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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

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Volume 15, Issue 1

# On Jordan mappings of inverse semirings

Sara Shafiq
Published Online: 2017-09-14 | DOI: https://doi.org/10.1515/math-2017-0088

## Abstract

In this paper, the notions of Jordan homomorphism and Jordan derivation of inverse semirings are introduced. A few results of Herstein and Brešar on Jordan homomorphisms and Jordan derivations of rings are generalized in the setting of inverse semirings.

MSC 2010: 16Y60; 16W25

## 1 Introduction

Let (S, +, ⋅) be a semiring with commutative addition and an absorbing zero 0. A semiring S is called an inverse semiring [1] if for every aS there exists a unique element a′ ∈ S such that a + a′ + a = a and a′ + a + a′ = a′. Throughout this paper, S will represent an inverse semiring which satisfies the condition that for every aS, a + a′ is in the center of S. This class of inverse semirings, known as MA-semiring [2], is useful in developing the theory of commutators and investigating certain additive mappings in semirings. In this connection, commuting maps [3], skew-commuting maps [4], centralizers [5,6], dependent elements and free actions [7] have been studied. However, the theory of Jordan homomorphism and Jordan derivation of inverse semirings has been unexplored. According to [2], a commutator [. , .] in inverse semiring is defined as [x, y] = xy + ýx = xy + $y\stackrel{´}{x}.$ We will make use of commutator identities [xy, z] = x[y, z] + [x, z]y and [x, yz] = [x, y]z + y[x, z] (see [2]). By [2], an additive map d : SS is a derivation if d(ab) = d(a)b + ad(b), ∀ a, bS. S is prime if aSb = (0) implies that a = 0 or b = 0 and is semiprime if aSa = (0) implies that a = 0. S is n-torsion free if nx = 0, xS implies that x = 0.

The idea of Jordan homomorphism of rings arose initially in Ancochea’s study of semi-automorphisms [8,9]. Later on, Kaplansky [10], Hua [11] and Jacobson and Ricket [12] made contribution and took the subject up. In 1950’s, Herstein studied Jordan homomorphisms [13,14] and Jordan derivations [15] in prime rings. His results have a notable impact on the study of Jordan structure and Jordan mappings. Brešar [16,17,18], Baxer and Martindale [19] generalized Herstein’s work on semiprime rings.

In this paper, we introduce Jordan homomorphism and Jordan derivation in inverse semirings and generalize a few results of Herstein [13,15] and Brešar [16,17] in the setting of inverse semirings. We define Jordan homomorphism between the two inverse semirings as follows. Let S and T be inverse semirings then an additive mapping φ : ST is called a Jordan homomorphism if φ(ab + ba) + φ(a)φ(b)′ + φ(b)φ(a)′ = 0, ∀ a, bS. In section 2, we generalize remarkable result of Herstein [13] for inverse semirings as follows: every Jordan homomorphism of inverse semiring onto prime inverse semiring is either homomorphism or anti-homomorphism. In section 3, we define Jordan triple homomorphism ρ between inverse semirings S and T as an additive mapping such that ρ(aba) + ρ(a)ρ(b)ρ(a)′ = 0, ∀ a, bS. Brešar [17] showed that a Jordan triple homomorphism ϕ of a ring R onto a prime ring R of characteristic different from 2 is of the form ϕ = ± θ, where θ is a homomorphism or an anti-homomorphism of R onto R. We generalize this result for inverse semirings. In section 4, we introduce the notion of Jordan derivation of inverse semirings thereby extend a classical result of Brešar [16] as follows. Every Jordan derivation of a 2-torsion free inverse semiring is a derivation. In the last section, Jordan triple derivation of inverse semiring is considered and a Brešar’s result [17] is extended in semirings.

We need the following lemmas in our arguments.

#### Lemma 1.1

(Lemma 1.1, [5]). Let S be an inverse semiring and a, bS. If a + b = 0 then a = b′.

#### Lemma 1.2

Let S be a 2-torsion free semiprime inverse semiring. If a, bS are such that axb + bxa = 0, ∀xS then axb = bxa = 0.

#### Proof

Using Lemma 1.1 in axb + bxa = 0 we have axb = bxa′, xS. Thus (bxa) y(bxa) = b(xby)axa = axbybxa. Hence, 2 (bxa)y(bxa) = (axb + bxa)ybxa = 0, since S is 2-torsion free, bxa = 0. Similarly, we can show axb = 0, ∀xS. □

The following lemma is an extension of Lemma 3.10 of [20], in a canonical fashion.

#### Lemma 1.3

Let S be a 2-torsion free prime inverse semiring. If a, bS are such that axb + bxa = 0, ∀xS then either a = 0 or b = 0.

#### Lemma 1.4

(Lemma 1.5, [5]). Let S be a semiprime inverse semiring and f, g : S × SS biadditive mappings. If f(x, y)wg(x, y) = 0, ∀x, y, wS then f(x, y)wg(s, t) = 0, ∀x, y, s, t, wS.

## 2 Jordan homomorphism

We begin this section by introducing the notion of Jordan homomorphism of inverse semirings. Let S and T be inverse semirings, an additive mapping φ : ST is called Jordan homomorphism if $φ(ab+ba)+φ(a)φ(b)′+φ(b)φ(a)′=0,∀a,b∈S$(1)

#### Lemma 2.1

Let φ be a Jordan homomorphism of inverse semiring S into 2-torsion free inverse semiring T. Then for all a, b, cS the following statements are true:

1. φ(a2) = φ(a)2

2. φ(aba) = φ(a)φ(b)φ(a)

3. φ(cba + abc) = φ(c)φ(b)φ(a) + φ(a)φ(b)φ(c)

#### Proof

Replacing b by a in (1) and then using Lemma 1.1 we obtain (i). For (ii); put b = ab + ba in (1), we have φ(a(ab + ba) + (ab + ba)a) + φ(a)φ(ab + ba)′ + φ(ab + ba)φ(a)′ = 0. In the view of Lemma 1.1 and (1), we can replace φ(ab + ba) by φ(a)φ(b) + φ(b)φ(a). Thus we have $φ(a2b+ba2)+2φ(aba)+φ(a)φ(a)φ(b)′+φ(a)φ(b)φ(a)′+φ(a)φ(b)φ(a)′+φ(b)φ(a)φ(a)′=0$

By (i) we have $φ(aba)+φ(a)φ(b)φ(a)′=0$ and hence by Lemma 1.1, we conclude (ii). Linearizing the above relation we obtain $φ(cba+abc)+φ(c)φ(b)φ(a)′+φ(a)φ(b)φ(c)′=0$(2) thus by Lemma 1.1, we arrive at (iii). □

Following [13], we fix some notations; ab = φ(ab) + φ(a)φ(b)′ and ab = φ(ab) + φ(b)φ(a)′. Thus (1) can be written as $ab+ba=0$

#### Lemma 2.2

If φ is a Jordan homomorphism from an inverse semiring S to 2-torsion free inverse semiring T then $abab=0=abab$

#### Proof

We have, $abab=φ(ab)φ(ab)+φ(ab)φ(b)φ(a)′+φ(a)φ(b)φ(ab)′+φ(a)φ(b)φ(b)φ(a)$

By Lemma 2.1 (i) we get $abab=φ(abab)+φ(ab)φ(b)φ(a)′+φ(a)φ(b)φ(ab)′+φ(a)φ(b2)φ(a)$

Using Lemma 2.1 (ii) and (2) we arrive at abab = 0, as desired. Similarly, we can compute abab = 0. □

#### Lemma 2.3

Let φ be a Jordan homomorphism from inverse semiring S to 2-torsion free inverse semiring T. Then for any a, b, rS $abφ(r)ab=abφ([a,b]r)andabφ(r)ab=φ([a,b]r)ab.$

#### Proof

Consider, φ(r)ab = φ(r)φ(ab) + φ(r)φ(a)φ(b)′.

Applying Lemma 1.1 in (2) and then using it in the above relation, we obtain $φ(r)ab=φ(r)φ(ab)+φ(b)φ(a)φ(r)+φ(rab+bar)′$

But φ(rab + bar)′ = φ(rab + (r + r′)ab + bar)′ = φ(rab + ab(r + r′) + bar)′ = φ(rab + abr)′ + φ(abr + bar′) = φ(r)φ(ab)′ + φ(ab)φ(r)′ + φ(abr + bar′). Thus we have, φ(r)ab = φ(r)(φ(ab) + φ(ab)′) + φ(b)φ(a)φ(r) + φ(ab)φ(r)′ + φ([a, b]r) = (φ(ab) + φ(ab)′)φ(r) + φ(b)φ(a)φ(r) + φ(ab)φ(r)′ + φ([a, b]r) = abφ(r) + φ([a, b]r). By Lemma 2.2 we obtain abφ(r)ab = abφ([a, b]r) and abφ(r)′ab + φ([a, b]r)ab = 0. □

#### Lemma 2.4

Let φ be a Jordan homomorphism from inverse semiring S to 2-torsion free inverse semiring T. If a, b, rS then $φ([a,b]r)=φ(r)ab+abφ(r)andφ(r[a,b])=abφ(r)+φ(r)ab.$

#### Proof

From (1), we can replace φ(ab + ba) by φ(a)φ(b) + φ(b)φ(a) ∀ a, bS. Using this together with Lemma 2.1 (iii), we have abφ(r) + φ(r)ab = φ(ab)φ(r) + φ(a)φ(b)φ(r)′ + φ(r)φ(ab) + φ(r)φ(b)φ(a)′ = φ(abr′ + rba′ + abr + rab) = φ(ab(r + r′) + rba′ + rab) = φ((r + r′)ab + rbá + rab) = φ(r[a, b]) Similarly, φ([a, b]r) = φ(r)ab + abφ(r). □

#### Theorem 2.5

Let φ be a Jordan homomorphism from inverse semiring S to 2-torsion free inverse semiring T. Then for all a, b, rS $abφ(r)ab+abφ(r)ab=0.$

#### Proof

Replacing r by [a, b]r in φ(r[a, b]) = abφ(r) + φ(r)ab we have $φ([a,b]r[a,b])=abφ([a,b]r)+φ([a,b]r)ab$

By Lemma 2.1 (ii) and Lemma 2.4, we get $φ([a,b])φ(r)φ([a,b])=abφ(r)ab+abφ(r)ab$

Now, φ([a, b]) = φ(ab + ab + ab′ + ba) = 2φ(ab) + φ(ab + ba)′ = 2φ(ab) + φ(a)φ(b) + φ(b)φ(a) = ab + ab.

Thus we have $(ab+ab)φ(r)(ab+ab)=abφ(r)ab+abφ(r)ab$ which implies that $abφ(r)ab+abφ(r)ab+abφ(r)ab+abφ(r)ab=abφ(r)ab+abφ(r)ab$

Adding baφ(r)ab + baφ(r)ab on both sides of the above equation and using the fact that ab + ba = 0 = ab + ba, we get the required result. □

#### Theorem 2.6

Every Jordan homomorphism φ of S onto 2-torsion free prime inverse semiring T is either a homomorphism or an anti-homomorphism.

#### Proof

By Theorem 2.5 and Lemma 1.3 we have either ab = 0 or ab = 0. Thus by Lemma 1.1 either φ is a homomorphism or an anti-homomorphism. □

## 3 Jordan triple homomorphism

Let S and T be inverse semirings. An additive map ρ : ST is called Jordan triple homomorphism if $ρ(aba)+ρ(a)ρ(b)ρ(a)′=0,∀a,b∈S$(3)

For the sake of convenience, we fix some notations $G1(a,b,c)=ρ(abc)+ρ(a)ρ(b)ρ(c)′ andG2(a,b,c)=ρ(abc)+ρ(c)ρ(b)ρ(a)′$

Linearization of (3) gives $ρ(abc+cba)+ρ(a)ρ(b)ρ(c)′+ρ(c)ρ(b)ρ(a)′=0$(4) or $G1(a,b,c)+G1(c,b,a)=0$

#### Lemma 3.1

If ρ is a Jordan homomorphism of inverse semiring S onto 2-torsion free prime inverse semiring T then $G1(a,b,c)ρ(x)G2(a,b,c)+G2(a,b,c)ρ(x)G1(a,b,c)=0,∀a,b,c,x∈S.$

#### Proof

Replacing a by abc, b by x and c by cba in (4), we obtain $ρ(abcxcba+cbaxabc)+ρ(abc)ρ(x)ρ(cba)′+ρ(cba)ρ(x)ρ(abc)′=0$(5)

According to Lemma 1.1 and (4) we have $ρ(cba)=ρ(abc)′+ρ(a)ρ(b)ρ(c)+ρ(c)ρ(b)ρ(a)$ thus from (5) we get $ρ(abcxcba+cbaxabc)+ρ(abc)ρ(x)ρ(abc)+ρ(abc)ρ(x)ρ(c)ρ(b)ρ(a)′+ρ(abc)ρ(x)ρ(a)ρ(b)ρ(c)′+ρ(abc)ρ(x)ρ(abc)+ρ(c)ρ(b)ρ(a)ρ(x)ρ(abc)′+ρ(a)ρ(b)ρ(c)ρ(x)ρ(abc)′=0$(6)

Also, by definition of triple homomorphism and Lemma 1.1, we have $ρ(abcxcba+cbaxabc)=ρ(a)ρ(b)ρ(c)ρ(x)ρ(c)ρ(b)ρ(a)+ρ(c)ρ(b)ρ(a)ρ(x)ρ(a)ρ(b)ρ(c)$(7)

From (6) and (7) we obtain $G1(a,b,c)ρ(x)G2(a,b,c)+G2(a,b,c)ρ(x)G1(a,b,c)=0$ as desired. □

#### Theorem 3.2

Let ρ be a Jordan triple homomorphism of an inverse semiring onto 2-torsion free prime inverse semiring. Then ρ = ϕ or ρ = ϕ′, where ϕ is either a homomorphism or an anti-homomorphism.

#### Proof

By Lemmas 3.1 and 1.3, we have either G1(a, b, c) = 0 or G2(a, b, c) = 0. If G1(a, b, c) = 0 then ρ(abc) = ρ(a)ρ(b)ρ(c). Replacing b by bxa and c by b we get $ρ(abxab)=ρ(a)ρ(b)ρ(x)ρ(a)ρ(b)$(8)

Also, by definition of triple homomorphism we have $ρ(abxab)+ρ(ab)ρ(x)ρ(ab)′=0$(9)

From (8) and (9) we get $ρ(a)ρ(b)ρ(x)ρ(a)ρ(b)+ρ(ab)ρ(x)ρ(ab)′=0$(10)

If ba = ρ(ab) + ρ(a)ρ(b)′ and ba = ρ(ab) + ρ(a)ρ(b) then using (10) we have $baρ(x)ba+baρ(x)ba={ρ(a)ρ(b)′ρ(x)+ρ(a)ρ(b)ρ(x)}ρ(ab)+ρ(ab){ρ(x)ρ(a)ρ(b)+ρ(x)ρ(a)ρ(b)′}$

But ρ(x)ρ(a)ρ(b) = ρ(xab), thus we get $baρ(x)ba+baρ(x)ba={ρ(abx)′+ρ(a)ρ(b)ρ(x)}ρ(ab)+ρ(ab){ρ(xab)+ρ(x)ρ(a)ρ(b)′}=0.$

Hence, by Lemma 1.3 either ba = 0 or ba = 0. Therefore, either ρ(ab) = ρ(a)ρ(b) or ρ(ab) = ρ(a)ρ(b)′ for all a, bS.

On similar lines, we can show that if G2(a, b, c) = 0 then ρ(ab) = ρ(b)ρ(a) or ρ(ab) = ρ(b)ρ(a)′ for all a, bS. Thus ρ = ϕ or ρ = ϕ′, where ϕ is either a homomorphism or an anti-homomorphism. □

## 4 Jordan derivation

We define Jordan derivation of an inverse semiring S as an additive map d : SS such that $d(x2)+d(x)x′+x′d(x)=0,∀x∈S$(11) holds. For example, if R is a commutative ring and I(R) is semiring of all two sided ideals of R with respect to ordinary addition and product of ideals, and T is subsemiring of I(R) then S = {(r, I) : rR, IT} is an inverse semiring with respect to ⊕ and ⊙ defined as (r1, I) ⊕(r2, J) = (r1 + r2, I + J) and (r1, I)⊙(r2, J) = (r1r2, IJ) (see [4]).

If we fix a = (r, {0}) ∈ S then d(x) = [a, x], ∀xS is a Jordan derivation.

#### Lemma 4.1

If d is a Jordan derivation of S then for all x, yS the following statements hold:

1. d(xyx) = d(x)yx + xd(y)x + xyd(x)

2. d(xyz + zyx) = d(x)yz + xd(y)z + xyd(z) + d(z)yx + zd(y)x + zyd(x).

#### Proof

Linearization of (11) gives $d(xy+yx)+d(y)x′+d(x)y′+x′d(y)+y′d(x)=0$(12)

Replacing y by xy + yx in (12) we get $d(x(xy+yx)+(xy+yx)x)+d(xy+yx)x′+d(x)(xy′+yx′)+x′d(xy+yx)+(xy′+yx′)d(x)=0$(13)

By (12),(13) and Lemma 1.1 we have $d(x2y+yx2)+2d(xyx)+2d(x)y′x+2x′d(y)x+2x′yd(x)+d(y)′x2+yd(x)x′+d(x)xy′+x′d(x)y+x2d(y)′+yx′d(x)=0$(14)

Replacing x by x2 in (12) we have $d(x2y+yx2)+d(y)′x2+d(x2)y′+x2d(y)′+y′d(x2)=0$

In the view of (11) and Lemma 1.1, we can replace d(x2) by d(x)x + xd(x) . Thus $d(x2y+yx2)+d(y)′x2+d(x)xy′+xd(x)y′+x2d(y)′+y′d(x)x+y′xd(x)=0$(15)

From (14) and (15) we have $d(xyx)+d(x)y′x+xd(y)x′+x′yd(x)=0$(16) which gives (i). To obtain (ii), linearize (16) we have $d(xyz+zyx)+d(x)yz+xd(y)′+xy′d(z)+d(z)yx′+zd(y)x′+zy′d(x)=0$(17) and hence (ii) follows by Lemma 1.1. This completes the proof. □

If we write xy for d(xy) + d(x)y′ + xd(y) then by (12) and Lemma 1.1, we have $xy=yx′$(18)

Moreover, $xy+z=xy+xz$(19) holds for all x, yS.

#### Lemma 4.2

Let S be a 2-torsion free inverse semiring and d : SS be a Jordan derivation. Then $xys[x,y]+[x,y]sxy=0.$(20)x, y, sS.

#### Proof

Replacing y by ysy in (16) we get $d(xysyx)+d(x)ysy′x+xd(ysy)x′+xys′yd(x)=0$(21)

Replacing x by y and y by xsx in (16) gives $d(yxsxy)+d(y)xsx′y+yd(xsx)y′+yxs′xd(y)=0$(22)

Adding (21) and (22), we have $d(xysyx+yxsxy)+d(x)ysy′x+xd(ysy)x′+xys′yd(x)+d(y)xsx′y+yd(xsx)y′+yxs′xd(y)=0$ By (i) of Lemma 4.1, we have $d(xysyx+yxsxy)+d(x)ysy′x+xd(y)sy′x+xyd(s)y′x+xysd(y)x′+xys′yd(x)+d(y)xsx′y+yd(x)sx′y+yxd(s)x′y+yxsd(x)y′+yxs′xd(y)=0$(23)

Replacing x by xy, y by s and z by yx in (17) we get $d(xysyx+yxsxy)+d(xy)sy′x+xyd(s)y′x+xy′sd(yx)+d(yx)sx′y+yxd(s)x′y+yx′sd(xy)=0$

From this, we have $d(xysyx+yxsxy)+xyd(s)y′x+yxd(s)x′y=d(xy)syx+xysd(yx)+d(yx)sxy+yxsd(xy)$(24) (24) (23) and (24) give $d(xy)syx+xysd(yx)+d(yx)sxy+yxsd(xy)+d(x)ysy′x+xd(y)sy′x+xysd(y)x′+xys′yd(x)+d(y)xsx′y+yd(x)sx′y+yxd(d)x′y+yxs′xd(y)=0$ which can be written as 0 = xysyx + xysyx + yxsxy + yxsxy = xysyx + xysxy + yxsxy + xysxy.

Hence, xys[x, y] + [x, y]sxy = 0. □

#### Theorem 4.3

A Jordan derivation d on 2-torsion free inverse semiring S is a derivation.

#### Proof

To prove the theorem, we have to show that xy = 0 ∀x, yS.

By Lemmas 1.2 and 4.2, we get $xys[x,y]=0=[x,y]sxy$(25)

Replacing y by y + z in xys[x, y] = 0 and using it again, we have $xys[x,z]+xzs[x,y]=0$(26) which gives xys[x, z] = xzś[x, y]. Thus (xys[x, z])t(xys[x, z]) = xzś[x, y]txys[x, z] = 0, ∀ tS.

Semiprimeness of S gives $xys[x,z]=0,∀x,y,z,s∈S$(27)

Similarly, [x, z]sxy = 0.

By using the above technique on (27), we get $xys[w,z]=0,∀x,y,z,s,w∈S$(28)

Pre multiplying (28) by [w, z] and post multiplying it by xy, we get $[w,z]xy=0,x,y,w,z∈S$(29)

Therefore, $[xy,t]r[xy,t]=xytr[xy,t]+t′xyr[xy,t]=0,∀r∈S$ which implies that [xy, t] = 0 or xyt + txy = 0. Adding txy on both sides of the last expression we have xyt + (t + t′)xy = xyt or xyt + xy(t + t′) = txy, ∀ tS. Thus xy is in the center of S.

From (18), we have $2(xy)2=xy(xy+yx′)=xy{d(xy)+d(yx)′+[y,d(x)]+[d(y),x]}$(30) or $2(xy)2=xy{d[x,y]+[y,d(x)]+[d(y),x]}$

Using (29) and the fact that xy is in center, we get $2(xy)2=xyd[x,y]$(31)

Also, xy[x, y] + [x, y]xy = 0 which gives $d(xy[x,y]+[x,y]xy)=0$(32)

Put x = xy and y = [x, y] in (12) and using (32), we obtain $d(xy)[x,y]+xyd[x,y]+d[x,y]xy+[x,y]d(xy)=0$

By (31), we have d(xy)[x, y] + [x, y]d(xy) + 4 (xy)2 = 0. Post multiplying last equation by xy, we have 4 (xy)3 = 0. Thus (xy)3 = 0.

Since xy is in the center of S so for tS, (xy)2r(xy)2 = (xy)3(xy)r = 0, rS. Thus (xy)2 = O. This implies that xy = 0, ∀x, yS, as required. □

## 5 Jordan triple derivation

An additive mapping d : SS is Jordan triple derivation if $d(aba)+d(a)ba′+ad(b)a′+abd(a)′=0$(33)

Linearization of (33) gives $d(abc+cba)+d(a)bc′+ad(b)c′+abd(c)′+d(c)ba′+cd(b)a′+cbd(a)′=0$(34)

Put F1(a, b, c) = d(abc) + d(a)bc′ + ad(b)c′ + abd(c)′ and F2(a, b, c) = abc + cba′ then $F1(a,b,c)+F1(c,b,a)=0$(35)

#### Lemma 5.1

Let S be an inverse semiring. Then $F1(a,b,c)xF2(a,b,c)+F2(a,b,c)xF1(a,b,c)=0$ for all a, b, c, xS.

Proof can be obtained by simple modification of the proof of Lemma 3.1.

#### Theorem 5.2

Every Jordan triple derivation d of a 2-torsion free semiprime inverse semiring S is a derivation.

#### Proof

By Lemmas 5.l and 1.3, we have either F1(a, b, c) = 0 or F2(a, b, c) = 0. If F1(a, b, c) = 0 then $d(abc)+d(a)bc′+ad(b)c′+abd(c)′=0$(36) which gives $d(abc)=d(a)bc+ad(b)c+abd(c)$(37)

Replacing b by bxa and c by b in (37), we get $d(abxab)=d(a)bxab+ad(b)xab+abd(x)ab+abxd(a)b+abxad(b)$(38)

Also, by definition of triple derivation we have $d(abxab)+d(ab)xab′+abd(x)ab′+abxd(ab)′=0$(39)

From (38) and (39) we get $(d(ab)+d(a)b′+ad(b)′)xab+abx(d(ab)+d(a)b′+ad(b)′)+abd(x)ab+abd(x)ab′=0$(40)

Applying Lemma 1.1 in (39) we get $abd(x)ab=d(abxab)+d(ab)xab′+abxd(ab)′$

Adding abd(x)ab′ on both sides we have abd(x)ab + abd(x)ab′ = 0. Thus from (40) we obtain $(d(ab)+d(a)b′+ad(b)′)xab+abx(d(ab)+d(a)b′+ad(b)′)=0$(41)

By Lemmas 1.2 and 1.4, we obtain $(d(ab)+d(a)b´+ad(b))xcd=0,∀a,b,c,d,x∈S$

Semiprimness of S implies that d(ab) + d(a)b′ + ad(b)′ = 0, this gives that d is a derivation.

If F2(a, b, c) = 0 then by Lemma 1.1, we have $abc=cba,∀a,b,c∈S$

Thus d(a)bc = cbd(a), ad(b)c = cd(b)a, abd(c) = d(c)ba. Hence from (34) $2d(abc)+2d(a)bc′+2ad(b)c′+2abd(c)′=0$

As concluded above, if d(abc) + d(a)bc′ + ad(b)c′ + abd(c)′ = 0 then d is a derivation. This completes the proof. □

## Acknowledgement

We are thankful to the referees for their useful comments and suggestions.

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Accepted: 2017-06-20

Published Online: 2017-09-14

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 1123–1131, ISSN (Online) 2391-5455,

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