Direct Product of BF-Algebras

Transcription

1 International Journal of Algebra, Vol. 10, 2016, no. 3, HIKARI Ltd, Direct Product of BF-Algebras Randy C. Teves and Joemar C. Endam Department of Mathematics College of Arts and Sciences Negros Oriental State University Dumaguete City 6200, Philippines Copyright c 2016 Randy C. Teves and Joemar C. Endam. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this paper, we introduce the direct product of BF-algebras and we obtain some properties of this concept. Mathematics Subject Classification: 06F35 Keywords: BF-algebras, direct product of BF-algebras, normal ideal, BFhomomorphism 1 Introduction In 2007, the concept of BF-algebras was introduced by A. Walendziak [3]. A BF-algebra is an algebra A = (A;, 0) of type (2, 0), that is, a nonempty set A together with a binary operation and a constant 0, satisfying the following axioms for all x, y A: (B1) x x = 0, (B2) x 0 = x, (BF) 0 (x y) = y x. In [3], Walendziak also introduced the notion of commutativity of BF-algebras. A BF-algebra A is commutative if x (0 y) = y (0 x) for all x, y A. In 2011, J.C. Endam and J.P. Vilela [2] characterized the commutativity of BF-algebras

2 126 Randy C. Teves and Joemar C. Endam and established the relationship of BF-algebras and groups. Walendziak also introduced the notions of subalgebras, ideals, and normality in BF-algebras, and established their properties. 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). A nonzero ideal I of A is said to be proper if I A. A nonempty subset N of A is called a subalgebra of A if x y N for any x, y N. It is known that an ideal need not be a subalgebra, and a subalgebra need not be an ideal. While a normal ideal is a subalgebra. Walendziak then used the concept of normality of BF-algebras to construct quotient BF-algebras. That is, given a normal ideal I of a BF-algebra A, the relation I is defined by x I y if and only if x y I for any x, y A. Then I is a congruence relation of A. For x A, write x/i for the congruence class containing x, that is, x/i = {y A: x I y}. We denote A/I = {x/i: x A} and define by x/i y/i = (x y)/i. Note that x/i = y/i if and only if x I y. Then the algebra A/I = (A/I;, 0/I) is a BF-algebra. The algebra A/I is called the quotient BF-algebra of A modulo I. The concept of BF-homomorphism was also introduced by A. Walendziak. A map ϕ : A B is called a BFhomomorphism if ϕ(x y) = ϕ(x) ϕ(y) for any x, y A. The kernel of ϕ, denoted by ker ϕ, is defined to be the set {x A : ϕ(x) = 0 B }. A BF-homomorphism ϕ is called a BF-monomorphism, BF-epimorphism, or BFisomorphism if ϕ is one-one, onto, or a bijection, respectively. In [3], A. Walendziak established the first isomorphism theorem for BF-algebras. In [1], J.C. Endam and J.P. Vilela established the second and third isomorphism theorems for BF-algebras. In this paper, we introduced the direct product of BF-algebras and established some of its properties. 2 Direct Product of BF-algebras We begin with some examples of BF-algebras. Example 2.1 [3] Let R be the set of real numbers and let A = (R;, 0) be the algebra with the operation defined by x if y = 0, x y = y if x = 0, 0 otherwise. Then A is a BF-algebra.

3 Direct product of BF-algebras 127 Example 2.2 [3] Let A = [0, ) = {x R : x 0}. Define the binary operation on A as follows: x y = x y for all x, y A. Then (A;, 0) is a BF-algebra. Example 2.3 [3] Let A = {0, 1, 2, 3} and be defined by the following table: Then (A;, 0) is a BF-algebra Let A = (A;, 0 A ) and B = (B;, 0 B ) be BF-algebras. Define the direct product of A and B to be the structure A B = (A B;, (0 A, 0 B )), where A B is the set {(a, b) : a A and b B} and whose binary operation is given by (a 1, b 1 ) (a 2, b 2 ) = (a 1 a 2, b 1 b 2 ). Note that the binary operation is componentwise. Thus, the properties (B1), (B2), and (BF) of A B follow from those of A and B. Hence, the following theorem easily follows. Theorem 2.4 The direct product of two BF-algebras is also a BF-algebra. Now, we extend this direct product to any finite family of BF-algebras. Let I n = {1, 2,..., n} and let {A i = (A i ;, 0 i ) : i I n } be a finite family of BF-algebras. Define( the direct product of BF-algebras A 1,..., A n to be the n ) structure A i = A i ;, (0 1,..., 0 n ), where A i = A 1 A n = {(a 1,..., a n ) : a i A i, i I n } and whose operation is given by (a 1,..., a n ) (b 1,..., b n ) = (a 1 b 1,..., a n b n ). Obviously, is a binary operation on A i. Remark 2.5 If {A i = (A i ;, 0 i ): i I n } is a family of BF-algebras, then A i is a BF-algebra.

4 128 Randy C. Teves and Joemar C. Endam Lemma 2.6 Let {A i = (A i ;, 0 i ): i I n } be a family of BF-algebras. Then each A i is commutative if and only if A i is commutative. Proof : Let each A i be commutative. If (a 1,..., a n ), (b 1,..., b n ) A i, then a i, b i A i and a i (0 i b i ) = b i (0 i a i ) for all i I n. Thus, (a 1,..., a n ) ((0 1,..., 0 n ) (b 1,..., b n )) = (a 1,..., a n ) (0 1 b 1,..., 0 n b n ) = (a 1 (0 1 b 1 ),..., a n (0 n b n )) = (b 1 (0 1 a 1 ),..., b n (0 n a n )) = (b 1,..., b n ) (0 1 a 1,..., 0 n a n ) = (b 1,..., b n ) ((0 1,..., 0 n ) (a 1,..., a n )). Therefore, A i is commutative. Conversely, let A i be commutative. If a i, b i A i for all i I n, then (a 1,..., a n ), (b 1,..., b n ) A i and (a 1,..., a n ) ((0 1,..., 0 n ) (b 1,..., b n )) = (b 1,..., b n ) ((0 1,..., 0 n ) (a 1,..., a n )). Thus, (a 1 (0 1 b 1 ),..., a n (0 n b n )) = (a 1,..., a n ) (0 1 b 1,..., 0 n b n ) = (a 1,..., a n ) ((0 1,..., 0 n ) (b 1,..., b n )) = (b 1,..., b n ) ((0 1,..., 0 n ) (a 1,..., a n )) = (b 1,..., b n ) (0 1 a 1,..., 0 n a n ) = (b 1 (0 1 a 1 ),..., b n (0 n a n )). This implies that a i (0 b i ) = b i (0 a i ) for all i I n. Therefore, each A i is commutative. Theorem 2.7 Let {ϕ i : A i B i : i I n } be a family of BF-homomorphisms. If ϕ is the map A i B i given by (a 1,..., a n ) (ϕ 1 (a 1 ),..., ϕ n (a n )), then ϕ is a BF-homomorphism with ker ϕ = ker ϕ i, ϕ( A i ) = ϕ i (A i ). Furthermore, ϕ is a BF-monomorphism (respectively, BF-epimorphism) if and only if ϕ i is.

5 Direct product of BF-algebras 129 Proof : Let {ϕ i : A i B i : i I n } be a family of BF-homomorphisms and let ϕ be the map A i B i given by (a 1,..., a n ) (ϕ 1 (a 1 ),..., ϕ n (a n )). If (a 1,..., a n ), (b 1,..., b n ) A i, then ϕ((a 1,..., a n ) (b 1,..., b n )) = ϕ((a 1 b 1,..., a n b n )) = (ϕ 1 (a 1 b 1 ),..., ϕ n (a n b n )) = (ϕ 1 (a 1 ) ϕ 1 (b 1 ),..., ϕ n (a n ) ϕ n (b n )) = (ϕ 1 (a 1 ),..., ϕ n (a n )) (ϕ 1 (b 1 ),..., ϕ n (b n )) = ϕ((a 1,..., a n )) ϕ((b 1,..., b n )). This shows that ϕ is a BF-homomorphism. Moreover, Thus, ker ϕ = (a 1,..., a n ) ker ϕ ϕ((a 1,..., a n )) = (0 1,..., 0 n ) (ϕ 1 (a 1 ),..., ϕ n (a n )) = (0 1,..., 0 n ) ker ϕ i. Let A = ϕ i (a i ) = 0 i for each i I n a i ker ϕ i for each i I n (a 1,..., a n ) ker ϕ i. A i. Then (b 1,..., b n ) ϕ(a) (a 1,..., a n ) A (b 1,..., b n ) = ϕ((a 1,..., a n )) (a 1,..., a n ) A (b 1,..., b n ) = (ϕ 1 (a 1 ),..., ϕ n (a n )) a i A i b i = ϕ i (a i ) ϕ(a i ) for each i I n (b 1,..., b n ) ϕ i (A i ). Thus, ϕ( A i ) = ϕ i (A i ). To prove the last statement, let ϕ be one-to-one. If ϕ i (a i ) = ϕ(b i ) for each i I n, then ϕ((a 1,..., a n )) = (ϕ 1 (a 1 ),..., ϕ n (a n )) = (ϕ 1 (b 1 ),..., ϕ n (b n )) = ϕ((b 1,..., b n )). Since ϕ is one-to-one, (a 1,..., a n ) = (b 1,..., b n ), that is, a i = b i for each i I n. Therefore, ϕ i is one-to-one for each i I n. Conversely, let ϕ i be

6 130 Randy C. Teves and Joemar C. Endam one-to-one for each i I n. If ϕ((a 1,..., a n )) = ϕ((b 1,..., b n )), then (ϕ 1 (a 1 ),..., ϕ n (a n )) = ϕ((a 1,..., a n )) = ϕ((b 1,..., b n )) = (ϕ 1 (b 1 ),..., ϕ n (b n )). Thus, ϕ i (a i ) = ϕ i (b i ) for each i I n. Since each ϕ i is one-to-one, a i = b i for each i I n and so (a 1,..., a n ) = (b 1,..., b n ). Therefore, ϕ is one-to-one. Finally, we show that ϕ is onto if and only if each ϕ i is. Let ϕ be onto. If b i B i for each i I n, then (b 1,..., b n ) B i. Since ϕ is onto, there exists (a 1,..., a n ) A i such that (b 1,..., b n ) = ϕ((a 1,..., a n )) = (ϕ 1 (a 1 ),..., ϕ n (a n )), that is, b i = ϕ i (a i ) for each i I n. Therefore, ϕ i is onto for each i I n. Conversely, let ϕ i be onto for each i I n. If (b 1,..., b n ) B i, then b i B i for each i I n. Since each ϕ i is onto, there exists a i A i such that b i = ϕ i (a i ) for each i I n so that (b 1,..., b n ) = (ϕ 1 (a 1 ),..., ϕ n (a n )) = ϕ((a 1,..., a n )). Therefore, ϕ is onto and so the theorem is finally proved. Remark 2.8 Let {A i = (A i ;, 0 i ): i I n }, {B i = (B i ;, 0 i ): i I n } be any two families of BF-algebras such that A i = Bi for each i I n. Then n A i = B i. Theorem 2.9 Let {A i = (A i ;, 0 i ): i I n } be a family of BF-algebras and let J i be a normal ideal of A i for each i I n. Then J i is a normal ideal of A i and A i / J i = n (A i /J i ). Proof : Let {A i = (A i ;, 0 i ): i I n } be a family of BF-algebras and let J i be a normal ideal of A i for each i I n. Then (0 1,..., 0 n ) J i since 0 i J i for each i I n and so J i is not empty. Let (a 1,..., a n ), (b 1,..., b n ) A i.

7 Direct product of BF-algebras 131 If (b 1,..., b n ), (a 1,..., a n ) (b 1,..., b n ) I n and (a 1 b 1,..., a n b n ) = (a 1,..., a n ) (b 1,..., b n ) J i, then b i J i for each i J i implies that a i b i J i for each i I n. By (I2), a i J i for each i I n and so (a 1,..., a n ) is an element of J i. Thus, J i is an ideal of A i. Let (c 1,..., c n ) A i. If (a 1,..., a n ) (b 1,..., b n ) (a 1 b 1,..., a n b n ) = (a 1,..., a n ) (b 1,..., b n ) J i, then J i so that a i b i J i for each i I n and so (c i a i ) (c i b i ) J i for each i I n. Moreover, ((c 1,..., c n ) (a 1,..., a n )) ((c 1,..., c n ) (b 1,..., b n )) = (c 1 a 1,..., c n a n ) (c 1 b 1,..., c n b n ) = ((c 1 a 1 ) (c 1 b 1 ),..., (c n a n ) (c n b n )) J i. Therefore, J i is a normal ideal of For simplicity, let J = A i. J i and A = A i. Define ϕ: A/J (A i /J i ) given by ϕ((a 1,..., a n )/J) = (a 1 /J 1,..., a n /J n ) for all (a 1,..., a n )/J A/J. Let (a 1,..., a n )/J, (b 1,..., b n )/J A/J. If (a 1,..., a n )/J = (b 1,..., b n )/J, then (a 1,..., a n ) J (b 1,..., b n ), that is, (a 1 b 1,..., a n b n ) = (a 1,..., a n ) (b 1,..., b n ) J. Thus, a i b i J i for all i I n, that is, a i Ji b i so that a i /J i = b i /J i. It follows that ϕ((a 1,..., a n )/J) = (a 1 /J 1,..., a n /J n ) = (b 1 /J 1,..., b n /J n ) = ϕ((b 1,..., b n )/J). This shows that ϕ is well-defined. If (a 1,..., a n )/J, (b 1,..., b n )/J A/J, then ϕ((a 1,..., a n )/J (b 1,..., b n )/J) = ϕ(((a 1,..., a n ) (b 1,..., b n ))/J) = ϕ((a 1 b 1,..., a n b n )/J) = ((a 1 b 1 )/J 1,..., (a n b n )/J n ) = (a 1 /J 1 b 1 /J 1,..., a n /J n b n /J n ) = (a 1 /J 1,..., a n /J n ) (b 1 /J 1,..., b n /J n ) = ϕ((a 1,..., a n )/J) ϕ((b 1,..., b n )/J).

8 132 Randy C. Teves and Joemar C. Endam This shows that ϕ is a BF-homomorphism. If ϕ((a 1,..., a n )/J) = ϕ((b 1,..., b n )/J), then (a 1 /J 1,..., a n /J n ) = ϕ((a 1,..., a n )/J) = ϕ((b 1,..., b n )/J) = (b 1 /J 1,..., b n /J n ). Thus, a i /J i = b i /J i for all i I n. Hence, a i Ji b i, that is, a i b i J i for all i I n so that (a 1,..., a n ) (b 1,..., b n ) = (a 1 b 1,..., a n b n ) J. Thus, (a 1,..., a n ) J (b 1,..., b n ) and so (a 1,..., a n )/J = (b 1,..., b n )/J. This shows that ϕ is one-to-one. If (a 1 /J 1,..., a n /J n ) (A i /J i ), then a i A i for all i I n, that is, (a 1,..., a n ) A. It follows that (a 1 /J 1,..., a n /J n ) = ϕ((a 1,..., a n )/J), where (a 1,..., a n )/J A/J. This shows that ϕ is onto. Therefore, ϕ is a BFisomorphism, that is, A i / J i n = (A i /J i ). References [1] J.C. Endam and J.P. Vilela, Homomorphism of BF-algebras, Mathematica Slovaca, 64 (2014), no. 1, [2] J.C. Endam and J.P. Vilela, The Relationship of BF-algebras and groups, Mindanawan Journal of Mathematics, 1 (2011), [3] A. Walendziak, On BF-algebras, Mathematica Slovaca, 57 (2007), Received: February 3, 2016; Published: April 12, 2016