Mappings of the Direct Product of B-algebras

Transcription

1 International Journal of Algebra, Vol. 10, 2016, no. 3, HIKARI Ltd, Mappings of the Direct Product of B-algebras Jacel Angeline V. Lingcong and Joemar C. Endam Department of Mathematics College of Arts and Sciences Negros Oriental State University Dumaguete City 6200, Philippines Copyright c 2016 Jacel Angeline V. Lingcong 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 two canonical mappings of the direct product of B-algebras and we obtain some of their properties. Mathematics Subject Classification: 06F35 Keywords: Direct product of B-algebras, canonical injections, canonical projections, B-homomorphism, normality 1 Introduction In [7], J. Neggers and H.S. Kim introduced the notion of B-algebras in A B-algebra A = (A;, 0) is an algebra 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, z A: (I) x x = 0, (II) x 0 = x, (III) (x y) z = x (z (0 y)). A B-algebra A is commutative if x (0 y) = y (0 x) for all x, y A. In [5], H.S. Kim and H.G. Park characterized commutativity of B-algebras. In [8], J. Neggers and H.S. Kim introduced the notions of subalgebras and normality in B-algebras, and established their properties. A nonempty subset N of A is called a subalgebra of A if x y N for any x, y N. By (I), 0 is always

2 134 Jacel Angeline V. Lingcong and Joemar C. Endam an element of a subalgebra. A nonempty subset N of A is called a normal subalgebra of A if (x a) (y b) N for any x y, a b N. A. Walendziak [10] characterized normality in B-algebras. J. Neggers and H.S. Kim used the concept of normality in B-algebras to construct quotient B-algebras. That is, given a normal subalgebra N of a B-algebra A, the relation N is defined by x N y if and only if x y N for any x, y A. Then N is a congruence relation of A. For x A, we write xn for the congruence class containing x, that is, xn = {y A : x N y}. Denote A/N = {xn : x A} and define on A/N by xn yn = (x y)n. Note that xn = yn if and only if x N y. The algebra A/N = (A/N;, N) is a B-algebra, and is called the quotient B- algebra of A modulo N. The concept of B-homomorphism was also introduced by J. Neggers and H.S. Kim. A map ϕ : A B is called a B-homomorphism 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 }. The ker ϕ is a normal subalgebra of A, and ker ϕ = {0 A } if and only if ϕ is one-one. A B-homomorphism ϕ is called a B-monomorphism, B-epimorphism, or B-isomorphism if ϕ is one-one, onto, or a bijection, respectively. In [4], J.A.V. Lingcong and J.C. Endam introduced and established the direct product of B-algebras. In this paper, we introduced and established two canonical mappings of the direct product of B-algebras. 2 Direct Product of B-algebras The results in this section are found in [4]. Example 2.1 The algebra (Z;, 0) is a B-algebra, where is defined by x y = x y for all x, y Z. Example 2.2 [7] Let A = {0, 1, 2, 3, 4, 5} be a set with the following table: Then (A;, 0) is a B-algebra Let A = (A;, 0 A ) and B = (B;, 0 B ) be B-algebras. Define the direct product of A and B to be the structure A B = (A B;, (0 A, 0 B )), where

3 Mappings of the direct product of B-algebras 135 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 (I), (II), and (III) of A B follow from those of A and B. Hence, the following theorem easily follows. Theorem 2.3 [4] The direct product of two B-algebras is also a B-algebra. Now, we extend this direct product to any finite family of B-algebras. Let I n = {1, 2,..., n} and let {A i = (A i ;, 0 i ): i I n } be a finite family of B-algebras. Define( the direct product of B-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. Corollary 2.4 [4] If {A i = (A i ;, 0 i ): i I n } is a family of B-algebras, then A i is a B-algebra. Theorem 2.5 [4] Let {ϕ i : A i B i : i I n } be a family of B-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 B-homomorphism with ker ϕ = ker ϕ i, ϕ( A i ) = ϕ i (A i ). Furthermore, ϕ is a B-monomorphism (respectively, B-epimorphism) if and only if each ϕ i is a B-monomorphism (respectively, B-epimorphism). Theorem 2.6 [4] Let {A i = (A i ;, 0 i ): i I n } be a family of B-algebras and let J i be a normal subalgebra of A i for each i I n. Then J i is a normal subalgebra of A i and A i / J i = n (A i /J i ).

4 136 Jacel Angeline V. Lingcong and Joemar C. Endam 3 Mappings of the Direct Product This section presents two canonical mappings of the direct product of any finite family of B-algebras and provides some of their properties. Theorem 3.1 Let {A i = (A i ;, 0 i ): i I n } be a family of B-algebras. Then f k : A i A k given by (a 1,..., a k,..., a n ) a k is a B-epimorphism of B-algebras for each k I n. Proof : For each k I n, define f k : for all (a 1,..., a k,..., a n ) be elements of A i A k by f k ((a 1,..., a k,..., a n )) = a k A i. Let (a 1,..., a k,..., a n ), (b 1,..., b k,..., b n ) A i. If (a 1,..., a k,..., a n ) = (b 1,..., b k,..., b n ), then a i = b i for each i I n. It follows that f k ((a 1,..., a k,..., a n )) = a k = b k = f k ((b 1,..., b k,..., b n )). Hence, f k is well-defined. If (a 1,..., a k,..., a n ), (b 1,..., b k,..., b n ) then f k ((a 1,..., a k,..., a n ) (b 1,..., b k,..., b n )) = f k ((a 1 b 1,..., a k b k,..., a n b n )) = a k b k = f k ((a 1,..., a k,..., a n )) f k ((b 1,..., b k,..., b n ). Thus, f k is a B-homomorphism. If c k A k, then (0 1,..., c k,..., 0 n ) A i, and f k ((0 1,..., c k,..., 0 n )) = c k. Therefore, f k is onto and so f k is a B- epimorphism. The maps f k in Theorem 3.1 are called the canonical projections of the direct product. The following theorem relates the direct product A i and its canonical projections. Theorem 3.2 Let {A i = (A i ;, 0 i ): i I n } be a family of B-algebras. Then there exists a B-algebra D, together with a family of B-homomorphisms A i

5 Mappings of the direct product of B-algebras 137 {f i : D A i : i I n } with the following property: for any B-algebra C and a family of B-homomorphisms {ϕ i : C A i : i I n }, there exists a unique B-homomorphism ϕ: C D such that f i ϕ = ϕ i for all i I n. Furthermore, D is uniquely determined up to B-isomorphism. Proof : Let {A i = (A i ;, 0 i ): i I n } be a family of B-algebras. Then a B-algebra by Corollary 2.4. Let D = A i is A i and let {f i : D A i : i I n } be the family of canonical projections. Suppose that C is any B-algebra and {ϕ i : C A i : i I n } a family of B-homomorphisms. Define ϕ : C D by ϕ(c) = (ϕ 1 (c),..., ϕ i (c),..., ϕ n (c)) for all c C. If c, d C, then ϕ(c d) = (ϕ 1 (c d),..., ϕ i (c d),..., ϕ n (c d)) = (ϕ 1 (c) ϕ 1 (d),..., ϕ i (c) ϕ i (d),..., ϕ n (c) ϕ n (d)) = (ϕ 1 (c),..., ϕ i (c),..., ϕ n (c)) (ϕ 1 (d),..., ϕ i (d),..., ϕ n (d)) = ϕ(c) ϕ(d). Hence, ϕ is a B-homomorphism. Moreover, f i ϕ = ϕ i for all i I n since (f i ϕ)(c) = f i (ϕ(c)) = f i ((ϕ 1 (c),..., ϕ i (c),..., ϕ n (c))) = ϕ i (c). To show that ϕ is unique, let ϕ : C D be another B-homomorphism such that f i ϕ = ϕ i for all i I n. If c C, then (f i ϕ)(c) = ϕ i (c) = (f i ϕ )(c). By the definition of ϕ, ϕ(c) = (ϕ 1 (c),..., ϕ i (c),..., ϕ n (c)) and assume that ϕ (c) = (a 1,..., a i,..., a n ). Thus, for each i I n, a i = f i ((a 1,..., a i,..., a n )) = f i (ϕ (c)) = (f i ϕ )(c) = (f i ϕ)(c) = f i (ϕ(c)) = f i ((ϕ 1 (c),..., ϕ i (c),..., ϕ n (c))) = ϕ i (c). Hence, ϕ(c) = (ϕ 1 (c),..., ϕ i (c),..., ϕ n (c)) = (a 1,..., a i,..., a n ) = ϕ (c). Therefore, ϕ is unique. Suppose that a B-algebra D has the same property as D with the family of B-homomorphisms {f i : D A i : i I n }. If we apply this property for D to the family of B-homomorphisms {f i : D A i : i I n } and also apply it for D to the family of B-homomorphisms {f i : D A i : i I n }, then we obtain unique B-homomorphisms α : D D and β : D D such that f i α = f i and f i β = f i for all i I n. Thus, α β : D D is a unique

6 138 Jacel Angeline V. Lingcong and Joemar C. Endam B-homomorphism such that f i (α β) = f i for all i I n. Since id D : D D is a B-homomorphism such that f i id D = f i for all i I n, α β = id D by uniqueness. A similar argument shows that β α = id D. Therefore, β is an B-isomorphism, that is, D is uniquely determined up to B-isomorphism. Theorem 3.3 Let {A i = (A i ;, 0 i ): i I n } be a family of B-algebras. Then g k : A k A i given by a k (0 1,..., 0 k 1, a k, 0 k+1,..., 0 n ) is a B- monomorphism of B-algebras for each k I n. Proof : Let {A i = (A i ;, 0 i ) : i I n } be a family of B-algebras. For each k I n, define g k : A k A i by g k (a k ) = (0 1,..., 0 k 1, a k, 0 k+1,..., 0 n ) for all a k A k. Let a k, b k A k. If a k = b k, then g k (a k ) = (0 1,..., 0 k 1, a k, 0 k+1,..., 0 n ) = (0 1,..., 0 k 1, b k, 0 k+1,..., 0 n ) = g k (b k ). Hence, g k is well-defined. If a k, b k A k, then g k (a k b k ) = (0 1,..., 0 k 1, a k b k, 0 k+1,..., 0 n ) = ( ,..., 0 k 1 0 k 1, a k b k, 0 k+1 0 k+1,..., 0 n 0 n ) = (0 1,..., 0 k 1, a k, 0 k+1,..., 0 n ) (0 1,..., 0 k 1, b k, 0 k+1,..., 0 n ) = g k (a k ) g k (b k ). Therefore, g k is a B-homomorphism. If g k (a k ) = g k (b k ), then (0 1,..., 0 k 1, a k, 0 k+1,..., 0 n ) = g k (a k ) = g k (b k ) = (0 1,..., 0 k 1, b k, 0 k+1,..., 0 n ). Hence, a k = b k. Thus, g k is one-to-one and so g k is a B-monomorphism. The maps g k in Theorem 3.3 are called the canonical injections. Theorem 3.4 Let {A i = (A i ;, 0 i ): i I n } be a family of B-algebras. For each k I n, if g k is the canonical injection, then g k (A k ) is a normal subalgebra of A i and A i. A i / g k (A k ) = i k

7 Mappings of the direct product of B-algebras 139 Proof : Note that g k (A k ) = {(0 1,..., 0 k 1, a k, 0 k+1,..., 0 n ) : a k A k }. It is easy to see that g k (A k ) is a normal subalgebra of A i. Define ϕ k : A i / g k (A k ) i k A i given by ϕ k ((a 1,..., a n )g k (A k )) = (a 1,..., a k 1, a k+1,..., a n ) for all (a 1,..., a n )g k (A k ) A i / g k (A k ). Let (a 1,..., a k,..., a n )g k (A k ), (b 1,..., b k,..., b n )g k (A k ) A i / g k (A k ). Suppose that (a 1,..., a k,..., a n )g k (A k ) = (b 1,..., b k,..., b n )g k (A k ). Then (a 1,..., a k,..., a n ) gk (A k ) (b 1,..., b k,..., b n ), that is, (a 1 b 1,..., a k b k,..., a n b n ) = (a 1,..., a k,..., a n ) (b 1,..., b k,..., b n ) g k (A k ) so that a i b i = 0 i for all i k. Hence, a i = b i for all i k and so ϕ k ((a 1,..., a k,..., a n )g k (A k )) = (a 1,..., a k 1, a k+1,..., a n ) = (b 1,..., b k 1, b k+1,..., b n ) = ϕ k ((b 1,..., b k,..., b n )g k (A k )). This shows that ϕ k is well-defined. Moreover, ϕ k ((a 1,..., a k,..., a n )g k (A k ) (b 1,..., b k,..., b n )g k (A k )) = ϕ k (((a 1,..., a k,..., a n ) (b 1,..., b k,..., b n ))g k (A k )) = ϕ k ((a 1 b 1,..., a k b k,..., a n b n )g k (A k )) = (a 1 b 1,..., a k 1 b k 1, a k+1 b k+1,..., a n b n ) = (a 1,..., a k 1, a k+1,..., a n ) (b 1,..., b k 1, b k+1,..., b n ) = ϕ k ((a 1,..., a k,..., a n )g k (A k )) ϕ k ((b 1,..., b k,..., b n )g k (A k )). This shows that ϕ k is a B-homomorphism. If ϕ k ((a 1,..., a k,..., a n )g k (A k )) = ϕ k ((b 1,..., b k,..., b n )g k (A k )), then (a 1,..., a k 1, a k+1,..., a n ) = ϕ k ((a 1,..., a k,..., a n )g k (A k )) = ϕ k ((b 1,..., b k,..., b n )g k (A k )) = (b 1,..., b k 1, b k+1,..., b n ). Thus, a i = b i for all i k so that a i b i = 0 i for all i k. Hence, (a 1,..., a k,..., a n ) (b 1,..., b k,..., b n ) = (a 1 b 1,..., a k b k,..., a n b n ) is an element of g k (A k ), that is, (a 1,..., a k,..., a n ) gk (A k ) (b 1,..., b k,..., b n ) so that (a 1,..., a k,..., a n )g k (A k ) = (b 1,..., b k,..., b n )g k (A k ). This shows that ϕ k is one-to-one.

8 140 Jacel Angeline V. Lingcong and Joemar C. Endam If (a 1,..., a k 1, a k+1,..., a n ) i k A i, then a i A i for all i k so that (a 1,..., a k 1, 0 k, a k+1,..., a n ) A i since 0 k A k. It follows that (a 1,..., a k 1, a k+1,..., a n ) = ϕ k ((a 1,..., a k 1, 0 k, a k+1,..., a n )g k (A k )), where (a 1,..., a k 1, 0 k, a k+1,..., a n )g k (A k ) A i / g k (A k ). Hence, ϕ k is onto. Therefore, ϕ k is an B-isomorphism, that is, References A i / g k (A k ) = i k [1] P.J. Allen, J. Neggers and H.S. Kim, B-algebras and groups, Scientiae Mathematicae Japonicae Online, 9 (2003), [2] J.R. Cho and H.S. Kim, On B-algebras and quasigroups, Quasigroups and Related Systems, 8 (2001), 1-6. [3] J.C. Endam and J.P. Vilela, The Second Isomorphism Theorem for B- algebras, Applied Mathematical Sciences, 8 (2014), no. 38, [4] J.A.V. Lingcong and J.C. Endam, Direct Product of B-algebras, International Journal of Algebra, 10 (2016), no. 1, [5] H.S. Kim and H.G. Park, On 0-commutative B-algebras, Scientiae Mathematicae Japonicae Online, (2005), [6] M. Kondo and Y.B. Jun, The Class of B-algebras Coincides with the Class of Groups, Scientiae Mathematicae Japonicae Online, 7 (2002), [7] J. Neggers and H.S. Kim, On B-algebras, Mat. Vesnik, 54 (2002), [8] J. Neggers and H.S. Kim, A fundamental theorem of B-homomorphism for B-algebras, Int. Math. J., 2 (2002), no. 3, [9] A. Walendziak, Some Axiomatizations of B-algebras, Math. Slovaca, 56 (2006), no. 3, [10] A. Walendziak, A note on normal subalgebras in B-algebras, Scientiae Mathematicae Japonicae Online, (2005), Received: February 3, 2016; Published: April 12, 2016 A i.