# On isomorphisms of SU-algebras.

[section] 1. Introduction

In 1966, K. Iseki introduced the notion of a BCI-algebras which is a generalization of BCK-algebras. He defined a BCI-algebra as an algebra (X, *, 0) of type (2,0) satisfying the following conditions: (BCI 1) ((x * y) * (x * z)) * (z * y) =0, (BCI 2) (x * (x * y)) * y = 0, (BCI 3) x * x = 0, (BCI 4) x * y = 0 = y * x imply x = y, (BCI 5) x * 0 = 0 imply x = 0, for all x, y, z [member of] X. If (BCI 5) is replaced by (BCI 6) 0 * x = 0 for all x [member of] X, the algebra (X, *, 0) is called BCK-algebra. In 1983, Hu and Li introduced the notion of a BCH-algebras which is a generalization of the notions of BCK-algebra and BCI-algebras. They have studied a few properties of these algebras and defined a BCH-algebra as an algebra (X, *, 0) of type (2,0) satisfying the following conditions: (BCH 1) x * x = 0, (BCH 2) (x * y) * z = (x * z) * y, (BCH 3) x * y = 0 = y * x imply x = y, for all x, y, z [member of] X. In 1998, Dudek and Zhang studied ideals and congruences of BCC-algebras. They gave the concept of homomorphisms and quotient of BCC-algebras. They presented some related properties of them. In 2006 Dar and Akram studied properties of endomorphism in BCH-algebra.

In this paper, we introduce a new algebraic structure, called SU-algebra and a concept of ideal and homomorphisms in SU-algebra. We also describe connections between ideals and congruences. We investigated some related properties of them. Moreover, this paper is to derive some straightforward consequences of the relations between quotient SU-algebras and isomorphisms and also investigate some of its properties.

[section] 2. The structure of SU-algebra

Definition 2.1. A SU-algebra is an algebra (X, *, 0) of type (2,0) satisfying the following conditions:

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

(2) x * 0 = x,

(3) if x * y = 0 imply x = y for all x, y, z [member of] X.

From now on, X denotes a SU-algebra (X, *, 0) and a binary operation will be denoted by juxtaposition.

Example 2.2. Let X = {0,1, 2, 3} be a set in which operation * is defined by the following:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Then X is a SU-algebra.

Theorem 2.3. Let X be a SU-algebra. Then the following results hold for all x,y, z [member of] X.

(1) xx = 0,

(2) xy = yx,

(3) 0x = x,

(4) ((xy)x)y = 0,

(5) ((xz)(yz))(xy) = 0,

(6) xy = 0 if and only if (xz)(yz) = 0,

(7) xy = x if and only if y = 0.

Theorem 2.4. Let X be a SU-algebra. A relation [less than or equal to] on X is defined by x [less than or equal to] y if xy = 0. Then (X, [less than or equal to]) is a partially ordered set.

Theorem 2.5. Let X be a SU-algebra. Then the following results hold for all x, y, z [member of] X.

(1) x [less than or equal to] y if and only if y [less than or equal to] x,

(2) x [less than or equal to] 0 if and only if x = 0,

(3) if x [less than or equal to] y, then xz [less than or equal to] yz.

Theorem 2.6. Let X be a SU-algebra. Then the following results hold for all x,y, z [member of] X.

(1) (xy)z = (xz)y,

(2) x(yz) = z(yx),

(3) (xy)z = x(yz).

Theorem 2.7. Let X be a SU-algebra. If xz = yz, then x = y for all x,y,z [member of] X.

Theorem 2.8. Let X be a SU-algebra and a [member of] X. If ax = x for all x [member of] X, then a = x.

[section] 3. Ideal and congruences in SU-algebra

Definition 3.1. Let X be a SU-algebra. A nonempty subset I of X is called an ideal of X if it satisfies the following properties :

(1) 0 [member of] I,

(2) if (xy) z [member of] I and y [member of] I for all x, y, z [member of] X, then xz [member of] I. Clearly, X and {0} are ideals of X.

Example 3.2. Let X be a SU-algebra as defined in Example 2.2. If A = {0,1}, then A is an ideal of X. If B = {0, 1, 2}, then B is not an ideal of X because (1 * 1) * 2 = 0 * 2 = 2 [member of] B but 1 * 2 = 3 [not member of] B.

Theorem 3.3. Let X be a SU-algebra and I be an ideal of X. Then

(1) if xy [member of] I and y [member of] I, then x [member of] I for all x,y [member of] X,

(2) if xy [member of] I and x [member of] I, then y [member of] I for all x, y [member of] X.

Proof. Let X be a SU-algebra and I be an ideal of X.

(1) Let xy [member of] I and y [member of] I. Since (xy)0 = xy [member of] I and y [member of] I, x = x0 [member of] I (by Definition 3.1).

(2) It is immediately followed by (1) and Theorem 2.3 (2).

Theorem 3.4. Let X be a SU-algebra and [A.sub.i] be ideal of X for i =1, 2, ..., n. Then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is an ideal of X.

Proof. Let X be a SU-algebra and [A.sub.i] be ideal of X for i = 1, 2, ..., n. Clearly, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Let x, y, z [member of] X be such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Then (xy)z [member of] [A.sub.i] and y [member of] [A.sub.i] for all i = 1, 2, ..., n. Since [A.sub.i] is an ideal, xz [member of] [A.sub.i] for all i = 1, 2, ..., n. Thus [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Hence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a ideal of X.

Definition 3.5. Let X be a SU-algebra. A nonempty subset S of X is called a SU-subalgebra of X if xy [member of] S for all x,y [member of] S.

Theorem 3.6. Let X be a SU-algebra and I be an ideal of X. Then I is a SU-subalgebra of X.

Proof. Let X be a SU-algebra and I be an ideal of X. Let x, y [member of] I. By Theorem 2.3 (4), ((xy)x) y = 0 [member of] I. Since I is an ideal and x [member of] I, (xy)y [member of] I. Since I is an ideal and y [member of] I, xy [member of] I.

Definition 3.7. Let X be a SU-algebra and I be an ideal of X. A relation ~ on X is defined by x ~ y if and only if xy [member of] I.

Theorem 3.8. Let X be a SU-algebra and I be an ideal of X. Then ~ is a congruence on X.

Proof. A reflexive property and a symmetric property are obvious. Let x,y,z [member of] X. Suppose that x ~ y and y ~ z. Then xy [member of] I and yz [member of] I. Since ((xz)(xy))(zy) =0 [member of] I and I is an ideal, (xz)(yz) = (xz)(zy) [member of] I. By Theorem 3.3 (1), xz [member of] I. Thus x ~ z. Hence ~ is an equivalent relation.

Next, let x,y,u,v [member of] X be such that x ~ u and y ~ v. Then xu [member of] I and yv [member of] I. Thus ux [member of] I and vy [member of] I. Since ((xy)(xv))(yv) = 0 [member of] I and Theorem 3.3 (1), (xy)(xv) [member of] I. Hence xy ~ xv. Since ((uv)(ux))(vx) = 0 [member of] I and Definition 3.1 (2), (uv)(vx) [member of] I. Since vx = xv, (uv)(vx) = (uv)(xv). Hence (uv)(xv) [member of] I and so xv ~ uv. Thus xy ~ uv. Hence ~ is a congruence on X.

Let X be a SU-algebra, I be an ideal of X and ~ be a congruence relation on X. For any x [member of] X, we define [[absolute value of x.sub.I] = {y [member of] X|x ~ y} = {y [member of] X|xy [member of] I}. Then we say that [absolute value of x].sub.I] is an equivalence class containing x.

Example 3.9. Let X be a SU-algebra as defined in Example 2.2. It is easy to show that I = {0,1} is an ideal of X, then [[0].sub.I] = {0,1}, [[1].sub.I] = {0,1}, [[2].sub.I] = {2, 3}, [[3].sub.I] = {2, 3}.

Remarks 3.10. Let X be a SU-algebra. Then

1. x [member of] [[x].sub.I] for all x [member of] X,

2. [[0].sub.I] = {x [member of] X|0 = x} is an ideal of X.

Theorem 3.11. Let X be a SU-algebra, I be an ideal of X and = be a congruence relation on X. Then [[x].sub.I] = [[y].sub.I] if and only if x = y for all x,y [member of] X.

Proof. Let [[x].sub.I] = [[y].sub.I]. Since y [member of] [[y].sub.I], y [member of] [[x].sub.I]. Hence x = y. Conversely, let x = y. Then y ~ x. Let z [member of] [[x].sub.I]. Then x ~ z. Since y ~ x and x ~ z, y ~ z, z [member of] [[y].sub.I]. Similarly, let w [member of] [[y].sub.I]. Then w [member of] [[x].sub.I]. Therefore [[x].sub.I] = [[y].sub.I].

The family {[[x].sub.I]|x [member of] X} gives a partition of X which is denoted by quotient SU-algebra X/I. For x,y [member of] X we define [[x].sub.I][[y].sub.I] = [[xy].sub.I]. The following theorem show that X/I is a SU-algebra.

Theorem 3.12. Let X be a SU-algebra and I be an ideal of X. Then X/I is a SU-algebra.

Proof. Let [[x].sub.I], [[y].sub.I], [[z].sub.I] [member of] X/I.

1) (([[x].sub.I][[y].sub.I])([[x].sub.I][[z].sub.I]))([[y].sub.I][[z].sub.I]) = ([[xy].sub.I][[xz].sub.I])[[yz].sub.I] = [[((xy)(xz))(yz)].sub.I] = [[0].sub.I].

2) [[x].sub.I][[0].sub.I] = [[x0].sub.I] = [[x].sub.I].

3) Suppose that [[x].sub.I][[y].sub.I] = [[0].sub.I]. Then [[xy].sub.I] = [[0].sub.I] = [[yx].sub.I]. Since xy [member of] [[xy].sub.I], 0 ~ xy. Hence xy [member of] [[0].sub.I]. Since [[0].sub.I] is an ideal, x ~ y. Hence [[x].sub.I] = [[y].sub.I]. Thus X/I is a SU-algebra.

[section] 4. Isomorphism of a SU-Algebra

In this section, we defined homomorphism and isomorphism of SU-algebras, then we show some consequences of the relations between quotient SU-algebras and isomorphisms.

Definition 4.1. Let (X, [*.sub.X], [0.sub.X]) and (Y, [*.sub.Y], [0.sub.Y]) be a SU-algebra and let f : X [right arrow] Y. We called f is a homomorphism if and only if f (xy) = f (x)f (y) for all x,y [member of] X.

The kernel of f defined to be the set ker(f) = {x [member of] X|f (x) = [0.sub.Y]}.

The image of f defined to be the set im(f) = {f(x) [member of] Y|x [member of] X}.

Definition 4.2. Let X and Y be a SU-algebra and let f : X [right arrow] Y be a homomorphism, then:

(1) f called a monomorphism if f is injective,

(2) f called an epimorphism if f is surjective,

(3) f called an isomorphism if f is bijective.

Definition 4.3. Let X and Y be a SU-algebra, then we say that X isomorphic Y (X [congruent to] Y) if we have f : X [right arrow] Y which f is an isomorphism.

Theorem 4.4. Let X be a SU-algebra, I be an ideal of X and = be a congruence on X. Then f : X [right arrow] X/I defined by f (x) = [[x].sub.I] for all x [member of] X is an epimorphism.

Proof. Let f : X [member of] X/I and defined f by f(x) = [[x].sub.I] for all x [member of] X. Let x,y [member of] X and x = y, then [[x].sub.I] = [[y].sub.I]. Thus f(x) = f(y). Hence f is a function. Let [[x].sub.I] [member of] X/I, then f(x) = [[x].sub.I]. Hence f is surjective. Since f(xy) = [[xy].sub.I] = [[x].sub.I][[y].sub.I] = f(x)f(y). Thus f is a homomorphism on X. Hence f is an epimorphism on X.

Theorem 4.5. Let (X, [*.sub.X], [0.sub.X]) and (Y, [*.sub.Y], [0.sub.Y]) be a SU-algebra and let f : X [right arrow] Y be a homomorphism, then:

(1) f ([O.sub.x]) = [0.sub.Y],

(2) im(f) is a SU-subalgebra,

(3) ker(f) = {[0.sub.X]} if and only if f is a injective,

(4) ker(f) is an ideal of X.

Proof. (1) Let x [member of] X, then f(x) [member of] Y. Since [0.sub.Y]f(x) = f(x) = f([0.sub.X]x) = f([0.sub.X])f( x), then by Theorem 2.7 we have [0.sub.Y] = f([0.sub.X]).

(2) Let a, b [member of] im(f), then there exists x, y [member of] X such that f(x) = a and f(y) = b. Thus ab = f(x)f(y) = f(xy) [member of] im(f). Hence im(f) is a SU-subalgebra.

(3) Suppose ker(f) = {[0.sub.X]}. Let x, y [member of] X and f(x) = f(y), then f(xy) = f(x)f(y) = [0.sub.Y]. Thus xy [member of] ker(f) = [0.sub.X]. Hence x = y. Therefore f is an injective. Conversely, it is obviously.

(4) By (1) we have f([0.sub.X]) = [0.sub.Y], Thus [0.sub.X] [member of] ker(f). Let (xy)z [member of] ker(f) and y [member of] ker(f), then f((xy)z) = [0.sub.Y] and f(y) = [0.sub.Y]. Since f is a homomorphism, f((xy)z) = f(xy)f(z) = (f (x)f (y))f (z). Since f ((xy)z) = [0.sub.y] and f(y) = [0.sub.x], [0.sub.y] = (f(x)[0.sub.y])f(z) = f(x)f(z) = f (xz). Thus xz [member of] ker(f). Hence ker(f) is an ideal of X.

Theorem 4.6. Let X and Y be a SU-algebra and let f : X [right arrow] Y be a homomorphism, then X/ker(f) [congruent to] im(f). In particular, if f is surjective, then X/ker(f) [congruent to] Y.

Proof. Consider the mapping g : X/ker(f) [right arrow] im(f) given by g([[x].sub.ker(f)]) = f(x)for all x [member of] X.

1) Let x,y [member of] X and [[x].sub.ker(f)] = [[y].sub.ker(f)], then x ~ y. Thus xy [member of] ker(f). Hence f(xy) = 0. Since f(y)f(x) = f(x)f(y) = f(xy) = 0, f(x) = f(y). Therefore g([[x].sub.ker(f)]) = f(x) = f(y) = g([[y].sub.fcer(f)]). Hence g is a function.

2) Let x,y [member of] X and g([[x].sub.fcer(f)]) = g([[y].sub.fcer(f)]), then f(x) = f(y). Thus [0.sub.x] = f(x)f(y) = f(xy). Hence xy [member of] ker(f). Since ker(f) is an ideal, x ~ y. Thus [[x].sub.ker(f)] = [[y].sub.ker(f)]. Hence g is a injective.

3) Let f(x) [member of] im(f). Since g([[x].sub.ker(f)]) = f(x), g is a surjective.

4) Let x,y [member of] X, then g([[x].sub.fcer(f)][[y].sub.ker(f)]) = g([[xy].sub.ker(f)]) = f(xy) = f(x)f(y) = g([[x].sub.fcer(f)]) g([[y].sub.ker(f)]). Hence g is a homomorphism. Therefore X/ker(f) [congruent to] im(f). In particular, let f be a surjective, then im(f) = Y. Hence X/ker(f) [congruent to] Y.

Theorem 4.7. Let H and K be an ideal of SU-algebra X and K [subset or equal to] H, then (X/K)/(H/K) [congruent to] X/H.

Proof. Consider the mapping g : X/K [right arrow] X/H given by g([[x].sub.K]) = [[x].sub.H] for all x [member of] X.

1) Let x,y [member of] X and [[x].sub.K] = [[y].sub.K], then x ~ y. Since K is a ideal , xy [member of] K. Since K [subset or equal to] H, xy [member of] H. Thus g([[x].sub.K]) = [[x].sub.H] = [[y].sub.H] = g([[y].sub.K]). Hence g is a function.

2) Let [[x].sub.H] [member of] X/H. Since g([[x].sub.K]) = [[x].sub.H], g is a surjective.

3) Let x,y [member of] X, then g([[x].sub.K][[y].sub.K]) = g([[xy].sub.K]) = [[xy].sub.H] = [[x].sub.H][[y].sub.H] = g([[x].sub.K])g([[y].sub.K]). Hence g is a homomorphism.

4) ker(g) = {[[x].sub.K]|g([[x].sub.K]) = [[0].sub.H]} = {[[x].sub.K]|[[x].sub.H] = [[0].sub.H]} = {[[x].sub.K]|x ~ 0} = {[[x].sub.K]|x = x0 [member of] H} = H/K. By Theorem 4.6 we have (X/K)/(H/K) [congruent to] X/H.

Let X be a SU-algebra and A,B be a subset of X defined AB by AB = {xy [member of] X|x G[member of] A, y [member of] B}.

Theorem 4.8. Let A and B be a subset of SU-algebra X, then AB is a SU-subalgebra of X.

Proof. Let a [member of] AB. By definition of AB we have a = xy for some x [member of] A,y [member of] B. Since A, B [subset or equal to], x, y [member of] X. Thus a = xy [member of] X. Hence AB [subset or equal to] X. Let m, n [member of] AB such that m = [a.sub.1] [b.sub.1], n = [a.sub.2][b.sub.2] for some [a.sub.1], [a.sub.2] [member of] A, [b.sub.1],[b.sub.2] [member of] B, then mn = ([a.sub.1][b.sub.1])([a.sub.2][b.sub.2]) = ([a.sub.1]([a.sub.2][b.sub.2]))[b.sub.1] = ([b.sub.2]([a.sub.2][a.sub.1]))[b.sub.1] = ([b.sub.2][b.sub.1])([a.sub.2][a.sub.1]) = ([a.sub.2][a.sub.1])([b.sub.2][b.sub.1]). Since [a.sub.2][a.sub.1] [member of] A and [b.sub.2][b.sub.1] [member of] B, mn G AB. Hence AB is a SU-subalgebra of X.

Remark 4.9. Let H and K be an ideal of SU-algebra X, then N is an ideal of HN .

Theorem 4.10. Let H and N be an ideal of SU-algebra X, then H/(H [intersection] N) [congruent to] HN/N.

Proof. Consider the mapping g : H [right arrow] HN/N given by g(x) = [[x].sub.N] for all x [member of] H.

1) Let x,y [member of] H and x = y, then [[x].sub.N] = [[y].sub.N]. Thus g(x) = g(y). Hence g is a function.

2) Let [[x].sub.N] [member of] HN/N, then g(x) = [[x].sub.N] Hence g is a surjective.

3) Let x,y [member of] H. Since g(xy) = [[xy].sub.N] = [[x].sub.N][[y].sub.N] = g(x)g(y). Hence g is a homomorphism on X.

4) Let x [member of] ker(g), then g(x) = [[0].sub.N]. Since g(x) = [[x].sub.N], [[x].sub.N] = [[0].sub.N]. Thus x ~ 0. Since N is an ideal, x = x0 [member of] N. Since ker(g) [subset or equal to] H, x [member of] H. Thus x [member of] H [intersection] N. Hence ker(g) [subset or equal to] H [intersection] N. Let x [member of] H [intersection] N, then x [member of] H and x [member of] N. Since g(x) = [[x].sub.N] = [[0].sub.N], x [member of] ker(g). Thus H [intersection] N [subset or equal to] ker(g). Hence ker(g) = H [intersection] N. By Theorem 4.6 we have H/(H [intersection] N) [congruent to] HN/N.

Acknowledgements

The authors are highly grateful to the referees for their valuable comments and suggestions for the paper. Moreover, this work was supported by a grant for Kasetsart University.

References

[1] Dar, K. H. and M. Akrem, On endomorphisms of BCH-algebras, Annals of University of Craiova, Math. Comp. Sci. Ser., 33(2006), 227-234.

[2] Dudek, W. A. and X. Zhang, On ideal and congruences in BCC-algebra, Czechoslovak Math. Journal, 48(1998), 21-29.

[3] Hu, Q. P. and X. Li, On BCH-algebras, Math. Seminar Notes, 11(1983), 313-320.

[4] Iseki, K., An algebra related with a propositiomal calculus, M. J. A., 42(1966), 26-29.

S. Keawrahun ([dagger]) and U. Leerawa ([double dagger])

([dagger])[double dagger]) Department of Mathematics, Kasetsart University, Bangkok, Thailand E-mail: eangg-11@hotmail.com fsciutl@ku.ac.th

In 1966, K. Iseki introduced the notion of a BCI-algebras which is a generalization of BCK-algebras. He defined a BCI-algebra as an algebra (X, *, 0) of type (2,0) satisfying the following conditions: (BCI 1) ((x * y) * (x * z)) * (z * y) =0, (BCI 2) (x * (x * y)) * y = 0, (BCI 3) x * x = 0, (BCI 4) x * y = 0 = y * x imply x = y, (BCI 5) x * 0 = 0 imply x = 0, for all x, y, z [member of] X. If (BCI 5) is replaced by (BCI 6) 0 * x = 0 for all x [member of] X, the algebra (X, *, 0) is called BCK-algebra. In 1983, Hu and Li introduced the notion of a BCH-algebras which is a generalization of the notions of BCK-algebra and BCI-algebras. They have studied a few properties of these algebras and defined a BCH-algebra as an algebra (X, *, 0) of type (2,0) satisfying the following conditions: (BCH 1) x * x = 0, (BCH 2) (x * y) * z = (x * z) * y, (BCH 3) x * y = 0 = y * x imply x = y, for all x, y, z [member of] X. In 1998, Dudek and Zhang studied ideals and congruences of BCC-algebras. They gave the concept of homomorphisms and quotient of BCC-algebras. They presented some related properties of them. In 2006 Dar and Akram studied properties of endomorphism in BCH-algebra.

In this paper, we introduce a new algebraic structure, called SU-algebra and a concept of ideal and homomorphisms in SU-algebra. We also describe connections between ideals and congruences. We investigated some related properties of them. Moreover, this paper is to derive some straightforward consequences of the relations between quotient SU-algebras and isomorphisms and also investigate some of its properties.

[section] 2. The structure of SU-algebra

Definition 2.1. A SU-algebra is an algebra (X, *, 0) of type (2,0) satisfying the following conditions:

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

(2) x * 0 = x,

(3) if x * y = 0 imply x = y for all x, y, z [member of] X.

From now on, X denotes a SU-algebra (X, *, 0) and a binary operation will be denoted by juxtaposition.

Example 2.2. Let X = {0,1, 2, 3} be a set in which operation * is defined by the following:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Then X is a SU-algebra.

Theorem 2.3. Let X be a SU-algebra. Then the following results hold for all x,y, z [member of] X.

(1) xx = 0,

(2) xy = yx,

(3) 0x = x,

(4) ((xy)x)y = 0,

(5) ((xz)(yz))(xy) = 0,

(6) xy = 0 if and only if (xz)(yz) = 0,

(7) xy = x if and only if y = 0.

Theorem 2.4. Let X be a SU-algebra. A relation [less than or equal to] on X is defined by x [less than or equal to] y if xy = 0. Then (X, [less than or equal to]) is a partially ordered set.

Theorem 2.5. Let X be a SU-algebra. Then the following results hold for all x, y, z [member of] X.

(1) x [less than or equal to] y if and only if y [less than or equal to] x,

(2) x [less than or equal to] 0 if and only if x = 0,

(3) if x [less than or equal to] y, then xz [less than or equal to] yz.

Theorem 2.6. Let X be a SU-algebra. Then the following results hold for all x,y, z [member of] X.

(1) (xy)z = (xz)y,

(2) x(yz) = z(yx),

(3) (xy)z = x(yz).

Theorem 2.7. Let X be a SU-algebra. If xz = yz, then x = y for all x,y,z [member of] X.

Theorem 2.8. Let X be a SU-algebra and a [member of] X. If ax = x for all x [member of] X, then a = x.

[section] 3. Ideal and congruences in SU-algebra

Definition 3.1. Let X be a SU-algebra. A nonempty subset I of X is called an ideal of X if it satisfies the following properties :

(1) 0 [member of] I,

(2) if (xy) z [member of] I and y [member of] I for all x, y, z [member of] X, then xz [member of] I. Clearly, X and {0} are ideals of X.

Example 3.2. Let X be a SU-algebra as defined in Example 2.2. If A = {0,1}, then A is an ideal of X. If B = {0, 1, 2}, then B is not an ideal of X because (1 * 1) * 2 = 0 * 2 = 2 [member of] B but 1 * 2 = 3 [not member of] B.

Theorem 3.3. Let X be a SU-algebra and I be an ideal of X. Then

(1) if xy [member of] I and y [member of] I, then x [member of] I for all x,y [member of] X,

(2) if xy [member of] I and x [member of] I, then y [member of] I for all x, y [member of] X.

Proof. Let X be a SU-algebra and I be an ideal of X.

(1) Let xy [member of] I and y [member of] I. Since (xy)0 = xy [member of] I and y [member of] I, x = x0 [member of] I (by Definition 3.1).

(2) It is immediately followed by (1) and Theorem 2.3 (2).

Theorem 3.4. Let X be a SU-algebra and [A.sub.i] be ideal of X for i =1, 2, ..., n. Then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is an ideal of X.

Proof. Let X be a SU-algebra and [A.sub.i] be ideal of X for i = 1, 2, ..., n. Clearly, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Let x, y, z [member of] X be such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Then (xy)z [member of] [A.sub.i] and y [member of] [A.sub.i] for all i = 1, 2, ..., n. Since [A.sub.i] is an ideal, xz [member of] [A.sub.i] for all i = 1, 2, ..., n. Thus [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Hence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a ideal of X.

Definition 3.5. Let X be a SU-algebra. A nonempty subset S of X is called a SU-subalgebra of X if xy [member of] S for all x,y [member of] S.

Theorem 3.6. Let X be a SU-algebra and I be an ideal of X. Then I is a SU-subalgebra of X.

Proof. Let X be a SU-algebra and I be an ideal of X. Let x, y [member of] I. By Theorem 2.3 (4), ((xy)x) y = 0 [member of] I. Since I is an ideal and x [member of] I, (xy)y [member of] I. Since I is an ideal and y [member of] I, xy [member of] I.

Definition 3.7. Let X be a SU-algebra and I be an ideal of X. A relation ~ on X is defined by x ~ y if and only if xy [member of] I.

Theorem 3.8. Let X be a SU-algebra and I be an ideal of X. Then ~ is a congruence on X.

Proof. A reflexive property and a symmetric property are obvious. Let x,y,z [member of] X. Suppose that x ~ y and y ~ z. Then xy [member of] I and yz [member of] I. Since ((xz)(xy))(zy) =0 [member of] I and I is an ideal, (xz)(yz) = (xz)(zy) [member of] I. By Theorem 3.3 (1), xz [member of] I. Thus x ~ z. Hence ~ is an equivalent relation.

Next, let x,y,u,v [member of] X be such that x ~ u and y ~ v. Then xu [member of] I and yv [member of] I. Thus ux [member of] I and vy [member of] I. Since ((xy)(xv))(yv) = 0 [member of] I and Theorem 3.3 (1), (xy)(xv) [member of] I. Hence xy ~ xv. Since ((uv)(ux))(vx) = 0 [member of] I and Definition 3.1 (2), (uv)(vx) [member of] I. Since vx = xv, (uv)(vx) = (uv)(xv). Hence (uv)(xv) [member of] I and so xv ~ uv. Thus xy ~ uv. Hence ~ is a congruence on X.

Let X be a SU-algebra, I be an ideal of X and ~ be a congruence relation on X. For any x [member of] X, we define [[absolute value of x.sub.I] = {y [member of] X|x ~ y} = {y [member of] X|xy [member of] I}. Then we say that [absolute value of x].sub.I] is an equivalence class containing x.

Example 3.9. Let X be a SU-algebra as defined in Example 2.2. It is easy to show that I = {0,1} is an ideal of X, then [[0].sub.I] = {0,1}, [[1].sub.I] = {0,1}, [[2].sub.I] = {2, 3}, [[3].sub.I] = {2, 3}.

Remarks 3.10. Let X be a SU-algebra. Then

1. x [member of] [[x].sub.I] for all x [member of] X,

2. [[0].sub.I] = {x [member of] X|0 = x} is an ideal of X.

Theorem 3.11. Let X be a SU-algebra, I be an ideal of X and = be a congruence relation on X. Then [[x].sub.I] = [[y].sub.I] if and only if x = y for all x,y [member of] X.

Proof. Let [[x].sub.I] = [[y].sub.I]. Since y [member of] [[y].sub.I], y [member of] [[x].sub.I]. Hence x = y. Conversely, let x = y. Then y ~ x. Let z [member of] [[x].sub.I]. Then x ~ z. Since y ~ x and x ~ z, y ~ z, z [member of] [[y].sub.I]. Similarly, let w [member of] [[y].sub.I]. Then w [member of] [[x].sub.I]. Therefore [[x].sub.I] = [[y].sub.I].

The family {[[x].sub.I]|x [member of] X} gives a partition of X which is denoted by quotient SU-algebra X/I. For x,y [member of] X we define [[x].sub.I][[y].sub.I] = [[xy].sub.I]. The following theorem show that X/I is a SU-algebra.

Theorem 3.12. Let X be a SU-algebra and I be an ideal of X. Then X/I is a SU-algebra.

Proof. Let [[x].sub.I], [[y].sub.I], [[z].sub.I] [member of] X/I.

1) (([[x].sub.I][[y].sub.I])([[x].sub.I][[z].sub.I]))([[y].sub.I][[z].sub.I]) = ([[xy].sub.I][[xz].sub.I])[[yz].sub.I] = [[((xy)(xz))(yz)].sub.I] = [[0].sub.I].

2) [[x].sub.I][[0].sub.I] = [[x0].sub.I] = [[x].sub.I].

3) Suppose that [[x].sub.I][[y].sub.I] = [[0].sub.I]. Then [[xy].sub.I] = [[0].sub.I] = [[yx].sub.I]. Since xy [member of] [[xy].sub.I], 0 ~ xy. Hence xy [member of] [[0].sub.I]. Since [[0].sub.I] is an ideal, x ~ y. Hence [[x].sub.I] = [[y].sub.I]. Thus X/I is a SU-algebra.

[section] 4. Isomorphism of a SU-Algebra

In this section, we defined homomorphism and isomorphism of SU-algebras, then we show some consequences of the relations between quotient SU-algebras and isomorphisms.

Definition 4.1. Let (X, [*.sub.X], [0.sub.X]) and (Y, [*.sub.Y], [0.sub.Y]) be a SU-algebra and let f : X [right arrow] Y. We called f is a homomorphism if and only if f (xy) = f (x)f (y) for all x,y [member of] X.

The kernel of f defined to be the set ker(f) = {x [member of] X|f (x) = [0.sub.Y]}.

The image of f defined to be the set im(f) = {f(x) [member of] Y|x [member of] X}.

Definition 4.2. Let X and Y be a SU-algebra and let f : X [right arrow] Y be a homomorphism, then:

(1) f called a monomorphism if f is injective,

(2) f called an epimorphism if f is surjective,

(3) f called an isomorphism if f is bijective.

Definition 4.3. Let X and Y be a SU-algebra, then we say that X isomorphic Y (X [congruent to] Y) if we have f : X [right arrow] Y which f is an isomorphism.

Theorem 4.4. Let X be a SU-algebra, I be an ideal of X and = be a congruence on X. Then f : X [right arrow] X/I defined by f (x) = [[x].sub.I] for all x [member of] X is an epimorphism.

Proof. Let f : X [member of] X/I and defined f by f(x) = [[x].sub.I] for all x [member of] X. Let x,y [member of] X and x = y, then [[x].sub.I] = [[y].sub.I]. Thus f(x) = f(y). Hence f is a function. Let [[x].sub.I] [member of] X/I, then f(x) = [[x].sub.I]. Hence f is surjective. Since f(xy) = [[xy].sub.I] = [[x].sub.I][[y].sub.I] = f(x)f(y). Thus f is a homomorphism on X. Hence f is an epimorphism on X.

Theorem 4.5. Let (X, [*.sub.X], [0.sub.X]) and (Y, [*.sub.Y], [0.sub.Y]) be a SU-algebra and let f : X [right arrow] Y be a homomorphism, then:

(1) f ([O.sub.x]) = [0.sub.Y],

(2) im(f) is a SU-subalgebra,

(3) ker(f) = {[0.sub.X]} if and only if f is a injective,

(4) ker(f) is an ideal of X.

Proof. (1) Let x [member of] X, then f(x) [member of] Y. Since [0.sub.Y]f(x) = f(x) = f([0.sub.X]x) = f([0.sub.X])f( x), then by Theorem 2.7 we have [0.sub.Y] = f([0.sub.X]).

(2) Let a, b [member of] im(f), then there exists x, y [member of] X such that f(x) = a and f(y) = b. Thus ab = f(x)f(y) = f(xy) [member of] im(f). Hence im(f) is a SU-subalgebra.

(3) Suppose ker(f) = {[0.sub.X]}. Let x, y [member of] X and f(x) = f(y), then f(xy) = f(x)f(y) = [0.sub.Y]. Thus xy [member of] ker(f) = [0.sub.X]. Hence x = y. Therefore f is an injective. Conversely, it is obviously.

(4) By (1) we have f([0.sub.X]) = [0.sub.Y], Thus [0.sub.X] [member of] ker(f). Let (xy)z [member of] ker(f) and y [member of] ker(f), then f((xy)z) = [0.sub.Y] and f(y) = [0.sub.Y]. Since f is a homomorphism, f((xy)z) = f(xy)f(z) = (f (x)f (y))f (z). Since f ((xy)z) = [0.sub.y] and f(y) = [0.sub.x], [0.sub.y] = (f(x)[0.sub.y])f(z) = f(x)f(z) = f (xz). Thus xz [member of] ker(f). Hence ker(f) is an ideal of X.

Theorem 4.6. Let X and Y be a SU-algebra and let f : X [right arrow] Y be a homomorphism, then X/ker(f) [congruent to] im(f). In particular, if f is surjective, then X/ker(f) [congruent to] Y.

Proof. Consider the mapping g : X/ker(f) [right arrow] im(f) given by g([[x].sub.ker(f)]) = f(x)for all x [member of] X.

1) Let x,y [member of] X and [[x].sub.ker(f)] = [[y].sub.ker(f)], then x ~ y. Thus xy [member of] ker(f). Hence f(xy) = 0. Since f(y)f(x) = f(x)f(y) = f(xy) = 0, f(x) = f(y). Therefore g([[x].sub.ker(f)]) = f(x) = f(y) = g([[y].sub.fcer(f)]). Hence g is a function.

2) Let x,y [member of] X and g([[x].sub.fcer(f)]) = g([[y].sub.fcer(f)]), then f(x) = f(y). Thus [0.sub.x] = f(x)f(y) = f(xy). Hence xy [member of] ker(f). Since ker(f) is an ideal, x ~ y. Thus [[x].sub.ker(f)] = [[y].sub.ker(f)]. Hence g is a injective.

3) Let f(x) [member of] im(f). Since g([[x].sub.ker(f)]) = f(x), g is a surjective.

4) Let x,y [member of] X, then g([[x].sub.fcer(f)][[y].sub.ker(f)]) = g([[xy].sub.ker(f)]) = f(xy) = f(x)f(y) = g([[x].sub.fcer(f)]) g([[y].sub.ker(f)]). Hence g is a homomorphism. Therefore X/ker(f) [congruent to] im(f). In particular, let f be a surjective, then im(f) = Y. Hence X/ker(f) [congruent to] Y.

Theorem 4.7. Let H and K be an ideal of SU-algebra X and K [subset or equal to] H, then (X/K)/(H/K) [congruent to] X/H.

Proof. Consider the mapping g : X/K [right arrow] X/H given by g([[x].sub.K]) = [[x].sub.H] for all x [member of] X.

1) Let x,y [member of] X and [[x].sub.K] = [[y].sub.K], then x ~ y. Since K is a ideal , xy [member of] K. Since K [subset or equal to] H, xy [member of] H. Thus g([[x].sub.K]) = [[x].sub.H] = [[y].sub.H] = g([[y].sub.K]). Hence g is a function.

2) Let [[x].sub.H] [member of] X/H. Since g([[x].sub.K]) = [[x].sub.H], g is a surjective.

3) Let x,y [member of] X, then g([[x].sub.K][[y].sub.K]) = g([[xy].sub.K]) = [[xy].sub.H] = [[x].sub.H][[y].sub.H] = g([[x].sub.K])g([[y].sub.K]). Hence g is a homomorphism.

4) ker(g) = {[[x].sub.K]|g([[x].sub.K]) = [[0].sub.H]} = {[[x].sub.K]|[[x].sub.H] = [[0].sub.H]} = {[[x].sub.K]|x ~ 0} = {[[x].sub.K]|x = x0 [member of] H} = H/K. By Theorem 4.6 we have (X/K)/(H/K) [congruent to] X/H.

Let X be a SU-algebra and A,B be a subset of X defined AB by AB = {xy [member of] X|x G[member of] A, y [member of] B}.

Theorem 4.8. Let A and B be a subset of SU-algebra X, then AB is a SU-subalgebra of X.

Proof. Let a [member of] AB. By definition of AB we have a = xy for some x [member of] A,y [member of] B. Since A, B [subset or equal to], x, y [member of] X. Thus a = xy [member of] X. Hence AB [subset or equal to] X. Let m, n [member of] AB such that m = [a.sub.1] [b.sub.1], n = [a.sub.2][b.sub.2] for some [a.sub.1], [a.sub.2] [member of] A, [b.sub.1],[b.sub.2] [member of] B, then mn = ([a.sub.1][b.sub.1])([a.sub.2][b.sub.2]) = ([a.sub.1]([a.sub.2][b.sub.2]))[b.sub.1] = ([b.sub.2]([a.sub.2][a.sub.1]))[b.sub.1] = ([b.sub.2][b.sub.1])([a.sub.2][a.sub.1]) = ([a.sub.2][a.sub.1])([b.sub.2][b.sub.1]). Since [a.sub.2][a.sub.1] [member of] A and [b.sub.2][b.sub.1] [member of] B, mn G AB. Hence AB is a SU-subalgebra of X.

Remark 4.9. Let H and K be an ideal of SU-algebra X, then N is an ideal of HN .

Theorem 4.10. Let H and N be an ideal of SU-algebra X, then H/(H [intersection] N) [congruent to] HN/N.

Proof. Consider the mapping g : H [right arrow] HN/N given by g(x) = [[x].sub.N] for all x [member of] H.

1) Let x,y [member of] H and x = y, then [[x].sub.N] = [[y].sub.N]. Thus g(x) = g(y). Hence g is a function.

2) Let [[x].sub.N] [member of] HN/N, then g(x) = [[x].sub.N] Hence g is a surjective.

3) Let x,y [member of] H. Since g(xy) = [[xy].sub.N] = [[x].sub.N][[y].sub.N] = g(x)g(y). Hence g is a homomorphism on X.

4) Let x [member of] ker(g), then g(x) = [[0].sub.N]. Since g(x) = [[x].sub.N], [[x].sub.N] = [[0].sub.N]. Thus x ~ 0. Since N is an ideal, x = x0 [member of] N. Since ker(g) [subset or equal to] H, x [member of] H. Thus x [member of] H [intersection] N. Hence ker(g) [subset or equal to] H [intersection] N. Let x [member of] H [intersection] N, then x [member of] H and x [member of] N. Since g(x) = [[x].sub.N] = [[0].sub.N], x [member of] ker(g). Thus H [intersection] N [subset or equal to] ker(g). Hence ker(g) = H [intersection] N. By Theorem 4.6 we have H/(H [intersection] N) [congruent to] HN/N.

Acknowledgements

The authors are highly grateful to the referees for their valuable comments and suggestions for the paper. Moreover, this work was supported by a grant for Kasetsart University.

References

[1] Dar, K. H. and M. Akrem, On endomorphisms of BCH-algebras, Annals of University of Craiova, Math. Comp. Sci. Ser., 33(2006), 227-234.

[2] Dudek, W. A. and X. Zhang, On ideal and congruences in BCC-algebra, Czechoslovak Math. Journal, 48(1998), 21-29.

[3] Hu, Q. P. and X. Li, On BCH-algebras, Math. Seminar Notes, 11(1983), 313-320.

[4] Iseki, K., An algebra related with a propositiomal calculus, M. J. A., 42(1966), 26-29.

S. Keawrahun ([dagger]) and U. Leerawa ([double dagger])

([dagger])[double dagger]) Department of Mathematics, Kasetsart University, Bangkok, Thailand E-mail: eangg-11@hotmail.com fsciutl@ku.ac.th

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Author: | Keawrahun, S.; Leerawat, U. |
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Publication: | Scientia Magna |

Date: | Jun 1, 2011 |

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