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, IsΓ©ki 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 (𝑋,βˆ—,0,1) 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], ξ€·ξ€Έπ‘₯βˆ”π‘¦π‘š=π‘₯π‘š+π‘¦π‘š,(1.1) for all π‘₯,π‘¦βˆˆπ‘‹ and π‘šβˆˆπ‘€ where π‘₯βˆ”π‘¦=(π‘₯βˆ—π‘¦)∨(π‘¦βˆ—π‘₯).

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 (𝑋,βˆ—,≀,0) satisfying for all π‘₯,𝑦,π‘§βˆˆπ‘‹(i)(π‘₯βˆ—π‘§)βˆ—(π‘₯βˆ—π‘¦)β‰€π‘¦βˆ—π‘§, (ii)π‘₯βˆ—(π‘₯βˆ—π‘¦)≀𝑦,(iii)0≀π‘₯,(iv)π‘₯≀π‘₯,(v)π‘₯≀𝑦 and 𝑦≀π‘₯ implies π‘₯=𝑦,(vi)π‘₯≀𝑦 if and only if π‘₯βˆ—π‘¦=0.

In addition, if there exists an element 1 in 𝑋 such that π‘₯≀1 for all π‘₯βˆˆπ‘‹, then 𝑋 is said to be bounded and we write 𝑁π‘₯ for 1βˆ—π‘₯. Also, if π‘₯βˆ—(π‘₯βˆ—π‘¦)=π‘¦βˆ—(π‘¦βˆ—π‘₯) for all π‘₯,π‘¦βˆˆπ‘‹, 𝑋 is said to be commutative. In addition, 𝑋 is called implicative if π‘₯βˆ—(π‘¦βˆ—π‘₯)=π‘₯ for all π‘₯,π‘¦βˆˆπ‘‹. As proved in [5, Theorem 10], implicative BCK-algebras are commutative. A subset 𝐼 of a BCK-algebra 𝑋 is called an ideal of 𝑋 if it satisfies (i) 0∈𝐼 and (ii) for every π‘₯,π‘¦βˆˆπ‘‹ such that π‘¦βˆˆπΌ and π‘₯βˆ—π‘¦βˆˆπΌ, then π‘₯∈𝐼.

As defined in [4, Definition 2.1], a left module over a bounded commutative BCK-algebra (𝑋,βˆ—,0,1) is an Abelian group (𝑀,+) with a multiplication (π‘₯,π‘š)↦π‘₯π‘š satisfying (i)(π‘₯βˆ§π‘¦)π‘š=π‘₯(π‘¦π‘š) for all π‘₯,π‘¦βˆˆπ‘‹ and π‘šβˆˆπ‘€,(ii)π‘₯(π‘š+𝑛)=π‘₯π‘š+π‘₯𝑛 for all π‘₯βˆˆπ‘‹ and π‘š,π‘›βˆˆπ‘€,(iii)0π‘š=0 for all π‘šβˆˆπ‘€,(iv)1π‘š=π‘š for all π‘šβˆˆπ‘€,

where π‘₯βˆ§π‘¦=π‘¦βˆ—(π‘¦βˆ—π‘₯).

If in addition, 𝑀 satisfy the axiom (v) below, we call 𝑀 an 𝑋-module of type 2.(v)(π‘₯βˆ”π‘¦)π‘š=π‘₯π‘š+π‘¦π‘š for all π‘₯,π‘¦βˆˆπ‘‹ and π‘šβˆˆπ‘€. Where π‘₯βˆ”π‘¦=(π‘₯βˆ—π‘¦)∨(π‘¦βˆ—π‘₯).

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

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

If 𝑀 is a left 𝑋-module, a subset 𝑆 of 𝑀 is a submodule if (𝑆,+) is a subgroup of (𝑀,+) such that π‘₯π‘šβˆˆπ‘† whenever π‘₯βˆˆπ‘‹ and π‘šβˆˆπ‘†.

Given two left 𝑋-modules 𝑀 and 𝑁, an 𝑋-module homomorphism from 𝑀 to 𝑁 is a map π‘“βˆΆπ‘€β†’π‘ satisfying(i)𝑓(π‘š+π‘šξ…ž)=𝑓(π‘š)+𝑓(π‘šξ…ž) for all π‘š,π‘šξ…žβˆˆπ‘€,(ii)𝑓(π‘₯π‘š)=π‘₯𝑓(π‘š) for all π‘₯βˆˆπ‘‹ and π‘šβˆˆπ‘€.

The set of all 𝑋-module homomorphisms from 𝑀 to 𝑁 is denoted by Hom𝑋(𝑀,𝑁) which has a natural structure of 𝑋-module via the multiplication (π‘₯𝑓)(π‘š)=π‘₯𝑓(π‘š).

We introduce the following definition.

Definition 2.1. A left 𝑋-module 𝑄 is weakly injective if for every left 𝑋-module 𝑀,𝑁 so that 𝑁 is of type 2, every injective homomorphism π‘“βˆΆπ‘€β†’π‘ and every homomorphism π‘”βˆΆπ‘€β†’π‘„, there exists a homomorphism πœ™βˆΆπ‘β†’π‘„ such that πœ™βˆ˜π‘“=𝑔.

Note that injective 𝑋-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 𝑋 be a bounded implicative BCK-algebra with unit 1. Then, for all π‘₯,𝑦,π‘§βˆˆπ‘‹, (i)π‘₯βˆ§π‘¦=π‘₯βˆ—π‘π‘¦,(ii)π‘₯βˆ—(π‘₯βˆ§π‘¦)=π‘₯βˆ—π‘¦, (iii)π‘₯∧(π‘¦βˆ—π‘§)=(π‘₯βˆ§π‘¦)βˆ—(π‘₯βˆ§π‘§), (iv)(π‘₯βˆ—π‘¦)βˆ”(π‘¦βˆ—π‘₯)=π‘₯βˆ”π‘¦,(v)(π‘₯βˆ”π‘¦)βˆ§π‘§=(π‘₯βˆ§π‘§)βˆ”(π‘₯βˆ§π‘§).

Proof. (i) From [5, Proposition 6], we have π‘₯βˆ§π‘π‘¦β‰€π‘₯βˆ—π‘¦. In addition, π‘₯βˆ—π‘¦β‰€π‘₯ and π‘₯βˆ—π‘¦β‰€1βˆ—π‘¦=𝑁𝑦, so π‘₯βˆ—π‘¦β‰€π‘₯βˆ§π‘π‘¦. Thus, π‘₯βˆ§π‘π‘¦=π‘₯βˆ—π‘¦. Therefore, π‘₯βˆ§π‘¦=π‘₯βˆ§π‘π‘π‘¦=π‘₯βˆ—π‘π‘¦.
For (ii), let π‘₯,π‘¦βˆˆπ‘‹. By (i), we do have π‘₯βˆ—(π‘₯βˆ§π‘¦)=π‘₯βˆ—(π‘₯βˆ—π‘π‘¦)=π‘π‘¦βˆ§π‘₯=π‘₯βˆ—π‘¦.(2.1) For (iii), let π‘₯,𝑦,π‘§βˆˆπ‘‹. We first prove that π‘₯∧(π‘¦βˆ—π‘§)=(π‘₯βˆ§π‘¦)βˆ—π‘§.(2.2) In fact, =π‘₯∧(π‘¦βˆ—π‘§)=(π‘¦βˆ—π‘§)βˆ—π‘π‘₯(π‘¦βˆ—π‘π‘₯)βˆ—π‘§=(π‘₯βˆ§π‘¦)βˆ—π‘§.(2.3)
Now, we use (ii) and (2.2) to show (iii)=(π‘₯βˆ§π‘¦)βˆ—(π‘₯βˆ§π‘§)=(π‘₯βˆ—π‘π‘¦)βˆ—(π‘₯βˆ§π‘§)(π‘₯βˆ—(π‘₯βˆ§π‘§))βˆ—π‘π‘¦=(π‘₯βˆ—π‘§)βˆ—π‘π‘¦by(ii)=(π‘₯βˆ—π‘π‘¦)βˆ—π‘§=(π‘₯βˆ§π‘¦)βˆ—π‘§=π‘₯∧(π‘¦βˆ—π‘§)by(2.2).(2.4)(iv) We have []∨[]ξ€Ίξ€»(π‘₯βˆ—π‘¦)βˆ”(π‘¦βˆ—π‘₯)=(π‘₯βˆ—π‘¦)βˆ—(π‘¦βˆ—π‘₯)(π‘¦βˆ—π‘₯)βˆ—(π‘₯βˆ—π‘¦)=((π‘₯βˆ—(π‘¦βˆ—π‘₯))βˆ—π‘¦)∨((π‘¦βˆ—(π‘₯βˆ—π‘¦))βˆ—π‘₯)by5,Equation(3)=(π‘₯βˆ—π‘¦)∨(π‘¦βˆ—π‘₯)since𝑋isimplicative=π‘₯βˆ”π‘¦.(2.5)(v) Recall that being a bounded implicative BCK-algebra, (𝑋,∨,∧) is a distributive lattice [5, Theorem 12]. We have ξ€·ξ€Έπ‘₯βˆ”π‘¦βˆ§π‘§=((π‘₯βˆ—π‘¦)∨(π‘¦βˆ—π‘₯))βˆ§π‘§=((π‘₯βˆ—π‘¦)βˆ§π‘§)∨((π‘¦βˆ—π‘₯)βˆ§π‘§)=((π‘₯βˆ§π‘§)βˆ—(π‘¦βˆ§π‘§))∨((π‘¦βˆ§π‘§)βˆ—(π‘₯βˆ§π‘§))by(ii)=(π‘₯βˆ§π‘§)βˆ”(π‘¦βˆ§π‘§).(2.6)

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

Proof. Consider the operational system (𝑋,βˆ”). Then, by [4, Proposition 2.5], (𝑋,βˆ”) is an Abelian group and together with the multiplication (π‘₯,𝑦)↦π‘₯π‘¦βˆΆ=π‘₯βˆ§π‘¦ satisfying (i), (ii), (iii), and (iv) of the definition of 𝑋-module above. It remains to verify that (𝑋,βˆ”) 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 𝑋 is a submodule, the argument is identical to the one of [2, Theorem 2.1].

Example 2.4. Consider the bounded implicative BCK, 𝑋=𝒫(β„€) with the standard operations. Consider 𝑀1=Maps(β„€,β„€) and 𝑀2=Maps(β„€,β„€2), then under the multiplication (𝐴,𝑓)↦1𝐴𝑓 (where 1𝐴 is the characteristic function of 𝐴), 𝑀1 is an 𝑋-module that is not of type 2 while 𝑀2 is an 𝑋-module of type 2.

Example 2.5. For every 𝑋-modules 𝑀 and 𝑁 so that 𝑁 is of type 2, the 𝑋-module Hom𝑋(𝑀,𝑁) is also of type 2. In particular, for every 𝑋-module 𝑀, Hom𝑋(𝑀,𝑋) is of type 2.

Remark 2.6. (𝑋,βˆ”) 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

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

Lemma 3.1. Let N be an 𝑋-module of type 2 and 𝑀 a submodule of 𝑁. For every π‘›βˆˆπ‘, define 𝐼𝑛={π‘₯βˆˆπ‘‹βˆ£π‘₯π‘›βˆˆπ‘€}.(3.1) Then, 𝐼𝑛 is an ideal of 𝑋 for all π‘›βˆˆπ‘.

Proof. Let π‘›βˆˆπ‘, then(i)0𝑛=0βˆˆπ‘€ as 𝑀 is a submodule; therefore, 0βˆˆπΌπ‘›.(ii)Let π‘₯,π‘¦βˆˆπ‘‹ such that π‘₯βˆ—π‘¦βˆˆπΌπ‘› and π‘¦βˆˆπΌπ‘›. As 𝑋 is implicative, π‘¦βˆ—π‘₯=π‘¦βˆ§π‘π‘₯=𝑁π‘₯βˆ§π‘¦ by Lemma 2.2 (i). Hence, (π‘¦βˆ—π‘₯)𝑛=(𝑁π‘₯βˆ§π‘¦)𝑛=𝑁π‘₯(𝑦𝑛) which is in 𝑀 as π‘¦π‘›βˆˆπ‘€ and 𝑀 is a submodule, thus π‘¦βˆ—π‘₯βˆˆπΌπ‘›. Therefore, π‘₯βˆ—π‘¦,π‘¦βˆ—π‘₯,𝑦 are all in 𝐼𝑛. Thus (π‘₯βˆ—π‘¦)𝑛,(π‘¦βˆ—π‘₯)𝑛 and 𝑦𝑛 are all in 𝑀, hence as 𝑀 is a submodule, then (π‘₯βˆ—π‘¦)𝑛+(π‘¦βˆ—π‘₯)𝑛+π‘¦π‘›βˆˆπ‘€. But from the axiom (v) of 𝑋-module, it follows that 𝑛=ξ€·ξ€Έ(π‘₯βˆ—π‘¦)𝑛+(π‘¦βˆ—π‘₯)𝑛+𝑦𝑛=(π‘₯βˆ—π‘¦)βˆ”(π‘¦βˆ—π‘₯)βˆ”π‘¦π‘₯βˆ”π‘¦βˆ”π‘¦π‘›byLemma2.2(iv)=π‘₯𝑛sinceπ‘¦βˆ”π‘¦=0.(3.2)Thus, π‘₯π‘›βˆˆπ‘€, consequently π‘₯βˆˆπΌπ‘› as desired.
Whence 𝐼𝑛 is an ideal of 𝑋 as stated.

Remark 3.2. Given an 𝑋-module 𝑀 and π‘šβˆˆπ‘€, then the set π‘‹π‘šβˆΆ={π‘₯π‘šβˆ£π‘₯βˆˆπ‘‹} is a submodule of 𝑀, the submodule generated by π‘š.

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

Proof. β‡’)∢ This direction is obvious as 𝑋 is an 𝑋-module of type 2, and every ideal of 𝑋 is an 𝑋-module [Proposition 2.3].
⇐)∢ Assume that for every ideal 𝐼 of 𝑋, every 𝑋-module homomorphism from 𝐼→𝑄 extends to a homomorphism from 𝑋→𝑄. Consider 0β†’π‘€π‘“βˆ’β†’π‘ and π‘€π‘”βˆ’β†’π‘„, where 𝑁 is type 2. Let X-Mod be the set of X-modules. Consider =ξ€½(||𝐢,πœ™)βˆΆπΆβˆˆπ‘‹-Mod,π‘€βŠ†πΆβŠ†π‘;πœ™βˆΆπΆβŸΆπ‘„;πœ™π‘€ξ€Ύ=𝑔.(3.3) First, note that βˆ‘β‰ βˆ… since βˆ‘(𝑀,𝑔)∈. Define on 𝑋 the relation β‰Ό by (𝐢1,πœ™1)β‰Ό(𝐢1,πœ™1) if 𝐢1βŠ†πΆ2 and πœ™2|𝐢1=πœ™1. Then, β‰Ό is easily verified to be an order on βˆ‘. The usual argument also show that every chain in (βˆ‘,β‰Ό) has an upper bound, and therefore, by the Zorn's lemma, (βˆ‘,β‰Ό) has a maximal element (𝐷,πœ‘).
We show that 𝐷=𝑁, and therefore, πœ‘ would be the required extension of 𝑔.
By definition, we have π·βŠ†π‘. Conversely, let π‘›βˆˆπ‘, then by Lemma 3.1, as 𝑁 is type 2, the set 𝐼𝑛={π‘₯βˆˆπ‘‹βˆ£π‘₯π‘›βˆˆπ·} is an ideal of 𝑋. Define, π›ΌβˆΆπΌπ‘›β†’π‘„ by 𝛼(π‘₯)=πœ‘(π‘₯𝑛). Then 𝛼(π‘₯βˆ”π‘¦)=πœ‘((π‘₯βˆ”π‘¦)𝑛)=πœ‘(π‘₯𝑛+𝑦𝑛)=πœ‘(π‘₯𝑛)+πœ‘(𝑦𝑛)=𝛼(π‘₯)+𝛼(𝑦). In addition 𝛼(π‘₯𝑦)=𝛼(π‘₯βˆ§π‘¦)=πœ‘((π‘₯βˆ§π‘¦)𝑛)=πœ‘(π‘₯(𝑦𝑛))=π‘₯πœ‘(𝑦𝑛)=π‘₯𝛼(𝑦). Hence, 𝛼 is an 𝑋-module homomorphism, and by hypothesis 𝛼 extends to π›½βˆΆπ‘‹β†’π‘„.
Define πœ‘ξ…žβˆΆπ·+𝑋𝑛→𝑄 by πœ‘ξ…ž(𝑑+π‘₯𝑛)=πœ‘(𝑑)+𝛽(π‘₯). We need to verify that πœ‘ξ…ž is a well-defined homomorphism (which clearly extends πœ‘).
For the well-definition, suppose 𝑑1+π‘₯1𝑛=𝑑2+π‘₯2𝑛, then 𝑑1βˆ’π‘‘2=π‘₯1𝑛+π‘₯2𝑛=(π‘₯1βˆ”π‘₯2)𝑛. So, (π‘₯1βˆ”π‘₯2)π‘›βˆˆπ·, hence (π‘₯1βˆ”π‘₯2)βˆˆπΌπ‘›. Now using the fact that πœ‘ is a homomorphism, we obtain πœ‘ξ€·π‘‘1ξ€Έξ€·π‘‘βˆ’πœ‘2ξ€Έπ‘₯=πœ‘ξ€·ξ€·1βˆ”π‘₯2𝑛π‘₯=𝛼1βˆ”π‘₯2ξ€Έξ€·π‘₯=𝛼1ξ€Έξ€·π‘₯+𝛼2ξ€Έξ€·π‘₯=𝛽1ξ€Έξ€·π‘₯+𝛽2ξ€Έ.(3.4) Hence, πœ‘(𝑑1)+𝛽(π‘₯1)=πœ‘(𝑑1)+𝛽(π‘₯1), because 𝛽(π‘₯1)+𝛽(π‘₯1)=0. Therefore, πœ‘ξ…ž(𝑑1+π‘₯1𝑛)=πœ‘ξ…ž(𝑑2+π‘₯2𝑛) and πœ‘ξ…ž is well defined. Next, we check that πœ‘ξ…ž is a homomorphism.
Let 𝑑,π‘‘β€²βˆˆπ· and π‘₯,π‘₯ξ…žβˆˆπ‘‹, then πœ‘ξ…žξ€·ξ€·π‘‘(𝑑+π‘₯𝑛)+ξ…ž+π‘₯ξ…žπ‘›ξ€Έξ€Έ=πœ‘ξ…žξ€·ξ€·π‘‘+π‘‘ξ…žξ€Έ+ξ€·π‘₯𝑛+π‘₯ξ…žπ‘›ξ€Έξ€Έ=πœ‘ξ…žξ€·ξ€·π‘‘+π‘‘ξ…žξ€Έ+ξ€·π‘₯βˆ”π‘₯ξ…žξ€Έπ‘›ξ€Έξ€·=πœ‘π‘‘+π‘‘ξ…žξ€Έξ€·+𝛽π‘₯βˆ”π‘₯ξ…žξ€Έξ€·π‘‘=πœ‘(𝑑)+πœ‘ξ…žξ€Έξ€·π‘₯+𝛽(π‘₯)+π›½ξ…žξ€Έ=πœ‘ξ…ž(𝑑+π‘₯𝑛)+πœ‘ξ…žξ€·π‘‘ξ…ž+π‘₯ξ…žπ‘›ξ€Έ,πœ‘β€²(π‘₯β€²(𝑑+π‘₯𝑛))=πœ‘ξ…žπ‘₯ξ€·ξ€·ξ…žπ‘‘ξ€Έ+π‘₯ξ…ž(ξ€Έπ‘₯𝑛)=πœ‘ξ…žξ€·π‘₯ξ…žξ€·π‘‘+π‘₯∧π‘₯ξ…žξ€Έπ‘›ξ€Έξ€·π‘₯=πœ‘ξ…žπ‘‘ξ€Έξ€·π‘₯+π›½ξ…žξ€Έβˆ§π‘₯=π‘₯ξ…žπœ‘(𝑑)+π‘₯ξ…žπ›½(π‘₯)=π‘₯ξ…žπœ‘ξ…ž(𝑑+π‘₯𝑛).(3.5) Thus, πœ‘ξ…ž is a homomorphism as needed. Whence (𝐷,πœ‘)β‰Ό(𝐷+𝑋𝑛,πœ‘ξ…ž) and by the maximality of (𝐷,πœ‘), we obtain 𝐷=𝐷+𝑋𝑛, so π‘›βˆˆπ· which shows that 𝐷=𝑁 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 π‘₯ of a BCK-algebra 𝑋 is called a zero-divisor if there exists a nonzero element 𝑦 in 𝑋 such that π‘₯βˆ§π‘¦=0. If 𝑋 has nontrivial zero-divisors, then 𝑋 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 {0,1} so that every module over such is always divisible.

Remark 3.6. Recall [7, Theorem 3] that if 𝑋 is a BCK-algebra (not necessarily implicative) and π‘Žβˆˆπ‘‹, the ideal of 𝑋 generated by π‘Ž is denoted by βŸ¨π‘ŽβŸ©is given by {π‘₯βˆˆπ‘‹βˆ£βˆƒπ‘›>0;π‘₯βˆ—π‘Žπ‘›=0}. In the case when 𝑋 is implicative, this simplifies to βŸ¨π‘ŽβŸ©={π‘₯βˆˆπ‘‹βˆ£π‘₯βˆ—π‘Ž=0}={π‘₯βˆˆπ‘‹βˆ£π‘₯β‰€π‘Ž}.

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

Example 3.8. (1) 𝑋={0,1,2,…,πœ”} as defined in [5, Example 1] is a principal BCK-algebra. In fact, it is easy to see that the only ideals of 𝑋 are 0,𝑋 and {0,1,2,…}. Note that 𝑋 is not implicative.
(2) The BCK-algebra 𝐡4βˆ’2βˆ’3 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 𝑀 be an 𝑋-module and suppose that 𝑋 is bounded implicative and principal. Then, 𝑀 is weakly injective. In particular, every bounded implicative and principal BCK-algebra is weakly injective as a module over itself.

Proof. Let 𝑀 be an 𝑋-module, with 𝑋 bounded implicative and principal, 𝐼 an ideal of 𝑋, and π‘“βˆΆπΌ=βŸ¨π‘ŽβŸ©β†’π‘€ an 𝑋-homomorphism. Define β„ŽβˆΆπ‘‹β†’π‘€ by β„Ž(π‘₯)=𝑓(π‘Žβˆ§π‘₯). It is clear that β„Ž is an 𝑋-homomorphism; in fact, using Lemma 2.2 (v), we obtain β„Žξ€·π‘₯1βˆ”π‘₯2ξ€Έξ€·ξ€·π‘₯=π‘“π‘Žβˆ§1βˆ”π‘₯2ξ€Έξ€Έ=π‘“ξ€·ξ€·π‘Žβˆ§π‘₯1ξ€Έβˆ”ξ€·π‘Žβˆ§π‘₯2ξ€·ξ€Έξ€Έ=π‘“π‘Žβˆ§π‘₯1ξ€Έξ€·+π‘“π‘Žβˆ§π‘₯2ξ€Έξ€·π‘₯=β„Ž1ξ€Έξ€·π‘₯+β„Ž2ξ€Έ,β„Ž(π›Ύβˆ§π‘₯)=𝑓(π‘Žβˆ§(π›Ύβˆ§π‘₯))=𝑓(π›Ύβˆ§(π‘Žβˆ§π‘₯))=𝛾𝑓(π‘Žβˆ§π‘₯)=π›Ύβ„Ž(π‘₯).(3.6) We claim that β„Ž extends 𝑓. In fact, let π‘₯∈𝐼=βŸ¨π‘ŽβŸ©. Then, π‘₯=π‘Žβˆ§π‘₯. So, β„Ž(π‘₯)=𝑓(π‘Žβˆ§π‘₯)=𝑓(π‘₯). Hence, 𝑀 is weakly injective by Baer's criterion.

Example 3.10. Consider 𝐡4βˆ’2βˆ’3 as above which is a bounded implicative and principal BCK-algebra. By Corollary 3.9, 𝐡4βˆ’2βˆ’3 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 𝑆 any infinite set and the BCK-algebra 𝕏=𝒫(𝑆) with the natural operations. Consider ξ€½||𝑋||𝐼=π‘‹βŠ†π‘†βˆΆ<∞.(4.1) Then, 𝐼 is clearly an ideal of 𝕏 and, therefore, an 𝕏-module by Proposition 2.3.
Claim 4.2. 𝐼 is not weakly injective.
To see this, it is enough to produce an 𝕏-module homomorphism πœ‘βˆΆπΌβ†’πΌ that does not extend to 𝕏.
For this, consider any finite complement subset 𝐴 of 𝑆 and πœ‘βˆΆπΌβ†’πΌ defined by πœ‘(𝑋)=π‘‹βˆ©π΄. Since ∩ distributes over β–΅ and ∩ is associative, it follows that πœ‘ is an 𝕏-module homomorphism.

We assert that there is no homomorphism πœ‘βˆΆπ•β†’πΌ such that πœ‘|𝐼=πœ‘. In fact, by contradiction, suppose there is such an extension. Then, for every π‘‹βŠ†π‘†, we have 𝑋=(π‘‹βˆ©π΄)β–΅(π‘‹βˆ©π΄πΆ); therefore, since πœ‘ is a homomorphism, then πœ‘(𝑋)=πœ‘(π‘‹βˆ©π΄)β–΅πœ‘ξ€·π‘‹βˆ©π΄πΆξ€Έ=ξ€·π‘‹βˆ©ξ€Έβ–΅ξ€·πœ‘(𝐴)π‘‹βˆ©πœ‘ξ€·π΄πΆξ€·ξ€Έξ€Έ=π‘‹βˆ©πœ‘(𝐴)β–΅πœ‘ξ€·π΄πΆξ€Έξ€Έ=π‘‹βˆ©ξ€·πœ‘(𝐴)πœ‘ξ€·π΄πΆξ€Έξ€·π΄=πœ‘πΆξ€Έξ€Έ.=βˆ…(4.2) Let 𝐡=πœ‘(𝐴), then since 𝐴∩𝐴𝐢=βˆ… and πœ‘ is a homomorphism, then 𝐴𝐢∩𝐡=βˆ…, so π΅βŠ†π΄. Note that 𝐡⊊𝐴, because if 𝐡=𝐴, then πœ‘(𝐴)=𝐴∩𝐴=𝐴 and π΄βˆ‰πΌ. Therefore, there exists an element π‘Ž of 𝐴 that is not in 𝐡. We have πœ‘({π‘Ž})=πœ‘({π‘Ž}), that is {π‘Ž}∩𝐴={π‘Ž}∩𝐡 which is a contradiction.

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

Example 4.3 (A weakly injective BCK-module that is not injective). 𝐡2βˆ’1βˆ’1 denotes the unique BCK-algebra with two elements: 0,1.
First, observe that every Abelian group (𝑀,+) has a natural structure of 𝐡2βˆ’1βˆ’1-module via 0β‹…π‘š=0 and 1β‹…π‘š=π‘š for all π‘šβˆˆπ‘€. We consider β„€ and β„š in this view as 𝐡2βˆ’1βˆ’1-modules. Consider the inclusion β„€β†ͺβ„š and π‘“βˆΆβ„€β†’π΅2βˆ’1βˆ’1 defined by 𝑓(2π‘˜)=0 and 𝑓(2π‘˜+1)=1. Then 𝑓 is a 𝐡2βˆ’1βˆ’1-module homomorphism. If 𝐡2βˆ’1βˆ’1 was injective, then, there would exist a homomorphism π‘“βˆΆβ„šβ†’π΅2βˆ’1βˆ’1 such that 𝑓(π‘š)=𝑓(π‘š) for all π‘šβˆˆβ„€. But such an extension would satisfy 1=𝑓(1)=𝑓(1)=𝑓(1/2)+𝑓(1/2)=0, which is a contradiction. Therefore, as a module over itself, 𝐡2βˆ’1βˆ’1 is not injective.
On the other hand, 𝐡2βˆ’1βˆ’1 is clearly weakly injective, as it has only two ideals making Baer's criterion obvious. Or more directly, use Corollary 3.9, since 𝐡2βˆ’1βˆ’1 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 𝑋-modules of type 2 so that injective objects in this category are characterized by the criterion.