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Kyungpook Mathematical Journal 2017; 57(1): 99-108

Published online March 23, 2017

On the Construction of Polynomial β-algebras over a Field

Keum Sook So1
Young Hee Kim2

Department of Mathematics, Hallym University, Chuncheon 24252, Korea1
Department of Mathematics, Chungbuk National University, Chongju 28644, Korea2

Received: September 22, 2016; Accepted: November 10, 2016

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.

Keywords: (linear, quadratic-linear, linear-quadratic) $eta$-algebra, (left, right)-zero semigroup, $BCK$-algebra

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.

A β-algebra([9]) is a non-empty set X with a constant 0 and two binary operations “+” and “−” satisfying the following axioms: for any x, y, zX,

• x −0 = x,

• (0 − x) + x = 0,

• (xy) − z = x − (z + y).

• Example 2.1.([9])

Let X := {0, 1, 2, 3} be a set with the following tables:

+0123
00123
11230
22301
33012
0123
00321
11032
22103
33210

Then (X,+,−, 0) is a β-algebra.

Given a β-algebra X, we denote x* := 0 − x for any xX.

Theorem 2.2.([6])

Let (K,+, ·, e) be a field (sufficiently large) and let x, yK. Then (K, ⊕, ⊖, e) is a β-algebra, where xy = x + ye and xy = xy + e for any x, yK.

We call such a β-algebra described in Theorem 2.2 a linear β-algebra.

If we let ϕ : KK be a map defined by ϕ(x) = e + bx for some bK. 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) := (xe)/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:

Proposition 2.3.([6])

The β-algebra (K, ⊖, ⊕, e) discussed in Theorem 2.2 is unique up to isomorphism.

A BCK/BCI-algebra is an important class of logical algebras introduced by K. Iséki and was extensively investigated by several researchers.

An algebra (X; *, 0) of type (2, 0) is called a BCI-algebra ([3, 4]) if it satisfies the following conditions: for all x, y, zX,

• (i) (((x * y) * (x * z)) * (z * y) = 0),

• (ii) ((x * (x * y)) * y = 0),

• (iii) (x * x = 0),

• (iv) (x * y = 0, y * x = 0 ⇒ x = y).

If a BCI-algebra X satisfies the following identity: for all xX,

• (v) (0 * x = 0),

then X is called a BCK-algebra ([3, 4]).

3. Constructions of Polynomial β-algebras

Let (K,+, ·, e) be a field (sufficiently large)and let x, yK. First, we consider the case of quadratic-linear β-algebras, i.e., xy is the polynomial of x, y with degree 2, and xy is the polynomial of x, y with degree 1. Define two binary operations “⊕, ⊖” on K as follows:

xy:=A+Bx+Cy+Dx2+Exy+Fy2,xy:=α+βx+γy

where α, β, γ, A, B, C, D, E, FK (fixed). Assume that (K, ⊕, ⊖, 0) is a β-algebra. It is necessary to find proper solutions for two equations. Since x = xe = α + βx + γe, we obtain (β − 1)x + (α + γe) = 0, and hence β = 1 and α = −γe. It follows that

xy=x+γ(y-e)

From (3.1) we obtain ex = e+γ(xe) = (1 − γ)e+γx. Since (ex)⊕x = e, we obtain

e=(ex)x=[(1-γ)e+γx]x=A+B[(1-γ)e+γx]+Cx+D[(1-γ)e+γx]2+E[(1-γ)e+γx]x+Fx2=A+B(1-γ)e+D(1-γ)2e2]+[Bγ+C+2Dγ(1-γ)e+E(1-γ)e]x+[Dγ2+Eγ+F]x2

If we assume that |K| ≥ 3, then we obtain

Dγ2+Eγ+F=0Bγ+C+2Dγ(1-γ)e+E(1-γ)e=0A+B(1-γ)e+D(1-γ)2e2=0

Case(i). γ ≠ 0. By formula (3.1) we obtain

(xy)z=xy+γ(z-e)=x+γ(y-e)+γ(z-e)=x+γ(y+z-2e)

and

x(zy)=x+γ(zy-e)

By (III), we obtain x + γ(y + z − 2e) = x + γ(zye). Since γ ≠ 0, we have zye = z + y − 2e. This shows that xy = x + ye for all x, yK. In this case, we obtain the linear case as described in Theorem 2.2.

Case (ii). γ = 0. By (3.1), we obtain xy = 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 = eBeDe2. This shows that

xy=(e-Be-De2)+Bx+(-Ee)y+Dx2+Exy=e+B(x-e)+D(x2-e2)+E(x-e)y=e+[B+D(x+e)+Ey](x-e)

which means that xy is of the form:

xy=e+(a+bx+cy)(x-e).

It is a quadratic form (not a linear form) and so (K, ⊕, ⊖, e) is a quadratic-linear β-algebra where xy = e + (a + bx + cy)(xe) and xy = x. We summarize:

Theorem 3.1

Let (K,+, ·, e) be a field (sufficiently large). If we define two binary operations “⊕, ⊖” on K by

xy:=e+(a+bx+cy)(x-e)xy:=x

for all x, yK, then (K,⊕, ⊖, e) is a quadratic-linear β-algebra.

Corollary 3.2

Let (K,+, ·, 0) be a field (sufficiently large) and let a, b, cK. If we define two binary operations “⊕, ⊖” on K by

xy:=ax+bx2+cxyxy:=x

for all x, yK, 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 xy := xx2−2xy and xy := x for all x, yR, then (R,⊕, ⊖, 0) is a quadratic-linear β-algebra.

Let (K,+, ·, e) be a field (sufficiently large) and let x, yK. Next, we consider the case of linear-quadratic β-algebras, i.e., xy is the polynomial of x, y with degree 1, and xy 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:

xy:=A+Bx+Cy,xy:=α+βx+γy+δx2+ɛxy+ξy2

where A, B, C, α, β, γ, δ, ɛ, ξK (fixed). Assume that (K, ⊕, ⊖, e) is a β-algebra and |K| ≥ 3. Then

x=xe=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

xy=x+γy+ɛxy+ξy2

It follows that 0 ⊖ x = 0+γx + ɛ0x + ξx2 = γx + ξx2 and hence

0=(0x)x=(γx+ξx2)x=A+B(γx+ξx2)+Cx=A+(Bγ+C)x+Bξx2

for all xK. This shows that A = 0, Bγ + C = 0, Bξ = 0.

Subcase (i-1). B = 0. Since C = −Bγ, we obtain C = 0 and hence xy = 0 for all x, yK. This shows that (0 ⊖ x) ⊕ x = 0. Using formula (3.4) we obtain

(xy)z=(x+γy+ɛxy+ξy2)z=(x+γy+ɛxy+ξy2)+γz+ɛ(x+γy+ɛxy+ξy2)z+ξz2=x+γy+γz+ɛxy+ξ(y2+z2)+ɛ(x+γy+ɛxy+ξy2)z

and

x(zy)=x0=x

By (III), we obtain γ = 0, ɛ = 0, ξ = 0, proving that xy = x. Hence xy = 0 and xy = 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 xy = x + γy + ɛxy. It follows that 0 ⊖ x = γx and hence 0 = (0 ⊖ x) ⊕ x = γxx = A + B(γx) + Cx = A + (Bγ + C)x for all xK. It follows that A = 0, Bγ + C = 0. Hence xy = B(xγy). It follows that

x(zy)=x+γ(zy)+ɛx(zy)=x+γB(z-γy)+ɛBx(z-γy)=x+γBz-γ2By+ɛBxz-ɛBγxy

and

(xy)z=(x+γy+ɛxy)z=x+γy+ɛxy+γz+ɛ(x+γy+ɛxy)z

By (III), we obtain ɛ = 0 and γ(1 + γ)B = 0. If γ = 0, then xy = Bx and xy = x. If B = 0, then it is the same case as subcase (i-1). If γ = −1, then xy = B(x + y) and xy = xy. This shows that (K, ⊕, ⊖, 0) is not a linear-quadratic β-algebra.

Case (ii). e ≠ 0. It follows from (3.3) that α = −γeξe2, β = 1− ɛe, δ = 0. Hence

xy=(γe-ξe2)+(1-ɛe)x+γy+ɛxy+ξy2

Subcase (ii-1). ξ ≠ 0. The formula (3.5) can be written as

xy=x+(γ+ɛx)(y-e)+ξ(y2-e2)

If y := e in (3.6), then xe = x + (γ + ɛx)(ee) + ξ(e2 e2) = x. If x := e and y := x in (3.6), then ex = x + (γ + ɛe)(xe) + ξ(x2 e2). It follows that

e=(ex)x=A+B(ex)+Cx=A+B[e+(γ+ɛe)(x-e)+ξ(x2-e2)]+Cx=[A+Be-B(γ+ɛe)e-Bξe2]+[B(γ+ɛe)+C]x+Bξx2

for all xK, and hence we obtain

Bξ=0B(γ+ɛe)+C=0A+Be-B(γ+ɛe)e-Bξe2=0

Since ξ ≠ 0, we have B = 0 and hence C = 0, A = 0, i.e., xy = 0 and xy = x+(γ+ɛx)(ye)+ξ(y2e2) does not form a β-algebra, since e⊖(ee) = ee(γ + ɛe) − ξe2 and (ee) ⊖ e = e.

Subcase (ii-2). ξ = 0. The formula can be written as

xy=x+(γ+ɛx)(y-e)

It follows that xe = x + (γ + ɛx)(ee) = x and ex = e + (γ + ɛe)(xe). By (II) we obtain the following.

e=(ex)x=[e+(γ+ɛe)(x-e)]x=A+B[e+(γ+ɛe)(x-e)]+Cx=[A+Be-B(γ+ɛe)e]+[B(γ+ɛe)+C]x

for all xK, and hence we obtain

A+Be-B(γ+ɛe)e=eB(γ+ɛe)+C=0

Hence A = e + Be[γ + ɛe − 1] and C = −B(γ + ɛe). It follows that

xy=[e+Be(γ+ɛe-1)]+Bx-B(γ+ɛe)y

Using formulas (3.7) and (3.8), we obtain

x(zy)=x+(γ+ɛx)(zy-e)=x+(γ+ɛx)[[e+Be(γ+ɛe-1)]+Bz-B(γ+ɛe)y-e]=x+(γ+ɛx)[[Be(γ+ɛe-1)]+Bz-B(γ+ɛe)y]=x+B(γ+ɛx)[e(γ+ɛe-1)+z-(γ+ɛe)y]

By formula (3.7) we obtain

(xy)z=xy+(γ+ɛ(xy))(z-e)=x+(γ+ɛx)(y-e)+[γ+ɛ(x+(γ+ɛx)(y-e))](z-e)=x+(γ+ɛx)[(y-e)+(1+ɛ(y-e))(z-e)]

By (III), we have

B(γ+ɛx)[e(γ+ɛe-1)+z-(γ+ɛe)y]=(γ+ɛx)[(y-e)+(1+ɛ(y-e))(z-e)]

Subcase (ii-2-a). γ = ε = 0. We obtain β = 1, α = 0 and hence xy = x, xy = (eBe) + Bx, a linear β-algebra.

Subcase (ii-2-b). γ + ɛx ≠ 0 for all xK. 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, zK. This shows that ɛ = 0, γ = −1, δ = 0, β = 1, α = e and A = −e and B = C = 1. Hence xy = x + ye and xy = xy + 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., xy and xy 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, yX. Similarly, a groupoid (X, *) is said to be a right-zero semigroup if x * y = y for all x, yX.

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 xX, and (xy) ⊖ z = x ⊖ (zy) = xz, i.e., (xy) ⊖ z = xz for all x, y, zX. It follows that xy = (xy) ⊖0 = x⊖0 = x, i.e., xy = x, for all x, yX, proving that (X, ⊖) is a left-zero semigroup.

Proposition 4.2

Let (X,⊕, 0) be an algebra with 0 ⊕ x = 0 for all xX. 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 xX 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 xX. 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 xy = (xy) ⊖ 0 = x ⊖ (0 ⊕ y) = x ⊖ 0 = x for all x, yX, 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 xX, 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 = xx for all xX. By (III), we have z = (xy) ⊖z = x⊖ (zy) = zy for all y, zX. It follows that x = xx = 0 for all xX, proving the proposition.

Theorem 4.7

Let (K,+,−, 0) be a field with |K| ≥ 4 and let eK. Define a binary operation “” on K by xy := p(x, y), i.e., a quadratic polynomial of x and y in K. If ex = e for all xK, then xy = e + (A + Bx + Cy)(xe) for all x, yK where A, B, CK.

Proof

Assume p(x, y) := A + Bx + Cy for all x, yK where A, B, CK. Since ex = e, we have e = ey = A + Be + Cy for all yK. It shows that A = e(1 − B), C = 0. Hence xy = e(1 − B) + Bx = e + B(xe). Assume p(x, y) := A + Bx + Cy + Dx2 + Exy + Fy2. Then

e=ey=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

p(x,y)=(e-Be-De2)+Bx-Eey+Dx2+Exy=e+[B+D(x+e)+Ey](x-e)=e+(B+De+Dx+Ey)(x-e),

i.e., p(x, y) is of the form p(x, y) = e + (A + Bx + Cy)(xe). This shows that xy = e + q(x, y)(xe) 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 eK. Define a binary operation “⊕” on K by

xy:=e+q(x,y)(x-e)

where q(x, y) is any polynomial of x and y in K. If xy := x for all x, yK, then (K,,, e) is a β-algebra.

+0123
00123
11230
22301
33012
0123
00321
11032
22103
33210
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