On multiplicative (generalized)-derivations in prime and semiprime rings

Aequat. Math. 86 (2013), 65–79c© Springer Basel 20130001-9054/13/010065-15published online May 29, 2013DOI 10.1007/s00010-013-0205-y Aequationes Mathematicae

On multiplicative (generalized)-derivationsin prime and semiprime rings

Basudeb Dhara and Shakir Ali

Abstract. Let R be a ring. A map F : R → R is called a multiplicative (generalized)-deriva-tion if F (xy) = F (x)y + xg(y) is fulfilled for all x, y ∈ R where g : R → R is any map (notnecessarily derivation). The main objective of the present paper is to study the followingsituations: (i) F (xy) ± xy ∈ Z, (ii) F (xy) ± yx ∈ Z, (iii) F (x)F (y) ± xy ∈ Z and (iv)F (x)F (y) ± yx ∈ Z for all x, y in some appropriate subset of R. Moreover, some examplesare also given.

Mathematics Subject Classification (2000). 16N60, 16U80, 16W25.

Keywords. Prime ring, semiprime ring, left ideal, derivation, multiplicative derivation,

generalized derivation, multiplicative (generalized)-derivation.

1. Introduction

Let R be an associative ring. The center of R is denoted by Z. For x, y ∈ R,the symbol [x, y] will denote the commutator xy − yx and the symbol x ◦ ywill denote the anticommutator xy + yx. For given x, y ∈ R, set [x, y]0 =x, [x, y] = xy − yx, and [x, y]k = [[x, y]k−1, y] for k > 1. We shall makeextensive use of the basic commutator identities [xy, z] = [x, z]y + x[y, z] and[x, yz] = [x, y]z+y[x, z]. Recall that a ring R is prime if for a, b ∈ R, aRb = (0)implies either a = 0 or b = 0 and is semiprime if for a ∈ R, aRa = (0)implies a = 0. An additive map d from R to R is called a derivation of Rif d(xy) = d(x)y + xd(y) holds for all x, y ∈ R. Let F : R → R be a mapassociated with another map g : R → R so that F (xy) = F (x)y + xg(y) holdsfor all x, y ∈ R. If F is additive and g is a derivation of R, then F is said to bea generalized derivation of R that was introduced by Bresar [8]. In [18], Hvalagave the algebraic study of generalized derivations of prime rings. We note

This research is partially supported by the research grants from UGC, India (Grant No. F.PSW-099/10-11 and 39-37/2010(SR), respectively).

66 B. Dhara and S. Ali AEM

that if R has the property that Rx = (0) implies x = 0 and h : R → R is anyfunction, and d : R → R is any additive map such that d(xy) = d(x)y + xh(y)for all x, y ∈ R, then d is uniquely determined by h and moreover h must be aderivation by ([8], Remark 1). Obviously, every derivation is a generalized deri-vation of R. Thus, generalized derivation covers both the concept of derivationand that of left multiplier map.

Following [10], a multiplicative derivation of R is a map D : R → R whichsatisfies D(xy) = D(x)y+xD(y) for all x, y ∈ R. Of course these maps are notadditive. To the best of our knowledge, the concept of multiplicative deriva-tions appears for the first time in the work of Daif [10] and it was motivated bythe work of Martindale [19]. Further, the complete description of those mapswere given by Goldmann and Semrl in [16]. Such maps do indeed exist inthe literature (viz. [10] and [16] where further references can be found). Moreprecisely, we consider R = C[0, 1], the ring of all continuous (real or complexvalued) functions and define a map D : R → R as follows:

D(f)(x) ={

f(x) log |f(x)|, when f(x) �= 00, otherwise.

Then, it is straightforward to check that D satisfies D(fg) = D(f)g+fD(g)for all f, g ∈ C[0, 1], but D is not additive. The notion of multiplicative deri-vation was extended in [12] as follows: a map F : R → R is called a multipli-cative generalized derivation if there exists a derivation d such that F (xy) =F (x)y + xd(y) for all x, y ∈ R. In this definition, if we consider that d isany map (not necessarily additive) then it is more reasonable to call F amultiplicative (generalized)-derivation. To give its precise definition, we makea slight generalization of Daif and Tammam-El-Sayiad’s definition for mul-tiplicative (generalized)-derivation (see [12]). Thus, a map F : R → R (notnecessarily additive) is said to be a multiplicative (generalized)-derivation ifF (xy) = F (x)y + xg(y) holds for all x, y ∈ R, where g is any map (not nec-essarily a derivation or an additive map). Hence, the concept of multiplicative(generalized)-derivation covers the concept of multiplicative derivation. More-over, multiplicative (generalized)-derivation with g = 0 covers the notion ofmultiplicative centralizers (not necessarily additive). The examples of mul-tiplicative (generalized)-derivation are multiplicative derivations and multi-plicative centralizers (see [15], Example 10). Obviously, every multiplicative(generalized)-derivation is a generalized derivation on R. However, the con-verse need not be true in general. The following example justifies the fact:

Example 1.1. Let S be any ring. Next, let R =

⎧⎨⎩

⎛⎝ 0 a b

0 0 c0 0 0

⎞⎠ ∣∣∣ a, b, c ∈ S

⎫⎬⎭ .

Define the maps d and F : R −→ R as follows: d

⎛⎝ 0 a b

0 0 c0 0 0

⎞⎠ =

⎛⎝ 0 0 a2

0 0 00 0 0

⎞⎠

Vol. 86 (2013) On multiplicative (generalized)-derivations 67

and F

⎛⎝ 0 a b

0 0 c0 0 0

⎞⎠ =

⎛⎝ 0 0 bc

0 0 00 0 0

⎞⎠. Then it is straightforward to verify that F is

a multiplicative (generalized)-derivation associated with a multiplicative deri-vation d, but F is not a generalized derivation of R.

Let S be a nonempty subset of R. A map F from R to R is said to becentralizing on S (resp. commuting on S) if [F (x), x] ∈ Z for all x ∈ S(resp. [F (x), x] = 0 for all x ∈ S). The first well-known result on com-muting maps is Posner’s second theorem in [21]. This theorem states thatthe existence of a nonzero commuting derivation on a prime ring R impliesR to be commutative. There has been considerable interest in commutingand centralizing maps in prime and semiprime rings (see for example [6] and[9], where further references can be found). During the past few years, sev-eral authors have been studying commutativity in prime and semiprime ringsadmitting derivations or generalized derivations. For example, we refer thereader to ([1,3–7,11,13,14,20,22,23], where further references can be found).In [5], Ashraf and Rehman proved that if R is a prime ring with a nonzeroideal I of R and d is a derivation of R such that either d(xy) − xy ∈ Zfor all x, y ∈ I or d(xy) + xy ∈ Z for all x, y ∈ I, then R is commu-tative. Being inspired by this result, recently Ashraf et al. [4] have studiedthe situations when derivation d is replaced with a generalized derivation F .More precisely, they proved that a prime ring R must be commutative, ifR satisfies any one of the following conditions: (i) F (xy) − xy ∈ Z for allx, y ∈ I, (ii) F (xy) + xy ∈ Z for all x, y ∈ I, (iii) F (xy) − yx ∈ Z forall x, y ∈ I, (iv) F (xy) + yx ∈ Z for all x, y ∈ I, (v) F (x)F (y) − xy ∈ Z forall x, y ∈ I, (vi)F (x)F (y) + xy ∈ Z for all x, y ∈ I; where F is a general-ized derivation of R associated with a nonzero derivation d and I is a nonzerotwo-sided ideal of R.

In the present paper, our main object is to discuss the commutativity ofprime and semiprime rings involving multiplicative (generalized)-derivations.More precisely, we study all the above mentioned situations in semiprime ringsby replacing the two-sided ideal I with a left-sided ideal L and generalized der-ivation with multiplicative (generalized)-derivation.

2. Main results

Now we begin with our first theorem:

Theorem 2.1. Let R be a semiprime ring, L be a nonzero left ideal of R andF : R → R be a multiplicative (generalized)-derivation associated with the mapg : R → R. If F (xy) ± xy = 0 for all x, y ∈ L, then Lg(L) = (0), F (xy) =F (x)y for all x, y ∈ L and F is a commuting map on L.


Proof. By the assumption, we have

F (xy) − xy = 0 (1)

for all x, y ∈ L. Putting y = yz, z ∈ L in (1) we get

F (xyz) − xyz = 0 (2)

for all x, y, z ∈ L. Since F (xy) = F (x)y+xg(y) for all x, y ∈ R, it follows that,

0 = F (xy)z + xyg(z) − xyz = (F (xy) − xy)z + xyg(z) (3)

for all x, y, z ∈ L. Application of (1) yields that

xyg(z) = 0 (4)

for all x, y, z ∈ L. Replacing y with g(z)ry, where r ∈ R, we get xg(z)ryg(z) =0. In particular, it follows that xg(z)Rxg(z) = (0) for all x, z ∈ L. Since R isa semiprime ring, the last expression forces that xg(z) = 0 for all x, z ∈ L,that is, Lg(L) = (0). Thus, for any x, y ∈ L,F (xy) = F (x)y + xg(y) = F (x)y.Then (1) implies that (F (x) − x)y = 0 for all x, y ∈ L. This implies thaty(F (x) − x)Ry(F (x) − x) = (0). Since R is semiprime, y(F (x) − x) = 0 for allx, y ∈ L. Thus for any x, y ∈ L, (F (x) − x)y = 0 and y(F (x) − x) = 0 togetherimplies [F (x), y] − [x, y] = 0. For x = y, we obtain [F (x), x] = 0 for all x ∈ L.That is, F is commuting on L.

In a similar manner, we can prove that the same conclusion holds forF (xy)+xy = 0 for all x, y ∈ L. Thereby the proof of the theorem is completed.

Corollary 2.2. Let R be a semiprime ring and F : R → R be a multiplicative(generalized)-derivation associated with the map g : R → R. If F (xy)±xy = 0for all x, y ∈ R, then g = 0 and F (x) = ∓x (respectively) for all x ∈ R.

Proof. In view of Theorem 2.11, we have g = 0, F (xy) = F (x)y for all x, y ∈ Rand F is a commuting map on R. Then by our hypothesis, we have for allx, y ∈ R that 0 = F (x)y±xy = (F (x)±x)y. That is, (F (x)±x)R = (0). SinceR is a semiprime ring, we conclude that F (x) ± x = 0 for all x ∈ R.

Theorem 2.3. Let R be a semiprime ring, L be a nonzero left ideal of R andF : R → R be a multiplicative (generalized)-derivation associated with the mapg : R → R. If F (xy) ± yx = 0 for all x, y ∈ L, then L[L,L] = (0), Lg(L) =(0), F (xy) = F (x)y for all x, y ∈ L and F is a commuting map on L.

Proof. First we consider the case

F (xy) − yx = 0 (5)

for all x, y ∈ L. Putting y = yz, z ∈ L in (5) we get

0=F (xyz) − yzx=F (xy)z+xyg(z) − yzx=(F (xy) − yx)z + y[x, z]+xyg(z)(6)


for all x, y, z ∈ L. By (5), it reduces to

y[x, z] + xyg(z) = 0. (7)

Now putting z = x in (7), we get xyg(x) = 0 for all x, y ∈ L. Since L is a leftideal of R, it follows that xRyg(x) = (0) which implies xg(x)Rxg(x) = (0) forall x ∈ L. Since R is semiprime, the last relation yields that xg(x) = 0 forall x ∈ L. Substituting z for y in (7) and then using the fact that xg(x) = 0for all x ∈ L, we obtain y[x, y] = 0 for all x, y ∈ L. Replace x by zx to get0 = y[zx, y] = yz[x, y] + y[z, y]x = yz[x, y] for all x, y, z ∈ L. This implies[x, y]z[x, y] = 0 for all x, y, z ∈ L. It follows that z[x, y]Rz[x, y] = (0) for allx, y, z ∈ L. The semiprimeness of R forces that z[x, y] = 0 for all x, y, z ∈ L.That is, L[L,L] = (0). Then (7) gives xyg(z) = 0 for all x, y, z ∈ L, whichimplies xRyg(z) = (0) and hence xg(z)Rxg(z) = (0) for all x, z ∈ L. Since R issemiprime, the above expression yields that xg(z) = 0 for all x, z ∈ L. That is,Lg(L) = (0). Then for any x, y ∈ L, we have F (xy) = F (x)y +xg(y) = F (x)y.Therefore, (5) becomes

F (x)y − yx = 0 (8)

for all x, y ∈ L. Now putting y = xy in (8), we obtain

F (x)xy − xyx = 0 (9)

for all x, y ∈ L. Left multiplying (8) by x and then subtracting it from (9), weget [F (x), x]y = 0 for all x, y ∈ L. Since L is a left ideal of R, we have

[F (x), x]Ry = (0) (10)

for all x, y ∈ L. Replacing y with x and then multiplying (10) from the rightby F (x) we find that

[F (x), x]RxF (x) = (0) (11)

for all x ∈ L. Replacing y by F (x)x in (10) we get

[F (x), x]RF (x)x = (0) (12)

for all x ∈ L. Subtracting (11) from (12), we find that

[F (x), x]R[F (x), x] = (0) (13)

for all x ∈ L. The semiprimeness of R gives that [F (x), x] = 0 for all x ∈ L.That is, F is commuting on L.

Using a similar approach with necessary variations, we can prove that thesame conclusion holds for the case F (xy) + yx = 0 for all x, y ∈ L.

Corollary 2.4. Let R be a semiprime ring and F : R → R be a multiplicative(generalized)-derivation associated with the map g : R → R. If F (xy)±yx = 0for all x, y ∈ R, then g = 0, R is commutative and F (x) = ∓x (respectively)for all x ∈ R.


Proof. By Theorem 2.3, we have g = 0, R is commutative, F (xy) = F (x)yfor all x, y ∈ R and F is a commuting map on R. Since R is commutative,F (xy) ± yx = 0 becomes F (xy) ± xy = 0 for all x, y ∈ R. Then by Corollary2.2, we conclude that F (x) = ∓x (respectively) for all x ∈ R.

Theorem 2.5. Let R be a semiprime ring, L be a nonzero left ideal of R andF : R → R be a multiplicative (generalized)-derivation associated with the mapg : R → R. If F (x)F (y) ± xy = 0 for all x, y ∈ L, then Lg(L) = (0), F (xy) =F (x)y for all x, y ∈ L and L[F (x), x] = (0) for all x ∈ L.

Proof. First we assume that

F (x)F (y) − xy = 0 (14)

for all x, y ∈ L. Substituting yz for y in (14) and then using the given hypoth-esis, we find that

0 = F (x)F (yz) − xyz = F (x)(F (y)z + yg(z)) − xyz

= (F (x)F (y) − xy)z + F (x)yg(z) (15)

for all x, y, z ∈ L. Using (14), it reduces to F (x)yg(z) = 0 for all x, y, z ∈ L.Replace x by ux, u ∈ L to get 0 = F (ux)yg(z) = (F (u)x + ug(x))yg(z) forall x, y, z ∈ L. Using the fact that F (x)yg(z) = 0 for all x, y, z ∈ L, it givesug(x)yg(z) = 0 for all x, y, z, u ∈ L. Since L is a left ideal, it follows thatLg(L)RLg(L) = (0). Since R is semiprime, we conclude that Lg(L) = (0).Thus for any x, y ∈ L, we obtain F (xy) = F (x)y + xg(y) = F (x)y. Replacingx by xy in (14), we get

F (xy)F (y) − xy2 = 0, (16)

that is,

F (x)yF (y) − xy2 = 0 (17)

for all x, y ∈ L. Right multiplying (14) by y and then subtracting it from (17),we get F (x)[F (y), y] = 0 for all x, y ∈ L. Putting xz for x in the last relation,we obtain F (x)z[F (y), y] = 0 for all x, y ∈ L. This implies [F (x), x]z[F (x), x] =0. That is, L[F (x), x]RL[F (x), x] = (0). Since R is semiprime, it follows thatL[F (x), x] = (0) for all x ∈ L.

A similar conclusion holds for the case F (x)F (y) + xy = 0 for all x, y ∈ L.

Corollary 2.6. Let R be a semiprime ring and F : R → R be a multiplicative(generalized)-derivation associated with the map g : R → R. If F (x)F (y) ±xy = 0 for all x, y ∈ R, then g = 0, F (xy) = F (x)y for all x, y ∈ R and F isa commuting map on R.

Proof. By Theorem 2.5, we conclude that g = 0, F (xy) = F (x)y for all x, y ∈ Rand F is a commuting map on R.


Theorem 2.7. Let R be a semiprime ring, L be a nonzero left ideal of R andF : R → R be a multiplicative (generalized)-derivation associated with the mapg : R → R. If F (x)F (y) ± yx = 0 for all x, y ∈ L, then Lg(L) = (0), L[L,L] =(0), F (xy) = F (x)y for all x, y ∈ L and L[F (x), x] = (0) for all x ∈ L.

Proof. We begin with the situation

F (x)F (y) − yx = 0 (18)

for all x, y ∈ L. Replacing y with yx in (18), we obtain

0 = F (x)F (yx) − yx2 = F (x)(F (y)x + yg(x)) − yx2

= (F (x)F (y) − yx)x + F (x)yg(x). (19)

Using (18), it reduces to F (x)yg(x) = 0 for all x, y ∈ L. Left multiplying byF (z) for z ∈ L,F (z)F (x)yg(x) = 0 for all x, y, z ∈ L. Using our hypothe-sis (18), the last expression yields xzyg(x) = 0 for all x, y, z ∈ L and hencexLRLg(x) = (0) for all x ∈ L. Since R is semiprime, it must contain a familyP = {Pα|α ∈ ∧} of prime ideals such that ∩Pα = (0) (see [2] for details). If Pis a typical member of P and x ∈ L, it follows that

xL ⊆ P or Lg(x) ⊆ P.

These two conditions together imply that xLg(x) ⊆ P for any P ∈ P. There-fore, xLg(x) ⊆ ⋂

α∈∧Pα = (0). That is, xLg(x) = (0) for all x ∈ L. Since L is a

left ideal of R, it follows that xRLg(x) = (0) for all x ∈ L. Again, let P be atypical member of P and x ∈ L. Then it follows that

x ∈ P or Lg(x) ⊆ P.

These two conditions together imply that xg(x) ∈ P for all x ∈ L and forany P ∈ P. Thus xLg(x) ⊆ ⋂

α∈∧Pα = (0). That is, xg(x) = 0 for all x ∈ L.

Now by our hypothesis, we have for any x, y, z ∈ L that

0 = F (x)F (yz) − (yz)x = F (x)(F (y)z + yg(z)) − yzx

= (F (x)F (y) − yx)z + y[x, z] + F (x)yg(z) = y[x, z] + F (x)yg(z). (20)

Replacing z with y, we obtain, by using yg(y) = 0 for any y ∈ L, that 0 =y[x, y] for any x, y ∈ L. Then by the same argument as given in Theorem 2.3,we have L[L,L] = (0). Then by (20), we have F (x)yg(z) = 0 for all x, y, z ∈ L.Now putting x = xz, we get 0 = F (xz)yg(z) = (F (x)z + xg(z))yg(z) =xg(z)yg(z) which yields xg(z)Ryg(z) = (0) for all x, y, z ∈ L. Since R is a semi-prime ring we get that Lg(L) = (0). Hence, we conclude that F (xy) = F (x)yfor all x, y ∈ L.

Now in (18) replacing x with xy and then using the above fact, we get

F (x)yF (y) − yxy = 0 (21)


for all x, y ∈ L. Now right multiplying (18) by y and then subtracting it from(21) we get F (x)[F (y), y] = 0 for all x, y ∈ L. Then replacing x by xz, it yieldsF (x)z[F (y), y] = 0 and hence [F (x), x]Rz[F (y), y] = (0) for all x, y, z ∈ L.Since R is semiprime, L[F (x), x] = (0) for all x ∈ L.

A similar conclusion holds for the case F (x)F (y) + yx = 0 for all x, y ∈ L.

Corollary 2.8. Let R be a semiprime ring and F : R → R be a multiplicative(generalized)-derivation associated with the map g : R → R. If F (x)F (y) ±yx = 0 for all x, y ∈ R, then g = 0, R is commutative, F (xy) = F (x)y for allx, y ∈ R and F is commuting on R.

Proof. By Theorem 2.7, we get our conclusions.

Theorem 2.9. Let R be a semiprime ring, L be a nonzero left ideal of R andF : R → R be a multiplicative (generalized)-derivation associated with the mapg : R → R. If F (xy) ± xy ∈ Z for all x, y ∈ L, then L[g(x), x] = (0) for allx ∈ L.

Proof. First we assume that

F (xy) − xy ∈ Z (22)

for all x, y ∈ L. Now we replace y with yz in (22), where z ∈ L and then weget

F (xyz) − xyz = F (xy)z + xyg(z) − xyz

= (F (xy) − xy)z + xyg(z) ∈ Z (23)

for all x, y, z ∈ L. Applying (22), (23) yields

[xyg(z), z] = 0 (24)

for all x, y, z ∈ L. Substituting rx for x in (24), where r ∈ R, we obtain

0 = [rxyg(z), z] = r[xyg(z), z] + [r, z]xyg(z) = [r, z]xyg(z). (25)

Taking x = g(z)x, we get

0 = [r, z]g(z)xyg(z), (26)

which implies (0) = [r, z]g(z)Rxyg(z) for all x, y, z ∈ L and r ∈ R. Interchang-ing x and y and then subtracting one from the other, we get (0) = [r, z]g(z)R[x, y]g(z) for all x, y, z ∈ L. In particular, (0) = [x, z]g(z)R[x, z]g(z) for allx, z ∈ L. The semiprimeness of R yields that 0 = [x, z]g(z) for all x, z ∈ L.Thus we have

[x, z]g(z) = 0 (27)

for all x, z ∈ L. Right multiplying (27) by z, we get

[x, z]g(z)z = 0 (28)

for all x, z ∈ L. Replace x by xz in (27) to get

[xz, z]g(z) = [x, z]zg(z) = 0 (29)


for all x, z ∈ L. Now (28) and (29) together imply that

[x, z][g(z), z] = 0 (30)

for all x, z ∈ L. Replacing x by g(z)x in the last expression, we obtain

[g(z), z]x[g(z), z] + g(z)[x, z][g(z), z] = 0

for all x, z ∈ L. Application of (30) yields that [g(z), z]x[g(z), z] = 0 for allx, z ∈ L that is L[g(z), z]RL[g(z), z] = (0) for all z ∈ L. Hence, the semiprime-ness of R forces that L[g(z), z] = (0) for all z ∈ L.

In a similar manner we can prove by assuming F (xy) + xy ∈ Z for allx, y ∈ L, that L[g(x), x] = (0) for all x ∈ L.

Corollary 2.10. Let R be a semiprime ring and F : R → R be a multiplicative(generalized)-derivation associated with the map g : R → R. If F (xy)±xy ∈ Zfor all x, y ∈ R, then [g(x), x] = 0 for all x ∈ R.

Theorem 2.11. Let R be a semiprime ring, L be a nonzero left ideal of R andF : R → R be a multiplicative (generalized)-derivation associated with themap g : R → R. If F (xy) ± yx ∈ Z for all x, y ∈ L, then x[x,L] ⊆ Z for allx ∈ L and L[g(x), x] = (0) for all x ∈ L. Moreover, if R is 3-torsion free, thenL[L,L] = (0).

Proof. First we consider that

F (xy) − yx ∈ Z (31)

for all x, y ∈ L. In the above relation, replacing x with xy and y with y2,respectively and then subtracting one from the other, we obtain

(F (xy2) − yxy) − (F (xy2) − y2x) ∈ Z. (32)

This implies that y[y, x] ∈ Z for all x, y ∈ L. Thus for all x ∈ L, x[x,L] ⊆ Z.Now substituting yz for y in (31), where z ∈ L, we get

F (xyz) − (yz)x = F (xy)z + xyg(z) − yzx

= (F (xy) − yx)z + y[x, z] + xyg(z) ∈ Z. (33)

Commuting both sides of (33) with z and then using (31), we obtain

[y[x, z], z] + [xyg(z), z] = 0 (34)

for all x, y, z ∈ L. Replacing x with xz in (34), we get

[y[x, z], z]z + [xzyg(z), z] = 0 (35)

for all x, y, z ∈ L. Right multiplying (34) by z and then subtracting it from(35), we get

[x[yg(z), z], z] = 0 (36)


for all x, y, z ∈ L. Replacing x with rx, r ∈ R in the above relation and thenusing (36), we have

0 = [rx[yg(z), z], z] = r[x[yg(z), z], z] + [r, z]x[yg(z), z]= [r, z]x[yg(z), z]. (37)

In particular, for r = yg(z), we have [yg(z), z]x[yg(z), z] = 0 for all x, y, z ∈ L.Since L is a left ideal of R, it follows that x[yg(z), z]Rx[yg(z), z] = (0) forall x, y, z ∈ L. Since R is semiprime, x[yg(z), z] = 0 for all x, y, z ∈ L. Nowreplacing y with g(z)y, we get

x[g(z)yg(z), z] = 0 (38)

that is,

x(g(z)yg(z)z − zg(z)yg(z)) = 0 (39)

for all x, y, z ∈ L. Now we put y = yg(z)u, where u ∈ L, and then obtain

x(g(z)yg(z)ug(z)z − zg(z)yg(z)ug(z)) = 0 (40)

for all x, y, z ∈ L. By (39), this can be written as

x(g(z)yzg(z)ug(z) − g(z)yg(z)zug(z)) = 0 (41)

that is,

xg(z)y[g(z), z]ug(z) = 0 (42)

for all x, y, z ∈ L. This implies x[g(z), z]y[g(z), z]u[g(z), z] = 0 for all x, y, z ∈ Land so (L[g(z), z])3 = (0) for all z ∈ L. Since a semiprime ring contains nononzero nilpotent left ideals (see [17]), it follows that L[g(z), z] = (0) for allz ∈ L, as desired.

Next assume that R is 3-torsion free. Then x[x,L] ⊆ Z for all x ∈ Lyields x[y, z] + y[x, z] ∈ Z for all x, y, z ∈ L. Replacing y with x2, it reduces tox[x2, z]+x2[x, z] ∈ Z for all x, z ∈ L. This implies, by using the fact x[x,L] ⊆ Zfor all x ∈ R, that 3x2[x,L] ⊆ Z. Since R is 3-torsion free, x2[x,L] ⊆ Z forall x ∈ L. Commuting both sides with L, we obtain (0) = [x2[x,L], L] thatgives (0) = [x,L]x[x,L] for all x ∈ L and so (0) = (x[x,L])2 for all x ∈ L.Since the center of a semiprime ring contains no nonzero nilpotent elements,(0) = x[x,L] for all x ∈ L. This yields (0) = x[x,L2] = xL[x,L] for all x ∈ L.Since L is a left ideal of R, it follows that (0) = L[x,L]RL[x,L] for all x ∈ L.The semiprimeness of R yields (0) = L[x,L] for all x ∈ L, as desired.

The same argument can be adapted if F (xy) + yx ∈ Z for all x, y ∈ L.

Corollary 2.12. Let R be a semiprime ring and F : R → R be a multiplicative(generalized)-derivation associated with the map g : R → R. If F (xy)±yx ∈ Zfor all x, y ∈ R, then x[x,R] ⊆ Z for all x ∈ R and [g(x), x] = 0 for all x ∈ R.Moreover, if R is 3-torsion free, then R is commutative.


Theorem 2.13. Let R be a semiprime ring, L be a nonzero left ideal of R andF : R → R be a multiplicative (generalized)-derivation associated with the mapg : R → R. If F (x)F (y) ± xy ∈ Z for all x, y ∈ L, then L[g(x), x] = (0) for allx ∈ L.

Proof. We begin with the situation

F (x)F (y) − xy ∈ Z (43)

for all x, y ∈ L. Replacing y with yz, z ∈ L, we have

F (x)F (yz) − x(yz) ∈ Z, (44)

which gives

F (x)(F (y)z + yg(z)) − xyz ∈ Z (45)

for all x, y ∈ L. Commuting both sides of (45) with z and then using (43), weget

[F (x)yg(z), z] = 0 (46)

for all x, y, z ∈ L. Putting y = zy in the above relation we obtain

[F (x)zyg(z), z] = 0 (47)

for all x, y, z ∈ L. Now putting x = xz in (46), we get

[(F (x)z + xg(z))yg(z), z] = 0 (48)

for all x, y, z ∈ L. Using (47), the above relation reduces to

[xg(z)yg(z), z] = 0 (49)

for all x, y, z ∈ L. In (49), we replace x with g(z)x and then using (49) weobtain

[g(z), z]xg(z)yg(z) = 0 (50)

for all x, y, z ∈ L. This implies [g(z), z]x[g(z), z]y[g(z), z] = 0 for all x, y, z ∈ L.That is, (L[g(z), z])3 = (0) for all z ∈ L. Since R is semiprime, it contains nononzero nilpotent left ideals, implying L[g(z), z] = (0) for all z ∈ L, as desired.

By the same argument, we may obtain the same conclusion when F (x)F (y)+ xy ∈ Z for all x, y ∈ L.

Corollary 2.14. Let R be a semiprime ring and F : R → R be a multiplicative(generalized)-derivation associated with the map g : R → R. If F (x)F (y) ±xy ∈ Z for all x, y ∈ R, then [g(x), x] = 0 for all x ∈ R.

Theorem 2.15. Let R be a semiprime ring, L be a nonzero left ideal of R andF : R → R be a multiplicative (generalized)-derivation associated with the mapg : R → R. If F (x)F (y) ± yx ∈ Z for all x, y ∈ L, then L[g(x), x] = (0) for allx ∈ L.


Proof. First we consider the case

F (x)F (y) − yx ∈ Z (51)

for all x, y ∈ L. Replacing y with yz, we get

F (x)(F (y)z + yg(z)) − yzx ∈ Z (52)

for all x, y, z ∈ L. This gives

(F (x)F (y) − yx)z + y[x, z] + F (x)yg(z) ∈ Z (53)

for all x, y, z ∈ L. Commuting both sides of (53) with z and then using (51),it reduces to

[y[x, z], z] + [F (x)yg(z), z] = 0 (54)

for all x, y, z ∈ L. Putting x = xz in the above relation we get

[y[x, z], z]z + [(F (x)z + xg(z))yg(z), z] = 0 (55)

for all x, y, z ∈ L. Putting y = zy in (54), we get

z[y[x, z], z] + [F (x)zyg(z), z] = 0 (56)

for all x, y, z ∈ L. Subtracting (56) from (55), we have

[[y[x, z], z], z] + [xg(z)yg(z), z] = 0 (57)

for all x, y, z ∈ L. Putting x = xz, the above relation yields

[[y[x, z], z], z]z + [xzg(z)yg(z), z] = 0 (58)

for all x, y, z ∈ L. Right multiplying (57) by z and then subtracting it from(58), we get

[x[g(z)yg(z), z], z] = 0 (59)

for all x, y, z ∈ L. Now we substitute g(z)yg(z)x for x in (59) and get

0 = [g(z)yg(z)x[g(z)yg(z), z], z]= g(z)yg(z)[x[g(z)yg(z), z], z] + [g(z)yg(z), z]x[g(z)yg(z), z]. (60)

Using (59), it reduces to

0 = [g(z)yg(z), z]x[g(z)yg(z), z] (61)

for all x, y, z ∈ L. Since L is a left ideal, it follows that x[g(z)yg(z), z]Rx[g(z)yg(z), z] = (0) and hence

x[g(z)yg(z), z] = 0 (62)

for all x, y, z ∈ L. This is the same as (38) in the proof of Theorem 2.11. Thususing the same arguments as we used in the last paragraph of the proof ofTheorem 2.11, we get the required result.


In the same manner the conclusion can be obtained when F (x)F (y)+yx ∈Z for all x, y ∈ L. Hence, the theorem is now proved.

Corollary 2.16. Let R be a semiprime ring and F : R → R be a multiplicative(generalized)-derivation associated with the map g : R → R. If F (x)F (y) ±yx ∈ Z for all x, y ∈ R, then [g(x), x] = 0 for all x ∈ R.

3. Examples

We can close this paper with some examples which show that the restrictionsin the hypothesis of several theorems are not superfluous.

Example 3.1. Consider R =

⎧⎨⎩

⎛⎝ 0 a b

0 0 c0 0 0

⎞⎠ |a, b, c ∈ S

⎫⎬⎭, where S is the set of

all integers. Note that⎛⎝0 1 1

0 0 00 0 0

⎞⎠ R

⎛⎝ 0 1 1

0 0 00 0 0

⎞⎠ = (0), but R is not a semiprime ring. We define

maps F, g : R → R, by F

⎛⎝ 0 a b

0 0 c0 0 0

⎞⎠=

⎛⎝0 0 b

0 0 00 0 0

⎞⎠ , g

⎛⎝ 0 a b

0 0 c0 0 0

⎞⎠ =

⎛⎝ 0 a2 b2

0 0 c0 0 0

⎞⎠ .

Then F is a multiplicative (generalized)-derivation associated with the map g.It is straightforward to verify that F (xy)−xy = 0 for all x, y ∈ R. We see thatg(R) �= (0), F (x) �= x for all x ∈ R and [g(x), x] �= 0 for all x ∈ R. Hence, thehypothesis of semiprimeness in Corollary 2.2 and in Corollary 2.10 can not beomitted.

Moreover, F satisfies F (xy) ± yx ∈ Z for all x, y ∈ R. Although R is 3-torsion free, R is not commutative. Hence, the hypothesis of semiprimeness inCorollary 2.12 is essential.

Example 3.2. Consider R =

⎧⎨⎩

⎛⎝ 0 a b

0 0 c0 0 0

⎞⎠ |a, b, c ∈ GF (2)

⎫⎬⎭. Here R is not

a semiprime ring. Now we define maps F, g : R → R, by F

⎛⎝ 0 a b

0 0 c0 0 0

⎞⎠ =

⎛⎝ 0 a 0

0 0 c0 0 0

⎞⎠ , g

⎛⎝ 0 a b

0 0 c0 0 0

⎞⎠ =

⎛⎝ 0 ab b2

0 0 c0 0 0

⎞⎠ . Then F is a multiplicative

(generalized)-derivation of R associated with the map g. It is very easy toverify that F (x)F (y) − xy = 0 for all x, y ∈ R. We see that g(R) �= (0) andF (xy) �= F (x)y for all x, y ∈ R. Hence, the hypothesis of semiprimeness inCorollary 2.6 is crucial.


Acknowledgements

The authors are very thankful to the referee for his/her careful reading of thepaper and valuable comments.

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Basudeb DharaDepartment of MathematicsBelda College, BeldaPaschim Medinipur 721424, Indiae-mail: basu [email protected]

Shakir AliDepartment of MathematicsAligarh Muslim UniversityAligarh 202002, Indiae-mail: [email protected]: http://ss1.amu.ac.in

Received: January 4, 2012

Revised: January 12, 2013

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On multiplicative (generalized)-derivations in prime and semiprime rings