proof is complete.

For the case of rings, these conditions are also necessary.

Theorem 3.17. A ring R is completely representable if and only if Spec(R) is Noetherian and for every non-zero ideal I of R, there exists minimal prime P of Ann(I) and rRP such that rI is P-secondary.

Proof. () It follows from Theorem 3.16.

() Let R be completely representable. Then R is representable and by Proposition 2.2 (iv), it is Laskerian. So by [10, Theorem 4], R has Noetherian spectrum. Now by Corollary 3.10, the proof is complete.

Remark 3.18. Note that by Corollary 3.10, Theorem 3.17, is the same as Proposition 2.10.

In [7], authors have considered representable linearly compact modules. In the next theorem we give a necessary and sufficient conditions for a finitely cogenerated linearly compact module to be completely representable.

Theorem 3.19. Let M be a finitely generated, linearly compact and finitely cogenerated R-module. Then M is completely representable if and only if Spec(R/Ann(M)) is Noetherian and for all non-zero submodule N of M, there exists a minimal prime ideal P of Ann(N) and rRP such that rN is P-secondary.

Proof. () Theorem 3.16.

() By Corollary 3.10, second condition is true. We show that R/Ann(M) has Noetherian spectrum. Let {x1,x2,...,xn} be a set of generators of M. Let φ:RMMM by r(rx1,rx2,,rxn). Then φ is an R-homomorphism, and hence R/Ann(M) is a submodule of MMM. But by Corollary 3.15, MMM is completely representable. So, R/Ann(M) is a representable ring. Therefore, by [10, Theorem 4], R/Ann(M) has Noetherian spectrum and the proof is complete.

In the following, we show that, if dim(R)=0, then the converse of Theorem 3.16 is true. For this purpose, we need the following lemmas.

Lemma 3.20. Let I be a primary ideal of a ring R such that every regular element of R/I is unit. Then R/I is a secondary ring.

Proof. By [23, Lemma 4.3], R/I is non-zero and every zero divisor in R/I is nilpotent. Thus, every element of R/I is unit or nilpotent. Hence R/I is a secondary ring.

Lemma 3.21. Let R be a ring with dim(R)=0. Then every regular element of R is unit.

Proof. Let r ∈ R be a regular element. By [26][Proposition 1.2], R satisfies (dccr). Thus, there exists integer n1 such that Rrn=Rrn+1. So rn=rn+1s, for some s ∈ R. Since r is regular, we have rs=1.

Proposition 3.22. Let M be a representable R-module and M=N1++Nt be a minimal secondary representation of M, with Ni, Pi-secondary (1it). Let Ii=Ann(Ni), (1it). Suppose, Pi's are pairwise comaximal and every regular element of R/Ii, (1it), is unit. Then Spec(R/Ann(M)) is Noetherian.

Proof. Since (Ii=)Pi's are pairwise comaximal, by [3, Proposition 1.16], Ii's are also pairwise comaximal. Therefore, by Chines Reminder Theorem, R/Ann(M)i=1tR/Ii. So, by Lemma 3.20, R/Ann(M) is representable. Thus, by Proposition 2.2 (v), Spec(R/Ann(M)) is Noetherian.

Corollary 3.23. Let R be a ring with dim(R)=0. Then R-module M is completely representable if and only if Spec(R/Ann(M)) is Noetherian and for every non-zero submodule N of M and every minimal prime ideal P over Ann(N), there exists rRP such that rN is P-secondary.

Proof. () Follows form Proposition 3.22 and Corollary 3.10.

() Follows form Theorem 3.16.

### 4. Strongly Representable Modules

By definition of representable modules, we can also consider the following definition.

Definition 4.1. Let M be an R-module. We say M is a strongly representable module if there exists secondary submodules N1,,Nt such that M=N1Nt.

Obviously, every strongly representable module is representable. But the converse is not true in general, see [21, Example 2.4]. We give some condition such that representable modules be strongly representable.

Theorem 4.2. Let M be a representable R-module such that the elements of Att(M) are pairwise comaximal. Then M is strongly representable.

Proof. Let M=N1++Nt be a minimal secondary representation of M with Ni, Pi-secondary (1it). Thus, Att(M)={P1,,Pt}. Let Ii=Ann(Ni) for i=1,,t. Since (Ii=)Pi's are pairwise comaximal, Ii's are also pairwise comaximal by [3, Proposition 1.16]. We show that Ni( jiNj)=0, for all i=1,,t. For simplicity, let i=1. Suppose xN1( j=2tNj). By [23, Proposition 3.59], I1+ j=2tIj=R. So, there exists αI1 and β j=2tIj such that α+β=1. Thus, x=1x=(α+β)x=αx+βx=0+0=0. Therefore, M=N1Nt. Hence M is strongly representable.

Corollary 4.3. Let M be a representable R-module such that Att(M)Max(R). Then M is strongly representable.

Proof. Every two distinct maximal ideals are comaximal. Hence Corollary follows form Theorem 4.2.

Corollary 4.4. Let R be a ring with dim(R)=0. Then every representable R-module is strongly representable. In particular, every representable ring is strongly representable.

Proof. If dim(R)=0, then Max(R)=Spec(R) and result follows from Corollary 4.3. The "in particular" statement follows from Proposition 2.2 (ii).

Remark 4.5. According to Proposition 3.3 and Corollary 4.4, representable, completely representable and strongly representable rings are the same.

Proposition 4.6. Let R be a domain with dim(R)=1 (e.g. R be a Dedekind domain) and M be a representable R-module such that contains no non-zero divisible submodule. Then M is strongly representable.

Proof. Over a domain, divisible modules and 0-secondary modules are the same. Since M contains no divisible submodule, Att(M)Spec(R){0}=Max(R). Hence, result follows from Corollary 4.3.

Corollary 4.7. Every representable module over a Dedekind domain is strongly representable.

Proof. By [16, Theorem 8], every module M over a Dedekind domain R can be decomposed as M=DE, where D is divisible and E has no non-zero divisible submodules. If M is representable, then E(M/D) is also representable (if non-zero). Hence result follows form Proposition 4.6.

Finally, we show that every finitely generated Artinian module is strongly representable. For this, we need the following lemma.

Lemma 4.8. Let P be a maximal ideal of a ring R and M be an R-module such that PnM=0, for some integer n1. Then M is a P-secondary R-module.

Proof. Let r ∈ R. If r ∈ P, then rnM=0. If rP, then since P is maximal, P+Rr=R. Hence 1=α+sr, for some αP and s ∈ R. Thus, 1=αn+γr, for some γR. So, for every x ∈ M, x=αnx+γrx=γrx. Hence rM=M.

Proposition 4.9. Every finitely generated Artinian module is strongly representable.

Proof. By [18, 6.3], if M is an Artinian R-module, then there exist (distinct) maximal ideals m1,,mt of R such that M=M(m1)M(mt), where M(I)= n=1(0:MIn)={xM|Inx=0,n1}, for every ideals I of R. If M is finitely generated, then every M(mi) is finitely generated and hence will be annihilate by some power of mi. So by Lemma 4.8, M(mi) is mi-secondary (1it). Therefore, M=M(m1)M(mt) is a secondary representation of M and M is strongly representable.

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