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Some Derivations on d-Algebras

Young Hee Kim

Department of Mathematics, Chungbuk National University, Cheongju, Korea
Correspondence to: Correspondence to: Young Hee Kim, (yhkim@chungbuk.ac.kr)
Received August 7, 2018; Revised September 20, 2018; Accepted December 20, 2018.
This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract

Using the notions of left and right derivations in d/BCK-algebras, we show that the set of all cycles forms a right ideal of a d-algebra, and show that the right translation maps become an (r, l)-derivation on BCK-algebras. Moreover, we discuss some properties on cycles and polynomial elements in d/BCK-algebras.

Keywords : d/BCK-algebra, (r,l)((l,r))-derivation, Excision, Cycle, Polynomial element
1. Introduction

The notion of derivations arising in analytic theory is extremely helpful in exploring the structures and properties of algebraic systems. Several authors [1, 2] studied derivations in rings and near rings. Jun and Xin [3] applied the notion of derivation in ring and near ring theory to BCI-algebras. In [4], the concept of derivation for lattices was introduced and some of its properties are investigated. Alshehri [5] introduced the notion of ranked bigroupoids and discussed (X, *, ω)-self-(co)derivations. Recently, Jun et al. [6] obtained further results on derivations of ranked bigroupoids, and Jun et al. [7] introduced the notion of generalized coderivations in ranked bigroupoids.

The notions of BCK-algebras and BCI-algebras were introduced by Iseki and his colleague [8, 9]. The class of BCK-algebras is a proper subclass of the class of BCI-algebras. We refer useful textbooks for BCK-algebras and BCI-algebras to [1012]. Neggers and Kim [13] introduced the notion of d-algebras which is another useful generalization of BCK-algebras, and then investigated several relations between d-algebras and BCK-algebras as well as several other relations between d-algebras and oriented digraphs.

In this paper we discuss left and right derivations in d/BCK-algebras, especially in excision-d-algebras, and cycles and polynomial elements in d/BCK-algebras related to (r, l)-derivations.

2. Preliminaries

A d-algebra [13] is a non-empty set X with a constant 0 and a binary operation “*” satisfying the following axioms:

1. (I) x * x = 0,

2. (II) 0 * x = 0,

3. (III) x * y = 0 and y * x = 0 imply x = y, for all x, yX. For more information on d-algebras we refer to [1417].

A BCK-algebra is a d-algebra X satisfying the following additional axioms:

1. (IV) ((x * y) * (x * z)) * (z * y) = 0,

2. (V) (x * (x * y)) * y = 0 for all x, y, zX.

### Theorem 2.1

If (X, *, 0) is a BCK-algebra, then

1. (VI) (x * y) * z = (x * z) * y,

2. (VII) xy implies x * zy * z,

3. (VIII) xy implies z * yz * x for all x, y, zX.

### Example 2.2 ([17])

Let X:= {0, 1, 2, 3, 4} be a set with the following table:

*01234
000000
110101
222030
333203
444110

Then (X, *, 0) is a d-algebra which is not a BCK-algebra.

A BCI-algebra is an algebra (X, *, 0) having the conditions (I), (III), (IV) and (V). It is well known that every BCK-algebra is a BCI-algebra. A BCI-algebra is said to be p-semisimple if x * (x * y) = y for all x, yX. Let X be a d-algebra and xX. X is said to be edge if for any xX, x*X = {x, 0}. It is known that if X is an edge d-algebra, then x *0 = x for any xX [17].

### Definition 2.3 ([17])

Let (X, *, 0) be a d-algebra and ∅︀ ≠ IX. I is called a d-subalgebra of X if x * yI whenever xI and yI. I is called a BCK-ideal of X if it satisfies:

1. (D0) 0 ∈ I,

2. (D1) x * yI and yI imply xI.

I is called a d-ideal of X if it satisfies (D1) and (D2) xI and yX imply x * yI, i.e., I * XI.

It is known that I is a right ideal of X if it satisfies the condition (D2).

3. Left and Right Derivations

Let (X, *, 0) be a d/BCI-algebra and let xy:= y *(y *x) for all x, yX. A map d: XX is said to be an (r, l)-derivation if d(x*y) = (x*d(y))∧(d(x) *y) for all x, yX. Similarly, a map d: XX is said to be an (l, r)-derivation if d(x * y) = (d(x) * y) ∧ (x * d(y)) for all x, yX.

### Proposition 3.1

Let (X, *, 0) be an edge d-algebra. If d: XX is an (r, l)-derivation (or an (l, r)-derivation), then d(0) = 0.

Proof

If d: XX is an (r, l)-derivation, then

$d(x*y)=(x*dy)∧(dx*y).$

If we let y:= x in (1), then

$d(0)=d(x*x)=(x*d(x))∧(d(x)*x).$

If we take x:= 0 in (2), since X is an edge d-algebra, we obtain

$d(0)=(0*d(0))∧(d(0)*0)=0∧d(0)=d(0)*(d(0)*0)=0.$

Similarly, we prove that d(0) = 0 for any (l, r)-derivation d.

### Corollary 3.2

Let (X, *, 0) be an edge d-algebra. If d: XX is an (r, l)-derivation, then (x * d(x)) ∧ (d(x) * x) = 0 for all xX.

Proof

It follows immediately from (2) and d(0) = 0.

Note that (a * b) ∧ (b * a) is not unusual in d-algebras.

### Proposition 3.3

Let (X, *, 0) be a d-algebra. If d: XX is an (r, l)-derivation, then d(x * d(x)) = 0 for all xX.

Proof

If we take y:= d(x) in (1), then

$d(x*d(x))=(x*d2(x))∧(d(x)*d(x))=(x*d2(x))∧0=0*(0*(x*d2(x))=0.$

Using the “idea” that d(x) represents the boundary of x, then x * d(x) is said to be x with its boundary removed and thus we may think of x * d(x) = y as an open element. Also, if d(x) = 0, then one usually thinks of x as being a cycle, i.e., we observe that, by Proposition 3.3, every open element in a d-algebra is a cycle.

### Proposition 3.4

Let (X, *, 0) be a d-algebra and let d: XX be an (r, l)-derivation. If x is a cycle, then x * y is also a cycle for any yX, i.e., if J is the set of all cycles in X, then J forms a right ideal of X.

Proof

If x is a cycle, then d(x) = 0. Since d is an (r, l)-derivation, we have d(x * y) = (x * d(y)) ∧ (d(x) * y) = (x * d(y)) ∧ (0 * y) = (x * d(y)) ∧ 0 = 0, proving that x * y is a cycle, for any yX.

Note that the set J of all cycles in Proposition 3.4 is a dsubalgebra of X.

Let (X, *, 0) be a d-algebra and let d: XX be an (r, l)-derivation (or an (l, r)-derivation). An element xX is said to be geometric if d2(x) = 0. Note that every cycle is a geometric element.

### Proposition 3.5

Let (X, *, 0) be a d-algebra and let d: XX be an (r, l)-derivation. Then any boundary of a geometric element is a cycle.

Proof

Straightforward.

If ZdX,GdX,BdX denotes the set of cycles, geometric elements and boundaries of geometric elements, respectively, then we have inclusions:

$BdX⊆ZdX⊆GdX.$
4. Excision d-Algebras

### Theorem 4.1

If (X, *, 0) is a BCK-algebra, then

$(x*y)*a=[x*(y*a)]∧[(x*a)*y]$

for any x, y, aX.

Proof

Given x, y, aX, we claim that (x*a)*yx*(y*a). By Theorem 2.1(VI), we obtain ((x * a) * y) * (x * (y * a)) = [(x*a) * (x*(y * a))] *y. By using (IV) and Theorem 2.1(VII), (VI), we have

$[(x*a)*(x*(y*a))]*y≤((y*a)*a)*y=((y*a)*y)*a=((y*y)*a)*a=0,$

which proves [(x * a) * y] * [x * (y * a)] = 0. Using the claim, we prove that

$[x*(y*a)]∧[(x*a)*y]=[(x*a)*y]*[((x*a)*y)*(x*(y*a))]=[(x*a)*y]*0=(x*a)*y=(x*y)*a.$

This proves the theorem.

### Corollary 4.2

Let (X, *, 0) be a BCK-algebra and let aX. If we define da: XX by da(x):= x * a, then it is an (r, l)-derivation on X.

Proof

Given x, yX, by (3) in Theorem 4.1, we have da(x * y) = (x * da(y)) ∧ (da(x) * y).

### Corollary 4.3

Let (X, *, 0) be a p-semisimple BCI-algebra and let aX. Then da(x * y) = x * da(y) for all x, yX.

Proof

Given x, yX, we have da(x * y) = (x * da(y)) ∧ (da(x)*y) = (da(x)*y)*[(da(x)*y)*(x*da(y))] = x*da(y), proving the corollary.

A d-algebra (X, *, 0) is said to be an excision-d-algebra if, for any aX, da is an (r, l)-derivation on X. We define a set dX by

$dX:={da∣da:an (r,l)-derivation X,a∈X}.$

By Corollary 4.2, every BCK-algebra is an excision-d-algebra. It is not yet known that there are examples of excision-d-algebras which are not BCK-algebras.

### Proposition 4.4

Let (X, *, 0) be aBCK-algebra and let a, bX. Then dadb = dbda where “∘” is the usual composition of functions.

Proof

Given xX, we have (dadb)(x) = da(db(x)) = da(x * b) = (x * b) * a = (x * a) * b = (dbda)(x), proving the proposition.

Given a groupoid (X, *), we define a set RLDer(X) by

$RLDer(X):={α∣α:an (r,l)-derivation on X}.$

### Proposition 4.5

If (X, *, 0) is an excision-d-algebra, then there is an injection ξ: XdX.

Proof

If we define ξ: XdX by ξ(a):= da, then it is injective. In fact, if ξ(a) = ξ(b) for any a, bX, then da = db and hence x * a = x * b for all xX. If we take x:= b, then b * a = b * b = 0. If we take x:= a, then a * b = a * a = 0. Since (X, *, 0) is a d-algebra, we obtain a = b.

Problem

Give some conditions for ξ to be a groupoid homomorphism from (X, *) to (dX, ∘).

If ξ is a groupoid homomorphism, then we obtain the following relation:

$X≃ξ(X)⊆dX⊆RLDer(X)$

An excision-d-algebra (X, *) is said to be complete if d is an (r, l)-derivation on X, then there exists aX such that d = dX.

Problems
1. What (if any) complete excision-d-algebras are there?

2. Are Boolean algebras complete excision-d-algebras?

5. Cycles and Polynomial Elements

Let (X, *, 0) be a d-algebra. An element xX is said to be polynomial if there exists nN such that dn(x) = 0.

### Proposition 5.1

Let (X, *, 0) be an edge d-algebra and let d be an (r, l)-derivation on X. If we define

$Ed(X):={x∈X∣d(x)=x}.$

Then Ed(X) is a d-subalgebra of (X, *, 0).

Proof

By Proposition 3.1, we have d(0) = 0, i.e., 0 ∈ Ed(X). If x, yEd(X), then d(x) = x and d(y) = y. It follows that

$d(x*y)=(x*d(y))∧(d(x)*y)=(x*y)∧(x*y)=x*y,$

proving that x * yEd(X).

### Proposition 5.2

Let (X, *, 0) be a d-algebra and let xX. If x is a cycle of an (r, l)-derivation da for some aX, then ax = 0.

Proof

If x is a cycle of an (r, l)-derivation da for some aX, then da(x) = 0. It follows that x * a = x and hence ax = x * (x * a) = x * x = 0.

### Proposition 5.3

Let (X, *, 0) be a d-algebra and let d be an (r, l)-derivation on X. If we define

$Pd(X):={x∈X∣x:polynomial element of X related to d}.$

Then Pd(X) ∩ Ed(X) = {0}.

Proof

If xPd(X) ∩ Ed(X), then there exists nN such that dn(x) = 0 and d(x) = x. It follows that 0 = dn(x) = dn−1(d(x)) = dn−1(x) = ⋯ = d(d(x)) = d(x) = x.

Proposition 5.3 shows that 0 is the only element which is both a cycle and a polynomial element in d-algebras.

### Proposition 5.4

Let (X, *, 0) be an edge d-algebra and let d be an (r, l)-derivation on X. If (x * y) * z = (x * z) * y for all x, y, zX and if d(y) ∧ xy for all x, yX, then (Ed(X), *) is a right ideal of (X, *, 0).

Proof

Given xEd(X) and yX, we have

$d(x*y)=(x*d(y))∧(d(x)*y)=(x*d(y))∧(x*y)=(x*y)*[(x*y)*(x*d(y))]=(x*y)*[(x*(x*d(y))*y]=(x*y)*[(d(y)∧x)*y]=(x*y)*0=x*y,$

proving that x * yEd(X).

### Proposition 5.5

Let (X, *, 0) be a BCK-algebra and let d be an (r, l)-derivation on X such that d(x) = x for some xX. Then (x * y) * (d(y) * y) ≤ d(x * y) for all yX.

Proof

Since (X, *, 0) is a BCK-algebra, we obtain (x * y) * (x * d(y)) ≤ d(y) * y. By applying Theorem 2.1(VIII), we obtain

$(x*y)*(d(y)*y)≤(x*y)*[(x*y)*(x*d(y))]=(x*d(y))∧(x*y)=(x*d(y))∧(d(x)*y)=d(x*y),$

proving the proposition.

6. Conclusion

In this paper, we proposed some properties on cycles and polynomial elements in d/BCK-algebra. We could get the results of 5 proposions in Section 5.

Conflict of Interest

TABLES
*01234
000000
110101
222030
333203
444110

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Biography

Young Hee Kim received her Ph.D. degree in the Department of Mathematics, Yonsei University, Korea in 1986. She is currently a professor in the Department of Mathematics, Chungbuk National University, Cheongju, Korea. Her current research interest includes the Fuzzy algebraic structure, BE-algebra, d-algebra.

E-mail: yhkim@chungbuk.ac.kr

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