On normal filters and congruence relations in BE-algebras

Andrzej Walendziak

On normal filters and congruence relations in BE-algebras

Abstract. In this paper we introduce the notion of a normal filter in BE-algebras (in transitive BE-algebras filters conicide with normal filters). We discuss some rela- tionships between congruence relations and normal filters of a BE-algebra A(ifAis commutative, then we show that there is a bijection between congruence rela- tions and filters inA). Moreover, we give the construction of quotient algebra A/F ofAvia a normal filterF ofA.

2010 Mathematics Subject Classification: 06F35, 03G25, 08A30.

Key words and phrases: BE-algebra, (Normal) Filter, Congruence relation, Transitive BE-algebra, Commutative BE-algebra.

1. Introduction. H.S. Kim and Y.H. Kim introduced in [3] the concept of BE- algebras. They also defined the notion of filters in these algebras. The filter theory of BE-algebras was established by B.L. Meng in [5] (see also [2], [4], and [7]).

In this paper we introduce the notion of a normal filter in BE-algebras and investigate its elementary properties. We prove that for transitive BE-algebras filters conicide with normal filters. We discuss some relationships between congruence relations and normal filters of BE-algebras. Moreover, we give the construction of quotient algebra A/F of a BE-algebra A via a normal filter F of A.

2. Preliminaries.

Definition 2.1 ([3]) An algebra (X; ∗, 1) of type (2, 0) is called a BE-algebra if for all x, y, z ∈ X the following identities hold:

(BE1) x ∗ x = 1;

(BE2) x ∗ 1 = 1;

(BE3) 1 ∗ x = x;

(BE4) x ∗ (y ∗ z) = y ∗ (x ∗ z).

We introduced a relation6 on X by x 6 y if and only if x ∗ y = 1.

Definition 2.2 ([1]) A BE-algebra X is said to betransitive if for any x, y, z ∈ X, y∗ z 6 (x ∗ y) ∗ (x ∗ z).

Definition 2.3 ([3]) We say that a BE-algebra X isself-distributive if x∗(y∗z) = (x ∗ y) ∗ (x ∗ z) for any x, y, z ∈ X.

Proposition 2.4 ([1]) If X is a self-distributive BE-algebra, then it is transitive.

Proposition 2.5 ([5]) If X is a transitive BE-algebra, then it satisfies the follo- wing condition:

(T) For all x, y, z ∈ X, if x 6 y, then z ∗ x 6 z ∗ y and y ∗ z 6 x ∗ z.

Definition 2.6 ([6]) We say that a BE-algebra X iscommutative if (C) (x ∗ y) ∗ y = (y ∗ x) ∗ x

for all x, y ∈ X.

Proposition 2.7 If X is a commutative BE-algebra, then it is transitive.

Proof From Theorem 3.4 of [6] we see that

(x ∗ y) ∗ ((y ∗ z) ∗ (x ∗ z)) = 1 for all x, y, z ∈ X. Applying (BE4) we conclude that

(y ∗ z) ∗ ((x ∗ y) ∗ (x ∗ z)) = 1,

and therefore X is transitive. _

Definition 2.8 ([3]) A subset F of a BE-algebra X is called a filter of X if it satisfies:

(F1) 1 ∈ F ;

(F2) x ∈ F and x ∗ y ∈ F imply y ∈ F .

We will denote by FilX the set of all filters in X. We have FilX 6= ∅, because X is a filter of X.

Lemma 2.9 Let F ∈ FilX. If x ∈ F and x 6 y, then y ∈ F .

Proof Trivial. _

Example 2.10 ([1]) Let X = {1, a, b, c} and ∗ be defined by the following table:

∗ 1 a b c

1 1 a b c

a 1 1 b c

b 1 1 1 1

c 1 a c 1

Then (X; ∗, 1) is a BE-algebra. It is easy to see that F¹ = {1}, F² = {1, a}, and F3 = X are filters of X. Let F be a filter of X and suppose that b ∈ F . Since b∗ a = b ∗ c = 1 ∈ F , (F2) shows that a, c ∈ F , and therefore F = X. Similarly, if c∈ F , then F = X. Thus FilX = {F¹, F2, F3}.

3. Normal filters and congruence relations.

Definition 3.1 A filter F of a BE-algebra X is said to be normal if it satisfies the following condition:

(NF) x ∗ y ∈ F ⇒ [(z ∗ x) ∗ (z ∗ y) ∈ F and (y ∗ z) ∗ (x ∗ z) ∈ F ] for all x, y, z ∈ X.

Remark 3.2 In Example 2.10, the filter F1 = {1} is not normal. Indeed, b ∗ a = 1 ∈ F¹, but (c ∗ b) ∗ (c ∗ a) = c ∗ a = a /∈ F¹. The filters F2and F3 are normal.

Proposition 3.3 If X is a transitive BE-algebra, then every filter of X is normal.

Proof Let F ∈ FilX and let x, y, z ∈ X. Suppose that x∗y ∈ F . By the transitivity of X,

x∗ y 6 (z ∗ x) ∗ (z ∗ y).

Since x ∗ y ∈ F , we have (z ∗ x) ∗ (z ∗ y) ∈ F by Lemma 2.9. Applying (BE4) we get (x ∗ y) ∗ [(y ∗ z) ∗ (x ∗ z)] = (y ∗ z) ∗ [(x ∗ y) ∗ (x ∗ z)] = 1 ∈ F.

From (F2) it follows that (y ∗ z) ∗ (x ∗ z) ∈ F . 

For a filter F of X we define the binary relation ∼^F in the following way:

x∼^F y⇔ x ∗ y ∈ F and y ∗ x ∈ F .

Lemma 3.4 If F is a normal filter of X, then ∼^F is an equivalence relation on X.

Proof By (BE1), x ∗ x = 1 ∈ F , that is, x ∼^F x for any x∈ X. This means that

∼^F is reflexive. From definition, ∼^F is symmetric. To prove that ∼^F is transitive, let x ∼^F y and y ∼^F z. Then x∗ y, y ∗ x, y ∗ z, z ∗ y ∈ F . Since F is normal, (y ∗ z) ∗ (x ∗ z) ∈ F . Hence x ∗ z ∈ F . Similarly, z ∗ x ∈ F . Consequently, x ∼^F z, and so ∼^F is transitive. Thus ∼^F is an equivalence relation onX. _

Theorem 3.5 If F is a normal filter of a BE-algebra X, then ∼^F is a congruence relation on X.

Proof By Lemma 3.4, ∼^F is an equivalence relation on X. Let x, y, z ∈ X and suppose that x ∼^F y. Then x∗y, y∗x ∈ F and hence (z∗x)∗(z∗y), (z∗y)∗(z∗x) ∈ F.

Therefore

z∗ x ∼^F z∗ y. (1)

Moreover, (y ∗ z) ∗ (x ∗ z) ∈ F and (x ∗ z) ∗ (y ∗ z) ∈ F . Thus

x∗ z ∼^F y∗ z. (2)

Let now x ∼^F y and u ∼^F v. From (2) it follows that x∗ u ∼^F y∗ u. By (1), y ∗ u ∼^F y∗ v. Since ∼^F is transitive, we have x ∗ u ∼^F y∗ v. Hence ∼^F is a

congruence relation on X. _

Let ConX denote the set of all congruence relations on X. For x ∈ X, we write x/θ for the congruence class containing x, that is, x/θ ={y ∈ X : yθx}.

Definition 3.6 Let θ ∈ ConX. We say that θ is normal if it satisfies the following condition:

(NC) x ∗ yθ1 ⇒ [(z ∗ x) ∗ (z ∗ y)θ1 and (y ∗ z) ∗ (x ∗ z)θ1]

for all x, y, z ∈ X.

Proposition 3.7 If θ is a normal congruence relation on X, then 1/θ is a normal filter of X.

Proof Obviously, 1 ∈ 1/θ. Let x, x∗y ∈ 1/θ. Then xθ1 and x∗yθ1. Hence x∗yθ1∗y, that is, x ∗ yθy. Consequently, yθ1, and therefore y ∈ 1/θ. Thus 1/θ is a filter of X.

From (NC) it follows that 1/θ satisfies (NF). Hence 1/θ is normal. _

Proposition 3.8 If X is a transitive BE-algebra, then every congruence relation on X is normal.

Proof Let θ ∈ ConX. By the proof of Proposition 3.7, 1/θ is a filter of X. From Proposition 3.3 we conclude that 1/θ is normal. To prove that θ is normal, let x∗ yθ1. Then x ∗ y ∈ 1/θ. Since the filter 1/θ satisfies (NF), we have (z ∗ x) ∗ (z ∗ y), (y ∗z)∗(x∗z) ∈ 1/θ. Therefore (z ∗x)∗(z ∗y)θ1 and (y ∗z)∗(x∗z)θ1. Consequently,

(NC) holds, and so θ is normal. _

Theorem 3.9 Let X be a BE-algebra. The following statements are equivalent:

(a) the filter {1} is normal;

(b) the identity congruence relation 0X= {(x, x) : x ∈ X} is normal;

(c) X satisfies condition (T).

Proof This is immediate from definitions. _

Proposition 3.10 Every normal filter of a BE-algebra is determined by some nor- mal congruence relation.

Proof Let X be a BE-algebra and let F be a normal filter of X. Observe that ∼^F satisfies condition (NC). Let x ∗ y ∼^F 1. Then x ∗ y = 1 ∗ (x ∗ y) ∈ F . Since the filter F is normal, we have (z∗ x) ∗ (z ∗ y), (y ∗ z) ∗ (x ∗ z) ∈ F . Hence (z ∗ x) ∗ (z ∗ y) ∼^F 1 and (y ∗ z) ∗ (x ∗ z) ∼^F 1. Therefore ∼^F is a normal congruence relation. Moreover, we have

x∈ 1/ ∼^F⇔ x ∼^F 1 ⇔ x ∗ 1 = 1, 1 ∗ x = x ∈ F ⇔ x ∈ F .

Then F = 1/ ∼^F. _

Let F be a normal filter of a BE-algebra X. For x ∈ X, the class containing x is denoted by x/F , that is, x/F = x/ ∼^F= {y ∈ X : y ∼^F x}. Set X/F = {x/F : x ∈ X}. We define a binary operation ∗ on X/F as follows:

x/F∗ y/F = x ∗ y/F.

The operation ∗ is well defined since ∼^F is a congruence relation on X. By the proof of Proposition 3.10, 1/F = 1/ ∼^F= F .

Proposition 3.11 (X/F ; ∗, F ) is a BE-algebra.

Proof Trivial. _

The BE-algebra X/F is called thequotient BE-algebra of X by F.

Definition 3.12 Let (X; ∗, 1) and (Y ; ∗, 1) be BE-algebras. A function ϕ : X → Y is called ahomomorphism from X into Y if ϕ(x ∗ y) = ϕ(x) ∗ ϕ(y) for all x, y ∈ X.

Proposition 3.13 Let ϕ : X → Y be a homomorphism from X into Y. Then:

(a) ϕ(1) = 1;

(b) for all x, y ∈ X, if x 6 y, then ϕ(x) 6 ϕ(y).

Proof (a) We have ϕ(1) = ϕ(x ∗ x) = ϕ(x) ∗ ϕ(x) = 1.

(b) Let x 6 y. Then x ∗ y = 1. Hence 1 = ϕ(x ∗ y) = ϕ(x) ∗ ϕ(y), and so

ϕ(x)6 ϕ(y). 

Definition 3.14 Let X and Y be BE-algebras and let ϕ : X → Y be a homo- morphism. The set Ker(ϕ) = {x ∈ X : ϕ(x) = 1} is called the kernel of ϕ.

Proposition 3.15 Let X be a BE-algebra and F be a normal filter of X. The func- tion π : X → X/F , given by π(x) = x/F , is a surjective homomorphism and Ker(π) = F.

Proof The function π is obviously surjective. Let x, y ∈ X. We have π(x∗ y) = x ∗ y/F = x/F ∗ y/F = π(x) ∗ π(y).

Hence π is a homomorphism. Moreover, Ker(π) = {x ∈ X : π(x) = F } = {x ∈ X : x/F = F} = {x ∈ X : x/F = 1/F } = {x ∈ X : x ∈ F } = F. 

Lemma 3.16 Let ϕ : X → Y be a homomorphism of BE-algebras. Then Ker(ϕ) is a filter of X.

Proof It is obvious that 1 ∈ Ker(ϕ). Let x, x∗y ∈ Ker(ϕ). Then ϕ(x) = ϕ(x∗y) = 1. Hence 1 = ϕ(x) ∗ ϕ(y) = 1 ∗ ϕ(y) = ϕ(y). Consequently, y ∈ Ker(ϕ). 

Proposition 3.17 Let ϕ : X → Y be a homomorphism of BE-algebras. The follo- wing statements are equivalent:

(a) Ker(ϕ) is a normal filter of X;

(b) ϕ(X) satisfies condition (T).

Proof We set F = Ker(ϕ).

(a) ⇒ (b): Let x, y, z ∈ ϕ(X) and suppose that x 6 y. Then x = ϕ(a), y = ϕ(b), and z = ϕ(c) for some a, b, c ∈ X. We have ϕ(a ∗ b) = ϕ(a) ∗ ϕ(b) = x ∗ y = 1.

Therefore a∗b ∈ F . Since F is normal, we see that (c∗a)∗(c∗b) ∈ F and (b∗c)∗(a∗c) ∈ F . Hence (z∗ x) ∗ (z ∗ y) = (ϕ(c) ∗ ϕ(a)) ∗ (ϕ(c) ∗ ϕ(b)) = ϕ((c ∗ a) ∗ (c ∗ b)) = 1, and so z ∗ x 6 z ∗ y. Similarly, y ∗ z 6 x ∗ z. Consequently, ϕ(X) satisfies (T).

(b) ⇒ (a): By Lemma 3.16, F is a filter of X. Let x, y, z ∈ X and suppose that x∗ y = 1. Then ϕ(x) ∗ ϕ(y) = 1, that is, ϕ(x) 6 ϕ(y). Since ϕ(X) satisfies (T), we get ϕ(z)∗ϕ(x) 6 ϕ(z)∗ϕ(y) and ϕ(y)∗ϕ(z) 6 ϕ(x)∗ϕ(z). Hence ϕ((z∗x)∗(z∗y)) = 1 = ϕ((y ∗ z) ∗ (x ∗ z)), and therefore (z ∗ x) ∗ (z ∗ y), (y ∗ z) ∗ (x ∗ z) ∈ F . Thus F

is normal. _

By Proposition 2.7, every commutative BE-algebra is transitive. From Proposi- tions 3.3 and 3.8 we have

Proposition 3.18 All filters and congruence relations of a commmutative BE- algebra are normal.

Theorem 3.19 Let X be a commutative BE-algebra. There is a bijection between congruence relations and filters of X.

Proof We consider two functions:

f : F → ∼^F and g : θ → 1/θ.

By Theorem 3.5, f maps FilX into ConX. From Proposition 3.7, we conclude that g maps ConX into FilX. Since F = 1/∼^F (see Proposition 3.10), we deduce that

g◦ f(F ) = F for all F ∈ FilX. (3)

Let θ ∈ ConX and let x, y ∈ X. Suppose that x ∗ yθ1 and y ∗ xθ1. Then (x ∗ y) ∗ yθ1∗ y = y and (y ∗ x) ∗ xθ1 ∗ x = x. Applying (C) we obtain xθy. Conversely, if xθy, then obviously x∗ yθ1 and y ∗ xθ1. Therefore

(x ∗ yθ1 and y ∗ xθ1) ⇔ xθy.

Now observe that ∼1/θ= θ. Indeed, x ∼1/θ y ⇔ x ∗ y, y ∗ x ∈ 1/θ ⇔ (x ∗ yθ1 and y∗ xθ1) ⇔ xθy. We have

f◦ g(θ) = f(1/θ) = ∼^1/θ= θ for all θ ∈ ConX. (4) From (3) and (4) it follows that f and g are inverse bijections between FilX and

ConX. _

References

[1] S.S. Ahn and K.S. So, On ideals and upper sets in BE-algebras, Scientiae Mathematicae Japonicae68 (2008), 279-285.

[2] K.H. Kim, A note on BE-algebras, Scientiae Mathematicae Japonicae72 (2010), 127-132.

[3] H.S. Kim and Y.H. Kim, On BE-algebras, Scientiae Mathematicae Japonicae66 (2007), 113- 116.

[4] H.S. Kim and K.J. Lee, Extended upper sets in BE-algebras, Bull. Malays. Math. Sci. Soc.34 (2011), 511-520.

[5] B.L. Meng, On filters in BE-algebras, Scientiae Mathematicae Japonicae71 (2010), 201-207.

[6] A. Walendziak, On commutative BE-algebras, Scientiae Mathematicae Japonicae69 (2009), 281-284.

[7] Y.H. Yon, S.M. Lee, and K.H. Kim, On congruences and BE-relations in BE-algebras, Inter- national Mathematical Forum5 (2010), 2263-2270.

Andrzej Walendziak

Institute of Mathematics and Physics, Siedlce University 3 Maja 54, 08–110, Siedlce, Poland

E-mail: walent@interia.pl

(Received: 28.11.2011)