A Structure of KK-Algebras and its Properties

Int. Journal of Math. Analysis, Vol. 6, 2012, no. 21, 1035-1044 A Structure of KK-Algebras and its Properties S. Asawasamrit Department of Mathematics, Faculty of Applied Science, King Mongkut s University of Technology North Bangkok, Bangkok 10800, Thailand Centre of Excellence in Mathematics, CHE, Sri Ayutthaya Road, Bangkok 10400, Thailand suphawata@kmutnb.ac.th A. Sudprasert University of the Thai Chamber of Commerce, Bangkok, Thailand aisuriya sud@utcc.ac.th Abstract In this paper, we define the notions of KK-algebras, quotient KKalgebras and investigate its properties. Moreover we show the relation between ideals and congruences on KK-algebras. Mathematics Subject Classification: 06B35, 03G10, 06B99 Keywords: KK-algebra, ideal, quotient KK-algebra, congruences 1 Introduction and Preliminaries By an algebra X =(X,, 0), we mean a non-empty set X together with a binary operation and a some distinguished 0. H. Yisheng [6] studied an algebraic structure called a BCI-algebra which is an algebra (X,, 0) with a binary operation such that for all x, y, z X, satisfies the following properties: (B-1) ((x y) (x z)) (z y) =0; (B-2) x x =0; (B-3) x y = 0 and y x = 0 imply x = y, for any x, y, z X. In this paper is introduction to the general theory of KK-algebra. We will first give the notion of KK-algebra, quotient KK-algebra and investigate elementary and fundamental properties, and we will show relation between ideals and congruences.

1036 S. Asawasamrit and A. Sudprasert 2 KK-algebras In this section, we do define some familiar concepts as KK-algebras, both for illustration and for review of the concept. First we give a few definitions and some notation. Definition 2.1. An algebra (X;, 0) with a binary operation and a nullary operation 0. Then X is called KK-algebra if it satisfies for all x, y, z X : (KK-1) (x y) ((y z) (x z))=0; (KK-2) 0 x = x; (KK-3) x y =0and y x =0if and only if x = y. First, give example of KK-algebra. Examples 2.2. Let be defined on an abelian group G by letting x y = x 1 y, where x, y in G, with e is unity element of G. Then (G;,e) is a KK-algebra. Examples 2.3. Let X = {0, 1} and let be defined by Then (G;, 0) is a KK-algebra. * 0 1 0 0 1 1 1 0 Theorem 2.3. Let (X;, 0) be a KK-algebra if and only if it satisfies the following conditions: for all x, y, z X, (1) (x y) ((y z) (x z))=0; (2) x ((x y) y) =0; (3) x x =0; (4) x y = 0 and y x = 0 if and only if x = y. Proof. Assume that (X;, 0) is a KK-algebra. From definition of KKalgebra, (1) and (4) holds. Then we see that x ((x y) y) =(0 x) ((x y) (0 y)) = 0, and x x =0 (x x) =(0 0) ((0 x) (0 x)) = 0, so (2) and (3) holds. Conversely, we need to show KK-2. By (1), (2) and (3), we see that ((0 x) x) 0 = ((0 x) x) (0 ((0 x) x)) = ((0 x) x) ((x x) ((0 x) x)) = 0. And since 0 ((0 x) x) =0. From (4), it follows that (0 x) x = 0 and x (0 x) =x ((x x) x) =0. Therefore 0 x = x, proving our theorem. Definition 2.4. Define a binary relation on KK-algebra X by letting x y if and only if y x =0.

A structure of KK-algebras and its properties 1037 Proposition 2.5. If (X;, 0) is a KK-algebra, then (X; ) is a partially order set. Proposition 2.6. If (X;, 0) be a KK-algebra and x 0, then x =0, for any x X. Moreover, 0 is called a minimal element in X. Proof. Let x 0, then 0 x =0. By KK-2, 0 x = x, and thus x =0. It is easy to show that the following properties are true for a KK-algebra. Theorem 2.7. Let (X;, 0) be a KK-algebra if and only if it satisfies the following conditions: for all x, y, z X, (1) ((y z) (x z)) (x y); (2) ((x y) y) x; (3) x y if and only if y x =0. Proposition 2.8. Let x, y, z be any element in a KK-algebra X. Then (1) x y implies y z x z. (2) x y implies z x z y. Proposition 2.9. Let x, y, z be any element in a KK-algebra X. Then x (y z) =y (x z). Proof. Since theorem 2.7(2), (x z) z x, and by proposition 2.8(1), we get that x (y z) ((x z) z) (y z). Putting x = y and y = x z in theorem 2.7(1), it follows that ((x z) z) (y z) y (x z). By the transitivity of gives x (y z) y (x z). And we replacing x by y and y by x, we obtain y (x z) x (y z). By the anti-symmetry of, thus x (y z) =y (x z) and finishing the proof. Corollary 2.10. Let x, y, z be any element in a KK-algebra X. Then (1) y z x if and only if x z y. (2) (z x) (z y) x y. (3) x y implies x z y z. Proposition 2.11. Let x, y, z be any element in a KK-algebra X. Then (1) ((x y) y) y = x y. (2) (x y) 0=(x 0) (y 0). Proof. (1) From theorem 2.3(2) and theorem 2.7(1), (((x y) y) y) (x y) x ((x y) y) =0. Thus (((x y) y) y) (x y) =0. Since (x y) (((x y) y) y) =((x y) y) ((x y) y) =0. So, by KK-3, (x y) y = x y. (2) Since (x 0) (y 0) = (x 0) (y ((x y) (x y))) = (x 0) ((x y) (y (x y))) = (x 0) ((x y) (x (y y))) = (x y) ((x 0) (x 0)) = (x y) 0. The proof is complete. In this paper we will denote N for the set of all nonnegative integers, i.e.,

1038 S. Asawasamrit and A. Sudprasert 0, 1, 2,..., and N for the set of all natural numbers, i.e., 1, 2, 3,..., and we will also use the following notation in brevity: y 0 x = x, y n x = y (... (y (y x))), }{{} ntimes where x, y are any elements in a KK-algebra and n N. Proposition 2.12. Let x, y be any element in a KK-algebra X. Then (1) ((y x) x) n x = y n x for any n N. (2) (x n 0) 0=(x 0) n 0 for any n N. Proof. Let X be a KK-algebra and x, y X and n, m N. (1) Proceed by induction on n and defined the statement P (n), ((y x) x) n x = y n x. We see that P (0) is true, ((y x) x) 0 x = x = y 0 x. Assume that P (k) is true for some arbitrary k 0, that is ((y x) x) k x = y k x. Since ((y x) x) k+1 x =((y x) x) (((y x) x) k x) =((y x) x) (y k x) = y k (((y x) x) x) =y k (y x) =y k+1 x. This show that P (k + 1) is true and by the principle of mathematical induction, P (n) is true for each n N. (2) Since (x n 0) 0=(x (x n 1 0)) 0=(x 0) ((x n 1 0) 0) = (x 0) ((x (x n 2 0)) 0) = (x 0) ((x 0) ((x n 2 0) 0)) = (x 0) 2 ((x n 2 0) 0) =... =(x 0) n 0 Given x X if it satisfies x 0=0, that is 0 x, the element x is called a positive element of X. By definition, the zero element 0 of X is positive. Proposition 2.12. Let x be any element in a KK-algebra X. Then ((x 0) 0) x is a positive element of X for every x X. Proof. Since (((x 0) 0) x) 0 = (((x 0) 0) 0)) (x 0) = (x 0) (x 0) = 0. Therefore ((x 0) 0) x is a positive element of X. 3 Ideals Definition 3.1. A non-empty subset A of a KK-algebra X is called a closed of X on condition that x y A whenever x, y A. Definition 3.2. A non-empty subset A of a KK- algebra X is called an ideal of X if it satisfies the following conditions: (I-1) 0 A (I-2) for any x, y X, x y A and x A imply y A. Examples 3.3. Let X = {0, 1, 2, 3} and let be defined by the table

A structure of KK-algebras and its properties 1039 * 0 1 2 3 0 0 1 2 3 1 0 0 3 3 2 3 3 0 0 3 3 2 1 0 Thus, it can be easily shown that X is a KK- algebra. I = {0, 1} and J = {0, 3} are closed ideals of X. And we see that Lemma 3.4. Let A be a closed of KK-algebra X. Then A is an ideal of X if and only if x A and z y A imply z (x y) A for all x, y, z X. Proof. Let A be an ideal of X and let x A whereas z y A. Suppose that z (x y) A. By proposition 2.9, we see that x (z y) A. Since A is an ideal of X and x A, z y A, a contradiction. So z (x y) A. Conversely, assume that if x A and z y A imply z (x y) A for all x, y, z X. Since A is a closed of X, then there is x A which 0 = x x A. That is, 0 A. Now, let x y A and x A. Assume that y A. We have that 0 y = y A. It follows that 0 (x y) A. Hence x y A, contradiction. Therefore A is an ideal of X. This completes the proof. Corollary 3.5. Let A be a closed of KK-algebra X. Then A is an ideal of X if and only if x A and y A imply x y A for all x, y X. Lemma 3.6. Let A be a closed of KK-algebra X. Then A is an ideal of X if and only if x (y z) A and x z A imply y A for all x, y, z X. Proof. Let A be an ideal of X and let x (y z) A, x z A. Suppose that y A. By proposition 2.9, we have y (x z) A. Since A is an ideal of X, thus x z A, contradiction, this shows that y A. Conversely, assume that x (y z) A and x z A imply y A for all x, y, z X. Since A is a closed of X, then there is y A which 0 = y y A. Then 0 A. Let y z A, y A and suppose that z A. By KK-2, 0 (y z) A and 0 z A. By assumption, so y A, a contradiction. This proves that A is an ideal of X. Corollary 3.7. Let A be a closed of KK-algebra X. Then A is an ideal of X if and only if x y A and y A imply x A for all x, y, z X. This Lemma gives some properties of ideal of KK-algebra. Lemma 3.8. If A is an ideal of KK-algebra X and B is an ideal of A, then B is an ideal of X. Proof. Since B is an ideal of A, then 0 B. Let x, y X such that x y B and x B. It follows that that x y A and x A. By assumption, A is an ideal of X, so y A and x B. From B is an ideal of A, so y B. Therefore, B is an ideal of X.

1040 S. Asawasamrit and A. Sudprasert Theorem 3.9. Let {J i : i N} be a family of ideals of a KK-algebra X where J n J n+1 for all n N. Then J n is an ideal of X. Proof. Let {J i : i N} be a family of ideals of X. It can be proved easily that J n X. Since J i is an ideal of X for all i, so 0 J n. Let x y J n and x J n. It follows that x y J j for some j N and x J k for some k N. Furthermore, let J j J k. Hence x y J k and x J k. By assumption, J k is an ideal of X, it follows that y J k. Therefore, y J n, proving that J n is an ideal of X. Theorem 3.10. Let {J i : i N} be a family of closed ideals of a KK-algebra X where J n J n+1 for all n N. Then J n is a closed ideal of X. Proof. Let {J i : i N} be a family of closed ideals of X. By theorem 3.9, J n is an ideal of X. We will show that J n is a closed of X. Let x, y J n. It follows that x J j for some j N and y J k for some k N. WLOG, we assume that j k, we obtain J j J k. That is, x J k and x J k. Since J k is a closed of X, we get x y J k J n. This proves that is a closed ideal of X. J n Theorem 3.11. Let {I j : j J} be a family of ideals of a KK-algebra X. Then I j is an ideal of X. Proof. Let {I j : j J} be a family of ideals of X. It is obvious that I j X. Since 0 I j for all j J, it follows that 0 I j. Let x y I j and x I j. We get that x y I j and x I j for all j J, then y I j for all j J. Because I j is an ideal of X. So y I j, proving our theorem. Theorem 3.12. Let {I j : j J} be a family of closed ideals of a KK-algebra X. Then I j is a closed ideal of X. Proof. Let {I j : j J} be a family of closed ideals of X. By theorem 3.11, I j is an ideal of X. We will show that I j is a closed of X. Let x, y I j. It follows that x, y I j for all j J. Since I j is a closed of X and x y I j for all j J, then x y I j. This show that I j is a closed ideal of X.

A structure of KK-algebras and its properties 1041 4 Quotient KK-Algebras In this section, we describe congruence on KK-algebras. Definition 4.1. Let I be an ideal of a KK-algebra X. Define a relation on X by x y iff x y I and y x I. Theorem 4.2. If I is an ideal of KK-algebra X, then the relation is an equivalence relation on X. Proof. Let I be an ideal of X and x, y, z X. By Theorem 2.3, x x =0 and assumption, x x I. That is, x x. Hence is reflexive. Next, suppose that x y. It follows that x y I and y x I. Then y x, so is symmetric. Finally, let x y and y z. Then x y, y x, y z, z y I and (y x) ((z y) (z x)) = 0 I. It follows that (z y) (z x) I, and since z y I, so z x I. Similarly, x z I. Thus is transitive. Therefore is an equivalence relation. Lemma 4.3. Let I be an ideal of KK-algebra X. For any x, y, u, v X, if u v and x y, then u x v y. Proof. Assume that u v and x y, for any x, y, u, v X, then u v, v u, x y, y x I and by K-1, we see that (u v) ((v x) (u x)) = 0 and (v u) ((u x) (v x)) = 0. From assumption and I is an ideal of X, these imply that (v x) (u x) I and (u x) (v x) I. This shows that v x u x. On the other hand, by corollary 2.10, we have that (y x) ((v y) (v x)) = 0 and (x y) ((v x) (v y)) = 0. From assumption and I is an ideal of X, these imply that (v y) (v x) I and (v x) (v y) I. Thus v x v y. Since is symmetric and transitive, so u x v y. Corollary 4.4. If I is an ideal of KK-algebra X, then the relation is a congruence relation on X. Proof. By theorem 4.2 and lemma 4.3. Definition 4.5. Let I be an ideal of a KK-algebra X. Given x X, the equivalence class [x] I of x is defined as the set of all element of X that are equivalent to x, that is, [x] I = {y X : x y}. We define the set X/I = {[x] I : x X} and a binary operation on X/I by [x] I [y] I =[x y] I. Theorem 4.6. If I is an ideal of KK-algebra X with X/I = {[x] I : x X} where a binary operation on a set X/I is defined by [x] I [y] I =[x y] I, then the binary operation is a mapping from X/I X/I to X/I. Proof. Let [x 1 ] I, [x 2 ] I, [y 1 ] I, [y 2 ] I X/I such that [x 1 ] I =[x 2 ] I and [y 1 ] I =

1042 S. Asawasamrit and A. Sudprasert [y 2 ] I. It follows that x 1 x 2 and y 1 y 2. By lemma 4.3, x 1 y 1 x 2 y 2, proving that [x 1 y 1 ] I =[x 2 y 2 ] I. Theorem 4.7. If I is an ideal of KK-algebra X, then (X/I;, [0] I ) is a KKalgebra. Moreover, the set X/I is called the quotient KK-algebra. Proof. Let [x] I, [y] I, [z] I X/I. Then ([z] I [x] I ) (([x] I [y] I ) ([z] I [y] I )) = [z x] I ([x y] I [z y] I )=[z x] I [(x y) (z y)] I =[(z x) ((x y) (z y))] I = [0] I. It is clear that [0] I [x] I =[0 x] I =[x] I. Now, let [x] I [y] I = [0] I and [y] I [x] I = [0] I. It follows that x y 0 and y x 0, that is 0 (x y), 0 (y x) I. Since I is an ideal of X and 0 I, we get that x y, y x I. Consequently, x y, proving that [x] I =[y] I. Therefore, (X/I;, [0] I ) is a KK-algebra. Example 4.8. According to example 3.3, we can get that X/I = {[0] I, [2] I }, where [0] I = [1] I = {0, 1} and [2] I = [3] I = {2, 3}. Let be defined on X/I by Then (X/I;, [0] I ) is a KK-algebra. [0] I [2] I [0] I [0] I [2] I [2] I [2] I [0] I Lemma 4.9. Let X be a KK-algebra and I,J be any sets such that I J X. Suppose that I is an ideal of X, then J is an ideal of X if and only if J/I is an ideal of X/I. Proof. Let I be an ideal of X with I J X. Suppose firstly that J is an ideal of X, then J/I = {[x] I : x J}, where [x] I = {y J : x y}, and X/I = {[x] I : x X}, where [x] I = {y X : x y}. Obviously, J/I X/I and [0] I J/I. Now, let [x] I [y] I J/I and [y] I J/I. Then [x y] I =[x] I [y] I J/I, it follows that x y J and x J. By assumption, y J. Accordingly, [y] I J/I, this shows that J/I is an ideal of X/I. On the other hand, suppose that J/I is an ideal of X/I and I is an ideal of X with I J X. Thus, 0 J. Let x y J and x J. It follows that [x y] I, [x] I J/I. Since [x y] I =[x] I [y] I, so [x] I [y] I J/I. By hypothesis, [y] I J/I implies y J, proving our lemma. Lemma 4.10. Let X be a KK-algebra and I,J be any sets such that I J X. Suppose that I is a closed ideal of X. Then J is a closed ideal of X if and only if J/I is a closed ideal of X/I. Proof. Similar to that of lemma 4.9 Lemma 4.11. Let I,J be ideals of a KK-algebra X with I J, then I is an ideal of X. Proof. Obvious.

A structure of KK-algebras and its properties 1043 Next, the basic properties of equivalence classes are considered are as the following Theorem. Theorem 4.12. Let I be a closed ideal of a KK-algebra X and a, b X. Then (1) [a] I = I iff a I. (2) [a] I =[b] I or [a] I [b] I =. Proof. Let I be a closed ideal of X and a, b X. (1) It is clear due to the fact that a a for all a X and a a =0 I, so we get that a [a] I = I. Conversely, let x [a] I. Then x a, it follows that x a, a x I. By hypothesis, x I. Hence, [a] I I. To show that I [a] I, choose x I. Since I is a closed of X, we have x a, a x I. Thus, x a, this means that x [a] I and shows that I [a] I. Consequently, [a] I = I. (2) Assume that [a] I [b] I. Then there is x [a] I [b] I such that x [a] I and y [a] I. It follows that x a and x b, so a b by the symmetric and transitive properties. Thus [a] I =[b] I. Theorem 4.13. If I is a closed ideal of a KK- algebra X and y I, then [y] I is closed ideal of X. Proof. Let I be a closed ideal of X and y I. It is clear that 0 [y] I. Now, suppose that a b [y] I and a [y] I. We will show that b [y] I. Then a b y and a y, it follows that y (a b) I and y a I. By assumption, a I. From proposition 2.9, a (y b) =y (a b) I, and I is a closed ideal of X and a I, therefore, y b I. By properties of X, we get that (a 0) (((a b) y) (b y)) = (a (b b)) (((a b) y) (b y)) = (b (a b)) (((a b) y) (b y)) = 0. By hypothesis, (a 0) (((a b) y) (b y)) I, and I is closed and a I, then a 0 I. Thus ((a b) y) (b y) I. From a b y and I is closed, then b y I. Hence, b y, this means a [y] I. Accordingly, [y] I is an ideal of X. Finally, let a, b [y] I. Then a y and b y, by lemma 4.3, a b y y. By theorem 2.3, it follows that a b 0. Thus a b [0] I. Now, we have 0,y I and I is closed, so 0 y I and y 0 I. That is, 0 y. Hence, [0] I =[y] I. By transitive, a b [y] I, proving our theorem. ACKNOWLEDGEMENTS. This paper is supported by Faculty of Applied Science, King Mongkut s University of Technology North Bangkok, Thailand, Grant No.5542109. References [1] S. Asawasamrit and U. Leerawat, On quotient binary algebras, Scientia Magna Journal, 6(2010), No.1, 82-88.

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