Generalized Jordan triple derivations associated with Hochschild 2–cocycles of rings

Abstract

In the present work, we introduce the notion of a generalized Jordan triple derivation associated with a Hochschild 2–cocycle, and we prove results which imply under some conditions that every generalized Jordan triple derivation associated with a Hochschild 2–cocycle of a prime ring with characteristic different from 2 is a generalized derivation associated with a Hochschild 2–cocycle.

Introduction

Let R denote an associative ring with center Z(R). A ring R is said to have characteristic n if n is the least positive integer such that nx=0 for all x∈R, and of characteristic not n if nx=0, x∈R, then x=0. An additive subgroup L of R is called a Lie ideal of R if [u,r]∈L for all u∈L, r∈R. A Lie ideal L is said to be a square-closed Lie ideal of R if u²∈L for all u∈L. An R-bimodule M is a left and right R-module such that x(my)=(xm)y for all m∈M and x,y∈R. Recall that a ring R is called prime if xRy=(0) implies that either x=0 or y=0, and R is called semiprime if xRx=(0) implies x=0. An additive mapping d:R→R is called a derivation if d(xy)=d(x)y+xd(y) for all x,y∈R. d is called a Jordan derivation in case d(x²)=d(x)x+xd(x) for all x∈R. Moreover, d is called a Jordan triple derivation if d(xyx)=d(x)yx+xd(y)x+xyd(x) for all x,y∈R. It is obvious to see that every derivation is a Jordan derivation and is a Jordan triple derivation but the converse is in general not true. A classical result of Herstein [1] asserts that any Jordan derivation of a prime ring with characteristic different from 2 is a derivation. In [2], Bre$\breve {s}$ar has proved Herstein’s result in the case of a semiprime ring. Also, he has shown in [3] that any Jordan triple derivation of a 2-torsion free semiprime ring is a derivation. An additive map f of a ring R is called a generalized derivation if there is a derivation d of R such that for all x, y in R, f(xy)=f(x)y+xd(y) and is called a generalized Jordan derivation if there is a Jordan derivation d such that f(x²)=f(x)x+xd(x) for all x∈R. Furthermore, f is said to be a generalized Jordan triple derivation if there is a Jordan triple derivation d of R such that for all x, y in R, f(xyx)=f(x)yx+xd(y)x+xyd(x). In [4], Jing and Lu have proved in a prime ring R of characteristic not two that every generalized Jordan derivation of R is a generalized derivation, and also every generalized Jordan triple derivation on R is a generalized derivation.

Let θ and ϕ be endomorphisms of a ring R. f is called a (θ,ϕ)−derivation if f(xy)=f(x)θ(y)+ϕ(x)f(y) for all x, y∈R. f is called a Jordan (θ,ϕ)–derivation if f(x²)=f(x)θ(x)+ϕ(x)f(x) for all x∈R. f is called a Jordan triple (θ,ϕ)−derivation if f(xyx)=f(x)θ(y)θ(x)+ϕ(x)f(y)θ(x)+ϕ(x)ϕ(y)f(x) for all x,y∈R. In [5], Liu and Shiue have proved that every Jordan triple (θ,ϕ)−derivation on a 2-torsion free semiprime ring R is a (θ,ϕ)−derivation, where θ and ϕ are automorphisms. An additive mapping f:R→R is said to be a left (right) centralizer, if f(xy)=f(x)y(f(xy)=xf(y)) for all x,y∈R. f is called a centralizer, if f is both a left and right centralizer. In [6], Vukman and Kosi-Ulbl have shown that if R is a 2-torsion free semiprime ring and f is an additive mapping of R such that 2f(xyx)=f(x)yx+xyf(x) for all x,y∈R, then f is a centralizer.

An additive mapping f:R→R is said to be a left (right) θ−centralizer associated with a function θ of R, if f(xy)=f(x)θ(y)(f(xy)=θ(x)f(y)) for all x,y∈R. f is called a θ−centralizer, if f is both a left and right θ−centralizer. Daif, El-Sayiad, and Muthana in [7] have proved that if R is a 2−torsion free semiprime ring and f is an additive mapping of R such that 2f(xyx)=f(x)θ(yx)+θ(xy)f(x) for all x,y∈R with θ(Z(R))=Z(R), where θ is a nonzero surjective endomorphism on R, then f is a θ−centralizer.

Now let R be a ring and M be an R-bimodule. A biadditive map α:R×R→M is called a Hochschild 2–cocycle, if xα(y,z)−α(xy, z)+α(x,yz)−α(x, y)z=0 for all x,y,z∈R, and α is called symmetric if α(x,y)=α(y,x) for all x,y∈R. Nakajima [8] has introduced a new type of generalized derivations and generalized Jordan derivations associated with Hochschild 2–cocycles in the following way. An additive map f:R→M is called a generalized derivation associated with a Hochschild 2–cocycle α if f(xy)=f(x)y+xf(y)+α(x,y) for all x,y∈R, and f is called a generalized Jordan derivation associated with α if f(x²)=f(x)x+xf(x)+α(x,x) for all x∈R. If α=0, then f means the usual derivation and Jordan derivation. He has given the following examples:

(1) If f is a generalized derivation associated with a derivation d, then the map α₁:R×R∋(x,y)↦x(d−f)(y)∈M is biadditive and satisfies the 2–cocycle condition. Hence, f is a generalized derivation associated with α₁.

(2) If f:R→M is a left centralizer, then by f(xy)=f(x)y+xf(y)+x(−f)(y), we have a 2–cocycle α₂:R×R→M defined by, α₂(x,y)=x(−f)(y), and hence, f is a generalized derivation associated with α₂.

(3) Let f be a (θ,ϕ)−derivation. Then, the map α₃:R×R∋(x,y)↦f(x)(θ(y)−y)+(ϕ(x)−x)f(y)∈M, is biadditive and satisfies the 2–cocycle condition. Since f(xy)=f(x)y+xf(y)+α₃(x,y), then f is a generalized derivation associated with α₃.

(4) In general, he has mentioned the following. Let f:R→M be an additive map and let α:R×R→M be a biadditive map. If f(xy)=f(x)y+xf(y)+α(x,y) holds, then by the associativity f((xy)z)=f(x(yz)), α satisfies the 2–cocycle condition. Thus f is a generalized derivation associated with α.

In his work, Nakajima [8] has shown the following result. Let R be a 2-torsion free ring. Then, every generalized Jordan derivation associated with a Hochschild 2–cocycle α is a generalized derivation associated with α in each of the following cases:

• (i) R is a noncommutative prime ring.

• (ii) There exist x,y∈R such that [x,y] is a nonzero divisor.

• (iii) R is commutative and α is symmetric.

Nawzad, et al. [9] have shown the following. Let R be a 2-torsion free ring. Then, every generalized Jordan derivation associated with a Hochschild 2–cocycle α is a generalized derivation associated with α in each of the following cases:

• (i) R is a noncommutative semiprime ring and α is symmetric.

• (ii) R is commutative.

In [10], Rehman and Hongan have proved the following result. Let R be a 2-torsion free ring and L a square-closed Lie ideal of R. Then, every generalized Jordan derivation associated with a Hochschild 2–cocycle α is a generalized derivation associated with α in each of the following cases.

• (i) R is a prime ring and L is noncommutative.

• (ii) R is a prime ring, L is commutative and α is symmetric.

• (iii) There exist x,y∈R such that [x,y] is a nonzero divisor in L.

In the present article, we introduce the notion of generalized Jordan triple derivations associated with Hochschild 2–cocycles in the following way. Let R be a ring and let M be an R-bimodule. An additive map f:R→M is called a generalized Jordan triple derivation associated with a Hochschild 2–cocycle α if f(xyx)=f(x)yx+xf(y)x+α(x,y)x+xyf(x)+α(xy,x) for all x,y∈R.

Examples (i) If f is a Jordan triple derivation, then the zero map α₁ is biadditive and satisfies the 2–cocycle condition. Therefore f is a generalized Jordan triple derivation associated with α₁.

(ii) If f is a generalized Jordan triple derivation associated with a Jordan triple derivation d, then α₂(x,y)=x(d−f)(y) is biadditive and satisfies the 2–cocycle condition and we can see that f(xyx)=f(x)yx+xf(y)x+α₂(x,y)x+xyf(x)+α₂(xy,x). Hence f is a generalized Jordan triple derivation associated with α₂.

Our aim in this work is to show that every generalized Jordan triple derivation associated with a Hochschild 2–cocycle α from a prime ring R with characteristic different from 2 to an R-bimodule M is a generalized derivation associated with α.

Preliminary results

The proof of our result is based on the following series of auxiliary lemmas.

Lemma 1

Let f be a generalized Jordan triple derivation from a ring R to an R- bimodule M associated with a Hochschild 2–cocycle map α from R×R into M. Then for all x,y,z∈R, f(xyz+zyx)=f(x)yz+xf(y)z+α(x,y)z+zyf(z)+α(xy,z)+f(z)yx+zf(y)x+α(z,y)x+zyf(x)+α(zy,x).

Proof

Let v=f((x+z)y(x+z)), we have for all x,y,z∈R

$${{} \begin{aligned} 0&=v-v\\ &=f(xyx)+f(xyz+zyx)+f(zyz)- \{f(x+z)y(x+z)\\ &\,\,\,\,\, +(x+z)f(y)(x+z)+\alpha(x+z,y)(x+z)\\& \quad+(x+z)yf(x+z)\\ &\,\,\,\,\, +\alpha((x+z)y,(x+z))\}. \end{aligned}} $$

Then,

$${{} \begin{aligned} 0&=f(xyx)+f(xyz+zyx)+f(zyz)-\{f(x)yx\\&\quad+f(x)yz+f(z)yx\\ &+f(z)yz+xf(y)x+xf(y)z+zf(y)x+zf(y)z\\&\quad+\alpha(x,y)x+\alpha(x,y)z\\ &+\alpha(z,y)x+\alpha(z,y)z+xyf(x)+xyf(z)+zyf(x)\\&\quad+zyf(z)+\alpha(xy,x)\\ &+\alpha(xy,z)+\alpha(zy,x)+\alpha(zy,z)\} \mathrm{\,\,\,for \,\,\, all\,\,\,} x,y,z\in R, \end{aligned}} $$

Therefore,

$${{} \begin{aligned} f(xyz+zyx)&=f(x)yz+xf(y)z+\alpha(x,y)z+zyf(z)\\&\quad+\alpha(xy,z)\\ &\,\,\,\, +f(z)yx+zf(y)x+\alpha(z,y)x+zyf(x)\\ &\,\,\,\, +\alpha(zy,x) \mathrm{\,\,\,for\,\,\, all\,\,\,} x,y,z\in R, \end{aligned}} $$

as required. □

For a generalized Jordan triple derivation f from a ring R to an R-bimodule M associated with a Hochschild 2−cocycle α, we denote by δ, F and β the maps from R×R×R into M defined by δ(x,y,z)=f(xyz)−f(x)yz-xf(y)z−α(x,y)z-xyf(z)−α(xy,z), F(x,y,z)=f(xyz)−f(x)yz-xf(y)z-xyf(z) and β(x,y,z)=xyz-zyx, respectively. Thus, δ(x,y,z)=F(x,y,z)−α(x,y)z−α(xy,z).

Lemma 2

For all x,y,z in a ring R, the following hold:

• (i) δ(x,y,z)=−δ(z,y,x), and

• (ii) δ(x,y,z) and β(x,y,z) are tri-additive.

Proof

(i) Follows easily from Lemma 1.

(ii) Replace x by a+b in the definition of δ, then (ii) is easily seen. □

Lemma 3

For any ring R and any a,b,c,x∈R,

δ(a,b,c)xβ(a,b,c)+β(a,b,c)xδ(a,b,c)=0.

Proof

Let v=f(abcxcba+cbaxabc), then 0=v−v=f((abc)x(cba)+(cba)x(abc))−f(a(bcxcb)a+c(baxab)c). By the definition of the generalized Jordan triple derivation f associated with a Hochschild 2-cocycle α and by Lemma 1, we get

$$ {\begin{aligned} 0&=f(abc)xcba+abcf(x)cba+\alpha(abc,x)cba+abcxf(cba)\\ &+\alpha(abcx,cba)+f(cba)xabc+cbaf(x)abc+\alpha(cba,x)abc\\ &+cbaxf(abc)+\alpha(cbax,abc)-\{f(a)bcxcba+af(b)cxcba\\ &+abf(c)xcba+abcf(x)cba+ab\alpha(c,x)cba+abcxf(c)ba\\ &+ab\alpha(cx,c)ba+a\alpha(b,cxc)ba+abcxcf(b)a+a\alpha(bcxc,b)a\\ &+\alpha(a,bcxcb)a+abcxcbf(a)+\alpha(abcxcb,a)+f(c)baxabc\\ &+cf(b)axabc+cbf(a)xabc+cbaf(x)abc+cb\alpha(a,x)abc\\ &+cbaxf(a)bc+cb\alpha(ax,a)bc+c\alpha(b,axa)bc+cbaxaf(b)c\\ &+c\alpha(baxa,b)c+\alpha(c,baxab)c+cbaxabf(c)+\alpha(cbaxab,c)\}. \end{aligned}} $$

(1)

Therefore, for all a,b,c,x∈R

$$ {\begin{aligned} 0&=F(a,b,c)xcba+abcxF(c,b,a)\\&+ \{\alpha(abc,x)-ab\alpha(c,x)\}cba+ \{\alpha(abcx,cba)-\alpha(abcxcb,a)\}\\ &-\{ab\alpha(cx,c)ba+a\alpha(b,cxc)ba+ a\alpha(bcxc,b)a+\alpha(a,bcxcb)a\}\\ &+F(c,b,a)xabc+cbaxF(a,b,c)\\ &+\{\alpha(cba,x)-cb\alpha(a,x)\}abc+ \{\alpha(cbax,abc)-\alpha(cbaxab,c)\}\\ &-\{cb\alpha(ax,a)bc+c\alpha(b,axa)bc+ c\alpha(baxa,b)c+\alpha(c,baxab)c\} \end{aligned}} $$

(2)

Since α is a 2–cocycle map, we obtain the following relations for all a,b,c,x∈R:

• (i) {α((ab)c,x)−(ab)α(c,x)}cba={α(ab,cx)−α(ab,c)x}cba.

• (ii) α(abcx,(cb)a)−α((abcx)(cb),a)=α(abcx,cb)a−(abcx)α(cb,a).

• (iii) {α((cb)a,x)−(cb)α(a,x)}abc={α(cb,ax)−α(cb,a)x}abc.

• (iv) α(cbax,(ab)c)−α((cbax)(ab),c)=α(cbax,ab)c−(cbax)α(ab,c).

Substituting from (i–iv) in (2), we get for all a,b,c,x∈R

$$ {\begin{aligned} 0&=F(a,b,c)xcba+abcxF(c,b,a)\\ &+\{\alpha(ab,cx)-\alpha(ab,c)x\}cba+ \{\alpha(abcx,cb)a-abcx\alpha(cb,a)\}\\ &-\{ab\alpha(cx,c)ba+a\alpha(b,cxc)ba+ a\alpha(bcxc,b)a+\alpha(a,bcxcb)a\}\\ &+F(c,b,a)xabc+cbaxF(a,b,c)\\ &+\{\alpha(cb,ax)-\alpha(cb,a)x\}abc+ \{\alpha(cbax,ab)c-cbax\alpha(ab,c)\}\\ &-\{cb\alpha(ax,a)bc+c\alpha(b,axa)bc+ c\alpha(baxa,b)c+\alpha(c,baxab)c\} \end{aligned}} $$

(3)

Since α is a 2–cocycle map, we conclude for all a,b,c,x∈R that

(i) α(ab,cx)=aα(b,cx)+α(a,b(cx))−α(a,b)(cx).

(ii) α(abcx,cb)a={−(abcx)α(c,b)+α((abcx)c,b)+α(abcx,c)b}a.

(iii) α(cb,ax)=cα(b,ax)+α(c,b(ax))−α(c,b)(ax).

(iv) α(cbax,ab)c={−(cbax)α(a,b)+α((cbax)a,b)+α(cbax,a)b}c.

Substituting from (i–iv) in (3), we obtain

$$ {\begin{aligned} 0&=\{F(a,b,c)-\alpha(ab,c)-\alpha(a,b)c\}xcba+abcx\{F(c,b,a)-\alpha(cb,a)\\ &-\alpha(c,b)a\}+\{a\alpha(b,cx)cba-ab\alpha(cx,c)ba-a\alpha(b,cxc)ba\}\\ &+\{\alpha(abcxc,b)a-a\alpha(bcxc,b)a-\alpha(a,bcxcb)a\}\\ &+\alpha(a,bcx)cba+\alpha(abcx,c)ba+\{F(c,b,a)-\alpha(cb,a)\\ &-\alpha(c,b)a\}xabc+cbax\{F(a,b,c)-\alpha(ab,c)-\alpha(a,b)c\}\\ &+\{c\alpha(b,ax)abc-cb\alpha(ax,a)bc-c\alpha(b,axa)bc\}\\ &+\{\alpha(cbaxa,b)c-c\alpha(baxa,b)c-\alpha(c,baxab)c\}\\ &+\alpha(c,bax)abc+\alpha(cbax,a)bc, \,\, \mathrm{for\,\, all}\,\ a,b,c,x\in R. \end{aligned}} $$

(4)

Again since α is a 2–cocycle map, we have

• (i) a{α(b,cx)c−bα(cx,c)−α(b,(cx)c)}ba=−aα(b(cx),c)ba.

• (ii) {α(a(bcxc),b)−aα(bcxc,b)−α(a,(bcxc)b)}a=−α(a,bcxc)ba.

• (iii) c{α(b,ax)a−bα(ax,a)−α(b,(ax)a)}bc=−cα(b(ax),a)bc.

• (iv) {α(c(baxa),b)−cα(baxa,b)−α(c,(baxa)b)}c=−α(c,baxa)bc.

Replacing (i–iv) into (4), we get, for all a,b,c,x∈R

$$ {\begin{aligned} 0&=\delta(a,b,c)xcba+abcx\delta(c,b,a)-a\alpha(bcx,c)ba-\alpha(a,bcxc)ba\\ &+\alpha(a,bcx)cba+\alpha(abcx,c)ba+\delta(c,b,a)xabc+cbax\delta(a,b,c)\\ &-c\alpha(bax,a)bc-\alpha(c,baxa)bc+\alpha(c,bax)abc+\alpha(cbax,a)bc. \end{aligned}} $$

(5)

Continuing in this manner, we obtain

• (i) {−aα(bcx,c)−α(a,(bcx)c)+α(a,bcx)c+α(a(bcx),c)}ba=0.

• (ii) {−cα(bax,a)−α(c,(bax)a)+α(c,bax)a+α(c(bax),a)}bc=0.

By (5), we conclude that 0=δ(a,b,c)xcba+abcxδ(c,b,a)+δ(c,b,a)xabc+cbaxδ(a,b,c) for all a,b,c,x∈R. By Lemma 2, we obtain 0=δ(a,b,c)xcba−abcxδ(a,b,c)−δ(a,b,c)xabc+cbaxδ(a,b,c) for all a,b,c,x∈R.

Therefore, δ(a,b,c)xβ(a,b,c)+β(a,b,c)xδ(a,b,c)=0 for all a,b,c,x∈R. This finishes the proof of the lemma. □

Lemma 4

If R is a prime ring of characteristic not 2, then for all a,b,c,x∈R,δ(a,b,c)xβ(a,b,c)=0,.

Proof

By Lemma 3 and Lemma 1.1 of Bre$\breve {s}$ar [3], we get the proof. □

Lemma 5

If R is a prime ring of characteristic not 2, then

δ(a₁,b₁,c₁)xβ(a₂,b₂,c₂)=0 for all a₁,b₁,c₁,a₂,b₂,c₂,x∈R.

Proof

From Lemma 2(ii), Lemma 4, and Lemma 1.2 of Bre$\breve {s}$ar [3], we get the proof. □

Lemma 6

Let R be a prime ring. Then, R is commutative iff β(a,b,c)=0 for all a,b,c∈R.

Proof

If R is commutative, then, by definition of β,β(a,b,c)=0 for all a,b,c∈R. Conversely, assume that β(a,b,c)=0 for all a,b,c∈R. Let Q be the Martindale right ring of quotients of R defined by Martindale [11]. Then Q is a prime ring with identity that contains the ring R. By Chuang [12], Q satisfies the same generalized polynomial identities as R. In particular abc−cba=0 for all a,b,c∈Q. Replacing c by the identity of Q yields the commutativity of Q, and hence R. □

Lemma 7

Let R be a prime ring of characteristic not 2. Then δ(a,b,c)=0 for all a,b,c∈R, in each of the following cases:

• (i) R is noncommutative.

• ii There exist x,y,z∈R such that β(x,y,z) is a nonzero divisor in M.

• iii R is commutative and α is symmetric.

Proof

(i) By Lemmas 5 and 6, we get our requirement.

(ii) By Lemma 5, we have δ(a,b,c)rβ(x,y,z)=0 for all a,b,c,r,x,y,z∈R. From our assumption δ(a,b,c)r=0 for all a,b,c,r∈R. Thus the primeness of R gives δ(a,b,c)=0 for all a,b,c∈R.

(iii) From Lemma 1 we have f(abc+cba)=f(a)bc+af(b)c+α(a,b)c+abf(c)+α(ab,c)+f(c)ba+cf(b)a+α(c,b)a+cbf(a)+α(cb,a) for all a,b,c∈R. Since R is commutative and α is symmetric, we get 0=2{f(abc)−f(a)bc−af(b)c−abf(c)}−α(a,b)c−α(ab,c)−aα(b,c)−α(a,bc) for all a,b,c∈R. Since α is 2–cocycle we have −aα(b,c)−α(a,bc)=−α(a,b)c−α(ab,c) for all a,b,c∈R. Therefore 0=2{f(abc)−f(a)bc−af(b)c−abf(c)−α(a,b)c−α(ab,c)} for all a,b,c∈R. Since R has characteristic not 2, then δ(a,b,c)=0 for all a,b,c∈R, as required. □

Main result

Theorem 1

Let R be a prime ring of characteristic not 2. Then every generalized Jordan triple derivation associated with a Hochschild 2–cocycle α is a generalized derivation associated with α in each of the following cases.

(i) R is noncommutative.

(ii) There exist x,y,z∈R such that β(x,y,z) is a nonzero divisor in M.

(iii) R is commutative and α is symmetric.

Proof

Suppose that f is a generalized Jordan triple derivation associated with a Hochschild 2−cocycle α. We denote by G(a,b) and a^b the elements of M defined by G(a,b)=f(ab)−f(a)b−af(b), and a^b=f(ab)−f(a)b−af(b)−α(a,b), respectively. Thus, a^b=G(a,b)−α(a,b). It is evident that a^b+c=a^b+a^c, and (a+b)^c=a^c+b^c. By Lemma 7, we have δ(a,b,c)=0 for all a,b,c∈R. Thus, for all a,b,c∈R

$$ f(abc)=f(a)bc+af(b)c+\alpha(a,b)c+abf(c)+\alpha(ab,c). $$

(6)

Now let v=f(abxab), then 0=v−v=f((ab)x(ab))−f(a(bxa)b). By (6), we have for all a,b,x∈R

$${\begin{aligned} 0&=f(ab)xab+abf(x)ab+\alpha(ab,x)ab+abxf(ab)+\alpha(abx,ab)\\ &-f(a)bxab-af(b)xab-abf(x)ab-a\alpha(b,x)ab-abxf(a)b\\ &-a\alpha(bx,a)b-\alpha(a,bxa)b-abxaf(b)-\alpha(abxa,b). \end{aligned}} $$

So, for all a,b,x∈R

$$ {{} \begin{aligned} 0&=G(a,b)xab+abxG(a,b)+ \{\alpha(ab,x)-a\alpha(b,x)\}ab\\ &+\{\alpha(abx,ab)-\alpha(abxa,b)\} -a\alpha(bx,a)b-\alpha(a,bxa)b. \end{aligned}} $$

(7)

Since α is 2-cocycle we have for all a,b,x∈R that

• (i) {α(ab,x)−aα(b,x)}ab={α(a,bx)−α(a,b)x}ab, and

• (ii) α(abx,ab)−α((abx)a,b)=α(abx,a)b−(abx)α(a,b).

Substituting from (i) and (ii) in (7), we get G(a,b)xab−α(a,b)xab+abxG(a,b)−abxα(a,b)+α(a,bx)ab+α(abx,a)b−aα(bx,a)b−α(a,bxa)b=0for all a,b,x∈R. But α is 2–cocycle, hence {α(a,bx)a+α(abx,a)−aα(bx,a)−α(a,bxa)}b=0. Therefore a^bx(ab)+(ab)xa^b=0 for all a,b,x∈R. By Lemma 1.1 of Bre$\breve {s}$ar [3], we get

$$ a^{b}x(ab)=(ab)xa^{b}=0\,\, \mathrm{for\,\, all}\,\, a,b,x\in R. $$

(8)

Replacing a by a+c in (8) and using (8), we obtain a^bxcb=−c^bxab for all a,b,c,x∈R, and then (a^bxcb)y(a^bxcb)=−a^bx(cbyc^b)xab=0 for all a,b,c,x,y∈R. Thus the primeness of R gives

$$ a^{b}xcb=0\,\, \mathrm{for\,\, all}\,\, a,b,c,x\in R. $$

(9)

Similarly replacing b by b+d in (9), we get

$$a^{b}xcd=0\,\, \mathrm{for\,\, all}\,\, a,b,c,d,x\in R. $$

(10)

Putting c=a^b and x=dx in (10) we have a^bdxa^bd=0 for all a,b,d,x∈R. Again, the primeness of R yields that a^bd=0 for all a,b,d∈R, and hence a^b=0 for all a,b∈R. Consequently, f is a generalized derivation associated with a Hochschild 2–cocycle α. □