ISRN Algebra

Volume 2011, Article ID 142403, 9 pages

http://dx.doi.org/10.5402/2011/142403

## Weakly Injective BCK-Modules

^{1}Department of Mathematical Sciences, Saginaw Valley State University, 7400 Bay Road, University Center, MI 48710-0001, USA^{2}Department of Mathematics, University of Oregon, Eugene, OR 97403, USA

Received 7 June 2011; Accepted 21 July 2011

Academic Editors: V. K. Dobrev and Y.-H. Quano

Copyright © 2011 Olivier A. Heubo-Kwegna and Jean B. Nganou. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We introduce the notion of weakly injective BCK-module and show that Baer's criterion holds for weakly injective BCK-modules but not for injective BCK-modules in general. We also provide examples and counterexamples of weakly injective BCK-modules.

#### 1. Introduction

Inspired by Meredith's BCK-systems, Iséki and Imai introduced the notion of BCK-algebra in 1966. These pioneers developed major aspects of the theory in the late 1960s and the 1970s. They were soon joined by many other researchers to develop various aspects of the BCK-algebra theory. Since then, BCK-algebras have been a subject of intense research. The main approach of this development has been trying to build a theory that is parallel to the standard ring theory. In this order of ideas, Noetherian and Artinian BCK-algebras [1], BCK-modules [2], injective and projective BCK-modules [2], and fractions BCK-algebras [3] have recently been treated. So far, the only articles on BCK-modules have been [2, 4]. Considering the topics covered by these two articles, it is quite clear that very little is known about the theory of modules over BCK-algebras. For instance, the notion of injective modules over BCK-algebras was introduced in [2], but not a single example was treated. In classical ring theory, injective modules are studied using Baer's criterion and divisible modules. Unfortunately, as we will show, this criterion does not hold for injective BCK-modules, and there are no natural notion of divisible modules over BCK-algebras.

The main goal of this work is to shed some light on the notion of injective modules over BCK-algebras. We do this by introducing a new class of modules (weakly injective modules) that strictly contains the above class and obtain a Baer's criterion for this class. In order to achieve this goal, we found ourselves imposing a new axiom to BCK-modules.

Recall that the notion of left module over a bounded commutative BCK-algebra was first introduced in 1994 by Abujabal et al. [4]. We consider the class of left BCK-modules that satisfy the following axiom in addition to the axioms of [4], for all and where .

We will refer to BCK-modules of this class as BCK-modules of type 2. The consideration of this class is motivated not only by the fact that it makes BCK-modules more in line with modules over rings, but also the fact that the main results obtained by the previous authors remain valid for this class. Using this class of BCK-modules, we introduce weakly injective BCK-modules. We prove that weakly injective BCK-modules are Characterized by Baer's criterion, which we use to prove that over principal bounded implicative BCK's, every module is weakly injective [Corollary 3.9]. We use these characterizations to build examples of (weakly) injective modules over BCK-algebras and also find examples that prove that our Baer's criterion is the sharpest we can get.

#### 2. Generalities on BCK-Modules

Recall that a BCK-algebra is an algebra satisfying for all (i), (ii),(iii),(iv),(v) and implies ,(vi) if and only if .

In addition, if there exists an element 1 in such that for all , then is said to be bounded and we write for . Also, if for all , is said to be commutative. In addition, is called implicative if for all . As proved in [5, Theorem 10], implicative BCK-algebras are commutative. A subset of a BCK-algebra is called an ideal of if it satisfies (i) and (ii) for every such that and , then .

As defined in [4, Definition 2.1], a left module over a bounded commutative BCK-algebra is an Abelian group with a multiplication satisfying (i) for all and ,(ii) for all and ,(iii) for all ,(iv) for all ,

where .

If in addition, satisfy the axiom (v) below, we call an -module of type 2.(v) for all and . Where .

Our terminology type 2 is motivated by the fact that every -module satisfying is as Abelian group, of exponent 2.

Recall [4, Lemma 2.4] that if is a bounded implicative BCK-algebra, then is a commutative ring. Therefore, -module of type 2 are modules over the ring .

If is a left -module, a subset of is a submodule if is a subgroup of such that whenever and .

Given two left -modules and , an -module homomorphism from to is a map satisfying(i) for all ,(ii) for all and .

The set of all -module homomorphisms from to is denoted by which has a natural structure of -module via the multiplication .

We introduce the following definition.

*Definition 2.1. *A left -module is weakly injective if for every left -module so that is of type 2, every injective homomorphism and every homomorphism , there exists a homomorphism such that .

Note that injective -modules as defined in [2] are clearly weakly injective. We have the following lemma whose some parts have been proved by other authors, but which we offer a proof here for the convenience of the reader.

Lemma 2.2. *Let be a bounded implicative BCK-algebra with unit 1. Then, for all , *(i)*,*(ii)*,
*(iii)*,
*(iv)*,*(v)*. *

*Proof. *
(i) From [5, Proposition 6], we have . In addition, and , so . Thus, . Therefore, .

For (ii), let . By (i), we do have
For (iii), let . We first prove that
In fact,

Now, we use (ii) and (2.2) to show (iii)(iv) We have
(v) Recall that being a bounded implicative BCK-algebra, is a distributive lattice [5, Theorem 12]. We have

Proposition 2.3. * Every bounded implicative BCK-algebra has a natural structure of -module of type 2. Furthermore, under this structure, every ideal of is a submodule of .*

*Proof. *Consider the operational system . Then, by [4, Proposition 2.5], is an Abelian group and together with the multiplication satisfying (i), (ii), (iii), and (iv) of the definition of -module above. It remains to verify that satisfies (v). But this is straightforward from the definition of the multiplication and Lemma 2.2 (v).

As for the proof that an ideal of is a submodule, the argument is identical to the one of [2, Theorem 2.1].

*Example 2.4. *Consider the bounded implicative BCK*, * with the standard operations. Consider and , then under the multiplication (where is the characteristic function of ), is an -module that is not of type 2 while is an -module of type 2.

*Example 2.5. *For every -modules and so that is of type 2, the -module is also of type 2. In particular, for every -module , is of type 2.

*Remark 2.6. * is an Abelian group of exponent 2; therefore, finite bounded implicative BCK-algebras have order a power of 2. This is not surprising as such BCKs are Boolean algebras [5, Theorem 12].

#### 3. Injective BCK-Modules and Baer's Criterion

will denote a bounded implicative BCK-algebra with unit 1. In addition, the term -module will refer to left -module. We start by the following lemma which is crucial for the proof of Baer's criterion.

Lemma 3.1. * Let be an -module of type 2 and a submodule of . For every , define
**
Then, is an ideal of for all . *

*Proof. *Let , then(i) as is a submodule; therefore, .(ii)Let such that and . As is implicative, by Lemma 2.2 (i). Hence, which is in as and is a submodule, thus . Therefore, are all in . Thus and are all in , hence as is a submodule, then . But from the axiom (v) of -module, it follows that
Thus, , consequently as desired.

Whence is an ideal of as stated.

*Remark 3.2. *Given an -module and , then the set is a submodule of , the submodule generated by .

Theorem 3.3 (Baer's Criterion). * Let be an - module.**Then, is weakly injective if and only if for every ideal of , every -module homomorphism from extends to a homomorphism from . *

*Proof. * This direction is obvious as is an -module of type 2, and every ideal of is an -module [Proposition 2.3].

Assume that for every ideal of , every -module homomorphism from extends to a homomorphism from . Consider and , where is type 2. Let X-Mod be the set of X-modules. Consider
First, note that since . Define on the relation by if and . Then, is easily verified to be an order on . The usual argument also show that every chain in has an upper bound, and therefore, by the Zorn's lemma, has a maximal element .

We show that , and therefore, would be the required extension of .

By definition, we have . Conversely, let , then by Lemma 3.1, as is type 2, the set is an ideal of . Define, by . Then . In addition . Hence, is an -module homomorphism, and by hypothesis extends to .

Define by . We need to verify that is a well-defined homomorphism (which clearly extends ).

For the well-definition, suppose , then . So, , hence . Now using the fact that is a homomorphism, we obtain
Hence, , because . Therefore, and is well defined. Next, we check that is a homomorphism.

Let and , then
Thus, is a homomorphism as needed. Whence and by the maximality of , we obtain , so which shows that as required.

The theory of modules over BCK-algebras displays some real pathologies as the remark below explains. Before the remark, a couple of definitions.

*Definition 3.4. *As defined in [6], an element of a BCK-algebra is called a zero-divisor if there exists a nonzero element in such that . If has nontrivial zero-divisors, then is called cancellative. These correspond to domains in ring theory.

*Remark 3.5. *The natural approach for understanding injective modules over rings consists of establishing the relationship with divisible modules. Unfortunately, what should be the natural equivalent of divisible modules over BCK-algebras turns out to be useless. In fact, it is straightforward to see that the only cancellative implicative BCK-algebra is so that every module over such is always divisible.

*Remark 3.6. *Recall [7, Theorem 3] that if is a BCK-algebra (not necessarily implicative) and , the ideal of generated by is denoted by is given by . In the case when is implicative, this simplifies to .

*Definition 3.7. *A BCK-algebra is principal if every ideal of is generated by one element.

*Example 3.8. * (1) as defined in [5, Example 1] is a principal BCK-algebra. In fact, it is easy to see that the only ideals of are and . Note that is not implicative.

(2) The BCK-algebra from [8, Appendix] is bounded implicative and principal.

We now deduce from the above Baer's criterion that all modules over principal bounded implicative BCK-algebras are weakly injective.

Corollary 3.9. * Let be an -module and suppose that is bounded implicative and principal. Then, is weakly injective. In particular, every bounded implicative and principal BCK-algebra is weakly injective as a module over itself.*

*Proof. *Let be an -module, with bounded implicative and principal, an ideal of , and an -homomorphism. Define by . It is clear that is an -homomorphism; in fact, using Lemma 2.2 (v), we obtain
We claim that extends . In fact, let . Then, . So, . Hence, is weakly injective by Baer's criterion.

*Example 3.10. *Consider as above which is a bounded implicative and principal BCK-algebra. By Corollary 3.9, is weakly injective as a module over itself.

#### 4. Examples

This section is devoted to constructing examples.

*Example 4.1 (A non weakly injective BCK-Module). *Consider any infinite set and the BCK-algebra with the natural operations. Consider
Then, is clearly an ideal of and, therefore, an -module by Proposition 2.3.*Claim 4.2. * is not weakly injective.

To see this, it is enough to produce an -module homomorphism that does not extend to .

For this, consider any finite complement subset of and defined by . Since distributes over and is associative, it follows that is an -module homomorphism.

We assert that there is no homomorphism such that . In fact, by contradiction, suppose there is such an extension. Then, for every , we have ; therefore, since is a homomorphism, then Let , then since and is a homomorphism, then , so . Note that , because if , then and . Therefore, there exists an element of that is not in . We have , that is which is a contradiction.

Whence, is an -module that is not weakly injective (much less injective).

*Example 4.3 (A weakly injective BCK-module that is not injective). * denotes the unique BCK-algebra with two elements: .

First, observe that every Abelian group has a natural structure of -module via and for all . We consider and in this view as -modules. Consider the inclusion and defined by and . Then is a -module homomorphism. If was injective, then, there would exist a homomorphism such that for all . But such an extension would satisfy , which is a contradiction. Therefore, as a module over itself, is not injective.

On the other hand, is clearly weakly injective, as it has only two ideals making Baer's criterion obvious. Or more directly, use Corollary 3.9, since is principal.

*Remark 4.4. *The existence of weakly injective modules that are not injective (see Example 4.3), shows that Baer's criterion does not characterize injective BCK-modules. That is, being able to extend homomorphisms from ideals to the whole BCK-algebra is weaker than being injective. Note also that the first example shows that not every type 2 module is weakly injective. Finally, the proof of Baer's criterion clearly works in the subcategory of -modules of type 2 so that injective objects in this category are characterized by the criterion.

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