HOMOMORPHISM OF BF-ALGEBRAS 1. Introduction

DOI: 10.2478/s12175-013-0182-6 Math. Slovaca 64 (2014), No. 1, 13–20

HOMOMORPHISM OF BF-ALGEBRAS Joemar C. Endam — Jocelyn P. Vilela (Communicated by Jiˇ r´ı Rach˚ unek ) ABSTRACT. In this paper, we provide some properties of homomorphism of BF-algebras. We also prove the Second and Third Isomorphism Theorems for BF2 -algebras. c 2014 Mathematical Institute Slovak Academy of Sciences

1. Introduction The concept of BF-algebras was introduced by A. Walendziak [2] together with BF1 -algebras and BF2 -algebras. He defined a BF-algebra as an algebra (A; ∗, 0) of type (2, 0) (i.e., a nonempty set A with a binary operation ∗ and a constant 0) satisfying the following axioms: (B1) x ∗ x = 0, (B2) x ∗ 0 = x, (BF) 0 ∗ (x ∗ y) = y ∗ x. A BF1 -algebra is a BF-algebra satisfying (BG) x = (x ∗ y) ∗ (0 ∗ y) and a BF2 -algebra is a BF-algebra satisying (BH) x ∗ y = 0 and y ∗ x = 0 imply x = y. 2010 M a t h e m a t i c s S u b j e c t C l a s s i f i c a t i o n: Primary 08A05. K e y w o r d s: BF-algebras, ideal, normal ideal, BF-algebra homomorphism, BF-algebra Isomorphism Theorems. This research is funded by the Department of Science and Technology-Philippine Council for Advanced Science and Technology Research and Development (DOST-PCASTRD).

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JOEMAR C. ENDAM — JOCELYN P. VILELA

He also introduced the notions of a subalgebra, an ideal and a normal ideal in BF-algebras and established some of their properties. Let A = (A; ∗, 0) be a BF-algebra. A nonempty subset N of A is called a subalgebra of A if x ∗ y ∈ N for any x, y ∈ N . A subset I of A is called an ideal of A if it satisfies the following for all x, y ∈ A: (I1) 0 ∈ I, (I2) x ∗ y ∈ I and y ∈ I imply x ∈ I. We say that an ideal I is normal if for any x, y, z ∈ A, x ∗ y ∈ I implies (z ∗ x) ∗ (z ∗ y) ∈ I. In this paper, we provide some properties of homomorphism of BF-algebras. We also prove the Second and Third Isomorphism Theorems for BF2 -algebras.

2. Some properties of BF-homomorphism Let A = (A; ∗, 0A ) and B = (B; ∗, 0B ) be BF-algebras. A mapping ϕ : A → B is called a BF-homomorphism from A into B if ϕ(x ∗ y) = ϕ(x) ∗ ϕ(y) for any x, y ∈ A. A BF-homomorphism ϕ is called a monomorphism, epimorphism, or isomorphism if ϕ is one-to-one, onto, or a bijection, respectively. A BF-homomorphism ϕ : A → A is called an endomorphism and an isomorphism ϕ : A → A is called an automorphism.

2.1

Let A be a BF-algebra. i. If Nα : α ∈ A is any nonempty collection of subalgebras of A, then Nα is a subalgebra of A. α∈A

Iα ii. If Iα : α ∈ A is any nonempty collection of ideals of A, then α∈A

is an ideal of A. iii. If Iα : α ∈ A is any nonempty collection of normal ideals of A, then Iα is a normal ideal of A. α∈A

P r o o f. Let A be a BF-algebra and let x, y, z ∈ A. i. Follows from the fact that a nonempty intersection of a system of subalgebras is a subalgebra. 14


HOMOMORPHISM OF BF-ALGEBRAS

ii. Let Iα : α ∈ A be any nonempty collection of ideals of A. Since each Iα is an ideal, 0 ∈ Iα for all α ∈ A . Hence, 0 ∈ Iα and so Iα is not empty. If x ∗ y ∈

Iα and y ∈

α∈A

α∈A

α∈A

Iα , then x ∗ y ∈ Iα and y ∈ Iα for each

α∈A

α ∈ A . Since each Iα is an ideal, x ∈ Iα for all α ∈ A . Therefore, x ∈

and so

Iα

α∈A

Iα is an ideal of A.

α∈A

iii. Let Iα : α ∈ A be any nonempty collection of normal ideals of A. By (ii), Iα is an ideal of A. If x ∗ y ∈ Iα , then x ∗ y ∈ Iα for all α ∈ A . α∈A

α∈A

x) ∗ (z ∗ y) ∈ I Since each Iα is normal, (z ∗ α for all α ∈ A and for all z ∈ A. Therefore, (z ∗ x) ∗ (z ∗ y) ∈ Iα and so Iα is normal in A. α∈A

α∈A

Since a subalgebra is also a BF-algebra contained in a given BF-algebra, the following remark easily follows. Remark 2.2 If I is a normal ideal of a BF-algebra A, then I is normal for every subalgebra of A containing I. For any two BF-algebras A and B, there exists always at least one BF-homomorphism ϕ : A → B, namely, the trivial BF-homomorphism defined by ϕ(x) = 0B for all x ∈ A. In [1], a BF-algebra A is commutative if x∗(0∗y) = y ∗(0∗x) for any x, y ∈ A.

2.3 Let ϕ : A → B be a BF-homomorphism from A into B. If A is commutative, then ϕ(A) is commutative. If ϕ is onto, then A is commutative implies B is commutative.

2.4 If ϕ : A → B and ψ : B → C are BF-homomorphisms, then ψ ◦ ϕ : A → C is also a BF-homomorphism. Assuming compatibility of functions so that composition is defined, the following corollary easily follows.

2.5

The composition of monomorphisms is a monomorphism, the composition of epimorphisms is an epimorphism, the composition of isomorphisms is an isomorphism, and the composition of automorphisms is an automorphism. 15



Clearly, the identity mapping idA : A → A of a BF-algebra A is an automorphism of A. If ϕ : A → B is a BF-homomorphism from A into B, then for all x ∈ A, ϕ(0A ∗ x) = ϕ(0A ) ∗ ϕ(x) = 0B ∗ ϕ(x). Given a BF-homomorphism, the following lemma also holds where the first two of which are routine in the study of algebra.

2.6

Let ϕ : A → B be a BF-homomorphism from A into B.

i. If N is a subalgebra of A, then ϕ(N ) is a subalgebra of B. ii. If L is a subalgebra of B, then ϕ−1 (L) is a subalgebra of A containing ker ϕ. iii. If I is an ideal of A and ϕ is one-to-one, then ϕ(I) is an ideal of ϕ(A). iv. If J is an ideal of B, then ϕ−1 (J) is an ideal of A containing ker ϕ. v. If I is a normal ideal of A and ϕ is one-to-one, then ϕ(I) is a normal ideal of ϕ(A). vi. If J is a normal ideal of B, then ϕ−1 (J) is a normal ideal of A. P r o o f. Let ϕ : A → B be a BF-homomorphism from A into B. iii. Let I be an ideal of A and ϕ be one-to-one. Then 0A ∈ I and ϕ(0A ) = 0B imply 0B ∈ ϕ(I). Thus, ϕ(I) is not empty. Let x, y ∈ ϕ(A). Then there exist a, b ∈ A such that ϕ(a) = x and ϕ(b) = y. If x ∗ y ∈ ϕ(I) and y ∈ ϕ(I), then there exist m, n ∈ I such that ϕ(m) = x ∗ y and ϕ(n) = y. Thus, ϕ(a ∗ n) = ϕ(a) ∗ ϕ(n) = x ∗ y ∈ ϕ(I) and so a ∗ n ∈ I since ϕ is one-to-one. By (I2), a ∈ I since n ∈ I which implies that x = ϕ(a) ∈ ϕ(I). Therefore, ϕ(I) is an ideal of ϕ(A). iv. Let J be an ideal of B. If x ∈ ker ϕ, then ϕ(x) = 0B ∈ J implies x ∈ ϕ−1 (J). Thus, ker ϕ ⊆ ϕ−1 (J). If x ∗ y ∈ ϕ−1 (J) and y ∈ ϕ−1 (J), then ϕ(x) ∗ ϕ(y) = ϕ(x ∗ y) ∈ J and ϕ(y) ∈ J. By (I2), ϕ(x) ∈ J. Therefore, x ∈ ϕ−1 (J) and so ϕ−1 (J) is an ideal of A. v. Let I be a normal ideal of A and ϕ be one-to-one. By (iii), ϕ(I) is an ideal of ϕ(A). Let x, y, z ∈ ϕ(A). Then there exist a, b, c ∈ A such that ϕ(a) = x, ϕ(b) = y, and ϕ(c) = z. If x ∗ y ∈ ϕ(I), then ϕ(a ∗ b) = ϕ(a) ∗ ϕ(b) = x ∗ y ∈ ϕ(I) and so ϕ is one-to-one implies a∗b ∈ I. Since I is normal in A, (c∗a)∗(c∗b) ∈ I. Thus, (z ∗ x) ∗ (z ∗ y) = (ϕ(c) ∗ ϕ(a)) ∗ (ϕ(c) ∗ ϕ(b)) = ϕ(c ∗ a) ∗ ϕ(c ∗ b) = ϕ((c ∗ a) ∗ (c ∗ b)) ∈ ϕ(I). 16



Therefore, ϕ(I) is a normal ideal of ϕ(A). vi. Let J be a normal ideal of B. By (iv), ϕ−1 (J) is an ideal of B. Let x, y, z ∈ A. If x ∗ y ∈ ϕ−1 (J), then ϕ(x) ∗ ϕ(y) = ϕ(x ∗ y) ∈ J. Since J is a normal ideal of B and ϕ(z) ∈ B, ϕ((z ∗ x) ∗ (z ∗ y)) = ϕ(z ∗ x) ∗ ϕ(z ∗ y) = (ϕ(z) ∗ ϕ(x)) ∗ (ϕ(z) ∗ ϕ(y)) ∈ J. Therefore, (z ∗ x) ∗ (z ∗ y) ∈ ϕ−1 (J) and so ϕ−1 (J) is a normal ideal of A.

2.7 If ϕ : A → B is a BF-homomorphism, then ϕ(A) is a subalgebra of B and ker ϕ is a subalgebra of A.

Remark 2.8 If ϕ : A → B is a monomorphism from a BF-algebra A into a BF-algebra B, then normal ideals of A correspond to normal ideals of ϕ(A).

3. Isomorphism Theorems for BF2 -algebras In [2], if I is a normal ideal of a BF-algebra A = (A; ∗, 0), then A/I = (A/I; ∗ , 0/I) where A/I = {x/I : x ∈ A} and ∗ is defined by x/I ∗ y/I = (x ∗ y)/I. For x ∈ A, x/I is the congruence class containing x, that is, x/I = {y ∈ A : x ∼I y}, where x ∼I y if and only if x ∗ y ∈ I for any x, y ∈ A. The algebra A/I is called the quotient BF-algebra of A modulo I. The following theorem is labeled as the First Isomorphism Theorem for BF2 -algebras and can be found in [2].

3.1

Let A and B be BF2 -algebras and let ϕ : A → B be a homomorphism from A onto B. Then A/ ker ϕ is isomorphic to B. In [2], if A = (A; ∗, 0) is a BF-algebra, then (P1) 0 ∗ (0 ∗ x) = x holds for all x ∈ A. If I is a normal ideal of A, then I is a subalgebra of A. Let A be a BF-algebra and I, J be normal ideals of A. Define the subset IJ of A to be the set x ∈ A : x = i ∗ (0 ∗ j) for some i ∈ I, j ∈ J . Remark 3.2 If I and J are normal ideals of a BF-algebra A, then IJ need not be equal to JI. If A is commutative, then IJ = JI.

3.3

Let I and J be normal ideals of a BF-algebra A. Then

i. I ⊆ IJ, JI and J ⊆ IJ, JI, ii. J ⊆ I implies IJ = JI = I, 17



iii. (i ∗ (0 ∗ j))/J = i/J for all i ∈ I and for all j ∈ J, iv. (SI) (i1 ∗ j1 ) ∗ (i2 ∗ j2 ) = (i1 ∗ i2 ) ∗ (j1 ∗ j2 ) for any i1 , i2 ∈ I, j1 , j2 ∈ J implies IJ is a subalgebra of A. P r o o f. Let I and J be normal ideals of a BF-algebra A. i. If x ∈ I, then by (B2) and (B1), x = x ∗ 0 = x ∗ (0 ∗ 0) ∈ IJ and by (P1), x = 0 ∗ (0 ∗ x) ∈ JI. Thus, I ⊆ IJ, JI. Similarly, J ⊆ IJ, JI. ii. If J ⊆ I, then conclusion easily follows since I is a subalgebra. iii. If i ∈ I and j ∈ J, then i ∗ i = 0 ∈ J, that is, i ∼J i. By (B2), (0 ∗ j) ∗ 0 = 0 ∗ j ∈ J, that is, 0 ∗ j ∼J 0. Since ∼J is a congruence relation, i ∗ (0 ∗ j) ∼J i ∗ 0 = i and so i ∗ (0 ∗ j)/J = i/J. iv. Let (i1 ∗ j1 ) ∗ (i2 ∗ j2 ) = (i1 ∗ i2 ) ∗ (j1 ∗ j2 ) for any i1 , i2 ∈ I, j1 , j2 ∈ J. If x, y ∈ IJ, then x = i1 ∗ (0 ∗ j1 ) and y = i2 ∗ (0 ∗ j2 ) for some i1 , i2 ∈ I, j1 , j2 ∈ J. Moreover, x ∗ y = (i1 ∗ (0 ∗ j1 )) ∗ (i2 ∗ (0 ∗ j2 )) = (i1 ∗ i2 ) ∗ ((0 ∗ j1 ) ∗ (0 ∗ j2 )) = i3 ∗ (j3 ∗ j4 ) = i3 ∗ (0 ∗ (j4 ∗ j3 )) = i3 ∗ (0 ∗ j5 ) ∈ IJ. where i3 = i1 ∗ i2 , j3 = 0 ∗ j1 , j4 = 0 ∗ j2 , and j5 = j4 ∗ j3 . Therefore, IJ is a subalgebra of A. Let A be a BF2 -algebra and let I and J be normal ideals of A. Then by Lemma 2.1(iii) and Remark 2.2, I/(J ∩ I) is defined. If A satisfies (SI), then by Lemma 3.3(i) and (iv) and Remark 2.2, IJ/J is defined.

3.4 (Second Isomorphism Theorem for BF2 -algebras) If I and J are normal ideals of a BF2 -algebra A such that A satisfies (SI), then I/(J ∩ I) ∼ = IJ/J. P r o o f. Let I and J be normal ideals of a BF2 -algebra A. Define the function ϕ : I → IJ/J by ϕ(i) = i/J for all i ∈ I. Let i1 , i2 ∈ I. If i1 = i2 , then i1 ∗ i2 = i1 ∗ i1 = 0 ∈ J, that is, i1 ∼J i2 . Thus, i1 /J = i2 /J and so ϕ(i1 ) = i1 /J = i2 /J = ϕ(i2 ). This shows that ϕ is well-defined. Moreover, ϕ(i1 ∗ i2 ) = (i1 ∗ i2 )/J = i1 /J ∗ i2 /J = ϕ(i1 ) ∗ ϕ(i2 ). This shows that ϕ is a homomorphism. If x/J ∈ IJ/J, then x = i ∗ (0 ∗ j) for some i ∈ I, j ∈ J. By Lemma 3.3(iii), 18



x/J = (i ∗ (0 ∗ j))/J = i/J = ϕ(i). Therefore, ϕ is onto. By Theorem 3.1, I/ ker ϕ ∼ = IJ/J. Furthermore, ker ϕ = i ∈ I : ϕ(i) = (0 ∗ (0 ∗ 0))/J = 0/J = {i ∈ I : i/J = 0/J} = {i ∈ I : i ∼J 0} = {i ∈ I : i = i ∗ 0 ∈ J} = J ∩ I. Therefore, I/(J ∩ I) ∼ = IJ/J.

If I and J are normal ideals of a BF-algebra A such that I ⊆ J, then by Remark 2.2, J/I is defined.

3.5

If I and J are normal ideals of a BF-algebra A such that I ⊆ J, then J/I is a normal ideal of A/I. P r o o f. Let I and J be normal ideals of a BF-algebra A such that I ⊆ J. Then J/I ⊆ A/I. Now, 0/I ∈ J/I since 0 ∈ J. Thus, J/I is not empty. If x/I ∗ y/I ∈ J/I and y/I ∈ J/I, then (x ∗ y)/I = x/I ∗ y/I ∈ J/I and y ∈ J. Hence, x ∗ y ∈ J. By (I2), x ∈ J and so x/I ∈ J/I. This shows that J/I is an ideal of A/I. Let x/I, y/I, z/I ∈ A/I. If x/I ∗ y/I ∈ J/I, then (x ∗ y)/I = x/I ∗ y/I ∈ J/I implies that x ∗ y ∈ J. Since J is normal in A, (z∗x)∗(z∗y) ∈ J. Thus, ((z∗x)∗(z∗y))/I ∈ J/I and so (z/I∗ x/I)∗ (z/I∗ y/I) = (z ∗ x/I) ∗ (z ∗ y/I) = (z ∗ x) ∗ (z ∗ y)/I ∈ J/I. Therefore, J/I is normal in A/I. If I and J are normal ideals of a BF2 -algebra A such that I ⊆ J, then by Lemma 3.5, (A/I)/(J/I) is defined.

3.6 (Third Isomorphism Theorem for BF2 -algebras) If I and J are normal ideals of a BF2 -algebra A such that I ⊆ J, then (A/I)/(J/I) ∼ = A/J. P r o o f. Let I and J be normal ideals of a BF2 -algebra A such that I ⊆ J. Define the function ϕ : A/I → A/J by ϕ(x/I) = x/J for all x/I ∈ A/I. Let x/I, y/I ∈ A/I. If x/I = y/I, then x ∼I y, that is, x ∗ y ∈ I ⊆ J. Thus, x ∼J y and so x/J = y/J. Hence, ϕ(x/J) = x/J = y/J = ϕ(y/I). Therefore, ϕ is welldefined. Moreover, ϕ is a homomorphism since ϕ((x/I ∗ y/I)) = ϕ((x ∗ y)/I) = (x ∗ y)/J = x/J ∗ y/J = ϕ(x/I) ∗ ϕ(y/I). If x/J ∈ A/J, then x/I ∈ A/I 19



and ϕ(x/I) = x/J. Thus, ϕ is onto. By Theorem 3.1, (A/I)/ ker ϕ ∼ = A/J. Furthermore, ker ϕ = x/I ∈ A/I : ϕ(x/I) = 0/J = {x/I ∈ A/I : x/J = 0/J} = {x/I ∈ A/I : x ∼J 0} = {x/I ∈ A/I : x = x ∗ 0 ∈ J} = J/I. Therefore, (A/I)/(J/I) ∼ = A/J.

REFERENCES [1] ENDAM, J. C.—VILELA, J. P.: The Relationship of BF-algebras and groups, Mindanawan J. Math. 1 (2011), 69–74. [2] WALENDZIAK, A.: On BF-algebras, Math. Slovaca 57 (2007), 119–128.

Received 1. 2. 2012 Accepted 24. 2 2012

Department of Mathematics and Statistics College of Science and Mathematics MSU-Iligan Institute of Technology Tibanga, Iligan City PHILIPPINES E-mail : ayungon [email protected] [email protected]

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