In this paper we construct quadratic β-algebras on a field, and we discuss both linear-quadratic β-algebras and quadratic-linear β-algebras in a field. Moreover, we discuss some relations of binary operations in β-algebras.
Y. Imai and K. Iséki introduced two classes of abstract algebras: BCK-algebras and BCI-algebras([3, 4]). We refer useful textbooks for BCK/BCI-algebra to [2, 7, 10]. J. Neggers and H. S. Kim([8]) introduced another class related to some of the previous ones, viz., B-algebras and studied some of its properties. They also introduced the notion of β-algebra([9]) where two operations are coupled in such a way as to reflect the natural coupling which exists between the usual group operation and its associated B-algebra which is naturally defined by it. P. J. Allen et al.([1]) gave another proof of the close relationship of B-algebras with groups using the observation that the zero adjoint mapping is surjective. H. S. Kim and H. G. Park([5]) showed that if X is a 0-commutative B-algebra, then (x * a) * (y * b) = (b * a) * (y * x). Using this property they showed that the class of p-semisimple BCI-algebras is equivalent to the class of 0-commutative B-algebras. Y. H. Kim and K. S. So ([6]) investigated some properties of β-algebras and further relations with B-algebras. Especially, they showed that if (X,−,+, 0) is a B*-algebra, then (X, +) is a semigroup with identity 0. They discussed some constructions of linear β-algebras in a field K.
In this paper we construct quadratic β-algebras on a field, and we discuss both linear-quadratic β-algebras and quadratic-linear β-algebras in a field. Moreover, we discuss some relations of binary operations in β-algebras.
Let (K,+, ·, e) be a field (sufficiently large) and let x, y ∈ K. Then (K, ⊕, ⊖, e) is a β-algebra, where x ⊕ y = x + y − e and x ⊖ y = x − y + e for any x, y ∈ K.
We call such a β-algebra described in Theorem 2.2 a linear β-algebra.
If we let ϕ : K → K be a map defined by ϕ(x) = e + bx for some b ∈ K. Then we have
ϕ(x+y)=e+b(x+y)=(e+bx)+(e+by)-e=ϕ(x)⊕ϕ(y)
and
ϕ(x-y)=e+b(x-y)=(e+bx)-(e+by)+e=ϕ(x)⊖ϕ(y),
so that ϕ(0) = e implies ϕ : (K,−,+, 0) → (K, ⊖, ⊕, e) is a homomorphism of β-algebras, where “−” is the usual subtraction in the field K. If b ≠ 0, then ψ : (K, ⊖, ⊕, e) → (K,−,+, 0) defined by ψ(x) := (x − e)/b is a homomorphism of β-algebras and the inverse mapping of the mapping ϕ, so that (K, ⊖, ⊕, e) and (K,−,+, 0) are isomorphic as β-algebras, i.e., there is only one isomorphism type in this case. We summarize:
Let (K,+, ·, e) be a field (sufficiently large)and let x, y ∈ K. First, we consider the case of quadratic-linear β-algebras, i.e., x ⊕ y is the polynomial of x, y with degree 2, and x ⊖ y is the polynomial of x, y with degree 1. Define two binary operations “⊕, ⊖” on K as follows:
x⊕y:=A+Bx+Cy+Dx2+Exy+Fy2,x⊖y:=α+βx+γy
where α, β, γ, A, B, C, D, E, F ∈ K (fixed). Assume that (K, ⊕, ⊖, 0) is a β-algebra. It is necessary to find proper solutions for two equations. Since x = x ⊖ e = α + βx + γe, we obtain (β − 1)x + (α + γe) = 0, and hence β = 1 and α = −γe. It follows that
x⊖y=x+γ(y-e)
From (3.1) we obtain e⊖x = e+γ(x−e) = (1 − γ)e+γx. Since (e⊖x)⊕x = e, we obtain
By (III), we obtain x + γ(y + z − 2e) = x + γ(z ⊕ y − e). Since γ ≠ 0, we have z ⊕ y − e = z + y − 2e. This shows that x ⊕ y = x + y − e for all x, y ∈ K. In this case, we obtain the linear case as described in Theorem 2.2.
Case (ii). γ = 0. By (3.1), we obtain x ⊖ y = x. Since |K| ≥ 3, the formula (3.2) can be represented as
F=0,C+Ee=0,A+Be+De2=e
It follows that F = 0, C = −Ee, A = e − Be − De2. This shows that
It is a quadratic form (not a linear form) and so (K, ⊕, ⊖, e) is a quadratic-linear β-algebra where x ⊕ y = e + (a + bx + cy)(x − e) and x ⊖ y = x. We summarize:
Theorem 3.1
Let (K,+, ·, e) be a field (sufficiently large). If we define two binary operations “⊕, ⊖” on K by
x⊕y:=e+(a+bx+cy)(x-e)x⊖y:=x
for all x, y ∈ K, then (K,⊕, ⊖, e) is a quadratic-linear β-algebra.
Corollary 3.2
Let (K,+, ·, 0) be a field (sufficiently large) and let a, b, c ∈ K. If we define two binary operations “⊕, ⊖” on K by
x⊕y:=ax+bx2+cxyx⊖y:=x
for all x, y ∈ K, then (K,⊕, ⊖, 0) is a quadratic-linear β-algebra.
Proof
It follows immediately from Theorem 3.1 by letting e := 0.
Example 3.3
Let R be the set of all real numbers. If we define x⊕y := x−x2−2xy and x ⊖ y := x for all x, y ∈ R, then (R,⊕, ⊖, 0) is a quadratic-linear β-algebra.
Let (K,+, ·, e) be a field (sufficiently large) and let x, y ∈ K. Next, we consider the case of linear-quadratic β-algebras, i.e., x ⊕ y is the polynomial of x, y with degree 1, and x ⊖ y is the polynomial of x, y with degree 2.
Theorem 3.4
There is no linear-quadratic β-algebras over a field (K,+,−, e).
Proof
Let (K,+, ·, e) be a field (sufficiently large). Define two binary operations “⊕, ⊖” on K as follows:
x⊕y:=A+Bx+Cy,x⊖y:=α+βx+γy+δx2+ɛxy+ξy2
where A, B, C, α, β, γ, δ, ɛ, ξ ∈ K (fixed). Assume that (K, ⊕, ⊖, e) is a β-algebra and |K| ≥ 3. Then
x=x⊖e=α+βx+γe+δx2+ɛex+ξe2=[α+γe+ξe2]+[β+ɛe]x+δx2
It follows that
α+γe+ξe2=0β+ɛe=1δ=0
Case (i). e = 0. By formula (3.3) we obtain α = 0, β = 1, δ = 0. It follows that
x⊖y=x+γy+ɛxy+ξy2
It follows that 0 ⊖ x = 0+γx + ɛ0x + ξx2 = γx + ξx2 and hence
for all x ∈ K. This shows that A = 0, Bγ + C = 0, Bξ = 0.
Subcase (i-1). B = 0. Since C = −Bγ, we obtain C = 0 and hence x ⊕ y = 0 for all x, y ∈ K. This shows that (0 ⊖ x) ⊕ x = 0. Using formula (3.4) we obtain
By (III), we obtain γ = 0, ɛ = 0, ξ = 0, proving that x⊖y = x. Hence x⊕y = 0 and x ⊖ y = x show that (K, ⊕, ⊖, 0) is not a linear-quadratic β-algebra.
Subcase (i-2). ξ = 0. We apply this condition to the formula (3.4), and obtain x ⊖ y = x + γy + ɛxy. It follows that 0 ⊖ x = γx and hence 0 = (0 ⊖ x) ⊕ x = γx ⊕ x = A + B(γx) + Cx = A + (Bγ + C)x for all x ∈ K. It follows that A = 0, Bγ + C = 0. Hence x ⊕ y = B(x − γy). It follows that
By (III), we obtain ɛ = 0 and γ(1 + γ)B = 0. If γ = 0, then x ⊕ y = Bx and x ⊖ y = x. If B = 0, then it is the same case as subcase (i-1). If γ = −1, then x ⊕ y = B(x + y) and x ⊖ y = x − y. This shows that (K, ⊕, ⊖, 0) is not a linear-quadratic β-algebra.
Case (ii). e ≠ 0. It follows from (3.3) that α = −γe − ξe2, β = 1− ɛe, δ = 0. Hence
x⊖y=(γe-ξe2)+(1-ɛe)x+γy+ɛxy+ξy2
Subcase (ii-1). ξ ≠ 0. The formula (3.5) can be written as
x⊖y=x+(γ+ɛx)(y-e)+ξ(y2-e2)
If y := e in (3.6), then x ⊖ e = x + (γ + ɛx)(e − e) + ξ(e2 − e2) = x. If x := e and y := x in (3.6), then e ⊖ x = x + (γ + ɛe)(x − e) + ξ(x2 − e2). It follows that
Since ξ ≠ 0, we have B = 0 and hence C = 0, A = 0, i.e., x ⊕ y = 0 and x⊖y = x+(γ+ɛx)(y−e)+ξ(y2−e2) does not form a β-algebra, since e⊖(e⊕e) = e − e(γ + ɛe) − ξe2 and (e ⊖ e) ⊖ e = e.
Subcase (ii-2). ξ = 0. The formula can be written as
x⊖y=x+(γ+ɛx)(y-e)
It follows that x ⊖ e = x + (γ + ɛx)(e − e) = x and e ⊖ x = e + (γ + ɛe)(x − e). By (II) we obtain the following.
Subcase (ii-2-a). γ = ε = 0. We obtain β = 1, α = 0 and hence x ⊖ y = x, x ⊕ y = (e − Be) + Bx, a linear β-algebra.
Subcase (ii-2-b). γ + ɛx ≠ 0 for all x ∈ K. We obtain the following formula.
B[e(γ+ɛe-1)+z-(γ+ɛe)y]=(y-e)+(1+ɛ(y-e))(z-e)
for all y, z ∈ K. This shows that ɛ = 0, γ = −1, δ = 0, β = 1, α = e and A = −e and B = C = 1. Hence x ⊕ y = x + y − e and x ⊖ y = x − y + e, i.e., (K, ⊕, ⊖, e) is a linear β-algebra which is not a quadratic-linear β-algebra. Hence there is no linear-quadratic β-algebra which is not a linear β-algebra.
Problem
Construct a complete quadratic β-algebra over a field, i.e., x ⊕ y and x ⊖ y are both polynomials of x and y of degree 2.
4. Some Relations of Binary Operations in β-algebras
In this section, we discuss some relations of binary operations in β-algebras. For example, given a groupoid (X, +), we want to know the structure of (X,−) if (X,+,−, 0) is a β-algebra. Given a non-empty set X, a groupoid (X, *) is said to be a left-zero semigroup if x * y = x for all x, y ∈ X. Similarly, a groupoid (X, *) is said to be a right-zero semigroup if x * y = y for all x, y ∈ X.
Proposition 4.1
If (X,⊕) is a left-zero semigroup and if (X,⊕, ⊖, 0) is a β-algebra, then (X,⊖) is also a left-zero semigroup.
Proof
Assume that (X, ⊕, ⊖, 0) is a β-algebra. Then x ⊖ 0 = x for all x ∈ X, and (x ⊖y) ⊖ z = x ⊖ (z ⊕ y) = x ⊖z, i.e., (x ⊖ y) ⊖ z = x ⊖ z for all x, y, z ∈ X. It follows that x⊖y = (x⊖y) ⊖0 = x⊖0 = x, i.e., x⊖y = x, for all x, y ∈ X, proving that (X, ⊖) is a left-zero semigroup.
Proposition 4.2
Let (X,⊕, 0) be an algebra with 0 ⊕ x = 0 for all x ∈ X. If (X, ⊖) is a left-zero semigroup, then (X,⊕, ⊖, 0) is a β-algebra.
Proof
Since (X, ⊖) is a left-zero semigroup, the conditions (I) and (III) hold. It follows from 0 ⊕ x = 0 for all x ∈ X that 0 = 0 ⊕ x = (0 ⊖ x) ⊕ x, proving the proposition.
Corollary 4.3
Let (X, *, 0) be a BCK-algebra. If (X, ⊖) is a left-zero semigroup, then (X, *,⊖, 0) is a β-algebra.
Proof
Straightforward.
Proposition 4.4
Let (X,⊕, 0) be an algebra with 0 ⊕ x = 0 for all x ∈ X. If (X,⊕, ⊖, 0) is a β-algebra, then (X, ⊖) is a left-zero semigroup.
Proof
Let (X,⊕, ⊖, 0) be a β-algebra. If we let z := 0 in (III), then x ⊖ y = (x ⊖ y) ⊖ 0 = x ⊖ (0 ⊕ y) = x ⊖ 0 = x for all x, y ∈ X, proving the proposition.
Proposition 4.5
Let (X,⊕) be a right-zero semigroup. If (X,⊕, ⊖, 0) is a β-algebra, then X = {0}.
Proof
Assume (X,⊕, ⊖, 0) is a β-algebra. Since (X,⊕) is a right-zero semigroup, by (II), we have 0 = (0 ⊖ x) ⊕ x = x for all x ∈ X, proving that X = {0}.
Proposition 4.6
Let (X, ⊖) be a right-zero semigroup. If (X,⊕, ⊖, 0) is a β-algebra, then X = {0}.
Proof
Assume (X,⊕, ⊖, 0) is a β-algebra. Since (X, ⊖) is a right-zero semigroup, by (II), we have 0 = (0 ⊖ x) ⊕ x = x ⊕ x for all x ∈ X. By (III), we have z = (x⊖y) ⊖z = x⊖ (z ⊕y) = z ⊕y for all y, z ∈ X. It follows that x = x⊕x = 0 for all x ∈ X, proving the proposition.
Theorem 4.7
Let (K,+,−, 0) be a field with |K| ≥ 4 and let e ∈ K. Define a binary operation “⊕” on K by x ⊕ y := p(x, y), i.e., a quadratic polynomial of x and y in K. If e ⊕ x = e for all x ∈ K, then x ⊕ y = e + (A + Bx + Cy)(x − e) for all x, y ∈ K where A, B, C ∈ K.
Proof
Assume p(x, y) := A + Bx + Cy for all x, y ∈ K where A, B, C ∈ K. Since e ⊕ x = e, we have e = e ⊕ y = A + Be + Cy for all y ∈ K. It shows that A = e(1 − B), C = 0. Hence x ⊕ y = e(1 − B) + Bx = e + B(x − e). Assume p(x, y) := A + Bx + Cy + Dx2 + Exy + Fy2. Then
e=e⊕y=A+Be+Cy+De2+Eey+Fy2=(A+Be+De2)+(C+Ee)y+Fy2
It follows that F = 0, C + Ee = 0, A + Be + De2 = e. Hence
i.e., p(x, y) is of the form p(x, y) = e + (A + Bx + Cy)(x − e). This shows that x ⊕ y = e + q(x, y)(x − e) where q(x, y) is a linear polynomial of degree ≤ 1.
Using Theorem 4.7 and Proposition 4.2, we obtain the following.
Corollary 4.8
Let (K,+,−, 0) be a field with |K| ≥ 4 and let e ∈ K. Define a binary operation “⊕” on K by
x⊕y:=e+q(x,y)(x-e)
where q(x, y) is any polynomial of x and y in K. If x⊖y := x for all x, y ∈ K, then (K,⊕, ⊖, e) is a β-algebra.
