ON RINGS WHOSE MODULES HAVE NONZERO HOMOMORPHISMS TO NONZERO SUBMODULES

We carry out a study of rings R for which HomR(M,N) 6= 0 for all nonzero N ≤MR. Such rings are called retractable. For a retractable ring, Artinian condition and having Krull dimension are equivalent. Furthermore, a right Artinian ring in which prime ideals commute is precisely a right Noetherian retractable ring. Retractable rings are characterized in several ways. They form a class of rings that properly lies between the class of pseudo-Frobenius rings, and the class of max divisible rings for which the converse of Schur’s lemma holds. For several types of rings, including commutative rings, retractability is equivalent to semi-Artinian condition. We show that a Köthe ring R is an Artinian principal ideal ring if and only if it is a certain retractable ring, and determine when R is retractable. 2010 Mathematics Subject Classification: Primary: 16D10; Secondary: 16E50,


Introduction
Throughout this paper rings will have unit elements and modules will be right unitary.Following [12], an R-module M is called retractable if Hom R (M, N ) = 0 for all nonzero submodules N of M .Semisimple modules and fully idempotent modules [21] are clearly retractable, and more generally self-projective modules with zero radical and essentially compressible modules are known to enjoy this property; see [5, 3.4] and [19,Theorem 3.1].Retractable modules have appeared in different situations.For example, in the study of nonsingular modules satisfying one of the properties: CS, continuous, quasi continuous or having a Baer endomorphism ring [16,Theorem 22].They have also been applied in the study of prime M -ideals that correspond to the isomorphism classes of indecomposable M -injective modules in σ[M ] [3, Theorems 2.10 and 6.7] and in the characterization of endomorphism rings of quasi-injective envelopes of polyform modules [5, 5.19]; see also, [9,Theorem 2.6], and [24,Section 2].In [21], it is shown that the commutative rings over which every module is fully idempotent are exactly the semisimple rings.
Rings with all finitely generated modules retractable are characterized in [8], and finitely generated retractable modules over right FBN rings are characterized in [18] where the term slightly compressible is used for retractable.
In the present work, we shall consider retractable rings which are rings with all nonzero module retractable.Recall from [7], R is a right CPF ring if for all proper ideals I of R, any faithful R/I-module is a generator in Mod-R/I.Artinian principal ideal rings are CPF [23, 56.9(c)].In Proposition 2.4 we show that the class of retractable rings properly lies between the class of right CPF rings and the class of divisible right max rings which are "CS" in the sense of Hirano and Park [11].These are rings for which the converse of Schur's Lemma holds; see also [10].Some equivalent conditions for a ring to be retractable are given in Theorem 2.2, where it is shown that retractable rings are precisely rings over which all torsion theories are hereditary.Over a retractable ring, a module is Artinian if and only if it is Noetherian and its second singular submodule is Artinian (Proposition 2.10).Retractable rings with Krull dimension and reduced retractable rings are characterized in Theorems 3.6 and 3.2.More generally, retractable rings R such that R/J(R) is reduced are shown to be left semi-Artinian, and they are precisely semi-Artinian if in addition J(R) ⊆ Cent(R) (Theorem 3.4 and Corollary 3.5).A result of Köthe states that over an Artinian principal ideal ring R every right (left) R-module is a direct sum of cyclic right (left) R-modules (i.e.R is a Köthe ring) [13].We investigate the converse of the Köthe theorem and as an application of our results, we show that a Köthe ring R is an Artinian principal ideal ring if and only if it is a retractable ring such that for any ring decomposition Mat n (S) × T R with local S, the ring S is Köthe (Theorem 3.10).The retractability of Köthe rings are then determined.Any unexplained terminology, and all the basic results on rings and modules that are used in the sequel can be found in [2], [5] and [14].

Retractability of modules
In this section we investigate the class of retractable rings in Theorem 2.2 and Propositions 2.4, 2.6 and study modules over retractable rings 0 for all essential submodules N of M ; see [22] for more information about essentially retractable modules.For an R-module M R , the injective hull of M is denoted by E(M R ) or simply E(M ).(i) M R is essentially retractable.(ii) There exists a nonzero f ∈ Hom R (M, E(M )) such that f (M ) is an essentially retractable R-module.
(ii) ⇒ (i).Suppose that (ii) holds.Let K be an essential submodule in M R .Then K is essential in E(M ), and hence K ∩ f (M ) is essential in f (M ).Thus there exists a nonzero homomorphism from f (M ) to Theorem 2.2.For a ring R, the following statements are equivalent.
(i) R is a retractable ring.
(ii) Every nonzero R-module is essentially retractable.
(iii) Every essential extension of a cyclic R-module is essentially retractable.(iv) Hom R (M, X) = 0 ⇔ Hom R (M, E(X)) = 0 for all R-modules M and X. (v) All torsion theories on R are hereditary.
(iii) ⇒ (ii).Note that if 0 = m ∈ M R , then mR essentially embeds in a suitable factor of M R [14, Proposition 6.18].
(ii) ⇒ (iv).If Hom R (M, E(X)) is nonzero, then similar to the proof of Lemma 2.1, we have Hom R (M, X) = 0.The converse is clear.
(iv) ⇒ (v).Let (T , F) be a torsion theory on R, N ≤ M R ∈ T and X ∈ F. Then Hom R (M, X) = 0, hence Hom R (M, E(X)) = 0 by (iv).It follows that Hom R (N, X) = 0, proving that N R ∈ T .(v) ⇒ (i).Let N ≤ M R and (T , F) be a torsion theory generated by M R .By (v) N R ∈ T and hence Hom R (M, N ) = 0.
In the following we collect more properties of modules over retractable rings.A module M R is called divisible if M c = M for any right regular element c ∈ R (i.e., r-ann R (c) = 0).The ring R is called right divisible if the module R R is divisible.It is well known that injective modules are divisible.If M is an R-module such that (M/N ) R and N R are divisible for some N ≤ M R , then it is easily seen that M R is divisible.Proposition 2.3.Let R be a retractable ring and let M and N be nonzero R-modules. (i) It is easy to verify that A R is also divisible and so it lies in N , a contradiction.Thus M = N and M R is divisible, as desired.Now let M R be nonzero and M I n = 0 for some ideal I of R and n ≥ 1.Since R/I is a retractable ring, by the first part, M I i /M I i+1 are divisible R/I-modules for i = 0, 1, . . ., n with I 0 = R.It follows that M c = M for every c ∈ R which is right regular modulo I.
(ii) By [23, 14.9], M R has a factor L such that Soc(L) = 0. Thus by the retractable condition on L, we can deduce that M has a maximal submodule, proving that J . Hence, by hypothesis, there exists a simple submodule S of M/J(M ) such that S embeds in N/f (J(M )).Now by retractable condition on N/f (J(M )), S N/K for some maximal submodule K of N .It follows that Hom R (N/J(N ), M/J(M )) = 0.
(iv) First note that if M R is nonsingular, then by hypothesis there is a nonzero map f : E(M ) → M .Thus Ker f is an essentially closed, and hence a direct summand of the injective R-module E(M ).It follows that Im f is a nonzero injective submodule of M R .Therefore, we can deduce that every nonzero nonsingular R-module contains a nonzero injective R-module.On the other hand, if m is any nonzero element of a nonsingular R-module M then r-ann R (m) is not an essential right ideal of R, and so there exists a right ideal A in R such that mA A. Consequently, if M R is nonsingular, then every nonzero submodule of M R contains a nonzero injective projective submodule.The proof is now completed by the fact that nonzero projective modules are not singular.To see this let P R = 0, P ⊕ K = F and F R be free with basis {e i | i ∈ I}.If P F/K is singular then for every i ∈ I, there exists an essential right ideal The following result together with Examples 3.9 show that the class of retractable rings properly lies between two classes of known rings.We first recall the necessary definitions.Following [7], a ring R is called right CPF if for all proper ideals I of R, any faithful R/I-module is a generator in Mod-R/I.Artinian principal ideal rings are known to be CPF [23, 56.9(c)].Also in [6], the ring R is called right HP (Hirano-Park) if for every non-zero R-module M , the converse of Schur's Lemma holds (i.e., if End R (M ) is a division ring, then M R is a simple module).More recent works on HP rings are cited in the references.Rings over which any non-zero module has a maximal submodule are called right max rings; see [20] for an excellent reference on the subject.Proposition 2.4.
(i) Right CPF rings are retractable.
(ii) Any retractable rings is a right divisible, right max and HP ring.
Proof: Part (i) follows from the definitions.For part (ii), note that R is a right max ring by Proposition 2.3(ii).Now if End R (M ) is a division ring and 0 = N ≤ M R , then the existence of a nonzero map M R → N R implies that N = M .Thus M R is simple and R is an HP ring.Applying Proposition 2.3(i) for M = R and I = 0, we have that R R is divisible. Lemma where e 1 and e 2 are central orthogonal idempotents in T such that e 1 R 2 = e 2 R 1 = 0 and e 1 + e 2 = 1 T .Clearly M e i is naturally an R i -module for i = 1, 2. Now let 0 = m ∈ M .We have m = m1 T = me 1 + me 2 .Hence there is i ∈ {1, 2} such that me i = 0.So by our assumption, there exists a nonzero is the natural projection.Hence M T is retractable, and T is a retractable ring.
In the following we investigate the retractability of the class σ[M R ] when M R is a locally Noetherian module.Recall from [23,15] that σ[M R ] is a full subcategory of the category Mod-R whose objects are submodules of modules which are generated by M R .Also a module M R is said to be polyform if Hom R (M/N, M ) = 0 for any N ≤ e M R .Here M is the M -injective envelope of M R in σ[M R ].Alternatively, M R is polyform if and only if End R ( M ) is a regular ring [5, 4.9].The class of polyform modules properly contains both the class of nonsingular and the class of semisimple modules.It is known that any submodule and any quasi-injective hull of a polyform module is again polyform.Proposition 2.7.Suppose that M R is polyform such that nonzero direct summands of M are retractable R-modules.If M R is locally Noetherian or it has acc (dcc) on direct summands, then M R is semisimple.
Proof: The first we show that every indecomposable submodule of M R is a simple M -injective R-module.Let U be an indecomposable submodule of M R and 0 = K ≤ U .Then Û , the M -injective hull of U , is a direct summand of M , and so by our assumption Û is retractable.Now similar to the proof of Proposition 2.3(iv), U contains a nonzero M -injective submodule of K. Therefore, K = U by the indecomposable condition on U , as desired.Now if M R has acc (dcc) on direct summands, then we are done by [ By Proposition 2.4 and the next lemma, we observe that if R is retractable, then every Artinian module is Noetherian.The converse will be investigated in Proposition 2.10.Lemma 2.9.Let M be a nonzero R-module and R is a right max ring.If every factor module of M has finite uniform dimension, then M R is Noetherian.
Proof: Just note that if N ≤ M R is not finitely generated then by [5, 5.11], there exists a finitely generated submodule K ≤ N such that N/K has no maximal submodule, a contradiction.Proposition 2.10.Over a retractable ring, a module is Artinian if and only if it is Noetherian and its second singular submodule is Artinian.
Proof: The necessity follows from Proposition 2.4(ii) and Lemma 2.9.Let M R be noetherian and Z 2 (M ) be Artinian.Note that L := M/Z 2 (M ) is nonsingular and hence a polyform module.Now apply Proposition 2.7 for the module L to deduce that L R is a semisimple noetherian module.Since now L and Z 2 (M ) are Artinian, M R is Artinian.

Characterization of some classes of rings
In this section, we give new characterizations for semisimple Artinian rings and certain semi-Artinian rings in terms of retractable rings.Two important subclasses of the class of Artinian rings are the class of Artinian principal ideal rings and the class of rings over which every right (and left) module is a direct sum of cyclic modules (Köthe rings).Let K (resp.AP) be the classes of Köthe (resp.Artinian principal ideal) rings.In [13], it is proved that AP ⊆ K and it is asked what the Köthe rings are; see also [17,Appendix B,Problem 2.48].Recently, in [4, Theorem 3.1], it is proved that normal Köthe rings are Artinian principal ideal rings.A restatement of our Corollary 3.8, gives AP ⊆ R where R is the class of retractable rings.Hence, if a Köthe ring is an Artinian principal ideal ring then it must be a retractable ring.We first characterize when a (semi-)Artinian ring is retractable and then determine when Köthe rings are Artinian principal ideal rings.
Recall from [14, 11.9] a ring R is said to be right Goldie if R has ascending chain condition on right annihilators and the uniform dimension of R R is finite.Left Goldie rings are defined similarly.Semiprime right Goldie rings are known to be right nonsingular.In [11,Proposition 11], it is shown that right nonsingular HP rings with finite uniform dimension are semisimple Artinian.Hence, by Proposition 2.4, retractable semiprime right Goldie rings are precisely semisimple Artinian rings.In the following, we obtain a similar result for retractable semiprime left (right) Goldie rings.Proposition 3.1.(i) Retractable domains are precisely division rings.
(ii) The ring R is a semiprime left (right) Goldie retractable ring if and only if R is a semisimple Artinian ring if and only if R is a right nonsingular retractable ring with acc (dcc) on direct summand right ideals.
(ii) Suppose that R is semiprime left Goldie and let I be an essential left ideal of R. Then I contains a regular element x and so Rx = R, by Proposition 2.3(i).It follows that R R has no proper essential left ideals, proving that R is a semisimple Artinian ring.The second equivalence is obtained by Proposition 2.7.
A ring R is said to be reduced if R has no nonzero nilpotent elements.A reduced ring which is a regular ring is called strongly regular ; see [23, 3.11] for more information.A ring R is said to be right (left) semi-Artinian if every nonzero right (left) R-module has a nonzero socle, and R is called semi-Artinian if it is right and left semi-Artinian.In [8,Theorem 2.7], it is shown that for a commutative ring, the semi-Artinian condition implies the retractable condition.The converse follows by [15,Theorem 8].We will give a generalization of this result in Theorem 3.4.Theorem 3.2.A ring is reduced and retractable if and only if it is a (right) semi-Artinian strongly regular ring.
Proof: For the sufficiency, note that since R is strongly regular, R is reduced, right ideals in R are two sided and cyclic R-modules are flat.Hence, all simple R-modules are injective by [14,Corollary 3.6A].It follows that the semi-Artinian ring R is retractable.Conversely, let R be a reduced retractable ring, 0 = a ∈ R, M = aR, and I = r-ann R (a).
Since R is a reduced ring, I is an ideal of R and so we have I ∩ aR = 0.It follows that a is right regular modulo I. Hence M a = M by Proposition 2.3, proving that R is a regular ring.Now R is strongly regular by the reduced condition on R.
To show that R is semi-Artinian, we will show that every cyclic R-module contains an injective R-module [5, 15.11].Suppose now B ≤ R R .Since R is a strongly regular ring, B is an ideal of R and the ring R/B is (right) nonsingular.By hypothesis, R/B is also a retractable ring and so it contains an injective R/B-module by Proposition 2.3(iv).On the other hand, R/B is a flat left R-module and so by [14, Corollary 3.6(A)], every injective right R/B-module is injective as a right R-module, as desired.
Following [2, p. 314], a non-empty subset Y of R is called left T-nilpotent provided for each sequence y 1 , y 2 , y 3 , . . . of elements of Y there exists a positive integer n such that y 1 y 2 . . .y n = 0. Proposition 3.3.Let R be a ring with J(R) ⊆ Cent(R).Then R is a retractable ring if and only if R/J(R) is a retractable ring and J(R) is a T-nilpotent ideal.(⇐) Let M R be a nonzero R-module and J = J(R).By Theorem 2.2, we shall show that M R is essentially retractable.If M J = 0 then M is an R/J-module and we are done.If M J = 0, then there exists r ∈ J such that M rJ = 0 but M r = 0. Since J ⊆ Cent(R), M r is an R/J-module and so it is essentially retractable as an R/J-module as well as R-module.Now multiplication by r defines a nonzero homomorphism Proof: Let R be a retractable ring and R/J(R) be reduced.By Theorem 3.2, R/J(R) is a (right) semi-Artinian strongly regular ring.By Proposition 2.4, R is a right max ring and so J(R) is a right T-nilpotent.It follows that R is left semi-Artinian [20, Lemma 2.12].The last statement is true because reduced rings are normal.ring.To see this let 0 = f ∈ Hom R (eR, I), then f (eR) = f (e)eR = 0, a contradiction.
(ii) Suppose that A and B are rings and A M B is a nonzero bimodule.Let R = [ A M 0 B ] , and e = [ 1 0 0 0 ] , then [ 0 M 0 0 ] lies in eR ∩ l-ann R (e).So by (i), R is not a retractable ring.Thus one may easily produce semi-Artinian rings which are not retractable.
(iii) For any ring R, the ring R[x] is never retractable (Proposition 2.4(ii)).
(iv) There exists a retractable ring which is not CPF.Suppose that S = Q[x i | i ∈ N] and I is the ideal of S generated by the subset {x i x j , x k+1 k | i = j, k ∈ N} and R = S/I.Then it is easy to verify that R is a local ring with J(R) = J = xi | i ∈ N .In view of Proposition 3.3, to show that R is retractable, we shall show that J is T-nilpotent.Let f i ∈ J and f 1 ∈ A := x1 , . . ., xn .Thus f 1 f 2 . . .f n+1 ∈ AJ n = 0. To proof that R is not a CPF ring, consider the faithful R-module M = i R/(⊕ j =i xj R).If M R is generator, then R must be embedded in M (k) R for some k ≥ 1, but every element in M (k) has nonzero annihilator, a contradiction.Hence M R is not generator, and so R is not a CPF ring.
(v) There exists a divisible, HP, max ring which is not retractable.Let R = Q N be the countable product of Q and I = Q (N) .Then it is well known that R is a self-injective regular ring such that Soc(R/I) R = 0. Thus R is a divisible, max ring and it is an HP ring by [11,Corollary 15], but R is not a retractable ring by Corollary 3.5.
A characterization of Artinian principal ideal rings in [23, 56.9], shows that a ring R is Artinian principal ideal ring if and only if such is Mat n (R).If S is a ring, we say that S is a matrix ring direct summand (matrix rds) of R whenever Mat n (S) × T R for some ring T and n ≥ 1.
Theorem 3.10.Let R be a Köthe ring.Then R is an Artinian principal ideal ring if and only if R is a retractable ring and every local ring which is a matrix rds of R is a Köthe ring.
Proof: (⇒) By Corollary 3.8 R is a retractable ring.Suppose that S is a matrix rds of R. By hypothesis and [23, 56.9], S is an Artinian principal ideal ring.Hence S is a Köthe ring by [13].
(⇐) Since R is a Köthe ring, it is Artinian.Hence, by Theorem 3.6, R i Mat ni (R i ) such that each R i is local and a matrix rds of R. Thus by hypothesis, each R i is a local Köthe ring.The proof is completed by [4,Theorem 3.1].

Lemma 2 . 1 .
The following statements are equivalent for a nonzero R-module M .

2 ,
Proposition 10.14].Let M R be a local noetherian.By the first part and [5, Corollary 5.2(2)], we can deduce Soc(M ) is an essential submodule of M .On the other hand, Soc(M ) is an M -injective submodule of M by [5, 2.5(c)].It follows that Soc(M ) = M .Corollary 2.8.Over a right Noetherian ring R, a nonzero module M R is semisimple if and only if it is polyform and the class σ[M R ] is retractable.Proof: By Proposition 2.7.
nonsingular if and only if every nonzero submodule of M R contains a nonzero injective projective submodule if and only if every nonzero submodule of M R contains a nonzero projective submodule.
Just note that in the definition of the (essentially) retractable modules, only categorical terms are used; see [2, Proposition 21.6].Proposition 2.6.The class of retractable rings is closed under homomorphic image, Morita equivalence and finite product.Proof: Let C be the class of all retractable rings.Clearly, C is closed under homomorphic image and Morita equivalence by Lemma 2.5.Now suppose R 1 and R 2 are retractable rings and set 2.5.Being (essentially) retractable is a Morita invariant property.Proof: