S-Injective Modules and Rings

Nasr A. Zeyada^1*

¹Department of Mathematics, Faculty of Science, Cairo University, Cairo, Egypt &Department of Mathematics, Faculty of Science, King Abdulaziz University, KSA.

DOI: 10.4236/apm.2014.41004                

Abstract

We introduce and investigate the concept of s-injective modules and strongly s-injective modules. New characterizations of SI-rings, GV-rings and pseudo-Frobenius rings are given in terms of s-injectivity of their modules.

Keywords

S-Injective Module; Kasch Ring; Pseudo Frobenius Ring

Keywords:

In this paper we introduce and investigate the notion of s-injective Modules and Rings. A right R-module M is called right s-N-injective, where N is a right R-module, if every R-homomorphism extends to, where K is a submodule of the singular submodule. M is called strongly s-injective if M is s-N-injective for every right R-module N. The connection between this new injectivity condition and other injectivity conditions has been established, and examples are provided to distinguish s-injectivity from other injectivity concepts such as mininjectivity, soc-injectivity. Several properties of this new class of injectivity are highlighted.

Throughout this paper all rings are associative with identity, and all modules are unitary R-modules. For a right R-module, we denoted, , and by the Jacobson radical, the socle and the singular submodule of, respectively., , , and are used to indicate the right socle, the left socle, the right singular ideal, the left singular ideal, and the Jacobson radical of, respectively. For a submodule of, the notations, and mean, respectively, that N is essential, maximal, and direct summand. If is a subset of a right R-module, right annihilators will be denoted by, with a similar definition of left annihilators. Multiplication maps and will be denoted by and, respectively. If M and N are right R-modules, then M is called N-injective if every R-homomorphism from a submodule of N into M can be extended to an R-homomorphism from N into M. Mod-R indicates the category of right R-modules. We refer to [1-3] for all the undefined notions in this paper.

2. Strongly S-Injective Modules

Definition 1 A right R-module M is called s-N-injective if every R-homomorphism extends to, where is a submodule of the singular submodule. M is called s-injective if is s-R-injective. is called strongly s-injective, if is s-N-injective for all right R-modules.

For example every nonsingular R-module is strongly s-injective. In particular, the ring of integers is strongly s-injective, but not injective.

Proposition 1

1) Let be a right R-module and a family of right R-modules. Then the direct product

is s-N-injective if and only if each is s-N-injective,

2) Let M, , and be right R-modules with If M is s-N-injective, then M is s-K-injective.

3) Let M, , and be right R-modules with If is s-M-injective, then is s-M-injective.

4) Let be a right R-module and a family of right R-modules. Then is injective

if and only if is injective,

5) A right R-module is s-injective if and only if is s-P-injective for every projective right R-module.

6) Let M, , and be right R-modules with If is s-K-injective, then is s-K- injective.

7) If, , and are right R-modules, and is s-A-injective, then is s-B-injective.

Proof. The proofs of 1) through 4) are routine.

5). This follows from 4).

6). Let be an R-homomorphism where is singular submodule of. Then the map can be extended to an R-homomorphism, where the inclusion map. Now, the map is an extension of, where the natural projection map into.

7). Let be an R-isomorphism, and an R-homomorphism where is a singular submodule of. The restriction of to induces an isomorphism By hypothesis, the map can be extended to an R-homomorphism Now, the map is an extension of. □

The next two corollaries are immediate consequences of the above proposition.

Corollary 1 Let be a right R-module. Then the following statements are true:

1) A finite direct sum of s-N-injective modules is again s-N-injective. In particular a finite direct sum of s-injective (strongly s-injective) modules is again s-injective (strongly s-injective).

2) A summand of s-N-injective (s-injective, strongly s-injective) module is again s-N-injective (s-injective, strongly s-injective) module.

Corollary 2

1) Let be a right R-module and in, where the are orthogonal idempotents. Then is s-injective if and only if is injective for each,

2) Assume that and are idempotents of, and is injective. Then is s-fN-injective.

Proposition 2 If is a finitely generated right R-module, then the following conditions are equivalent:

1) Any direct sum of s-N-injective modules is s-N-injective.

2) Any direct sum of injective modules is s-N-injective.

3) is noetherian.

Proof. 1) 2). Clear.

2) 3). Consider a chain of singular submodules of and. Let be the injective hull of, and be a map defined by. Since, is s-N-injective, can be extended to an R-homomorphism Since is finitely generated, for some then and for

every. Hence is noetherian.

3) 1). Let be a direct sum of s-N-injective modules, and be a homomorphism of

right R-modules where. Since is noetherian, for some finite subset

Since finite direct sums of s-N-injective modules is s-N-injective, can be extended to an R-homomorphism □

The second singular submodule of a right R-module, denoted by, is defined by the equality. We see that a is closed submodule of and is non-singular. A right R-module is Goldie torsion if.

Lemma 1 Let and be right R-modules such that is injective. Then every homomorphism where extends to.

Proof. Let where is injective and. If is a homomor- phism where such that, then is singular. So extendable

to. Now suppose that, so is singular which is a contradiction. Thus

. Suppose that where and,. So and the kernal of the map by is essential in which is a contradiction. Then every homomorphism where extends to. □

Proposition 3 The following statements are equivalent:

1) is strongly s-injective.

2) is injective, where is the injective hull of.

3), where is nonsingular and is injective with.

4) is injective.

5) is G-injective for every Goldie torsion module.

6) is I-injective, where is the injective hull of.

Proof. 1) 2). Clear.

2) 3). If, we are done. Assume that and consider the following diagram:

where and are inclusion maps and is injective closure of in. Since is injective, is s-D-injective. So there exists an R-homomorphism which extends Since, is an embedding of in. If we write then for some submodule of because is injective. Finally is nonsingular because.

3) 4). Since is singular and is singular submodule of, so and for some submodule of. Then and is injective.

4) 5). Let be a Goldie torsion right R-module and is a submodule of. Using the above Lemma, every homomorphism extends to.

5) 6). If is the injective hull of, then and is Goldie torsion.

6) 1). Let be a right R-module and be a singular submodule of. Consider the diagram

where, , and are the inclusion maps. Since is I-injective and is injective. So, there exist R- homomorphisms and such that and. Thus. Hence is strongly s-injective. □

Corollary 3 Let a Goldie torsion right R-module. Then is injective if and only if is strongly s-injective.

Proposition 4 For a ring, the following conditions are equivalent:

1) is strongly s-injective.

2) is s-I(R_R)-injective, where is the injective hull of.

3) where is injective and is nonsingular. Moreover, if, then, and in this case and are relatively injective.

4) is injective.

5) is G-injective for every Goldie torsion right R-module.

6) is G-injective, where is the injective hull of.

7) Every finitely generated projective right R-module is strongly s-injective.

Proof. The equivalence between 1), 2), 3), 4), 5) and 6) is from Proposition 3.

1) 7) Since a finite direct sum of s-N-injective is s-N-injective for every right R-module (Corollary 1), so every finitely generated free right R-module is strongly s-injective. But a direct summand of strongly s-injective is strongly s-injective (Corollary 1). Therefore every finitely generated projective module is strongly s-injective. The converse is clear. □

The following examples show that the two classes of strongly s-injective rings and of soc-injective rings are different.

Example 1 Let be the field of two elements, for, , If

is the subring of generated by and then is a von Neumann regular ring with, and hence and is strongly s-injective. However, the map given by cannot be extended to an R-homomorphism from into (suppose that for some. Then for every, which is impossible), and so is not a soc-injective ring.

Example 2 Let where for all for all and for all and Then is a commutative, semiprimary, local ring with and has simple essential socle. It is not difficult to see that is right soc-injective. However the R-homomorphism defined by for all can not be extended to an endomorphism of, and so is not s-injective ring.

Definition 2 A ring is called a right generalized V-ring (right GV-ring) if every simple singular right R-module is injective.

Proposition 5 A ring is right GV-ring if and only if every simple right R-module is strongly s-injective.

Proof. Let be a right GV-ring and be a simple right R-module. The module is either projective or singular, so is strongly s-injective. Conversely, if is a simple singular right R-module, then is strongly s-injective. Thus is injective by Proposition 3. □

Lemma 2 For a right R-module the following conditions are equivalent:

1) satisfies on essential submodules.

2) is noetherian.

Proof. Assume that has on essential submodules. Then is noetherian for every submodule where is an intersection complement of and is noetherian). In particular, every uniform submodule of is noetherian. Let be an intersection complement of (see Kasch [2] p.112). Then is noetherian. So, to prove that is noetherian it is enough to show that is noetherian. Assume that contains an infinite direct sum of nonzero submodules. Since, each contains a proper essential submodule and is essential in. But this gives that is noetherian which is impossible because with each nonzero. Then contains independent uniform submodules such that is essential in. Thus and are noetherian. Hence is noetherian. □

It is well-known that, a ring is right noetherian if and only if all direct sums of injective right R-modules are injective. In the next Proposition we obtain a characterization of ring which has on essential right ideals in terms of strongly s-injective right R-modules.

Proposition 6 The following conditions on a ring are equivalent:

1) Every direct sum of strongly s-injective right R-modules is strongly s-injective.

2) Every direct sum of injective right R-modules is strongly s-injective.

3) Every finitely generated right R-module has on essential submodules.

4) is noetherian.

Proof. 1) 2). Clear.

2) 3). Consider a chain of essential submodules of a finitely generated right R-module

and. Let be the injective hull of and be a map defined by. Since is strongly s-injective and has an essential singular submodule, so is injective and can be extended to an R-homomorphism Then for some and. Thus

for every. Hence has on essential submodules.

3) 4). Above Lemma.

4) 1). Let be a direct sum of injective Goldie torsion modules. Then, , may be

considered as an injective R/S_r-module. Thus is an injective R/S_r-module. Now, suppose that

is a nonzero map where is simple right ideal, so is a singular right ideal and where is finite. Thus extends to. Then which is a contradiction

with. Hence any homomorphism where is a right ideal of induces a map with. The map extends to a homomorphism. Then the map where π is the natural epimorphism, extends. Hence is injective. Therefore, every direct sum of strongly s-injective right R-modules is strongly s-injective. □

If is an ideal of, is called I-semiperfect if for every right ideal, there is a decomposition such that and [4].

Lemma 3 If is Z_r-semiperfect, then the following statements hold:

1) A module is s-injective if and only if is injective.

2) for all singular right ideals of if and only if for all right ideals of.

Proof. 1) Let be s-injective, and be an R-homomorphism where is a right ideal of. Then where Let be an extension of the restriction map Define by for all Clearly, is an extension of, and so is injective by the Baer's Criterion.

2) Let be a right ideal of. Since is right -semiperfect, then where

and. So and. If

, then and so

. The last equality is because that is a singular right ideal of. Write where. Then. Therefore,. □

Proposition 7 Let be a right R-module. is semisimple summand of if and only if every right R-module is s-M-injective.

Proof. If is semisimple summand of, then every right R-module is s-M-injective. Conversely, if every right R-module is s-M-injective, then every identity map where is singular submodule of extends to. Thus is a summand of. Hence is a semisimple summand of. □

Corollary 4 A ring is right nonsingular if and only if every right R-module is s-injective.

A ring is called a right (left) ring if every singular right (left) R-module is injective. rings were initially introduced and investigated by Goodearl [5]

Theorem 1 The following statements are equivalent:

1) is right ring.

2) Every right R-module is strongly s-injective.

3) Every singular right R-module is strongly s-injective.

Proof. Clear from Proposition 3. □

A module is said to satisfy the C2-condition, if and are submodules of, and then We also say satisfies the C3-condition if and are submodules of with and then is a summand of. It is a well-know fact that the C2-condition implies the C3-condition. In the next proposition we show that s-quasi-injective modules inherit a weaker version of these conditions.

Proposition 8 Suppose is a s-quasi-injective right R-module.

1) If and are singular submodules of, and then

2) Let and be singular submodules of with If and then is a summand of.

Proof. 1) Since and is s-injective, being a summand of the s-quasi-injective right R-module is s-injective. If is the inclusion map, the identity map has an extension such that and so is a summand of.

2) Since and are summands of, and is s-quasi-injective, both and are s-M- injective. Thus the singular module is s-M-injective, and so is a summand of. □

It is a well-known fact that the C2-condition implies the C3-condition.

Proposition 9 If a module has s-C2-condition, then has s-C3-condition.

Proof. Consider singular summands and of such that. Write and let π denote the projection. Then. If, where

and and, and. Then is a mono-

morphism; so is a summand of by -. As,.

Proposition 10 Let and be Morita-equivalent rings with category equivalence Let, , and be right R-modules. Then

1) is singular if and only if is singular.

2) is s-N-injective if and only if is s-injective.

Proof. There is a natural isomorphisms and. This means that for each there is an isomorphism in such that for each, in and each in, the following diagram commutes

1). The right R-module is singular if and only if there is an exact sequence of right R-modules with essential monomorphism. But using [6, Proposition 21.4 and Proposition 21.6] the sequence is exact with essential monomorphism if and only if the sequence of right S-modules is exact with essential monomorphism. So is singular if and only if is singular.

2). Let be a s-N-injective and be a singular submodule of. Let be a homomorphism. Since is singular, is a monomorphism and the maps and are isomorphisms ( we may assume that is a submodule of), then we have the commutative diagram

So the following diagram

is commutative where is injective. The converse is similarly. □

As for right self-injectivity, right strongly s-injectivity turns out to be a Morita invariant.

Theorem 2 Right strong s-injectivity is a Morita invariant property of rings.

Proof. Let and be Morita-equivalent rings with category equivalences and. Let, and be right R-modules. is finitely generated projective R-module if and only if is finitely generated projective S-module [6, Propositions and]. Also is s-N-injective if and only if is s-F(N)-injective (Proposition 10) and then is strongly s-injective if and only if is strongly s-injective. Then, every finitely generated projective right R-module is strongly s-injective if and only if every finitely generated projective right S-module is strongly s-injective. Therefore right strong s-injectivity is a Morita invariant property of rings. □

Proposition 11 For a projective right R-module, the following conditions are equivalent:

1) Every homomorphic image of a s-M-injective right R-module is s-M-injective.

2) Every homomorphic image of an injective right R-module is s-M-injective.

3) Every singular submodule of is projective.

Proof. 1) 2) Obvious.

2) 3) Consider the following diagram:

Where is a singular submodule of, and are right R-modules with injective, an R-epimorphism, and an R-homomorphism. Since is s-injective, can be extended to an R-homomo- rphism Since is projective, can be lifted to an R-homomorphism such that Now, define by Clearly, and hence is projective.

3) 1) Let and be right R-modules with an R-epimorphism, is a singular submodule of and is s-M-injective. Consider the following diagram:

Since is projective, can be lifted to an R-homomorphism such that Since is s-injective, can be extended to an R-homomorphism Clearly, extends. □

Corollary 5 The following conditions are equivalent:

1) Every quotient of s-injective right R-module is s-injective.

2) Every quotient of an injective right R-module is s-injective.

3) Every singular right ideal is projective.

Proposition 12 The following conditions are equivalent:

1) Every strongly s-injective right R-module is injective.

2) Every nonsingular right R-module is semisimple injective.

3) For every right R-module, where semisimple injective.

4) is semiperfect.

Proof. 1) 2) If is nonsingular right R-module, then is strongly s-injective. Thus is semisimple injective.

2) 3) Let be a right R-module. If is Goldie torsion, we are done. Now suppose that is not essential in and be a maximal submodule with respect to. Then is semisimple injective and. It is clear that.

3) 4). Let be a right ideal of. Then where is semisimple injective. We have and is a summand of and generated by an idempotent. Hence is semiperfect.

4) 1). Let be a strongly s-injective and be an R-homomorphism where is a right ideal of By where and Using Proposition 3 let be an extension of the restriction map Define by for all Clearly, is an extension of, and so is injective by the Baer's Criterion. □

Theorem 3 The following are equivalent for a ring:

1) is a right PF-ring.

2) is Z_r-semiperfect, right strongly s-injective ring with essential right socle.

3) is semiperfect, right min-C2, right strongly s-injective ring with essential right socle.

4) is semiperfect with, right strongly s-injective ring with essential right socle.

5) is right finitely cogenerated, right min-C2, right strongly s-injective ring.

6) is a right Kasch, right strongly s-injective ring.

7) is a right strongly s-injective ring and the dual of every simple left R-module is simple.

Proof. 1) 2). Clear.

2) 1). Clear by Lemma 2.

1) 3). Clear.

3) 4). Suppose that is a non singular simple right ideal in, so for some simple right ideal with. Thus and is a summand which is a contradiction. Then. The other inclusion is clear.

4) 1). Let be semiperfect and right strongly s-injective ring with essential right socle and. Then where is injective and. Thus and. The right ideal may be considered as an R/J-module. Let be a map where is a right ideal of, induces a map given by. Since is injective as an R/J-module so extends to. The map, where π is the natural epimorphism, extends and is injective. Therefore is right selfinjective and is right.

1) 5). Clear.

5) 1). Since is right strongly s-injective ring, it follows from Proposition 3 that, where is injective and is nonsingular. If is simple right ideal in such that, then, by the proof of 3) 1) and every simple right ideal in is a summand of. Since is right finitely cogenerated, has a finitely generated essential socle. Thus is semisimple. Hence is injective and is right ring.

1) 6) Proposition 3 and [7, Theorem 5]

1) 7). Assume that is right strongly s-injective and the dual of every simple left R-module is simple. If is a nonsingular simple right ideal, then for some simple right ideal with. But is, so and is a summand. Thus is right min-CS, so by [3, Theorem 4.8] is semiperfect with essential right socle. Hence is right by 3). □

The following is an example of a right perfect, left Kasch ring and right strongly s-injective which is not right self-injective ring.

Example 3 Let be a field and let be the ring of all upper triangular, countably infinite square matrices over with only finitely many off-diagonal entries. Let be the K-subalgebra of generated by and Then is a right perfect, left Kasch (has only one simple left R-module up to isomorphism. So) such that whereas is neither left perfect nor right self-injective because it is not right finite dimensional.

Remark 1 Note that the ring of integers is an example of a commutative noetherian strongly s-injective ring which is not quasi-Frobenius.

Definition 3 A ring is called right CF-ring (FGF-ring) if every cyclic (finitely generated) right R-module embeds in a free module. It is not known whether right CF-rings (FGF-rings) are right artinian (quasi- Frobenius). In the next result we provide a positive answer if we assume in addition that the ring is strongly right s-injective.

Proposition 13 Every right right strongly s-injective ring is quasi-Frobenius.

Proof. Theorem 3 and [7, Theorem 5] □

3. S-CS Modules and Rings

A module is said to satisfy C1-condition or called module if every submodule of is essential in a direct summand of.

Definition 4 A right R-module is called s-CS module if every singular submodule of is essential in a summand of.

For example, every nonsingular module is s-CS. In particular, the ring of integers is s-CS but not

Proposition 14 For a right R-module, the following statements are equivalent:

1) The second singular submodule is and a summand of.

2) is s-CS.

Proof. 1) 2). If the second singular submodule of is and a summand of, then every singular submodule of is a summand of and a summand of.

2) 1). Let be s-CS and is a submodule of. Then where is a sum-

mand of and. But is closed, so. Since and is closed in, so and is. In particular, is the only closure of. Thus is a summand of. □

A module is called s-continuous if it satisfies both the s-C1- and s-C2-conditions, and a module is called quasi-s-continuous if it satisfies the s-C1- and s-C3-conditions, and is called a right s-continuous ring (right quasi-s-continuous ring) if has the corresponding property. Clearly every strongly s-injective is s-continuous.

Proposition 15 If every singular simple right R-module embeds in and is s-CS, then is finitely cogenerated.

Proof. Let be a s-CS and every singular simple right R-module embeds in. Then is a and summand of by above Proposition. Also cogenerats every simple quotient of then by [3, Theorem 7.29], is finitely cogenerated.

Proposition 16 Let be a ring. Then is a right PF-ring if and only if is a cogenerator and is.

Proof. Every right PF-ring is right self-injective and is a right cogenerator by [3, 1.56]. Conversely, if is a and is cogenerator then has finitely generated, essential right socle by Proposition 15. Since is right finite dimensional and is a cogenerator, let and be the injective hull of, then there exists an embedding for some set. Then for some projection, so and hence is monic. Thus is monic, and so where is nonsingular. So is a right PF-ring by Theorem 3. □

Proposition 17 If every simple singular right R-module embeds in and is continuous, then is semiperfect.

Proof. Let be continuous and every simple singular right R-module embeds in. Then has a finitely generated essential socle by Proposition 15. Thus, by hypothesis, there exist simple submodules

of such that is a complete set of representatives of the isomorphism classes of

simple singular right R-modules. Since is, there exist submodules of such that is a direct summand of and for. Since is an indecomposable continuous R-module, it has a local endomorphism ring; and since is projective, is maximal and small in by [3, 1.54]. Then is a projective cover of the simple module. Note that clearly implies; and the converse also holds because every module has at most one projective cover up to isomorphism. But it is clear that if and only if if and only if. Moreover, every is singular. Thus, is a complete set of representatives of the isomorphism classes of simple singular right R-modules. Hence every simple singular right R-module has a projective cover. Since every non-singular simple right R-module is projective, we conclude that is semiperfect. □

References

Conflicts of Interest

The authors declare no conflicts of interest.

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