Abstract. In this paper, we introduce the concept of multiplier of BE-algebras ... if x
= b. Then it is easy to check that f is a multiplier of BE-algebra X. Example 3.3.
International Mathematical Forum, Vol. 6, 2011, no. 17, 815 - 820
Multipliers in BE-Algebras Kyung Ho Kim Department of Mathematics, Chungju National University Chungju 380-702, Korea
[email protected] Abstract In this paper, we introduce the concept of multiplier of BE-algebras and obtained some properties of BE-algebras. Also, we introduce the simple multiplier and characterized the kernel of multipliers of BEalgebras.
Mathematics Subject Classification: 06F35, 03G25, 08A30 Keywords: BE-algebra, filter, multiplier, simple multiplier, self-distributive, commutative, kernel
1
Introduction
In [2] a partial multiplier on a commutative semigroup (A, ·) has been introduced as a function F from a nonvoid subset DF of A into A such that F (x) · y = x · F (y) for all x, y ∈ DF . In this paper, we introduce the concept of multiplier of BE-algebras and obtained some properties of BE-algebras. Also, we introduce the simple multiplier and characterized the kernel of multipliers of BE-algebras.
2
Preliminaries
In what follows, let X denote an BE-algebra unless otherwise specified. By an BE-algebra we mean an algebra (X; ∗, 1) of type (2, 0) with a single binary operation “∗” that satisfies the following identities: for any x, y, z ∈ X, (BE1) x ∗ x = 1 for all x ∈ X, (BE2) x ∗ 1 = 1 for all x ∈ X, (BE3) 1 ∗ x = x for all x ∈ X,
Kyung Ho Kim
816 (BE4) x ∗ (y ∗ z) = y ∗ (x ∗ z) for all x, y, z ∈ X.
We introduce a relation “≤” on X by x ≤ y imply x ∗ y = 1. An BE-algebra (X, ∗, 1) is said to be self-distributive if x ∗ (y ∗ z) = (x ∗ y) ∗ (x ∗ z) for all x, y, z ∈ X. A non-empty subset S of an BE-algebra X is said to be a subalgebra of X if x ∗ y ∈ S whenever x, y ∈ S. In an BE-algebra, the following identities are true: (p1) x ∗ (y ∗ x) = 1. (p2) x ∗ ((x ∗ y) ∗ y)) = 1. Let (X, ∗, 1) be an BE-algebra and F a non-empty subset of X. Then F is said to be a filter of X if (F1) 1 ∈ F, (F2) If x ∈ F and x ∗ y ∈ F, then y ∈ F.
3
Multipliers in BE-algebras
Definition 3.1. Let (X, ∗, 1) be an BE-algebra. A self-map f is called a multiplier of X if f (x ∗ y) = x ∗ f (y) for all x, y ∈ X. Example 3.2. Let X = {1, a, b, c} in which “∗” is defined by ∗ 1 a b c
1 a 1 a 1 1 1 1 1 1
b b a 1 a
c c a a 1
It is easy to check that (X, ∗, 1) is an BE-algebra. Define a map f : X → X by 1 f (x) = a
if x = 1, a, c if x = b
Then it is easy to check that f is a multiplier of BE-algebra X. Example 3.3. Let X = {1, a, b} in which “∗” is defined by
Multipliers in BE-algebras
817 ∗ 1 a b
1 1 1 1
a a 1 1
b b 1 1
Then X is an BE-algebra. Define a map f : X → X by 1 if x = 1, b f (x) = b if x = a Then it is easy to check that f is a multiplier of BE-algebra X. Example 3.4. The identity mapping , the unit mapping ι : a −→ 1 and for p ∈ X, the mappings αp (a) = p ∗ a are multipliers Lemma 3.5. Let f be a multiplier in BE-algebra X. Then we have f (1) = 1. Proof. Substituting f (1) for x and 1 for y in Definition 3.1, we obtain f (1) = f (f (1) ∗ 1) = f (1) ∗ f (1) = 1. Proposition 3.6. Let f be a multiplier in BE-algebra. Then x ≤ f (x) for all x ∈ X, Proof. Putting x = y in Definition 3.1, we get 1 = f (1) = f (x ∗ x) = x ∗ f (x), that is, x ≤ f (x). Let X be an BE-algebra and f1 , f2 two self maps. We define f1 ◦f2 : X → X by (f1 ◦ f2 )(x) = f1 (f2 (x)) for all x ∈ X. Proposition 3.7. Let X be an BE-algebra and f1 , f2 two multipliers of X. Then f1 ◦ f2 is also a multiplier of X. Proof. Let X be an BE-algebra and f1 , f2 two multipliers of X. Then we have (f1 ◦ f2 )(a ∗ b) = f1 (f2 (a ∗ b)) = f1 (a ∗ f2 (b)) = a ∗ f1 (f2 (b)) = a ∗ (f1 ◦ f2 )(b). This completes the proof.
Kyung Ho Kim
818 We define x y by
x y = (y ∗ x) ∗ x
for all x, y ∈ X. Let X be an BE-algebra and f1 , f2 two self maps. We define f1 f2 : X → X by (f1 f2 )(x) = f1 (x) f2 (x) for all x ∈ X. Proposition 3.8. Let X be an BE-algebra and f1 , f2 two multipliers of X. Then f1 f2 is also a multiplier of X. Proof. Let X be an BE-algebra and f1 , f2 two multipliers of X. Then we have (f1 f2 )(a ∗ b) = f1 (a ∗ b) f2 (a ∗ b) = a ∗ f1 (b) a ∗ f2 (b) = ((a ∗ f2 (b)) ∗ (a ∗ f1 (b))) ∗ (a ∗ f1 (b)) = (a ∗ (f2 (b) ∗ f1 (b))) ∗ (a ∗ f1 (b)) = a ∗ ((f2 (b) ∗ f1 (b)) ∗ f1 (b)) = a ∗ (f1 (b) f2 (b)) = a ∗ (f1 f2 )(b). This completes the proof. Proposition 3.9. Let f be a multiplier in BE-algebra and Ff = {x ∈ X | f (x) = x}. Then Ff is a subalgebra of X. Proof. Let x, y ∈ Ff . Then we have f (x) = x and f (y) = y. Hence f (x ∗ y) = x ∗ f (y) = x ∗ y, and so x ∗ y ∈ Ff . This proves that Ff is a subalgebra of X. Proposition 3.10. Let X be an BE-algebra and f a multiplier. If x ∈ F, then x y ∈ F. Proof. Let x ∈ F. Then we have f (x) = x, and so f (x y) = f (y ∗ x) ∗ x) = (y ∗ x) ∗ f (x) = (y ∗ x) ∗ x = x y. This completes the proof. An BE-algebra X is said to be commutative if for all x, y ∈ X, (y ∗ x) ∗ x = (x ∗ y) ∗ y, i.e., x y = y x.
Multipliers in BE-algebras
819
Proposition 3.11. Let X be a commutative BE-algebra and f a multiplier. If x ∈ F and x ≤ y, then y ∈ F. Proof. Let X be a commutative BE-algebra. If x ∈ F and x ≤ y, then we have f (y) = f (1 ∗ y) = f ((x ∗ y) ∗ y) = f ((y ∗ x) ∗ x) = (y ∗ x) ∗ f (x) = (y ∗ x) ∗ x = (x ∗ y) ∗ y = 1 ∗ y = y. This completes the proof. Denote by M(X) the set of all multipliers of X. Definition 3.12. For any ϕ ∈ M(X), we define the kernel of ϕ as follows: Kϕ := {x ∈ X | ϕ(x) = 1}. Proposition 3.13. Let ϕ be a multiplier of X. Then Kϕ is a subalgebra of X. Proof. Clearly, 1 ∈ Kϕ and so Kϕ is nonempty. For any x, y ∈ Kϕ , we have ϕ(x ∗ y) = x ∗ ϕ(y) = x ∗ 1 = 1, and so x ∗ y ∈ Kϕ . Hence Kϕ is a subalgebra of X. Theorem 3.14. Let ϕ be a multiplier of X. If x ∈ Kϕ and x ≤ y, then y ∈ Kϕ . Proof. Let x ∈ Kϕ and x ≤ y. Then ϕ(x) = 1. By Proposition 3.7, we have 1 = ϕ(x) ≤ ϕ(y), and so 1 = ϕ(y), that is, y ∈ Kϕ . This completes the proof. Theorem 3.15. Let ϕ be a multiplier and a homomorphism of X. Then Kϕ is a filter of X. Proof. Clearly, 1 ∈ Kϕ since ϕ(1) = 1. Let x ∈ Kϕ and x ∗ y ∈ Kϕ . Then 1 = ϕ(x ∗ y) = ϕ(x) ∗ ϕ(y) = 1 ∗ ϕ(y) = ϕ(y), and so y ∈ Kϕ . This completes the proof. We call the multiplier αp (a) = p ∗ a of Example 3.3 as simple multiplier. Proposition 3.16. For every p ∈ X, the simple self-distributive multiplier αp is an endomorphism of X. Proof. Let a, b ∈ X. Then we have αp (a ∗ b) = p ∗ (a ∗ b) = (p ∗ a) ∗ (p ∗ b) = αp (a) ∗ αp (b). Hence αp is an endomorphism of X..
Kyung Ho Kim
820
Proposition 3.17. The simple multiplier α1 is an identity function of X. Proof. For every a ∈ X, α1 (a) = 1 ∗ a = a. This completes the proof. Proposition 3.18. Let X be an BE-algebra. Then, for each p ∈ X, we have αp (x p) = 1. Proof. For each p ∈ X, we have αp (x p) = αp ((p ∗ x) ∗ x) = p ∗ ((p ∗ x) ∗ x) = (p ∗ x) ∗ (p ∗ x) = 1. This completes the proof. Acknowledgements. The research on which this paper is based was supported by a grant from Chungju National University Research Foundation, 2011.
References [1] H. S. Kim and Y. H. Kim, On BE-algebras, Sci. Math. Japo. 66 (1) (2007), 113-116. [2] R. Larsen, An Introduction to the Theory of Multipliers, Berlin: Springer-Verlag, 1971. [3] S. S. Ahn and K. S. So, On ideals and uppers in BE-algebras, Sci. Math. Japo. Online e-2008, 351-357. Received: October, 2010
