A dissertation presented to. the faculty of. the College of Arts and Sciences of Ohio University. In partial fulfillment

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1 The Subprojectivity and Pure-Subinjectivity Domains of a Module A dissertation presented to the faculty of the College of Arts and Sciences of Ohio University In partial fulfillment of the requirements for the degree Doctor of Philosophy Joseph Lawrence Mastromatteo May Joseph Lawrence Mastromatteo. All Rights Reserved.

2 2 This dissertation titled The Subprojectivity and Pure-Subinjectivity Domains of a Module by JOSEPH LAWRENCE MASTROMATTEO has been approved for the Department of Mathematics and the College of Arts and Sciences by Sergio López-Permouth Professor of Mathematics Robert Frank Dean, College of Arts and Sciences

3 Abstract 3 MASTROMATTEO, JOSEPH LAWRENCE, Ph.D., May 2014, Mathematics The Subprojectivity and Pure-Subinjectivity Domains of a Module (59 pp.) Director of Dissertation: Sergio López-Permouth A module M and is said to be N-subprojective if for every epimorphism g : B N and homomorphism f : M N, there exists a homomorphism h : M B such that gh = f. The subprojectivity domain of M is defined to be the collection of all modules N such that M is N-subprojective. The subprojective profile of a ring R is the class of all subprojectivity domains for R modules. A module M is called N-pure-injective if every homomorphism from a pure submodule of N to M can be extended to a homomorphism from N to M. The pure-injectivity domain of M is defined to be the collection of all modules N such that M is N-pure-injective. The pure-injective profile of a ring R is the class of all pure-injectivity domains for R modules. In chapter 2 of this dissertation, we study subprojectivity domains. We show that the subprojective profile of R is a semilattice, and consider when this structure has coatoms or a smallest element. Modules whose subprojectivity domain is smallest as possible will be called subprojectively poor (sp-poor) or projectively indigent (p-indigent) and those with co-atomic subprojectivy domain are said to be maximally subprojective. The existence of sp-poor modules and maximally subprojective modules is determined for various families of rings. For example, we determine that artinian serial rings have sp-poor modules and attain the existence of maximally subprojective modules over the integers and for arbitrary V-rings. In chapter 3 of this dissertation, we turn our attention to the study of pure-subinjectivity domains. In particular, we give characterizations of right pure hereditary rings, von Neumann regular rings, and semisimple rings, by comparing the

4 4 pure-subinjectivity domains with the other domains of pure-injectivity and injectivity. We also study when the pure-subinjectivity domain of a module is closed under pure quotients.

5 Table of Contents 5 Page Abstract List of Symbols Preliminaries Ring and Module Theory Background An Alternative Perspective on Projectivity of Modules Introduction Subprojectivity and the subprojectivity domain of a module Subprojectivity domains and torsion-free classes Bounds in the subprojectivity domain of a module Pure-Injectivity from a Different Perspective Introduction The pure-subinjectivity domain of a module When the pure-subinjectivity domain of a module is closed under (pure) quotients Pure-subinjectively poor modules References

6 List of Symbols 6 Mod-R rad(m) Soc(M) t(m) pr.dim(m) J(R) E(M) PE(M) Ext 1 R(K, N) ip(r) pp(r) sip(r) spp(r) Pr 1 (M) In 1 (M) ppr 1 (M) pin 1 (M) Pr 1 (M) In 1 (M) ppr 1 (M) pin 1 (M) N Z Q Z n right R modules Jacobson radical of a module M socle of a module M torsion part of a module M projective dimension of a module M Jacobson radical of a ring R injective hull of a module M pure-injective hull of a module M extensions of N by K injective profile of a ring R projective profile of a ring R subinjective profile of a ring R subprojective profile of a ring R projectivity domain of a module M injectivity domain of a module M pure-projectivity domain of a module M pure-injectivity domain of a module M subprojectivity domain of a module M subinjectivity domain of a module M pure-subprojectivity domain of a module M pure-subinjectivity domain of a module M natural numbers integers rational numbers integers modulo n

7 1 Preliminaries Ring and Module Theory Background This chapter provides the necessary ring and module theory notations and definitions used in this dissertation. Throughout this dissertation, R will denote an associative ring with identity and modules will be unital right R-modules, unless otherwise explicitely stated. We denote by Mod-R the category of right R-modules. If M is an R-module, then rad(m), Soc(M), pr.dim(m), E(M), and PE(M) will respectively denote the Jacobson radical, socle, projective dimension, injective hull, and pure-injective hull of M. The Jacobson radical of a ring R will be denoted by J(R). N, Z, Q, and Z n, denote the natural numbers, integers, rational numbers, and the ring of integers modulo n, respectively. A ring R is called a right V-ring if every simple R-module is injective; a right hereditary ring if submodules of projective modules are projective or, equivalently, if quotients of injective modules are injective; a right perfect ring if every module has a projective cover; and a right coherent ring if every finitely generated right ideal is finitely presented or, equivalently, if products of flat left R-modules are flat. A torsion theory T is a pair of classes of modules (T, F ) such that (i) Hom(M, N) = 0 for every M T, N F ; (ii) if Hom(A, N) = 0 for all N F, then A T ; and (iii) if Hom(M, B) = 0 for all M T, then B F. In this situation, T and F are called the torsion class and torsion-free class of T, respectively. A class of modules T is the torsion class of some torsion theory if and only if it is closed under quotients, extensions and arbitrary direct sums. Likewise, a class of modules F is the torsion-free class of some torsion theory if and only if it is closed under submodules, extensions and arbitrary direct products. If C is a class of modules, then T C := (T C, F C ), where F C = {N Mod-R : Hom(C, N) = 0 for every C C} and

8 8 T C = {M Mod-R : Hom(M, N) = 0 for every N F C } is said to be the torsion theory generated by C. Likewise, T C = (T C, F C ) where T C = {M Mod-R : Hom(M, C) = 0 for every C C} and F C = {N Mod-R : Hom(M, N) = 0 for every M T C } is said to be the torsion theory cogenerated by C, [24, Chapter VI]. The torsion theory T C (resp. T C ) can also be characterized as the smallest torsion theory such that every object in C is torsion (resp. torsion-free). If M Mod-R, we write T M and T M for T {M} and T {M}, respectively. A module M is said to be quasi-projective if it is projective relative to itself. Over a right perfect ring R, every quasi-projective module M satisfies the following conditions: (D1) For every submodule A of M, there is a decomposition M = M 1 M 2 such that M 1 A and A M 2 M. (D2) If A M is such that M/A is isomorphic to a direct summand of M, then A is a direct summand of M. (D3) If M 1 and M 2 are direct summands of M with M 1 + M 2 = M, then M 1 M 2 is a direct summand of M. Modules satisfying (D1) are called lifting, see [10]. Modules satisfying (D1) and (D2) are called discrete, while modules satisfying (D1) and (D3) are called quasi-discrete. Every discrete module is quasi-discrete, as it is the case that (D2) (D3). [21, Lemma 4.6]. It is not the case that every projective module is lifting, as, for example Z is not a lifting Z-module. However, if R is right perfect, then every projective module is discrete, cf. [21, Theorem 4.41]. Every quasi-discrete module decomposes as a direct sum of modules whose every submodule is superfluous, see [21, Theorem 4.15]. Let P be a submodule of a right R-module M. Then P is called a pure submodule of M if for any left R-module X, the natural induced map on tensor products i 1 X : P X M X is injective, equivalently, if for any finite system of equations over P which is solvable in M, the system is also solvable in P. A submodule K of a module M

9 9 is said to be pure-essential in M if K is pure in M and for any nonzero submodule N of M, either K N 0 or (K N)/N is not pure in M/N. Let M and N be R-modules. M is called N-pure-injective if every homomorphism from a pure submodule of N to M can be extended to a homomorphism from N to M, and M is called pure-injective if it is N-pure-injective for every module N. A ring R is called right pure-semisimple if every right R-module is a direct sum of finitely generated modules. For additional concepts and results not mentioned here, we refer the reader to [4]. [5] and [17].

10 2 An Alternative Perspective on Projectivity of Modules 2.1 Introduction In this chapter, we initiate the study of an alternative perspective on the analysis of the projectivity of a module, as we introduce the notions of relative subprojectivity and assign to every module its subprojectivity domain. A module is projective if and only if its subprojectivity domain consists of all modules. Therefore, at this extreme, there is no difference in the role played by the projectivity and subprojectivity domains. Interesting things arise, however, when we focus on the subprojectivity domain of modules which are not projective. It is easy to see that every module is subprojective relative to all projective modules, and one can show (Proposition 2.2.8) that projective modules are the only ones sharing the distinction of being in every single subprojectivity domain. It is thus tempting to ponder the existence of modules whose subprojectivity domain consists precisely of only projective modules. We refer to these modules as sp-poor or, to keep in line with [6], we sometimes use the expression p-indigent. This dissertation is inspired by similar ideas and notions studied in several papers. On the one hand, relative injectivity, injectivity domains and the notion of a poor module (modules with smallest possible injectivity domain) have been studied in [1], [11] and [19]. Dually, relative projectivity, projectivity domains and the notion of a p-poor module have been studied in [16] and [19]. On the other hand, in [19] the authors name a class of modules an i-portfolio (resp. p-portfolio) if it coincides with the injectivity (resp. projectivity) domain of some module. Then, they proceed to define the injective profile (resp. projective profile) of a ring R, an ordered structure consisting on all the i-portfolios (resp. p-portfolios) in Mod-R. In this dissertation, we study these concepts in the context of subinjectivity and subprojectivity domains, thus obtaining the ordered invariants sip(r) and spp(r), the subinjective and subprojective profile of R, respectively. We study some 10

11 11 of its properties, such as the existence of coatoms and their relations with the lattice of torsion theories in Mod-R. One of the first things that comes to the surface in this type of study is the potential existence of modules which are least injective or projective possible with respect to whichever measuring approach one may be using. Injectively and projectively poor modules have been studied in [1], [11], [19] and [16]. In [6], Aydoğdu and López-Permouth modify in a subtle yet significant way the notion of relative injectivity to obtain relative subinjectivity. They also study subinjectivity analogs of poor modules, calling them indigent. Here, we study the projective analog of relative subinjectivity and indigent modules. In order to emphasize the analogy between poor and indigent modules, we also call indigent modules subinjectively poor, or si-poor for short. In this same spirit, we call the subprojective analog of indigent modules either p-indigent or sp-poor. We depict the different analogies between the different ways of measuring the injectivity and projectivity of a module in the following diagram. Relative injectivity i-poor modules i-portfolios ip(r) Relative projectivity p-poor modules p-portfolios pp(r) Relative subinjectivity indigent (= si-poor) modules si-portfolios sip(r) Relative subprojectivity p-indigent (= sp-poor) modules sp-portfolios spp(r)

12 Subprojectivity and the subprojectivity domain of a module Definition Given modules M and N, M is said to be N-subprojective if for every epimorphism g : B N and for every homomorphism f : M N, then there exists a homomorphism h : M B such that gh = f. The subprojectivity domain, or domain of subprojectivity, of a module M is defined to be the collection Pr 1 (M) := { N Mod-R : M is N-subprojective }. The domain of subprojectivity of a module is a measure of how projective that module is. Just as with projectivity domains, a module M is projective precisely when Pr 1 (M) is as large as possible (i.e. equal to Mod-R.) Before we proceed, we need to introduce two additional notions. Definition Let C Mod-R. We say that C is a subprojective-portfolio, or sp-portfolio for short, if there exists M Mod-R such that C = Pr 1 (M). The class spp(r) := {C Mod-R : C is an sp-portfolio} will be named the subprojective profile, or sp-profile, of R. Our first lemma says that, in order for M to be N-subprojective, one only needs to lift maps to projective modules that cover N, to free modules that cover N or even to a single projective module that covers N. Lemma Let M, N Mod-R. Then, the following conditions are equivalent. 1. M is N-subprojective. 2. For every f : M N and every epimorphism g : P N with P projective, there exists h : M P such that gh = f. 3. For every f : M N and every epimorphism g : F N with F free, there exists h : M F such that gh = f.

13 4. For every f : M N there exists an epimorphism g : P N with P projective and a morphism h : M P such that gh = f. 13 Proof. The implications (1) (2) (3) (4) are clear. To show (4) (1), assume (4) and let f : M N be a morphism and g : B N be an epimorphism. By (4), there exist an epimorphism g : P N and a morphism h : M P such that gh = f. Since P is projective, there exists a morphism h : P B such that g = gh. Then, hh : M B and ghh = gh = f. Hence, M is N-subprojective. Using the preceding lemma, we can show that a module M is projective if and only if it is M-subprojective, thus ruling out the possibility of a non-trivial subprojective analogue to the notion of quasi-projectivity. Proposition For any module M, the following are equivalent; 1. M is projective. 2. M Pr 1 (M). Proof. The implication (1) (2) is clear. For (2) (1), put M = N and f = 1 M, the identity morphism on M in the condition of Lemma (4), to see that M is a direct summand of a projective module. Hence, M is projective. Some modules can be shown easily to belong to a subprojectivity domain. Proposition If Hom R (M, A) = 0, then A Pr 1 (M). Proof. If Hom R (M, A) = 0, then given any epimorphism g : C A if we let h : M C be the zero mapping then gh = 0. Whence A Pr 1 (M). As an easy consequence of Proposition 2.2.5, we have the following. Corollary Let M and A be right R-modules. Then,

14 14 1. If rad(m) = M and rad(a) = 0, then M is A-subprojective. 2. If M is singular and A is nonsingular, then M is A-subprojective. 3. If M is semisimple and Soc(A) = 0, then M is A-subprojective. Proposition is also instrumental in figuring out the next example, where we see that sometimes the conditioned study in that Proposition actually characterizes certain subprojectivity domains. This subject will be picked up again later in Proposition Example We see that in the category of Z-modules, Pr 1 (Q) consists precisely of the abelian groups granted by Proposition for, if there is a nonzero morphism f : Q M, let π : F M be an epimorphism with F free. Since there are no nonzero morphisms from Q to F, we cannot lift f to a morphism Q F, so M Pr 1 (Q). Consequently, the subprojectivity domain of Q consists precisely of the class of reduced abelian groups. It is a natural question to ask how small Pr 1 (M) can be. The next proposition shows that the domain of subprojectivity of any module must contain at least the projective modules, and the projective modules are the only ones that belong to all sp-portfolios. Proposition The intersection Pr 1 (M), running over all R-modules M, is precisely {P Mod-R : P is projective}. Proof. To show the containment, suppose M is a module which is subprojective relative to all R-modules. Then in particular, M Pr 1 (M). So by Proposition 2.2.4, M is projective. To show the containment, let P be a projective module and M be any R-module. Let g : B P be an epimorphism and f : M P be a homomorphism. Now since P is projective g splits and so there exists a homomorphism k : P B such that gk = 1 P. Then

15 15 g(k f ) = (gk) f = f and so by definition P Pr 1 (M). Since M was arbitrary the result follows. Proposition provides a lower bound on how small the domain of subprojectivity of a module can be. If a module does achieve this lower bound, then we will call it suprojectively-poor. Definition A module M is called subprojectively poor, sp-poor, or p-indigent, if its subprojectivity domain consists of only the projective modules. Notice that it is not clear whether sp-poor modules over a ring R must exist. Section 4 will be devoted to this problem, but first we go deeper into our study of subprojectivity. The following several propositions show that subprojectivity domains behave nicely with respect to direct sums. Proposition Let {M i } i I be a set of R-modules. Then, Pr ( 1 i I M ) i = i I Pr 1 (M i ), that is, the subprojectivity domain of a direct sum is the intersection of the subprojectivity domains of the summands. Proof. To show the containment, let N be in the subprojectivity domain of M i I i and fix j I. Let g : B N be an epimorphism and f : M N be a homomorphism. Let p j : M i I i M j denote the projection map and e j : M j M i I i denote the inclusion map. Since N Pr ( 1 i I M i), then there exists a homomorphism h : M i I i B such that gh = f p j. Letting h := h e j : M j B, then it is straightforward to check that gh = f. Whence, N Pr ( ) 1 M j. To show the containment, let N be in the subprojectivity domain of M i for every i I, let g : B N be an epimorphism and let f : M i I i N be a homomorphism. Since for each j I, N Pr ( ) 1 M j then h j : M j B such that f e j = gh j. Letting

16 16 h := h i I i : M i I i B, then gh = gh i = f e i = f e i = f. Hence, N Pr 1 ( i I M i ). i I i I i I Note that Proposition tells us that spp(r) is a semilattice with a biggest element, namely Mod-R, the subprojectivity domain of any projective R-module. In view of Proposition 2.2.8, spp(r) has a smallest element if and only if R has an sp-poor module. Proposition If N Pr 1 (M), then every direct summand of N is in Pr 1 (M). Proof. Suppose A is a direct summand of N, and let g : C A be an epimorphism and f : M A be a homomorphism. Consider the epimorphism g 1 : C N/A A N/A N, where 1 : N/A N/A is the identity map. Since N Pr 1 (M), then there exists a homomorphism ĥ : M C N/A such that (g 1)ĥ = e f, where e : A N is the inclusion map. Therefore, g(pĥ) = p(g 1)ĥ = p(e f ) = f, where p : N A denotes the projection map. Hence, A Pr 1 (M). Proposition If A i Pr 1 (M) for i {1,, m}, then m i=1 A i Pr 1 (M). Proof. By induction, it is sufficient to prove the proposition when m = 2. Let g : C A B be an epimorphism and f : M A B be a homomorphism. Since A Pr 1 (M), then there exists a homomorphism h 1 : M C such that p A gh 1 = p A f, where p A : A B A is the projection map. So p A (gh 1 f ) = 0 and hence Im(gh 1 f ) 0 B B. Since B Pr 1 (M), then there exists a homomorhism h 2 : M g 1 (0 B) C such that gh 2 = gh 1 f. Let h := h 1 h 2. Then gh = gh 1 gh 2 = gh 1 gh 2 f + f = gh 2 gh 2 + f = f.

17 17 Hence, A B Pr 1 (M). Proposition If M is finitely generated and A i -subprojective for every i I, then M is i I A i-subprojective. Proof. Let f : M A i I i be a homomorphism and g : C A i I i be an epimorphism. Let X := {m 1,..., m k } be a set of generators for M. Then there exists a finite index set J I such that f (X) A j J j. By Proposition , there exists a homomorphism h : M C such that gh(m i ) = f (m i ) for all i {1,..., k}. Since X generates M, then gh = f, as hoped. We don t know if the subprojectivity domain of a module is, in general, closed under arbitrary direct sums. However, we have some information regarding when it is closed under arbitrary direct products. If the subprojectivity domain of every module is closed under products, then, by Proposition 2.2.8, the class of projective modules is closed under products. By [9, Theorem 3.3], this means that R is a right perfect, left coherent ring. This condition is enough to ensure that subprojectivity domains are closed under products. Proposition Let R be a ring. The following conditions are equivalent. 1. R is a right perfect, left coherent ring. 2. The subprojectivity domain of any right R-module is closed under arbitrary products. Proof. (2) (1) follows from the discussion on the preceding paragraph. For (1) (2), let M Mod-R and let {N λ } λ Λ be a set of modules in Pr 1 (M). Let f = ( f λ ) λ Λ : M λ Λ N λ. For every λ Λ, be g λ : P λ N λ be an epimorphism with P λ projective. By hypothesis, there exists h λ : M P λ such that f λ = g λ h λ. Let h = (h λ ) λ Λ : M λ Λ P λ, and g : λ Λ P λ λ ΛN λ be defined by

18 18 g((x λ ) λ Λ ) = (g λ (x λ )) λ Λ. It is routine to check that g is an epimorphism and that gh = f. Note that, since R is right perfect and left coherent, λ Λ P λ is projective. By Lemma (4), M is λ Λ N λ -subrojective. Similarly, recall that R is said to be right perfect if submodules of projective modules are projective. If every sp-portfolio is closed under submodules, then R must be right hereditary. The next proposition tells us that the converse of this statement is also true. Proposition Let R be a ring. The following conditions are equivalent. 1. R is right hereditary. 2. The subprojectivity domain of any right R-module is closed under submodules Proof. (2) (1). If the subprojectivity domain of any right R-module is closed under submodules then, by Proposition 2.2.8, the class of projective modules is closed under submodules. Then, R is right hereditary. For (1) (2), let M be a right R-module, K Pr 1 (M) and N K. Let f : M N. We can consider f as a morphism from M to K with image in N. Let g : P K be an epimorphism with P projective. Then, there exists h : M P such that gh = f. Now let P = g 1 (N) P. Since R is right hereditary, P is projective. Note that h(m) P, and g(p ) = N. By Proposition (4), N Pr 1 (M). In general, the subprojectivity domain of a module is not closed with respect to quotients. Consider for example the Z-modules, M = Z/(2), A = Z, and B = 2Z. Since A and B are projective, then by Proposition A, B Pr 1 (M). But M = A/B is not projective, so by Proposition A/B Pr 1 (M). Using similar arguments, note that Pr 1 (M) is closed under quotients if and only if M is projective.

19 Subprojectivity domains and torsion-free classes. Hereditary pretorsion classes are an important tool in the study of the injective and projective profile of a ring R, see [19]. For this reason, it seems reasonable to see if torsion-theoretic notions or techniques may help in the study of spp(r). Our next result tells us that it is torsion-free classes thay play a role in the study of this semilattice. Proposition tells us that, for every module M, the torsion-free class generated by M, F M, is contained in Pr 1 (M). In Example we found that, in the category of Z-modules one actually has that F Q = Pr 1 (Q). Our next goal is to characterize those subprojective portfolios for which this phenomenon happens. First, we give a definition. Definition Let C spp(r). We say that C is a basic sp-portfolio if there exists M Mod-R such that C = Pr 1 (M) = {N Mod-R : Hom(M, N) = 0}. As a quick example, notice that Mod-R is always a basic sp-portfolio, as Mod-R = Pr 1 (0). Moreover, if R is a cogenerator for Mod-R (e.g. a QF-ring) then the only basic subportfolio is Mod-R. It is clear that if C is a basic sp-portfolio and M is a module as in Definition then Hom(M, P) = 0 for every projective module P. The following proposition tells us that this condition is indeed sufficient for Pr 1 (M) to be basic. Proposition Let C Mod-R be an sp-portfolio. The following conditions are equivalent. 1. C is basic. 2. There exists a module M such that C = Pr 1 (M) and Hom(M, R) = 0. Proof. (1) (2) is clear. We show (2) (1). Let N be a module such that Hom(M, N) 0, and let f : M N be a nonzero morphism. Let p : F N be an

20 20 epimorphism with F free. By (2), Hom(M, F) = 0, so f cannot be lifted to a morphism M F, so N Pr 1 (M). Hence, Pr 1 (M) is basic. Note that if C = Pr 1 (M) is basic, then it does not follow that Hom(M, R) = 0. For example, by Proposition, if M Mod-R then M and M R have the same subprojectivity domain, while it is always the case that Hom(M R, R) 0. As a consequence of Proposition 2.3.2, we can list the subprojectivity domain of some classes of modules. 1. Pr 1 ( Z p n) = {M Mod-Z : M does not have elements of order p}. 2. Pr 1 ( p prime Z p) = {M Mod-Z : Soc(M) = 0} = {M Mod-Z : t(m) = 0}. 3. Pr 1 ( Z p ) = {M Mod-Z : Zp is not isomorphic to a submodule of M}. As we have said, for every module M, the class {N Mod-R : Hom(M, N) = 0} is a torsion-free class, that is, it is closed under submodules, extensions and arbitrary direct products. Then, we have the following consequence of Proposition Corollary Let M be a module such that Hom(M, R) = 0. Then, Pr 1 (M) is closed under arbitrary products, submodules, and extensions. The module Z p exhibits an interesting behaviour. It is not projective, but its subprojectivity domain is in some sense large. We formalize this in the following definition. Definition Let M be an R-module. We say that M is maximally subprojective if Pr 1 (M) is a coatom in spp(r). Proposition Let M be a module such that Pr 1 (M) = {K Mod-R : K does not have a direct summand isomorphic to M}. Then, M is maximally subprojective.

21 21 Proof. Assume N is a module with Pr 1 (M) Pr 1 (N). If N M K, then Pr 1 (M) Pr 1 (N) = Pr 1 (M) Pr 1 (K) Pr 1 (M), a contradiction. Hence, N does not have direct summands isomorphic to M. By our assumptions, N Pr 1 (M) Pr 1 (N), so N is N-subprojective, that is, N is projective. Hence, M is maximally subprojective. Corollary Z p is a maximally subprojective Z-module. 2. For any ring R, if S is a simple injective nonprojective module over any ring R, then S is maximally subprojective. Proof. (1) is clear from the examples after Proposition For (2), if S is a simple injective nonprojective module, then Hom(S, R) = 0, for otherwise S would be a summand of a projective module. Then, Pr 1 (S ) is basic. Finally, note that those modules M for which Hom(S, M) = 0 are precisely the modules that do not contain a direct summand isomorphic to S. Then, S is maximally subprojective. Corollary Let R be a non-semisimple right V-ring. Then, R has maximally subprojective modules. The following proposition tells us that every sp-portfolio is basic if and only if every sp-portfolio is a torsion free class and describes precisely for which rings these conditions hold. Proposition Let R be a ring. The following are equivalent. 1. R is a right hereditary, right perfect, left coherent ring. 2. Every sp-porfolio is basic. 3. Every sp-portfolio is a torsion free class.

22 22 Proof. (2) (3) is clear. (3) (1) follows because torsion free classes are closed under arbitrary intersections, so the class of projective modules is a torsion free class (Proposition 2.2.8). Finally, for (1) (2), let M be a right R-module. Since R is right hereditary, right perfect, left coherent, the class of projective modules is closed under direct products and submodules. Let N = {Ker( f ) : f : M P, and P is projective}. M/N is projective and N is the smallest submodule of M that yields a projective quotient. Now, M M/N K. If Hom(K, R) 0 then K has a projective quotient and we can find a submodule of M properly contained in N that yields a projective quotient of M, a contradiction. Hence, Hom(K, R) = 0, so Pr 1 (K) is basic, and Pr 1 (M) = Pr 1 (M/N) Pr 1 (K) = Pr 1 (K). Note that if we ignore (2) in Proposition 2.3.8, the equivalence (1) (3) can be easily obtained from propositions and The use of torsion-theoretic techniques can also be applied to study the notion of subinjectivity, as defined in [6]. Definition Let I Mod-R. We say that I is a subinjective-portfolio, or si-portfolio for short, if there exists M Mod-R such that I = In 1 (M). The class sip(r) := {I Mod-R : I is an si-portfolio} will be called the subinjective profile, or si-profile of R. By [6, Proposition 2.4(1)], sip(r) is a semilattice with biggest element. Analog to Proposition 2.2.5, we have the following result, that tells us that, for every module M, the torsion class cogenerated by M T M is contained in In 1 (M). Proposition Let M, N Mod-R. If Hom(N, M) = 0, then N In 1 (M). Motivated by Proposition , we have the following definition. Definition We say that an si-portfolio I is basic if there exists a module M for which I = In 1 (M) = {N Mod-R : Hom(N, M) = 0}.

23 23 Proposition Let I Mod-R be an si-portfolio. The following conditions are equivalent. 1. I is basic. 2. There exists M Mod-R such that I = In 1 (M) and Hom(E, M) = 0 for every injective module E. If R is right noetherian, this happens if and only if Hom(E, M) = 0, where E = {E : E is an indecomposable injective}. Proof. (1) (2) is clear from [6, Proposition 2.3]. Now assume (2), and let N Mod-R be such that Hom(N, M) 0. Then, there exists a nonzero f : N M that cannot be extended to a morphism f : E(N) M. Hence, N In 1 (M). The last assertion is clear. It is again worth noticing that if I = In 1 (M) is basic, it does not follow that Hom(E, M) = 0 for every injective module E. For example, if E 0 is any injective module, it follows from [6, Proposition 2.4 (2)] that M and M E 0 have the same subinjectivity domain, while it is always the case that Hom(E 0, E 0 M) 0. Example The subinjectivity domain of Z is In 1 (Z) = {N Z - Mod : Hom(N, Z) = 0} = {N Z - Mod : Z is not isomorphic to a direct summand of N}. Clearly, Z is not an injective (= divisible) Z-module. However, it is proved in [19] that In 1 (Z) is a coatom in the injective profile of Z. Interestingly enough, In 1 (Z) is also a coatom in the subinjective profile of Z. Definition Let M Mod-R. We say that M is maximally subinjective if In 1 (M) is a coatom in the subinjective profile of R.

24 24 Proposition Let M be an R-module such that In 1 (M) = {N Mod-R : M is not isomorphic to a direct summand of N}. Then, M is maximally subinjective. Proof. Let K be a module such that In 1 (M) In 1 (K). Note that, if K L M, then In 1 (K) = In 1 (L) In 1 (M) In 1 (M), a contradiction. Hence, M is not isomorphic to a direct summand of K, so K In 1 (M) In 1 (K), that is K is K-subinjective. Therefore, K is injective and M is maximally subinjective. Again, note that Mod-R = In 1 (0) is always a basic si-portfolio. If Mod-R has an injective generator (e.g. a right self-injective ring) then Mod-R is the only basic si-portfolio. The next proposition tackles the extreme opposite case, that is, when every si-portfolio is basic. Proposition Let R be a ring. The following conditions are equivalent. 1. R is a right hereditary, right noetherian ring. 2. Every si-portfolio is basic. 3. Every si-portfolio is a torsion class. Proof. (2) (3) (1) is clear. For (1) (2), assume R is right hereditary, right noetherian and let M Mod-R. Then, M = d(m) r(m), where d(m) is the divisible part of M and r(m) its reduced part. Since r(m) does not have injective submodules, Hom(E, r(m)) = 0 for every injective module E. Then, In 1 (M) is basic and In 1 (M) = In 1 (d(m)) In 1 (r(m)) = In 1 (r(m)). Now, assume R is a right hereditary, right noetherian ring. Then, the class E of injective modules is a torsion class. Let F denote its corresponding torsion free class. Then, M F if and only if Hom(E, M) = 0 for every E E, that is, if and only if

25 25 In 1 (M) = {N Mod-R : Hom(N, M) = 0}. Then, a module K F is indigent if and only if {N Mod-R : Hom(N, K) = 0} = E, if and only if K is a cogenerator of the torsion theory (E, F). With these observations, we have proved the following result. Proposition Let R be a right hereditary right noetherian ring, and let T = (E, F) be the torsion theory where E is the class of injective modules. Then, a module M is si-poor if and only if M/d(M) is a cogenerator of T. In particular, R has an si-poor module if and only if T can be cogenerated by a single module. As an application of Proposition , we show that si-poor modules exist over a hereditary finite dimensional algebra over an algebraically closed field k = k. Recall that, over such an algebra A, in mod-a we have the Auslander-Reiten translate τ, where τm is the k-dual of the transpose of M. The Auslander-Reiten translate has several interesting properties, a number of which can be found in [5, Chapter IV]. Corollary Let A be a hereditary finite dimensional algebra over an algebraically closed field k, and let E be the direct sum of indecomposable injectives. Then, τe is an si-poor module, where τ denotes the Auslander-Reiten translate. Proof. Since E is injective, Ext 1 A(E, E) = 0. Since A is hereditary, pr.dim(e) 1. The number of non-isomorphic indecomposable summands of E equals the rank of the Grothendieck group K 0 (A). Then, E is a tilting A-module, see [5, Corollary VI.4.4]. It follows that (Gen(E), Cogen(τE)) is a torsion theory [5, Theorem VI.2.5]. Note that Gen(E) is the class of injective modules. Hence, τe is an indigent module. Example Let A be the path algebra of the quiver The indecomposable projectives are P(1) = K K 0 0, P(2) = 0 K 0 0, P(3) = 0 K K K, and P(4) = K; and

26 26 the indecomposable injectives I(1) = K 0 0 0, I(2) = K K K 0, I(3) = 0 0 K 0 and P(4) = 0 0 K K. The Auslander-Reiten quiver of A, Γ(A), is 1100 τ τ τ 0111 τ τ 0110 τ 1000 where we represent each module by its dimension vector. Now, the sum of indecomposable injectives is E = , so τe = is si-poor. 2.4 Bounds in the subprojectivity domain of a module. In this section, we investigate both upper and lower bounds one may impose on the subprojectivity domain of a module. Recall that a module is said to be sp-poor, or p-indigent, if its subprojectivity domain consists precisely of the projective modules. We will show next that this condition may be softened with equivalent results. Proposition Consider the following conditions on a module M. 1. Pr 1 (M) = {P Mod-R : P is projective}. 2. Pr 1 (M) {P Mod-R : P is quasi-projective}. 3. Pr 1 (M) {P Mod-R : P is discrete}. 4. Pr 1 (M) {P Mod-R : P is quasi-discrete}.

27 27 Then, (1) and (2) are equivalent, and the four conditions are equivalent if R is a right perfect ring. Proof. (1) (2) is clear. Assume (2). Let K Pr 1 (M), and let P be a projective module that covers K. By Proposition , K P Pr 1 (M). Then, K P is quasi-projective, so K is P-projective and hence projective. If R is right perfect, then every quasi-projective module is discrete (cf. [21, Theorem 4.41]), so we have (1) (2) (3) (4). To show (4) (1), assume (4) and let K Pr 1 (M). Then, K is quasi-discrete, so by [21, Theorem 4.15] there exists a decomposition K = K i I i, where each K i is a quasi-discrete hollow module. We show that each K i is projective. Indeed, let P be the projective cover of K i. Since R is perfect, P is quasi-discrete, so P = P j J j, where each P j is a projective hollow module. By Proposition , each P j is in Pr 1 (M). Then, P j K i Pr 1 (M), so P j K i is quasi-discrete. Then, by [21, Theorem 4.48] K i is P j -projective. Since R is right perfect, projectivity domains are closed under arbitrary direct sums ([4, Exercise 7.16]), which implies that K i is P-projective. Then, K i is projective. Note that if R is not right perfect then the implication (2) (3) does not hold as, by [21, Theorem 4.41], a ring is right perfect if and only if every quasi-projective right module is discrete. Definition A ring R is called right manageable if there exist a set S of non-projective right R-modules such that for every non-projective R-module M, there exists A S such that A is isomorphic a direct summand of M. For convenience, we refer to the set S as the manageable set associated with R. Recall that a ring R is said to be Σ-cyclic if every right R-module is a direct sum of cyclic modules. See [12, Chapter 25]. Example If R is a right Σ-cyclic ring, then R is right manageable. In particular, an artinian serial ring is both right and left manageable.

28 28 Proposition Every manageable ring R has an sp-poor module. Proof. Let S be the manageable set of modules associated with R. Let X = A S A. We claim that X is sp-poor. To see this, let B Pr 1 (X). If B is not projective, then there exists C S such that C is isomorphic to a direct summand of B. By Proposition C Pr 1 (X). By Proposition , C Pr 1 (C) and by Proposition 2.2.4, C is projective, a contradiction. Then, B is projective and X is sp-poor. If R is an artinian chain ring then Proposition implies that the direct sum of nonprojective cyclic right R-modules is sp-poor. The next proposition tells us that, in fact, for such a ring every non-projective right R-module is sp-poor. This is an interesting discovery giving us a glance into the phenomenon of a ring R having no subprojective middle class. It should be noted that artinian chain rings also fail to have a subinjective middle class [6]. The study of rings without a projective or injective middle class has been undertaken in [1], [11], [19] and [16]. Proposition If R is an artinian chain ring, then every non-projective module is sp-poor. Proof. Since R is an artinian chain ring, every R-module is a direct sum of cyclic uniserial modules. Consequently, it suffices to consider cyclic modules by Proposition and Proposition Because R is an artinian chain ring, the ideals of R are zero or the powers J(R) n of J(R), the Jacobson radical of R. Moreover, if p J(R) but p J(R) 2, then J(R) n = p n R for every n 0. Hence, we have the finite chain for some positive integer n: R pr p 2 R... p n R = 0. Therefore, it is enough to show that p k R is sp-poor for every positive integer k. Let A = p k R, where k 0 and let g : R p k R be the quotient map. If k > m, then let f : A p m R be the inclusion map. Assume there exists h : A R such that gh = f.

29 29 Since R is a chain ring either Ker g Im h or Im h Ker g. If Im h Ker g then gh = 0, a contradiction. Hence, Ker g Im h. But since g is not monic, then Ker g 0. Hence, there is a nonzero element x A such that 0 = gh(x) = f (x), a contradiction. Thus, p m R Pr 1 (A). If k < m, then consider the homomorphism f : A p m R, where f (p k ) = p m. Assume there exists h : A R such that gh = f. But then p m = f (p k ) = gh(p k ) = g(1)p m p (2m) R, a contradiction. Thus, p m R Pr 1 (A). Finally, we note that by Proposition 2.2.4, p k R Pr ( 1 p k R ), as p k R is not projective. Example If p is a prime, then Z/p i Z is a p-indigent ( Z/p k Z ) -module for every i < k. Now we investigate the existance of sp-poor modules over a right hereditary, right perfect, left coherent ring. We choose this class of rings because it is precisely when the projective modules form a torsion-free class, P. Proposition Let R be a right hereditary, right perfect, left coherent ring. Let P be the torsion-free class consisting of the projective R-modules, and let T be the corresponding torsion class. Then, R has an sp-poor module if and only if there exists a module M that generates (T, P). Proof. Assume there exists a module M that generates (T, P). Then, since (T, P) is a torsion theory, the class {N : Hom(M, N) = 0} = P. Then, M is sp-poor. Now, assume M is an sp-poor module. Let M be the smallest module that yields a projective quotient, so M M/M N and Pr 1 (M) = Pr 1 (N), with N T. Now Pr 1 (N) is basic, and Pr 1 (N) = P. Therefore, N is a generator of (T, P). Now we investigate the existence of an sp-poor Z-module. Note that Z is not perfect, so we cannot apply the preceding proposition. In fact, we have the following.

30 30 Proposition Let R be a ring which is not right hereditary, or not right perfect or not left coherent. If there exists an sp-poor module M, then Hom(M, R) 0. Proof. If Hom(M, R) = 0, then Pr 1 (M) is a torsion-free class. But this cannot happen, as the class of projective modules is either not closed under submodules or not closed under arbitrary direct products. Then, Hom(M, R) 0. Corollary If M is an sp-poor Z-module, then Hom Z (M, Z) 0 and, consequently, Hom Z (M, N) 0 for every abelian group N. Since Z is a principal ideal domain, the last corollary tells us that if M is an sp-poor Z-module, then M Z N, and Pr 1 (M) = Pr 1 (Z) Pr 1 (N) = Pr 1 (N), so N is also an sp-poor Z-module. Iterating the process and taking a direct limit, we have the following result. Corollary Let M be a p-indigent Z-module. Then, there exist a submodule N M such that N Z (N). Our next goal is to show that the Z-modules T = i(z /p i Z) and S = ( i(z /p i Z) / ( i Z /p i Z) ) are not sp-poor, where p 1 < p 2 <... are the rational primes in increasing order. To do so, we will need the following result, which can be found in [13]. For each i N, let e i Z N be the standard unit vectors in Z N, that is, e i ( j) = δ i j, the Kronecker delta. Proposition Every homomorphism f : Z N Z is completely determined by its action on Z (N). In particular, if f (e i ) = 0 for all i, then f = 0. Proposition Hom(T, Z) = 0. Proof. Let f Hom(T, Z). Define P = Z N and g : P T by [g(α)](i) := α(i) + p i Z Z /p i Z; that is (α 1, α 2,...) (α 1 + p 1 Z, α 2 + p 2 Z,...).

31 31 Then g is epic and f g Hom(P, Z). Fix k N. Then g(p k e k ) = 0. Hence 0 = ( f g)(p k e k ) = p k ( f g)(e k ) Z. Hence ( f g)(e k ) = 0. So what we have shown is that ( f g)(e i ) = 0 for all i, and so by the last statement of Proposition , f g = 0. Since g is epic, it follows that f = 0, which concludes the proof. Proposition Hom(S, Z) = 0. Proof. Let f Hom(S, Z). Let h : T S be the epic mapping each element to its equivalence class. Then f h Hom(T, Z). By , f h = 0. Since h is epic, it follows that f = 0, which concludes the proof. From Propositions and , we conclude that both S and T are not sp-poor Z-modules. In view of Corollary , another natural candidate for an sp-poor Z-module is the Baer-Specker group Z N. However, we don t know if this is the case. Also note that, by [17, Lemma 2.8 ], the group Z N / Z (N) is also not p-indigent. Now we consider a lower bound on Pr 1 (M) which is inspired by [3], where they define a module M to be strongly soc-injective if, for every N Mod-R, every morphism f : Soc(N) M can be extended to a morphism f : N M. It is not hard to see that the requirement of M to be strongly soc-injective is equivalent to saying that SSMod -R In 1 (M). Definition Let M Mod-R. We say that M is strongly soc-projective if SSMod -R Pr 1 (M). Of course, a strongly soc-projective module need not be projective, as the Z-module Q Z shows. However, in the category of abelian groups, a finitely generated strongly-soc projective module is projective.

32 32 Proposition Let M be a finitely generated abelian group such that every semisimple module is in Pr 1 (M). Then, M is projective. Proof. By the Fundamental Theorem of Finitely Generated Abelian groups, M Z n Z p α 1 1 Then, M Z n is projective. Z p αn. If α n i 0 for some i, then Z pi Pr 1 (M), a contradiction. If R is a semiperfect ring, then every simple module has a projective cover, which has to be a local module, that is, with only one maximal submodule. In this case, we have the following proposition. Proposition Let R be a semiperfect ring and let M be a right R-module. Assume S is a simple module such that S Pr 1 (M). Then, either Hom(M, S ) = 0 or M = P(S ) K, where P(S ) stands for the projective cover of S. Proof. Assume Hom(M, S ) 0, and let f : M S be a nonzero morphism. By hypothesis, we can lift this morphism to a morphism f : M P(S ). Now, f (M) cannot be contained in the unique maximal ideal of P(S ), which is the kernel of the epimorphism P(S ) S. Then, f (M) = P(S ) and, by the projectivity of P(S ), we conclude that M = P(S ) K. Corollary Let R be a semiperfect ring and let M be a finitely generated right R-module of finite uniform dimension that is subprojective with respect to every simple (equivalently, with respect to every semisimple) module. Then, M is projective. Proof. Iterating the process of Proposition , and using the fact that M has finite uniform dimension, we have that M = P K, with P projective and K has no nonzero morphisms to a simple module. If K 0 then K has maximal submodules, a contradiction. Hence, M = P

33 33 Note that the conclusion of Corollary may hold even if R is not semiperfect. For example, by Proposition , it holds for the ring of integers Z. If we assume that R is right perfect, we can drop the finitely generated assumption in Corollary , as the existence of a maximal submodule of K is guaranteed by the conditions on R. Then, we have the following. Corollary Let R be a right perfect ring and let M be a right R-module of finite uniform dimension that is subprojective with respect to every semisimple module. Then, M is projective. If, moreover, we assume that our ring is right hereditary, right perfect, left coherent, we can also remove the finite uniform dimension assumption in Corollary Proposition Let R be a right perfect, right hereditary, left coherent ring. Then, a right module M is projective if and only if it is strongly soc-projective. Proof. As we ve seen, in this case every module has a smallest module that yields a projective quotient. Then, we can decompose every module M as M P X, with P projective and X a module without projective quotients. Moreover, since R is right perfect, every nonzero module has maximal submodules, cf. [8]. Then, if X 0 there exists a simple module S such that Hom(X, S ) 0. Since Pr 1 (M) = Pr 1 (X), S Pr 1 (X). This implies, by Proposition , that P(S ) is a direct summand of X, a contradiction. Hence, X = 0 and M P is projective.

34 3 Pure-Injectivity from a Different Perspective Introduction In [14], the pure-injective profile of a ring is introduced as an analog to the injectivity profile of [19]. The aim of this chapter is to investigate the viability of obtaining valuable information about a ring R from the pure-subinjectivity profile (also called sub pure-injectivity profile). The following remarks summarize various well-known results so that they may be easily referenced in this chapter. Remark ([27, 53.6]) A ring R is right pure-semisimple if and only if every right R-module is pure-injective. Remark ([27, 34.13(2)]) Every Z-module K, with nk = 0 for some n N, is pure injective. Remark ([26, Proposition 6]) If A is a module and f : A B is an embedding of A as a pure submodule in a pure-injective module B, then f extends to a homomorphism PE(A) B, imbedding PE(A) as a pure submodule of B. Remark ([14, Lemma 3.6]) Let A be a pure submodule of a module B. Then A/AJ can be embedded in B/BJ as a pure submodule for any left ideal J of R. Remark ([27, 34.6]) For any module M, the map ϕ M : M Hom Z (Hom Z (M, Q/Z), Q/Z) defined by ϕ M (m)(α) = α(m) for any m M and α Hom Z (M, Q/Z) is a pure monomorphism. In [20], a module M is said to be absolutely pure if each embedding in Mod-R with domain M is a pure embedding. Remark ([22, Theorem 2.8, Example 3.4]) For any module M, the following are equivalent.

35 35 1. M is absolutely pure. 2. M is a pure submodule of an injective module. 3. PE(M) = E(M). 3.2 The pure-subinjectivity domain of a module Definition Let M and N be modules. M is called N-pure-subinjective if for every pure extension K of N, every homomorphism from N to M can be extended to a homomorphism from K to M. The pure-subinjectivity domain of M is pin 1 (M) which consists of those modules N such that M is N-pure-subinjective. Lemma Let M and N be modules. Then the following are equivalent. 1. N pin 1 (M). 2. For every pure-essential extension K of N, every homomorphism from N to M can be extended to a homomorphism from K to M. 3. Every homomorphism from N to M can be extended to a homomorphism from PE(N) to M. Proof. (1) (2) (3) Clear. (3) (1) Let K be a pure extension of N and f : N M a homomorphism. By (3), there exists a homomorphism g : PE(N) M such that g N = f. By Remark 3.1.3, PE(N) is a direct summand of PE(K). Hence, g is extended to g 0 : PE(K) M. Since (g 0) K i = f where i : N K is inclusion, the proof is completed. Corollary A module M is pure-injective if and only if pin 1 (M) = Mod-R if and only if M pin 1 (M).

36 Proposition The intersection of pure-subinjectivity domains of all R-modules is the class of all pure-injective R-modules. 36 Proof. Since a pure-injective module is a direct summand of its pure extensions, the class of all pure-injective modules is contained in the intersection of pure-subinjectivity domains of all modules. Let M be a module which every module is M-pure-subinjective. Since M is pure in PE(M) and M is M-pure-subinjective, M is a direct summand of PE(M). This implies that M is pure-injective. It is easy to see that the subinjectivity domain In 1 (M) of a module M is contained in pin 1 (M), but it need not be equal to pin 1 (M) as the following example shows. Example Let M denote the submodule (2 + 4Z)Z of the Z-module Z/4Z. By Remark 3.1.2, M is pure-injective and so by Corollary 3.2.3, M pin 1 (M). On the other hand, since M is not a direct summand of Z/4Z, it does not belong to its subinjectivity domain. It is well known that the character module Hom Z (M, Q/Z) of a module M is pure-injective, so it is N-pure-subinjective for every module N. In the next theorem, we determine the intersection of subinjectivity domains of character modules. In addition, we see that N-pure-subinjectivity and N-subinjectivity coincide for an absolutely pure module N, moreover this condition is a characterization of being absolutely pure module for N. Theorem The following are equivalent for a module N. 1. N is an absolutely pure module. 2. Every module M is N-pure-subinjective if and only if it is N-subinjective. 3. Hom Z (M, Q/Z) is N-subinjective for every module M. 4. Hom Z (Hom Z (N, Q/Z), Q/Z) =: N is N-subinjective.

37 37 5. PE(N) is N-subinjective. Proof. (1) (2) Since N is absolutely pure, by Remark 3.1.6, PE(N) = E(N). The rest is clear. (2) (3) (4) It is obvious from the fact that the character module of any module is pure-injective and Corollary (4) (1) For the homomorphism ϕ N mentioned in Remark 3.1.5, by (4), there exists f : E(N) N such that f i N = ϕ N where i N : N E(N) is inclusion. Since i N is an essential monomorphism, f is monomorphism. Since ϕ N (N) is pure submodule of N and ϕ N (N) = f (N) f (E(N)) N, f (N) is pure submodule of f (E(N)). Thus N is pure in E(N). Therefore N is absolutely pure by Remark (2) (5) Obvious by Corollary (5) (1) Let i 1 : N E(N) and i 2 : N PE(N) be inclusions. By (5), there exists a homomorphism f : E(N) PE(N) such that f i 1 = i 2. Since i 1 is an essential monomorphism, f is monomorphism. It follows that N is pure in E(N). So Remark completes the proof. Corollary The following are equivalent for a ring R. 1. R is von-neumann regular. 2. For every R-module M, pin 1 (M) = In 1 (M). 3. For every R-module M, pin 1 (M) In 1 (M). 4. For every R-module M, pin 1 (M) = In 1 (M). 5. For every R-module M, pin 1 (M) In 1 (M). Proof. (1) (2), (1) (3), (1) (4) and (1) (5) They are easy by the fact that every extension of a module over a von-neumann regular ring is a pure extension.

38 38 (2) (3) and (4) (5) are obvious. (3) (1) It is easy consequent of Theorem (5) (1) Let M be a pure-injective R-module. Then Mod-R = pin 1 (M) and by hypothesis, Mod-R = In 1 (M). Hence M is injective and so R is von-neumann regular. The pure-subinjectivity domain of a module need not be closed under pure submodules, for example, consider the Z-module Z. Since Z is not pure-injective, by Corollary 3.2.3, Z does not belong to pin 1 (Z) although PE(Z) pin 1 (Z). Proposition The following are equivalent for a module M. 1. M is pure-injective. 2. pin 1 (M) is closed under pure submodules. 3. pin 1 (M) = pin 1 (M). 4. pin 1 (M) pin 1 (M). Proof. (1) (2) and (1) (3) are clear since pin 1 (M) = Mod-R = pin 1 (M). (2) (1) Since PE(M) pin 1 (M), by (2), M is also in pin 1 (M). Then Corollary completes the proof. (3) (4) Is obvious. (4) (1) By hypothesis, PE(M) belongs to pin 1 (M). This implies that M is a direct summand of PE(M), and so M is pure-injective. Corollary The following are equivalent for a ring R. 1. R is right pure-semisimple. 2. For every R-module M, pin 1 (M) pin 1 (M).

39 39 3. For every R-module M, pin 1 (M) = pin 1 (M). Proof. Clear from Proposition and Remark Theorem The following are equivalent for an R-module M. 1. M is injective R module. 2. pin 1 (M) = In 1 (M). 3. pin 1 (M) In 1 (M). 4. In 1 (M) = In 1 (M). 5. In 1 (M) In 1 (M). Proof. (1) (2), (1) (3), (1) (4), (1) (5), (1) (6), (4) (5) and (2) (3) are clear. (3) (1) Since E(M) is pure-injective, E(M) pin 1 (M). By hypothesis, E(M) In 1 (M), and hence M is injective. Therefore R is semisimple. (5) (1) Since E(M) is injective, E(M) In 1 (M) and so by hypothesis E(M) In 1 (M). Therefore M is injective and hence R is semisimple. Corollary The following are equivalent for a ring R. 1. R is semisimple. 2. For every R-module M, pin 1 (M) = In 1 (M). 3. For every R-module M, pin 1 (M) In 1 (M). 4. For every R-module M, In 1 (M) = In 1 (M). 5. For every R-module M, In 1 (M) In 1 (M).

40 40 6. For every R-module M, pin 1 (M) = In 1 (M). Proof. (1) (2) (3) (4) (5) (1) are clear by Theorem (6) (1) Let M be an R-module. Then Mod-R = pin 1 (PE(M)) and by hypothesis, Mod-R = In 1 (PE(M)). Thus PE(M) is injective and so PE(M) In 1 (M) pin 1 (M). Therefore by Corollary 3.2.3, M is (pure-)injective and hence R is smisimple. Remark Let R be a ring. If R pin 1 (M) for a module M, then pin 1 (M) need not be equal to Mod-R. Consider a right pure-injective ring R which is not right pure-semisimple (for example, an infinite direct product of the ring Z/2Z is injective as ring but not pure-semisimple because it is regular). Then R pin 1 (M) for every module M by Proposition On the other hand, by Remark 3.1.1, there exists a module M which is not pure-injective. Recall that a module M is called cotorsion if Ext 1 R(F, M) = 0 for every flat module F. It is known that every pure-injective module is cotorsion. Proposition Let M and N be modules. Consider the following conditions. 1. N pin 1 (M). 2. There exists a pure-injective extension K of N such that every homomorphism from N to M can be extended to a homomorphism from K to M. 3. There exists a cotorsion extension K of N such that every homomorphism from N to M can be extended to a homomorphism from K to M. 4. For every extension K of N with K/N flat, every homomorphism from N to M can be extended to a homomorphism from K to M. Then (1) (2) (3) (4). Also, (4) (1) if PE(N)/N is flat.

41 41 Proof. (1) (2) Obvious by Lemma (2) (1) Let L be a pure extension of N and f : N M a homomorphism and i : N L inclusion. By hypothesis, there exists a pure-injective module K with N K and a homomorphism g : K M such that gi N = f where i N : N K is inclusion. Also, there exists a homomorphism h : L K with hi = i N because of the pure-injectivity of K. Hence ghi = f, and so f is extended to gh. (2) (3) Since every pure-injective module is cotorsion, the proof is clear. (3) (4) Let L be an extension of N with L/N flat and f : N M a homomorphism. Hence 0 N L L/N 0 is a pure-exact sequence. If we apply the functor Hom R (, K) to the pure-exact sequence where K is a cotorsion module in (3), then we obtain the exact sequence Hom R (L, K) Hom R (N, K) Ext 1 R(L/N, K) = 0. Since Hom R (L, K) Hom R (N, K) is surjective, by a similar discussion in the proof of the implication (2) (1), f is extended to a homomorphism L M. (4) (3) Let C(N) denote the cotorsion envelope of N. By Wakamatsu s Lemma (see [28, Lemma 2.1.2]), C(N)/N is flat. The rest is clear by (4). (4) (1) Let f : N M be a homomorphism. Since PE(N)/N is flat, f is extended a homomorphism PE(N) M. This implies that N pin 1 (M) due to Lemma Proposition Let R be a ring, {M i } i I a class of R-modules for any index set I and ( ) let N be an R-module. Then N pin 1 M i if and only if N pin 1 (M i ) for all i I. i I ( Proof. Let N pin 1 M i ), i I and f : N M i be a homomorphism. Then there i I exists a homomorphism g : PE(N) M i such that gi N = i Mi f where i N : N PE(N) i I and i Mi : M i M i are inclusions. Let π Mi denote the natural projection M i M i. i I i I Since π Mi gi N = f, f is extended to π Mi g. Therefore N pin 1 (M i ). Conversely, let N pin 1 ((M i )) for all i I and f : N M i be a homomorphism. Hence for each i I

42 42 i I, there exists g i : PE(N) M i with g i i N = π Mi f. Now define g : PE(N) M i by ( i I g : x (g i (x)). Since gi N = f, g extends f. Thus N pin 1 M i ). i I Corollary Let N be a module. Then the following hold. 1. Every finite direct sum of N-pure-subinjective modules is N-pure-subinjective. 2. Every direct summand of an N-pure-subinjective module is also N-pure-subinjective. We can characterize a ring to be N-pure-subinjective for a module N as the following. Theorem The following are equivalent for a ring R and an R-module N. 1. R is N-pure-subinjective as an R-module. 2. Any direct product of copies of R is N-pure-subinjective. 3. Every cyclic projective R-module is N-pure-subinjective. 4. Every finitely generated projective R-module is N-pure-subinjective. 5. Every finitely generated free R-module is N-pure-subinjective. Proof. (1) (2) Clear by Proposition (4) (3) (1) are obvious. (1) (5) Let F be a finitely generated free module. Since N pin 1 (R), by Proposition , N pin 1 (F). (5) (4) Let P be a finitely generated projective module. Then it is a direct summand of a finitely generated free module F. Hence N pin 1 (F). By Corollary (2), N pin 1 (P). Proposition Let M, N 1 and N 2 be modules. Then N 1 N 2 pin 1 (M) if and only if N i pin 1 (M) for i = 1, 2.

43 43 Proof. Let N 1 N 2 pin 1 (M), and f : N 1 M a homomorphism. Since N 1 is pure in N 1 N 2, by Remark 3.1.3, PE(N 1 ) is a direct summand of PE(N 1 N 2 ). Let π N1 : N 1 N 2 N 1 and i N1 N 2 : N 1 N 2 PE(N 1 N 2 ) be the natural projection and inclusion, respectively. Then there exists a homomorphism g : PE(N 1 N 2 ) M such that gi N1 N 2 = f π N1. It is easy to check that gi PE(N1 )i N1 = f where i PE(N1 ) : PE(N 1 ) PE(N 1 N 2 ) is inclusion. Thus N 1 pin 1 (M). Similarly, N 2 pin 1 (M), as desired. Now let f : N 1 N 2 M be a homomorphism, i N1 N 2, i N j and i N j denote the inclusions N 1 N 2 PE(N 1 N 2 ), N j N 1 N 2 and N j PE(N j ), respectively, for each j = 1, 2. Since N j pin 1 (M), there exist g j : PE(N j ) M with g j i N j = f i N j for j = 1, 2. Since PE(N 1 ) and PE(N 2 ) are direct summands of PE(N 1 N 2 ), consider the homomorphism g 1 π PE(N1 ) + g 2 π PE(N2 ) : PE(N 1 N 2 ) M where π PE(N1 ) and π PE(N2 ) are natural projections. Due to (g 1 π PE(N1 ) + g 2 π PE(N2 ))i N1 N 2 = f, M is N 1 N 2 -pure-subinjective. As a consequence of Proposition and Corollary , we have the next result. Corollary Let {M i } i I and {N i } i I be classes of modules for an index set I = {1,, n} with n a positive integer. Then i I if M i is N j -pure-subinjective for all i, j I. M i is i I N i -pure-subinjective if and only The following examples show that the Propositions and are not hold for infinite direct sum. Examples Let M i = Z Pi and N = i N Z P i. Since every M i is pure injective, N pin 1 (M i ) and M i pin 1 (N) for all i N. But N is pure submodule of i N Z Pi and N isn t direct summand of it, hence N pin ( 1 i N M i) by Corollary Let R be a right pure-injective ring which is not Σ-pure-injective over itself, then R pin ( 1 R (I)) and R (I) pin 1 (R), but R (I) pin ( 1 R (I)) for some infinite index set I.

44 Proposition Let M be an N/K-pure-subinjective module with Hom R (K, M) = 0. Then M is N-pure-subinjective. 44 Proof. Let f : N M be a homomorphism. Since f i K Hom R (K, M) = 0 where i K : K N is inclusion, K Ker f. By the Factor Theorem, there exists f : N/K M with f π N = f where π N : N N/K is natural projection. Also, N/K is pure in PE(N)/K. By hypothesis, there exists h : PE(N)/K M with hi N/K = f where i N/K : N/K PE(N)/K is inclusion. Since hπ PE(N) i N = f, N pin 1 (M). 3.3 When the pure-subinjectivity domain of a module is closed under (pure) quotients The following Theorem is generalization of [28, Theorem 3.5.1]. Theorem Let N be an R-module and C(N) be a cotorsion hull of N, then the following are equivalent. 1. PE(N)/N is flat. 2. Every Cotorsion R-module is N-pure-subinjective. 3. C(N) is N-pure-subinjective. 4. C(N) is pure injective (i.e C(N) = PE(N)). 5. For every exact sequence 0 P P P 0 with P pure injective and (a) P pure injective, or (b) P = PE(N), or (c) P = C(N), P is N-pure-subinjective.

45 45 6. For every exact sequence 0 P P P 0 with P pure injective and P, N-pure-subinjective, P is N-pure-subinjective. Proof. (1) (2) Let M be a cotorsion module. By applying the functor Hom R (, M) to the exact sequence 0 N PE(N) PE(N)/N 0, we obtain Hom R (PE(N), M) Hom R (N, M) Ext 1 R (PE(N)/N, M) Since PE(N)/N is flat, Ext 1 R (PE(N)/N, M) = 0. Hom R(PE(N), M) Hom R (N, M) being surjective implies N pin 1 (M). (2) (3) Obvious. (3) (4) Let i 1 : N C(N) and i : N PE(N) be natural homomorphism. By (3), C(N) is N-pure-subinjective, hence by Lemma there exists a homomorphism φ : PE(N) C(N) such that φi = i 1. Since every pure injective is cotorsion and C(N) is cotorsion hull of N, there exists a homomorphism α : C(N) PE(N) such that αi 1 = i. Thus αφi = i and φαi 1 = i 1, by the definition of injective hull and cotorsion hull (see [28, Definition 1.2.1]) αφ and φα are automorphism and hence α and φ are isomorphism. (4) (2) Let M be a cotorsion R-module. Since C(N) is cotorsion hull of N, every homomorphism from N to M can be extend to homomorphism from C(N) to M. Thus by Lemma 3.2.2, M is N-pure-subinjective. (2) (5) Let 0 P P P 0 be an exact sequence with P and P cotorsion R-modules. Since the class of cotorsion modules is closed under extension, P is also cotorsion and hence by (2), P is N-pure-subinjective. (5)(c) (3) Let P = C(N) P by hypothesis P is N-pure-subinjective hence by Corollary , C(N) is N-pure-subinjective. (6) (5)(a) (5)(b) are obviously. (5)(b) (6) Let 0 P P α P 0 be an exact sequence with P pure injective and P, N-pure-subinjective. Since P is N-pure-subinjective, for every homomorphism

46 46 f : N P (by Lemma 3.2.2) there exists a homomorphism ϕ : PE(N) P such that α f = ϕi where i : N PE(N). Now, consider following pullback diagram 0 P L ψ PE(N) P P α g ϕ P 0 by pullback diagram property there exists homomorphism φ : N L such that gφ = f. By hypothesis L is N-pure-subinjective, so by Lemma there exists a homomorphism λ : PE(N) L such that λi = φ. Thus gλi = gφ = f and so by Lemma 3.2.2, P is N-pure-subinjective. (5)(b) (1) By [28, Lemma 3.4.1] it is enough to show that Ext 1 R (PE(N)/N, K) = 0 for every pure injective R-module K. Let K be a pure injective R-module. For every exact sequens 0 K H f PE(N)/N 0 consider the pullback diagram of f : H PE(N)/N and π : PE(N) PE(N)/N, 0 K L α PE(N) K H and since the following diagram commute N g i f π PE(N)/N 0 PE(N) 0 H f π PE(N)/N so by the property of pullback diagram there exists a homomorphism σ : N L such that ασ = i and gσ = 0. By hypothesis L is N-pure-subinjective, thus by Lemma there exists a homomorphism φ : PE(N) L such that φi = σ and so αφi = i. By the property of pure injective hull, αφ is automorphism of PE(N) hence i = (αφ) 1 i and gφ(αφ) 1 i = gφi = gσ = 0. Therefore, π = f gφ(αφ) 1 and it implies that i PE(N)/N = f ψ

47 47 where ψ = gφ(αφ) 1 : PE(N)/N H and i PE(N)/N : PE(N)/N PE(N)/N. Consequently Ext 1 R(PE(N)/N, K) = 0. Corollary For an R-module M, if PE(M) is flat, then C(N) is pure injective flat R-module where N is pure submodule of M and C(N) is cotorsion envelop of N. 2. For a cotorsion R-module M, PE(M)/M is flat if and only if M = PE(M). 3. Let A be a class of all cotorsion Z-modules. Z i N p i and Z i N pi belong to pin 1 (M) for every Z-module M in A. Proof. (1) and (2) Are easy consequent of Theorem (3) It is consequent of Theorem by the facts that PE( Z i N p i ) = i N Z pi and i N Z pi / Z i N p i is divisible torsion free Z-module. We recall that a ring R is called left coherent provided that every finitely generated left ideal is finitely presented. Proposition Let R be a left coherent. If for every exact sequence 0 P P P 0 with P pure injective and P, N-pure-subinjective, P is N-pure-subinjective then PE(N)/N is flat. Proof. Let 0 P P PE(N) 0 be a exact sequence with P pure injective, we shall show that P is N-pure-subinjective then by Theorem 3.3.1, PE(N)/N is flat. Since R is left coherent by [28, Lemma 3.1.5], PE(N) has a flat cover F. Consider the following pullback

48 K K 0 P L τ F f 0 1 σ 0 P P ρ g PE(N) by [28, Lemma 3.2.4], F and K are pure injective. Since P is pure injective, L F P and so L is pure injective, by hypothesis P is N-pure subinjective. Proposition For an R-module N the following statements are equivalent, 1. The class of N-pure-subinjective is closed under extension. 2. Every exact sequence 0 P P C(N) 0 with P, N-pure-subinjective implies P is N-pure-subinjective. 3. Every exact sequence 0 P P PE(N) 0 with P, N-pure-subinjective implies P is N-pure-subinjective. Proof. (1) (2) Let 0 P P C(N) 0 be a exact sequence with P, N-pure-subinjective. By Theorem 3.3.1, C(N) is N-pure-subinjective and so by hypothesis P is N-pure-subinjective. (2) (3) By Theorem 3.3.1, C(N) is pure injective therefore C(N) = PE(N). (3) (1) It is similar to the proof (5)(b) (6) of Theorem The next lemma is analog of the characterization of the projectivity in terms of the injective modules (see [23, Lemma 4.22]).

49 49 Lemma A module M is pure-projective if and only if it has the projective property relative to every pure-exact sequence 0 A B C 0 of modules with B pure-injective. Proof. The necessity is clear. For the sufficiency, let 0 K L N 0 be a pure-exact sequence of modules and consider the following diagram. 0 M 0 K i L τ N f K σi σ PE(L) π Q 0 where Q = PE(L)/Im(σi). Note that ρ exists by diagram chasing with πσ = ρτ and the second row is pure-exact. By hypothesis, there exists g : M PE(L) such that πg = ρ f. ρ Since Img L, M is pure-projective. Proposition Let N be a module and consider the following conditions. 1. N is pure-projective. 2. Every pure quotient of an N-pure-subinjective module is N-pure-subinjective. 3. Every pure quotient of a cotorsion module is N-pure-subinjective. 4. Every pure quotient of a pure-injective module is N-pure-subinjective. Then (1) (2) (4) and (3) (4). If PE(N) is pure-projective, then (4) (1). Also, (2) (3) if and only if PE(N)/N is flat. Proof. (1) (2) Let M be an N-pure-subinjective module, K a pure submodule of M and f : N M/K a homomorphism. Let π : M M/K denote the natural projection. By the pure-projectivity of N, there exists a homomorphism g : N M such that f = πg. Since M is N-pure-subinjective, g = hi N for some homomorphism h : PE(N) M where

50 50 i N : N PE(N) is inclusion. Due to πhi N = f, M/K is N-pure-subinjective. Similarly, if we take into account of the projectivity of N, then M/K is N-pure-subinjective for every submodule K of M. (2) (4) and (3) (4) are obvious. (4) (1) Let M be a pure-injective module and consider a pure-exact sequence 0 K M π M/K 0 where K is a pure submodule of M. Let f : N M/K be a homomorphism. By (4), N pin 1 (M/K). Then gi = f for some g : PE(N) M/K where i : N PE(N) is inclusion. By the pure-projectivity of PE(N), there exists h : PE(N) M such that πh = g. Hence we have πhi = f. Therefore N is pure-projective by Lemma (2) (3) It is proved by Theorem Corollary Consider the following conditions for a module N. 1. Every flat R-module is N-pure-subinjective. 2. Every projective R-module is N-pure-subinjective. 3. Every free R-module is N-pure-subinjective. Then (1) (2) (3). If N is pure-projective, then (3) (1). Proof. (1) (2) (3) Obvious. (3) (2) Clear by Corollary (2). (3) (1) Assume that N is pure-projective. Let M be a flat module. Then M is a pure quotient of a free module F. Since F is N-pure-subinjective, by Proposition 3.3.6, M is N-pure-subinjective. Proposition Let N be a module and consider the following conditions. 1. N is projective. 2. Every quotient of an N-pure-subinjective module is N-pure-subinjective.

51 51 3. Every quotient of a cotorsion module is N-pure-subinjective. 4. Every quotient of a pure-injective module is N-pure-subinjective. 5. Every quotient of an injective module is N-pure-subinjective. Then (1) (2) (4) (5) and (3) (4). If PE(N) is projective, then all of them are equivalent. Moreover consider 6. Every quotient of a flat cotorsion module is N-pure-subinjective. 7. Every cotorsion module is N-pure-subinjective. Then (3) (6) (7). Also, together (1) and (7) imply (3). Proof. (1) (2) and (1) & (7) (3) are similar to the proof of (1) (2) in Proposition (2) (4) (5), (3) (4) and (3) (6) are obvious. (5) (1) It is proved as (4) (1) in Proposition (2) (3) PE(N) being projective implies that PE(N)/N is flat. So Theorem completes the proof. (6) (7) Let M be a cotorsion module. Then M has a flat cover F f M 0. By Wakamatsu s Lemma (see [28, Lemma 2.1.1]), Ker f is cotorsion. It follows that F is also cotorsion. By (6), M F/Ker f is N-pure-subinjective. Corollary Consider the following conditions for a module N. 1. R is N-pure-subinjective as an R-module. 2. Every finitely presented R-module is N-pure-subinjective. 3. Every finitely generated R-module is N-pure-subinjective. Then (3) (2) (1). If N is projective, then (1) (3).

52 52 Proof. (3) (2) Is clear. (2) (1) Since every finitely generated projective module is finitely presented, Theorem completes the proof. (1) (3) Assume that N is a projective module and let M be a finitely generated module. Then M is a quotient of a finitely generated free module F. By Theorem , F is N-pure-subinjective. So by Proposition 3.3.8, N pin 1 (M). Corollary A module M is R-pure-subinjective if and only if every finitely generated projective R-module belongs to pin 1 (M/K) for each submodule K of M. Proof. Let R pin 1 (M). By Proposition , every finitely generated free R-module belongs to pin 1 (M), then so is every finitely generated projective R-module. For any (pure) submodule K of M and any finitely generated projective R-module P, due to Proposition (Proposition 3.3.6), P pin 1 (M/K). The sufficiency is clear. In [2], a ring R is called right pure hereditary if every pure right ideal of R is projective. The next result is a characterization of right pure hereditary rings whose pure-injective hulls are projective in terms of the notion of pure-subinjectivity. Theorem Let R be a ring and S denote the set {I R R : I is pure in R}. Consider the following conditions. 1. R is right pure hereditary. 2. For every I S, every quotient of an I-pure-subinjective module is I-pure-subinjective. 3. For every I S, every quotient of a cotorsion module is I-pure-subinjective. 4. For every I S, every quotient of a pure-injective module is I-pure-subinjective. 5. For every I S, every quotient of an injective module is I-pure-subinjective.

53 53 Then (1) (2) (4) (5) and (3) (4). If PE(R R ) is projective, then all of them are equivalent. Proof. Let I S. By [26, Proposition 6], PE(I) is a direct summand of PE(R R ). If PE(R R ) is projective, then so is PE(I). The rest is clear by Proposition Proposition Every pure submodule N of a module M with N pin 1 (M) is pure-essential in a direct summand of M, also this direct summand is PE(N). Proof. Let M be a module, N a pure submodule of M and N pin 1 (M). Then there exists f : PE(N) M such that f N = i where i : N M is inclusion. We claim that f is monic. To see this, let p PE(N) with f (p) = 0. If p N, then p = 0. Assume that p N. Since f N = i, clearly, pr N = 0. Let n x i r i j = n j + pr be a finite system of equations i=1 over (pr + N)/pR where r i j R, n j N for j = 1,, m and {a i + pr : i = 1,, n} a solution in PE(N)/pR. Then n (a i + pr)r i j = n j + pr, and so n a i r i j = n j + pr j for some i=1 r j R and j = 1,, m. Since f N = i and f (p) = 0, we obtain n i=1 i=1 f (a i )r i j = n j for j = 1,, m. Due to purity of N in M, there exists {b i N : i = 1,, n} such that n b i r i j = n j for j = 1,, m. It follows that n (b i + pr)r i j = n j + pr for j = 1,, m. i=1 i=1 Hence (pr + N)/pR is pure in PE(N)/pR. Since N is pure-essential in PE(N), we have pr = 0, and so p = 0. Thus PE(N) M. On the other hand, since N is pure in M, PE(N) is a direct summand of PE(M). By the modularity condition, the pure-essential extension PE(N) of N is a direct summand of M. The next result is a consequence of Proposition and Proposition Corollary Let M be a module and N a submodule of M. Then the following are equivalent. 1. N is pure in M and N pin 1 (M). 2. M = PE(N) K for some submodule K of M and N pin 1 (K).

54 54 Proposition Let M be a nonsingular module and N pin 1 (M). Then every essential pure extension of N belongs to pin 1 (M). Proof. Let K be an essential pure extension of N, L a pure extension of K, and f : K M a homomorphism. Then N is pure in L. Since N pin 1 (M), there exists g : L M with g N = f N. We claim that r R ( f (k) g(k)) is essential in R for any k K. Let k K and 0 a R. If ka = 0, then a r R ( f (k) g(k)). If ka 0, then there exists b R such that 0 kab N by the essentiality of N in K. Hence f (kab) = g(kab), and so 0 ab r R ( f (k) g(k)). Thus we proved the assertion. This implies f (k) g(k) Z(M) = 0. It follows that g extends f. Therefore K pin 1 (M). Proposition Let R be a ring, I an ideal of R, M an R/I-module and N an R-module. If M is an N/(NI)-pure-subinjective R/I-module, then it is an N-pure-subinjective R-module. Proof. Let N/(NI) pin 1 ( M R/I ), an R-module K be a pure extension of N and f : N M an R-homomorphism. By Remark 3.1.4, N/NI can be embedded in K/KI as a pure submodule via g : N/NI K/KI defined by g(n + NI) = n + KI for any n N. Since NI Ker f, by Factor Theorem, there exists f : N/(NI) M such that f π N = f where π N : N N/NI is natural projection. By assumption, there exists an R/I-homomorphism h : K/KI M such that hg = f. Since h is also an R-homomorphism and hπ K i N = f where π K : K K/KI is natural projection and i N : N K is inclusion, hπ K extends f. Thus N pin 1 (M R ). Corollary Let R be a ring, I an ideal of R. The following hold. 1. Let M and N be R/I-modules. Then M is an N-pure-subinjective R-module if and only if it is an N-pure-subinjective R/I-module. 2. ([14, Proposition 3.7]) Let M be an R/I-module. Then M is a pure-injective R-module if and only if it is a pure-injective R/I-module.

55 55 3. Let M be an R-module. Then PE(M/MI) has an R/I-module structure. Proof. (1) The necessity is clear. The sufficiency holds by Proposition (2) This known result is a consequence of (1) by using Corollary (3) Due to (2), PE(M/MI) R/I is pure-injective as an R-module. By the definition of the pure-injective hull of a module, PE(M/MI) R is contained in PE(M/MI) R/I. This implies that PE(M/MI) R has an R/I-module structure. Theorem Let M be a right R-module and N a left S - right R-bimodule. If M is N-pure-subinjective, then Hom R (N, M) is an S -pure-subinjective right S -module. Proof. Let K be a pure extension of S as a right S -module. We need to show that Hom S (K, Hom R (N, M)) Hom S (S, Hom R (N, M)) is epic. For any right R-module A, since S is pure in K as a right S -module, we have 0 S S (N R A) K S (N R A) and so 0 (S S N) R A (K S N) R A. Hence S S N is pure in the right R-module K S N. Since S S N N as a right R-module and N pin 1 (M), we obtain Hom R (K S N, M) Hom R (S S N, M) 0. By the Adjoint Isomorphism, we have the exact sequence Hom S (K, Hom R (N, M)) Hom S (S, Hom R (N, M)) 0 as desired. Therefore S pin 1 (Hom R (N, M)). Corollary (2) is known from [25, Lemma 33.4], also it is obtained as a consequence of Corollary and Theorem Corollary The following hold. 1. Let M and N be right R-modules. If N pin 1 (M), then End R (N) pin 1 (Hom R (N, M)) as a right End R (N)-module.

56 2. Let M be a module and N a submodule of M with N pin 1 (M). If f (N) N for every homomorphism f : N M, then End R (N) is right pure-injective Endomorphism ring of a pure-injective module is right pure-injective. Proof. (1) Clear from Theorem (3) The proof follows from (2). (2) By (1), the right End R (N)-module End R (N) belongs to pin 1 (Hom R (N, M)). On the other hand, Hom R (N, M) = End R (N) by hypothesis. Hence End R (N) is right pure-injective due to Corollary Pure-subinjectively poor modules Definition A module M is called pure-subinjectively poor (or simply, ps-poor) if the pure-subinjectivity domain of M consists of only pure-injective modules. By using Corollary (2), we have the next result. Remark Let M be a module and N a direct summand of M. If N is ps-poor, then so is M. Theorem The following are equivalent for a ring R. 1. R is a right pure-semisimple ring. 2. Every R-module is ps-poor. 3. There exists a pure-injective ps-poor R-module is a ps-poor R-module. 5. R has a ps-poor module and every ps-poor R-module is pure-injective. 6. R has a ps-poor module and every quotient of a ps-poor R-module is ps-poor.

57 57 7. R has a ps-poor module and every direct summand of a ps-poor R-module is ps-poor. Proof. (1) (2) and (1) (5) by Remark (2) (4) and (2) (6) (7) Obvious. (4) (2) Clear from Remark (2) (3) For any module M, PE(M) satisfies the mentioned properties in (3). (3) (1) Let M be a module and P a pure-injective ps-poor module. Since P is M-pure-subinjective, M is pure-injective. By Remark 3.1.1, R is right pure-semisimple. (5) (1) Let M be a module. By (5), there exists a ps-poor module N. Then M N is also ps-poor by Remark 3.4.2, and so it is pure-injective due to (5). Also M is pure-injective. Thus R is right pure-semisimple. (7) (2) Let M be a module. Similar to the proof of (5) (1), M N is a ps-poor module for some ps-poor module N. The rest is clear by hypothesis. Proposition A module M is ps-poor if and only if i I M i is ps-poor for any index set I where M i = M for all i I. Proof. Clear from Proposition Theorem Let R be a ring. Then the following are equivalent. 1. R is ps-poor as an R-module. 2. Any direct product of copies of R is ps-poor. 3. Every free R-module is ps-poor. 4. There exists a cyclic projective ps-poor R-module.

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