International Journal of Scientific and Research Publications, Volume 6, Issue 10, October 2016 ISSN f -DERIVATIONS ON BP-ALGEBRAS

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1 ISSN f -DERIVATIONS ON BP-ALGEBRAS N.Kandaraj* and A.Arul Devi**, *Associate Professor in Mathematics **Assistant professor in Mathematics SAIVA BHANU KSHATRIYA COLLEGE ARUPPUKOTTAI TAMILNADU. INDIA. ABSTRACT. Motivated by some results on derivations on rings, and the generalizations of BCK and BCI algebras, in this paper, we define f -derivations on BP-algebras and investigate some important results. Key Words: BP-algebras, derivations on BP-algebras, f -derivations on BP-algebras. Subject Classification: AMS(2000) 06F35, 03G25, 06D99, 03B47 1. Introduction BCK and BCI algebras are two new classes of algebras based on propositional calcului or logic introduced by Imai and Isaki[5].In[6] K.Isaki and K.Tanaka introducd the theory of BCK-algebras. In[3,4] Q.P.Hu and X.Li and introduced a wider class of abstract algebras BCH-algebras. The class of BCI-algebras is a proper subclass of the class BCHalgebras. J.Neggers and H.S.Kim [9] introduced the notion of d-algebras which is another generalization of BCK-algebras. In S.S.Ahn and J.S.Han [1] introduce the notion of a BP-algebras. In 2004 Y.B.Jun and X.L..Xin [7] inroduced the notion of derivations of BCI-algebras, which was motivated from a lot of workdone on derivations of rings. Since then many authors worked on the notion of derivations on several algebras such as d-algebras and TM-algebras [2,8] motivated by this paper introduce the notion of f-derivations on BP-algebras. 2. Preliminaries In this section we recall some basic definitions that are required in our work. Definition 2.1. [6] Let X be a set with a binary operation and a constant 0.Then ( X,,0) is called a BCK -algebras if it satisfies the following axioms: (1) x x =0 (2) 0 x = 0 (3) (( x y ) (x z) ) (z y) =0 (4) (x (x y)) y =0 (5) x y =0 and y x =0 imply x = y x, y, z X Definition 2.2. [7] Let X be a set with a binary operation and a constant 0.Then ( X,,0) is called a BCI-algebra if it satisfies the following axioms:

2 ISSN (1) ((x y) (x z)) (z y) =0 (2) (x (x y)) y =0 (3) x x =0 (4) x y =0 and y x =0 x = y x, y, z X Definition 2.3. Let x be a BCI-algebra. Two elements x and y in X are said to be comparable if x y or y x. Here x y if and only if x y =0.Also we define y ( y x ) by x y. Definition 2.4. [9] A d -algebra is a non-empty set X with a constant 0 and binary operation satisfying the following axioms: (1) x x =0 (2) 0 x = 0 (3) x y =0 and y x =0 x = y. Definition 2.5. [1] Let X be a set with a binary operation and a constant 0. Then (X,, 0) is called a BP-algebra if it satisfies the following axioms. (1) x x = 0 (2) x (x y) = y (3) (x z) (y z) = x y for any x, y, z X. Definition 2.6. [9] Let X be a d -algebra and I be a subset of X,then I is called d -ideal of X if it satisfies the following conditions. (1) 0 I (2) x y I and y I = x I (3) x I and y X = x y I (ie) I X I Example 2.7. Let X = {0, 1, 2, 3},( X,,0) be a set with the following cayley table Then (X,, 0) is a BP-algebra Definition 2.8. Let X be a d - algebra. A map θ :X X is a left - right derivation (briefly, (l, r) - derivation ) on X, if it satisfies the identity θ ( x y ) = ( θ(x) y) (x θ(y)) for all x, y X. If θ satisfies the identity θ ( x y ) = (x θ(y)) (θ(x) y)) for all x, y X,then θ is called a right-left derivation (briefly, (r, l) -derivation) on X. If θ is both an (l, r) and an (r, l) -derivation, then θ is called a derivation on X. 3. f-derivations ON BP-ALGEBRA In this section, we define the notion of f-derivations and regular of f-derivations on BP-algebras and prove some results. Throughout this section we assume that f is an endomorphism of the BP-algebra ( X,,0)

3 ISSN Definition 3.1. Let X be a BP-algebra. By a left - right f - derivation (briefly, (l, r) - f -derivation ) on X, we mean a self map θ f of X satisfies the identity θ f ( x y ) = ( θ f (x) f(y)) (f(x)) θ f (y) ) for all x, y X. If θ f satisfies the identity θ f ( x y ) =( f(x) θ f (y)) (θ f (x) f(y) ) for all x, y X, then it is said that θ f is a right - left f - derivation (briefly, (r, l) - f -derivation) of X. If θ f is both an (r, l) - and an (l, r) - f -derivation, then θ f is said to be a f - derivation. Example 3.2. Let X = {0, 1, 2, 3} be a BP-algebra with the following cayley table (1) Define an endomorphism f of X by f(0) =0, f(1) =3, f(2) =2, f(3) =1. and a self map θ f : X X by θ f (0)=1, θ f (1)=0, θ f (2)=3, θ f (3)=2. Then it is easily checked that θ f is a (l, r) - f -derivation of X. (2) Define an endomorphism f of X by f(0) =0, f(1) =3, f(2) =2, f(3) =1. and a self map θ f : X X by θ f (0)=2, θ f (1)=1, θ f (2)=0, θ f (3)=3. Then it is easily checked that θ f is a f -derivation of X. Definition 3.3. θ f (0) = 0. An f -derivation θ f on a BP-algebra X is said to be regular if Proposition 3.4. Every (r, l) - f -derivation ((l, r) - f -derivation) of a BP-algebra is regular. Let X be a BP-algebra and θ f be a (r, l) - f -dreivation on X.Then for all x X, we have θ f (0) = θ f (x x) = (f(x) θ f (x)) (θ f (x) f(x)) = (θ f (x) f(x)) ((θ f (x) f(x)) (f(x) θ f (x) = f(x) θ f (x) ( x x = 0) = 0. Let θ f be a (l, r) f -derivation on X. Then for all x X.we have θ f (0) = θ f (x x) = (θ f (x) f(x)) (f(x) θ f (x)) = (f(x) θ f (x)) ((f(x) θ f (x)) (θ f (x) f(x))) = θ f (x) f(x) = 0. 3

4 ISSN One can easily prove that the following result gives a necessary and sufficient condition for the derivation θ f to be regular. Proposition 3.5. Let θ f be a self map of a BP-algebra on X, then the following hold: (1) If θ f is an (l, r) f -derivation on X, then θ f (x) = θ f (x) f(x) for all x X if and only if θ f (0)=0. (2) If θ f is an (r, l) f -derivation on X, then θ f (x) = f(x) θ f (x) for all x X if and only if θ f (0)=0. Proposition 3.6. Let θ f be a (l, r) f -derivation on a BP-algebra X. Then θ f (a) = θ f (0) (0 f(a) ), for all a X. proof: Now, Let θ f be an (l, r) f -derivation on a BP-algebra X. θ f (a) = θ f (0) (0 f(a)). θ f (a) = θ f (0 (0 a)) ( 0 (0 x) = x) = (θ f (0) f(0 a)) (f(0) θ f (0 a)) = (f(0) θ f (0 a)) ((f(0) θ f (0 a)) (θ f (0) f(0 a))) = θ f (0) f(0 a) = θ f (0) (f(0) f(a)) = θ f (0) (0 f(a)) Proposition 3.7. Let θ f be a self map on a BP-algebra X and θ f be an (r, l) f - derivation on X.Then θ f (x) = f(x),for all x X if and only if θ f (0)=0. Let θ f be an (r, l) f -derivation on X. Assume that θ f (0)=0. Now, θ f (x) = θ f (x 0) ( x 0 = x) = (f(x) θ f (0)) (θ f (x) f(0)) = (θ f (x) f(0)) ((θ f (x) f(0)) (f(x) θ f (0))) = f(x) θ f (0) = f(x).

5 ISSN Coversely, assume that θ f (x) = f(x). Now, Definition 3.8. f ( A ) A. θ f (0) = 0. θ f (0) = θ f (x x) = (f(x) θ f (x)) (θ f (x) f(x)) = (θ f (x) f(x)) ((θ f (x) f(x)) (f(x) θ f (x))) = f(x) θ f (x) = f(x) f(x) ( θ f (x) = f(x)) = 0. An ideal A on a BP-algebra X is said to be an f -ideal if Example 3.9. Let X = {0, 1, 2, 3} be a BP-algebra with the following cayley table. Consider the ideal A ={0,3} of X If θ f : X X is defined by θ f (0)=0, θ f (1)=2, θ f (2)=1, θ f (3)=3 and define an endomorphism f of X by θ f (x) = f(x). Since f(0) = 0, f(3) = 3, f(a) A proving that A is an f -ideal on X. Definition Let θ f be a self map of a BP-algebra X. An f -ideal on X is said to be θ f -invariant if θ f ( A ) A. Example Example(3.9), θ f (0) A and θ f (3)=3 A. Hence θ f (A) A,showing that A is θ f invariant. Theorem Let θ f be a regular (r, l) f -derivation on a BP-algebra X. Then f -ideal A on X is θ f invariant. proof: Now, Let θ f be a regular (r, l) f -derivation on X. θ f (x) = θ f (x 0) = (f(x) θ f (0)) (θ f (x) f(0)) = (f(x) 0) (θ(x) 0) = f(x) θ f (x) = θ f (x) (θ f (x) f(x)) = f(x), x X. 5

6 ISSN Let y θ f (A) then y = θ f (x) for some x A. It follows that y f(x) = θ f (x) f(x) = 0 A. Since x A,then f(x) f(a) A as A is an f -ideal. It follows that y A since A is an ideal on X. Hence θ f (A) A. Thus A is θ f -invariant. 4. Composition of f-derivation Definition 4.1. f be two self maps on X. We define θ f θ f :X X as ( θ f θ f )( x )= θ f ( θ f ( x )) for all x X. Proposition 4.2. f are the (l, r) f -derivations on X. Let f 2 = f f = f, then ( θ f θ f )is also a (l, r) f -derivation on X. Let X be a BP-algebra, and θ f and θ f are the (l, r) - f -derivations on X. (θ f θ f )(x y) = θ f (θ f (x y)) = θ f [(θ f (x) f(y)) (f(x) θ f (y))] = θ f [(f(x) θ f (y)) ((f(x) θ f (y)) θ f (x) f(y)))] = θ f (θ f (x) f(y)) ( y (y x) = x) = (θ f (θ f (x)) f 2 (y)) (f(θ f (x)) θ f (f(y))) = θ f (θ f (x)) f 2 (y) (θ f θ f )(x y) = (θ f (θ f (x)) f(y)) = (f(x) θ f (θ f (y))) [(f(x) θ f (θ f (y))) ((θ f (θ f (x)) f(y))] = (f(x) (θ f θ f )(y) [(f(x) (θ f θ f )(y)) (θ f θ f (x) f(y))] = (θ f θ f )(x) f(y)) (f(x) (θ f θ f )(y)). Which implies that ( θ f θ f ) is a (l, r) f -derivation on X. One can easily prove that the following proposition. Proposition 4.3. Let X be a BP-algebra, θ f and θ f are the (r, l) f -derivations on X such that f 2 = f f = f. Then θ f θ f is also a (r,l) - f -derivation on X.

7 ISSN Theorem 4.4. f be two f -derivations on X such that f 2 = f. Then θ f θ f is also a f -derivation on X. One can easily prove that the following proposition that the composition of derivations is commutative. Proposition 4.5. f, be two f-derivations on X such that f θ f = θ f f, θ f f = f θ f. Then θ f θ f = θ f θ f. f,be the f-derivations on X. Since θ f is a (l,r) - f-derivation on X, then for all x, y X. (θ f θ f )(x y) = θ f (θ f (x y)) But θ f is a (r,l) - f -derivation on X. = θ f ((θ f (x) f(y)) (f(x) θ f (y))) = θ f (θ f (x) f(y)) (θ f θ f )(x y) = θ f ((θ f (x) f(y)) = (f(θ f (x)) θ f (f(y))) (θ f (θ f (x)) f 2 (y)) = (f(θ f (x)) θ f (f(y))) = (f θ f )(x) (θ f f)(y) Thus we have for all x, y X, (θ f θ f )(x y) = (f θ f )(x) (θ f f)(y) Also since θ f is a (r,l) - f -derivation on X then for all x, y X. But θ f (θ f θ f )(x y) = θ f (θ f (x y)) is a (l,r) - f-derivation on X = θ f ((f(x) θ f (y)) (θ f (x) f(y))) = θ f ((f(x) θ f (y)) (θ f θ f )(x y) = (θ f (f(x)) f(θ f (y))) (f 2 (x) θ f (θ f (y))) Thus we have for alll x, y X = (θ f (f(x)) f(θ f (y))) = (θ f f)(x) (f θ f )(y) = (f θ f )(x) (θ f f)(y) (θ f θ f )(x y) = (f θ f )(x) (θ f f)(y) From (1) and (2) we get for all x, y X, (θ f θ f )(x y) = (θ f θ f )(x y) By putting y = 0 we get for all x X, which implies that (θ f θ f ) = (θ f θ f ). (θ f θ f )(x) = (θ f θ f )(x) 7

8 ISSN Definition 4.6. θ f θ f : X X as f (θ f θ f )x = θ f (x) θ f (x) for all x X. be two self maps on X. We define Proposition 4.7. f are f -derivations on X. Then ( f θ f ) ( θ f f ) = ( θ f f ) ( f θ f ) f be two derivations on X. Since θ f - f -derivation on X. Then for all x, y X. (θ f θ f )(x y) = θ f (θ f (x y)) But θ f is a (r,l) - f -derivation on X. = θ f [θ f (x) f(y)) (f(x) θ f (y))] = θ f (θ f (x) f(y)) θ f (θ f (x) f(y)) = (f(θ f (x)) θ f (f(y))) (θ f (θ f (x)) f 2 (y)) = (f(θ f (x) θ f (f(y))) = (f θ f )(x) (θ f f)(y) ( θ f θ f )( x y ) = ( f θ f )( x ) ( θ f f)( y ) for all x, y X. (1) Also we have that θ f is a (r,l) - f -derivation on X, then for all x, y X. (θ f θ f )(x y) = θ f (θ f (x y)) But θ f is a (l,r) - f -derivation on X. = θ f [(f(x) θ f (y)) (θ f (x) f(y))] = θ f (f(x) θ f (y)) θ f (f(x) θ f (y)) = (θ f (f(x)) f(θ f (y))) (f 2 (x) θ f (θ f (y))) = (θ f (f(x)) f(θ f (y))) (θ f θ f )(x y) = (θ f f)(x) (f θ f )(y), x, y X (2). From (1) and (2) we get for all x X (By putting y = x ) (f θ f )(x) (θ f f)(x) = (θ f f)(x) (f θ f )(x) (f θ f ) (θ f f)(x) = (θ f f) (f θ f )(x) which implies that (f θ f ) ( θ f f) = ( θ f f) (f θ f ) Notation: Der f ( X ) denotes the set of all f-derivations on X. is a (l,r)

9 ISSN Definition 4.8. Let θ f, θ f Der f ( X ). Define the binary operation as ( θ f θ f )( x ) = θ f ( x ) θ f ( x ). Proposition 4.9. f is also a (l,r) - f -derivation on X. f (θ f θ f )(x y) = θ f (x y) θ f (x y) are (l,r) - f -derivations om X. Then θ f θ f are (l,r) - f -derivations on X. We have = {(θ f (x) f(y)) (f(x) θ f (y))} {(θ f (x) f(y)) (f(x) θ f (y))} = (θ f (x) f(y)) (θ f (x) f(y)) = θ f (x) f(y) = (θ f (x) (θ f (x) θ f (x))) f(y) = (θ f (x) θ f (x)) f(y) = (θ f θ f )(x) f(y) = (f(x) (θ f θ f )(y)) (f(x) (θ f θ f )(y)) ((θ f θ f )(x) f(y)) = ((θ f θ f )(x) f(y)) (f(x) (θ f θ f )(y)) This shows that ( θ f θ f ) is a (l,r) - f -derivation on X. This completes the proof. In the similar fashion, we can establish the following. Proposition f is also a (r,l) - f -derivation on X. are (r,l) - f -derivations on X. Then θ f θ f Theorem If θ f, θ f Der f ( X ), θ f θ f Der f ( X ). Also ( θ f ( θ f θ f ))( x y ) = (( θ f θ f ) θ f )( x y ). If θ f, θ f Der f ( X ), then θ f is both a (l,r) and a (r,l) derivation.similarly θ f is both a (l,r) and a (r,l) derivation. By proposition (4.9) and (4.10), we observe that θ f θ f is both a (l,r) and a (r,l) derivation. Hence θ f θ f Der f ( X ). 9

10 ISSN To shows the associativity, choose θ f, θ f, θ f Der f ( X ). ((θ f θ f ) θ f )(x y) = (θ f θ f )(x y) (θ f )(x y) = ((θ f )(x y)) ((θ f )(x y)) ((θ f θ f )(x y)) = (θ f θ f )(x y) = (θ f )(x y) (θ f )(x y) = [(θ f (x) f(y)) (f(x) θ f (y))] [(θ f (x) f(y)) (f(x) θ f (y))] = (θ f (x) f(y)) (θ f (x) f(y)) = (θ f (x) f(y)) Also, (θ f (θ f θ f )(x y) = (θ f )(x y) (θ f θ f )(x y) = θ f (x y) [(θ f (x y) θ f (x y)] = θ f (x y) (θ f )(x y) (x (x y) = y) = [(θ f (x) f(y)) (f(x) θ f (y))] [(θ f (x) f(y)) (f(x) θ f (y))] This shows that, = (θ f (x) f(y)) (θ f (x) f(y)) = (θ f (x) f(y)) (( θ f θ f ) θ f )( x y ) = ( θ f ( θ f θ f )( x y ) which implies that (( θ f θ f ) θ f ) = ( θ f ( θ f θ f ) From the above theorem, we conclude that Der f ( X ) is closed under the binary composition defined in (4.8) which is also associative. Hence we have the following theorem. Theorem Der f ( X ) us a semigroup under the binary composition. ACKNOLEDGEMENT Authors are thankful to Dr.M.Chandramouleeswaran, Associate Professor and Head of Department of Mathematics, Saiva Bhanu Kshatriya College, Aruppukottai, for their suppported and encouragement of our Research work.

11 ISSN References [1] Ahn.S.S and Han.J.S: On BP-algebras, Hacettepe Journal of Mathematics and Statistics, Vol 42 (5) (2013), [2] Ganeshkumar T and Chandramouleeswaran M : Derivations on TM-algebras,International Journal of Mathematical Archive, 3 (11), 2012, [3] Hu Q.P and Li.X : On BCH-algebras,Math.seminar Notes,Kobe univ., 11(1983), [4] Hu Q.P and Li.X: On proper of BCH-algebras, math, Japan 30 ), (1985) [5] Imai.Y and Iseki.K: On axiom systems of Propositional calculi, XIV, Proc. Japan Acad. Ser A, Math Sci., 42 (1966), [6] Iseki and Tanaka.S: An introduction to theory of BCK-algebras, Math Japan 23,(1978), 1-26 [7] Jun Y.B and Xin X.L: On derivations of BCI-algebras Inform Sci, 159,(2004), [8] Kandaraj N and Chandramouleeswaran M: On Left derivations of d -algebras, International Journal of Mathematical Archive, 3 (6), 2012, [9] Neggers J and Kim H.S: On d-algebrasmath. Slovaca, Co.,49 (1999), AUTHORS First Author Kandaraj N, Associate Professor, Department of Mathematics, Saiva Bhanu Kshatriya College, Aruppukottai. address:n.kandarajsbkc1998@gmail.com Second Author Arul Devi A, Assistant Professor, Department of Mathematics, Saiva Bhanu Kshatriya College, Aruppukottai. address:aruldevika.22@gmail.com 11